Is 'charged black hole' an oxymoron?

[stevendaryl]

Start with the photons dropped in the freely falling box. We assume that the collisions of the photons with the walls of the box are perfectly elastic (the box's inner walls are perfectly reflecting mirrors). That means that, in a local inertial frame that is free-falling with the box, there is no net force on the box--the pressure of the photon gas against the walls is isotropic, so it all cancels out. Also, there is no energy exchange anywhere, so the energy at infinity of the box stays the same as it falls, and the measured photon temperature inside the box stays the same as well. Finally, we also assume that nothing about the box itself is changed by this process; this is an idealization as well, but it basically amounts to the total energy at infinity of the box being negligible compared to the total energy at infinity of the photons inside it. That is certainly possible in principle, though of course we would not be able to come anywhere near realizing it in practice with our current technology.

Now, as the box approaches an observer who is static at a constant radius r, where r is much less than the radius from which the box was dropped, a small slot opens in the bottom of the box and some photons escape. If the observer compares the frequency of those photons to the frequency of photons generated locally from an identical source, he will find that the photons from the box are blueshifted. This is because, by assumption, the photons behave identically to photons from the same source that are simply "dropped" in free fall from the same much higher radius where we dropped the box from. (In a real experiment, of course, this would not be true; the photons would exchange energy with the walls of the box. But we've idealized that away. The only real assumption we're making is that the change in gravitational potential energy between two fixed heights is independent of the particular free-falling path taken: the null path that the photons would have followed without the box is equivalent to the timelike path the box actually follows. But that assumption follows easily from the fact that the metric is time-independent.)

Okay. I was skeptical, but I think you're right. Another way to see the same result is to use the relativistic Doppler shift. After the elevator drops in freefall,
it's going to be traveling downward at some speed v by the time it reaches the observer. If we imagine a photon bouncing up and down inside the elevator,
and its frequency is \nu, as measured by the observer in the elevator, its frequency as measured by the observer at "rest" will be \nu' = \nu \sqrt{(1-v/c)/(1+v/c)} when it is heading up, and \nu \sqrt{(1+v/c)/(1-v/c)} when it is heading down. The average frequency will be \nu/√(1-(v/c)²)

Now, consider a second box, which we set up at the same higher radius where the first one was dropped from, in an identical fashion to the first, so that both start out identical. We attach a rope to this box and lower it very slowly to the same lower radius where our observer is. We assume (another idealization) that we can lower the box in such a way as to extract the maximum possible work from this process; it is easy to show that that work will be equal to the difference in gravitational potential energy of the box between the two altitudes; i.e., it will be equal to the kinetic energy possessed by the first box just as it falls past the observer at the lower radius. We also make the same assumption about the box as before, that its total energy at infinity is negligible compared to that of the photons.

Now compare the two boxes when they are both at the observer's location at the lower radius. Box #1 is freely falling with some kinetic energy; box #2 is at rest. They both have the same gravitational potential energy, because they are both at the same height and they both have identical numbers of identical atoms in the walls, and identical numbers of photons contained inside (because we prepared them identically). The only difference between them is their kinetic energy. So what difference will that make to our observations?

It seems obvious, put that way, that the photons coming out of the second box will have to have the *same* frequency as photons coming from an identical source locally. That is, they will *not* be blueshifted. After all, the work we extracted as we lowered the box, which is equal to the kinetic energy of the first box + its photons, had to come from somewhere, and the only place it could have come from, by hypothesis, was the photons in the second box (remember the box's energy itself is negligible). So the photons in the second box must be redshifted, compared to those in the first box (which must have all of the kinetic energy of the system, since again the box's energy itself is negligible), by exactly the amount by which the photons in the first box were blueshifted in the process of falling.

I think that's right. I'll have to think about it more, though.

 

[PeterDonis]

Okay. I was skeptical, but I think you're right. Another way to see the same result is to use the relativistic Doppler shift...

Yes, this is a good way to describe how things would look from the perspective of a static observer watching the box fall.

 

[jartsa]

You are correct that force "blue shifts" in the following sense: if I attach a rope to the box, and exert a constant force on the rope as measured by a force gauge at my height, then as the box lowers, the force measured by a force gauge at the box's end of the rope will show an increasing force. However, this...
is *not* correct. The lower force meter reads a larger force because there is a larger force; it takes more force to hold the box at the box's height than it does to hold the upper end of the rope at the higher height. If you put force gauges all along the rope, you would measure the tension in the rope to be steadily decreasing as you went up from the box to the upper end. It has nothing to do with the weakening of the spring of the force meter; if you ran all the standard calibration tests on the force meter at the lower height, it would pass them just the same as at the higher height. (Of course, every force meter has an upper limit to the force it can measure; we're assuming that that limit is not exceeded anywhere in this scenario. In fact, another way of saying that the force meter doesn't weaken is that the upper limit of the force it can measure doesn't change; if you do a test to measure the limit at the higher height, and again at the lower height, the two tests will give the same results.)

The reason it takes more force at the lower height is simple: the proper acceleration required to hold static at a constant height increases as you go lower in the field. It's the same reason why it takes more rocket thrust to "hover" closer to a massive body than further away: the "acceleration due to gravity" is higher at a lower altitude.

Usually we think that a hanging non-massless rope has higher tension at upper parts, both in homogeneous or inhomogeneous gravity field. Right?

A hanging massless rope, with a weight on the lower end then again ... usually we think that the tension is constant along the rope, and it doesn't matter what the gravity field is like. Usually we think the massless rope doesn't know there is a gravity field, it only knows there's some force pulling both ends. Right?

 

[PeterDonis]

Usually we think that a hanging non-massless rope has higher tension at upper parts, both in homogeneous or inhomogeneous gravity field. Right?

Usually we don't deal with relativistic gravity fields, where the redshift/blueshift is large enough to affect the tension in a rope. Inhomogeneous fields can still be very weak, like the Earth's. What's required for the redshift/blueshift of force to be observed is not inhomogeneity, but strong enough gravity.

If you're talking here about a rope that's just hanging loose, with nothing on the end of it, then in the case of weak gravity where the redshift/blueshift is negligible, yes, I would expect the tension in the rope to be higher at the upper end, since that end is supporting the entire weight of the rope, whereas the lower end is only supporting itself. But that's a different case than the scenario we've been discussing, because the lower end of the rope is not constrained: it's not holding anything and can move freely. You seem to recognize this, because you go on to describe a case with a weight on the end of the rope.

If the redshift/blueshift is not negligible (meaning the rope is very long and/or the field is very strong), then I'm not sure how the tension in the rope would vary, precisely because the lower end of the rope is unconstrained; so it can move vertically to adjust itself as the tension above it changes. I don't think one could make a definite prediction without a more detailed material model of the rope and how it would respond to forces in a relativistic gravity field. Greg Egan has a web page on relativistic elasticity which at least gives some idea of what such a model would have to look like (but it's still a very rudimentary model, as he says):

http://gregegan.customer.netspace.net.au/SCIENCE/Rindler/SimpleElasticity.html

A hanging massless rope, with a weight on the lower end then again ... usually we think that the tension is constant along the rope, and it doesn't matter what the gravity field is like. Usually we think the massless rope doesn't know there is a gravity field, it only knows there's some force pulling both ends. Right?

As I said above, usually we don't deal with gravity fields that are strong enough to have a redshift/blueshift factor that's large enough to affect measurable forces and tensions. If the redshift/blueshift is negligible then yes, the tension should be constant along the rope if there is a weight on the end and the rope is idealized as massless. But the scenario we've been discussing requires the redshift/blueshift to not be negligible. In that case the force is redshifted/blueshifted. The key thing here that makes this prediction doable, by contrast with the above case, is not just that there is a weight on the end of the rope, but that we specified the weight's motion: either it's static at a constant height, or it's being slowly lowered in such a way as to extract maximum work. Those are huge constraints; without them, as I said above, you would need a detailed material model of the rope to make predictions about the observed tensions. But any such model would still have to take into account the redshift/blueshift factor changing with height.

