Transformation of velocity exceeding light speed

  • #1
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Main Question or Discussion Point

In discussion with my friend, we reached a conclusion that transformation formula of velosity v to another IFR moving V, i.e.
[tex]v'=\frac{v+V}{1+vV/c^2}[/tex]
is valid even if v is hypothetical velocity,i,e,
[tex]v=\frac{x_2-x_1}{t_2-t_1}[/tex]
[tex]v'=\frac{x'_2-x'_1}{t'_2-t'_1}[/tex]
where interval of ##(t_1,x_1)\rightarrow (t'_1,x'_1)## and ##(t_2,x_2)\rightarrow (t_2',x_2')## are time-like, space-lile, null, it doen't matter.
For example when ##t_2-t_1=0## ,##v=\pm \infty## is trandformed to
[tex]\pm \infty \rightarrow \frac{\pm \infty + V}{1+\pm \infty V/c^2}=\frac{c^2}{V}[/tex]
Of course it is over c but it seems to work describing change of synchronicity.
I have never thought of such an application of the law so appreciate your comment whether it is OK or No Good.
 

Answers and Replies

  • #2
PeterDonis
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i,e,

$$
v=\frac{x_2-x_1}{t_2-t_1}
$$
$$
v'=\frac{x'_2-x'_1}{t'_2-t'_1}
$$

where interval of ##(t_1,x_1)\rightarrow (t'_1,x'_1)## and ##(t_2,x_2)\rightarrow (t_2',x_2')## are time-like, space-lile, null, it doen't matter.
There are two problems here. First, "velocity" is ##dx / dt##, not ##(x_2 - x_1) / (t_2 - t_1)##. It's a derivative, not a ratio.

Second, the Lorentz transformation, which is where you're getting all this from, is only valid if ##dx / dt < 1## (or ##c## in conventional units). There is no such thing as a Lorentz transformation with ##v \ge 1##, because there is no such thing as an inertial frame with a null or spacelike "time axis".
 
  • #3
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Thanks. I will give some more details of the discussion.

----------------------
A stationary wave
[tex]\Psi/A=sin[\omega t]sin[k z]=\frac{1}{2} cos[kz-\omega t]-\frac{1}{2} cos[-kz-\omega t][/tex]
is transfomed to moving IFR of velocity v in z direction
[tex]\Psi/A=\Psi'/A'= sin[\omega \gamma (t'+vz'/c^2)]sin[k \gamma(z'+vt')]=\frac{1}{2} cos[\gamma(k-\omega v/c^2)z'-\gamma(\omega-kv)t']-\frac{1}{2} cos[-\gamma(k+\omega v/c^2)z'-\gamma(\omega+kv)t'][/tex]
where
dispersion relation [tex]\omega=u k[/tex],
for a wave going out
[tex]\omega'_1=\gamma(\omega-kv)[/tex]
[tex]k'_1=\gamma(k-\omega v/c^2)[/tex] 
for a wave coming in
[tex]\omega'_2=\gamma(\omega+kv)[/tex]
[tex]k'_2=-\gamma(k+\omega v/c^2)[/tex]
-------------------

We can easily confirm that velocity ##\omega'_1/k'_1## ##\omega'_2/k'_2## follow the addition rule.
Furthermore as for ##sin[\omega t]## where ##sin[0\cdot z + \omega t]## means velocity ##-\omega/0=\pm\infty## ,and ##sin[\omega \gamma (t'+vz'/c^2)]## where velocity is ##c^2/v > c##, these "velocities" seem satisfying the addition rule as mentioned in OP.
Is the velocity addition rule applicable also for such velocities exceeding c?

More clearly, as for the formula
[tex]x'=\frac{x+V}{1+xV/c^2}[/tex]
where x' is a quantity in moving IFR by velocity V, which corresponds to x in the original IFR, it is sure that this formula stands for x of ordinary velocity <c so this formula becomes the velocity addition rule in that case.
I am suggested that this formula also stands for x of extraordinary velocities exceeding c or even infinity as exemplified above. Is it all right?

PS The formula suggest that in all the IFRs, -c< ordinary speed <c and |extraordinary speed |>c . They are in the different regions and no contamination take place.
 
Last edited:
  • #4
PeterDonis
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Is the velocity addition rule applicable also for such velocities exceeding c?
I've already answered this--the answer is no--and explained why.
 
  • #5
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There are two problems here. First, "velocity" is dx/dt, not (x2−x1)/(t2−t1). It's a derivative, not a ratio.
As for the ratio , say, two events
[tex]A(t_A,z_A),B(t_B,z_A)[/tex] be expressed as [tex]A(t'_A,z'_A),B(t'_B,z'_A)[/tex] in change of IFRs where
[tex]t'_A=\gamma(t_A+Vz_A/c^2), z'_A=\gamma(z_A+Vt_A)[/tex]
[tex]t'_B=\gamma(t_B+Vz_B/c^2), z'_B=\gamma(z_B+Vt_B)[/tex]
Let us see how the ratio would be transformed
[tex]\frac{z'_B-z'_A}{t'_B-t'_A}=\frac{\gamma(z_B-z_A)+\gamma V(t_B-t_A)}{\gamma(t_B-t_A)+\gamma V(z_A-z_B)/c^2}=\frac{\frac{z_B-z_A}{t_B-t_A}+V}{1+V/c^2 \frac{z_B-z_A}{t_B-t_A}}[/tex]
It is same as the velocity addition rule. There is no condition put between A and B, e.g. interval is time-like, space-like or null.

This includes the case if we take both A and B is on the same world line and very close to infinity zero then the ratio turns out to be velocity dz/dt.
 
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  • #6
PeterDonis
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There is no condition put between A and B, e.g. interval is time-like, space-like or null.
You are incorrect. The transformation you are using is a Lorentz transformation, which is only valid if the relative velocity is less than the speed of light. I have already explained why.

Since your question has been answered and you are simply repeating incorrect statements at this point, this thread is closed.
 

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