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Thread: Does Newton's Third Law apply to electrostatic forces in Curved Space-Time?

  1. #1 Does Newton's Third Law apply to electrostatic forces in Curved Space-Time? 
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    I have been unable to work my way through a seemingly simple problem: If you have two charges, q1 and q2 above one another in a gravitational field (say at r1 > r2), is the force on q1 equal and opposite to the force on q2?

    Certainly this is true in flat space-time, but I have reason to believe it is not true in curved space-time. Can anyone answer this with authority? Can you help me figure out how to do this calculation using covariant formulation of EM? (That is the harder question...)

    Rich L.
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    Curved or flat I would have thought that the distance from q1 to q2 would still be the same as the distance from q2 to q1. What reason would lead you to believe that the symmetry is broken?
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    I have a thought experiment that suggests otherwise. It is based on the thought experiment that Einstein used to deduce the gravitational red shift:

    Consider two positions in a gravitational field, A and B, where A is directly above B. Supported at B is a vertical rod that extends up to A. On this rod are two levers, one at A and the other at B. The lever at A is connected to a generator, so that if the lever is turned energy is harvested with 100% efficiency.

    Now suppose there is an experimenter at B who turns the lever one full turn. We will not make any assumption what torque B must exert to do this, but will call it TB. The work done by B is
    W_B=2πT_B, where T_B is the torque at B

    The work done at A is likewise
    W_A=2πT_A

    Now lets suppose the experimenter at A collects this energy via the generator and converts it to a photon directed back down to B. As a result of the gravitational red shift, when this photon arrives at B it will have a higher energy than W_A:
    W_A (at B)=2πT_A √((g_00 (at A))/(g_00 (at B) )), (Sorry about the equation, but LaTeX is not working for me...)

    But conservation of energy requires that the energy of the photon be equal to the work that B did on the lower lever. Therefore
    W_B=2πT_B=2πT_A √((g_00 (at A))/(g_00 (at B) ))

    T_B=T_A √((g_00 (at A))/(g_00 (at B) ))

    If each lever is the same length, the forces applied to the end of each lever must follow this same ratio.

    We can generalize this a little bit further by considering a threaded vertical rod with frictionless nuts at A and B. Since the force on the nut vertically will be translated into torque by the pitch of the threads, a very similar argument involving force times distance will again require that the forces applied to the nuts must also follow this proportionality when the torques are balanced.

    For example, if two positive charges are attached to the nuts, Newtons Third Law states that the force on each charge should be equal. However weve shown that if that were the case the rod would turn with a net torque at the upper end and energy could be harvested. However if the pitch of the threads were equal at top and bottom the distance between the nuts would not change, and thus the distance between the charges would not change. Therefore the charges would not be doing any work and energy would not be conserved.


    The idea I'm working on is that in curved spacetime, the electromagnetic potential function should not be calculated as 1/r but as 1/t, where t is the time between the field point and the retarded position of the source charge, as seen from the field point frame. This definitely works, but I'm trying to determine if the conventional calculation using Covariant EM gives the same result.

    Rich L.
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    Erm, are taking energy out if the EM file here or the gravitational field? Or both?
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    Neither. The conclusion is that a joule of energy at one level in the gravitaional field does not represent a joule at a different level. If you take a Kg mass at A and drop it to B below A, when that mass arrives at B it has both the rest mass energy of the mass and also the kinetic energy gained during the fall. From the point of view of an observer at A the moving mass at B has a total energy equal to its rest mass at A.

    Another way to see this is to consider a photon directed down from A to B below. When the photon arrives at A it will be gravitationally blue shifted and have a higher energy (from the point of view of B) than it did at A. From the point of view of A however, that photon has the same energy as it did when it left. If B reflects the photon off a mirror and sends it back up to A again, it will arrive with the same energy as it had originally at A.

    Energy depends on the observer. Likewise force. My thought experiment shows that a force deep in a gravitational field is "red shifted" as it is transmitted up a gravitational field just like photons are red shifted.

