Graduate Mandelstam Variables for 2->2 Scattering with Equal Masses in CoM Frame

[spaghetti3451]

For $2\rightarrow 2$ scattering with equal masses in the centre-of-mass frame, are all the four three-momenta equal to each other, or is it that the incoming three-momenta and the outgoing three-momenta sum to 0 separately?

 

[PeterDonis]

Have you tried to use conservation laws to obtain the answer? Remember that energy must be conserved as well as momentum.

 

[spaghetti3451]

I have. I get my second answer.

 

[PeterDonis]

I have. I get my second answer.

So what's the problem? Do you think that answer is wrong? Can you show your work?

 

[spaghetti3451]

This is about calculating one of the incoming momenta and the angle of scattering in terms of the Mandelstam variables. I get

$|\vec{k}| = \frac{1}{2}\sqrt{2s+t+u}$

which is one of the incoming three-momenta $|\vec{k}|$ in terms of the Mandelstam variables and

$t-u = 4\ |\vec{k}|\ |\vec{p}|\ \text{cos}\ \theta,$

where $|\vec{p}|$ is one of the final momenta and $\theta$ is the angle of scattering. How do I proceed with the second equation to remove $|\vec{p}|$ and express $\cos\ \theta$ in terms of the Mandelstam variables?

 

[PeterDonis]

Ok, after seeing your work, you weren't doing what I suggested: using conservation laws to obtain the answer. In the center of momentum frame, it should be obvious that the following two facts are true:

(1) The magnitudes of all four 3-momentum vectors are the same (since the masses are equal and both momentum and energy are conserved).

(2) The directions of the incoming pair of 3-momentum vectors are opposite, and the directions of the outgoing pair are opposite; but the two pairs can be pointing in different directions (the difference in the incoming and outgoing directions is of course the scattering angle).

Does that help?

 

[spaghetti3451]

Why should all the four three-momenta have the same magnitude just because their masses are the same?

Conservation law gives us $E_{1}+E_{2}=E_{\text{cm}}=E_{3}+E_{4}$ and $\vec{k}_{1}+\vec{k}_{2}=0=\vec{k}_{3}+\vec{k}_{4}$, where $1$ and $2$ are incoming and $3$ and $4$ are outgoing, so that $\vec{k}_{1}=-\vec{k}_{2}$ and $\vec{k}_{3}=-\vec{k}_{4}$, but not $|\vec{k}_{1}|=|\vec{k}_{3}|$.

 

[PeterDonis]

Why should all the four three-momenta have the same magnitude just because their masses are the same?

The conservation laws you wrote down should tell you. Try writing the energies in terms of the (squared) magnitudes of the momentum vectors.

 

[spaghetti3451]

Right, I get it.

 

[spaghetti3451]

Do you also have some suggestion for managing this calculation:

$\displaystyle{\frac{d\sigma}{d\theta} = \frac{|\mathcal{M}|^{2}}{64\pi^{2}E_{\text{cm}}^{2}}}$

$\displaystyle{= \frac{g^{4}}{64\pi^{2}E_{\text{cm}}^{2}}\left|\frac{i}{s^{2}-m^{2}} + \frac{i}{t^{2}-m^{2}} + \frac{i}{u^{2}-m^{2}}\right|^{2}}$

$\displaystyle{= \frac{g^{4}}{64\pi^{2}E_{\text{cm}}^{2}}\bigg[\frac{1}{(s^{2}-m^{2})^{2}}+\frac{1}{(t^{2}-m^{2})^{2}}+\frac{1}{(u^{2}-m^{2})^{2}}+\frac{2}{(s^{2}-m^{2})(t^{2}-m^{2})}+\frac{2}{(s^{2}-m^{2})(u^{2}-m^{2})}+\frac{2}{(t^{2}-m^{2})(u^{2}-m^{2})}\bigg]}$

$\displaystyle{=\frac{g^{4}}{64\pi^{2}E_{\text{cm}}^{2}}\bigg[\frac{[(t^{2}-m^{2})(u^{2}-m^{2})]^{2}+[(s^{2}-m^{2})(u^{2}-m^{2})]^{2}+[(s^{2}-m^{2})(t^{2}-m^{2})]^{2}}{[(s^{2}-m^{2})(t^{2}-m^{2})(u^{2}-m^{2})]^{2}}+\frac{2[(u^{2}-m^{2})+(t^{2}-m^{2})+(s^{2}-m^{2})]}{(s^{2}-m^{2})(t^{2}-m^{2})(u^{2}-m^{2})}\bigg]}$

 

[PeterDonis]

Do you also have some suggestion for managing this calculation

This is for the same problem?

 

[spaghetti3451]

Yes, for $2\rightarrow 2$ scattering, but now specialised to the case of $\phi^{3}$ theory.

