Momentum transfer in electron-proton collision

[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?

 

[PeroK]

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

 

[PeroK]

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

 

[TheMercury79]

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 $\theta=0$

 

[PeroK]

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.

And I instead got $Q^2=-4\frac{EE'}{c^2}sin^2\frac{\theta}{2}$

 

[vela]

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

 

[TheMercury79]

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

 

[PeroK]

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?

 

[TheMercury79]

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.

 

[TheMercury79]

anyway, thanks for the help