 

[stevendaryl]

Yes, this is a good way to describe how things would look from the perspective of a static observer watching the box fall.

Okay, now I'm having second thoughts about your conclusion. From the point of view of Doppler shift, the "blueshift" is symmetric: The photons produced by the "falling" observer will be blueshifted, when measured by the observer at "rest", and the photons produced by the the observer at rest will be blueshifted, as measured by the falling observer.

 

[PeterDonis]

From the point of view of Doppler shift, the "blueshift" is symmetric: The photons produced by the "falling" observer will be blueshifted, when measured by the observer at "rest", and the photons produced by the the observer at rest will be blueshifted, as measured by the falling observer.

You have to be careful with the Doppler shift analysis. You derived the "average" Doppler shift of photons in the box, as seen by a "static" observer, and showed that it is a blueshift. If we consider instead an observer free-falling with the box, that observer would not see the photons to be blueshifted "on average". If that free-falling observer, just as he and the box are reaching the lower altitude, were to look at photons emitted by an identical source at that altitude, yes, he would see those photons to be blueshifted, due to his velocity relative to the "static" observer. But if you work it out, you will see that the blueshift he sees is *different* than the "average" blueshift you calculated for the photons in the box, as seen by the "static" observer!

From the viewpoint of the "static" observer, the free-falling observer is moving at velocity v, and vice versa. So the Doppler shift between them (which is symmetric, as you note) is (in units where c = 1)

\sqrt{\frac{1 + v}{1 - v}}

whereas the "average" shift of the photons in the box is

\sqrt{\frac{1}{1 - v^{2}}}

But the first Doppler shift does not apply directly to the photons in the box; it would apply, as I said, to photons emitted by the "static" observer and seen by the falling observer, or to photons emitted by an identical source that was falling with the box, just as the box reaches the "static" observer's height. The second Doppler shift is for the "average" of the photons in the box, as a result of them falling inside the box, which is a different thing.

This does point out, though, that I left out one key specification in my thought experiment: when the photons are let out of the box, they are released without any interaction with the box itself. That is to make sure that the box does not act as a "source" of the photons and add any additional Doppler shift to them as they are emitted; they are emitted just as if they had fallen freely by themselves from the upper height. This should be OK for the "average" of the photons, which is what we have been idealizing.

 

[Q-reeus]

When did I agree that a physically measured E field is frame dependent? What I agreed to is that the E field of a capacitor as seen by a charge q--"seen" meaning "as measured by the actual, observable force on the charge"--is larger when the charge is moving relative to the capacitor, than when the charge is at rest relative to the capacitor. If that is what you mean by "the physically measured field is frame dependent", then fine.

Of course that's what I meant. The usual and only sensible interpretation. And my perfectly clear and undeviating position from the start.

But I would much rather stick to less ambiguous terminology; "frame dependent" could just as easily mean that you think the field somehow changes when I calculate it in one frame vs. another, which is false.

You have'nt defined here what 'the field changes' actually means. Presumably a straw-man definition: that it's conceivable an observer stationary in some frame S notices a change in some 'fixed' field, measured in S, just because another observer originally stationary in S', moves relative to S and thus sees a different field *value* in S', which is then bizarrely 'projected' back into S! No-one sane and sober believes such a lunatic interpretation. From the start I have stuck to the usual understanding - value of 'observables' *seen/actually measured* in one frame will be generally *seen/actually measured* different in another. That's what those field and force and momentum and energy and time etc. transformation rules are all about!

I don't understand this at all. Once again, I'm talking about actual, observable readings on strain gauges and scales. I'm not saying anything about "frames". Actual, observable readings can always be directly compared.

Only generally if made in the same frame! Your insistence O in frame S = O' in S' violates that rule. it outright denies validity of Lorentz force law F = q(E+uxB), as stated time and again previously. It inevitably follows from your insistence the proper value of F' as measured by scales stationary in S' must be identically the value F measured by scales stationary in S. Wrong. Apply that logic to the carousel arrangement described earlier. You already claimed earlier the orbiting charges experience a greater axial force, measured in the non-spinning frame, owing to their circular motion relative to that frame. Contrary to Lorentz expression which explicitly forbids a velocity dependent change - as measured in that one frame.

Ironically in a way it is charge invariance here that shows just how wrong that position is. if you want force balance to hold then obviously those orbiting charges must then exert an equally greater velocity dependent electric force back on the non-spinning capacitor plate charges. Won't happen. Mutual electrostatic net forces between the spinning carousel charges and stationary capacitor charges are not only equal and opposite, but also velocity independent - as measured in the non-spinning frame. Same values as when carousel has zero spin. What does change is the greater E field seen/actually measured in the frame of an orbiting charge. And that higher E field E' = γE measured there as a higher F' = qE', transforms back into the non-spinning frame as F = F'/γ = qE - as measured there. How many times do we have to thrash that one out?

I don't understand this at all. Once again, I'm talking about actual, observable readings on strain gauges and scales. I'm not saying anything about "frames". Actual, observable readings can always be directly compared.
In this particular case, you are claiming that I can have a strain gauge and a scale, both registering a force that is purely in the x direction, both motionless in the x direction, yet showing different magnitudes for the force.

Ad nauseam - because they have relative motion in the y direction. Think of Doppler-shift or something as vague parallel here.

That makes no sense to me at all; it means you have a situation that's static but with unbalanced forces; the arm has one force, O', on one end of it (what the strain gauge is reading), but another, smaller force, O, on the other end of it (what the scale is reading). How can the arm stay motionless?

Because the measurement in one frame does *not* transform into the same measurement determined in the frame with differing relative y-axis motion. Apply the force transformation rules as per link given before.

Once again, why all this hoopla about "frames"? Why not just describe, in plain English, how you would go about building a perpetual motion machine, given that what I have said about the forces is true?

Pardon me being in a rush last post. What I had in mind was not directly related to the charge moving horizontally on rails thing, but rather the implication for a charge in general free motion between the cap plates. You have consistently claimed a larger charge proper force is what also is measured in the cap frame - i.e. F' = qE' = qγE = F (transverse relative motion only). Which as I've insisted a number of times now already, explicitly denies validity of the usual 3-force transformation expressions, and Lorentz force F = q(E+uxB). Only the latter are consistent with energy/momentum conservation in this non-radiative scenario. To generalize your position though, recall in #75 you referenced to a Wiki article giving the 'covariant' expression
F = γq(E+uxB) - (1),
as being that giving the 'relativistically corrected' measured force on a charge with any relative velocity gamma factor applying in that frame. My view expressed in #86 was that it was a puzzling, seeming expression for proper force but made no sense as shown. Nevertheless you continued to be happy with it as correct.

So let's apply your logic to that expression (1). The capacitor plates have small-bore holes in them, lined-up in your defined x direction, such that a speeding charge of gamma factor in lab frame can enter one hole, pass through between the plates parallel to the applied E field, and exit the hole in the other plate, experiencing the very nearly uniform E field essentially all the way between the plates. You insist the charge's proper force F' as per (1) will also be that measured in the capacitor (lab) frame - i.e. F' = qE' = qγE = F. Well, hooray if it were only so. Energy gain is force times distance. Directly measurable force *in the lab frame* by your logic goes up with speed u *in the lab frame*. Traversed distance d between plates in lab frame is of course completely independent of charge speed u. We can all do the simple math. Need I elaborate? And the apparent energy gains will apply for oblique motion also if your rules are applied, but a bit more messy to work through. But sadly, energy/momentum conservation will hold - proper application of 3-force transformation rules and Lorentz force will show that to be the case. And yes, it is the those dp/dt expressions that give the actually measured force on q in the lab frame (determined by acceleration and energy change), not the 'remote' dp/dtau applying to q's rest frame only.