 

[spaghetti3451]

I computed

$|\vec{k}_{1}| = \frac{1}{2}\sqrt{2s+t+u}$

and

$\text{cos}\ \theta = \frac{t-u}{2s+t+u}$.

They don't help, do they?

 

[PeterDonis]

Yes

Have you obtained expressions for $s$, $t$, and $u$ taking into account the conservation laws? For example, $s = E_{cm}^2$, correct?

 

[spaghetti3451]

Yes, they are

$s=E_{\text{cm}}^{2}$

$t=-2|\vec{k}_{1}|^{2}(1-\cos\theta)$

$u=-2|\vec{k}_{1}|^{2}(1+\cos\theta)$.

 

[PeterDonis]

they are

So each one has units of energy squared--if so, it seems to me that the formula in post #10 should have factors like $s - m^2$, $t - m^2$, $u - m^2$, instead of having the squares of $s$, $t$, and $u$.

 

[spaghetti3451]

Yes, they do. But how do you proceed?

 

[PeterDonis]

You can also use the relativistic energy-momentum relation to write $E_{cm}^2 - m^2 = s - m^2 = |\vec{k}|^2 = |\vec{p}|^2$. You can use that to get a relation between $s$ and $t$ (and/or $s$ and $u$).

 

[spaghetti3451]

I see. So, you do have to expand the brackets in the formula and simplify using the relations for $s+t$, $s+u$, $t+u$, etc?

 

[PeterDonis]

you do have to expand the brackets in the formula and simplify using the relations for $s+t$, $s+u$, $t+u$, etc?

That seems like a good thing to try, yes.

 

[spaghetti3451]

Well, I have this so far:

$\displaystyle{\frac{d\sigma}{d\theta} = \frac{|\mathcal{M}|^{2}}{64\pi^{2}E_{\text{cm}}^{2}}}$

$\displaystyle{= \frac{g^{4}}{64\pi^{2}E_{\text{cm}}^{2}}\left|\frac{i}{s-m^{2}} + \frac{i}{t-m^{2}} + \frac{i}{u-m^{2}}\right|^{2}}$

$\displaystyle{=\frac{g^{4}}{64\pi^{2}E_{\text{cm}}^{2}}\bigg[\frac{1}{(s-m^{2})^{2}}+\frac{1}{(t-m^{2})^{2}}+\frac{1}{(u-m^{2})^{2}}+\frac{2}{(s-m^{2})(t-m^{2})}+\frac{2}{(s-m^{2})(u-m^{2})}+\frac{2}{(t-m^{2})(u-m^{2})}\bigg]}$

$\displaystyle{=\frac{g^{4}}{64\pi^{2}E_{\text{cm}}^{2}}\bigg[\frac{[(t-m^{2})(u-m^{2})]^{2}+[(s-m^{2})(u-m^{2})]^{2}+[(s-m^{2})(t-m^{2})]^{2}}{[(s-m^{2})(t-m^{2})(u-m^{2})]^{2}}+\frac{2[(u-m^{2})+(t-m^{2})+(s-m^{2})]}{(s-m^{2})(t-m^{2})(u-m^{2})}\bigg]}$

$\displaystyle{= \frac{g^{4}}{64\pi^{2}E_{\text{cm}}^{2}}\bigg[\frac{[tu-(t+u)m^{2}+m^{4}]^{2}+[su-(s+u)m^{2}+m^{4}]^{2}+[st-(s+t)m^{2}+m^{4}]^{2}}{[(s-m^{2})(t-m^{2})(u-m^{2})]^{2}}+\frac{2[s+t+u-3m^{2}]}{(s-m^{2})(t-m^{2})(u-m^{2})}\bigg]}$

Expanding out the next set of brackets now seems like a foolish thing to try.

 

[PeterDonis]

I'm not sure what your goal is. Sometimes these expressions are complicated. The only things I can see that you haven't done are to write $s$ instead of $E_{cm}^2$ in the denominator of the factor in front, and to try to express $t$ and $u$ in terms of $s$ by using the energy-momentum relations, per one of my previous posts.

 

[spaghetti3451]

I'm not sure what your goal is. Sometimes these expressions are complicated. The only things I can see that you haven't done are to write $s$ instead of $E_{cm}^2$ in the denominator of the factor in front, and to try to express $t$ and $u$ in terms of $s$ by using the energy-momentum relations, per one of my previous posts.

How is $\displaystyle{\frac{d\sigma}{d\theta}}$ usually written?

Is it usually written in terms of $s$?

You can also use the relativistic energy-momentum relation to write $E_{cm}^2 - m^2 = s - m^2 = |\vec{k}|^2 = |\vec{p}|^2$. You can use that to get a relation between $s$ and $t$ (and/or $s$ and $u$).

