Feynmann diagrams, bhabha scattering

[physics_fun]

There is nowone who can help me on the homework forum, so i'll try it here...

I'll try a little LATEX to make the problem more clear:

$$p_{1} + p_{2} = p_{3} + p_{4}$$
These are four vectors of the in- and going momenta

You take the frame where the threemomentum $$p_{2}=0$$

Questions:
1) Why does then defining threemomentum $$p_{4}=p_{3}$$ imply that $$p_{4}=p_{3}=0$$? (threemomenta!)
And why does defining $$p_{3}*p_{4}=0$$ (three vectors) imply that $$p_{3}=0 or p_{4}=0$$?

3. The Attempt at a Solution

I tried: 4-vectors: $$p_{1} + p_{2} = p_{3} + p_{4}$$
square this: $$(p_{1})^(2) + (p_{2})^(2) + 2p_{1}p_{2}= (p_{3})^(2) + (p_{4})^(2) + 2p_{3}p_{4}<br /> //<br /> (p_{1})^(2)=(p_{3})^(2)=m (electron mass)//<br /> (p_{2})^(2)=(p_{4})^(2)=M (positron mass)$$
So: $$p_{1}p_{2}=p_{3}*p_{4}$$
The lab frame condition gives:
$$p_{1}^{0}p_{2}^{0}=p_{3}^{0}p_{4}^{0}-p_{3}*p_{4}(three-vectors)$$

But what are the next steps?

 

[physics_fun]

Thanks for your reply!

I still don't get it exactly

First Question:

If p3=p4, then p3-p4=0 and the moment of q=p1+p2=p3+p4, I don't get why it should be 0...

Second Question:

The '*' is the innerproduct (inproduct), so I don't know if then the rest of your reply holds?

 

[Hans de Vries]

There is nowone who can help me on the homework forum, so i'll try it here...

First Question: (edited for the annihilation-creation diagram)

The answer follows from the on-shell condition.
E1+m = E3+E4, p1 = p3+p4
This means that E1 and p1 can only be on-shell if p3=p4=0

Second Question:

General advice: First try to simplify things with the given relations before
starting to take squares and things like that. Then use the on-shell
relations to prove the given restrictions.

I suppose the '*' is the cross product here. The cross product means that
p3 and p4 are in the same direction, thus (p4x,p4y,p4z) = a(p3x,p3y,p3z)

This gives us:

$$E_4+E_3\ =\ \sqrt{ a^2p_3^2 + m^2} + E_3$$

Since p2= (m,0,0,0) we have q = (E1+m, p1x, p1y, p1z)

It follows that (p1x,p1y,p1z) = (1-a)(p3x,p3y,p3z)

This gives us:

$$E_1 = \sqrt{ (1-a)^2p_3^2 + m^2) }$$

Thus:

$$E_3 = m + \sqrt{ (1-a)^2p_3^2 + m^2} - \sqrt{ a^2p_3^2 + m^2) }$$

This means that either $a$ must be 0 and thus p4=0 or p3=0.
There is another solution and that is $a=1$. In this trivial case p4=p3
and they can be non-zero...Regards, Hans

 

[Hans de Vries]

Thanks for your reply!

I still don't get it exactly

First Question:

If p3=p4, then p3-p4=0 and the moment of q=p1+p2=p3+p4, I don't get why it should be 0...

To which of the two diagrams contributing to Bhabha scattering are you
referring to. The electron positron scattering or the annihilation/creation one?

Second Question:

The '*' is the innerproduct (inproduct), so I don't know if then the rest of your reply holds?

It holds for the x product... (for the annihilation/creation diagram)Regards, Hans

 

[physics_fun]

To which of the two diagrams contributing to Bhabha scattering are you
referring to. The electron positron scattering or the annihilation/creation one?

Annihahilition: p1+p2(incoming momenta)=p3+p4(outgoing momenta)
Scattering: p1-p3=p4-p2 (toch?)

It holds for the x product... (for the annihilation/creation diagram)

Doesn't it hold for the innerproduct?

 

[Hans de Vries]

To which of the two diagrams contributing to Bhabha scattering are you
referring to. The electron positron scattering or the annihilation/creation one?

Annihahilition: p1+p2(incoming momenta)=p3+p4(outgoing momenta)
Scattering: p1-p3=p4-p2 (toch?)

OK, The answer then follows from the on-shell condition.

E1+m = E3+E4, p1 = p3+p4

This means that E1 and p1 can only be on-shell if p3=p4=0Regards, Hans

 

[physics_fun]

Mmm...
I still don't get it completely

 

[Hans de Vries]

Doesn't it hold for the innerproduct?

So p3 and p4 are under 90 degrees. It follows that:

$$p_{3\bot} = -p_{4\bot}$$

$$p_{3\|} + p_{4\|} = q = p_1$$

The latter is the same as

$${p_1 \over |p_1|} \cdot p_3\ +\ {p_1 \over |p_1|} \cdot p_4\ =\ p_1$$

The first left hand term is zero if the second one is at a maximum
and visa versa. For the energy we have:

$$E_1+m = E_3+E_4$$

Now, p1 should be on shell so you have to prove that this is only
true in the two cases given above.


Regards, Hans

 

[physics_fun]

Ok, I think I'm getting it now:

E1, E2, E3 are the rest-energies, p1, p3, p4 the three vectors (p2 is defined 0)

 

[physics_fun]

You have:

E1 = E3 + E4 - m = 2E3 - m

and

p1 = p3 + p4 = 2p3

now

$$E_1^2 = p_1^2 + m^2$$

thus

$$(2E_3 - m)^2 = (\ 2p_3\ )^2 + m^2$$

It follows that p3=p4=0


Regards, Hans

I'm understanding this one!

 

[Hans de Vries]

I'm understanding this one!

The factor 2 in 2p3 is true because p3 and p4 must be parallel to q and p1
otherwise it would be a variable.Regards, Hans

 

[physics_fun]

About the second: is it true that (threevectors:)
if p2=0, p1=p3+p4
so (p1)^2=(p3)^2+(p4)^2+2(p3)(p4)

now (p3)(p4)=0, so (p1)^2=(p3)^2+(p4)^2, and this can only be true if p4=0 or p3=0?

 

[physics_fun]

The complete proof is a bit more involved since the factor 2 in 2p3 is true only if
p3 and p4 are parallel to q but you can replace the 2 in 2p3 with a variable and follow
the same logic.


Regards, Hans

small angle approximation

 

[Hans de Vries]

small angle approximation

no, the 2 is true anyway , since p3 and p4 must be parallel to q and p1.
(I modified my post)

Regards, Hans.

 

[Hans de Vries]

About the second: is it true that (threevectors:)
if p2=0, p1=p3+p4
so (p1)^2=(p3)^2+(p4)^2+2(p3)(p4)

now (p3)(p4)=0, so (p1)^2=(p3)^2+(p4)^2, and this can only be true if p4=0 or p3=0?

You've got an expression for p1 now, as well as E1+m = E3+E4.
Next thing is to prove that

$$E_1^2 = p_1^2 + m^2$$

is true only in the two given cases.

Regards, Hans

 

[physics_fun]

It's completely clear to me now!

Thank you very much for your help!