Can an image be smaller than an atom?

[sorgoth]

I was thinking about those mirror rooms that reflect off of each other (you know, the image in the mirror that reflects the other mirror that relfects the other mirror...) and this goes on, apparently, to infinity. So since it just gets smaller and smaller, apparently without end, would it be possible that the image ends up being smaller than an atom?

[WooBot]

Originally posted by sorgoth
I was thinking about those mirror rooms that reflect off of each other (you know, the image in the mirror that reflects the other mirror that relfects the other mirror...) and this goes on, apparently, to infinity. So since it just gets smaller and smaller, apparently without end, would it be possible that the image ends up being smaller than an atom?

No.

But did you know that advanced civilizations live in the dirt under your fingernails?

Originally posted by sorgoth
I was thinking about those mirror rooms that reflect off of each other (you know, the image in the mirror that reflects the other mirror that relfects the other mirror...) and this goes on, apparently, to infinity. So since it just gets smaller and smaller, apparently without end, would it be possible that the image ends up being smaller than an atom?

I think it is possible.

-Who

[Fade]

I was always under the impression that light had a certain sort of size, and that you couldn't magnify past a certain point (which we are far beyond in terms of optics) because of this. There literally isn't enough light bouncing off things of such small size to give any sort of image at all, and thus you have to turn to electron tunneling microscopes.

[WooBot]

Originally posted by Whodini


I think it is possible.

-Who

Do you think the little teeny atoms have little teeny mirrors in their little teeny bathroom closets that are small enough to reflect the little teeny photons the little teeny image is made of?

[zakur]

Originally posted by WooBot


Do you think the little teeny atoms have little teeny mirrors in their little teeny bathroom closets that are small enough to reflect the little teeny photons the little teeny image is made of? There should be a law limiting the number of times one can use the phrase "little teeny" in one sentence. ;)

Originally posted by WooBot


Do you think the little teeny atoms have little teeny mirrors in their little teeny bathroom closets that are small enough to reflect the little teeny photons the little teeny image is made of?

So you said "No" origianllly in rseponse to the poster. Now is your chance to explain. :)

-Who

[jj]

An image is made of light.

The wavelength of (visible) light is 400-700 nanometers, give or take.
If you get smaller than that, you can't represent anything in terms of visible light.

[WooBot]

Originally posted by jj
An image is made of light.

The wavelength of (visible) light is 400-700 nanometers, give or take.
If you get smaller than that, you can't represent anything in terms of visible light.

What he said. :)

Originally posted by jj
An image is made of light.

The wavelength of (visible) light is 400-700 nanometers, give or take.
If you get smaller than that, you can't represent anything in terms of visible light.


So were do the images go?

Do the images keep getting smaller and smaller?

What is the smallest that they can get?

-Who

[WooBot]

Originally posted by Whodini



So were do the images go?

Do the images keep getting smaller and smaller?

What is the smallest that they can get?

-Who

Where does the picture on your TV go when you are close enough to see the dots it's made of?

Can you have a TV picture smaller than those dots?

When you understand the answers to these questions you'll get it. A picture is made of photons of a certain range of wavelengths; a picture made of them must of course be larger than that.

Same reason you can't sign your name with a 40-foot-wide crayon.

[jj]

Originally posted by WooBot


Where does the picture on your TV go when you are close enough to see the dots it's made of?

Can you have a TV picture smaller than those dots?

When you understand the answers to these questions you'll get it. A picture is made of photons of a certain range of wavelengths; a picture made of them must of course be larger than that.

Same reason you can't sign your name with a 40-foot-wide crayon.

Bot, don't waste your time with Whodini. He is an expert at asking questions that imply specific, often absurd, and sometimes insulting things, without actually taking such positions, and he has yet to offer support for any of the implications he makes.

