Say we had a %100 efficient spring controlled by an internal computer. We compress it and leave it alone and measure from afar.
The spring system has a certain mass-energy which will cause it to have a certain gravitational pull. If the spring is sprung, it will start to move faster and so will “weigh” more since it is increasing its mass energy.
But a previous thread (about the same sort of situation but w/r/t gravitational potential rather than spring potential) led me to believe that the spring’s gravitational effects from afar will not change because its stress-energy tensor will still “add up” to the same.
So, when the spring is at rest, the “potential energy” rests in the stress portion of the stress-energy tensor, thus producing the same gravity from afar as would happen once the spring has sprung and is thus moving faster?
There seem tio several parts to your question. The energy in a distorted spring is stored in the stretched or compressed electrostatic fields between the atoms.
You are also asking about the relativistic mass of a compressed spring versus a relaxed spring + kinetic energy why I am not qualified to really answer
The spring does not move. Some parts of the spring will travel faster than other bits, but because the Center of Mass is stationary, relativistic mass change (assuming that your spring has stored enough energy to allow this) in one direction is counterbalanced by the same mass moving in the other direction. From outside the system, nothing changes.
*two suitably large masses with a spring (100% efficient) compressed between them. The spring has enough energy stored to accelerate the masses to relativistic velocity on release. The system is isolated.
At this point, the system has a mass, and thus a gravitational field.
The spring is released. The OP then believes that the system will gain a net relativistic velocity, which affects the observed mass/gravitational field, and wishes to know where the energy that produced the mass change was stored.*
This is wrong, of course. While the two individual masses may achieve relativistic velocities, they are in opposite directions, and thus the effects cancel out. The center of mass does not move, and there is no change in the system.
Let’s say the momentum were transferred to the two halves, instead of from a 100% efficient spring, from a 100% efficient fusion reactor. There would definitely be less fermionic mass after the two halves were accelerated.
So does the gravitation of the fusion-based system decrease, because the velocities are exactly counterbalanced from the two halves, but containing less matter, even if the system is 100% efficient? Or, rather, observing from the outside, is the mass-energy of the missing matter transferred to the mass-energy of the separate halves of the system somehow, thus creating the same gravitation as observed from outside?
Oblique, but FWIW there’s a classic experiment where two identical springs are dissolved in acid, but one of them is compressed and the other is not during their immersion. The acid gets a little warmer in the compressed case than in the relaxed case.
IANAP, but I think the compressed spring weighs more (the increase could not be measured, but it is there). The reason is that energy is mass and mass is energy and there is energy in the compressed spring. Where is that energy stored? In the inter-atomic bonds that are not in their lowest energy state. I would like to hear a real physicist confirm this, though.
Hari, I think your statement is correct. I’m a real physicist. But, Einsteinian relativity (the kind you mean, as opposed to Gallillean or Newtonian or others) isn’t my area and I haven’t studied it since the '70s.
I’m going to talk about the situation described by si_blakely: two equal masses at the ends of a spring, initially compressed and at rest, then released so that the masses vibrate back and forth.
The energy of a compressed spring is stored in the interatomic electrostatic forces within the spring, as scm1001 and Hari Seldon say. This additional energy appears in the electromagnetic stress-energy tensor, which is a component of the full general-relativity stress-energy tensor.
When the spring is released, the energy stored in the spring is transferred to kinetic energy in the two masses (and gravitational potential energy, to separate the two masses; of course this is typically very small). So energy is transferred back and forth between the electromagnetic and kinetic parts of the stress-energy tensor. But the total energy of the system is conserved (mostly; see below), so the gravitational field as seen from far away is (mostly) unchanged.
The caveat is that this pair of vibrating masses has a varying quadrupole moment, so it will emit gravitational radiation. This removes energy from the system, so the amplitude of the vibrations will eventually die down; and the gravitational radiation travels outward at c and can be detected far away from the system.
Thanks U3. I knew that neutron-star-level pressure could squeeze electrons into different orbits but did not know that the same could occur for any old spring.