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Ziggurat · 20th June 2007, 08:12 AM

Originally Posted by Paulhoff

OK, let’s try this if it hasn’t been done all ready.

Not explicitly, but the key difference here is the distinction between what you see and what you observe, and if you understand that distinction, these questions are easy.

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We have two spacemen and three clocks and they are big clocks that can be seen from a 100 feet with no problem. The speed of light for this experiment is 100 ft. per second. They are all somewhere in deep space, who cares where. The clocks are all reading the same time.

I am going to assume furthermore that the clocks and the spacemen also start at the same place and at rest with respect to some specified reference frame (which I'll refer to as the start frame).

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1. Now one of the spacemen takes one of the clocks and travels at 99.9999% speed of light for one second and stops and looks back at the other spaceman and the two clocks, he is 100 ft away.

The one second time must be refering to the time in the start frame, not the time he experiences. I will refer to this spaceman as spaceman 1, and his clock as clock 1.

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Is the time on his clock is the same as the other two clocks.
Is the time on his clock is one second behind the other two clocks.
Is the time on his clock is two seconds behind the other two clocks.
Is the time on his clock is one second ahead of the other two clocks.
Is the time on his clock is two seconds ahead of the other two clocks.

The time on clock 1 is observed to be different than the other two clocks. The proper time he experiences while moving with respect to the start frame is very short, so clock 1 will be observed to be almost 1 second behind the other clocks. However, spaceman 1 will see the other clocks as being about the same time as clock 1, while spaceman 2 will see clock 1 being two seconds behind the other clocks, since there's an approximately 1 second delay for the signal to get transmitted from one location to another. They don't see the same thing, but they both observe the same thing: clock 1 is ~1 second behind the other clocks.

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2. Now the other spaceman takes one of the two remaining clocks

I'll call this spaceman 2 and clock 2, with the clock left behind being clock 3.

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and travels very slowly, about one inch a second to the other spaceman.

You can go slower than that, the point is that we take the limiting behavior of slow travel time.

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When he gets to the other spaceman he looks at the other spaceman’s clock.

Clock 1 and clock 2 are now in the same location. This is important.

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Is the time on his clock is the same as the other spaceman’s clock.
Is the time on his clock is one second behind the other spaceman’s clock.
Is the time on his clock is two seconds behind the other spaceman’s clock.
Is the time on his clock is one second ahead of the other spaceman’s clock.
Is the time on his clock is two seconds ahead of the other spaceman’s clock.

Clock 1 will be observed to be about 1 second behind clock 2, as before. Because they are now in the same location, all observers will also see this same ~1 second difference, since any time delay in signal propagation to any location is the same from both clocks.

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3. Now the other spaceman looks back at the other clock left behind.

Is the time on his clock is the same as the other clock.
Is the time on his clock is one second behind the other clock.
Is the time on his clock is two seconds behind the other clock.
Is the time on his clock is one second ahead of the other clock.
Is the time on his clock is two seconds ahead of the other clock.

Spaceman 1 and 2 both observe that clock 2 is reading approximately the same time as clock 3. Spaceman 1 and 2 both see clock 2 reading approximately 1 second ahead of clock 3 because of the time delay for the signal from clock 3 to reach them.

Thabiguy · 20th June 2007, 08:29 AM

Originally Posted by Gertrude

When the two twins meet, they do have different ages, but it is not because of SR and its time dilation. The effect is a result of GR: the change occurs when the traveling twin decelerates, changes direction and accelerates back towards the earth.

Not quite. The twin paradox is indeed because of SR and time dilation, and nothing else. GR needn't be invoked at all.
Explanation.

ponderingturtle · 20th June 2007, 08:39 AM

Originally Posted by ynot

The rock only exists in its own rest frame and falls in a straight line toward the force of gravity when dropped. That it is perceived to be moving on an angle when observed from another rest frame is an illusion. To use the movie analogy used earlier, the single event is being separated in to a strip of many events and this creates the illusion.

Is it me, or does this seem to disprove orbital mechanics?

Gertrude · 20th June 2007, 08:47 AM

Originally Posted by Thabiguy

Not quite. The twin paradox is indeed because of SR and time dilation, and nothing else. GR needn't be invoked at all.
Explanation.

Actually there had been a debate among writers in the late 50's regarding the need to introduce GR to explain the twin paradox but nowadays, most physicists agree that SR is sufficient. And yes, SR can deal with acceleration. But we must be clear. SR deals only with acceleration as measured in inertial reference frames. This is not quite the case in the twin paradox, where one of the twins himself undergoes acceleration. In this sense, it is a problem of GR. We are appealing to the reality of this acceleration to explain the paradox.

The reason most writers agree that SR is sufficient in explaining the paradox is that we can use methods such as the Doppler effect (ie. pretend the twins send each other signals and count these) as described by SR and obtain the same result. But this is not to say that SR can handle acceleration as experienced by an observer.

Ziggurat · 20th June 2007, 08:57 AM

Originally Posted by Gertrude

Do not confuse this with the twin paradox. When the two twins meet, they do have different ages, but it is not because of SR and its time dilation. The effect is a result of GR: the change occurs when the traveling twin decelerates, changes direction and accelerates back towards the earth. The twin paradox imho is the worst example in teaching SR.

This is simply wrong. Let me expand upon Thai's response: special relativity can accomodate acceleration. It would be a crap theory if it couldn't. The math can get ugly and so most introductory treatments don't address it, but there is absolutely NO impediment to using acceleration within special relativity. It's mathematically exactly equivalent to the problem of curved lines in Euclidean space. You don't have to abandon Euclidean geometry or the pythagorean theorem in order to calculate the length of a curved line, and you don't need to go to general relativity to handle acceleration.

But it is a common misconception that you need it, for two reasons: first, because it's often not taught (because you don't need to do it to understand the basics of SR), and second, because of a misunderstanding of the equivalence principle (which I could go into in more detail but isn't really important at the moment). So you're far from alone in your mistake, but the twin paradox really can be handled completely within the confines of special relativity, including the bits where one twin (or even both) accelerates. The ONLY thing you cannot handle correctly within the confines of special relativity is gravity. If you think the twin paradox gets resolved because of some GR effect, then you don't understand the problem.

Ziggurat · 20th June 2007, 09:27 AM

Originally Posted by Gertrude

Actually there had been a debate among writers in the late 50's regarding the need to introduce GR to explain the twin paradox but nowadays, most physicists agree that SR is sufficient. And yes, SR can deal with acceleration. But we must be clear. SR deals only with acceleration as measured in inertial reference frames. This is not quite the case in the twin paradox, where one of the twins himself undergoes acceleration. In this sense, it is a problem of GR. We are appealing to the reality of this acceleration to explain the paradox.

The reason most writers agree that SR is sufficient in explaining the paradox is that we can use methods such as the Doppler effect (ie. pretend the twins send each other signals and count these) as described by SR and obtain the same result. But this is not to say that SR can handle acceleration as experienced by an observer.

This is also incorrect. There's generally little reason to treat special relativity problems from within a non-inertial frames (and plenty of reason not to - the math can get ugly), but there are no actual impediments. All the mathematical framework is there already. It's no different than Newtonian physics in that respect - using a rotating coordinate system introduces coriolis and centripidal pseudo-forces, but that's it. The reason that GR is different is not that it can handle non-inertial reference frames and SR can't, it's that GR can handle non-uniformity of that non-inertiality. That requires a whole new (and much more complicated) mathematical framework, but just non-inertiality can be handled within special relativity alone.

Gertrude · 20th June 2007, 09:29 AM

Originally Posted by Ziggurat

This is simply wrong. Let me expand upon Thai's response: special relativity can accomodate acceleration. It would be a crap theory if it couldn't. The math can get ugly and so most introductory treatments don't address it, but there is absolutely NO impediment to using acceleration within special relativity. It's mathematically exactly equivalent to the problem of curved lines in Euclidean space. You don't have to abandon Euclidean geometry or the pythagorean theorem in order to calculate the length of a curved line, and you don't need to go to general relativity to handle acceleration.

But it is a common misconception that you need it, for two reasons: first, because it's often not taught (because you don't need to do it to understand the basics of SR), and second, because of a misunderstanding of the equivalence principle (which I could go into in more detail but isn't really important at the moment). So you're far from alone in your mistake, but the twin paradox really can be handled completely within the confines of special relativity, including the bits where one twin (or even both) accelerates. The ONLY thing you cannot handle correctly within the confines of special relativity is gravity. If you think the twin paradox gets resolved because of some GR effect, then you don't understand the problem.

Like I said, SR can very well deal with accelerated motion as measured in inertial frames. But to deal with accelerated frames, we must use approximations (and ingenuity). There is no rigorous treatment (ie. without the need for approximations) that I know of. If there is, please do point me to some reference.

The reason I stated that it was a problem of GR (and you are right that I should have weighed my words more as it does come accross as a 'miracle' of GR) is that we must appeal to the reality of this acceleration to explain the paradox. Some treatment that does not involve GR can very well explain the paradox (for instance, the relativistic Doppler effect), but the turnaround is the key.

I stated this because it appears that many people take the example of the twin paradox to illustrate and understand SR. In this sense, they will think that time dilation/length contraction is still valid when the observers meet. But the twin paradox appears at first sight to be in contradiction with the tenets of SR and imho should not be used to understand the basics of SR.

Gertrude · 20th June 2007, 09:37 AM

Originally Posted by Ziggurat

This is also incorrect. There's generally little reason to treat special relativity problems from within a non-inertial frames (and plenty of reason not to - the math can get ugly), but there are no actual impediments. All the mathematical framework is there already. It's no different than Newtonian physics in that respect - using a rotating coordinate system introduces coriolis and centripidal pseudo-forces, but that's it. The reason that GR is different is not that it can handle non-inertial reference frames and SR can't, it's that GR can handle non-uniformity of that non-inertiality. That requires a whole new (and much more complicated) mathematical framework, but just non-inertiality can be handled within special relativity alone.

When I mention 'acceleration', I certainly refer to "all kinds" of accelerations, not only uniform acceleration...

But I'd be curious to see a rigorous treatment of (uniform) accelerating frames within SR. If you have any reference, I'd be highly interested.

boooeee · 20th June 2007, 09:55 AM

Originally Posted by Ziggurat

This is simply wrong. Let me expand upon Thai's response: special relativity can accomodate acceleration. It would be a crap theory if it couldn't. The math can get ugly and so most introductory treatments don't address it, but there is absolutely NO impediment to using acceleration within special relativity. It's mathematically exactly equivalent to the problem of curved lines in Euclidean space. You don't have to abandon Euclidean geometry or the pythagorean theorem in order to calculate the length of a curved line, and you don't need to go to general relativity to handle acceleration.

