Your original source had something like:
dt = \frac{d\tau}{\sqrt{1-2M/x}}
and
dx = dS \sqrt{1-2M/x}
This is a consistent definition of the quantities, because the quantities on the LHS of both equations are coordinate measurements made by an observer at infinity. Your preferred method of defining the ratios is dtau/dt and dx/dS but this is inconsistent because the former is the ratio (local)/(coord) and the second is the ratio (coord)/(local). Now it is not important whether you invert the ratios or not, but if you wish to compare the ratios you should be consistent and use either (local)/(coord) for both or (coord)/(local) for both, to make a meaningful comparison.
Q-reeus said:
Yes, as per above, and that raises another issue. In #40 you said "This hints at what I said earlier that length contraction is a function of potential gradient rather than the value of the potential itself like gravitational force." Given that the two curves (inverting the length curve as graphed) match precisely everywhere outside the shell, there is no choice but to have both, or neither, to be functions of potential or potential gradient. But if functions of potential gradient (like gravitational acceleration), they would both, just as for 'gravity', disappear inside the shell! Welcome to the Twilight Zone - or not!
As I said before, if you use a consistent definition of ratios, length contraction and time dilation are not the same outside the shell.
Q-reeus said:
This is the fundamental issue that can't just be redefined like the exterior curves can. I can find nothing amiss re your derivation in #35 of dt/dtau, dr/dS, given the usage of [PLAIN]https://www.physicsforums.com/latex_images/30/3050819-0.png. But that expression bears closer scrutiny. How was it derived, or sourced from where?
That equation is based on the (fairly) well known and documented interior Shwarzschild solution:
<br />
c^2d\tau^2 = \left(\frac{3}{2}\sqrt{1-\frac{2M}{R}}-\frac{1}{2}\sqrt{1-\frac{2M x^{2}}{R^{3}}}\right)^2 c^2 dt^2 - \left(1-\frac{2M x^2}{R^3}\right)^{-1}dx^2 -x^2(d\theta^2+\sin(\theta)^2d\phi^2)<br />
where I am using x instead of r to represent the Schwarzschild radial displacement coordinate. This solution is for a solid sphere with uniform mass density and no cavity.
See Gron page 255 for example
http://books.google.co.uk/books?id=IyJhCHAryuUC&pg=PA255#v=onepage&q&f=false
Even using the raw metric in its unadulterated form, the metric shows that length contraction is greatest at the outer surface (x=R) and reduces to unity at the centre (x=0), contrary to your instincts that length contraction does not reduce as you transfer from the exterior vacuum solution to the interior solution.
Now a closer look at how I derived my equations from the standard solution.
First, look at this expression that appears twice in the metric:
<br />
\left(1-\frac{2M x^{2}}{R^{3}}\right)<br />
Multipying the top and bottom of the fraction by the density (p) and x(4/3)pi the following is obtained:
<br />
\left(1-\frac{2M}{x} \frac{ (4/3)\pi x^{3}p}{(4/3)\pi R^{3}p}\right)<br />
Now (4/3)\pi R^{3}p = M so the expression above can be written as:
<br />
\left(1-\frac{2}{x} (4/3)\pi x^{3}p}}\right)<br />
and (4/3)\pi x^{3}p is the mass enclosed (M_x) within a sphere of radius x and the expression can now be written as:
<br />
\left(1-\frac{2M_x}{x}\right)<br />
Substituting this rearranged expression back into the origianl metric gives the form:
<br />
c^2d\tau^2 = \left(\frac{3}{2}\sqrt{1-\frac{2M}{R}}-\frac{1}{2}\sqrt{1-\frac{2M_x }{x}}\right)^2 c^2 dt^2 - \left(1-\frac{2M_x }{x}\right)^{-1}dx^2 -x^2(d\theta^2+\sin(\theta)^2d\phi^2)<br />
which is the equation you were suspicious about.
This is still the metric for a solid sphere of uniform density expressed in a different way.
To obtain the hollow shell metric we need to calculate the mass enclosed with a radius of x.