[Captain_Snort]

the reflected light will eventually run out of energy and no longer be reflected, how far can you see say a LED from?

just with (drunken) thinking if it had enough energy to keep on going down to such a level, the image would start blurring (probably a lot longer before this with focussing etc.) but arguably down to the wavelength of light and it would depend a lot on what you would term an image, or merely a number of photons hitting the target.

[TillEulenspiegel]

The answer is no and yes.
NO: If the image is composed of photons , no. You cannot make an object which has detail smaller the it's constituent parts. Besides we don't have the heisenberg compensators that Star Trek enjoys. There a corollary in digital encoding of signals ( audio recording, video recording) called the Nyquist limit, it basically states that the sampling rate of the encoding process, that is changing the analog signal to a digital snapshot, must be at least twice the rate of the highest frequency of the signal or the information is lost.

YES: IBM a few years ago produced it's corporate logo in individual atoms. If and when it becomes technological possible to manipulate sub-atomic particles ( neutrons, protons,haydrons and spins and flavors of quarks =) you'll be able to view the Mona Lisa thru an electron microscope.

Originally posted by Captain_Snort
the reflected light will eventually run out of energy and no longer be reflected, how far can you see say a LED from?

just with (drunken) thinking if it had enough energy to keep on going down to such a level, the image would start blurring (probably a lot longer before this with focussing etc.) but arguably down to the wavelength of light and it would depend a lot on what you would term an image, or merely a number of photons hitting the target.

Captain-Snort,

Your words make sense to me. Although, I am wondering then at what point, say how many reflections, until there is no more picture? Can one put a number on this?

-Who

[EdipisReks]

Originally posted by Whodini


Captain-Snort,

Your words make sense to me. Although, I am wondering then at what point, say how many reflections, until there is no more picture? Can one put a number on this?

-Who

well, it would depend on the strength of the light, the prevailing conditions around the light and the mirrors, and how reflective the mirrors are, wouldn't it?

Originally posted by EdipisReks


well, it would depend on the strength of the light, the prevailing conditions around the light and the mirrors, and how reflective the mirrors are, wouldn't it?

I'm not sure. :confused:

Say the strength of the light is S, the number of mirrors is M, and the reflectiveness of the mirrors is R. Is it possible to come up with a general formula that involves S, M, and R?

-Who

[Uncertainty]

If S is in number of photons, M is the number of mirrors, and R is the % of photons lost in each reflection then the number of photons you have TOTAL after each reflection is
S * M * R

If you want to figure out what you could see...

Your eye needs about a hundred photons in a second to see something (I think). Your eye can only see what light intersects it. This is the area of your eye/the area covered by a sphere with a radius = the distance from your eye to the sounce. You can include it in your calculation if you also divide by (4 *Pi * D) / A if D is the distance from your eye to the source and A is the cross sectional area of your eye. In this case, treat S as number of photons per unit area and the units all work out! :)

This would make a good test question.


edit: added / A because I forgot it earlier

[evildave]

Originally posted by TillEulenspiegel
YES: IBM a few years ago produced it's corporate logo in individual atoms. If and when it becomes technological possible to manipulate sub-atomic particles ( neutrons, protons,haydrons and spins and flavors of quarks =) you'll be able to view the Mona Lisa thru an electron microscope.

Actually, they did it with a really fine needle and an electrical field, and scanned with it to "see". They in fact made the image with the same needle.

Anything made of single atoms (or other such particles) that you bombarded with electrons would generally come apart like a sculpture made of pinballs being hit by a continuous rain of shotgun blasts.

http://www.research.ibm.com/atomic/nano/roomtemp.html
http://focus.aps.org/story/v11/st19

A google search for things like:

IBM atom needle

Will yield up more links about STM and AFM microscopes.

As for an image smaller than an atom, perhaps if you used something other than "light" to see it... certainly atoms and subatomic things wouldn't work. They'd tend to destroy what you were attempting to project on. Like trying to observe the solar system by bombarding it with planets. You could never see it directly, only infer from some observable side-effects (debris hitting a surface) that something had been there.

