How Can You Derive the Lorentz Equations from Basic Assumptions?

[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.

 

[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}}}

 

[stevmg]

Perfect!

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

Thanks,

Steve G
Melbourne, FL