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Preface

In classic physics the potential energy of a compressed (or stretched) spring is

$E_{\rm sp} = (\frac{1}{2})k_{\rm sp}(\Delta l)^2$

where $\Delta l$ is the length that the spring is compressed by, $k_{\rm sp}$ is spring constant.

Also in classic physics the potential energy of two charged particles is

$E_{\rm el} = k_{\rm el}\frac{q_1q_2}{r}$

where $q_1$ is charge of particle 1, $q_2$ is charge of particle 2, $r$ – distance between the particles, $k_{\rm el}$ – Coulomb constant.

The main part

Now, suppose we have 2 observers: one is at rest with the spring and the two charged particles (lets call this observer Ralf); and the other is moving relative to these objects with a speed close to speed of light (lets call this observer Maria).

For Maria the formulas above will be different due to length contraction:

$l’=\frac{l}{\gamma} $

where $l$ – proper length, $\gamma$ – Lorentz factor.

As I understand, for Maria the potential energies of the spring and the two particles will be $\gamma$ times the potential energies measured by Ralf, that is

$E_{\rm sp}' = \gamma E_{\rm sp}$

$E_{\rm el}' =\gamma E_{\rm el}$

where $E_{\rm el}'$, $E_{\rm sp}'$ - energies measured by Maria, $E_{\rm el}$, $E_{\rm sp}$ – by Ralf.

Here I have two questions.

  1. If we try to calculate potential energy by lengths or distances, then potential energy turns out to be frame dependent. I can barely believe that, because I always thought that any type of internal energy (including potential one) is invariant both in classic physic and in special relativity. Is this true? If it is, then Ralf and Maria should get the same values for potential energies.

  2. If potential energy is frame dependent, then how can energy conservation law hold in special relativity? Examine a situation of a cart at the base of a hill and it is being pushed up to the hill by a compressed spring. Let cart start its motion with a smallest possible speed just sufficient to pass the peak of the hill. All the observers (let them move horizontally) must agree on that the cart succeeds in passing the peak of the hill. This event requires that spring gives some energy ($E=mgh$, where $h$ – height of the hill) to the cart, otherwise the cart can’t get to the peak.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Buzz
    Commented Jan 17, 2023 at 19:14

2 Answers 2

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The value of any energy measured by different observers (relativistic or non-relativistic) might be different. The conservation of energy states that whatever the energy measured by one observer remains constant for him/her. For example, if you have an object of mass $1kg$ moving at speed $v=2m/s$, its kinetic energy in your frame is $2J$. If there are no forces it will remain $2J$ forever. For someone moving with the object, the speed is 0, hence kinetic energy is $0J$. Similarly, the value of potential energy will depend on the observer. Even a given observer can measure different values of potential energy depending on where they put their reference point. (center of earth/surface of the earth)

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  • $\begingroup$ I understand that 0 value for potential energy is arbitrary, but lets thinks that 2 observes always can agree on the same condition for 0 (like a spring is not stretched nor compressed). In a textbook by Eric Mazur there is the following extract. “Remember that the internal energy of an object or system is a quantitative measure of the state of that object or system. Because the state of an object cannot depend on the motion of the observer relative to the object, internal energy also must be an invariant.” $\endgroup$
    – Alexandr
    Commented Jan 14, 2023 at 13:18
  • $\begingroup$ IMO, the potential energy of a system is always internal for that system, so it should be invariant (if Eric Mazur is not mistaken) $\endgroup$
    – Alexandr
    Commented Jan 14, 2023 at 13:22
  • $\begingroup$ I think by "internal energy," he means rest mass, which is invariant. Total energy generally, transform like $0th$ component of 4 vectors so it can not be invariant. Electrostatic potential "$\phi$" for example is actually $0th$ component of vector potential {$\phi$,$\vec{A}$}. Potential energy is charge times the potential $\phi$. The charge is invariant but $\phi$ is not. Thus electrostatic potential energy is not invariant. $\endgroup$
    – bhoutik
    Commented Jan 14, 2023 at 13:55
  • $\begingroup$ You are definitely right about “rest mass” to be invariant, but I am afraid that Mazur means exactly what he writes. Here is another extract in the following page: “The work done on the spring during the compression increases the spring’s potential energy and therefore the spring’s internal energy.” $\endgroup$
    – Alexandr
    Commented Jan 15, 2023 at 7:30
  • $\begingroup$ I quote the extract above so that you can see how Eric Mazur uses term “internal energy”. So internal energy is internal energy (anything except energy of motion of the whole system). $\endgroup$
    – Alexandr
    Commented Jan 15, 2023 at 8:12
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Examine a situation of a cart at the base of a hill and it is being pushed up to the hill by a compressed spring. Let cart start its motion with a smallest possible speed just sufficient to pass the peak of the hill.

Moving observer sees this:

The extra small internal potential energy of the spring becomes the extra small internal kinetic energy of the cart-planet system, and that kinetic energy becomes the extra small internal gravitational potential energy of the cart-planet system.

You have a correct idea: If some of the internal energies is not extra small, then there is a problem.

So we can conclude that all types of internal energies of a moving object are equally small.

Now, if we consider a moving arrow hitting a tree, we now know that there occurs an increase of all internal energies of the arrow, which we probably have not thought about until now.

So there is extra energy involved in the collision. So the arrow had extra kinetic energy.

Se we know what becomes of the internal energies of an object whose internal energies decrease as the object's speed is increased. Kinetic energy.

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  • $\begingroup$ I am sorry, but I don’t understand your following comments: “You have a correct idea: If some of the internal energies is not extra small, then there is a problem.” “So we can conclude that all types of internal energies of a moving object are equally small.” I didn’t say anything about smallness of potential energies, I did just indicate that any two observer must agree on that the cart got specific amount of energy from the spring (in the example I described); and my thought was that any observers should get same value of that energy (equal to $mgh$). Maybe I am wrong. $\endgroup$
    – Alexandr
    Commented Jan 15, 2023 at 9:54
  • $\begingroup$ @Alexandr A planet moves to direction that its north pole points to. At the pole there is a hill, whose height is contracted, and a vertical spring, that is length contracted. Contraction has decreased the energy required to climb the hill, by 1/gamma. And contraction has decreased the energy of the spring, by 1/gamma. See, same transformation, and no problem. Do you know how forces transform? $\endgroup$
    – stuffu
    Commented Jan 15, 2023 at 10:28
  • $\begingroup$ Ok I see the point of your example, but I described the other one. I repeat: a horizontal spring is compressed and attached to a cart at the base of a hill, the spring pushes the cart, so that it moves up to the hill. For an observer at rest with the hill the cart reaches the peak of the hill and stops (i.e. the cart gets the minimal energy required to get to the peak - $mgh$). What would see an observer who moves horizontally at very high speed? Note that in this example, the height of the hill is not contracted, because height h is perpendicular to the velocity of the 2nd observer. $\endgroup$
    – Alexandr
    Commented Jan 15, 2023 at 10:29
  • $\begingroup$ So these 2 observers see the same $mgh$ energy but different energies of the spring (before it starts pushing the cart), because the second observer sees length contraction (he/she moves horizontally and the spring is horizontal). $\endgroup$
    – Alexandr
    Commented Jan 15, 2023 at 10:33
  • $\begingroup$ As a consequence, these observers see different amounts of the energy that the cart gains. It is very weird to me. $\endgroup$
    – Alexandr
    Commented Jan 15, 2023 at 10:35

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