Electron/positron annihilation/momentum

[Nick89]

Exactly. Even if momentum can increase without bound, so can the energy of the photons, so there is no problem at all.

 

[benk99nenm312]

Gamma does not stop increasing when v = c. Gamma is undefined when v = c. And when v > c, gamma is imaginary so you can no longer speak of increasing, since complex numbers have no 'order'. You cannot say (a + bi) > (c + di).


Of course, v must always be less than c. But you can make it go as near to c as you want (theoretically of course). In fact, the closer you get to c, the larger the momentum increase becomes.

Ahhh... I see. Interesting.

 

[granpa]

gamma increases without limit as v approaches c

 

[Nick89]

gamma increases without limit as v approaches c

Yes, we decided on that. But so can the energy of the photons. So is there still a problem for you, or is this question resolved?

 

[granpa]

an electron and positron annihilate producing 2 0.5 Mev photons. these 2 photons possesses a certain amount of momentum.
the sum of the 2 momentums possessed by the photons can be regarded therefore as an upper limit to the net momentum that the original
electron positron pair possessed.

let us suppose that there exists another electron positron pair with twice that net momentum.
I would assume that their net velocity would be tiny so when they annihilate they should produce 2 0.5 Mev photons.

these photons can't carry twice the momentum of the first photons.

 

[Nick89]

I don't really understand your question (it's getting late), but don't forget that momentum is a VECTOR! It has direction. By 'adjusting' the angle at which the two photons are emitted, you can increase or decrease the net momentum. Two photons moving in the same direction don't have the same momentum as two photons moving away from each other (those would have a net momentum of zero).

 

[Nabeshin]

an electron and positron annihilate producing 2 0.5 Mev photons. these 2 photons possesses a certain amount of momentum. if all of that momentum were transferred to another electron positron pair (scratch that. let's just create a pair and make them move with twice that net momentum)[/color] then what would the net velocity of that pair be? I suspect it would be very tiny. if a pair were moving at double that velocity their total energy would be only very very slightly over the total energy of a stationary pair so when they annihilate they should produce 2 0.5 Mev photons. these photons can't carry twice the momentum of the first photons.

You realize that when you double very tiny you still sometimes get something very tiny... doesn't mean it didn't double.

That's all I can grok from your paragraph, otherwise it's pretty much gibberish to me. Try citing a specific example?

 

[benk99nenm312]

if a pair were moving at double that velocity their total energy would be only very very slightly over the total energy of a stationary pair so when they annihilate they should produce 2 0.5 Mev photons. these photons can't carry twice the momentum of the first photons.

I think I see your point, but the solution is that our assumptions are not quite accurate. We are assuming that the net momentum is equal to the sum of the individual particle's momentum. This is wrong. Also, we haven't considered angles. Here's how you come to the wrong conclusion:

The relativistic equation for total energy is E=ymc^2. If we consider two velocities: 1/4 of c, and 1/2 of c, then we can show the change in energy.

When v=1/4 c, E= 1/\sqrt{1- v^2/c^2}mc^2 becomes
E=1/\sqrt{1- (c/4)^2/c^2}mc^2 = 1/\sqrt{1- c^2/16 /c^2}mc^2
= 1/\sqrt{1- 1/16}mc^2 = 1/\sqrt{15/16}mc^2 = 1.0328mc^2.

Now, when v=1/2 c:
E=1/\sqrt{1- (c/2)^2/c^2}mc^2 = 1/\sqrt{1- c^2/4 /c^2}mc^2
= 1/\sqrt{1- 1/4}mc^2 = 1/\sqrt{3/4}mc^2 = 1.1547mc^2.

1.0328 compared to 1.1547 isn't huge increase in energy by any means.

This means that the momentum of a photon holding the same amount of energy as an electron traveling at 1/4th the speed of light is
\frac{1.0328mc^2}{c} = 2.82 x 10^-22 kg m/s

and the photon holding the same amount of energy as an electron traveling at 1/2 the speed of light is
\frac{1.1547mc^2}{c} = 3.15 x 10^-22 kg m/s.

As you may have noticed, 2.82 is not half of 3.15.

Therefore, the only way this works is to consider what everyone else has already mention: momentum is a vector. It has direction. A perfect head on collision would have no net momentum, so the photons would zoom away from each other at 180 degrees. If there is momentum, the only way to make it add up is by adjusting the angle in which the photons zoom away, and by the angle at which the particles collide.

Also, we must consider that the momentum can never just be the sum of the two momentums of the individual particles that annihalate, because that would infer that they are traveling in exactly the same direction. Then, they wouldn't ever annihalate. It all depends on angles.

 

[benk99nenm312]

You realize that when you double very tiny you still sometimes get something very tiny... doesn't mean it didn't double.

That's all I can grok from your paragraph, otherwise it's pretty much gibberish to me. Try citing a specific example?

As I have shown above, it actually didn't double. granpa was right in his assumption, but there was just something we forgot (momentum is a vector).

 

[Vanadium 50]

strange.

https://www.physicsforums.com/showthread.php?t=310644

photon -> e+ e- : that violates momentum conservation

but the reverse process does not.