But it is a common misconception that you need it, for two reasons: first, because it's often not taught (because you don't need to do it to understand the basics of SR), and second, because of a misunderstanding of the equivalence principle (which I could go into in more detail but isn't really important at the moment). So you're far from alone in your mistake, but the twin paradox really can be handled completely within the confines of special relativity, including the bits where one twin (or even both) accelerates. The ONLY thing you cannot handle correctly within the confines of special relativity is gravity. If you think the twin paradox gets resolved because of some GR effect, then you don't understand the problem.

I know this issue has probably been covered ad nauseum in this forum, but can somebody provide an SR interpretation of the travelling twin's point of view?

In other words, the travelling twin is the one that accelerates away from his twin and then accelerates back. The travelling twin is sitting in his spaceship and observing the age of his twin. How does the travelling twin use Special Relativity to explain that time is moving faster for his twin?

Ziggurat · 20th June 2007, 10:02 AM

Originally Posted by Gertrude

When I mention 'acceleration', I certainly refer to "all kinds" of accelerations, not only uniform acceleration...

That's not what I mean (though I phrased it poorly). I don't mean that your acceleration is uniform over time (there's no requirement for that, just as Euclidean geometry can handle curves with varying curvature), I mean that you're adopting a frame in which all of space-time adapts to your single acceleration. This goes back to the equivalence principle: standing still in a gravitational field is locally equivalent to accelerating on a flat space-time background. If you can handle acceleration with special relativity, why can't you handle gravity? Because gravity isn't local, and it isn't uniform. If we look at two people sitting on either side of a planet, their "accelerating" frames are accelerating away from each other (because gravity is not uniform, but pulls them in different directions), and yet the distance between them isn't changing. So any attempt to simply use a special relativity non-inertial reference frame to describe these two guys is going to fail, but not because special relativity can't handle accelerated frames.

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But I'd be curious to see a rigorous treatment of (uniform) accelerating frames within SR. If you have any reference, I'd be highly interested.

Not at my fingertips, sorry.

Ziggurat · 20th June 2007, 10:15 AM

Originally Posted by Gertrude

The reason I stated that it was a problem of GR (and you are right that I should have weighed my words more as it does come accross as a 'miracle' of GR) is that we must appeal to the reality of this acceleration to explain the paradox. Some treatment that does not involve GR can very well explain the paradox (for instance, the relativistic Doppler effect), but the turnaround is the key.

Depends what you mean by that.

Let's look at the metric:
ds² = dx² + dy² + dz² - (ct)²
To find the proper time of any path, we integrate this quantity along that path. This is EXACTLY like taking the integral of a Euclidean metric:
ds² = dx² + dy² + dz²
to find the length of a curve, it's just a different metric. Now, we can make our traveling twin's trajectory curved (finite acceleration), or we can make it kinked (infinite acceleration, instant velocity change), it doesn't matter: the process of taking that path integral is the same either way. Saying that it's the turnaround that matters is equivalent to saying that (say) a right angle in a Euclidean path is what makes it longer than the straight line. That may be true in one sense, but the extra length in the bent path isn't found in the bend itself. Likewise, the "missing" time isn't subtracted from the turnaround of the twin either. Each leg of the traveling twin's path really is shorter than half the stationary twin's path.

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I stated this because it appears that many people take the example of the twin paradox to illustrate and understand SR. In this sense, they will think that time dilation/length contraction is still valid when the observers meet.

I'm not even sure what you mean by this anymore.

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But the twin paradox appears at first sight to be in contradiction with the tenets of SR and imho should not be used to understand the basics of SR.

That's the whole point, though: it's only an apparent contradiction if you don't really understand what those tenets actually are.

Ziggurat · 20th June 2007, 10:31 AM

Originally Posted by boooeee

I know this issue has probably been covered ad nauseum in this forum, but can somebody provide an SR interpretation of the travelling twin's point of view?

In other words, the travelling twin is the one that accelerates away from his twin and then accelerates back. The travelling twin is sitting in his spaceship and observing the age of his twin. How does the travelling twin use Special Relativity to explain that time is moving faster for his twin?

Recall that your space axis (which marks all points in space at the same time for your reference frame) tilts when you change reference frame. If you're accelerating, that axis is continually tilting. So as the traveling twin accelerates back towards the stationary twin, the intersection of his space axis with the stationary twin's world-line move upwards (in time). During the acceleration phase, then, the traveling twin will indeed observe the stationary twin's clock to be moving faster, in his accelerated frame. If he changes velocity instantaneously, then his space axis will tilt instantly, and he will observe that the stationary twin's current time will have instantaneously jumped forward. An important caveat, though: this is what he oberves, not what he sees. Even in the case of instantaneous velocity change, there's no discontinuity in what the twin sees.

If you want to ask what the twins see (which is also interesting, BTW), that's actually pretty straight forward. While moving away from each other, each twin sees the other twin's clock slowed by time dilation, plus additional slowing from doppler shift. When the traveling twin turns around and heads back, he will see the stationary twin's clock slowed by time dilation, but also speeded up by the doppler shift. So for half of his journey, the doppler shift increases the clock rate, and for half of it the doppler shift decreases the clock rate.

The stationary twin, however, doesn't see the traveling twin turn around at the moment he turns around, because the signal is time-delayed. But when he does see the traveling twin heading back towards him, he will see the clock slowed by time dilation, plus speeded up because of the doppler shift. Unlike the traveling twin, however, he sees the doppler shift slowing the clock for MORE than half the time interval, and speeding up the clock for LESS than half the time interval (because it took time for the light from the turnaround to get back to him). Which is how the twins end up with different times on their clocks: what they see over the length of the whole journey isn't the same, even if instantaneous snapshots on the outward and return journeys look superficially symmetric.

Gertrude · 20th June 2007, 11:18 AM

Originally Posted by Ziggurat

Depends what you mean by that.

Let's look at the metric:
ds² = dx² + dy² + dz² - (ct)²
To find the proper time of any path, we integrate this quantity along that path. This is EXACTLY like taking the integral of a Euclidean metric:
ds² = dx² + dy² + dz²
to find the length of a curve, it's just a different metric. Now, we can make our traveling twin's trajectory curved (finite acceleration), or we can make it kinked (infinite acceleration, instant velocity change), it doesn't matter: the process of taking that path integral is the same either way. Saying that it's the turnaround that matters is equivalent to saying that (say) a right angle in a Euclidean path is what makes it longer than the straight line. That may be true in one sense, but the extra length in the bent path isn't found in the bend itself. Likewise, the "missing" time isn't subtracted from the turnaround of the twin either. Each leg of the traveling twin's path really is shorter than half the stationary twin's path.

Absolutely. The point here is that, neglecting accelerations, two inertial frames cannot be brought together and still experience different durations. I could put it differently: the lack of symmetry in the paradox is key. The fact that one twin definately feels an acceleration (be it a constant acceleration as in a circle or a sudden turnaround) while the other one doesn't makes up for the difference.

In SR, time dilation works both ways. In all cases we can say: "A is travelling at X m/s relative to A" in the same way as "B is travelling at X m/s relative to A". If the only effect that we considered was velocity, then there indeed would be a paradox, as the effects of SR are symmetric for both observers, ie. both observers would see the other aging slower. It is the idea that the situation is not symmetrical that resolves the paradox.

In this sense, I think the subtlety may lead to an incorrect understanding that time dilation is not only apparent and occurs to the 'fastest traveler' only(whatever that means).

Yllanes · 20th June 2007, 11:32 AM

Originally Posted by Gertrude

But I'd be curious to see a rigorous treatment of (uniform) accelerating frames within SR. If you have any reference, I'd be highly interested.

Do you mean references as in web pages or books? Because if it is the latter, most textbooks discuss them. For example, let us see how uniform accelerated motion looks in SR.

First of all we must define it, because acceleration is relative. I'm going to find the trajectory of a particle with constant acceleration in its instantaneous rest frame. This rest frame is the inertial reference frame that happens to have the same velocity as the particle at each moment, and so it is different at each point of the movement. Let a be the ordinary acceleration of Newtonian physics, and v the ordinary velocity. As you may now, we can define a four velocity if we differentiate the position x with respect to the proper time

$\footnotesize \begin{align*} x^\mu&=(ct,\ x,\ y,\ z),\\ u^\mu&=\frac{\mathrm{d} x^\mu}{\mathrm{d} \tau} = \left(\gamma c,\ \gamma {\vec v}\right),\qquad \tau =t \sqrt{1-v^2/c^2} = t/\gamma,\\ w^\mu&=\frac{\mathrm{d} x^\mu}{\mathrm{d} \tau^2} \end{align*} $

The important thing about four vectors is that their magnitude is the same in all reference frames. We write that magnitude as u² and we have to remember it is not calculated with the Euclidean metric, but that the time component is subtracted. For example

$\footnotesize \[u^2 = (u^1)^2+(u^2)^2+(u^3)^2-(u^0)^2 = -c^2 \] $

In our accelerated motion, we now that w^m = (0, a, 0,0) in the rest frame. So w² = a² in every reference frame. We are interested in the acceleration as seen from a 'fixed' frame. So we go to the value of w^m in that system and force it to have w²=a². This leads to the equation (v is the velocity of the particle in the fixed frame).

$\footnotesize \[\frac{\mathrm{d}}{\mathrm{d}t} \frac{v}{\sqrt{1-\frac{v^2}{c^2}}} = a \] $

It's just a matter of integrating now. If we set x = 0 at t = 0 the movement is

$\footnotesize \begin{align*} v&=\frac{at}{\sqrt{1+\frac{a^2t^2}{c^4}}}, & x &= \frac{c^2}{a}\left(\sqrt{1+\frac{a^2t^2}{c^2}}-1\right) \end{align*} $

And it is inmediate to see that in the limit at << c we recover the Newtonian equations for the uniform accelerated movement ( v ~ at, x ~ at²/2). As you see, it is very simple to handle accelerated motion in SR.

Ziggurat · 20th June 2007, 11:33 AM

Originally Posted by Gertrude

It is the idea that the situation is not symmetrical that resolves the paradox.

Correct.

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In this sense, I think the subtlety may lead to an incorrect understanding that time dilation is not only apparent and occurs to the 'fastest traveler' only(whatever that means).

I'm confused by what you mean here. If you're claiming that time dilation is only an apparent effect, that is wrong. It's a very real effect. But it can be wrongly applied to the traveling twin paradox if the issue of simultaneity is not handled correctly as well.

Gertrude · 20th June 2007, 12:17 PM

Originally Posted by Yllanes

First of all we must define it, because acceleration is relative. I'm going to find the trajectory of a particle with constant acceleration in its instantaneous rest frame. This rest frame is the inertial reference frame that happens to have the same velocity as the particle at each moment, and so it is different at each point of the movement.

