I finally figured out how to ask my question

[PaulRacer]

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

 

[Dale]

It would run slower because there is a gravitational redshift.

 

[PaulRacer]

It would run slower because there is a gravitational redshift.

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?

 

[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.

 

[Dale]

The gravitational effect is much larger than the velocity effect.

 

[PaulRacer]

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

 

[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.

 

[russ_watters]

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?

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

 

[russ_watters]

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.

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.

 

[russ_watters]

It would run slower because there is a gravitational redshift.

Redshift or time dilation?

 

[sylas]

Redshift or time dilation?

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

 

[PaulRacer]

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

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?

 

[sylas]

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?

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 anyone 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

 

[PaulRacer]

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 anyone 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

Time dilation plus relative movement is what I am saying.

 

[PaulRacer]

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

 

[sylas]

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

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.

 

[George Jones]

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.

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

https://www.physicsforums.com/showthread.php?p=1543402#post1543402.

 

[atyy]

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

https://www.physicsforums.com/showthread.php?p=1543402#post1543402.

There is also a little bit about this in section 5 of Ashby's http://relativity.livingreviews.org/Articles/lrr-2003-1/ :"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. "

 

[Dale]

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?

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

I am not sure what you are referring 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.

 

[Austin0]

Redshift or time dilation?

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

 

[Dale]

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 received 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.

 

[atyy]

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

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.

 

[Austin0]

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 received 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.

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 I am my mind.
Thanks again

 

[Austin0]

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.

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

 

[atyy]

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

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.

 

[Dale]

it would seem to mean the signal itself was initially emitted at .99MHz and did not change during transit.

I am not sure what you mean by this. Let's say that you have an array of identical receivers at different altitudes. A receiver at the same altitude of the transmitter would receive a 1 MHz signal and receiver at progressively higher altitudes would receive progressively lower frequencies. This is what is observed experimentally, but I don't know if this is what you are describing.

 

[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.

I cheated. Here there is no such thing as the frequency of a photon. To assign photons frequencies, we have a photon stream shoot out from one atomic clock at some proper time interval of that clock, and define the inverse of that interval to be the frequency of the photon stream at the shooting clock. The frequency of the photon stream at the receiving clock is similarly defined.

More correctly, one should write the Einstein field equations, and the equation of state for the electromagnetic field (Maxwell's equations), and the equation of state for the atomic clock fields - then solve them together with some initial/boundary conditions that correspond to the particular physical situation you have in mind. In this formalism, there are no photons, but there are electromagnetic waves, and we can assign them frequencies. The cheated version is the ray limit of this formalism (at least in some cases - I don't think anyone has written an equation of state for the atomic clock fields).

 

[PaulRacer]

I am not sure what you are referring 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.

Twisting due to rotation and time rate differences within a single object. How could you view a clock moving at a conflicting rate in the center and not have a "twist". Whether time dilation is caused by gravitational redshift or angular velocities doesn't really matter.
Thanks for the replies.

 

[PaulRacer]

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

https://www.physicsforums.com/showthread.php?p=1543402#post1543402.

Thanks George, but your "low level of mathematical rigor" is still way above my highest level of mathematical understanding.

 

[Dale]

How could you view a clock moving at a conflicting rate in the center and not have a "twist".

It sounds like you think there should be some twist, even in the case of a non-rotating planet, purely due to the gravitational time dilation. This is not correct, any twist is entirely due to the rotation of the planet and would occur even for a hollow (no significant mass) rotating planet.

 

[PaulRacer]

It sounds like you think there should be some twist, even in the case of a non-rotating planet, purely due to the gravitational time dilation. This is not correct, any twist is entirely due to the rotation of the planet and would occur even for a hollow (no significant mass) rotating planet.

Without rotation, obviously there would be no twist. My whole question has to do with the rotation plus time dilation.
Thanks.

 

[Dale]

Then the twisting is entirely due to fictitious forces that are not purely radial (e.g. coriolis force).

 

[PaulRacer]

No, I see what you are getting at though. Let me put it this way, if you had a string tied between two people, One person remains stationary but rotating in the center and one walks around the person in the center with the string taught doing laps. The person in the center makes one rotation every x seconds the person on the circumference makes one lap every x seconds. If their watches were out of sync the string would eventually wrap itself around the center person.

 

[Dale]

That is not how it works. The person at the center of the Earth will measure the length of a day to be slightly less than the person at the surface of the earth.

 

[PaulRacer]

Ok, so you are saying that if I was traveling in a car at .5c and I started counting off mile markers to calculate speed compared to my watch that I would conclude that the speedometer was wrong even though it is correct. Thanks Dale!