Is 'charged black hole' an oxymoron?

[Q-reeus]

So, in this scenario the differential equations of interest are the Einstein field equations and Maxwell's equations. The boundary conditions could be a spherically symmetric E-field at some specified radius outside of the EH (represented by Q), and the spherically symmetric spacetime curvature at some specified radius outside of the EH (represented by M). The differential equations can then be solved together with those boundary conditions to determine the curvature and E-field at other locations.

I have no problem with that in general - the problem is in deciding if the correct underlying physics has been applied; i.e is charge invariance in curved spacetime imposed as axiom or can it be proven as utterly inevitable? Tricky one I would say.

I will see if I can dig up a derivation and then we can take a direct look at the assumptions and see if any are indeed improper or doubtful.

That could be very illuminating - provided a succinct interpretation is uncoverable.

Once we agree on the physics outside a charged planet then we can discuss the BH case.

Yes probably the best place to start - or rather continue, as e.g. my charged mat/balloon example was an oblique poke in that direction.

 

[Q-reeus]

If I interpret the two equations as you've just said, they say the same thing:
dp/dtau = dt/dtau dp/dt = gamma dp/dt
Which is fine; if by "force" you just mean the "coordinate force" dp/dt, then you're correct, the equation you wrote is always valid, because it's just a different way of writing the equation I wrote, a way which makes the relativistic covariance more obscure.

And I will respond that relativistic covariance obscures here what was said back in #61 - sticking with the lab frame, relative motion of charge q+arm has no effect on the actually measurable 3-force that scales placed under the tracks the q+arm is moving on will observe! And if that seems stretched in some way to you, rearrange the setup. Multiple charge+cantilever-arms mounted evenly on a spinning carousel subject to a uniform axial E field from coaxially arranged charged fixed circular-plates capacitor. Do we agree no matter how fast the carousel spins, scales placed under the carousel will register no changed weight owing to electrical forces on those orbiting charges? And that at some speed, the arms will break nevertheless? It goes without saying we discount centripetal effects here (and the greater the carousel radius, the less those centripetal forces become for a given rim speed). I stand fully by the legitimacy of 'relativistic weakening' in that context and as specifically related in #61 and following. And notwithstanding current position, the gist of your earlier counter arguments in #69, #72 seemed clear enough.

The question is, what do we actually measure? The answer is, we actually measure dp/dtau. That's what, for example, a strain gauge attached to the arm that is holding the charge q in place would read. And that observable has to be invariant; it can't be frame dependent, because it's a direct observable. The reading on the strain gauge attached to q can't depend on which frame we decide to calculate it in. I thought that by "force" you meant the actual measured force, the force the strain gauge would read, which is why I thought your version of the force law was incorrect. You're right, it's not incorrect; it just doesn't mean what you think it means.

Oh I think it means exactly what I think it does - as explained above. To repeat, I did mean the actual measured force - but the 3-force charge-velocity-independent one as measured in the capacitor rest frame. The one scales under the tracks/carousel directly physically measures. Yet the arm breaks owing to velocity. But let's call this one a day. Rest of your lengthy expose in #89 is no doubt good for picking up the procedure of going from one form to another, but somewhat over my head - I just like to get basic concepts dead right. :zzz:

 

[PeterDonis]

And I will respond that relativistic covariance obscures here what was said back in #61 - sticking with the lab frame, relative motion of charge q+arm has no effect on the actually measurable 3-force that scales placed under the tracks the q+arm is moving on will observe!

Well, obviously we disagree, but I'm not sure what we disagree about. We seem to agree that the actual observable is what "scales placed under the tracks the q+arm is moving on will observe"--which I take to be equivalent to what strain gauges attached to the arm would observe. Let's call this observable O. Your claim is that, in the lab frame, relative motion of q and the capacitor has no effect on observable O. But I'm not sure which of the two following positions you are taking about what happens in the rest frame of q (in which the capacitor is moving):

(1) Are you saying that, in the rest frame of q, observable O somehow has a different value, when the relative motion of q and the capacitor is the same? This obviously violates relativistic invariance; the actual observable has to be the same regardless of which frame we calculate it in.

(2) Or are you saying that, in the rest frame of q, observable O has the same value, and it's the lower value you calculated (without gamma), not the higher value I calculated (with gamma)? But if that's the case, then the equation you wrote down for the Lorentz force law is only valid in the rest frame of the capacitor; in the rest frame of q it predicts a number which is larger by gamma, because the E field in the rest frame of q is larger by gamma.

Of course the position I am taking is that both of the above are wrong: observable O has the same value regardless of which frame we calculate it in, and its value is the value I calculated, with the gamma in it. In other words, the correct relativistic equation predicts that relative motion between q and the capacitor *does* affect the observed force.

And if that seems stretched in some way to you, rearrange the setup. Multiple charge+cantilever-arms mounted evenly on a spinning carousel subject to a uniform axial E field from coaxially arranged charged fixed circular-plates capacitor. Do we agree no matter how fast the carousel spins, scales placed under the carousel will register no changed weight owing to electrical forces on those orbiting charges?

