Prove 4*s*k^2 = (s-m^2)^2 | Mandelstam Variables

[Manojg]

Hi,

For collision: k + p1 -> q + p2, where k, p1, q and p2 are 4 Vectors of the colliding particles.

Any body show me how to prove this expression:

4*s*k*k = (s - m*m)^2, where s = (k+p1)^2 and m^2 = p1^2: mass^2 of the particle p1.Thanks.

 

[malawi_glenn]

give it a try yourself first, then we help you.

 

[Manojg]

I tried it.
For an example, if k is photon then
K*k = 0 and it gives
s - m*m = 0;
k*k + p1*p1 + 2kp1 - m*m = 0;
0 + m*m + 2kp1 - m*m = 0;
kp1 = 0;
E_photon * E_p1 - P_photon * P_p1 = 0.

if p1 is at rest then
E_photon * m = 0
which is wrong.

 

[Meir Achuz]

s equals (k+p1)^2 is the 4-momentum squared. Then
$$s=(k+E_1)^2-{\bf k}^2$$.
This assumes the particle of mass m is initially at rest (lab system).

 

[humanino]

By Mandelstam's definition, $$s=(k+p_{1})^2$$ in $$k p_{1}\rightarrow q p_{2}$$ (total energy squared in the CM frame)

$$(s-m^{2})^{2}=\left[(k+p_{1})^{2}-p_{1}^{2}\right]^{2}=\left[k^{2}+2kp_{1}\right]^{2}=4k^{2}\left[\frac{k}{2}+p_{1}\right]^{2}\neq4k^{2}s$$

Why would that supposedly be right ?

 

[Manojg]

Sorry for late reply.

I got this equation from papers Phy. Rev. 129, 2264(1963) (equation # 5) and Phy. Rev. 123, 1882(1961) (equation # 2.8).

 

[humanino]

I got this equation from papers Phy. Rev. 129, 2264(1963) (equation # 5) and Phy. Rev. 123, 1882(1961) (equation # 2.8).

Both of them being photoproduction : $$k^{2}=0$$ which is kind of useful in this context.
I understand your confusion now. Look at first row of equations in 2.4. in the CM in Phys. Rev. 123, 1882 - 1887 (1961)
$$k=(|\vec{k}|,\vec{k})$$ , $$p_1=(\epsilon_1,-\vec{k})$$
Now it is easy to see that
$$s=(k+p_{1})^{2}=_{\text{CM}}(|\vec{k}|+\epsilon_1)^2-\vec{0}^2$$
since the total momentum in the CM is zero (I explicitly wrote the norme squared of the null vector).
$$(s-m^{2})^{2} = \left[(k+p_{1})^{2}-p_{1}^{2}\right]^{2} = \left[k^{2}+2kp_{1}+p_{1}^{2}-p_{1}^{2}\right]^{2} = \left[2kp_{1}\right]^{2} = 4\left[|\vec{k}|\epsilon_1+\vec{k}^2\right]^{2} = 4\vec{k}^{2} \left[\epsilon_1+|\vec{k}| \right]^{2} = 4\vec{k}^2s$$

So if you followed, the $$k^{2}$$ drops by itself in the expansion of $$s^{2}$$, the $$p_1^{2}$$ cancels with $$m^{2}$$ which we put in, and we end up only with the double cross product, which we then write in the CM. This is the CM momentum you have in your mysterious 2.8 ! I would say it is confusing, because the paper should have taken care to write a bold-face k, which is really $$|\vec{k}|_{\text{CM}}$$ instead of $$k^{2}=\frac{(s-m^{2})^{2}}{4s}$$, which when interpreted as a 4-vector squared and plugged $$k^{2}=0$$ in 2.8 would give $$s=m^{2}$$, which is not true.

Note something important : we have obtained $$(s-m^2)^2=4|\vec{k}|^2s$$ in the CM but this does not depend on the reference frame, so is valid anywhere.

 

[humanino]

BTW, I'm glad to see how people read my posts

$$\left[k^{2}+2kp_{1}\right]^{2}=4k^{2}\left[\frac{k}{2}+p_{1}\right]^{2}$$

Wrong !

That should have been :
$$\left[k^{2}+2kp_{1}\right]^{2}=4k^{2}\left[\frac{k}{2\sqrt{k^2}}+\frac{1}{\sqrt{k^2}}kp_{1}\right]^{2}$$
which obviously I can't do anymore.

 

[Manojg]

Thanks a lot humanino.