Redshifting Through Hole in Earth

[ClamShell]

I have been wondering what to expect if there was a hole,
from pole to pole, that inside which we could measure the
Einstein red and blue shifts due to gravity. At first, I suspect
a blue-shift as light travels from the surface of the Earth to
the center; just less and less as the center is approached.

But, and here is where I become ashamed even to have
imagined...isn't this the same as light going from large
gravity to smaller gravity? A red-shift in other words, like
a photon leaving a massive body?

All I believe for sure is that a photon would be restored
to its original wavelength if it was measured at the poles.

 

[PeterDonis]

If there was a long tunnel to the center of the Earth, and one observer, B, was at the bottom (i.e., at the center of the Earth) and another, T, was at the top (i.e., at the surface of the Earth), then light from B would be redshifted when observed by T, and light from T would be blueshifted when observed by B.

isn't this the same as light going from large
gravity to smaller gravity?

No; "gravity" with respect to redshift means gravitational potential, not the "acceleration due to gravity". The "acceleration due to gravity" is zero at the center of the Earth, but the center is still deeper in the Earth's potential well than the surface is.

 

[WannabeNewton]

Just to add on, consider first the general case of a stationary gravitational field. This is a gravitational field that's independent of time so we can write the metric as $ds^2 = g_{00}dt^2 + g_{0i}dt dx^i + g_{ij}dx^i dx^j$. An observer at rest at some location $R$ in this gravitational field has a clock that ticks at the rate $d\tau = \sqrt{-g_{00}|_R}dt$. Then two clocks at rest in this gravitational field at respective locations $R_1$ and $R_2$ differ in their tick rates by $\frac{d\tau_2}{d\tau_1} = \frac{\sqrt{-g_{00}|_{R_2}}}{\sqrt{-g_{00}|_{R_1}}}$. Then the gravitational redshift of frequency of a light wave emitted at $R_1$ by an observer at rest there and received at $R_2$ by an observer at rest here is given by $\frac{\nu_2}{\nu_1} = \frac{\sqrt{-g_{00}|_{R_1}}}{\sqrt{-g_{00}|_{R_2}}}$ (we measure the frequencies by using the periodic ticking of the respective clocks).

In the Newtonian limit, $g_{00} = -(1 + 2\Phi)$ where $\Phi$ is the gravitational potential. Let's model the Earth as a sphere of uniform mass density and radius $a$. Then a basic application of Gauss's law gives $\Phi(r) = -GM\frac{3a^2 - r^2}{2a^3}$ for $r \leq a$ i.e. inside the Earth so we have $\sqrt{-g_{00}} = \sqrt{1 -\frac{3}{2}\frac{GM}{a}(1 - \frac{r^2}{3a^2})}$. If we let $R_1 = 0$ correspond to an observer hovering at the center of the Earth and $R_2 = a$ correspond to an observer standing on the surface of the Earth then $\frac{\nu_2}{\nu_1} = \frac{\sqrt{1 -\frac{3}{2}\frac{GM}{a}}}{\sqrt{1 - \frac{GM}{a}}} < 1$ so a light ray emitted by the observer at the center will be redshifted when received by the observer at the surface. Conversely, a light ray emitted by the observer at the surface will be blueshifted when received by the observer at the center as Peter stated.

 

[ClamShell]

Well, thankyou very much...my right-brain has problems with
mathematics, but ole lefty can usually figure out the direction
of change pretty well. Right-brain still is skeptical though;
thinks there is "similarity" between going from a location
where clocks run slow(Earth's surface), to a location where
clocks run faster(Earth's center OR deep space). Have we
ever done this experiment with holes in the ground instead
of towers and satellites? IE, am I accepting an untested
conclusion even if the conclusion is correct?

 

[Nugatory]

Well, thankyou very much...my right-brain has problems with
mathematics, but ole lefty can usually figure out the direction
of change pretty well. Right-brain still is skeptical though;
thinks there is "similarity" between going from a location
where clocks run slow(Earth's surface), to a location where
clocks run faster(Earth's center OR deep space). Have we
ever done this experiment with holes in the ground instead
of towers and satellites? IE, am I accepting an untested
conclusion even if the conclusion is correct?

Ahhh... Why does that right-brain believe that clocks run faster at the center of the Earth than at the surface of the earth? (always doing our surface work at one of the poles, as in your original post).

