Photon-photon collision - pair creation

[WrongMan]

Homework Statement

Two photon (wavelengths 1pm, 2pm) collide, decaying into a pair positron-electron find the velocities of these particles
Everything happens in the x axis

Homework Equations

The system
hc(1*10¹²+0.5*10¹²)=E₊+E_-
h(1*10¹²-0.5*10¹²)=P₊-P_-
E=sqrt(c²P²+m²c⁴)

The Attempt at a Solution

Well this yields a 2 page attempt at solving this system so i won't bother pasting it here. I have a feeling that like for finding the compton wave length, there is a specific path i should take that would make everything go smoother. I just can't find it

 

[vela]

Your relevant equations are wrong. It usually helps to refrain from plugging numbers in right from the start.

Try solving the problem in the center-of-mass frame and then transforming back to the lab frame.

 

[WrongMan]

Your relevant equations are wrong. It usually helps to refrain from plugging numbers in right from the start.

Try solving the problem in the center-of-mass frame and then transforming back to the lab frame.

Ok so i understand the momentum part might be wrong because i don't know if the pair goes in opposite directions, I haven't plugged in any numbers... But how do i use a center of mass frame with photons?

 

[vela]

Ok so i understand the momentum part might be wrong because i don't know if the pair goes in opposite directions, I haven't plugged in any numbers...

Where did the 1*10^12 and 2*10^12 come from then?

But how do i use a center of mass frame with photons?

In the case of photons, you're looking for the case where the total momentum of the system is zero.

 

[WrongMan]

Where did the 1*10^12 and 2*10^12 come from then?In the case of photons, you're looking for the case where the total momentum of the system is zero.

Ah those are 1/λ the second one should be 0.5 and not 2.

Im not sure if i covered CM collisions in class but il give it a shot.
So i have to figure out the velocity frame in which the sum of momentums of the photons is 0 right? Then the velocities of the pair would be the same?

 

[vela]

The electron and positron would move with the same speed but in opposite directions so that their momentum would sum to 0.

 

[WrongMan]

The electron and positron would move with the same speed but in opposite directions so that their momentum would sum to 0.

So P₁ is the momentum of photon 1, P₂ photon 2, P_cm momentum of the CM; so
(P₁-P_cm)+(-P₂-P_cm) =0?
And then for the pair i substitute the new momentum of the photons in the equations and i get the energy of both photons and from the energy i get the velocity in the cm frame? I am confused

 

[WrongMan]

Ok so here is what i did; i used the formula in my last post to find the momentum of the cm; then used that momentum to find the momentum of both photons in that frame, (same momentum opposite directions);
And that momentum i transformed in energy so i get the same energy with same "signal" used the photon energy to find the energy of positron (electron is the same energy because of the CM frame) used that energy to calculate momentum;
Subtracted/added momentum of the CM to find momentum of the lab frame;
Used that momentum to find energy then velocity and got that both of them were at aprox. 0.65c
Did i do this right? I can provide a pic of resolution if you want.
Thank you!

Edit: ok this seems right! Momentum was conserved: original photon resulting momentum was 3.315*10^-22 and final pair was 3.2*10^-22 I am checking energy conservation(disregard that velocity i wrote above i made an aproximation error I am recalculating that also)

Edit2: ok energy also conserved, inicial 2.98 *10^-13 j ; final 2.99*10^-13 j.
Electron/positron velocities -0.9c and 0.66c

Since everything was conserved I am assuming my answer is correct. Thank you very much! Much easier this way!

 

[vela]

The description of your calculations doesn't sound right, but your answers are pretty close to what I found, the differences perhaps due to rounding error.

You shouldn't transform between reference frames by just adding and subtracting vectors, like you did with the momenta. You should use the Lorentz transformations. On the other hand, the speed of the center of mass is pretty low, only 1/3 c, so maybe it was a decent approximation.

Recall Newtonian mechanics. The momentum of the center of mass is just the sum of the individual momenta of the particles making up the system. The same holds true in relativistic mechanics, so $p_{cm} = p_1 - p_2$. It looks like you put in a factor of two.

 

[WrongMan]

The description of your calculations doesn't sound right, but your answers are pretty close to what I found, the differences perhaps due to rounding error.

You shouldn't transform between reference frames by just adding and subtracting vectors, like you did with the momenta. You should use the Lorentz transformations. On the other hand, the speed of the center of mass is pretty low, only 1/3 c, so maybe it was a decent approximation.

Recall Newtonian mechanics. The momentum of the center of mass is just the sum of the individual momenta of the particles making up the system. The same holds true in relativistic mechanics, so $p_{cm} = p_1 - p_2$. It looks like you put in a factor of two.

I found the momentum of the CM so that the momentums of the photons would be the same in the CM reference. I will redo the problem with that CM and compare results
How would i use the lorentz transforms to find the momentum of the pair?
What other parts of my resolution don't seem right to you?

 

[vela]

You'd need to show your work. Verbal description are often too vague to clearly describe what you're doing. For example, you wrote, "I found the momentum of the CM so that the momentums of the photons would be the same in the CM reference." If you the calculation correctly, then great, but if you didn't, I don't know what you actually did.

