# Scattering an electron off a proton.

1. Mar 21, 2009

### Alpha&Omega

Protons and neutrons have a diameter of around $$10^{-15}$$m. Despite this small size, physicists were able first to probe inside these particles in the 1970s, by scattering high energy electrons off them.

i). explain briefly why it is important that the electrons have high energy.

Since momentum is conserved in a collision the resulting particles will also have a high energy. This makes them easier to be detected by a detector.

ii). Assuming that the protons and neutrons were at rest, calculate the minimum momentum to which the electrons should be accelerated, in order to perform such experiments successfully

$$\lambda=\frac{h}{\rho} \Rightarrow \ \rho=\frac{h}{\lambda}=\frac{h}{10^{-15}}=6.63 \times 10^{-19}Jsm^{-1}$$

iii). Considering such scattering collisions in the laboratory frame of reference (where the target is stationary), write formulae for the energies of the electrons and target, and evaluate then in Gev (you may ignore the rest mass of the electron)

I found this part of the question pretty difficult. Here's my attempt:

Initial state: 0---> O
I have an electron (in the "picture" above as "0") and a proton ("O"). The electron has an energy $$E_1$$ and a momentum $$\underline{\rho}$$. I wrote this as $$(E_1,\underline{\rho})$$.

Using the equation $$E=\rho c$$, $$E_1=\underline{\rho}c$$

Final state: <-----0 O----->
Using the (energy, momentum) co-ordinates, the electron has $$(E_2, -\underline{\rho})$$ and the proton has $$(E_3, \underline{\rho})$$.

The initial energy I calculated using $$E=\sqrt{(\rho c)^2+(m_0 c^2)^2}$$
$$\sqrt{(\frac{6.63 \times 10^{-19} c}{1.6 \times 10^{-10}})^2+(0.938)^2}$$

(the mass of a proton is $$0.938\frac{Gev}{c^2}$$)

This gives E=1.557 GeV.

Hence I found the initial energy of the proton to be 0.938Gev and the energy of the electron to be 1.557-0.938=0.619Gev.

I'm unsure whether this is correct. I may have completely misunderstood the entire situation!

iv). The total energy in the centre-of-mass reference frame in such electron-nucleon collisions corresponds to the maximum possible mass of all particles in the final state and is a Lorentz-invariant quantity. Ignoring the rest mass of the electron, calculate this quality

The "maximum possible mass" suggests that all the particles are stationary. Hence I should be calcuating their rest mass.

However, since I only have a proton and an electron at the end would the answer by just the rest mass of a proton- $$0.938\frac{GeV}{c^2}$$?

2. Mar 22, 2009

### turin

Not always. The idea behind part ii) is what they are probably looking for here. Think in terms of scattering and coherence.

You should think of part iv) as a hypothetical: what if there were a group of particles with masses that add up to $\Sigma{}m$? Then, you need to determine what is the largest that $\Sigma{}m$ could be for this hypothetical group of particles to be produced. You need to understand the concept of invariant mass, which is not necessarily the rest mass of any particle. The invariant mass is simply $\frac{1}{c^2}\sqrt{E^2-c^2\vec{p}.\vec{p}}$.

Actually, I would have done this calculation before part iii), but I'm not quite sure what part iii) is asking.

Last edited: Mar 22, 2009
3. Mar 23, 2009

### Alpha&Omega

Ah, I understand what you mean.

For part ii) you can use $$E=\rho c$$. Therefore $$\rho=\frac{E}{c}$$.
There's also the equation for De Broglie wavelength (it won't let me put it here for some reason).

Hence if you make the energy larger the wavelength of the electron decreases. A smaller wavelength gives better resolving power.

I'm also not sure about part iii). I'm not sure if it's a collision or not since I thought the electrons were just diffracting around the proton? =S

I'll give part iv) another go but any hints anyone could tell me would be super!!

Watch this space!

4. Mar 23, 2009

### nathan12343

Part iv is actually pretty easy. Find the total CM energy-momentum 4-vector, and calculate the invariant mass (<E,E> = $$\frac{1}{c^2}\sqrt{E^2 - (pc)^2}$$). You can also do part iii using this information as well as some Lorentz transformations... might save you a bit of tedious algebra, although I haven't tried this.

5. Mar 23, 2009

### Alpha&Omega

Right, I have a few questions.

Firstly, what is a "CM energy-momentum 4 vector"? Does this have 4 dimensions so I should be including time as well?

Secondly, is there any reason why it's "<E,E>" in the invariant mass equation?

Here's my go at finding the invariant mass:

$$\frac{1}{c^2}\sqrt{E^2-(\rho c)^2}$$

$$\frac{1}{c^2} \sqrt{(1.67 \times 10^{-27} c^2)^2-(6.63 \times 10^{-19} c)^2$$

$$\frac{1}{c} \sqrt{(1.67 \times 10^{-27} c^2)^2-(6.63 \times 10^{-19})^2}$$

$$=5.01 \times 10^{-19}$$kg

Is this part right?

