Electron and Proton Collision (relativity)

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Homework Statement


An electron is accelerated through a potential of 10^9V to the right. A proton is heading to the left towards the electron at .9c.
Find E,K,p for the electron and proton from the rest frame, and from each others frames.


Homework Equations





The Attempt at a Solution


Since the potential is 10^9V, the kinetic energy of the electron is K=10,000MeV, and the rest mass is mc^2=.51099MeV. So E=10000.51 MeV.

Using E=γmc^2, I get γ=19570.9, and β=.999999999

Which gives me values of E=10000.53 MeV, K=10000.02 MeV, and p=E/c=10000.53 MeV/c.


For the proton, I do similar math to get E=2152MeV, K=1212.3 MeV and p=2152 MeV/c.

Now to find the E,K,p of the electron from the reference from of the PROTON, I use the combination of velocities equation u'=-ve-vp/(1+vevp/c^2) and I always get the speed of the electron u'=c. How do I get around this? if u=c, then λ=∞, and all the energies and momenta are infinite.. :( Am I doing something wrong? :(
 

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  • #2
vela
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Homework Statement


An electron is accelerated through a potential of 10^9V to the right. A proton is heading to the left towards the electron at .9c.
Find E,K,p for the electron and proton from the rest frame, and from each others frames.


Homework Equations





The Attempt at a Solution


Since the potential is 10^9V, the kinetic energy of the electron is K=10,000MeV, and the rest mass is mc^2=.51099MeV. So E=10000.51 MeV.

Using E=γmc^2, I get γ=19570.9, and β=.999999999

Which gives me values of E=10000.53 MeV, K=10000.02 MeV, and p=E/c=10000.53 MeV/c.
As I mentioned in the other thread, you're off by a factor of 10 for K. Note you already had K and E. You didn't need to recalculate them after you found γ.

For the proton, I do similar math to get E=2152MeV, K=1212.3 MeV and p=2152 MeV/c.
E and K are right, though you should watch the sig figs. The momentum isn't correct though. The electron was ultrarelativistic, so you can use the approximation that it's massless and use E=p/c. That's not true for the proton.

Now to find the E,K,p of the electron from the reference from of the PROTON, I use the combination of velocities equation u'=-ve-vp/(1+vevp/c^2) and I always get the speed of the electron u'=c. How do I get around this? if u=c, then λ=∞, and all the energies and momenta are infinite.. :( Am I doing something wrong? :(
Are you familiar with the Lorentz transformations?
 
  • #3
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Okay I fixed all that, and got B=.9999999 for the electron (Sorry I decided to start a thread with the entire problem.. But thank you so much for all your help!)

And for the proton, I'm guessing the energies are so much higher just due to the fact that it is more massive than the electron? And I don't know how I missed this but p should be: p=Eβ/c, correct? So p=1936.8 MeV/c!

And I have seen lorentz transforms before, is that what I should be using here? So the momentum four-vector (in this case only has 2 dimensions) would be:
p=(E/c,p_x) And I would multiply it by:
(γ, -βγ)
(-βγ, γ)
to get:
(Eγ/c-βγp_x/c)
(-βγE/c+p_x)

Or is that wrong? I'm not quite sure how to use it for this.. :/
 
  • #4
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I think I got it! This time (with the new value for the speed of the electron that you helped me with) I got an actual value for u'=.999999995c which makes γ=9746.48. Then I just used:
E=γmc^2=4980.35 MeV
K=(γ-1)mc^2=4979.84 MeV
p=βE/c=4980.35 MeV/c

Then it would be the exact same from the ELECTRONS reference frame, but with a slightly lower value for u'

Are these correct?
 
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  • #5
vela
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For which particle are those quantities and in which frame?
 
  • #6
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Those quantities of E,K, and p are for the electron, from the protons frame of reference.
 
  • #7
vela
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Hmm, I get different results.
 
  • #8
vela
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Okay I fixed all that, and got B=.9999999 for the electron (Sorry I decided to start a thread with the entire problem.. But thank you so much for all your help!)

And for the proton, I'm guessing the energies are so much higher just due to the fact that it is more massive than the electron? And I don't know how I missed this but p should be: p=Eβ/c, correct? So p=1936.8 MeV/c!
Yes, this matches what I got.

And I have seen lorentz transforms before, is that what I should be using here? So the momentum four-vector (in this case only has 2 dimensions) would be:
p=(E/c,p_x) And I would multiply it by:
(γ, -βγ)
(-βγ, γ)
to get:
(Eγ/c-βγp_x/c)
(-βγE/c+p_x)

Or is that wrong? I'm not quite sure how to use it for this.. :/
Yes, this would work.
 
  • #9
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Awesome! I think I've got it now. Thank you so much for your help Vela; I really appreciate it!
 
  • #10
vela
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Just thought I'd mention...
I think I got it! This time (with the new value for the speed of the electron that you helped me with) I got an actual value for u'=.999999995c which makes γ=9746.48. Then I just used:
E=γmc^2=4980.35 MeV
K=(γ-1)mc^2=4979.84 MeV
p=βE/c=4980.35 MeV/c

Then it would be the exact same from the ELECTRONS reference frame, but with a slightly lower value for u'

Are these correct?
I assume you found u' using the velocity-addition formula. Your method is probably fine, but when the speeds get this close to c, even a small change can cause the value of γ to change wildly. Your answers are probably off just because of rounding errors.
 
  • #11
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I assume you found u' using the velocity-addition formula. Your method is probably fine, but when the speeds get this close to c, even a small change can cause the value of γ to change wildly. Your answers are probably off just because of rounding errors.

Yeah I did use the velocity addition formula and I noticed the huge changes for the smallest errors; I think it should be alright for this problem though (hopefully,) the next part of it definitely needs lorentz transform (find frame where momenta are equal.)
 

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