Actually, I get

$s=E_{\text{cm}}^{2}$

$s-m^{2}=E_{\text{cm}}^{2}-m^{2}$

$s-m^{2}=E_{1}^{2}+E_{2}^{2}-m^{2}$

$s-{m^{2}}=\vec{k}^{1}_{1}+m^{2}+\vec{k}^{2}_{1}+m^{2}-m^{2}$

$s-m^{2}=2\vec{k}^{2}_{1}-m^{2}$,

where $\vec{k}_{1}$ and $\vec{k}_{2}$ are the incoming momenta.

 

[PeterDonis]

I get

You're right, I had misstated the result. But I think your last line should be

$$
s - m^2 = 2 \vec{k}_1^2 + m^2
$$

In the line before you have $+ m^2$ twice and $- m^2$ once, which adds up to $+ m^2$. This would imply

$$
s = 2 \left( \vec{k}_1^2 + m^2 \right)
$$

which is the correct relativistic energy-momentum relation for the incoming pair (or the outgoing pair since all of the squared momenta are the same).

 

[spaghetti3451]

Yes, thanks.

Now,

$t=-2|\vec{k}_{1}|^{2}(1-\cos\theta)$ so that $t = -2s(1-\cos\theta)$

and $u=-2|\vec{k}_{1}|^{2}(1+\cos\theta)$ so that $u=-2s(1+\cos\theta)$.

Therefore,

$\displaystyle{\frac{d\sigma}{d\theta} = \frac{|\mathcal{M}|^{2}}{64\pi^{2}E_{\text{cm}}^{2}}}$

$\displaystyle{= \frac{g^{4}}{64\pi^{2}E_{\text{cm}}^{2}}\left|\frac{i}{s-m^{2}} + \frac{i}{t-m^{2}} + \frac{i}{u-m^{2}}\right|^{2}}$

$\displaystyle{= \frac{g^{4}}{64\pi^{2}E_{\text{cm}}^{2}}\left(\frac{1}{s-m^{2}} + \frac{1}{t-m^{2}} + \frac{1}{u-m^{2}}\right)^{2}}$

$\displaystyle{= \frac{g^{4}}{64\pi^{2}s}\left(\frac{1}{s-m^{2}} + \frac{1}{2s(1-\cos\theta)-m^{2}} + \frac{1}{2s(1+\cos\theta)-m^{2}}\right)^{2}}$

Is this a good form to in which to leave the final answer?

 

[PeterDonis]

Now,

$t=-2|\vec{k}_{1}|^{2}(1-\cos\theta)$ so that $t = -2s(1-\cos\theta)$

and $u=-2|\vec{k}_{1}|^{2}(1+\cos\theta)$ so that $u=-2s(1+\cos\theta)$.

These don't look right since we just got through figuring out that $s = 2 \left( \vec{k}_1^2 + m^2 \right)$, which would mean $2 \vec{k}_1^2 = s - 2 m^2$.

Is this a good form to in which to leave the final answer?

I don't know. I don't know what you want to use the answer for, or what your personal preferences are for what kinds of expressions look neater than others.

 

[spaghetti3451]

Actually, I get

$s=E_{\text{cm}}^{2}$

$s-m^{2}=E_{\text{cm}}^{2}-m^{2}$

$s-m^{2}=E_{1}^{2}+E_{2}^{2}-m^{2}$

$s-{m^{2}}=\vec{k}^{1}_{1}+m^{2}+\vec{k}^{2}_{1}+m^{2}-m^{2}$

$s-m^{2}=2\vec{k}^{2}_{1}-m^{2}$,

where $\vec{k}_{1}$ and $\vec{k}_{2}$ are the incoming momenta.[/QUOTE]

There's a mistake here, I think:

$E_{cm}^{2} \neq E_{1}^{2}+E_{2}^{2}$.

Am I correct?

 

[PeterDonis]

There's a mistake here, I think:

$E_{cm}^{2} \neq E_{1}^{2}+E_{2}^{2}$.

Am I correct?

Yes, there should be an extra $2 E_1 E_2$ term, so $s = \left( E_1 + E_2 \right)^2 = E_1^2 + 2 E_1 E_2 + E_2^2$.

 

[vanhees71]

It's not as difficult as it looks. For elastic $2 \rightarrow 2$ scattering of particles of the same mass, $m$ we have with $p_1,p_2$ being the four-momenta in the initial and $p_3,p_4$ those of the final state
$$s=(p_1+p_2)^2=(p_3+p_4)^2, \quad t=(p_1-p_3)^2=(p_2-p_4)^2, \quad u=(p_1-p_4)^2=(p_2-p_3)^2.$$
Then we have the on-shell conditions $p_1^2=p_2^2=p_3^2=p_4^2=m^2$.

