Do Gravitating Bodies Warp the Fabric of Space?

[Passionflower]

Not proper acceleration, that can only have one value, the one measured by an accelerometer. An accelerometer at the center of the Earth reads 0, and accelerometer at the surface of the Earth reads 9.8 m/s² upwards. A.T.'s counterexample is correct, gravitational time dilation is not due to differences in proper acceleration the way you suggest.

Really? So when a ball explodes what single value is the proper acceleration of parts of the shell?
A clock is not a point mass. Are you suggesting there are no stresses on an object placed in the core of a massive object like the earth?

An accelerometer at the center of the Earth reads 0

How do you know?
It is likely an accelerometer could not even operate properly at the center of the earth. By the way the fact that an accelerometer only measures an acceleration in one direction is not a limitation of nature but a limitation of the device.

The uniform field is also a good counter example. Suppose you have an ideal gravitational field where the proper acceleration of a stationary particle is everywhere constant. In such a field a light pulse going "up" would be gravitationally red-shifted and therefore there would be gravitational time dilation despite the fact that the proper acceleration is constant.

You mean a gravitational field without tidal forces right?
Could you show me a physical configuration where we have such a "field"? Or is this good example in the same category as "when a chicken travels at the speed of light he cannot lay any eggs because time stands still"? Also please define "gravitational well" and "up" as well, not by using Newtonian gravity but by using GR.

Yes. E.g. as a photon goes up it gains potential energy, loses kinetic energy, and therefore becomes redshifted. This indicates that time is slower lower in the potential well.

A emitted photon is absorbed redder or bluer not because something on its path affected it as you seem to suggest, it is simply because the clocks at the emission and reception run at a different rate.

 

[Dale]

Passionflower, you have a strong habit of making factually wrong statements and then when you are corrected by someone who actually knows what they are talking about making it worse by persisting in your error and trying to support it with completely irrelevant statements. You did it in the thread where you were confusing path length with distance and now in this thread.

Really? So when a ball explodes what single value is the proper acceleration of parts of the shell?
A clock is not a point mass. Are you suggesting there are no stresses on an object placed in the core of a massive object like the earth?

Irrelevant. Proper acceleration is defined along a worldline, stresses are not.

How do you know?
It is likely an accelerometer could not even operate properly at the center of the earth. By the way the fact that an accelerometer only measures an acceleration in one direction is not a limitation of nature but a limitation of the device.

Also irrelevant. As you yourself mention, measuring acceleration in one direction is a limitation of a specific type of accelerometer. Similarly, not operating properly at the center of the Earth is a limitation of a specific device, not a limitation of nature. In relativity when we talk about clocks, rods, or accelerometers we are always speaking of ideal devices.

You mean a gravitational field without tidal forces right?
Could you show me a physical configuration where we have such a "field"? Or is this good example in the same category as "when a chicken travels at the speed of light he cannot lay any eggs because time stands still"? Also please define "gravitational well" and "up" as well, not by using Newtonian gravity but by using GR.

Sure, no problem. A scalar gravitational potential can be defined in a static spacetime, such as the Schwarzschild solution. "Up" is the direction of the gradient of the potential, and the region around the minimum of a potential is described as a "well".

A emitted photon is absorbed redder or bluer not because something on its path affected it as you seem to suggest, it is simply because the clocks at the emission and reception run at a different rate.

And how would you make an experiment that would test this distinction? As far as I can tell it is two different ways of saying the same thing.

Again, all of this is in response to your factually incorrect statement that "The reason a clock attached to a ceiling runs faster than a clock on the floor is that the clock on the floor has a greater proper acceleration than the clock near the ceiling." A.T. provided two correct counter examples and none of your posts since then have addressed anything relevant, which is the exact same behavior as in the other thread.

 

[yuiop]

Start with the Schwarzschild solution in the weak field approximation:

(cd\tau)^2=(1-\frac{2\Phi}{c^2})(cdt)^2+(1-\frac{2\Phi}{c^2})^{-1}(dr)^2+...

