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Finally logged in; moment of inertia of the electric field.

  1. Jul 25, 2008 #1

    After some persistent trouble, I finally succeeded in registering and logging in. Hurrah.

    I cannot often use internet, but I will try to return on this forum at least once a week, ususlly in nor near the weekend.

    I have a question about electromagnetism and I searched first using the words "moment of inertia of the electric field". The search returned nothing, so here is my conundrum. (It is not my homework assignment, or anything like that, as I am 53 years old; it is something I recenty thought of myself.)

    Consider a spherical conductor of radius R, carrying an electric charge Q. There will be an electric field around it, and this field will contain some energy. Therefore, by Eintein's familiar formula it will contain some mass. This mass will increase the work needed to change the velocity of the sphere. (mvv/2) It will also, by its moment of inertia, increase the work needed to change the rotation of the sphere. (Iww/2)

    The size of these phenomena should be calculable from Q, R and some constants of nature. There will be no electric field inside the conductor, and the field outside can be subdivided into an infinite number of equally thin spherical shells concentric with the sphere. By Coulomb's Law, the field strength in every shell will be proportional to the charge and inversely proportional to the square of the shell's radius r. Proportional, that is, to Q/rr. The energy density will be proportional to the square of the field strength, hence to QQ/rrrr. But the volume of the shell will be proportional to rr, so the energy contained in it will be proportional to to QQ/rr.

    Adding up the energies contained in all the shells yields a total energy which is proportional to QQ/R. This is, of course, as it should be. It is the energy one had to expend in gathering together the charge from infinity onto the surface of the sphere. This energy had to go somewhere; it has gone into the field surrounding it.

    But this energy has mass, and therefore adds to the moment of inertia relative to any axis passing through the center of the sphere. Here, too, every shell has its own contribution to make. Its moment of inertia is proportional to its mass multiplied by rr. Its mass, however, is proportional to QQ/rr. Therefore, its moment of inertia is proportional to QQ and independent of r. Every one of the infinite number of shells contributes the same amount. Adding up these amounts yields an infinite moment of intertia.

    How can I escape from this crazy result? I can think of only three counterarguments.

    1. The situation has spherical symmetry; therefore rotating the sphere does not change the field strength anywhere, which makes it meaningless to say that the field is rotating with the sphere.

    But any scratch or bump on the sphere's surface would create an inhomogeneity in the field, which would rotate with the sphere. Moreover, rotating the sphere might set up a circular current, causing a magnetic field to form, and changing the situation "outside" from what it would be for a nonrotating sphere. (Such a current would certainly exist if the charge had been sitting on the surface of an insulating sphere.)

    2. The field does not propagate at infinite speed; therefore charging the sphere cannot set up a field extending to infinity in finite time.

    But this suggests that the moment of inertia would steadily, and linearly, increase without bounds. One would be able to ascertain how long ago the sphere was charged by measuring the torque needed to set it spinning.

    3. It is impossible to gather a charge from infinity; therefore the charge on the sphere came from somewhere else, where an equal but opposite charge must now exist. The fields from the two charges will cancel at large distances from both, and anything contained in these fields will therefore have a finite value.

    But this would still allow one to create a device carrying a huge moment of inertia relative to the axis separating the two charges while expending just a tiny amount of energy. If one considers two charges Q and -Q, both sitting on a sphere of radius R with the midpoints of these spheres separated by a distance D, the energy would for large values of D/R approach a value proportional to QQ/R, but the moment of inertia would be roughly proportional to QQD.

    I wonder whether there is another counterargument.

    (I also wonder whether, if I log out, I will again be able to log in; so if you never hear from me again, I probably wasn't able to. But I'll try anyway tomorrow.)

  2. jcsd
  3. Jul 25, 2008 #2
    The electric field will not really rotate. If you think in terms of fieldlines you can say that field lines can never move. They can get stronger or weaker but they will never move. It's similar to a water wave. The water isn't really moving with the wave. The water only moves up and down. So if you move a charged sphere from point A to point B the field at A will get weaker and the field at B will get stronger but it isn't actually moving.
    However the electrons that produce the field are moving. The inertia depends on the speed of those excess electrons which are all on the surface of the sphere.
  4. Jul 26, 2008 #3

    I have been able to log in again, so the trouble I experienced earlier seems to have become a thing of the past.