Now let's specialize to the CM frame, where $\vec{p}_2=-\vec{p}_1=-\vec{p}$ and $\vec{p}_4=-\vec{p}_3=-\vec{p}'$. In this frame we have
$$s=(E_1+E_2)^2=4 E_1^2=4 E_3^2 \; \Rightarrow \; |\vec{p}'|=|\vec{p}|$$
and
$$s=4(m^2+\vec{p}^2) \; \Rightarrow \; |\vec{p}|=|\vec{p}'|=\frac{1}{2} \sqrt{s-4m^2}.$$
The only other parameter of the scattering in the CM frame is the scattering angle $\vartheta$, defined by
$$\vec{p} \cdot \vec{p}'=|\vec{p}|^2 \cos \vartheta.$$
This angle must be related to the Mandelstam variable $t$ (or $u$, but $s+t+u=4m^2$).Indeed we have
$$t=(p_1-p_3)^2=2 m^2-2 p_1 \cdot p_3=2m^2-2(E_1 E_3-\vec{p} \cdot \vec{p}')=2 (m^2-E_1^2+|\vec{p}|^2 \cos \vartheta) = -2 |\vec{p}|^2 (1-\cos \vartheta)=-\sqrt{s-4m^2} (1-\cos \vartheta).$$
The manifestly covariant form of the differential cross section, expressed in Mandelstam variables only reads
$$\frac{\mathrm{d} \sigma}{\mathrm{d} t}=\frac{1}{64 \pi s |\vec{p}|^2} \mathcal{M}(s,t).$$
For the more general case of $2 \rightarrow 2$ scattering (and more kinematics), see
http://pdg.lbl.gov/2016/reviews/rpp2016-rev-kinematics.pdf

 

[spaghetti3451]

It's not as difficult as it looks. For elastic $2 \rightarrow 2$ scattering of particles of the same mass, $m$ we have with $p_1,p_2$ being the four-momenta in the initial and $p_3,p_4$ those of the final state
$$s=(p_1+p_2)^2=(p_3+p_4)^2, \quad t=(p_1-p_3)^2=(p_2-p_4)^2, \quad u=(p_1-p_4)^2=(p_2-p_3)^2.$$
Then we have the on-shell conditions $p_1^2=p_2^2=p_3^2=p_4^2=m^2$.

Now let's specialize to the CM frame, where $\vec{p}_2=-\vec{p}_1=-\vec{p}$ and $\vec{p}_4=-\vec{p}_3=-\vec{p}'$. In this frame we have
$$s=(E_1+E_2)^2=4 E_1^2=4 E_3^2 \; \Rightarrow \; |\vec{p}'|=|\vec{p}|$$
and
$$s=4(m^2+\vec{p}^2) \; \Rightarrow \; |\vec{p}|=|\vec{p}'|=\frac{1}{2} \sqrt{s-4m^2}.$$
The only other parameter of the scattering in the CM frame is the scattering angle $\vartheta$, defined by
$$\vec{p} \cdot \vec{p}'=|\vec{p}|^2 \cos \vartheta.$$
This angle must be related to the Mandelstam variable $t$ (or $u$, but $s+t+u=4m^2$).Indeed we have
$$t=(p_1-p_3)^2=2 m^2-2 p_1 \cdot p_3=2m^2-2(E_1 E_3-\vec{p} \cdot \vec{p}')=2 (m^2-E_1^2+|\vec{p}|^2 \cos \vartheta) = -2 |\vec{p}|^2 (1-\cos \vartheta)=-\sqrt{s-4m^2} (1-\cos \vartheta).$$
The manifestly covariant form of the differential cross section, expressed in Mandelstam variables only reads
$$\frac{\mathrm{d} \sigma}{\mathrm{d} t}=\frac{1}{64 \pi s |\vec{p}|^2} \mathcal{M}(s,t).$$
For the more general case of $2 \rightarrow 2$ scattering (and more kinematics), see
http://pdg.lbl.gov/2016/reviews/rpp2016-rev-kinematics.pdf

Thanks.

So, my goal is to try and simplify the following expression:

$\displaystyle{\frac{d\sigma}{d\theta} = \frac{|\mathcal{M}|^{2}}{64\pi^{2}E_{\text{cm}}^{2}}}$

$\displaystyle{= \frac{g^{4}}{64\pi^{2}E_{\text{cm}}^{2}}\left|\frac{i}{s-m^{2}} + \frac{i}{t-m^{2}} + \frac{i}{u-m^{2}}\right|^{2}}$

$\displaystyle{= \frac{g^{4}}{64\pi^{2}E_{\text{cm}}^{2}}\left(\frac{1}{s-m^{2}} + \frac{1}{t-m^{2}} + \frac{1}{u-m^{2}}\right)^{2}}$

and then to find out which angles dominate the cross-section in the high-energy limit $\sqrt{s} \gg m$.

How do you simplify the expression first? I think the trick is to express $t$ and $u$ in terms of $s$. Is this so?