# Understanding how to derive the cross-section formula in the CoM frame

• I

## 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.

## Main Question or Discussion Point

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.

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

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

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nrqed
Homework Helper
Gold Member
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.

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

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.
If the book has $|\mathbf p'_1|^2 d|\mathbf p'_1|^2 d|\mathbf p'_1| d \Omega'_1 = d^3 \mathbf p'_1$, then there is a typo in the book. It should be
$|\mathbf p'_1|^2 d|\mathbf p'_1| d \Omega'_1 = d^3 \mathbf p'_1$. This is the equivalent of $d^3 \mathbf r = r^2 dr d \Omega$ in momentum space.

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

$$v^2_{rel} = \frac{|p_1|^2}{(E_1 E_2)^2}(E^2_1 + E^2_2)$$
There is a mistake there. Can you show your steps? You know that $(p_1 p_2)^2$ means $(E_1 E_2 + \mathbf p^2)^2$, right?

JD_PM
vanhees71
Gold Member
2019 Award
In view of this thread, I've slightly revised the derivation of the said cross-section formula in my QFT manuscript:

https://itp.uni-frankfurt.de/~hees/publ/lect.pdf

There's no need to think about any other specific reference frame than the laboratory frame, where in the initial state one particle ("the target") is at rest (in the manuscript that's particle 2), because that's how the cross section is defined. This is for historical reasons, I guess, because in the early era of particle physics one had usually one particle beam and a fixed target. Then you simply write everything in manifestly covariant form, and finally you get a manifestly covariant cross-section formula, i.e., you can calculate the result in any reference frame you like. Usually it's most simple in the center-of-momentum frame of course, where you have $\vec{p}_2=-\vec{p}_1$ by definition.

JD_PM
I'll reply soon in this thread; I am still dealing with the proof (I'll post my work in the QP thread)

$$\lim_{y \rightarrow \infty} \frac{\sin^2(xy)}{yx^2}=\pi \delta(x)$$

If the book has $|\mathbf p'_1|^2 d|\mathbf p'_1|^2 d|\mathbf p'_1| d \Omega'_1 = d^3 \mathbf p'_1$, then there is a typo in the book. It should be
$|\mathbf p'_1|^2 d|\mathbf p'_1| d \Omega'_1 = d^3 \mathbf p'_1$. This is the equivalent of $d^3 \mathbf r = r^2 dr d \Omega$ in momentum space.
You're right, it is my typo. Thanks!

There is a mistake there. Can you show your steps? You know that $(p_1 p_2)^2$ means $(E_1 E_2 + \mathbf p^2)^2$, right?
You're right, I actually missed a term.

We get the following 5 terms out of $v^2_{rel} = \frac{(p_1 p_2)^2}{(E_1 E_2)^2} - \frac{(E^2_1 - |p_1|^2)(E^2_2 - |p_2|^2)}{(E_1 E_2)^2}$

$$v^2_{rel} = \frac{(\mathbf p_1 \mathbf p_2)^2}{(E_1 E_2)^2} -1 -\frac{(\mathbf p_1 \mathbf p_2)^2}{(E_1 E_2)^2} + \frac{E_1^2 |\mathbf p_2|^2}{(E_1 E_2)^2}+ \frac{E_2^2 |\mathbf p_1|^2}{(E_1 E_2)^2}$$

Simplifying and using the CoM condition we get

$$v^2_{rel} = |\mathbf p_1|^2 \Big( \frac{E_1^2+E_2^2-(E_1 E_2)^2/|\mathbf p_1|^2}{(E_1 E_2)^2}\Big)$$

So I made a mistake somewhere, as

$$E_1^2+E_2^2-(E_1 E_2)^2/|\mathbf p_1|^2 \neq E_1+E_2$$

I still do not see it...

PeroK
Homework Helper
Gold Member
For equation (3), we have:
$$v_1 = \frac{\gamma m_1 v_1}{\gamma m_1} = \frac{|\mathbf {p_1}|}{E_1}$$
And
$$v_2 = \frac{\gamma m_2 v_2}{\gamma m_2} = \frac{|\mathbf {p_2}|}{E_2}$$
So:
$$v_1 + v_2 = \frac{|\mathbf {p_1}|E_2 + |\mathbf {p_2}|E_1}{E_1E_2}$$
Then set $|\mathbf {p_1}| = |\mathbf {p_2}|$ for the CoM frame.

JD_PM
nrqed
Homework Helper
Gold Member
You're right, it is my typo. Thanks!

You're right, I actually missed a term.

