# Basic derivation of Lorentz Transformation

1. Sep 26, 2011

### ben.g95

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

2. Sep 26, 2011

### atyy

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.

Last edited: Sep 26, 2011
3. Sep 26, 2011

### ben.g95

How did he plug in constants a and b to obtain formulas (5)?

4. Sep 26, 2011

### atyy

From a=(λ+μ)/2 and b=(λ-μ)/2, we get a+b=λ and a-b=μ.

5. Sep 26, 2011

### ben.g95

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.

6. Sep 26, 2011

### atyy

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?

Last edited: Sep 26, 2011
7. Sep 26, 2011

### ben.g95

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?

8. Sep 27, 2011

### atyy

I'm a biologist.

I don't follow his reasoning either! Is it correct?

9. Sep 27, 2011

### ben.g95

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. Sep 28, 2011

### atyy

OK, it's correct. Will post an explanation later.

11. Sep 29, 2011

### ben.g95

Good, I couldn't get anywhere with it.

12. Oct 10, 2011

### ben.g95

I don't mean to sound like a jerk here, but could you please post your progress? Thanks.

13. Oct 10, 2011

### atyy

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. Oct 12, 2011

### atyy

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.

Last edited: Oct 13, 2011
15. Oct 13, 2011

### ben.g95

Where did you get xo=xo'/a and L=L'/a? I follow you completely minus those two.

16. Oct 13, 2011

### atyy

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. Oct 13, 2011

### zaybu

From x' = ax → Δx' = a Δx

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

Hope it helps.