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politas
10th March 2005, 02:19 AM
There are some people on these forums who have a much better understanding of relativistic effects than myself, so I was wondering if anyone can answer these questions, which might help me to understand it.

Imagine a spaceship 189,000 miles wide, with a light source on one side and a mirror on the other side, arranged so that light from the light source can be seen in the mirror.

If the spaceship is at rest relative to the observer, the light from the light source should appear one second later in the mirror, right?

So, if D is the difference in seconds between the light reaching an observer from the light source and the mirror, at rest, D=1.

What happens to the value of D under the following circumstances, where speeds and accelerations are sufficient for measurable relativistic effects? (I'm looking for answers like "D<1","D>1","D=1"):

1) Light source and mirror are in a line perpendicular to a line drawn from ship to observer, and

a) Spaceship is accelerating away from the observer, and travelling away from the observer.

b) Spaceship is accelerating away from the observer and travelling towards the observer.

c) Spaceship is accelerating towards the observer and travelling towards the observer.

d) Spaceship is accelerating towards the observer and travelling away from the observer.

e) Spaceship is travelling towards the observer at a fixed velocity relative to the observer.

f) Spaceship is travelling away from the observer at a fixed velocity relative to the observer.

2) Light source and mirror are in a line parallel to a line drawn from ship to observer, with the mirror only half the above stated distance further away from the observer than the light source (so that at rest, D=1), and

a) Spaceship is accelerating away from the observer, and travelling away from the observer.

b) Spaceship is accelerating away from the observer and travelling towards the observer.

c) Spaceship is accelerating towards the observer and travelling towards the observer.

d) Spaceship is accelerating towards the observer and travelling away from the observer.

e) Spaceship is travelling towards the observer at a fixed velocity relative to the observer.

f) Spaceship is travelling away from the observer at a fixed velocity relative to the observer.

Ziggurat
10th March 2005, 08:25 AM
Originally posted by politas

What happens to the value of D under the following circumstances, where speeds and accelerations are sufficient for measurable relativistic effects? (I'm looking for answers like "D<1","D>1","D=1"):

Working out the details for each scenario takes some time, and I'm not actually sure how general an answer you can really get (it's easier to work with specific numbers). But I will point out that solving case #2 with acceleration is actually highly non-trivial. The problem is that accelerating large objects at relativistically significant speeds will necessarily run into the problem of the speed of sound in the ship. For example, let's say you've got thrusters on the back side of this giant ship which is one light-second long. You turn the thrusters on, and the ship starts accelarating. OK, but which PART of the ship? Well, the speed of sound in the ship cannot exceed c (in real materials, of course, it's not even close), which means that the front end of the ship CANNOT start accelerating until at least one second AFTER the back end of the ship has started to accelerate. So even assuming a perfectly stiff ship (speed of sound = c), calculating the trajectories of the front and back of the ship becomes quite difficult.

Say we want to try to simplify this. We stick thrusters on the front and back of the ship, with sufficient power such that both ends of the ship start accelerating together (let the middle part of the ship be stretchy for this part). Suppose they each accelerate for 1 second, and obtain some significant fraction of c. In the reference frame they started in, the front and back ends of the ship are both moving at the same speed, and are still the same distance apart. But in the ship's reference frame, you'll find that the distance between the front and back end of the ship has gotten larger. We've stretched the ship by accelerating both ends at once. While their trajectories are now easy to calculate, we've fundamentally changed the problem you were asking.

In other words, once you throw in acceleration, particularly for such a large object, you're forced to address a whole host of complicated questions, and figuring out the answers isn't easy. It's not so bad for case 1, since you can accelerate both the light and mirror side without worrying about length contraction between them, but I'm still not sure how general an answer you can give without more details of the speeds and the position of the observer. The observer can be placed exactly between the light and the mirror when there's no relative motion and no acceleration, but there isn't such an obvious place to position the observer in the other cases, particularly once you add acceleration, and the position of the observer is going to make a big difference in your answer (even with the ship at rest).

politas
10th March 2005, 08:52 AM
The size of the ship isn't really important here, at least, not for what I'm trying to find out. I'm wondering about how time dilation works. I don't understand when things appear to speed up and down. The idea was to have some time-based event occur that is measurable and visible to an observer. Treat the spaceship as a point. D is the duration of a measurable, timed event that has a value of 1 when the ship is at rest relative to the observer. The light/mirror thing is an artificial construct to create such a measurable event, since it is based on a handy constant that doesn't require any masses moving relative to each other. Shrinking the distance between light and mirror reduces the base value of D; as long as variations are still measurable, it can be far smaller, if that helps.

What I want to know is under what conditions does D reduce, and under what conditions does it increase? (to the observer, obviously; it is always 1 for the moving object)

Case 2 I expected to be more complex than case 1; I didn't realise just how complex. I'm primarily interested in case 1.

See, I was reading some of the discussions about relativity, and the "twin paradox", and wondering about the revolutions. How fast are they occurring on the trips out and back, both while accelerating and "decelerating? I wanted to reduce the question to a simpler case, as a thought experiment to try to get a better grasp (if not a complete understanding; the maths is beyond my grasp without serious study I just don't have the time or inclination for)

Ziggurat
10th March 2005, 09:36 AM
Originally posted by politas
What I want to know is under what conditions does D reduce, and under what conditions does it increase? (to the observer, obviously; it is always 1 for the moving object)

If I understand you correctly, then what you're asking is basically when does it look like a clock on the ship is slowing down (you can turn the bouncing light problem into a clock problem if you ask how long it takes the reflected light to get back to the source, and not to some external observer whose position is then relevant). Clocks will always be *observed* to slow down for any moving object, regardless of direction or acceleration. But they can *look* like they're moving faster if the object is moving towards you, and *look* like they are slowed down even more than they are if the object is moving away from you: Doppler shift effects are added on top of time dilation when you want to know what you actually see.

Note that there's a distinction in relativity between what you see and what you observe. An example from everyday life is airplanes flying overhead at a significant fraction of the speed of sound. You hear the noise of the plane coming from some position behind where it is currently located, but knowing the speed of sound, you know that you're hearing sound that was emitted some time ago, and if you've been watching and calculate things correctly, you *observe* that the sound came from where the plane was, not from some point behind the plane. Observations in special relativity are independent of such optical illusions, but what you actually see can have very strong optical illusion effects. A famous example from astronomy is objects close to the speed of light moving towards you at an angle can *look* like they are moving sideways faster than the speed of light. But correctly accounting for their motion towards you (which can be hard to measure at intergallactic distances) will correctly produce an *observation* of the object moving at less than c.

rwguinn
10th March 2005, 09:42 AM
Originally posted by Ziggurat
If I understand you correctly, then what you're asking is basically when does it look like a clock on the ship is slowing down (you can turn the bouncing light problem into a clock problem if you ask how long it takes the reflected light to get back to the source, and not to some external observer whose position is then relevant). Clocks will always be *observed* to slow down for any moving object, regardless of direction or acceleration. But they can *look* like they're moving faster if the object is moving towards you, and *look* like they are slowed down even more than they are if the object is moving away from you: Doppler shift effects are added on top of time dilation when you want to know what you actually see.

Note that there's a distinction in relativity between what you see and what you observe. An example from everyday life is airplanes flying overhead at a significant fraction of the speed of sound. You hear the noise of the plane coming from some position behind where it is currently located, but knowing the speed of sound, you know that you're hearing sound that was emitted some time ago, and if you've been watching and calculate things correctly, you *observe* that the sound came from where the plane was, not from some point behind the plane. Observations in special relativity are independent of such optical illusions, but what you actually see can have very strong optical illusion effects. A famous example from astronomy is objects close to the speed of light moving towards you at an angle can *look* like they are moving sideways faster than the speed of light. But correctly accounting for their motion towards you (which can be hard to measure at intergallactic distances) will correctly produce an *observation* of the object moving at less than c.

I believe (that word again) that your reference to Doppler effect on time dialation is incorrect. I was under the impression that the value is a scalar, not a vector quantity.

Roger

Ziggurat
10th March 2005, 09:52 AM
Originally posted by rwguinn
I believe (that word again) that your reference to Doppler effect on time dialation is incorrect. I was under the impression that the value is a scalar, not a vector quantity.

Roger

I'm using the term "added" rather loosely, and should have said something more like "you calculate doppler shift in addition to time dilation" (although both the Doppler effect and your time dilation numbers will both be scalar quantities, and I think you'll actually need a multiplication). What I meant was that you first calculate the time dilation using relativity to figure out how much a moving clock is truly slowed down in reference frame. This calculation only depends on the magnitude of the relative velocity of the clock. After you have calculated the time dilation, you then calculate a Doppler shift based upon not only the magnitude of the velocity, but also relative direction. If the object is moving away from you, it is red-shifted, if it is moving towards you, it is blue-shifted, if it is moving perpendicular to you there is no Doppler shift. But in all cases there is still a time dilation effect. It's the fact that you have to do the Doppler shift calculation as well as ("in addition to") the time dilation if you want to figure out what you would see (and not just observe) that I was trying to get at. Hope that clarifies it.

rppa
10th March 2005, 10:37 AM
[QUOTE]Originally posted by politas
The size of the ship isn't really important here, at least, not for what I'm trying to find out.

But such things as the fact that the front and back of the
ship will be moving at different velocities, does come into play in resolving any paradoxes. As a general rule, when people create relativity paradoxes, apparent contradictions, there's usually something hidden which violates causality in some way (for instance, infinitely rigid sticks for which the far end moves at the exact same instant that the near end is moved).

BTW, one small nitpick: the number is 186,000, not 189,000.

I'm wondering about how time dilation works. I don't understand when things appear to speed up and down. The idea was to have some time-based event occur that is measurable and visible to an observer. Treat the spaceship as a point.

A red flag is going up in my mind: It's not a point, and the physical length may make a difference in what you're about to say. But let's reserve judgement...

D is the duration of a measurable, timed event that has a value of 1 when the ship is at rest relative to the observer. The light/mirror thing is an artificial construct to create such a measurable event, since it is based on a handy constant that doesn't require any masses moving relative to each other.

I'm confused about the setup, and one of the things that makes a big difference is how you measure D.

Here's how things might be phrased when you're being careful about stating the conditions: "Two clocks are synchronized and then separated and moved to the positions of the source and the mirror. When the light leaves the source, a time signal is sent from the source. When the light arrives at the mirror, another signal is sent from the mirror. How far apart is the time of arrival of these two signals at the observer? What is the difference between the timestamps sent in the two signals?"

Here are at least two important undefined things in your setup:
(1) Which direction is the motion, relative to light-mirror axis and position of observer?

(2) How does the observer measure what time the light left the (distant) source and what time it arrives at the (distant) mirror?

You can't just talk about things happening at exactly the same time at physically separated places, because different observers won't agree on which events are simultaneous.

I can say some general things, though. People have already mentioned the relativistic Doppler effect and that will come into play here. There are two parts to relativistic doppler: One is the standard doppler, which causes the times between events to appear shorter when an object is getting closer (blue shift) and longer when the object is receding (red shift). The other part is a smaller time dilation effect, which causes times to appear shorter and is independent of direction of motion.

