Are the Electron & Positron Created from a Photon Scattered at the Same Angles?


by Puky
Tags: angles, electron, photon, positron, scattered
Puky
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#1
Nov6-12, 02:44 PM
P: 22
Hello. Firstly, I hope this is the right sub-forum for this question. Here's a quick question that I have.

After an electron-positron pair production, do the angles that the paths of the electron and the positron make with the original path of the photon have to be equal? Another way to ask this is, do the velocities of the created electron and positron have to be equal? If they do have to be equal, I'd really like to know how this can be proved. Thanks.
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tom.stoer
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#2
Nov6-12, 07:50 PM
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try energy- and momentum-conservation
Puky
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#3
Nov7-12, 07:15 AM
P: 22
Okay, let the photon's movement be in the +x direction. Then, if the resultant particles have different scattering angles with the horizontal, the one with the greater angle should have greater momentum, so that the sum of the y-components of their momenta will be 0.

Some of the photon's momentum in the +x direction will be transferred to a nucleus, so I'm not sure if we can express the conservation of the x-component of momentum with an equation here.

We still have the energy conservation. Assuming the speeds are much smaller than the speed of light, we can use the non-relativistic momentum formula [itex]\bf{P} = m\bf{v}[/itex]. So the relationship between the velocities will be linear ([itex]v_1sin(\theta1) = v_2sin(\theta2)[/itex]). Then we can just write [itex]\frac{1}{2}mv_1^2 + \frac{1}{2}m(v_1\frac{sin(\theta_1)}{sin(\theta_2)})^2 = hν-2mc^2[/itex].

I'm just asking this becase I've heard some people say that the two angles have to be equal, but none of the things I have said above require that. Is there something I am missing? If they indeed do have to be equal, how would we show it? I'd be glad if anyone could explain it briefly or just point me to a relevant website.

qinglong.1397
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#4
Nov7-12, 12:24 PM
P: 107

Are the Electron & Positron Created from a Photon Scattered at the Same Angles?


They should be equal in magnitude but opposite in direction, because of symmetry. Note: that one of the products is particle while the other anti-particle plays no role here. What matters to this problem is the mass.
tom.stoer
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#5
Nov7-12, 05:11 PM
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The initial state is a photon γ and a target T. The final state is an electron-positron pair e- and e+

The initial 4-momenta are (with c=1)

[tex]p^\mu_\gamma = (p_\gamma,\vec{p}_\gamma)[/tex]
[tex]p^\mu_T = (M,0)[/tex]

The final 4-momenta are

[tex]p^\mu_{e^-} = (\sqrt{m^2+\vec{p}_{e^-}^2},\vec{p}_{e^-}) = (\sqrt{m^2+\vec{p}_{e^-}^2},p_x,\vec{p}_\perp) [/tex]
[tex]p^\mu_{e^+} = (\sqrt{m^2+\vec{p}_{e^+}^2},\vec{p}_{e^+}) = (\sqrt{m^2+\vec{p}_{e^+}^2},p_x,-\vec{p}_\perp) [/tex]

The y- and z-components (perpendicular) add to zero b/c of momentum conservation.

One immediately sees that the angles must be identical, and one can calculate this explicitly

[tex]\cos\alpha_{\gamma,e^\pm} = \frac{\vec{p}_\gamma\,\vec{p}_{e^\pm}}{p_\gamma\,p_{e^\pm}} = \frac{p_\gamma\,p_x}{p_\gamma\,p_{e^\pm}} = \frac{p_x}{p_{e^\pm}} [/tex]
Puky
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#6
Nov8-12, 06:07 AM
P: 22
Quote Quote by tom.stoer View Post
The initial state is a photon γ and a target T. The final state is an electron-positron pair e- and e+

The initial 4-momenta are (with c=1)

[tex]p^\mu_\gamma = (p_\gamma,\vec{p}_\gamma)[/tex]
[tex]p^\mu_T = (M,0)[/tex]

The final 4-momenta are

[tex]p^\mu_{e^-} = (\sqrt{m^2+\vec{p}_{e^-}^2},\vec{p}_{e^-}) = (\sqrt{m^2+\vec{p}_{e^-}^2},p_x,\vec{p}_\perp) [/tex]
[tex]p^\mu_{e^+} = (\sqrt{m^2+\vec{p}_{e^+}^2},\vec{p}_{e^+}) = (\sqrt{m^2+\vec{p}_{e^+}^2},p_x,-\vec{p}_\perp) [/tex]

The y- and z-components (perpendicular) add to zero b/c of momentum conservation.