For the record, I guess I should state that of course I don't see any violation of energy conservation. It is true that q+arm has a larger kinetic energy in #2 than in #1, as seen from the rest frame of the capacitor; but it has that larger KE because it was set in motion in the y direction, so some energy had to be added to it from some source to cause that motion. So if I then remove the arm and let q hit the capacitor and release its larger KE, the larger KE it releases is because I added some energy to it from a source. Extra energy captured = extra energy from source. No violation of conservation anywhere.

All explained above - understandably you drew the wrong inference and have argued against a straw man.

Edit: I guess I should also add that I'm not sure I understand why the reading on the scale matters anyway. You've already agreed that the force on the charge q increases if q is moving relative to the capacitor; and that's the only force that matters in determining how much KE the charge will gain if it is allowed to "fall" in the capacitor's field.

No I hav'nt agreed with that view ever. It's your view and it leads to violations of energy conservation as per above.

So when you were talking earlier about just using capacitors with holes in them instead of cyclotrons, you were talking about a scenario that you had already agreed was true. If so, why haven't you gone ahead and patented your capacitor substitute for a cyclotron?

See above.

Edit again: After looking at that "sciencebits" page you linked to, what it calls "force" is what I called "coordinate force", i.e., dp/dt, not dp/dtau. So what they are calling "force" is not a direct observable.
Only dp/dtau is a direct observable. You are confusing coordinate quantities with direct observables.

Disagree - it is dp/dt that is *the* direct observable that matters - that applying to the frame we measure it in. It's the force measured by the scales under the tracks in lab frame. Improperly applying dp/dtau, valid only in charge rest frame, to lab frame, without applying force transformation formula, is what gives erroneous results. Only if u = 0 do the two coincide, and I expect you agree with that much.

This bizarre episode makes me feel like having fallen through some spacetime warp into an almost-the-same parallel universe! This 'PeterDonis' just can't be the real PD from back there. Help. I want back. And btw that bit in #114 is also a bizarre interpretation; that dp/dt = 0 always - or something?! Just what I would expect of 'PeterDonis' residing in an evil parallel universe.

 

[Q-reeus]

In a way, all this is a matter of "interpretation"; we agree on the physical observables.

Agreed - unnecessary nitpicking over inconsequentials is not productive. I kept things 'compact' and expected reasonable interpretation.

So most people would say the redshift of the energy if it's radiated upward is due to something about the spacetime in between and has nothing to do with the force between the balls, which is a purely local thing.

Tug on the balls remotely (net forces case) via rope and pulleys, and redshifted forces will be observed. One then can make a reasonable inference from F = qE or F = mg (g being here that due to the other ball, not of planet), that we have depressed values for field or mass/charges or both. That's the point.

Ok here; there's no net force, so there's no net work. But that also means there's nothing to redshift. See below.
How are you going to measure the nonexistent net work in order to show that it's redshifted?

You didn't get the intended angle obviously. We already deduced redshift for uncharged situation. So now there is precisely equal and opposite work being done, and 'force neutral' condition actually makes the close analogy here between gravitational and EM effects very evident - imho.

It seems to me that a better experiment would be to have two oppositely charged objects where the Coulomb force is much larger than the Newtonian gravity force, so the latter can be neglected. Then let them roll together and capture the work done and let it be radiated as photons. In that case, yes, the energy will be redshifted if it is captured "at infinity", but will not if it is captured locally. So the same issue of interpretation arises as for the gravity case, above.

I thought about doing it like that (have done so before), but preferred the 'force neutral' arrangement as giving a better feel for things. Minor preferences are no big deal imo.

To start talking about this, you need to first specify something that you left out of the scenario above. How are you measuring "m"? If you are measuring "m" by a local Cavendish experiment--basically the same thing you described, separate the two objects by a known distance d, measured locally, then measure the work done, locally, when they come together--then you are correct, the contribution of the object with mass "m", measured locally, to the total mass M, measured "at infinity", will be redshifted.

Answered your own question well - great. Moving on now...

Again, how are you measuring the q's? I'll assume you're measuring them locally by the same "Cavendish" method as above--take two objects with same magnitude of charge but opposite signs, separate them by a known distance, then measure, locally, the work done when they come together.

Why...yes!

In this case, I'm not sure how those two q's, measured locally, would contribute to the total charge Q observed "at infinity" for the spacetime as a whole. I see your argument as to why it ought to be redshifted just as the masses are, but I also see the interpretational issues that I raised above, and I'm not sure what the math actually says. I'll have to take a deeper look.

Now we're down to the nitty-gritty. I anxiously await further feedback!

 

[PeterDonis]

Q-reeus, I'm going to respond first to just the last part of the first of your two recent posts, for reasons which should be clear in a moment.

This bizarre episode makes me feel like having fallen through some spacetime warp into an almost-the-same parallel universe! This 'PeterDonis' just can't be the real PD from back there. Help. I want back. And btw that bit in #114 is also a bizarre interpretation; that dp/dt = 0 always - or something?! Just what I would expect of 'PeterDonis' residing in an evil parallel universe.

No, it's still the same PeterDonis here. I don't see why you're so surprised; finding highly provocative ways to express my disagreements with your positions has always been my trademark, hasn't it?

Disagree - it is dp/dt that is *the* direct observable that matters - that applying to the frame we measure it in. It's the force measured by the scales under the tracks in lab frame. Improperly applying dp/dtau, valid only in charge rest frame, to lab frame, without applying force transformation formula, is what gives erroneous results. Only if u = 0 do the two coincide, and I expect you agree with that much.

No, I don't; at least, not as you apply it here. Apparently you didn't get the real point of my previous post, and it's fundamental, so let me restate it, carefully.

First of all, to clear up one point: if an object is at rest in a given frame, dp/dtau for that object equals dp/dt in that frame, because tau = t for an object at rest. So I have no problem saying that dp/dt is a "direct observable" for an object at rest in the frame, since it's the same, in that frame, as dp/dtau for that object, which is the direct observable expressed in covariant form. I prefer thinking about dp/dtau because it's covariant; I can calculate dp/dtau in any frame, and it must give the same number. That's not true for dp/dt. But that's a minor point for this discussion.

The key point I was making is: *what* observable is dp/dt? What does it mean? Well, look at it: "p" is the object's 4-momentum, and dp/dt is the rate of change of 4-momentum with respect to time. Also, it's vectorial, so there are three components: dp_x/dt, dp_y/dt, and dp_z/dt. It's only one component, dp_x/dt, that we are concerned with in this problem; the other rates of change are stipulated to be zero already by the statement of the problem.

So, what is dp_x/dt? It is the rate of change of the x component of 4-momentum, with time, in a given frame. But as I pointed out, *all* of the objects in this scenario, in both "situations", #1 and #2, are *motionless* in the x-direction. That means their x momentum is *not changing with time*. So dp_x/dt is *zero* for all objects in this scenario, by simple inspection. And this is true in both frames, the rest frame of the capacitor and the rest frame of the charge q, because the relative motion of the frames is in the y direction only.

(Strictly speaking, what I've said is true only on the assumption that the apparatus as a whole is floating freely in flat spacetime. But I had assumed from the outset that that was what we were talking about, since we have been strictly talking about SR and its predictions.)

Once you recognize this, you realize that everything we've both been saying about dp_x/dt, and how it transforms, etc., etc., is *irrelevant* to this scenario. Zero transforms to zero, so "force transformation" is irrelevant to this problem.