When an object is moving perpendicular to the line of sight, there is no classical doppler, just the time dilation effect. This is called "transverse doppler", and is a purely relativistic prediction.

politas
10th March 2005, 04:20 PM
Originally posted by rppa
[QUOTE]Originally posted by politas
[B]The size of the ship isn't really important here, at least, not for what I'm trying to find out.

But such things as the fact that the front and back of the
ship will be moving at different velocities, does come into play in resolving any paradoxes. As a general rule, when people create relativity paradoxes, apparent contradictions, there's usually something hidden which violates causality in some way (for instance, infinitely rigid sticks for which the far end moves at the exact same instant that the near end is moved).

BTW, one small nitpick: the number is 186,000, not 189,000.

Whoops. I really didn't want to get into the physics of the moving object for this. Fascinating as it is, what I'm trying to do is get a grasp on the individual effects, before trying to understand how they interact.

A red flag is going up in my mind: It's not a point, and the physical length may make a difference in what you're about to say. But let's reserve judgement...

D is the duration of a measurable, timed event that has a value of 1 when the ship is at rest relative to the observer. The light/mirror thing is an artificial construct to create such a measurable event, since it is based on a handy constant that doesn't require any masses moving relative to each other.

I'm confused about the setup, and one of the things that makes a big difference is how you measure D.

Here's how things might be phrased when you're being careful about stating the conditions: "Two clocks are synchronized and then separated and moved to the positions of the source and the mirror. When the light leaves the source, a time signal is sent from the source. When the light arrives at the mirror, another signal is sent from the mirror. How far apart is the time of arrival of these two signals at the observer? What is the difference between the timestamps sent in the two signals?"

Ok, maybe you could phrase it that way, but then I'm no longer sure what's being measured.

Here are at least two important undefined things in your setup:
(1) Which direction is the motion, relative to light-mirror axis and position of observer?

perpendicular

(2) How does the observer measure what time the light left the (distant) source and what time it arrives at the (distant) mirror?

By tracking it very carefully with a really big telescope? :-)

You can't just talk about things happening at exactly the same time at physically separated places, because different observers won't agree on which events are simultaneous.

I can say some general things, though. People have already mentioned the relativistic Doppler effect and that will come into play here. There are two parts to relativistic doppler: One is the standard doppler, which causes the times between events to appear shorter when an object is getting closer (blue shift) and longer when the object is receding (red shift). The other part is a smaller time dilation effect, which causes times to appear shorter and is independent of direction of motion.

Ok, so does the doppler effect act in much the same way as it does for sound?

When an object is moving perpendicular to the line of sight, there is no classical doppler, just the time dilation effect. This is called "transverse doppler", and is a purely relativistic prediction.

I think my set up was flawed for what I'm trying to understand. How about we forget about the enormous spaceship and try strobe lights?

Take two strobe lights, A and B, each producing a flash once per second. Start at time T0. Accelerate A away from B until time T1, then towards B until they are at rest relative to each other, at time T2. Continue accelerating A towards B until time T3, then accelerate A away from B until A and B are back together at time T4. Acceleration has a constant scalar value throughout the trip, only changing direction.
Scalar value of acceleration is such that at time T4, A has flashed half as many times as B.

If D1 is the duration between flashes emitted by A, as viewed from B, and D2 is the duration between flashes emitted by B, as viewed from A, how do D1 and D2 change throughout the journey?

RussDill
10th March 2005, 04:33 PM
Originally posted by politas

Take two strobe lights, A and B, each producing a flash once per second. Start at time T0. Accelerate A away from B until time T1, then towards B until they are at rest relative to each other, at time T2. Continue accelerating A towards B until time T3, then accelerate A away from B until A and B are back together at time T4. Acceleration has a constant scalar value throughout the trip, only changing direction.
Scalar value of acceleration is such that at time T4, A has flashed half as many times as B.

If D1 is the duration between flashes emitted by A, as viewed from B, and D2 is the duration between flashes emitted by B, as viewed from A, how do D1 and D2 change throughout the journey?

However, in the twin paradox, general relativity makes a back seat, and i think part of your question relates directly to general relativity. I think general relativity would just slow time further for the accelerator in relation to the one that does not accelerate.

[edited to add: I think the answers depend on what the rate of acceleration is. If you are just traveling away from a strobe source, the beats will slow down. If you are accelerating away from something, the beats will speed up. So, which net effect they add up to depends.

Ziggurat
10th March 2005, 04:43 PM
Originally posted by RussDill
However, in the twin paradox, general relativity makes a back seat, and i think part of your question relates directly to general relativity. I think general relativity would just slow time further for the accelerator in relation to the one that does not accelerate.

General relativity has nothing to do with any of this. You don't need it unless you deal with gravity, and nobody brought up gravity. Special relativity can handle acceleration quite fine on its own.

RussDill
10th March 2005, 04:51 PM
Originally posted by Ziggurat
General relativity has nothing to do with any of this. You don't need it unless you deal with gravity, and nobody brought up gravity. Special relativity can handle acceleration quite fine on its own.

acceleration and gravity are equivelent.

politas
10th March 2005, 04:58 PM
Originally posted by RussDill

Ok, the diagram on that page shows this big gap between the simultaneity planes for the stationary twin. What happens in between those times? What is seen?

Atlas
10th March 2005, 05:11 PM
I've got a question too for you smart guys.

If the ship is at rest I think a strobe goes straight and hits the mirror 1 light second away.

But if the ship is traveling at c would a laser strobe have to lead its target. That is, shoot along the hypotenuse of a right triangle to where the refector target will be in, what would it be, the square root of 2 light seconds?

Does that calculation need to be part of the math to compute the answer?

RussDill
10th March 2005, 05:11 PM
Originally posted by politas
Ok, the diagram on that page shows this big gap between the simultaneity planes for the stationary twin. What happens in between those times? What is seen?

Basically, if you had a side view from the stationary point (B), as it's headed out, you'd see strobe spaceing like this:

B | | | | | | A->

So both parties would be receiving strobes at intervals longer than once per second.

When A turns around, you have a situation like this (strobes from A, B's frame of reference)

B | | | |||||A<-

As you can see, they are stacking up. The diagram on the wikipedia page represents the three frames of reference. It is trying to show the two different paths a and b traveled through spacetime to arrive at their final destination.

RussDill
10th March 2005, 05:15 PM
Originally posted by Atlas
I've got a question too for you smart guys.

If the ship is at rest I think a strobe goes straight and hits the mirror 1 light second away.

But if the ship is traveling at c would a laser strobe have to lead its target. That is, shoot along the hypotenuse of a right triangle to where the refector target will be in, what would it be, the square root of 2 light seconds?

Does that calculation need to be part of the math to compute the answer?

first of all, if the ship is traveling at c, you are screwed anyway, because by the time you see it, it's too late to fire a laser in order to hit it.

Second, I assume in this example, the ship is traveling past you, ie:

ship->

you

You need to do some math to calculate its current position and trajectory based on two observations. Once you do that, you can calculate where it would really be, since light takes time to reach you. Then, fire a laser beam so it intersects with the path of the ship, taking into account, again, the speed of light.

In this case, you don't need any relativity, just the speed of light.

Atlas
10th March 2005, 05:51 PM
/|
/ |
/ |
<------------------< |------------------
\ | <=== direction
\ | speed = c
\|
mirror>- 1 light second to near wing tip
2 light seconds to far wing tip

O
me

I was thinking something like this. The original ship was 1 light second across. If the blink occurs where its right in my line of sight I'll see it 2 seconds after the flash but it will have missed the mirror which has moved forward by 1 light second by the time the light reaches it.

The flash would have to be pointed forward at 45 degrees for it to hit the mirror if it were traveling at c. Am I thinking about that right?

69dodge
10th March 2005, 11:33 PM

Stimpson J. Cat
11th March 2005, 03:23 AM
Originally posted by Atlas
I've got a question too for you smart guys.

If the ship is at rest I think a strobe goes straight and hits the mirror 1 light second away.

But if the ship is traveling at c would a laser strobe have to lead its target. That is, shoot along the hypotenuse of a right triangle to where the refector target will be in, what would it be, the square root of 2 light seconds?

Does that calculation need to be part of the math to compute the answer?
It's not just a part of the math to compute the answer, in effect, it is the math for computing the answer. That is, you can derive the formulas for time dilation and length contraction from simple geometry, using the assumption that the speed of light is the same for all observers.

Consider the following triangle:

***
*** *
z *** *
*** * y
*** *
*** *
*********************
x

Here x is the distance traveled by the light from the POV of the person on the ship, y is the distance traveled by the ship, as measured by the stationary observer, and z is the distance traveled by the light, as measured by the stationary observer. This gives us

z^2 = x^2 + y^2, or equivalently x^2 = z^2 - y^2

If we let t be the time as measured by the stationary observer, and t' be the time as measured by the person on the ship, then this gives us

(ct')^2 = (ct)^2 - (vt)^2

where v is the velocity of the ship. Solving for t' we get

t' = t sqrt(1-(v/c)^2)

which is the equation for time dilation. Keeping in mind that both observers agree on the relative velocity v, this means that

y' = y sqrt(1-(v/c)^2)

where y' is the distance traveled by the ship from the POV of the person on the ship. Since less time passes for the person on the ship, the distance he has traveled must be shorter by the same factor.

Dr. Stupid

Ziggurat
11th March 2005, 06:50 AM
Originally posted by RussDill
acceleration and gravity are equivelent.

No. Acceleration is equivalent to a UNIFORM gravitational field. Special relativity can handle a uniform gravitational field exactly the same as it can handle acceleration. You're just constantly changing reference frames. However, since every gravitational field of any interest is NOT uniform, you cannot treat gravity globally as a simple acceleration, and special relativity falls apart. Two objects sitting on opposite sides of a planet cannot be treated as if they are accelerating away from each other when they clearly aren't. It is this nonuniformity of gravity which requires general relativity. Absent that, you can do everything you want with special relativity.

rppa
11th March 2005, 08:17 AM
How does the observer measure what time the light left the (distant) source and what time it arrives at the (distant) mirror?

By tracking it very carefully with a really big telescope? :-)

That's perfectly legitimate, but you have to realize that you don't see events when they happen. You see them only when some sort of signal reaches you. In this case it's the light from the event, which travels to you at speed c and arrives at your telescope d/c seconds after the event happened.

If you further model some way of knowing d, then you can backtime your observation to say "from my point of view, that event happened at distance d, at time d/c ago". Now you have a way of assigning a time in your frame, to something that happened far away.

The point is that in all relativistic discussions, paradoxes are created by being vague about the information being measured (or by postulating faster than light communication), and paradoxes are resolved by acknowledging that information can't get from A to B faster than c.

There are two parts to relativistic doppler: One is the standard doppler, which causes the times between events to appear shorter when an object is getting closer (blue shift) and longer when the object is receding (red shift).

Ok, so does the doppler effect act in much the same way as it does for sound?

What I called the "standard doppler" above is exactly the same as the doppler effect for sound, except that in sound you can have a situation where the observer is moving relative to the medium, changing the apparent sound speed. That can't happen to light.

But the analysis of what happens to signals leaving a moving source is exactly the same as for sound. For a source moving toward you, emitting signals once per second, later signals travel less distance and get to you less than one second after earlier signals.

