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Combining gravitational time dilation components.

  1. Nov 22, 2013 #1
    Consider a sphere with outer radius r2 and a centred inner cavity of radius r1, forming a constant density shell with density p.

    Let's say the time dilation on a clock on the inner surface of the shell is ta.
    Now the shell with filled with a material of the same constant density as the shell such that the time dilation factor due to the material that fills the cavity is tb = √(1-2*p*4/3*pi*r12). If we ignore the time dilation due to the material in the original shell.

    Would the total time dilation factor at r1 due to the material in the filled inner cavity and the material in the outer shell, be equal to ta*tb?

    Starting again with original hollow shell and a clock at r1, an additional shell with the same constant density and inner radius of r2 and outer radius r3 now encloses the original shell. If the time dilation on the clock at r1, due to the new outer shell alone is tc, would the combined time dilation acting on the clock at r1 due to both shells be ta*tc or does time dilation combine in some other way?

    From my initial investigations, time dilation does not seem to combine as a simple product (or sum) and the time dilation due to an outer shell, does not seem to have a time dilation component that is independent of material inside the cavity.

    Can anyone shed any light (or equations) on these issues?
     
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  3. Nov 22, 2013 #2

    bcrowell

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    In a static spacetime, we can define the notion of a gravitational time dilation factor. It equals the square root of the time-time component of the metric, or [itex]e^\Phi[/itex], where Φ is the gravitational potential. In the weak field approximation, potentials add, so in that approximation, time-dilation factors multiply.

    There is no general rule for this that avoids the need for the weak field approximation, because the Einstein field equations are nonlinear. The notion of gravitational time dilation can't even be defined in an arbitrary spacetime.
     
  4. Nov 23, 2013 #3

    tom.stoer

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    It's not clear what you mean. Consider a local reference frame with coordinate time t and two curves i=1,2 representing two observers moving through spacetime w.r.t. the reference frame. Then we have for the proper times of the two observers

    ## \tau_i = \int_{C_i} d\tau = \int_0^T dt \,\sqrt{g_{ab}\,\dot{x^a}\,\dot{x}^b} ##

    This expression summarizes differential aging between the two observers, as well a time dilation w.r.t. the (artificial) reference frame. The physically relevant and observable phenomenon is the difference of the proper times.

    I guess what you mean is that in general you cannot separate the "gravitational effects" due to the metric g from the "velocity-dependent effects" due to dx/dt. I would agree to that statement.
     
  5. Nov 23, 2013 #4

    bcrowell

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    Yes, that's what I meant.
     
  6. Nov 24, 2013 #5
    In the original question I was only considering gravitational time dilation as all observers/clocks were intended to be stationary with respect to the gravitational source.

    As for how gravitational time dilation combines, here is one simple way to demonstrate that the relationship is definitely not linear.

    Consider a shell with outer radius R and inner radius r and mass M. The time dilation at the surface of the shell is ##t1 = \sqrt(1-2M/R)##. If we have another sphere with mass m and an outer radius of r in otherwise empty space, then the time dilation due to this sphere at a radius R is ##t2 = \sqrt(1-2m/R)##. Now consider a solid sphere with total mass (M+m) and outer radius R, then the time dilation at the surface of this sphere ##t3 = \sqrt(1-2(M+m)/R)## and this is not equal to ##t1*t2 ## and there does not seem to be any simple way to relate t3 to t1 and t2.

    =============================================

    My original question stemmed form trying to figure out what is going on inside and outside a spherical shell. In this post by pervect he gives a definition of J(r) as:

    ##J(r) = K \, exp \, \left( \int_{x=R1}^{x=r}\frac{2\,m(x)\,dx}{x\left(x-2\,m(x)\right)}\right)##

    where ##m(r) = \frac{M \left( r^3 - R1^3\right)}{R2^3 - R1^3}##

    Is it possible to obtain an explicit expression for J(r) using the above information?

    If I substitute the full expression for m(r) for m(x) in the integral all hell breaks loose and the integral is almost impossible to determine. Should m(x) be treated as constant equal to m(r) when carrying out the integration?
     
  7. Nov 24, 2013 #6

    Mentz114

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    If the antiderivative of ## \frac{2\,m(x)\,dx}{x\left(x-2\,m(x)\right)}## were ##j(x)## then ##J(r) =K (exp( j(r))-exp(j(R1))##.

    There doesn't appear to be a closed form of the integral. So that isn't very helpful.