But what about the Lorentz force law, you ask? I addressed that too, but let me address it again. The Lorentz force law, as it needs to be applied in this problem, does not predict any nonzero dp_x/dt, for any object; it can't, because as we saw above, dp_x = 0 for all objects. Instead, what the Lorentz force law predicts is how much stress there will have to be in the arm to hold the charge q motionless. That stress will show up as pressure in the arm, and that pressure times the cross sectional area of the arm will have to equal the force given by the Lorentz force law for the charge q. But none of this leads to any nonzero dp_x/dt for anything; everything is held motionless, by hypothesis, so it's in static equilibrium, and the Lorentz force law just contributes a term to the equation determining static equilibrium.

I'll defer responding to other parts of your post; the above point is so critical that it deserves a discussion all on its own.

 

[PeterDonis]

Ok, I said I would defer responding to the rest of Q-reeus' post, but there are a couple of other things I should comment on without waiting:

because they have relative motion in the y direction. Think of Doppler-shift or something as vague parallel here.

Vague, and irrelevant. Relative motion in the y direction does not affect *static equilibrium* in the x direction.

Improperly applying dp/dtau, valid only in charge rest frame

NO! Sorry for shouting, but you have it backwards. The covariant observable, valid in *any* frame, is dp/dtau. In a frame in which the object in question is not at rest, you have to apply the gamma factor, for that frame, to get dp/dtau, the frame-independent quantity, from dp/dt, the frame-dependent quantity. The equations I wrote down with dp/dtau in them are valid in any frame. (They are also irrelevant to this problem, as I discussed in my last post; but the point about which expressions are invariant is important too and is worth stressing.)

 

[Q-reeus]

No, it's still the same PeterDonis here. I don't see why you're so surprised; finding highly provocative ways to express my disagreements with your positions has always been my trademark, hasn't it?

True - enough trademarks there - still in the same universe. Phew.

However, the key point I was making is: *what* observable is dp/dt? What does it mean? Well, look at it: "p" is the object's 4-momentum, and dp/dt is the rate of change of 4-momentum with respect to time. Also, it's vectorial, so there are three components: dp_x/dt, dp_y/dt, and dp_z/dt. It's only one component, dp_x/dt, that we are concerned with in this problem; the other rates of change are stipulated to be zero already by the statement of the problem.

You've just dropped a bombshell. Thought we were working with p representing 3-momentum. That's what I've been working from in writing many times 3-force F = dp/dt etc. Hmmm. Well as long as we just use spatial component projections appropriately, probably won't change anything of substance. Scales will read the 3-force component of that 4-force dp/dtau. [Just now caught your #130. Given this new revelation I agree there that dp/dtau is not just valid in proper frame, but then I was using it to mean proper value of 3-force]

So, what is dp_x/dt? It is the rate of change of the x component of 4-momentum, with time, in a given frame. But as I pointed out, *all* of the objects in this scenario, in both "situations", #1 and #2, are *motionless* in the x-direction. That means their x momentum is *not changing with time*. So dp_x/dt is *zero* for all objects in this scenario, by simple inspection. And this is true in both frames, the rest frame of the capacitor and the rest frame of the charge q, because the relative motion of the frames is in the y direction only.

Once you recognize this, you realize that everything we've both been saying about dp_x/dt, and how it transforms, etc., etc., is *irrelevant* to this scenario. Zero transforms to zero, so "force transformation" is irrelevant to this problem.

But what about the Lorentz force law, you ask? I addressed that too, but let me address it again. The Lorentz force law, as it needs to be applied in this problem, does not predict any nonzero dp_x/dt, for any object; it can't, because as we saw above, dp_x = 0 for all objects. Instead, what the Lorentz force law predicts is how much stress there will have to be in the arm to hold the charge q motionless. That stress will show up as pressure in the arm, and that pressure times the cross sectional area of the arm will have to equal the force given by the Lorentz force law for the charge q. But none of this leads to any nonzero dp_x/dt for anything; everything is held motionless, by hypothesis, so it's in static equilibrium, and the Lorentz force law just contributes a term to the equation determining static equilibrium.

I see your point in this but not agreeing with the overall inference. There is zero net dp_x/dt yes, but it's the result of, as you were saying previously also, a balance of opposing 3-forces - electric one F_e = qE vs mechanical one |F_m| = σA (σ the average stress, A the cross-sectional area of arm or other support structure). That these two are in static equilibrium tells us nothing however about whether these equal and opposite magnitudes, in a given frame, may be a function of relative motion u in that frame, or how they co-vary as measured in another frame. So the dp_x/dt of interest is that applying to the 'active' electric part, and we simply assume automatic balance by 'reactive' mechanical part. So it's how this 'partial' dp_x/dt = F = q(E+uxB) calculated in lab frame values, transforms and is measured in the proper frame of q, and vice versa, that very much matters imo. To repeat - scale measurement of 3-forces in a given frame is what matters. Bottom line is still this - I say that, for e.g. spinning carousel scenario, Lorentz force ensures measured weight in non-spinning frame is indifferent to spin-rate, whereas you have claimed it will rise with spin-rate. A matter of observable physics.

 

[PeterDonis]

Thought we were working with p representing 3-momentum.

If you are writing a covariant equation, you can't use 3-momentum; 3-vectors are never covariant in 4-D spacetime. So all the covariant quantities and equations I've written, including dp/dtau, use 4-momentum.

The Lorentz force equation you wrote down does use p to mean 3-momentum, I agree; that's because it's written in a particular frame, where we split 4-momentum into energy (a scalar) and momentum (a 3-vector).

Well as long as we just use spatial component projections appropriately, probably won't change anything of substance. Scales will read the 3-force component of that 4-force dp/dtau.

Yes, as long as you define "3-force component" appropriately. See further comments below.

There is zero net dp_x/dt yes, but it's the result of, as you were saying previously also, a balance of opposing 3-forces - electric one F_e = qE vs mechanical one |F_m| = σA (σ the average stress, A the cross-sectional area of arm or other support structure).

Yes.

That these two are in static equilibrium tells us nothing however about whether these equal and opposite magnitudes, in a given frame, may be a function of relative motion u in that frame, or how they co-vary as measured in another frame.

True, the fact of static equilibrium by itself doesn't tell us how the magnitudes will be represented in different frames. But it does tell us that the magnitudes must be *equal and opposite*. That, by itself, is enough to show that the two observables, O and O', are equal. I gave the chain of equal and opposite magnitudes that shows it. That chain of reasoning works regardless of which frame you choose to evaluate the magnitudes.

Re whether the magnitudes are a function of relative motion, see below. [Edit: actually, see separate post, forthcoming. ]

To repeat - scale measurement of 3-forces in a given frame is what matters.

You were OK until you added the qualifier "in a given frame". Once again, the two observables, O and O', are actual observed numbers; they are readings on a scale. The readings on a scale cannot depend on what frame you use to calculate them. Nor can they depend on the state of motion of the "observer" relative to the scale. So they are not numbers "in a given frame"; I must be able to calculate the numbers using quantities represented in *any* frame, and come out with the same answer. You've agreed to this before, but you keep on using language that just begs to be interpreted as contradicting it.

Let's step back for a moment and look at a simpler case. Suppose I have a capacitor whose field in its rest frame is E in the x direction, and a small object with charge q. Consider two scenarios:

(1) At some instant of time, the small object is at rest with respect to the capacitor (and between its plates, so it "sees" the field). The Lorentz force law, in covariant form, gives (the only nonzero component of the general version I posted before):

\frac{dp^{1}}{d\tau} = q F^{1}_{0} u^{0}

Substituting F^{1}_{0} = E and u^{0} = 1 gives us the "coordinate" form in the rest frame of the capacitor, which is the version you wrote down:

\frac{dp^{x}}{dt} = q E

All good so far. Now try this scenario:

(2) At some instant of time, the small object is moving solely in the y direction, relative to the capacitor (and it is between the plates). First work the problem in covariant form:

\frac{dp^{1}}{d\tau} = q F^{1}_{0} u^{0}

Looks the same as above, doesn't it? But note that, since this is written in *covariant* form, it is valid in any frame. So we can generate two different "coordinate" forms:

In the rest frame of the charge, F^{1}_{0} = E' = \gamma E, and u^{0} = 1, so we have (I'll use primed quantities for this frame):

\frac{dp^{x'}}{dt'} = q E' = q \gamma E

This matches what we would write down if we wrote down the Lorentz force law, in the form you gave it, in this frame, since E' is the E field "as seen" in this frame (the rest frame of the charge).

