# Velocity and kinetic energy, violating galilean relativity

I have a toy car and a battery. The barrery has a screen that shows how much energy it has left. Since kinetic energy is proportional to velocity squared, I need 1J of energy to go from 0m/s=>1m/s, but 3J of energy to go from 1m/s =>2m/s.

So I can surely use this to tell what absolute reference frame I am in! If the toy car takes 1J to accelerate by 1m/s then I am in the rest frame.

Where's the catch?

• Energy isn't invariant under transformation of reference frames, the total energy gain and lost in a process will be different in different reference frames. Commented Aug 9, 2021 at 9:54
• @OfekGillon the energy in a battery is definitely invariant under reference frame transformation, but the kinetic energy is not. How can one side of the equation change with reference frame, but the other remain the same?
– Matt
Commented Aug 9, 2021 at 9:55
• @MattTobias you have assumed that all the energy for the acceleration comes from the battery. In the frame where the ground is at rest, this is correct. In other frames, it is not (because the reaction mass is changing speed). Commented Aug 9, 2021 at 10:01
• The energy of the acceleration does come from the battery. It's just that the work done is different in a different frame. So there is no contradiction with requiring different changes in KE in different frames and Galilean relativity. I have tried to show this, to some extent, in my answer. Commented Aug 9, 2021 at 11:43
• @BrainStrokePatient OK, so the 'fuel efficiency' of the battery becomes lower and lower, the faster the car is going relative to the ground, since more and more KE is getting dumped into the earth? And I guess there's some equivalent insight about the efficiency of electric motors or internal combustion engines changing with speed?
– Matt
Commented Aug 10, 2021 at 23:50

In all your figures, you are referring to the speed of the vehicle with respect to the surface of the earth. Since you cannot tell the speed of the earth, you do not have access to an absolute reference frame. You do have access to the relative speed of the vehicle, but of course that could be done as a speedometer as well.

If you instead mean you could determine the speed of the car relative to the ground, then that is true. Since the car accelerates by pushing on the ground, the relative velocity of the ground matters. If you had a car that operated as a pure rocket and didn't interact with the surface, this wouldn't be possible.

How can you ever truly calculate this without thinking about the earth?

Because the earth is so massive that in any human interaction, we cannot measure the acceleration or the resultant velocity. Yes, the earth will accelerate backward when you drive your car, but the energy that takes is many orders of magnitude less than the energy given to your car in that frame. Ignoring this is fine.

It is only in frames where the earth is already moving that we have a concern. The work done on the earth (or by the earth) is equal to $$F \times d$$ or $$Fvt$$. In the center of mass at rest frame, $$v$$ is so near to zero that this term disappears.

• That's because it's usually appropriate (and simpler) to work only in the frame with the earth at rest. Working in other frames (or worse, switching frames) normally only increases the complexity without helping things along. Commented Aug 9, 2021 at 10:06
• Added a little bit. Commented Aug 9, 2021 at 16:02
• @matt With a rocket, think in terms of the COM (centre of mass) frame of the rocket plus the propellant. The engine imparts equal & opposite momentum to the rocket and the exhaust, so the COM of the combined system doesn't move. Commented Aug 9, 2021 at 23:37
• Note that the tweet says the energy needs go as the square. That's not the same as saying the fuel needs go as the square. Rocket efficiency changes with speed. Commented Aug 10, 2021 at 21:49
• Maybe look at physics.stackexchange.com/questions/186587/… Commented Aug 10, 2021 at 21:54