    Putting ##m(x) = \frac{M \left(x^3 - R1^3\right)}{R2^3 - R1^3}## into the integrand gives ##-\frac{2\,M\,\left( R1-x\right) \,\left( {R1}^{2}+x\,R1+{x}^{2}\right) }{x\,\left( x\,{R2}^{3}+2\,M\,{R1}^{3}-x\,{R1}^{3}-2\,{x}^{3}\,M\right) }##, in case you want a check.
     
    Last edited: Nov 24, 2013
  8. Nov 25, 2013 #7
    Could someone help me reconcile these two equations by Pervect and Peter Donnis.

    =============================================
    In this post by pervect he gives a definition of J(r) as:

    ##J(r) = K \, exp \, \left( \int_{x=R1}^{x=r}\frac{2\,m(x)\,dx}{x\left(x-2\,m(x)\right)}\right)##

    where ##m(r) = \frac{M \left( r^3 - R1^3\right)}{R2^3 - R1^3}##

    =============================================

    In this blog by PeterDonnis he gives a defintion of J(r) as:

    ##\frac{dJ}{dr} = 2J \left( \frac{m(r) + 4 \pi r^3 p(r)}{r^2 (1 - 2 m(r) / r)} \right)##

    (See also this post by Peter).

    ============================

    Now pervect is using a dust model with no stress so he is using p=0. If I set p=0 in Peter's equation I get:

    ##\frac{dJ}{dr} = J \left( \frac{2m(r) }{r^2 (1 - 2 m(r) / r)} \right)##

    Changing the variable from r to x to align with pervect's notation and integrating wrt r this becomes:

    ##J(r) = \left( \int_{x=R1}^{x=r}J \frac{2\,m(x)\,dx}{x\left(x-2\,m(x)\right)}\right)##

    I am not too bothered about the K in pervect's solution as I understand that is a scaling factor but I am not sure why he has an exp function that does not appear in Peter's and I am not sure why Peter has a J in the integrand, that does not appear in pervect's, unless J is scaling factor like K?

    So where does the exp come from?

    Also, can anyone confirm the equation is definitely not meant to be:

    ##J(r) = K \, exp \, \left( \int_{x=R1}^{x=r}\frac{2\,m(r)\,dx}{x\left(x-2\,m(r)\right)}\right)##?

    That form would have a closed solution unlike the other integrals.
     
  9. Nov 25, 2013 #8

    PeterDonis

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    In the expression you derived from my blog post, there is a ##J## in the integrand on the RHS. Before integrating, you need to first move the ##J## to the LHS; then when you integrate you get:

    $$
    \int \frac{dJ}{J} = \int \left( \frac{2 m(r)}{r \left( r - 2m(r) \right)} dr \right)
    $$

    The integral on the LHS gives ##\ln J##, so you need to exponentiate to get ##J## itself. That's where the ##\exp## in pervect's expression comes from.
     
  10. Nov 25, 2013 #9
    Thanks for that! :approve: I understand now.

    What about my last proposition:

    Can that be ruled out?
     
  11. Nov 25, 2013 #10

    PeterDonis

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    Is this supposed to be the same as pervect's equation? Or different?
     
  12. Nov 25, 2013 #11
    Different. I have replaced m(x) with m(r) in the integrand, so that that the enclosed mass itself is not integrated with respect to x. My reasoning is that formula for m(r) given by pervect already calculates the enclosed mass, so integrating it again by x seems like overkill.
     
  13. Nov 25, 2013 #12

    PeterDonis

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    Then it's ruled out; pervect's equation is correct. See below.

    But the integral isn't calculating ##m##, it's calculating ##J##. What the equation for ##(1 / J) dJ / dr## says is that, as you move radially outward from the center, ##\ln J## increases by the ratio of ##2m(x)## to ##x (x - 2m(x))## *at the value of ##x## you are currently at*. If you instead take the ratio using the value of ##2 m## at the outer surface of the object (i.e., at ##x = r##), you will get the wrong answer for ##J##. (For one thing, if you use the value of ##2 m## at the surface, then at sufficiently small values of ##x##, ##x - 2m(r)## will be negative and the sign of ##dJ / dr## will change, so the minimum of ##J## will not be at the inner surface of the shell--it will be somewhere within the shell. That's obviously wrong physically; the minimum of ##J## has to be at a point where there is no mass closer to the center, and that's only possible at the inner surface of the shell.)
     
  14. Nov 25, 2013 #13
    I think I follow your arguments here and I agree that it is physically wrong in Newtonian terms, but I would prefer to see the calculations in GR terms, rather than assume them.

    Does the following work?

