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Are the directions of electric fields lines affected by Gravity?

  1. Jul 27, 2012 #1
    I recall that the path of light itself can be altered by gravity, then, being part of the electromagnetic force, then is it safe to assume that the paths of electric fields lines can also be warped?

    I would imagine that the consequences would be enormous for electric flux in curved space.

    For instance, the flux about a point charge is given by (1/4)(1/pi)(1/Eo)Q/r^2 by Gauss's law INTEGRAL[E * dA] take a sphere so the E field is tanget to the sphere and decreases at a uniform rate from the center, then you get Q/Eo = E(4pir^2), then solve for E.

    However, if we're in a region of highly curved space, then the E field no longer retains its uniform rate of change from the center of a point charge. The flux lines would appear something like the picture attached below.

    Eventually all of the flux lines would concentrate themselves at a singular point, as if the field of the point charge originated from two or more locations.

    Or conversely, in a rapidly expanding area of space, which would be negatively curved, the E-field would then diminish even faster than 1/r^2, much like a dipole, which decreases at a rate of 1/r^3. In a region with a strong enough negative curve, the electric field about a point charge could reach zero over a finite distance (or possibly have reverse effects, an electric field under time reversal? Such as electrons attracting other electrons?)

    Of course this sounds like a bunch of nonsense, but the Gaussian laws would suggest this given my current understanding, and thus my understanding must be severely flawed!
     

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  3. Jul 27, 2012 #2
    Yep, electric field lines are warped since the space itself is warped. One of the more interesting ideas I've seen is the idea that electrons themselves are a kind of microscopic wormhole. Electric field lines converge on the wormhole giving the sort of point particle behavior we've come to expect. Positrons would be the other end of the wormhole. :D The probelm I see with this view is that you still have a sort of electromagnetic field filling all of spacetime separate from gravity. Which I suppose isn't all that different from quantum field theory in which every fundamental particle has a unique field associated with it that fills the universe. Seems conceptually icky to me. Ideally, the electromagnetic field would itself be understood in terms of geometry.
     
  4. Jul 27, 2012 #3

    pervect

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    Field lines work fine with General Relativity, and are a fine way to visualize electromagnetism. The tubes of the field lines are representations of anti-symmetric tensors, also known as a "differential forms", of rank 2, referred two as two-forms.

    Elelctromagnetism turns out to be represented by a rank 2 anti-symmetric tensor, so it's a perfect match.

    See for instance http://125.71.228.222/wlxt/ncourse/DCCYDCB/web/condition/9.pdf , "Teaching Electromagnetic Field Theory Using Differential Forms", IEEE TRANSACTIONS ON EDUCATION, VOL. 40, NO. 1, FEBRUARY 1997. This doesn't specifically talk about the usefulness of two forms with GR, but it does talk about using two forms to teach E&M.

    Gravity turns out to be much harder - in general it takes a rank 4 tensor to represent it, and while it has a high degree of symmetry, it's not a differential form.
     
  5. Jul 27, 2012 #4
    It may be easier to talk about EM fields not as "differential forms" but in terms of bivectors. The EM field is a vector field in spacetime but a field of oriented planes. The most natural way to think of this iis to take the electric field lines you know and extend them in the time dimension to draw out sheets. This is the natural object that describes the EM field, and boosting these planes is what gives the relationship between electric and magnetic fields.

    This also explains why gravity isn't usually said to have field lines. The Riemann tensor isn't a bivector field like the EM field; rather, it's a linear operator on bivectors.
     
  6. Jul 27, 2012 #5
    Maybe an occasion to at last get some kind of resolution to what was proposed but never settled here: https://www.physicsforums.com/showthread.php?p=3946413
    Seems to be from above entries a general agreement E field will be warped by gravitation. That makes good sense to me, yet how is that squared with the generally accepted as mathematically sound Reissner-Nordstrom charged black hole, where at large r the E field, owing to charge infalling at the infinitely redshifted BH EH (event horizon) is asymptotically close to the flat spacetime expression q/(4πε0r2)? Does it make any sense to say E field lines could in general be gravitationally bent somehow (as suggested by illustration in #1) yet with absolutely no change in field strength - which the mere existence of a RN BH surely implies? To put this concretely in a very simple and less extreme context, suppose we have a uniformly charged thin spherical mass shell. What does the marriage of EM with GR predict for the dependence of radial E field at large r on the mass M of that shell? Or suppose instead of the surface charge, a 'point' electric dipole is placed at the center of that mass shell. Will the field lines be altered at large r owing to M (assuming, apart from any purely gravitational effects, shell electromagnetic transparency - i.e. εr=1)?
     