We get the following 5 terms out of $v^2_{rel} = \frac{(p_1 p_2)^2}{(E_1 E_2)^2} - \frac{(E^2_1 - |p_1|^2)(E^2_2 - |p_2|^2)}{(E_1 E_2)^2}$

$$v^2_{rel} = \frac{(\mathbf p_1 \mathbf p_2)^2}{(E_1 E_2)^2} -1 -\frac{(\mathbf p_1 \mathbf p_2)^2}{(E_1 E_2)^2} + \frac{E_1^2 |\mathbf p_2|^2}{(E_1 E_2)^2}+ \frac{E_2^2 |\mathbf p_1|^2}{(E_1 E_2)^2}$$

Simplifying and using the CoM condition we get

$$v^2_{rel} = |\mathbf p_1|^2 \Big( \frac{E_1^2+E_2^2-(E_1 E_2)^2/|\mathbf p_1|^2}{(E_1 E_2)^2}\Big)$$

So I made a mistake somewhere, as

$$E_1^2+E_2^2-(E_1 E_2)^2/|\mathbf p_1|^2 \neq E_1+E_2$$

I still do not see it...
Watch out, you used $(p_1 p_2) = \mathbf p_1 \cdot \mathbf p_2$. This is incorrect. As I mentioned earlier, $p_1 p_2$ here means the contraction of the four-momenta, $p_1 p_2 = E_1 E_2 - \mathbf p_1 \cdot \mathbf p_2$ which is how I got (in the COM) $p_1 p_2 = E_1 E_2 + \mathbf p^2$.

JD_PM
In view of this thread, I've slightly revised the derivation of the said cross-section formula in my QFT manuscript:

https://itp.uni-frankfurt.de/~hees/publ/lect.pdf
I went through it and I have specific questions

1) Isn't there a typo in the primed factors of the Delta Functions of Eqs (6.113) and (6.114)? I think the Delta Functions should be (respectively):

$$\delta^{(4)} (p_1' + p_2' - p_1 - p_2)$$

$$\delta^{(4)} (\omega_1' + \omega_2' - \omega_1 - \omega_2)$$

2) I do not understand Eq. (6.116)

I understand that $\vec p'^2 d|\vec p'| = d^3 \vec p_1'$ (I used $\vec p_1' =-\vec p_2' := \vec p$) because we use polar coordinates for $\vec p_1'$, but why $\vec p'^2 d|\vec p'|=|\vec p'| \omega_1' d \omega_1'$?

To match the answer I used

$$\omega_1'^2 = \vec p_1'^2 + m_1^2$$

And set $m_1^2=0$. But why all mass is converted to energy after the collision? (Maybe this was not your reasoning. How did you get Eq. (6.116) then?).

3) I do not understand Eq. (6.117) (solved while typing the question! I'll type it for completeness, you can simply ignore 3))

We assume

$$\vec p'^2 d|\vec p'^2| = |\vec p| \omega_1' d \omega_1' \ \ \ \ (*)$$

$$\vec p'^2 d|\vec p'^2| = |\vec p| \omega_2' d \omega_2' \ \ \ \ (**)$$

We multiply Eq. (*) by $\omega_2'$ and Eq. (**) by $\omega_1'$, add them up, solve for $\vec p'^2 d|\vec p'^2|$ and we get Eq. (6.117).

4) Why is the relativistic-invariant flux $I=|\vec p| \omega$ in the CoM frame (Eq. 6.119))?

At Eq. (6.100), you defined $I$ as follows

$$I := \sqrt{(p_1p_2)^2 -(m_1m_2)^2}=\sqrt{|p|^2 -(m_1m_2)^2}$$

The relative velocity is given to be (Eq. 2 in this thread; Eq. 8.9, in Mandl & Shaw's book):

$$\omega_1 \omega_2 v_{rel} = [(\vec p_1 \vec p_2)^2 - m_1^2 m_2^2]^{1/2} \ \ \ \ (2)$$

In the center-of-mass frame we get

$$v_{rel}= |\vec p_1|\frac{\omega_1 + \omega_2}{\omega_1\omega_2} \ \ \ \ (3)$$

So I do not know how to combine the definition of $I$ and the relative velocity in the CoM frame to get it.

OTHER IDEA: I've been thinking that we may not need (3) at all to show $I=|\vec p| \omega$ in the CoM frame. If that is the case, we'd only need to use $I := \sqrt{(p_1p_2)^2 -(m_1m_2)^2}=\sqrt{|p|^2 -(m_1m_2)^2}$; but how to show it only based on this Eq.?

5) How to obtain Eq. (6.122)? (didn't you miss the square on it; $t=(p_1-p_1')^2=...$)?

