Rotation of a photograph of an object due to relativity

[user1139]

Homework Statement

Understanding the rotation of a photograph of an object due to relativistic effects.

Relevant Equations

See below.

I came across an interesting question in the Hartle's textbook, "An Introduction to Eisntein's General Relativity". The question is as follows:

Explain why a photograph of an object moving uniformly with a speed approaching the speed of light, parallel to the plane of the film appears not contracted, but rather, rotated. (Assume that the object subtends a small angle from the camera lens.)

Hartle argues that if we consider a rectangular object moving parallel to the plane of the film with speed $V$, whose length is of rest length $a$ and breadth of rest length $b$, then the effect can be attributed to the fact that light coming from the far side takes a longer time of $b/c$ than the short side to get to the film.

My question is, what exactly does does Hartle meant by far and near side? And, why does light need to travel an extra distance $b$?

 

[.Scott]

Let's say that the letter "T" is moving from left to right with the top of the T farthest from the camera.
The top of the T will be seen on the left of the image, because as light from that part of the letter crossed to the bottom of the T, the rest of the T move to the right. So the T will appear about 90 degrees rotated - with the top of the T on the left.

That extra distance b is the height of the letter.

 

[ergospherical]

To elaborate, consider an observer sufficiently far away from the rectangle such that the light rays can be considered to be parallel upon reaching the observer. If the rectangle has rest length $a$ parallel to the velocity $v$ then in this frame the rectangle is of length $a\sqrt{1-v^2}$.

If you imagine the rectangle as moving to the right (with the observer positioned below) then a photon emitted vertically downward from the top left corner at $t=0$ becomes vertically level with a photon emitted vertically downward from the bottom left corner at $t = b$. The horizontal separation between these two photons is $L_1 = vb$. Meanwhile the horizontal separation between two photons emitted from the bottom left and right corners at $t=b$ is simply $L_2 = a\sqrt{1-v^2}$. Putting $\varphi = \sin^{-1} v$ it follows that $L_1 = b\sin{\varphi}$ and $L_2 = a\cos{\varphi}$, i.e. the image is identical to that of a rectangle at rest with respect to the observer, but rotated by angle $\varphi$.

 

[PeroK]

And, why does light need to travel an extra distance $b$

Because the far side of the object is further from the camera than the near side.

The far side is the side furthest away from the camera and the near side is the side nearest the camera.