No. But until we clarify the question I asked above, I think we should hold off on other scenarios.

It goes without saying we discount centripetal effects here (and the greater the carousel radius, the less those centripetal forces become for a given rim speed).

As an approximation, I think this is OK; I'd have to run some numbers to be sure. But as I said above, one scenario at a time.

 

[jartsa]

In this case you are extracting energy from the system--you have to, to keep it from freely falling. Since you're modeling the photons as a gas, that means you are decreasing the temperature of the gas. There's no "weakening" of the atoms in the elevator; they stay the same, because the forces that hold them together don't depend on temperature.

(Strictly speaking, the temperature of the elevator walls will decrease as the photon gas temperature decreases, which means the atoms will be jiggling around less energetically. But that only affects the motion of the atoms as whole atoms; it doesn't affect the internal binding energy of the atoms that holds them together.)



Here you are not extracting any energy from the system, so of course the temperature of the photon gas stays the same. Again, that has nothing to do with whether or not the atoms are "weakening".

Now you made an error! What you really think to happen is that photon gas in a free falling elevator experiences a blueshift. And in the elevator that descends at constant speed there's no temperature drop, because gravity adds as much energy as is extracted.

Is there some law that says that you can't know inside a closed elevator, if the elevator is ascending or descending?

 

[Dale]

I have no problem with that in general - the problem is in deciding if the correct underlying physics has been applied

Well, if you accept the laws and the boundary conditions as correctly describing the physics then everything else follows. That is why it is so important to make sure that you pick good boundary conditions. Since this is Physics Forums, and since the EFE and Maxwell's equations are accepted mainstream physics, we should just use those as givens. But that still leaves a lot of freedom to choose good boundary conditions.

; i.e is charge invariance in curved spacetime imposed as axiom or can it be proven as utterly inevitable?

I don't know the answer to that. I will look and see if I can find a derivation of that also.

 

[PeterDonis]

Now you made an error!

If I did then so did you; everything I said was based on the premise that your description of the two scenarios was correct. You said:

1: Let's consider photon gas in a mirror lined elevator that is going down at constant speed. Photons redshift when bouncing from the floor, and blueshift when bouncing from the ceiling. A photon that moves upwards along a gently sloping path will be deflected downwards by the gravity, before it hits the ceiling. Therefore there is a net redshift of the photon gas, and the force that the gas exerts on the walls is decreasing.

The force the gas exerts on the walls is its pressure, and so you are saying its pressure is going down, which means its temperature must be going down as well, as I said.

You also said:

2: Let's consider photon gas in a mirror lined elevator that is free falling. No gravity is felt in the elevator, so no net blue or redshift of the photon gas happens, and the force that the gas exerts on the walls does not change.

Here you are saying the gas pressure doesn't change, so its temperature can't change either, as I said.

What you really think to happen is that photon gas in a free falling elevator experiences a blueshift.

Did I say that? Are you now saying your original description of the scenario, quoted above, was wrong?

And in the elevator that descends at constant speed there's no temperature drop, because gravity adds as much energy as is extracted.

Again, did I say that? And are you now saying your original description, quoted above, was wrong?

Is there some law that says that you can't know inside a closed elevator, if the elevator is ascending or descending?

There are principles in GR which say two things:

(1) Locally, inertial motion (free fall, no force felt, weightless) is indistinguishable from being rest in free space.

(2) Locally, accelerated motion (feeling a force, feeling weight) is indistinguishable from being held at rest in a gravitational field.

I'll leave it to you to work out how, if at all, those statements relate to elevators.

 

[stevendaryl]

I have no problem with that in general - the problem is in deciding if the correct underlying physics has been applied; i.e is charge invariance in curved spacetime imposed as axiom or can it be proven as utterly inevitable? Tricky one I would say.

I thought you were claiming that the combination of General Relativity and electromagnetism was inconsistent. That's a stronger claim than the claim that some of its claims haven't been proved.

 

[jartsa]

If I did then so did you; everything I said was based on the premise that your description of the two scenarios was correct. You said:



The force the gas exerts on the walls is its pressure, and so you are saying its pressure is going down, which means its temperature must be going down as well, as I said.

You also said:



Here you are saying the gas pressure doesn't change, so its temperature can't change either, as I said.



Did I say that? Are you now saying your original description of the scenario, quoted above, was wrong?



Again, did I say that? And are you now saying your original description, quoted above, was wrong?

Let's consider a photon gas container falling at great speed, breaking apart near the ground. Observers on the ground will say that the photons coming out are blue shifted.

According to me that happens because the observers are weakened. According to you that happens because the photons are gravitationally blue shifted. Right?

I tried to do a favor and add a blue shift into your description of the events in the elevator.

 

[PeterDonis]

Let's consider a photon gas container falling at great speed, breaking apart near the ground. Observers on the ground will say that the photons coming out are blue shifted.