Based on $\omega^2=|\vec p|^2 + m^2$ I get

$$t:= (p_1-p_1')^2 = \Big( \sqrt{\omega_1^2 - m_1^2} - \sqrt{\omega_1'^2 - m_1'^2} \Big)^2=\omega_1^2 - m_1^2+\omega_1'^2 - m_1'^2 - 2\sqrt{\omega_1^2 - m_1^2}\sqrt{\omega_1'^2 - m_1'^2}$$

PS: I want to thank you again for sharing your manuscript. Your derivation of the cross-section in the CoM frame is more detailed compared to the nice Mandl & Shaw's QFT book.

vanhees71
Gold Member
2019 Award
ad 1: In (6.113) it's indeed $\delta^{(4)}(p_1+p_2-p_1'-p_2')$ and that's what's in the manuscript (note that $\delta^{(4)}(k)=\delta^{(4)}(-k)$). In (6.114) it must be a single $\delta(\omega_1+\omega_2-\omega_1'-\omega_2')$, and that's what's in the manuscript.

ad 2: For an on-shell momentum $k$ you have for the radial part of the integral measure $\mathrm{d} |\vec{k}| k^2=\mathrm{d} \omega_k |\vec{k}|$, because $\omega^2=m^2+|\vec{k}|^2$ and thus $2 \omega \mathrm{d} \omega=\mathrm{d} |\vec{k}| |\vec{k}|$.

ad 4: $I^2=(p_1 \cdot p_2)^2-m_1^2 m_2^3$. Let's write $P=|\vec{p}_1|=|\vec{p}_2|$ in the CoM frame, where $\vec{p}_1=-\vec{p}_2$. Now we have
$$s=(p_1+p_2)^2=\omega^2=m_1^2+m_2^2+2 p_1 p_2 \; \Rightarrow \; p_1 p_2=\frac{1}{2} ( \omega^2-m_1^2-m_2^2)$$
and thus
$$I^2=(P^2+\omega_1 \omega_2)^2-m_1^2 m_2^2=P^4+2\omega_1 \omega_2 P^2 + \omega_1^2 \omega_2^2 -m_1^2 m_2^2 = P^4 + 2 \omega_1 \omega_2 P^2 + (m_1^2+P^2)(m_2^2+P^2)-m_1^2 m_2^2=2 P^4+P^2(m_1^2+m_2^2+2 \omega_1 \omega_2)=2 P^4 +P^2 (\omega_1^2+\omega_2^2-2P^2+2 \omega_1 \omega_2)=P^2 (\omega_1+\omega_2)^2=P^2 \omega^2.$$
Your Eq. (2) is wrong. It's an invariant and thus you have the Minkowski product p_1 p_2 in the first term under the square root and not $\vec{p}_1 \cdot \vec{p}_2$.

Eq. (3) cannot be right either, because the relative speed is covariantly defined as
$$v_{\text{rel}}=\frac{I}{p_1 p_2}=\frac{P\omega}{\omega_1 \omega_2+P^2}.$$
Indeed the general formula for the relative velocity, as derived in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

Eq. (1.6.5), confirms the above formula since or collinear velocities $\vec{\beta}_1=\vec{p}_1/E_1=\vec{p}/E_1$ and $\vec{\beta}_2=\vec{p}_2/E_2=-\vec{p}/E_2$ the formula simplifies to
$$v_{\text{rel}}=\frac{1}{1-\vec{\beta}_1 \cdot \vec{\beta}_2} |\vec{\beta}_1-\vec{\beta}_2|.$$
In your formula (3) the factor in front of the "naive" formula for the relative velocity. Unfortunately this is wrong in some textbooks. I cannot check Mandl and Shaw, whether it's correct in there, but I guess not, given your Eq. (3).

ad 5) Indeed I forgot a square (I've corrected the typo; thanks for pointing it out), but we indeed have
$$t=(p_1-p_1')^2=m_1^2 + m_1^{\prime 2}-2 \omega_1 \omega_1' + 2 \vec{p} \cdot \vec{p}'.$$
I don't know, how you come to your equation for Mandelstam $t$.

nrqed, JD_PM and dextercioby
ad 1: In (6.113) it's indeed $\delta^{(4)}(p_1+p_2-p_1'-p_2')$ and that's what's in the manuscript (note that $\delta^{(4)}(k)=\delta^{(4)}(-k)$). In (6.114) it must be a single $\delta(\omega_1+\omega_2-\omega_1'-\omega_2')$, and that's what's in the manuscript.
Oh so, if I am not mistaken, both ways are correct due to the Dirac-Delta distribution property $\delta^{(4)}(k)=\delta^{(4)}(-k)$. Now it is clear, thank you.

