# Derivation of an old-time formula

1. Jul 3, 2010

### stevmg

In certain physics textbooks, one starts with the assumption (in a one linear and one time dimension) that

1) x2 - c2t2 = x'2 - c2t'2

I don't want to go into that. Let us start from there.

Now, if you assume that

2) x = vt, then

3) x' = 0 always because the origin is moving at v along the x-axis so that x' is always zero.

Using those three bits of information one can derive the Lorentz equations:

x' = $$\gamma$$(x - vt)
t' = $$\gamma$$(t - xv/c2)

where $$\gamma$$ = SQRT[1 - v2/c2] (I can't get the square root to come out in LATEX)

I have tried, tried, tried to do that but I cannot.

Anyone can help?

If I substitute the Lorentz equations back into the above three bits of information, it will work out but that's backwards. I want to do it forwards.

2. Jul 3, 2010

### espen180

$$c^2\text{d}\tau^2=c^2\text{d}t^2-\text{d}x^2$$

This is a usual way to express the minowski metric, or SR metric. $$\tau$$ is proper time, measured by an observer at rest relative to the frame being measured.

From here:

$$\left(\frac{\text{d}\tau}{\text{d}t}\right)^2=1-\frac{1}{c^2}\left(\frac{\text{d}x}{\text{d}t}\right)^2$$

We use the symbol $$v$$ for $$\frac{\text{d}x}{\text{d}t}$$.

$$\left(\frac{\text{d}\tau}{\text{d}t}\right)^2=1-\frac{v^2}{c^2}$$

$$\frac{\text{d}\tau}{\text{d}t}=\sqrt{1-\frac{v^2}{c^2}}$$

In the case of constant velocity, the right hand side is a constant. We can integrate wrt. $$t$$, letting $$\tau=t=0$$ initially and obtain

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

or the more familiar

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

Last edited: Jul 4, 2010
3. Jul 3, 2010

### stevmg

Perfect!

Actually, that makes it clearer than the text I was reading from.

Thanks,

Steve G
Melbourne, FL