You can certainly "imagine" something smaller than an atom... say by standing an a plastic army man figure next to a basket ball, pointing at the ball and saying "that's an ATOM". Of course, you could imagine pixies and leprechauns wrestling while you're at it for all this exercise accomplishes.

The answer seems to lie in someone inventing a "better mousetrap" for imaging.


Any amound of light signals may pass through the same point in space without interfering with each other.

Another exercise might be to imagine we're projecting an image from a 10000x10000 grid of atoms to another 10000x10000 grid of atoms through a construct of atoms acting as a lens. At some point, before the source image was translated to the target, it would all pass through a very small amount of space. Theoretically, a small enough surface placed at that point would have a very small copy of the image on it... except that the light would be like tidal-wave sized impulses on a BB, so if it were a single atom in space, it would be rocketed off at ridiculous speed in a random direction as soon as the first wave of light hit it.

[jj]

Originally posted by Uncertainty
If S is in number of photons, M is the number of mirrors, and R is the % of photons lost in each reflection then the number of photons you have TOTAL after each reflection is
S * M * R


Actually, if S is the original number of photons, R is the reflectivity (not lost photons), and M the number of mirrors...

Then

s(M) = S * ( R ^ M )

[Uncertainty]

Actually, if S is the original number of photons, R is the reflectivity (not lost photons), and M the number of mirrors...

ACK!
I guess thats one test question I would have gotten wrong.

-5 points!

[Soapy Sam]

Here is a photograph of an oil painting of a single photon of pure white light, emitted by a single photon pure white laser. It is in jpg format (640 x 480). Sorry the res isn't better, it's due to quantum camera shake. (Hand held , 1/4 sec at f8)

Its here somewhere....
just a minute...

Bloody quantum tunneling.

I'll have to go take it again.

[COCT]

I was thinking about those mirror rooms that reflect off of each other (you know, the image in the mirror that reflects the other mirror that relfects the other mirror...) and this goes on, apparently, to infinity.

The problem, as I see it, is this: Unless your head is centered in the mirror, it won't look like a long straight tunnel, but rather a long curved tunnel. You won't be able to see the "end" of it, because it would appear to curve around a corner. If your head is centered, you won't be able to see the end because your head will be in the way.

I think the best you could do would be to make a hole in one mirror, and peak through it from outside the mirror-room. But even then, the mirrors would need to be perfectly flat and parallel for you to really see to the "end" and even so, the "end" would be blocked by the reflection of the hole you made.

Wouldn't it?

COCT

[BillyJoe]

Originally posted by sorgoth
I was thinking about those mirror rooms that reflect off of each other (you know, the image in the mirror that reflects the other mirror that relfects the other mirror...) and this goes on, apparently, to infinity. So since it just gets smaller and smaller, apparently without end, would it be possible that the image ends up being smaller than an atom? I was waiting for someone to correct this, but after about 20 replies no one has, so I guess I must be wrong, but can some one explain why the image gets smaller and smaller?

Thanks,
BillyJoe.

[sorgoth]

Quote:The wavelength of (visible) light is 400-700 nanometers, give or take.
If you get smaller than that, you can't represent anything in terms of visible light.


Hmm...I had never really thought about the how 'big' light is...

I love this board. Learn something new every day.

[Skeptical Greg]

Originally posted by Whodini

quote:
--------------------------------------------------------------------------------
Originally posted by jj
An image is made of light.

The wavelength of (visible) light is 400-700 nanometers, give or take.
If you get smaller than that, you can't represent anything in terms of visible light.
--------------------------------------------------------------------------------

So were do the images go?

Do the images keep getting smaller and smaller?

What is the smallest that they can get?

-Who


Urrh, umm.. That would be about 400 nanometers..