One immediately sees that the angles must be identical, and one can calculate this explicitly

[tex]\cos\alpha_{\gamma,e^\pm} = \frac{\vec{p}_\gamma\,\vec{p}_{e^\pm}}{p_\gamma\,p_{e^\pm}} = \frac{p_\gamma\,p_x}{p_\gamma\,p_{e^\pm}} = \frac{p_x}{p_{e^\pm}} [/tex]
Thank you very much for your explanation. The only problem is that I'm not familiar with 4-momentum yet. Can this be shown using the classical 3-momentum, or will I have to wait until I learn about 4-momentum?
tom.stoer
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#7
Nov8-12, 09:34 AM
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The μ=0 component is the energy; the μ=1,2,3 components are the usual 3-momentum; forget about the μ=0 component, you don't need it for the scattering angle
Puky
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#8
Nov10-12, 07:22 AM
P: 22
Quote Quote by tom.stoer View Post
The μ=0 component is the energy; the μ=1,2,3 components are the usual 3-momentum; forget about the μ=0 component, you don't need it for the scattering angle
Okay, I'm able to follow your calculation now but I still don't understand why the x-components of [itex]\vec{p}_{e^+}[/itex] and [itex]\vec{p}_{e^+}[/itex] have to be equal. If they are, the result easily follows from there but I don't see why they have to be. If you think of the photon as a bomb which explodes into two particles of equal mass (I am aware that it's an entirely different example but momentum is conserved in both of them), the x-components of the momenta of the new particles need not be equal, as long as they add up to the original momentum of the bomb (photon).
tom.stoer
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#9
Nov10-12, 08:01 AM
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I see.

1)
Start with different values px and p'x which results in different energies (!) for electron and positron and try to find a contradiction with energy-momentum-conservation.

or 2)
Start in the c.o.m. frame with vanishing total momentum. The initial momenta (in x and -x direction) are carried by photon and the target and add up to zero. So the total momentum in x-direction is zero. Now make a Lorentz transformation from c.o.m to rest frame of M; the Lorentz trf. introduces one velocity so both particles must have same momentum in x direction.
Puky
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#10
Nov10-12, 08:59 AM
P: 22
That's what I was trying to do in my second post, but I couldn't come to the conclusion that the two angles have to be equal. I'll try to make it clearer what I did there so you can point out what I'm missing.

Say the photon is moving in the +x direction and the resultant particles move in the x-y plane, for simplicity (so no z-component of momenta). Let's say the electron is scattered at an angle [itex]\theta_1[/itex] and the positron at [itex]\theta_2[/itex]. By the conservation of momentum in the y direction,

[itex]\gamma_1mv_1sin\theta_1=\gamma_2mv_2sin\theta_2[/itex]

[itex]\gamma_1v_1sin\theta_1=\gamma_2v_2sin\theta_2[/itex]


The condition for the x-components of the momenta to be different is:

[itex]\gamma_1mv_1cos\theta_1\neq\gamma_2mv_2cos\theta_2[/itex]

[itex]\frac{\gamma_1v_1}{\gamma_2v_2}\neq\frac{cos\theta_2}{cos\theta_1}[/itex]

which is satisfied as long as the two angles are different because:

[itex]\frac{\gamma_1v_1}{\gamma_2v_2}=\frac{sin\theta_2}{sin\theta_1}\neq\fra c{cos\theta_2}{cos\theta_1}[/itex]


Also, some of the momentum of the photon will be transferred to a nucleus, so the initial momentum of the photon will be greater that the sum of the x-components of the final momenta, but as far as I know, we can't write an exact equality for it:

[itex]\frac{hν}{c}>\gamma_1mv_1+\gamma_2mv_2[/itex]
[itex]hν>\gamma_1mv_1c+\gamma_2mv_2c[/itex]


And for the energy conservation, the energy transferred to the nucleus will be insignificant compared to the momentum transferred, so:

[itex]hν=\gamma_1mc^2+\gamma_2mc^2[/itex] (approximately)

It's clear that
[itex]hν=\gamma_1mc^2+\gamma_2mc^2>\gamma_1mv_1c+\gamma_2mv_2c[/itex]
so there is no contradiction with what we've found using the energy conservation and the momentum conservation in the x direction.


I don't see anything here that violates one of the conservation laws if the angles are not equal. It is even easier to see that it should be allowed if we let the speeds be much less than c, so [itex]\gamma_1=\gamma_2=1[/itex]. That way, there will be a linear relationship between the speeds, and you can easily determine the speeds for a given [itex]ν[/itex] and two angles you pick, as I explained in my second post.

Quote Quote by tom.stoer View Post
or 2)
Start in the c.o.m. frame with vanishing total momentum. The initial momenta (in x and -x direction) are carried by photon and the target and add up to zero. So the total momentum in x-direction is zero. Now make a Lorentz transformation from c.o.m to rest frame of M; the Lorentz trf. introduces one velocity so both particles must have same momentum in x direction.
In that case, in the CoM frame, the electron and the positron could move in any two opposite directions, since the total momentum is zero and any two opposite directions would be equally good. So, the x components of their momenta would add up to 0. Let's call the electron's speed in the x direction (in the CoM frame) [itex]u'[/itex] and the positron's [itex]-u'[/itex].

Now the CoM is moving in the +x direction with respect to the frame where M is at rest. Let's call this speed [itex]v[/itex]. Then using the Lorentz transformation, the speeds in the x direction [itex]u_1[/itex] (of the electron) and [itex]u_2[/itex] (of the positron) in the rest frame are:

[itex]u_1=\frac{u'+v}{1+\frac{u'v}{c^2}}[/itex]

[itex]u_2=\frac{-u'+v}{1-\frac{u'v}{c^2}}[/itex]

which cannot be equal unless [itex]u'=0[/itex] or [itex]v=c[/itex]. [itex]u'=0[/itex] is the case where the particles' scattering angles are the same in the rest frame.
Puky
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#11
Nov16-12, 04:15 PM
P: 22
Innocent bump.


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