ad 2: For an on-shell momentum $k$ you have for the radial part of the integral measure $\mathrm{d} |\vec{k}| k^2=\mathrm{d} \omega_k |\vec{k}|$, because $\omega^2=m^2+|\vec{k}|^2$ and thus $2 \omega \mathrm{d} \omega=\mathrm{d} |\vec{k}| |\vec{k}|$.
Mmm please let me check if I understand why $\vec p'^2 d|\vec p'|=|\vec p'| \omega_1' d \omega_1'=|\vec p'| \omega_2' d \omega_2'$.

We start from the relativistic energy-momentum equation (let me neglect primes from here on)

$$\omega_p^2 = m^2+|\vec{p}|^2$$

Taking the derivative on both sides we get

$$2\omega_p d \omega_p = 2|\vec{p}| d|\vec{p}|$$

Multiplying both sides by $|\vec{p}|$ we get

$$|\vec{p}|\omega_p d \omega_p = |\vec{p}|^2 d|\vec{p}|$$

And as we are in the CoM frame (i.e. $\vec p_1 =-\vec p_2$) we thus get

$$\vec p'^2 d|\vec p'|=|\vec p'| \omega_1' d \omega_1'=|\vec p'| \omega_2' d \omega_2'$$

ad 4: $I^2=(p_1 \cdot p_2)^2-m_1^2 m_2^3$. Let's write $P=|\vec{p}_1|=|\vec{p}_2|$ in the CoM frame, where $\vec{p}_1=-\vec{p}_2$. Now we have
$$s=(p_1+p_2)^2=\omega^2=m_1^2+m_2^2+2 p_1 p_2 \; \Rightarrow \; p_1 p_2=\frac{1}{2} ( \omega^2-m_1^2-m_2^2)$$
and thus
$$I^2=(P^2+\omega_1 \omega_2)^2-m_1^2 m_2^2=P^4+2\omega_1 \omega_2 P^2 + \omega_1^2 \omega_2^2 -m_1^2 m_2^2 = P^4 + 2 \omega_1 \omega_2 P^2 + (m_1^2+P^2)(m_2^2+P^2)-m_1^2 m_2^2=2 P^4+P^2(m_1^2+m_2^2+2 \omega_1 \omega_2)=2 P^4 +P^2 (\omega_1^2+\omega_2^2-2P^2+2 \omega_1 \omega_2)=P^2 (\omega_1+\omega_2)^2=P^2 \omega^2.$$
Naive question: why $(p_1+p_2)^2=(\omega_1 +\omega_2)^2$ holds? This is my reasoning: in your notation $p$ is the four-momentum vector $p:=(\omega, \vec p)$. Thus we get (using the CoM condition $\vec{p}_1=-\vec{p}_2$, $\omega:=\omega_1+\omega_2$ and the on-shell condition $p^2=m^2$)

$$s=(p_1+p_2)^2=(\omega_1+\vec p_1+\omega_2+\vec p_2)^2=\omega^2=(m_1+m_2)^2=m_1^2+m_2^2+2 p_1 \cdot p_2$$

Where $p_1 \cdot p_2$ is the 4-momentum inner product. Then we end up with

$$p_1 \cdot p_2=\frac{1}{2} ( \omega^2-m_1^2-m_2^2) \ \ \ \ (12)$$

Do you agree at this point? I think you will not because it seems that I shouldn't get the 4-momentum inner product $p_1 \cdot p_2$. Why I think so? Because

$$p_1 \cdot p_2=P^2+\omega_1 \omega_2 \neq \frac{1}{2} ( \omega^2-m_1^2-m_2^2)$$

If I shouldn't get the 4-momentum inner product at Eq. 12, what did you mean by $p_1 p_2$?

Your Eq. (2) is wrong. It's an invariant and thus you have the Minkowski product p_1 p_2 in the first term under the square root and not $\vec{p}_1 \cdot \vec{p}_2$.

