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The first sentence in the article claims: https://en.wikipedia.org/wiki/Length_contraction Length contraction is the phenomenon of a decrease in length of an object as measured by an observer who is traveling at any non-zero velocity relative to the object. As I understand, If I move relative to my friend and have a photo camera and I will make a picture of my friend, he will appear contracted. He will appear of the same height, but will appear “very slim”, is it correct? But if my friend will take picture, he will see that I am contracted. Is it correct? But, doesn't a film in a camera Lorentz contract? If the film contracts, my friend will appear stretched. I will appear stretched too. How come? If my friend holds two Einstein - synchronized clocks in left and right hands - will they show the same time on the picture?

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2 Answers 2

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Surprisingly, Lorentz contraction is not perceived by an observer as a contraction, but as a rotation. This is because the light rays that arrive at the observer simultaneously have not left the object at the same time, whereas Lorentz contraction is associated to the distance of the ends of moving objects at the same time in the observers frame of reference.

I quote from [Terrell,1959]

A sphere will photograph with precisely the same circular outline, whether stationary or in motion with respect to the camera. An object with less symmetry than a sphere, such as a meter stick, will appear, when in rapid motion with respect to an observer, to have undergone rotation, not contraction.

See also [Penrose,1959] and the Wikipedia article on Terrell rotation.

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  • $\begingroup$ I understand that a sphere may appear rotated. But a measuring rod is very thin. Well, will it's length will be the same, shorter or longer? $\endgroup$
    – user139020
    Dec 14, 2016 at 19:21
  • $\begingroup$ Ok. Camera is "at rest"". Rod is "in motion" in the reference frame of the camera. Rod Lorentz contracts and appears contracted? But motion is relative. Now rod is at rest and camera is in "motion" in a reference frame of the rod. Now film Lorentz contracts. The rod appears stretched. Outcome is different. How come? $\endgroup$
    – user139020
    Dec 14, 2016 at 19:31
  • $\begingroup$ Imagine there are two synchronized clocks on the end on the rod. A flash on camera doesn't work. Clockfaces break out simultaneously in the reference frame of the rod. What distance will be between the clocks? Shorter? Longer? On the picture I mean. $\endgroup$
    – user139020
    Dec 14, 2016 at 19:39
  • $\begingroup$ The source of the paradox of the different outcome depending on the description of choice (rod moves-camera static or camera moves-rod static) is the implicit change of frames of reference in the second case. To see this clearly, let us deconstruct your argument. $\endgroup$
    – zap
    Dec 14, 2016 at 23:54
  • $\begingroup$ 1) In the rod's frame of reference, the film is contracted -> 2)The rod is imprinted on a contracted film. -> 3)In the camera's frame of reference the film is not contracted, therefore the rod should appear stretched. The problem here is that from 2->3 you change frames of reference, while talking about the same event. But change of frame of reference is not permitted for an inertial observer. If you limit yourself to describing the event in a single frame of reference, you should find that there is no contradiction. For a better understanding, read Terrell's paper (link in my answer above) $\endgroup$
    – zap
    Dec 15, 2016 at 0:07
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So the first problem here is that you're trying to say "I make a picture of my friend" and the article is saying "length of an object as measured." Photographs are not a great way to measure distances. Go ahead and take a picture with your camera and then look very carefully at the picture and try to ask, "how would I say how long this thing is?". You will probably need to make reference to the things you see in the photograph, 'hey, this thing I know is 7 feet tall' and that will be difficult.

The second problem here is that your camera collects light waves, which move at the speed of light. When you start moving relative to your friend at speeds approaching the speed of light, your naive expectation of what's going to happen with those light waves is not necessarily correct. So to do this calculation you really need to say:

  1. I have this friend showing me a picture of dots at some $x_i$ coordinates, she is at a closest-approach distance $y=L$ from me: in her coordinates, with me passing through the origin, their world-lines are "$(w,x,y,z) = (c\tau, x_i, L, 0),$ for all $\tau.$
  2. Now I do a Lorentz transformation to these lines in the $x$-direction and find the lines $(\gamma~(c\tau - \beta x_i), \gamma~(x_i - \beta c \tau), L,0) $ for all $\tau$ for each $i$. Furthermore if each point has a clock at it, $\tau$ is the time that appears on that clock.
  3. Now I know that in my reference frame I am at the origin and some light emitted by points on these two lines travels forward in time until it hits the aperture of my camera at some time $t$. If I say that the distance it travels divided by the time it travels must be $c$, via the Pythagorean theorem that amounts to: $$[c t - \gamma~(c\tau - \beta x_i)]^2 = L^2 + [\gamma~(x_i - \beta c \tau)]^2,$$ letting me find $\tau$ as a function of $t$ and $x_i.$
  4. Now that I know $\tau(t, x_i)$, I have to evaluate what the angles are between the 2D vectors $[L, \gamma (x_i - \beta c ~ \tau(t, x_i))]$ at constant $t$. Those angles correspond to the distances between these points that I'll see in the photograph. Alternatively I can try to project these light rays through an "image plane," probably one that is perpendicular to the light ray corresponding to $x_i = 0.$

Only after doing all of these steps can you be fully sure what the people in the spaceships see.

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  • $\begingroup$ Thank you very much! The first problem can be resolved. We can make a picture of rectangle. By aspect ratio, we can determine the proportion. $\endgroup$
    – user139020
    Dec 14, 2016 at 20:56
  • $\begingroup$ @Albert If I know as much as I know about relativity and would be highly uncertain about "relativity preserves proportions", what makes you so confident? $\endgroup$
    – CR Drost
    Dec 14, 2016 at 21:01
  • $\begingroup$ Let's say that proper height is 2 feet and proper length is 10 feet. We put marks every feet at each edge. The rectangle Lorentz contracts. We can determine magnitude of contraction, because it contracts only in the direction of travel. ?By looking at picture I mean. It doesn't preserve, sure. $\endgroup$
    – user139020
    Dec 14, 2016 at 21:08
  • $\begingroup$ Well, maybe the best thing to say is that some people have done the above 4-step process and made videos out of it, e.g. youtube.com/watch?v=JQnHTKZBTI4 , and so if you want to know what things look like, there you go. There's also an interesting paper here: arxiv.org/pdf/0708.3454.pdf . And I have some answers around this site about how to use spinor calculus to derive the distortion directly and its complement, the "relativistic beaming" effect for outbound light rays. $\endgroup$
    – CR Drost
    Dec 14, 2016 at 21:52
  • $\begingroup$ Thank you very much! It seems the article is interesting and the problem too. Do you think so? $\endgroup$
    – user139020
    Dec 14, 2016 at 22:02

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