    Start with your original equation and invert it

    ##\frac{dr}{dJ} = \frac{1}{J} \left( \frac{r^2 (1 - 2 m(r)/r)}{2m(r) + 8 \pi r^3 p(r)} \right)##,

    Integrate wrt J

    ##r = \int{\frac{1}{J} \left( \frac{r^2 (1 - 2 m(r)/r)}{2m(r) + 8 \pi r^3 p(r)} \right)} dJ = \log{(J)} \left( \frac{r ^2(1- 2 m(r)/r)}{2m(r) + 8 \pi r^3 p(r)} \right)##,

    Solve for J

    ##J(r) = \exp{ \left( \frac{2m(r) + 8 \pi r^3 p(r)}{r (1 - 2 m(r) / r)} \right)}##

    where ##m(r) = \frac{M \left( r^3 - R1^3\right)}{R2^3 - R1^3}##

    Here, J(r) has the property that it is minimal at the inner surface of the shell where m(r)=0.
     
    Last edited: Nov 25, 2013
  15. Nov 25, 2013 #14

    PeterDonis

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    Pervect's calculation *is* a calculation "in GR terms". I'm not sure I understand what you're looking for here.

    It doesn't look correct to me. Inverting the equation makes ##r## a function of ##J##, so your "integral with respect to J" is a *lot* more complicated than you appear to think it is.
     
  16. Nov 26, 2013 #15
    I was hoping to find some closed form that can be evaluated, studied/plotted, but it appears that is impossible. It seems that the only way to resolve the integral in numerically. Consider a shell with an inner radius R1=0 and outer radius R2=9/4 (so effectively a solid sphere). Assume constant density, and zero stress and using units of G=c=1, a total mass of 1. After carrying out the numerical integration I get the following results:

    r=9/4, J(r) = 3
    r=9/8, J(r) = 1.13389
    r=0, J(r) = 1

    Now if I apply the same parameters to the interior Schwarzschild solution I get:

    r=9/4, J(r) = 1/9 = 0.1111
    r=9/8, J(r) = 0.00348589
    r=0, J(r) = 0

    There does not appear to be any constant K that can scale the first set of results to match the second set by multiplication or addition. So why the difference? I have assumed constant density in both cases. Does the interior Schwarzschild solution automatically include an unspecified stress parameter that is a consequence of assuming a perfect fluid.
     
    Last edited: Nov 26, 2013
  17. Nov 26, 2013 #16

    PeterDonis

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    Unfortunately, yes, this is the case.
     
  18. Nov 26, 2013 #17

    PeterDonis

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    This is not a good case because it's a limiting case: the central pressure goes to infinity. Numerical methods don't give good results when one of your parameters is increasing without bound. I would advise picking a larger starting radius.

    Pervect's equation doesn't include the effect of pressure, so it isn't a good one to use for any situation where pressure is non-negligible, such as this one. You would have to pick a *much* larger starting radius to get a meaningful answer using his equation.

    It includes pressure, so you need to solve for the pressure numerically as well as for J(r). If you do you will, as I said above, find that the pressure increases without bound as r = 0 is approached.
     
  19. Nov 26, 2013 #18
    I was imagining a sphere of dust particles that has not compacted, so there is no pressure. Presumably the interior Schwarzschild metric assumes a hydrostatic pressure that is in equilibrium, but I am not sure about that.

    If by starting radius you mean the inner surface of the shell, then the reason I chose R1=0 is so that the shell metric can be directly compared to the interior Schwarzschild metric. As I understand it, the interior solution is only valid for a sphere of constant density, so it would not be valid to consider a sphere with a vacuum cavity in the centre, when using the interior Schwarzschild metric. If you mean the outer radius, then the choice of R2=9/4 is possibly a poor choice because that is when J(0) goes to zero, which is an extreme.

    Can you provide any information how p(r) would be calculated in the case of an ideal fluid? I assume it not as simple as calculating m(r) which is just a function of enclosed volume and density. Then again, it should not be too complicated, as it is included in interior solution and that is fairly straightforward.
     
    Last edited: Nov 26, 2013
  20. Nov 27, 2013 #19

    PeterDonis

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    Such a configuration can't be static; it will implode.

    Yes, it does. Otherwise, as above, the solution would not be static. (Note that my blog post only treats the static case; the equations there are not valid if the system is not static. The same is true for the Schwarzschild interior solution.)

    That part is fine: pervect's solution with R1 = 0 would give the same results as the interior Schwarzschild metric, *if* pervect's solution included pressure.

    Yes, that's what I was referring to.

    My blog post gives the equation for dp/dr. You need to numerically integrate both dp/dr and dJ/dr, either inward from the shell's outer radius or outward from r = 0.
     
