Calculating Photon Energy with Proton Momentum | Special Relativity Homework

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Homework Help Overview

The discussion revolves around calculating the energy of a photon that shares momentum with a proton having a kinetic energy of 10 MeV. The subject area includes concepts from special relativity, particularly relating to energy, momentum, and the Lorentz factor.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss using the Lorentz factor and momentum equations to relate the kinetic energy of the proton to the photon energy. There are attempts to clarify the calculations involving the Lorentz factor and its implications on the problem setup.

Discussion Status

Some participants provide guidance on calculation methods and tools, while others question the assumptions made regarding the values of the Lorentz factor. Multiple interpretations of the calculations are being explored, with no explicit consensus reached.

Contextual Notes

There are mentions of potential confusion regarding the squaring of the Lorentz factor and its approximation, as well as the need for accurate computational tools to handle small values. The original poster's approach is noted to be potentially complicated, with alternative methods suggested.

lockedup
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Homework Statement


What is the energy of a photon whose momentum is the same as that of a proton with a kinetic energy of 10 MeV?

Homework Equations


K = mc^{2}(\gamma-1)
p = \gammamv
E = pc

The Attempt at a Solution


I figured I would go at it like this. I used the first equation listed to obtain a value for gamma and v. I put these into the second equation along with the mass of a proton. Then, I would put that value into the third equation. Is that even close?
 
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Seems ok if I understood it right. Your equations might help more.
 
As far as I can understand, the procedure you describe is correct. As long as you have done the calculations properly, you should end up with the correct answer.

Torquil
 
Ok, so I have \gamma = 1 + 1.18 \times 10^{-19} So, um, how am I supposed to square that (so I can find beta)? My calculator won't go that far... :*)
 
use an online calculator
 
or any computer program will do it
 
or you can just square the 1.18 and then double 19
 
dacruick said:
use an online calculator
They all give me 1
 
the answer is 1.3924e-38
 
  • #10
so for next time this is what you will do. you should square 1.18. then you should double the exponent. and voila
 
  • #11
dacruick said:
so for next time this is what you will do. you should square 1.18. then you should double the exponent. and voila
You're misreading the problem. If it was just 1.18e-19, I wouldn't have a problem. I need to square 1+1.18e-19. BTW, I found a calculator that will do it. Google can be tricksy sometimes...
 
  • #12
if \gamma=1+1.18\times10^{-19} then you should just approximate \gamma\simeq1.

Also, use www.wolframalpha.com It's like having Mathematica available to you with 0 cost
 
  • #13
Your value for \gamma is way too small. The proton's kinetic energy is about 1% of the rest energy, so \gamma should be about 1.01.
 
  • #14
lockedup said:

Homework Statement


What is the energy of a photon whose momentum is the same as that of a proton with a kinetic energy of 10 MeV?



Homework Equations


K = mc^{2}(\gamma-1)
p = \gammamv
E = pc



The Attempt at a Solution


I figured I would go at it like this. I used the first equation listed to obtain a value for gamma and v. I put these into the second equation along with the mass of a proton. Then, I would put that value into the third equation. Is that even close?

The speed of light = 299,792,458 m/s;
Proton mass: 1.67262158*10^{-27} kg;
K= 1.602176487*10^{−15} J.

So with a simple mind-based calculation you at least can get that the Lorentz factor is around 1. The following shows its exact value up to 9 decimal digits:

\gamma=1.000010658.

AB
 
  • #15
This is such a complicated way to do this problem. The energy of a proton is its kinetic energy plus its mass, and that squared is p^2 + m^2 (with c=1). Find p. E(photon) = p. Done.
 

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