[slimshady2357]

Originally posted by BillyJoe
I was waiting for someone to correct this, but after about 20 replies no one has, so I guess I must be wrong, but can some one explain why the image gets smaller and smaller?

Thanks,
BillyJoe.

If you are 5 feet from the mirror you are facing, your reflection appears at the size it would if a 'mirror' you was actually just 10 feet away from you. That first reflection is then reflected in the mirror behind you (say it is 5 feet behind you).

So 10 feet seperates the two mirrors and your first reflection is the size it would be if it were another 5 feet behind the mirror you are facing. Therefore the reflection in the mirror behind you is the size it would be if you were 30 feet away! (15 feet doubled).

Does that make sense? Your images get smaller and smaller because the 'mirror space' between them gets larger and larger....

Hmmmmm I know what I mean, but I'm not sure that will help you.

Adam

[BillHoyt]

Originally posted by Diogenes



Urrh, umm.. That would be about 400 nanometers..

You have to keep in mind that the question was about an "image". That requires two things: a collection of light rays large enough to carry a representation of the object being "imaged" and a receiver with a sufficiently fine resolution to "see" it.

Cheers,

[MRC_Hans]

Originally posted by BillyJoe
I was waiting for someone to correct this, but after about 20 replies no one has, so I guess I must be wrong, but can some one explain why the image gets smaller and smaller?

Thanks,
BillyJoe. The image gets smaller because it is at a steadily increasing distance.

Suppose the room is 4 yards with a mirror at each end, you stand in the middle facing mirror A (the other mirror is B).

Now, the first "image" is your direct view of mirror A, 2 yards away. In that mirror, you see mirror B reflected at an effective distance of 6 yards 8your distande to mirror A + the AB distance. It will appear correspondingly smaller.

In the reflection of mirror B, you see a reflection of miror A, perspectively only 4 yards further away, because your distance to the initial mirror is only added once. In this reflection you see -- etc. etc.

As COCT notes, we have an observation problem: As always, we cannot observe a system without interefering with it. In this case, we need to take out a bit of the picture which must be situated over the "end" of the apparant tunnel.

If we choose to disregard this, what would happen as the images became smaller. Assuming that the mirrors were sufficiently loss-free, this would equal looking at a single mirror, and pulling it backward to indreasing distance. As the image size approaches light wavelength, the picture becomes increasingly coarse like a picture with ever larger pixels (but they won't be square), until it is just one single "pixel", at which point all image information except average luminosity and average color is lost.

Obviously, we could never observe this with the naked eye, because the maximum resolution of the eye is much coarser.


Hans

[slimshady2357]

Originally posted by MRC_Hans

If we choose to disregard this, what would happen as the images became smaller. Assuming that the mirrors were sufficiently loss-free, this would equal looking at a single mirror, and pulling it backward to indreasing distance. As the image size approaches light wavelength, the picture becomes increasingly coarse like a picture with ever larger pixels (but they won't be square), until it is just one single "pixel", at which point all image information except average luminosity and average color is lost.

Hans

Good answer Mark.

That's pretty much what I was thinking too.

And as Bill said, you must consider when you would no longer call the small collection of 'pixels' an image any longer....

Adam

[arcticpenguin]

None of your formulas are relevant if they do not account for wavelength. The smallest object you can resolve with light (electromagnetic readiation) is 1/2 x the wavelength.

You could resolve smaller objects by using a shorter wavelength. With the technique of X-ray crystallography, distances smaller than 0.1 nm can currently be resolved. I will not say imaged, because focusing optics are not adequate for that wavelength of 'light'.

[Skeptical Greg]

Not sure how this fits in, but I read soewhere that the eye can detect a single photon..

I know that is not the same as an image, but if it is true, it is interesting...:)



I liked MRC_Hans' post also....

[Skeptical Greg]

Originally posted by BillHoyt


You have to keep in mind that the question was about an "image". That requires two things: a collection of light rays large enough to carry a representation of the object being "imaged" and a receiver with a sufficiently fine resolution to "see" it.