    Dr. Zoidberg, your reply is a variant of my counterargument number 1; the field as such does not rotate. But if we think in terms of field lines, then the strength of the field at a certain point in space is proportional to the number of field lines piercing a unit surface perpendicular to these lines. (They are in a way a book keeping device, keeping track of how the field strength can vary in a volume withouth sources; i.e. without charge. They do this by not being allowed to begin or end anywhere but on a charge.) If the rotating sphere is perfectly homogeneous, no field line need move to keep track of the field strength. But if there is a tiny scratch or bump, some field lines will have to move, and there will always be some field lines moving with the rotating sphere. This means than some mass (or energy) is orbiting the sphere, and this mass must carry a moment of momentum. A moment of inertia must therefore exist, which is found by dividing the moment of momentum J by the rotation frequency w. I = J/w. It would seem strange if only an asymmetric field has such a moment of inertia. Because the symmetric part could then be subtracted from the total field, and that might be done in several ways. (The symmetric part might be a weak as the weakest region of the total field, or as strong as the strongest region.)

    As regards the electrons, their movement (if constrained to move with the sphere) would of course have its own moment of momentum, quite apart from the field. The same is true of the protons and neutrons making up the rest of the sphere. That is; the sphere itself has a moment of inertia. But I feel that this is a different problem.

    Am I right in understanding that a completely spherically symmetrical field would have no moment of inertia at all?

    I will try to return in the beginning of next week.
  5. Jul 26, 2008 #4
    Even if there is a bump in the sphere the field will not move. The field changes. But changing just means it gets stronger at one location and weaker at another.
    You can say it's a basic law of physics that electric and magnetic fields can never move. They can only get stronger or weaker.
    In particle physics an electric field is said to consist of virtual photons. That means a charged sphere will constantly send out virtual photons. Those photons then cause a force on other charges. Of course the virtual photons won't move with the sphere.
    A flashlight for example will send out real photons. If you move the flashlight, the photons that have already left the flashlight don't move with it.
  6. Jul 29, 2008 #5
    However, if I move the flashlight, any new photons will be sent into a new direction, so the same might apply to the virtual photons in an electromagnetic field. Here I would like to compare the rotating sphere with a lighthouse, sending rotating beams of photons out over the sea. Of course the beams will be slightly bent, for the photons travel at a finite speed. And they will not be infinitely long if the lighthouse was built a finite time ago. That is just my counter-argument 2: nothing infinite can be set up in a finite time, But one might imagine a charged sphere which has been rotating for a very long time, and my reasoning suggests that it would be very hard to stop such a sphere.

    I have been thinking about the electrons. I feel that their individual mass, charge and other properties cannot be relevant here, because the same reasoning would apply to a negatively charged sphere (which has an excess of electrons) and a positively charged sphere (which has a deficit). The spheres themselves would have slightly different masses, assuming that they were identical before being charged. But the field around them would carry the same mass for a certain amount of charge.

    I have also been thinking about magnetism. It felt safe not to include magnetism because including it could only increase the mass everywhere in the field by an amount proportional to the square of the magnetic field strength. If it could not decrease the total mass, it could not decrease the total moment of inertia, which had already been found to be infinite.

    However, if I could include magnetism, this would give me another way to calculate the moment of inertia, by first calculating the angular momentum (moment of momentum) in the field. There is something called the Poynting Vector, which purports to give the local momentum density in any electromagnetic field. It is the vector product of the electric field strength E and the magnetic field strength B, scaled by a suitable constant. Summing all the several contributions from all locations would give a total angular momentum, which could then be divided by the rotation rate to give the moment of inertia.

    Now I can't find the magnetic field for a rotating charged sphere, but at large distances it must be nearly equal to the field of a magnetic dipole, whose strength is proportional to 1/rrr, or smaller. (B[x] with 3xz/rrrrr, B[y] with3yz/rrrrr, and B[z] with (rr-3zz)/rrrrr, for a rotation around the Z-azis.) I know that everywher E[x] is proportional to x/rrr, E[y] to y/rrr and E[z] to z/rrr. It is therefore possible to approximate the Poynting vector at large distances. The vector product of the radius vector with this Poynting vector would approximate the local contribution to the angular momentum. And because the infinity arises at the large distances, a finite result out there would be enough to contradict the earlier reasoning..

    The total angular momentum will be in the Z-direction, so I only want to consider J[z], or its local density j[z]. This would be found from xP[y]-yP[x]. P[x] would be proportional to B[y]E[z]-B[z]E[y], and P[y] to B[z]E[x]-B[x]E[z]. Therefore j[z] will be proportional to B[z](xE[x]+yE[y]+zE[z])-E[z](xB[x]+yB[y]+zB[z]). The first term is proportional to B[z]/r, or (rr-3zz)/rrrrrr, the second term to (z/rrr)*(z/rrr), and the sum over a shell with radius r and surface rr must be proportional to rr/rrrr or 1/rr. Hence the total angular omentum J[z] would be roughly proportional to 1/R, and therefore finite, even if nothing can be calculated exactly.

    But there is something fishy about this, isn't there? Some mass seems to be orbiting the rotating sphere alright, but not at the same angular speed as the sphere itself, or the result would be infinite.

    I will try to return next weekend.
    Last edited: Jul 29, 2008
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