What caused time to run slower in a field of gravity

[Bjarne]

What cause time to run slower in a field of gravity?

1.) So far I have understood is caused by space deformation ?

2.) According to the dilation equation 2GM is dividen with: c^2 - why ?

http://upload.wikimedia.org/math/1/2/b/12b6af8d31abe31378245988a0e74f66.png"

 

[rrogers]

1) first space-time distortion:) Time doesn't really run slower, but the relative time does. That is, when an outside observe intercepts subsequent null geodisics they are separated by larger time intervals than measured by the subject in the gravitational field.
The inverse experiment (done) is to take a precise oscillator to the top of a tower and watch it from down below. The oscillator frequency appears to be faster than you "know" it is. The physics reasoning would be that the photons increase in momentum/energy as they fall just like anything else. The mathematical explanation is that spreading of space-time manifold as you move away from the source causes geodisics to diverge and separate in terms of proper distance (time).
2) You should have been walked through this in dimensional analysis type of reasoning.
The particular common system you are normalizing to yields c=1. If you walk through the dimensions of G and M you will see that you have to scale GM by (ft/sec)^2/(f1*sec1)^2 and so on. Alternately look at GM/r=energy and identify energy with the time coordinate in [energry,momentum] in the tangent space. Then is you change the time and energy measurement units you are forced to scale GM like so.
Be aware I probably misstated something above; I usually do in off the cuff comments.
As always you don't know if something works until you understand it yourself; there is a lot of BS around.
Ray

 

[A.T.]

What cause time to run slower in a field of gravity?
1.) So far I have understood is caused by space deformation ?

It is spacetime deformation:
http://www.physics.ucla.edu/demoweb..._and_general_relativity/curved_spacetime.html

2.) According to the dilation equation 2GM is dividen with: c^2 - why ?

That is just unit conversion, between time and length units. You can use geometric units where c=1.

 

[meopemuk]

Hi Bjarne,

You may find it interesting that gravitational time dilation can be explained without any reference to space-time distortion/curvature, etc. This explanation requires only classical Newtonian theory of gravity, Einstein's formula E=mc^2 and a bit of quantum mechanics.

Consider a physical system (e.g., a clock) whose behavior depends on time. Quantum mechanics tells us that in this case the system is not in a stationary state. The wave function of the system is a linear superposition of energy eigenstates. To make things simple, assume that our system has only two discrete energy levels E_i and E_f, so that E_i > E_f. The frequency of any time-dependent process in our system is directly proportional to the energy separation between the levels. For example, the frequency of a photon emitted in the transition E_i \to E_f is

\omega = (E_i - E_f)/\hbar...(1)

Next use the formula E=mc^2, which tells that the mass of the system in the excited state m_i = E_i/c^2 is higher than the mass in the ground state m_f = E_f/c^2. Next it follows from the Newtonian gravity theory that attraction to the Earth is stronger in the excited state. The energies of the two states in the gravitational field of the Earth (radius=R, mass=M, gravitational constant=G) are

E_i' = E_i - GMm_i/R = E_i(1-GM/(c^2R))...(2)
E_f' = E_f - GMm_f/R = E_f(1-GM/(c^2R))...(3)

This means that the energy level structure of our clock on the Earth surface becomes uniformly compressed. According to quantum mechanics this should result in universal slowdown of all time-dependent processes in the clock. Indeed, inserting (2) and (3) in formula (1) we obtain that the characteristic frequency of a time-dependent process in our system becomes (1-GM/(c^2R)) lower

\omega' = (E_i' - E_f')/\hbar = \omega(1-GM/(c^2R))

 

[feynmann]

What cause time to run slower in a field of gravity?

1.) So far I have understood is caused by space deformation ?

