Relating classical and relativistic energy&work

Wox

Can work and energy in special relativity be described by drawing the analogy with classical physics as shown below?

$\bar{F}$: four force
$\bar{v}$: four velocity
$\tilde{F}$: classical three force
$\tilde{v}$: classical three velocity
$\Psi$: electromagnetic tensor

A. Classical
The work done by the classical force $\tilde{F}$ as derived in classical physics
$W(t)=\int_{t_{0}}^{t} \tilde{F}\cdot \tilde{v} dt=m\int_{t_{0}}^{t} \frac{d\tilde{v}}{dt}\cdot \tilde{v} dt=m\int_{\tilde{v}(t_{0})}^{\tilde{v}(t)} \tilde{v}d\tilde{v}=\frac{m\tilde{v}(t)^{2}}{2}-\frac{m\tilde{v}(t_{0})^{2}}{2}$

Furthermore if $\tilde{F}$ is conservative then (using the gradient theorem)
$W(t)=\int_{t_{0}}^{t} \tilde{F}\cdot \tilde{v} dt=-\int_{t_{0}}^{t} \tilde{\nabla}E_{pot}\cdot \tilde{v} dt=-\Delta E_{pot}$

From this we define the total energy of an object in a force field as
$E_{tot}(t)=\frac{m\tilde{v}(t)^{2}}{2}+E_{pot}(t) \equiv E_{kin}(t)+E_{pot}(t)$

A. Relativistic
I will try to do the same thing as in classical physics, but now using these relativistic relations:
1. Relation between four and three force:
$\bar{F}=(mc\gamma\frac{d\gamma}{dt},m\gamma\frac{d\gamma\tilde{v}}{dt})$
$\bar{F}=q\Psi \bar{v}$
$\Leftrightarrow \bar{F}=(mc\gamma\frac{d\gamma}{dt},\gamma\tilde{F})$ where $\tilde{F}=q(\tilde{E}+\tilde{v}\times\tilde{B})$
2. Four force and four velocity are orthogonal:
$\bar{v}=(c\gamma,\gamma\tilde{v})$
$<\bar{F},\bar{v}>=0\Leftrightarrow \tilde{F}\cdot \tilde{v}=mc^{2}\frac{d\gamma}{dt}$

The work done by the classical force $\tilde{F}$ as derived in special relativity
$W(t)=\int_{t_{0}}^{t} \tilde{F}\cdot \tilde{v} dt=mc^{2}\int_{\gamma(t_{0})}^{\gamma(t)}d\gamma=m\gamma(t)c^{2}-m\gamma(t_{0})c^{2}$

Furthermore if $\tilde{F}$ is conservative then (using the gradient theorem)
$W(t)=\int_{t_{0}}^{t} \tilde{F}\cdot \tilde{v} dt=-\int_{t_{0}}^{t} \tilde{\nabla}E_{pot}\cdot \tilde{v} dt=-\Delta E_{pot}$

From this we define the total energy of an object in a force field as
$E_{tot}(t)=m\gamma(t)c^{2}+E_{pot}(t) \equiv mc^{2}+E_{kin}(t)+E_{pot}(t)$

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pervect

Staff Emeritus
I didn't spot any obvious errors (doesn't mean that I didn't miss some). Note that if you try to apply this approach to electromagnetic forces, it will work only insofar as you have an unchanging electromagnetic field that isn't affected by the motion of the test particle, as there is no concept yet of "field energy" in the approach you outlined.

You can go a bit further with a Lagrangian approach, see for instance Goldstein "Classical Mechanics".

Wox

My main problem is that the relativistic $E_{tot}$ doesn't converge to the classical $E_{tot}$ for low speeds, meaning that they aren't describing the same thing. I think it has something to do with taking internal energy into account or not. But why does it pop-up in the relativistic $E_{tot}$ and not in the classical $E_{tot}$?

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Chestermiller

Mentor
My main problem is that the relativistic $E_{tot}$ doesn't converge to the classical $E_{tot}$ for low speeds, meaning that they aren't describing the same thing. I think it has something to do with taking internal energy into account or not. But why does it pop-up in the relativistic $E_{tot}$ and not in the classical $E_{tot}$?
The classial Etot usually omits the internal energy, since it doesn't change in lots of problems. It does change in inelastic collisions, however.

Chet

Wox

But how is it that the internal energy pops-up in the relativistic analogue to the classical definition of energy, while it doesn't in the classical definition? I mean, I can see that it does pop up, but why? The difference between the classical and the relativistic is the presence of $\gamma$ in $\tilde{F}=m\frac{d\gamma\tilde{v}}{dt}$. Is there a way to understand that if we omit $\gamma$, we're "neglecting the internal energy" in some way.

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Chestermiller

Mentor
But how is it that the internal energy pops-up in the relativistic analogue to the classical definition of energy, while it doesn't in the classical definition? I mean, I can see that it does pop up, but why? The difference between the classical and the relativistic is the presence of $\gamma$ in $\tilde{F}=m\frac{d\gamma\tilde{v}}{dt}$. Is there a way to understand that if we omit $\gamma$, we're "neglecting the internal energy" in some way.
No. The classical treatment frequently omits the internal energy in mechanics problems because it often does not change. However, in general thermodynamics analyses, the internal energy, the kinetic energy, and the potential energy are all typically included in the energy balance. However, the thermodynamic analyses only look at changes in total energy, and not its absolute value. It took Einstein to realize that, for a given frame of reference, the total energy can be regarded as an absolute quantity like absolute temperature, and that the absolute internal energy is equal to mc2.

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