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

Question

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:

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

Answer 1

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/ .

Answer 2

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 $(dE/dt,dp/dt)\cdot(dt,dx)=0$.