A Common Misunderstanding Of Special Relativity

Bombadil · October 10, 2014

Well the opening post situation is based on the idea that if some photons enter the two holes of the rest ship in particular angle, then those two photons would also enter the two identical holes of the moving ship - and that angle would be the same if you correct for the aberration. Typically this would mean correcting to the rest frame of the source. If you don't do the correction, you get where the star was in your frame when the light was emitted. If you do the correction, you get where the star is "now", i.e. where it is if you stop into the frame of the star.

I assume that when you are correcting for aberration of light you are also correcting for Lorentz contraction, if this is the case then I will agree with you, but if you are doing as it sounds like you are purposing and that is only correcting for adoration of light then I have to differ with you, and think that you will get the length given by the Lorentz contraction.

If on the other hand you don't correct for Lorentz contraction or aberration of light you might still be able to convince me that at least in some cases both observers will agree on the distances measured. Actually its kind of funny one minuet I can convince myself that there is no way that the same distance is measured and every now and then I start thinking about Terrell rotation and think of course both a moving observer and a stationary observer must say that a star is the same distance away.

So there are these websites that are saying that Penrose demonstrated that you can't photograph length contraction, but thinking about it a bit I would like to see what type of argumentation he used. I just can't find it from anywhere.

I think that what you are looking for is this

http://www.astro.gla.ac.uk/honours/labs/relativistic_camera/terrell.pdf

to me this is still a local theorem of a sort, that is it is assumed that the angles involved are small, which is an assumption that I always wonder about in cases like this. Second it seems to me that it is more of a case of proportions changing just so that it looks like the orientation has changed and not the length.

As an example in the case of passing by a wall at relativistic speeds, if we think that the wall is short enough that it won't appear to bend towards the road then it will still appear to point towards the road depending on how far from passing it we are, if we combine this with the added time that the light will take to travel from the close end verses the far end and where we will be when this happens then we can begin to understand that the wall will appear to be rotating as we pass it and the length will appear to be the same as if we were not moving.

The problem that I have with this is that we must assume that the wall is sufficiently short for the needed approximation to be valid. And I think that the actual case of what we see will be different then this because at some point the wall will take up a large portion of our view, especially if it appears to maintain its original length.

Thinking about it though, in the case of measuring the distance to a star, the angle between the ship where we are standing and the point of interest, the star is effectively zero and as a result it would make sense that both a stationary and a moving observer might measure the same distance to the star.

Add another identical train moving at the same speed but to the opposite direction on the next lane. And make the trains line up as seen from the platform, just as the lightning hits. The observer on the platform thinks the trains are the same length. The passenger in the newly added train thinks his train is longer than the markings, and furthermore both passengers symmetrically think their train is longer than the other train, as long as they both think their one-way C is the same as two-way C.

While I know that the idea that the one-way C is the same as two-way C is the way that you always seem to be thinking of this, I can't help but think of this as saying that the speed of C is velocity invariant by definition. I am wondering though would you agree that these are the same idea just stated in different ways?

Now let's switch the two trains for the glass tube and metal rod passing through each others.

Let's look at the pictures that any observers standing on the platform would be taking of the situation. All the observers anywhere on the platform would make the same conclusions about the simultaneity, and thus about the plotted lengths of things. But the pictures they are taking would be quite different.

An observer in the middle of the platform would receive information about both lightning strikes simultaneously, and that is what would display on his picture; the ends of the rod and tube neatly matching.

I have to disagree here, the observer standing in the middle of one of the pipes watching the other pass by and lets remember it really doesn't matter which one is passing by, will see the one passing by as shorter then the one he is stationary with. This is due to the fact that as far as he is concerned he is not moving and so there is no aberration of light taking place only Lorentz contraction and so he will see only the effect of Lorentz contraction and so will see the one moving past as shorter then the one that he is stationary with.

To say as I did above that no aberration of light is taking place is of course not true as the light has to travel a different distance to reach the observer depending on if it travels from the front or the back of the train but if the only time that we are interested in is when the observer is standing in the middle of the train then of course even this effect will vanish at that particular instant seeing as by definition the light will have to travel for the same amount of time to reach the observer standing in the middle of the platform.

You seem to be assuming that both ends of the train can get hit by the lightning. I see no reason that this should be true if the rest length of the train is the same as the distance between the lightning strikes. On the other hand I see no reason that if the train is the correct length to get hit by both lightning strikes while it is passing to then assume that this is also its rest length.

An observer at one end of the platform could take a picture where the lightning strike at that end is visible, but the strike at the other end is not. The further away something is from the observer, the older states he is photographing.

Thus, if you take a photograph while standing at one end of the platform, with such timing that the end of the tube and the rod near you neatly overlap in your photograph, the far ends are still approaching each others in your picture (you photograph older states).

I will agree, that in this situation, that the moving object will appear even shorter then in the last situation, but this still won't change my opinion on the last situation. This situation just adds one more issue of simultaneity and the finite speed of light that must be considered in the equations.

The end that is moving towards you looks elongated, and the end moving away from you looks shrunk in your picture. Symmetrically, this is exactly what you expect via aberration analysis as well (think of the rod or the tube as at rest, and the platform as moving).

So it's clear that any photograph is distorted by the fact that the events appearing on the picture are not simultaneous. However, this time the distortion is a function of your location, not a function of your speed!

Well, what you are saying here could easily be misunderstood to be that you are suggesting that the Lorentz contraction is a result of the aberration effects, but lets not forget that the aberration effects are a consequence of using a finite speed of light and a preferred reference frame, while the Lorentz contraction is a consequence of our definitions resulting in a velocity invariant speed of light.

In particular I don't like the way you say “ Symmetrically, this is exactly what you expect via aberration analysis as well”. It just sounds to much like you are blaming aberration of light for the Lorentz contraction.

So it's clear that any photograph is distorted by the fact that the events appearing on the picture are not simultaneous. However, this time the distortion is a function of your location, not a function of your speed!

If you add a cameras to this setup such that they are at rest with the tube or with the rod, and they will neatly co-incide with a camera on the platform just as that camera takes a picture, all the cameras co-inciding must display the same events in their pictures (only aberration distorted).

Thus, the observer in the middle of the platform taking a picture of neatly matching tube and rod, will match the pictures of the cameras moving along with the tube and with the rod, as long as those cameras just happen to be in the middle of the platform when the picture is taken.

This might be the kind of argument Penrose made to conclude length contraction can't be photographed, but it is actually an erroneous conclusion. The above doesn't mean that length contraction cannot be photographed, it just means length contraction of objects moving in opposite directions is the same.

I'm sorry but I just don't follow what you are saying, the more that I read this the more it seems like you are saying that the aberration of light is the source of the Lorentz contraction, but I know that these are two vary different things. One is the result of our definitions resulting in a speed of C that is velocity invariant (the Lorentz contraction). The other is a result of saying that we prefer the others coordinate system and so, doing calculations as though we where in it and using their speed of light and simultaneity definitions.

As for what Penrose was doing I suggest that you read the link I gave above. The augment there is that the Lorentz contraction and the aberration of light have more or less opposite effects and will tend to cancel each other out over small areas of view.

AnssiH · November 15, 2014

I assume that when you are correcting for aberration of light you are also correcting for Lorentz contraction, if this is the case then I will agree with you, but if you are doing as it sounds like you are purposing and that is only correcting for adoration of light then I have to differ with you, and think that you will get the length given by the Lorentz contraction.

No I wasn't correcting for Lorentz contraction, but you might be right. Even if the front plates in the original thought experiments are set to co-incide at the moment of detection, the implied positions of the rear holes at the moments of the photons entering in each case are different. Meaning, the same photons that are considered in the rest ship case, could not have entered the moving ship. We are looking at photons that entered the moving ship before the rear plates co-incided; at a larger angle even if we do the correction for aberration.

If on the other hand you don't correct for Lorentz contraction or aberration of light you might still be able to convince me that at least in some cases both observers will agree on the distances measured.

