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Possibly solved the metric without field equations

  1. Aug 12, 2014 #1
    Error below :grumpy:

    For a couple of years now, I have been attempting to solve for the values in GR of the time dilation z and the radial and tangent length contractions, L and L_t respectively, which form the metric

    [tex]c^2 dτ^2 = c^2 z^2 dt^2 - dr^2 / L^2 - d_θ^2 r^2 / L_t^2[/tex] (along a plane)

    using only two main principles in GR that give rise to the single equation

    [tex]m L_t^2 / r^2 = z' L[/tex]

    as demonstrated in this thread. With this equation, we can make a coordinate choice for one of the variables and find a relationship between the other two, or likewise, make a coordinate choice for the relationship between two of the variables and solve precisely for the third, but we are still left to determine the precise values between the two variables we only have a relationship for. I gave up a few times, thinking it impossible to solve precisely for all three variables without applying an additional principle which would give another mathematical relationship between them. I tried various methods, attempting to apply relationships involving circular orbits, angular momentum, the equivalence principle, etc., all in vain and usually just adding even more unknown variables to determine. Einstein applied field equations, but I wanted to find and use something much simpler, some basic principle that would solve for all three. As it turns out, I may have stumbled across just the principle I need to solve for all three precisely, the low Newtonian approximation. The word 'approximation' threw me off for a long time as I wanted precise values and so I was reluctant to use it as a guiding principle, but it appears it may have indeed provided me with the final relationship I have been searching for.

    Finding the derivatives for the previous equation produces the 00 component of the Ricci tensor as also demonstrated in that thread. If one prefers, we can solve for the three values starting with only the 00 component of the Ricci tensor and reverse engineer it to gain the previous relationship, which can also be demonstrated simply by finding the indefinite integrals of the 00 component which directly gives us the previous equation along with an unknown constant which we find to be m in the low Newtonian approximation. So applying the low Newtonian approximation has already helped us. Let's see what else it can do.

    The energy relationship in GR as shown in the thread is

    [tex]z / sqrt(1 - (v/c)^2) = k[/tex]

    where k is a constant for all r. In the low Newtonian limit as r approaches infinity, for an object falling from rest at r = infinity, we have v^2 = 2 m c^2 / r. Applying this directly to the energy relationship, we would have z = 1 and v = 0 at r = infinity, so K = 1, giving

    [tex] z = sqrt(1 - 2 m / r) [/tex]

    Now, we could apply this directly to GR as a coordinate choice for the time dilation at any r, in other words by placing each spherical shell with an invariant value for the time dilation z at the corresponding r defined by the equation, and we can then gain a relationship between L and L_t by doing so, but we cannot solve for each of the contractions separately, whereas we want to solve for all three. So instead, we will define the relationship between z and r as

    [tex] z(r) = sqrt(1 - 2 m f(r) / r)[/tex]

    where f is some unknown function of r. This allows z to have any arbitrary relation to r that we choose by using different functions for f, so we have not yet applied a coordinate choice. It is close to the actual relationship in the low Newtonian limit, but all we can say at this point is that f must work toward unity at r = infinity, as must also z, L , and L_t.

    So we now have four variables to solve for, seemingly making things worse, but let's go with it. By introducing f, we still have a coordinate choice available, so whereas L_t = 1 is generally applied, we will instead use z = L. We assume that this is possible, that whatever the value for z at a particular shell, we can place the shells at each r in such a manner that z = L. It can also be demonstrated weakly how this can be performed, by beginning randomly with a spherical shell at a random r, and then placing consecutive spherical shells both toward r = 0 and r = infinity such that the invariant locally measured distance between two shells is L smaller according to the distant observer, placed such that L is equal to the invariant time dilation z of a particular shell. The entire map can then be slid inwards or outwards together at will without affecting the values of z and L since the distances between the shells remain the same so L remains the same and z is invariant for each shell. Only the value of L_t at each shell will be affected, and we will slide the map to the place where L_t is precisely unity at r = infinity, no more and no less.