Compared to what?

 

[jartsa]

Compared to what?

Let's say only one type of flash light exists. The photons in the container were produced by these flash lights at some high altitude compared to ground level.

The ground dwellers know all this. And they say the light became blue shifted when the container was free falling, blue shifted compared to light from flash lights at the ground level.

 

[PeterDonis]

Let's say only one type of flash light exists. The photons in the container were produced by these flash lights at some high altitude compared to ground level.

The ground dwellers know all this. And they say the light became blue shifted when the container was free falling, blue shifted compared to light from flash lights at the ground level.

So basically, your scenario is identical to just having the photons fall in the gravitational field, correct? Then what's the point of having the box with the mirrors inside it?

 

[Q-reeus]

Well, obviously we disagree, but I'm not sure what we disagree about. We seem to agree that the actual observable is what "scales placed under the tracks the q+arm is moving on will observe"--which I take to be equivalent to what strain gauges attached to the arm would observe. Let's call this observable O.

Immediate disagreement. There are two observables here, not one. Call O the 3-force F observed by scales under the tracks in the lab frame. Then O is independent of charge q's velocity u in that frame - as per usual Lorentz force law. Call O' the proper 3-force F' experienced by q in it's rest frame. That O' is greater by factor γ than O - and the strain-gauges attached to and moving with q+arm will register that increase. O' strain-gauge reading is of course then also observed in the lab frame - but in the lab frame we attribute the higher strain-gauges *readings* to a weakening of the cantilever arm. Instead of strain-gauges on the arm, substitute spring scales weighing the charge. In the lab frame, we say the higher *reading* of the spring scales moving with the charge is owing to RW (relativistic weakening - I'm using that often enough to give it an abbreviation) of the scale's springs - since the scales stationary in lab frame read no such increased force. Same goes for carousel arrangement.

This answers and sets right all your later points (1) and (2) in #93. And really the RW phenomenon is nothing more than consistently applying and interpreting the usual SR 3-force transformation relation - in particular the transverse force component. In the simple case there is only a transverse force, the proper force is just γ times that observed in the lab frame. It is then only the proper force that is a function of lab frame measured relative speed. The RW interpretation has much wider application than just charges moving through an E field and whenever there is transverse forces acting together with relative motion we have that all relevant parameters such as elastic constant, mechanical and dielectric strength are subject to this RW effect.

For instance, a spinning hollow cylinder, subject to axial compression by a force F measured in the non-spinning frame, will undergo axial strain by factor γ greater than if there is no spin (we assume an average peripheral speed u in cylinder wall applied to γ expression). In the proper frame of an element in the cylinder wall, that axial force becomes γF. In this situation, axial strain is identical in both frames, but attributed differently. And many other examples can be given - e.g. bursting stress of a rotating hoop will be less than naively applying only γ factor increased mass in the otherwise strictly Newtonian calculations - the correct factor to apply is γ² - one γ for mass, another γ for RW. In the orbiting proper frame of a hoop element this is attributed to centripetal acceleration being γ² times greater than measured in the lab frame. [There are extra complications from e.g. that SR length contraction effects the hoop radius that enter the overall calculations but are not pertinent to the current issue.] Hope this about exhausts this lengthy side-track!

 

[jartsa]

So basically, your scenario is identical to just having the photons fall in the gravitational field, correct? Then what's the point of having the box with the mirrors inside it?

Well, I don't know.

So let's consider a physicist lowering a photon gas filled box into a gravity well, using a rope. The physicist says that "I'm extracting energy from the photon gas and from the box, the radiation in the box cools down, and the box weakens."

When the box is really low the physicist says: "If I tug this rope just slightly, the weakened box is about to break, and the temperature of the cooled radiation rises by a large fraction almost melting the box"

When the physicist is doing this experiment, the idea of weakening of things when lowered down, becomes quite natural.

Do you see something wrong in this scenario?

 

[stevendaryl]

So let's consider a physicist lowering a photon gas filled box into a gravity well, using a rope. The physicist says that "I'm extracting energy from the photon gas and from the box, the radiation in the box cools down, and the box weakens."

When the box is really low the physicist says: "If I tug this rope just slightly, the weakened box is about to break, and the temperature of the cooled radiation rises by a large fraction almost melting the box"

When the physicist is doing this experiment, the idea of weakening of things when lowered down, becomes quite natural.

I don't get why you consider that lowering a box weakens the box. That doesn't make any sense at all to me.

I also don't think it makes sense to say that lowering the box cools down the radiation. The temperature as measured by a thermometer inside the box at rest relative to the box (and that's the only meaningful notion of temperature) does not decrease when the box is lowered.

Einstein's equivalence principle states that, for a small enough elevator, and for a uniform enough gravitational field, there is no measurable difference between an elevator at rest in a gravitational field and an elevator accelerating upward in gravity-free space. In either case, giving the elevator a constant velocity doesn't change anything; You can't tell the difference between an elevator that is moving at a constant velocity in a uniform gravitational field and an elevator that is at rest in a uniform gravitational field.