Nah, I'm pretty conviced that doesn't happen. Those effects just never co-incide, if you think about it carefully... (Referring to my earlier post about this)

I think that what you are looking for is this
http://www.astro.gla.ac.uk/honours/labs/relativistic_camera/terrell.pdf

The link doesn't work... :I

But they refer to it for instance in here;

http://math.ucr.edu/home/baez/physics/Relativity/SR/penrose.html

But I am right now quite convinced that Penrose is wrong about that. It would be nice to get some feedback from someone who actually knows what Penrose's argument exactly was. Maybe he even means something else than I think he does?

to me this is still a local theorem of a sort, that is it is assumed that the angles involved are small, which is an assumption that I always wonder about in cases like this. Second it seems to me that it is more of a case of proportions changing just so that it looks like the orientation has changed and not the length.

As an example in the case of passing by a wall at relativistic speeds, if we think that the wall is short enough that it won't appear to bend towards the road then it will still appear to point towards the road depending on how far from passing it we are, if we combine this with the added time that the light will take to travel from the close end verses the far end and where we will be when this happens then we can begin to understand that the wall will appear to be rotating as we pass it and the length will appear to be the same as if we were not moving.

The problem that I have with this is that we must assume that the wall is sufficiently short for the needed approximation to be valid. And I think that the actual case of what we see will be different then this because at some point the wall will take up a large portion of our view, especially if it appears to maintain its original length.

Thinking about it though, in the case of measuring the distance to a star, the angle between the ship where we are standing and the point of interest, the star is effectively zero and as a result it would make sense that both a stationary and a moving observer might measure the same distance to the star.

I wasn't really able to follow what you meant here. :(

While I know that the idea that the one-way C is the same as two-way C is the way that you always seem to be thinking of this, I can't help but think of this as saying that the speed of C is velocity invariant by definition. I am wondering though would you agree that these are the same idea just stated in different ways?

Yeah that is what I mean. The reason I brought up one-way speed vs two-way speed is because it is very important to remember that we really are talking about a property that cannot logically be measured at all. The world is full of physicians who have somehow forgotten this fact. Maybe it is just the poor state of education; today everybody "knows" that the validity of relativity "proves" that light moves exactly at C always. When in actual fact relativity doesn't prove such thing at all, relativity is just a working convention, and it was originally proposed that way. Very very important difference.

I have to disagree here, the observer standing in the middle of one of the pipes watching the other pass by and lets remember it really doesn't matter which one is passing by, will see the one passing by as shorter then the one he is stationary with. This is due to the fact that as far as he is concerned he is not moving and so there is no aberration of light taking place only Lorentz contraction and so he will see only the effect of Lorentz contraction and so will see the one moving past as shorter then the one that he is stationary with.

If you follow carefully what I was describing, we were at the case where there were two trains moving at the opposite directions at equal speeds, so that they would co-incide when they pass each others by the platform. And then I replaced the trains with the tube and the rod; still moving at opposite directions. That is why the observer in the middle of the platform sees (and plots) them as same length; they are both length contracted by the same amount.

To say as I did above that no aberration of light is taking place is of course not true as the light has to travel a different distance to reach the observer depending on if it travels from the front or the back of the train but if the only time that we are interested in is when the observer is standing in the middle of the train then of course even this effect will vanish at that particular instant seeing as by definition the light will have to travel for the same amount of time to reach the observer standing in the middle of the platform.

You seem to be assuming that both ends of the train can get hit by the lightning. I see no reason that this should be true if the rest length of the train is the same as the distance between the lightning strikes. On the other hand I see no reason that if the train is the correct length to get hit by both lightning strikes while it is passing to then assume that this is also its rest length.

I will agree, that in this situation, that the moving object will appear even shorter then in the last situation, but this still won't change my opinion on the last situation. This situation just adds one more issue of simultaneity and the finite speed of light that must be considered in the equations.

Well, what you are saying here could easily be misunderstood to be that you are suggesting that the Lorentz contraction is a result of the aberration effects, but lets not forget that the aberration effects are a consequence of using a finite speed of light and a preferred reference frame, while the Lorentz contraction is a consequence of our definitions resulting in a velocity invariant speed of light.

In particular I don't like the way you say “ Symmetrically, this is exactly what you expect via aberration analysis as well”. It just sounds to much like you are blaming aberration of light for the Lorentz contraction.

I'm sorry but I just don't follow what you are saying, the more that I read this the more it seems like you are saying that the aberration of light is the source of the Lorentz contraction, but I know that these are two vary different things. One is the result of our definitions resulting in a speed of C that is velocity invariant (the Lorentz contraction). The other is a result of saying that we prefer the others coordinate system and so, doing calculations as though we where in it and using their speed of light and simultaneity definitions.

All these comments are arising from that same mis-interpretation of what I said. I re-read what I had written and could not find any errors, maybe just try to re-read it with the correction I just provided and see if it makes more sense this time.

As for what Penrose was doing I suggest that you read the link I gave above. The augment there is that the Lorentz contraction and the aberration of light have more or less opposite effects and will tend to cancel each other out over small areas of view.

Do you have that .pdf still in your cache? I would love to see it because what you are describing sounds completely wrong. Drop me a PM or something if you have it and can e-mail it?

-Anssi

Bombadil · November 17, 2014

Nah, I'm pretty conviced that doesn't happen. Those effects just never co-incide, if you think about it carefully... (Referring to my earlier post about this)

I think that I will hold off on commenting on this until you have read about Terrell rotation. As the other option is for me to redo a derivation that I did when this thread was just starting, and didn't post because I thought that it was what this thread was about. Now I am starting to wish that I had just wrote it all out then so that at least we would be on the same page. Although it was nothing more then a special case of Terrell rotation, and I seem to remember my approach being a little bit messy.

I wasn't really able to follow what you meant here. :(

Well until you read that document it just sounds like I am making stuff up any way.

All these comments are arising from that same mis-interpretation of what I said. I re-read what I had written and could not find any errors, maybe just try to re-read it with the correction I just provided and see if it makes more sense this time.

I haven't got to it yet but I will reread that post in the mean time.

Do you have that .pdf still in your cache? I would love to see it because what you are describing sounds completely wrong. Drop me a PM or something if you have it and can e-mail it?

Actually I made a copy of the document when I found it, so yes I have a copy, if there is no objection I might be able to upload a copy for any one else that is trying to follow the thread, I think that hypography would support this although I have never tried to do something like that, otherwise if you can PM me an e-mail I will send you a copy.

AnssiH · November 19, 2014

Actually I made a copy of the document when I found it, so yes I have a copy, if there is no objection I might be able to upload a copy for any one else that is trying to follow the thread, I think that hypography would support this although I have never tried to do something like that, otherwise if you can PM me an e-mail I will send you a copy.

Thanks, I received it.

For everyone else, the paper is by James Terrell, written at 1959. I could not find it by googling it, but I put it here;

https://drive.google.com/file/d/0B-26v8xdJjFrX0J5cUVIUGl1aU0/view?usp=sharing

There's a similar argument by Penrose, which I still have not found. But I believe it is the paper called "The Apparent shape of a relativistically moving sphere".

So I read this Terrell paper, and yeah there appears to be a rather obvious error in there. It's pretty unbelievable that this paper is still considered valid... Let me explain.

Basically he is running similar analysis as I was running in post #49

http://www.scienceforums.com/topic/27564-a-common-misunderstanding-of-special-relativity/?p=332951

Interestingly, he idealized a camera the same way as I did; an infinitely small pinhole, infinite shutter speed, and photons passing the hole define the picture. After which I can't believe he does such a complicated proof of the fact that cameras in the same position, but with different velocity, will take effectively the same picture. Of course they do; the way we just defined the pinhole means they catch the same photons; how could the pictures be any different (except for the apparent directions of the photons).

Anyway, in the first two pages, he is analyzing a very specific case in terms of observer positions, and the error he is making is he assumes his conclusion is a general conclusion to all observer positions. This is not true at all.

To quote myself;

An observer at one end of the platform could take a picture where the lightning strike at that end is visible, but the strike at the other end is not. The further away something is from the observer, the older states he is photographing.

Thus, if you take a photograph while standing at one end of the platform, with such timing that the end of the tube and the rod near you neatly overlap in your photograph, the far ends are still approaching each others in your picture (you photograph older states).

The end that is moving towards you looks elongated, and the end moving away from you looks shrunk in your picture. Symmetrically, this is exactly what you expect via aberration analysis as well (think of the rod or the tube as at rest, and the platform as moving).

