# Electric car in different inertial frames

Assume there are two inertial frames $A$ and $B$ where $B$ is moving with speed $v$ relative to $A$. Suppose an electric car is stationary in frame $A$. The car is accelerated to speed $v$ to be stationary relative to frame $B$.

An observer in frame $A$ will conclude the electric car gains KE of $\frac{1}{2}mv^2$. He will also conclude that the battery loses chemical energy $\frac{1}{2}mv^2$ by principle of energy conversation.

However, an observer in frame $B$ will conclude the electric car loses KE of $\frac{1}{2}mv^2$. But this observer will notice the battery loses chemical energy $\frac{1}{2}mv^2$. So where does total energy $mv^2$ go?

Please assume it is frictionless. There is no lose of energy to thermal.

• The car can't accelerate without static friction between the tire and the road, but this doesn't dissipate any energy. So I assume by "frictionless" you simply mean "there's no energy dissipation." Aug 1, 2018 at 1:54

I admit this looked confusing at first, but I think you can resolve the issue if you take the work done by the car when its wheels pushes against the Earth (or the propellant if you swap out the car with a rocket). Basically, if you consider the (car + battery) as a closed system, then we break the conservation of momentum, even though everything stays translationally invariant. So (car + battery) can't be a closed system.

Let's say the car is on a board. In frame A, both are stationary. The battery has $\frac{1}{2} mv^2$ worth of energy. Let's assume for simplicity that the mass of the board is same as that of the car. Now the car wheels push against the board, and by conservation of momentum, they must travel in opposite direction. The observer in frame A will conclude that they move $\pm \frac{mv}{\sqrt{2}}$ by conservation of energy and momentum.

Now let's go to frame B. Since it is defined such that the final velocity of the car is zero, frame B is moving at $\frac{v}{\sqrt{2}}$ with respect to frame A. When the car comes to a stop, the battery has drained, and the board moves at the velocity of $-2\frac{v}{\sqrt{2}}$. So the final momentum is still $-2\cdot \frac{mv}{\sqrt{2}}$, and the final energy is $mv^2$, which is equivalent to the initial energy of $\left(2 \cdot \frac{mv^2}{4}\right) + \frac{mv^2}{2} = mv^2$.

The situation you consider is the limit of the mass of the board going to infinity. I am certain that if you obtain the general expressions when you set the mass of the board to arbitrary value, you will still satisfy the conservation of energy and momentum.

Okay, I'm adding the general expressions.

Let's denote the mass of the car as $m$ and the mass of the board as $M$. Let's assume the battery has energy worth $\frac{1}{2} m v_a^{2}$.

After the battery drains out, calculate the resulting velocities using conservation of momentum and energy.

In frame A, after the battery drains out,

The board travels at $v_{board} = - \sqrt{\frac{m^2}{M\cdot m + M^2}}v_a$.

The car travels at $v_{car} = + \sqrt{\frac{M^2}{M\cdot m + M^2}}v_a$.

Again, since we define frame B to be the frame in which the final velocity of the car is zero, frame B is moving at $\sqrt{\frac{M^2}{M\cdot m + M^2}}v_a$ with respect to frame A. The initial kinetic energy is then $\frac{1}{2}(M+m)\frac{M^2}{M\cdot m + M^2} v_a^2 = \frac{1}{2} M v_a^2$. So the total energy is $\frac{1}{2} (M+m)v_a^2$ in frame B. When the car stops in frame B, the board will be traveling with velocity $-\sqrt{\frac{M+m}{M}}v_a$.

Now take the limit in which $M$ approaches infinity.

• Thanks. You are correct. My reasoning is flawed as electric car cannot accelerate alone as it requires the earth to provide the force. I guess one cannot accelerate a rocket using battery.
– PCur
Aug 1, 2018 at 2:26
• @PCur You can definitely accelerate a rocket using a battery - just emit electromagnetic radiation out of the tail in the same way that a rocket would normally emit exhaust. Electromagnetic fields carry momentum, so they can propel you (very slowly) forward. The general principle is that to propel yourself forward, you must have something to push backward (this is just conservation of momentum). That something doesn't necessarily have to be the Earth - but it happens to be the Earth in the case of electric cars, which move forward by pushing the Earth backward with their tires. Aug 1, 2018 at 3:31

What you're forgetting about is the kinetic energy of the Earth below you. The car gains speed by exerting force on the Earth, and the kinetic energy is distributed differently in each frame.