In the rest frame of the capacitor, F^{1}_{0} = E and u^{0} = \gamma, but we also have dt / d\tau = \gamma, so the gammas cancel, as we established way back many posts ago, and we have

\frac{dp^{x}}{dt} = q E

So the Lorentz force law, in the form you gave it, is correct in both frames. But the two versions refer to two *different* things! The first version, in the rest frame of the charge, where dp/dtau = dp/dt', refers to the actual observable, what would be measured by an accelerometer attached to the charge; that's because dp/dtau *is* that observable. But the second version, in the rest frame of the capacitor, does *not* directly represent that observable; that should be obvious, because the RHS is a different number, q E instead of q \gamma E. The same actual observable, the actual reading on an accelerometer (or a scale), can't be two different numbers; it has to be one or the other. In this case, it's q \gamma E (as we've agreed before). That means the actual observable, the reading on the accelerometer, is equal to dp^{x'} / dt', but is *not* equal to dp^{x} / dt. The "coordinate" form only directly represents the actual observable in one particular frame, while the "covariant" form represents it in a way that's valid in any frame. That's why you have to be very careful in specifying how the "3-force component" you are interested in is defined. Defining it using the covariant equation works in any frame; defining it as dp/dt, the "coordinate" form, only works in one particular frame.

This post is getting long so I'll continue with the part about static equilibrium and relative motion in a separate post.

 

[PeterDonis]

Ok, follow-up post on static equilibrium and relative motion. Basically, there are two distinct questions at issue, and we seem to agree on one while not agreeing on the other.

That these two are in static equilibrium tells us nothing however about whether these equal and opposite magnitudes, in a given frame, may be a function of relative motion u

True. I think we have established that the magnitudes *are* a function of relative motion; that is, when the charge q is moving in the y direction relative to the capacitor, the force on the charge, as measured by an accelerometer attached to it, is larger than when the charge is at rest relative to the capacitor, by the gamma factor associated with the relative motion of charge and capacitor.

in that frame,

A minor point but I've been repeating it so I'll repeat it again: please be careful to distinguish frame-dependent from frame-independent quantities. Here, "in that frame" is *not* correct. Relative motion is relative motion; it's an invariant thing, so it's there regardless of which frame you choose to describe it. That the charge and the capacitor in our scenario are in relative motion, with relative velocity v, is an invariant, frame-independent statement.

or how they co-vary as measured in another frame.

False: static equilibrium is static equilibrium, regardless of relative motion in directions orthogonal to the equilibrium. That means the magnitudes that are equal and opposite stay equal and opposite regardless of that relative motion. So O and O' have to remain equal when the charge is moving relative to the capacitor, because the motion is orthogonal to the direction of the equilibrium. Their actual magnitude changes with relative motion; it goes up, by the first point above. But they are still equal and opposite; both magnitudes change in sync.

Consider another scenario, not involving electromagnetism. I put weights on a scale and set the scale on a trolley which sits at rest on a track. There is another scale under the track, measuring the weight of the track and whatever is on it. The scales, trolley, and track all have negligible mass compared to the weights. I define two observables: O, the weight shown on the scale under the track, and O', the weight shown on the scale on the trolley.

To the best of my knowledge, all of the following are true:

(1) When the trolley sits at rest on the track, O = O'. Call this reading A.

(2) When the trolley is rolling along the track at a constant velocity v, O = O'. (We assume the track is perfectly horizontal, so the trolley's motion is perfectly orthogonal to the direction of gravity.) Call this reading B.

(3) In the non-relativistic approximation (v << c), reading A will be the same as reading B. However, if we allow v to be relativistic, then the gravitational "force" on the weights is no longer just the standard Newtonian force; there is an additional velocity-dependent force which can be viewed as due to "gravitomagnetism". (In the limit as v -> c, the total force goes to twice the Newtonian value; another way of saying this is that "acceleration due to gravity" is higher on a relativistically moving object. This is why light grazing the Sun, for example, bends by twice the amount you would predict by just doing a Newtonian calculation for a mass "falling" in the Sun's field, where the mass is the light's energy divided by c^2.) So in the case of relativistic v, reading B will be larger than reading A. (I don't think anyone has actually run this experiment, of course; a real experiment like this done on Earth could only last a small fraction of a second at relativistic velocities without violating the "perfectly horizontal" requirement.) But O = O' will still hold even in this case.

 

[PeterDonis]

One other note: after looking at some references, my use of the term "static equilibrium" is not standard. All the references I've looked at use that term only to describe a situation where there is *no* motion whatsoever. There doesn't seem to be a standard term for describing a situation where there is motion in one direction but no forces in that direction, and a balance of forces resulting in no net motion in an orthogonal direction. Since there doesn't seem to be one, I'll just use "force balance", and state the claim I've been defending as "relative motion in the y direction does not affect force balance in the x direction".

 

[Austin0]

Start with the photons dropped in the freely falling box. We assume that the collisions of the photons with the walls of the box are perfectly elastic (the box's inner walls are perfectly reflecting mirrors). That means that, in a local inertial frame that is free-falling with the box, there is no net force on the box--the pressure of the photon gas against the walls is isotropic, so it all cancels out. Also, there is no energy exchange anywhere, so the energy at infinity of the box stays the same as it falls, and the measured photon temperature inside the box stays the same as well.

Now, as the box approaches an observer who is static at a constant radius r, where r is much less than the radius from which the box was dropped, a small slot opens in the bottom of the box and some photons escape. If the observer compares the frequency of those photons to the frequency of photons generated locally from an identical source, he will find that the photons from the box are blueshifted. This is because, by assumption, the photons behave identically to photons from the same source that are simply "dropped" in free fall from the same much higher radius where we dropped the box from. (In a real experiment, of course, this would not be true; the photons would exchange energy with the walls of the box. But we've idealized that away. The only real assumption we're making is that the change in gravitational potential energy between two fixed heights is independent of the particular free-falling path taken: the null path that the photons would have followed without the box is equivalent to the timelike path the box actually follows. But that assumption follows easily from the fact that the metric is time-independent.)

Now, consider a second box, which we set up at the same higher radius where the first one was dropped from, in an identical fashion to the first, so that both start out identical. We attach a rope to this box and lower it very slowly to the same lower radius where our observer is. We assume (another idealization) that we can lower the box in such a way as to extract the maximum possible work from this process; it is easy to show that that work will be equal to the difference in gravitational potential energy of the box between the two altitudes; i.e., it will be equal to the kinetic energy possessed by the first box just as it falls past the observer at the lower radius. We also make the same assumption about the box as before, that its total energy at infinity is negligible compared to that of the photons.

Now compare the two boxes when they are both at the observer's location at the lower radius. Box #1 is freely falling with some kinetic energy; box #2 is at rest. They both have the same gravitational potential energy, because they are both at the same height and they both have identical numbers of identical atoms in the walls, and identical numbers of photons contained inside (because we prepared them identically). The only difference between them is their kinetic energy. So what difference will that make to our observations?

