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I Volume vs Circumference of Massive Body

  1. Aug 16, 2016 #1
    Volume vs Circumference of a massive body.

    In the spacetime diagrams it is often shown that the space around massive bodies is warped in a way that seems to indicate the actual volume of a massive body is larger than 4/3 pi r^3. If you measure the radius from the circumference by traversing it, and then measured the diameter by traversing it, and then calculated the respective radius’s, then it appears that the radius as measured from traversing the diameter would be larger than the radius as derived from traversing the circumference.

    It’s kinda as if the massive body is like the Tardis from Doctor Who, more space on the inside.


    Is this indeed correct?
     
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  3. Aug 16, 2016 #2

    Orodruin

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    I suggest that you do not take images like these too literally for several reasons:
    1. It is the space-time that is curved, not only space.
    2. The images are often artists' impressions.
    3. It is not obvious what you would call "volume" in relativity. It generally depends on the space-time foliation.
    That being said, for some such coordinate dependent definitions of volume, volume can be larger than in flat space. However, this is completely negligible for a body like the Earth.
     
  4. Aug 17, 2016 #3

    pervect

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    The well known phenomenon of Lorentz contraction helps illustrate why volume is observer dependent even in special relativity. A box with sides of 1 meter, proper length, has a volume of 1m^3. But to a moving observer, one of the sides will Lorentz contract, and the volume of the same box will be less in a different frame.
     
  5. Aug 17, 2016 #4

    A.T.

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    Yes.
     
  6. Aug 17, 2016 #5

    PAllen

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    Well, just as one can talk about the proper length of a rod in SR (length in its rest frame), one can talk about the proper volume of a body in SR or GR, at least in simplified models. A body in SR or GR can be said to be defined by a congruence of world lines. In the simplest case of a rigid body, it is a congruence with zero expansion tensor. Such can be defined for exact, spherically symmetric solutions for a body in GR. Then, the proper volume is as measured by the unique hypersurface orthogonal foliation defined by the congruence. The hypersurface orthogonal condition generalizes the SR notion of the "the body's rest frame".
     
    Last edited: Aug 18, 2016
  7. Aug 17, 2016 #6

    PeterDonis

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    There is a caveat to this, which is that in the most general case of a rigid congruence, such a foliation might not exist. For example, any congruence with nonzero vorticity (i.e., a congruence describing a rotating body) cannot be hypersurface orthogonal. A rotating body will also not be spherically symmetric, however (at best it will be axially symmetric), so for the spherically symmetric case a hypersurface orthogonal foliation can be defined.
     
  8. Aug 18, 2016 #7

    DrGreg

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    However, in this case there is the "quotient manifold" construction (that has been discussed in other threads quite some time ago) to obtain a 3D space in which volume is defined. It's not a flat space, though, (even for a lightweight rotating object where gravity is negligible) and it works only for rigid congruences.
     
  9. Aug 18, 2016 #8
    Thanks. With this idea, I get the image that there is more space at my feet than at my head (however so small), and its as though I am falling into that larger space, or pushed into that larger space, because of this space gradient.

    So instead of thinking as gravity as a force pulling me down, I can think in terms of say a space gradient where mass/energy has a tendency to move to a larger spacial volume, or such?
     
  10. Aug 18, 2016 #9

    PeterDonis

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    No, that's not correct. It's not even meaningful; what does "more space" mean? How do you measure "the amount of space"?

    All that is going on is that the geometry of space is non-Euclidean. Non-Euclidean geometry is perfectly well-defined and doesn't imply any of the other things you are thinking.

    There is no "space gradient". See above.

    No. See above.
     
  11. Aug 18, 2016 #10
    Thanks Peter for your reply and I her your "No's". I am trying to reconcile some information in Kip Throne's videos on youtube. In the following two videos..

    from 22:05 to about 23:30


    from 12:05 to about 12:48


    In these videos he explains the warping of space, one for a black hole and a second one for the sun, as the diameter is larger than the circumference would imply. If you can measure the length across any possible related diameter/circumference and they all have the same warping where the diameter is larger than the respective circumference's expected diameter, then the space inside the sphere would be larger than indicated by normal geometric ideas. In addition, it appears as though the length is warped more toward the center of the sphere in a continual manner, thus each step closer to the center, the longer the distance than the previous step. This seems to mean more and more space toward the center.

    This seem to imply a gradient in a spacial context, where there is more and more space, meaning we could fit more and more 1m^3 cubes toward the center, because of the warping of space made more space in the center.

    He makes this argument in many of his videos and lectures. What is he trying to say if he is not saying "there is more space in the regions toward the center"?

    In one of his videos, he even says if he put a meter stick through the center along a diameter, it would have to be much longer than expected before it came out the other side.
     
    Last edited: Aug 18, 2016
  12. Aug 18, 2016 #11

    A.T.

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    The spatial geometry has nothing to do with this effect of gravity. It's the time geometry: Free falling stuff tends towards areas of slower time passage:

     
  13. Aug 18, 2016 #12

    Orodruin

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    Do we really want to go there? The separation in space and time is arbitrary. The relevant issue is the space-time geometry. I think we can be that precise in an I level thread.

    I also would not call it "areas of slower time passage". This is convention and observer dependent.
     
  14. Aug 18, 2016 #13

    PeterDonis

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    In the case of the sun, this is fine. In the case of a black hole, it's not quite right, because a black hole doesn't have a diameter. But you can still see the non-Euclideanness by looking at the radial distance between two 2-spheres, both of which are outside the hole's horizon, and comparing that distance with what you would expect based on Euclidean geometry given the areas of the 2-spheres. The actual distance will be larger.

    No, it would be larger than indicated by Euclidean geometry (with the caveat above for the black hole case). But calling Euclidean geometry "normal" is not physics; it's just words.

    That the geometry of space is non-Euclidean. The only reason you are using the words "more space" is that you are comparing it with Euclidean geometry, which you are claiming is the "normal" amount of space. Once again, that's not physics, it's just words.
     
  15. Aug 20, 2016 #14
    Thanks again.

    Space in my post is more properly the idea of volume. I don't know if that makes a difference or not, but that was my intended meaning.

    I am still trying to understand the consequences of the larger diameter so bear with me please.

    Say we use the sun example so we don't have to be concerned with black hole singularity issues. I get the image that if we replaced the sun with an imaginary empty shell, leaving in place all the curvature of space that was created by the mass of that sun, and then fill that shell with a non compressible fluid of no mass/energy etc, would that empty shell hold more of the fluid than say the same shell at a location that did not have the space warping?
     
  16. Aug 20, 2016 #15

    PeterDonis

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    That doesn't change any of what I said. There is still no justification for saying that the Euclidean volume is "normal" and any different volume is a change from the "normal" one.

    This is not a meaningful question in at least two ways:

    (1) You can't just replace the sun with an empty shell and not change the curvature of spacetime. The shell is a different solution to the Einstein Field Equation and will have different properties.

    (2) You can't have "the same shell that did not have the space warping". The shell is what it is; changing the curvature of spacetime changes the shell.

    The meaningful thing you can say is that the interior volume of the Sun is larger than

    $$
    \frac{1}{6 \sqrt{\pi}} A^{3/2}
    $$

    where ##A## is the Sun's surface area. (All I've done here is to rewrite the standard Euclidean formula ##4 \pi R^3 / 3## by substituting ##R = \sqrt{A / 4 \pi}##, so as to put it in terms of surface area, which is the thing we are "holding constant", so to speak.) Computing exactly how much larger, and computing the actual physical radius (or diameter) of the Sun, what you would measure if you could take a ruler and extend it all the way through the Sun, is more complicated since it requires evaluating an integral. But you can also meaningfully say that the radius will be larger than ##\sqrt{A / 4 \pi}##.
     
  17. Aug 20, 2016 #16

    pervect

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    If I might suggest an analogy, consider the curved surface of the globe from a flatlander perspective, i.e. consider the geometry as measured by inhabitants who never leave the surface of the globe.

    The surface are of the globe is 4 pi r^2, for a half-sphere, a hemisphere, the surface area is 2 pi r^2, while on a plane, the surface area would be pi r^2. The circumference in both cases is 2 pi r. So the ratio of the area over the circumference is different.

    The amount of difference depends on the size of the circle - for small circles, the curvature is barely noticable and there is no difference in the ratio, , for the largest possible circle, there is a 2:1 difference.

    I wouldn't describe this as the globe "making more area on the inside". It may or may not be a valid way of thinking about a globe, I'm not sure. What I can say is that I don't think Kip Thorne was suggesting that sort of interpretation.
     
  18. Aug 20, 2016 #17
    Thanks again for the input/s.

    I understand the concept of non Euclidean spaces (although not in enough detail to do volumetric calculation) and I have worked through 3D vector calculus, although many many years ago now. What caught my attention is that he clearly calls out that the diameter of a massive body vs the circumference is distorted do to space time warping according to Einsteins field equations, especially when he said if he put a ruler of the expected diameter in flat space through a diameter of the sphere that it would not come out the other side.

    I also understand we have to look at the math to answer such a question and I don't know how to do the differential geometric calculations to get the answer myself to this question.

    Keeping it simple, given a homogeneous mass distribution in the shape of a sphere, with out regard to size as even the smallest objects would have their respective field distortions however small, that if indeed the presence of mass distorts space in the way he described which seems pretty clear in his videos, that the area of a crossection would be much larger than the corresponding circumference. I had the image that all the little deltas of the integral for volume would follow the same pattern and sum to a larger volume inside the sphere than would be expected in classical flat space with out the distortions. Thus I drew the conclusion that the volume would also be larger inside the spherical object than one would expect in flat space.

    Subsequently, I was pondering if this is indeed so, and posted this question to see if I was interpreting this correctly. My interpretation was that there is more 'space' /'volume' inside the shell of the sphere due to the mass distorting the space and that you could fit more arbitrary objects into that space as there was more room to put stuff; kinda like the Tardis from the Dr Who shows.

    And that if this was true, one could interpret this as a volumetric graident that increases as you approach the center of the sphere, as the distortion increases as you move toward the center point of the sphere, thus there is more 'volume'/'space' etc. at your feet than at your head and that you kinda sink into that extra volume as it is less, umm... volumetrically dense, at your feet than at your head.

    Many videos (from reputable sources such as PBS SpaceTime) abound that state that 'gravity is an illusion', and is a consequence of warped spacetime and is not a force field where some particle field such as the EM field is responsible for the gravitational force, thus I imagined this image of gravity that was due to spacetime warpage and am just bouncing it off this forum for as a sanity check.

    As of now, I still am not sure how to interpret the whole idea :)
     
  19. Aug 20, 2016 #18

    PeterDonis

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    More compared to the Euclidean volume of a sphere with the same surface area, yes.

    No. Nothing in this paragraph corresponds to anything in the actual math or physics.

    This is the most common interpretation of classical GR, yes. But it is not the only possible interpretation. See below.

    There is actually a way to derive GR starting with the theory of a spin-2 field in flat spacetime. (The EM field is a spin-1 field, so it is not quite the same, but the theories are very similar.) The original flat spacetime turns out to be unobservable, but there is nothing inconsistent in interpreting the resulting theory, GR, as saying that gravity is caused by the effects of the spin-2 field.

    What this is really telling you is that "curved spacetime" is ultimately an interpretation of GR, not GR itself. The curved spacetime interpretation is the one most relativists use most of the time, but it's not the only possible one. Kip Thorne talks about this in one of the later chapters of his book Black Holes and Time Warps.
     
  20. Aug 30, 2016 #19
    I had another though on this... In the common examples of time and length warping, it is commonly explained that if a car wizzed past me at relativistic speeds, we would see the car greatly shortened and the clocks inside the car running slow. To the person inside the car, he would see his space as normal (local) yet he sees our space is contracted and our clocks running slow.

    Say I am outside the car and the person in the car is holding a meter stick. I see his meter stick as shortened in the direction of travel. The person in the car sees his meter stick as he would generally think of a meter stick. Subsequently, to the passenger in the car, there is 'normal' space where a meter is a meter is a meter. I however, see his space warped. If I put my meter stick next to his as he wizzes by, mine would be longer. He could have maybe two meter sticks to my one.

    In a way, its like there is 'more space' in the car than I would other wise assume. Again, it's like the tardis idea where there is more space on the inside in a way. Thus, I can kinda see the idea where there is 'more space' inside the object in a seemingly tangible way

    Just thinking.
     
  21. Aug 31, 2016 #20

    PeterDonis

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    It should be emphasized that the word "see" here does not refer to what you actually would see with your eyes (or a telescope or other actual measuring device). A better word would be "calculate": when you take what you actually see (observe, measure, etc.), and correct it for the travel time of the light from the object to you, you calculate that the object is shortened and that its clocks are running slow.

    Again, you would calculate this from your observations, but it's not what you would actually observe. You would observe his meter stick to be rotated, not contracted (look up Penrose-Terrell rotation), because the light reaching your eyes from his meter stick at a given instant would have traveled different distances to get to you and so would have been emitted at different times. But if you set up a mechanism to record where the two ends of his meter stick were against your meter stick at the same instant of your time, the results would indicate that his meter stick was shortened.

    No. First, the car is shortened, so if you just naively calculated its volume in your coordinates, the volume would be less than its volume if it were at rest relative to you.

    But more important, the space inside the car is still Euclidean. If you tried to pack blocks inside the moving car, by putting each block in motion at the same speed as the car and then pushing it inside, you would find that the blocks would shorten relative to you, so exactly as many would fit inside the car as if it were at rest relative to you. So this is not the same kind of thing that is going on near a massive body.
     
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