# Kinetic energy with respect to different reference frames

I'm having problems understanding the following situation. Suppose two 1-tonne cars are going with the same orientations but opposite senses, each 50 km/h with respect to the road. Then the total energy is

$$\begin{eqnarray}E=E_1+E_2&=&\frac{1\mathrm t\times(50\mathrm{km}/\mathrm h)^2}2+\frac{1\mathrm t\times(50\mathrm{km}/\mathrm h)^2}2\\&=&1\mathrm t\times(50\mathrm{km}/\mathrm h)^2\\&=&2500\frac{\mathrm t\times\mathrm{km}^2}{\mathrm h^2}.\end{eqnarray}$$

Now if we look at it from the point of view of one of the cars, then the total energy is

$$\begin{eqnarray}E=E_1+E_2&=&\frac{1\mathrm t\times(0\mathrm{km}/\mathrm h)^2}2+\frac{1\mathrm t\times(100\mathrm{km}/\mathrm h)^2}2\\&=&\frac{1\mathrm t\times(100\mathrm{km}/\mathrm h)^2}2\\&=&5000\frac{\mathrm t\times\mathrm{km}^2}{\mathrm h^2}.\end{eqnarray}.$$

I know that kinetic energy is supposed to change when I change the frame of reference. But I understand that then there must be some other kind of energy to make up for it so that the energy in the system stays unchanged. But I don't see any other kind of energy here. I only see two total energies of the same system that seem to be different. Could you explain this to me?

Please note that while I don't understand any physics, I do understand college level mathematics, so if necessary please use it. (I doubt anything more than high school maths should be needed here, but I want to say this just in case.)

-
Hi Bartek, and welcome to Physics Stack Exchange! I'm curious, what did you read on Physics Meta that says elementary questions are not allowed here? I'd like to know if it's something that needs to be changed. The true policy is that questions of any level are allowed, but they have to be insightful and original and show research effort and prior thought on the part of the poster. It tends to be harder to come up with a question like that about a low-level topic. – David Z Jan 15 '13 at 2:00
@DavidZaslavsky: He may have seen the homework policy, which does put restrictions on homeworky problems, especially the basic ones. But this question is OK as far as the HW policy goes :) – Manishearth Jan 15 '13 at 5:41

You have successfully discovered that the kinetic energy depends on the reference frame.

That is actually true. What is amazing, however, is that the fact that kinetic energy is conserved is NOT reference frame-dependent. So, when you balance your conservation of energy equation in the two frames, you'll find different numbers for the total energy, but you will also see that the energy before and after an elastic collision will be that same number.

So, let's derive the conservation of energy in two reference frames. I'm going to model an elastic collision between two particles. In the first reference frame, I am going to assume that the second particle is stationary, and we have:

\begin{align} \frac{1}{2}m_{1}v_{i}^{2} + \frac{1}{2}m_{2}0^{2} &= \frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^{2}\\ m_{1}v_{i} &= m_{1}v_{1}^{2} + m_{2}v_{2}^{2} \end{align}

to save myself time and energy, I'm going to call $\frac{m_{2}}{m_{1}} = R$, and we have:

$$v_{i}^{2} = v_{1}^{2} + Rv_{2}^{2}$$

Now, what happens if we shift to a different reference frame, moving to the right with speed v? This is essentially the same thing as subtracting $v$ from all of these terms. We thus have:

\begin{align} (v_{i}-v)^{2} + R(-v)^{2} &= (v_{1}-v)^{2} + R(v_{2}-v)^{2}\\ v_{i}^{2} -2v_{i}v + v^{2} + Rv^{2} &= v_{1}^{2} - 2 vv_{1} + v^{2} + Rv_{2}^{2}-2Rv_{2}v + Rv^{2}\\ v_{i}^{2} -2v_{i}v &= v_{1}^{2}- 2vv_{1} + Rv_{2}^{2}-2Rv_{2}v\\ v_{i}^{2} &= v_{1}^{2} + Rv_{2}^{2} + 2v(v_{i} - v_{1} - R v_{2}) \end{align}

So, what gives? It looks like the first equation, except we have this extra $2v(v_{i} - v_{1} - R v_{2})$ term? Well, remember that momentum has to be conserved too. In our first frame, we have the conservation of momentum equation (remember that the second particle has initial velocity zero:

\begin{align} m_{1}v_{i} + m_{2}(0) &= m_{1}v_{1} + m_{2}v_{2}\\ v_{i} &= v_{1} + Rv_{2}\\ v_{i} - v_{1} - Rv_{2} &=0 \end{align}

And there you go! If momentum is conserved in our first frame, then apparently energy is conserved in all frames!

-
I've deleted my previous comment because I think I misunderstood your answer. I will need to think more about this. – Bartek Jan 14 '13 at 23:17
@Bartek: I've written a greatly expanded answer that actually walks through the proof that energy is conserved in all frames. – Jerry Schirmer Jan 14 '13 at 23:34

In newtonian mechanics, kinetic energy is reference frame dependent.

If this were a relativistic description, the rest mass of the system is invariant under boosts and rotations.

-

As you say, energy is not invariant under reference frame change.

Imagine a moving ball. It has kinetic energy, but if I move in its reference frame, it doesn't. It's as simple as this.

There's no need to make up for the missing energy.

-
But here's what Wikipedia says: "The kinetic energy of any entity depends on the reference frame in which it is measured. However the total energy of an isolated system, i.e. one which energy can neither enter nor leave, does not change in whatever reference frame it is measured." Doesn't this mean we need something to make the energies equal? – Bartek Jan 14 '13 at 23:01
you are misinterpreting the statement. The statement is what Jerry Schirmer is saying. That is, the total energy of an isolated system, whatever value it might be in this reference frame, will not change a.k.a conservation of energy. of course the value of this energy in different frames is in general different. in fact, the article also goes on to say "Different observers moving with different reference frames disagree on the value of this conserved energy." – nervxxx Jan 14 '13 at 23:13
@nervxxx OK, so the total energies can be different in different frames of reference, but whatever they are, they don't change over time if we look at them from the same point of view all of that time. Am I understanding it correctly? If so, this is mind-blowing to me, and I will need to think about this a lot! – Bartek Jan 14 '13 at 23:21
@Bartek yup that is right! – nervxxx Jan 15 '13 at 0:24