 

[TrickyDicky]

I thought you were claiming that the combination of General Relativity and electromagnetism was inconsistent. That's a stronger claim than the claim that some of its claims haven't been proved.

Maybe expressed like that it sounds like a strong claim, anyway in those vague (GR vs EM) terms it could mean a lot of different things, but if one specifies a bit and refers to the EFE and Maxwell equations it seems as one must logically infer that at some point those two set of equations lead to inconsistent (between them) results. If only for one thing, we know that the linear Maxwell equations don't account for vacuum polarization and are no longer valid in the non-linear limit of QED. However the nonlinear EFE are considered valid at all regimes except (supposedly) at the more remote Planckian scale that is not yet reachable experimentally.

 

[PeterDonis]

Call O the 3-force F observed by scales under the tracks in the lab frame. Then O is independent of charge q's velocity u in that frame - as per usual Lorentz force law.

I can see a sort of logic to this: if we *assume* that the force exerted on q itself is independent of q's velocity, then the force exerted on the scales under the tracks will be independent of q's velocity as well. For example, suppose that instead of a charge q we had a weight placed on the arm; then the weight would not change if the arm was moving horizontally (i.e., perpendicular to the direction of the force) vs. being at rest.

However, the assumption is wrong for the actual scenario we've been discussing: the force on the charge q is *not* independent of its velocity. See further comments below.

Call O' the proper 3-force F' experienced by q in it's rest frame. That O' is greater by factor γ than O - and the strain-gauges attached to and moving with q+arm will register that increase.

So now you are saying this: I start with q+arm at rest relative to the capacitor. I measure the two observables, O and O', and find that they are the same: the reading on the strain gauge is the same as the reading on the scales under the tracks.

Now I put q+arm in motion relative to the capacitor at velocity v, directed perpendicular to the E field of the capacitor. In this case, you are saying that O and O' now *differ*? The reading on the strain gauge goes up, but the reading on the scales under the tracks stays the same?

That doesn't make sense. The force read by the strain gauge will also be the force exerted by the arm on the tracks; which means you are saying that the force exerted by the arm on the tracks is larger than the force exerted by the tracks on the scales. But there is no motion of any of those parts in the direction of the force; the arm's motion is perpendicular to it and so it doesn't affect the force balance in that direction at all. So the entire force read by the strain gauge has to be transmitted to the scales.

I won't bother commenting on the rest of your post since it just carries forward this same mistake.

 

[PeterDonis]

So let's consider a physicist lowering a photon gas filled box into a gravity well, using a rope. The physicist says that "I'm extracting energy from the photon gas and from the box, the radiation in the box cools down, and the box weakens."

Why would he say the box weakens? I understand the part about extracting energy: he can measure that by extracting work from the system. But how does he determine that the box weakens? What measurement does he make to show that?

When the box is really low the physicist says: "If I tug this rope just slightly, the weakened box is about to break, and the temperature of the cooled radiation rises by a large fraction almost melting the box"

How are you deducing that this is what would happen? Are you using GR? If so, please show your work. If you are using some other personal theory, please explain what it is.

 

[Q-reeus]

So now you are saying this: I start with q+arm at rest relative to the capacitor. I measure the two observables, O and O', and find that they are the same: the reading on the strain gauge is the same as the reading on the scales under the tracks.

So far so good.

Now I put q+arm in motion relative to the capacitor at velocity v, directed perpendicular to the E field of the capacitor. In this case, you are saying that O and O' now *differ*? The reading on the strain gauge goes up, but the reading on the scales under the tracks stays the same?

Yes.

That doesn't make sense. The force read by the strain gauge will also be the force exerted by the arm on the tracks;...

Yes - but only as determined in proper frame of the arm.

...which means you are saying that the force exerted by the arm on the tracks...

This is O' force F' = γF seen in the moving frame of q+arm

...is larger than the force exerted by the tracks on the scales.

Yes, because we have implicitly shifted reference frame! Latter is now O force F = F'/γ

But there is no motion of any of those parts in the direction of the force;...

True, and we are not discussing energy balances, so what does that matter?

...the arm's motion is perpendicular to it and so it doesn't affect the force balance in that direction at all. So the entire force read by the strain gauge has to be transmitted to the scales.

Yes - but one frame determines that force differently to the other!

I won't bother commenting on the rest of your post since it just carries forward this same mistake.

Good grief! Peter, maybe we should conduct a poll!

 

[Q-reeus]

Getting back to the title subject proper, best to give a refresher as to what imo is the key argument. There are two basic situations - 'hovering' and free-fall charge. Latter is strictly the only one applying to an otherwise neutral BH that acquires charge via infall. It's by definition an extreme strong gravity case, so for now I'd rather just summarise the argument for a weak gravity hovering charge situation.

