# A few questions from introduction to sr by rindler.

superluminal non physical motion

The question is not looking for the speed at which light spreads from a flash point; this is what's confusing about the question. $$F = \left( t, \frac{c^2}{v} t, B, C \right) \right}.$$

Now, consider two times, $t_1$ and $t_2$, for S, with $t_1 < t_2.$ At time $t_1$, a bunch of flashes go off simultaneously in the spatial plane $x = c^2/v t_1;$ at time $t_2$, a bunch of flashes go off simultaneously in the spatial plane $x = c^2/v t_2.$ The spatial distance between the planes divided by difference in times gives that the "plane of flashes" propagates with speed $c^2/v.$
The speed cc/V is superluminal but the supposed motion is not physical. A Lorentz transformation relates the spece-time coordinates of two events which take place at the same point in space and performing it the point in space does not move!
Regards

we have two inertial frames, S and S' where S' is moving with speed v along the x axis.
here are a few questions about these frames:
..............
3. in the inertial frame S' the standard lattice clocks all emit a 'flash' at noon. prove that in S this flash occurs on plane orthogonal to the x-axis and travlling in the positive x direction at speed c^2/v.
..............
but in three, the beam of light from S' obviously travels at speed c, and according to einstein's postulate the speed of light is constant to all observers, so shouldnt the flash travel at c?
Of course you're right that the flashes cannot possibly travel at speed c.c/v. If you have quoted the problem correctly it is clearly misleading since as v approaches zero the velocity would become infinite. Rindler is either confused or being confusing. Two different simultaneous flashes along x' in S' would appear separated in S by a time x.v/(c.c) over a distance x, so if one 'imagined' they were the same flash propagating along x the velocity might 'appear' to be c.c/v. The illusion would be spoiled by the arrival of the earlier flash at normal light speed a moment later.
It's trivially like a "mexican wave" at a football match, which can "travel" at any speed you like, even superluminal.

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quotation

it's called "Introduction to Special Relativity", i think have the second edition of this book, and thus far the questions were from chapter one.
I remind you that the standard way to quote a book is
Author Title (Editor year) page
Helping me please quote the source in that way. The problem has a high pedagogical potential illustrating a superluminat but not physical motion.
Thanks

Gold Member
Wolfgang Rindler, Introduction to Special Relativity, second edition,page 21.( i think the editor year is 1995).

I have found in a paper I studied long time ago that in order to solve a problem it is advisable to start in the reference frame where it is the simplest and to find out there the significant events. Transform them via the LT to another inertial reference frame. So we start in S where the events generated by the two photons (light signals) are E(1)[a,a/c) and E(2)[a+L,(a+L)/c). Detected from S' the space coordinates of the two events are
x'(1)=ga(1-b) and x'(2)=g(a+L)(1-b)
and so the distance between the two events is
x'(2)-x'(1)=Lsqrt[(1-b)/(1+b)]
probably the desired result? Do you see some analogy with the formula whixh accounts for the Doppler Effect? Please give me the exact statement of the problem proposed by Rindler and its quotation, because my edition is quite old.
Hi,

I know this is a very old post, but the answer x'(2)-x'(1)=Lsqrt[(1-b)/(1+b)] is obviously wrong, as indicated by the correct solution: [..]prove that in S' the distance between them is L*(c+v)^0.5/(c-v)^0.5[..]

To obtain that, you have to pick a rest frame at first. If we take S' as a rest frame the correct Lorentztransformation is

$$x' = \gamma (x + vt)$$

and not

$$x' = \gamma (x - vt)$$ since S is moving.

Therefore we get:

$$\Delta x' = \gamma \Delta x (1 + \frac{v}{c})$$

with t=c*x.

Now we have
$$\Delta x = L = const.$$ in S and this gives the desired result in S':

$$\Delta x' = L\sqrt{\frac{c+v}{c-v}}$$.

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