Cheers,

Of course I was ribbing Whodini..:)

But isn't this kind of like the tree falling in the forrest?

Is there an ' image ' if there is no one/nothing to see it?

And couldn't that ' collection of light rays ' be just ' one ' if that was all the object could reflect ?

[jj]

Originally posted by BillHoyt


You have to keep in mind that the question was about an "image". That requires two things: a collection of light rays large enough to carry a representation of the object being "imaged" and a receiver with a sufficiently fine resolution to "see" it.

Cheers,

Well, I was assuming that superman was the viewer, i.e. I was addressing the smallest resolution available from the image in visible light.

I said 1 wavelength, too, using the idea that you had to have at least a 2x2 set of resolvable pixels to make an "image" instead of a spot.

And we're both leaving out the fact that if it's that small, it's going to acquire color due to its limited range of reflectivity....

But I thought that was maybe a bit much detail to start with. :)

Originally posted by Diogenes


Of course I was ribbing Whodini..:)

But isn't this kind of like the tree falling in the forrest?

Is there an ' image ' if there is no one/nothing to see it?

And couldn't that ' collection of light rays ' be just ' one ' if that was all the object could reflect ?

Ribbed for my pleasure, how curteous of you!

-Who

[pgwenthold]

Originally posted by jj
An image is made of light.

The wavelength of (visible) light is 400-700 nanometers, give or take.
If you get smaller than that, you can't represent anything in terms of visible light.

But an image doesn't have to be created from visible light. To see it visually, yes. But consider x-ray diffraction. The "image" doesn't look like a visible image, but the diffraction pattern is certainly a unique property of the object that you are "looking at." It takes a different combinatorial approach to put the "image" into something we are used to seeing, but it is still an image, just in a different vector space.

The number 1001 in base 2 is still a number, it just doesn't look like the number 9 that we are used to seeing in base 10.

[ceptimus]

Originally posted by Diogenes
Not sure how this fits in, but I read soewhere that the eye can detect a single photon..In his book, "The fabric of reality", David Deutsch says that a frog's eye can detect single photons, but that human eyes are several times less sensitive, and can't.

ceptimus.

So we have a:

No
No and yes

etc.

So I'm curious still. What is the answer? :)

-Who

[jj]

Originally posted by pgwenthold


But an image doesn't have to be created from visible light. To see it visually, yes.

We were, however, talking about visible-light images in context.

I have used an electron microscope, too.

[jj]

Originally posted by pgwenthold

It takes a different combinatorial approach to put the "image" into something we are used to seeing, but it is still an image, just in a different vector space.


Well, let's talk about holograms then, optical Wiener Spectra and DFT's, etc.

All of them can be used to produce images, indeed, and all of them use different basis vectors, but in order to see them, as spectra or whatever, we must still turn them into visible light.

It seems to me that you're confusing the issue for people who aren't familiar with linear algebra, basis functions, etc, and who are thinking purely in terms of visible light reflected from a series of mirrors...

[pgwenthold]

Originally posted by jj


Well, let's talk about holograms then, optical Wiener Spectra and DFT's, etc.

All of them can be used to produce images, indeed, and all of them use different basis vectors, but in order to see them, as spectra or whatever, we must still turn them into visible light.

It seems to me that you're confusing the issue for people who aren't familiar with linear algebra, basis functions, etc, and who are thinking purely in terms of visible light reflected from a series of mirrors...

Well, I apologize for trying to be technical in a science forum. However, the concept of "image" and "to see" are very broad concepts, and leave a lot of room for discussion.

To answer Whodini's question, it depends on what you mean by "image"?

Originally posted by pgwenthold

To answer Whodini's question, it depends on what you mean by "image"?

Yikes pgwenthold, I'm not entirely sure here. :(

-Who

[sorgoth]

Okay, let's talk in terms of NONvisible light.