  7. Jul 27, 2012 #6

    PeterDonis

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    Hi, Q-reeus, yes, we never really got closure on the questions raised in that other thread. I actually did work through a lot of the computations but never got to a point where I felt ready to post them. If I have time I'll try to go back and look again.

    Regarding the specific point in your quote above, in a R-N BH the E field lines are not bent; they all stick straight out, radially, from the BH, all the way out to infinity. That has to be true by spherical symmetry. So that particular case may not be the best one to investigate whether and how gravity can bend EM field lines in a more general scenario that doesn't have that special symmetry.
     
  8. Jul 27, 2012 #7
    Hi Peter. It would be good to see things resolved there (and/or here) if possible (big if!).
    Agreed and I probably could have clarified slightly better that E field bending was not implied in as you say spherically symmetric RN case. However the point there was that imo field warpage of any kind is inconsistent with RN implied total absence of coupling between source charge gravitational potential and far field radial E field strength (radial was specified). Logically *static* EM field (no disputing EM radiation does couple to gravity) is either effected or not by gravity as a package - can't in general admit to directional warpage without logically field strength also being effected. And you know my arguments re finite RN E field. :rolleyes:
     
  9. Jul 27, 2012 #8

    PeterDonis

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    I'm not sure I would describe such a coupling, if it is present in the spherically symmetric RN case, as "field warpage". But I think by "warpage" you really mean "some observable effect on the field", which seems more general to me and would cover the RN case.
     
  10. Jul 28, 2012 #9
    Two identical charges side by side in a gravity field might repel each other in these alternative ways:
    1: both experience some lift
    2: both experience some down force
    3: for some very odd reason the identical charges experience a different vertical force
    4: the forces are opposite and horizontal

    Well quite obviously number 4 is the only reasonable of these alternatives.

    We know that two observers side by side in a gravity field must look upwards, if they want to see each other, but electric charges are felt as being in the direction where they really are.
     
  11. Jul 28, 2012 #10
    Yes it just means here 'any departure from flat spacetime field configuration' - field pattern or field strength.

    It may be worth reminding at this stage of a finding you made in another thread that isotropic Minkowski spacetime spatial metric components interior to a spherical mass shell are identical to asymptotic exterior Schwarzschild values at infinity. Consequently if one collapses together the plates of a charged parallel plate capacitor lying inside such a shell, it's coordinate determined travel distance is identical to that when done at infinity. Alternately simply discharge the plates at constant separation - the capacitor field effective volumes are identical inside and at infinity. Yet we know that the energy release when operation is performed within the shell is depressed wrt infinity by the redshift factor √-gtt = √(1-2GM/(rc2)).

    And that imo inescapable fact presents a rather difficult dilemma for RN supporters (i.e. practically whole of GR community). I raised this scenario several times in the other thread linked to in #5, but worth raising it here again. Cuts right through all the mathematical elegance and sophistication of the standard picture. Which basically wants to have it's cake (RN metric with E field unaffected by infinitely depressed grav redshift) and eat it too (standard grav redshift applies to emitted radiation etc.).

    Could splitting between an invariant 'active' charge |E| = qa/(4πε0r2) and potential effected 'passive' charge F = qpE, with qp = q√-gtt, allow just that? It certainly works for the collapsing capacitor plates discharge scenario energy-wise. And for the static plates discharge too if we identify gravitationally depressed electrostatic energy density with W = 1/2√-gttε0|E|2. Even works if applied to output of a dipole oscillator where field output is reduced solely owing to redshifted frequency - both input and output power drops as -gtt (radiation field strength drops proportional to frequency and thus as √-gtt), as required. Alas there is at least one fatal failing.

    Suppose we separate two charged spheres apart via a dielectric rod, equal and opposite charges but with one difference - one sphere is more massive than the other. If RN is true, qa's are identical and hence magnitude of E field induced by one charged sphere on the other. However we had to have qp dependent on potential, so action and reaction are unequal, with a greater force exerted on the less massive sphere. Newton not happy.

    A better approach imo that has no such inconsistency at least for our static scenario was suggested in #248 in the other thread linked in #5. Leave charge as just plain q but with coordinate values for vacuum permittivity and permeability given by: ε,μ = (1/√-gtt)(ε00). We now have that coordinate determined light speed c, EM energy and power densities, and Newton's 3rd law for electrostatics (and similarly for magnetostatics) all work out good. I do not claim this can be automatically extended to more complex situations involving general motion in non-static, non-symmetric spacetimes, but then it does not fall badly at the first hurdle either!
     
  12. Jul 31, 2012 #11
    They're going for gold in London, but a different type of gold applies to this thread - as one golden oldie goes 'Silence is golden, golden...' Or another: 'The kiss of death, from Mr Gold Finger...' Gee, hate to think that's my label here. Anyway just in case someone wants to step out of the shadows, here's another thought following on from #10.

    Say we have a steady current loop as magnetic dipole of moment m, arranged coaxial with and centred about an electric dipole p of the same energy (as determined by the mutual mm and pp forces of attraction/repulsion between other identical such dipoles). Two such mp pairs, but one pair with opposite relative orientation between m and p, will, provided separations are not too close to invalidate 'point dipole' approximation, experience no net interaction forces or torques. That's in flat spacetime. What about if these pairs are within the spherical mass shell of #10? Still flat spacetime there, but potential depressed by √-gtt. Locally, non-interaction between dipole pairs must still hold. Which is interesting, since coordinate value of drift speed of magnetic dipole circulating currents is depressed wrt infinity by √-gtt, and so therefore the magnitude of the magnetic dipole moments m - assuming as per RN metric implies, charge itself as a source of E field is invariant wrt potential.

    By RN metric reckoning, the coordinate value for electric dipole moments p are unaffected by the surrounding mass shell (charge magnitude and displacement distance both indifferent to gtt). Locally then, within the shell, net interaction-free mp pairs, but by coordinate measure there is net interaction owing to dominance by the electric dipoles?! Just doesn't add-up imo. And this coordinate imbalance persists regardless of whether one applies the split into 'active' and 'passive' charge done in #10 or not. Oddly perhaps the force imbalance problem between charged spheres as per #10 cancels out when applied to an equivalent situation involving current loops of differing mass and therefore locally differing √-gtt factors. Small comfort.

    Now try the suggested cure: depressed values for ε,μ = (1/√-gtt)(ε00). From the Wiki article here: http://en.wikipedia.org/wiki/Dipole, we have expressions for magnetic dipole:
    b7d9fcf7464a06bec4fb084517b2927a.png
    and electric dipole:
    134903c8e2bdf92a4cadf1a7205c57ab.png
    At first sight the inverse relative locations of ε00 in the two expressions appears in conflict with having each directly modified by 1/√-gtt factor - seemingly implying an increased B field in conflict with an expected reduction in E field. Recall however that the magnetic moment m is that owing to circulating charges, which in coordinate measure not only circulate slower by √-gtt, but with reduced effective charge by the same factor. Hence m by itself is reduced by the factor -gtt, the field B then further modified by μ = 1/√-gttμ0 whereas electric moment p is unaffected in the above expression for E, it's reduced effective charge appearing outside of p itself - in the single modifier 1/√-gtt for ε0. Properly interpreted and applied then, there is imo a harmony here that works. That a well-known 94 yo solution to EFE's is directly under challenge seems to be a matter of sheer indifference here at PF. Or maybe Mr GF just generates too much perplexity/fear/loathing for comfort. :confused:
     
    Last edited: Jul 31, 2012
  13. Jul 31, 2012 #12

    PeterDonis

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    Yes, I did make such a finding. But it doesn't mean what you think it means. The fact that the spatial metric coefficients are the same does not mean you can directly compare distances inside the shell to distances at infinity the way you are trying to. See further comments below.

    However, your capacitor scenario is irrelevant to the questions you've raised about RN spacetime, because in the capacitor scenario, unless I'm misunderstanding something, the EM field is supposed to be zero everywhere except between the capacitor plates. So the EM fields of both capacitors are purely local, and a purely local analysis is all that can be applied to them.

    You're not stating this precisely enough. The correct statement is that the *locally measured* capacitor plate separation is the same inside the shell and at infinity. See below.