So it's clear that any photograph is distorted by the fact that the events appearing on the picture are not simultaneous. However, this time the distortion is a function of your location, not a function of your speed!

Think about that. The further away something is from you, the older states of it your photograph displays. In case of moving objects, this leads into an optical "lag" distortion that is a function of your location; meaning different observers sharing the same velocity but different position, see different geometrical lengths in their pictures;

An object moving towards you will look elongated; its further end displays older states (when the whole object was further away) than its closer end.

And as a flip-side, an object moving away from you will look shortened for the same reason; its further end displays older states (when it was still closer to you)

If you understand that distortion, you also understand there is a very specific single moment for each observer, where that distortion will co-incide with Lorentz contraction for any passing object. But only for that passing moment.

The whole two first pages, Terrell is not realizing that he is analyzing exactly a momentary circumstance where that distortion effect (not aberration effect) happens to cancel Lorentz contraction. The reason he fell into that trap is pretty obvious; he first analyzed a picture taken fully at rest, and then rotated the situation around the observer; effectively moving the "moving object" exactly into the position where it must look the same (the photons he is catching are by definition the same). That rotation amount happens to be exactly the same amount as what you get from aberration formula; of course, it is the direction where the "same photons" appear to be coming from for the moving observer who takes the same picture...

It is very likely Penrose did the same exact error in his analysis with spheres. You know, this is another plain and simple "let's see if water flows downhill... But first, I define downhill as where water flows"-situation.

The most amazing thing about this Terrell paper is that in the third page he actually points out Lorentz contraction can be photographed - contrary to his concluding argument! Look at figure 1, and the discussion associated with it. He is correctly analyzing that part, and he has even drawn (valid) illustrations of photographs taken at circumstances where length contraction is in fact fully visible!

He also correctly points out the same thing that I did in my post #46, that any camera catching the same photons will display the same picture regardless of its velocity, the only difference being the picture is interpreted as being a result of different types of lag distortions.

So he was basically inches away from catching his own error already.

-Anssi

Rade · November 20, 2014

see this 1960 paper by Weisskopf where he discusses Terrell paper and what it means:

https://stuff.mit.edu/afs/athena/course/8/8.20/www/weisskopf.pdf

==

See also this 1989 comment by Terrell:

http://henry.pha.jhu.edu/terrell.effect.pdf

==

From a journal used by teachers of physics to help students understand Terrell paper:

http://engagedscholarship.csuohio.edu/cgi/viewcontent.cgi?article=1149&context=sciphysics_facpub&sei-redir=1&referer=http%3A%2F%2Fscholar.google.com%2Fscholar%3Fstart%3D91%26q%3Drelated%3A_bAKW9kV81_coM%3Ascholar.google.com%2F%26hl%3Den%26as_sdt%3D0%2C39#search=%22related%3A_bAKW9kV81_coM%3Ascholar.google.com%2F%22

Edited November 21, 2014 by Rade

AnssiH · November 24, 2014

see this 1960 paper by Weisskopf where he discusses Terrell paper and what it means:

https://stuff.mit.edu/afs/athena/course/8/8.20/www/weisskopf.pdf

==

See also this 1989 comment by Terrell:

http://henry.pha.jhu.edu/terrell.effect.pdf

==

From a journal used by teachers of physics to help students understand Terrell paper:

http://engagedscholarship.csuohio.edu/cgi/viewcontent.cgi?article=1149&context=sciphysics_facpub&sei-redir=1&referer=http%3A%2F%2Fscholar.google.com%2Fscholar%3Fstart%3D91%26q%3Drelated%3A_bAKW9kV81_coM%3Ascholar.google.com%2F%26hl%3Den%26as_sdt%3D0%2C39#search="related%3A_bAKW9kV81_coM%3Ascholar.google.com%2F"

Thank you for these.

They appear to be propagating the same error. I'm a little puzzled as to why, because especially the last one points out at one point of its analysis quite correctly that the apparent (or photographed) length of the front facing part of the fast moving object is exactly the same as Lorentz length contraction when the moving object is photographed at its closest point of pass-by. And before that it appears elongated, and after that it appears shrunk even more, because of simple "lag distortion" (my term).

I created a quick and dirty visualization that makes it very simple to think about;

There's a green rod and a cyan tube, that are the same length at rest.

In the lab frame the cyan tube is at rest, but the green rod is moving at ~.86c, meaning it must be plotted Lorentz contracted into half of its rest length.

There's two cameras, one at rest in the lab frame, and one moving at the same speed as the green rod; they see each others at rest.

The animation represents the simple situation of a camera taking a photograph (=catching photons) where the cyan tube is directly in front of the camera, and the green rod is seen just passing through the cyan tube. I'm simply plotting the path of the photons that define the photograph.

The circle represents a particular radius around the camera pinhole; The photographed photons start propagating simultaneously in the lab frame.

Note how these photons define the lengths of the photographed objects; the front face of the green rod is going to be seen in the photograph exactly at half its rest length.

However, notice now that the moving camera will catch the exact same photons; by definition it must also photograph a picture where the green rod appears shorter than the cyan tube; even though it is the cyan tube that is moving. This is completely expected result.

Note that this camera must be situated diagonally from the green tube (i.e. in lab frame it is trailing behind), and it must see the green rod at rest, sitting "up ahead" in diagonal angle.

1.

We can already say that that is exactly the angle where the moving camera will see the photons as coming from (this is aberration of light)

2.

In its photograph, the green rod is seen at its rest length, and the cyan tube is seen elongated because those four photons are not taken as simultaneous; the ones from the back are interpreted as representing older states (this is "lag distortion", I don't know if there's a better name)

And last but not least;

3.

The moving camera must photograph the green rod (which it sees as at rest) from rearward angle at all times; thus also in the lab frame camera the moving rod must be seen as from rearward angle. This is Terrel Effect and it can be derived in all kinds of ways (as has been done in above papers). The last author you linked correctly points out that it's not so much rotation as it is shearing.

How much that "shearing" would increase the apparent length of the object is entirely a function of the thickness of the object, is it not?

So, I really don't see it as a justified notion to say that Terrell Effect or aberration or something would always exactly compensate for Lorentz length contraction.

Rade · November 25, 2014

So, I really don't see it as a justified notion to say that Terrell Effect or aberration or something would always exactly compensate for Lorentz length contraction.

It does not always exactly compensate for the set of all possible objects...but it does always compensate for some...see this paper by Mary Boas:

http://www-zeus.desy.de/~wing/AJP000283.pdf

AnssiH · November 29, 2014

It does not always exactly compensate for the set of all possible objects...but it does always compensate for some...see this paper by Mary Boas:

http://www-zeus.desy.de/~wing/AJP000283.pdf

Heh, yeah, isn't it interesting that we've gone from papers implying that objects will generally get compensated, to a paper saying only perfect spheres will get compensated...

It reminds me of that old joke about the physicist who had to create a mathematical formula to predict the outcome of horse races, and sure enough he came up with absolutely bullet proof solution "provided the horse is perfectly spherical and moving in a vacuum".

And this sphere case doesn't seem general to me either. I mean, it seems to only work when you analyze the case where the moving observer happens to cross the stationary observer at the same moment that he is at his closest point of approach to the sphere.

Take a look at Fig 3 on that Terrel's paper;

http://api.viglink.com/api/click?format=go&jsonp=vglnk_14172514686229&key=6afc78eea2339e9c047ab6748b0d37e7&libId=34d275ae-1484-4b4d-9491-f4047367c0f2&loc=http%3A%2F%2Fwww.scienceforums.com%2Ftopic%2F27564-a-common-misunderstanding-of-special-relativity%2Fpage-4&v=1&out=https%3A%2F%2Fdrive.google.com%2Ffile%2Fd%2F0B-26v8xdJjFrX0J5cUVIUGl1aU0%2Fview%3Fusp%3Dsharing&ref=http%3A%2F%2Fwww.scienceforums.com%2Ftopic%2F27564-a-common-misunderstanding-of-special-relativity%2F&title=A%20Common%20Misunderstanding%20Of%20Special%20Relativity%20-%20Page%204%20-%20Philosophy%20of%20Science%20-%20Science%20Forums&txt=https%3A%2F%2Fdrive.google...iew%3Fusp%3Dsharing

First note that he is immediately approximating the situation into a "spherical horse" (never actually possible) by tracing light rays in parallel; light rays that never hit a single observer.

