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Transfer of Energy through Forces like gravity

  1. Oct 10, 2013 #1
    In a Newtonian Universe where gravitational and electric forces are instantly experienced, Energy is exchanged between 2 objects acting on each other in order to affect their motion. However in a relativistic universe, these forces travel at the speed of light and so given distance d between these same 2 objects, it take d/c seconds in order for the force to be experienced.

    Assuming I have this correct, what happens to the energy to be transferred over this time that the force is travelling before it's actually transferred to the other object. I don't think it could continue being held by the object for this time and then instantly jumping over afterwards but I can't think of a better solution and haven't had any luck researching. Anyone know? Thanks.
     
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  3. Oct 10, 2013 #2

    WannabeNewton

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    The information propagates via a field e.g. the electromagnetic field.
     
  4. Oct 10, 2013 #3

    PeterDonis

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    Gravity travels instantaneously in Newtonian theory, but electromagnetism does not; it travels at the speed of light.

    In the case of electromagnetism, the energy is stored in the electromagnetic field. In fact, the electromagnetic force felt by a charged object is standardly viewed as being caused by the EM field at the object's location, not by the distant sources of charge that ultimately cause the field. Note that this is true even if we are using Newtonian mechanics, not relativistic mechanics (see above).

    In the case of gravity, in relativity gravity is not a force; it's spacetime curvature. However, in certain situations, gravity can be viewed similarly to electromagnetism: the motion of freely falling objects can be viewed as being due to the gravitational field in their vicinity, and the field is ultimately caused by distant sources (objects with mass-energy). On this view, the energy transferred between objects by gravity is stored, while being transferred, in the gravitational field.
     
  5. Oct 10, 2013 #4
    Thanks!

    This is what I thought first, but wouldn't this require that only the necessary amount of energy is being transferred that will be transferred to all other objects, which I can't see quantum mechanics allowing? Or does it require constantly giving off energy in order to maintain the gravitational field as it propagates out and therefore to bend space-time? And if this is the case, does that mean that it will slowly disappear from giving off all its energy?
     
  6. Oct 10, 2013 #5

    PeterDonis

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    If you mean, does this viewpoint satisfy conservation of energy, yes, it does. Remember that, in the case of gravity (general relativity), this viewpoint is limited to certain situations; it is not completely general. The completely general answer is that gravity is not a force and does not transfer energy; it's spacetime curvature.

    Why do you think this?

    In the situations where this viewpoint can be used with gravity, the gravitational field is static, so it does not need to be "maintained", and the object that is the source of the field is in a steady state.
     
  7. Oct 10, 2013 #6
    It doesn't seem like Quantum Mechanics would allow it because it would require being certain of the position of the object when the energy is transferred.

    What I got from your answer is that Space-time is bent by the presence of Energy and this is what causes the change in motion. But a transfer of Energy is necessary from one object to another is necessary for this and so the energy is stored in space-time over the duration of time the transfer is made. I just don't see how this transfer is made so that just the right amount of energy is exchanged. Unless you're saying that Space-time transfers energy to object based on the curvature there and then to maintain its energy to comply with the law of conservation of Energy takes energy from other surrounding objects making them appear to be a source?
     
  8. Oct 10, 2013 #7

    PeterDonis

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    The energy doesn't get transferred directly between objects; energy is exchanged between the object moving in the field, and the field itself. The source of the field doesn't gain or lose energy at all. See below.

    Yes, this part is OK.

    No, it isn't. The object that is the source of the gravitational field (e.g., the Earth) does not gain or lose energy at all; it is static, as is the field it produces.

    The object that is moving in the field exchanges energy with the field at its location; for example, a ball that is thrown upwards stores energy in the field, and then gains it back once it reaches the top of the trajectory and starts moving downwards again. But this all happens locally; the Earth's energy stays constant through all of this. The only role the Earth plays is to determine the global (static) state of the field.
     
  9. Oct 10, 2013 #8
    Awesome, Thanks! Just to check one thing, cases of gravitational waves are an exception to the object being static, right? I know that these are energy being emitted from the system as waves, just making sure it is the actual objects in the system emitting it and not an effect on the field caused by them.
     
  10. Oct 10, 2013 #9

    PeterDonis

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    Gravitational waves can't be emitted from a static system; in fact even some non-static systems can't emit them.