    Let's restate this scenario more precisely. We have two capacitors: C1 is at infinity, C2 is inside a spherical mass shell. The proper distance between the plates, measured locally, is D for both C1 and C2. Both capacitors are stipulated to start in the same state of charge, again as measured locally. That means that the same energy E, measured locally, will be released when we discharge each capacitor. So measured locally, the capacitors are identical.

    It is true that an observer at infinity will see energy E coming from C1, but energy fE coming from C2, where f < 1 is the "redshift factor" inside the spherical mass shell. What accounts for the difference? Obviously that the energy was redshifted as it climbed out of the gravity well. It doesn't imply any local change in C2 as compared to C1; as we saw above, locally the two capacitors are identical. That means the energy redshifting has nothing to do with the local behavior of the EM field. (If you want, we can transport the energy from C2 to infinity by some means other than EM radiation.) The difference in observed energy received at infinity is purely due to the effects of the intervening spacetime; it has nothing to do with the local properties of C2 vs. C1.
     
  14. Jul 31, 2012 #13

    PeterDonis

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    I've only given your latest scenario a quick read-through, but you are making an awful lot of claims about what GR and the RN metric supposedly say, without backing them up with math. This has happened before. You also continue to insist on focusing on coordinate-dependent quantities instead of invariants, which has also happened before. And you insist on comparing coordinate-dependent quantities in different parts of spacetime (e.g., at infinity vs. inside the shell) and then insisting that these comparisons have some physical meaning. This has happened before as well. All of these are reasons why your proposed challenges are so often received with what you term "sheer indifference".

    Furthermore, it's not clear to me exactly what you think you are challenging. Are you trying to claim that the R-N metric is not a solution of the EFE? That's ludicrous; it's easy to check, and thousands of physics undergraduates probably check it every year as a homework assignment. Are you trying to claim that the EFE is wrong? Good luck with that; within its domain of validity it has a lot of experimental support (to put it mildly). Are you trying to claim that the R-N solution, while mathematically correct, is somehow "unphysical"? Well, yes, it is, in at least two ways. One way, which is IMO not that big an issue, is that it assumes exact spherical symmetry; but that's easy to justify as an idealization which is often pretty well approximated by real objects. The other way, which is more of an issue, is that the interior of the R-N solution has a number of features that make it not very reasonable physically. But neither of those issues are relevant to your proposed challenge; you are also assuming perfect spherical symmetry, and your challenge can be applied purely in the exterior region of R-N spacetime. So I'm not sure exactly what you think your challenge is supposed to prove.
     
  15. Jul 31, 2012 #14
    Sure it does - as obviously implied in #10 and explicitly stated in that other thread linked to in #5. We can, especially obviously for collapsing plates scenario, use ropes & pulleys, rods & bell cranks etc. and make a 1:1 correspondence that is entirely physically meaningful. Do you challenge that? How so exactly if you do?
    Asserting my examples are irrelevant does not make them so - see my last comments.
    A trivial true statement which ignores what's going on here. Go back and check context of that entire first main para in #10, esp. preceeding sentence to above. Actually, don't bother; here it is in full:
    No reasonable excuse for taking it other than how it is clearly meant to be taken. And how it then subsequently leads on.
    Another 'precise' and equally trivially true statement.
    And via ropes & pulleys etc. we can tie that down to equivalent fields and forces acting 'down there'. As per earlier comments. And it's not just 'absolutes' - this time as per your opening comments in #13, think carefully about the example given in #11 - we have a number of ratios 'paradoxes' of RN making there. If it's all so easy to resolve within standard picture, offer your own full explanation/resolution please, up to your own standards of 'precision'. I maintain logic behind RN metric inevitably leads to paradox.
    Please do not keep repeating that red herring - you aught to know perfectly well it is a false representation of what I have been saying all along. And I don't like going around in circles.
    Well the intervening spacetime doesn't have any effect on charge as source of E if RN metric is true. Do you disagree? How so if so?
    Having read your #13 best I don't respond to it because it's style and tone invites a slanging match we can do without. what would impress in a positive way wold be for you to apply that undoubted prowess with 'the math' to at the very least the simple scenario of #5, and actually commit to a definite prediction. Why was that not done quite some time ago in that other thread? Here I have given an increased choice of alternative configurations - all laughably simple for someone of your grasp of GR surely. The one time you tentatively presented what passed as a proper math solution, not for for charged shell but RN exterior E field was here:https://www.physicsforums.com/showpost.php?p=3969524&postcount=345
    My somewhat delicate response was in #348, and you came back, very briefly, in #366 with what looked to be a quiet disowning of the original. Fair enough I suppose, but time to properly settle at minimum the charged shell case, wouldn't you say? Then we might come back to the highly relevant scenarios here and see it all in a new light.