I lost you right there! If the rest frame has the same velocity as the particle, it cannot be inertial since the particle undergoes acceleration!?

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In our accelerated motion, we now that wm = (0, a, 0,0) in the rest frame. So w2 = a2 in every reference frame. We are interested in the acceleration as seen from a 'fixed' frame. So we go to the value of wm in that system and force it to have w2=a2. This leads to the equation (v is the velocity of the particle in the fixed frame).

Ok. So you are analysing the motion as an observer in an inertial frame. Actually, I have seen this treatment before. What interests me is looking at events from an accelerating frame. If you have any suggestion...

Ziggurat · 20th June 2007, 12:42 PM

Originally Posted by Gertrude

I lost you right there! If the rest frame has the same velocity as the particle, it cannot be inertial since the particle undergoes acceleration!?

No. Rather, acceleration is always measured with respect to some specific reference frame. This doesn't sound significant until you realize that 1) the same accelerating trajectory will have a different acceleration when measured from different reference frames, and 2) this is not the same as classical mechanics where velocity is relative but acceleration is absolute. To adopt constant proper acceleration, you basically find a trajectory where the instantaneous acceleration in the inertial reference frame defined by the instantaneous velocity is always constant. Which frame you use has to keep changing, which means that in any fixed reference frame the acceleration will not appear constant, but each of those tangent frames is still inertial.

Gertrude · 20th June 2007, 12:46 PM

Originally Posted by Ziggurat

I'm confused by what you mean here. If you're claiming that time dilation is only an apparent effect, that is wrong. It's a very real effect. But it can be wrongly applied to the traveling twin paradox if the issue of simultaneity is not handled correctly as well.

I am quite good at expressing things in the most confusing possible manner. I'll give an example that may be clearer. The statement "moving clocks run slower" is confusing because it does imply that time slows down with speed (relative to any observer). The reality is that moving clocks measure events occuring in a slower inertial frame as having longer duration than what a clock at rest relative to said event would measure.

The confusion is only increased with the twin paradox. It could be interpreted that velocity indeed slows time down for every observer (as 'confirmed' by the fact that when they meet, the two twins no longer have the same age). But this is not the case. If A and B are in two different inertial frames, B would see the clock of A running slower because it is moving relative to A and A would see the clock of B running slower because it is moving relative to B. There is no 'absolute' time dilation and it is in this sense that I say time dilation is apparent.

Paulhoff · 20th June 2007, 01:08 PM

OK, let’s try this again.

We have two spacemen and three clocks and the clocks are big so that they can be seen and read from a 100 feet with no problem. The speed of light for this experiment is 100 ft. per second. The spacemen and clocks are all together somewhere in deep space, who cares where, and who cares about their speed. The clocks are all reading the same time.

1. Now spaceman (1) takes one of the clocks and travels in any direction at 99,9999999% speed of light for one second and stops. Spaceman (1) now is 100 ft away from spaceman (2) and the two clocks remanding clocks, and spaceman (1) looks at the two clocks with spaceman (2).

A. Is the time on spaceman's (1) clock is the same as the other two clocks.
B. Is the time on spaceman's (1) clock is one second behind the other two clocks.
C. Is the time on spaceman's (1) clock is two seconds behind the other two clocks.
D. Is the time on spaceman's (1) clock is one second ahead of the other two clocks.
E. Is the time on spaceman's (1) clock is two seconds ahead of the other two clocks.

2. Now spaceman (2) takes one of the two remaining clocks and travels very slowly, about one inch a second to spaceman (1). When he gets to spaceman (1) he looks at spaceman's (2) clock.

A. Is the time on spaceman's (2) clock is the same as spaceman’s (1) clock.
B. Is the time on spaceman's (2) clock is one second behind spaceman’s (1) clock.
C. Is the time on spaceman's (2) clock is two seconds behind spaceman’s (1) clock.
D. Is the time on spaceman's (2) clock is one second ahead of spaceman’s (1) clock.
E. Is the time on spaceman's (2) clock is two seconds ahead of spaceman’s (1) clock.

3. Now spaceman (2) looks back at the clock left behind.

A. Is the time on his clock is the same as the clock left behind.
B. Is the time on his clock is one second behind the left clock.
C. Is the time on his clock is two seconds behind the left clock.
D. Is the time on his clock is one second ahead of the left clock.
E. Is the time on his clock is two seconds ahead of the left clock.

Paul

Yllanes · 20th June 2007, 01:11 PM

Originally Posted by Gertrude

I lost you right there! If the rest frame has the same velocity as the particle, it cannot be inertial since the particle undergoes acceleration!?

Ziggurat answered this. There is a different rest frame at each moment of time.

Quote:

Ok. So you are analysing the motion as an observer in an inertial frame. Actually, I have seen this treatment before. What interests me is looking at events from an accelerating frame. If you have any suggestion...

Ah, well it's just the same. For example, t is time as measured by the stationary observer. To switch to the accelerated observer we have to use his proper time, which is just a matter of integrating

$\footnotesize \[ \tau=\int_0^t \sqrt{1-\tfrac{v^2}{c^2}}= \frac{c}{a} \mathrm{arcsinh}\left(\frac{at}{c}\right)\] $

so the distance from its starting point, as measured by the accelerated particle, is

$\footnotesize \[ x(\tau)=\left(\cosh \tfrac{at}{c} - 1\right)\frac{c^2}{a} \] $

Incidentally, such a uniformly accelerated observer can outrun a photon, even though his velocity never reaches c.

Thabiguy · 20th June 2007, 01:14 PM

Originally Posted by Paulhoff

OK, let’s try this again.

We have already answered you, here and here.

Why do you keep asking the same question?

Paulhoff · 20th June 2007, 01:20 PM

Originally Posted by Thabiguy

We have already answered you, here and here.

Why do you keep asking the same question?

It was reworded......

Paul

Ziggurat · 20th June 2007, 01:25 PM

Originally Posted by Gertrude

There is no 'absolute' time dilation and it is in this sense that I say time dilation is apparent.

Time dilation is relative, that's certainly true, but I think the word "apparent" is a poor choice, precisely because the effect is still quite real and definite.

Here's a nice little problem involving time dilation which might help out a bit. We set up a "rest" reference frame, stick an observer at the origin and let him stay there (along with a clock) for the duration of our experiment. Now we take several volunteers with clocks, sync all the clocks together, and have them start moving in circular trajectories which all start at the origin (and hence return there periodically), which they will travel along at constant speed (with respect to our rest frame). Now one of the nice features of using circular orbits is that the acceleration and the speed are now independent parameters. We can have high-speed trajectories with small accelerations by using large circles, we can have small speed trajectories with large accelerations by using small circles, or any other combination. Now we ask, when these traveling clocks swing by the origin, how do their times compare with our stationary clock?

The answer is quite simply that the travelling clocks are slowed down compared to the stationary clock because of time dilation, by a factor given by their speed. Now, just like the twin paradox, there's an asymmetry in the problem because the moving clocks experience acceleration and the stationary clock doesn't. But here's the interesting bit which becomes clear because of the setup of the problem: the acceleration never actually enters into the calculation! Remember, we can make trajectories which have the exact same speed but different accelerations, or different speeds but the exact same acceleration, but the ONLY thing we need to know to figure out how much time dilation a trajectory experiences is the speed. The answer is totally independent of the value of the acceleration, and so acceleration is, in one sense, completely irrelevant to that observed time dilation.

Ziggurat · 20th June 2007, 01:28 PM

Originally Posted by Paulhoff

It was reworded......

Paul

My answers in post 81 still apply. Do you have any questions about those answers?

slyjoe · 20th June 2007, 01:34 PM

Nice illustration Zig - never seen that one before!

slyjoe · 20th June 2007, 01:36 PM

ETA: But I'm not sure if it will help those who don't want to use the math.

ETA: Not sure what just happened to cause 2 posts here.

Paulhoff · 20th June 2007, 01:56 PM

Originally Posted by Yllanes

Incidentally, such a uniformly accelerated observer can outrun a photon, even though his velocity never reaches c.

Really, then how do his atoms not fall apart because the virtual photons in the atom, as you say, can't keep up?

Paul

Paulhoff · 20th June 2007, 01:58 PM

Originally Posted by boooeee

How does the travelling twin use Special Relativity to explain that time is moving faster for his twin?

In a few words, there is only so much speed to go around for all 3 dimansions. The more you used in one direction and/or dimension the less you have for the other two.

Paul

Paulhoff · 20th June 2007, 02:15 PM

Originally Posted by Ziggurat

My answers in post 81 still apply. Do you have any questions about those answers?

Apparently my question was not clear enough since you had such a long answer.

Paul

Ziggurat · 20th June 2007, 02:26 PM

Originally Posted by Paulhoff

Really, then how do his atoms not fall apart because the virtual photons in the atom, as you say, can't keep up?

He didn't make this explicit, but what he's refering to is photons which you have a head start on. In other words, for constant proper acceleration and a sufficient head start, you can stay ahead of a photon forever (but you can't ever stop accelerating). More details here:
http://en.wikipedia.org/wiki/Event_h...rated_particle

Paulhoff · 20th June 2007, 03:26 PM

Originally Posted by Ziggurat

He didn't make this explicit, but what he's refering to is photons which you have a head start on. In other words, for constant proper acceleration and a sufficient head start, you can stay ahead of a photon forever (but you can't ever stop accelerating).

This most likely would work a lot easier if space itself is expanding, which looks like it is now happening.

Paul

Art Vandelay · 20th June 2007, 07:27 PM

Originally Posted by Yllanes

The important thing about four vectors is that their magnitude is the same in all reference frames.

I think that for clarity's sake, you should write "four-vectors" and "four-velocities".

Originally Posted by Gertrude

I lost you right there! If the rest frame has the same velocity as the particle, it cannot be inertial since the particle undergoes acceleration!?

It's like the tangent lines in the derivative. Each point has its own tangent line, so even though each line is straight, the curve is not.

Originally Posted by Gertrude

I am quite good at expressing things in the most confusing possible manner. I'll give an example that may be clearer. The statement "moving clocks run slower" is confusing because it does imply that time slows down with speed (relative to any observer). The reality is that moving clocks measure events occuring in a slower inertial frame as having longer duration than what a clock at rest relative to said event would measure.

I think that's still not quite right. Rather, if you have events proceeding, in one reference frame, at the same place, but at different places, then in another reference frame, they will be assigned different spatial corrdinates, and temporal coordinates that increment at a lower rate than those in the other reference frame.

A big problem in understanding relativity is that people don't understand how relative it is. When they're told that moving clocks run slower, they think that there is some absolute "time" through which the clock moves more slowly. Rather, the clock runs at normal speed in its own reference frame. That is, there is a quantity called "time" which increases at a normal rate. In another reference frame, the clock is measured with a different quantity, and found to have less of that quantity. That this other quantity is given the same name as the other one is what causes the confusion.

