# Mandelstam variable equality derivation for $s$-channel process

I am supposed to show that the Mandelstam variable

$s=(p_1^2 + p_2^2) = 4(P_{cm}^2 + m^2)$

for an electron/positron interaction, where $p_1$ is the incident 4-momentum of the electron and $p_2$ is the incident 4-momentum of the positron, $m$ is the electron/positron mass, and $P_{cm}$ is the magnitude of the 3-momenta of the incident and scattered particles in the center of mass frame. The derivation that my professor gave is as follows:

$(p_1^2 + p_2^2) = (E_1 + E_2)^2 - (\vec p_1 + \vec p_2)^2$

In the center of mass frame, $\vec p_1 + \vec p_2 =0$ (since each are the regular 3-momenta of the particles), so

$=(E_1 + E_2)^2 = (2E)^2$ since the energy of each particle is the same in the center of mass frame we just express each energy as $E$.

But then the step I don't understand is the following:

$E^2 = P_{cm}^2 + m^2$

which gets plugged in to $(2E)^2$ to get the final answer. I believe the equation in the line above comes from the relativistic energy-momentum relation but I don't see how we can change from the energy $E$ in the center of mass frame to $P_{cm}^2$, which generally is not in the center of mass frame. Is $E^2$ a Lorentz invariant? If so, I can understand how it works in to the derivation.

Edit: as pointed out, $P_{cm}$ is not the momentum of the cneter of mass but the momenta int he center of mass frame. This means that the above paragraph is irrelevant. Thus, the energy-momentum relation holds because $P_{cm}$ is the momentum of each particle in the center of mass frame and so there is no jumping between frames as I thought previous.

• I think, by $P_{cm}$ your professor might mean momentum of the two particles in the COM frame $P_{cm} = \frac{\vec{p}_i + \vec{p}_2}{2}$ rather than momentum of center of mass. In that case, in such a frame $E_{cm}^2 = P_{cm}^2 + m^2$ – Siva Jan 20 '14 at 23:18
• By golly, I think you're right. I still don't understand the step that I didn't understand before, so my question still stands. – NeutronStar Jan 20 '14 at 23:20
• My comment is slightly incoherent. I've expanded in the answer below. – Siva Jan 20 '14 at 23:30
• Joshua: "I am supposed to show that the Mandelstam variable $s=(p^2_1+p^2_2)=$ ..." -- If you're using instead the definition of the Mandelstam variable $s := (p_1 + p_2)^2$, as indicated for instance here, then showing what you're supposed to show might become more ... compelling. – user12262 Jan 21 '14 at 17:22

$(p_1^2 + p_2^2) = (E_1 + E_2)^2 - (\vec p_1 + \vec p_2)^2$

This equation should hold in any inertial frame, so let's boost into the COM frame. In the center of mass frame (by definition), $\vec p_1 + \vec p_2 =0$ (since each are the regular 3-momenta of the particles), so

$=(E_1 + E_2)^2 = (2E)^2$ since the energy of each particle is the same in the center of mass frame we just express each energy as $E$.

Note that all the calculations so far have been in the COM frame. So we need the energies of the particles, in this frame. We know their masses, so we need the momenta of the particles, in this frame.

Momentum of one of the particles in the COM frame $P^1_{cm} \equiv p_1 - \dfrac{\vec{p}_1 + \vec{p}_2}{2}$ rather than momentum of center of mass. The other particle will have the opposite 3-momentum, which means that they'll both have the same energy in this frame: $$E_{cm}^2 = P_{cm}^2 + m^2$$

But then the step I don't understand is the following:

$E^2 = P_{cm}^2 + m^2$

which gets plugged in to $(2E)^2$ to get the final answer.