In each frame, we must clarify one thing: whether your reference frame is at rest relative to the Earth before the car begins to accelerate, but not actually attached to Earth's surface (let's call these frames A1 and B1), or whether it's fixed to the Earth (frame A2 and B2).

In frame A1, before the car accelerates, the Earth is stationary in your frame, but as the car accelerates, the Earth begins to move, very slowly (let's call its final speed after the acceleration $\Delta v_{Earth}$). Now the car's speed relative to the Earth's surface is $v$, but, since the Earth is now moving in frame A1, the speed that the observer measures is slightly less, $v-\Delta v_{Earth}$. In line with energy conservation, the sum of the kinetic energy imparted to the Earth and the car is the same as the chemical potential energy lost by the battery, which we observe, in each frame, to be $\frac{1}{2}mv^2$. Therefore, we can solve for the speed of the Earth under us, using $M$ for the Earth's mass:

$$\frac{1}{2}M\Delta v_{Earth}^2+\frac{1}{2}m(v-\Delta v_{Earth})^2=\frac{1}{2}mv^2$$

Solving for $\Delta v_{Earth}$ and ignoring the zero solution gives

$$\Delta v_{Earth}=\frac{2mv}{m+M}$$

which, you'll notice, is a pretty small number, given that $M\gg m$. So, in the inertial reference frame that was initially at rest relative to the Earth, the Earth begins to imperceptibly move, stealing some kinetic energy from the car. Frame A1 will not measure the car's final kinetic energy to be $\frac{1}{2}mv^2$; rather, the observer will measure the car's kinetic energy to be

$$\frac{1}{2}m(v-\Delta v_{Earth})^2=\frac{1}{2}mv^2\left(1-\frac{2m}{m+M}\right)$$

which is slightly less than the kinetic energy you would measure if you were moving along with the Earth (but, if you were in an inertial frame moving along with the Earth now, then you would see the car and the Earth moving imperceptibly backward before the car accelerated with speed $\Delta v_{Earth}$. This should give you some indication that something is weird with frame A2...).

In frame A2, which is fixed to the Earth, the Earth remains stationary the whole time. But, as we saw in frame A1, the Earth accelerates when the car exerts a force on it. This means that frame A2 is not inertial. As such, we can expect some fictitious forces in this frame that will change the physical situation, including the particular way that energy is conserved. Since the Earth remains stationary, and energy is still conserved, we will, as you said, measure the car to accelerate from being stationary to moving at speed $v$. This increase in kinetic energy is provided by the corresponding decrease in potential energy from the battery. Here's the strange part: in this frame, the car exerts a force on the Earth, but the Earth doesn't accelerate. This means that there must be a fictitious force acting on the Earth in this frame that exactly counterbalances the force the car exerts. This fictitious force acts on all objects in this frame, including the car. So the car actually accelerates faster in this frame, which makes sense, because its total change in speed is greater than in the inertial frame. What's interesting here is that energy is not quite conserved until you account for the fictitious work done by the fictitious force. Remember that the original force on the car was only enough to accelerate it to $v-\Delta v_{Earth}$. If only that force was doing work on the car, then the car wouldn't be able to accelerate to a speed of $v$. The ficitious force did some work on the car, in this situation, transferring energy from the Earth to the car, which accounts for the extra kinetic energy the car has that "comes from nowhere" in this frame. So the car receives energy from both the battery and the Earth. This will become more important once we analyze frame B2.

In frame B1, we will assume that the observer is not fixed to the Earth (as in frame A1). The observer initially sees both the car and the Earth moving backward at speed $v$. As such, the total kinetic energy in this frame initially is $\frac{1}{2}(m+M)v^2$. After the car accelerates, it is stationary in the observer's frame, but the Earth is now moving faster than it was before, at $\frac{1}{2}M(v+\Delta v_{Earth})^2$. The battery loses $\frac{1}{2}mv^2$ of potential energy in this frame to effect this change. Rewriting the final kinetic energy, after some long algebra, should lead you to get exactly $\frac{1}{2}(m+M)v^2+\frac{1}{2}mv^2$; in other words, the energy of the battery went into accelerating the Earth.

