# Relativity question, photon and proton

1. Dec 8, 2011

### Prodigium

Relativity question, photon and proton....

1. The problem statement, all variables and given/known data
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?

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

3. 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.

Last edited: Dec 8, 2011
2. Dec 8, 2011

### vela

Staff Emeritus
Re: Relativity question, photon and proton....

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.

Last edited: Dec 8, 2011
3. Dec 8, 2011

### Prodigium

Re: Relativity question, photon and proton....

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

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 cant seem to get it , did I do this correctly?

4. Dec 8, 2011

### vela

Staff Emeritus
Re: Relativity question, photon and proton....

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

5. Dec 8, 2011

### Prodigium

Re: Relativity question, photon and proton....

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 $t_2$.

Last edited: Dec 8, 2011