- #1
yuiop
- 3,962
- 20
In various other threads we have been kicking around various equations for a spherical shell and discussing the implications. In this thread I would like to present what (I think) I have worked out about how the shell metric relates to the vacuum metric inside and outside the shell.
I hope to show here that ##g_{tt}## or ##J(r)## inside the vacuum cavity must always be unity and that the http://preposterousuniverse.com/grnotes/grnotes-seven.pdf can be derived from the shell equation derived by PeterDonnis and that the metric for the shell can connect the metric inside and outside without any 'patching' factors.
For reference I am assuming spherical symmetry with no rotation, no charge and a momentarily stationary dust shell.
First consider this post by WannabeNewton
.. concludes that ##g_{rr}=1## inside the cavity. If we accept the above conclusions by WBN and Peter, then we must accept that ##g_{tt}=J(r) =1## inside the cavity and cannot take any value, other than unity, as long as the vacuum cavity exists. <EDIT: Now I have looked again, I see that WBN had already concluded that ##g_{tt}## must be unity inside the vacuum cavity, in his elegant post above>.
If we require the metric of the shell to smoothly connect ##g_{tt}=1## at the inside surface to ##g_{tt} = (1-2M/r)## at the outside surface of the shell then that requirement is met by Sean Carroll's metric that has ##g_{tt} = J(r) = (1-2m(r)/r)##. I know this is not a popular idea but I will try to show here that it can be obtained from Peter's equation. We start with:
##\frac{1}{J} \frac{dJ}{dr} = \frac{2(m(r) - 4 \pi r^3 \rho (r))}{r \left( r - 2 m(r) \right)}##
Initially I will consider the constant density case so ##\rho = \frac{M}{(4/3)*\pi(R2^3-R1^3)}## where M is the total mass of the shell, R1 is the inner radius of the shell and R2 is the outer radius. Similarly we can replace ##m(r)## with ##M*(r^3-R1^3)/(R2^3-R1^3)## as shown by pervect here. When these substitutions are made, the equation becomes:
##\frac{1}{J} \frac{dJ}{dr} = \frac{2M(-2r^3-R1^3)}{r^2(R2^3-R1^3)-r2M(r^3-R1^3)}##
Now when we carry out the definite integration from r=R1 to r=R0 where R0 is the location where we wish to determine ##g_{tt}##, we get a closed form solution:
##{J(R0)} = \left(\frac{R1^3(1-2M/R0)+2MR0^2-R2^3}{R1^3(1-2M/R1)-2MR1^2-R2^3}\right)##
The online calculation of the indefinite integral can be seen here.
This algebraically simplifies to:
##J(R0) = \left(1-\frac{2M}{R0}\left(\frac{R0^3-R1^3}{R3^3-R1^3}\right)\right)##
The algebraic manipulation is not obvious, but the equivalence can be confirmed by subtracting the expressions from each other and getting zero, as shown here.
It is easy to see that the last term is the enclosed volume and that the equation can be expressed as:
##J(r) = \left(1-\frac{2M(r)}{r}\right)##
where it should be understood that M(r) is the mass enclosed within a sphere of radius r. This is the result quoted by Sean Carroll and has been obtained directly from Peter's equation. It also has the pleasant result that the integral has a closed form and directly connects ##g{tt}=1## of the vacuum cavity with ##g_{tt}=(1-2M/r)## of the Schwarzschild metric outside the shell without any scaling factors.