But there's this time dilation effect also. It's quite small. The size of the sound doppler effect, and the first-order light doppler, is v/c. The size of the time-dilation effect is v^2/c^2, much smaller. You don't see it unless there is no first-order effect (because the motion is perpendicular to line of sight). In that case for sound, there's no effect at all. Zero. But for relativity, there is a transverse doppler.

Take two strobe lights, A and B, each producing a flash once per second. Start at time T0. Accelerate A away from B until time T1, then towards B until they are at rest relative to each other, at time T2. Continue accelerating A towards B until time T3, then accelerate A away from B until A and B are back together at time T4. Acceleration has a constant scalar value throughout the trip, only changing direction.
Scalar value of acceleration is such that at time T4, A has flashed half as many times as B.

If D1 is the duration between flashes emitted by A, as viewed from B, and D2 is the duration between flashes emitted by B, as viewed from A, how do D1 and D2 change throughout the journey?

As somebody has already said, this is the twin paradox. It's actually an easy way to show the twins don't have identical experiences.

A is the accelerating twin. What B sees is that the flashes get farther and farther apart as A recedes. As A starts to decelerate, the corresponding flashes get closer together, but they're still red-shifted: D1 > 1.

When A turns around, the corresponding pulses will be blue-shifted, D1<1. But A is very far away. Those pulses will take awhile to get back to B. By the time B sees the pulses change from D1 > 1 to D1 < 1, A is already enroute back.

Meanwhile, A also sees a red-shift D2>1 as he recedes, and a blue-shift D2<1 when he turns around. But he sees the change as soon as he turns around, because the pulses that are red-shifted/blue-shifted are already nearby, have already travelled out to where he is.

RussDill
11th March 2005, 12:08 PM
Originally posted by Ziggurat
No. Acceleration is equivalent to a UNIFORM gravitational field. Special relativity can handle a uniform gravitational field exactly the same as it can handle acceleration. You're just constantly changing reference frames. However, since every gravitational field of any interest is NOT uniform, you cannot treat gravity globally as a simple acceleration, and special relativity falls apart. Two objects sitting on opposite sides of a planet cannot be treated as if they are accelerating away from each other when they clearly aren't. It is this nonuniformity of gravity which requires general relativity. Absent that, you can do everything you want with special relativity.

The two reference frames are experiencing different strengths of gravitational fields, yes?

RussDill
11th March 2005, 12:15 PM
I was thinking something like this. The original ship was 1 light second across. If the blink occurs where its right in my line of sight I'll see it 2 seconds after the flash but it will have missed the mirror which has moved forward by 1 light second by the time the light reaches it.

The flash would have to be pointed forward at 45 degrees for it to hit the mirror if it were traveling at c. Am I thinking about that right?

Don't forget that photons have momentum, for conservation of momentum to occur, the photons leave the strobe at what seems to be a 45 degree angle.

Ziggurat
11th March 2005, 12:58 PM
Originally posted by RussDill
The two reference frames are experiencing different strengths of gravitational fields, yes?

Which two frames are you refering to? The two that I gave in my example, with two objects sitting on opposite sides of a planet? In that case the strength of the gravitational field is identical, only the direction changes. But in any case, the distinction between special relativity and general relativity is still that the former cannot handle non-uniform gravitational fields. That is its ONLY limit compared to general relativity.

RussDill
11th March 2005, 01:23 PM
Originally posted by Ziggurat
Which two frames are you refering to? The two that I gave in my example, with two objects sitting on opposite sides of a planet? In that case the strength of the gravitational field is identical, only the direction changes. But in any case, the distinction between special relativity and general relativity is still that the former cannot handle non-uniform gravitational fields. That is its ONLY limit compared to general relativity.

I'm refering to the stationary twin, vs the twin accelerating away. Certainly, there are really 6 frames of reference in the twin paradox.

a) stationary twin
b) twin accelerating away (Actualy, a contium of reference frames)
c) twin moving away at a constant velocity
d) twin turning around (again, a continum of reference frames)
e) twin moving back at a constant velocity
f) twin slowing to a stop

b, d, and f, would seem to require integration, and I think general relativity, set me straight here.

Stimpson J. Cat
11th March 2005, 01:57 PM
RussDill,

I'm refering to the stationary twin, vs the twin accelerating away. Certainly, there are really 6 frames of reference in the twin paradox.

a) stationary twin
b) twin accelerating away (Actualy, a contium of reference frames)
c) twin moving away at a constant velocity
d) twin turning around (again, a continum of reference frames)
e) twin moving back at a constant velocity
f) twin slowing to a stop

b, d, and f, would seem to require integration, and I think general relativity, set me straight here.

They do require integration, but special relativity is sufficient. Anyway, when it comes to the Twin Paradox, the acceleration is kind of a red herring. After all, you can always take the limit in which the period of acceleration is so short compared to the total time of the trip that it makes no difference.

The appearance of the paradox comes from the misconception that the situation is symmetric for the two twins. Of course it is not. But the breaking of the symmetry is not in the fact that one observer is accelerating and the other isn't. After all, from the other frame of reference the opposite is true. The break in symmetry comes from the fact that the two evens which mark the beginning and ending of the experiment are in the same inertial frame that one of the twins is in for the duration of the experiment, while the other twin goes from that frame of reference to a different one, and back again.

If you think about it, the question "how much time passes between event A and event B for the two twins?" is only well defined for some specific reference frame. In this case, the frame which the two events are defined in, is the frame of the twin on Earth. You could do the experiment from any frame, and even try to make the experiment symmetric between the two twins by observing the whole thing from a 3rd reference frame. In that case, the relative ages of the two twins at the end of the experiment would depend on their velocities relative to that frame. For example, if we send both twins away from the Earth in opposite directions, and have them come back after going the same distance at the same speed, then they will be the same age when they return, even though from each of their points of view the other twin was moving relative to them. In this case the experiment is symmetric, so they both age the same amount and there is no paradox. The paradox only appears to be there when the situation is not symmetric, but superficially appears to be.

Dr. Stupid

Ziggurat
11th March 2005, 02:15 PM
Originally posted by RussDill
I'm refering to the stationary twin, vs the twin accelerating away. Certainly, there are really 6 frames of reference in the twin paradox.

a) stationary twin
b) twin accelerating away (Actualy, a contium of reference frames)
c) twin moving away at a constant velocity
d) twin turning around (again, a continum of reference frames)
e) twin moving back at a constant velocity
f) twin slowing to a stop

b, d, and f, would seem to require integration, and I think general relativity, set me straight here.

b,d, and f all require integration to do correctly, and the math can get considerably more complicated than the simple Lorenz transformations people usually deal with (although they're still the basis for the calculus involved. But none of those cases requires general relativity.

I think I might see what's tripping you up. First, take it as a complete given that special relativity can handle acceleration, any kind of acceleration, including everything you need to solve the twin problem. And it can handle acceleration because nothing about acceleration breaks special relativity (if it did, the theory would not have gotten much attention). But gravity does break special relativity, in a way that acceleration does not.

The acceleration = gravity is in some senses secondary: what is perhaps more fundamental is FALLING in a uniform gravitational field is exactly equivalent to being in an inertial reference frame. If you let your coordinate system fall, it is an inertial coordinate system. But of course, you can't make such a reference frame globally with special relativity once the gravitational field is non-uniform, because different parts of your reference frame then need to "fall" at different rates and in different directions, so it "falls" apart (ha ha!... sorry). In the twin problem, we've got various inertial reference frames, and accelerations which constantly change WHICH inertial reference frame any one twin is in at any one moment. But each of the infinite possible inertial reference frames we can choose is still itself constant, and they all maintain their relative velocities at all times, so special relativity has no problems with any of this. In short: acceleration is just changing inertial reference frames, which special relativity has no problem with, but non-uniform gravity means you can no longer even describe an inertial reference frame with special relativity on anything more than a local level.

So falling reference frames are inertial reference frames. That means if you're standing still on the surface of the earth, you're constantly accelerating compared to a reference frame that is falling. Hence gravity = acceleration, but only to the extent that locally the gravitational field is pretty damned close to uniform. And that's actually always the case locally, if you look at a small enough patch of space, except in the case of singularities, just as the surface of a sphere looks like a plane if you zoom up close enough, but what's true for a plane (parallel lines never cross) is no longer true for your curved surface on a global scale. The relationship of general relativity to special relativity is much the same: aside from singularities, special relativity still describes the tangent space for general relativity.

Hope that clarifies things a bit.

Atlas
11th March 2005, 02:17 PM
Originally posted by RussDill
Don't forget that photons have momentum, for conservation of momentum to occur, the photons leave the strobe at what seems to be a 45 degree angle. Now there's a crazy idea. I wouldn't have thought of that. I still am wondering though... because photons are massless would they really have an X directional component if they were being aimed in the Y coordinate.

I'm asking. I learn something new each time these ideas are presented here.

Just one more comment on the diagram I presented earlier. I mentioned that if the craft was moving at speed c along the x coordinate that I thought the strobe should be aimed forward at 45 degrees to lead the target. But at speed c the hypotenuse of the triangle is longer than the the distance along the x coordinate. With the craft flying at speed c there is no angle at which a flash could be projected such that it would be received at the opposite wing tip.

Is that true or is there some relativistic weirdness that makes me wrong once more?

(edit: I know at any speed the hypotenuse of a right triangle is longer than either leg but what I meant was that because of that longer distance the light traveling at top speed c will miss the target which is traveling a shorter distance at the same speed c. )

RussDill
11th March 2005, 03:10 PM
Originally posted by Ziggurat
Hope that clarifies things a bit.

Doesn't time dialation occur in an accelerating reference frame? (vs a non accelerating reference frame)? How does special relativity address this dialation?

RussDill
11th March 2005, 03:12 PM
Originally posted by Atlas

Just one more comment on the diagram I presented earlier. I mentioned that if the craft was moving at speed c along the x coordinate that I thought the strobe should be aimed forward at 45 degrees to lead the target. But at speed c the hypotenuse of the triangle is longer than the the distance along the x coordinate. With the craft flying at speed c there is no angle at which a flash could be projected such that it would be received at the opposite wing tip.
I

right, if some hypothetical spaceship was traveling at c, then you could only hit targets behind you. Of course, spaceships can't travel at c, so you don't have this problem. (something similar does occur in a blackhole though).

Ziggurat
11th March 2005, 03:49 PM
Originally posted by RussDill
Doesn't time dialation occur in an accelerating reference frame? (vs a non accelerating reference frame)? How does special relativity address this dialation?

ds^2 = dx^2 + dy^2 + dz^2 - dt^2

That's your metric. If there's no acceleration, all you need are start and end coordinates, and you can calculate time along a path with very simple algebra. If the path is curved (acceleration), then you need to do some integration, and that can get messy very fast, but it's still just special relativity, regardless of how twisty you make that trajectory. The difference between calculating it for an accelerating trajectory versus a non-accelerating trajectory is simply the difference between calculating the length of a curved line and a straight line. In this sense, there's nothing special about time dilation for an accelerating object compared to a non-accelerating object.

