Is 'charged black hole' an oxymoron?

[Q-reeus]

If you are right that there is higher E in frame S, so that the force on q is higher in frame S, then it must be higher in frame S' as well. The force on q, meaning the proper acceleration it experiences times its rest mass, must be invariant. Instead of waving your hands, you should actually do the calculation.

Easy - 3-force F = qE in both frames (using appropriately transformed values for E naturally). The magnetic field seen in S exerts no force on q there, and there is no B field in S'. It is a fact motion of charge at any speed through a given fixed E field of any orientation has no effect on the 3-force on moving charge in that frame. Accelerations are a different matter. And ipso facto, transverse 3-force component is frame variant. You should be very familiar with SR 3-force transformation relations - only the force component parallel to relative velocity is frame invariant wrt v. Now when it gets to 4-vectors etc I get crossed eyes, but the 3-force stuff I do know here. I stand by previous 'relativistic weakening' comment as a legitimate viewpoint on that basis.

 

[PeterDonis]

Easy - 3-force F = qE in both frames (using appropriately transformed values for E naturally).

Exactly--*transformed* values for E. Which means your notation is obscuring the physics.

Adopting a better notation, define "E" as the electric field in the rest frame of the capacitor. Then, if gamma is the Lorentz gamma factor due to velocity v, the velocity of the charge q relative to the capacitor, then the electric force on q is F = q * gamma * E in the rest frame of the capacitor. If we transform to the rest frame of q, then E' = gamma * E and the force on q is F' = q * E' = q * gamma * E, i.e., the same actual, physically measured force calculated in both frames.

So the reason the arm breaks when it's in motion relative to the capacitor, instead of at rest, is that the electric force on q, and hence the force on the arm, is higher when q is in motion relative to the capacitor. No weakening of the arm at all.

 

[Q-reeus]

Exactly--*transformed* values for E. Which means your notation is obscuring the physics.

Adopting a better notation, define "E" as the electric field in the rest frame of the capacitor. Then, if gamma is the Lorentz gamma factor due to velocity v, the velocity of the charge q relative to the capacitor, then the electric force on q is F = q * gamma * E in the rest frame of the capacitor. If we transform to the rest frame of q, then E' = gamma * E and the force on q is F' = q * E' = q * gamma * E, i.e., the same actual, physically measured force calculated in both frames.

So the reason the arm breaks when it's in motion relative to the capacitor, instead of at rest, is that the electric force on q, and hence the force on the arm, is higher when q is in motion relative to the capacitor. No weakening of the arm at all.

Sorry Peter - your turn to experience a brain snap it seems. I've given up finding a website explicitly dealing with the type of scenario discussed here, but not needed. Lorentz force law F = qE + vxB is quite explicit - if there is no applied B field present, the purely electric force in a given frame is thus totally independent of the charge velocity in that frame. Basic stuff. Yes as we both agree the force is higher in the rest frame of q when set in motion relative to the capacitor. But transverse 3-force and transverse E field transformations cancel each other out such that, in keeping with vanilla Lorentz force expression, force in a given frame has no input from v there (zero applied B field) - period. To ease your pain a little, let me make a confession. Remember that ongoing thread about shell metric transition? Well guess what - I lately came to realize I was mistaken and standard SC's do not after all imply any unphysical jumps in radial metric - at least to first order locally observed. But then it was a certain someone's earlier erroneous claim that got me started on that wrong track in the first place... Must go. :zzz:

 

[Dale]

As you will be aware, my real focus is on logical reason for any external E for a BH.

I am not sure exactly what the source of the logical concern is, but maybe this will help clarify the real point of disagreement. I know that you are not into the math, but do you understand what is meant by the term "boundary condition"?

If you do, then you can simply take the EM field just outside the event horizon (on the edge of the Scwarzschild coordinate chart) as a spherically symmetric boundary condition. Then the question is: what is the EM field and the spacetime curvature in the rest of the chart? The EM field is given by Maxwell's equations and the curvature by the Einstein field equations. The two sets of equations are coupled due to the fact that the EM field has its own stress-energy tensor. The solution to the EFE is the RN metric which is described by two free parameters.

