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Say you want to store hot coffee in a container surrounded by a vacuum. To remove all sources of conductive energy loss the container is suspended in the vacuum by a magnetic field and does not have a physical connection to the sides of the vacuum chamber,

My question is would the magnetic field be a path for energy to be conducted out of the suspended container?

Another way to look at this question would be two magnets are suspended in a vacuum with their poles aligned. A heat source is attached to one of the magnets. Would the second magnet show a corresponding increase in temperature, excluding radiated heat transfer?

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You should say that you're ignoring the radiative energy transfer that would always be present. –  endolith Jun 6 '11 at 23:55

4 Answers 4

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Heat consists of random vibrations in a material. If a magnetic field connects two objects, then it creates a mechanical coupling between the two objects. Such a coupling will couple vibrations, therefore heat will be conducted by the magnetic field.

In practice, the effect will be very small, but given enough time, heat will be conducted by the magnetic field. The same applies to suspension by a combination electric and magnetic field.

This is not entirely incompatible with some of the other (wrong) answers. For example, since the suspended object is not at absolute zero, it is not possible for the magnetic field suspending it to be entirely static.

As an example, let's consider a superconducting metal container of gas suspended in a magnetic field:
enter image description here

Now suppose that the gas is at some non-zero temperature. Then there is a non-zero probability that the gas will concentrate onto one side:
enter image description here

In order for this to happen, assuming that the center of mass of the container remained constant, the metal part of the container had to have shifted in the opposite direction:
enter image description here

Assuming that the levitation is stable, a motion of the superconducting container must be opposed by the magnetic field. Thus the above motion is opposed by a force from the magnetic field. However, for every action there is an opposite and equal reaction (Newton's 3rd law), therefore there will be a corresponding opposing force on the suspension:
enter image description here

Therefore thermal motion in the suspended container will induce thermal motion in the base, hence there will be conduction of heat.

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A magnetic field is created and sustained by moving charges. In those terms, could you explain how a vibrating material lattice structure will conduct heat by the magnetic field? How would this be consistent with the claim that the fields are static? Do you mean to say that a permanent magnet really has minute fluctuations that allow the heat transfer? What would be the nature of such fluctuations? –  AlanSE Jun 3 '11 at 22:38
If there is a temperature involved, it's impossible for the fields to be static. I should probably edit in a calculation. –  Carl Brannen Jun 3 '11 at 23:11
Actually the fields could be static if the suspended object were so arranged that its thermal fluctuations resulted in no change in the suspension force, but this would require impossibly exact design. –  Carl Brannen Jun 3 '11 at 23:53
I think you are right that some small transfer of energy will take place. You are examining the question really: can a magnetic field transfer energy through a vacuum, and the answer is yes. If we go back to the question asked, it is not clear that it will be in the form of heat. In the two magnets case, a vibration might be set up. If one vibrates one of the magnets (50herz for example) the vibration will be transfered. I suppose that considering heat vibrations because the frequency is thermal one could call it heat transfer. –  anna v Jun 4 '11 at 4:12
@anna v; What I've described is only a single mode of vibration. In fact, it's insignificant compared to the (almost infinite) number of possible vibration modes. That these modes are not coherent is why we call the result "heat" rather than "vibration". An actual calculation is possible. One follows the outline given above (for example, the rate of heat flow will depend on the form of the restoring force). If I recall, the techniques for the calculation is covered in graduate physics statistical mechanics classes under the title "fluctuation dissipation theory". –  Carl Brannen Jun 4 '11 at 17:30

No, a static magnetic field does not conduct heat.

The only way for thermal energy to cross a vacuum is by radiation. Of course, radiative heat transfer actually does consist of electric and magnetic fields: hot objects give off electromagnetic waves, which travel through the vacuum and are absorbed by, say, the walls of the vacuum chamber. But those are waves, not static fields, and besides, they are actually emitted by the hot object - it's not like it takes advantage of preexisting fields to move the energy.

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This would be correct if there were a really static magnetic field. Since it is built up by magnetic fields collectively, which come from individual molecules, then just by the heisenberg uncertainty principle the field would not be static. –  anna v Jun 4 '11 at 8:20
continued: Think of the two magnets in the question above. If I impose a 50 herz vibration on one, the other will vibrate too. Thermal vibration of the individual molecules will also, in my opinion, introduce vibrations in the magnetic field and will affect the second magnet, as I commented on Carl's answer too. I would not exclude this without an experiment. –  anna v Jun 4 '11 at 8:23

I would say the first point to stress here is that heat by definition is the transfer of energy. 'Heat loss' can occur via convection, conduction or radiation. Clearly, the first two can be essentially reduced to zero by surrounding your coffee (or whatever) with a perfect vacuum. Creating a perfect vacuum is pretty much impossible though.

All objects at temperatures above absolute zero will radiate heat via emission of electromagnetic waves. (For more details, see this wiki article on black-body radiation.) This is a matter-of-fact, with or without a vacuum, hence why flasks don't keep your coffee warm indefinitely. A reflective layer (normally some sort of foil) is used to contain some of this radiated energy but it works better for some wavelengths, and worse for others.

To answer your question directly: no - The magnetic field used to suspend the vessel will not interfere with the energy being radiated, nor indeed does it constitute a medium through which heat can excape via any other means.

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The various comments on the earlier Answers to this Question point to its moderately fundamental misconception. Because of special relativity there is no such thing as a magnetic field separately from the electromagnetic field. Thermal fluctuations of a body, including vibrations, are transferred to a remote body by interactions between the bodies, the chief of which is likely to be electromagnetism at scales at which magnetism has been engineered to be a specific constraint on movement.

The first part of the Question obscures matters by introducing a magnetic field without introducing its sources --permanent magnets or electromagnets, say. I think this idealization is unhelpful. If there were really a magnetic field without any source for the magnetic field, then thermodynamics would be inapplicable unless there are other heat sources; it would further have to be specified whether there are any interactions between the various heat sources, and what those interactions might be. EDIT: An electromagnetic field is dynamical, in contrast to a magnetic field, so the electromagnetic field itself can be considered to be a separate thermodynamic system, in which case the heat of the confined container will dissipate into the electromagnetic field or vice versa, depending on the relative temperatures of the electromagnetic field and of the container.

The final sentence, "Would the second magnet show a corresponding increase in temperature, excluding radiated heat transfer?"[my emphasis] sets things up so that the answer is no by definition. If radiated heat transfer is excluded --an impossible idealization according to experience-- then there will be no change of temperature. This, however, is again not a helpful idealization.

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I see your points, trying to oversimplify the question by making to many idealizations and restrictions on important factors results in answers with very narrow and specific relevance which do not translate well to more general assertions about the fundamental nature of the system. –  Rob Jun 6 '11 at 19:06
That's a reasonable enough summary. What can and cannot be omitted from a model can ultimately only be learned from experience of the relative scales at which different types of detailed models can be used. To some extent that experience can be either theoretical or experimental, though most Physicists prefer there to be a mix of both. –  Peter Morgan Jun 6 '11 at 19:53

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