2.) According to the dilation equation 2GM is dividen with: c^2 - why ?

http://upload.wikimedia.org/math/1/2/b/12b6af8d31abe31378245988a0e74f66.png"

We can learn something from Einstein. Did he ask himself the question what caused time to run slower in a field of gravity? The answer is no, he only asked himself how much slower would time run in a field of gravity. He figured it out using a thought experiment

 

[Fredrik]

Do clocks run slower in a field of gravity? Does that question even make sense? Let's say that it does, and that the answer is yes. I can only interpret that as saying that a clock in a field of gravity runs slower than it would be running if the field of gravity wasn't there. I don't know how to compare those two situations, and I'm not at all convinced that it makes sense in general. (As I was writing this post, I figured out one way to make sense of it in a specific case. See the stuff at the end).

The relevant comparison is between two clocks held at different but constant altitudes in the same gravitational "field". I don't think "gravitational field" is a well-defined concept in GR, so let's be more specific. I'm talking about e.g. two clocks held at different but constant r coordinates in a Schwarzschild spacetime. In this case, the reason why their ticking rates are different is that they are accelerating by different amounts.

The situation is similar to the case of two clocks attached to opposite ends of the same solid rod in SR. Accelerate the rod gently, and it will get shorter in the original rest frame, because of Lorentz contraction (or maybe I should say "because solids behave in a way that's consistent with the Lorentz contraction formula when they are accelerated gently along a geodesic"). This means that the rear is accelerating faster than the front. What a clock measures is the proper time of the curve in spacetime that represents its motion. In an inertial frame, proper time can be expressed as the integral of \sqrt{dt^2-dx^2} along the curve, and because of the greater acceleration at the rear, there will be a greater contribution from dx along that curve. That makes the proper time smaller, so the clock at the rear ticks slower.

Here's a way to make sense of the phrase "a clock ticks slower in a gravitational field", at least in the specific case of a Schwarzschild spacetime. If we go far away from the spherical mass distribution, a Newtonian description of gravity gets more accurate. In particular, the concept of gravitational field makes more sense at greater distances from the mass. In the limit where the distance goes to infinity, the description in terms of the "gravitational field" becomes exact, and in that limit there's no field at all. So what we're really talking about is the relative ticking rates of two clocks at different altitudes, but one of them is at a finite altitude, and the other is infinitely far away.

 

[A.T.]

In this case, the reason why their ticking rates are different is that they are accelerating by different amounts.

I don't think, that different proper acceleration is connected to gravitational time dilation. You can have two clocks with the same proper acceleration, but different clock rates. For example: one in the center of the earth, and one very far away.

 

[Fredrik]

Hm, my argument should at least hold when the clocks are close together, e.g. when they are on different floors of the same building. In this case, they are experiencing different proper accelerations because they are attached to the same solid object (the building).

Not sure about the example with a clock at the center of Earth and one far away. I don't know what the metric is like inside a spherical distribution of mass. My first guess (based only on some very naive spacetime diagrams that I'm drawing in my mind) is that the ticking rates of those two clocks will be the same, but it's really no more than a guess, so I could be completely wrong.

If my argument is completely wrong for clocks at very different altitudes, then I'm back to having absolutely no idea what it means for "clocks to run slower in a gravitational field. (And I don't mean that I have no idea why they do. What I mean is that I wouldn't know what the sentence "clocks run slower in a gravitational field" means).

 

[v2kkim]

... If my argument is completely wrong for clocks at very different altitudes, then I'm back to having absolutely no idea what it means for "clocks to run slower in a gravitational field. (And I don't mean that I have no idea why they do. What I mean is that I wouldn't know what the sentence "clocks run slower in a gravitational field" means).

One situation to experience the slow clock due to gravitation may be as follow:
Suppose you travel from far away to a very heavy star. After reaching the star surface stay there for a year in your hand carrying clock, and came back. Another guy makes a similar travel but visits very light mass star. Now after the travel they found a time difference, say delta_T, then we can conclude this time difference comes from the time dilation under gravity.

 

[neopolitan]

Hm, my argument should at least hold when the clocks are close together, e.g. when they are on different floors of the same building. In this case, they are experiencing different proper accelerations because they are attached to the same solid object (the building).

