Understanding the Lorentz Transformation Equation for Time

Click For Summary

Homework Help Overview

The discussion revolves around the Lorentz transformation equations in the context of special relativity, specifically focusing on the transformation of time between two inertial frames of reference, S and S’. The original poster attempts to derive the Lorentz transformation equation for time based on given expressions for displacement.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the derivation of the inverse Lorentz transformation and how to manipulate the equations to isolate the time variable. There are attempts to substitute variables and simplify expressions, with some participants expressing confusion about the algebra involved.

Discussion Status

Several participants have provided guidance on how to approach the problem, including suggestions to substitute variables and collect terms. There is an acknowledgment of confusion regarding the algebraic manipulation, but no consensus has been reached on a final solution.

Contextual Notes

Some participants express uncertainty about their algebra skills and the complexity of the transformations, indicating that this is a challenging topic for those new to the subject.

cr41g
Messages
10
Reaction score
0
Question
Suppose an inertial frame of reference S’ moves at a constant velocity v in the
positive x-direction with respect to a second inertial frame S. The Lorentz
transformation from S to S’ for the x coordinate of displacement is given by:
x′ =γ (x − vt)
Write down a corresponding expression for the inverse transformation, i.e. from
S’ to S, giving x in terms of x’ and t’.
Use these two expressions to derive the Lorentz transformation equation for time:

t'=γ(t-vx/c^2)


I think I have the first part, I answered x =γ (x' + vt). But the second part I have no idea I have been looking online and even watching lectures on youtube.
Thanks in advance.
 
Physics news on Phys.org
vt? Are you sure?
 
Sorry vt'

Do you know the second part or even where I should start?
 
You've got x' in terms of x and t, and x in terms of x' and t'. Which variable don't you want to appear in the final expression?
 
X' I think. Sorry if I sound a bit stupid. I'm in my first year and this is only my second question.
 
Take your answer for the first part, substitute in for x' in terms of x and t, then solve for t' in terms of x and t.
 
Ok I've done what you have said and got-

t'=(x+γ(γx-γvt))/γv

Now I am competent lost
 
Completely*
 
This is correct, you just need to simplify it. What is γ equal to?
 
  • #10
Expand the Lorentz gammas and collect terms. Courage! This one always looks a mess to me until it all clicks into place at the end.
 
  • #11
Can anyone actually show me this step as I am completely thrown. I just can't make sense of it.
 
  • #12
You have a sign wrong in your expression for t', I just noticed. That might be your problem.
 
  • #13
Yeah I found that when I started from scratch. But I'm still at a loss. It just looks a mess. Do you expand all the gammas?
 
  • #14
Collect your x and t terms. The t term should fall out straight away. That leaves the x term. I'd suggest that if in doubt, expand, is a good rule of thumb here. You might want to use [itex]\beta=v/c[/itex] to save ink.
 
  • #15
Ok. Well just got into bed so I will give it a go before I go to university tomrrow. Thanks for your help guys.
 
  • #16
When you studied algebra, did they teach you how to solve 2 linear algebraic equations in two unknowns?
 
  • #17
Yeah as far as I'm aware no hate when algebra is explained in words. But yeah I think I did.
 
  • #18
cr41g said:
Yeah as far as I'm aware no hate when algebra is explained in words. But yeah I think I did.
If that is the case, that's all you have to do in this problem. Your two unknowns are x and t. Solve for 'em.

Chet
 
  • #19
Thank you
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 35 ·
2
Replies
35
Views
5K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 40 ·
2
Replies
40
Views
3K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K