    Rich L.
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    Rich L, your thought experiment is making an assumption that because the energy is force x distance the forces must be different. General Relativity would say that it is the distances that are different.
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    Quote Originally Posted by Jilan View Post
    Curved or flat I would have thought that the distance from q1 to q2 would still be the same as the distance from q2 to q1.
    This may surprise you, but that is not necessarily true. For example, in the exterior Kerr space-time the length of world-lines from (1) to (2) may be different than the one going from (2) to (1), even though they are the same points in space.
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    Markus said: "This may surprise you, but that is not necessarily true. For example, in the exterior Kerr space-time the length of world-lines from (1) to (2) may be different than the one going from (2) to (1), even though they are the same points in space."

    I'm not too familiar with the Kerr metric yet, but I can believe this might be the case. My thought experiment is specifically for a stationary metric (i.e. Schwartzschild), and between two physical points that are stationary wrt themselves and the source of the gravitational field.

    In anycase, the effect I'm describing is despite the distances measured from each charge being the same measured from each side. This is why I think I can make an argument that more generally the potential should be 1/t instead of 1/r.
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    Are you trying to describe both fields as a single one? If so mega respect to you!
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    No, definitely not. I'm just trying to understand how gravity affects electromagnetism. I noticed that the 1/r potential we were all taught could just as well be a 1/t potential if t is taken to be the retarded time to the source charge. In Minkowski spacetime these would be indistinguishable. In curved spacetime these can be different things. If you consider two experimenters, one at the top of a ladder and one at the bottom, because of the gravitational red shift, the one at the top will say that light takes n seconds to reach the bottom and come back again. The experimenter at the bottom will say that light takes m<n seconds to make a round trip up and back. Yet both experimenters would measure the same distance if they laid out rulers along the ladder. I believe my thought experiment shows that electrostatic potential is actually 1/t, and I'm trying to solidify the argument for this, or else get someone to convince me that I'm missing something.
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    Rich, I commend your efforts but I am unclear as to how's this would differ from the current theory that describes pretty well how light is "deflected" by a gravitational field.
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    It should not affect propagation of light at all. Light travels on null paths (aka light cones). I think the understanding that light is red/blue shifted as it rises/falls in a gravitational field is compatible with the idea. The primary effect it would have is on the intensity of electromagnetic fields very near a black hole. At the moment I'm not even sure that the conventional covariant electromagnetic theory does not already account for this effect. That is really the point of my original question.
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    Quote Originally Posted by Rich L. View Post
    At the moment I'm not even sure that the conventional covariant electromagnetic theory does not already account for this effect.
    I wasn't following the discussion up until now - can you quickly summarise in a sentence or two what effect it is you are referring to ?
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    Rich, it might be worth considering the charges inside an elevator in free-fall in a gravitational field. If we consider opposite charges that are attracting each other, in the elevator frame they would meet in the middle and we would conclude that the forces on each are equal. However for an external observer one would appear to have travelled further than he other one and we might conclude that the forces were different.
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    Quote Originally Posted by Jilan View Post
    Rich, it might be worth considering the charges inside an elevator in free-fall in a gravitational field. If we consider opposite charges that are attracting each other, in the elevator frame they would meet in the middle and we would conclude that the forces on each are equal. However for an external observer one would appear to have travelled further than he other one and we might conclude that the forces were different.
    You have to be very careful here, because electric charges in "free fall" do not exactly follow geodesics in space-time; the equivalence principle applies only to uncharged particles. The reason for this is conceptually quite simple - such a charge consists of both a particle carrying it ( which can be considered point-like and local ) and the accompanying electromagnetic field, which spatially extends into infinity and is hence most definitely not local. In a curved space-time, the spatially extended field will be distorted according to the covariant formulation of electrodynamics, and that distortion affects a back-reaction on the particle itself, introducing an additional force that induces a deviation from "free" fall under the influence of gravity only. The magnitude of this effect will vary according to where the particle is located at a given time. There is a name for this effect, but for the life of it I can't recall it at the moment ! This effect is undetectably small for weak gravitational fields, but it becomes important in the strong field domain. Actually calculating this is very difficult apparently, and even among experts there seems to be some disagreements on even the fundamentals, such as whether or not a charge in free fall will radiate, and who and how such radiation could be detected.