If we look at a car going down a road, one person might describe it as going 40 miles north in an hour. That is, after an hour, its ditance from the South Pole increases 40 miles. Another person might observe that its distance from the Prime Meridian increases by 50 miles. If the first just says it traveled 40 miles, and the second says that it traveled 50 miles, it may seen that they are contradicting each other. But if we realize that the two instances of the word "mile" refer to different things (North-miles vs. East-miles), there is no contradiction.

One way of thinking about it (which, I suppose, might inspire its own misconceptions) is that each tick of the clock has a certain magnitude, so to speak, and that magnitude stays the same no matter what. A stationary clock has all of that magnitude going into time, so it runs at full speed. For a moving clock, each tick moves through space and time, so some of its magnitude is "used up" by the movement through space, and the amount that it travles through time is less. But the whole issue of whether the stuff it goes through is "space" or "time" is simply a labeling issue; one person calls the stuff its going through "time" and another person calls it "some time and some space".

Quote:

If A and B are in two different inertial frames, B would see the clock of A running slower because it is moving relative to A and A would see the clock of B running slower because it is moving relative to B.

It's important to note that they don't just disagree about the results of the measurement of the clocks. They disagree as to what to measure. That's why they're getting different answers. If they both measured the same thing, they would get the same answer. Remember how earlier, I talked about East being the distance to the Prime Meridian? The Prime Meridian is a set of points that is considered to all be “zero East”, and one’s distance East is then measured from the closest one of those points. For A, there is a collection of points which he considers to all be at “time zero”. He measures time as the distance to the closest of those points. When he looks at his own clock, that’s no problem, because the closest is always the same (his starting point). But when he looks at B’s clock, the closest point in the “time zero” set is constantly changing. (For those are getting lost: asking what the “closest point” in A’s “time zero” is the same thing as asking “According to A, where would B have been at the beginning, if he had remained at rest and ended up where he is now?” or “What, according to A, was, at time zero, the same place as B?”) Now, both A and B agree on what the distance to this point is. The trouble is that, according to B, this point that A is measuring from is the wrong point. B doesn’t disagree with the result of A’s measuring, he just says that A is measuring the wrong thing. According to B, he has remained still, so his “closest point” is always the same: where B started from.

So, really, they don’t disagree about the speed of their clocks. They both agree that B’s clock is proceeding through “A time” more slowly than A’s clock. It’s just B says that “A time” isn’t “real” time, it’s a confabulation of time and space.

Originally Posted by RussDill

Einstein says rubbish. Time and space are inseperable, we live in a 4 dimensional world.

[Annoying pedant]According to Einstein, time and space are inseparable[/Annoying pedant]

Quote:

Kinda like if you take slices of a cylinder. The flat slice will show a circle. Other slices will show an ellipse. You are saying that it can't be a circle and an ellipse at the same time. It isn't. Its a cylinder. You just can't visualize the cylinder.

Actually, the issue of visualization isn’t really the issue. We can visualize two dimensional space just fine, but the same issues come up. The real issue is that we want to deal with real numbers. In other words, we want to decompose space into one-dimensional coordinates. And once we do that, we’re going to get different answers depending on what coordinate system we use.

69dodge · 20th June 2007, 09:29 PM

Originally Posted by Ziggurat

Saying that it's the turnaround that matters is equivalent to saying that (say) a right angle in a Euclidean path is what makes it longer than the straight line. That may be true in one sense, but the extra length in the bent path isn't found in the bend itself. Likewise, the "missing" time isn't subtracted from the turnaround of the twin either. Each leg of the traveling twin's path really is shorter than half the stationary twin's path.

I think the traveling twin would agree only if he is permitted to include in "half the stationary twin's path" that portion of the path simultaneous, according to himself, with half his period of acceleration. He would not agree that the stationary twin ages more rapidly than he does, during any part of the unaccelerated portion of his trip.

BillyJoe · 20th June 2007, 09:35 PM

Originally Posted by Paulhoff

It was reworded......

But you still didn't get it right...

Quote:

1. Now spaceman (1) takes one of the clocks and travels in any direction at 99,9999999% speed of light for one second and stops. Spaceman (1) now is 100 ft away from spaceman (2) and the two clocks remanding clocks.

If Spaceman (1) travels for one second, he is going to be travelling a WHOLE lot further than 100 ft.
I think you mean "for one second in frame of reference of Spaceman (2)"

Ziggurat · 20th June 2007, 09:53 PM

Originally Posted by 69dodge

I think the traveling twin would agree only if he is permitted to include in "half the stationary twin's path" that portion of the path simultaneous, according to himself, with half his period of acceleration.

This isn't actually necessary. Pick a frame, any frame, and you can calculate the halfway point of the stationary twin's worldline. Furthermore, every observer, regardless of their frame, will agree on the calculation of the proper time experienced over this half journey. The only reason to sync the halfway point of the traveling twin's acceleration with the halfway point of the stationary twin's path is if we want them to be simultaneous, but there's no reason we need to pick a frame where the two halfway points are simultaneous.

Quote:

He would not agree that the stationary twin ages more rapidly than he does, during any part of the unaccelerated portion of his trip.

Well, sure. I thought this was already clear.

Alkatran · 20th June 2007, 11:49 PM

http://casa.colorado.edu/~ajsh/sr/paradox.html

This website is what I was looking at when I finally understood what special relativity was getting at. Read down the page and think about it for a minute, then go to the next page for Einstein's solution.

The Grave · 21st June 2007, 02:48 AM

Originally Posted by Alkatran

http://casa.colorado.edu/~ajsh/sr/paradox.html

This website is what I was looking at when I finally understood what special relativity was getting at. Read down the page and think about it for a minute, then go to the next page for Einstein's solution.

Looked on this link of yours...pretty good...BUT...take this situation here...

Time dilation

But now suppose Cerulean goes off at velocity v relative to Vermilion, in a direction perpendicular to the direction of the mirror.
A far as Cerulean is concerned, his clock tick-tocks at the same rate as before, a tick at the mirror, a tock on return.
But from Vermilion's point of view, although the distance between Cerulean and his mirror at any instant remains the same as before, the light has further to go. And since the speed of light is constant, Vermilion thinks it takes longer for Cerulean's clock to tick-tock than her own. Thus Vermilion thinks Cerulean's clock runs slow relative to her own.

Let's say I am red {stationary} and my twin 'goes for a fast ride'...I want to keep things descriptive rather than mathematical because... as Einstein said to Bohr..."Q.M. just doesn't ''feel'' right"...

By the way...having never really studied GR/SR I got the slopey turning of the cones thing pretty easily... before they told me. So what 'we' want is a physical description... so back to it....

Now my twin, Blue, moves off and I see his clock showing a slower time, but he will see the same 'normal' time on his clock, as I do with my clock... that's what that diagram tells me. It most certainly doesn't imply that Blue's clock is 'actually' slowing down and he is getting younger, or conversely that I myself {Red} am (not) getting older.

"thinks" is the word! Even 'observes'... but not actual time change and so not actual age difference.

What do you guys think... I teach, and I know it's very frustrating 'not to be able to get the point across'... and I'm not blaming your efforts or 'our' ignorance of GR/SR, I'm just asking/learning....

Thanks, Griff....Interesting...spell check can't live with a tock!

The Grave · 21st June 2007, 03:04 AM

So Lorentz was nearly there before Mr.E! Well all this comes from some easy 'O' level-GCSE maths.... and just add a 'c' to get (v/c) and Bob's' your uncle...

12 + (

v)2 =

2

from which it follows that the Lorentz gamma factor

is related to Cerulean's velocity v by

which is Lorentz's famous formula.

"While Vermilion thinks events happen simultaneously along horizontal planes in this diagram, Cerulean thinks events occur simultaneously along skewed planes. Thus Vermilion thinks her clock

ticks when Cerulean is at the point

, before Cerulean's clock

ticks. Conversely, Cerulean thinks his clock

ticks when Vermilion is at the point

, before Vermilion's clock

ticks."

Now this paragraph above, correct me if I'm wrong (asking for it!)... but how can Red or Blue 'see' something that hasn't happened?

Isn't this all just a bunch of 'observed' effects due to the wave properties of light and in no way reflects physical reality?

I can hear clogs turning....

Griff...Please, no maths... not because I'm afraid of it, I nearly chose it for degree, but instead I want true physical explanations....

eg. to explain diffusion I would say the balls wiggle past each other to get 'somewhere' that's no more important than where they are; because there's nothing to stop them.

No maths needed. So come you Physicists...who's up for it? Trust me, if you can make 'us' understand, you'll be doing science a great favour!

The Grave · 21st June 2007, 03:16 AM

Sorry but what does x=vt mean?

"Step 3

Draw a rectangle, sides 45o from vertical, with the origin at the centre and Cerulean at one corner. The rectangle represents the path of light rays which Cerulean uses to define a hypersurface of simultaneity, as described on the simultaneity page, Spacetime diagram illustrating simultaneity from Cerulean's point of view. Draw the extra diagonal across this rectangle. The diagonal is a hypersurface (reduced to a line) of simultaneity, a `now' line, for Cerulean. The now line lies along t = vx in Vermilion's frame, and along t´ = 0 in Cerulean's frame. "

Earlier they said "x=vt" which is fine; displacement = vel x time.

But Time = vel x displacement ? I don't get that!

MortFurd · 21st June 2007, 03:17 AM

Originally Posted by polaristutoring@tisc

Looked on this link of yours...pretty good...BUT...take this situation here...

Time dilation

http://casa.colorado.edu/~ajsh/sr/timev.gif
But now suppose Cerulean goes off at velocity v relative to Vermilion, in a direction perpendicular to the direction of the mirror.
A far as Cerulean is concerned, his clock tick-tocks at the same rate as before, a tick at the mirror, a tock on return.
But from Vermilion's point of view, although the distance between Cerulean and his mirror at any instant remains the same as before, the light has further to go. And since the speed of light is constant, Vermilion thinks it takes longer for Cerulean's clock to tick-tock than her own. Thus Vermilion thinks Cerulean's clock runs slow relative to her own.

Let's say I am red {stationary} and my twin 'goes for a fast ride'...I want to keep things descriptive rather than mathematical because... as Einstein said to Bohr..."Q.M. just doesn't ''feel'' right"...

By the way...having never really studied GR/SR I got the slopey turning of the cones thing pretty easily... before they told me. So what 'we' want is a physical description... so back to it....

Now my twin, Blue, moves off and I see his clock showing a slower time, but he will see the same 'normal' time on his clock, as I do with my clock... that's what that diagram tells me. It most certainly doesn't imply that Blue's clock is 'actually' slowing down and he is getting younger, or conversely that I myself {Red} am (not) getting older.

