I was faced with a situation where I suddenly realized that although Kinetic Energy in Special Relativity is defined as

$KE=\gamma m_0 c^2 - m_0 c^2$

The work energy theorem says

$\Delta KE +\Delta PE_{atomic,electric,…}=W$

Potential energy can’t be gravitational in special relativity. The work energy theorem is expressed in the lab frame, so if enough of the initial or final values are known in the lab frame, simple. If the total is known in the moving frame and the velocity and position and time are also known, the solution in the lab frame is also straightforward and well-defined.

What if we want to use conservation of energy to compute something about the motion of the moving frame? If work is done, that means a force is applied and the moving frame is accelerated. If we know the force in the lab frame, we can integrate the work and find the KE in the lab frame. Then the straightforward relativistic formula applies.

But in practice, do we usually know forces in the lab frame??? I feel like it may be more normal to know them in the accelerated frame.

None of forces, work, nor KE are components of a 4-vector and they are not Lorentz covariant. Consider for example Forces due to E and B fields. The magnetic vector potential A is a 4-vector, but the Force has mixed dependencies (Maxwells equations) due to induction.

I’m thoroughly surprised by how complex this transformation, in pursuit of an equation of motion based on the work, appears to be. To find the equations of motion, it is necessary to have the velocity of the moving (accelerated) frame as a function of its time $\tau$ or the lab time $t$, from which one can find the time $t(\tau)$ and the position $x(\tau)$, not discussed here.

Consider Newton’s second law in the accelerated frame, in an attempt to characterize this.

Define time to always be in the accelerated frame, so time is $\tau$. Allow mass to change with time

$F(\tau)=m(\tau) a(\tau)$

$\frac{dv(\tau)}{d\tau} =a(\tau)=\frac{d^2 x(\tau)}{d\tau^2} $

What is the time $t$ in the lab frame, if the accelerated frame is moving at speed $v$? It is necessary to be careful about who is measuring the relative velocity $v$ of the two frames, since the times are not the same in the two frames and $v$ is a function of time. So if the current velocity is $v(\tau)$ or $v(t)$ then

$t=\gamma(v(\tau)) (\tau-\frac{v(\tau)x(\tau)}{c^2})$ so


But this is actually a total derivative isn’t it? So this is where the first problem is

Moving on, rather than correcting this issue, to show the structure of the argument.


Again, the total derivative

Take the total second derivative to get the total second derivative of the acceleration and use these quantities to relate it to the force $F(\tau)$ and the mass $m(\tau)$. But I believe this yields a differential equation in $v(t)$ or $x(t)$, or alternatively in $v(\tau)$ and $x(\tau)$.



Where $\gamma$ depends implicitly on $v(\tau)$ Would be the answer if acceleration were a partial derivative but since it is a total derivative it is not. I have to admit I haven’t finished that calculation. But this is sufficient demonstration to ask my question.

So it’s a differential equation to obtain the equation of motion in the lab frame, while usually while using the Kinetic Energy and Work we can simply solve a quadratic equation. I am wondering the utility of the work energy theorem and if there is something else it is used for and how and why? Also, is it really this complex to get the equations of motion if the force and mass are allowed to vary in the accelerated frame?

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    $\begingroup$ Please edit your post using MathJax for formulas. $\endgroup$
    – Hyperon
    Commented Apr 12 at 7:57
  • $\begingroup$ Okay hold on I’ll try to. This will be my first time learning it. $\endgroup$ Commented Apr 12 at 8:00
  • $\begingroup$ Of course I know LaTeX. I don’t know much about bulletin boards. $\endgroup$ Commented Apr 12 at 8:03
  • $\begingroup$ It is not even half a day ago that I just wrote about how Work Energy Theorem is nonsense in Special Theory of Relativity, and that no textbook author even wants to present it. Technically, you can still make it work, but it is so ugly that nobody wants to. Work is not a well-defined concept, especially if you consider statistical thermodynamics with magnetism, so you probably should take a hint and just move on. The fundamental facts of our universe is that energy and momentum are conserved, and we use them to understand our universe. $\endgroup$ Commented Apr 12 at 8:12
  • $\begingroup$ Please clarify what exactly you are trying to do. In SR the equation $F = \mathrm{d}E/\mathrm{d}x$ is still true in inertial frames, as long as the correct relativistic expressions for $F$ and $E$ are used. Conservation of energy itself is ill-defined in accelerating frames, even in newtonian mechanics. $\endgroup$ Commented Apr 12 at 8:53

3 Answers 3


But in practice, do we usually know forces in the lab frame??? I feel like it may be more normal to know them in the accelerated frame.

Yes, we do. The "lab frame" is really an inertial frame, where Maxwell's equations hold, so we know how EM fields $\mathbf E,\mathbf B$ behave, and also the Lorentz force formula holds:

$$ \mathbf F = q\mathbf E + q\mathbf v \times \mathbf B. $$

Expressing force and equations of motion is harder in accelerated frames, because fields are weird (do not obey Maxwell's equations) and in such frames, conservation of energy does not hold in any usual sense, because of the additional inertial forces (sometimes called "fictitious forces") supplying whatever energy is needed from nothing.

