# Mandelstam variables

1. Jun 20, 2008

### 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.

Last edited: Jun 20, 2008
2. Jun 21, 2008

### malawi_glenn

3. Jun 22, 2008

### 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.

4. Jun 27, 2008

### clem

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).

5. Jun 27, 2008

### 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 ?

Last edited: Jun 27, 2008
6. Jul 14, 2008

### Manojg

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

7. Jul 14, 2008

### humanino

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.

Last edited: Jul 14, 2008
8. Jul 14, 2008

### humanino

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.

Last edited: Jul 14, 2008
9. Jul 16, 2008

### Manojg

Thanks a lot humanino.