Is there a minimum size for a wave? Or can it just keep on getting smaller?

[pgwenthold]

Originally posted by sorgoth
Okay, let's talk in terms of NONvisible light.

Is there a minimum size for a wave? Or can it just keep on getting smaller?

Is there any meaning to be shorter than the Planck length?

[arcticpenguin]

Originally posted by sorgoth
Okay, let's talk in terms of NONvisible light.

Is there a minimum size for a wave? Or can it just keep on getting smaller?
There's a practical limit: Shorter wavelength means higher energy. Get the energy too high and it will penetrate those mirrors instead of bouncing off them.

[jj]

Originally posted by arcticpenguin

There's a practical limit: Shorter wavelength means higher energy. Get the energy too high and it will penetrate those mirrors instead of bouncing off them.

Hmm, yeah, how much energy in that photon when you get to planck length?

Hadn't thought about that in particular. Somebody got numbers handy?

I've had a similar discussion with audiophiles who claim "but analog signals have infinite bandwidth"... I keep telling them "that last photon will get ya"... (this leaving aside real-world issues aplenty, too)

But somehow they keep saying the same old stuff...

[arcticpenguin]

I guess you could construct the mirrors out of 'unobtanium'...

Originally posted by arcticpenguin
I guess you could construct the mirrors out of 'unobtanium'...

LOL

I guess I've never thought what that 'last photon' would do the mirrors.

Hmm..

I'll try and think of how I can better phrase what I am trying to get at.

-Who

[Dancing David]

No genius me,
but i seem to recall that the images in the mirrors get very dim as they progress to smaller and smaller.
Given the loss of reflection, you might loose the light before you loose the image.

[WooBot]

Originally posted by Dancing David
No genius me,
but i seem to recall that the images in the mirrors get very dim as they progress to smaller and smaller.
Given the loss of reflection, you might loose the light before you loose the image.

Which is of course exactly the correct answer in real life. Every reflection must go through an additional pane of glass and before long, no light left to form an image.

[LaserCool]

Ahh, quantum optics. My field of research.

Hmm, yeah, how much energy in that photon when you get to planck length?

Assuming Planck length l = sqrt[Gh/c^3] = 4.05 x 10^-35m, and E=pc = hf= hc/l then:

E= (6.626x10^-34 J-s)(3.00 x 10^8 m/s)/(4.05x10^-35 m) = 4.91x10^9 J

1eV = 1.60 x10^-19 J

E = 3.06 x10^ 28 eV

Okay, let's talk in terms of NONvisible light.

Light, theoretically, can have a wavelenght as small as you like, at the expense of more and more energy to provide it. See above.

One of the ways to get around this is to use massive objects rather than massless photons. Electron microscopes, which have much smaller wavelenghts than light ones, capitalize on the de Broglie wavelengths of massive electrons (pc=hc/l). In theory, more massive particles (protons, neutrons, etc.) could be used. in fact, neturon scattering is one way to produce diffraction images.



I guess the question about imaging can be answered this way:

Is it physically possible to use a lens of some sort to focus an image down to sub-atomic sizes? Technically, yes, it could be done in a sense, but encounters a few problems.

One is that he resolving power is limited to a half-wavelength of the signal, so you have to have a light source at least half the size of an atom in wavelength (i.e. the Bohr radius or 0.529Angstroms). That energy is in the range of ~2.40 KeV. Reachable, but energy expensive.

Part of the problem here is that an "image" should be more than a spot, so for more detail, you are going to need better resolution. Shorter wavelengths are required, and budget that in your energy equation.

The issue lies more in what you're going to put the image *on*.

