# How to derive this relativistic Energy-Momentum relation?

## Plz explain in detail , how to derive this relation : ${E}^{2} = {p}^{2} {c}^{2} + {m}^{2} {c}^{4}$ where all the symbols have their usual meanings.

Apr 19, 2018

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

#### Explanation:

${p}^{2} = {m}^{2} {v}^{2} = {\gamma}^{2} {m}_{o}^{2} {v}^{2}$

$\implies {p}^{2} \cdot {c}^{2} + {m}_{o}^{2} {c}^{4}$

$= \frac{{m}_{o}^{2} {v}^{2} {c}^{2}}{1 - {v}^{2} / {c}^{2}} + {m}_{o}^{2} {c}^{4}$

$= {m}_{o}^{2} {c}^{2} \left(\frac{{v}^{2}}{1 - {v}^{2} / {c}^{2}} + {c}^{2}\right)$

$= {m}_{o}^{2} {c}^{2} \left(\frac{{v}^{2}}{1 - {v}^{2} / {c}^{2}} + {c}^{2} \cdot \frac{1 - {v}^{2} / {c}^{2}}{1 - {v}^{2} / {c}^{2}}\right)$

$= {m}_{o}^{2} {c}^{2} \left(\frac{{v}^{2}}{1 - {v}^{2} / {c}^{2}} + \frac{{c}^{2} - {v}^{2}}{1 - {v}^{2} / {c}^{2}}\right)$

$= {m}_{o}^{2} {c}^{4} \left(\frac{1}{1 - {v}^{2} / {c}^{2}}\right) = {\gamma}^{2} {m}_{o}^{2} {c}^{4} = {E}^{2}$

Apr 19, 2018

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 $\left(\frac{E}{c} , {p}_{x} , {p}_{y} , {p}_{z}\right) \equiv \left(\frac{E}{c} , \vec{p}\right)$ form a 4-vector, called, rather unimaginatively, the energy-momentum 4-vector . This is a generalization to four dimensions of the notion of ordinary, or 3-vectors. Just like the components of all 3-vectors (like force, momentum, velocity, ...) transform like the coordinates are rotated, components of all 4-vectors transform just like the prototype 4-vector $\left(c t , x , y , z\right)$ under a Lorentz transformation - the rule which describes how measurements change from one inertial observer to another.

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 $\left({A}^{0} , {A}^{1} , {A}^{2} , {A}^{3}\right)$ this is given by

${\left({A}^{0}\right)}^{2} - {\left({A}^{1}\right)}^{2} - {\left({A}^{2}\right)}^{2} - {\left({A}^{3}\right)}^{3}$

By the way, the 0 in ${A}^{0}$, for example, is not a power - the index is written as a superscript instead of the customary subscript. Why? That's a whole new story - which we do not need to go into here.

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

${E}^{2} / {c}^{2} - {p}^{2}$

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 $E = m {c}^{2} , \vec{p} = 0$, so that in this frame the invariant is

${\left(\frac{m {c}^{2}}{c}\right)}^{2} - {0}^{2} = {m}^{2} {c}^{2}$

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

${E}^{2} / {c}^{2} - {p}^{2} = {m}^{2} {c}^{2}$

from which the relation follows trivially!