For the case dr=d\theta=d\phi=0 you get the well known relationship:

d\tau=\sqrt{1-\frac{2\Phi}{c^2}}dt

Writing the above for two different gravitational potentials \Phi_1 and \Phi_2 you obtain the well-known time dilation relationship:

\frac{d\tau_1}{d\tau_2}=\sqrt{\frac{1-\frac{2\Phi_1}{c^2}}{1-\frac{2\Phi_2}{c^2}}}

At the Earth surface :

\Phi_1=-\frac{GM}{R}

At the Earth center:

\Phi_2=-3/2\frac{GM}{R}

Now, due to the fact that \frac{\Phi}{c^2}<<1 you can obtain the approximation:

\frac{d\tau_1}{d\tau_2}=1-\frac{\Phi_1-\Phi_2}{c^2}=1-\frac{GM}{2Rc^2}<1

So, f_1>f_2 where f_1 is the clock frequency on the Earth crust and f_2 is the frequency of the clock at the center of the Earth.

In addition, the time dilation depends on the difference in the gravitational field \Phi_1-\Phi_2.

Generalization:

At a distance r<R from the center of the sphere, inside the sphere, the gravitational potential is:

\Phi_2(r)=-\frac{GM}{R}(\frac{3}{2}-\frac{r^2}{2R^2})

The above gives:

\frac{d\tau_1}{d\tau_2}=1-\frac{\Phi_1-\Phi_2}{c^2}=1-\frac{GM}{2Rc^2}(1-\frac{r^2}{R^2})<1

For r=0 (clock2 at the center of the Earth) you recover the results from above.

For r=R you get the expected:

\frac{d\tau_1}{d\tau_2}=1

Here is a slightly better equation based on the interior Schwarzschild solution for dr=d\theta=d\phi=0 :

\frac{d\tau}{dt}=\frac{3}{2}\sqrt{1-\frac{2GM}{c^2R}}-\frac{1}{2}\sqrt{1-\frac{2GMr^2}{c^2R^3}}

where d\tau is the proper clock rate of a clock located at r, dt is the clock rate of a clock located very distant from the gravitational body and R is the the radius of the massive body with mass M, valid for r<=R. This equation has the slightly curious result that when R <(9/4)GM/c^2 the proper time of a clock at the centre runs backwards compared to the distant clock.

The equation assumes an even density distribution within the massive body and can be be expressed more generally for uneven density distributions (using units of G=c=1) as:

\frac{d\tau}{dt}=3/2\sqrt{1-2M/R}-1/2\sqrt{1-2 p (4/3)\pi r^2}

where p is the average density of the mass enclosed within a sphere of radius r where the clock is located. To compare the clock rates of two clocks located at a radius less than or equal to the radius of the gravitational body the following equation can be used:

\frac{d\tau_1}{d\tau_2}= \frac{3\sqrt{1-2M/R}-\sqrt{1-2 p_1 (4/3)\pi r_1^2}}{3\sqrt{1-2M/R}-\sqrt{1-2 p_2 (4/3)\pi r_2^2}}

For the case that one clock is located at the surface (dt_1 = dt_s) of the massive body (r_1 = R) and the other clock is located at the centre (dt_2= dt_c) of the massive body (r_2 = 0) the equation simplifies to:

\frac{d\tau_s}{d\tau_c}= \frac{2 \sqrt{1-2M/R}}{3\sqrt{1-2M/R}-1}

which has a value of greater than unity for all 2M<R<\infty

 

[Passionflower]

Dalespam, if the center of the Earth would not accelerate it would simply collapse into a black hole.

 

[starthaus]

\frac{d\tau_s}{d\tau_c}= \frac{2 \sqrt{1-2M/R}}{3\sqrt{1-2M/R}-1}

which has a value of greater than unity for all 2M<R<\infty

You must have made an error somewhere. If you did your calculations correctly, you should have gotten:\frac{d\tau_s}{d\tau_c}<1.

The clock at the center of the Earth (deeper in the gravitational well) should tick slower. You got the reverse.

 

[Passionflower]

The clock at the center of the Earth (deeper in the gravitational well) should tick slower. You got the reverse.

That is certainly true.

However, in order to calculate such situations one has to start with some kind of perfect fluid solution as the Earth is not hollow. Geodesic paths, due to Ricci curvature, tend to get in in each others way in the center and create havoc to say the least.

 

[Dale]

Dalespam, if the center of the Earth would not accelerate it would simply collapse into a black hole.

You just insist on making wrong statement after wrong statement. The interior Schwarzschild solution does not have a singularity nor an event horizon. Also the worldline defined by a constant r=0 is a geodesic in the interior Schwarzschild solution, meaning that it has no proper acceleration.