Atoms aren't billiard balls. To reflect a signal off a single atom in any meaningful way is not really possible. Reflection of a signal is a complex process, but it does require a bulk of atoms. Incoming light that interacts with an atom (i.e. is resonant with the electron energy levels) is re-emitted after a short time in a random direction. This is maximally absorbtive. Less absorbtive, but still strongly is a material which absorbs and re-emits semi-randomly, but which is diffuse enough to allow the light to pass into its bulk and die after multiple moderately absorbtive interactions. Most materials fall into this category.

Shiny objects like metals have relatively free electrons on thier surface, that may be excited by nonresonant energy (since they're unbound), and re-emitted easily. Such actions are low energy, short time frame and don't allow for enough time for "randomization" of the re-emitted photon. This time frame is much shorter than the non-reflective object.

So clearly to be "reflective" a bulk has to support realtively free electrons which can occupy unbound energy states that allow for very small excitations and de-excitations. This can't happen in a single atom.

Also, remember that when you shine a light that interacts with that atom, you impart the energy of that photon's momentum into the structure of the atom. Basically, you give it a kick. So sending such a signal can displace a free atom so far you won't be able to find the reflected signal. In fact, this is a layman's way of describing the Heisenberg uncertianty principle.

BTW my field of research is in electromagnetially induced transparency in resonant atom structures. This is how I approached the problem.

[popeStephen]

Kay, so... I just found this site... and this is the first thread I came across...

NEWayz-- I think LC has the best answer so far--

quote:
--------------------------------------------------------------------------------
Assuming Planck length l = sqrt[Gh/c^3] = 4.05 x 10^-35m, and E=pc = hf= hc/l then:

E= (6.626x10^-34 J-s)(3.00 x 10^8 m/s)/(4.05x10^-35 m) = 4.91x10^9 J

1eV = 1.60 x10^-19 J

E = 3.06 x10^ 28 eV
--------------------------------------------------------------------------------

Yet... and excuse me for being mathmatically illiterate (from your perspective) but... huh? (E = 3.06 x10^ 28 eV? I'm still trying to figure out E=MC2...)

Nevertheless please allow me to attempt to distill the various learned answers provided above into a single response. (All you more learn-ed folks--please feel free to correct me if I'm wrong because, who the hell am I? What do I know?)

So, IF you assume a light source of infinate energy, and IF you assume a mirror that reflects with infinate resolution, and IF you assume a camera with infinate resolution then----------

the resolution of the reflected reflections should get down to the level of individual atoms (eventually) but my question is still--

Where do you place the camera? Do we also have to assume a camera smaller than an atom to make this thing work? Maybe it would just be better to say that the image keeps getting smaller untill either--

The light source poops out of power or,

The mirror runs out of resolution or,

The camera runs out of resolution, or

The camera gets in the way.

What if we had an invisible camera? What then? HUH?

[Yahweh]

Originally posted by sorgoth
I was thinking about those mirror rooms that reflect off of each other (you know, the image in the mirror that reflects the other mirror that relfects the other mirror...) and this goes on, apparently, to infinity. So since it just gets smaller and smaller, apparently without end, would it be possible that the image ends up being smaller than an atom?
I used to think about things like that alot. The answer is No... unless you try to imagine every photon reflected from the mirrors to become constrained like into a laser beam so all the photons are so close that they... you get the idea.

[popeStephen]

No?

quote: (Yahweh)
--------------------------------------------------------------------------------
I used to think about things like that alot. The answer is No... unless you try to imagine every photon reflected from the mirrors to become constrained like into a laser beam so all the photons are so close that they... you get the idea.
--------------------------------------------------------------------------------

So the photons get closer and closer untill... what?

[BillyJoe]

Originally posted by popeStephen
So the photons get closer and closer untill... what? .....until....they...can't..get.anycloser

regards,
BillyJoe.
(BTW, Slimshady and Hans, thanks for the answer to my question
.....rather obvious wasn't it!)

[LaserCool]

Originally posted by popeStephen
Kay, so... I just found this site... and this is the first thread I came across...