"thinks" is the word! Even 'observes'... but not actual time change and so not actual age difference.
What do you guys think... I teach, and I know it's very frustrating 'not to be able to get the point across'... and I'm not blaming your efforts or 'our' ignorance of GR/SR, I'm just asking/learning....

Thanks, Griff....Interesting...spell check can't live with a tock!

Wrongo. Very much a real age difference.
Take a look here for data on an experiment in which this was acutally tested.

Remember, these clocks traveled only at the speed of a normal commercial jet. The difference was present and measurable, and in line with the predictions made by relativity theory.

Here's a more prosaic application of relativity. Your GPS receiver depends on compensation for relatavistic effects for it's accuracy.

20th June 2007, 08:12 AM	#81
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 26,285	Originally Posted by Paulhoff OK, let’s try this if it hasn’t been done all ready. Not explicitly, but the key difference here is the distinction between what you see and what you observe, and if you understand that distinction, these questions are easy. Quote: We have two spacemen and three clocks and they are big clocks that can be seen from a 100 feet with no problem. The speed of light for this experiment is 100 ft. per second. They are all somewhere in deep space, who cares where. The clocks are all reading the same time. I am going to assume furthermore that the clocks and the spacemen also start at the same place and at rest with respect to some specified reference frame (which I'll refer to as the start frame). Quote: 1. Now one of the spacemen takes one of the clocks and travels at 99.9999% speed of light for one second and stops and looks back at the other spaceman and the two clocks, he is 100 ft away. The one second time must be refering to the time in the start frame, not the time he experiences. I will refer to this spaceman as spaceman 1, and his clock as clock 1. Quote: Is the time on his clock is the same as the other two clocks. Is the time on his clock is one second behind the other two clocks. Is the time on his clock is two seconds behind the other two clocks. Is the time on his clock is one second ahead of the other two clocks. Is the time on his clock is two seconds ahead of the other two clocks. The time on clock 1 is observed to be different than the other two clocks. The proper time he experiences while moving with respect to the start frame is very short, so clock 1 will be observed to be almost 1 second behind the other clocks. However, spaceman 1 will see the other clocks as being about the same time as clock 1, while spaceman 2 will see clock 1 being two seconds behind the other clocks, since there's an approximately 1 second delay for the signal to get transmitted from one location to another. They don't see the same thing, but they both observe the same thing: clock 1 is ~1 second behind the other clocks. Quote: 2. Now the other spaceman takes one of the two remaining clocks I'll call this spaceman 2 and clock 2, with the clock left behind being clock 3. Quote: and travels very slowly, about one inch a second to the other spaceman. You can go slower than that, the point is that we take the limiting behavior of slow travel time. Quote: When he gets to the other spaceman he looks at the other spaceman’s clock. Clock 1 and clock 2 are now in the same location. This is important. Quote: Is the time on his clock is the same as the other spaceman’s clock. Is the time on his clock is one second behind the other spaceman’s clock. Is the time on his clock is two seconds behind the other spaceman’s clock. Is the time on his clock is one second ahead of the other spaceman’s clock. Is the time on his clock is two seconds ahead of the other spaceman’s clock. Clock 1 will be observed to be about 1 second behind clock 2, as before. Because they are now in the same location, all observers will also see this same ~1 second difference, since any time delay in signal propagation to any location is the same from both clocks. Quote: 3. Now the other spaceman looks back at the other clock left behind. Is the time on his clock is the same as the other clock. Is the time on his clock is one second behind the other clock. Is the time on his clock is two seconds behind the other clock. Is the time on his clock is one second ahead of the other clock. Is the time on his clock is two seconds ahead of the other clock. Spaceman 1 and 2 both observe that clock 2 is reading approximately the same time as clock 3. Spaceman 1 and 2 both see clock 2 reading approximately 1 second ahead of clock 3 because of the time delay for the signal from clock 3 to reach them.
	__________________ "As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law

20th June 2007, 08:29 AM	#82
Thabiguy Muse Join Date: Feb 2007 Posts: 783	Originally Posted by Gertrude When the two twins meet, they do have different ages, but it is not because of SR and its time dilation. The effect is a result of GR: the change occurs when the traveling twin decelerates, changes direction and accelerates back towards the earth. Not quite. The twin paradox is indeed because of SR and time dilation, and nothing else. GR needn't be invoked at all. Explanation.
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20th June 2007, 08:39 AM	#83
ponderingturtle Orthogonal Vector Join Date: Jul 2006 Location: Tarrytown, NY Posts: 26,557	Originally Posted by ynot The rock only exists in its own rest frame and falls in a straight line toward the force of gravity when dropped. That it is perceived to be moving on an angle when observed from another rest frame is an illusion. To use the movie analogy used earlier, the single event is being separated in to a strip of many events and this creates the illusion. Is it me, or does this seem to disprove orbital mechanics?
	__________________ Sufficiently advanced Woo is indistinguishable from Parody "There shall be no poofing in science" Paul C. Anagnostopoulos Force ***** on reasons back" Ben Franklin

20th June 2007, 08:47 AM	#84
Gertrude Scholar Join Date: May 2007 Posts: 106	Originally Posted by Thabiguy Not quite. The twin paradox is indeed because of SR and time dilation, and nothing else. GR needn't be invoked at all. Explanation. Actually there had been a debate among writers in the late 50's regarding the need to introduce GR to explain the twin paradox but nowadays, most physicists agree that SR is sufficient. And yes, SR can deal with acceleration. But we must be clear. SR deals only with acceleration as measured in inertial reference frames. This is not quite the case in the twin paradox, where one of the twins himself undergoes acceleration. In this sense, it is a problem of GR. We are appealing to the reality of this acceleration to explain the paradox. The reason most writers agree that SR is sufficient in explaining the paradox is that we can use methods such as the Doppler effect (ie. pretend the twins send each other signals and count these) as described by SR and obtain the same result. But this is not to say that SR can handle acceleration as experienced by an observer.
Gertrude Scholar Join Date: May 2007 Posts: 106

20th June 2007, 08:57 AM	#85
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 26,285	Originally Posted by Gertrude Do not confuse this with the twin paradox. When the two twins meet, they do have different ages, but it is not because of SR and its time dilation. The effect is a result of GR: the change occurs when the traveling twin decelerates, changes direction and accelerates back towards the earth. The twin paradox imho is the worst example in teaching SR. This is simply wrong. Let me expand upon Thai's response: special relativity can accomodate acceleration. It would be a crap theory if it couldn't. The math can get ugly and so most introductory treatments don't address it, but there is absolutely NO impediment to using acceleration within special relativity. It's mathematically exactly equivalent to the problem of curved lines in Euclidean space. You don't have to abandon Euclidean geometry or the pythagorean theorem in order to calculate the length of a curved line, and you don't need to go to general relativity to handle acceleration. But it is a common misconception that you need it, for two reasons: first, because it's often not taught (because you don't need to do it to understand the basics of SR), and second, because of a misunderstanding of the equivalence principle (which I could go into in more detail but isn't really important at the moment). So you're far from alone in your mistake, but the twin paradox really can be handled completely within the confines of special relativity, including the bits where one twin (or even both) accelerates. The ONLY thing you cannot handle correctly within the confines of special relativity is gravity. If you think the twin paradox gets resolved because of some GR effect, then you don't understand the problem.
	__________________ "As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law

20th June 2007, 09:27 AM	#86
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 26,285	Originally Posted by Gertrude Actually there had been a debate among writers in the late 50's regarding the need to introduce GR to explain the twin paradox but nowadays, most physicists agree that SR is sufficient. And yes, SR can deal with acceleration. But we must be clear. SR deals only with acceleration as measured in inertial reference frames. This is not quite the case in the twin paradox, where one of the twins himself undergoes acceleration. In this sense, it is a problem of GR. We are appealing to the reality of this acceleration to explain the paradox. The reason most writers agree that SR is sufficient in explaining the paradox is that we can use methods such as the Doppler effect (ie. pretend the twins send each other signals and count these) as described by SR and obtain the same result. But this is not to say that SR can handle acceleration as experienced by an observer. This is also incorrect. There's generally little reason to treat special relativity problems from within a non-inertial frames (and plenty of reason not to - the math can get ugly), but there are no actual impediments. All the mathematical framework is there already. It's no different than Newtonian physics in that respect - using a rotating coordinate system introduces coriolis and centripidal pseudo-forces, but that's it. The reason that GR is different is not that it can handle non-inertial reference frames and SR can't, it's that GR can handle non-uniformity of that non-inertiality. That requires a whole new (and much more complicated) mathematical framework, but just non-inertiality can be handled within special relativity alone.
	__________________ "As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law

20th June 2007, 09:29 AM	#87
Gertrude Scholar Join Date: May 2007 Posts: 106	Originally Posted by Ziggurat This is simply wrong. Let me expand upon Thai's response: special relativity can accomodate acceleration. It would be a crap theory if it couldn't. The math can get ugly and so most introductory treatments don't address it, but there is absolutely NO impediment to using acceleration within special relativity. It's mathematically exactly equivalent to the problem of curved lines in Euclidean space. You don't have to abandon Euclidean geometry or the pythagorean theorem in order to calculate the length of a curved line, and you don't need to go to general relativity to handle acceleration. But it is a common misconception that you need it, for two reasons: first, because it's often not taught (because you don't need to do it to understand the basics of SR), and second, because of a misunderstanding of the equivalence principle (which I could go into in more detail but isn't really important at the moment). So you're far from alone in your mistake, but the twin paradox really can be handled completely within the confines of special relativity, including the bits where one twin (or even both) accelerates. The ONLY thing you cannot handle correctly within the confines of special relativity is gravity. If you think the twin paradox gets resolved because of some GR effect, then you don't understand the problem. Like I said, SR can very well deal with accelerated motion as measured in inertial frames. But to deal with accelerated frames, we must use approximations (and ingenuity). There is no rigorous treatment (ie. without the need for approximations) that I know of. If there is, please do point me to some reference. The reason I stated that it was a problem of GR (and you are right that I should have weighed my words more as it does come accross as a 'miracle' of GR) is that we must appeal to the reality of this acceleration to explain the paradox. Some treatment that does not involve GR can very well explain the paradox (for instance, the relativistic Doppler effect), but the turnaround is the key. I stated this because it appears that many people take the example of the twin paradox to illustrate and understand SR. In this sense, they will think that time dilation/length contraction is still valid when the observers meet. But the twin paradox appears at first sight to be in contradiction with the tenets of SR and imho should not be used to understand the basics of SR.
Gertrude Scholar Join Date: May 2007 Posts: 106

