The formulas of Special Relativity do not always tell what things really would look like for our eyes when they travel at a speed near the speed of light. The formula I have in mind is the length contraction formula. When we say that objects get contracted when they travel with speeds near light speed, this does not mean that we (considering us as being at rest) actually see fast objects as shorter then they would be at rest, but that the distance we will measure between two points where the ends of an object are located simultaneously wrt us (but these are not simultaneous events wrt the moving object) is smaller than the rest length of the object. The length contraction formula is therefore highly dependent upon the time synchronization procedure used in SR (as are other SR formulas).(adsbygoogle = window.adsbygoogle || []).push({});

But if we want to know what we actually will see, we must remember that when we see a distant object (or a part of it) we see it not as it looks right now, but as it looked when the light from it which reach us right now was sent, so if the object is moving with a high speed we will not see it at the location where it is right now, but where it was when the light was sent.

To calculate what we actually will see, let an observer O be located at the origin of some (inertial) reference system. Another reference system moves with velocity v to the

right along the x-axis of the first system (or relative to it). The systems have parallel axes and in both, the clocks are synchronized so that the origins of the systems meet when t=0 in both systems. We also imagine that the systems are made as clearly visible, indeed shining, scaffolds.

Now, at t=0, O looks at the point (x,y,z) in O's own, stationary, system. What will then the coordinates (x',y',z') be in the moving system for the event O looks at? This can be calculated by applying the Lorentz transformation (LT) for the velocity v to the coordinates (x,y,z,t), where t is the time (wrt the stationary)system) when the light beam that O received at the origin at t=0 left (x,y,z).

We will actually make two calculations, one for the classical case, where we use the Galilei transformation (GT), and one for the relativistic case, using LT. This way we can see which of the effects are caused by SR and which are (to a greater extent) caused by the high speeds. In the classical case, we assume that the stationary system is stationary wrt the "ether", so that the light speed is constant in that system. (In other systems, the light speed will be different in different directions.)

Clearly, the time (wrt the stationary system) when the light leaves (x,y,z) is [itex]t=-\frac{\sqrt{x^2+y^2+z^2}}c[/itex]. Therefore, in the classical case, GT gives [tex]x'=x+\frac vc\sqrt{x^2+y^2+z^2},[/tex] and in the relativistic case, LT gives [tex]x'=\frac {x+\frac vc\sqrt{x^2+y^2+z^2}}{\sqrt{1-\frac {v^2}{c^2}}}.[/tex]

Also, y'=y and z'=z in both cases.

The following images, the first for the classical and the second for the relativistic case, shows how the moving system seems distorted from the viewpoint of observer O, for the relative velocity v=0.5 c. We see in both images the plane z=0.

http://img851.imageshack.us/img851/7448/classical1.png [Broken]

http://img202.imageshack.us/img202/3971/relativistic1.png [Broken]

The red curves are O:s images of the originally vertical lines of the moving system, and they are seen against the background of the stationary system, with green vertical and blue horizontal lines, the latter coinciding with the originally horizontal lines of the moving system, since y'=y. Every region bounded by red curves and blue lines is seen as a square by an observer in the moving system.

Here are two similar images, now for v=0.9 c.

http://img202.imageshack.us/img202/1802/classical2.png [Broken]

http://img684.imageshack.us/img684/669/relativistic2.png [Broken]

From the images, we see immediately that O will see no simple length contraction. If O looks along the x-axis, horizontal lengths in the moving system will seem contracted to the right of the origin (in the direction of the motion) yes, but to the left of the origin, horizontal lengths will seemdilated. These effects weaken the farther away from the x-axis O looks. Vertical lengths seems unchanged, since y'=y.

These contraction and dilation effects are present in both the classical and the relativistic case.

This is easily explained for the classical case: when the light which is received by O at the origin leaves the point with coordinates (x,0) wrt the stationary system, the point with the same coordinates wrt the moving system has not yet reached the first point but is located to the left of it. This first point must therefore have coordinates (x',0), with x'>x, wrt the moving system. This difference will be greater the farther away from the origin the point is located. It means that |x|<|x'| if x>0, so to the right of the origin, lengths will appear contracted, while |x|>|x'| if x<0, so lengths will appear dilated to the left of the origin.

I think this effect is sometimes called a Doppler effect.

It is basically the same in the relativistic case. But here, the contraction effect on the right side is enhanced and the dilation effect on the left side is weakened, compared to the classical case. This is of course due to ordinary length contraction. But this length contraction can never cancel out the dilating effect on the left side, far from it.

Those apparent contractions and dilations and the distortions of shapes will always be more striking for an observer than the contribution from ordinary length contraction, and these effects are not really relativistic but present in the classical case too. They are in both cares caused by the high relative velocity.

If O looks along the y-axis (more generally: perpendicular to the direction of the motion), he/she will see shapes that were originally square be distorted to parallelograms (if the dimensions of the square is small compared to the distance to the origin). The reason is that when the light from the back side of the square reaches the front side, and the light seen by O leaves the front side, the square has moved a bit to the right from its position when the light left the back side, and therefore O will se the front side lying to the right of the back side. Thus the square will seem sheared. This can, of course be appropriately generalized to all kinds of shapes. It is valid both in the classical and the relativistic case.

But in the relativistic case, it turns out that this effect, combined with ordinary length contraction, gives rise to the effect that the image of the square on O:s retina or on a photographic plate (2-dimensional) looks the same as it would look if the square was rotated a certain angle. (But O would actually perceive the square as rotated only if the scaffolds of the stationary system are removed, so that O has nothing to compare with to perceive depth.) Again, this can be generalized to all kinds of shapes.

This is not true in the classical case.

My main point of this post is that length contraction, as it is usually presented, is more of a mathematical abstraction than something that is actually perceived by an observer.

I hope you could follow this post, despite that my English isn't perfect (it isn't my mother tongue), for which I apologize.

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# What relativity really looks like

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