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## Homework Statement

Two photon of energy ##E_1 ## and ## E_2## collide with their trajectory at an angle $\theta$ with respect to each other.

a) Total four-momentum before collision?

b) square length of 4-momentum in lab frame (LB)and in center of momentum frame (CM)?

c) 4-momentum of two photon system in CM?

d) After photons annihilate they form two identical particle with mass m. what can we say about the spatial components of their 4-momentum in CM?

e) requirement of this process is that ##\sqrt{E_1E_2}\geq E_{min}##. find ##E_{min}## in terms of rest energy of particles and angle ##\theta##.

## Homework Equations

-Conservation of Energy

-Conservation of Momentum

-For a photon square of 4-momentum is zero (light like particle)[/B]

## The Attempt at a Solution

**Attempt at Answer:(assuming without loss of generality that particle 1 is moving in + x direction and the whole motion is in x-y plane)**

a)

- The initial momentum for each particle is:

$$p^{\mu}_1=(E_1, E_1,0,0) \\ p^{\mu}_2=(E_2,-E_2cos\theta,E_2sin\theta,0) \\$$

hence the total initial momentum is: $$p^{\mu}_T=(E_1+E_2,E_1-E_2cos\theta,E_2sin\theta,0)$$

b)

- squared length of 4-momentum in lab frame (LB): $$(p_{T\mu}p^{\mu}_T)_{LB}=(E_1+E_2)^2- (E_1-E_2cos\theta)^2-(E_2sin\theta)^2=2E_1E_2(1+cos\theta)$$

- square 4-momentum in COM = square length of 4-momentum in LB (** is it right??!!):

$$(p_{T\mu}p^{\mu}_T)_{LB}=(p_{T\mu}p^{\mu}_T)_{COM}=2E_1E_2(1+cos\theta)$$

c)

Now to find the momentum in COM I use the above fact that square of the total four momentum is invariant under frame transformation (again is it right!!) and combining this with the fact that the square of 4-momentum of photon is zero we get:

$$p^{\mu}_{1(COM)}=(E',E'_x,E'_y,0)\\p^{\mu}_{2(COM)}=(E',-E'_x,-E'_y,0)$$

-So the total 4-momentum in COM is:

$$p^{\mu}_{T(COM)}=(2E',0,0,0)$$

squaring it and equating to square of total 4-momentum in COM, gives: $$E'^2=\frac{E_1E_2}{2}(1+cos\theta)=E_1E_2cos^2{\frac{\theta}{2}}$$

As I have the magnitude and I know that $$E'^2_x+E'^2_y=E_1E_2cos^2{\frac{\theta}{2}}$$

its evident that the two components will have the following form:$$E'_x=\sqrt{E_1E_2}\cos{\frac{\theta}{2}}\cos{\theta'}\\E'_y=\sqrt{E_1E_2} \cos{\frac{\theta}{2}} \sin{\theta'}$$

but here I face a problem. I still don't know $$\theta'$$

Any help or hint is appreciated. Thank you so much for your time.

a)

- The initial momentum for each particle is:

$$p^{\mu}_1=(E_1, E_1,0,0) \\ p^{\mu}_2=(E_2,-E_2cos\theta,E_2sin\theta,0) \\$$

hence the total initial momentum is: $$p^{\mu}_T=(E_1+E_2,E_1-E_2cos\theta,E_2sin\theta,0)$$

b)

- squared length of 4-momentum in lab frame (LB): $$(p_{T\mu}p^{\mu}_T)_{LB}=(E_1+E_2)^2- (E_1-E_2cos\theta)^2-(E_2sin\theta)^2=2E_1E_2(1+cos\theta)$$

- square 4-momentum in COM = square length of 4-momentum in LB (** is it right??!!):

$$(p_{T\mu}p^{\mu}_T)_{LB}=(p_{T\mu}p^{\mu}_T)_{COM}=2E_1E_2(1+cos\theta)$$

c)

Now to find the momentum in COM I use the above fact that square of the total four momentum is invariant under frame transformation (again is it right!!) and combining this with the fact that the square of 4-momentum of photon is zero we get:

$$p^{\mu}_{1(COM)}=(E',E'_x,E'_y,0)\\p^{\mu}_{2(COM)}=(E',-E'_x,-E'_y,0)$$

-So the total 4-momentum in COM is:

$$p^{\mu}_{T(COM)}=(2E',0,0,0)$$

squaring it and equating to square of total 4-momentum in COM, gives: $$E'^2=\frac{E_1E_2}{2}(1+cos\theta)=E_1E_2cos^2{\frac{\theta}{2}}$$

As I have the magnitude and I know that $$E'^2_x+E'^2_y=E_1E_2cos^2{\frac{\theta}{2}}$$

its evident that the two components will have the following form:$$E'_x=\sqrt{E_1E_2}\cos{\frac{\theta}{2}}\cos{\theta'}\\E'_y=\sqrt{E_1E_2} \cos{\frac{\theta}{2}} \sin{\theta'}$$

but here I face a problem. I still don't know $$\theta'$$

Any help or hint is appreciated. Thank you so much for your time.