The reason "acceleration" comes into the twin problem is that one of the twins has a curved trajectory, while one of them doesn't (the form of the curve doesn't matter). In Euclidean geometry, the shortest distance between two points is a straight line. In special relativity, the longest, rather than shortest, interval between two events is a straight line (play around with the metric a little bit, draw out line segment paths and calculate distances and you'll see what I mean). The earth-bound twin experiences no acceleration, and so traces out a straight line in space-time, while the traveling twin traces out a curved (or segmented, however you want to deal with it) trajectory. So people like to say that the time difference is because one of the tins accelerates, and in a sense that's correct. But it's not because acceleration has to be treated fundamentally any differently when calculating time dilation, any more than the length of curves in Euclidean geometry is any different than the length of straight lines. The calculus is still the same.

RussDill
11th March 2005, 03:57 PM
Again, from what I understand, time dialation occurs to a reference frame in a gravitational well (not free falling). On earth, it is the same time dialation that would occur to a reference frame accelerating at 9.8m/s^2. Is this not correct?

Ziggurat
11th March 2005, 07:43 PM
Originally posted by RussDill
Again, from what I understand, time dialation occurs to a reference frame in a gravitational well (not free falling). On earth, it is the same time dialation that would occur to a reference frame accelerating at 9.8m/s^2. Is this not correct?

Ah, now I see what you're getting at.

The difficulty though is you always need to ask, time dilation compared to what? With the earth, you could ask, for example, time dilation compared to a distant spot far from the earth. But the answer depends on the entire gravitational well geometry, not just the local gravitational field. If you hollowed out a spot at the center of the earth, where gravity was zero, the time dilation there would still be "stronger" than the time dilation at the surface of the earth. To get the right answer, there's a sort of integration process you need to carry out along the path - which also means that if your path length is essentially zero, there's no relevant time dilation regardless of your acceleration. And you can't pick a path length between two inertial reference frames in special relativity, since all inertial reference frames are infinite. It never makes sense to talk about the time dilation due to a given acceleration, because it doesn't exist. There's no special time dilation needed to account for acceleration in special relativity.

The reason gravitational time dilation pops up has to do with making a reference frame that doesn't move with respect to your gravitational source, and hence is not really an inertial reference frame. It gets messy, and I don't think it's really worth getting into now.

69dodge
12th March 2005, 10:07 PM
Originally posted by Stimpson J. Cat
The appearance of the paradox comes from the misconception that the situation is symmetric for the two twins. Of course it is not. But the breaking of the symmetry is not in the fact that one observer is accelerating and the other isn't. After all, from the other frame of reference the opposite is true.Well. The other frame of reference is not inertial, so it doesn't count.

That's what people mean when they say that one observer is accelerating: he is accelerating relative to an inertial reference frame. Obviously, he is not accelerating relative to the reference frame in which he is motionless.The break in symmetry comes from the fact that the two evens which mark the beginning and ending of the experiment are in the same inertial frame that one of the twins is in for the duration of the experiment, while the other twin goes from that frame of reference to a different one, and back again.Events aren't in one reference frame or another; they're just events. The difference between the twins is that one doesn't accelerate relative to an inertial reference frame and the other one does.If you think about it, the question "how much time passes between event A and event B for the two twins?" is only well defined for some specific reference frame. In this case, the frame which the two events are defined in, is the frame of the twin on Earth.Huh? The travelling twin is just as involved in the events of his departure and arrival as the Earthbound twin is. More involved, arguably. Events aren't defined in any particular reference frame. They can be referred to any reference frame.

The time between two events does depend on which reference frame that time is measured in. Between the departure and arrival events, more time passes in the inertial reference frame in which the Earthbound twin is motionless than in the noninertial reference frame in which the travelling twin is motionless.You could do the experiment from any frame, and even try to make the experiment symmetric between the two twins by observing the whole thing from a 3rd reference frame. In that case, the relative ages of the two twins at the end of the experiment would depend on their velocities relative to that frame.Observing the whole thing from a different reference frame won't change the relative ages of the twins when they reunite. Every observer will agree that the twin who left Earth is younger on his return than the twin who stayed on Earth, regardless of the observer's own motion.For example, if we send both twins away from the Earth in opposite directions, and have them come back after going the same distance at the same speed, then they will be the same age when they return, even though from each of their points of view the other twin was moving relative to them. In this case the experiment is symmetric, so they both age the same amount and there is no paradox.This experiment is symmetric, but it's a different experiment, not merely the same experiment viewed from a different reference frame. Before, one twin's motion was inertial; now, neither's is.

epepke
13th March 2005, 01:13 AM
Originally posted by politas
The size of the ship isn't really important here, at least, not for what I'm trying to find out. I'm wondering about how time dilation works. I don't understand when things appear to speed up and down.

In that case, you're overcomplicating the problem.

Instead of picking D as the time for the light to get to the mirror, pick D' as the time for the light to get to the mirror and back again, which would be 2 seconds. Then it's easy. D>2 as seen by an observer in all cases other than rest (provided that the measurements take into account the change in distance that the spaceship travels with respect to the observer.)

The reason that it's overcomplicating is that in the case where the path of the light is even partially along the axis of travel of the spaceship, the time for light to get to the mirror will be different for another observer than the time it takes to get back.

As for how time dilation works, it's extremely simple. Have the light going at right angles to the direction of travel of the ship. At rest WRT an observer, the light will appear to bounce side to side in the spaceship. When the spaceship is moving, the light will go in a zig-zag pattern. A zig-zag pattern is longer. Because the speed of light is constant, if the light is going in a longer (zig-zag) pattern, it will take longer to get there. Therefore, that clock made with the light will appear to go slower.

All clocks have to go slower, too, because if not, you could determine something's absolute motion by comparing clocks, which would violate relativity.

epepke
13th March 2005, 01:21 AM
Originally posted by RussDill
Doesn't time dialation occur in an accelerating reference frame? (vs a non accelerating reference frame)? How does special relativity address this dialation?

Basic time dilation occurs in a moving reference frame. Another aspect of time dilation occurs in an accelerating reference frame.

Special relativity can be used to handle lots of accelerating cases, but historically, this wasn't done much until general relativity. And the so-called "twin paradox" can be resolved with SR ignoring the acceleration entirely.

As for your other question, the acceleration-equivalent time effects of gravitation work out to be exactly what you would get if you just dropped a clock (no air resistance, of course). So as long as you know what speed the clock is going when it passes you, you can work it out without using GR in lots of cases.

Stimpson J. Cat
13th March 2005, 02:18 AM
69Dodge,

I think you are misunderstanding what I am saying.

Consider this alternative formulation of the "paradox":

A spaceship is coming from a very distant star towards Earth at 0.8c. An observer on Earth figures that the ship will pass very close to Earth, so he sets up a marker in space 4 light minutes away in the direction the ship is coming from, and 4 light minutes away in the direction the ship will moving away from the Earth in. Note that there will be no acceleration in this experiment.

He then looks at a clock in the alien ship, using a telescope, and the alien does likewise.

When he sees the ship pass marker A, he looks at the clock in the ship, and the clock in his office. Let's say his clock says 12:04PM, and the ship clock says 2:00PM. He then subtracts 4 minutes from the time on his clock, because he knows it took 4 minutes for the light from marker A to reach him. So his correct starting time is 12:00.

When the ship passes marker B, he looks again. This time his clock will clearly say 12:14PM (12:10, when he subtracts the 4 minute light-travel time), because at 0.8c it will take 10 minutes to travel 8 light minutes. Because of time dilation the spaceship clock will say 2:06PM, because (1-sqrt(0.8)^2)=0.6.

But what will the alien see? From his POV he is stationary. An inertial frame containing both markers, and the Earth, is moving past him at 0.8c. Because of length contraction, he will see the distance from marker A to marker B as being not 8 light minutes, but instead 4.8 light minutes. At 0.8c he will therefore think the trip takes 6 minutes. And of course he will agree with observer on Earth as to what times his clock says when he passes the markers (2:00PM and 2:06PM).

But he can not agree with observer on Earth that the amount of time which passes on Earth is 10 minutes. On the contrary, because of time dilation, he must say that only 3.6 minutes pass on Earth. Thus we have an apparent paradox, similar to that of the twin paradox, but with no acceleration to save it (as I said, though, acceleration is not the issue in the twin paradox either).

The resolution comes from the fact that, while both observers agree on what the alien's clock said at the beginning and ending of the experiment, they do not agree on what the Earth observer's clock said. Remember that the way the Earth observer knows what the alien's clock said, is by looking at the alien's clock when he passes the marker. The alien must do the same.

So the alien crosses marker A and is looking at the Earth. When the marker passes him, he knows that the Earth is 2.4 light minutes away (length contraction). Of course, at that time he sees the Earth further away, but he can correct for the time it takes the light to reach him. This correction tells him that the light emitted by the Earth when the marker passed him will reach him in 2.4 minutes, just 0.6 minutes before the Earth reaches him. So at that time, the Earth is only 0.48 light minutes away.

So now we can figure out what his clock will say. When that light from Earth reaches the alien, in the frame of Earth, the alien will be 0.8 light minutes away. So that means that the light from Earth currently reaching the alien will have departed only 0.8 minutes ago. Furthermore, in the frame of the Earth, the alien will have travelled 3.2 light minutes since the beginning of the experiment, which means that 4 minutes have already passed. Thus the clock on Earth will say 12:04, which means that the alien will calculate that the Earth clock said it was 12:03.2 when he passed marker A.

Likewise when marker B passes the alien, the Earth will be 2.4 light minutes away. Again, it will take 2.4 minutes for that light to reach the alien. By that time the earth will have moved away another 1.92 light minutes, giving a total distance of 4.32 light minutes (in the alien's frame). In the Earth frame, the alien will therefore be 7.2 light minutes from the Earth. This means that the light then reaching the alien left the Earth 7.2 minutes ago. And since, in the Earth frame, the alien passed marker B at 12:10, and has travelled 3.2 light minutes since then, the current time will be 12:14. Subtracting 7.2 minutes we get 12:06.8.

Thus from the alien's POV, the clock on Earth said 12:03.2 when the experiment started, and 12:06.8 when it finished. Total time: 3.6 minutes, just as time dilation predicts.

The point of all this is that the apparent contradiction arises from the problem that the two observers cannot agree on which events occurred simultaneously. They both agree that the start of the experiment occurred simultaneously with the alien's clock saying 2:00, and that the end of the experiment occurred simultaneously with the alien's clock saying 2:06. But the alien says that the beginning occurred simultaneously with the Earth clock saying 12:03.2, and the Earth observer says that the beginning occurred simultaneously with the Earth clock saying 12:00. Likewise they disagree on when the experiment ended in the Earth frame.

Again, there is no acceleration here. In the twin paradox, acceleration is not the issue either. In that frame, they can agree on what the clocks both said at the beginning and ending of the experiment, because they are both in the same frames at those times. Acceleration is only a factor in that one or both of them must accelerate for motion to occur. But the amount of time which passes during acceleration can be negligible in both frames. In fact, you could have the entire Earth-star frame do the accelerating instead of the ship, and it wouldn't make any difference. The twin is younger not because he did the accelerating, but because the beginning and end points of the trip are the Earth and star, and that distance is always longest in the frame of the Earth and star. Do the math, and you'll see. You get exactly the same answer either way.