You might be able to placate that right-brain by asking it to think about how much work it takes to move from one location to another, instead of just how strong gravity is at a particular location. It takes more work to lift a weight from the center of the Earth to deep space than it does to lift a weight from the surface of the Earth to deep space.

We haven't done the experiment with holes drilled in the ground for that specific purpose, but the GPS system tests General Relativity from underground every day. The GPS system assigns times to points on the surface of the Earth as if it were a smooth surface, but of course the Earth isn't smooth - it's lumpy with mountains and valleys, and anything at the bottom of a valley is effectively at the bottom of a hole.

 

[WannabeNewton]

IE, am I accepting an untested
conclusion even if the conclusion is correct?

Thankfully no. See here: http://en.wikipedia.org/wiki/Pound–Rebka_experiment

 

[ClamShell]

Started again thinking about this concept when
Righty read a short paper by Stephen Hawking
claiming that he no longer thinks event horizons
exist. Righty immediately informed Lefty that
the mathematical singularity at 2GM/c^2 should
extend closer to the black-hole now; which has
to mess-up the mathematics somehow. Righty
also "reminded" that clocks run "faster" at
the center of the Earth and has "reminded"
of that "fact" over-and-over-again until Lefty
actually started to believe it as "essentially
obvious." So we came back here to get some
real facts so that Righty would shut-up.

Righty always wanted the "well" to bottom-out
at a zero of some sort, but instead of admitting
being wrong...has now started researching the
"best" material to under-water-basket-weave in
the bathtub with its' rubber-duckies.

Thanks again everyone...then you say,
"glad to be of help."

 

[ClamShell]

I've pondered the posts by PeterDonis and WannaBeNewton
and am proud to say that I totally believe that a light
beam transmitted from the center of the Earth would be
redshifted when received at the surface. But now I
wonder if the radial distribution of matter density
would have any affect on the total redshift. I'm
thinking that the gravitational potential at the center
would be the same if the total mass of the Earth was the
same for any possible radial distribution of matter;
including the case where the density would increase
from the center to the surface.

 

[PeterDonis]

I wonder if the radial distribution of matter density would have any affect on the total redshift.

If the total mass is the same, the density distribution won't affect the total redshift from the center to the surface. But it will affect the details of how the redshift changes as you go from the center to the surface. The redshift at any radius from the surface depends on the total mass inside that radius; changing the density distribution will change the total mass inside a given radius in the Earth's interior.

 

[ClamShell]

And what about if the Earth was a hollow spherical shell?
Would all clocks float inside and all run the same speed
identical with the speed they would have run at the center
of a real Earth? I'm thinking yes.

 

[WannabeNewton]

Would all clocks float inside and all run the same speed
identical with the speed they would have run at the center
of a real Earth? I'm thinking yes.

Yes for the most part however I'm unsure about whether the tick rate would be identical to that of a clock placed at the center of a solid sphere of uniform density-I'll have to check and get back to you on that. Also, the clocks will run at different rates when compared to clocks outside the shell.

 

[PeterDonis]

If the total mass is the same, the density distribution won't affect the total redshift from the center to the surface.

Actually, on thinking this over, I'm no longer sure it's true. The redshift factor depends on the pressure as well as the density, and changing the density distribution will change the pressure distribution as well. It's not clear to me from looking at the equations (which are in this post on my PF blog) that the changes will always work out to keep the redshift factor at the center the same.

 

[PeterDonis]

And what about if the Earth was a hollow spherical shell?

Same comment here as in the post I just made on the solid sphere case.

 

[ClamShell]

Yes, now we're making some progress. If the hollow sphere contained
a gas, there might be enough pressure at the center to liquefy or
even solidify the gas.

 

[PeterDonis]

Yes, now we're making some progress. If the hollow sphere contained a gas, there might be enough pressure at the center to liquefy or
even solidify the gas.

If the hollow sphere contains a gas, it's not vacuum inside, so spacetime inside will not be flat. For spacetime inside the hollow sphere to be flat (i.e., for clocks everywhere inside to go at the same rate), it has to be vacuum inside.

 

[ClamShell]

If the hollow sphere contains a gas, it's not vacuum inside, so spacetime inside will not be flat. For spacetime inside the hollow sphere to be flat (i.e., for clocks everywhere inside to go at the same rate), it has to be vacuum inside.