Eq. (3) cannot be right either, because the relative speed is covariantly defined as
$$v_{\text{rel}}=\frac{I}{p_1 p_2}=\frac{P\omega}{\omega_1 \omega_2+P^2}.$$
Indeed the general formula for the relative velocity, as derived in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

Eq. (1.6.5), confirms the above formula since or collinear velocities $\vec{\beta}_1=\vec{p}_1/E_1=\vec{p}/E_1$ and $\vec{\beta}_2=\vec{p}_2/E_2=-\vec{p}/E_2$ the formula simplifies to
$$v_{\text{rel}}=\frac{1}{1-\vec{\beta}_1 \cdot \vec{\beta}_2} |\vec{\beta}_1-\vec{\beta}_2|.$$
In your formula (3) the factor in front of the "naive" formula for the relative velocity. Unfortunately this is wrong in some textbooks. I cannot check Mandl and Shaw, whether it's correct in there, but I guess not, given your Eq. (3).
Thank you for sharing another Manuscript of yours! . I have to say I was confused by Mandl & Shaw's definition of relative velocity. This deserves a separated thread

ad 5) Indeed I forgot a square (I've corrected the typo; thanks for pointing it out), but we indeed have
$$t=(p_1-p_1')^2=m_1^2 + m_1^{\prime 2}-2 \omega_1 \omega_1' + 2 \vec{p} \cdot \vec{p}'.$$
I don't know, how you come to your equation for Mandelstam $t$.
Sorry, I was sloppy. Let me start from scratch. We know that the following equation holds

$$t:=(p_1-p_1')^2=(\omega_1 + \vec p_1 - \omega_1' - \vec p_1')^2 = (m_1-m_1')^2$$

Thus I get

$$t:=(p_1-p_1')^2=m_1^2 + m_1'^2 - 2p_1 \cdot p_1'=m_1^2 + m_1'^2 - 2\omega_1\omega_1' - 2\vec p_1 \cdot \vec p_1'$$

Mmm but why did you get the term $2\vec p_1 \cdot \vec p_1'$ with a positive sign?

nrqed
Homework Helper
Gold Member
Oh so, if I am not mistaken, both ways are correct due to the Dirac-Delta distribution property $\delta^{(4)}(k)=\delta^{(4)}(-k)$. Now it is clear, thank you.

Mmm please let me check if I understand why $\vec p'^2 d|\vec p'|=|\vec p'| \omega_1' d \omega_1'=|\vec p'| \omega_2' d \omega_2'$.

We start from the relativistic energy-momentum equation (let me neglect primes from here on)

$$\omega_p^2 = m^2+|\vec{p}|^2$$

Taking the derivative on both sides we get

$$2\omega_p d \omega_p = 2|\vec{p}| d|\vec{p}|$$

Multiplying both sides by $|\vec{p}|$ we get

$$|\vec{p}|\omega_p d \omega_p = |\vec{p}|^2 d|\vec{p}|$$

And as we are in the CoM frame (i.e. $\vec p_1 =-\vec p_2$) we thus get

$$\vec p'^2 d|\vec p'|=|\vec p'| \omega_1' d \omega_1'=|\vec p'| \omega_2' d \omega_2'$$

Naive question: why $(p_1+p_2)^2=(\omega_1 +\omega_2)^2$ holds? This is my reasoning: in your notation $p$ is the four-momentum vector $p:=(\omega, \vec p)$. Thus we get (using the CoM condition $\vec{p}_1=-\vec{p}_2$, $\omega:=\omega_1+\omega_2$ and the on-shell condition $p^2=m^2$)

$$s=(p_1+p_2)^2=(\omega_1+\vec p_1+\omega_2+\vec p_2)^2=\omega^2=(m_1+m_2)^2=m_1^2+m_2^2+2 p_1 \cdot p_2$$
Watch out. It is incorrect to write $(p_1+p_2)^2=(\omega_1+\vec p_1+\omega_2+\vec p_2)^2$. Note that one can never add scalars to vectors, this is not well defined mathematically. Here, $p_1+p_2$ is the sum of the two four-momenta, so it is
$$(\omega_1, \mathbf p_1)+ (\omega_2 , \mathbf p_2) = (\omega_1+ \omega_2, \mathbf p_1 + \mathbf p_2 )$$
which in the COM gives $(\omega_1 + \omega_1, \mathbf 0)$ which is what we must then square.

It seems that you get the correct results at the end, but it is important to not write things incorrectly in intermediate steps.

vanhees71 and PeroK
nrqed
Homework Helper
Gold Member
Thus I get

$$t:=(p_1-p_1')^2=m_1^2 + m_1'^2 - 2p_1 \cdot p_1'=m_1^2 + m_1'^2 - 2\omega_1\omega_1' - 2\vec p_1 \cdot \vec p_1'$$

Mmm but why did you get the term $2\vec p_1 \cdot \vec p_1'$ with a positive sign?
The definition of the product between two four vectors is $p_1 \cdot p_2 = \omega_1 \omega_2 - \vec p_1 \cdot \vec p_2$. Note the negative sign.

vanhees71