    I don't often reference other forum discussions, but in this case I think it is appropriate, due to the variety of good sources given here :

    http://www.physicsforums.com/showthread.php?t=160533
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    Thar's yer problem!

    Quote Originally Posted by Rich L.
    ...If you take a Kg mass at A and drop it to B below A, when that mass arrives at B it has both the rest mass energy of the mass and also the kinetic energy gained during the fall.
    No! Imagine you drop a 1kg bowling ball into a black hole. The mass of the black hole increases by 1kg. Not 1000000kg. The kinetic energy of a falling body comes from that falling body. The mass-equivalence of the rest-mass-energy plus the kinetic energy always adds up to 1kg.

    Quote Originally Posted by Rich L.
    ...Another way to see this is to consider a photon directed down from A to B below. When the photon arrives at A it will be gravitationally blue shifted and have a higher energy (from the point of view of B) than it did at A. From the point of view of A however, that photon has the same energy as it did when it left. If B reflects the photon off a mirror and sends it back up to A again, it will arrive with the same energy as it had originally at A.
    The descending photon doesn't gain any energy. If you direct a 511keV photon into a black hole, the black hole mass increases by 511keV/c. Descending photons don't gain any energy, and nor do descending bowling balls, because gravity is not a force in the Newtonian sense. Yes, you added energy when you lifted the bowling ball. Its rest mass is higher as a result. We call it gravitational potential energy. When you drop the bowling ball, that extra rest-mass energy is converted into kinetic energy. But photons are *just* kinetic energy, and you can't lift a photon like you can lift a bowling ball.

    Quote Originally Posted by Rich L.
    My thought experiment shows that a force deep in a gravitational field is "red shifted" as it is transmitted up a gravitational field just like photons are red shifted.
    Descending photons don't gain any energy. Ascending photons don't lose any.
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    It's a curious thing. I would say there are 3 forms of energy involved here. The rest mass, the kinetic energy and the gravitational potential energy. The second two will cancel out.
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    I have a lengthy response to Farsight below. He clearly did not follow what I meant, but I accept responsibility for that. In GR we must be very clear about who (and where) is the observer when talking about energy and momentum, etc. I was not sufficiently clear about that. I believe we are actually in agreement on this point, I just didn't state my argument clearly enough. My comments below:

    Farsight said:
    Thar's yer problem!

    Quote Originally Posted by Rich L.
    ...If you take a Kg mass at A and drop it to B below A, when that mass arrives at B it has both the rest mass energy of the mass and also the kinetic energy gained during the fall.

    Farsight: No! Imagine you drop a 1kg bowling ball into a black hole. The mass of the black hole increases by 1kg. Not 1000000kg. The kinetic energy of a falling body comes from that falling body. The mass-equivalence of the rest-mass-energy plus the kinetic energy always adds up to 1kg.

    Rich L. response: From the point of view of the person dropping the ball from radius R, and remaining there, the ball has constant total energy as it falls, as JiLan points out. It is only from the point of view of an observer at lower altitude that the ball has rest mass 1kg*c^2 plus a kinetic energy (=mgh if the acceleration of gravity was essentially constant during the fall). However, if the person at R does not drop the ball, but lowers it slowly on a string, then the energy of the ball decreases as it is lowered (as assessed by the observer at R), while the lower observer will see the stationary ball as having only the mass energy.

    Quote Originally Posted by Rich L.
    ...Another way to see this is to consider a photon directed down from A to B below. When the photon arrives at A it will be gravitationally blue shifted and have a higher energy (from the point of view of B) than it did at A. From the point of view of A however, that photon has the same energy as it did when it left. If B reflects the photon off a mirror and sends it back up to A again, it will arrive with the same energy as it had originally at A.