 

[yuiop]

You must have made an error somewhere. If you did your calculations correctly, you should have gotten:

\frac{d\tau_s}{d\tau_c}<1.

The clock at the center of the Earth (deeper in the gravitational well) should tick slower. You got the reverse.

The clock at the centre does tick slower than the clock at the surface and as a result the ratio should be:

\frac{d\tau_s}{d\tau_c}>1

So, f_1>f_2 where f_1 is the clock frequency on the Earth crust and f_2 is the frequency of the clock at the center of the Earth.

From this you should get:

\frac{f_1}{f_2} > 1

For example the clock rate (d\tau_s) of a clock on the surface of a body with a radius R=8M is 0.8666 seconds for every second that passes on a clock at infinity.

The clock rate (d\tau_c) of a clock at the centre of the same body is 0.7990 seconds for every second that passes on a clock at infinity (and slower than the surface clock).

The ratio is this example is:

\frac{d\tau_s}{d\tau_c} = \frac{0.8666}{0.7990} = 1.084 >1

Maybe there is a mistake in your calculations?

 

[Passionflower]

Dalespam, if the center of the Earth would not accelerate it would simply collapse into a black hole.

You just insist on making wrong statement after wrong statement. The interior Schwarzschild solution does not have a singularity nor an event horizon. Also the worldline defined by a constant r=0 is a geodesic in the interior Schwarzschild solution, meaning that it has no proper acceleration.

Really now, who is talking about particular solutions you or I?

Do you realize that the EM forces in all atoms inside the Earth's have to resist the tendency of gravity to collapse the Earth into a black hole. That resistance is called proper acceleration.

 

[starthaus]

The clock at the centre does tick slower than the clock at the surface and as a result the ratio should be:

\frac{d\tau_s}{d\tau_c}>1

The above signifies that the period of the clock at the surface is greater than the one at the ceneter of the Earth. This is incorrect.

From this you should get:

\frac{f_1}{f_2} > 1

Nope, you got this backwards:

\frac{d\tau_s}{d\tau_c}>1

implies:

\frac{f_1}{f_2} < 1

which is clearly wrong. Remember, clock frequency is the inverse of clock period.

 

[A.T.]

Dalespam, if the center of the Earth would not accelerate it would simply collapse into a black hole.

If you think that the proper acceleration of the Earth's center is not zero, then please tell us the direction & magnitude of the vector.

 

[Dale]

That resistance is called proper acceleration.

No, it is not. Proper acceleration is what is measured by an accelerometer, what you are describing is stress which is measured by a stress transducer.

And, going back to your post 54, yes, if there were no stress at the center of the Earth it would collapse into a black hole.

 

[yuiop]

Nope, you got this backwards:

\frac{d\tau_s}{d\tau_c}>1

implies:

\frac{f_1}{f_2} < 1

which is clearly wrong. Remember, clock frequency is the inverse of clock period.

I have consistently referred to d\tau as clock rate which is the same as frequency. This is the normal meaning for d\tau in relativity.

 

[Passionflower]

No, it is not. Proper acceleration is what is measured by an accelerometer, what you are describing is stress which is measured by a stress transducer.

So you think when you stand with your feet on the Earth's surface there is no proper acceleration but instead stress?
Perhaps you might think a little and perhaps realize that both on a microscopic scale are actually the same thing.

And, going back to your post 54, yes, if there were no stress at the center of the Earth it would collapse into a black hole.

Oh I see, good! So let me ask you this: does stress dilate clocks? Or in different terms: does an object under, as you call it, stress, travel on a geodesic?

Also while we are at it, could you describe a clock or accelerometer that could even in principle be the size of a zero dimensional point?

 

[A.T.]

So you think when you stand with your feet on the Earth's surface there is no proper acceleration but instead stress?

On the surface there is both of them. In the center there is only stress, no proper acceleration.

Also while we are at it, could you describe a clock or accelerometer that could even in principle be the size of a zero dimensional point?

Irrelevant. My example works fine with a normally sized clock & accelerometer floating in a cavity at the center. They will measure zero proper acceleration and a slower clock rate than on the surface, according to General Relativity.

If your gravitational theory predicts a non-zero proper acceleration at the center, then I'm really curious about the direction of the vector.

 

[Passionflower]

a normally sized clock & accelerometer floating in a cavity at the center.

Oh I see, now it is a cavity, you are moving the goalposts!
Did you come to realize that Ricci curvature spoiled your original claim?

