# Momentum transfer in electron-proton collision

• TheMercury79
In summary: 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.
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?

Last edited:
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.

PeroK said:
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##

TheMercury79 said:
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.

TheMercury79 said:
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
PeroK said:
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.

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

TheMercury79 and PeroK
vela said:
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

TheMercury79 said:
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
PeroK said:
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

## 1. What is momentum transfer in electron-proton collision?

Momentum transfer in electron-proton collision refers to the exchange of momentum between an electron and a proton during a collision. This transfer is a result of the interaction between the two particles and can be described by the laws of conservation of momentum.

## 2. How is momentum transfer calculated in electron-proton collision?

Momentum transfer in electron-proton collision can be calculated by subtracting the initial momentum of the electron and proton before the collision from their final momentum after the collision. This calculation takes into account the direction and magnitude of the momentum vectors.

## 3. What is the significance of momentum transfer in electron-proton collision?

Momentum transfer in electron-proton collision is significant because it provides information about the interaction between the two particles. It can be used to study the structure and properties of the particles and can also be used in particle accelerators to create high-energy collisions.

## 4. How does momentum transfer affect the outcome of an electron-proton collision?

The amount of momentum transfer in an electron-proton collision can affect the outcome of the collision. A higher momentum transfer can result in a more energetic collision, leading to the creation of new particles or the breaking apart of existing particles.

## 5. Can momentum transfer in electron-proton collision be controlled?

Yes, momentum transfer in electron-proton collision can be controlled by adjusting the energy and trajectory of the particles involved in the collision. This allows scientists to study specific interactions and phenomena in a controlled manner.

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