So there is an ideal spherical and non-rotating planet as central mass M. Lying horizontal (defined as an equipotential region wrt to M's influence) on the planet's surface there are two small spheres each of proper mass m, separated a small distance d apart. Gravity of M is weak and that of masses m much weaker again. I will take the position to very good accuracy we can then treat the proper mutual gravitational attraction
|F_g| = Gmm/d² - (1),
as Newtonian situation at least locally. Let balls gravitationally roll together so at point of impact, an amount W_g of initial mutual gravitational potential energy is converted to KE. Locally there is no effect from residing in the potential well of M, but in SC coordinate measure, W_g is reduced by redshift factor f = √(1-2GM/(cr²)), and since d is impervious to redshift, one ascribes the same reduction to the Newtonian force F_g in (1). Suppose now equal charges q are added to each ball, just enough to electrostatically cancel F_g. Bring the balls together now and no net work is involved, as both Newtonian gravitational and Coulombic influences fall off at the same inverse r² rate. Which means in coordinate measure mutual electrostatic field energy and repulsive forces |F_q| = kqq/d² have redshifted precisely the same as for Newtonian gravitational case.

Now here's the thing. The gravitational contribution of m's to overall mass M is not that given by proper values m, but the redshifted values fm. Let someone explain precisely why one should not make the obvious connection - given above conclusions, same redshift applies to remotely observed electrostatic field owing to q's. It's that simple imo. Too bumpkin simple you may say. Fine - just point out why in clear terms. No argument from authority here please.

 

[PeterDonis]

Yes, because we have implicitly shifted reference frame! Latter is now O force F = F'/γ

You're missing the point. I'm talking about actual observables: the readings on the strain gauge and the scale. I thought you'd already agreed that actual observables, like readings on strain gauges and scales, are the same regardless of what frame you calculate them in. I'm not even talking about "frames" at all. See below.

True, and we are not discussing energy balances, so what does that matter?

We are talking about force balances, which means forces have to, you know, balance. Otherwise the system won't stay fixed in the direction of the forces. See below.

Yes - but one frame determines that force differently to the other!

Again, you're missing the point. What I'm saying has nothing to do with frames. Let me run through it again:

(1) We have a capacitor, an object with charge q, an arm, and a track. We start out with all of them at rest relative to each other. We have a scale under the track; its reading is O. We have a strain gauge on the arm; its reading is O'. With this setup, reading O = reading O'. I think we agree on that, at least.

(2) Now we put q+arm in motion relative to capacitor+track. You are saying that O' increases, but O does not. That means O' > O. The actual, observable reading on the strain gauge is *higher* than the actual, observable reading on the scale. This is how I'm reading what you've said. Am I reading you right?

Notice that I said nothing whatever about "frames" in (2). I only made statements about actual, observable readings on devices. Those readings are independent of "frames"; they have nothing to do with "frames". They are direct observables, and either they are the same in situation #2 or they are not. You are saying they are not; I am saying they are.

Here's why I'm saying they are. (Do I really need to go through this? Apparently I do.) Construct free body diagrams for each object in the scenario, one after the other. In the direction of all the forces in question, which I called the "1" direction before but we can also call the "x" direction if it's easier to remember, all the objects are static. This is true in both situations, #1 and #2, above. In situation #2 two objects, q+arm, are moving in the "y" direction relative to the other two, but that has nothing to do with the force balance in the "x" direction. All objects are motionless in the "x" direction, therefore a free body diagram for each object must have all forces in the "x" direction (which is all forces in the problem) cancel, for both situations, #1 and #2.

Now take the objects one at a time:

Charge q: There is a force on q in what we'll call the positive "x" direction due to the field of the capacitor; call this force +F. (Note that we aren't specifying the exact magnitude of F; that will vary between situations #1 and #2. But for a given situation, F is fixed, and that's all we need to know to determine the force balance.) There is also a force on q in the opposite, or negative "x" direction, due to the arm, which is holding q in place. That force must be -F, equal and opposite to the force from the field of the capacitor, so that q is motionless in the "x" direction.

Arm: By Newton's Third Law, there is a force +F from q on the arm, equal and opposite to the force -F that the arm exerts on q. This is the force measured by the strain gauge, which we are calling observable O'. So O' = +F.

There is also a force on the arm from the track, which must be -F, equal and opposite to the force on the arm from q, so that the arm is motionless in the "x" direction.

Track: There is a force +F from the arm on the track, equal and opposite to the force -F that the track exerts on the arm (Newton's 3rd Law again). This is the force that the scale measures, which we are calling observable O. So O = +F = O'.

So in both situations, #1 and #2, O = O'; that has to be true because of the force balances as just described. The only difference between the situations is that in #1, O = O' = +F = qE, while in #2, O = O' = +F = q gamma E.

 

[PeterDonis]

So there is an ideal spherical and non-rotating planet as central mass M. Lying horizontal (defined as an equipotential region wrt to M's influence) on the planet's surface there are two small spheres each of proper mass m, separated a small distance d apart. Gravity of M is weak and that of masses m much weaker again.

