Sir, in some answer on the main site I noticed a neutrino on the product side when transferred to the reactant side it becomes an antineutrino and vice versa. So is it something like $\nu=-{\nu}^{-}$? And is the case same for a positron and an electron?

I don't remember where I saw that kind of transformation.

Yes

In effect you add an antineutrino to both sides and the $\nu$ and $\bar\nu$ cancel.

Ok sir. So both an electron capture reaction and beta plus reaction are merely the same?

Or do they take place simultaneously?

Some care is needed ...

Let's start with the electron capture:

$$ p + e \to n + \nu_e $$

Yes sir.

Now we can add a positron to both sides and allow the electron and positron to cancel on the left side. Then we get:

$$ p + 2\gamma \to n + e^+ + \nu_e $$

Because the electron and positron annihilate to two photons so they can be treated as equaivalent to two photons.

Where did $\gamma$ come from?

@JohnRennie I see. I have to learn about quantum "annihilation" before I question this further. I remember, some time ago, we came to the same concept.

@GuruVishnu the electron and positron both have the same energy of $m_e c^2$ so they can't just vanish or energy wouldn't be conserved.

@JohnRennie Ok sir. So $e^-+e^+=\text{energy}$ instead of $0$. Did I get it?

Yes

Although writing energy is a poor way to put it. The electron and positron convert into two new particles i.e. two photons.

Photons are not energy - they are massless particles.

@JohnRennie Two photons of equal energy?

Yes.

The reason two photons are produced is because momentum cannot be conserved otherwise.

Consider the reaction in the centre of momentum frame i.e. total momentum is zero.

@JohnRennie Ok sir. Now I can see why they must be of the same energy.

So initially the electron and positron have equal and opposite momenta.

The photon momentum is $h/\lambda$

So the only way for momentum to be conserved is if we get two photons with equal and opposite momenta $+h/\lambda$ and $-h/\lambda$.

Ok sir. If the initial momentum is non-zero, say we hit an electron on a fixed positron, again, will two photons be emitted? If yes, I think the two photons will have different energies and hence different momenta.

There is always a centre of momentum frame, so there must always be two photons produced.

In a frame that is not the COM frame the photons will not have equal and opposite momenta.

The photon emitted in the direction of light will be blue shifted and the other be red shifted. Is this conclusion correct?

In general the motion relative to the COM frame will change the direction of the photon momenta, and it will blue shift one of the photons and red shift the other.

@GuruVishnu yes

@JohnRennie Ok sir. I'm slightly confused, why not the opposite take place - red shifted in the prograde direction and blue shifted in the retrograde direction?

Let's consider an example. I'll draw a diagram:

Suppose we have a stationary positron (red) being hit my a moving electron (blue).

Ok sir.

So if we take the positive x direction to be right we the total momentum is $+P_x$

Yes sir.

On collision the particles annihilate. We'll assume the photons get produced along the line of motion - they can be produced in any direction but we'll assume along the line of motion for simplicity.

Ok sir.

Then the momentum of the right moving photon is $+h/\lambda_r$ and the momentum of the left moving photon is $-h/\lambda_l$.

So we require $+h/\lambda_r - h/\lambda_l = P_x$

@JohnRennie Yes sir. This shows why is $\lambda_l>\lambda_r$

That immediately tells you $\lambda_r < \lambda_l$ so the right moving photon is blue shifted amd the left moving photon is red shifted.

Fine sir. I understood this.

Though it could look like this:

i.e. the photons emitted at equal angles to the direction of travel. In that case their wavelengths would be equal.

@GuruVishnu how did you get confused?

@JohnRennie Ok sir. I understood everything we discussed so far. But is it necessary to have two photons after collision? Is this a property of "annihilation" (quantum mechanics)?

You can't have one photon because with one photon there is no way to conserve momentum. So the question is could you have three, four, five, etc photons. Yes?

@JohnRennie Yes sir.

> the photons emitted at equal angles

I understand why the photons must be emitted at equal angles, else the momentum along the vertical will not be conserved.

But, in the final case, why can't the equal angles be equal to zero. Are there any other values, these angles can't take?

And the answer is that yes you could have three or more photons produced from an $e \bar{e}$ annihilation, but when you calculate the probability of this happening it turns out to be much, much smaller than the probability for two photons.

With massive particles the momentum is proportional to $v$ and the energy to $v^2$, so in general any angles are allowed because you can always find a combination of angle and velocity that satisifes both conditions.

But for photons $E = pc = hc/\lambda$ so the energy and momentum are both inversely proportional to the wavelength.

This restricts the possibilities for the angle of emission.

@JohnRennie Ok sir. But I don't see any violation of conservation of momentum if only one photon is emitted along the direction of initial momentum. Anyway, if two photons must be emitted, I've thought of another reason based on change of reference frame. If we imagine ourselves to be in an inertial frame where the momentum of the two particles is initially zero, two photons must be emitted and they must have equal energies.

Emission of two photons is invariant of the reference frame. So even in the original reference frame two photons must be emitted, but they will be blue and red shifted accordingly.

Could you tell whether my reasoning is correct?

> Ok sir. But I don't see any violation of conservation of momentum if only one photon is emitted along the direction of initial momentum.

So what would this look like in the centre of momentum frame?

Yes sir. I realise my mistake. So our observation (emission of two photons) must not depend on the reference frame we choose - something like, one photon in one frame and two in the other frame is not permissible. Am I right?

Correct. Changing the frame is just a choice of coordinates i.e. just mathematics. It cannot change things like the number of photons created.

Ok sir. Could you comment on my reasoning in my previous message? Now I feel it's also correct.

"...if two photons must be emitted, I've thought of another reason... ...and red shifted accordingly."

Yes sir. The energies get redistributed when we change our frame of reference.

Is this correct?

Yes

Just curious, in which reference frame will the total energy $E$ be distributed as $0$ and $E$ for the retrograde and prograde photons? Is that the frame which travels in the speed of light in the direction of the initial momentum?

Or is that the frame travelling with the speed of light in the direction opposite to the initial momentum?

@GuruVishnu there is no frame in which the energy of the red shifted photon is exactly zero because that would require travelling at the speed of light, and in that frame our electron and positron would have to be travelling at the speed of light. And massive particles cannot travel at the speed of light.

But it possible to travel arbitrarily close to speed of light, and in that frame the energy of the red shifted photon can come arbitrarily close to zero.

@JohnRennie Ah. Nice. And this also explains why there must necessarily be two photons. A proof by contradiction.

It's good fun this physics thing :-)

Thank you for your help and time sir :-) Colliding quantum particles is more fun than I thought initially. People at particle accelerators probably enjoy their lives.

I would guess the people working at CERN really enjoy it. It has to be a dream job :-)

:-)