Zwiebach Section 12.4 Homework: M^2 = -p^2

[ehrenfest]

Homework Statement

On page 221 Zwiebach uses the "definition" M^2 = -p^2. I am not sure where this comes from since normally

m^2c^4 + p^2 c^2 = E^2

and even dropping the c's does not reduce to that.

EDIT: I see. It is the light-cone Lorentz generator of section 11.6. How is he getting the square of it though? Working it out with 11.76?

EDIT 2: Wrong again. It is the just the energy-momentum invariant since the p I wrote above was only a three-vector. So, why is it a capital M?

Homework Equations

The Attempt at a Solution

 

[nrqed]

Homework Statement

On page 221 Zwiebach uses the "definition" M^2 = -p^2. I am not sure where this comes from since normally

m^2c^4 + p^2 c^2 = E^2

and even dropping the c's does not reduce to that.

Don't confuse four-momenta with three-momenta. In M^2 = - P^2, P is a four-momentum. In your seond equation p is the three-momentum.

In any case, depending on the metric used, the square of the four-momentum may either be taken as P^2 = E^2-p^2 OR as p^2-E^2 which gives either +m^2 or -m^2 (setting c=1 everywhere)