Basic derivation of Lorentz Transformation

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Discussion Overview

The discussion revolves around the derivation of the Lorentz Transformation in the context of special relativity. Participants are exploring the mathematical relationships and constants involved in the transformation, seeking clarification on specific equations and their derivations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about the transition between equations (x'+ct')=μ(x+ct) and a=(λ+μ)/2, indicating difficulty in understanding the derivation process.
  • Another participant suggests determining the constants λ and μ indirectly through a and b, proposing that a=(λ+μ)/2 and b=(λ-μ)/2 are foundational to the derivation.
  • There is a question about how constants a and b are plugged into the equations to derive further formulas, with participants attempting to clarify this process.
  • One participant mentions getting stuck on the equations x'=ax-act and ct'=bx+bct, seeking guidance on the next steps.
  • Another participant confirms the correctness of a derived equation but expresses uncertainty about the origin of dx=1/a, prompting further inquiry into its derivation.
  • A later contribution discusses the relationship between the coordinates of a moving rod and the derived equations, suggesting a connection between the transformations and the physical interpretation of motion.
  • Participants are also trying to reconcile different forms of equations and their implications for the transformation process.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and confusion regarding specific steps in the derivation. There is no consensus on the clarity of Einstein's explanation or the derivation of certain equations, indicating that multiple competing views and uncertainties remain.

Contextual Notes

Some participants note that they are still working through Einstein's original text and have not been able to simplify the explanations provided. There are unresolved questions about the derivation of specific constants and equations, as well as dependencies on the definitions used in the transformation process.

Who May Find This Useful

This discussion may be useful for students and enthusiasts of physics, particularly those interested in special relativity and the mathematical foundations of the Lorentz Transformation.

ben.g95
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Hey, I'm just learning the ins and outs of special relativity and I'm having a little trouble with the derivation of the Lorentz Transformation. I bought Einstein's book 'Relativity: the special and general theories' and this is described in an appendix. For background, the appendix is linked to at the bottom of this post. (It was posted on marxists.org, I found that pretty funny, even though I knew Einstein was a socialist.) I get lost between
(x'+ct')=μ(x+ct) and
a=(λ+μ)/2

Also, looking ahead, I've gotten lost before (this isn't my first time looking over the appendix, I lost my work from last time, though) between
dx=1/a
and x'=a(1-(v2)/c^2))x <-- this one I really can't get (how does he get there?)

If anyone could be of any help, thanks!
http://www.marxists.org/reference/archive/einstein/works/1910s/relative/ap01.htm
 
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ben.g95 said:
I get lost between
(x'+ct')=μ(x+ct) and
a=(λ+μ)/2[/url]

In defining the mathematical forms of Eq 3 & 4, he sets up the unknown constants λ and μ. The problem then is to determine them.

Instead of determining λ and μ directly, he will determine them indirectly by determining a=(λ+μ)/2 and b=(λ-μ)/2 first.
 
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How did he plug in constants a and b to obtain formulas (5)?
 
ben.g95 said:
How did he plug in constants a and b to obtain formulas (5)?

From a=(λ+μ)/2 and b=(λ-μ)/2, we get a+b=λ and a-b=μ.
 
When plugging in a and b, I keep getting x'=ax-act and ct'=bx+bct and getting stuck there. I'm a beginner at this, could you walk me through it (thanks for being patient)? I feel like I'm close but I don't know where to go next.
 
By definition:
a=(λ+μ)/2
b=(λ-μ)/2

Adding the two equations gives: a+b=λ, ie. λ=a+b
Subtracting the two equations gives: a-b=μ, ie. μ=a-b
Now we know λ and μ in terms of a and b.

So x'-ct'=λ(x-ct) becomes:
x'-ct'=(a+b)(x-ct)
x'-ct'=ax+bx-act-bct (Eq I)

And x'+ct'=μ(x+ct) becomes:
x'+ct'=(a-b)(x+ct)
x'+ct'=ax-bx+act-bct (Eq II)

If I add Eq I and Eq II, the left hand side (LHS) is x'-ct' + x'+ct' = 2x'.
If I add Eq I and Eq II, the right hand side (RHS) is ax+bx-act-bct + ax-bx+act-bct = 2ax-2bct.
Equating LHS and RHS gives 2x'=2ax-2bct.

So x'=ax-bct, which is the first of Einstein's Eq 5.

Is that ok so far?
 
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That was great! You are a physics god among men.

Now on to the tougher stuff (for me at least).

Where does he get dx=1/a from?
 
ben.g95 said:
That was great! You are a physics god among men.

I'm a biologist.

ben.g95 said:
Now on to the tougher stuff (for me at least).

Where does he get dx=1/a from?

I don't follow his reasoning either! Is it correct?
 
I'm not really sure... it seems to come up in the next term (lost on that one as well), but I don't know what hat he pulled it out of.
 
  • #10
OK, it's correct. Will post an explanation later.
 
  • #11
Good, I couldn't get anywhere with it.
 
  • #12
I don't mean to sound like a jerk here, but could you please post your progress? Thanks.
 
  • #13
I haven't been able to simplify Einstein's explanation, am still working on it.

But there are many others here who presumably can.
 
  • #14
Let's say we have a rod of length L' moving with constant velocity u with respect to [x',t'] spacetime coordinates. Its left and right ends will trace out parallel lines of spacetime points described by equations of the form x'=xo'+ut' and x'=xo'+L'+ut'. In the particular case that the rod is stationary with respect to [x',t'] spacetime coordinates, u=0, and the parallel lines are x'=xo' and x'=xo'+L'.

Substituting Einstein's Eq 5 into parallel lines x'=xo' and x'=xo'+L', we get xo'=ax-bct and xo'+L'=ax-bct, which we can rearrange to x=(xo'+bct)/a and x=(xo'+L'+bct)/a to see more clearly that they are parallel lines in [x,t] coordinates. These parallel lines have the form x=xo+vt and x=xo+L+vt, as expected for the left and right ends a rod of length L moving with velocity v with respect to [x,t] coordinates, with xo=xo'/a, L=L'/a and v=bc/a. The equation L=L'/a is Einstein's Eq 7.

Let us note that since the rod was stationary in [x',t'] coordinates, and moving with velocity v=bc/a in [x,t] coordinates, v is also the velocity of [x,'t'] coordinates with respect to [x,t] coordinates. Hence our equation v=bc/a is Einstein's Eq 6.
 
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  • #15
Where did you get xo=xo'/a and L=L'/a? I follow you completely minus those two.
 
  • #16
Compare these two equations from post #14.

1) The equation for the spacetime line traced out by the right end of the rod obtained by getting the equation in [x',t'] coordinates then transforming to [x,t] coordinates:
x=(xo'+L'+bct)/a

2) The equation for the spacetime line traced out by the right end of the rod in [x,t] coordinates, obtained by the definition that it is a rod moving with velocity v with respect to [x,t] coordinates:
x=xo+L+vt

These two equations describe the same thing, so they must be the same equation, which will be the case only if xo=xo'/a, L=L'/a and v=bc/a
 
  • #17
ben.g95 said:
Where does he get dx=1/a from?

From x' = ax → Δx' = a Δx

But I = Δx', therefore I = a Δx, or Δx = I/a

Hope it helps.
 

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