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Masses interaction.

  1. May 5, 2005 #1
    There is an analogy of the phenomenon of curvature of space by mass. It is a sheet of rubber with the some mass placed on it. Then the deepening on a sheet is formed and all surrounding objects with only smaller (?) mass should sliding down to the mass lying on a deepening bottom. This model is intended for evident demonstration of an origin of a gravitational field existing around of mass.
    However this model does not take into account a mutual character of attraction between masses.
    If consider such masses as the Earth and a tennis ball then their interaction is entered excellently in a drawn picture.
    But if consider interaction of two equal, for example, masses then this model does not work completely. Each of masses are in the own deepening of a sheet and there is not any reason for their mutual attraction.
    Normally, they should meet on the middle of a distance between them. Even to admit one mass is placed in a deepening of a sheet and another one is placed on a flat part of a sheet on the edge of a slope, it is impossible to present a picture then mass is rolled out from a deepening and meet another one on the middle of a slope.
    It turns out so, that the curvature of spacetime in the mass area really exist, but this curvature doesn’t acts directly as attraction force between masses.
    So, how this contradiction can be explained?
  2. jcsd
  3. May 5, 2005 #2

    Andrew Mason

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    The problem is with the analogy. It is only intended to give a general concept of space curvature. I have never liked the analogy because it only works if you assume there is gravity that draws the masses down into the depression - so it is somewhat circular.

    The concept of space curvature can only be really explained mathematically. We cannot physically conceive of a curved three dimensional space. The mathematical definition of curvature in 2 dimensions is applied to three dimensions and is treated as curvature. A geodesic or shortest path through space is a straight line in the absence of gravity. With gravity present, the geodesic becomes a curved line. So we say that gravity curves space.

  4. May 5, 2005 #3
    Well I think you never tried it for real by placing 2 equals mass (metal balls) on an elastic surface...

    You can explain this as : the energy the curvature has when the 2 balls are apart is greater than the energy when the 2 balls are at the same place : the stable solution is just the extremal one...

    The problem is : I've never heard of an analytical solution of Einstein field equation with 2 masses....

    But we can compute the curvature with the equivalence principle :

    Then it's known that the curvature of space-time comes from the local lorentz frame, depending on the equivalent speed at this point. This speed corresponds in GR to the classical speed. Example :

    a) Schwarzschild : [tex] \frac{1}{2}mv^2=\frac{GMm}{r} [/tex]

    Hence the squared time dilatation factor from the Lorentz transformation with this speed depending on r is : [tex] B(r)=1-\frac{v^2}{c^2}=1-\frac{2GM}{rc^2} [/tex]

    which is the time coefficient in the Schwarzschild metric. The radial coefficient is just -1/B(r).

    b) In your case we have : 2 masses M at a distance 2L apart, the origin of the polar referential being the middle of them.

    [tex] \frac{1}{2}mv^2=GMm\left(\frac{1}{L^2+r^2-2rL\cos(\theta)}+\frac{1}{L^2+r^2+2rL\cos(\theta)\right)} [/tex]

    In this case we should have coefficients of the metric for dt, dr,dtheta.... The time dilatation is just :

    [tex] g_{00}(r,\theta)=1-\frac{v^2}{c^2}=1-\frac{2GM}{c^2}\left(\frac{1}{L^2+r^2-2rL\cos(\theta)}+\frac{1}{L^2+r^2+2rL\cos(\theta)\right)} [/tex] with the speed obtained from the previous equation.

    Then you get the spatial parts [tex] g_{11},g_{22} [/tex] by the symmetry condition : g11=g22<0 and g00*g11*g22=1...

    When you get the coefficient of the metric, just use Einstein's equation in the reverse way :

    plug the obtained metric g into G(g) and you get the stress-energy tensor out....take the energy component (00) of the tensor, and just look what is the dependence towards L.....I cannot make a guess about the answer.
  5. May 5, 2005 #4
    Does it means that calculation had never been executed?
    If the trajectories of Apollo or Kassini space ships was calculated in a way which you have suggested, they never would make their travel.
    I guess also, that these calculations have been executed with use of classical Newtonian equation .
    In the result we have the theory which cannot make a simple calculation.
  6. May 5, 2005 #5
    I didn't know that classicl mechanics was calculating the curvature of space-time..??

    There is no motion calculated here, but only the coefficients of the metric given two masses...If you want the equation of motion with the given metric you have to compute the geodesics, but I won't do that for you.....

    What I mean is : this way of calculating is not solving Einstein's field equation, but computing the curvature of space-time with the equivalence principle.

    You agree that the case of 1 mass gives the same result as the Schwarzschild solution ?
  7. May 5, 2005 #6
    I think there does not exist a curvature of spacetime which surrounded mass because it exist only inside mass.
    This curvature of spacetime is mass itself.
  8. May 6, 2005 #7
    In my opinion the main lack of existing model is that it does not show an action of mass at spacetime indeed . Seems that a model was invented by someone when he sat on a sofa and then could not find out something that should be beside. This something was found in a deepening formed in a sofa under weight of this person some time later.
    But what attitude to this home experiment with a sofa has spacetime itself?
    And...what we are waiting to find out as a mass?
  9. May 6, 2005 #8
    I have some problem in understanding your last message :

    a) Visual model : 2 balls under gravitation on an elastic surface (classical motion with constraints)
    b) GR : gravitation is curvation of space-time....the motions are the geodesics of this space-time.

    For your note on curvature :

    It's clear that the Ricci scalar (intrisic curvature) vanishes everywhere in the Scwharzschild solution, because it's the solution of Einstein's equ. in vacuum...but this does not mean that space-time has no curvature....(take a cylinder, it's Ricci scalar is 0 (Gaussian curvature), but all but one direction, are curved...
  10. May 6, 2005 #9
    a) If assume that mass has some distinction from spacetime then the principle of uniformity will be violated. The elementary mass (mass of quarks) is just a curved ( compressed, COLLAPSED) spacetime and does not contain any "special stuffing" inside.
    b) The reason of masses interaction is the TENSION OF SPACETIME. It spread from mass as a RADIAL STRINGS into all direction. Therefore the gravitation force is inversely proportional to square of sphere.

    Gravitation force= {G (4pi)}* M1*M2/(4pi)r^2
    Here (4pi)r^2 is a square of sphere with radius r.

    c) There are evidences that spacetime is a set of enclosed spheres.
  11. May 6, 2005 #10
    What you do is classical physics...because in GR the Gravitational potential (you recover the force with the gradient) :

    [tex] V_{effective GR}(r)=-\frac{GM}{r}-\frac{GMl^2}{c^2r^3} [/tex]

    If you don't take into account the force in r^{-4} then you have no perihelion precession for Mercury...
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