    Yes; for example, the Hulse-Taylor binary pulsar system has been steadily emitting them for the last few decades, and astronomers have measured the resulting energy loss from the system.

    Actually, if you look at the source objects in the system in isolation, they don't change; for example, each individual pulsar in the binary pulsar system doesn't change. What changes is the relationship between them: the orbital parameters. (These are actually what astronomers measure in order to see the energy loss from the system.) This changes the externally measured energy of the system as a whole (for example, the mass that would be measured if you placed a satellite in orbit about the binary pulsar system at a distance very much larger than the distance between the pulsars), but does not change the masses of the individual objects. So if anything, this looks more like a change in the "field" than a change in the source objects.
     
    Last edited: Oct 11, 2013
  11. Oct 10, 2013 #10
    Alright, thanks again! This was really confusing me.
     
  12. Oct 11, 2013 #11

    PeterDonis

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    You're welcome! Glad I could help. :smile:
     
  13. Oct 11, 2013 #12
    Isn't the whole thing with energy transfer on interactions the quantum mechanics in action?

    I have always seen it this way:
    Let's consider electromagnetism. There is some emitter antenna that emits radio waves. Let's say it emitted a very tiny amount of energy, equal to a single photon. The radio wave propagates radially, in all directions. One may get an impression that the energy is distributed equally throughout space. But when we put a receiver antenna in some neighbourhood and it happens to absorb the wave, then all the energy goes to it. Now it looks like a discrete photon flied between the emitter and the receiver. This is exactly the particle-wave duality.

    When we see the radio signal as a wave, there is a problem with the energy smeared everywhere and not in a single point.
    When we see the radio signal as a photon, there is a problem how the photon knew the place of the receiver in the first place.

    Of course, the receiver antenna has only certain probability of absorbing the photon. Exactly equal to the amount of energy it would receive in the classical case.

    So, waves expand in all directions, but they only carry probability of interaction, not the energy.
    Then, when the bodies have interacted, it looks like a discrete quantum of energy passed between them, knowing their position in advance.

    This not only applies to electromagnetic waves (excitations), but also to static sources. A static charged body spreads a static field around, that doesn't hold energy, but only probability. Only when another charged body happens to interact with it, it looks like it interchanged energy with the field. But again, if we wanted to say that the energy was stored in the field from the beginning, then we would have to admit that spread energy suddenly concentrated in one point.

    I think the case of gravity is similar. I don't know if I'm completely correct.
     
  14. Oct 11, 2013 #13

    PeterDonis

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    In this case, the emission can't really be viewed as a wave. To get observed wavelike behavior you need large numbers of photons. The single photon has a wave *function*, but that's not really the same thing.

    Not really. The photon's wave *function* has a part that looks like a spherical wavefront expanding radially at c, yes. But it also has longitudinal and timelike components that do *not* propagate at c. In the case of an ordinary EM wave, these components cancel out when summed over large numbers of photons.

    Sort of, but there are complications. See above, and further comments below.

    No, when we see the radio signal as a wave, it's because there are large numbers of photons, each of which can get absorbed at a different point in the receiving antenna. The pattern of absorptions looks like a wave.

    If that's a problem, it's a problem for the wave case too, because, as above, the observed "wave" is composed of a large number of individual photon absorption events.

    Then where is the energy while they are traveling? Is conservation of energy violated? If so, how? All of the quantum mechanical equations are consistent with energy conservation.

    Yes.

    Not really.

    Quantum mechanically, static sources interact by means of virtual particles (at least, this is one common view, which also has limitations). For example, a static charged body interacts with other charged bodies by exchanging virtual photons. (The main component of these photons that contributes to the interaction is the timelike one, IIRC: the components that appear in an observed EM wave don't contribute much to the static interaction amplitude.)

    On the virtual particle view, there is no "field"; there are just amplitudes for different possible virtual particle exchanges. The field is something that appears when you sum over a large number of virtual particle interactions. See below.

    If we sum over a large number of virtual particle interactions, yes. But as you say, the body interchanges energy *with the field*. In other words, at this level of description, it no longer looks like the charged bodies are interchanging energy directly; instead, each charged body only exchanges energy with the field at its location. The field can change if its source changes, but those changes are propagated at the speed of light, so there's no causality violation.

    No, we wouldn't, because the energy exchange is not between the objects; it's between each object and the field at its location. Everything is local. See above and my previous posts.
     
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