The work done by the battery is frame dependent. In order to make my point and to simplify the math, I'm assuming that the battery does work by applying the force on a single point, instead of applying forces on multiple charges to move the current. I'll also be doing things in 1d. You can think in this particular case of the battery as something like a compressed spring, if it helps. Assuming that initially the potential energy of the battery was $$V$$ and finally it was $$0$$, and that the car was initially at rest you can say $$V = \frac{1}{2}mv_f^2$$ where $$v_f$$ is the final velocity of the car after all the battery energy has been expended. Now for a frame moving with respect to the ground frame at a velocity $$v_{rel}$$, you might be tempted to say that the energy equation in that frame would be $$\frac{1}{2}mv_{rel}^2 + V = \frac{1}{2}mv_f'^2$$ where $$v_f' = v_f - v_{rel}$$ But this is wrong. To see this substitute the Galilean tranformation equation into it $$\frac{1}{2}mv_{rel}^2 + V = \frac{1}{2}m(v_f - v_{rel})^2$$ $$V = \frac{1}{2}mv_f^2 - mv_{rel}v_f$$ which clearly contradicts with the first equation. The answer lies in the work energy principle. Before applying this in the moving frame, remember that $$x$$ and $$x'$$ are related by the Galilean transformation rule $$x' = x - v_{rel}t$$ and so $$dx' = dx - v_{rel}dt$$ So when integrating for computing the work, there are two contributions. The point of application of force moves because of the force applied ($$dx$$) and also because the point itself is moving relative to the moving frame ($$v_{rel}dt$$). Now applying work energy, we get $$\frac{1}{2}mv_f'^2 - \frac{1}{2}mv_{rel}^2 = \int F dx'$$ $$= \int F dx - v_{rel} \int F dt$$ $$= V - mv_{rel}(v_f' + v_{rel})$$ The first term is just the work done by the battery in the stationary frame, i.e. ,V. The second term has the impulse of the force, and so it's just the difference in final and initial momentum. The plus sign is there because the initial momentum in the moving frame is $$-mv_{rel}$$. This is the actual energy equation that you should have in the moving frame. To check it's consistency, replace all the $$v_f'$$ with $$v_f - v_{rel}$$ to get $$\frac{1}{2}m(v_f - v_{rel})^2 - \frac{1}{2}mv_{rel}^2 = V - mv_{rel}v_f$$ $$\implies \frac{1}{2}mv_f^2 = V$$ which agrees with the first equation.

In essence, a similar thing is happening in 3 dimensions with all the complicated forces in your battery. The screen only shows the work it can do in the ground frame. The work it can do in any other moving frame is different. The correct way to think about these sorts of energy problems is always through the work energy principle.

• Thanks. OK, so the battery really does do less work per unit energy as the car goes faster? And each additional unit of energy produces less and less delta v?
– Matt
Commented Aug 11, 2021 at 5:03

You're right that this is a subtle trick, because the amount by which the battery drains - in effect, how many molecules of battery chemicals react - is not something trivial enough to simply be transformed away by a reference frame change, and you can indeed use this to single out a "preferred" reference frame for the problem.

In particular, this special reference frame will be that of the Earth - and that should be a clue. Think about how the vehicle actually accelerates up to speed in the first place. Can it just do this were it simply floating around out in empty space? No! It needs to be physically contacting something to move across - and when you're on Earth, that thing is the Earth's surface (in a general sense). And this is what selects the reference frame you get. If you run it in empty space, the battery will still drain, but the wheels will simply spin uselessly and no problematic kinetic energy changes will occur. If you put the vehicle on another planet, like the Moon or Mars, then its battery will likewise "select" that planet as its "preferred" frame.

Of course, this is just one component of the problem. The other component is how do we reconcile the 4 joules, say, as being burnt by the battery in every reference frame, with yielding more or less joules in another reference frame? That is, if we shift to a reference frame where that the kinetic energy gain is only 2 joules, where did the other 2 joules go? And it turns out that if you work it out, the answer is they went into the Earth. They boosted its kinetic energy by 2 joules. Likewise, if we're in a frame where the kinetic energy gain is, say, 6 joules, the excess 2 joules beyond what the battery put in came from the Earth: in such a frame, the Earth looks like it slowed down a bit - note that with a suitably-moving reference frame, any "speeding up" object can be made to look like it's "slowing down", at least for a time, and conversely.

(Finally, note that you won't actually see any of these changes in Earth's motion in any reasonable sense. At a mass of $$6 \times 10^{21}$$ tonnes, a two-joule change in kinetic energy can at most change the Earth's orbit by about $$10^{-29}\ \mathrm{m/s}$$, which is roughly equivalent to a motion of about the size of a proton, over the entire elapsed span of existence of the human species!)

• > the wheels will simply spin uselessly and no problematic kinetic energy changes will occur< Doesn't the same problem exist for wheels rotating? 0=>1 rad/s vs 100=>101 rad/s? I can't believe this is actually true. Surely the world would look very different, high speeds would be almost unattainable etc.
– Matt
Commented Jul 8 at 16:28
• Is there anything in culture/film etc about high speeds taking enormously more energy etc?
– Matt
Commented Jul 8 at 16:52