Calculating Photon Energy with Proton Momentum | Special Relativity Homework

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SUMMARY

The discussion centers on calculating the energy of a photon with momentum equivalent to that of a proton possessing a kinetic energy of 10 MeV. Participants utilized key equations from special relativity, including K = mc²(γ-1), p = γmv, and E = pc, to derive the Lorentz factor (γ) and ultimately the photon energy. A critical point raised was the approximation of γ, which should be around 1.01 due to the proton's kinetic energy being approximately 1% of its rest energy. The use of online calculators, such as Wolfram Alpha, was recommended for complex calculations.

PREREQUISITES
  • Understanding of special relativity concepts, particularly Lorentz transformations.
  • Familiarity with the equations for kinetic energy and momentum in relativistic physics.
  • Basic knowledge of photon energy calculations and the speed of light.
  • Proficiency in using scientific calculators or online computational tools.
NEXT STEPS
  • Learn how to derive the Lorentz factor (γ) for various kinetic energies in relativistic contexts.
  • Explore the relationship between kinetic energy and momentum in special relativity.
  • Investigate the use of online tools like Wolfram Alpha for complex physics calculations.
  • Study the implications of relativistic effects on particle physics and energy calculations.
USEFUL FOR

Students and educators in physics, particularly those focusing on special relativity, as well as anyone involved in calculations related to particle momentum and energy in high-energy physics contexts.

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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