Does work differ between inertial frames?

Question

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 $dr$.
4. Additional question, is, what about potential energy, is it frame-dependent? This should follow from the answer of "is work frame-dependent?", right?

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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?

Answer 6

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 $W'=\int_{Γ'} m\frac {d^2\mathbf r'}{dt^2}\cdot d\mathbf r'=\int_{α}^{β}m\frac {d^2 \mathbf r'}{dt^2} \mathbf (\mathbf γ'(t))\cdot \frac {d\mathbf γ'}{dt}(t)dt=\int_{α}^{β} m\frac {d^2 (\mathbf r -\mathbf a)}{dt^2} (\mathbf γ(t))\cdot \frac {d(\mathbf γ -\mathbf a )}{dt}(t)dt=\int_{α}^{β} m\frac {d^2 \mathbf r}{dt^2} \mathbf (\mathbf γ(t))\cdot \frac {d\mathbf γ}{dt}(t)dt= \int_{Γ} m\frac {d^2\mathbf r}{dt^2}\cdot d\mathbf r=W.$

Answer 7

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