And what if there was a gas inside? Would it become a Bose condensate...just sayin'?

 

[PeterDonis]

And what if there was a gas inside? Would it become a Bose condensate...just sayin'?

It would depend on the details. Off the top of my head, I would say that if the spherical shell is exerting enough inward pressure on the gas inside to turn it into a Bose-Einstein condensate, the shell probably can't support itself against its own weight and it will implode. In fact, even if the shell exerts enough pressure to liquefy the gas, it probably can't support itself against its own weight, because if the shell is exerting inward pressure on the gas inside at all, that means it's not in equilibrium. But I would have to look at the detailed math to be sure.

 

[ClamShell]

In the case of a vacuum, is the rate of ticking of clocks
inside of the shell the same as it would have been for
a clock ticking at the center of real Earth? This would
be both a fascinating and elegant result...wouldn't you say?

 

[WannabeNewton]

Even if the Earth is modeled as a solid sphere of uniform density that would not be true. For a thin spherical shell, clocks at rest inside of the shell tick at the same rate as a clock at rest on the shell itself because the gravitational potential is everywhere constant inside of the shell, fixed by the gravitational potential of the shell itself. For a solid sphere of uniform density in the weak field limit on the other hand, a clock at rest at the center ticks at a different rate than a clock at rest on the surface, as shown in post #3.

 

[ClamShell]

Even if the Earth is modeled as a solid sphere of uniform density that would not be true. For a thin spherical shell, clocks at rest inside of the shell tick at the same rate as a clock at rest on the shell itself because the gravitational potential is everywhere constant inside of the shell, fixed by the gravitational potential of the shell itself. For a solid sphere of uniform density in the weak field limit on the other hand, a clock at rest at the center ticks at a different rate than a clock at rest on the surface, as shown in post #3.

I feel that "fascinating and elegant" just got thrown out with the
wash-water. Do you mean that the potential well for a hollow-
shell , the same mass as the Earth, and the same outer surface
radius, would be shallower than the potential well of a solid
Earth? That dense thin shells don't redshift as much as less-
dense thick shells do? I guess I was just hoping that the work
to go through a dense thin shell would be equivalent to the work
to go through a less-dense thick shell. I'm thinking I need some
"hand-holding" again.

 

[PeterDonis]

Do you mean that the potential well for a hollow-
shell , the same mass as the Earth, and the same outer surface
radius, would be shallower than the potential well of a solid
Earth?

The potential at the outer surface (of the shell or the solid Earth) will be the same because it only depends on the total mass and the outer surface radius. But once you go below the surface, the potential depends on the details of the density and pressure inside; change those details and you change the potential. In other words, the potential at the outer surface is not sufficient to determine the potential everywhere inside, which is what you appear to be assuming.

 

[ClamShell]

Then the redshift from the center to the surface IS affected by the radial distribution
of mass density. Right?

 

[PeterDonis]

Then the redshift from the center to the surface IS affected by the radial distribution of mass density. Right?

Yes, and the radial distribution of pressure as well. (If we assume everything stays spherically symmetric, then the density and pressure can only be functions of radius.)

 

[ClamShell]

Even if the Earth is modeled as a solid sphere of uniform density that would not be true. For a thin spherical shell, clocks at rest inside of the shell tick at the same rate as a clock at rest on the shell itself because the gravitational potential is everywhere constant inside of the shell, fixed by the gravitational potential of the shell itself. For a solid sphere of uniform density in the weak field limit on the other hand, a clock at rest at the center ticks at a different rate than a clock at rest on the surface, as shown in post #3.

Does this mean that at the limit of thinness of a shell,
there would be no redshift? Total work is a summing
of force times distance, so if the distance is zero,
wouldn't the work be zero as well? The well would be
zero as well as the well of the shell? Well, what do you say?

 

[PeterDonis]

Does this mean that at the limit of thinness of a shell,
there would be no redshift?

Not if the shell has mass; an infinitely thin shell with finite mass would get modeled as a spherical shell of zero thickness and "infinite" density, which is not physically realistic, but it would still have an effect on the redshift--there would be a discontinuous "jump" in the redshift factor at the shell.

Total work is a summing of force times distance

Only if the "force" is finite. For an infinitely thin shell with finite mass, the "force" required to go from inside to outside the shell is "infinite". What that really means is that a finite amount of work is required to go from inside to outside the shell even though the shell is of zero thickness. As I said above, that's not physically realistic, but that's what you get when you make physically unrealistic assumptions like a shell of zero thickness but with finite mass.

Of course if the shell is of zero thickness because it has zero mass, that's different: then the shell simply isn't there at all and of course it won't have any effect on the redshift.

 

[pervect]

Only if the "force" is finite. For an infinitely thin shell with finite mass, the "force" required to go from inside to outside the shell is "infinite".

Are you sure? As I recall, the Newtonian limit of a flat sheet with a surface density $rho_s$ is finite - the easiest way to calculate it is Gauss' law (for an infinite sheet). The Newtonian gravity field winds up perpendicular to the sheet. and the total flux escaping at the box (at top and bottom) is proportional to the finite mas enclosed in the box.

Offhand, I don't see that the relativistic case should have an infite force.

 

[PeterDonis]

the Newtonian limit of a flat sheet with a surface density $rho_s$ is finite

A flat sheet is different from a spherical shell; going from one side of the flat sheet to the other takes zero work, but going from the inside of the shell to the outside takes finite work.

Also, I'm not saying anything measurable is infinite; see below.

I don't see that the relativistic case should have an infite force.

I put "force" in scare-quotes because you can't actually measure it. What you can measure is the work done in going from the inner to the outer surface of the shell. That work is finite; but if you define "force" in terms of work done as "work = force times distance", then since "distance" is zero in the case of a shell of zero thickness (but finite mass), "force" must be infinite to give a finite product. Really what this is saying is that the "force" can't be defined this way in the case of a zero thickness shell; but that's just another way of saying that a zero thickness shell with finite mass is unphysical.

 

[ClamShell]

Isn't an "event horizon"(scare-quotes) something like an "infinitely" thin
shell with "zero" mass at R = 2GM/c^2, that requires an infinite amount
of work to escape across? I'm trying to "jury-rig" why Hawking should
now dismiss Schwarzschild's analysis. I'm next expecting Hawking to
tell us that black holes don't gobble each other up.

 

[PeterDonis]

Isn't an "event horizon"(scare-quotes) something like an "infinitely" thin shell with "zero" mass at R = 2GM/c^2, that requires an infinite amount of work to escape across?

No. First of all, I said the "force" was infinite for an infinitely thin shell (if the shell has *nonzero* mass, btw, not zero mass--a zero mass shell doesn't exist, as I also said), but the work done to go from inside to outside the shell is finite.

Second, the reason you can't go from inside an event horizon to outside it has nothing to do with requiring an infinite amount of "work"; it has to do with not being able to go faster than light. The event horizon is an outgoing lightlike surface; you can't go from inside it to outside for the same reason that you can't catch a light ray that's moving in the same direction as you and starts out ahead of you. The thin shell in your scenario is not a lightlike surface; it's an ordinary spherical surface, and the "world-sheet" it generates in spacetime is timelike (more precisely, its third dimension, other than the 2 dimensions of the spherical surface, is timelike), not lightlike.

I'm trying to "jury-rig" why Hawking should now dismiss Schwarzschild's analysis.

When did he do that? Schwarzschild's analysis was of a classical black hole; Hawking's analysis is of a quantum black hole. Nothing in Hawking's analysis (or any of the other work that has been done on how quantum effects change the behavior of black holes) invalidates any of Schwarzschild's analysis; in fact, all of the quantum treatments of black holes assume that Schwarzschild's analysis gives the correct classical behavior of black holes.

 

[ClamShell]

First of all, I am dissapointed that mathematically moving
atoms and molecules from the center to the surface of the
Earth DOESN'T yeild a hollow space where floating clocks
tick at the same rate as they would have ticked at the
center of a solid Earth. I'm thinking that what you have
taught me is that, as more and more atoms and molecules
are moved to the surface, the redshift of a photon
traveling from the center to the surface would lessen as
more and more matter was moved. Can you show me this?
Is there a limiting redshift approached asymtotically?
Is this limiting redshift not zero? Is there an assumption
we could make, like inclusion of some cosmological
constant, that might recover a constant redshift for
any radial distribution of matter?

Secondly, am I being too skeptical in the judging of
theories that have inconsistant assumptions and/or
predictions? I'm thinking that bad predictions mean
bad theories, and that Hawking has falsified
Schwarzschild's prediction.