Dr. Stupid

69dodge
13th March 2005, 06:24 AM
Originally posted by Stimpson J. Cat
[...]

The point of all this is that the apparent contradiction arises from the problem that the two observers cannot agree on which events occurred simultaneously. They both agree that the start of the experiment occurred simultaneously with the alien's clock saying 2:00, and that the end of the experiment occurred simultaneously with the alien's clock saying 2:06. But the alien says that the beginning occurred simultaneously with the Earth clock saying 12:03.2, and the Earth observer says that the beginning occurred simultaneously with the Earth clock saying 12:00. Likewise they disagree on when the experiment ended in the Earth frame.

Again, there is no acceleration here.Right. Relativity of simultaneity at a distance.In the twin paradox, acceleration is not the issue either. In that frame, they can agree on what the clocks both said at the beginning and ending of the experiment, because they are both in the same frames at those times. Acceleration is only a factor in that one or both of them must accelerate for motion to occur. But the amount of time which passes during acceleration can be negligible in both frames. In fact, you could have the entire Earth-star frame do the accelerating instead of the ship, and it wouldn't make any difference. The twin is younger not because he did the accelerating, but because the beginning and end points of the trip are the Earth and star, and that distance is always longest in the frame of the Earth and star. Do the math, and you'll see. You get exactly the same answer either way.Star? What star? One twin stays on Earth; the other leaves and returns. The trip starts and ends on Earth. There's no distance to mess up simultaneity. Both at the beginning of the trip and at the end, the two twins are in the same place so they can look directly at each other's clocks. Yet their clocks still disagree.

From the point of view of the travelling twin, a great deal of Earth time passes while he's turning around (i.e., accelerating)---more than enough time to make up for all the rest of his trip, during which, again from his point of view, Earth's clocks go slower than his.

Stimpson J. Cat
13th March 2005, 07:37 AM
69dodge,

Star? What star? One twin stays on Earth; the other leaves and returns. The trip starts and ends on Earth.
Sorry, I should have been more specific. I was referring to the standard formulation of the twin paradox, where the twin travels to some nearby star and back again. Substitute "turning point" for star.

There's no distance to mess up simultaneity. Both at the beginning of the trip and at the end, the two twins are in the same place so they can look directly at each other's clocks. Yet their clocks still disagree.
There is a distance. The distance to whatever point the twin travels to, before he turns around and comes back. In fact, since you can synchronize clocks on the Earth and whatever point in space he turns around at, you can break the experiment into two parts. The first half of the trip takes less time for the twin, as does the second half. When the twin stops (relative to Earth) at the turning point, he can check the clock there which he knows is synchronized (in the frame of the Earth) to the clock on Earth. Sure enough, that clock will say more time has passed than has passed for the twin. The same thing happens on the return half of the trip.

From the point of view of the travelling twin, a great deal of Earth time passes while he's turning around (i.e., accelerating)---more than enough time to make up for all the rest of his trip, during which, again from his point of view, Earth's clocks go slower than his.
This is incorrect. The accelerating and decelerating time can be made as small as we want, so that it accounts for a negligible portion of the total trip time. For example, we could easily construct the experiment so that only 1 second of time passes (in the Earth frame) during all the combined acceleration and deceleration periods. Clearly the time difference is accumulated over the entire trip, during uniform motion, and not during that brief acceleration period.

Again, it doesn't even matter who does the accelerating. Imagine this experiment:

A huge ruler is built in space, which is 4 light minutes long. One twin is on one end of the ruler. The other in a spaceship next to his twin. He takes off for the other end at 0.8c, stops there and turns around to come back, also at 0.8c. Assume that all acceleration times are negligibly small.

From twin A's POV (on the ruler), the trip takes 10 minutes (8 light minutes / 0.8c). From twin B's POV, the total distance traveled by the ruler during the first part is 2.4 light minutes, and then 2.4 more during the second part, giving a total of 4.8 light minutes, for a total trip time of 6 minutes.

But you can switch things around. Leave twin B stationary, and move the ruler instead. You will get exactly the same result. Twin B will still only experience 6 minutes, while twin A experiences 10, even though twin A did the accelerating.

The reason is simple. The distance traveled is always 8 light minutes from the POV of twin A, and always 4.8 light minutes from the POV of twin B, regardless of who is actually doing the accelerating.

The acceleration is only important in the respect that somebody has to accelerate for the experiment to happen. It is not the case that the time discrepancy accumulates during the acceleration period. Of course, if the acceleration period is not negligible, some time difference will accumulate there too, but it does not all accumulate there.

Dr. Stupid

69dodge
13th March 2005, 11:06 PM
Originally posted by Stimpson J. Cat
But you can switch things around. Leave twin B stationary, and move the ruler instead. You will get exactly the same result. Twin B will still only experience 6 minutes, while twin A experiences 10, even though twin A did the accelerating.This cannot be right. An observer's proper time depends on whether he's dragging a ruler behind him? What if neither A nor B is attached to a ruler? That's the original question anyway. The only difference between the twins is that one accelerates and one doesn't. It obviously doesn't matter which one we call A and which one we call B.

Stimpson J. Cat
14th March 2005, 02:34 AM
69Dodge,

This cannot be right. An observer's proper time depends on whether he's dragging a ruler behind him? What if neither A nor B is attached to a ruler? That's the original question anyway. The only difference between the twins is that one accelerates and one doesn't. It obviously doesn't matter which one we call A and which one we call B.
That is not the only difference, nor is it the relevant difference. The relevant difference is what events mark the beginning and end of the trip, and how they are defined.

Note that in the twin paradox, what you really have are two separate trips. One to the destination, and one back. For both of these trips, the events which mark the beginning and ending of the trip are defined to be when the twin leaves Earth, arrives at some destination, leaves that destination, and arrives back at Earth. Both the Earth and the destination locations are in the same inertial frame (we are assuming that they are not moving relative to each other). In the Earth-destination inertial frame, the distance between the Earth and the destination is the largest it will be in any inertial frame. That is why the twin who spends the duration of the trip in the Earth frame experiences a longer passage of time than the other one. It really has nothing to do with who does the accelerating.

You asked what if neither one is attached to the ruler. I am not sure what you mean. In the original thought experiment there is no "ruler", but the Earth-bound twin is still "attached" to the Earth-destination frame, just as twin A is "attached" to the frame of the ruler in my variant. That is why he experiences more time passing.

Do you mean what happens if both twins are moving relative to the Earth-destination frame? If that is the case, then the results are going to be different. The obvious example being the triplets variant of the experiment. In this case, Triplet A stays on the Earth, and triplets B and C each head off in opposite directions at the same speed, and for the same distance, and come back likewise.

In such a case, both B and C will be younger than A, but they will both be the same age as each other in spite of the fact that they were moving relative to each other. In this case the experiment really is symmetric with respect to B and C, so they must experience the same passage of time. Again the relevant issue is that the events which determine the beginning and endings of the trips are defined in the frame of the Earth, so the distance travelled is longest in that frame.

Dr. Stupid

69dodge
14th March 2005, 04:04 AM
Originally posted by Stimpson J. Cat
But you can switch things around. Leave twin B stationary, and move the ruler instead. You will get exactly the same result. Twin B will still only experience 6 minutes, while twin A experiences 10, even though twin A did the accelerating.I still disagree.

Can you describe this scenario in more detail? What does it mean exactly to move the ruler? If B is stationary and A accelerates, how is that different---apart from simply swapping labels---from when A is stationary and B accelerates?

The proper time between two events is longest for an inertial observer, i.e., one who does not accelerate. So, if one twin leaves the other and returns, he will experience less time than his inertial twin, because he accelerated and his twin didn't. I do not believe you can construct a scenario where this isn't true.

Stimpson J. Cat
14th March 2005, 05:06 AM
69dodge,

Can you describe this scenario in more detail? What does it mean exactly to move the ruler? If B is stationary and A accelerates, how is that different---apart from simply swapping labels---from when A is stationary and B accelerates?
That's just the point. It isn't any different. It does not make any difference whether the ruler moves or the guy in the spaceship does. What matters is that the events which mark the beginning and ending of the trip are when the spaceship and the ends of the ruler reach each other. The length of the ruler is always longest in the frame of the ruler, so a person in the frame of the ruler will always measure the time required for something travelling at a specific speed to cross that distance as being more than the observer which is crossing the distance.

The only difference in the amount of time experienced that the acceleration makes is in the amount of time that passes for each observer during the period of acceleration. And we can make that time as small compared to the total time of the experiment as we want. As I mentioned before, if from the frame of the person on the ruler, the entire acceleration time amounts to only 1 second, then even if no time passed for the accelerating ship at all, that could only account for 1 second difference in total elapsed time. Where does the other 3 minutes and 59 seconds go? Clearly the 4 minute time difference between the 10 minutes experienced by the observer on the ruler, and the 6 minutes experienced by the observer in the spaceship, is accumulated over the course of the entire trip.

The proper time between two events is longest for an inertial observer, i.e., one who does not accelerate. So, if one twin leaves the other and returns, he will experience less time than his inertial twin, because he accelerated and his twin didn't. I do not believe you can construct a scenario where this isn't true.
This is incorrect. Different inertial observers will measure different amounts of time between two events.

And that is really what we have in this thought experiment. Since we can make the acceleration times negligibly small, what we really have are two separate sets of events.

Event 1 is when the spaceship leaves the Earth. Event 2 is when it arrives at the destination. During the time between these events, we have two inertial frames. One is the frame of the Earth and destination. The other is the frame of the spaceship. The distance between these places is the longest in the frame of the Earth and destination. Likewise the length of the spaceship is longest in the frame of the spaceship. Since the distance travelled is longest in the frame of the Earth, and observer in that frame measures the longest duration of time.

Note that for this portion of the experiment, there is no acceleration. This is what I was trying to illustrate with my example of an alien flying past the Earth. Even without any acceleration on the part of any of the participants, the fact remains that the time required for the spaceship to go from the Earth to the destination, will be less in the frame of the spaceship than in the frame of the Earth and destination. Likewise for the trip back.

We can work out how much time passes for each observer during just the periods of uniform motion, and then add on the times which pass during acceleration and deceleration separately. The relative amounts of time which pass for each twin during the acceleration periods will clearly be different for each twin, and will depend on who is accelerating, but the relative amounts of time which pass for each twin during the periods of uniform motion do not, in any way, depend on who did the accelerating to get them to that relative speed.

Do you agree that we can break the thought experiment down into six parts?

1) Acceleration of the ship towards the destination.
2) Travel at uniform speed to the destination.
3) Deceleration of the ship at the destination.
4) Acceleration of the ship towards the Earth.
5) Travel at uniform speed to the Earth.
6) Deceleration at the Earth.

If so, when do you think the time difference between the twins accumulates? I don't see how it could accumulate only during the acceleration periods. If that were the case, the time difference would not even depend on the total trip time. It would depend only on the time spent accelerating. But of course, that isn't the case.