    So may I humbly suggest we put off vague accusations of 'imprecision' etc. Once the long awaited kosher GR solution for charged spherical mass shell materializes, discussions of objective substance will be possible. Of course I do not accept there is anything lacking in logical rigor to that already given here in #10 & #11, but I understand the imo underlying Sacred Cow mentality that is denied in word but adhered to in practice. Don't think I'm singling you out on that score - far from it. Must go. :zzz:
     
  16. Jul 31, 2012 #15

    PeterDonis

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    And any such linking mechanism will be affected by the curvature of spacetime in between, just as the energy being transmitted back upward is.

    I know no such thing. I wasn't making a statement about what you were saying; I was making a statement about what GR says. Did you read the part where I said I was assuming that the EM field is zero everywhere except inside the capacitor plates?

    If you'll remember, in one of those other threads I said I needed to go back and re-evaluate what I said about the charge integral being unchanging, since it didn't seem to be consistent with other "obvious" things about the R-N metric.

    Because such things take time and effort. First, before even constructing a math model, I have to be sure I understand your scenario correctly. Your way of stating scenarios does not always make that easy. Please understand that I'm not saying that as a criticism, and I apologize for coming across as critical in my previous post. But it is a fact that I (and apparently others here on PF) sometimes find it difficult to understand your scenarios. That makes it difficult to model them in the math. Obviously *you* understand your own scenarios, but you can't model them in the math, so there has to be a translation step involved, and often it doesn't appear to work very well.

    Second, please understand that I am participating in these discussions because I find them interesting and fun, but that doesn't mean they are at the top of my priority list. (I suspect this goes for others as well who have said things about trying to model these things in math.) I understand that you believe you have a knock-down refutation of GR, and I also understand that nobody here on PF has yet presented a counterargument that you consider valid. That doesn't change the fact that I, and probably others, are applying a heuristic that says that, when presented with a scenario like those you have proposed, which is claimed to refute GR, it is far more likely that the person who is proposing the scenario has made a mistake somewhere, than that it actually refutes GR. That doesn't mean we have actually found the mistake; nor is it a *proof* that there must be a mistake. It's just a heuristic judgment in order to set our priorities for how much time and effort we are willing to spend in trying to either verify your claims, or find your mistake.

    I understand that you think it *should* be at the top of all our priority lists (or at least higher than it is now), because you think you are right; you think you *have* found a refutation of GR. You even think it is obvious. But if you are really, really convinced that you have found a refutation of GR, then you should not be depending on us to provide the proof for you. *You* should learn the math yourself; *you* should learn the physics yourself. *You* should be able to write up your arguments and proofs in the standard language of the field. Yes, that's a lot of work; but if you're right, there's probably a Nobel Prize at the end of it. Of course part of the reason I am telling you this is that I think there's a very low probability that you are right; so I think that if you actually do all that work, you will end up just finding out for yourself where you have made mistakes in the scenarios you have proposed here. But I do think the effort is worthwhile quite independently of where you end up with regard to those particular scenarios. Of course that's another heuristic judgment of mine, with which you may disagree.
     
  17. Jul 31, 2012 #16

    pervect

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    Gauss's law says that we can define the charge enclosed by a surface as the integral of the surface area multiplied by the electric field normal to the surface. There's a very similar theorem in two forms, when the integration process is carried out appropriately.

    The easiest way to define the apropriate sense of integration is to use the idea of counting field lines. You can also envision it as a more traditional integral carried out by dividing the whole area into a large number of small pieces, by insisting that you use observers with unit diagonal metrics - i.e. Lorentzian observers - to carry out the integration of each piece.

    Consider a space-like sphere surrounding a black hole. This can easily be visualized for any point outside the event horizon by a sphere of constant schwarzschild radius r. The surface area of said 3-sphere will be 4 pi r^2, by he definition of the Schwarzschild r coordinate.

