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JD_PM

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- TL;DR Summary
- I am studying how to get the differential cross-section formula (in the CoM frame)

##\Big( \frac{d \sigma}{d \Omega'_1} \Big)_{CoM} = \frac{1}{64 \pi^2 (E_1 + E_2)^2} \frac{|\mathbf p'_1|}{|\mathbf p_1|} \Pi_l \Big( 2 m_l \Big) |\mathscr{M}|^2##

As explained in Quantum Field Theory's book by Mandl and Shaw and I basically got lost in the Mathematics of the derivation.

We work in natural units.

Let's assume in this post that the differential cross-section of two particles that, after collision, yield ##N## particles is given by the following formula:

$$d \sigma = (2\pi)^4 \delta^{(4)} (\sum p'_f - \sum p_i) \frac{1}{4 E_1 E_2 v_{rel}} \Pi_l (2m_l) \Pi_f \Big(\frac{d^3 \mathbf p'_f}{(2\pi)^3 2 E'_f}\Big) |\mathscr{M}|^2 \ \ \ \ (1)$$

Where ##v_{rel}## refers to the relative velocity of the two colliding particles and ##\mathscr{M}## to the Feynman amplitude.

Eq. (1) holds in any Lorentz frame in which the colliding particles move collinearly. In such a frame the relative velocity ##v_{rel}## is given by

$$E_1 E_2 v_{rel} = [(p_1 p_2)^2 - m_1^2 m_2^2]^{1/2} \ \ \ \ (2)$$

Where ##m_1## and ##m_2## are the rest masses of the colliding particles.

The formula for the relative velocity in the CoM frame is as follows

$$v_{rel}=\frac{|\mathbf p_1|}{E_1}+\frac{|\mathbf p_2|}{E_2}=|\mathbf p_1|\frac{E_1+E_2}{E_1E_2} \ \ \ \ (3)$$

The formula for the relative velocity in the LAB frame is as follows

$$v_{rel}=\frac{|\mathbf p_1|}{E_1} \ \ \ \ (4)$$

Because of conservation of energy and momentum, the final-state-3D momentum ##\mathbf p'_1,..., \mathbf p'_N## are not all independent variables. In order to fix this we must integrate Eq. (1) with respect to the remaining variables.

Then Mandl and Shaw chose to work out the two-body state as an example. In such a case, Eq (1) becomes

$$d \sigma=f(p'_1, p'_2) \delta^{(4)} (p'_1 + p'_2 - p_1 - p_2)d^3 \mathbf p'_1 d^3 \mathbf p'_2 \ \ \ \ (5)$$

Where

$$f(p'_1, p'_2) := \frac{1}{64 \pi^2 v_{rel} E_1 E_2 E'_1 E'_2} \Pi_l (2m_l) |\mathscr{M}|^2 \ \ \ \ (6)$$

Integrating Eq. (5) wrt ##\mathbf p'_2## yields

$$d \sigma = f(p'_1, p'_2) \delta (E'_1 + E'_2 - E_1 - E_2) |\mathbf p'_1|^2 d|\mathbf p'_1|^2 d|\mathbf p'_1| d \Omega'_1 \ \ \ \ (7)$$

Where ##\mathbf p'_2 =\mathbf p_1 + \mathbf p_2 -\mathbf p'_1##. Integrating Eq. (7) wrt ##\mathbf p'_1## we get

$$d \sigma = f(p'_1, p'_2) |\mathbf p'_1|^2 d \Omega'_1 \Big[ \frac{\partial(E'_1 + E'_2)}{\partial |\mathbf p'_1|}\Big]^{-1} \ \ \ \ (8)$$

Where we have used the following trick to solve the integral

$$\int f(x,y) \delta [g(x,y)]dx = \int f(x,y) \delta [g(x,y)] \Big( \frac{\partial x}{\partial g} \Big) dg = \Big[\frac{f(x,y)}{(\partial g / \partial x)_y}\Big]_{g=0} \ \ \ \ (9)$$

We know that the energy-momentum equation must be safisfied:

$$(E'_f)^2 = (m'_f)^2 + |\mathbf p'_f|^2 \ \ \ \ (10)$$

As we are in the CoM frame we get that the following equation is satisfied

$$\frac{\partial(E'_1 + E'_2)}{\partial |\mathbf p'_1|} = |\mathbf p'_1|\frac{E_1 + E_2}{E_1 E_2} \ \ \ \ (11)$$

Combining Eqs. (8), (6), (3) and (11) we finally get the desired result.

$$\Big( \frac{d \sigma}{d \Omega'_1} \Big)_{CoM} = \frac{1}{64 \pi^2 (E_1 + E_2)^2} \frac{|\mathbf p'_1|}{|\mathbf p_1|} \Pi_l \Big( 2 m_l \Big) |\mathscr{M}|^2$$

1) How to compute the integral of Eq. (5) wrt ##\mathbf p'_2##.

