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lambdadandbda

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- 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.

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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