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Are gravitational fields additive?

  1. Jul 17, 2007 #1
    Are gravitational "fields" additive?

    By this I mean, if you are placed directly between two solid massful objects, that are somehow held still for this example(maybe binary system revolving around the center of mass).: You would not move because of an unstable equillibrium, true. But does your equations for fields at that point ADD, or equate to zero?

    I guess what I'm asking is, I know there are some effects that are affected by gravitation fields, so by being in the presences of two equally stong fields, is that effect doubled? or cancelled out?

    I would assume doubled. Seems that just because a sum of the force vectors due to potentials zero out in 3D, from that observation point, doesnt mean that there isnt still a displacement of curvature when compared to a flat plane.
    Though I understand this curvature would be relative to the observer at the point, or distant outside the group.

    Or WOULD it be so locally flat, at that exact point, that it WOULD be zero?
    I'm imagining a maxima of some sort, where you could either say zoomed in it approaces a flat plane for its derivative, or looking a little larger you see that there is curvature.
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  3. Jul 17, 2007 #2


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    In Newtonian gravitation, the gravitation fields are additive.
    In Einsteinian gravitation, they're not in general.
  4. Jul 17, 2007 #3


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    You seem to mix up different concepts. In Newtonian Gravity, potential, forces (acceleration), and tidal forces ("tidal accelerations", curvature) all are linear and add up.
    If you add two balancing Forces, they will cancel each other, the Potential at this point will decrease (get more negative), and tidal forces will add but are not specified and irrelevant in your post.
  5. Jul 17, 2007 #4
    Well, I see what you're saying. I just have a hard time imagining it.
    I'll try to clarify the picture I have in my head:
    Firstly to compare flat 1D(simple) space-time:
    Now imagine two massful objects placed so that


    I assume it would curve space to be such that:


    where the straight lines are spacetime and the "," are just spacers.

    Now the way I imagine it is, if massful enough,


    Look at B. See how the curvature due to each makes it "lower"? If you were to approximate the curved space with a flat line, the difference between the flat and curved models would be like the second drawing right? Where points A and C at some larger distance away would be a closer approximation to flat and any possible point B. As if they ADD in some way? I'm asking "in what way" am I describing?

    Think about it as a potential well. You have one well, and overlap it with another well. So the point in the middle has some potential, though it can be an unstable equillibrium. Does this fact, that even though you're there at equillibrium, since youre in SOME potential well, make any modifications to reality? To measurement? To anything? Or is it since it's at equillibrium and the slope is 0 at that point its as if it were a flat plane that was at the same potential as a regular flat-spacetime.

    Is that more clear what I'm asking?
    In my mind its like the letter W but where the middle of the W is lower than the outer points.

    Seems like in my head I can only think of spacetime as 4-spacial dimensions. Maybe thats the problem, I'm extending the curvature into another dimension.
  6. Jul 17, 2007 #5


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    Newtonian Potential (GR doesn't add new insight here) go like M/r, and you can add them. When you do that, you arrive at your drawings (but avoid speaking of "curvature", this only adds confusion.
    The point in the middle is exactly equivalent to a point in an empty universe. The region around this point is gravity-free, depending on you measurement accuracy. High accuracy - small region, low accuracy - large region.
    So: Yes to
  7. Jul 17, 2007 #6


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    I think you are attempting to draw an "embedding diagram", like the one in

    http://www.bun.kyoto-u.ac.jp/~suchii/embed.diag.html [Broken]

    The embeding diagram is just a diagram, isn't even necessarily unique, and has no physical significance. It's just a visual aid. So if you're asking is there any physical significance to the "height" in your embedding diagram, the answer is "NO".

    If you want to understand GR more fully, you need to study the notion of "intrinsic curvature", sometimes known as "Gaussian curvature" as opposed to your embedding diagram approach.

    The simplest thing I could find on this was http://physics.tamuk.edu/~hewett/Mo...ativity/CurvedSpaces/Intrinsic/Intrinsic.html

    The point of intrinsic curvature is that it is something that can be measured and defined in terms of measurements we can make in 4-d space-time. There are not any "other" dimensions in which space-time is "embedded" in this approach, there is simply a 4-d geometry which is curved.
    Last edited by a moderator: May 3, 2017
  8. Jul 18, 2007 #7
    Hmm, Thank you for the link. I think I have a grasp on intrinsic curvature, though I didn't know the definition before.

    And yes, I guess I was trying to think outside the given dimensions, which is though visually an aide, not correct.
    Whats funny is when I picture a 1D or 2D world I do the embedded method in my head, but when imagining a 3D world I think of it as intrinsic curvature. There is no dimension into which it bends, but rather almost a condensing and stretching in 3D, locally to the mass.

    So I guess my questions end with "No difference." between my examples and a flat spacetime.

    I really need to learn this stuff at a more advanced level. I seem to have an easier time visualizing and understanding higher-level physics when I can see the math. Does anyone have any resources that would start with basic GR? Maybe with an intro to tensor analysis? Or should I just till I get that far in grad school (think that class is still a year off). We never talked about GR in undergrad. Anything I understand of it so far is from PF.com :)
  9. Jul 18, 2007 #8
    Last edited by a moderator: Apr 22, 2017
  10. Jul 19, 2007 #9
    Thanks for the link, it is helping me also.

  11. Jul 21, 2007 #10
    Here is a situation that is similar to the one originally posed.

    A test particle anywhere inside a spherical shell (with constant thickness and mass-density) experiences no gravitational force. This can be explained in terms of cancelling forces, because the gravitational force from any portion of the spherical wall is compensated by a portion of the opposite wall. Imagine two equal solid angles extending in opposite directions from the particle: the area subtended at the wall in each direction is proportional to the square of the distance to the wall, which, from Newton's equation, means that the gravitational forces are equal and opposite at the particle.

    Let there be a small hole in the sphere. If the particle moves from infinity toward the sphere, it starts from zero force and increasingly experiences the force due to the mass of the sphere. Then, when it enters the hole, the force returns to zero. Thus, the gravitational potential is lower inside the sphere than at infinity, even though the force is zero at both locations.
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