    The event horizon isn't space-like , but null. But you can't have a static observer at the event horizon anyway, so it makes no sense to ask what the E-field would be there.

    So it is necessary and sufficient that (4/3 pi r^2) * E_normal = constant for charge to be conserved. Which is in fact the prediction of the RN metric when you convert the metric using the usual transformation rules to that of a locally Lorentzian observer.

    Note that this is a different behavior than what a local measure of gravitation will give. The integral of local normal force * area (defined in the special way we did above) isn't constant for gravity.
     
    Last edited: Aug 1, 2012
  18. Jul 31, 2012 #17

    PeterDonis

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    This should be "2-sphere" and surface area 4 pi r^2, correct? (Vs. "3-sphere" and 4/3)

    This was my thought, too, in the other thread that Q-reeus linked to; it's why I later said that the calculation in the specific post he linked to didn't look right to me, because it didn't match other things. The calculation was:

    [tex]Q = \frac{1}{4 \pi} \int{F_{ab} n^{a} u^{b} dS}[/tex]

    where [itex]dS[/itex] is the surface element of a 2-sphere, [itex]n^{a}[/itex] is the outward-pointing normal to the 2-sphere, and [itex]u^{b}[/itex] is the 4-velocity of a static observer--or, more relevant for this calculation, it's the 4-velocity of an observer whose worldline is orthogonal to the spacelike hypersurface in which the 2-sphere lies. Since none of the quantities in the integrand depend on the angular coordinates, the integral just gives a factor of [itex]4 \pi r^2[/itex] and we end up with

    [tex]Q(r) = r^2 F_{ab} n^{a} u^{b}[/tex]

    I inserted [itex]u^{b} = (1, 0, 0, 0)[/itex], [itex]n^{a} = (0, \sqrt{1 - 2M / r + Q^2 / r^2}, 0, 0)[/itex], and [itex]F_{10} = Q / r^{2}[/itex] to obtain

    [tex]Q(r) = Q \sqrt{1 - \frac{2M}{r} + \frac{Q^2}{r^2}}[/tex]

    which, as I noted in that old post, could be interpreted as the "charge at radius r" being "redshifted" relative to the "charge at infinity". However, by your argument, we should get [itex]Q(r) = Q[/itex] for all r, which would require [itex]u^{b} = (1 / \sqrt{1 - 2M /r + Q^2 / r^2}, 0, 0, 0)[/itex]; i.e., the timelike vector would have to be normalized using the metric to be a unit vector. The latter seems more physically reasonable to me, but I haven't been able to convince myself definitely one way or the other, if I don't use the fact that I already know the answer ought to be constant, Q(r) = Q at all r.
     
  19. Jul 31, 2012 #18

    PAllen

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    Pervect did say:

    " You can also envision it as a more traditional integral carried out by dividing the whole area into a large number of small pieces, by insisting that you use observers with unit diagonal metrics - i.e. Lorentzian observers - to carry out the integration of each piece."

    This alone would rule out ub being (1,0,0,0) in RN coordinates. Also, I have never seen this symbol used for anything other than a timelike unit vector.
     
    Last edited: Jul 31, 2012
  20. Aug 1, 2012 #19

    pervect

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    Yes, I wanted the surface of the 3 sphere, which is a 2 sphere, and it's obviously 4 pi r^2

    As far as your calculation goes, I haven't looked it over in great detail, but if u^b = (1,0,0,0) g_00 u^b u^b is not equal to 1, so it's not a unit length vector, because g_00 is not unity. At least if you're using the coordinates I think you are (Schwarzschild).

    A four-velocity should have unit length, it's only (1,0,0,0) if g_00 = 1.
     
  21. Aug 1, 2012 #20

    PeterDonis

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    Ok, good.

    Yes, I'm using Schwarzschild coordinates. I agree the u^b vector I wrote down is not a timelike unit vector (and so I shouldn't have called it a "4-velocity", as PAllen pointed out); however, it is the timelike Killing vector, correct? So the question is, which is appropriate for the formula I wrote down: the timelike unit vector (in which case the integral gives Q at any radius r), or the timelike Killing vector (in which case the integral gives Q(r), which includes a "redshift factor" at any finite r)?
     
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