Secondly note that he is only analyzing the specific situation where the considered light rays are perpendicular to the velocity of the sphere in the observer's coordinate system. This is to say, to an observer who would take the photograph at its closest point of approach to the sphere.

On Mary's paper about the sphere's, notice she does the same thing. First she says;

"If we want to ask what O sees at other times, we need only consider a whole set of O' observers each at rest with respect to the sphere, and ask, as O passes each of them, what O sees; Each O' sees a circular outline as above"

But then when she goes to actually analyze the situation, she does so as a function of a unit vector in the direction to the sphere. She is using the components of the unit vector to set the coordinate system, where velocity only occurs in x direction. That sets v to always be perpendicular to the direction of the considered light cone, no matter where you place O'. Which is to say, where-ever you place O´, you are still only analyzing the situation where the observer is at its closest point of approach to the sphere.

And this always works with a sphere, because sphere looks spherical to O' from all directions... Also works with spherical horses, cows, and chickens.

Note that in any other case - even with a sphere - the "lag distortion" would still cause elongation when approaching (at a specific point matching Lorentz contraction), and shrinking when receding the observer.

-Anssi

Bombadil · December 11, 2014

AnssiH before I go into responding to your posts I want to see if I can explain a few things that I think are giving you a different view of these proofs then is intended by their author as so far I don't have the same opinion of these papers as you seem to be having, but I think that it is not so much a case of incorrect analysis of the problem, by you are as put forth by Terrell, but rather I think that it is a question of how the problem is thought about and what the question is.

I suspect that the first thing to consider is that the papers are mathematical proofs and so when they were done the point was not to be understandable but rather to be mathematically strict about something that the person doing it already knew and simply wanted to confirm with strict definitions and tell other mathematicians about, we are looking at a finished product here and not the first draft. Notice that I say strict not precise, not to say that he was not using precise definitions but rather to say that I don't think that Terrell was at all concerned with any body being able to follow his argument that didn't have a degree in physics and in mathematics. Not to say that he was right or wrong just to say on some level he knew what he wanted to prove and he knew who he wanted to prove it to. At least on some level.

Also I get the impression that you are perhaps not overly used to reading mathematical proofs like this, and by that I mean that you are not used to some of the things that mathematicians do and only mention in passing, for instance.

They love to say this is the origin, it's true more then once I have seen entire proofs where the whole point was that they decided to choose a new representation where the desired result was trivial even if the original problem was not and this often involves cleverly choosing the origin. I think that they are doing this to some extant in all of the above papers. On this point just remember that just because they say this is the origin of a coordinate system doesn't mean that it is not just some random point. The key question is what property must a point have in order to call it a origin and can we just call any point the origin and use it as such, but perhaps the more important question is, was the point that was chosen picked out of thin air.

In fact I think that mathematicians get so used to doing this sorts of thing that they don't even bother to tell you when they do it, rather they assume that you understand what they did and can figure out the justification on your own.

And the second thing I will point out springs form a joke that I have heard from time to time. And that is, A mathematician can't tell the difference between his donut and his coffee cup. You have probably heard this one before but have you really thought about what it means or that it might be more then just a joke and actually consider using it as an insight into a mathematicians mind.

In truth I think that this joke springs from the field of topology, as in this case you really can't tell the difference between a coffee cup and a donut, or at least at some point you stop caring about the difference. As at some point you only are interested in a continuous mapping from one object to some other object and if such a mapping exists then we might as well work with the simpler object whichever one that is. I won't go into details about what I just said unless you really want some gory details, but there are very precise mathematical definition that I am referring to when I say continuous mapping and if I want to be precise I should use the term homomorphism. If you want to know what I mean just try looking up a precise mathematical definition of ether of these terms.

My point here is that when a mathematician says something we must ask the question is he trying to be precise or is he trying to be readable and who is going to read this.

I think that you should try to keep both of these ideas in mind when reading all of the papers in question.

Interestingly, he idealized a camera the same way as I did; an infinitely small pinhole, infinite shutter speed, and photons passing the hole define the picture. After which I can't believe he does such a complicated proof of the fact that cameras in the same position, but with different velocity, will take effectively the same picture. Of course they do; the way we just defined the pinhole means they catch the same photons; how could the pictures be any different (except for the apparent directions of the photons).

I can't get this out of the proof, yes he idealizes a camera the same way that you do but then I don't think that he says both pictures will be the same, except on an infinitely small scale, the way that you put it though you seem to expect that both pictures will be identical, which I have a hard time grasping. If it were then neither Lorenz counteraction nor any form of abortion effects could be visible and I think that both must be combined.

Further more I think that he plainly states, or at least as plainly as he is going to, that what you say is not what he is showing when he says

the property of conformality is this sense, which is intrinsic to relativistic aberration, is sufficient to ensure that observers O and O' will obtain pictures which are identical, except for a magnification factor, over comparable regions of small subtended solid angle.

First, try to understand that he is talking about a Conformal map, remember when I said that “A mathematician can't tell the difference between his donut and his coffee cup.” This is just the sort of thing that I meant. In this case though it's more like he can't tell the difrence between a straight line and an ark as long as any line that intercepts them is doing it at the same angel. Don't assume that this makes the pictures look the same.

Next notice that he is only interested in small angles. This is not a global property that he is talking about here, he really can't distinguish between a strait line and an ark (he is looking at it too closely) further more what he is doing is only true when the cosine of the angle in interest is close to 1 in fact this seems to be why he is interested in this case he wants to set cosine to 1. In fact I think that it is safe to say that the only case that he is interested in is the case where you are looking at an infinitesimally small object, I actually think that he right up says this in as plain of language as we can expect from him.

Finally, notice that what he is using as a constant he gives a value to that is not a constant at all, although this is vary cleverly disguised. Again he is only interested in objects that are small enough that, that magnification factor is constant over them. We might call this a clever move, we might also say that this is impossible. Actually from a philosophical stance we could argue that it is impossible, but I suggest that we don't as this idea is embedded vary deeply in mathematics and if we go into it here I suggest that we take a very definition based approach to the question of if this is true as we very quickly will be discussing just what continuity and differentiation really are, in the mathematical sense of the words. In any case I think that this is almost off topic.

Think about that. The further away something is from you, the older states of it your photograph displays. In case of moving objects, this leads into an optical "lag" distortion that is a function of your location; meaning different observers sharing the same velocity but different position, see different geometrical lengths in their pictures;

If you think closely about what I just said and consider small enough angles in the proof of Terrell I think that we can agree that the value of M in his proof is not a constant over the whole field of view and so there will be distortion of the picture that is taken just not the type that might be expected.

It is very likely Penrose did the same exact error in his analysis with spheres. You know, this is another plain and simple "let's see if water flows downhill... But first, I define downhill as where water flows"-situation.

No I think this is what happens when you let mathematicians do physics and then try to read it from any other point of view, something is just lost is translation. They say things like this and then you have to understand that when they say over sufficient small areas they mean like infinitesimal areas.

Just out of curiosity though if you where going to describe your argument of what the picture looks like for a moving observer in a mathematically strict setting how would you do it. I ask because when I think about it Terrell's argument really doesn't seem that complicated anymore, although I think it is more complicated then needed actually I am pretty sure that I have seen a simpler proof by some one else, I can try to find it if you want. Further more your simple animation really doesn't convince me due to the fact of how many things must be left out or assumed to be true in order to except it as useful.

An observer at one end of the platform could take a picture where the lightning strike at that end is visible, but the strike at the other end is not. The further away something is from the observer, the older states he is photographing.

But that is not what we are interested in, we are interested in the length of the objects. Think about this the faster that the train is moving the shorter that the train is and so the light from the far end will have to travel for less time to make it to us, when the lightning strikes, is really of very little interest to me, what I care about is where it hits the ground and how much of the train is past that point.

