Relativity question, photon and proton

[Prodigium]

Relativity question, photon and proton...

Homework Statement

Proton of gamma factor 10^12 and a photon set off from one side of our galaxy to the other take the distance to be 9.3x10^20m.
What is the time interval that the photon precedes the proton?

Homework Equations

\gamma = sqrt(c/2epsilon). (epsilon being c-v)

The Attempt at a Solution

Worked out epsilon to be 1.5x10^(-18)Not sure where to go from there anyhelp/hints would be most appreciated.

 

[vela]

You should try to form ratios that are dimensionless. In this case, you could write
$$v = c - \Delta v = c\left(1 - \frac{\delta v}{c}\right) = c(1-\varepsilon)$$where $\varepsilon = \Delta v/c$. Why? Because it avoids any complication due to units, for one thing. Regardless of which set of units you use, $\varepsilon$ defined this way will always turn out to have the same value. Moreover, an expression is often more naturally expressed in terms of these unitless ratios.

You can show that
$$\gamma = \frac{1}{\sqrt{1-(v/c)^2}} = \frac{1}{\sqrt{1-(1-\varepsilon)^2}} \cong \frac{1}{\sqrt{2\varepsilon}}$$

Now write down expressions for the time each takes to travel the given distance.

 

[Prodigium]

$\Delta t = \frac{(1- \epsilon )x-x}{c (1- \epsilon) }$


after putting $t_1 = \frac{x}{c}$ and $t_2 = \frac{x}{c(1- \epsilon )}$, then to find $\Delta t = t_1 - t_2$

but I can't seem to get it , did I do this correctly?

 

[vela]

$\Delta t = \frac{(1- \epsilon )x-x}{c (1- \epsilon) }$


after putting $t_1 = \frac{x}{c}$ and $t_2 = \frac{x}{c(1- \epsilon )}$, then to find $\Delta t = t_1 - t_2$

but I can't seem to get it , did I do this correctly?

Because ε is so small, you can approximate t₂ well with a series expansion to first order.

 

[Prodigium]

Because ε is so small, you can approximate t₂ well with a series expansion to first order.

so $t_2= \frac{x}{c} \cdot (1- \epsilon )^{-1} = \frac{x}{c} \cdot (1)$

or, because that completely gets rid of $\epsilon$ :-

$t_2= \frac{x}{c} \cdot (1- \epsilon )^{-1} = \frac{x}{c} \cdot (1+ \epsilon )$

That after finding $\Delta t$ gave me and answer of $-1.55 \cdot 10^{-12} s$

only problem is you have to find the time as being in the reference frame of the proton later and need $t_2$.