Photon emission opening angle

  • #1
Homework Statement
Let a particle of mass M travelling at speed ##\beta = 1/2## (##\gamma = 2/\sqrt 3 \ \ c=1##) decay in to two photons: ##A \rightarrow 1+2##
1) Calculate energy and moment of the photons in the reference frame of the center of mass (CM)
2) Calculate energy and moment of the photons in the reference frame of the lab (LAB)
3) Let ##\theta_1## be the angle between the x axis and the first photon (counter clock wise) and ##\theta_2## the angle the second photon x axis angle (clock wise), discuss the relation between this angles and ##\theta^*##
4) The opening angle is ##\alpha = \theta_1+\theta_2##, find the condition required for ##\alpha## to be minimized and calculate it in this condition.
Relevant Equations
$$\alpha=2arctan \left( \frac{1}{\gamma\beta} \right) $$
I'm doing special relativity in undergrad and I have the following problem:

Let a particle of mass M travelling at speed ##\beta = 1/2## (##\gamma = 2/\sqrt 3 \ \ c=1##) decay in to two photons: ##A \rightarrow 1+2##

1) Calculate energy and moment of the photons in the reference frame of the center of mass (CM):

$$ E_1^*=E_2^*=|P_1^*|=|P_2^*|=E^*=M/2 $$



2) Calculate energy and moment of the photons in the reference frame of the lab (LAB):

$$ P_1^* =(E^*, E^*cos\theta^*, E^*sin\theta^*, 0) $$

$$ P_2^* =(E^*, -E^*cos\theta^*, -E^*sin\theta^*, 0) $$

applying Lorentz transformation

$$ |P_1|=E_1 =\gamma(E^* +\beta cos\theta^*) $$

$$ |P_2|=E_2 =\gamma(E^* -\beta cos\theta^*) $$

3) Let ##\theta_1## be the angle between the x axis and the first photon (counter clock wise) and ##\theta_2## the angle the second photon x axis angle (clock wise), discuss the relation between this angles and ##\theta^*##

$$ P_1 =(E_1, E_1cos\theta_1, E_1sin\theta_1, 0) $$

$$ P_2 =(E_2, E_2cos\theta_2, -E_2sin\theta_2, 0) $$

$$ E_1cos\theta_1 = E^*\gamma(\beta+cos\theta^*)$$

$$ E_2cos\theta_2 = E^*\gamma(\beta-cos\theta^*)$$

So for both photons to be emitted forward we need ##\beta+cos\theta^*>0## and ##\beta-cos\theta^*>0## resulting in ##|cos\theta^*|<\beta=1/2## so ##\pi/3<\theta^*<2/3\pi##, for one photon to be emitted along y axes we need ##\theta^*=\pm 2/3\pi## or ##\theta^*=\pm 1/3\pi## and if ##\theta^*=0## the photons are emitted along x axis one forward and the other backward.

4) The opening angle is ##\alpha = \theta_1+\theta_2##, find the condition required for ##\alpha## to be minimized and calculate it in this condition.

Here I get stuck. I know that $$tan\theta_1=\frac{sin\theta^*}{\gamma(\beta + cos\theta^*)}\quad tan\theta_2=\frac{sin\theta^*}{\gamma(\beta - cos\theta^*)} $$ and I found out on some books that the wanted condition is ##\theta^*=\pi/2## in that case $$\alpha=2arctan \left( \frac{1}{\gamma\beta} \right) $$

but how can I prove that ##\theta^*=\pi/2## is the wanted condition? Are the other results right?

Thanks, have a nice day.
 
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  • #2
lambdadandbda said:
4) The opening angle is ##\alpha = \theta_1+\theta_2##, find the condition required for ##\alpha## to be minimized and calculate it in this condition.

Here I get stuck. I know that $$tan\theta_1=\frac{sin\theta^*}{\gamma(\beta + cos\theta^*)}\quad tan\theta_2=\frac{sin\theta^*}{\gamma(\beta - cos\theta^*)} $$
how can I prove that ##\theta^*=\pi/2## is the wanted condition?
I couldn't think of a short way to show that the minimum of ##\alpha## occurs for ##\theta^* = \pi/2##. I did get it with some tedious :oldfrown: manipulations .

Starting with ##\tan\theta_1= \large \frac{\sin\theta^*}{\gamma(\beta + \cos\theta^*)}##, you can take the derivative of both sides with respect to ##\theta^*## and then manipulate to solve for ##\large \frac{d \theta_1}{d \theta^*}##. With the help of the relation ##\gamma = \large \frac 1 {\sqrt{1-\beta^2}}##, I found that it could be simplified finally to $$\frac{d \theta_1}{d \theta^*} = \frac 1 {\gamma (\beta \cos \theta^*+ 1)}$$ Of course, this should be checked. I get a similar expression for ##\large \frac{d \theta_2}{d \theta^*}##. These results can be used to derive the condition on ##\theta^*## corresponding to ##\large \frac{d \alpha}{d \theta^*} = 0## and to show that ##\alpha## takes on its minimum value here.

Aside: Without any calculation you can argue that ##\alpha## must have a local extremum at ##\theta^* = \pi/2##. By symmetry of the physics, the function ##\alpha(\theta^*)## must be an even function about ##\theta^* = \pi/2##.

lambdadandbda said:
Are the other results right?
They look right to me.
 
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  • #3
thanks! I got confused using the tangent because of the asyntotes at ##\pi/2## so I would prefer a proof without it. Anyhow I cant find a clear way to state the result using symmetry, for sure we can see that ##\alpha## has maxima at ##\theta^*= 0,\pi## but I don't know how to formulate the symmetry around ##\theta^*=\pi/2##...
 
  • #4
lambdadandbda said:
I don't know how to formulate the symmetry around ##\theta^*=\pi/2##...

In the CM frame, consider a pair of photons, shown in red below, for which one of the photons is emitted at the angle ##\theta^*## to the x-axis.

1707689231018.png

As long as one of the photons is emitted in any direction that lies on the cone, then we will get the same value for ##\alpha## back in the lab frame. Consider the blue pair of photons on the cone as shown below.

1707689480463.png

The blue pair will have the same value for ##\alpha## in the lab frame as the red pair. But the red pair and the blue pair make the same angle relative to the plane corresponding to ##\theta^* = \frac{\pi}{2}##. This plane is the y-z plane. The z-axis comes out of the page and is not shown in the figure.

Start with the red pair initially at the angle ##\theta^*## in the CM as shown and then continuously increase ##\theta^*## until the red pair is coincident with the blue pair. The starting and ending values of ##\alpha## back in the lab frame must be equal. So, ##\alpha## as a function of ##\theta^*## must be symmetric about ##\theta^* = \frac {\pi}{2}##. That is, it must be an even function about ##\theta^* = \frac {\pi}{2}##.
 
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  • #5
Beautiful! Having it written like this it's easy to see that ##\alpha## is a periodic function of period ##\pi## with maximum at 0 and ##\pi## and symmetric about 0, ##\pi/2## and ##\pi/2## and hence with minimum at ##\pi/2##.
Thank you very much, have a nice day!
 

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