# Does work differ between inertial frames?

I have a question regarding the frame dependence of both kinetic energy and work.

1. I was able to prove that if I have a system of particles with masses $$m_i$$ and momenta $$p_i$$, then the change in kinetic energy is invariant under Galilean transformation only if the total momentum is conserved.
2. If momentum is not conserved, then the change in kinetic energy is not invariant, and hence (by the work-energy theorem), the work done is different between frames.
3. My best thought till now is that work is $$\text{power}\times\text{time}$$ and power is $$F \cdot v$$. Since inertial frames do agree on $$F$$, but not $$v$$, power is frame-dependent. Hence, work is also frame-dependent, but this doesn't make sense to me! If I exerted some work to move things apart for example in one frame, everyone else should see me doing the same work, right? we both agree on the force $$F$$ and the path $$dr$$.

Take a simple example of one ball of mass 1 kg pushed to go from 0 m/s to 1 m/s in one frame and from 9 m/s to 10 m/s in another frame. What is the work done on it? And what is the meaning of each number in that frame?

• Given a body moves in a vector field from point A to point B the work done is any frame is the same. Instead of changing frames, imagine a e.g charge moving at different speeds. The work done is the same as it moves the same distance, but because the faster object spends less time in the field, it has a lower change of momentum. Indicative of the fact that the same amount of kinetic energy added to a faster body changes the velocity less than if it were moving slower. As its proportional to v^2 Jun 29 at 13:22
• Thanks for your contribution. What you're saying here is "Same amount of work done (change in kinetic energy) changes the speed less if the speed is higher". I totally agree with this, but I can't see how this is related to the question. The question says, that two inertial frames see different changes in kinetic energy, so which one is correct regarding the work spent? Since this work should be the same as it might indicate for example an energy release from a chemical interaction or so. I think this is what we should think about, is it really the case that work should be the same?! Jun 30 at 11:31
• The 2 different speed can be thought of as seeing the object in 2 difference frames. The work is the same regardless. Will need to think abit more before I answer, but it appears in a different frame, the power is different. Power is different, total work isn't. Jun 30 at 13:15
• Note that energy is invariant under coordinate transformations. Jun 30 at 16:49
• Aha, correct, I think I found the answer finally. Thanks for your contribution. It's about defining the system. I will accept and edit (a bit) Frank's answer soon. Jun 30 at 20:27

The conclusion: Work done is dependent on different frames. Let's look at a simple example. Say there is an object(1kg) that is initially at rest and we apply a constant force to move it for a certain distance to reach 2m/s in the rest frame(call it O). The KE change in frame O is 2J. Now consider a moving frame O' moving with a velocity=1m/s to the left with respect to frame O. In frame O', the object is initially moving to the right with a speed of 1m/s(NOT at rest), and the final speed is 3m/s, the KE change in frame O' = 4J, which is different from that in O. We know that in this simple case the KE change must equal to the work done on the object, and this must be true in both frames. Thus, we conclude the work done is different in different frames. This is not difficult to understand as in frame O', the object is moving with a greater average speed when the force is acting on it. Since the time that the force applied on the object must be the same in both frames, the displacement of the object in frame O' is indeed larger and the resulting work done is also larger.

• I totally agree with your setup and thanks for putting it this clear. However, my question is: Say, I exerted this work from my muscles, shouldn't its value be definite? Like, everyone should agree that I used up 20 calories doing that! Jun 29 at 9:37
• Your metabolism operates in a single frame, so it doesn't seem like any ambiguity is possible. Jun 29 at 18:39
• That's an interesting question. But remember the person who is acting the force is also moving, a moving object of course has more ability to do work. Or we can put it in this way, imagine the object and the person are on a moving train. People on the ground are the frame O' and people on the train are the frame O. Jun 30 at 14:01
• Now a person exerts a force on the object and the object will exert a reaction force back to slow down the person. If he wants to keep his speed, then he has to gain energy from the train to keep moving with the train. So the extra energy gain is indeed from the train(or any other source that keeps the person moving.) Jun 30 at 14:01

Yes, work can differ in inertial frames as others have discussed. You also asked about the chemical energy of your body in numerous comments.

The energy from your muscles is a change in internal energy (burning food), not work. That change in internal energy can be used to do work. For example, a car engine burning gas (change in internal energy) can increase the rotational kinetic energy of the driven wheels (do work) which friction converts partly to translational kinetic energy to move the car forward.

Also, in non-inertial frames fictitious forces contribute to work.

