Does the concept of work make kinetic energy Galilean invariant?

Question

I have read at this site that the work is defined in such a way that kinetic energy will be Galilean invariant.

Could someone explain a little more about what is meant by this?

Answer 1

Kinetic energy is not invariant: its value can change from one system to another. However the relationship connecting the work done by a force and the change in kinetic energy must hold in every inertial frame of reference!

That kinetic energy is not the same in all systems is clear: if you are in a car moving at constant speed $v_0$ your energy is $1/2mv_0^2$ (m being your mass). But you, inside the car and at rest, if you computed your kinetic energy, would say it is 0 - because you are at rest! So kinetic energy changes across frames of reference but in intertial frame the work-kinetic energy relationship still holds.

Let's state what I mean: I am in a reference system (inertial). I see an object moving. I measure the work done $W_1$. I measure the change in kinetic energy due to that work $\Delta K_1$. The two quantities must be the same or physics is wrong! The I go to another inertial reference frame. Again, I measure $W_2$ and $\Delta K_2$ in the new frame. These two new quantities can (and will) be different from the ones before, but $\Delta K_2 = W_2$ must still hold, because we are in an inertial frame where $F=ma$ has not changed.

This is because work is defined in a way that $W=\Delta K$ holds in all inertial frames of reference but the individual value of $W$ can change. If it does, also $\Delta K$ must change to stay equal to $W$. The definition of work stems directly from $F=ma$, so if the work-kinetic energy relationship breaks, so does $F=ma$ i.e. so does newtonian mechanics! However, notice that this only holds for intertial references.

Let's see a (long-ish) example [at the end there is a more general further comment]:

Suppose you have a constant force $F_1$ pushing an object, initially at rest in the origin, in a reference system which is at rest for a length $x_1$ in a time $t$. The work done will be $$W_1(t)=F_1x_1(t)$$ Because with constant force and no initial speed, setting the acceleration to be $F_1=ma_1$ (Newton's law) and using the equation of motion $$x_1(t)={1\over 2}at^2={1\over 2}{F_1\over m}t^2$$ we get $$W_1(t)={1\over 2}{F^2_1\over m}t^2$$

This quantity must be the same as the change in kinetic energy $$\Delta K_1(t)={1/2}mv_1^2(t)$$ Indeed, because the force is constant, the object will have moved according to (again, equation of motion) $$v_1(t)=a_1t={F_1\over m}t$$ [remember this relationship for later] so that $$\Delta K_1(t)={1\over 2}mv_1^2(t)={1\over 2}m({F/m}t)^2$$ so that $$\Delta K_1(t)={1\over 2}{F^2_1\over m}t^2$$ i.e. $$\Delta K_1=W_1$$

Physics is safe so far!

Now, let's go into another frame of reference which is moving with speed $v_0$ in the same direction. This means that the speed of the object you observe in this frame at the beginningis going to be $v_2(t)=v_1(t)-v_0$, which means that at the beginning (when $v_1=0$) $v_2=v_0$. This also means that if after a time $t$ the observer would measure the change in kinetic energy (final minus initial) she would get $$\Delta K_2={1\over2}mv_2(t)^2-{1\over2}m(-v_0)^2$$ which is $${1\over2}m(v_1^2-2v_1v_0)$$ so $$\Delta K_2=\Delta K_1-mv_1v_0$$ which is different than what we measured in the first frame by the factor $mv_1v_0$. Oh no: where is the problem?!

The "problem" is that also the work $W_2$ in this frame of reference will be different. Let's see... let's measure the force the observer would measure in his system of reference.

Because the observer is moving, it would measure a speed $v_2(t)=v_1(t)-v_0$, as we said. Additionally, at the beginning, because the object was a rest in the first frame, the initial speed of the object in the second frame would be $-v_0$ (the observer would see the object moving in the negative direction: but it is not, it's actually the observer moving!). Additionally, as the observer sees a speed increasing linearly in time, she can assume there is a constant force $F_2$ acting on the object. By using Newton's law, we can write that $$v_1(t)-v_0=v_2(t)=v_2(0)+a_2t=-v_0+{F_2\over m}t$$ This means $$F_2=v_1(t)m/t$$ which is the same expression we found before for $F_1$ so that $$F_2=F_1$$

The force is the same in both systems - and that makes sense, because we are moving at constant speed and the acceleration the object sees must be the same, hence also the force. However, while the force is the same, the displacement of the object won't be the same because we are looking at it from another (moving) reference. This already suggests that that factor in the kinetic energy that was missing is coming from the fact that in this new reference system I see a smaller displacement i.e. a smaller work, so the "speed" must compensate for that. Let's check the two quantities are indeed equal!

We know $$x_2(t)=x_1(t)-v_0t$$ so that the work is [I am now dropping the $(t)$] $$W_2=F_2x_2=F_2(x_1-v_0t)$$ now, using $F_2=F_1$ we get

$$W_2=F_1x_1-F_1v_0t$$ and using $W_1=F_1x_1$ and $F_1=mv_1/t$ [told you to remember this one]:

$$W_2=W_1-mv_0v_1$$ which now satisfies $$\Delta K_2=W_2$$ despite $$W_1\ne W_2$$ and $$\Delta K_2 \ne \Delta K_1$$

Summing up, in a reference frame the absolute value of the kinetic energy can change but if I measure forces, speeds and displacements in the same reference system then $\Delta K= W$ always hold.

This is because the definition of work comes directly from $F=ma$ and $F=ma$ holds in all intertial reference systems by definition.

p.s. if the 2nd reference system were accelerating or rotatig, stuff would have been a bit more complex, involving apparent forces but the results would have been the same: $\Delta K = W$ always hold, but you need to include "special forces" due to the system not being inertial.

p.p.s. this also means that to apply conservation of energy, you need to do it in the same frame of reference.

Answer 2

I'm pretty sure that the guy intended Energy and not just kinetic energy. As you can see in the example he made, it seems that energy is not conserved in some reference systems. By defining the work done by a force in this way:

\begin{align} W = \int _\gamma \mathbf F \cdot d\mathbf r, \end{align}

you're implicitly saying that $\gamma$, the trajectory made by an object, depends on the reference system, and so does the work done by a force on that object. Changing the reference system is going to change the work done by the force, but also the kinetic energy of the object itself. The change in these two quantities (kinetic energy and work) in different reference systems is going to cancel out only by defining work as we said. This implies that energy is invariant for Galilean boosts.