    Farsight: The descending photon doesn't gain any energy. If you direct a 511keV photon into a black hole, the black hole mass increases by 511keV/c. Descending photons don't gain any energy, and nor do descending bowling balls, because gravity is not a force in the Newtonian sense. Yes, you added energy when you lifted the bowling ball. Its rest mass is higher as a result. We call it gravitational potential energy. When you drop the bowling ball, that extra rest-mass energy is converted into kinetic energy. But photons are *just* kinetic energy, and you can't lift a photon like you can lift a bowling ball.

    Rich L. response: As for the photon, I am in agreement, but I stand by what I said, that for the observer at lower level the energy of the photon DOES increase per gravitational red shift. There is no absolute energy scale, so you have to be careful who is doing the assessment of energy. This is actually no different for the ball than for the photon. If you direct a photon into a black hole, or drop a ball into the hole, in both cases the energy represented by each at the starting radius R is constant as they fall freely. Directing a photon upwards is no different than launching the ball upwards. The total energy is constant for the person doing the launching, but the total energy gets smaller and smaller for observers high and higher in altitude.

    Quote Originally Posted by Rich L.
    My thought experiment shows that a force deep in a gravitational field is "red shifted" as it is transmitted up a gravitational field just like photons are red shifted.

    Farsight: Descending photons don't gain any energy. Ascending photons don't lose any.

    Rich L. Response: How do you account for gravitational red shift then, other than the clarification I've given above?
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    Quote Originally Posted by Rich L. View Post
    At the moment I'm not even sure that the conventional covariant electromagnetic theory does not already account for this effect.

    Markus said: I wasn't following the discussion up until now - can you quickly summarize in a sentence or two what effect it is you are referring to ?

    Rich L. Response: Markus, my question is: If you have two equal charges separated vertically in a gravitational field (e.g. Schwartzschild metric), is the electric field at each charge due to the other the same? It seems to me that even in Covariant EM in curved spacetime, that the electric fields are still calculated as the spatial gradient of the potential (using covariant derivatives, of course). For the simplest case of two stationary charges in a stationary field, this would necessarily give equal electric field magnitude at each charge for the case I'm asking about. Can you confirm that that is true?

    Rich L.
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    Quote Originally Posted by Rich L. View Post
    is the electric field at each charge due to the other the same?
    What quantity exactly do you want to compare ?
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    Markus,

    I want to calculate the electric field at each charge due to the other charge.

    If I have two equal charges, q, one at radius r1 and another at radius r2, at the same theta and phi coordinates, where r1 > r2, is the electric field at r1 due to the charge at r2 equal to the electric field at r2 due to the charge at r1?
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    Quote Originally Posted by Rich L.
    From the point of view of the person dropping the ball from radius R, and remaining there, the ball has constant total energy as it falls, as JiLan points out. It is only from the point of view of an observer at lower altitude that the ball has rest mass 1kg*c^2 plus a kinetic energy (=mgh if the acceleration of gravity was essentially constant during the fall). However, if the person at R does not drop the ball, but lowers it slowly on a string, then the energy of the ball decreases as it is lowered (as assessed by the observer at R), while the lower observer will see the stationary ball as having only the mass energy.
    Sounds reasonable. I think the thing to appreciate is that the observer loses energy when he's lowered on a string. So to him it looks like something has gained energy when actually it hasn't.

    Quote Originally Posted by Rich L.
    As for the photon, I am in agreement, but I stand by what I said, that for the observer at lower level the energy of the photon DOES increase per gravitational red shift.
    The observer will claim that the photon has increased in energy. But so will the observer who accelerates towards a photon in gravity free space. In both cases the photon energy didn't change. Instead the observer changed.

    Quote Originally Posted by Rich L.
    There is no absolute energy scale, so you have to be careful who is doing the assessment of energy.
    Agreed. But in similar vein there is no magical mysterious action-at-a-distance mechanism by which an object acquires energy. There is no surge of energy flying through space into the descending photon.

