# Momentum transfer in electron-proton collision

• TheMercury79
Because I knew the answer is not zero. Why should I check and see if the answer i zero? That's just ridiculous, of course it's not zero.anyway, thanks for the help.f

#### TheMercury79

Homework Statement
What is the squared 4-momentum transfer between the particles
Relevant Equations
4-momentum vector P = (E/c, ##\vec p##)
In a head-on collision between the proton and electron, what is the squared 4-momentum transfer between the two particles.

Starting with the difference in momentum of the electron with the 4-vectors before and after the event: $$(P-P')^2=P^2+P'^2-2P\cdot P'$$
The circumstances are such that the energies are large enough for the masses of the electron and proton to be ignored, thus ##E\sim pc## and ##E'\sim p'c##. Where ##E## is the energy of the electron before collision, ##E'## the energy of the electron after collision and ##\theta## is the angle between the the electron before and after the collision. Then:$$(P-P')^2=m_e^2c^2+m_e^2c^2-2(\frac{EE'}{c^2}-\vec p \cdot \vec p')\sim -2\frac{EE'}{c^2}(1-cos\theta)$$

The squared momentum transer would accordingly be ##Q^2=-2\frac{EE'}{c^2}(1-cos\theta)##

But I was told it is ##Q^2=-2\frac{EE'}{c^2}(1+cos\theta)##. These differ only by a plus and minus sign.

I think it is a bit weird that in my calculation above, I didn't have to include the proton at all. But then again I figure the momentum transfer is what the electron transfers to the electron (correct me if I'm not using the term momentum transfer correctly).

I am thinking I could perhaps use Mandelstam variables and include the proton in the calculation and see if I somehow end up with a positive cos-term instead of the negative.

Or is my calculation ##Q^2=-2\frac{EE'}{c^2}(1-cos\theta)## correct?

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Are you using the same definition of ##\theta##?

With your definition, if ##\theta = 0## there is no collision and no momentum transfer. So the ##-## looks right.

With your definition, if ##\theta = 0## there is no collision and no momentum transfer. So the ##-## looks right.
maybe I formulated it badly but the original statement was"##\theta## is the angle between the outgoing electron and the incoming proton". I figured the collision was head on, so this is the same as what I said "##\theta## is the angle between the electron before and after the collision"
I don't see where I said ##\theta=0##

maybe I formulated it badly but the original statement was"##\theta## is the angle between the outgoing electron and the incoming proton".
Exactly, their ##\theta## is the opposite of yours. Both equations are correct, with a different definition of ##\theta## is each case.

I don't see where I said ##\theta=0##
Choosing a simple value of ##\theta## to test your equation is a common idea. Your equation works with ##\theta = 0##.

• TheMercury79
Exactly, their ##\theta## is the opposite of yours. Both equations are correct, with a different definition of ##\theta## is each case.

Choosing a simple value of ##\theta## to test your equation is a common idea. Your equation works with ##\theta = 0##.
But the correct answer is supposed to be ##Q^2=-4\frac{EE'}{c^2}cos^2\frac{\theta}{2}## I just put in it the other form to illustrate the plus and minus difference.

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Your ##\theta## isn't the same as the one specified in the problem statement.

• TheMercury79 and PeroK
Your ##\theta## isn't the same as the one specified in the problem statement.
Now I see it! Pero said the same thing but he threw me off by insisting we should set ##\theta## to zero

Now I see it! Pero said the same thing but he threw me off by insisting we should set ##\theta## to zero
I didn't insist! It was just the obvious example to sanity check your answer.

How have you got this far in your studies without trying simple values to check a formula makes sense? • vela
I didn't insist! It was just the obvious example to sanity check your answer.

How have you got this far in your studies without trying simple values to check a formula makes sense? Because I knew the answer is not zero. Why should I check and see if the answer i zero? That's just ridiculous, of course it's not zero.

• • vela and PeroK
anyway, thanks for the help