The relativistic energy–momentum relation describes how an object’s total energy is determined by both its momentum (motion through space) and its invariant mass (rest mass).
To derive the relation, consider the relativistic definition of momentum in terms of proper time,
where 
is the rest mass and 
is the proper time measured by a clock moving with the object.
According to time dilation, 
, where 
. Since the object’s velocity in the observer’s frame is 
, we have 
. Therefore,
From the definition of work, 
, or equivalently, 
, since 
and 
. Differentiating eq254a throughout yields
=md\biggr\(\frac{v}{\sqrt{1-\frac{v^2}{c^2}}}\biggr\))
Question
Show that =\gamma^3dv)
and 
.
Answer
}{dv}&=\biggr\(1-\frac{v^2}{c^2}\biggr\)^{-\frac{1}{2}}+v\biggr\(-\frac{1}{2}\biggr\)\biggr\(1-\frac{v^2}{c^2}\biggr\)^{-\frac{3}{2}}\biggr\(-\frac{2v}{c^2}\biggr\)\\&=\biggr\(1-\frac{v^2}{c^2}\biggr\)^{-\frac{3}{2}}\biggr\(1-\frac{v^2}{c^2}+\frac{v^2}{c^2}\biggr\)=\frac{1}{\biggr\(1-\frac{v^2}{c^2}\biggr\)^{\frac{3}{2}}}=\gamma^3\end{align})
Therefore, . Also,
\biggr\(1-\frac{v^2}{c^2}\biggr\)^{-\frac{3}{2}}\biggr\(-\frac{2v}{c^2}\biggr\)=\frac{\gamma^3v}{c^2})
Substituting 
and into results in 
, or its integrated form
where 
is a constant.
To satisfy Einstein’s mass-energy equivalence relation 
, where is the rest mass, eq254b must give when 
. Since 
at , this requires 
. Thus, eq254b becomes
From eq254a, 
. Substracting this from the square of eq254c gives
Since 
,
=m^2c^4)
Therefore,
This is the relativistic energy–momentum relation.
Interestngly, 
has two solutions. The negative-energy solution seems problematic because ordinary particles are expected to have positive energy. However, it cannot consistently be discarded. This eventually led Paul Dirac to interpret the negative solution as corresponding to antiparticles.