Also I think that you might consider looking closely at the equations

[math]
x'=\frac{1}{\sqrt{1-(\frac{v}{c})^2}}[x-vt]
\;\;\;\;
and
\;\;\;\;
t'=\frac{1}{\sqrt{1-(\frac{v}{c})^2}}[t-\frac{vx}{c^2}]
[/math]

notice the x dependence in both equations. Can you tell me the exact effect that it is going to have because I just have a hard time picturing it in my mind and the only other option I have is setting up a rather difficult set of equations to solve. One that everywhere I find someone trying to solving a equivalent set of equations, it says that the objects appear to be rotating as they pass us.

Thus, if you take a photograph while standing at one end of the platform, with such timing that the end of the tube and the rod near you neatly overlap in your photograph, the far ends are still approaching each others in your picture (you photograph older states).

I'm not so sure of this, the thing is, the light from the far end is older, implying that the object will look longer then it is, however the rod is not as long as in a rest frame due to the Lorenz transformation and both of these effects must be taken into consideration in order to form the photograph. In fact I don't think that there will be anything trivial about calculating either the apparent length of the train, or equivalently how long the light will travel from the far end of the train before it is to the close end of the train. Just think about it how would you calculate that time if you had to without the use of a cleverly designed thought experiment.

I will admit that I think that length contraction is just a consequence of our definitions however I also think that it is every bit as real as the rest length of an object, which is also nothing more then a consequence of the definitions that we use.

So, I really don't see it as a justified notion to say that Terrell Effect or aberration or something would always exactly compensate for Lorentz length contraction.

I don't think that any one has outright said that, although that was my first imposition to, and I may have at the least tried to imply it at some point, and I also don't think that is the intent of any of these papers, although again that was my first impression to seeing the Terrel paper.

Rather I think that the idea is that the object appears to be rotating as it passes by. Think about that possibility for a second, as a object passes by it always appears to have the same dimensions as when it was at rest however it appears to have rotated and so if you have a 1 foot by six inch box going by you with the six inch side of it pointing in the direction of motion, then if it reached close to the speed of light you would only ever see the six inch side of it and it would always be facing you and if you tried to measure it, it would measure six inches.

Finally try to ignore the contaminator as much as possible, but as weird as it may seem I think that this is what Terrell is trying to describe.

not saying its right or wrong but its seems worth consideration.

LaurieAG · December 11, 2014

I can't get this out of the proof, yes he idealizes a camera the same way that you do but then I don't think that he says both pictures will be the same, except on an infinitely small scale, the way that you put it though you seem to expect that both pictures will be identical, which I have a hard time grasping. If it were then neither Lorenz counteraction nor any form of abortion effects could be visible and I think that both must be combined.

Hi Bombadil and AnssiH,

Surely any photons that could be captured at any discrete location and time would arrive at that discrete location and time regardless of whether there was an observer there or not unless this particular observer was causing the aberration?

In Øyvind Grøn's paper 'Space geometry in rotating reference frames: A historical appraisal', ( http://areeweb.polito.it/ricerca/relgrav/solciclos/gron_d.pdf ) figure 9 part C shows a relativistic rolling wheel solution. If the wheel was pinned and the road moved the wheel as it moves past at relativistic velocity (disregarding Born rigidity etc as per the original proof) surely the camera would capture the same image if the emission locations and timings were the same for both tests?

The following link points to a SR solution from another forum below. I asked the author to add the 'Axle velocity between events' column and order the events in their firing order to cross check that the axle was moving at the same velocity as the angular velocity of all the points moving along the rim of the wheel.

http://www.thephysicsforum.com/special-general-relativity/5577-relativistic-rolling-wheel-ii-3.html#post12639

The absence of aberration in this example may be due to all the z axis elements being zero to prevent the wheel from distorting as it rotated (as Gron states in his papers Conclusion below).

4. If the disk is regarded as a 2-dimensional surface it can be put into rotation in a Born rigid way, that is without any displacements in the tangential plane of the disk, by bending for example upwards so that it obtains the shape of a cup.

LaurieAG · December 13, 2014

When you have a stationary observer and a rotating observer in the same plane you can use SR time dilation to get the emission location and time as shown in the relativistically rolling wheel solution.

Additionally you can effectively triangulate two points on one such plane without time dilation by placing a third observer at a distance away from the plane along the axis of rotation. At any time during one complete rotation the first and second observers are at equal distances from the third observer.

In the image below C shows the third observer in the same plane of rotation while A shows the third observer at 90 degrees to the plane of rotation. Note that despite the angles A, B, C and D shown, all of the light paths are very similar and the first photons that would arrive from all 4 (starting from 1,0 or 3,0) would also arrive at the same time.

AnssiH · December 13, 2014

AnssiH before I go into responding to your posts I want to see if I can explain a few things that I think are giving you a different view of these proofs then is intended by their author as so far I don't have the same opinion of these papers as you seem to be having, but I think that it is not so much a case of incorrect analysis of the problem, by you are as put forth by Terrell, but rather I think that it is a question of how the problem is thought about and what the question is.

I was thinking about that quite a bit when I was reading the paper, but there are so many comments that imply even Terrell himself interprets the result as somehow more general than they are, and I decided to go with that interpretation. Even if Terrell meant his results just as a specific case result, it is clear almost all the references to his (and Penrose's) paper interpret it as a general result, which is clearly wrong.

But it does seems to me that the reason this problem has not been clearly called out is that the math itself probably has got no errors in it, it is just the semantical interpretation of the whole thing that is completely wrong, and then people don't even bother trying to fix it.

I think a lot of the comments in those articles Rade linked to also imply the same interpretation.

And the second thing I will point out springs form a joke that I have heard from time to time. And that is, A mathematician can't tell the difference between his donut and his coffee cup. You have probably heard this one before but have you really thought about what it means or that it might be more then just a joke and actually consider using it as an insight into a mathematicians mind.

In truth I think that this joke springs from the field of topology, as in this case you really can't tell the difference between a coffee cup and a donut, or at least at some point you stop caring about the difference. As at some point you only are interested in a continuous mapping from one object to some other object and if such a mapping exists then we might as well work with the simpler object whichever one that is. I won't go into details about what I just said unless you really want some gory details, but there are very precise mathematical definition that I am referring to when I say continuous mapping and if I want to be precise I should use the term homomorphism. If you want to know what I mean just try looking up a precise mathematical definition of ether of these terms.

I don't remember ever hearing that joke but I can see what you are getting at.

The only sensible way I can think of to define length contraction in a photograph is to photograph two meter sticks at the same distance but at different velocities. If you have a reference stick that is at rest, you can see whether or not other meter sticks are longer or shorter than the reference within the same picture.

A meter stick that is at rest in the coordinate system of the camera serves as the reference, and then you have a defined meaning to something being elongated, or shrunk, inside the captured picture.

And if those paper define "same length" as just different cameras taking conformally the same pictures, then they amount to a very complex proof of a very trivial thing; the same photons will represent the same events by any camera that captures them.

On that note;

I can't get this out of the proof, yes he idealizes a camera the same way that you do but then I don't think that he says both pictures will be the same, except on an infinitely small scale, the way that you put it though you seem to expect that both pictures will be identical, which I have a hard time grasping. If it were then neither Lorenz counteraction nor any form of abortion effects could be visible and I think that both must be combined.

What I mean by identical is that they display the same exact events.

You can easily describe a situation where you take pictures of trains passing various mile markers, and it must be plainly true that any cameras that take pictures at the same location at the same moment, will capture the same order of things inside their pictures regardless of their own velocities. If a train fits neatly between two markers in one cameras picture, it will be so in all of their pictures.

Likewise, if any picture can be show to see some object from such and such angle, then all the cameras capturing the same photons will obviousy see the same thing; exactly what Terrell Effect is describing.

If you want to find the circumstance where a camera takes a picture of a moving meter stick that appears to be exactly the same length as a co-inciding meter stick at rest in its frame, consider the following;

Take an observer watching two meter sticks approaching from opposite directions, and neatly passing each others right in front of the observer. That is to say, the length contraction of the sticks is exactly equal so their lengths appear to match.

The ends of the sticks passing each others will be taken as simultaneous, and capturing a picture with those photons will also display this so.

Any moving camera capturing the same exact photons, will also display the ends neatly matching in its picture.