And what about my example of the alien flying past the Earth? In that case, there is no acceleration. And yet the alien experiences less time travelling from the marker to the Earth than the Earth observer does. Clearly this example is identical to having the alien remain stationary in space, while the marker and the Earth fly past. Clearly it makes absolutely no difference how that relative velocity was initially attained. All that matters is that the marker and Earth are both in the same inertial frame. The distance between them is thus greater in the frame of the Earth than in the frame of the alien. Incidentally, this example has been confirmed experimentally by observing the percentage of short-lived particles which reach the Earth after being produced in the upper atmosphere.

Dr. Stupid

RussDill
14th March 2005, 10:31 AM
Originally posted by epepke
Special relativity can be used to handle lots of accelerating cases, but historically, this wasn't done much until general relativity. And the so-called "twin paradox" can be resolved with SR ignoring the acceleration entirely.

I know you can solve the twin paradox and ignore acceleration, but one of the questions of the OP specifically involves acceleration. Heck, in the twin paradox, you could say that the twin accelerates at a rate that makes his time dilation due to acceleration equal to the time dialation that the other twin experiences due to gravity.

Ziggurat
14th March 2005, 12:14 PM
Originally posted by RussDill
I know you can solve the twin paradox and ignore acceleration, but one of the questions of the OP specifically involves acceleration. Heck, in the twin paradox, you could say that the twin accelerates at a rate that makes his time dilation due to acceleration equal to the time dialation that the other twin experiences due to gravity.

There IS no time dilation due to acceleration. I've been trying to tell you that for a while now. You can calculate a sort of pseudo-time dilation from acceleration which is distance-dependent quantity, but it's really the equivalent of coriolis or centripedal forces: it's a way to make a non-inertial reference frame look like an inertial reference frame, but it's a mathematical slight-of-hand. As long as you stay in an inertial reference frame, such time dilation never comes up.

RussDill
14th March 2005, 12:39 PM
Originally posted by Ziggurat
There IS no time dilation due to acceleration. I've been trying to tell you that for a while now. You can calculate a sort of pseudo-time dilation from acceleration which is distance-dependent quantity, but it's really the equivalent of coriolis or centripedal forces: it's a way to make a non-inertial reference frame look like an inertial reference frame, but it's a mathematical slight-of-hand. As long as you stay in an inertial reference frame, such time dilation never comes up.

so...what happened to the whole equivelency of gravitational and accelerating reference frames in GR?

Ziggurat
14th March 2005, 01:52 PM
Originally posted by RussDill
so...what happened to the whole equivelency of gravitational and accelerating reference frames in GR?

Nothing.

Here's the deal: in Newtonian mechanics, if you make your coordinate system rotate, it is not inertial. To account for this, you need to introduce fictitious forces, in particular coriolis and centrifugal forces. But these forces are NOT real.

In special relativity, if you want your reference frame to accelerate, it is NO LONGER an inertial reference frame. To account for this, you need to introduce what LOOKS like a position-dependent time dilation. All it comes from is the fact that as you accelerate, the axis marking your "current" time keeps tilting, shifting what you consider "now" for locations displaced from you. That makes clocks in the direction you're accelerating seem to run slower, and clocks in the opposite direction seem to run faster, than they would simply based on relative speeds. You can treat this like position-dependent time dilation, but it is NOT. Just like coriolis forces, it's completely due to the fact that you're in a non-inertial reference frame, and its strength is completely position-dependent: at the origin there's no effect. And it can be completely calculated with special relativity. But just like fictitious forces, there's no bloody point in doing so if you have an inertial reference frame which is easier to work with. And if you're using an inertial coordinate system, there is NO acceleration-dependent time dilation, period.

Now in GR, you get exactly the same sort of thing: locally, gravity LOOKS like acceleration, but not globally, so you also get what looks like a distance-dependent time dilation depending on your gravitational field. But you can't change coordinate systems to get rid of gravity, since you can't have your entire coordinate system fall in a non-uniform gravitational field (the only fields of interest). It's there regardless, which means, unlike the acceleration-dependent time dilation you're refering to, gravitational time dilation IS real. There is never any coordinate system you can adopt which gets rid of it.

Ziggurat
14th March 2005, 07:08 PM
Originally posted by Stimpson J. Cat
"The proper time between two events is longest for an inertial observer, i.e., one who does not accelerate. So, if one twin leaves the other and returns, he will experience less time than his inertial twin, because he accelerated and his twin didn't. I do not believe you can construct a scenario where this isn't true."

This is incorrect. Different inertial observers will measure different amounts of time between two events.

I think you're kind of talking past each other on this one. 69dodge's statement is exactly correct, if what you're looking at is the time experienced along a trajectory: if the two events in question both occur at the same location in the oberver's reference frame, then yes, an accelerated observer will observe a shorter time interval. But 69dodge's statement is comparing different possible observers of two events who all see the events happen in the same location (which automatically means there's only one inertial reference frame for this, all other frames must be accelerated). This is a different question than asking the time interval between two events for different inertial observers when the two events need NOT occur in the same location, which (if I'm not mistaken) is what you are addressing.

Stimpson J. Cat
15th March 2005, 05:32 AM
I think you're kind of talking past each other on this one. 69dodge's statement is exactly correct, if what you're looking at is the time experienced along a trajectory: if the two events in question both occur at the same location in the oberver's reference frame, then yes, an accelerated observer will observe a shorter time interval. But 69dodge's statement is comparing different possible observers of two events who all see the events happen in the same location (which automatically means there's only one inertial reference frame for this, all other frames must be accelerated). This is a different question than asking the time interval between two events for different inertial observers when the two events need NOT occur in the same location, which (if I'm not mistaken) is what you are addressing.

Yes, that may be what he had in mind.

If you want to look at the Twin Paradox from the point of view of space-time events, and the distances and time intervals between them, then it is vital to recognize what those events are.

For the first half of the experiment, we have 2 events.

Event one occurs at Earth, at the time when the twin leaves. The relative velocity between the twin and the Earth is 0.8c, or whatever other speed you want to use, so this is just the instance after the twin starts. We are assuming the limit that the acceleration time is negligible.

Event two occurs at the destination, at the time when the twin arrives. The relative velocity is again 0.8c, so this is just the instant before the twin stops.

Now, in special relativity there are two types of relationships between pairs of events. Spacelike and timelike.

For spacelike events, there will be some inertial frame in which the two events occur at the same time, but with some distance between them. The distance in this frame will be the shortest distance between the events in any inertial frame. In any other frame, the events will occur further apart, and with some time delay between them. Naturally this time delay between them will always be less than the time required for light to travel from the location of one event to the other.

For timelike events, there will be some inertial frame in which the two events happen at the same location, but at different times. In this frame the time difference will be the smallest possible. In all other frames the events will occur at different positions, and with a longer time delay. In all cases the time delay will be long enough for light to travel the distance between them.

Now the two events which define the beginning and end of the first part of the twin experiment are clearly timelike events. In fact, the frame of the spaceship is exactly the frame in which the two events occur at the same location. That means that the time which passes in this frame must be smaller than the time between those events as measured from any other frame.

And that is the key point to the twin experiment. It does not make any difference who did the accelerating, or even if they both do. What matters is that in one frame the events marking the beginning and ending of the experiment happen in the same place, and in the other, they do not.

The fact that we then combine the first part of the experiment with an identical return trip, tends to obscure this fact. It then appears that the beginning and ending events are at the same time and place for both observers. But that is a mistake. You have to calculate for each part of the trip separately. When you do this, you will find that time experienced by the twin is smaller. The question of who did the accelerating never even enters into the calculations. All that matters is the relative velocities, and the space-time events which define the experiment.

Dr. Stupid

politas
18th March 2005, 06:16 AM
Originally posted by 69dodge
Ah, thanks. That is an excellent site, and gives exactly the information I was searching for.

Thanks also to everyone who has responded. There's far more in this thread than I can absorb at one time. I think I'll be dealing with it all for some time.

69dodge
19th March 2005, 06:09 AM
Originally posted by Stimpson J. Cat
Do you agree that we can break the thought experiment down into six parts?

1) Acceleration of the ship towards the destination.
2) Travel at uniform speed to the destination.
3) Deceleration of the ship at the destination.
4) Acceleration of the ship towards the Earth.
5) Travel at uniform speed to the Earth.
6) Deceleration at the Earth.

If so, when do you think the time difference between the twins accumulates?The twins disagree about when the time difference accumulated. The stationary twin thinks it accumulated gradually over the entire trip: from his point of view, the travelling clock ran slowly because it was moving. But from the traveller's point of view, it is the Earth twin who was moving and whose clock therefore ran slowly. So how does he explain the fact that, when he returns, the Earth clock clearly shows more elapsed time than his, even though it was running slowly during the whole trip? He explains it by saying that it ran slowly during almost the whole trip; around the midpoint of his trip, however, while he accelerated towards Earth (parts 3 and 4), the Earth clock ran much faster than his, just as if it were at the top of a giant gravity well and his clock were at the bottom. Which, from his point of view, was precisely the case.I don't see how it could accumulate only during the acceleration periods. If that were the case, the time difference would not even depend on the total trip time. It would depend only on the time spent accelerating. But of course, that isn't the case.It depends also on the distance between the two twins during the acceleration, just as gravitational time dilation depends on the (vertical) distance between two clocks being compared.And what about my example of the alien flying past the Earth? In that case, there is no acceleration. And yet the alien experiences less time travelling from the marker to the Earth than the Earth observer does.The alien experiences less time starting from when it thinks it passed the marker, than the Earthling experiences starting from when he thinks the alien passed the marker. (They both agree about when the alien passed the Earth.) The alien still thinks the Earthling's clock is running slower than its own, however. But it also thinks that the Earthling started his stopwatch too early, before the alien actually passed the marker; that's why, in its opinion, the Earthling measured a longer period of time.

The reason the twin paradox seems more paradoxical is that the twins are together at the beginning of the trip and also at the end, so they can directly compare their stopwatches; and they agree that both were started and stopped at the right times. Even so, the two stopwatches end up showing different amounts of elapsed time.For the first half of the experiment, we have 2 events.

Event one occurs at Earth, at the time when the twin leaves. [ ... ] Event two occurs at the destination, at the time when the twin arrives. [ ... ] the frame of the spaceship is exactly the frame in which the two events occur at the same location. That means that the time which passes in this frame must be smaller than the time between those events as measured from any other frame.One could just as easily define an "event two" that occurs on Earth. Suppose the round trip takes two hours, as measured by an Earth clock. Then, let "event two" occur at the location of the Earth clock when it shows that one hour has elapsed. Now, the travelling twin ought to think that more than one hour passes for each one-way trip. Oops. (In other words, relatively moving inertial observers each think the other's clock runs slowly.)

The problem is the assumption that we can ignore the acceleration. We can't ignore it, because that's when all the interesting stuff happens; it's the only thing that breaks the symmetry between the two twins.

Each twin thinks the acceleration takes a small amount of his own time; the Earth twin thinks it also takes a small amount of spaceship time, but the spaceship twin thinks it takes a lot of Earth time.

Suppose the round trip takes two hours of Earth time and one hour of spaceship time, and suppose it starts at 10:00. Around the midpoint of the trip, as he is still moving away from Earth but right before he turns around, what does the spaceship twin think the time on Earth is? Well, his clock shows that 30 minutes have passed, but since---according to him---the Earth was moving and so its time slowed down, he says the Earth time is only 10:15.

