# Redshifting Through Hole in Earth

1. Jan 31, 2014

### 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.

2. Jan 31, 2014

### Staff: Mentor

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.

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.

3. Jan 31, 2014

### 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.

4. Feb 1, 2014

### 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?

5. Feb 1, 2014

### Staff: Mentor

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.

6. Feb 1, 2014

7. Feb 1, 2014

### ClamShell

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
being wrong...has now started researching the
the bathtub with its' rubber-duckies.

Thanks again everyone...then you say,

8. Feb 7, 2014

### 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.

9. Feb 8, 2014

### Staff: Mentor

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.

10. Feb 8, 2014

### 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.

11. Feb 8, 2014

### WannabeNewton

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.

12. Feb 8, 2014

### Staff: Mentor

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.

13. Feb 8, 2014

### Staff: Mentor

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

14. Feb 8, 2014

### 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.

15. Feb 8, 2014

### Staff: Mentor

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.

16. Feb 8, 2014

### ClamShell

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

17. Feb 9, 2014

### Staff: Mentor

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.

18. Feb 9, 2014

### 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?

19. Feb 9, 2014

### 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.

20. Feb 9, 2014

### ClamShell

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.