The above derivation is for the constant density case. It turns out there is alternative way to obtain the same result that implies the result has a more general application. This time I start with Peter's equation and replace ##(m(r) - 4 \pi r^3 \rho (r))##with M(x), where M(x) is a constant that represents the total effective gravitational mass enclosed by a volume with r=x, taking into account all factors such as energy, stress, pressure that contribute to the gravitational mass, just as M does the same job in the external Schwarzschild metric. By this definition we have
##\frac{1}{J(r)}\frac{dJ(r)}{dr} = \frac{2M(x)}{r \left( r - 2M(x) \right)}##
The indefinite integral is then:
##J(r) = (1-2M(x)/r) +C##
This is the same result as obtained earlier if the constant of integration is discarded. Normally an indefinite integral does not have defined boundaries but it works here as the boundaries are implicitly defined when M(x) is calculated. A more general form of M(x) can given as:
##M(x) = M\frac{x^d-R1^d}{R2^d-R1^d}##
For constant density, d=3 but other density profiles can be defined by using different values of d. For example when d=(-1), the density near the inner surface of the shell is considerably higher than near the outer surface. Changing the density profile change the shape of the potential gradient within the shell, but the curve always neatly connects the inner Minkowski metric to the outer Schwarzschild metric.
See these plots for J(r) where d=1,-1,3 with M=G=c=1 and R1=1/2 and R2=7.
##g_{tt}## or ##J(r)## can represent the gravitational potential and examination of the slope of the potential in the graphs linked above, produces some surprising conclusions that might offend our Newtonian based intuitions, but we should not be too surprised that GR can differ significantly.
An interesting aspect of this solution is that one metric (with no 'patching' factors) can continuously cover the entire region inside the vacuum of the cavity, through the shell and on through the external vacuum, if it is clearly understood that M(x) is the gravitational mass enclosed within a volume of radius r=x.
I hope to show here that ##g_{tt}## or ##J(r)## inside the vacuum cavity must always be unity and that the http://preposterousuniverse.com/grnotes/grnotes-seven.pdf can be derived from the shell equation derived by PeterDonnis and that the metric for the shell can connect the metric inside and outside without any 'patching' factors.
For reference I am assuming spherical symmetry with no rotation, no charge and a momentarily stationary dust shell.
First consider this post by WannabeNewton
that concludes that ##g_{tt} = g_{rr}^{-1}## inside the cavity and this quote by PeterDonnisWannabeNewton said:You have to consider a thin spherically symmetric shell configuration wherein the source matter is a dirac delta distribution, or at the least confined to a region ##R_1 < r < R_2##. Then the region exterior to ##R_2## will of course have to be Schwarzschild as a consequence of Birkhoff's theorem but note that the region interior to ##R_1## is also spherically symmetric vacuum by hypothesis. Hence Birkhoff's theorem should apply in the interior as well implying that the interior metric is static.
Now the general form of the metric in a static spherically symmetric space-time can be written in appropriate coordinates as ##ds^{2} = -A(r)dt^{2} + B(r)dr^2 + r^{2}d\Omega^{2}##. One can then use the vacuum Einstein equations to deduce that ##B(r) = A(r)^{-1}, A(r) = 1 + \frac{C}{r}##. Now in the interior case ##A(r)## must be well behaved as ##r\rightarrow 0## because the metric has to be non-degenerate, so it must be the case that ##C = 0## otherwise ##A(r)## will necessarily blow up at ##r =0## and this will make the metric degenerate at ##r =0## but by definition the metric must be non-degenerate everywhere.
PeterDonis said:However, at the *inner* surface of the object (i.e., the outer edge of the hollow cavity), ... ##m(r) = 0##--there is no mass left inside that radius. ... there is *no* "space distortion" any more, even in this global metric; the coefficient of ##dr^2## is ##1##
.. concludes that ##g_{rr}=1## inside the cavity. If we accept the above conclusions by WBN and Peter, then we must accept that ##g_{tt}=J(r) =1## inside the cavity and cannot take any value, other than unity, as long as the vacuum cavity exists. <EDIT: Now I have looked again, I see that WBN had already concluded that ##g_{tt}## must be unity inside the vacuum cavity, in his elegant post above>.
