# Relativistic energy = four-force times displacement four-vector?

In mechanics energy is $$E = \frac{m v^2}{2}$$

The corresponding relativistic equation is $$E = m (\gamma -1) c^2$$ which for v<<c is appoximately $$\frac{m v^2}{2}$$

I know that the equation above is correct because I have seen the derivation at Wikipedia.

But energy can also be calculated by $$E = f d$$

The corresponding relativistic equation would be four-force times displacement four-vector (i.e. four-position)

$$E = \left(\gamma {\mathbf{f}\cdot\mathbf{v} \over c},\gamma{\mathbf f}\right) \cdot \left(ct, \mathbf{r}\right)$$

Is there a way to show that this second relativistic equation gives a value for energy that doesnt contradict the first equation above?

($$ct$$ has units of distance. $$\frac{v}{c}$$ is dimensionless and so is $$\gamma$$)

f is the rate of change of proper momentum (mass times proper velocity)

$${\mathbf f}={\mathrm{d} \over \mathrm{d}t} \left(\gamma m {\mathbf v} \right)={\mathrm{d}\mathbf{p} \over \mathrm{d}t}$$

and

$${\mathbf{f}\cdot\mathbf{v}}={\mathrm{d} \over \mathrm{d}t} \left(\gamma mc^2 \right)$$

The derivative of gamma is:

$$\dot\gamma = \frac{d \gamma}{d t} = \frac{d \gamma}{dv} \frac{dv}{dt} = \frac{v \gamma^3 a}{c^2}$$

• There is no term "proper velocity". If it would exist then it would be the zero 3-vector. If you mean proper 4-velocity vector then it would be $\mathbf U_0 \boldsymbol{=}(c, \boldsymbol{0})$. Dec 21, 2020 at 14:15
• @Frobenius sometimes “celerity” is called “proper velocity”: en.m.wikipedia.org/wiki/Proper_velocity But I think “proper velocity” is terrible terminology and would always use the term “celerity” instead.
– Dale
Dec 21, 2020 at 14:32
• @Dale : Thanks for the info. Dec 21, 2020 at 14:39

It's pretty easy to see that if you take the inner product of the force four-vector with a displacement four-vector, you don't get a correct expression for the mechanical work done by the force. This is because the inner product of two four-vectors is a scalar, which is the same in all reference frames. But energy obviously depends on your frame of reference. A more compact way of expressing this is that in terms of three-vectors, $$\textbf{F}\cdot\textbf{v}$$ is an expression for the power, while in terms of four-vectors, this expression vanishes identically.

It is true, however, that if you express work in terms of the force and displacement three-vectors, the result is relativistically valid, and you don't need to introduce factors of gamma or anything like that. There is a compact proof of this fact (here given in one dimension):

$$\frac{ d E}{ d x} = \frac{ d E}{ d p}\frac{ d p}{ d t} \frac{ d t}{ d x} = \frac{ d E}{ d p} \frac{F}{v}$$

The desired result follows from application of the identity $$dE/dp=v$$.

For a more detailed discussion of this kind of thing, see ch. 4 of my SR book, http://lightandmatter.com/sr/ .

To reconcile the 4D calculation, you need $$(dE/dt,dp/dt)\cdot(dt,dx)$$ for infinitesimal 4-displacement

Take the units where $$c=1$$. It's better to regard $$E=m\gamma$$ instead of $$E=m(\gamma-1)$$ although it's just a constant shift. The Minkowski-norm $$(E,p)\cdot (E,p)=E^2-p^2=m^2\gamma^2-m^2 v^2\gamma^2=m^2$$ is a constant, the static mass squared as expected. So the differential of that is identically 0, which gives $$0=2(dE/dt,dp/dt)\cdot(E,p)=2(E\;dE/dt-dp/dt\cdot p).$$ So you arrive at $$dE/dt=dp/dt \cdot (p/E)=dp/dt\cdot dx/dt,$$ which is consistent with $$(dE/dt,dp/dt)\cdot(dt,dx)=0$$.