I suspect that the sifting property of the Dirac Delta function (i.e. ##\int f(t) \delta (t-T) dt = f(T)##) has been applied and that's actually why we go from ##\delta^{(4)} (p'_1 + p'_2 - p_1 - p_2)## to ##\delta^{(1)} (E'_1 + E'_2 - E_1 - E_2)##. Besides, why ##|\mathbf p'_1|^2 d|\mathbf p'_1|^2 d|\mathbf p'_1| d \Omega'_1 = d^3 \mathbf p'_1##? I know, based on what I learned in Calculus, that the solid angle satisfies ##d \Omega = \sin \theta d\theta d \phi## (where ##\theta## is the scattering angle and ##\phi## is the azimuthal angle) but I do not see why it shows up here.

2) I do not see how to prove Eq. 3 (I am a bit ashamed to be honest )

I started with Eq. (2)

$$v^2_{rel} = \frac{(p_1 p_2)^2}{(E_1 E_2)^2} - \frac{(m_1 m_2)^2}{(E_1 E_2)^2}$$

Using the famous energy-momentum relativistic equation we get

$$v^2_{rel} = \frac{(p_1 p_2)^2}{(E_1 E_2)^2} - \frac{(E^2_1 - |p_1|^2)(E^2_2 - |p_2|^2)}{(E_1 E_2)^2}$$

We know that in the CoM frame ##\mathbf p_1=-\mathbf p_2##. Applying such an idea we get 4 terms of which 2 cancel each other out, so we end up getting these two

$$v^2_{rel} = \frac{|p_1|^2}{(E_1 E_2)^2}(E^2_1 + E^2_2)$$

Mmm but my mistake is that I get ##\sqrt{E^2_1 + E^2_2}## instead of ##E^2_1 + E^2_2##... Where did I go wrong?

PS: I'd like to discuss the Physics of the differential cross section formula once I understand how to derive it. This could be done in this thread or I could simply open another. I'll do it as you all wish. Any help is appreciated.

Thank you

Let's assume in this post that the differential cross-section of two particles that, after collision, yield ##N## particles is given by the following formula:

$$d \sigma = (2\pi)^4 \delta^{(4)} (\sum p'_f - \sum p_i) \frac{1}{4 E_1 E_2 v_{rel}} \Pi_l (2m_l) \Pi_f \Big(\frac{d^3 \mathbf p'_f}{(2\pi)^3 2 E'_f}\Big) |\mathscr{M}|^2 \ \ \ \ (1)$$

Where ##v_{rel}## refers to the relative velocity of the two colliding particles and ##\mathscr{M}## to the Feynman amplitude.

Eq. (1) holds in any Lorentz frame in which the colliding particles move collinearly. In such a frame the relative velocity ##v_{rel}## is given by

$$E_1 E_2 v_{rel} = [(p_1 p_2)^2 - m_1^2 m_2^2]^{1/2} \ \ \ \ (2)$$

Where ##m_1## and ##m_2## are the rest masses of the colliding particles.

The formula for the relative velocity in the CoM frame is as follows

$$v_{rel}=\frac{|\mathbf p_1|}{E_1}+\frac{|\mathbf p_2|}{E_2}=|\mathbf p_1|\frac{E_1+E_2}{E_1E_2} \ \ \ \ (3)$$

The formula for the relative velocity in the LAB frame is as follows

$$v_{rel}=\frac{|\mathbf p_1|}{E_1} \ \ \ \ (4)$$

Because of conservation of energy and momentum, the final-state-3D momentum ##\mathbf p'_1,..., \mathbf p'_N## are not all independent variables. In order to fix this we must integrate Eq. (1) with respect to the remaining variables.

