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Four-Momenta derivation: is this mathematical step even valid?

  • Thread starter whyamihere
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  • #1

Homework Statement



Hey all, this is my first post to the forum, but I've been reading for years. Anyways, I was reviewing a solution to a Compton Scattering derivation we learned in class, and am still trying to make sense out of it.

Homework Equations


Given, we have the initial and final four-vectors of the photon and electron:
Pphoton,i = (Ei, Ei ni)
Pphoton,f = (Ef, Ef nf)
Pelectron,i = (me, 0)
Pelectron,f = (gamma me, gamma mevi)


The Attempt at a Solution


Using conservation of momentum, we get this expression:
Pphoton,i + Pelectron,i = Pphoton,f + Pelectron,f

This is where I get confused. Our tutor solved for the final electron term and then squared both sides like so:
|Pelectron,i|2 = |Pphoton,f + Pelectron,f - Pphoton,i|2

Is this action legal? Don't you have to square them first, then solve for the final electron term? Please, someone help clarify this. Everyone I ask seems to not even think twice about it. Is it really just a trivial step?
 

Answers and Replies

  • #2
CompuChip
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If you first square, you get


|Pphoton,i + Pelectron,i|² = |Pphoton,f + Pelectron,f

How are you going to get |Pelectron,f|² from that?

(Note that |a + b|² = |a|² + |b|² + (some very unpleasant cross terms) )
 
  • #3
Thanks CompuChip.

Yes, that makes sense mathematically. It's the physics that I still am not getting. What is the reason for squaring both sides? Does that give us the square of the rest mass?
 
  • #4
Actually, now I am convinced that I am right and my tutor made a mistake.

In order for this to be Lorentz invariant, the threshold energy s = |Pphoton,i + Pelectron,i|² = |Pphoton,f + Pelectron,f

So we MUST square each side first, otherwise it is not invariant.

Can anyone verify/refute this?
 
  • #5
vela
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The square of any four-vector is an invariant, and more generally, the product of any two four-vectors is an invariant.
 
  • #6
That really clarifies everything. Thanks vela!
 
  • #7
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Homework Equations


Given, we have the initial and final four-vectors of the photon and electron:
Pphoton,i = (Ei, Ei ni)
Pphoton,f = (Ef, Ef nf)
Pelectron,i = (me, 0)
Pelectron,f = (gamma me, gamma mevi)


You will want the units of photons and electons to match. For n unitless, and all vector components in units of energy:

Pphoton,i = (Ei, Ei ni)
Pphoton,f = (Ef, Ef nf)
Pelectron,i = (mec2, 0)
Pelectron,f = (γmec2, γmecvf)

Squaring things loses information.

Isn't Pphoton,i + Pelectron,i = Pphoton,f + Pelectron,f
four linear equations in four unknowns, γ and γvf?

However, you haven't actually stated what your givens and unknowns are.
 
Last edited:
  • #8
vela
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You will want the units of photons and electons to match. For n unitless, and all vector components in units of energy:

Pphoton,i = (Ei, Ei ni)
Pphoton,f = (Ef, Ef nf)
Pelectron,i = (mec2, 0)
Pelectron,f = (γmec2, γmecvf)
It's common to use units where c=1 and drop the explicit factors of c from expressions. For example

[tex](mc^2)^2 = E^2 - (pc)^2[/tex]

becomes

[tex]m^2 = E^2 - p^2[/tex].
Squaring things loses information.
Sometimes you don't care about that information. :wink: A common trick in relativity calculations is to arrange the four-vectors so that a well-placed 0 will wipe out terms you don't really care about when you square the equation.
Isn't Pphoton,i + Pelectron,i = Pphoton,f + Pelectron,f
four linear equations in four unknowns, γ and γvf?

However, you haven't actually stated what your givens and unknowns are.
 

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