• Hey, thanks for contributing to this question. I think you have a point here by interpolating the internal energy in the context since it has a kinetic energy component, and hence, its change is variant as well. This seemed to solve it EXCEPT that we can speak about sources of work like a gravitational field for example. I mean, should the work done by the field be the same or not between frames? Jun 30 at 11:38
• Work done by gravity is different in different frames. For object initially at rest the change in KE is different than for object initially moving as seen by a moving observer. Same as answer by @Frank earlier, where here force is gravity. This does not violate conservation of energy; work is also required to increase the KE of the stationary frame to the KE of the moving frame. Jun 30 at 14:39
• Yeah, I see. Yet, you have to be careful here about the kinetic energy of the "frame". There is nothing like that in mechanics. A frame is a "considered" coordinate system, no more. It's not something physical, it has no mass and it's not there, we attach it to bodies. Though, I think you meant that frame attached to the person who throws the ball. This is totally correct and it's my final answer to the question. Thanks! Jun 30 at 20:41
• Yes. Also, the train's acceleration matters; if train moving at constant velocity (not accelerating or rotating) there are no fictitious forces. If the train is accelerating, in that frame there are fictitious forces and they contribute to work. Simple example: body at rest to observer on ground, no force no work, no change in KE. For train moving at constant velocity body, in that frame, constant KE so no change in KE no work. For train accelerating linearly, in that frame. fictitious force -ma acts accelerating the body doing work and KE changes. Jun 30 at 21:30

Regarding the last part of your question, the thing that you are neglecting is the part of equal and opposite reaction. Suppose I am standing in a train carriage and throw a ball. Then not only do I accelerate the ball forward, but I also accelerate the train carriage backwards by a very tiny amount. In a frame where the train is standing still, this means that I use chemical energy both to impart kinetic energy mostly onto the ball but also a tiny bit of int onto the train.

Now do the same thing in a freely rolling, moving train (or a respective frame), such that the train is moving in direction throw (respectively the frame is moving in opposite direction). Then on one hand, the ball gains more kinetic energy, but on the other hand, the speed and thus the kinetic energy of the train now gets reduced. So what happens is that you use chemical energy to impart kinetic energy on the ball, but you also convert some of the kinetic energy of the train into kinetic energy of the ball. If you take that into account, you will find that the chemical energy needed in both cases is the same. (As will be the case for any potential energy if defined properly, only the kinetic part changes.)

In case you don't want to be on the train, you can also do the same thing standing on the ground. If you throw a ball, then not only are there forces between you and the ball but also between you and the ground. In a non-moving frame, the latter perform no work, as they don't cover any distance. In a moving frame however, the same force covers a distance and thus in some sense the ground performs work, which again precisely produces the missing energy needed for the change in kinetic energy of the ball.

This is irrespective of the source of the force, even if it was not a contact force like a gravitational or electric field. At the end of the day, these fields have a source! If this source was taken into consideration, we would get the same work done.(Btw, contact force is electromagnetic, so it's no different)

Then the problem was raised from:

1. attaching the frame to one of the interacting bodies and this made it non-inertial.
2. asking about the work on the ball and not noticing there are other works as well that should be taken into consideration.

What this says really is that you need to take into account all the interacting bodies so that all forces are internal, and conservation of momentum applies, and if this is the case, the change in total kinetic energy would be invariant and total work would be invariant as well.

• Thanks for your answer. Does this raise the claim that the work-energy theorem is only meaningful in a frame where momentum is conserved? Jun 30 at 11:47
• @IbrahimAlmetwale It actually holds in every frame, however in a non-inertial frame you will have to account for work done by pseudo-forces. See the example in the last paragraph, the frame fixed to the ground (or moving with a constant velocity with respect to it) is not an inertial frame, as the throw will actually have accelerated the earth by a tiny amount. In that frame there is a force coming from you acting on the ground that is not canceled or results in acceleration, so it performs work without taking energy from somewhere, to account for the acceleration of the frame.
– mlk
Jun 30 at 13:59
• Yeah, I see! thank you, this is super clear, I will just another paragraph to the answer to we accept it. Jun 30 at 20:47

If I exerted some work to move things apart for example in one frame, everyone else should see me doing the same work, right?

Without realizing it, here you are describing a situation where an observer in a different frame considers your frame's point of view. It's really just the fact that you're imagining them thinking all this from their point of view that gives you the false impression that it's "their reference frame". It's not.

When you say "they see me doing X work", thats actually them watching you and imagining your frame.

If they truly used the velocities relative to their frame, the work they calculate would indeed be different.

• Yep, That's the problem! The work calculated in both frames shouldn't be different, right? My body used a specific amount of chemical energy. Jun 29 at 9:39

Train moves at speed 100 m/s.

Bob stands on train floor, train moves at constant speed. Bob does not do any kind of work.

Joe stands on train floor, Joe is exerting a force on the train, because the train is slowing down. Joe does a large amount of work on the train, Joe does not do any muscle work.

Alice stands on train floor, train moves at constant speed. Then Alice starts walking. Alice is exerting a force on the train. Alice does a small amount of muscle work. Alice does a large amount of work on the train.

Joe is just standing. And doing a lot of work. Is it possible to understand this strange mystery? Well, Joe uses his kinetic energy to do the work.

Now if we understand how Joe does lot of work with zero effort, perhaps we can understand how Alice does lot of work work with small effort?