If we require the metric of the shell to smoothly connect ##g_{tt}=1## at the inside surface to ##g_{tt} = (1-2M/r)## at the outside surface of the shell then that requirement is met by Sean Carroll's metric that has ##g_{tt} = J(r) = (1-2m(r)/r)##. I know this is not a popular idea but I will try to show here that it can be obtained from Peter's equation. We start with:
##\frac{1}{J} \frac{dJ}{dr} = \frac{2(m(r) - 4 \pi r^3 \rho (r))}{r \left( r - 2 m(r) \right)}##
Initially I will consider the constant density case so ##\rho = \frac{M}{(4/3)*\pi(R2^3-R1^3)}## where M is the total mass of the shell, R1 is the inner radius of the shell and R2 is the outer radius. Similarly we can replace ##m(r)## with ##M*(r^3-R1^3)/(R2^3-R1^3)## as shown by pervect here. When these substitutions are made, the equation becomes:
##\frac{1}{J} \frac{dJ}{dr} = \frac{2M(-2r^3-R1^3)}{r^2(R2^3-R1^3)-r2M(r^3-R1^3)}##
Now when we carry out the definite integration from r=R1 to r=R0 where R0 is the location where we wish to determine ##g_{tt}##, we get a closed form solution:
##{J(R0)} = \left(\frac{R1^3(1-2M/R0)+2MR0^2-R2^3}{R1^3(1-2M/R1)-2MR1^2-R2^3}\right)##
The online calculation of the indefinite integral can be seen here.
This algebraically simplifies to:
##J(R0) = \left(1-\frac{2M}{R0}\left(\frac{R0^3-R1^3}{R3^3-R1^3}\right)\right)##
The algebraic manipulation is not obvious, but the equivalence can be confirmed by subtracting the expressions from each other and getting zero, as shown here.
It is easy to see that the last term is the enclosed volume and that the equation can be expressed as:
##J(r) = \left(1-\frac{2M(r)}{r}\right)##
where it should be understood that M(r) is the mass enclosed within a sphere of radius r. This is the result quoted by Sean Carroll and has been obtained directly from Peter's equation. It also has the pleasant result that the integral has a closed form and directly connects ##g{tt}=1## of the vacuum cavity with ##g_{tt}=(1-2M/r)## of the Schwarzschild metric outside the shell without any scaling factors.
The above derivation is for the constant density case. It turns out there is alternative way to obtain the same result that implies the result has a more general application. This time I start with Peter's equation and replace ##(m(r) - 4 \pi r^3 \rho (r))##with M(x), where M(x) is a constant that represents the total effective gravitational mass enclosed by a volume with r=x, taking into account all factors such as energy, stress, pressure that contribute to the gravitational mass, just as M does the same job in the external Schwarzschild metric. By this definition we have
##\frac{1}{J(r)}\frac{dJ(r)}{dr} = \frac{2M(x)}{r \left( r - 2M(x) \right)}##
The indefinite integral is then:
##J(r) = (1-2M(x)/r) +C##
This is the same result as obtained earlier if the constant of integration is discarded. Normally an indefinite integral does not have defined boundaries but it works here as the boundaries are implicitly defined when M(x) is calculated. A more general form of M(x) can given as:
##M(x) = M\frac{x^d-R1^d}{R2^d-R1^d}##
For constant density, d=3 but other density profiles can be defined by using different values of d. For example when d=(-1), the density near the inner surface of the shell is considerably higher than near the outer surface. Changing the density profile change the shape of the potential gradient within the shell, but the curve always neatly connects the inner Minkowski metric to the outer Schwarzschild metric.
See these plots for J(r) where d=1,-1,3 with M=G=c=1 and R1=1/2 and R2=7.
##g_{tt}## or ##J(r)## can represent the gravitational potential and examination of the slope of the potential in the graphs linked above, produces some surprising conclusions that might offend our Newtonian based intuitions, but we should not be too surprised that GR can differ significantly.
An interesting aspect of this solution is that one metric (with no 'patching' factors) can continuously cover the entire region inside the vacuum of the cavity, through the shell and on through the external vacuum, if it is clearly understood that M(x) is the gravitational mass enclosed within a volume of radius r=x.
Last edited: