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Force as measured at different heights in a gravitational field.

  1. May 28, 2012 #1
    Say we have identical masses side by side low down in a strong gravitational field. We attach ideal massless cords to each mass and suspend them from spring balances higher in the field. One spring balance (with a longer cord) is at a much higher altitude (h2) than the other at h1. Will the spring balances at h1 and h2 read the same or is there a force transformation?

    P.S. I probably should of stated that the same two spring balances read the same as each other when they are at the same height as each other and the masses are at the same height as each other, that the massive body is non rotating and uncharged and that the test masses are much smaller than the massive body.
    Last edited: May 28, 2012
  2. jcsd
  3. May 28, 2012 #2


    Staff: Mentor

    Assuming that everything is in static equilibrium, and that all objects are at a constant height, yes, the force measured at h2 will be less than that measured at h1. The force is "redshifted" based on the difference in heights.

    The limiting case of this, where the height of the object being suspended above the horizon approaches zero, and the height at which the spring balance is located approaches infinity, is used to define the "surface gravity" of the hole. The Wikipedia entry on surface gravity talks about this some:


    However, my initial Googling hasn't found a more detailed online reference. This is talked about in Wald's textbook on black hole thermodynamics; based on the Wiki entry's reference list it is also talked about in Wald's GR textbook, but I don't have it handy to give a specific chapter/page.
  4. May 28, 2012 #3


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    Staff Emeritus
    Science Advisor

    Force does decrease according to the redshift factor, aka the time dilation factor. It turns out that the surface integral of the "force at infinity" is equal to the enclosed Komar mass for a static system. You'll find the relevant discussion in Wald in the chapter on energy. I could dig up the exact reference, but unless you have Wald there's not a lot of point to it.
  5. May 28, 2012 #4
    Thanks for the replies. You have given me some useful pointers of where to start looking. On thing that has always puzzled me is that the surface gravity (per unit mass of the test object) of a black hole is sometimes stated as being GM/r^2 and at other times as being 1/(4M). The last quoted expression does not even have the right units. I found this article that seems to clarify things a bit for me. http://www.physics.uoguelph.ca/poisson/research/agr.pdf in the first paragraph on page 142. It gives an equation for the acceleration at radius r, as measured at infinity, as:

    [tex]a(\infty) = f^{1/2} a(r) = f'(r) [/tex]

    It turns out when f = (1-2GM/(rc^2)) the expression in the middle is:

    [tex] a(r)*\sqrt{1-2GM/(rc^2)}[/tex]

    and the expression on the right is half the derivative of (1-2GM/rc^2)) which works out as the familiar Newtonian expression:

    [tex] \frac{GM}{r^2}[/tex]

    Now if r (where the test mass is) is equal to the Schwarzschild radius (2GM/c^2) the expression becomes:

    [tex]a(\infty) = \frac{GM}{4G^2M^2/c^4} = \frac{c^4}{4GM} [/tex]

    which is the frequently quoted a(∞) = 1/(4M) when we use units of G=c=1.

    Glad I finally got that sorted out :approve:

    For the slightly more general situation, where the test mass is at r and the spring balance is at r2, the equation is:

    [tex]a(r_2) = \frac{GM}{r^2}*\frac{1}{\sqrt{1-r_S/r_2}}[/tex]

    where rS is the Schwarzschild radius.
    Last edited: May 29, 2012
  6. May 29, 2012 #5
    Take the correct 'force redshift factor' is as per #3 just the usual relative frequency redshift factor f' = √(1-2GM/(r1c2))/√(1-2GM/(r2c2)), i.e. the fractional relative change in redshift between h1 and h2, measured at h2. As a corollary this implies dr1 = dr2, where dr is the radial displacement measured locally at h1, h2 respectively, given an incremental radial movement in the 'rigid' string attached to the scales at h2. This can be easily deduced from the simple energy relation dE = F.dr applied locally at h1, h2. Since dE2 = f'dE1 and F2 = f'F1, we must have dr1 = dr2.