    Okay, so here we go. So far we have made the coordinate choice

    [tex]z = L = sqrt(1 - 2 m f / r)[/tex]

    WolframAlpha gives the derivative of z, then, as

    [tex]z' = m (f - r f') / (r^2 sqrt(1 - 2 m f / r))[/tex]

    and applying that to our previous relationship, we gain

    [tex]m L_t^2 / r^2 = m (f - r f') / r^2[/tex]

    [tex]L_t^2 = f - r f'[/tex]

    and the derivative of that gives

    [tex]2 L_t L_t' = - r f"[/tex]

    Starting back with the original relationship again, we gain

    [tex]L_t^2 = z z' r^2 / m[/tex]

    and taking the derivative of that,

    [tex]2 L_t L_t' = r (r z'^2 + z (r z" + 2 z')) / m[/tex]

    So we now have the new relationship,

    [tex]- r f" = r (r z'^2 + z (r z" + 2 z')) / m[/tex]

    [tex]f" = -(r z'^2 + z (r z" + 2 z')) / m[/tex]

    Taking the indefinite integral for that using WolframAlpha also, we get

    ERROR HERE. Wolfram found for the integral of m rather than r for some reason

    [tex]f' = - log(m) (r z'^2 + z (r z" + 2 z')) + j[/tex], where j is some constant

    and again to find f,

    [tex]f = j r - r log(m) z z' - log(m) z^2 / 2 + p[/tex], where p is another constant

    Starting back with what we originally had for z,

    [tex]z = sqrt(1 - 2 m f / r)[/tex]

    [tex]2 m f / r = 1 - z^2[/tex]

    [tex]f = r (1 - z^2) / (2 m)[/tex]

    and then substituting,

    [tex]r (1 - z^2) / (2 m) = j r - r log(m) z z' - log(m) z^2 / 2 + p[/tex]

    and plugging this equation into Wolframalpha, it finds the solution for z to be

    [tex]z = sqrt(1 + (c1 / r) e^(r / (m log(m))) + m (1 - 2 j m) log(m) / r - 2 m (j r + p) / r)[/tex], where c1 is yet another constant

    Now, we want z = 1 precisely at r = infinity. (c1 / r) e^(r / (m log(m))) will become infinite at r = infinity unless c1 = 0, so c1 = 0 and then for any finite value of r, this part drops out and we are left with

    [tex]z = sqrt(1 + m (1 - 2 j m) log(m) / r - 2 m j - 2 m p / r)[/tex]

    The value m(1 - 2 j m) log(m) - 2 m p is all constant, so to simplify we will denote it with the variable q, now giving

    [tex]z = sqrt(1 + q / r- 2 m j)[/tex]

    At r = infinity, this equation will reduce to

    [tex]z = sqrt(1 - 2 m j)[/tex]

    where 2 m j is a constant whereas we want z to be precisely unity, so j = 0. This leaves us with the equation

    [tex]z = sqrt(1 + q / r)[/tex]

    and we originally had

    [tex]z = sqrt(1 - 2 m f / r)[/tex]

    so

    [tex]q = 2 m f[/tex]

    [tex]f = q / (2 m)[/tex]

    q / (2 m) is a constant, which makes f constant also, so f' = 0. Going back to our original equation for L_t^2, then, we gain

    [tex]L_t^2 = f - r f'[/tex]

    [tex]L_t^2 = f[/tex]

    Since we have determined that f is a constant for all r, then L_t is a constant also for all r when z = 1 at r = infinity. We want L_t = 1 at r = infinity as well, so if L_t = 1 at r = infinity and L_t is a constant for all r, then L_t = 1 for all r. We have now solved for the variable L_t. Now let's solve for the other two. As our coordinate choice, we had

    [tex]z = L = sqrt(1 - 2 m f / r)[/tex]

    and we just gained

    [tex]L_t^2 = f = 1[/tex]

    giving

    [tex]z = L = sqrt(1 - 2 m / r)[/tex]

    We now have exact solutions for all three variables :)

    Of course, everybody check my math. I just found this solution today and was so excited I had to go ahead and post it. I've gone back through it a couple of times though and it seems pretty straightforward and error free. I never would have thought, however, that all three variables could be determined without applying some other underlying principle, as determined by the full Einstein field equations or otherwise, rather than just using the low Newtonian approximation. Yet apparently, the best that I can tell so far anyway, they can.
     
    Last edited: Aug 12, 2014
  2. jcsd
  3. Aug 12, 2014 #2

    PeterDonis

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    You are already assuming a particular solution of the full Einstein field equations, simply by writing the metric in the form you've written it; that form of the metric only applies to spherically symmetric solutions, and the particular exact solution you derive for the three variables only applies to a spherically symmetric vacuum solution, i.e., to the Schwarzschild geometry (written in any coordinate chart in which the metric assumes the form you use). None of what you're doing generalizes beyond those particular special cases.
     