I guess the original context is that I’m trying to understand why anyone would use the concept of kinetic energy at all in special relativity. The work-energy theorem doesn’t seem to help

The work-energy theorem is fine in special relativity, provided we use the correct kinetic energy $K= (\gamma - 1)mc^2$. Work of net force on point particle equals change of its kinetic energy:

$$ \int_{\mathbf r_1}^{\mathbf r_2} \mathbf F \cdot d\mathbf x = K_2 - K_1. $$

  • $\begingroup$ I meant if a force was a thrust of a rocket, for example, although there could be other kinds of forces. That might not be a priori known in the lab frame. $\endgroup$ Commented Apr 14 at 1:36
  • $\begingroup$ @StevenDorsher OK but force and kinetic energy of a relativistic rocket is a specific problem, better ask about it explicitly. $\endgroup$ Commented Apr 14 at 2:13
  • $\begingroup$ I don’t think there’s anything specific to the question of “is work well defined in special relativity if the force or mass in the accelerated frame is allowed to vary” $\endgroup$ Commented Apr 14 at 5:14
  • $\begingroup$ Basically, this is a question that should have predictive power in its own right, because it is mathematically possible to formulate, whatever the source of the force. How that can be applied to other problems is then a matter of experiment or engineering, or maybe observational astronomy, but it is something that should be possible to address theoretically without any specific named force or reason the mass may change. This is just as Newton’s second law applies regardless of what the force is. $\endgroup$ Commented Apr 14 at 5:18
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    $\begingroup$ Work is well defined in inertial frame, where the rocket moves. In the accelerated frame of the rocket, the rocket does not move, so there is no point in using the concept of work in this accelerated frame - it is zero. $\endgroup$ Commented Apr 14 at 11:49

... returning to the motivation of the post, what is the kinetic energy useful for?

If you are designing a next generation super collider and you know what sort of collision velocities are you are aiming for, it is useful to know how much energy you will have to supply to the collider coils to achieve those velocities. If you are designing a inter stellar rocket and know what velocity you require to reach a target star system in a given time period, then you need to know the maximum kinetic energy to determine how much fuel the rocket will have to carry. An example of a fuel calculation which allows for the fact that rocket fuel adds to the total load to be accelerated, is given in this document: "The relativistic rocket" which has many useful equations for constant (proper) acceleration, which you will find very enlightening.

The work done on a particle is equal to (Final Total Energy - Initial Total Energy) = $\Delta$(Kinetic Energy) which hints at the usefulness and meaning of relativistic kinetic energy.

  • $\begingroup$ My question wasn’t about constant proper acceleration. Nor do rockets experience that. Their masses change and so does their thrust. I was curious how one would even begin to set up an equation for the work in that circumstance if the final kinetic energy was unknown (because the final velocity was unknown and was the desired quantity to be computed). $\endgroup$ Commented Apr 14 at 1:39
  • $\begingroup$ Also… your link about the Relativistic Rocket may be useful to others, although it is not a full answer to my question. I can’t follow the link though, it doesn’t seem to work. Could you please edit and fix that? $\endgroup$ Commented Apr 14 at 1:41
  • $\begingroup$ You will be surprised how often the proper acceleration is constant. For example if you are accelerating a particle in a particle collider with a constant maximum force the due to the fact there is a limit to how much current the coils can handle, then the acceleration is initially fast but steadily reduces in the the reference frame for the collider, but in the reference frame of the particle, the (proper ) acceleration is constant. It would not be too hard to determine the final velocity in this case. $\endgroup$
    – KDP
    Commented Apr 14 at 4:12
  • $\begingroup$ Also fixed the link. I am sure you will find it useful and educational. $\endgroup$
    – KDP
    Commented Apr 14 at 4:13
  • $\begingroup$ Maybe thanks. I’ve seen the derivation before and it can’t hurt to review it. I do appreciate the link. But there most definitely do exist situations on this planet or elsewhere where the acceleration is not constant and my question was about those. Is work even well-defined in that context? $\endgroup$ Commented Apr 14 at 5:13

I believe this is much clearer

I don't believe the question I asked here was particularly clear. The question I linked above is a much clarified approach to the underlying physics I was discussing here. I do not believe energy is the essential issue, as others pointed out. I believe the differential equations of motion, and possibly the geodesic equation, are much more relevant to obtaining velocities in the presence of a time dependent force, and that energy may not be a particularly useful concept at all in this context.

I believe it is most clear that Kinetic Energy is most useful in relativity during impulse-like changes in velocity, such as during atomic transitions. Problems that require integration over time or ongoing accelerations may not have equations for work that are particularly useful as highly relativistic speeds are approached, since it is pretty obvious that it is not possible to maintain a constant acceleration throughout this time in the lab frame.


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