If you place camears that are at rest with each stick, and position them so that they will happen to capture the same photons as the first camera, both of those pictures also will display a moving meter stick that appears exactly the same length as the meter stick that is at rest with the camera.

Position the camera anywhere else, and it will display varying length; either elongated or shrunk moving stick.

Likewise, if the original camera had a meter stick at rest for reference in its picture, it would plainly be visible in the picture that the other sticks are shorter. For the two "moving" cameras, that meter stick appears elongated (and it appears to be approaching in the picture).

Sorry I'm using quick and sloppy languge but I think this is simple enough for you to be able to think it through easily.

Just out of curiosity though if you where going to describe your argument of what the picture looks like for a moving observer in a mathematically strict setting how would you do it. I ask because when I think about it Terrell's argument really doesn't seem that complicated anymore, although I think it is more complicated then needed actually I am pretty sure that I have seen a simpler proof by some one else, I can try to find it if you want. Further more your simple animation really doesn't convince me due to the fact of how many things must be left out or assumed to be true in order to except it as useful.

Hmm, I would have to think about it, but roughly I would probably just describe two meter sticks superimposed, one at rest and one moving, and then define a convenient position where a camera would capture some relevant photons to define the picture.

In the video, notice that the speed of things was selected so that they'd be described as length contracted to half of their rest length, and I chose to represent a set of photons that define some events that will be visible in the photographs. As of the assumptions, those would be assumptions embedded into relativity itself, and it's relativity we are talking about so... What other assumptions did you run into?

But that is not what we are interested in, we are interested in the length of the objects. Think about this the faster that the train is moving the shorter that the train is and so the light from the far end will have to travel for less time to make it to us,

The shorter the train is, the less additional time lag there is from the far end, but the far end is always going to be further away from us than the closer end, so there always is more lag from the far end no matter which way it's put.

Also I think that you might consider looking closely at the equations

[math]
x'=\frac{1}{\sqrt{1-(\frac{v}{c})^2}}[x-vt]
\;\;\;\;
and
\;\;\;\;
t'=\frac{1}{\sqrt{1-(\frac{v}{c})^2}}[t-\frac{vx}{c^2}]
[/math]
notice the x dependence in both equations. Can you tell me the exact effect that it is going to have because I just have a hard time picturing it in my mind and the only other option I have is setting up a rather difficult set of equations to solve.

The x dependence on the t transformation represents relativity of simultaneity; the events across x-axis in unprimed coordinate system that are simultaneous (share the same t value), are not simultaneous in the primed coordinate system (their t value varies).

See for instance;

http://en.wikipedia.org/wiki/Relativity_of_simultaneity#Lorentz_transformations

And just remember the whole thing arises from the ignorance of the one-way speed of light, and how it allows us to define simultaneity against isotropic C. If each coordinate system have their own notion of one-way C, then they all must have their own notion of simultaneity; meaning they all have their own notion of t as a function of x.

That transformation is pretty easy to visualize actually. See the animation;

http://en.wikipedia.org/wiki/Lorentz_transformation#mediaviewer/File:Lorentz_transform_of_world_line.gif

It's basically animating lorentz transformation in a spacetime diagram; the dots are not objects, they are events.

Imagine a horizontal line, and you see what events across that line share the same "t"; what events are considered simultaneous in terms of C of that coordinate system.

Watch how that notion of simultaneity transforms with Lorentz transformation, and notice especially how this particular transformation preserves the speed of light (the diagonal lights) as isotropic, in the expense of the events sliding into the future or into the past as necessary to set the one-way speed of C to remain the same all the time.

That's all there is to it.

You may also notice that Lorentz transformation is in fact merely a scale transformation. Take a spacetime diagram where speed of light is represented in 45 degree angle, and you can scale the picture along those angles so that it shrinks in one axis by the same amount that it elongates on another axis. That skews the picture in exactly the same way as that animation displays, but the topology of events remains the same; for any natural observer everything plays out the same way.

One that everywhere I find someone trying to solving a equivalent set of equations, it says that the objects appear to be rotating as they pass us.

Well I think skewing is a better term. It's also related to the comment I made about "all the cameras in the same position will take the same picture", i.e. picture the same events. So, imagine a camera that is at rest with the moving object, and coinciding with you taking the picture of the moving object. The moving camera must be trailing the moving object far behind, to catch the same photons as you catch where the object appears to be directly in front of you.

The angle the moving camera has towards the object is clearly such that it sees the back end of the object it is trailing (like in the animation I created), and that is exactly what must be visible also in the pictures both cameras take. If not, then you would have to say the same exact photons represent two different sets of events; that would obviously imply an inconsistency in your theory.

I'm not so sure of this, the thing is, the light from the far end is older, implying that the object will look longer then it is, however the rod is not as long as in a rest frame due to the Lorenz transformation and both of these effects must be taken into consideration in order to form the photograph.

In the comment you quoted, remember it was describing a situation where there was two identical objects moving at the same speed to opposite directions.

I went through various versions of picture taking to point out different aspects of the problem, which can make it little bit hard to read... But I think with careful enough reading, it all seems to be in order there...

I will admit that I think that length contraction is just a consequence of our definitions however I also think that it is every bit as real as the rest length of an object, which is also nothing more then a consequence of the definitions that we use.

Yeah exactly.

Rade · December 29, 2014

I will admit that I think that length contraction is just a consequence of our definitions however I also think that it is every bit as real as the rest length of an object, which is also nothing more then a consequence of the definitions that we use.

I have a question concerning your statement. Do you also think that our definitions that we use for the rest length of an object (e.g., the definitions we attach as pointers to the concepts) are also just a consequence of something ? I would appreciate your thoughts about this.

Bombadil · January 2, 2015

In Øyvind Grøn's paper 'Space geometry in rotating reference frames A historical appraisal', http://areeweb.polito.it/ricerca/relgrav/solciclos/gron_d.pdf ) figure 9 part C shows a relativistic rolling wheel solution. If the wheel was pinned and the road moved the wheel as it moves past at relativistic velocity (disregarding Born rigidity etc as per the original proof) surely the camera would capture the same image if the emission locations and timings were the same for both tests?

I'm not going to even try to make a serious endeavor into this for the simple reason that I simply don't know much of anything about general relativity or its application, and clearly special relativity no longer applies in the situation of a rotating wheel, maybe AnssiH has a more useful comment on this.

When you have a stationary observer and a rotating observer in the same plane you can use SR time dilation to get the emission location and time as shown in the relativistically rolling wheel solution.

If you say so but I really don't even know what the question to ask in this situation is. As for even trying to prove what you just said, well it is just outside of my experience to even fully understand the consequences, and I don't have the time right now to dive into the topic.

If you want my advice try to get doctor dick to respond to you, the only problem there is that you probably won't like the answer as I think that you are looking for a mainstream view on the topic and I don't think that is what you would get, maybe I'm wrong.

I was thinking about that quite a bit when I was reading the paper, but there are so many comments that imply even Terrell himself interprets the result as somehow more general than they are, and I decided to go with that interpretation. Even if Terrell meant his results just as a specific case result, it is clear almost all the references to his (and Penrose's) paper interpret it as a general result, which is clearly wrong.

Well I know what you mean there. Some of the pages that I have found, seem to say that objects appear the same as if they where at rest and don't even bother saying that they appear to be rotated. That objects will appear to be rotated in some way though makes a lot of scenes to me, just think about this thought experiment.

Lets suppose that a picture is passing by us so that the picture is on edge to its direction of travel, lets further suppose that the picture is infinitively thin, perhaps this is unneeded but I don't care what the picture looks like from the side and I would just as soon you cant see it from the side.

Now as the picture passes I ask what will it look like. Well you might say that when we are in perfect line with the side that we can't see it since I did say that it is infinity thin, but this is over looking an important fact and that is that the speed that the picture is moving is a large fraction of the speed of light, think about that for a moment, that means that a light ray can bounce off of the far side of the picture then the picture can move almost as far as it is long before it will line up with the light that will arrive at the same time from the close end of it. That means that even thought the picture is on edge and we shouldn't be able to see the picture, we can see the picture nearly as though it where facing us. In fact there is only going to be one location in which we can't see it the rest of the time it will be vary nearly facing up.

