I finally figured out how to ask my question

1. Sep 2, 2009

PaulRacer

What would a clock do at the center of the earth compared to a clock on the surface and why?

2. Sep 2, 2009

Staff: Mentor

It would run slower because there is a gravitational redshift.

3. Sep 2, 2009

PaulRacer

Thanks Dale. So if you drilled a hole to observe it, wouldn't that hole at some point become twisted sort of like a yoyo string?

4. Sep 2, 2009

PaulRacer

What about time dilation on the surface due to angular velocities within the spinning sphere? (I think I got my terminology right) Does the gravitational redshift make the entire object time neutral? Sorry, I meant to ask this first.

5. Sep 2, 2009

Staff: Mentor

The gravitational effect is much larger than the velocity effect.

6. Sep 2, 2009

PaulRacer

What is the answer to the hole question? Is the earth wound like a clockspring inside?

7. Sep 2, 2009

v2kkim

Something is going strange in this dialog. I think the gravitation effect is stronger on the earth surface and minimum at the center of the earth. Therefore the surface clock is slower than center one.

8. Sep 2, 2009

Staff: Mentor

No. Every part of this setup is stationary wrt each other.

9. Sep 2, 2009

Staff: Mentor

It isn't. What is strongest at the surface is the force one feels. The gravitational potential (the strength of the field) is strongest at the center. The reason no force is felt at the center is that every particle in your body is being pulled in all directions at once.

10. Sep 2, 2009

Staff: Mentor

Redshift or time dilation?

11. Sep 2, 2009

sylas

You can use the redshift to measure time dilation, and as a thought experiment to verify that there is a time dilation effect.

Imagine source of electromagnetic radiation at the center of the Earth. The signal measured at the surface will have a redshift, corresponding to the change in gravitational potential. Therefore the clock at the radiation source is running slow with respect to the clock at the surface.

Cheers -- sylas

12. Sep 2, 2009

PaulRacer

How is that possible with the object spinning? If the spinning is calculated in distance/sec., and the seconds are longer in the center......? Let's say the center clock is slower by 1 second per x amount of years, then it should take the center x years times 86400 to be one rotation behind the surface right?

13. Sep 2, 2009

sylas

No. As Russ says, every part is stationary with respect to every other part. They ALL spin and in a way that there's no twist introduced. As measured from any one point, all parts spin at the same rate.

However, the rate of spin as measured from different locations is different.

The proper inference of your thought experiment is that the rate of spin in radians per second is greater when measured from the middle than when measured from the surface. But no matter where you make the measurement, that spin the same at all levels. There's no "twist", because the difference is from time dilation, not from any relative movement.

Cheers -- sylas

14. Sep 2, 2009

PaulRacer

Time dilation plus relative movement is what I am saying.

15. Sep 2, 2009

PaulRacer

Okay, if what you are saying is true, then is it possible for it to be percieved as twisted but actually not be physically?

16. Sep 2, 2009

sylas

I don't think so; I can't see what you mean by a twist. You can see moving particles curving to one side if they are moving in freefall, if that is what you mean; but that doesn't apply for the Earth itself, which is not moving in its own rotating frame.

17. Sep 2, 2009

George Jones

Staff Emeritus
The only way to find out is to do some sort of a calculation. For example, a calculation might show that, as DaleSpam states, one effect dominates, or a calculation that includes both effects, as I have done (using a low level of mathematical rigor) in

18. Sep 2, 2009

atyy

There is also a little bit about this in section 5 of Ashby's http://relativity.livingreviews.org/Articles/lrr-2003-1/ [Broken] :"Figure 2 shows the net fractional frequency offset of an atomic clock in a circular orbit, which is essentially the left side of Eq. (35) plotted as a function of orbit radius a, with a change of sign. Five sources of relativistic effects contribute in Figure 2. The effects are emphasized for several different orbit radii of particular interest. For a low earth orbiter such as the Space Shuttle, the velocity is so great that slowing due to time dilation is the dominant effect, while for a GPS satellite clock, the gravitational blueshift is greater. The effects cancel at 9545 km. The Global Navigation Satellite System GALILEO, which is currently being designed under the auspices of the European Space Agency, will have orbital radii of approximately 30,000 km. "

Last edited by a moderator: May 4, 2017
19. Sep 2, 2009

Staff: Mentor

I am not sure what you are refering to here. Do you mean a twisting due to gravity or a twisting due to the rotation? Any twisting will be due to the rotation rather than the gravity.

20. Sep 3, 2009

Austin0

Hi perhaps you can clear up a question I have regarding the difference [if any?] between redshift and time dilation in this situation.

As per Sylas's scenario: there is an EM source at the center . an electron (1) with a low resonant frequency.
At the surface that photon is absorbed by an electron with a lower frequency than an electron on the surface comparable to electron (1)
By comparison the photon is considered red shifted .

But this scenario could be explained two ways.

The electron frequencies at the center were dilated by location and the photon was emited at that frequency and was unchanged by the translation to the surface.
No redshift. Only Time dilation affecting the electrons.

The photon itself was lowered in frequency through transit ,up the well so to speak.
Redshift.

Both of the above.

I hope you can shed some light on this question.
Thanks

21. Sep 3, 2009

Staff: Mentor

Gravitational redshift implies gravitational time dilation. Consider a transmitter that transmits a perfect 1 MHz sinusoidal wave. That is a clock. If the signal goes up a gravitational potential it may be recieved at .99 MHz due to gravitational redshift. Unlike in the normal Doppler case the time between transmission and reception is not changing, so there are no transmission delay effects to correct for wrt the frequency. This implies that the frequency of the transmitter is lower and its time must therefore be dilated.

22. Sep 3, 2009

atyy

See Cliiford Will's comments on p49/50 of http://books.google.com/books?id=9ZuP9JQzc00C&dq=clifford+will+einstein&source=gbs_navlinks_s . Basically, the distinction is not a concept with an operational meaning.

23. Sep 3, 2009

Austin0

If I am reading this right you mean the transmitter would put out a 1 MHz signal if located at the potential altitude of the receiver but at the transmission potential location it put out a .99MHz signal???
If this is correct then it is exactly what I thought. ANd in that case it would seem to mean the signal itself was initially emitted at .99MHz and did not change during transit.
That the effect is totally due to the relative dilation of the transmitter and receiver.
Is this correct?
Then maybe the question is only one of semantics. To me the term gravitational redshift implies that gravity had a direct effect on the photons. I have no problem with this concept per se but I do want to get it clear im my mind.
Thanks again

24. Sep 3, 2009

Austin0

Hi thanks for the link, although I tried it and didnt get through.

I understand that it does not have an operational meaning. It is a phenomenon and practically speaking the reality behind it is not relevant to how we deal with it.
But conceptually I think it might have some relevance regarding understanding gravity and its interaction with photons.
Thanks for the input

25. Sep 3, 2009

atyy

I meant that there is no conceptual relevance because a concept is defined by an experiment. The underlying concepts are (i) metric which defines null, timelike, spacelike geodesics (ii) atomic clocks tick proper time (iii) photons follow null geodesics.