From this answer for the question Is the energy conserved in a moving frame of reference?, I learnt that the work-energy theorem is independent of the frame of reference. But, is the theorem valid even for non inertial frames? I know that in non inertial frames we need to include inertial (pseudo or fictional) forces. Is the work done by all forces including the inertial forces equal to the change in kinetic energy?

Please note: According to this question and answer - Work associated with pseudo force, the work done by pseudo forces must be included in order to determine the total work done by all the forces. But the answer doesn't discuss about the validity of the theorem itself.


3 Answers 3


Of course the theorem is still valid! To see why, let's review what goes into the proof of the work-energy theorem, which I'll state as $$\Delta K = \int \mathbf{F} \cdot d \mathbf{x}, \quad K = \frac12 m v^2.$$ The easiest way to prove this is to work differentially, $$dK = \mathbf{F} \cdot d \mathbf{x}.$$ By the definition of the differential, $$dK = m \mathbf{v} \cdot d \mathbf{v} = m \frac{d\mathbf{x}}{dt} \cdot d \mathbf{v} = m \, d \mathbf{x} \cdot \frac{d \mathbf{v}}{dt} = m \mathbf{a} \cdot d \mathbf{x}.$$ Each of the steps here requires no physical input at all, besides the definitions of $K$ and $\mathbf{v}$, and some basic calculus like the product rule and the chain rule. So, we see the only physical assumption needed is $$\mathbf{F} = m \mathbf{a}.$$ This is of course true in an inertial frame.

Now recall why fictitious forces are used. Going from an inertial reference frame to a noninertial one changes the acceleration. Therefore, naively it makes $\mathbf{F} = m \mathbf{a}$ stop working. The whole point of introducing fictitious forces is to adjust $\mathbf{F}$ so that $\mathbf{F} = m \mathbf{a}$ is true again. Then, as long as this holds, the proof of the work-energy theorem goes through exactly as above, so the theorem holds in non-inertial reference frames if you count the work done by the fictitious forces.


In the following papers you can see the derivation and analysis of the work-energy theorem in noninertial reference frames

Work and energy in inertial and noninertial reference frames https://doi.org/10.1119/1.3036418

An extension to rotating reference frames is also given in

Work and energy in rotating systems https://doi.org/10.1119/1.4807897

The work energy theorem is valid even for noninertial frames.

Best regards,

Diego Manjarrés


Yes why not, work energy theorem for a system of particles accelerating w.r.t an inertial frame is given as:-For an $n$ particle system let the inertial force on $i^{th}$ particle be$\vec{F_{inertial}}^i$ then,$$\sum_{i=0}^n \int (\vec{F_{inertial}}^i+\vec{F_{pseudo}}).d\vec{s}=\Delta K_{system}$$ $$W_{inertial}+W_{pseudo}=\Delta K_{system}$$ wheres change in mechanical energy of the system is given as $$\Delta E_{mechanical}=W_{ext}+W_{int,non cons}+W_{pseudo}$$ $W_{int,non cons}$ is the work done by internal non conservative forces of the system.

  • $\begingroup$ With respect to your first equation, I think inertial and pseudo forces are the same. Further, from where did you get the first equation of work-energy theorem? Till now, I've only encountered only this form - $W_{conservative}+W_{non-conservative}+W_{external}=K_f-K_i$. $\endgroup$
    – Vishnu
    Nov 17, 2019 at 7:32
  • $\begingroup$ @Intellex I would not provide a proof of the 1st equation as it would be a homework like proof but I would give you a hint that total work done by all forces is equal to change in kinetic energy which includes pseudo and coriolis forces as well.Remeber to calculate the work done by inertial forces from heliocentric reference frame.Is it fine now? $\endgroup$ Nov 17, 2019 at 8:03
  • $\begingroup$ Thanks. So we are proving this under the assumption "...total work done by all forces is equal to change in kinetic energy which includes pseudo and coriolis forces as well" which is same as saying the work energy theorem is valid in non inertial frames. Further, why "heliocentric reference"?, I'm not doing celestial mechanics. Even this frame is non-inertial. Am I right? $\endgroup$
    – Vishnu
    Nov 17, 2019 at 8:18

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