Do you see any of that as illogical? If not, then it seems that the only real question is under what physical conditions the boundary conditions can arise. Basically I am trying to "localize" the disagreement to inside or outside the EH.

 

[PeterDonis]

Lorentz force law F = qE + vxB is quite explicit

That's the non-relativistic approximation; it only holds if v << c. The correct relativistic law can be found in many places, for example the Wikipedia page:

http://en.wikipedia.org/wiki/Lorentz_force#Relativistic_form_of_the_Lorentz_force

As you will see, there is a gamma in there, just as I posted.

- if there is no applied B field present, the purely electric force in a given frame is thus totally independent of the charge velocity in that frame.

Nope--see above. Also, after looking at how the EM field tensor transforms, I'm no longer sure the induced B field due to the motion of the charge q relative to the capacitor has no effect on the force observed, but I'll save that for a separate post.

Remember that ongoing thread about shell metric transition? Well guess what - I lately came to realize I was mistaken and standard SC's do not after all imply any unphysical jumps in radial metric - at least to first order locally observed.

Thanks for the feedback, but as you can see from the above, I don't need any consolation for this discussion.

Also, that "to first order locally observed" is not strong enough: there are no "unphysical jumps" to any order. Seems like you still have a bit of thinking to do.

 

[PeterDonis]

after looking at how the EM field tensor transforms, I'm no longer sure the induced B field due to the motion of the charge q relative to the capacitor has no effect on the force observed

False alarm--I was forgetting that, even though in the rest frame of the charge there is an induced B field (because that frame is moving relative to the capacitor), the charge is not moving in that frame, so the v x B term is zero anyway. Move along, nothing to see here.

 

[jartsa]

Imagine light from some event (say, a charged particle accelerating) comes to you from two different directions; for example, suppose there is a very massive star, or black hole between the event and you, and the light can go around in one direction, or the other. When the light gets to you, the two images can have different amounts of redshift. You can't explain that in terms of weaker or stronger electrical forces down in gravitational well. The way to explain it is to realize that light has to travel from the event to your eyes. Depending on the path it takes, the light is changed by its journey.

Nope. There is a certain potential difference between two points. You plug that into some equation, and out comes certain redshift. I mean, redshift is determined by gravitational potential difference.

 

[jartsa]

I've read that thread, and I don't see how you got the following out of it:



There was no talk at all in that thread about the planet moving in the same direction as the light. You said the light losing energy when falling was your own idea, and indeed that's not in the thread either. Nor is the energy staying the same "to an extremely good approximation"; Jonathan Scott said that the frequency of the light stays the same, but he meant something different by "frequency", and he specified that that was relative to a particular coordinate system, the global Schwarzschild coordinates.

Let me try to summarize what the technical posts in that thread were actually saying. A photon emitted at a particular event in any spacetime will have a 4-momentum vector $k^{a}$ associated with it. Since photons travel on null geodesics, that 4-momentum vector will be parallel transported along the photon's worldline; this is the sense in which the photon "does not change" as it travels.

However, the observables associated with the photon are determined, not just by the photon's 4-momentum, but by geometric objects, vectors and tensors, associated with the observer. For example, the energy the photon is measured to have by that observer is the contraction of the photon's 4-momentum with the observer's 4-velocity $u^{b}$:

$$E = g_{ab} k^{a} u^{b}$$

So even if $k^{a}$ is unchanged as the photon travels, its observed energy can still change if either the metric $g_{ab}$ or the observers' 4-velocity $u^{b}$ changes. (We actually measure photon frequency, not energy, but the latter is just Planck's constant times the former.)

In the case of the standard Doppler shift, the measured energy (frequency) changes because the 4-velocity of the observer $u^{b}$ changes relative to that of the emitter, which determines the photon's 4-momentum $k^{a}$.

In the case of a photon falling into or climbing out of a gravity well, the energy (frequency) measured by static observers--observers who are "hovering" at a constant radius r--changes with r because the metric $g_{ab}$ changes. (The 4-velocity of "hovering" observers is the same at all r--all their 4-velocity vectors point "in the same direction".)