 

[WrongMan]

You'd need to show your work. Verbal description are often too vague to clearly describe what you're doing. For example, you wrote, "I found the momentum of the CM so that the momentums of the photons would be the same in the CM reference." If you the calculation correctly, then great, but if you didn't, I don't know what you actually did.

thank you for your patience, here is my work:
for P_cm
(P₁-P_cm)+(-P₂-P_cm)=0 ; this is "the momentum of the CM so that the momentums of the photons would be the same in the CM reference"
then :
P₁+P_cm=P_1cm; P₂+P_cm=P_2cm
and so P_1cm+P_2cm=0.
found the energies of the photons now E₁=cP_1cm and E₂=E₁ so:
2E₁=2E_p
2E_p is the sum of the energy of the pair, since we are in the CM the velocities are equal.
since E²=c²p²+m²c⁴
found P of pair, and P+/-P_cm is the momentum of the pair turned this back into energy and used:
E=(mc²)/(sqrt(1-(V²/C²)) to find each velocity

 

[vela]

If a particle has energy E and momentum p in one frame, then in the other frame, its energy and momentum are
\begin{align*}
E' &= \gamma[E - (v/c)(pc)] \\
p'c &= \gamma[pc - (v/c)E].
\end{align*} where $v$ is the speed of the second frame relative to the first.

 

[WrongMan]

If a particle has energy E and momentum p in one frame, then in the other frame, its energy and momentum are
\begin{align*}
E' &= \gamma[E - (v/c)(pc)] \\
p'c &= \gamma[pc - (v/c)E].
\end{align*} where $v$ is the speed of the second frame relative to the first.

And the speed of the second frame is P_cm/2m_e?

 

[WrongMan]

doing this did not give correct results
i got 0.9c and 0.72c this does not conserve energy...

i got vcm =0.30c; i used the previously calculated energy and momentum, and used the formula you stated to calculate the new energy; γ[E−(v/c)pc]., with γ= 1/sqrt(1-v²/c²) E and p of the particle and v of the cm
what am i doing wrong?

 

[vela]

The speed's not right for one thing. Assume the photons have momentum $p_1$ and $p_2$ in the lab frame. Then calculate what their momenta would be in the CM frame, set those two expressions equal, and solve for $v$.

 

[WrongMan]

The speed's not right for one thing. Assume the photons have momentum $p_1$ and $p_2$ in the lab frame. Then calculate what their momenta would be in the CM frame, set those two expressions equal, and solve for $v$.

i don't get it
photons have no mass how do i find "v" of the cm with photons? am i to use:
γ(p_1cmc−(v_cm/c)E)= γ(p_2cmc−(v_cm/c)E) ?

 

[vela]

I'm not sure what you mean by p_1cm, etc.

 

[WrongMan]

I'm not sure what you mean by p_1cm, etc.

momentum of photon 1 in CM frame

 

[vela]

In the CM frame, you want p_1cm = p_2cm, right?

 

[WrongMan]

In the CM frame, you want p_1cm = p_2cm, right?

yes so i used :

(P₁-P_cm)+(-P₂-P_cm)=0

to find pcm

 

[vela]

Which is wrong.

 

[WrongMan]

Which is wrong.

ok but

$p_{cm} = p_1 - p_2$.

doesnt give |p2cm| = |p1cm|

 

[vela]

You need to put aside your preconceptions. When changing from one frame to another, it doesn't work the same way as it did in Newtonian mechanics.

Reread posts 13 and 16.

 

[WrongMan]

You need to put aside your preconceptions. When changing from one frame to another, it doesn't work the same way as it did in Newtonian mechanics.

Reread posts 13 and 16.

ok so

\begin{align*}
E' &= \gamma[E - (v/c)(pc)] \\
p'c &= \gamma[pc - (v/c)E].
\end{align*}

since p' is the same for both photons i get:
p₁c+p₂c=v/c(-E₂+E₁) this doesn't work

 

[vela]

That's close. Check your algebra.

 

[WrongMan]

so for photon 1
p1c= E1 ; (v/c)E= vp;
p'1=γ[E1−(v)p1];
then
p'2=-p'1;
E1-vp1=-E2+vp2;
E1+E2=v(p1+p2);
v=c

 

[vela]

You have the right idea, but you need to watch the signs more carefully. You should find
$$\frac vc = \frac{p_1-p_2}{E_1+E_2},$$ where $p_1$ and $p_2$ are the magnitudes of the photon momentum. If you think about it, this makes sense. If $p_1 = p_2$, then you find v=0 since the momenta are already equal.

 

[WrongMan]

ok so i took a break from this, i was obviously tired of studying this, and moved on to other things. Today i came back to this.
i quickly got the same expression for vcm and i got Vcm=0.045c
and in the new frame got E_photon=1.039*10^-13J
and, E_e^-=E_e⁺=E_photon,
P_e⁺=-P_e^-=1.87*10^-22mkg/s

ok i must say upfront i got worse results. Inicial photon energy was 2.98*10^-13J and final pair energy i got 2.88*10^-13J.
what i did:
used E²=p²c²+me²c⁴ to find p, then used:

If a particle has energy E and momentum p in one frame, then in the other frame, its energy and momentum are
\begin{align*}
E' &= \gamma[E - (v/c)(pc)] \\
p'c &= \gamma[pc - (v/c)E].
\end{align*}

as a system for each one, where E' and p' are the energy and momentum in the new frame and E and p in the lab frame, and v is the speed of the new frame,
in the case in which the particle was moving along the frame direction i added the v/c term instead of subtracting. i might have made some mistake wth the calculus I'm redoing this from scratch tommorow.

 

[WrongMan]

Ok have redone this one last time and i think i got it finally...
Vcm i got exactly 1/3 c
And final energy i got 2.95*10^-13j (comparing to initial of 2.98 and accounting for rounding mistakes i think this is pretty good!) I resolved the problem like i described in my last post, difference was yesterday i used nanometers instead of picometers for wavelengths, my bad... Thank you for all your help!

Just need clarification on using the transform formula i quoted above, the sign on v/c should be changed according to the direction of the Vcm correct? If its moving in the same direction as the particle the particle's speed on that frame would be lower therefore i must subtract v/c otherwise i should add, correct?