# Find the energy of a photon after this annihilation process

Tags:
1. Apr 2, 2017

### leo.

1. The problem statement, all variables and given/known data

The problem is as follows: in a reference frame there is one electron at rest and one incoming positron which annihilates with the electron. The positron energy is $E$ and two gamma rays are produced. Find first the energy of the photons in the center of mass frame as function of the energy $E$ in the lab frame. Then supposing one photon is detected in the lab frame at opposite direction than the proton's find the photon's energy.

2. Relevant equations

Of course I believe the relevant equations are: independence of reference frame of the metric tensor inner product and the definition of the center of mass frame as the one on which the total momentum is zero.

3. The attempt at a solution

The first part I think I solved right. The situation before, in the electron's frame of reference, we have one electron at rest with four-momentum $p_{1}^\mu = (m,0)$ and one positron with four-momentum $p_2^\mu = (E,\mathbf{p}_2)$. Using invariance of the inner product, we can compute $p_2^\mu {p_2}_{\mu}$ in the positron's frame, and comparing the results in both frames we have $E^2 - |\mathbf{p}_2|^2 = m^2$ from which we determine $|\mathbf{p}_2|^2 = E^2 + m^2$.

After the process, we have two photons with four-momentum now in the center of mass frame $p_3^\mu = (E_1,\mathbf{p}_\gamma)$ and $p_4^\mu = (E_2,-\mathbf{p}_\gamma)$ so that the total momentum zero. Now since these are photons, their four-momentums are lightlike, hence $p_3^\mu {p_3}_\mu = 0$ from which we get $E_1^2 + |\mathbf{p}_\gamma|^2 = 0$ and the same to the other photon. This gives that $E_1 = E_2 = E_\gamma$.

Hence we have: the total four momentum before the annihilation in the electron's frame is $p^\mu = (E+m, \mathbf{p}_2)$. The total four momentum after the annihilation in the center of mass frame is $\tilde{p}^\mu = (2E_\gamma, 0)$. Now considering the abstract four-vectors $p$ and $\tilde{p}$ conservation of four-momentum indicates we should have $p = \tilde{p}$. In that case, since the metric tensor inner product doesn't depend on reference frame we can equate the one computed on the electron's frame and the one computed on the center of mass frame, giving $p^\mu p_\mu = \tilde{p}^\mu \tilde{p}_\mu$ which is

$$(E+m)^2 - (E^2 + m^2) = 4E_\gamma^2$$

and hence $E_\gamma = \sqrt{mE/2}$.

The problem is being the other part. I mean let now $p_\gamma^\mu = (E', \mathbf{p}')$ be the four momentum of that photon in the lab reference frame. We only know $\mathbf{p}'$ has opposite direction to $\mathbf{p}_2$, hence $\mathbf{p}'\cdot \mathbf{p}_2 = -E'|\mathbf{p}_2|$.

This lead me to the idea to take the inner product between $p_2^\mu$ the positron's four-momentum and $p_\gamma^\mu$ this photon's four momentum. This yields

$$p_2^\mu {p_\gamma}_\mu = EE' - E' |\mathbf{p}_2|$$

but now I don't know anything else I can use. So how can I finish this computation? And also, is my approach up to this point correct?

Last edited: Apr 2, 2017
2. Apr 2, 2017

### Staff: Mentor

I don't understand your first part, and it is confusing how you switch between reference frames.

The 4-vector in different frames won't be the same, only the invariant mass is the same between frames.
Your result has an "e" in it, what is that? If you mean me, then the photon energy would not depend on the incoming positron energy at all? That is wrong. If e is the positron energy, it is wrong as well.

The second part will need a correct answer to the first part.

3. Apr 2, 2017

### leo.

It was a typo that I corrected, it should be $E$ the energy of the incoming positron. I denote $m$ the common mass of the electron and positron.

I'll try to explain my reasoning in a more detailed way: I agree that the 4-vector components are different in different frames, but the four vector itself doesn't depend on frame at all. Spacetime is a four-dimensional lorentzian manifold $M$ which in this flat case happens to be $M=\mathbb{R}^4$. A four vector is an element of $TM$, which in this flat case can be thought of as just one element of $\mathbb{R}^4$ also. A reference frame is then a choice of orthonormal basis containing one timelike and three spacelike vectors. So given $p\in \mathbb{R}^4$ a four vector, we can pick a reference frame (that is a basis) $e_\mu$, and write components $p^\mu$ so that $p = p^\mu e_\mu$, but I can also pick another reference frame $\overline{e}_\mu$ which gives components $\overline{p}^\mu$ with expansion $p = \overline{p}^\mu \overline{e}_\mu$.

What I mean is: I can represent the total four momentum in both the lab frame and the center of mass frame. By the problem statement, the total four momentum in the lab frame is $p^\mu = (E+m, \mathbf{p}_2)$. On the other hand the total four momentum after the annihilation is in the center of mass frame $\tilde{p}^\mu = (2E_\gamma, 0)$. Incidentely, the total four momentum is conserved, hence the total four momentum vector equality $p = \tilde{p}$. Obviously to write down this equality in components, we need to use the same reference frame for both of sides, but here I represent the equality between the abstract four vectors in coordinate independent way.