The "frequency staying the same" that Jonathan Scott was talking about was a different sense of "frequency": if I have a blinker, say, emitting flashes of light deep in a gravity well, such that it emits N flashes between Schwarzschild coordinate times t = 0 and t = 1, then an observer much higher up in the gravity well will also count N flashes between coordinate times t = 0 and t = 1. The two observers will differ in how much *proper* time they experience between those two coordinate times, so they will assign a different proper frequency (flashes per second of proper time) to the blinker; but the frequency relative to *coordinate* time is the same. This is all consistent with what I said above.

Well, "frequencies" and frequencies must stay proportional. Probably that was the point of the talk about "frequencies" there in the other thread. So frequencies stay constant, because "frequencies" stay constant.

 

[PeterDonis]

Well, "frequencies" and frequencies must stay proportional.

But the constant of proportionality changes as the metric changes. In the case of a gravity well, the constant of proportionality changes as you go deeper into the gravity well. So "frequencies" in the sense Jonathan Scott used the term in the particular post I referred to stay constant, but frequencies in the sense of local measurements do not.

 

[jartsa]

But the constant of proportionality changes as the metric changes. In the case of a gravity well, the constant of proportionality changes as you go deeper into the gravity well. So "frequencies" in the sense Jonathan Scott used the term in the particular post I referred to stay constant, but frequencies in the sense of local measurements do not.

Well, anyway frequency of a falling photon cannot increase, because it decreases slightly. When a fast moving object (photon) hits a still standing object (black hole) the kinetic energy of the still standing object increases, so the kinetic energy of the photon must decrease.

 

[PeterDonis]

Well, anyway frequency of a falling photon cannot increase, because it decreases slightly.

The reasoning you are about to give is incorrect (see below), but even if it were correct, it would certainly not apply while the photon is still above the hole's horizon. As it "falls" in that region, its frequency as observed by static observers (observers who maintain a constant altitude) increases. This has been experimentally measured; see, for example, here:

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

When a fast moving object (photon) hits a still standing object (black hole) the kinetic energy of the still standing object increases, so the kinetic energy of the photon must decrease.

The above reasoning is incorrect. A photon will not "hit" a BH; it just falls into it. A BH is not a "still standing object" in the sense you are trying to use the term here. It is better to think of the BH as a region of spacetime with properties very different from what we can intuitively picture.

 

[stevendaryl]

Nope. There is a certain potential difference between two points. You plug that into some equation, and out comes certain redshift. I mean, redshift is determined by gravitational potential difference.

No, that is not true. It is true in a static, symmetric case (black hole), but it isn't true in general. If a photon passes near a massive object that is itself moving, the photon can pick up additional energy, in exactly the same way that the slingshot effect can send a space probe shooting out of our solar system.

In general, two different photons that start off identical, and take different paths can have different frequencies.

 

[jartsa]

No, it happened because the photon was blueshifted as it dropped into the gravity well. A visible light photon emitted locally, at the same altitude as the atom, won't ionize it, so the field of the atom is not "weakened" at all according to local measurements. The difference that lowering the atom gently makes is that it is at rest deeper inside the well, so it "sees" the blueshift of the photon. To see why that's important, consider an alternate experiment where you let both a hydrogen atom and a visible light photon free-fall into the gravity well, in such a way that they meet up somewhere much deeper into the well than where you released them (you time the release of the atom and the photon from your much higher altitude to ensure this). Will the photon ionize the atom? No, because the atom is not at rest in the field; it is falling inward at a high speed, so there is a large Doppler redshift when it absorbs the photon that cancels the gravitational blueshift.

Hey now think I have thought about it plenty enough.

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.

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.

In case 1 the radiation weakens, so we can conclude that the atoms of the elevator weaken equally.

In case 2 radiation does not change, so we can conclude that the atoms of the elevator don't change either.

 

[PeterDonis]

1: Let's consider photon gas in a mirror lined elevator that is going down at constant speed.

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.)

2: Let's consider photon gas in a mirror lined elevator that is free falling.

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".

 

[Q-reeus]

I am not sure exactly what the source of the logical concern is, but maybe this will help clarify the real point of disagreement. I know that you are not into the math, but do you understand what is meant by the term "boundary condition"?

Generally. Like in EM where we say specify the vanishing of tangent E field component at a conductor surface etc. But impression is 'boundary condition, can get rather abstract in GR.

If you do, then you can simply take the EM field just outside the event horizon (on the edge of the Scwarzschild coordinate chart) as a spherically symmetric boundary condition. Then the question is: what is the EM field and the spacetime curvature in the rest of the chart? The EM field is given by Maxwell's equations and the curvature by the Einstein field equations. The two sets of equations are coupled due to the fact that the EM field has its own stress-energy tensor. The solution to the EFE is the RN metric which is described by two free parameters.

I understand in broad outline your point. My point is that maybe improper assumptions have been made in coupling the two, and gedanken experiments in #1 and subsequently are intended to be taken as potential counterexamples to be carefully thought through. Apart from dealing with a side issue here, what currently looks like a confusing mess getting full consistency working in SC's needs sorting out and haven't devoted much time to that.

Do you see any of that as illogical? If not, then it seems that the only real question is under what physical conditions the boundary conditions can arise. Basically I am trying to "localize" the disagreement to inside or outside the EH.

Well if we take the view charge has fallen onto the EH of a previously neutral BH, there it asymptotically exactly resides in 'frozen free-fall' from coordinate perspective. And it's from coordinate perspective 'charged BH' is supposed to make sense ('Inside' EH is essentially a meaningless concept imo). Which boils down to, as said times earlier, implying infinite redshift at EH has no effect - exterior E is as though no BH is swallowing the infalling charged matter. Hence my take in example 3: in #1, and the others there. Maybe that clarifies a little more. Before it was moved, I argued in the earlier thread this one forked off from, on the matter of whether from virtual photon view things made any better sense, but probably best to entirely avoid that angle here.

 

[Q-reeus]

That's the non-relativistic approximation; it only holds if v << c. The correct relativistic law can be found in many places, for example the Wikipedia page:
http://en.wikipedia.org/wiki/Lorentz_force#Relativistic_form_of_the_Lorentz_force
As you will see, there is a gamma in there, just as I posted.

You are referring no doubt to the *covariant* expression for Lorentz force in vector form shown on that Wiki page as:
dp/dtau = qγ(E+uxB) - (1),
and claiming it represents the 'relativistically corrected' version of the usual 'low velocity' expression:
dp/dt = F = q(E+uxB) - (2),
that is somehow only an approximation. I'm flabbergasted! Check the vector expressions earlier in that Wiki page, or any reliable articles such as:
http://arxiv.org/abs/1205.1080
web.mit.edu/sahughes/www/8.022/lec10.pdf

The usual 'low velocity' expression (2) is in fact exact and good for any velocities whatsoever. Provided it is applied as intended. Forces, fields and velocities all evaluated in the one frame they appear. Notice I used dp/dt = F in (2) - not dp/dtau. Important difference. No wiggling out that non-covariant expression (2) was clearly the form being discussed earlier!
My sole error in #73 was of the typo type - forgetting to use parenthetic enclosure around both E and vxB rather than the qE+vxB as shown. Pardon my oversight, but the pertinent issue is whether your 'relativistically corrected' version with the added gamma factor is right as you interpret it - to be applied directly to the E field and charge velocity appearing in the capacitor rest frame. Stop and think what that would mean. Action and reaction would be out the window, as would conservation of energy. Why bother to build huge superconducting magnet array monsters like at Cern LHC, when a simple bank of capacitors with holes in them would accomplish the task far better! Do I need to supply yet another gedanken arrangement or two to illustrate the truth of that? To again ease your pain, I promise not to use here the following smarty smiley's - :tongue2: :tongue::rofl::uhh:. Promise!