In frame B2, the observer is fixed to be moving along Earth's surface at speed $v$. Just as in frame A2, frame B2 is non-inertial, and as such, we can expect to see fictitious forces here too. Borrowing the same logic as in frame A2, the car decelerates faster in this frame than in frame B1, due to a presence of a fictitious force that is responsible for accelerating the frame. But this frame might be even stranger than frame A2: though the battery has lost $\frac{1}{2} mv^2$ of potential energy, the kinetic energy of the car is required to be zero, and the Earth doesn't gain any kinetic energy after the deceleration. So where does this energy go? Well, remember that to ensure energy conservation, we have to account for the work done by the fictitious force. In frame B1, kinetic energy was added to the Earth and removed from the car by the force provided by the battery. But the fictitious force of this frame specifically prevents the Earth from moving any faster. Therefore, the energy of the battery was spent doing work on the Earth against the fictitious force in this frame.

Summary:

Frame A1 (initially at rest, floating):

• Battery does work on both car and Earth
• Car ends acceleration not quite at speed $v$

Frame A2 (initially at rest, fixed):

• Battery only does work on car
• Fictitious force only does work on car
• Forces are aligned, so work done on car adds
• Car ends acceleration at speed $v$

Frame B1 (initially at speed $v$, floating):

• Battery does work on both car and Earth
• Car ends acceleration not quite at rest

Frame B2 (initially at speed $v$, fixed):

• Battery only does work on Earth
• Fictitious force only does work on Earth
• Forces are anti-aligned, so work done cancels
• Car ends acceleration at rest

Instead of an electric car (!) we can consider a mass $M$ on a horizonal spring. The sping is fixed by one end to the Earth (or to a RF A). In FR A the total pendulum energy $E=mv^2/2+kx^2/2$ is conserved, but in a moving RF B it is not. It is explainable with the fact that in the moving RF B the potential energy $kx^2/2=k(x'\pm Vt)^2)/2$ depends explicitly on time , that's it. There is no need to consider a "closed system" to make ends meet: - in a moving RF the energy is not obliged to conserve.

The simplest example is a tennis ball bouncing from a wall. If the wall is at rest, the tennis ball energy is concerved all the time. If the wall is moving, no ball energy conservation may be required ;-)

• In frame B, moving with velocity $V$, the total energy of the mass on the spring is $E=mv^2/2+k(\Delta x)^2/2+mV^2/2$, which is still conserved, because all we've done is add a constant term to the energy (if frame B is inertial). It's not enough to be moving, as energy conservation holds in all inertial frames, even the moving ones. The potential energy term depends only on the distance between the mass's current position and its equilibrium position. Both of those positions are moving, so no, the potential energy term is not time-dependent if frame B is inertial. Aug 1, 2018 at 3:26
• @probably_someone: I understand your desire to "extend" the original system in order to prove that the total energy of the extended system is conserved. However, I still consider the original system (a mass $M$ on a massless spring) in both inertial systems. The energy being non-conserved is explained with explicit time dependence of the potential energy in a moving RF. Aug 1, 2018 at 3:54
• Your potential energy term is wrong. I'm not suggesting an extension, I'm making a correction. A spring with equilibrium position $x_0$ has potential energy $k(x-x_0)^2/2$. When you transition to a moving frame, you have to apply both $x\to x+Vt$ and $x_0\to x_0+Vt$, so that the potential energy is $k(x+Vt-x_0-Vt)^2/2=k(x-x_0)^2/2$, which is exactly what it was in the rest frame. This is still true if you now take $x_0=0$ in the rest frame (which would imply that $x_0=Vt$ in the moving frame). Aug 1, 2018 at 3:59
• In a moving RF the potential energy becomes dependent explicitly on time due to $Vt$. If you arrive at another conclusion, you are wrong. Aug 1, 2018 at 4:10
• Where is the spring's equilibrium position in the moving frame? And where is it in the at-rest frame? Aug 1, 2018 at 4:35