Not sure about the example with a clock at the center of Earth and one far away. I don't know what the metric is like inside a spherical distribution of mass. My first guess (based only on some very naive spacetime diagrams that I'm drawing in my mind) is that the ticking rates of those two clocks will be the same, but it's really no more than a guess, so I could be completely wrong.

If my argument is completely wrong for clocks at very different altitudes, then I'm back to having absolutely no idea what it means for "clocks to run slower in a gravitational field. (And I don't mean that I have no idea why they do. What I mean is that I wouldn't know what the sentence "clocks run slower in a gravitational field" means).

Don't clocks on GPS satellites have to be modified for 1) orbital velocity and 2) the difference between the GR effect on clocks due to gravity on the satellite and the Earth receiver (nominally at sea level)?

One effect will make the satellite clock slower (orbital velocity) and the other will make the Earth receiver clock run slower (gravity is greater since r is lower). I don't know how much the correction is, but it seems like a potential accuracy error that could be sensibly removed.

Any satellite engineers out there who can testify?

cheers,

neopolitan

Found a link: http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps.html

 

[Al68]

The situation is similar to the case of two clocks attached to opposite ends of the same solid rod in SR.

I'd say the word "similar" is an understatement. The word equivalent comes to mind. After all, this is in principle the scenario used by Einstein to derive gravitational time dilation from SR to begin with.

 

[Bjarne]

Don't clocks on GPS satellites have to be modified for 1) orbital velocity and 2) the difference between the GR effect on clocks due to gravity on the satellite and the Earth receiver (nominally at sea level)?

One effect will make the satellite clock slower (orbital velocity) and the other will make the Earth receiver clock run slower (gravity is greater since r is lower). I don't know how much the correction is, but it seems like a potential accuracy error that could be sensibly removed.Any satellite engineers out there who can testify?

That's right.

http://en.wikipedia.org/wiki/Global_Positioning_System"

quotation:
Clocks in GPS orbital altitudes will tick more rapidly, by about 45.9 microseconds (μs) per day, because they have a higher gravitational potential than atomic clocks on Earth's surface.

Special relativity predicts that atomic clocks moving at GPS orbital speeds will tick more slowly than stationary ground clocks by about 7.2 μs per day.

When combined, the discrepancy is about 38 microseconds per day

I begin to get a bit confused according to what causes the GR dilation difference ?
I mean is it a simple way to understand the cause ot the dilation?
Is it due to that distances "out there" are relative larger than at the surface of the earth?

 

[Fredrik]

A clock measures the proper time of the curve in spacetime that represents its motion, so the difference is always due to the world lines having different proper times.

In some special cases, such as two clocks on different floors of the same building, the difference can be described in other terms. (That particular case is equivalent to the accelerating solid rod in SR).

The equivalence with the accelerating solid rod scenario in SR can also explain the difference between the ticking rates of a clock on the ground and a clock held stationary at the altitude of a GPS satellite, at least approximately, but the approximation may not be very good in this case, since the clocks are not as close to each other. (The equivalence principle is only exact in the limit where the region of spacetime that's being considered goes to zero).

When the clock is in orbit, things get more tricky. We could try to use the equivalence with the accelerating solid rod in SR to find the ticking rate of a clock held stationary at the correct altitude, and then adjust that result by using the standard time dilation formula from SR with the orbital speed of the satellite plugged in. I'm guessing that this would give us a pretty accurate result actually, it would be even more accurate to just calculate the proper times along the relevant curves in a Schwarzschild spacetime.

 

[A.T.]

Not sure about the example with a clock at the center of Earth and one far away. I don't know what the metric is like inside a spherical distribution of mass. My first guess (based only on some very naive spacetime diagrams that I'm drawing in my mind) is that the ticking rates of those two clocks will be the same, but it's really no more than a guess, so I could be completely wrong.

The clock in the center of the Earth is the slowest one, with the biggest rate difference to the clock far away. But you don't need such extreme examples. For every radial coordinate outside of the earth, there is a corresponding radial coordinate inside the earth, with the same proper acceleration but slower clock rate.