Right after he turns around, as he is moving toward Earth, what does he think the time on Earth is? Well, he is going to reach Earth in 30 of his own minutes, which again correspond---according to him---to 15 Earth minutes. But when he does reach Earth, he will find that clocks there show 12:00. So, now they must be showing 11:45.

And there you go. According to the twin on the spaceship, a full hour and a half of Earth time passed as he was turning around, even though the turnaround was practically instantaneous from his own point of view.

Stimpson J. Cat
19th March 2005, 12:05 PM
69dodge,

The twins disagree about when the time difference accumulated. The stationary twin thinks it accumulated gradually over the entire trip: from his point of view, the travelling clock ran slowly because it was moving. But from the traveller's point of view, it is the Earth twin who was moving and whose clock therefore ran slowly. So how does he explain the fact that, when he returns, the Earth clock clearly shows more elapsed time than his, even though it was running slowly during the whole trip? He explains it by saying that it ran slowly during almost the whole trip; around the midpoint of his trip, however, while he accelerated towards Earth (parts 3 and 4), the Earth clock ran much faster than his, just as if it were at the top of a giant gravity well and his clock were at the bottom. Which, from his point of view, was precisely the case.

Sorry, but this is simply incorrect. The travelling twin will see the Earth's clock going very slow while he is moving away from the Earth, and very quickly while returning to the Earth. But the speed at which he sees the Earth's clock moving at any given time is strictly a function of their relative velocity at that time. It does not depend on acceleration at all.

Of course, most of the difference in the speed the Earth's clock seems to be moving will be due to the doppler affect. After correcting for that, the travelling twin will always calculate that the Earth's clock is running slow by a factor of sqrt(1-(v/c)^2), where v is whatever their relative velocity is at that time.

Either way, he will not see the Earth clock suddenly speed up during the acceleration period. That just isn't how it works.

I don't see how it could accumulate only during the acceleration periods. If that were the case, the time difference would not even depend on the total trip time. It would depend only on the time spent accelerating. But of course, that isn't the case.
It depends also on the distance between the two twins during the acceleration, just as gravitational time dilation depends on the (vertical) distance between two clocks being compared.
This is incorrect. Nothing about either special relativity, or general relativity, suggests that the time dilation during periods of acceleration depends on the relative distance of the two objects. And in GR, the gravitational time dilation depends on the difference in gravitational potential, not on distance. In the case of planets, it just so happens that gravitational potential varies with distance.

And what about my example of the alien flying past the Earth? In that case, there is no acceleration. And yet the alien experiences less time travelling from the marker to the Earth than the Earth observer does.
The alien experiences less time starting from when it thinks it passed the marker, than the Earthling experiences starting from when he thinks the alien passed the marker. (They both agree about when the alien passed the Earth.) The alien still thinks the Earthling's clock is running slower than its own, however. But it also thinks that the Earthling started his stopwatch too early, before the alien actually passed the marker; that's why, in its opinion, the Earthling measured a longer period of time.
I know. That's the whole point. The twin flying back and forth between the Earth and his destination will, while he is moving, say that the twin back on Earth stopped his stopwatch at the wrong time (when the twin arrived at the destination), and likewise started it at the wrong time (when the twin heads back for Earth). Again, you have to recognize that we've got essentially two different experiments here.

Imagine there is a clock at the destination, which is (in the frame of the Earth and destination) synchronized with the clock on Earth. When the twin arrives at the destination, its clock will read (using the 4 light minutes and 0.8c example from before) 5 minutes later then the Earth clock read when he left. Let's assume that the acceleration period was very small (say 1 second).

When the twin leaves the Earth, he will see that the clock at the destination says -4 minutes. When he arrives, it will say +5 minutes. During the trip (which he says takes 3 minutes), he will see the clock advance 9 minutes. He will see the clock going 3 times its normal speed. He will not see it running slow all the way until the end, and then suddenly go through the rest of the 9 minutes just during deceleration. This is clear from the fact that even if he does not decelerate at all, and flies right past the destination clock, he will still see that it says +5 when he passes it! So the extra time can't be elapsing during the deceleration time. My calculations for the flyby alien prove this, because if it were happening during the deceleration, the results for the flyby alien would have to be different.

For the first half of the experiment, we have 2 events.

Event one occurs at Earth, at the time when the twin leaves. [ ... ] Event two occurs at the destination, at the time when the twin arrives. [ ... ] the frame of the spaceship is exactly the frame in which the two events occur at the same location. That means that the time which passes in this frame must be smaller than the time between those events as measured from any other frame.
One could just as easily define an "event two" that occurs on Earth. Suppose the round trip takes two hours, as measured by an Earth clock. Then, let "event two" occur at the location of the Earth clock when it shows that one hour has elapsed. Now, the travelling twin ought to think that more than one hour passes for each one-way trip. Oops. (In other words, relatively moving inertial observers each think the other's clock runs slowly.)
You are forgetting about the simultineaty problem. If the Earth observer says that the instant just before the ship stops and turns around is simultaneous with event 2, then it necessarily follows that the observer on the ship will not, and vice-versa. Your experiment is thus rendered impossible, because the two observers will not be able to agree on what events mark the beginning and ending of the experiment.

In your experiment, we actually have 4 events.

Event 1) The Earth's clock says t=0.
Event 2) The spaceship leaves.
Event 3) The Earth's clock says t=2 hours.
Event 4) The spaceship stops.

Events 1 and 2 happen at the same location, so they are simultaneous in all frames. Thus they are effectively the same event.
Events 3 and 4 happen at different locations, so they are not simultaneous in all frames. If they are simultaneous in the Earth frame, then the twin on the spaceship will not agree that he turned around when the Earth clock said 2 hours. If the twin on the spaceship thinks they are simultaneous, then the Earth twin will say that the spaceship twin did not turn around at the right time.

The problem is the assumption that we can ignore the acceleration. We can't ignore it, because that's when all the interesting stuff happens; it's the only thing that breaks the symmetry between the two twins.
It's not. What breaks the symmetry is the fact that the events defining the beginning and ending of each part of the experiment happen at the same place for the spaceship, and at different places for the Earth. If you did design the experiment in such a way as to reverse this, you would get the opposite effect. What you have to consider is how such an experiment would be defined.

To see this better, imagine two huge rulers in space. One right next to the other like this ||. Now we could move ruler A up, until its middle is flush with the top end of ruler B, or we could move ruler B down, until its middle is flush with the bottom of ruler A. It seems like these experiments should be equivalent, since when the middle of A is flush with the tip of B, the middle of B will be flush with the bottom of A. So there is no broken symmetry. The experiment looks the same to an observer on the middle of ruler A and to an observer on the middle of ruler B. By symmetry, they must experience the same duration of time, or there is a logical contradiction, a true paradox.

And yet we can think of the middle of B as being the Earth, and its top end being the destination, and the middle of A as being the spaceship. Then we've got the twin paradox, right? Likewise if we let the middle of A be the Earth, and the bottom of A be the destination, and the middle of B be the spaceship.

The problem is, you can't do it. If, from the POV of an observer in the middle of B (the Earth), the middle of A (the spaceship) arrives at the top of B (the destination) at the same time that the bottom of A arrives at the middle of B, then it will absolutely not be the case that, from the POV of an observer at the middle of A, the middle of B arrives at the bottom of A at the same time that the top of B arrives at the middle of A.

I realize that is a bit confusing. Please try drawing it out on paper, and you will see what I mean. The point is that there is no way to make the experiment symmetrical. If we define the beginning and end points of the experiment (in this case, each half of the experiment) in such a way that they happen at the same location for one observer, then the beginning and end points of the experiment cannot possibly happen at the same location for the other observer. One of the two people is going to experience less time passing, and in any case, it is going to be the guy for whom both events happened at the same place. This would only create a paradox if it was possible to have the beginning and end of the experiment happen at the same place for both observers, but it isn't.

Each twin thinks the acceleration takes a small amount of his own time; the Earth twin thinks it also takes a small amount of spaceship time, but the spaceship twin thinks it takes a lot of Earth time.
The alien flyby example clearly shows that this is not the case. The alien, who has not decelerated yet at all, will see exactly the same time on the clock at the destination that the twin who stops there will. The second before the spaceship stops (from the spaceship's POV), the clock at the destination will say 4 minutes 57 seconds (using the 4 light minute and 0.8c example I gave). After he stops, it will say 5 minutes. It is very easy to verify that this is the case.

Suppose the round trip takes two hours of Earth time and one hour of spaceship time, and suppose it starts at 10:00. Around the midpoint of the trip, as he is still moving away from Earth but right before he turns around, what does the spaceship twin think the time on Earth is? Well, his clock shows that 30 minutes have passed, but since---according to him---the Earth was moving and so its time slowed down, he says the Earth time is only 10:15.
Sure, but a clock which is synchronized with the Earth's (in the Earth's frame), which is at the turn around point, will say 11:00. It will not say 10:15 just before he stops, and then suddenly advance to 11:00 over the deceleration period.

Right after he turns around, as he is moving toward Earth, what does he think the time on Earth is? Well, he is going to reach Earth in 30 of his own minutes, which again correspond---according to him---to 15 Earth minutes. But when he does reach Earth, he will find that clocks there show 12:00. So, now they must be showing 11:45.
Yes, when he changes speed at a distance far from the Earth, his calculation of what time it currently is on the Earth changes. That is why it is useful to have a clock at the destination which is in the same frame as the clock on Earth. From the Earth frame, the two clocks are synchronized. From the frames moving relative to the Earth, they are not. While the ship is moving towards the destination, it sees the clock on at the destination as being 45 minutes ahead of the one on Earth. When the ship stops at the destination, it sees them as being synchronized again. Likewise going back it sees the clock on Earth as being 45 minutes ahead of the clock at the destination, but when it stops back on Earth they are again synchronized.

This may be where you are getting the idea that the time difference suddenly accumulates when the velocity changes. But it doesn't. All we have here are disagreements about what time it currently is at some distant location, due to relative velocity.

If the spaceship is looking back at a clock on Earth, he does not see the clock suddenly speed up and go through the missing 45 minutes as he decelerates. In fact, the time he sees pass on the Earth clock will be less than what passes on his own during the deceleration period. All that happens is that his calculation of how long ago that light from Earth left, based on how far away the Earth is, has changed.

Just before he stops, he will look back at the clock on Earth. From my calculations, what he will see should be about 10:08. Note that from his POV, the earth is moving away from him at nearly light speed. So he is seeing where the Earth was some time ago. When he corrects for how far away the Earth was when this light was emitted, he will determine that right now the Earth clock says 10:15.

When he stops, he will still see that the clock on Earth says 10:08. But now the Earth will appear to be much further away than before, so his calculations of how long ago the Earth clock said 10:08 will be different. Thus he now determines that the clock on Earth says 11:00.

And there you go. According to the twin on the spaceship, a full hour and a half of Earth time passed as he was turning around, even though the turnaround was practically instantaneous from his own point of view.
Wrong. During the turn around, his calculation of what time it currently is on Earth will change by an hour and a half. But he will still only see a small amount of time pass on the Earth frame during this period.