6. Mar 24, 2009

### turin

A 4-vector is a fundamental object of special relativity; it is the basis of the fundatmental representation of the Lorentz Group, which is the group of all Lorentz transformations. It has four components; one of the components is the time component, and the other three components are the space components. The prototype 4-vector is the vector whose time component is the time coordinate of an event, and whose space components are the space coordinates of the event. An energy-momentum 4-vector (usually referred to simply as 4-momentum) has energy as the time component and momentum components as the space components. CM means center-of-mass. You can treat the CM of a system of particles as a massive object whose energy is the total energy of the system and whose momentum is the total momentum of the system. So, now you tell us, what do you think a CM energy-momentum 4-vector is?

It looks like you're using the proton mass for the CM energy and the electron momentum for the CM momentum. That would be incorrect.

7. Mar 24, 2009

### Alpha&Omega

Okay, so the CM energy-momentum 4-vector is $$(E, \rho_x, \rho_y, \rho_z)$$.

Since the proton is stationary relative to the electron, it's CM energy-momentum 4-vector is $$(1.67 \times 10^{-27}c^2, 0,0,0)$$.

In the case of the electron, i'll assume it's travelling along the x-axis towards the proton. Hence it's CM energy-momentum 4-vector is $$(6.63 \times 10^{-19}c,6.63 \times 10^{-19},0,0)$$. (I got $$6.63 \times 10^{-19}$$ from part ii). I also used the equation $$E=\rho c$$ for the energy component).

I'll also have to calculate the co-ordinates for each particle after the collision. Should I assume that the electron travels the opposite way down the x axis?

For my invariant mass, should I be taking $$\rho=0$$ since the proton has no momentum before the collision?

8. Mar 24, 2009

### turin

You need to be clear about exactly what physical object has this 4-momentum.

You are half-right. These are trivial CM 4-momenta for each particle. What you want is the CM 4-momentum for the entire system (proton and electron together as a single object). To get this, simply add the vectors (that's one of the beauties of this formalism, you can add 4-vectors just like you add any other kind of vectors). So, pCM=pe+pp (as four-component vectors).

I don't think so. Certainly not for part iv.

The final state is completely undetermined. This problem is hinting that the proton is composed of quarks, so that the electron basically sees three randomly positioned scattering centers, and it diffracts through them. Don't worry about the details. All you need to realize is that, unless you are given a final state, you cannot know what it is. (there may not even be a proton or electron after the collision! - but don't worry about that either.)

Invariant mass of what? rho of what?

9. Mar 24, 2009

### Alpha&Omega

So the CM energy momentum 4-vector of the system is $$\rho_{CM}=(1.67 \times 10^{-27}c^2+6.63 \times 10^{-19}c, 6.63 \tims 10^{-19},0,0)$$ (it's probably simpler using it in this form).

That's a little weird! Slowly but surely i'm starting to get a better grasp of this question.

I meant that the proton before the collision has no momentum since it's stationary. Hence $$\rho=0$$.

Would the invariant mass be this:

$$\frac{1}{c^2} \sqrt{E^2-(\rho c)^2}$$

So we need to find this for both the electron and the proton:

$$\frac{1}{c^2} \sqrt{1.67 \times 10^{-27} c^2-0}+\frac{1}{c^2} \sqrt{(6.63 \times 10^{-19}c)^2}$$

$$\frac{1}{c^2}\sqrt{1.67 \times 10^{-27}c^2}+\frac{1}{c^2}6.63 \times 10^{-19}$$

10. Mar 24, 2009

### turin

So, now just apply the definition of invariant mass to this 4-vector, where the first component is E and the other three components are the components of p.

edit: BTW, it is probably better to stick with units of GeV and c, rather than writing it in terms of kg, J, m, s etc.. GeV/c^2 is a perfectly acceptable unit of mass, and is preferred by many in the high energy physics community.

Last edited: Mar 24, 2009
11. Mar 24, 2009

### Alpha&Omega

YAY!!!

BYW?

$$m=\frac{1}{c^2}\sqrt{E^2-(pc)^2}$$

$$m=\frac{1}{c^2}\sqrt{(1.67 \times 10^{-27}c^2+6.63 \times 10^{-19}c)^2-(6.63 \times 10^{-19}c)^2}$$

$$m=3.18 \times 10^{-27}kg$$ (sorry about the units, I worked out everything in kg =S).

12. Mar 24, 2009

### turin

Sorry. I corrected this typo. I meant BTW.

That looks correct. Do you know how to interpret this result? For example, would you have been surprised if m<1.67E-27kg?

13. Mar 25, 2009

### Alpha&Omega

Oh, I thought this was an actual physics term =S

If m< 1.67E-27 it would imply that the least amount of energy that both the electron and proton have is less than the rest mass energy of the proton. This can't happen!

14. Mar 25, 2009

### turin

Correct (after I replaced "least" with "initial").

I don't want to confuse you, so if you are just working on this stuff at a superficial level, then ignore the following. However, if you really want to get into this particle physics stuff, then you should be aware of bound states, which can have an energy that is less than the energy of the sum of the masses!