It seems obvious, put that way, that the photons coming out of the second box will have to have the *same* frequency as photons coming from an identical source locally. That is, they will *not* be blueshifted. After all, the work we extracted as we lowered the box, which is equal to the kinetic energy of the first box + its photons, had to come from somewhere, and the only place it could have come from, by hypothesis, was the photons in the second box (remember the box's energy itself is negligible). So the photons in the second box must be redshifted, compared to those in the first box (which must have all of the kinetic energy of the system, since again the box's energy itself is negligible), by exactly the amount by which the photons in the first box were blueshifted in the process of falling.



For example, we could confirm our analysis of the first box by asking where the photons' energy would go as the first box falls--it can't go into the box, since the box's energy is negligible, and it can't go anywhere else, since the box won't let any photons out (until the slot is opened at the very end). But in the second box, there *is* a place the energy can go: into the rope that is holding up the box and keeping it from freely falling. The photons do work on the second box, which is transmitted up the rope and extracted at the top. The photons can't do any net work on the first box.

There seem to be two main concepts here. Energy conservation and the kinetic theory of frequency shift.

Energy:

Obviously both boxes had to originate from hovering systems. Under constant acceleration requiring energy. The box that descends slowly does so through the upward acceleration transmitted through the rope. The energy does not go from the box to the platform but in the other direction.

As far as I can see the only energy that goes up the rope is the kinetic energy of the box. This results in a reduction of total system potential energy. But this energy comes from spatial translation not from intrinsic energies of the photons.

And how could the photons possibly be doing work to effect an upward acceleration??

how could they have any net effect on the motion of the box whatsoever?

Kinetic energy;

I have said that photons *do* "gain energy" (that's one way of interpreting what's happening, anyway) when *freely falling*.

You are saying that the blueshift occurring with the freefall box due to the kinetic energy gained through freefall would not occur with the lowered box because of the lack of gained kinetic energy in that case. Correct??

I agree completely with your description here

The only real assumption we're making is that the change in gravitational potential energy between two fixed heights is independent of the particular free-falling path taken: the null path that the photons would have followed without the box is equivalent to the timelike path the box actually follows. But that assumption follows easily from the fact that the metric is time-independent.)

This seems to contradict your conclusion;That the lowered box photons would be redshifted.

As far as the photons in the lowered box are concerned they are just as much in freefall as the inertial box photons.
The only difference is the velocity of the mirrors they are reflecting off of.

If we isolate a single reflection, say off the lower end of a box. The mirror is moving away from the photon. This is the case for both boxes. Likewise for a reflection off the upper mirror when the mirror is approaching the photon , which in both cases should compensate for any effect at the other end.
In this context how do you distinguish between the conditions in the two boxes?

We assume that the collisions of the photons with the walls of the box are perfectly elastic (the box's inner walls are perfectly reflecting mirrors). That means that, in a local inertial frame that is free-falling with the box, there is no net force on the box--the pressure of the photon gas against the walls is isotropic, so it all cancels out. Also, there is no energy exchange anywhere,

So having defined a perfect atiabatic isolation how do you reconcile this with your conclusion of reduced temperature /frequency?

 

[jartsa]

There seem to be two main concepts here. Energy conservation and the kinetic theory of frequency shift.

Energy:

Obviously both boxes had to originate from hovering systems. Under constant acceleration requiring energy. The box that descends slowly does so through the upward acceleration transmitted through the rope. The energy does not go from the box to the platform but in the other direction.

As far as I can see the only energy that goes up the rope is the kinetic energy of the box. This results in a reduction of total system potential energy. But this energy comes from spatial translation not from intrinsic energies of the photons.

And how could the photons possibly be doing work to effect an upward acceleration??

how could they have any net effect on the motion of the box whatsoever?

Kinetic energy;



You are saying that the blueshift occurring with the freefall box due to the kinetic energy gained through freefall would not occur with the lowered box because of the lack of gained kinetic energy in that case. Correct??

I agree completely with your description here


This seems to contradict your conclusion;That the lowered box photons would be redshifted.

As far as the photons in the lowered box are concerned they are just as much in freefall as the inertial box photons.
The only difference is the velocity of the mirrors they are reflecting off of.

If we isolate a single reflection, say off the lower end of a box. The mirror is moving away from the photon. This is the case for both boxes. Likewise for a reflection off the upper mirror when the mirror is approaching the photon , which in both cases should compensate for any effect at the other end.
In this context how do you distinguish between the conditions in the two boxes?


So having defined a perfect atiabatic isolation how do you reconcile this with your conclusion of reduced temperature /frequency?

A photon in a box, bouncing up and down, in a perfectly vertical direction, in a gravity field, is weightless. That means, there is no momentum exchange with the gravitating body.

That's because, as I have explained, a falling photon gains no energy, so it does not gain any momentum either.

Magnetic field never does any work on moving charges. Gravity field never does any work on photons.

 

[Dale]

as I have explained, a falling photon gains no energy

Nature disagrees: http://en.wikipedia.org/wiki/Pound–Rebka_experiment

 

[PeterDonis]

Austin0, good questions. I'll respond to them slightly out of order.

So having defined a perfect atiabatic isolation how do you reconcile this with your conclusion of reduced temperature /frequency?

Just to clarify, only the first box, the one that's freely falling, is perfectly adiabatically isolated. And the photon temperature inside it, measured relative to the box, does *not* change.

The second box has a connection to the environment: the rope. All the stuff about mirror walls and so on just ensures that the rope is the *only* connection the second box has to its environment. And because of that connection, the photon temperature inside the second box, relative to the box, *does* change.

Sorry if the above wasn't fully clear. See further comments below on why the photon temps behave as I've stated.

Obviously both boxes had to originate from hovering systems. Under constant acceleration requiring energy.

But the "energy" could come from static equilibrium; the boxes could have started out sitting on a platform at the higher height, which is held up by columns, for example. The scenario does not specify how any "static" objects--objects at a constant height--are held there, since it's not relevant to the question the scenario was intended to answer. If it matters, assume all objects are held in static equilibrium; no rocket engines or other "hovering" methods that require energy expenditure.

The box that descends slowly does so through the upward acceleration transmitted through the rope.

Yes.

The energy does not go from the box to the platform but in the other direction.

No. The direction of "transmission of acceleration", so to speak, does not have to be the same as the direction of energy flow, or work done. Work is being extracted from the box during the lowering process. I haven't specified how that's being done, because again it's irrelevant to the question the scenario was intended to answer; but here's one way it could be done. On the platform at the higher height there is a large pulley; the rope is wound around the pulley. The weight of the box slowly unwinds the rope, causing the pulley to turn. As the pulley turns, it spins up a flywheel by means of a belt system that allows the speed of the flywheel to increase while the speed of the pulley remains constant (something like a continuously variable transmission, that's found in some cars). So work is done on the flywheel by the pulley, and therefore work is done on the pulley by the box. That means energy is extracted from the box and stored in the flywheel.

From the box's point of view, the reason energy is being extracted from it is that it is being lowered in a gravity field, so its potential energy is decreasing; and it's traveling at a (slow) constant velocity, so its kinetic energy is unchanged. So on net its energy is decreasing.

As far as I can see the only energy that goes up the rope is the kinetic energy of the box.

Nope; see above. The KE of the box is unchanged during the lowering process (except for brief periods at the start and end, where the box is put into slow motion, and then stopped again.)

And how could the photons possibly be doing work to effect an upward acceleration??

They're not. The upward acceleration comes from the rope, which holds the box in place and prevents it from freely falling.

how could they have any net effect on the motion of the box whatsoever?

See further comments below.

You are saying that the blueshift occurring with the freefall box due to the kinetic energy gained through freefall would not occur with the lowered box because of the lack of gained kinetic energy in that case. Correct??

That's one way of looking at it, yes.

As far as the photons in the lowered box are concerned they are just as much in freefall as the inertial box photons.

During the time they are not reflecting, yes, that's true. But the reflection events are *not* the same. See below.

The only difference is the velocity of the mirrors they are reflecting off of.