    This seems to create an odd situation. From the Schwarzschild line element we get dt2/dt1 = 1/f' - the usually understood relative redshift of frequency. But we also formally get dr2/dr1 = f'. Applying the same interpretational procedure as for the temporal components, radial displacement should be measured greater at the lower potential. Which is inconsistent with the previous finding dr1 = dr2. How is this reconciled, given this is a static spacetime scenario with purely radial displacement involved?
  7. May 29, 2012 #6


    Staff: Mentor

    "Radial displacement" in terms of *coordinate* differentials, yes. But *not* measured in terms of actual physical distance. When you derived dr1 = dr2, your "dr"s were representing actual physical distances; they had to be to satisfy the equation you wrote down. But when you derived dr2/dr1 = f' from the Schwarzschild line element, your "dr"s were representing coordinate differentials. So the "dr"s in the two equations represent different things; that's why the equations are different.

    Edit: I think you are also not quite clear about the interpretation of "dE" and "F" in your equation "applied locally". An equation that is truly "applied locally" should be written in a local inertial frame, and should not have any "redshift factors" incorporated into any of the terms in it.
    Last edited: May 29, 2012
  8. May 30, 2012 #7
    Meaning proper values - i.e. locally measured as I had clearly stated. Sure. But you are saying there is no 1:1 linkage between coordinate determined ratio and that measured locally? See below.
    Nice try Peter but this imo fails to deal with the main issue pointed to in #5 - your same logic should apply to the temporal component, also expressed in coordinate measure. The coordinate temporal ratio is precisely reflected also in the proper, local measure. A light pulse of duration dt2 at height h2 is measured as of duration dt1 = f'dt2 at height h1. No puzzling mismatch between coordinate determined ratio and that locally measured here. Yet it all falls apart for the spatial counterpart. There needs to be a clear explanation as to why the formalism works in the temporal but not the spatial case. There is of course in standard Schwarzschild coordinates no issue for the transverse spatial component since f' = 1 there.
    Not inertial frame - we do not assume free-fall but stationary in a gravitational field - hence there is weight, force, string tension etc., as per OP scenario.
    Depending on how you take the above point, redshift factors as used there are appropriate surely. We are comparing measurement outcomes at the different potentials.

    [Concrete example of where both temporal and spatial components are propogated from one potential to another: a radially oriented and notionally rigid rod undergoing harmonic axial motion so as to transfer power at a steady rms rate between generator at h2 and load at h1. Assuming correctness of F2 = f'F1 and requiring locally measured power has ratio W2/W1 = f'2, it's easy to find that given temporal rate propagates as per coordinate calculations, radial spatial component does not.]
    Last edited: May 30, 2012
  9. May 30, 2012 #8


    Staff: Mentor

    I think we've been here before. :wink:

    Neither the temporal nor the spatial "coordinate ratio" "propagates". Certain global coordinates happen to have certain relationships to actual local physical quantities in this particular spacetime, and the relationship varies from event to event. You are interpreting the global Schwarzschild coordinates as if they had some inherent direct physical meaning. They don't.

    I said local inertial frame--an inertial frame in which a "hovering" observer is momentarily at rest. That's how you write the laws of physics in invariant form. A correct equation corresponding to your "dE = F * dr" would involve actual physical quantities only; but you are trying to interpret the quantities as coordinate quantities in the global coordinates, which have no actual physical meaning. But if you write the law in terms of quantities in a local inertial frame, then you can assign direct physical meaning to them; what you are calling "dr" would be an actual physical differential displacement, what you are calling "F" would be an actual physical force, and what you are calling "dE" would be an actual physical energy change corresponding to the actual physical work done.

    If you write the equation that way, you can then "translate" it into global coordinates and see what it looks like. You may find that it doesn't look like you expect. I'm out of time right now but if I have time later I will try to write down how I think this would work.
  10. May 30, 2012 #9
    There is a whiff of deja vu about it, yes. But just a whiff. :smile:
    Pardon then my terminology, but vibrating rod example does not give the sense of what 'propagate' means? Perhaps I should have specified high frequency motion implying wave propagation to drive home the sense of 'propagate'. At any rate there is physical transfer from a to b involving time and length, and certain transformation rules are meant to apply to that process. There is some frequency and stroke length applying at h2, which at h1 measures differently at least for frequency - but according to what consistent transformation criteria? Some kind of 'supplementary' maths or selection rule seems to have to come into play, apart from straightforward use of SC's.
    My interpretation is they are meant to provide a consistent framework for determining relationships - transformations, from one spacetime locale to another. And as said over and over previously, formal application gives this seemingly inconsistent finding of temporal component match-up but spatial (radial) component mismatch. All based on an assumed force relation being true.
    That is still free-fall environment before the fall part gathers pace. In that environment there is no weight on the scales, no sense of 'g'. I suspect you will be angling at that only tidal effects should count re physical quantities here.
    Careful - next you will be telling me SC's are meaningless! Seriously, what is misplaced in using SC's to determine, as I have done, the *ratios* between physical quantities at different potentials? You have not so far disputed this works perfectly in the case of temporal redshift. And that is hardly a coincidence.
    All of these quantities I have given as locally measured (really real surely) - as *determined* via coordinate measure using SC's.
    Can barely wait! :tongue2:
    Last edited: May 30, 2012
  11. May 30, 2012 #10
    The derivative is [tex]\frac{Gm}{r^2c^2}[/tex]
    If r=rs, it becomes [tex]a(\infty) = \frac{Gm}{4G^2m^2/c^2} = \frac{c^2}{4Gm}[/tex]
    [tex]\frac{Gm}{c^2}[/tex] is taken as M by definition, so a(∞)=1/4M
  12. May 30, 2012 #11
    You are right that there are some small errors in my last post, but your correction also fails in that the result does not have units of acceleration (meters per second2) but has units of length (meters). This is my revised version that tries to get the units right:


    The equation for the acceleration a(r) at radius r, as measured at infinity a(∞), is given as:

    [tex]a(\infty) = f^{1/2} a(r) = (c^2/2)* f'(r) [/tex]

    where f = (1-2Gm/(rc^2)) and f' is the derivative of f, so the expression on the right is:

    [tex] \frac{Gm}{r^2}[/tex]

    If r = rs where rs is the Schwarzschild radius (2Gm/c^2) = 2M the expression becomes:

    [tex]a(\infty) = \frac{Gm}{4G^2m^2/c^4} = \frac{c^4}{4Gm} = \frac{c^2}{4M}[/tex]

    which is the frequently quoted a(∞) = 1/(4M) when we use units of c=1.

  13. May 30, 2012 #12
    Looks reasonable. The derivative was wrong.
  14. May 30, 2012 #13


    Staff: Mentor

    Q-reeus, I still don't have enough time to post a detailed description, but I wanted to respond to a few things in your post:

    I don't disagree that *something* is propagating in the vibrating rod. I just would not describe what is propagating as "the spatial coordinate ratio" or "radial spatial component" (which I'll call K, just to increase the deja vu sensation :wink:, so K > 1 is the ratio between radial coordinate differentials and actual physical distances). I also would not say that the "temporal coordinate ratio" or "temporal metric component" (which I'll call J, so J < 1 is the "time dilation factor", the ratio between time coordinate differentials and actual physical proper time for a "hovering" observer) is propagating. Metric coefficients don't propagate. Different parts of spacetime may have different coefficients; that's all.

    In this particular case, yes, you can use the metric coefficients and how they change from point to point in spacetime to do this. But you have to be careful about interpreting what that means. See below.

    No, you are misunderstanding. A free-falling observer at rest in the local inertial frame is *momentarily* at rest with respect to the hovering observer. But we can still describe the hovering observer's motion, at least in that small patch of spacetime, using the local inertial frame. The "hovering" observer's worldline in that frame won't be a straight line, as the free falling observer's is. But we are still describing the "hovering" observer's motion. We're just using an inertial frame centered on the free-faller that is momentarily at rest with respect to the hovering observer to do it. That's so we can apply all the usual rules of SR, within the local inertial frame, including how SR describes accelerated observers, like the "hovering" observer. We can also apply all the usual rules of SR for force, energy, etc.

    Nope; see above. The proper acceleration of the "hovering" observer's worldline is a perfectly good physical quantity, and we can describe it in the local inertial frame.
  15. May 30, 2012 #14
    OK terminology was bad and agree that 'propagate' is not the right wording. So instead I would say that using SC's the physically propagating power and associated time and distance quantities transform locally with potential by those K, J factors.
    Thanks for the clarification - had a feeling you may have meant it that way. So I have no problem with that position; just not sure what advantage it brings to the table.

    [retracted claim about radial motion - coordinate distortion of geometry cancels out other effects and one still has dr2 = dr1 locally measured]

    ...a spring-plunger-dashpot arrangement, oriented for radial motion of spring-plunger. By coordinate measure energy release upon unlatching of spring is redshifted by f, as is the radial displacement. Hence the radial acting spring force is by this measure invariant wrt gravitational potential. Yet if this corresponds to that an observer at any height locally measures, it obviously conflicts with dr2/dr1 = 1 determined earlier requiring measured force redshift.

    [Later: Tied myself in knots earlier over a relative motion scenario best forgotten. What now comes out as odd but apparently ok is this: For lateral orientation of spring-plunger arrangement, coordinate determined transverse force is 'straightforwardly' redshifted by f (energy redshifted but not spring displacement), whereas as per before coordinate radial force is unaffected by redshift (energy and spring displacement redshifted). Convey the transverse spring force to a distant observer via rope and right-angle pulley such that the tug is radial down to the pulley radial entrance, then transverse to the spring. Locally there can be no difference in the observed rope tension - also true whether radial or transverse allignment of spring. Which must tally for an observer holding the rope, at any height greater than the spring. Yet for a given radial tug movement, coordinate transverse spring displacement is greater by factor 1/f than for radial. Evidently in coordinate measure the pulley acts effectively as a 'gear reduction unit' - transforming a redshifted lateral rope tension into an unredshifted radial tension and speed reducing in the process. Well ok it's really the action of metric doing the coordinate 'gear reduction', but one gets the picture.]
    Last edited: May 31, 2012
  16. May 31, 2012 #15


    Staff: Mentor

    It makes reasoning easier, because you can use SR directly in a local inertial frame, so you can reason about what's happening locally without having to worry about the effects of curvature.

    For example, take your scenario of a mass m at height h1 suspended by a cable which is attached to a motor at a higher height h2. The motor pulls the cable up by some small distance; what force does it have to exert, and how much work does it do?

    First, we need to make sure that the problem is completely specified. To do that, we need to specify how the displacements of the cable are related at h1 and h2. Your previous statement of the scenario appeared, to me, to assume that the actual, physical displacements were the same; i.e., that the cable's motion was "rigid". We'll see what that means mathematically below.

    Second, we need to clarify what it is we are actually trying to analyze. We are *not* analyzing the dynamics of the movement: the tension in the cable, the time it takes to move, time delay between the motor starting the cable moving at its end and the mass actually moving at its end, etc. We are only comparing two static equilibrium configurations: the "before" configuration, with mass at h1, and the "after" configuration, with mass at h1 + some small distance, and some additional cable retracted at h2. We are essentially assuming that the time it takes to transition from one static equilibrium to the other is very fast compared to everything else in the problem. This seems reasonable if the black hole is large enough compared to the difference in heights.

    Given the above, we can write down a simple "work equation" in the local inertial frame of the mass m at height h1, centered around the event at which the displacement takes place:

    [tex]dW_{1} = F_{1} dx_{1}[/tex]

    where dW1 is the work done on the mass m, F1 is the force applied, and dx1 is the displacement. Note that dx1 is a differential of the x coordinate in the local inertial frame, which we are assuming is pointed radially outward; and since it's a local inertial frame, dx1 directly represents the actual, physical displacement (because all the metric coefficients in a local inertial frame are 1--there is no "distortion" due to curvature).

    We can also write down a simple work equation for the motor at height h2 when it retracts a length of cable:

    [tex]dW_{2} = F_{2} dx_{2}[/tex]

    where dW2, F2, and dx2 are defined similarly.

    The "rigidity" condition above lets us write:

    [tex]dx_{1} = dx_{2} = dx[/tex]

    That is, the actual physical displacements are the same. So we can rearrange the two equations above to both be equations for dx, and set them equal:

    [tex]\frac{dW_{1}}{F_{1}} = \frac{dW_{2}}{F_{2}}[/tex]

    Or, rearranging to obtain the quantity we want:

    [tex]dW_{2} = \frac{F_{2}}{F_{1}} dW_{1}[/tex]

    That is, the work done by the motor is "redshifted" relative to the work done on the mass, by the same factor as the force is "redshifted".

    Note that I didn't have to say *anything* about how the displacements in the local inertial frames related to coordinate differentials dr1, dr2 in the global Schwarzschild coordinates. That's simply irrelevant, given that you have already specified that the actual, physical displacements are the same. The local inertial frame analysis gives the simple answer directly, without having to deal with any such complications.

    I'll refrain from commenting yet on the rest of your post, since it kind of seems like a work in progress... :wink:
  17. May 31, 2012 #16
    Well fine Peter nice derivation but that result as you say just falls out on the assumption of 'rigid' motion, and is what I obtained earlier working from a different angle - deriving 'rigid' motion by taking known redshift energy relation and assuming redshift of forces. But what I have wanted is a formal explanation of why temporal and radial spatial derivative ratios via SC's have divergent applicability. I have some thoughts expressed below.
    More like a descent into madness at one stage - but still cogent - best as can self-assess! :cry: Too late to further edit there but anyway my handwavy resolution for why dt2/dt1 works but dr2/dr1 doesn't, now comes down to this:
    In the temporal case redshift of a signal can be properly interpreted as purely the effect of locally varying clock-rate, there being no coordinate perceived change in frequency with potential - i.e. a wave emitted at some local frequency maintains that frequency forever after but a local observer's clock at some other potential measures a different time rate and hence relative frequency. So just one parameter is 'really' varying. In the spatial case both the local ruler length and say the transmission rod length are equally effected by potential. Two parameters that effectively cancel out re measured effect. So although from a coordinate perspective a differential radial movement at some higher potential is determined to grow progressively smaller (by redshift factor) further down, decreased ruler length there cancels any locally perceived difference.

    Hence although my finding via spring-plunger scenario in #14 that radial force is invariant wrt potential can be considered 'true', it is cancelled out by the coordinate measured 'gear reduction' from spatially varying rod length scale - greater coordinate motion further up implies leverage advantage! So I'd say just applying dr2/dr1 via SC's as done before is tacitly excluding the transmission medium (the rod say) but not local rulers, a physical absurdity that implied the increase in locally observed displacement which does not in fact apply. And physically that cancellation of coordinate radial spatial effects is what matters - measured force is redshifted 'as expected'. Just not up with the formal means of expressing this - as said before it doesn't come out from treating radial spatial derivative ratio the same as for temporal, via straightforward use of SC line element. The 'supplementary maths' is some metric operation applying to the rod and obviously numerically it is the exact inverse of that 'straightforward' application of coordinate determined dr2/dr1. Enough rambling for now.
  18. May 31, 2012 #17


    Staff: Mentor

    A clarification: the "redshift of forces" isn't "assumed". It's *required* for the work-energy relationships I wrote down to hold in the given scenario, given the known redshift of energy.

