# Issue with deriving the Lorentz Factor

1. Jan 1, 2013

### guillefix

Hello,

I've seen several derivations of it and I understand them, specially the one in two dimensions with the right angled triangle, but when I try it in 1 dimension, I always find that the condition that the speed of light is the same in all the inertial frames, isn't enough to specify what the Lorentz Factor is. Here's my problem:

Take the Lorentz Transformation eq.:

x'=y(x-vt)
t'=y(t-vx/c^2)

Now focus on a light signal, so that in S, x/t=c. We then want x'/t'=c in S' too:

x'/t'=(x-vt)/(t-vx/c^2)=c

As you can see the Lorentz Factor always drops out of the eq...so it seems like it doesn't matter what its value is. But then using other methods and the same assumptions, you do get a definite expression. I think there must be something wrong in my derivation above, but I don't know what it can be. Any ideas?

2. Jan 1, 2013

### Staff: Mentor

You haven't used all the constraints that follow from assuming a constant speed of light; so far you've only required that both observers agree about the speed of a right-moving light signal for which (x-ct)=0. Try adding another constraint for a left-moving signal for which (x+ct) and (x'+ct') must also be zero.

(I'm assuming that you don't want to just be pointed at a successful derivation - if you want to see it worked, the derivation in the first appendix of Einstein's book "relativity: the special and general theory" is pretty good)

Last edited: Jan 1, 2013
3. Jan 1, 2013

### guillefix

In the case for a left-moving signal I will just have x/t=-c so:

x'/t'=(-c*t-vt)/(t+v*t/c)=-(c+v)/(1+v/c)=-c

So again, no Loretnz factor involved no?

4. Jan 1, 2013

### Staff: Mentor

Yes, but think about why the Lorentz factor is disappearing in both cases. You're looking at sets of points x and t that satisfy (x-vt)=γ(x'-vt')=0; as long as x-vt and x'-vt' are both zero the equation will hold for all values of γ.

Try writing (x-vt)=A(x'-vt') and (x+vt)=B(x'+vt') and then adding and subtracting the two equations. That will give you two equations for two unknowns, A and B. Select some points off the lines of the two light signals for which you know both coordinates in both systems (like the origins at times other than zero) and solve for A and B. You'll have to use the symmetry of the situation (however the unprimed coordinates transform into the primed coordinates is also how the primed coordinates transform into the unprimed coordinates with v replaced by -v).

Or you could just look it up in one of the many online versions of Einstein's book, but that's not as much fun

5. Jan 1, 2013

### Staff: Mentor

I should add: the previous post is taking c=1 and that's why it's v not v/c.

6. Jan 1, 2013

### guillefix

Sorry, but I don't understand where (x-vt)=γ(x'-vt')=0 and (x-vt)=A(x'-vt') come from. I've seen them in wikipedia, but I don't get them. I mean why (x-vt)=γ(x'-vt') if x=γ(x'-vt') ?

7. Jan 1, 2013

### Darwin123

Because (t-vx/c^2)=γ(t'-vx'/c^2)

I didn't even know that Wikipedia showed mathematical derivations. Wikipedia often shows the final result without derivations. However, Wikipedia often provides references where one can read about the derivation.

As another poster pointed out, one has to include all the constraints. The separate parts of the Lorentz transformation do not demonstrate Lorentz invariance while separated.

There are a lot of derivations of the Lorentz transformation in the literature. There is more than one way to derive them. However, all derivations all start out with physical hypotheses. One can't just look at the Lorentz transformation and "get them".

8. Jan 1, 2013

### Staff: Mentor

My mistake, I fumbled the 'v' and 'c' variables in my last post. The two equations should be (x-t)=A(x'-t') and (x+t)=B(x'+t').
(Replace t and t' with ct and ct' if you don't want to use units in which c is equal to one).

These equations are valid for the same reason that γ keeps on disappearing in your calculations: we are considering sets of points (x,t) that describe the path of a light signal so the terms in the parentheses are equal to zero and the equations are true no matter what the values of A and B are.

Add the two equations and we get: 2x = (A+B)x' + (B-A)t'
Subtract them and we get: -2t = (A-B)x' - (A+B)t'

These two equations are pretty clearly transformations between the primed and unprimed coordinate systems. All we need now are some interesting (x,t)<=>(x',t') correspondences that we can use to supply x,t,x',t' values that will turn these into equations in the unknowns, A and B. For example, we know that the point (vt,t) corresponds to the point (0',t') - it's the origin of the primed system at any given time.

Einstein's derivation along these lines is at http://www.bartleby.com/173/a1.html

9. Jan 2, 2013

### guillefix

Thank you for the answers. I finally realized my problem is that I wasn't solving the two equations simultaneously. As with each the light ray moving to the right and left, gamma could be anything, but when you put the constraints together simultaneously, it constraints the answer. BTW, Wikipedia does _sometimes_ give quite good derivations, and for the Lorentz Transformations it's the case: http://en.wikipedia.org/wiki/Lorentz_transformation#From_physical_principles

10. Jan 4, 2013

### stevendaryl

Staff Emeritus
I don't think that invariance of lightspeed uniquely determines the transforms.

If you have a general transformation:

$x' = Ax + Bt$​
$t' = Cx + Dt$​

Then you have the constraints:
1. $x=vt \Rightarrow x'=0$
2. $x=ct \Rightarrow x' = c t'$
3. $x=-ct \Rightarrow x'=-c t'$

then we can conclude:
1. $B = -vA$
2. $D = A$
3. $C = -vA/c^2$

So the equations become:
$x' = A(x - vt)$​
$t' = A(t - vx/c^2)$​

with the inverse equations:
$x = \dfrac{1}{A(1-\dfrac{v^2}{c^2})} (x' + vt')$​
$t = \dfrac{1}{A(1-\dfrac{v^2}{c^2})} (t' + vx'/c^2)$​

I think that to determine $A$, you need to impose the assumption that the inverse transformation must look the same as the forward transformation, if you replace $v$ by $-v$. With that additional assumption, it must be that

$A = \dfrac{1}{A(1-\dfrac{v^2}{c^2})}$​

which gives

$A = \dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}$​

11. Jan 4, 2013

### Staff: Mentor

I agree. I mentioned this back in #4, probably should have repeated it.