So what is the correct usage for expression (1)? To be honest, not being conversant with covariant formulations, I'm at a loss. Appears to quite simply yield the proper force dp/dtau - as seen in the moving charge's rest frame, in terms of the lab frame E, B, and u parameters. Trouble is, it just can't work in general on that basis. For example no proper force gamma factor applies for an E field component parallel to u, and that's simply not represented in (1), where implicitly all components are operated on equally. Hmmm.

Thanks for the feedback, but as you can see from the above, I don't need any consolation for this discussion.

Still sure about that?

Also, that "to first order locally observed" is not strong enough: there are no "unphysical jumps" to any order. Seems like you still have a bit of thinking to do.

Maybe - but enough said for now.

 

[stevendaryl]

That's the non-relativistic approximation; it only holds if v << c. The correct relativistic law can be found in many places, for example the Wikipedia page:

http://en.wikipedia.org/wiki/Lorentz_force#Relativistic_form_of_the_Lorentz_force

It's true that dP/dt = q (E + 1/c v x B) is the nonrelativistic form of the Lorentz force law, but that doesn't mean that it's an approximation.

The relativistic form is

dP_α/dτ = q F^αβ U_β

where τ is proper time.

Using U₀ = γ
U_j = γ v_j
F^0k = E^k
F_ij = B_k

this implies

dP/dτ = q γ (E + v x B)

But d/dτ = γ d/dt, so this becomes

γ dP/dt = q γ (E + v x B)

or

dP/dt = q (E + v x B)

So the latter is exactly equivalent to the relativistic form, not an approximation. The factors of γ cancel.

 

[Dale]

Generally. Like in EM where we say specify the vanishing of tangent E field component at a conductor surface etc. But impression is 'boundary condition, can get rather abstract in GR.

Well, in general physics is described by a class of equations known as differential equations. The differential equations, by themselves, cannot be solved. In order to solve them you must also include a set of conditions, known as boundary conditions. The boundary conditions describe the physical situation that is being modeled, and the differential equations then describe how the physical laws make that situation evolve.

So, for example, Newton's 2nd law: $f=mx''(t)$ and Newton's law of gravitation: $f=G m_1 m_2/r^2$ govern the motion of projectiles in vacuum. However, by themselves they cannot tell you where a given projectile will land. For that, you need to specify the initial velocity and initial position, which are the boundary conditions describing a particular projectile. Once you do that you can solve the system of differential equations and boundary conditions together to make specific predictions about a given projectile.

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.

My point is that maybe improper assumptions have been made in coupling the two

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.

'Inside' EH is essentially a meaningless concept imo

I understand that you want to work in Schwarzschild coordinates, and I agree that inside the EH is meaningless in those coordinates. That is a known limitation of Schwarzschild coordinates, but there is nothing wrong with using them as long as you stick to regions of the manifold where they are not meaningless. To do so, we can just specify the field at some radius outside the EH as the boundary condition and not worry too much about how it got there. E.g. we can suppose, if we like, that we are simply dealing with the fields outside a uniformly charged spherical planet rather than a BH.

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

 

[PeterDonis]

The usual 'low velocity' expression (2) is in fact exact and good for any velocities whatsoever. Provided it is applied as intended. Forces, fields and velocities all evaluated in the one frame they appear. Notice I used dp/dt = F in (2) - not dp/dtau. Important difference. No wiggling out that non-covariant expression (2) was clearly the form being discussed earlier!

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.

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.

So what is the correct usage for expression (1)? To be honest, not being conversant with covariant formulations, I'm at a loss.

Indeed. Try the most general covariant version, using the EM field tensor, which appears in the Wiki article (I'm giving it in a slightly different form which will be easier to work with):

$$\frac{dp^{a}}{d\tau} = q F^{a}_{b} u^{b}$$

This equation is valid in any frame, so you can just read off the frame-dependent versions by transforming the tensor components and the 4-velocity appropriately.

For example, in the rest frame of the capacitor, we have $F^{0}_{1} = F^{1}_{0} = E$, with all other EM field tensor components zero. We also have for the 4-velocity of the charge q $u^{b} = (\gamma, 0, \gamma v, 0)$. We have oriented the spatial axes so the capacitor's field is in the "1" direction and the charge is moving at velocity v in the "2" direction (perpendicular to the capacitor field, as was my understanding).