This is perhaps a quick and dirty thought experiment but I think that you can get the idea. By the way what did you think of the video that I found and put in my last post?

And if those paper define "same length" as just different cameras taking conformally the same pictures, then they amount to a very complex proof of a very trivial thing; the same photons will represent the same events by any camera that captures them.

I have to wonder here about your use of the term “conformal mapping” as I really don't think that any one would consider a conformal mapping to preserve length. If we go to Wikipedia they describe it as

In mathematics, a conformal map is a function that preserves angles locally.

I think that part of the point is that the Lorenz transformation wont preserve length, but it is suppose to be a conformal mapping at least that is what Terrell said in his paper and I suspect that this is actually a true statement about the Lorentz transformation although this perhaps says nothing about what will be seen.

You can easily describe a situation where you take pictures of trains passing various mile markers, and it must be plainly true that any cameras that take pictures at the same location at the same moment, will capture the same order of things inside their pictures regardless of their own velocities. If a train fits neatly between two markers in one cameras picture, it will be so in all of their pictures.

I will agree with this statement but I think that it is worth noting that we must be very careful about what we consider to be the same moment and I think that we must also be careful to consider what we mean by order of events, in that after something has reflected light we have no way that we can change that event, however some observers at other points need not receive the light in the same order as we do as long as causality is preserved.

If you place camears that are at rest with each stick, and position them so that they will happen to capture the same photons as the first camera, both of those pictures also will display a moving meter stick that appears exactly the same length as the meter stick that is at rest with the camera.

I think that the symmetry of what you are describing makes your result a necessity of not being able to tell who is moving and who is stationary, but I think that it is somewhat hiding the question at the same time. And that is, how will the pictures compare afterworlds when the two pieces of film are compared.

Hmm, I would have to think about it, but roughly I would probably just describe two meter sticks superimposed, one at rest and one moving, and then define a convenient position where a camera would capture some relevant photons to define the picture.

In the video, notice that the speed of things was selected so that they'd be described as length contracted to half of their rest length, and I chose to represent a set of photons that define some events that will be visible in the photographs. As of the assumptions, those would be assumptions embedded into relativity itself, and it's relativity we are talking about so... What other assumptions did you run into?

Well for one thing it is not entirely obvious that you have accurately represented all of the photons as moving at the same speed or represented all of the relevant photons, or that you have the rod moving at the correct speed, but maybe this is a detail as symmetry would suggest that you have the correct arrival times for the photons that you do represent. More to the point, you are assuming that just because two cameras cache the same photons they will take the same picture, while I think that there is more to the question then this, in my mind this is not the case, but I think that this has more to do with what a picture is and how it has been defined then anything else, I will come back to this issue shortly.

The x dependence on the t transformation represents relativity of simultaneity; the events across x-axis in unprimed coordinate system that are simultaneous (share the same t value), are not simultaneous in the primed coordinate system (their t value varies).

I think that I was perhaps not entirely clear as to what I was referring to as it is the other equation that I was looking at and not from the question of what effect that this has on measurements, as I am well aware that it is more or less a rotation in a sort of hyperbolic space, just look into the hyperbolic sine and cosine functions. Rather my question is how will it effect what is actually seen. Currently I don't think that it will have any effect at all, as what is important is the actual length which is only going to be scaled and is independent of location.

Well I think skewing is a better term. It's also related to the comment I made about "all the cameras in the same position will take the same picture", i.e. picture the same events. So, imagine a camera that is at rest with the moving object, and coinciding with you taking the picture of the moving object. The moving camera must be trailing the moving object far behind, to catch the same photons as you catch where the object appears to be directly in front of you.

I suspect that we are actually very close to agreeing on what is actually seen, and that the real disagreement here is not on what is seen, but when it is seen, how simultaneity is defined and what a picture is. Actually I think it all boils down to the definitions that we are using and we really agree on the rest of it.

To try and explain what I mean I will use the thought experiment as you put forward in your video. Firstly the obvious, clearly both ships will see the side of the rod inside of the pipe. You have said several times that this is the case and I agree with you. Now for a few things that are not so obvious and I think could be pointed out better in the video.

Firstly the observer that is at rest with the pipe will see the rod pass though the pipe directly ahead of him and it will be possible for him to see the trailing side of the rod at this time, the thing to realize though, is that the observer that is moving will see a slightly different picture of the event. He will see that pipe as ahead and to the right of him, as you have it in the video, he will also be able to see the trailing side of the rod, the important part in my mind though is the location that they will see the rod inside of the pipe, as your video seems to be hiding this and rather assuming that both pictures will be the same because they are capturing the same photons. I will admit that the pictures must be conformal mappings of each other and that they must preserve casualty but the pictures will show the rod inside of the pipe in different directions from the front of the crafts.

I think that it is worth pointing out one or two other things, although I think that you already realize them. Firstly the observer at rest with the pipe will see the rod as longer when it is in the pipe then the observer that is moving. Why, well because he will see the hole thing taking place closer to him then the moving observer. And because of how you have conveniently chosen the lengths of the pipe and rod to be the same when at rest, we know that when the pipe is seen directly ahead of him he will see it to be the same length as the observer at rest with the pipe sees the rod. Actually the only way that I have found to convince myself of this is by symmetry.

Also the more that I think about it, the more that I have to conclude that the observer at rest with the rod will have to see the pipe as stretched when the rod is inside of it. Since if your video is correct and this event took place in front of the moving observer he would see the rod and pipe lengths reversed.

If you are having a hard time understanding what I am talking about just consider for a moment that when the observer at rest with the rod sees the pipe directly in front of him, he will see it as the same length as the observer at rest with the pipe see the rod to be when it is directly in front of him (inside of the pipe).

In my mind this is very counterintuitive and I really can't fully explain its cause although I am sure that it must be the case.

I have to wonder at this point if you have miss identified the path of the photons to the rest observer and that in fact the rest observer will see both the tube and the pipe as the same length from his vantage point. Put simply, when any object is seen at a right angel to the direction of travel it is seen as having the same length no matter how fast it is moving. How you have it right now Lorentz contraction would be visible.

I have to ask at this point how did you generate the velocities in your picture?

If the above seems somewhat confusing try to understand I am trying to picture the video from the prospective of the other observer, and I would be very interested in a video from that prospective if you think you can make one, if you do try to make one remember that in that video the pipe is half the length of the rod.

I really would be interested in such a video as I think that it is more or less what is being assumed when I watch the video that you made.

I have a question concerning your statement. Do you also think that our definitions that we use for the rest length of an object (e.g., the definitions we attach as pointers to the concepts) are also just a consequence of something ? I would appreciate your thoughts about this.

My first answer to this question would be that they are a convenient set of definitions that result in useful consequences, but somehow I have to wonder if this is the question that you are asking.

I think that the question that you are asking is if there is some profound reason that we have chosen these definitions out of all of the different possibilities that we could have used. To which I would respond that you must try to understand that no matter what set of definitions that you use they must at some fundamental level be the same if you are going to use them for the same thing.

Also I suggest that you think very carefully about what you mean by rest length as it is not a trivial idea.

Rade · January 2, 2015

I think that the question that you are asking is if there is some profound reason that we have chosen these definitions out of all of the different possibilities that we could have used. To which I would respond that you must try to understand that no matter what set of definitions that you use they must at some fundamental level be the same if you are going to use them for the same thing.

Also I suggest that you think very carefully about what you mean by rest length as it is not a trivial idea.

Hello. It does not appear that you understood my question. First it does not matter what I mean by the term rest length, my question concerns what you mean when you say....."rest length of an object, which is also nothing more then a consequence of the definitions that we use". OK, if as you say rest length of an object is a consequence of the definitions we use, then what are the definitions you say we use a consequence of ? Sorry if i am not making myself clear.

LaurieAG · January 3, 2015

Hi Bombadil,

I'm not going to even try to make a serious endeavor into this for the simple reason that I simply don't know much of anything about general relativity or its application, and clearly special relativity no longer applies in the situation of a rotating wheel, maybe AnssiH has a more useful comment on this.

The proof for the rotating wheel was entirely SR based and, although Gron didn't show his workings, the calculations from the SR proof on the forum gave identical results so it would appear that SR is applicable to rotating wheels in certain circumstances (i.e. external and rotating observers are in the same plane).