It's not just the velocity of the mirrors that's different; it's the *acceleration* of the mirrors, as in proper acceleration. The free-falling box's mirrors feel no force; they are weightless. The lowered box's mirrors feel a force, from the rope, so they are accelerated. That makes a difference when the photons reflect off the mirrors. See below.

If we isolate a single reflection, say off the lower end of a box. The mirror is moving away from the photon. This is the case for both boxes.

No, it's not. Look at it in the momentarily comoving inertial frame (MCIF) of the box. The freely falling box will be at rest in this frame, and will remain at rest for the time of flight of a photon from one wall to the other. (Strictly speaking, we need to assume that the length of the box is small enough that light can cross it before tidal effects become measurable; but that is a pretty easy condition to meet.) So the walls won't be moving relative to the photon, and the two impacts will just cancel each other out, as you say.

The lowered box, however, is accelerated upward; so if the box is momentarily at rest when the photon leaves one wall, it will be moving when the photon reaches the other wall. If the photon is going from the upper to the lower wall, the wall will actually be moving *towards* the photon in the MCIF when the photon reaches the lower wall. The wall will be moving away from the photon if the photon is moving from the lower towards the upper wall. So here the two impacts do *not* cancel out; the downward impact, on the lower wall, exerts more force than the upward impact, on the upper wall, does. So on net the photons exert a downward force on the box; this force does work that is transmitted up the rope.

(Note that we can ignore the constant downward velocity of the box; we simply adjust the MCIF so the box is at rest in it at the instant a photon leaves one wall. That means the MCIF will have a small downward boost relative to the MCIF of a static observer at the same altitude. That has no effect on the analysis I just gave, but if you're worried about it having some other effect, we can make it negligibly small by lowering the box slowly enough.)

 

[jartsa]

Nature disagrees: http://en.wikipedia.org/wiki/Pound–Rebka_experiment

Well this experiment is something I should explain. And as I said, I did explain. Let's argue when you have read the whole thread

 

[Dale]

The statement was wrong*, regardless of what else you may have said wrong or right previously.

*as was the statement about a photon in a box being weightless

 

[jartsa]

The statement was wrong*, regardless of what else you may have said wrong or right previously.

*as was the statement about a photon in a box being weightless

If you know how photons in a box behave, then educate us, because we have a small problem here. We all know about gravitational redshift of photons already.EDIT: Here's a short course of Jartsa's photon redshift theory:

Electric charges are deposited on the clock hands of a clock. Now the clock produces an EM-wave. The clock is dropped into a gravity well. Now clock hands move more slowly, and the frequency of the EM-wave is lower.

 

[Austin0]

A photon in a box, bouncing up and down, in a perfectly vertical direction, in a gravity field, is weightless. That means, there is no momentum exchange with the gravitating body.

That's because, as I have explained, a falling photon gains no energy, so it does not gain any momentum either.

Magnetic field never does any work on moving charges. Gravity field never does any work on photons.

What did I write that you could interpret as my saying a falling photon gains energy or momentum?

 

[jartsa]

What did I write that you could interpret as my saying a falling photon gains energy or momentum?

Well actually, I don't know what you said. Something about photons in a box ...

So I added my two cents about photons in a box.

 

[Q-reeus]

If you are writing a covariant equation, you can't use 3-momentum; 3-vectors are never covariant in 4-D spacetime. So all the covariant quantities and equations I've written, including dp/dtau, use 4-momentum.
The Lorentz force equation you wrote down does use p to mean 3-momentum, I agree; that's because it's written in a particular frame, where we split 4-momentum into energy (a scalar) and momentum (a 3-vector).

Fine with that.

True, the fact of static equilibrium by itself doesn't tell us how the magnitudes will be represented in different frames. But it does tell us that the magnitudes must be *equal and opposite*. That, by itself, is enough to show that the two observables, O and O', are equal. I gave the chain of equal and opposite magnitudes that shows it. That chain of reasoning works regardless of which frame you choose to evaluate the magnitudes...

You were OK until you added the qualifier "in a given frame". Once again, the two observables, O and O', are actual observed numbers; they are readings on a scale. The readings on a scale cannot depend on what frame you use to calculate them. Nor can they depend on the state of motion of the "observer" relative to the scale. So they are not numbers "in a given frame"; I must be able to calculate the numbers using quantities represented in *any* frame, and come out with the same answer. You've agreed to this before, but you keep on using language that just begs to be interpreted as contradicting it.

Trouble is you have here that O and O' are readings on one and the self-same scale - which moves with the charge q. In which case I obviously agree that when someone holds up a placard reading '2', an observer moving at u relative to the placard holder still sees '2'. But that is not the definition of O and O' I gave back in first para of #102 and to which seemingly we both agreed to work from. That definition had O the reading on one scale static wrt capacitor (call it lab frame S), whereas O' was the reading on a different scale - that moving with the charge q in frame S' having relative speed u wrt S. Both O and O' are perfectly legitimate physical observables. Both detecting the same phenomenon - force on a given charge q. And one reads differently to the other! Now if working definitions have transmogrified along the way, little wonder there is continued disagreement. it impacts on the rest of your #132 the same way.

 

[Q-reeus]

Q-reeus: "or how they co-vary as measured in another frame."
False: static equilibrium is static equilibrium, regardless of relative motion in directions orthogonal to the equilibrium. That means the magnitudes that are equal and opposite stay equal and opposite regardless of that relative motion.

Quite so. Opened myself to attack on that one but it was just down to an inapt use of terminology. By "co-vary as measured in another frame" it was meant as "both varying together equally as measured in another frame". No real disagreement there.

So O and O' have to remain equal when the charge is moving relative to the capacitor, because the motion is orthogonal to the direction of the equilibrium. Their actual magnitude changes with relative motion; it goes up, by the first point above. But they are still equal and opposite; both magnitudes change in sync.

Last bit true but the confusion and disagreement is over your current definition of O and O', as discussed last post. But then it changes below...

Consider another scenario, not involving electromagnetism. I put weights on a scale and set the scale on a trolley which sits at rest on a track. There is another scale under the track, measuring the weight of the track and whatever is on it. The scales, trolley, and track all have negligible mass compared to the weights. I define two observables: O, the weight shown on the scale under the track, and O', the weight shown on the scale on the trolley.

Great so far - you have moved back to original definition in #102 of what O and O' refer to.

To the best of my knowledge, all of the following are true:
(1) When the trolley sits at rest on the track, O = O'. Call this reading A.

There are two readings involved here so I would prefer to put it as situation (1): O_A = O'_A

(2) When the trolley is rolling along the track at a constant velocity v, O = O'. (We assume the track is perfectly horizontal, so the trolley's motion is perfectly orthogonal to the direction of gravity.) Call this reading B.

My rewording; situation (2) has O_B = O'_B.

(3) In the non-relativistic approximation (v << c), reading A will be the same as reading B.

However, if we allow v to be relativistic, then the gravitational "force" on the weights is no longer just the standard Newtonian force; there is an additional velocity-dependent force which can be viewed as due to "gravitomagnetism". (In the limit as v -> c, the total force goes to twice the Newtonian value; another way of saying this is that "acceleration due to gravity" is higher on a relativistically moving object. This is why light grazing the Sun, for example, bends by twice the amount you would predict by just doing a Newtonian calculation for a mass "falling" in the Sun's field, where the mass is the light's energy divided by c^2.) So in the case of relativistic v, reading B will be larger than reading A. (I don't think anyone has actually run this experiment, of course; a real experiment like this done on Earth could only last a small fraction of a second at relativistic velocities without violating the "perfectly horizontal" requirement.) But O = O' will still hold even in this case.

I have a different take. Treating above scenario (2) as straight-line equivalent of particles circulating in a synchrotron storage ring, we assume that transferring energy from some source (e.g. bank of batteries) to the circulating particles of relativistic energy γm (m the rest energy), no overall change in gravitating mass has occurred.