The density (p) of the shell with inner radius r and outer radius R is :
p = \frac{M}{(4/3)\pi (R^3-r^3)}
The mass enclosed within a radius of x is then:
M_x = M\frac{(4/3)\pi (x^3-r^3)}{(4/3)\pi (R^3-r^3)} = M\frac{(x^3-r^3)}{ (R^3-r^3) }
Substituting this equation into the uniform density solid sphere metric gives the uniform density shell metric as:
<br />
c^2d\tau^2 = \left(\frac{3}{2}\sqrt{1-\frac{2M}{R}}-\frac{1}{2}\sqrt{1-\frac{2M}{x}\frac{(x^3-r^3)}{ (R^3-r^3) }}\right)^2 c^2 dt^2 - \left(1-\frac{2M}{x}\frac{(x^3-r^3)}{ (R^3-r^3) }\right)^{-1}dx^2 -x^2(d\theta^2+\sin(\theta)^2d\phi^2)<br />
Q-reeus said:
In particular the abrupt gradient reversal for dS/dx (or the inverse as I prefer) shown in graphs in #40, 42 seem wholely unphysical. Think of the shell as divided into a number of concentric sub-shells, each separated by a small gap. After traversing the first sub-shell, a tiny test particle has entered a new region of slightly lower gravitational potential but otherwise as for outside the shell proper. Continuing on through successive sub-shells, the cumulative change in potential becomes progressively less, until finally entering the constant potential interior. One must expect smooth transitional curves, tangent at outer and inner radii to the respective values, both for potential and it's related functions of space and time. What physical reason can there be for the abrupt change at the outer radius?
The physical reason for the abrupt change is that you are moving from a vacuum to a region that has a non zero mass density. This is a physical change! There is nothing unusual about this. For example the Newtonian gravitational force behaves in exactly the same way. As you get approach Earth from infinity the force of gravity increases to a peak at the surface, but if you go down a mineshaft into the interior of the Earth, the force of gravity abruptly changes and reduces as you get closer to the centre of the Earth until at the centre of the Earth the force of gravity is zero. Length contraction and the force of gravity only depend on the enclosed mass and ignore all the mass above, while time dilation and gravitational potential take all the mass above and below into account. That is what the equations are telling us.
Q-reeus said:
In searching for someone else who has tackled or at least discussed the shell problem, could only find the following, and yes it does support my view:
http://www.bautforum.com/showthread.php/108079-Shell-Theorem-in-General-relativity?p=1796618" #2 "Ah, the interior of the shell is completely flat spacetime, Minkowski up to the boundary of the shell. The outside of the shell will be Schwarzschild from spatial infinity down to the shell. To the far, outside Schwarzschild observer, clocks inside the shell are ticking slow and radial rulers are short by just the value of the Schwarzschild factor at the shell -- this corresponds to the Newtonian potential difference. But as far as observers inside are concerned, nothing has happened, their clocks and rulers are just fine."
Only quibble here is why he specified radial rulers - being flat spacetime then by definition circumferential rulers contract by just the same. Probably just a slip up in expression.
Becoming a bit of a saga, yes?
Circumferential (or horizontal) rulers cannot length contract anywhere in the cavity, in the shell or outside the shell. The reason is simple. Outside the shell we are certain that circumferential rulers do not length contract. Continuity at the boundary means that circumferential rulers cannot length contract as you cross from the exterior to the material of the shell and as you pass into the cavity. For example if the exterior solution predicts that the circumference of the Earth is 40,000km it would be silly if the interior solution predicted the circumference of the Earth was 6,000,000km. Metrics have to agree at boundaries to avoid contradictions.
Now that we have determined that circumferential rulers cannot length contract inside the cavity, we must conclude that radial rulers cannot length contract inside the the cavity either, if the geometry inside the cavity is to be Euclidean. This is exactly what my model predicts, but it contradicts what publius in the other forum is saying. I would go so far as to say publius is wrong, but his statements are a bit vague. He says "radial rulers are short by just the value of the Schwarzschild factor at the shell". Does he mean the same as the inner surface of the shell or at the outer surface of the shell or some average? If he means the inner surface then he is sort of correct, but spoils it by saying rulers are "short" inside the cavity which is false. Rulers inside the cavity have length contraction ratio of unity (inverted or not) and are exactly the same length as rulers at infinity. Saying rulers are "short" inside the cavity demonstrates that publius does not understand that. To be fair, I did not understand that either earlier in this thread, until I did the actual calculations.
and ... yes
To be fair to passionflower, he may be right about the perceived length increase, but it needed to be specified the distant observer was aligned with both the in-falling object and the shell center. If on the other hand the distant observer, in-falling object and shell center form a right-angle (or near enough allowing for curvature), I would expect the 'optics' to be quite different, and length contraction as per your sqrt(1-2m/r) would apply. It's a matter of specifying things carefully. What bugs me is how to reconcile the Schwarzschild situation near a BH event horizon with the free-falling observer who experiences typically only a brief moment before oblivion at the singularity. How much outside time has passed during this one-way trip? I would think forever, in which case will the BH still be there - for the infaller at any distance past the event horizon in fact? But that's for another thread I guess.
If an observer inside the hollow shell measures radii and circumferences, he will find that Circ < 2*pi*r which is a bit weird in flat space!
