Consider a rigid inertial coordinate system $K$. A photo camera is located on the point $O=(0,0,0)$ and in $t=0$ it captures a picture from the light rays of the physical objects at rest wrt $K$. Let's suppose that the capture takes an infinitesimal time. Consider the two cases:

  1. at $t=0$ the camera is in $O$ with zero velocity wrt $K$

  2. at $t=0$ the camera is in $O$ with non-zero velocity wrt $K$ in the direction where the camera is pointing.


According to special relativity do we expect to find differences between the photos in case 1 and 2? What kind of differences? (Size of the objects? Angle of view?) How can these differences be explained in terms of the dynamics of light rays going inside the camera?


3 Answers 3


This is actually a really interesting question and you will probably be surprised by some of the findings: for example, you might be surprised to know that as you accelerate faster and faster, the stars all "crowd into" the part of the sky in front of you. Even more spectacular, if I throw a beach ball past you, at no point does it present anything other than a circular profile to you: its length contraction is always "hidden" by other effects! These are called Terrell effects or "Terrell rotations" in general; the world as-you-see-it looks surprisingly sensible even though if you literally tried to calculate where everything was in your coordinate system you would have to ultimately conclude that things were squished in weird ways.

As for proving it, I only know of one good way, and that is the spinor calculus.

Adjoin the identity matrix $I$ to the usual Pauli matrices as $\sigma_w$ (where $w=ct$ as usual) and take the 4-vector $v^\mu$ of your choice in the $(w,~x,~y,~z)$ coordinates of your choice; now form the Hermitian matrix $V = v^\mu \sigma_\mu:$ now it turns out that $\det V = v^\mu v_\mu$ in the $(+~-~-~-)$ metric.

This has two consequences. First off null 4-vectors have $\det V = 0$ and therefore they are projections; $V = \phi~\phi^\dagger$ for some "spinor" $\phi$ living in $\mathbb C^2$. This is great because null vectors are precisely the light rays that are coming into your camera! To get a better sense of where they are, chart them starting from whenever they pass through a sphere of radius $R$ around you, then their 4-vector components are $(v^w,~v^x,~v^y,~v^z) = (-R, x, y, z),$ and the above procedure assembled them into some $\phi = [\alpha;\;\beta]$ according to: $$ V = \begin{bmatrix}v^w + v^z & v^x - i v^y\\ v^x + i v^y & v^w - v^z\end{bmatrix} = \begin{bmatrix}-R + z & x - i y\\ x + i y & -R - z\end{bmatrix} = \begin{bmatrix}\alpha~\alpha^* & \alpha~\beta^*\\ \beta~\alpha^* & \beta~\beta^*\end{bmatrix}.$$ Note that therefore $\zeta = -\beta/\alpha = -\beta\alpha^*/(\alpha\alpha^*) = -(x + i y)/(-R + z) = (x + i y)/(R - z),$ which is the stereographic projection of this point $(x, y, z)$ on this sphere of radius $R$ onto the complex plane $\mathbb C.$

So it's actually just fallen out of our maths that we want to describe these images by stereographically projecting whatever we see onto the complex plane!

I said above that there was also a second consequence; it is that the Lorentz transforms are linear transforms of $V$ which preserve $\det V$ and Hermitian-ness: aside from the parity transforms $V \to -V$ and $V \to (\det V)~V^{-1},$ you have the special Lorentz transforms $V \to L~ V~ L^\dagger$ for some complex matrix $L$ such that $\det L = 1.$

One such matrix is $$ L = \begin{bmatrix} e^{r/2} & 0 \\ 0 & e^{-r/2} \end{bmatrix}.$$ This has the effect on our 4-vectors of $$ V \to \begin{bmatrix} e^{r/2} & 0 \\ 0 & e^{-r/2} \end{bmatrix} \begin{bmatrix}v^w + v^z & v^x - i v^y\\ v^x + i v^y & v^w - v^z\end{bmatrix} \begin{bmatrix} e^{r/2} & 0 \\ 0 & e^{-r/2} \end{bmatrix} = \begin{bmatrix}e^r(v^w + v^z) & v^x - i v^y\\ v^x + i v^y & e^{-r}(v^w - v^z)\end{bmatrix}, $$ but since $v^w = (V_{00} + V_{11})/2$ and $v^z = (V_{00} - V_{11})/2$ you can see that this is straightforwardly the Lorentz boost:$$\begin{array}{rl} v^w\to&v^w \cosh r + v^z \sinh r\\ v^z\to&v^w \sinh r + v^z \cosh r, \end{array}$$making this the Lorentz boost in the $z$ direction by rapidity $-r$ or velocity $v = -c \tanh r.$ In other words this is the transform you use when you transfer to a reference frame moving relative to you in the $-z$ direction.

Similarly we find that $L \phi = [e^{r/2}~\alpha;\; e^{-r/2} \beta]$ and so this maps the images that we're seeing on the complex plane by $\zeta \to e^{-r} \zeta,$ so they all crowd together towards the origin point $\zeta = 0.$ But looking at "what is $\zeta = 0$ on the original sphere?" we have that $x, y \to 0$ while $R - z \to \text{nonzero},$ hence it is the point $(0, 0, -R),$ proving that indeed when you change these coordinates the stars all crowd in the direction that you're going.

Similarly you can use this to calculate the angle-distortion in your photographs; it's as simple as "stereographically project around the direction you're going; scale $\zeta \to k \zeta$ for some constant $k < 1$, then stereographically project back."

(It is also instructive to consider future-pointing null vectors, in which case the formula for the stereographic projection is different and you derive a similar effect; the light that you are emitting in all directions is concentrated in the direction you're seen as going from some third-party perspective. This is somewhat more well-known than the Terrell effects and is called relativistic beaming.)

As for that ball? Well, stereographic projections map circles on the sphere to circles and straight lines (for circles passing through $(0, 0, R)$) and vice versa. The action of any such matrix $L$ upon $\zeta$ is a bilinear or "Mobius" transform which also maps circles and lines to circles and lines. So therefore every Lorentz transform preserves every circle that you visually see, even though they may distort a pattern of uniformly nested circles (producing the Terrell rotation effect and other apparent skews).

  • $\begingroup$ I forgot to mention the relatively intuitive other consequence that the unitary matrices generate rotations under $V \to U V U^\dagger$ since rotations preserve $v^{w}$ and hence $\operatorname{Tr} V,$ and due to the cyclic property of trace $\operatorname{Tr}(UVU^\dagger) = \operatorname{Tr}(VU^\dagger U) = \operatorname{Tr}V.$ The reason this gets interesting is that rotations similarly pick up a factor of 1/2 just like scale factors; e.g. $\exp(i\theta\sigma_z/2)$ being a rotation by $\theta$ about $z$. Choosing $\theta=2\pi$ we get $\phi\to-\phi$ which is why I called it a spinor above. $\endgroup$
    – CR Drost
    Commented Nov 29, 2016 at 19:30

Both these references deal with moving objects and a stationary camera, but that is of course equivalent to stationary objects and a moving camera.

  • 1
    $\begingroup$ I am downvoting this because your second point suggests that if I throw a ball past you you might be able from some angle to see it as anything other than circular in outline, which turns out to be incorrect. $\endgroup$
    – CR Drost
    Commented Nov 29, 2016 at 18:05
  • $\begingroup$ Ah, yes, I was neglecting the path from the light to the camera. Thank you, I will edit. $\endgroup$
    – user121664
    Commented Nov 29, 2016 at 18:13
  • $\begingroup$ What about the size of the a sphere pictured in the photo? How would it change in case 1 and 2? $\endgroup$ Commented Nov 29, 2016 at 18:16

The previous answers mainly explain optical appearance of an object like Terrell – Penrose rotation. If you make pictures by pinhole camera, Terrell – Penrose rotation manifests when you take a picture of the WHOLE object and the object is highlighted by external light source, so it shines by reflected light.

If you want to measure relativistic contraction by pinhole camera and to get rid of optical illusions, you have to follow some rules.

We can consider two different cases:

1) A physical object moves in the reference frame of photo camera. In this case a physical object will appear CONTRACTED gamma times. If there are two clocks, attached at light and right end of material object, the clocks will show different time.

2) A photo camera moves in the reference frame of the physical object. In this case a physical object will appear STRETCHED gamma times because photo – film Lorentz - contracts. If there are two clocks, attached at light and right end of material object, the clocks will show the same time.

If we make a photo by means of pinhole camera, we can evaluate magnitude of relativistic distortions (contraction) by looking at the picture, but there are some details.

Let’s consider a picture of a square. By distortions of sides on the picture (vertical sides vs horizontal ones) we can measure magnitude of Lorentz contraction.

We do like that: we attach lamps into each corner of the square. Then we release flashes simultaneously.

If the camera is “at rest”, we release flashes simultaneously in the camera frame. The camera and the physical object (center of a square) are at points of the closest approach at this moment.

If the camera is “in motion” and the physical object is at rest, we release flashes simultaneously in the physical objects frame. When the light rays reach aperture, camera and physical object are ay points of closest approach. Since photo film in camera Lorentz – contracts, the square will appear stretched gamma times.

If the square is “at rest” and we release flashes simultaneously in the square’s frame, THE ONLY picture we can get is the STRETCHED square.

It should be noted, that path of photons are hypotenuses of equal triangles and photons will come to aperture simultaneously and next moment will simultaneously hit the photo film.


By the way – this article explains Terrell – Penrose rotation in very simple way. The article considers pictures taken by a photo camera, which was in motion in the reference frame of a sphere.



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.