20th June 2007, 09:37 AM	#88
Gertrude Scholar Join Date: May 2007 Posts: 106	Originally Posted by Ziggurat This is also incorrect. There's generally little reason to treat special relativity problems from within a non-inertial frames (and plenty of reason not to - the math can get ugly), but there are no actual impediments. All the mathematical framework is there already. It's no different than Newtonian physics in that respect - using a rotating coordinate system introduces coriolis and centripidal pseudo-forces, but that's it. The reason that GR is different is not that it can handle non-inertial reference frames and SR can't, it's that GR can handle non-uniformity of that non-inertiality. That requires a whole new (and much more complicated) mathematical framework, but just non-inertiality can be handled within special relativity alone. When I mention 'acceleration', I certainly refer to "all kinds" of accelerations, not only uniform acceleration... But I'd be curious to see a rigorous treatment of (uniform) accelerating frames within SR. If you have any reference, I'd be highly interested.
Gertrude Scholar Join Date: May 2007 Posts: 106

20th June 2007, 09:55 AM	#89
boooeee Dart Fener Join Date: Aug 2002 Location: The Lando System Posts: 2,403	Originally Posted by Ziggurat This is simply wrong. Let me expand upon Thai's response: special relativity can accomodate acceleration. It would be a crap theory if it couldn't. The math can get ugly and so most introductory treatments don't address it, but there is absolutely NO impediment to using acceleration within special relativity. It's mathematically exactly equivalent to the problem of curved lines in Euclidean space. You don't have to abandon Euclidean geometry or the pythagorean theorem in order to calculate the length of a curved line, and you don't need to go to general relativity to handle acceleration. But it is a common misconception that you need it, for two reasons: first, because it's often not taught (because you don't need to do it to understand the basics of SR), and second, because of a misunderstanding of the equivalence principle (which I could go into in more detail but isn't really important at the moment). So you're far from alone in your mistake, but the twin paradox really can be handled completely within the confines of special relativity, including the bits where one twin (or even both) accelerates. The ONLY thing you cannot handle correctly within the confines of special relativity is gravity. If you think the twin paradox gets resolved because of some GR effect, then you don't understand the problem. I know this issue has probably been covered ad nauseum in this forum, but can somebody provide an SR interpretation of the travelling twin's point of view? In other words, the travelling twin is the one that accelerates away from his twin and then accelerates back. The travelling twin is sitting in his spaceship and observing the age of his twin. How does the travelling twin use Special Relativity to explain that time is moving faster for his twin?
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20th June 2007, 10:02 AM	#90
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 26,285	Originally Posted by Gertrude When I mention 'acceleration', I certainly refer to "all kinds" of accelerations, not only uniform acceleration... That's not what I mean (though I phrased it poorly). I don't mean that your acceleration is uniform over time (there's no requirement for that, just as Euclidean geometry can handle curves with varying curvature), I mean that you're adopting a frame in which all of space-time adapts to your single acceleration. This goes back to the equivalence principle: standing still in a gravitational field is locally equivalent to accelerating on a flat space-time background. If you can handle acceleration with special relativity, why can't you handle gravity? Because gravity isn't local, and it isn't uniform. If we look at two people sitting on either side of a planet, their "accelerating" frames are accelerating away from each other (because gravity is not uniform, but pulls them in different directions), and yet the distance between them isn't changing. So any attempt to simply use a special relativity non-inertial reference frame to describe these two guys is going to fail, but not because special relativity can't handle accelerated frames. Quote: But I'd be curious to see a rigorous treatment of (uniform) accelerating frames within SR. If you have any reference, I'd be highly interested. Not at my fingertips, sorry.
	__________________ "As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law

20th June 2007, 10:15 AM	#91
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 26,285	Originally Posted by Gertrude The reason I stated that it was a problem of GR (and you are right that I should have weighed my words more as it does come accross as a 'miracle' of GR) is that we must appeal to the reality of this acceleration to explain the paradox. Some treatment that does not involve GR can very well explain the paradox (for instance, the relativistic Doppler effect), but the turnaround is the key. Depends what you mean by that. Let's look at the metric: ds² = dx² + dy² + dz² - (ct)² To find the proper time of any path, we integrate this quantity along that path. This is EXACTLY like taking the integral of a Euclidean metric: ds² = dx² + dy² + dz² to find the length of a curve, it's just a different metric. Now, we can make our traveling twin's trajectory curved (finite acceleration), or we can make it kinked (infinite acceleration, instant velocity change), it doesn't matter: the process of taking that path integral is the same either way. Saying that it's the turnaround that matters is equivalent to saying that (say) a right angle in a Euclidean path is what makes it longer than the straight line. That may be true in one sense, but the extra length in the bent path isn't found in the bend itself. Likewise, the "missing" time isn't subtracted from the turnaround of the twin either. Each leg of the traveling twin's path really is shorter than half the stationary twin's path. Quote: I stated this because it appears that many people take the example of the twin paradox to illustrate and understand SR. In this sense, they will think that time dilation/length contraction is still valid when the observers meet. I'm not even sure what you mean by this anymore. Quote: But the twin paradox appears at first sight to be in contradiction with the tenets of SR and imho should not be used to understand the basics of SR. That's the whole point, though: it's only an apparent contradiction if you don't really understand what those tenets actually are.
	__________________ "As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law

20th June 2007, 10:31 AM	#92
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 26,285	Originally Posted by boooeee I know this issue has probably been covered ad nauseum in this forum, but can somebody provide an SR interpretation of the travelling twin's point of view? In other words, the travelling twin is the one that accelerates away from his twin and then accelerates back. The travelling twin is sitting in his spaceship and observing the age of his twin. How does the travelling twin use Special Relativity to explain that time is moving faster for his twin? Recall that your space axis (which marks all points in space at the same time for your reference frame) tilts when you change reference frame. If you're accelerating, that axis is continually tilting. So as the traveling twin accelerates back towards the stationary twin, the intersection of his space axis with the stationary twin's world-line move upwards (in time). During the acceleration phase, then, the traveling twin will indeed observe the stationary twin's clock to be moving faster, in his accelerated frame. If he changes velocity instantaneously, then his space axis will tilt instantly, and he will observe that the stationary twin's current time will have instantaneously jumped forward. An important caveat, though: this is what he oberves, not what he sees. Even in the case of instantaneous velocity change, there's no discontinuity in what the twin sees. If you want to ask what the twins see (which is also interesting, BTW), that's actually pretty straight forward. While moving away from each other, each twin sees the other twin's clock slowed by time dilation, plus additional slowing from doppler shift. When the traveling twin turns around and heads back, he will see the stationary twin's clock slowed by time dilation, but also speeded up by the doppler shift. So for half of his journey, the doppler shift increases the clock rate, and for half of it the doppler shift decreases the clock rate. The stationary twin, however, doesn't see the traveling twin turn around at the moment he turns around, because the signal is time-delayed. But when he does see the traveling twin heading back towards him, he will see the clock slowed by time dilation, plus speeded up because of the doppler shift. Unlike the traveling twin, however, he sees the doppler shift slowing the clock for MORE than half the time interval, and speeding up the clock for LESS than half the time interval (because it took time for the light from the turnaround to get back to him). Which is how the twins end up with different times on their clocks: what they see over the length of the whole journey isn't the same, even if instantaneous snapshots on the outward and return journeys look superficially symmetric.
	__________________ "As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law

20th June 2007, 11:18 AM	#93
Gertrude Scholar Join Date: May 2007 Posts: 106	Originally Posted by Ziggurat Depends what you mean by that. Let's look at the metric: ds² = dx² + dy² + dz² - (ct)² To find the proper time of any path, we integrate this quantity along that path. This is EXACTLY like taking the integral of a Euclidean metric: ds² = dx² + dy² + dz² to find the length of a curve, it's just a different metric. Now, we can make our traveling twin's trajectory curved (finite acceleration), or we can make it kinked (infinite acceleration, instant velocity change), it doesn't matter: the process of taking that path integral is the same either way. Saying that it's the turnaround that matters is equivalent to saying that (say) a right angle in a Euclidean path is what makes it longer than the straight line. That may be true in one sense, but the extra length in the bent path isn't found in the bend itself. Likewise, the "missing" time isn't subtracted from the turnaround of the twin either. Each leg of the traveling twin's path really is shorter than half the stationary twin's path. Absolutely. The point here is that, neglecting accelerations, two inertial frames cannot be brought together and still experience different durations. I could put it differently: the lack of symmetry in the paradox is key. The fact that one twin definately feels an acceleration (be it a constant acceleration as in a circle or a sudden turnaround) while the other one doesn't makes up for the difference. In SR, time dilation works both ways. In all cases we can say: "A is travelling at X m/s relative to A" in the same way as "B is travelling at X m/s relative to A". If the only effect that we considered was velocity, then there indeed would be a paradox, as the effects of SR are symmetric for both observers, ie. both observers would see the other aging slower. It is the idea that the situation is not symmetrical that resolves the paradox. In this sense, I think the subtlety may lead to an incorrect understanding that time dilation is not only apparent and occurs to the 'fastest traveler' only(whatever that means).
Gertrude Scholar Join Date: May 2007 Posts: 106	Last edited by Gertrude; 20th June 2007 at 11:20 AM.

20th June 2007, 11:32 AM	#94
Yllanes Muse Join Date: Sep 2005 Location: Madrid Posts: 826	Originally Posted by Gertrude But I'd be curious to see a rigorous treatment of (uniform) accelerating frames within SR. If you have any reference, I'd be highly interested. Do you mean references as in web pages or books? Because if it is the latter, most textbooks discuss them. For example, let us see how uniform accelerated motion looks in SR. First of all we must define it, because acceleration is relative. I'm going to find the trajectory of a particle with constant acceleration in its instantaneous rest frame. This rest frame is the inertial reference frame that happens to have the same velocity as the particle at each moment, and so it is different at each point of the movement. Let a be the ordinary acceleration of Newtonian physics, and v the ordinary velocity. As you may now, we can define a four velocity if we differentiate the position x with respect to the proper time $\footnotesize<br /> \begin{align}<br /> x^\mu&=(ct,\ x,\ y,\ z),\\<br /> u^\mu&=\frac{\mathrm{d} x^\mu}{\mathrm{d} \tau} = \left(\gamma c,\ \gamma {\vec v}\right),\qquad \tau =t \sqrt{1-v^2/c^2} = t/\gamma,\\<br /> w^\mu&=\frac{\mathrm{d} x^\mu}{\mathrm{d} \tau^2} <br /> \end{align}<br />$ The important thing about four vectors is that their magnitude is the same in all reference frames. We write that magnitude as u² and we have to remember it is not calculated with the Euclidean metric, but that the time component is subtracted. For example $\footnotesize<br /> \[u^2 = (u^1)^2+(u^2)^2+(u^3)^2-(u^0)^2 = -c^2<br /> \]<br />$ In our accelerated motion, we now that w^m = (0, a, 0,0) in the rest frame. So w² = a² in every reference frame. We are interested in the acceleration as seen from a 'fixed' frame. So we go to the value of w^m in that system and force it to have w²=a². This leads to the equation (v is the velocity of the particle in the fixed frame). $\footnotesize<br /> \[\frac{\mathrm{d}}{\mathrm{d}t} \frac{v}{\sqrt{1-\frac{v^2}{c^2}}} = a<br /> \]<br />$ It's just a matter of integrating now. If we set x = 0 at t = 0 the movement is $\footnotesize<br /> \begin{align}<br /> v&=\frac{at}{\sqrt{1+\frac{a^2t^2}{c^4}}}, &<br /> x &= \frac{c^2}{a}\left(\sqrt{1+\frac{a^2t^2}{c^2}}-1\right)<br /> \end{align}<br />$ And it is inmediate to see that in the limit at << c we recover the Newtonian equations for the uniform accelerated movement ( v ~ at, x ~ at²/2). As you see, it is very simple to handle accelerated motion in SR.