Finally, I use the fact that the inner product itself is invariant under change of reference frame. Thus given $p$ I can compute $p^\mu p_\mu$ in both the lab and center of mass frame and they are the same. Since $p = \tilde{p}$ I compute $p^\mu p_\mu$ in the lab frame and $\tilde{p}^\mu \tilde{p}_\mu$ in the center of mass frame and they must be the same since the vectors are the same, and the inner product doesn't change with reference frame.

This is what I used to arrive at this result.

Now if this is wrong, what have I missed and what would be the best way to do it? Thanks for the help

4. Apr 2, 2017

### Staff: Mentor

It does. At least in the way 4-vectors are used in physics.

You throw around with buzzwords where the problem can actually be solved in 2 lines and without knowing about manifolds at all. Calculate the magnitude of the 4-vector in the lab frame. There is your center of mass energy. That directly leads to the photon energy in the center of mass frame.

To check your answer, think of the case E<<me. What do you expect as photon energy for the annihilation of positron and electron at rest?

5. Apr 2, 2017

### vela

Staff Emeritus

Your reasoning seems to be okay, but you made a mistake somewhere because your final expression for the photon energy is wrong. It was a bit strange to see $p=\tilde{p}$ because most of us don't go into the gory detail you did here since it tends to obscure the physics.

Let 1=electron, 2=positron, 3=photon 1, and 4=photon 2. In the lab frame, conservation of four-momentum gives you
$$p_1 + p_2 = p_3 + p_4.$$ Similarly, in the COM frame, you have
$$p'_1 + p'_2 = p'_3 + p'_4.$$
You're saying that $(p_1+p_2)^2 = (p'_3+p'_4)^2$ where $p_1+p_2 = (E+m, \vec{p}_2)$ and $p'_3+p'_4 = (2E'_\gamma,0)$.

6. Apr 2, 2017

### TSny

A mistake occurs here in rearranging.

7. Apr 2, 2017

### leo.

Indeed I found a mistake @vela and I believe after I resolved it I got this right. It was exactly the mistake in rearranging that @TSny pointed out. I missed a minus sign.

I understand. I believe it is all just a question of notation really. I learned mostly in the differential geometry way where vectors elements of the tangent spaces to manifolds, defined usually as derivations, independently of choice of basis. Only when we pick a basis we get components. So when I state the equality, I'm not equating components in two different frames, actually I'm equating two vectors, that may be resolved in different frames.

Let me just re do my reasoning in what I believe is here being considered a more standard notation. I'll let a tilde means the components in the center of mass frame. In that case, we have $p_1^\mu = (m,0)$ and $p_2^\mu = (E,\mathbf{p}_2)$. Let $p^\mu$ be the total 4-momentum before the colision in the lab frame, so that $p^\mu = (E+m,\mathbf{p}_2)$. Now, by conservation of 4-momentum, after the colision the total 4-momentum is the same $p^\mu = {p'}^\mu$.

Now the photons have 4-momenta $\tilde{p}_3^\mu = (E_3, \mathbf{p}_3)$ and $\tilde{p}_4^\mu = (E_4,\mathbf{p}_4)$ where now $\mathbf{p}_3+\mathbf{p}_4=0$. Thus since these 4-momenta are lightlike we gain $E_3 = E_4 = E_\gamma$. In that way, in the center of mass frame, the total 4-momentum after the colision is ${\tilde{p}'}^\mu = (2E_\gamma,0)$.

But now ${p'}^\mu$ is the total 4-momentum after the colision in the lab frame and ${\tilde{p}}^\mu$ is the total 4-momentum after the colision in the center of mass frame. Taking the magnitude we find ${p'}^\mu{p'}_\mu = {\tilde{p}'}^\mu {\tilde{p}'}_\mu$ which is
$$(E+m)^2 - |\mathbf{p}_2|^2 = 4E_\gamma^2$$and upon using the correct expression $|\mathbf{p}_2|^2 = E^2 - m^2$ we find
$$E_\gamma= \sqrt{\dfrac{mE+m^2}{2}}$$

Is it right now? I mean, it is the same reasoning, it was really the sign mistake the problem. I just shifted notation here, to avoid confusion.

8. Apr 2, 2017

### leo.

As for the second part, I did the following: first in the reference frame of the lab, we have $p_2^\mu = (E,\mathbf{p}_2)$ as before, and now the photon has 4-momentum $p_3^\mu = (E_3,\mathbf{p}_3)$ and we want to find $E_3$.

I compute the following inner product, to be able to use the fact that the directions are opposite
$$p_3^\mu {p_2}_\mu = E_3E - \mathbf{p}_2\cdot \mathbf{p}_3 = E_3E+ |\mathbf{p}_2| |\mathbf{p}_3|$$.

But for the photon $|\mathbf{p}_3| = E_3$ and thus $p_3^\mu {p_2}_\mu = E_3(E+|\mathbf{p}_2|)$.