    Quote Originally Posted by Rich L.
    This is actually no different for the ball than for the photon. If you direct a photon into a black hole, or drop a ball into the hole, in both cases the energy represented by each at the starting radius R is constant as they fall freely. Directing a photon upwards is no different than launching the ball upwards. The total energy is constant for the person doing the launching, but the total energy gets smaller and smaller for observers high and higher in altitude.
    Because the higher observers have more energy. The height they attain depends on the energy they were launched with.

    Quote Originally Posted by Rich L.
    How do you account for gravitational red shift then, other than the clarification I've given above?
    It's an observer effect. The ascending photon isn't actually red-shifted, but people say it is. They say it has lost energy, but it hasn't. Instead they gained it.
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    Quote Originally Posted by Farsight View Post
    Sounds reasonable. I think the thing to appreciate is that the observer loses energy when he's lowered on a string. So to him it looks like something has gained energy when actually it hasn't.

    The observer will claim that the photon has increased in energy. But so will the observer who accelerates towards a photon in gravity free space. In both cases the photon energy didn't change. Instead the observer changed.

    Agreed. But in similar vein there is no magical mysterious action-at-a-distance mechanism by which an object acquires energy. There is no surge of energy flying through space into the descending photon.

    Because the higher observers have more energy. The height they attain depends on the energy they were launched with.

    It's an observer effect. The ascending photon isn't actually red-shifted, but people say it is. They say it has lost energy, but it hasn't. Instead they gained it.
    (All of a sudden the Quote button works for me! Great!)

    By taking this point of view you are, in effect, adopting a preferred reference frame. Doing this is your right, I suppose, but is counter to the consensus opinion about how to look at GR. That opinion, as I understand it, is that all reference frames are equally valid, just like in SR where any inertial frame is equally valid. In effect you are saying the "correct" frame to use is the one very far from any massive objects.
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    The quote button doesn't work for me! I have to copy and paste and put quote= in square brackets.

    Quote Originally Posted by Rich L.
    By taking this point of view you are, in effect, adopting a preferred reference frame. Doing this is your right, I suppose, but is counter to the consensus opinion about how to look at GR.
    Maybe. But I like to think I'm in line with Einstein's on this. Take a look at the quotes in this post. Year after year from 1911 thru 1916 he talked about how the special relativity postulate no longer applied.

    Quote Originally Posted by Rich L.
    That opinion, as I understand it, is that all reference frames are equally valid, just like in SR where any inertial frame is equally valid. In effect you are saying the "correct" frame to use is the one very far from any massive objects.
    It's more like the correct thing to do is to take a step back and see the big picture. By not limiting yourself to any one frame. Think of it as the "global" view rather than the "local view". And then remind yourself that one viewpoint is just as valid as any other.
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    Quote Originally Posted by Farsight View Post
    The quote button doesn't work for me! I have to copy and paste and put quote= in square brackets.
    I get the same problem. The first time I click on "quote" it turns into a spinner animation and nothing more happens. As a workaround, I found that if I then click on the spinner animation it opens the advanced edit form.

    I'm using Chrome version 35. I haven't tested the site with other browsers.
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    Quote Originally Posted by Farsight View Post
    Maybe. But I like to think I'm in line with Einstein's on this. Take a look at the quotes in this post. Year after year from 1911 thru 1916 he talked about how the special relativity postulate no longer applied.
    I hate to break it to you, Farsight, but after those quotes, he wrote a whole theory, one that appears to work to great approximation, that also rejects preferred reference frames. You might be interested in reading about "The General Theory of Relativity".
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    Quote Originally Posted by Rich L. View Post
    If I have two equal charges, q, one at radius r1 and another at radius r2, at the same theta and phi coordinates, where r1 > r2, is the electric field at r1 due to the charge at r2 equal to the electric field at r2 due to the charge at r1?
    I can't just pull this out of my hat, so I went to do some research, and it turns out that this is a highly non-trivial problem that was only solved in 1975 ( I think ); I won't even attempt to do this myself, not even for something comparatively simple as Schwarzschild. In any case, the answer seems to be no, the electric fields aren't the same. This makes sense if one just looks at the form of the Maxwell equations :