You don't want to make M's gravity too weak, or the redshift factor will be unmeasurable. But I assume that by "weak" you just mean that M << R, where R is the radius at the surface of the planet; i.e., the planet is much larger than the Schwarzschild radius associated with its mass.

I will take the position to very good accuracy we can then treat the proper mutual gravitational attraction
|F_g| = Gmm/d² - (1),
as Newtonian situation at least locally.

This looks fine to me.

Let balls gravitationally roll together so at point of impact, an amount W_g of initial mutual gravitational potential energy is converted to KE.

Actually, the PE gets converted to KE continuously as the balls roll together. (I assume they're rolling on an ideal frictionless surface so the only thing driving their motion is their mutual gravity.) What happens when they collide is that their mutual KE gets converted to something else. Normally it would be heat, raising the temperature of the two balls; but they could, for example, re-radiate that heat as photons directed upward. See below.

in SC coordinate measure, W_g is reduced by redshift factor f = √(1-2GM/(cr²)),

How do you measure this? It seems to me that the way you would measure it is by capturing the energy produced somehow; say by having the heat generated in the collision re-radiated upward as photons, and captured at a very large radius where the redshift factor due to M is negligible. In that case, yes, the energy captured would be redshifted, compared to what you would capture if you captured the photons locally. But this may not mean what you think it means; see below.

and since d is impervious to redshift, one ascribes the same reduction to the Newtonian force F_g in (1).

Why? The only way to measure the redshift is to measure the difference between the energy captured locally and the energy captured "at infinity". But if you're trying to tell whether the Newtonian force is changed, why would you privilege the energy captured "at infinity" over the energy captured locally? The latter is a much better measure of the force, since it's captured locally; the energy captured "at infinity" is affected by the spacetime curvature in between.

In a way, all this is a matter of "interpretation"; we agree on the physical observables. If you insist on saying that the Newtonian force is "redshifted", I can't stop you. But you'll have to cover yourself with a lot of caveats because the "redshift" won't be observed locally; local experiments will all show the same Newtonian force as given by your equation above, with no "redshift". 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.

Sppose now equal charges q are added to each ball, just enough to electrostatically cancel F_g. Bring the balls together now and no net work is involved, as both Newtonian gravitational and Coulombic influences fall off at the same inverse r² rate..

Ok here; there's no net force, so there's no net work. But that also means there's nothing to redshift. See below.

Which means in coordinate measure mutual electrostatic field energy and repulsive forces |F_q| = kqq/d² have redshifted precisely the same as for Newtonian gravitational case.

How are you going to measure the nonexistent net work in order to show that it's redshifted?

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.

Now here's the thing. The gravitational contribution of m's to overall mass M is not that given by proper values m, but the redshifted values fm.

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.

same redshift applies to remotely observed electrostatic field owing to q's

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.

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.

 

[Q-reeus]

Again, you're missing the point. What I'm saying has nothing to do with frames. Let me run through it again:

(1) We have a capacitor, an object with charge q, an arm, and a track. We start out with all of them at rest relative to each other. We have a scale under the track; its reading is O. We have a strain gauge on the arm; its reading is O'. With this setup, reading O = reading O'. I think we agree on that, at least.

(2) Now we put q+arm in motion relative to capacitor+track. You are saying that O' increases, but O does not. That means O' > O. The actual, observable reading on the strain gauge is *higher* than the actual, observable reading on the scale. This is how I'm reading what you've said. Am I reading you right?

Yes - again.

Notice that I said nothing whatever about "frames" in (2). I only made statements about actual, observable readings on devices. Those readings are independent of "frames"; they have nothing to do with "frames".

Of course they do! How on Earth do you interpret meaning and application of the standard 3-force transformation expressions here: http://www.sciencebits.com/node/176 (last two groupings). You accept that a physically measured E field will be frame dependent. The transformations for E are of exactly the same form as for force (B field excluded).

They are direct observables, and either they are the same in situation #2 or they are not. You are saying they are not; I am saying they are.

Unfortunately, another yes to that last bit.

Here's why I'm saying they are. (Do I really need to go through this? Apparently I do.) Construct free body diagrams for each object in the scenario, one after the other. In the direction of all the forces in question, which I called the "1" direction before but we can also call the "x" direction if it's easier to remember, all the objects are static. This is true in both situations, #1 and #2, above. In situation #2 two objects, q+arm, are moving in the "y" direction relative to the other two, but that has nothing to do with the force balance in the "x" direction. All objects are motionless in the "x" direction, therefore a free body diagram for each object must have all forces in the "x" direction (which is all forces in the problem) cancel, for both situations, #1 and #2.

True if applied rightly - there must be force balance - as determined in each frame. But not if we mix and try to match across frames.

