Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Functional independence and energy

  1. Dec 4, 2005 #1
    I put together two questions :

    a) suppose there is a point mass with mass M..if it is moving, then from a certain oberver, the total energy is higher, via E=Mc^2...hence, following the generaly relativity qualitatively, where the energy density defines the curvature, the gravitation should be different ?

    b) let's take two coordinate systems, linked by a transformation, how are the metric obtained as solution linked together : [tex] x^\mu=f^\mu(x'^\nu)[/tex] how are the [tex]g_{\mu\nu}[/tex] linked to the [tex]g'_{\mu\nu}[/tex]...since the tensor is symmetric, there are only 6 degree of freedom, and hence there should exist 6 degree of functional degree of freedom for the coordinates. Those should hence describe the same space-time. (e.g. Shwarzschild(singular)<->Kruskal(non singular))..Or the question can be stated as : how to know that 2 metrics cannot be obtained by a change of coordinates ?
     
  2. jcsd
  3. Dec 4, 2005 #2

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    Note that your set of functions function [itex]x^v = f^{v}(x^{'u})[/itex] must be invertible to be a transformation, i.e. [itex]x^{'v'} = g^{v}(x^{u})[/itex]

    The upper indices of the tensor transform like this:

    [tex]
    T^{'u'} = \sum_u T^{u} \frac{\partial x^{'u'}}{\partial x^u}
    [/tex]

    The lower indices of a tensor transfom like this

    [tex]
    T_{'u'} = \sum_u T_{u} \frac{\partial x^u}{\partial x^{'u'}}
    [/tex]

    These transformation properties are sometimes used to define tensors.

    If this seems involved or abstract, there is another route involving the line elment that may make the process seem clearer, though it really isn't any different than what I wrote above.

    If you have a line element

    [itex]ds^2 = g_{00} dt^2 + g_{11} dx^2[/itex]

    and you let t = p(t1,x1) and x = q(t1,x1) then you can write, by the chain rule

    [tex]
    dt = \frac{\partial p}{\partial t1} dt1 + \frac{\partial p}{\partial x1} dx1
    [/tex]

    [tex]
    dx = \frac{\partial q}{\partial t1} dt1 + \frac{\partial q}{\partial x1} dx1
    [/tex]

    Then you just algebraically substitute in the expression for the line elelment in terms of dx and dt to get it in terms of dx1 and dt1. After the algebraic manipulation, you "read off" the new metric coefficients.

    If you happen to have an explicit defintion of the new coordinates, you either have to invert the transformation to put it in the implicit form above, or add a step where you do some algebra to compute dt and dx as a function of dt1 and dx1 (basically a matrix inversion).

    Going back to your original question, in GR we do not have point masses, instead we have a stress-energy tensor at any point in space. (One component of this tensor is the mass density, usually called T_00).

    Because the stress-energy tensor is a tensor, it transforms according to the above rules. Note that in general you can't transform the mass density alone because it isn't a tensor - you need the extra information in the stress energy tensor to find the mass-density component of this tensor under a coordinate transformation.

    The curvature is also a tensor which transforms like any other tensor.

    I'm not aware of any foolproof way to recognize when two metrics are equivalent under a change of coordinates, but to prove they are not, people often look at invariants of the metrics (Petrov invariants, I think is the right term).
     
    Last edited: Dec 4, 2005
  4. Dec 6, 2005 #3
    I don't know exactly the Kerr-metric, but I think, like in Schwarzschild case, it is computed in Vaccuum, where T=0 which is always 0 whatever the change of coordinates.

    You mean that if the stress-energy tensor of a massive sphere is taken, then the rotational case is obtained by a certain coordinate tranformation from the static one...(But then if there are more sphere rotating with different axes ?)
     
  5. Dec 6, 2005 #4

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    The Kerr solution is not the same as the Schwarzschild one - it's a physically different solution. It is not obtained via a coordinate transform from the Schwarzschild solution.

    You have to solve Einstein's equaitons from scratch for every different configuration. There are only a few exact solutions known. Because the equations are non-linear, you cannot add solutions together, except as an approximation in the weak field limit.
     
  6. Dec 6, 2005 #5
    Suppose you have a non linear equ K : for example [tex]f'(x)^2=f(x) [/tex].
    You can compute the function [tex] \phi(f(x))[/tex] such that it is again a solution of the equation : for the example : [tex] \phi(x)=x+c\sqrt{x}+\frac{c^2}{4}[/tex]
    Then, suppose the set [tex] f_n(x)=\phi(f_{n-1}(x))[/tex] is complete, you can derive the operation : [tex] f_n +_{K} f_m=f_{m+n}[/tex] as a generalized superposition : suppose [tex] h(x)=\sum a_i f_i(x)\quad k(x)=\sum b_i f_i(x)[/tex] the spectral decompostion of two other solutions at first not generated by this principle. Then
    [tex] h(x) plus_{K}k(x)[/tex] is a solution...or am i doing a mistake here ?
    Maybe one has to pass through the fixpoint of phi...but i don't see the trick yet...
     
  7. Dec 6, 2005 #6

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    Think partial differential equations, not ordinary differential equations.

    The electromagnetic anology would be solving Poisson's equation in an electrostatics problem. (That would apply only to static gravitational systems, of course). [itex]\nabla^2 \Phi = \frac{\rho}{\epsilon}[/itex]

    What is important to both problems is both the distribution of charge (mass) and the boundary conditions.

    In the electrostatic case, because the equations are linear, the solutions always add together linearly. It can be shown that the differential equations that model GR formulate a well-posed initial value problem. Because of this, it is known that solutions do exist for any well-posed initial conditions.

    However, if you want to find the solution for two masses, said solution is not the algebraic sum of two one-mass solutions except in the weak field limit. It think it's possible to do a sort of series expansion to model cases near the weak field limit, but I don't recall the details offhand.
     
  8. Dec 7, 2005 #7
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Functional independence and energy
  1. Background independent (Replies: 6)

  2. Independent Continuum (Replies: 8)

Loading...