I want to consider the visual appearance of a moving rod with relativistic speed. The rod moves with velocity $v$ with respect to the laboratory.

Let $L$ be the length of the rod measured in the laboratory, so $L=L_0/\gamma$ where $L_0$ is the proper length and $\gamma$ the Lorentz factor.

The photograph records photons that arrive simultaneously, in such a way that at a photon coming from the right tip covers a distance $d$ from the rod to the camera, at a right angle to the rod (figure below).

The simultaneous photon coming from the left tip must have left when that tip was a distance $x$ to the left. It covers a distance $\sqrt{d^2+(L+x)^2}$ when going to the camera.

The difference between the travel times of those two photons must equal the time it takes for the rod to move a distance $x$.

So, the equation describing this situation is $$ \frac{\sqrt{d^2+(L+x)^2}}{c}=\frac{d}{c}+\frac{x}{v}$$

My question is: is this correct?

I read someplaces that the Lorentz contraction should be invisible, in the sense that the length of the rod as registered by the photograph (which I think should be $L+x$) equals the proper length, $L_0$. But this is not the solution to the above equation.

enter image description here

  • 1
    $\begingroup$ No. The Lorentz Transform does not tell you what you actually see. To see/photograph something the light rays have to enter your eye/camera. To do this you will have to learn about en.wikipedia.org/wiki/Relativistic_aberration. It is not that hard but might take some digging so here's a better link mathpages.com/rr/s2-05/2-05.htm $\endgroup$
    – m4r35n357
    Commented Sep 13, 2018 at 11:23
  • $\begingroup$ Also see en.wikipedia.org/wiki/Terrell_rotation $\endgroup$
    – PM 2Ring
    Commented Sep 13, 2018 at 12:45
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    $\begingroup$ I will look into aberration, but the page on Terrell rotation is not helpful at all. $\endgroup$
    – thedude
    Commented Sep 13, 2018 at 15:47
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    $\begingroup$ @thedude how is this pagebe "not helpful at all"? It even has a very clear animation that predicts the answer to your question. Some folks are hard to please. $\endgroup$
    – my2cts
    Commented Sep 13, 2018 at 18:20
  • $\begingroup$ @my2cts The picture is cute but without the corresponding calculation I don't think it can be considered an answer. $\endgroup$
    – thedude
    Commented Sep 13, 2018 at 18:31

1 Answer 1


The picture depends on the spatial position of the camera wrt to the measuring rod. In your case it is not quite clear at what position the camera is when it takes picture. It seems that it takes picture at that moment, when its “front” end is just opposite to aperture.

I believe that good photographer aims his camera straight into the center of the object, ot into the center of the rod and makes time of exposure as short as possible to avoid blurred picture.

So, the shutter of the aperture must be open for infinitely small amount of time.

For the sake of simplification, we can swap frames (relativity allows this) and think about the rod “at rest” and the camera moving.

We will take a picture exactly at that moment, when the aperture of the camera is at closest approach (just opposite) to the center of the rod.

Let’s photons from the ends of the “stationary” rod travel to the aperture. The photons have been emitted simultaneously in the rod’s frame. These photons travel equal distances to the aperture (hypotenuses of right-angled triangles) and meet each other straight in the pinhole. Shutter makes a click at this moment* and lets them go through the aperture.

Then these photons travel equal distances from the aperture to the photo emulsion (the film) and hit it simultaneously (in the rod’s frame).

Photons patches from the ends of the rod to the aperture and from the aperture to the film form similar triangles.

Due to aberration of light the picture will be shifted on the film “to the left” if the camera moves “to the right”.

Obviously, since photo emulsion (the film) Lorentz contracts, distance between the ends of the rod will be $\gamma$ times stretched, i.e. the rod will appear $\gamma$ times longer than it “originally” is.

If that was a rectangle, we could even measure Lorentz contraction (of the film) by measuring ratio of its sides.

Interesting to note, that the different parts of the rod will be of different color on the picture (as a rainbow), even though the rod's proper color was even, for example green. This is the effect of relativistic Doppler shift.

The center of the rod on the picture will be of intense violet color, if the camera moves close to the speed of light. This is the Transverse Doppler effect - blueshift of frequency. That means, that the clock "at rest" is ticking $\gamma$ times faster than the cameraman's own clock.

However, the cameraman may say, that these photons from the ends of the rod have been emitted not simultaneously in his frame and hit the film also not simultaneously.**

The cameraman may wish to take picture of these photons, that have been emitted from the ends of the rod simultaneously in his frame. In this case the rod will appear $\gamma$ times contracted and the camera will not be just opposite to the center of the rod (at the moment it makes a click), but in the other spatial position.

In the latter case the rod will appear $\gamma$ times contracted.



*when the aperture is at closest approach to the center of the rod

**If the cameraman synchronizes clocks in his frame by Einstein.


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