That thread is about a completely different process.

yes. its about the reverse process

No, it's not. Having one photon convert to an electron-positron pair (in the presence of a nucleus) is not the reverse process to having an electron and a positron annihilate into two photons. 1 \neq 2. More generally,

\gamma + N \rightarrow e^+ + e^- + N

is not the reverse of

e^+ + e^- \rightarrow \gamma + \gamma

 

[diazona]

Just to address a point that I don't think has been answered yet:

I have been thinking myself and I have a question too (I'm probably over-thinking this):
If you have a 'simple' annihilation, one electron moving to the right, and one positron to the left, annihilating in the center, then the net momentum is zero. For that reason, it is impossible that less than two photons are created, as only one photon would not be able to have zero momentum and thus conserve momentum.

However, assuming the electron and positron already have momentum (like in the OP's question, they could be orbiting around each other, and their center of mass could be moving relative to some observer, no?), then in my reasoning it is perfectly allowed to have only one photon be created, since that can conserve momentum.

What if we have this process, but then decide to co-move with the e-p pair? Net momentum, relative to us, would be zero, but only one photon would be emitted. What happens here? Where does relativity save us? :p

The process needs to be consistent in all reference frames. If there exists any reference frame in which momentum is not conserved, it isn't allowed. So an e+ e- annihilation can't produce a single photon because there exists a frame, namely the center of mass frame, in which momentum will not be conserved.

 

[Vanadium 50]

I don't understand what you're crapping your pants with such stubbornness about, so I'm just going to answer your OP:

The momentum goes into the 2 photons that are produced. They have momentum p = h/lambda.

Exactly.

an electron and positron annihilate producing 2 0.5 Mev photons. these 2 photons possesses a certain amount of momentum.

Close. An electron and positron at rest annihilate producing 2 0.5 Mev photons.

the sum of the 2 momentums possessed by the photons can be regarded therefore as an upper limit to the net momentum that the original
electron positron pair possessed.

And here is where things start to go pear shaped. Momentum is conserved. Absolutely. The vector sum of the momenta (the plural of momentum is momenta) of the photons is the same as the vector sum of the momenta of the electron and positron.

let us suppose that there exists another electron positron pair with twice that net momentum. I would assume that their net velocity would be tiny so when they annihilate they should produce 2 0.5 Mev photons.

these photons can't carry twice the momentum of the first photons.

This is just wrong. You are confusing so many things here that it's difficult to tell where one error ends and the next begins. A few facts:

• Electrons and positrons can annihilate at any relative velocity.
• Changing the (lab frame) momentum of the electron-positron system changes the (lab frame) momentum of the photons.

 

[Vanadium 50]

PS Why is this Classical physics?

 

[granpa]

Close. An electron and positron at rest annihilate producing 2 0.5 Mev photons.

that is what I meant.

the sum of the 2 momentums possessed by the photons can be regarded therefore as an upper limit to the net momentum that the original
electron positron pair possessed.

And here is where things start to go pear shaped. Momentum is conserved. Absolutely. The vector sum of the momenta (the plural of momentum is momenta) of the photons is the same as the vector sum of the momenta of the electron and positron.

that is why we can be sure that the original pair of particles can't have had more net momentum than the sum of the momentums of the photons. (of course we know that they had no net momentum but what I'm saying becomes important down below)

let us suppose that there exists another electron positron pair with twice that net momentum. I would assume that their net velocity would be tiny so when they annihilate they should produce 2 0.5 Mev photons.

these photons can't carry twice the momentum of the first photons.

Electrons and positrons can annihilate at any relative velocity.

Yes, I know.
I meant that their annihilation should produce 2 approximately[/color] 0.5 Mev photons. in other words not much more energetic than the original pair of photons.

Changing the (lab frame) momentum of the electron-positron system changes the (lab frame) momentum of the photons.

does it change their energy too. down below people have been telling me that the momentum of the photon depends only on its energy. in particular that it is proportional to the energy.

 

[Bob S]

This reaction, positronium annihilation in flight, is nearly exactly the same as positron annihilation in flight, except in the latter case the electron is at rest (in a thin target). All the kinematic expressions were derived by Kendall and Deutsch in 1956:
http://prola.aps.org/abstract/PR/v101/i1/p20_1
Positron annihilation in flight by accelerator produced positrons was later observed in 1961:
http://prola.aps.org/abstract/PR/v121/i3/p866_1
http://prola.aps.org/abstract/PR/v121/i2/p605_1

 

[benk99nenm312]

0.5 Mev photons. in other words not much more energetic than the original pair of photons.

does it change their energy too. down below people have been telling me that the momentum of the photon depends only on its energy. in particular that it is proportional to the energy.

You were correct about the changing of energy.

The momentum of a photon is dependent on energy E=pc (p=E/c).
The net momentum of two photons does however depend on the angle of separation between them, as well as the energy of the pair.

 

[Dale]

In the center of momentum frame the electron and positron have four-momenta of
\textbf{p}_{e^\pm} = ( \gamma m_0 c, \pm \gamma m_0 \textbf{v})

After annihilation the photons have four-momenta of
\textbf{p}_p = (\gamma m_0 c, \pm \gamma m_0 \textbf{c})

Where m_0 is the rest mass of an electron, \textbf{v} is the velocity of the positron in the center of momentum frame, \gamma is the time dilation factor for that velocity, and \textbf{c} is a velocity vector of magnitude c in a random direction.

To get the photons in the lab frame, simply boost the result in the center of momentum frame using a standard Lorentz transform. It is easy to verify that momentum and energy are conserved in all reference frames.

 

[protonchain]

PS Why is this Classical physics?

My thoughts exactly. This has to do with quantum.

 

[benk99nenm312]

My thoughts exactly. This has to do with quantum.

I was wondering that too, but I guess the original question was about momentum. Momentum seems more classical to me, but.. oh well.