    The Hodge dual depends explicitly on the metric, and the differentials in the exterior derivative become covariant derivatives, so the resulting system of differential equations will depend explicitly on the Schwarzschild r coordinate. Even though the charges as described by J are identical, the resulting electrostatic fields are position-depended, since they are distorted by the curvature of space-time, which is not constant.
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    Quote Originally Posted by Markus Hanke View Post
    I can't just pull this out of my hat, so I went to do some research, and it turns out that this is a highly non-trivial problem that was only solved in 1975 ( I think ); I won't even attempt to do this myself, not even for something comparatively simple as Schwarzschild. In any case, the answer seems to be no, the electric fields aren't the same. This makes sense if one just looks at the form of the Maxwell equations :



    The Hodge dual depends explicitly on the metric, and the differentials in the exterior derivative become covariant derivatives, so the resulting system of differential equations will depend explicitly on the Schwarzschild r coordinate. Even though the charges as described by J are identical, the resulting electrostatic fields are position-depended, since they are distorted by the curvature of space-time, which is not constant.
    That might fit with what I'm calculating by a more intuitive (and less general) approach. I still don't have any sense of what the Hodge dual means or what effect it has on the electric fields. By my calculation, which is what I'm trying to validate, I get that the ratio of the electric fields should be the square root of the ratio of the time metric components, g_00 at each location. (This, of course, only applies to a restricted class of coordinates since it is based on the actual metric components. But the Schwarzschild metric is in this class.) I get that the electric field at the lower charge will be larger (by this ratio) than that at the upper charge. Does that seem reasonable based on your knowledge of this?

    Can you point me to a reference, such as the 1975 solution you mention, that might help me get a handle on this?
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    I'm kind of an "amateur relativist", but I think there's a massive chicken-and-egg problem with all this. One Einstein struggled with. Because he approached it from the wrong end. Think about an electron. It's got an electromagnetic field, field interactions result in force, and that force can be titanic. See hyperphysics. A metal ion is somewhat similar. But in the current-in-the-wire the forces almost cancel one another. There's a small "trace" force left, and as a result two wires move together. Or apart. OK, stop the current. There's still a small "trace" force left. Only we don't call it electromagnetism.
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    Quote Originally Posted by Rich L. View Post
    Markus,

    I want to calculate the electric field at each charge due to the other charge.

    If I have two equal charges, q, one at radius r1 and another at radius r2, at the same theta and phi coordinates, where r1 > r2, is the electric field at r1 due to the charge at r2 equal to the electric field at r2 due to the charge at r1?
    It might be a good strategy to start by considering the much simpler case of two "stationary" world-lines (i.e. field lines of the timelike Killing vector) that are extremely close together, and very close to the event horizon of the Schwarzschild solution, and see what you get. The advantage of that is that the small "tube" of spacetime containing the two world-lines (in coordinates which keep the particles at fixed positions) looks like a piece of Rindler metric (i.e. Minkowski space in accelerated coordinates), and so we have effectively reduced the problem to that of solving Maxwell's equations for two uniformly accelerated charges in ordinary Minkowski space (with slightly differing proper accelerations), such that the radar distance between the two charges is kept constant.

    So, consider two uniformly-accelerated particles in Minkowski space in inertial coordinates (t, x, y, z), both moving along the z-axis, which are very close and stationary at t = 0 and have very slightly different positive accelerations in the z-direction. The two world-lines are given by the two following equations:




    where the accelerations satisfy g2 > g1 > 0, and g2 is only very slightly greater than g1, and x = y = 0 for both world-lines. Note that the two world-lines described are indeed separated by a constant radar distance (which you can figure out as an exercise), just as we require.

    We now need the electromagnetic field of a uniformly accelerated charge. This is harder than it sounds, and involves some nasty subtleties, so I will just give the result without derivation (if you are curious, search for Electrodynamics of Hyperbolically Accelerated Charges by E. Eriksena and . Grn, and papers that cite it). The key result for our purposes is that if particle 1 has charge q, the E-field due to it on points along the z-axis has vanishing x and y components (which is obvious from the symmetry), and a z component given by




    The plus sign is required for z2 > t2 + 1/g12 (i.e. to the "right" of the particle). The minus sign is required to the "left" of the particle. All components of the B field vanish along the z-axis, so we don't have to worry about them.