    Last edited: Nov 27, 2013
  21. Nov 27, 2013 #20

    WannabeNewton

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    This is mathematically impossible. Dust particles cannot generate static solutions. Assume we have a dust field with 4-velocity field ##u^{\mu}## that generates a static solution ##g_{\mu\nu}## to Einstein's equation. Then we can always find coordinates ##\{x^{\mu}\}## adapted to the static killing field ##\xi^{\mu}## so that ##g_{it} = 0## and ##\partial_t g_{\mu\nu} = 0##. Contracting both sides of Einstein's equation with ##\xi^{\mu}##, we have ##R_{\mu\nu}\xi^{\nu} = 8\pi \rho(u_{\mu}u_{\nu} + \frac{1}{2}g_{\mu\nu})\xi^{\nu}## hence for ##\mu = i## this reduces to ##R_{it} = 8\pi \rho u_{i}u_{\nu}\xi^{\nu} ##.

    Standard formulas give us ##R_{it} = \partial_{\mu}\Gamma ^{\mu}_{it} + \Gamma ^{\mu}_{it}\Gamma^{\nu}_{\mu\nu} - \Gamma^{\mu}_{i\nu}\Gamma^{\nu}_{t\mu}##. Note that ##g^{\mu\nu}g_{\nu\alpha} = \delta^{\mu}_{\alpha}## implies ##g^{i t}g_{t t} = 0## i.e. ##g^{it} = 0##. Using this and the standard formula for the connection coefficients one can show that ##R_{it} = 0##. Hence ##u_{i} = 0## or ##u_{\mu}\xi^{\mu} = 0## but the latter of the two is impossible since this would imply that ##u^{\mu}## is space-like. Therefore ##u^{i} = g^{i\nu}u_{\nu} =g^{it}u_{t} + g^{ij}u_{j} = 0##. What this means is that ##u^{\mu} = \alpha \xi^{\mu}## for some smooth scalar field ##\alpha## that is determined by the normalization condition and is thus a function of the metric tensor.

    The next step is to use Raychaudhuri's equation for the dust field ##u^{\mu}## which reads ##u^{\mu}\nabla_{\mu}\theta = -R_{\mu\nu}u^{\mu}u^{\nu} + \omega_{\mu\nu}\omega^{\mu\nu} - \sigma_{\mu\nu}\sigma^{\mu\nu} - \frac{1}{3}\theta^{2} + \nabla_{\mu}(u^{\nu}\nabla_{\nu}u^{\mu})## where ##\theta = \nabla_{\mu}u^{\mu}## is the expansion, ##\omega_{\mu\nu}## is the rotation, and ##\sigma_{\mu\nu}## is the shear. A dust field is necessarily geodesic so ##u^{\nu}\nabla_{\nu}u^{\mu} = 0##. Also, since ##u^{\mu} = \alpha \xi^{\mu}##, we have ##\nabla_{\mu}u^{\mu} = \alpha \nabla_{\mu}\xi^{\mu}+ \xi^{\mu}\nabla_{\mu}\alpha = 0 ## where the first term vanishes due to Killing's equation ##\nabla^{(\mu}\xi^{\nu)} = 0## and the second term vanishes because ##\xi^{\mu}\nabla_{\mu}\alpha = \partial_{t}\alpha = 0##. Finally, since ##\xi^{\mu}## is static, it satisfies (by definition) the hypersurface orthogonality condition ##\xi_{[\alpha}\nabla_{\mu}\xi_{\nu]} = 0## which implies ## \epsilon^{\mu\nu\alpha\beta}\xi_{\nu}\nabla_{\alpha}\xi_{\beta} = 0## (i.e. a static killing field is twist-free). Thus ##\omega^{\mu} = \epsilon^{\mu\nu\alpha\beta}u_{\nu}\nabla_{\alpha}u_{\beta} = \alpha^2 \epsilon^{\mu\nu\alpha\beta}\xi_{\nu}\nabla_{\alpha}\xi_{\beta} + \alpha\epsilon^{\mu[\nu|\alpha|\beta]}\xi_{(\nu}\xi_{\beta)}\nabla_{\alpha}\alpha = 0## therefore ##\omega_{\mu\nu} = 0##.

    This finally leaves us with ##R_{\mu\nu}u^{\mu}u^{\nu} + \sigma_{\mu\nu}\sigma^{\mu\nu} = 0## but ##\sigma_{\mu\nu}\sigma^{\mu\nu} \geq 0## since it is purely space-like and ##R_{\mu\nu}u^{\mu}u^{\nu} > 0## for non-vanishing mass density due to the strong energy condition, leaving us with a contradiction. This means that a dust field cannot generate a static solution to Einstein's equation.
     
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