Relativity question, photon and proton

AI Thread Summary
The discussion revolves around calculating the time interval that a photon precedes a proton, given a proton with a gamma factor of 10^12 and a distance of 9.3x10^20m across the galaxy. The user attempts to derive the time difference using relativistic equations and dimensionless ratios, specifically focusing on the relationship between the velocities of the photon and proton. They express the time taken by each particle to travel the distance and attempt to find the difference in time, Δt. Despite initial calculations leading to a result of -1.55 x 10^(-12) seconds, the user struggles with the need to adjust the time calculation to the proton's reference frame. The discussion emphasizes the importance of approximations in relativistic calculations when dealing with small values of ε.
Prodigium
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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.
 
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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.
 
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\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?
 


Prodigium said:
\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.
 


vela said:
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.
 
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