    And the short answer to that is, why not? I understand that you have an intuitive sense that they shouldn't have "divergent applicability", but that intuitive sense is simply wrong. I know it *seems* compelling, but it's still wrong. That happens with plenty of intuitions we have from our everyday experience, because relativity covers scenarios far outside our everyday experience. You just have to deal with it.

    A better way of stating this would be: when a "wave", which I'll specify here as a light ray for concreteness, is emitted, it has a definite (null) 4-momentum [itex]k^{a}[/itex]. This 4-momentum is parallel transported along the ray's worldline; in this sense the 4-momentum "never changes", since parallel transport *defines* what it means to "not change" along a geodesic. But the energy (or frequency, if we divide by Planck's constant) the ray will be seen to have by a particular observer is given by the inner product of the ray's 4-momentum with the observer's 4-velocity [itex]u^{b}[/itex]: [itex]E = g_{ab} k^{a} u^{b}[/itex]. So if [itex]g_{ab}[/itex] changes, the measured energy E will change as well, even if we consider observers for whom [itex]u^{b}[/itex] is constant, such as "hovering" observers, for whom [itex]u^{b} = (1, 0, 0, 0)[/itex] .

    Strictly speaking, they are affected by [itex]g_{rr}[/itex]; the "potential" is related to [itex]g_{tt}[/itex]. In the particular case of the static gravitational field around a massive body (or a black hole), it just so happens that [itex]g_{rr} = \frac{1}{g_{tt}}[/itex] in Schwarzschild coordinates. But that's not true in general; in fact it's not even true for other coordinate charts on the same spacetime (such as Painleve).

    Also, you have to be clear on what you mean by "local ruler length", and on the physical assumptions you are making about "transmission rod length". "Local ruler length" is length measured by a free-falling ruler in the local inertial frame that is momentarily at rest relative to the length you want to measure. I specify "free-falling" because the ruler needs to be unstressed for us to be sure that it is really a good measure of "true length" or "proper length"; stresses in the ruler will change its length and invalidate its measurement.

    Similarly, for "transmission rod length" to be a good way to "transmit" spatial displacements, the rod's motion has to be "rigid" in the sense I defined in an earlier post. But physically, that's actually a very restrictive condition, which in fact can't be realized exactly, though it can to a good approximation under the conditions I described. In a real transmission rod, force exerted at one end is not instantaneously transmitted to the other; different pieces of the rod start moving at different times, so the rod's instantaneous "length" changes, but with a proper setup the rod should "settle" back into a state with the same overall length when all motion has been completed, at least to a good approximation, as long as the change in height is very small compared to the size of the massive object (as I said in an earlier post). But in general, moving any real object from one height to another in the field will induce changing stresses in the object, which will change its overall length.

    With all those caveats, yes, the "effect" of the change in [itex]g_{rr}[/itex] with height will be the same for "local ruler length" and "transmission rod length", so it cancels out.
  19. May 31, 2012 #18
    A circular situation seems to me, since you acknowledged assumption of rigidity which only then in conjunction with energy redshift forces the force redshift to hold. But admittedly it's logical to do so.
    You may have noticed a certain dealing with it last post of mine.
    Thanks for putting it professionally and formally that way - but I guess you are basically saying yes.
    Sure agree with all these caveats and as extended further in your post, but OP specified the static spacetime SC setting and further I think it quite clear material elastic effects were not under consideration here.
    Phew - Peter do you realize we are in agreement on at least two points in all this!? :approve: On that high note I retire for some much needed R and R. :zzz:
  20. May 31, 2012 #19


    Staff: Mentor

    What's circular about it? Redshift of energy + assumption of rigidity => redshift of force.

    Oh, no, I must have missed something. :surprised
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