Constraints on Lorentz Velocity Transformation

[alhuebel]

TL;DR Summary

I'm doing some undergraduate research related to the Lorentz velocity transformation. I've done a basic derivation of the Lorentz transformation in the attached document. I got some feedback from my professor but I am not understanding what he is looking for. Comments from my professor are in the details below. Any insight would be appreciated as I do not see where or in what ways the transformation is invalid for ux < c.

1. The 2nd line on the 3rd page of your notes, you have x=ct and
x'=ct', thus ux=dx/dt and ux'= dx'/dt' =c according to Einstein's
assumptiuon.
2. But near the end of the last page, you wrote dx'/dt' = (ux
-v)/(1-vux/c2) . Compare with 1. This equation can be valid only for ux=c
and it is not valid when ux <c.
3. But in the book ux can be to be less than c, how come?

 

[malawi_glenn]

Seems to me that the professor is worried about your derived transformation is only valid for the case where ux = ux' = c, but that is how you define the transformation and thus come up with the approriate value of the constant a, namely 1/sqr(1-(v/c)^2) which guarantees that ux = c <=> ux' = c aswell.

No idea what your professor is referring to.

 

[alhuebel]

Thank you. Are you aware of any inconsistencies, breakdowns, singularities etc. involved in the lorentz velocity transformation?

 

[malawi_glenn]

Thank you. Are you aware of any inconsistencies, breakdowns, singularities etc. involved in the lorentz velocity transformation?

If v = c you could get into trouble.

Which book is the professor referring to?

 

[alhuebel]

The book is the Serway / Moses / Moyer Modern Physics 3rd edition. He got back to me with the following:
"The Lorenz-Transformation is based on c²t²-x² -y² -z² =c²t'²-x'² -y'² -z'² (A) where (x,y,z,t) and (x',y',z',t') are coordinates of a light signal (photon) with speeds u_x = u_x' = c in systems O and O' respectively;

and O' is moving with speed v respect to O along +x direction. Can you show that (A) is also valid for a material particle which moves with speed ux and ux' < c?"

My response at this point is as follows:
The equation [A] above is the invariant spacetime interval and is a fundamental component of general relativity. If we simply replace c with velocities of matter u_x and u_x' then those would be isolated observations within each reference frame and not translations of relative velocities between inertial reference frames. That is the whole purpose of the lorentz transformation and the gamma factor.

Of course I may be missing something completely here so I would appreciate any input and perspective you can provide.

 

[malawi_glenn]

you did derive ux' as delta x'/delta t' and then you used the Lorentz transformations for delta x' and delta t' to get an expression for ux' in terms of c, v and ux

Let me drop the y and z now, its not important for your boost in the x-direction.

I guess you could use a space-time diagram and draw the worldline (blue line) for a particle traveling with constant speed. In S (black) it would be u = x/t and in S' (red) it would be u' = x'/t'

In S calculate (ct)^2 - x^2 = (ct)^2 - (ut)^2 = (c^2 - u^2)t^2
In S' calculate (ct')^2 - x'^2 = (c^2 - u'^2)t'^2
and show that you get the same value by using the formula for u' in terms of u and v.
Nice numbers to use are v = 0.6c which gives gamma = 1.25
Then use some values for u expressed in c (s.t. u < 1c)

I am taking a summer class on introductory relativity now, given that I have not been doing physics for about 7 years and wanted to fresh my memory. Anyway, last week I did an excersie similar to this, I calculated the invariant space-time interval using Lorentz-transformation of x,t into x' and t' and then I show that I got the same result using the addition of velocity formula. It was sweet to see the numbers actually work out instead of just cruching algebra for 2 pages :)

 

[alhuebel]

you did derive ux' as delta x'/delta t' and then you used the Lorentz transformations for delta x' and delta t' to get an expression for ux' in terms of c, v and ux

Let me drop the y and z now, its not important for your boost in the x-direction.

I guess you could use a space-time diagram and draw the worldline (blue line) for a particle traveling with constant speed. In S (black) it would be u = x/t and in S' (red) it would be u' = x'/t'
View attachment 303510
In S calculate (ct)^2 - x^2 = (ct)^2 - (ut)^2 = (c^2 - u^2)t^2
In S' calculate (ct')^2 - x'^2 = (c^2 - u'^2)t'^2
and show that you get the same value by using the formula for u' in terms of u and v.
Nice numbers to use are v = 0.6c which gives gamma = 1.25
Then use some values for u expressed in c (s.t. u < 1c)

I am taking a summer class on introductory relativity now, given that I have not been doing physics for about 7 years and wanted to fresh my memory. Anyway, last week I did an excersie similar to this, I calculated the invariant space-time interval using Lorentz-transformation of x,t into x' and t' and then I show that I got the same result using the addition of velocity formula. It was sweet to see the numbers actually work out instead of just cruching algebra for 2 pages :)

nice thank you very much. I will look at this more closely this evening. I just recently took Modern physics in the spring for fun. I'm more or less a perpetual student. I got an undergrad degree in accounting in 1999, an MS in telecommunications in 2002 and I've been taking math and physics courses off and on ever since.