# Derivation of Lorentz transformation (time)

1. Aug 24, 2014

### rem45

Hi all,

I'm trying to derive that t=δ(t'+vx'/c^2)

Using x'=δ(x-vt) then substituting for x=δ(x'+vt') I should be able to isolate t and solve the problem but I am getting to the following point after simplification and can't figure out where to go next...?

x'= δ[δ(x'+vt')-vt] (isolate "vt")

vt = -x'/δ + δ(x'+vt') (divide by "v")

t = -x'/δv + (δ/v)(x'+vt')

2. Aug 24, 2014

### Fredrik

Staff Emeritus
If you're trying to solve the system of equations
\begin{align}
&t'=\gamma(-vx/c^2+t)\\
&x'=\gamma(x-vt)
\end{align} for t, it's really weird to use the result for x in terms of t' and x', because that's no easier to find than the result for t.

I recommend that you start by finding a linear combination of x' and t' that doesn't involve x. (I would also use units such that c=1, to avoid having to keep track of factors of c in every step, but that's optional).

3. Aug 24, 2014

### rem45

I agree but the assumption is that I do not know the equation for t or t', only x and x'. I see this approach on numerous websites including the wikipedia derivation but everyone seems to get to where I am the just write that this becomes the familiar t=(gamma)(t'+vx'/c^2)

4. Aug 24, 2014

### Fredrik

Staff Emeritus
OK, in that case, the trick is to rewrite $\gamma^2-1$ in a clever way. Can you see it? Use the definition of $\gamma$.
$$t=-\frac{x'}{\gamma v}+\frac{\gamma}{v}(x'+vt') =\frac{x'}{\gamma v}(\gamma^2-1)+\gamma t'.$$

5. Aug 24, 2014

### rem45

Yes, it took me min but now I see it! Very sneaky. I don't see how this becomes t=δ(t'+vx'/c^2)....

Apparently δ(vx'/c^2) = [x'/δ(v)](δ^2-1) but there must be some other trick to reduce to the standard formula

6. Aug 24, 2014

### Fredrik

Staff Emeritus
Not sure why there's an x' in your result for $\gamma^2-1$, but you can cancel it from that equation. Then solve for $\gamma^2-1$, and insert the result into the right-hand side of the calculation in post #4.

Why are you using the symbol δ instead of γ. Is it just because the latter looks too much like a y?

Edit: OK, now I see what you're doing. The reason you didn't solve for $\gamma^2-1$ is that you wanted to rewrite the first term on the right-hand side in post #4. You found that it's equal to $\gamma v x'/c^2$. This turns the result in post #4 into
$$t=\frac{\gamma v x'}{c^2}+\gamma t' =\gamma\left(t'+\frac{vx'}{c^2}\right).$$

Last edited: Aug 24, 2014
7. Aug 24, 2014

### rem45

Got it! Sorry i didn't have my reading glasses with me but now i can see now that i've been using delta instead of gamma.

8. Aug 24, 2014

### Fredrik

Staff Emeritus
OK, cool. I think the easiest way to do this calculation is like this (in units such that c=1):

\begin{align}
&x'=\gamma(x-vt) =\gamma\big(\gamma(x'+vt')-vt\big) =\gamma^2(x'+vt')-\gamma vt\\
&\gamma vt =\gamma^2(x'+vt')-x' =(\gamma^2-1)x'+\gamma^2vt'\\
&\hspace{20mm}\bigg[\gamma^2-1=\frac{1}{1-v^2}-1 =\frac{1-(1-v^2)}{1-v^2}=\gamma^2v^2\bigg]\\
&\gamma vt =\gamma^2v^2x'+\gamma^2vt'\\
&t=\gamma v x'+\gamma t' =\gamma(vx'+t')
\end{align}