Now my idea was to compute this inner product on the center of mass frame and equate the two. On the center of mass frame the photon has the previously computed 4-momentum ${\tilde{p}_3}^\mu = (E_\gamma, \tilde{\mathbf{p}}_3)$ while the positron has components $\tilde{p}_2^\mu = (\tilde{E}_2, \tilde{\mathbf{p}}_2)$. If the electron has 4 momentum in the center of mass frame given by $\tilde{p}_1^\mu = (\tilde{E}_1, \tilde{\mathbf{p}}_1)$ we deduce the following from the center of mass condition
$$\tilde{E}_1^2 - |\tilde{\mathbf{p}}_1|^2 = m^2 = \tilde{E}_2^2 - |\tilde{\mathbf{p}}_2|^2 \Longrightarrow \tilde{E}_1 = \tilde{E}_2$$
Now with this we find $\tilde{E}$ by using the total 4-momentum magnitude:
$$(E+m)^2 - |\mathbf{p}|^2 = \tilde{E}^2$$
which leads to $\tilde{E} = E_\gamma$ the photon energy computed at the first part.

With this, the inner product on the center of mass frame is

$$\tilde{p}_3^\mu {\tilde{p}_2}_\mu = E_\gamma^2 - \tilde{\mathbf{p}}_2\cdot \tilde{\mathbf{p}}_3$$

We can equate this to the inner product computed in the lab frame. The problem is that I don't know what to do with this dot product between the momenta of the photon and the positron in the center of mass frame. Is this all right? How can I deal with this last detail?

Last edited: Apr 3, 2017
9. Apr 2, 2017

### TSny

Your result and method of reasoning look good.

10. Apr 2, 2017

### vela

Staff Emeritus
The photon and positron momenta will be parallel in the COM frame too, so you should be able to do the same calculation you did for the lab frame.

A more straightforward way to solve the problem, though, is to find the velocity of the lab frame relative to the COM frame and then apply the appropriate Lorentz transformation to the photon's four-vector.

BTW, is $\vec{p}_3$ the momentum of the forward-moving photon? If not, I think you have a sign mistake when calculating the inner product of the four-momenta in the lab frame.

11. Apr 2, 2017

### TSny

Think about how the energy of the positron in the cm frame is related to the energy of one of the photons in the cm frame. This should allow you to express $|\tilde{\mathbf{p}}_2|$ of the positron in terms of $E_\gamma$ and $m$. You can also relate $|\tilde{\mathbf{p}}_3|$ of the photon to $E_\gamma$.

But this way is quite tedious. Instead, stay in the lab frame and compare the components of the total four-momentum before the reaction to the components of the total four-momentum after the reaction.

12. Apr 3, 2017

### leo.

Indeed, the directions are opposite so $\mathbf{p}_2\cdot \mathbf{p_3}$ should read $-|\mathbf{p}_2||\mathbf{p}_3|$, and this shifts the minus sign. I corrected on the post.

I applied this, so that the inner product on the lab frame becomes $$p_3^\mu {p_2}_\mu = E_3(E+|\mathbf{p}_2|)$$ and on the center of mass frame $$\tilde{p}_3^\mu {\tilde{p}_2}_\mu = E_\gamma^2 + |\tilde{\mathbf{p}}_2||\tilde{\mathbf{p}}_3|$$

but then $|\tilde{\mathbf{p}}_3| = E_\gamma$ and $E_\gamma^2 - |\tilde{\mathbf{p}}_2|^2 = m^2$ so that $|\tilde{\mathbf{p}}_2|^2 = E_\gamma^2 - m^2$. Thus in the end we get

$$E_3(E+|\mathbf{p}_2|)=E_\gamma^2 + E_\gamma \sqrt{E_\gamma^2 - m^2}$$

or else

$$E_3 = \dfrac{E_\gamma^2 + E_\gamma \sqrt{E_\gamma^2 - m^2}}{E+\sqrt{E^2-m^2}}$$

is it the correct result? I mean, it is quite unpleasant to tell the truth. Anyway, why the three momentum being paralel in the lab frame implies being paralel in the other frame?

13. Apr 3, 2017

### vela

Staff Emeritus
Yes, that simplifies to the correct result. It should be expressed in terms of $E$ and $m$ only.

Try TSny's suggestion of solving the problem solely in the lab frame.

Assume the positron is moving along the x-axis. To transform from the lab frame to the COM frame, you need a boost along the x-axis. If you have no y- or z- components in one frame, can you get non-zero y- and z- components in the other from a boost along the x-direction?

14. Apr 3, 2017

### leo.

I did express it like that. It becomes: $$E_3 = \dfrac{mE+m^2+\sqrt{m^2E^2-m^4}}{2(E+\sqrt{E^2-m^2})}$$ which happens to be approximately $m/2$ when $E$ is much greater than $m$, so the result gets a lot better in this limit.

Fine, I got your point, viewing it like this makes it quite simple.

Thanks for all the help by the way

15. Apr 3, 2017

### vela

Staff Emeritus

It simplifies even further down to
$$E_3 = \frac{E+m-\sqrt{E^2-m^2}}{2}.$$