# A interesting problem in relativity!

1. Dec 30, 2004

### Doctordick

Hurklyl, you are the resident mathematician, you might check this out carefully.

Relativity is really a very complex issue seldom examined in detail by anyone. For example, I suspect almost everyone here who has ever taken a course in relativity has seen the exposition on looking at the back side of an object before it gets past you via relativistic effects. For those who are ignorant of what I am talking about, I will put the issue forward here in detail.

Let us do the following thought experiment:

1. We are an observer at rest at the origin of our personal Euclidian coordinate system (this will all be special relativity so there is no need to go into curved space).

2. One kilometer away from us (in the direction of our y axis) is a railroad track running parallel to our x axis.

3. On that track is a four wheeled vehicle with two taillights (positioned as they would be on a standard automobile).

4. That vehicle is proceeding (from a large positive x position) at a velocity near the speed of light towards our y axis.

5. The driver of that vehicle has his own Euclidian coordinate system attached to his vehicle in which his vehicle is at rest. His x axis (which we can call x') is parallel to the track and his y axis (which we can call y') is parallel to the original observer's y axis.

Now arises the question as to what the observer actually sees! You can obtain the transformation between events in the two coordinate systems from any common text on relativity. If we set t=t'=0 to the event of his taillights crossing the original x axis, those transformation equations are particularly simple (we can ignore z as it does not bear on the problem):

$$y'=y$$

$$x'=\frac{x-vt}{\sqrt{1-\beta^2}}$$

$$t'=\frac{1}{\sqrt{1-\beta^2}}\left[t-\beta\left(\frac{x}{c}\right)\right]$$

(Only given here so that one can see that all lines parallel to the y axis in one coordinate system are parallel to the y axis in the other coordinate system at all times. And all lines parallel to the x axis are the same in both systems: i.e., y'=y everywhere.)

Now, let us examine the light coming from the right rear taillight of the moving vehicle. Suppose that taillight stands a fraction of a centimeter to the rear of the actual vehicle (so it can be seen when the eye is looking right down the line defining the rear of the vehicle).

In the driver's frame, a beam of photons running exactly parallel to the rear of the vehicle (emanating from that taillight) will just miss the rear of the vehicle and be entirely free to propagate to the eye of the rest observer on a line parallel to the driver's y axis.

From the rest observer's perspective, that beam of photons will not be traveling parallel to the y axis but will have a component of its velocity in the negative x direction (so that it just keeps up with the vehicle as it proceeds towards the observer. It follows that the observer will see that light proceeding towards the origin of his coordinate system on a line diagonal to that rectilinear coordinate system. He will thus be able to see light from the right rear taillight before he sees the rear of the vehicle cross the line defined by x=0 (the moment at which he could have seen the light if the vehicle were moving at a low classical velocity.)

The conclusion is that he can see the taillight of the moving vehicle before he sees the vehicle pass him. A common deduction from this (very correct analysis) is that, since special relativity is completely reversible (with regard to who is moving and who is standing still), if we could move a camera on the earth fast enough, we could take a picture of the back side of the moon without actually going out there.

Now that is a ridiculous statement on the face of it as, in order to take that picture, the EM fields associated with the light would have to exist and if those fields existed, we could measure them and have our picture without moving our camera (just do the calculations).

Now, the easiest way to see the error in the speculation above is with another simple thought experiment. Since these are thought experiments (and we are not concerned with general relativity) we can attach a rigid rod to the rear of that moving vehicle perfectly parallel to the line defining the rear which reaches all the way over to the x axis (i.e., the line y=0).

Now, let us take a picture of that right rear taillight! We have a slight problem! That rigid rod is going to hit us up the side of the head a fraction of a moment before we can see the light so we can't get the picture. It is also interesting to note that the definitive event occurs at exactly the same moment that the rear of the vehicle crosses the x=0 line (in both coordinate systems no less!)

This little discussion points out a very important issue. The transformation equations above do not at all tell us what things "look like" at relativistic velocities. That problem is a far more complex problem than it is generally conceived to be.

That is the problem facing some people trying to present a computer model of what the universe would look like if one could freely move around at relativistic velocities. Check out the forum at

http://www.shatters.net/forum/

A few of the people there are interested in correcting that software for relativistic effects. The piece of software they are talking about can be downloaded from:

http://sourceforge.net/project/showfiles.php?group_id=21302

I have downloaded it and it's a lot of fun, but it is not relativistically correct. The pictures presented are exactly what one would see if the universe were covered with a collection of cameras at rest with the solar system and, as you traveled through the universe, instead of looking with your eyes, you were instead were given pictures taken with a rest camera from your current position. Not a realistic perspective at all.

As far as I am aware, there is no piece of software which yields even a special relativistically correct picture of what things would look like if one could travel at relativistic velocities, much less one which takes into account general relativity.

One of the things shown above is that a rectilinear moving coordinate system will not appear to be rectilinear to a rest observer. Note that rigid rod is exactly parallel to the x=0 line in the vehicles coordinate system above and yet will appear to lie along the path of the light from the taillight.

It is interesting to look at the relativistic distortions produced in a polar coordinate system. Consider two polar coordinate systems moving at a relativistic velocity with respect to one another. For simplicity (as things get complex quickly here) let us examine how the universe appears to a moving observer at the instant his origin coincides with the origin of the rest observer.

