How to Derive the Lorentz Factor from Pythagoras?

  • Context: Undergrad 
  • Thread starter Thread starter MathiasArendru
  • Start date Start date
  • Tags Tags
    Variable
Click For Summary

Discussion Overview

The discussion revolves around deriving the Lorentz factor using the Pythagorean theorem in the context of a light clock on a train. Participants explore the mathematical steps involved in manipulating the equation and isolating variables, with a focus on understanding the relationship between time, distance, and velocity in special relativity.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant expresses confusion about deriving the Lorentz factor from the equation \( (ct)^2 = (cx)^2 + (vt)^2 \) and seeks help in starting the derivation.
  • Another participant points out that the expression for \( \gamma \) is a definition rather than something that can be derived in the way the original poster suggests.
  • A participant corrects their earlier misunderstanding regarding the variable being solved for, clarifying that they were looking for \( t \) instead of \( \gamma \).
  • Several participants provide hints on how to isolate \( t^2 \) in the equation, suggesting to bring all terms involving \( t \) to one side and isolate it.
  • One participant expresses frustration at being unable to manipulate the equation effectively, questioning if there is a method they have not learned that could help solve it.
  • Another participant draws a parallel to a simpler algebraic equation to illustrate how to isolate terms, suggesting that similar techniques could be applied to the current problem.
  • Participants discuss the challenge of dealing with terms like \( v^2t^2 \) and whether they can be treated similarly to constants in algebraic manipulations.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to derive the Lorentz factor, and there are multiple competing views on how to manipulate the equation effectively. The discussion remains unresolved regarding the specific steps needed to complete the derivation.

Contextual Notes

Participants express uncertainty about their mathematical skills and the techniques required for isolating variables in the context of the Lorentz factor derivation. There are indications of missing assumptions or methods that could aid in the solution.

MathiasArendru
Messages
17
Reaction score
0
Hey guys, this is a little silly question but it bothers me. I am not a math genius (yet i hope) and I am still in elementary school so there's a lot to learn. But i just read about the lorenz factor in this example he basically used pythagoras of this light clock in a train, so it started of as

(ct)^2 = (cx)^2 + (vt)^2

and he derived it into:

t = \frac{x}{\sqrt{1-\frac{v^2}{c^2}}}

I would have posted an attemp to solve it but i really just don't know how to crack it and get started

Pleeeaaase help it would be really nice :D
 
Last edited:
Physics news on Phys.org
That is nothing you can derive (at least not in the way you ask for here), that is a definition of γ.
 
right its me god I am stupid! he solved for t not gamma don't really know what went through my head while i wrote it. i corrected it in the post now
 
OK, I'll help you get started: bring all terms which have a ##t## to one side of the equation and the other terms on the other side. Isolate ##t^2## so you have ##t^2 = \text{something}##. Then take roots.
 
ok ill try:

c^2t^2 = c^2x^2 + v^2t^2
t^2 = \frac{c^2x^2 + v^2t^2}{c^2}
Dividing both sides by t^2
1 = \frac{c^2x^2 + v^2t^2}{c^2t^2}

im stuck... lol
normally i don't really have trouble when solving for variables but this one irritates me.. can i have another hint ? :)
 
MathiasArendru said:
ok ill try:

c^2t^2 = c^2x^2 + v^2t^2
t^2 = \frac{c^2x^2 + v^2t^2}{c^2}

I can see a ##t^2## on the LHS and on the RHS. The idea is to have all occurences of ##t^2## on the LHS.
 
Exacly that's the though part, because normally i would just divide out the t^2 but that won't help in this example as it would leave me with a 1 on the LHS.. and that wouldn't help much,, is there some mechanism or method that i am missing that could solve this? i feel like there's something i haven't learned that could allow this to be solved.. or is it just me that's blind?
 
MathiasArendru said:
Exacly that's the though part, because normally i would just divide out the t^2 but that won't help in this example as it would leave me with a 1 on the LHS.. and that wouldn't help much,, is there some mechanism or method that i am missing that could solve this? i feel like there's something i haven't learned that could allow this to be solved.. or is it just me that's blind?

If you have ##2x^2 = 5 - 3x^2##, can you solve that? You just need to do the same thing here. Isolate ##x##.
 
Yea no problem but there i can just add 3x^2 to both sides, in my example its v^2t^2 so i can't do the same thing here?
 
  • #10
MathiasArendru said:
Yea no problem but there i can just add 3x^2 to both sides, in my example its v^2t^2 so i can't do the same thing here?

You could try subtracting ##v^2t^2## from both sides.
 
  • #11
MathiasArendru said:
Yea no problem but there i can just add 3x^2 to both sides, in my example its v^2t^2 so i can't do the same thing here?

Why do you think v2 is different from 3?
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
7K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 19 ·
Replies
19
Views
5K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 20 ·
Replies
20
Views
2K