# Relativistic spring

Imagine a cart at the base of the hill. The cart is next to a spring. You push the cart into the spring until the spring compresses $x$ meters. In your frame of reference, the cart has a potential energy of $\frac{1}{2} k x^2$ from the spring. You release the spring, and the cart goes over the hill.

Before releasing the spring, an alien passed the scene going very close to the speed of light. From the alien's frame of reference, the spring only compresses $x / \gamma$ meters, so the cart only gains an energy of $\frac{1}{2}k\frac{x^2}{\gamma^2}$. We can assume that the alien was flying so close to the speed of light that $\gamma$ was large enough so that the cart did not have enough energy to cross the hill. How does the cart make it over the hill in your frame of reference, but not the alien's?

Note that in this example one could replace the cart with a charged particle and the hill with a change in voltage (or even just replace the cart with an arbitrary body and the hill with an arbitrary potential barrier) and have a similar outcome, so gravitational potential and the gravitational mass of the cart aren't important.

Also, though it doesn't alter the essence of the question, analyzing how $k$ might change relativistically seems to make the discrepancy only a bit better: as detailed below, $k$ increases proportionally to $\gamma$, so the cart still gains less energy, namely $\frac{1}{2}k\frac{x^2}{\gamma}$.

I initially thought that $k$ must increase by a factor of $\gamma^2$ to cancel the length contraction. However, I think it actually increases only by a factor of $\gamma$: for an elastic material (which will behave as a spring), the spring constant is given by $k = \frac{EA}{L}$, where $E$ is Young's modulus, $A$ is the cross sectional area of the material, and $L$ is the initial length of the material. In turn, Young's modulus is given by $E = \frac{stress}{strain} = \frac{F / A}{\Delta L / L}$, where $F$ is a force acting on the body, $A$ is the cross-sectional area upon which $F$ acts, $\Delta L$ is the amount the length changes due to the force $F$, and $L$ is the initial length. Considering Young's modulus in a relativistic frame (with $v$ parallel to $F$ and $L$), the contraction of $\Delta L$ and $L$ cancel, and $A$ does not change because it has no component in the direction of motion. $F$ does not change either— according to the relativistic Newton's second law, the component of force parallel to velocity (in this case the only component) is given by $\gamma^3 m a$, where $a$ is the acceleration of the body in the same frame of reference as the force we're considering (see edit). That acceleration would be $\frac{1}{\gamma^3}$ times the "rest" acceleration of the body (i.e. the acceleration of the body in a frame where the body's velocity is not close to the speed of light, for example the ground's frame of reference). The factors $\frac{1}{\gamma^3}$ and $\gamma^3$ cancel, so $F$ is unchanged. Returning to the spring constant, considering that in a relativistic frame would give $k = \frac{E A}{L / \gamma} = \gamma k_0$, where $k_0$ is the value for the spring constant in the original frame of reference. This doesn't substantially change the paradox.

EDIT: stuffu is correct in that I used the wrong acceleration when calculating how much force should increase. I initially assumed that in $F = \gamma^3 ma$, the acceleration referred to some sort of rest acceleration, which doesn't make sense. It actually refers to the acceleration in the same frame of reference in which we are considering $F$ and $\gamma$ (in the example that would be the alien's frame of reference). That changes my calculation for $k$ changing, but it does not substantially affect the main idea of the paradox.

• Isn't the hill shortened too as observed by the moving alien?
– Deep
Commented Aug 24, 2017 at 6:06
• Yes, but the length of the hill doesn't matter— only the energy required to cross it. In the case of an actual hill, the height is the only dimension that affects the energy, and the height doesn't change. Commented Aug 24, 2017 at 21:35

If the hill moves and pushes on something, the hill does work.

If the spring is pushed by somethings that move, the moving things do work.

If we take the above mentioned things into account, we will probably notice that the net effect of those things is zero.

If in an electrostatic potential well a length contracted spring gives a negative charge some kinetic energy, then the magnetic field, that exists in the frame where the system has large velocity, can help the charge to leave the well. (the charge just has to deflect 90 degrees after being shot)

In the case of a gravity well, it's a gravito-magnetic effect that helps the projectile to escape.

In case of another spring, it's electro-magnetism inside the spring that helps spring A to compress spring B. (Spring B is at 90 degrees angle to spring B)

In case of a magnet being shot away from a chunk of iron, it's the electric field, that exists in the frame where the system has large velocity, that helps the magnet to leave the chunk of iron.

Your force calculations are wrong because you are using wrong acceleration. Acceleration transforms like this: a'=a/$\gamma^3$

From relativistic velocity addition it follows that the $\Delta v$ of the cart is divided by $\gamma ^2$. And from time dilation it follows that the duration of the acceleration is multiplied by $\gamma$. So the acceleration that the spring produces is divided by $\gamma^3$

Lengths transform like this:
longitudinal length: L'=L/$\gamma$
transverse length: L'=L

Forces transform like this:
longitudinal force: F'=F
transverse force: F'=F/$\gamma$

From those two transformation laws it follows that force multiplied by length, where force is aligned with the length, and which product we call here E, transforms like this in all directions: E'=E/$\gamma$

• Is there a reason why all of these different types of potential wells should each have some specific quality forcing the apparent energy decrease to exactly cancel? I didn't use any specific qualities of the potential well to ask the question, so it seems unsatisfactory that specific qualities of the potential well are necessary to resolve it— unless there is some law forcing those specific qualities to exist for any arbitrary potential well. Commented Aug 24, 2017 at 21:38
• You are correct about the acceleration; I misunderstood what I had been reading. I edited the question to fix that, but it doesn't make a qualitative difference. Commented Aug 24, 2017 at 21:40
• I don't know the inner workings of potential wells, do you? If we create a force field by shooting a continuous stream of carts, and the whole system moves fast in x-direction, there is large $\Delta$ p when a cart is shot in x-direction, 'normal' $\Delta$ p when a cart is shot in y-direction. If we say the carts are force carrying particles, then we have a force field, whose behavior is explained by the behavior of the carts. Commented Aug 25, 2017 at 6:12

I investigated what happens with the spring. The result was a bit surprising.

In the alien's frame of reference the spring in fact extends by the length $$x/\gamma$$ and the force increases from $$F=0$$ to $$F=kx.$$ You would then expect the work being $$W = F_{\text{mean}} x/\gamma = \frac12 kx^2/\gamma.$$

But there are two things that we have not considered:

1. the ends are moving,
2. relativity of simultaneity.

The work done at the front end of the spring is $$F_{\text{mean}} vt_{\text{front}} + \frac12 kx^2/\gamma$$ and the work done at the back end of the spring (assuming that this end is fixed in your frame of reference) is $$-F_{\text{mean}} vt_{\text{back}},$$ where $$t_{\text{front}}$$ and $$t_{\text{back}}$$ are the time lengths during which the forces from expanding the spring are acting on the front and back ends of the spring in the alien's frame of reference. The total work thus is $$W = F_{\text{mean}} v (t_{\text{front}} - t_{\text{back}}) + \frac12 kx^2/\gamma$$.

Now, because of relativity of simultaneity, $$t_{\text{front}} - t_{\text{back}} = \gamma xv/c^2$$ giving $$W = \frac12 \gamma kx^2v^2/c^2 + \frac12 kx^2/\gamma = \gamma \frac12 kx^2$$ which is in accordance with what should be expected from the formula for "relativistic mass" which can also be applied on energy.