20th June 2007, 11:33 AM	#95
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 26,285	Originally Posted by Gertrude It is the idea that the situation is not symmetrical that resolves the paradox. Correct. Quote: In this sense, I think the subtlety may lead to an incorrect understanding that time dilation is not only apparent and occurs to the 'fastest traveler' only(whatever that means). I'm confused by what you mean here. If you're claiming that time dilation is only an apparent effect, that is wrong. It's a very real effect. But it can be wrongly applied to the traveling twin paradox if the issue of simultaneity is not handled correctly as well.
	__________________ "As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law

20th June 2007, 12:17 PM	#96
Gertrude Scholar Join Date: May 2007 Posts: 106	Originally Posted by Yllanes First of all we must define it, because acceleration is relative. I'm going to find the trajectory of a particle with constant acceleration in its instantaneous rest frame. This rest frame is the inertial reference frame that happens to have the same velocity as the particle at each moment, and so it is different at each point of the movement. I lost you right there! If the rest frame has the same velocity as the particle, it cannot be inertial since the particle undergoes acceleration!? Quote: In our accelerated motion, we now that wm = (0, a, 0,0) in the rest frame. So w2 = a2 in every reference frame. We are interested in the acceleration as seen from a 'fixed' frame. So we go to the value of wm in that system and force it to have w2=a2. This leads to the equation (v is the velocity of the particle in the fixed frame). Ok. So you are analysing the motion as an observer in an inertial frame. Actually, I have seen this treatment before. What interests me is looking at events from an accelerating frame. If you have any suggestion...
Gertrude Scholar Join Date: May 2007 Posts: 106	Last edited by Gertrude; 20th June 2007 at 12:27 PM.

20th June 2007, 12:42 PM	#97
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 26,285	Originally Posted by Gertrude I lost you right there! If the rest frame has the same velocity as the particle, it cannot be inertial since the particle undergoes acceleration!? No. Rather, acceleration is always measured with respect to some specific reference frame. This doesn't sound significant until you realize that 1) the same accelerating trajectory will have a different acceleration when measured from different reference frames, and 2) this is not the same as classical mechanics where velocity is relative but acceleration is absolute. To adopt constant proper acceleration, you basically find a trajectory where the instantaneous acceleration in the inertial reference frame defined by the instantaneous velocity is always constant. Which frame you use has to keep changing, which means that in any fixed reference frame the acceleration will not appear constant, but each of those tangent frames is still inertial.
	__________________ "As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law

20th June 2007, 12:46 PM	#98
Gertrude Scholar Join Date: May 2007 Posts: 106	Originally Posted by Ziggurat I'm confused by what you mean here. If you're claiming that time dilation is only an apparent effect, that is wrong. It's a very real effect. But it can be wrongly applied to the traveling twin paradox if the issue of simultaneity is not handled correctly as well. I am quite good at expressing things in the most confusing possible manner. I'll give an example that may be clearer. The statement "moving clocks run slower" is confusing because it does imply that time slows down with speed (relative to any observer). The reality is that moving clocks measure events occuring in a slower inertial frame as having longer duration than what a clock at rest relative to said event would measure. The confusion is only increased with the twin paradox. It could be interpreted that velocity indeed slows time down for every observer (as 'confirmed' by the fact that when they meet, the two twins no longer have the same age). But this is not the case. If A and B are in two different inertial frames, B would see the clock of A running slower because it is moving relative to A and A would see the clock of B running slower because it is moving relative to B. There is no 'absolute' time dilation and it is in this sense that I say time dilation is apparent.
Gertrude Scholar Join Date: May 2007 Posts: 106

20th June 2007, 01:08 PM	#99
Paulhoff You can't expect perfection. Join Date: May 2005 Location: South Florida Posts: 12,575	OK, let’s try this again. We have two spacemen and three clocks and the clocks are big so that they can be seen and read from a 100 feet with no problem. The speed of light for this experiment is 100 ft. per second. The spacemen and clocks are all together somewhere in deep space, who cares where, and who cares about their speed. The clocks are all reading the same time. 1. Now spaceman (1) takes one of the clocks and travels in any direction at 99,9999999% speed of light for one second and stops. Spaceman (1) now is 100 ft away from spaceman (2) and the two clocks remanding clocks, and spaceman (1) looks at the two clocks with spaceman (2). A. Is the time on spaceman's (1) clock is the same as the other two clocks. B. Is the time on spaceman's (1) clock is one second behind the other two clocks. C. Is the time on spaceman's (1) clock is two seconds behind the other two clocks. D. Is the time on spaceman's (1) clock is one second ahead of the other two clocks. E. Is the time on spaceman's (1) clock is two seconds ahead of the other two clocks. 2. Now spaceman (2) takes one of the two remaining clocks and travels very slowly, about one inch a second to spaceman (1). When he gets to spaceman (1) he looks at spaceman's (2) clock. A. Is the time on spaceman's (2) clock is the same as spaceman’s (1) clock. B. Is the time on spaceman's (2) clock is one second behind spaceman’s (1) clock. C. Is the time on spaceman's (2) clock is two seconds behind spaceman’s (1) clock. D. Is the time on spaceman's (2) clock is one second ahead of spaceman’s (1) clock. E. Is the time on spaceman's (2) clock is two seconds ahead of spaceman’s (1) clock. 3. Now spaceman (2) looks back at the clock left behind. A. Is the time on his clock is the same as the clock left behind. B. Is the time on his clock is one second behind the left clock. C. Is the time on his clock is two seconds behind the left clock. D. Is the time on his clock is one second ahead of the left clock. E. Is the time on his clock is two seconds ahead of the left clock. Paul
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20th June 2007, 01:11 PM	#100
Yllanes Muse Join Date: Sep 2005 Location: Madrid Posts: 826	Originally Posted by Gertrude I lost you right there! If the rest frame has the same velocity as the particle, it cannot be inertial since the particle undergoes acceleration!? Ziggurat answered this. There is a different rest frame at each moment of time. Quote: Ok. So you are analysing the motion as an observer in an inertial frame. Actually, I have seen this treatment before. What interests me is looking at events from an accelerating frame. If you have any suggestion... Ah, well it's just the same. For example, t is time as measured by the stationary observer. To switch to the accelerated observer we have to use his proper time, which is just a matter of integrating $\footnotesize<br /> \[<br /> \tau=\int_0^t \sqrt{1-\tfrac{v^2}{c^2}}= \frac{c}{a} \mathrm{arcsinh}\left(\frac{at}{c}\right)\]<br />$ so the distance from its starting point, as measured by the accelerated particle, is $\footnotesize<br /> \[<br /> x(\tau)=\left(\cosh \tfrac{at}{c} - 1\right)\frac{c^2}{a}<br /> \]<br />$ Incidentally, such a uniformly accelerated observer can outrun a photon, even though his velocity never reaches c.
	Last edited by Yllanes; 20th June 2007 at 01:15 PM.

20th June 2007, 01:14 PM	#101
Thabiguy Muse Join Date: Feb 2007 Posts: 783	Originally Posted by Paulhoff OK, let’s try this again. We have already answered you, here and here. Why do you keep asking the same question?
Thabiguy Muse Join Date: Feb 2007 Posts: 783

20th June 2007, 01:20 PM	#102
Paulhoff You can't expect perfection. Join Date: May 2005 Location: South Florida Posts: 12,575	Originally Posted by Thabiguy We have already answered you, here and here. Why do you keep asking the same question? It was reworded...... Paul
	__________________ For our money "IN WHICH GOD DO YOU TRUST" Much worse than the Question not asked, is the Answer not Given Don't accept an answer that can't be questioned - God is Surperfluous A society fails when ignorance outweighs knowledge Science doesn’t know everything, but religion doesn’t know anything Life is so horrent and also so beautiful, but without it there is nothing

20th June 2007, 01:25 PM	#103
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 26,285	Originally Posted by Gertrude There is no 'absolute' time dilation and it is in this sense that I say time dilation is apparent. Time dilation is relative, that's certainly true, but I think the word "apparent" is a poor choice, precisely because the effect is still quite real and definite. Here's a nice little problem involving time dilation which might help out a bit. We set up a "rest" reference frame, stick an observer at the origin and let him stay there (along with a clock) for the duration of our experiment. Now we take several volunteers with clocks, sync all the clocks together, and have them start moving in circular trajectories which all start at the origin (and hence return there periodically), which they will travel along at constant speed (with respect to our rest frame). Now one of the nice features of using circular orbits is that the acceleration and the speed are now independent parameters. We can have high-speed trajectories with small accelerations by using large circles, we can have small speed trajectories with large accelerations by using small circles, or any other combination. Now we ask, when these traveling clocks swing by the origin, how do their times compare with our stationary clock? The answer is quite simply that the travelling clocks are slowed down compared to the stationary clock because of time dilation, by a factor given by their speed. Now, just like the twin paradox, there's an asymmetry in the problem because the moving clocks experience acceleration and the stationary clock doesn't. But here's the interesting bit which becomes clear because of the setup of the problem: the acceleration never actually enters into the calculation! Remember, we can make trajectories which have the exact same speed but different accelerations, or different speeds but the exact same acceleration, but the ONLY thing we need to know to figure out how much time dilation a trajectory experiences is the speed. The answer is totally independent of the value of the acceleration, and so acceleration is, in one sense, completely irrelevant to that observed time dilation.
	__________________ "As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law

20th June 2007, 01:28 PM	#104
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 26,285	Originally Posted by Paulhoff It was reworded...... Paul My answers in post 81 still apply. Do you have any questions about those answers?
	__________________ "As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law

20th June 2007, 01:34 PM	#105
slyjoe Muse Join Date: Mar 2007 Location: Near Harmonica Virgins Posts: 969	Nice illustration Zig - never seen that one before!