Now take the objects one at a time:

Charge q: There is a force on q in what we'll call the positive "x" direction due to the field of the capacitor; call this force +F. (Note that we aren't specifying the exact magnitude of F; that will vary between situations #1 and #2. But for a given situation, F is fixed, and that's all we need to know to determine the force balance.) There is also a force on q in the opposite, or negative "x" direction, due to the arm, which is holding q in place. That force must be -F, equal and opposite to the force from the field of the capacitor, so that q is motionless in the "x" direction.

Yes to this much.

Arm: By Newton's Third Law, there is a force +F from q on the arm, equal and opposite to the force -F that the arm exerts on q. This is the force measured by the strain gauge, which we are calling observable O'. So O' = +F.

Well, I would prefer you added a prime to +F, but ok per se.

There is also a force on the arm from the track, which must be -F, equal and opposite to the force on the arm from q, so that the arm is motionless in the "x" direction.

Yes, all true.

Track: There is a force +F from the arm on the track, equal and opposite to the force -F that the track exerts on the arm (Newton's 3rd Law again). This is the force that the scale measures, which we are calling observable O. So O = +F = O'.

And here's where things go awry - comparing apples with oranges.

So in both situations, #1 and #2, O = O'; that has to be true because of the force balances as just described. The only difference between the situations is that in #1, O = O' = +F = qE, while in #2, O = O' = +F = q gamma E.

Trivially true for #1, wrong for #2. If it were so - rush to patent - you have solved the world's energy needs forever! Think about the energetics in each frame.
[sorry, but must go - will respond to your #111 considerably later]

 

[PeterDonis]

Yes - again.

Ok, so we have a disagreement about an actual, observable experimental result. In which case discussion about theory is rather pointless. But still:

Of course they do! How on Earth do you interpret meaning and application of the standard 3-force transformation expressions here: http://www.sciencebits.com/node/176 (last two groupings). You accept that a physically measured E field will be frame dependent. The transformations for E are of exactly the same form as for force (B field excluded).

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. 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. In fact, it seems to me that confusing "frames" with actual physical observables is contributing to much of what you are saying here. See further comments below.

True if applied rightly - there must be force balance - as determined in each frame. But not if we mix and try to match across frames.

...

And here's where things go awry - comparing apples with oranges.

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. 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?

Think about the energetics in each frame.

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?

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.

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. 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?

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.

 

[PeterDonis]

You are confusing coordinate quantities with direct observables.

Actually, on re-reading and further consideration, the confusion is about even more than that. The equations on the sciencebits page talk about "force" as dp/dt, as I said, which is a coordinate-dependent quantity. But that page also completely fails to talk about the fact that, in a static situation, what we have been calling "force" does *not* correspond to any dp/dt at all, because in a static situation, dp/dt is *zero*.

Let me unpack this. All of the objects in both situations, #1 and #2, are motionless in the x direction, as I said. So there is *no* change in the x momentum, p_x, of *any* object, even though "forces" are being applied and measured. The reason, of course, is that the forces are balanced, so there is no *net* force on any object. And the equations on the sciencebits page are talking about *net* force, *not* about what we have been calling "force", as in "force measured by the strain gauge on the arm", or "force measured by the scale". As far as the equations on the sciencebits page are concerned, dp/dt is *zero* in all cases, so that force equation in any "frame" is just 0 = 0.

So what are these "forces" that the strain gauge and the scale are actually measuring? They are pressures times cross sectional areas. And they all have to balance as I described because the pressures times cross sectional areas have to balance across any plane normal to the x direction in any situation where all objects are motionless in the x direction. It has nothing to do with "force transformation", or indeed even with the Lorentz force law. The only way the Lorentz force law comes in is to determine the initial pressure times cross sectional area, between the charge q and the arm, that is required to hold q motionless. That pressure times cross sectional area must be equal to the Lorentz force on the charge, which, as Q-reeus has already agreed, is larger when the charge is moving relative to the capacitor. But once we have that answer in hand, everything else is determined by static equilibrium as I described; there are no transformations or frames or anything like that involved, because no object is moving in the x direction, so there's nothing to transform; dp_x/dt is *zero* for every object in the problem, regardless of which frame you pick to do the analysis.

The only other physical assumption is that static equilibrium in the x direction is unchanged by relative motion in the y direction. If anyone wants to try and refute that assumption, go ahead.

 

[jartsa]

I don't get why you consider that lowering a box weakens the box. That doesn't make any sense at all to me.

I also don't think it makes sense to say that lowering the box cools down the radiation. The temperature as measured by a thermometer inside the box at rest relative to the box (and that's the only meaningful notion of temperature) does not decrease when the box is lowered.

Einstein's equivalence principle states that, for a small enough elevator, and for a uniform enough gravitational field, there is no measurable difference between an elevator at rest in a gravitational field and an elevator accelerating upward in gravity-free space. In either case, giving the elevator a constant velocity doesn't change anything; You can't tell the difference between an elevator that is moving at a constant velocity in a uniform gravitational field and an elevator that is at rest in a uniform gravitational field.

Well, Peter Donis said that the radiation cools. And I could understand that sentence.