Then Mandl and Shaw chose to work out the two-body state as an example. In such a case, Eq (1) becomes

$$d \sigma=f(p'_1, p'_2) \delta^{(4)} (p'_1 + p'_2 - p_1 - p_2)d^3 \mathbf p'_1 d^3 \mathbf p'_2 \ \ \ \ (5)$$

Where

$$f(p'_1, p'_2) := \frac{1}{64 \pi^2 v_{rel} E_1 E_2 E'_1 E'_2} \Pi_l (2m_l) |\mathscr{M}|^2 \ \ \ \ (6)$$

Integrating Eq. (5) wrt ##\mathbf p'_2## yields

$$d \sigma = f(p'_1, p'_2) \delta (E'_1 + E'_2 - E_1 - E_2) |\mathbf p'_1|^2 d|\mathbf p'_1|^2 d|\mathbf p'_1| d \Omega'_1 \ \ \ \ (7)$$

Where ##\mathbf p'_2 =\mathbf p_1 + \mathbf p_2 -\mathbf p'_1##. Integrating Eq. (7) wrt ##\mathbf p'_1## we get

$$d \sigma = f(p'_1, p'_2) |\mathbf p'_1|^2 d \Omega'_1 \Big[ \frac{\partial(E'_1 + E'_2)}{\partial |\mathbf p'_1|}\Big]^{-1} \ \ \ \ (8)$$

Where we have used the following trick to solve the integral

$$\int f(x,y) \delta [g(x,y)]dx = \int f(x,y) \delta [g(x,y)] \Big( \frac{\partial x}{\partial g} \Big) dg = \Big[\frac{f(x,y)}{(\partial g / \partial x)_y}\Big]_{g=0} \ \ \ \ (9)$$

We know that the energy-momentum equation must be safisfied:

$$(E'_f)^2 = (m'_f)^2 + |\mathbf p'_f|^2 \ \ \ \ (10)$$

As we are in the CoM frame we get that the following equation is satisfied

$$\frac{\partial(E'_1 + E'_2)}{\partial |\mathbf p'_1|} = |\mathbf p'_1|\frac{E_1 + E_2}{E_1 E_2} \ \ \ \ (11)$$

Combining Eqs. (8), (6), (3) and (11) we finally get the desired result.

$$\Big( \frac{d \sigma}{d \Omega'_1} \Big)_{CoM} = \frac{1}{64 \pi^2 (E_1 + E_2)^2} \frac{|\mathbf p'_1|}{|\mathbf p_1|} \Pi_l \Big( 2 m_l \Big) |\mathscr{M}|^2$$

**What I do not understand is:**1) How to compute the integral of Eq. (5) wrt ##\mathbf p'_2##.

I suspect that the sifting property of the Dirac Delta function (i.e. ##\int f(t) \delta (t-T) dt = f(T)##) has been applied and that's actually why we go from ##\delta^{(4)} (p'_1 + p'_2 - p_1 - p_2)## to ##\delta^{(1)} (E'_1 + E'_2 - E_1 - E_2)##. Besides, why ##|\mathbf p'_1|^2 d|\mathbf p'_1|^2 d|\mathbf p'_1| d \Omega'_1 = d^3 \mathbf p'_1##? I know, based on what I learned in Calculus, that the solid angle satisfies ##d \Omega = \sin \theta d\theta d \phi## (where ##\theta## is the scattering angle and ##\phi## is the azimuthal angle) but I do not see why it shows up here.

2) I do not see how to prove Eq. 3 (I am a bit ashamed to be honest )

I started with Eq. (2)

$$v^2_{rel} = \frac{(p_1 p_2)^2}{(E_1 E_2)^2} - \frac{(m_1 m_2)^2}{(E_1 E_2)^2}$$

Using the famous energy-momentum relativistic equation we get

$$v^2_{rel} = \frac{(p_1 p_2)^2}{(E_1 E_2)^2} - \frac{(E^2_1 - |p_1|^2)(E^2_2 - |p_2|^2)}{(E_1 E_2)^2}$$

We know that in the CoM frame ##\mathbf p_1=-\mathbf p_2##. Applying such an idea we get 4 terms of which 2 cancel each other out, so we end up getting these two

$$v^2_{rel} = \frac{|p_1|^2}{(E_1 E_2)^2}(E^2_1 + E^2_2)$$

Mmm but my mistake is that I get ##\sqrt{E^2_1 + E^2_2}## instead of ##E^2_1 + E^2_2##... Where did I go wrong?

PS: I'd like to discuss the Physics of the differential cross section formula once I understand how to derive it. This could be done in this thread or I could simply open another. I'll do it as you all wish. Any help is appreciated.

Thank you