20th June 2007, 01:36 PM	#106
slyjoe Muse Join Date: Mar 2007 Location: Near Harmonica Virgins Posts: 969	ETA: But I'm not sure if it will help those who don't want to use the math. ETA: Not sure what just happened to cause 2 posts here.

20th June 2007, 01:56 PM	#107
Paulhoff You can't expect perfection. Join Date: May 2005 Location: South Florida Posts: 12,575	Originally Posted by Yllanes Incidentally, such a uniformly accelerated observer can outrun a photon, even though his velocity never reaches c. Really, then how do his atoms not fall apart because the virtual photons in the atom, as you say, can't keep up? Paul
	__________________ For our money "IN WHICH GOD DO YOU TRUST" Much worse than the Question not asked, is the Answer not Given Don't accept an answer that can't be questioned - God is Surperfluous A society fails when ignorance outweighs knowledge Science doesn’t know everything, but religion doesn’t know anything Life is so horrent and also so beautiful, but without it there is nothing

20th June 2007, 01:58 PM	#108
Paulhoff You can't expect perfection. Join Date: May 2005 Location: South Florida Posts: 12,575	Originally Posted by boooeee How does the travelling twin use Special Relativity to explain that time is moving faster for his twin? In a few words, there is only so much speed to go around for all 3 dimansions. The more you used in one direction and/or dimension the less you have for the other two. Paul
	__________________ For our money "IN WHICH GOD DO YOU TRUST" Much worse than the Question not asked, is the Answer not Given Don't accept an answer that can't be questioned - God is Surperfluous A society fails when ignorance outweighs knowledge Science doesn’t know everything, but religion doesn’t know anything Life is so horrent and also so beautiful, but without it there is nothing

20th June 2007, 02:15 PM	#109
Paulhoff You can't expect perfection. Join Date: May 2005 Location: South Florida Posts: 12,575	Originally Posted by Ziggurat My answers in post 81 still apply. Do you have any questions about those answers? Apparently my question was not clear enough since you had such a long answer. Paul
	__________________ For our money "IN WHICH GOD DO YOU TRUST" Much worse than the Question not asked, is the Answer not Given Don't accept an answer that can't be questioned - God is Surperfluous A society fails when ignorance outweighs knowledge Science doesn’t know everything, but religion doesn’t know anything Life is so horrent and also so beautiful, but without it there is nothing

20th June 2007, 02:26 PM	#110
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 26,285	Originally Posted by Paulhoff Really, then how do his atoms not fall apart because the virtual photons in the atom, as you say, can't keep up? He didn't make this explicit, but what he's refering to is photons which you have a head start on. In other words, for constant proper acceleration and a sufficient head start, you can stay ahead of a photon forever (but you can't ever stop accelerating). More details here: http://en.wikipedia.org/wiki/Event_h...rated_particle
	__________________ "As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law

20th June 2007, 03:26 PM	#111
Paulhoff You can't expect perfection. Join Date: May 2005 Location: South Florida Posts: 12,575	Originally Posted by Ziggurat He didn't make this explicit, but what he's refering to is photons which you have a head start on. In other words, for constant proper acceleration and a sufficient head start, you can stay ahead of a photon forever (but you can't ever stop accelerating). This most likely would work a lot easier if space itself is expanding, which looks like it is now happening. Paul
	__________________ For our money "IN WHICH GOD DO YOU TRUST" Much worse than the Question not asked, is the Answer not Given Don't accept an answer that can't be questioned - God is Surperfluous A society fails when ignorance outweighs knowledge Science doesn’t know everything, but religion doesn’t know anything Life is so horrent and also so beautiful, but without it there is nothing

20th June 2007, 07:27 PM	#112
Art Vandelay Illuminator Join Date: May 2004 Posts: 4,790	Originally Posted by Yllanes The important thing about four vectors is that their magnitude is the same in all reference frames. I think that for clarity's sake, you should write "four-vectors" and "four-velocities". Originally Posted by Gertrude I lost you right there! If the rest frame has the same velocity as the particle, it cannot be inertial since the particle undergoes acceleration!? It's like the tangent lines in the derivative. Each point has its own tangent line, so even though each line is straight, the curve is not. Originally Posted by Gertrude I am quite good at expressing things in the most confusing possible manner. I'll give an example that may be clearer. The statement "moving clocks run slower" is confusing because it does imply that time slows down with speed (relative to any observer). The reality is that moving clocks measure events occuring in a slower inertial frame as having longer duration than what a clock at rest relative to said event would measure. I think that's still not quite right. Rather, if you have events proceeding, in one reference frame, at the same place, but at different places, then in another reference frame, they will be assigned different spatial corrdinates, and temporal coordinates that increment at a lower rate than those in the other reference frame. A big problem in understanding relativity is that people don't understand how relative it is. When they're told that moving clocks run slower, they think that there is some absolute "time" through which the clock moves more slowly. Rather, the clock runs at normal speed in its own reference frame. That is, there is a quantity called "time" which increases at a normal rate. In another reference frame, the clock is measured with a different quantity, and found to have less of that quantity. That this other quantity is given the same name as the other one is what causes the confusion. If we look at a car going down a road, one person might describe it as going 40 miles north in an hour. That is, after an hour, its ditance from the South Pole increases 40 miles. Another person might observe that its distance from the Prime Meridian increases by 50 miles. If the first just says it traveled 40 miles, and the second says that it traveled 50 miles, it may seen that they are contradicting each other. But if we realize that the two instances of the word "mile" refer to different things (North-miles vs. East-miles), there is no contradiction. One way of thinking about it (which, I suppose, might inspire its own misconceptions) is that each tick of the clock has a certain magnitude, so to speak, and that magnitude stays the same no matter what. A stationary clock has all of that magnitude going into time, so it runs at full speed. For a moving clock, each tick moves through space and time, so some of its magnitude is "used up" by the movement through space, and the amount that it travles through time is less. But the whole issue of whether the stuff it goes through is "space" or "time" is simply a labeling issue; one person calls the stuff its going through "time" and another person calls it "some time and some space". Quote: If A and B are in two different inertial frames, B would see the clock of A running slower because it is moving relative to A and A would see the clock of B running slower because it is moving relative to B. It's important to note that they don't just disagree about the results of the measurement of the clocks. They disagree as to what to measure. That's why they're getting different answers. If they both measured the same thing, they would get the same answer. Remember how earlier, I talked about East being the distance to the Prime Meridian? The Prime Meridian is a set of points that is considered to all be “zero East”, and one’s distance East is then measured from the closest one of those points. For A, there is a collection of points which he considers to all be at “time zero”. He measures time as the distance to the closest of those points. When he looks at his own clock, that’s no problem, because the closest is always the same (his starting point). But when he looks at B’s clock, the closest point in the “time zero” set is constantly changing. (For those are getting lost: asking what the “closest point” in A’s “time zero” is the same thing as asking “According to A, where would B have been at the beginning, if he had remained at rest and ended up where he is now?” or “What, according to A, was, at time zero, the same place as B?”) Now, both A and B agree on what the distance to this point is. The trouble is that, according to B, this point that A is measuring from is the wrong point. B doesn’t disagree with the result of A’s measuring, he just says that A is measuring the wrong thing. According to B, he has remained still, so his “closest point” is always the same: where B started from. So, really, they don’t disagree about the speed of their clocks. They both agree that B’s clock is proceeding through “A time” more slowly than A’s clock. It’s just B says that “A time” isn’t “real” time, it’s a confabulation of time and space. Originally Posted by RussDill Einstein says rubbish. Time and space are inseperable, we live in a 4 dimensional world. [Annoying pedant]According to Einstein, time and space are inseparable[/Annoying pedant] Quote: Kinda like if you take slices of a cylinder. The flat slice will show a circle. Other slices will show an ellipse. You are saying that it can't be a circle and an ellipse at the same time. It isn't. Its a cylinder. You just can't visualize the cylinder. Actually, the issue of visualization isn’t really the issue. We can visualize two dimensional space just fine, but the same issues come up. The real issue is that we want to deal with real numbers. In other words, we want to decompose space into one-dimensional coordinates. And once we do that, we’re going to get different answers depending on what coordinate system we use.

20th June 2007, 09:29 PM	#113
69dodge Illuminator Join Date: Nov 2002 Posts: 3,607	Originally Posted by Ziggurat Saying that it's the turnaround that matters is equivalent to saying that (say) a right angle in a Euclidean path is what makes it longer than the straight line. That may be true in one sense, but the extra length in the bent path isn't found in the bend itself. Likewise, the "missing" time isn't subtracted from the turnaround of the twin either. Each leg of the traveling twin's path really is shorter than half the stationary twin's path. I think the traveling twin would agree only if he is permitted to include in "half the stationary twin's path" that portion of the path simultaneous, according to himself, with half his period of acceleration. He would not agree that the stationary twin ages more rapidly than he does, during any part of the unaccelerated portion of his trip.
69dodge Illuminator Join Date: Nov 2002 Posts: 3,607

20th June 2007, 09:35 PM	#114
BillyJoe Penultimate Amazing Join Date: Aug 2001 Location: MOOROOLBARK Posts: 12,539	Originally Posted by Paulhoff It was reworded...... But you still didn't get it right... Quote: 1. Now spaceman (1) takes one of the clocks and travels in any direction at 99,9999999% speed of light for one second and stops. Spaceman (1) now is 100 ft away from spaceman (2) and the two clocks remanding clocks. If Spaceman (1) travels for one second, he is going to be travelling a WHOLE lot further than 100 ft. I think you mean "for one second in frame of reference of Spaceman (2)"
	__________________ A secular society is one in which no one loses any liberty as a consequence of someone else's religious beliefs. NB Allowing yourself to get led around the nose by a person like Craig is a losing strategy. SH Morality is a social coating around a Darwinian core. JC My joke about freewill: There is no basis for it.

20th June 2007, 09:53 PM	#115
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 26,285	Originally Posted by 69dodge I think the traveling twin would agree only if he is permitted to include in "half the stationary twin's path" that portion of the path simultaneous, according to himself, with half his period of acceleration. This isn't actually necessary. Pick a frame, any frame, and you can calculate the halfway point of the stationary twin's worldline. Furthermore, every observer, regardless of their frame, will agree on the calculation of the proper time experienced over this half journey. The only reason to sync the halfway point of the traveling twin's acceleration with the halfway point of the stationary twin's path is if we want them to be simultaneous, but there's no reason we need to pick a frame where the two halfway points are simultaneous. Quote: He would not agree that the stationary twin ages more rapidly than he does, during any part of the unaccelerated portion of his trip. Well, sure. I thought this was already clear.
	__________________ "As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law