    Great! Now we just calculate the field at the position of particle 2. The calculation is a piece of cake:




    As we'd expect, the value is independent of t. If we now go through the same process and calculate the field due to particle 2 at the position of particle one, we get




    Note the minus sign, and the appearance of 1/g22 in the denominator instead of 1/g12. We're not quite done, because to find out the electric field as experienced by the particles, we need to transform to each of their instantaneous rest frames in turn. As luck would have it, though, at t = 0 in the current coordinates they are both instantaneously at rest, and the transformation is trivial! For other times, we just take advantage of the fact that the component of the E-field parallel to a Lorentz boost doesn't change. Therefore, at all times the ratio of field strengths experienced by each particle is just




    i.e. it is just minus the inverse of the square of the ratio of the proper accelerations, in line (I think) with what your intuition suggested.

    By incredibly strange coincidence, this preprint was published just a few days ago. You might find it interesting:

    The Fields of a Charged Particle in Hyperbolic Motion, Joel Franklin & David J. Griffiths
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    btr,

    Thank You! That was extremely useful, especially the references, but also the derivation. I did something similar using Rindler coordinates and got a similar result, but by a somewhat different path. I need to go back to my notes to check that it is exactly the same result.

    As for the Covariant formulation, I would still like to understand what it predicts. How about an even simpler calculation. For my purposes we don't need a solution for all space, just in two small volumes local to each charge. For the two equal charges at r1 and r2, can we just do a calculation of the electric field locally at each charge? That is, very near r1 we can define a locally flat (Minkowski) space and take the gradient of the electrostatic potential, taking the derivatives wrt the local coordinates. If the electrostatic potential is really 1/r then this gradient should give the same result at r1 as it does at r2. The field tensor is defined in terms of these potentials and their gradients. Is there some modification to the potential function in curved spacetime that invalidates this approach?
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    You can pick a point on the first particle's world-line and work in its instantaneous local rest frame around that point (Riemann normal coordinates). However, the particle will have non-zero acceleration in this frame (in the scenario we've been considering), so you don't get a spherically symmetric 1/r2 field. Instead, you will get something that locally is very similar to the field of a uniformly accelerated charge in Minkowski space (like the one depicted in Figure 2 of the article I mentioned earlier):


    (The diagram shows the field in Rindler coordinates; the particle is accelerating to the right.)

    Incidentally, I happened across this the other day:

    The Motion of Point Particles in Curved Spacetime, Poisson et al.

    There's a lot in there which is relevant to this topic. For example, in equations 18.26--18.28, there are expressions for the field of a charged particle, averaged over a small sphere of radius s in Fermi normal coordinates. If the field were locally spherically symmetric, these expressions should be exactly zero, but in general you can see that they do not vanish (even for unaccelerated particles, in fact, since there is a contribution to the average electric field proportional to the Ricci tensor).
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    btr,

    Thank you again. I have some studying to do...
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    Is there some modification to the potential function in curved spacetime that invalidates this approach?
    Locally, in the rest frames of your individual charges, there is no difference whatsoever. However, if the charges are far enough apart that the curvature cannot be neglected, the field tensor of the EM between them becomes :



    with the double pipe denoting covariant differentiation. If you expand this expression you will find that the components of the field tensor now contain terms that are explicitly dependent on derivatives of the metric. These derivatives are functions of r, and because , I would expect the E fields in the vicinity of the charges to not be the same.

    Globally, the 4-potential satisfies the de Rham wave equation :



    So again, the global 4-potential is explicitly dependent on the metric and its derivatives - it is in fact directly connected to Ricci curvature, which, if there are charges present, does not vanish.

    Sorry I cannot immediately find the original solution I was referring to. I will keep looking and post the link when I come across it.
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