Deriving the Energy-Momentum Formula

[Rulesby]

Homework Statement

Show that the energy-momentum relationship, E^2 = p^2 * c^2 + (m*c^2)^2, follows from the expressions E = (gamma)*m*c and p = (gamma)*m*u

where

(gamma) = 1 / sqrt(1 - (u^2)/(c^2)) the lorentz transformation factor.
m is the rest mass.
c is the speed of light
u is the velocity of the particle
E is the total energy
p is the momentum

The book does not teach about relativistic mass, so I think I supposed to derive this without making a distinction between m and m0.

Homework Equations

(1): E^2 = p^2 * c^2 + (m*c^2)^2
(2): E = (gamma)*m*c
(3): p = (gamma)*m*u

The Attempt at a Solution

When the chapter introduces the formula E^2 = p^2 * c^2 + (m*c)^2, it does not show how it derived this equation. Instead it says that it just says "By squaring [equations (2) and (3)] and subtracting, we can eliminate u. The result after some algebra is [equation (1)]."

My first attempt was to start by squaring both sides of equation (3).

p^2 = (gamma)^2*m^2*u^2

Then get it in terms of m^2.

m^2 = (p^2)/(u^2) - (p^2)/(c^2)

In equation (2), square both sides, then substitute m^2 to get:

E^2 = (gamma)^2 * [(p^2)/(u^2) - (p^2)/(c^2)] * c^2

After some algebra I got:

E^2 = [(gamma)^4 * m^2 * c^4] - [(gamma)^2 * p^2 * c^2]

This is nearly what I'm trying to derive, however, the (gamma) terms are still there and I don't know how to get rid of them.

 

[Rulesby]

Nevermind, I found the solution.

https://www.physicsforums.com/showthread.php?t=64683