Let the moving observer be coming from minus infinity along the $\theta =0$ line. If the rest observer is looking at a star along the radial line $\theta = \theta_1$, at the instant the two observers are in the same place, the moving observer will see that same star at an angle in his coordinate system given by

$$\theta = arctan \left(\frac{\sqrt{1-\beta^2}sin\theta_1}{\beta + cos\theta_1} \right)$$

A plot of this functional relationship is quite strange. The moving observer will find the angular position of the stars to appear to be displaced in the direction he is traveling. There is also another subtle illusion not as easily calculated (not by me anyway; maybe a decent mathematician can calculate it - Hurkyl, can you help???)

The illusion I am speaking of is the radial position the moving observer will see that same star. It is easy to show that the apparent radial distance to the star will be different from that distance as seen by the rest observer. Presume that the two observers have sufficiently accurate knowledge of the apparent $\theta$ of the star to calculate the radial distance by triangulation based on the distance between their eyes (i.e., they have perfect stereoscopic vision).

Because their eyes are in a slightly different position, each eye must look in a very slightly different direction to focus on the star. Given the direction the two eyes of the rest observer must look, it is possible to calculate the direction the two eyes of the moving observer must look. The star will appear to be at the intersection of those two lines.

That eye which is downstream from the other (with regard to the motion) has to look at a smaller $\theta$ than does the upstream eye (otherwise the two lines don't cross). That eye which is looking at the greater angle (the upstream eye) has to shift its view by a larger amount than does the down stream eye. Thus it follows that the difference between the angles looked at by the two eyes becomes smaller (or the apparent radial distance to the star increases).

That is in severe contradiction to Lorenz contraction. Now, understand that I am not contradicting the transformation equations between coordinate systems at all; I am talking about how things look, a totally different and severely neglected issue.

If anyone out there can establish the exact radial relationship as a function of $\theta_1$ and $\beta$ I would certainly like to see his algebra.

Come on Hurkyl; you can do it!

Have fun everybody -- Dick

2. Dec 30, 2004

### Hurkyl

Staff Emeritus
Let me get this straight -- you're just interested in how the Lorentz transforms look when the spatial coordinates are given in polar form? (particularly, the radial coordinate?)

3. Dec 30, 2004

### Staff: Mentor

Penrose-Terrell "rotation"

No, I believe he is interested in the literal appearance of a rapidly moving object. That the Lorentz contraction would be invisible (and appear as a rotation, not a contraction) was pointed out in papers by Penrose and Terrell in 1959. (See: http://math.ucr.edu/home/baez/physics/Relativity/SR/penrose.html)

4. Dec 30, 2004

### pervect

Staff Emeritus
The particular phenomenon you are talking about has a name - "Terrell rotation". This name is helpful in finding materials that talk about the problem. I've posted a number of links that talk about the topic in detail, some of the better ones are:

http://www.anu.edu.au/Physics/Searle/paper2.pdf

which presents the basic equations, and the overview page

http://www.anu.edu.au/Physics/Searle/

which presents some downloadable movies (excerpts from an educational film that's being sold commercially, I believe).

The later link, from the Australian National university, has movies that show what everyday scenes would look like if the speed of light were very slow. This scaling is done primarily to make the movies visually interesting, as the author points out the alternative would be to consider these as photographs of Very Large objects with a normal speed of light. (It's not very interesting to look at a real-time picture of something that happens in a few tens of nanoseconds).

Sorry that I haven't looked over your equations in more detail, but I'm more interested in some other problems at the moment. You can compare your answers to the results in the paper I posted if you're looking for a "sanity check".

5. Dec 31, 2004

### Andrew Mason

The problem here is that the light will never reach the observer in the rest frame. The light will be highly collimated in the x direction.

This is not correct. The reason is: time dilation prevents the photon from moving in the y direction any farther away from the car than the x direction in a given time interval.

Consider two photons emitted at the same moment and moving at 90 degrees to each other in the source frame: one moving in the x direction and another moving in the y direction. In the source frame the ratio of y to x speeds is unity:
$$c = dy/dt = dx/dt; dy/dx=1$$

In the rest frame dx'/dt' = c by the second postulate of SR. But the y direction is complicated because the time of observations are made at different x positions and the Lorentz transformations for dy'/dt' contain an x component and as a result:

[tex]\frac{dy'/dt'}{dx'/dt'} \propto \frac{1}{\gamma}[/itex]

Since $1/\gamma \rightarrow 0 \text{ as } v/c \rightarrow 1$, the angle that the light takes is very sharply in the x direction: $dy'/dx' = tan\theta \rightarrow 0$. Where $\gamma = 1000$, the light emitted from the taillight in the y direction will be at y = 1 metre (ie. one metre from the car by the time the car reaches the origin). It never reaches the distant observer.

This phenomenon is observable in synchrotons, in which light is emitted by relativistic electrons. The light is highly directional with very narrow beam spread, which makes it very useful. It is all relativity. Incidentally, I commented on this a while ago at the end of the SR Quiz thread: https://www.physicsforums.com/showthread.php?p=288161

AM