The only nonzero component of the above equation is the "1" component of the 4-acceleration:

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

which is what I wrote. And, of course, we can obtain what you wrote by just writing dp/dt = dtau/dt dp/dtau, which removes the factor of gamma.

Now do the same thing in the rest frame of the charge q. Obviously the 4-velocity is just $u^{0} = 1$ in this frame, all other components zero. But perhaps it's worth writing out the Lorentz transform of the EM field tensor explicitly, since you actually have to transform twice because the tensor has two indexes. And because we're using the "mixed" form of the tensor, with one upper and one lower index, we need two versions of the transform, one for each kind of index. I haven't found a good online reference for all this, but it's in the textbooks (I'm working from MTW).

A given EM field tensor component transforms like this (indexes are primed or unprimed according to which frame they refer to):

$$F^{a'}_{b'} = F^{c}_{d} \Lambda^{a'}_{c} \Lambda^{d}_{b'}$$

where repeated indexes are summed over, and the two transformation matrices are (you do have to keep careful track of which indexes are primed and which are unprimed):

"Forward" matrix (primed upper index):

$$\Lambda^{0'}_{0} = \Lambda^{2'}_{2} = \gamma$$

$$\Lambda^{0'}_{2} = \Lambda^{2'}_{0} = - \gamma v$$

$$\Lambda^{1'}_{1} = \Lambda^{3'}_{3} = 1$$

"Inverse" matrix (unprimed upper index):

$$\Lambda^{0}_{0'} = \Lambda^{2}_{2'} = \gamma$$

$$\Lambda^{0}_{2'} = \Lambda^{2}_{0'} = \gamma v$$

$$\Lambda^{1}_{1'} = \Lambda^{3}_{3'} = 1$$

If you take each of the above matrices individually, and work out how it acts on a vector, you will see that it's saying the same thing as you're used to with Lorentz transformations: x' = gamma (x - vt), t' = gamma(t - vx), etc., with the sign of the v terms switched for the inverse matrix.

Now we need to apply the matrices to the tensor components as shown above. This looks like a lot, but if you work through the possible combinations of indexes, you will see that there are only four nonzero components of the primed EM field tensor: $F^{0'}_{1'} = F^{1'}_{0'} = \gamma E$, and $F^{1'}_{2'} = - F^{2'}_{1'} = \gamma v E$. The first two terms correspond to an electric field in the "1" direction, as before, but now you can see that it is stronger, in "coordinate" terms, than it was in the rest frame of the capacitor, by a factor gamma. The last two terms correspond to a magnetic field in the "3" direction, i.e., perpendicular to both the electric field and the relative velocity of the charge and the capacitor. This is what you're used to seeing written out, in 3-vector form, as v x E.

Now read off the force in the primed frame from the index equation above; you will see that there is again only one nonzero component, that for the "1" component of the 4-acceleration:

$$\frac{dp^{1'}}{d\tau} = q F^{1'}_{0'} u^{0'} = q \gamma E$$

which is the same number as in the other frame. So the observable quantity is the same in both frames, as required.

But you could "interpret" the two equations differently. You could say that, in the rest frame of the capacitor, the "E field" is just E, and the "coordinate force" on the charge q is qE, but to obtain the observable number, the "4-force", you have to multiply by gamma to convert dp/dt to dp/dtau. Whereas, in the rest frame of the charge, the "E field " is already gamma E, so dp/dt and dp/dtau are the same (because t = tau for the charge in its own rest frame). This is all just "interpretation" and doesn't change the physics.

What you *can't* do is say that somehow the "real force" is different in the two frames, because that leads you to a false prediction: it leads you to predict that the reading on the strain gauge would differ depending on which frame you calculated it in.

 

[PeterDonis]

So the latter is exactly equivalent to the relativistic form, not an approximation. The factors of γ cancel.

Yes, I saw that when I realized that by "force" he meant the "coordinate force", dp/dt, not the actual observed force, dp/dtau. See my post in response.

 

[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.