[My turn to clarify better here. 'Overall' is referring to that of source+circulating particles taken together - not that the circulating particles show no net change in gravitating mass - they clearly do do! Hence if a quantity W of battery PE energy is converted into particle KE, gravitating mass lost in batteries is precisely added to particles.]

Which requires by analogy that in (2), (3),:
O_B = γO_A, O'_B = γ²O'_A, so O_B = O'_B/γ.

The matter of that doubling the Newtonian deflection has intrigued me when it comes to applying it to a spinning flywheel with spin axis oriented somewhere between radial and transverse to say Earth's gravitational field. I had at one time thought radial orientation of axis might lead to flywheel weighing more than with transverse orientation. Suggesting oblique orientation yielding a transverse net acceleration. Which no doubt is equivalence principle violating - and I presume coincidentally checked along with the precessional stuff re gravity probe B experiment? Or there have been Eotvos style experiments specifically checking that spin/orbital motion, either macroscopic (gyro) or microscopic (magnetized material) has no orientational gravitational/inertial effect re 'weight'?
And, umm, we are sort of getting pretty sidetracked from OP topic I believe...HINT!

 

[PeterDonis]

Fine with that.

Trouble is you have here that O and O' are readings on one and the self-same scale - which moves with the charge q.

No, they're not. My definition matches yours, as given just a bit later in your post:

That definition had O the reading on one scale static wrt capacitor (call it lab frame S), whereas O' was the reading on a different scale - that moving with the charge q in frame S' having relative speed u wrt S.

This is my definition too, and that's what I was using when I made my post. I see now that I worded it as "a reading on a scale", which doesn't make it clear that O and O' were supposed to refer to the readings on two *different* scales, as you describe. Sorry if that wasn't clear.

Both O and O' are perfectly legitimate physical observables. Both detecting the same phenomenon - force on a given charge q.

I would say that they are both *determined* by the same phenomenon, the Lorentz force on the charge q. One, O', is determined by that force more directly than the other (fewer pieces in between), but they both ultimately are determined by the same force. But that means that this:

And one reads differently to the other!

Makes no sense! The two scales are in force balance in the x direction; the relative motion in the y direction does not change that. They have to read the same, or they won't be motionless in the x direction; the forces in that direction won't be in balance and something will move. As you have it, with O < O' (force measured at the arm greater than force measured at the track), the arm will push the track in the negative x direction.

 

[PeterDonis]

I would say that they are both *determined* by the same phenomenon, the Lorentz force on the charge q. One, O', is determined by that force more directly than the other (fewer pieces in between), but they both ultimately are determined by the same force.

Maybe it's worth expanding on this a bit to see if it clears up an issue. Both O and O' are *determined*, ultimately, by the Lorentz force on the charge q, but neither one "detects" that force directly. That is, if I write down the most fundamental local equation that determines O and O', neither one will have the Lorentz force in it! The only fundamental local equation that has the Lorentz force in it is the force balance directly on the charge q: that balances the Lorentz force on q, q E' = q gamma E (E' is in the rest frame of q, E in the rest frame of the capacitor), against the force exerted on q by the arm, which must be equal and opposite for q to be motionless.

But the force exerted by q on the arm is not what the scale attached to the arm measures; it measures the force exerted on the arm by q. That force is determined by Newton's Third Law, not the Lorentz force law. So observable O' is not directly "detecting" the force on the charge q; it's only "detecting" it indirectly, through the force exerted on the arm by q.

Similarly, observable O, the force exerted on the track by the arm, is determined by Newton's Third Law; it must be equal and opposite to the force exerted on the arm by the track. As I said before, the latter force must be equal and opposite to O', for the arm to be motionless; so again observable O is determined by Newton's Third Law, not the Lorentz force law. Observable O "detects" the Lorentz force on q indirectly, just as O' does; it just detects it more indirectly since there are more intermediate steps in the chain that links them.

 

[PeterDonis]

Great so far - you have moved back to original definition in #102 of what O and O' refer to.

See my previous two posts; I never changed the definition, but my wording may not have made that clear. Hopefully it's clarified now.

There are two readings involved here so I would prefer to put it as situation (1): O_A = O'_A

My rewording; situation (2) has O_B = O'_B.

I have no real problem with this notation; I'm just too lazy to type a lot of tags, so I tend to prefer notation that doesn't require me to do that.

I have a different take. Treating above scenario (2) as straight-line equivalent of particles circulating in a synchrotron storage ring, we assume that transferring energy from some source (e.g. bank of batteries) to the circulating particles of relativistic energy γm (m the rest energy), no overall change in gravitating mass has occurred. Which requires by analogy that in (2), (3),:
O_B = γO_A, O'_B = γ²O'_A, so O_B = O'_B/γ.

I'm not sure this analogy holds; I'll have to take a bit of time to consider the rotating scenario. My only comment right now is that I would not use "gravitating mass" here; I would instead say that, if the system as a whole is isolated, then its total 4-momentum and total angular momentum must be conserved. That makes it clear that we are only considering flat spacetime; we are not considering the self-gravity of the system, only the flat spacetime effects of rotation, internal energy/momentum exchange, etc.

The matter of that doubling the Newtonian deflection has intrigued me...we are sort of getting pretty sidetracked from OP topic I believe...HINT!

It intrigues me too, but I agree it's getting too far off topic for this thread. There was at least one other recent thread on the spinning flywheel in a gravity field scenario; I don't remember offhand where that one ended up.

 

[Q-reeus]

This is my definition too, and that's what I was using when I made my post. I see now that I worded it as "a reading on a scale", which doesn't make it clear that O and O' were supposed to refer to the readings on two *different* scales, as you describe. Sorry if that wasn't clear.

OK thanks for clarifying, but wish my earlier misinterpretation on that was right, as it's back to disagreeing on reading O, which you maintain is identical to O'.

Makes no sense! The two scales are in force balance in the x direction; the relative motion in the y direction does not change that. They have to read the same, or they won't be motionless in the x direction; the forces in that direction won't be in balance and something will move. As you have it, with O < O' (force measured at the arm greater than force measured at the track), the arm will push the track in the negative x direction.

No it won't. Relative motion of scales generates a 'lever advantage' situation of sorts. Scales in S' are closer to the lever 'fulcrum point' so to speak and record a larger applied force than scales in S - in fact necessarily so if the 'lever' (static equilibrium in x direction) is to stay in balance. I see no greater paradox here than the more familiar one involving clocks and rulers in different frames - in particular with one frame being a rotating one so the differing readings are not reciprocally observed. None of these imperfect analogies may impress you and it's getting to be a last ditch effort at seeking accord.

So I ask you to just clarify please your take on the mechanical example of spinning hollow cylinder given in first part of last paragraph in #102. What do you believe will be observed, and what explains it as seen in the two frames. I completely skip your #147 as answer to this problem may help change one of our outlooks fundamentally - cross fingers!

 

[Q-reeus]

I'm not sure this analogy holds; I'll have to take a bit of time to consider the rotating scenario. My only comment right now is that I would not use "gravitating mass" here; I would instead say that, if the system as a whole is isolated, then its total 4-momentum and total angular momentum must be conserved. That makes it clear that we are only considering flat spacetime; we are not considering the self-gravity of the system, only the flat spacetime effects of rotation, internal energy/momentum exchange, etc.

Sure I agree it's all analyzable using SR principles. Please note my edit in #145 - can't afford any more misunderstandings!

It intrigues me too, but I agree it's getting too far off topic for this thread. There was at least one other recent thread on the spinning flywheel in a gravity field scenario; I don't remember offhand where that one ended up.

OK - and best move back to center as we're not out to break the record for longest running thread. :zzz: