# How to derive this relativistic Energy-Momentum relation?

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Plz explain in detail , how to derive this relation :

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

where all the symbols have their usual meanings.

Plz explain in detail , how to derive this relation :

where all the symbols have their usual meanings.

##### 2 Answers

#### Answer:

The algebra is below. In terms of explanation, the point to get an equation in momentum, so it is really just manipulation.

#### Explanation:

#### Answer:

If you happen to know about more advanced stuff like 4-vectors, the proof of this relation becomes a lot simpler.

#### Explanation:

The four quantities

One important consequence is that just like the length of a 3-vector that does not change when you rotate your coordinate system, there is an invariant "length" that does not change when you change the inertial observer. The only difference is that the square of this "length", instead of the sum of squares, is a **signed sum**. For a 4-vector

By the way, the 0 in

Now, for the energy-momentum 4-vector, this invariant is

Being invariant, this is the same in all inertial frames. In particular, its value is the same in the frame in which the particle is (at least instantaneously) at rest. In this frame

Since the invariant is the **same** in all frames, we must have

from which the relation follows trivially!