  4. Aug 12, 2014 #3
    Thanks PeterDonis. The metric is in a general form with the three unknown variables and one can always for a spherically symmetric solution, but it turns out that I was too excited and too quick to post after all. Running back through the derivation once again, I found when finding the indefinite integral from f" to f', Wolfram used dm rather than dr (or dx as it is put in, substituting r for x) for some reason, not sure why when dr gives a solution and the variables were all in terms of r, whereas m was supposed to be a constant. The correct integral, however, just gives me back what I started with, which is what is what I originally expected in the first place when doing it this way. I'm devastated.
     
  5. Aug 12, 2014 #4

    PeterDonis

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    No, it isn't. The form in which you have written the metric is only valid for a spherically symmetric solution. MTW gives a derivation that shows why this is the case.
     
  6. Aug 12, 2014 #5
    Right, I am using spherical symmetry. :smile: Or was. Now it looks like I'm back to the drawing board just minutes after I posted. :rolleyes: I still need another underlying principle. Is there one main basic principle that the Einstein field equations give us that can be easily visualized and expressed mathematically in algebraic form?
     
  7. Aug 12, 2014 #6
    The assumption of spherical symmetry does not have much to do with the Einstein equation itself, does it?
    I, personally, find offering different viewpoints important.

    OP, may I ask why you are writing this here and not in a tex file? Obviously you're putting a lot of work into it and even if whatever you're doing gets rejected by the journals you can still publish it on the arxiv and see whether someone likes it and uses it. Also putting everything in a proper form usually let's one question some of the things one puts in and makes the work better.

    Btw, if you're looking for postulates of GR I can recommend Kriele's book "spacetime", which is a great work of mathematical physics. On a conceptual level I can tell you that GR is a natural synthesis of analytical mechanics and special relativity.
     
  8. Aug 12, 2014 #7
    Thanks, it's already been rejected. Postulates or principles of GR is exactly what I need, thank you. Something general that be expressed in a mathematically simple way.
     
  9. Aug 12, 2014 #8

    robphy

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    This might be a useful reference
    for trying to study aspects of general relativity (in nice cases)
    without using the full Einstein field equations and the full machinery of differential geometry.
    http://www.eftaylor.com/general.html
     
  10. Aug 12, 2014 #9

    PeterDonis

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    Once you assume spherical symmetry, the most you can get out of the Einstein Field Equation is to reduce the number of free functions in the metric to two. You can't reduce it to one, which is basically what you are trying to do. The MTW derivation that I referred to previously shows why two free functions is the minimum you can achieve; basically, getting to that point uses up all of the components of the EFE that are not either identically zero or just restatements of other components, given the constraint of spherical symmetry.

    (Using your notation, the easiest way to see that two free functions are required is to use the Schwarzschild chart, where ##L_t = 1##. Then you can show that ##L = 1 - 2 m / r##, where ##m## is the "mass inside radius ##r##", and is one of the free functions. The other free function is ##z##, which can't be simplified any further unless you assume vacuum--see below. Note also that, in general, ##m## and ##z## can be functions of both ##r## and ##t##, the coordinate time; there is no need to assume a static metric. In the vacuum case, staticity is not an assumption, it's a theorem, part of the conclusion of Birkhoff's theorem.)

    If you assume spherical symmetry plus vacuum, then you can reduce the number of free parameters in the metric to one, the mass ##M##, which is a constant, not a function--i.e., you can prove that the geometry must be the Schwarzschild geometry, so the relationship between your ##z##, ##L##, and ##L_t## is fixed once you choose your coordinate chart (Schwarzschild chart, isotropic chart, etc.). This is just Birkhoff's theorem.
     
    Last edited: Aug 12, 2014
  11. Aug 13, 2014 #10
    I assume a vacuum solution sets certain quantities to zero? If so, what quantities would those be?
     
  12. Aug 13, 2014 #11

    PeterDonis

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    Vacuum means the stress-energy tensor is identically zero, so the RHS of each component of the Einstein Field Equation is zero. (I'm assuming a zero cosmological constant.)
     
  13. Aug 13, 2014 #12

    robphy

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    In the vacuum case, the Ricci tensor is zero.
    So, you need to formulate your equations accordingly.

    http://en.m.wikipedia.org/wiki/Ricci-flat_manifold
     
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