Conservation of momentum in compton scattering

  • Thread starter Avatrin
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  • #1
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Hi

I recently got some weird feedback on homework. It said that "..momentum is only conserved componentwise..". The problem involved Compton scattering, and I had used [itex] p_{\lambda,1} = p_{e,2} + p_{\lambda,2}[/itex] to find [itex]p_{e,2}[/itex] for collision between an electon and a photon. I had used the absolute values for the momenta because that was all I had.

So, in the general case, how is momentum conserved? Also, in the specific case, how should it be used in problems like the one above?
 

Answers and Replies

  • #2
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The feedback is a little confusing. Especially the "only" part.
It's not that it does not conserve otherwise but the momentum is a vector and so the conservation law is a vector expression.
So in your expression above, all three terms should be vectors. Then you can split it by components - this is what they say.
If you want to look at the absolute values, than you should realize that the absolute value of the sum of two vectors (as yo have on the right-hand side) is not the sum of the absolute values. So the equation does not hold as you wrote it, if these are absolute values.
 
  • #3
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Momentum is a vector quantity, so what is conserved is a vector. The sum of the magnitudes may not be conserved. If you only have the magnitudes then you don't really have enough information. You also need to know the directions to use momentum.

EDIT: and its nasu for the win!
 
  • #4
jtbell
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I had used [itex] p_{\lambda,1} = p_{e,2} + p_{\lambda,2}[/itex]
More explicitly, you should have started with $$p_{\lambda,1,x} = p_{e,2,x} + p_{\lambda,2,x} \\ p_{\lambda,1,y} = p_{e,2,y} + p_{\lambda,2,y}$$ For convenience, we usually choose the coordinate system so the incoming photon enters along the x-axis which makes ##p_{\lambda,1,y} = 0##. You don't need to consider a third component because the three vectors define a plane.
 
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  • #5
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Thank you all!
 

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