NEWayz-- I think LC has the best answer so far--

quote:
--------------------------------------------------------------------------------
Assuming Planck length l = sqrt[Gh/c^3] = 4.05 x 10^-35m, and E=pc = hf= hc/l then:

E= (6.626x10^-34 J-s)(3.00 x 10^8 m/s)/(4.05x10^-35 m) = 4.91x10^9 J

1eV = 1.60 x10^-19 J

E = 3.06 x10^ 28 eV
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Yet... and excuse me for being mathmatically illiterate (from your perspective) but... huh? (E = 3.06 x10^ 28 eV? I'm still trying to figure out E=MC2...)

I'm using something known as the de Broglie hypothesis to compute how much energy is contained in a photon with a given wavelength of light. The length in question was the Planck length, a distance related to black hole production and quantum gravity. It turns out tha this length is the smallest size a massive body (e.g. a black hole) could be and whose information could still be accessable to the outside universe. Essentially, it's detectability smears out in quantum uncertianty below the Planck length

The Planck length is on the order of 10^-35m. The diameter of a proton is ~10^-15m. Hence, it's a really small wavelength. It represents the amount of energy a worst-case scenario a photon would have to have in order to resolve anything onto an object of the Planck length.

So, after a little hand-waving and some algebra, I computed the energy necissary to produce a photon with a wavelength short enough to be "seen" at the Planck length. That turned out to be ~2.4KeV. Most photons you encounter in the visible range are in the 1-5eV band. You'd need 10^28 or 10,000,000,000,000,000,000,000,000,000 times as much energy in a photon as is in a photon of visible light in order to resolve the Planck length.

This is rather overkill, since an atom's diameter is waaaay larger than the Planck length, but it's fun to work out as a limiting case.

[jj]

Arg bargelit.

Whodini dested, so now I can't have his stuff on ignore.

Grrrr!

[Larspeart]

Sure it can!

Just make a smiley face out of the protons and neutrons!

[popeStephen]

Originally posted by LaserCool



So, after a little hand-waving and some algebra, I computed the energy necissary to produce a photon with a wavelength short enough to be "seen" at the Planck length. That turned out to be ~2.4KeV. Most photons you encounter in the visible range are in the 1-5eV band. You'd need 10^28 or 10,000,000,000,000,000,000,000,000,000 times as much energy in a photon as is in a photon of visible light in order to resolve the Planck length.

This is rather overkill, since an atom's diameter is waaaay larger than the Planck length, but it's fun to work out as a limiting case.

So, given sufficient energy the wavelength can get within the diameter of an atom, but we still have the problem of where to put the camera--

What if the camera moved in and out of position fast enough to allow light to pass from behind the camera yet was back in position quickly enough to catch (some of) it on the return trip?

[rwguinn]

Originally posted by LaserCool
Ahh, quantum optics. My field of research.



Assuming Planck length l = sqrt[Gh/c^3] = 4.05 x 10^-35m, and E=pc = hf= hc/l then:

E= (6.626x10^-34 J-s)(3.00 x 10^8 m/s)/(4.05x10^-35 m) = 4.91x10^9 J

1eV = 1.60 x10^-19 J

E = 3.06 x10^ 28 eV



Major Snippage


BTW my field of research is in electromagnetially induced transparency in resonant atom structures. This is how I approached the problem.

Does your head hurt? Mine certainly does. Thanks for the post, though. Very informative!

[COCT]

What if we had an invisible camera? What then? HUH?


Then you end up ruining the film.

It's like the problem of being an Invisible Man. If light passes through all parts of you, you're blind... right? The light has to be affected by the lens of your eye and fall on your retina properly for you to see it, doesn't it?

COCT

[popeStephen]

QUOTE
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It's like the problem of being an Invisible Man. If light passes through all parts of you, you're blind... right? The light has to be affected by the lens of your eye and fall on your retina properly for you to see it, doesn't it?
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There goes my invisible in the girls locker room fantasy...