• Thanks for your trial. Your example is a bit ambiguous to me. What is the difference between Bob and Joe? Why does Joe exert force on the train? It's just the contact force that he exerts on the train and the train exerts it back on him, and it does NO work since it's not in the direction of the displacement. Jun 30 at 11:56

Let a point moving on a curve $$Γ=Γ(t)\subset \Bbb R^2$$ parametrized by time $$t\in [α.β]$$ . We suppose two inertial frames $$xOy$$ and $$x'O'y'$$, so the position vector of the point is $$\mathbf r$$ and $$\mathbf r'$$ respectively. Furthermore $$\mathbf r-\mathbf r'=\mathbf {OO'}$$, which yields $$\frac {d^2\mathbf r}{dt}=\frac {d^2\mathbf r'}{dt}\implies m\frac {d^2\mathbf r}{dt}=m\frac {d^2\mathbf r'}{dt}\implies \mathbf F=\mathbf F'\implies \mathbf F\cdot d\mathbf r =\mathbf F'\cdot d \mathbf r'\implies \int_{Γ(t)}\mathbf F\cdot d\mathbf r =\int_{Γ(t)}\mathbf F'\cdot d \mathbf r'\implies W=W',$$ because $$d(\mathbf r-\mathbf r')=d(\mathbf {OO'})\implies d\mathbf r=d\mathbf r'.$$

The answer above is about inertial frames with zero velocity. If we are about the Gelilean Transformation, we have as usually $$x=x', y=y', z=z'+vt$$. Since $$v$$ is constant, the second time derivative of three coefficients is invariant, so the acceleration is the the same, $$a=a'$$, which means that $$F=F'$$. Then $$W=Fd$$ and $$W'=F'd'=F(d+vt)=Fd+Fvt=W+Fvt$$, which means that the work has a $$Fvt$$ change compared to the first one.

*About the invariane of above line integral: Let a curve parametrized as $$Γ=Γ(t)$$ in $$xΟy$$ and $$Γ'=Γ'(t)$$ in $$x'O'y'$$, $$t\in [α,β]$$. Then, using the previous symbols, we get $$W'=\int_{Γ'} m\frac {d^2\mathbf r'}{dt^2}\cdot d\mathbf r'=\int_{α}^{β}m\frac {d^2 \mathbf r'}{dt^2} \mathbf (\mathbf γ'(t))\cdot \frac {d\mathbf γ'}{dt}(t)dt=\int_{α}^{β} m\frac {d^2 (\mathbf r -\mathbf a)}{dt^2} (\mathbf γ(t))\cdot \frac {d(\mathbf γ -\mathbf a )}{dt}(t)dt=\int_{α}^{β} m\frac {d^2 \mathbf r}{dt^2} \mathbf (\mathbf γ(t))\cdot \frac {d\mathbf γ}{dt}(t)dt= \int_{Γ} m\frac {d^2\mathbf r}{dt^2}\cdot d\mathbf r=W.$$

• Thanks for your contribution! What you have indicated is showing that the "since the force is invariant between the frames, and space intervals are so as well, work must be the same". I am fine with this, and pretty much everyone. However, this doesn't answer the question. The question is "Since the change in kinetic energy is variant under Galilean transformation so long as momentum is not conserved, by the work-energy theorem, work is also variant in such circumstances." Jun 29 at 12:01
• I will enrich my answer considering your comment. I hope I 'll help.
– SK_
Jun 29 at 18:49
• There is some ambiguity in what you have provided in the last paragraph. What is small gamma? and is it a vector or a scalar? because you removed a vector infinitesimal and replaced with gamma in strange way, could you illustrate that please? thanks! Jun 30 at 11:55
• Firstly, small gamma is the symbol of curve parametrization; consequently it i s a vector (bold in text). It could be a bold r=r(t) as well, but I set a different symbol in order to show the parametrization of the curve. I cannot see some "strange removal". I 've used the definition of line integral: dr=γ΄dt.
– SK_
Jun 30 at 13:04

Maybe $$dr$$ is the same between frames, but the limits of integration are different in different frames.

• We go from point 1 to point 2 on a path, which is a specific set of points in space. So, neither the path nor the limits of integration should change, you are summing the same pieces of space intervals dr. Jun 29 at 5:22
• @IbrahimAlmetwale: In the example you give at the end of your question, if we assume a constant force occurs over the course of 1 second, then the ball travels $\frac{1}{2}$ m in one of the frames and $9 \frac{1}{2}$ m in the other frame. This is why the limits of integration are different. Or, to put it another way, if the end-point is moving in one frame and not in the other, then the distance travelled to get to that endpoint will be different in the two frames. Jun 29 at 18:36
• Thanks for clarifying your answer. I am still thinking about the interpretation of the result. What does it mean for work to be variant? think about the source of it? (have a look at other comments, please). Jun 30 at 11:58
• if you want to clarify it, write an accelerating trajectory, say $x(t) = t^2$ from $t=0$ to $t=1$, compute the work. Now transform to a moving frame and do it again.
– JEB
Jul 2 at 0:08