What I mean is that I wouldn't know what the sentence "clocks run slower in a gravitational field" means).

Well "slower" is a relative term. It should be "... slower than far away from the mass" or "clocks run at different rates in a gravitational field"

 

[Fredrik]

The clock in the center of the Earth is the slowest one, with the biggest rate difference to the clock far away. But you don't need such extreme examples. For every radial coordinate outside of the earth, there is a corresponding radial coordinate inside the earth, with the same proper acceleration but slower clock rate.

I can't rule that out since I don't know the metric on the inside, but it sounds pretty strange to me. For example, if you're right, then a clock held 1 m from the center of the Earth is ticking faster than one at the center, even though the one at the center is the only one that's doing geodesic motion (maximizing proper time).

 

[feynmann]

The clock in the center of the Earth is the slowest one, with the biggest rate difference to the clock far away. But you don't need such extreme examples. For every radial coordinate outside of the earth, there is a corresponding radial coordinate inside the earth, with the same proper acceleration but slower clock rate.

Supposed there is a cavity at the center of the earth, Would a clock held 1 m from the center of the Earth is ticking faster than one at the center?

 

[A.T.]

I can't rule that out since I don't know the metric on the inside

Only in German, sorry:
http://de.wikipedia.org/wiki/Schwarzschild-Metrik#Innere_L.C3.B6sung

For example, if you're right, then a clock held 1 m from the center of the Earth is ticking faster than one at the center, even though the one at the center is the only one that's doing geodesic motion (maximizing proper time).

That rule doesn't hold in curved space time.

Supposed there is a cavity at the center of the earth, Would a clock held 1 m from the center of the Earth is ticking faster than one at the center?

No. Both would tick at the same rate, slower than a clock outside of the empty shell.

 

[stevebd1]

Schwarzschild interior solution-

https://www.physicsforums.com/showpost.php?p=1543402&postcount=8

 

[rrogers]

Supposed there is a cavity at the center of the earth, Would a clock held 1 m from the center of the Earth is ticking faster than one at the center?

A crude method is to compute the Energy difference between the different positions. Similiar to what was done in the QM example.
The proper method in all cases is to integrate the "proper" distance as mentioned. In GR this would always be the proper time elapsed; this can also included null geodesics (which contribute nothing). The integration path doesn't need to be a geodesic. Typically intuitive type of reasoning can be fooled since we have little experience in these realms. GR is mathematically consistent, so we need to bring our intuition into line with what the mathematics says. Mathematical consistency says nothing about whether it matches reality though.

Ray

 

[rrogers]

Schwarzschild interior solution-

https://www.physicsforums.com/showpost.php?p=1543402&postcount=8

Looks more or less okay, but are you sure about M/R ? I don't see the density term and I think you need it. For instance if M is constant then your last expression blows up.
Presuming that your original expression is right then the density should be the constant; OTOH constant density is a stretch for the earth. Although not for a meteor.

Ray

 

[stevebd1]

In the metric M is geometric (M=Gm/c^2), 2M being the schwarzschild radius of the object of mass and R being the overall radius of the object of mass. Based on this, as r tends to zero, R remains constant and none of the expressions appear to blow up. Which expression are you referring to?

 

[rrogers]

Okay, I will reexamine the result. It still seems strange to me that the expression uses Schwarzschild terms instead of density. The longer I think about the stranger it seems; probably a shortcoming on my part.


Ray

 

[K.J.Healey]

Okay, I will reexamine the result. It still seems strange to me that the expression uses Schwarzschild terms instead of density. The longer I think about the stranger it seems; probably a shortcoming on my part.


Ray

It comes from making the boundary conditions at the surface; going to the exterior solution (Pressure at surface goes to zero). By using the boundary conditions all of the unknwons will be solved for in terms of M enclosed and R boundary. (given the EOS)

 

[rrogers]

I presume you are using Florides solution. I will have to wait until I move into my house and have a real computer to do a real analysis. This little laptop running Vista is really trying.
Ray