Maybe Anssih could let us know if he is aware of any other similar situations/solutions where SR is applicable?

If you want my advice try to get doctor dick to respond to you, the only problem there is that you probably won't like the answer as I think that you are looking for a mainstream view on the topic and I don't think that is what you would get, maybe I'm wrong.

I would appreciate any constructive comments from doctordick or Rade as this approach seems to be a valid (but little used) extension of mainstream SR that should provide some insights into the issues of rotating sources.

Incidentally there are some interesting structural connections between relativistically rotating wheels and cycloid arcs that may be more relevant to SR than GR due to their pure geometric derivation. For instance, over one complete cycle, (1) the area above the cycloid arc equals the area of the wheel that produced the arc, (2) the area below the arc equals 3 times the area of the wheel and (3) the length of the arc equals 4 times the diameter of the wheel.

I find this interesting, structurally and mathematically, as the total plot area equals the surface area of a sphere of the same radius as the wheel. It seems that any mapping between the area of a cycloid arc plot and such a sphere would be a 1:1 mapping (not Cartesian) and the area of a circle with the same radius of the sphere on the surface of the sphere would not contain an area that was equal to the area of the circle due to the curvature of the sphere.

AnssiH · January 4, 2015

Responding in two posts because the system told me there are too many quotation blocks. (meh)

I'm not going to even try to make a serious endeavor into this for the simple reason that I simply don't know much of anything about general relativity or its application, and clearly special relativity no longer applies in the situation of a rotating wheel, maybe AnssiH has a more useful comment on this.

Oh... Hmm, I didn't realize there was a question there. In fact, I'm still struggling to understand what is the question.

Surely any photons that could be captured at any discrete location and time would arrive at that discrete location and time regardless of whether there was an observer there or not unless this particular observer was causing the aberration?

In Øyvind Grøn's paper 'Space geometry in rotating reference frames: A historical appraisal', ( http://areeweb.polit...clos/gron_d.pdf ) figure 9 part C shows a relativistic rolling wheel solution. If the wheel was pinned and the road moved the wheel as it moves past at relativistic velocity (disregarding Born rigidity etc as per the original proof) surely the camera would capture the same image if the emission locations and timings were the same for both tests?

Yes. The appearance of the ground is not being analyzed so obviously doing something with it changes nothing. So I presume by "moving ground" you in fact imply to changing to the coordinate system of the non-rotating part of the axle; inertial frame where the camera is moving (i.e. the camera is attached to the ground).

So in effect you are just talking about describing the same situation from two different frames. After taking a quick peek, the result in the paper looks right to me.

Bombadil, this part of the analysis is simply SR since it just involves tracing back the photon emission events. In the picture taken by the camera (which is flat against the ground btw), the wheel is going to look distorted the way it is described in the paper. But if you place a camera into the coordinate system of the axle, the wheel must look symmetrical. In fact taking the wheel as rotating only complicates the picture; you get the same distortion without rotation, just with linear motion of the wheel. Relativistic spinning wheel problems are just a massive headache... :I

Well I know what you mean there. Some of the pages that I have found, seem to say that objects appear the same as if they where at rest and don't even bother saying that they appear to be rotated. That objects will appear to be rotated in some way though makes a lot of scenes to me, just think about this thought experiment.

Well yeah, and a lot of the references seem to assume length contraction is just completely compensated out from the photographs once you trace back the emission events to get the appearance in the photograph. But the result you get from that analysis varies depending on where in the coordinate system you place the camera (e.g. whether the object is receding or approaching, and in what angle).

Lets suppose that a picture is passing by us so that the picture is on edge to its direction of travel, lets further suppose that the picture is infinitively thin, perhaps this is unneeded but I don't care what the picture looks like from the side and I would just as soon you cant see it from the side.

Now as the picture passes I ask what will it look like. Well you might say that when we are in perfect line with the side that we can't see it since I did say that it is infinity thin, but this is over looking an important fact and that is that the speed that the picture is moving is a large fraction of the speed of light, think about that for a moment, that means that a light ray can bounce off of the far side of the picture then the picture can move almost as far as it is long before it will line up with the light that will arrive at the same time from the close end of it. That means that even thought the picture is on edge and we shouldn't be able to see the picture, we can see the picture nearly as though it where facing us. In fact there is only going to be one location in which we can't see it the rest of the time it will be vary nearly facing up.

Well it's exactly the same mechanism as what is visible in that video I did. The camera A that is stationary in the pictured frame, will see the tube directly from the side. The camera B that is moving will compose its picture from the same exact photons; the glass tube must appear to be in the same angle - which is also the result you get from tracing back the emission events. Even though the tube appears to be somewhere up front in camera B's coordinate system; that's where the emission events occurred in its coordinate system. That is also where the metal pipe is at rest in B's coordinate system (which will be pictured inside the tube)

This is perhaps a quick and dirty thought experiment but I think that you can get the idea. By the way what did you think of the video that I found and put in my last post?

Looks about right to me.

I have to wonder here about your use of the term “conformal mapping” as I really don't think that any one would consider a conformal mapping to preserve length.

I wrote the "same length" in quotation marks exactly because, like you, I also don't think it is good semantics to take conformal mapping as meaning "same length". I was wondering if they do in these papers. But if they don't, then conformal mapping just has got nothing to do with whether or not you can photograph length contraction, under the sensible definition for "length". So... It just seems like the semantical interpretation of these papers is all over the place.

I think that part of the point is that the Lorenz transformation wont preserve length, but it is suppose to be a conformal mapping at least that is what Terrell said in his paper and I suspect that this is actually a true statement about the Lorentz transformation although this perhaps says nothing about what will be seen.

Yeah, if it wasn't, it would not be self-consistent transformation; your expectations would be coordinate system dependent.

You can easily describe a situation where you take pictures of trains passing various mile markers, and it must be plainly true that any cameras that take pictures at the same location at the same moment, will capture the same order of things inside their pictures regardless of their own velocities. If a train fits neatly between two markers in one cameras picture, it will be so in all of their pictures.
I will agree with this statement but I think that it is worth noting that we must be very careful about what we consider to be the same moment and I think that we must also be careful to consider what we mean by order of events, in that after something has reflected light we have no way that we can change that event, however some observers at other points need not receive the light in the same order as we do as long as causality is preserved.

"...any cameras that take pictures at the same location at the same moment...", means the same moment is well defined; the cameras are at the same location. This is of course a thought experiment only, so you can imagine camera superimposed into the same location to analyze what picture they would take.

So, as long as their positions co-incide, they must also agree on simultaneity, and they must also compose their picture out of the same exact collection of photons (being how I idealized the shutter as infinitely small hole)

I think that the symmetry of what you are describing makes your result a necessity of not being able to tell who is moving and who is stationary, but I think that it is somewhat hiding the question at the same time. And that is, how will the pictures compare afterworlds when the two pieces of film are compared.

All the photographs are composed out of the same collection of photons, so they must all display all the objects in same relative sizes with each others. Of course different cameras see all the objects in different angles in their fields of view; at the angle where the emission events are in their own coordinate systems.

Well for one thing it is not entirely obvious that you have accurately represented all of the photons as moving at the same speed or represented all of the relevant photons, or that you have the rod moving at the correct speed, but maybe this is a detail as symmetry would suggest that you have the correct arrival times for the photons that you do represent.

Yeah, I suppose you would just have to take my word for it. Of course I approximated the speeds and distances, but the result is well within the error of margin. I just chose a speed of light, and then set the massive objects to move at .86c, because it allows me to represent length contraction at exactly 50%.

So pay attention to the two inner photons. If there is room for photons to pass through the glass tube, but miss the length contracted metal rod, that means that the length contraction of the rod must be visible in the picture composed out of those photons. Notice I also chose emission events that are taken as simultaneous in the pictured frame, purely out of convencience.

In other words, that is the result you would get in terms of special relativity; the rod would be seen as length contracted in the picture.

More to the point, you are assuming that just because two cameras cache the same photons they will take the same picture, while I think that there is more to the question then this, in my mind this is not the case, but I think that this has more to do with what a picture is and how it has been defined then anything else, I will come back to this issue shortly.

Well they both see the angles of the photons differently.

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