And I think Peter Donis and I did agree that the truth value of that sentence is false, if photons gain energy when falling. Now I'm saying the radiation cools, because photons don't gain energy when falling.

Oh yes, I agree that the reading of the thermometer in the box does not change.

 

[jartsa]

Why would he say the box weakens? I understand the part about extracting energy: he can measure that by extracting work from the system. But how does he determine that the box weakens? What measurement does he make to show that?

He sees the box deforming when he's tugging the rope.

How are you deducing that this is what would happen? Are you using GR? If so, please show your work. If you are using some other personal theory, please explain what it is.

So I thought hard and long how to deduce the blue shift of tugs on a rope, and this I came up with:

An accelerating spacecraft tows an asteroid, using a long massless rope. Force meters at the two ends of the rope show diffrent readings. I know this, because I read in this forum that accelerometers at the two ends of an accelerating rocket show different readings, and in this case the force meters can be thought to be also accelerometers.

Here is the thread: https://www.physicsforums.com/showthread.php?t=608495

Now let's use the equivalence principle: Force meters at the two ends of a massless rope, supporting a weight in a homogeneous gravity field, show different readings.

So here we have the blueshift of force, which is actually the weakening of the spring of the lower force meter.

 

[PeterDonis]

Well, Peter Donis said that the radiation cools. And I could understand that sentence.

And I think Peter Donis and I did agree that the truth value of that sentence is false, if photons gain energy when falling. Now I'm saying the radiation cools, because photons don't gain energy when falling.

Oh yes, I agree that the reading of the thermometer in the box does not change.

You have managed to contradict yourself and (at least possibly) misinterpret what I was saying, all in one short post.

You say the radiation cools, but you say the reading of the thermometer in the box does not change. Those statements contradict each other.

Also, you seem to be implying that I agreed that photons don't gain energy when falling. I didn't agree to that statement as it stands; I have said that photons *do* "gain energy" (that's one way of interpreting what's happening, anyway) when *freely falling*. But I'm not sure what you meant by "falling", so I'm not sure whether you misinterpreted me or not.

Rather than continue to try to get you to say something that is not either vague or inconsistent, let me go ahead and actually analyze the scenario from scratch. This is really more for stevendaryl's sake than for yours, since he posed some good questions. But you, and anyone else of course, are welcome to comment further.

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.)

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.

But how could that happen, you ask? Doesn't the equivalence principle say that the photons can't change while being slowly lowered? Actually, no, it doesn't. The equivalence principle says that acceleration in free space can't be distinguished, locally, from being at rest in a gravitational field; and the principle of relativity says that moving at a constant velocity *in free space* can't be distinguished, locally, from rest. But you can't put those two together and say that moving at a constant velocity *in a gravitational field* (or in an accelerated rocket) is indistinguishable from being at rest in the field. It isn't. The two cases are different, because in the first case (constant velocity in free space), nothing feels any force; everything is weightless. But in the second case (constant velocity relative to accelerating rocket/gravitational field), everything *does* feel a force. And that makes a difference; something has to be supplying/transmitting that force, and you have to include that something in your analysis.

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.

I haven't said anything about thermometer readings yet, but I think I'll let the above stand for now.

 

[PeterDonis]

So I thought hard and long how to deduce the blue shift of tugs on a rope

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...

So here we have the blueshift of force, which is actually the weakening of the spring of the lower force meter.

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.

 

[PeterDonis]

Now let's use the equivalence principle: Force meters at the two ends of a massless rope, supporting a weight in a gravity field, show different readings.

The equivalence principle doesn't apply here, because the measurement isn't local. The force meters will only show different readings if the difference in heights is enough to measurably change the "acceleration due to gravity"; i.e., that the effects of tidal gravity (change in field strength with height) are detectable. But the equivalence principle only applies within a small enough region of spacetime that tidal effects are not detectable.

 

[jartsa]

Why would he say the box weakens? I understand the part about extracting energy: he can measure that by extracting work from the system. But how does he determine that the box weakens? What measurement does he make to show that?
How are you deducing that this is what would happen? Are you using GR? If so, please show your work. If you are using some other personal theory, please explain what it is.

The equivalence principle doesn't apply here, because the measurement isn't local. The force meters will only show different readings if the difference in heights is enough to measurably change the "acceleration due to gravity"; i.e., that the effects of tidal gravity (change in field strength with height) are detectable. But the equivalence principle only applies within a small enough region of spacetime that tidal effects are not detectable.

Force meters at different altitudes in a homogeneous gravity field will show different readings, because force meters (accelerometers) at different heights in a rocket with constant proper acceleration show different readings.

That's very simple, and sounds quite plausible to me.

If we are in this rocket, and the the accelerometers show different readings, we still don't know if we are in a rocket or in a gravity field. So in a gravity field force meters must show different readings, otherwise we would know.

Oh yes an asteroid may have a gravity field. So let's make it a very small asteroid, and let's make the acceleration very large. Is the situation more clear now? We ignore the gravity field of the asteroid.