Again, you have to remember that what he sees happening on the Earth is largely due to the doppler effect. During his trip outward, he sees very little time pass on Earth (I figure about 8 minutes). During his trip back he sees the other hour and 52 minutes pass on Earth. He does not see the actual rate at which time is passing on the Earth at all. He can only calculate it. And what he calculates depends on their relative speed. During his outward trip he calculates a passage of 15 minutes on Earth. Likewise during his return trip. He just can't add these two numbers together to get the total time, because his calculations of what time it was at the beginning of and ending of those periods don't match up.

Dr. Stupid

Ziggurat
19th March 2005, 01:24 PM
Originally posted by Stimpson J. Cat
Sorry, but this is simply incorrect. The travelling twin will see the Earth's clock going very slow while he is moving away from the Earth, and very quickly while returning to the Earth. But the speed at which he sees the Earth's clock moving at any given time is strictly a function of their relative velocity at that time. It does not depend on acceleration at all.

Actually, 69dodge is correct, although that's not usually how we think of it, and it's easy to get tripped up. But in order to understand what he means, it's absolutely crucial to understand that what he says ONLY applies to what the twin "oberves", not what he sees.

Wrong. During the turn around, his calculation of what time it currently is on Earth will change by an hour and a half. But he will still only see a small amount of time pass on the Earth frame during this period.

But I think this is exactly his point: he's talking about what the traveling twin would calculate, and hence "observe", NOT what the traveling twin would see.

Stimpson J. Cat
19th March 2005, 03:30 PM
Ziggurat,

Sorry, but this is simply incorrect. The travelling twin will see the Earth's clock going very slow while he is moving away from the Earth, and very quickly while returning to the Earth. But the speed at which he sees the Earth's clock moving at any given time is strictly a function of their relative velocity at that time. It does not depend on acceleration at all.
Actually, 69dodge is correct, although that's not usually how we think of it, and it's easy to get tripped up. But in order to understand what he means, it's absolutely crucial to understand that what he says ONLY applies to what the twin "oberves", not what he sees.

It may be that I have misunderstood what Dodge69 was saying. If what he meant by this

So how does he explain the fact that, when he returns, the Earth clock clearly shows more elapsed time than his, even though it was running slowly during the whole trip? He explains it by saying that it ran slowly during almost the whole trip; around the midpoint of his trip, however, while he accelerated towards Earth (parts 3 and 4), the Earth clock ran much faster than his, just as if it were at the top of a giant gravity well and his clock were at the bottom. Which, from his point of view, was precisely the case.

was that once the travelling twin calculates what the Earth clock really said at each part of the trip, he comes up with a gap between what he estimated it said just before he stopped at the destination, and what he estimated that it said just after he started back for Earth, then I agree. Like I said above, during the period of acceleration his calculation of what time it is on Earth changes by a huge amount.

But that is not the same as saying that during the period of acceleration a huge amount of time passes in the reference frame of the Earth. It doesn't, as my point about the synchronized clock at the turning point illustrates.

I think that at this point we are mostly suffering from problems of terminology. The real issue here is that when the twin reaches the turning point, just before he slows down, he calculates that only 15 minutes has passed on Earth. After stopping, he recalculates and finds that 1 hour has passed on Earth. But this is just because what he considers to be simultaneous with his reaching the turning point, at the location of the Earth, is different before and after he stops. It is not because his acceleration somehow causes time to pass extremely slowly for him, compared to the other frame. Perhaps this is what dodge69 meant when he said that the time dilation depended on the distance travelled. If so, then I misunderstood him. I thought he was saying that the relative passage of time between the Earth frame and the accelerating frame depends on the distance travelled before deceleration, which is clearly wrong.

Wrong. During the turn around, his calculation of what time it currently is on Earth will change by an hour and a half. But he will still only see a small amount of time pass on the Earth frame during this period.
But I think this is exactly his point: he's talking about what the traveling twin would calculate, and hence "observe", NOT what the traveling twin would see.
You may be right. If so, then I misunderstood what he was saying, and we are not really in disagreement.

Dr. Stupid

69dodge
20th March 2005, 03:02 AM
Originally posted by Stimpson J. Cat
If what [69dodge] meant [...] was that once the travelling twin calculates what the Earth clock really said at each part of the trip, he comes up with a gap between what he estimated it said just before he stopped at the destination, and what he estimated that it said just after he started back for Earth, then I agree. Like I said above, during the period of acceleration his calculation of what time it is on Earth changes by a huge amount.

But that is not the same as saying that during the period of acceleration a huge amount of time passes in the reference frame of the Earth. It doesn't, as my point about the synchronized clock at the turning point illustrates.Not "in the reference frame of the Earth." Just "on Earth."

While the moving twin is moving, he doesn't agree that the clock at the turning point is in fact synchronized with clocks on Earth, so why should he care about what it says when trying to determine what the clocks on Earth are doing?

I'm not saying that the twin on Earth thinks that a lot of Earth time passes during the acceleration; I'm saying that the twin on the spaceship thinks that a lot of Earth time passes during the acceleration.

In what sense is the "calculated" Earth time not the "real" Earth time? There's no absolutely real Earth time; there are different Earth times in different reference frames. But it seems to me that what the moving twin calculates, based on his current reference frame, is as real for him as anything is.I think that at this point we are mostly suffering from problems of terminology. The real issue here is that when the twin reaches the turning point, just before he slows down, he calculates that only 15 minutes has passed on Earth. After stopping, he recalculates and finds that 1 hour has passed on Earth. But this is just because what he considers to be simultaneous with his reaching the turning point, at the location of the Earth, is different before and after he stops. It is not because his acceleration somehow causes time to pass extremely slowly for him, compared to the other frame.What's the difference between the twin in the spaceship saying, "at the start of my deceleration it was 10:15 on Earth, and at the end of my deceleration it was 11:00 on Earth" and him saying, "during my deceleration, 45 minutes passed on Earth"? Seems the same to me. I mean, the Earth clocks didn't just jump from 10:15 to 11:00.I thought [69dodge] was saying that the relative passage of time between the Earth frame and the accelerating frame depends on the distance travelled before deceleration, which is clearly wrong.I don't think it's even meaningful to talk about the relative passage of time between the Earth frame and the accelerating frame, because according to the accelerating twin, different clocks in the Earth frame are running at different rates. I'm just talking about what, according to the accelerating twin, is happening on Earth itself.

Anyway, to back up a bit, this whole business of splitting the trip into two parts is merely something that we can do, if we want to, to help with the calculations. An actual trip simply is what it is; it doesn't have to be to a particular star and back. If two twins start out together in empty space, and one of them just coasts inertially while the other uses his engine to zoom around in any complicated path he likes, he will be younger than his coasting twin if and when he returns to him. So even if one's favorite explanation doesn't explicitly involve acceleration, it doesn't seem wrong to say that the important thing is in fact the acceleration, because the twin who accelerated is always the one who ends up younger.

(If space is not empty, but instead there are gravitating masses around, it's more complicated (http://www.mathpages.com/rr/s6-05/6-05.htm).)

Ziggurat
20th March 2005, 04:25 AM
Originally posted by 69dodge

In what sense is the "calculated" Earth time not the "real" Earth time? There's no absolutely real Earth time; there are different Earth times in different reference frames. But it seems to me that what the moving twin calculates, based on his current reference frame, is as real for him as anything is.

The problem with taking acceleration-based time "dilation" as anything more than an artifact is that it can lead to clocks at large distances away being observed to run backwards when you're accelerating away from said clock. You never run into such problems from within an inertial reference frame, and that clock which the accelerator observes is running backwards won't think that the accelerator's clock is running backwards (there's no symmetry, unlike for inertial observers). That doesn't make the calculation wrong, but it's easy for someone who doesn't fully grasp the whole thing to get really, really confused. As mentioned, for any inertial observer, nothing weird ever happens to an accelerating clock, regardless of the distance or the degree of acceleration. So I don't think you and stimpy and I actually disagree at all about the physics, this is mostly about how to frame the problem.

Stimpson J. Cat
20th March 2005, 06:11 AM
69dodge,

Not "in the reference frame of the Earth." Just "on Earth."

While the moving twin is moving, he doesn't agree that the clock at the turning point is in fact synchronized with clocks on Earth, so why should he care about what it says when trying to determine what the clocks on Earth are doing?

I'm not saying that the twin on Earth thinks that a lot of Earth time passes during the acceleration; I'm saying that the twin on the spaceship thinks that a lot of Earth time passes during the acceleration.

OK. Apparently I did misunderstand what you were saying. Of course I agree with this.

Dr. Stupid

69dodge
22nd March 2005, 02:22 AM
Originally posted by Ziggurat
The problem with taking acceleration-based time "dilation" as anything more than an artifact is that it can lead to clocks at large distances away being observed to run backwards when you're accelerating away from said clock.Oh my. That is weird.You never run into such problems from within an inertial reference frame, and that clock which the accelerator observes is running backwards won't think that the accelerator's clock is running backwards (there's no symmetry, unlike for inertial observers). [ ... ] As mentioned, for any inertial observer, nothing weird ever happens to an accelerating clock, regardless of the distance or the degree of acceleration.To someone who is used to relativity, symmetry between inertial observers in relative motion is normal, and asymmetry between them is weird. But, really, if I see your clock running slowly, it's pretty weird that you would see my clock running slowly too; if yours is slower than mine, how could mine not be faster than yours?

So is that an artifact too?

How weird does something have to be, to be considered "just an artifact"?

These are not rhetorical questions. I don't know.

Stimpson J. Cat
22nd March 2005, 03:10 AM
69dodge,

To someone who is used to relativity, symmetry between inertial observers in relative motion is normal, and asymmetry between them is weird. But, really, if I see your clock running slowly, it's pretty weird that you would see my clock running slowly too; if yours is slower than mine, how could mine not be faster than yours?

So is that an artifact too?
It's simply an effect of the fact that they don't agree on which events are simultaneous. The example of the alien flyby that I gave earlier illustrates this effect in detail. Both the alien, and an observer on Earth, each think that the other one's clock is running slowly. But when we look at the events which determine those intervals, it all works out.

The alien observes both his own clock, and the Earth clock (compensating for the time required for light to travel, and time dilation), when he passes some marker which is 4 light minutes from the Earth. So does the Earth observer. Again they look at each other's clocks when the alien passes the Earth.

If the alien is moving at 0.8c, then the trip will take him 3 minutes. Both the Earth observer and the alien will agree on what his clock said at both times, and that three minutes passed on the clock during the trip.

For the Earth observer, the trip will take 5 minutes. He will observe that his clock reads 5 minutes later when the alien arrives, than when the alien passed the marker.

But the alien will say that only 1.8 minutes passed on Earth. He will observe (after compensating for time-dilation and distance), that the Earth's clock only advanced 1.8 minutes between when he passed the marker, and when he passed the Earth.

The Earth observer claims that the alien passing the marker was simultaneous with the Earth clock reading t-5 minutes, but the alien claims that his passing the marker was simultaneous with the Earth clock reading t-1.8 minutes. This difference accounts for why they each measure time as going more slowly for the other one.

Dr. Stupid