If your gravitational theory predicts a non-zero proper acceleration at the center, then I'm really curious about the direction of the vector.

First of all it is not my gravitational theory, I am simply applying the equivalence principle.

For instance consider an atom in the center of the Earth. From all directions there will be a tendency to interfere with the EM forces due to the curvature of spacetime, the resistance to that will result in a proper acceleration in all directions.

 

[Al68]

For instance consider an atom in the center of the Earth. From all directions there will be a tendency to interfere with the EM forces due to the curvature of spacetime, the resistance to that will result in a proper acceleration.

Proper acceleration certainly is the result of net force applied, in this case it's equal to zero, since the net force applied to the '"center of earth" atom is equal to zero.

The individual components of force that result in a net force of zero are irrelevant. Proper acceleration depends only on the net force applied, and is equal to zero in this case.

 

[Passionflower]

You just insist on making wrong statement after wrong statement. The interior Schwarzschild solution does not have a singularity nor an event horizon. Also the worldline defined by a constant r=0 is a geodesic in the interior Schwarzschild solution, meaning that it has no proper acceleration.

Oh wait I missed that.

So let me get this right you claim that r=0 in the int. Schwarzschild solution is the center of the ball and there can never be a singularity?

 

[A.T.]

Oh I see, now it is a cavity, you are moving the goalposts!

No, I just gave the third counter example that disproves your wrong statement: "The reason a clock attached to a ceiling runs faster than a clock on the floor is that the clock on the floor has a greater proper acceleration than the clock near the ceiling."

While you (as usual) fail to address the counter example.

Did you come to realize that Ricci curvature spoiled your original claim?

No. My first counter example is still valid. GR predicts zero proper acceleration for the center of a solid sphere, because the worldline of this point is a geodesic.

First of all it is not my gravitational theory

It clearly contradicts GR.

a proper acceleration in all directions.

Ohhh... so the non-zero proper acceleration vector is pointing in all directions simultaneously? Fascinating...

So, what is its magnitude then?

 

[Dale]

Perhaps you might think a little and perhaps realize that both on a microscopic scale are actually the same thing.

No, they are not. For one thing they have completely different units. In SI units proper acceleration is in m/s² and stress is in kg/(s²m). Additionally, stress is a tensor field with 9 elements at each event in spacetime, and proper acceleration is a vector with 3 elements at each event along a single worldline. It doesn't matter if you are looking at a microscopic scale or not, they are not the same thing at all.

So let me ask you this: does stress dilate clocks?

Stress is part of the stress-energy tensor which curves spacetime according to the EFE.

Or in different terms: does an object under, as you call it, stress, travel on a geodesic?

Whether or not an object is under stress is irrelevant to whether or not it is traveling on a geodesic. What is relevant is if the sum of the real forces is non-zero.

So let me get this right you claim that r=0 in the int. Schwarzschild solution is the center of the ball and there can never be a singularity?

Yes, that is correct. The singularity at r=0 is for the exterior Schwarzschild solution, not the interior one.

 

[yuiop]

You must have made an error somewhere. If you did your calculations correctly, you should have gotten:

\frac{d\tau_s}{d\tau_c}<1.

The clock at the center of the Earth (deeper in the gravitational well) should tick slower. You got the reverse.

The error is yours. I have edited your post below to show where your errors are (my corrections in red):

Start with the Schwarzschild solution in the weak field approximation:

(cd\tau)^2=(1-\frac{2\Phi}{c^2})(cdt)^2+(1-\frac{2\Phi}{c^2})^{-1}(dr)^2+...

That should be:

{\color{red}(cd\tau)^2=(1+\frac{2\Phi}{c^2})(cdt)^2-(1+\frac{2\Phi}{c^2})^{-1}(dr)^2-...}

See post 185 of this different thread https://www.physicsforums.com/showthread.php?t=397403&page=12 where you gave the correct equation. This error in the signs propagates all the way through the rest of your calculations.

For the case dr=d\theta=d\phi=0 you get the well known relationship:

d\tau=\sqrt{1{\color{red}+}\frac{2\Phi}{c^2}}dt

Writing the above for two different gravitational potentials \Phi_1 and \Phi_2 you obtain the well-known time dilation relationship:

\frac{d\tau_1}{d\tau_2}=\sqrt{\frac{1{\color{red}+}\frac{2\Phi_1}{c^2}}{1{\color{red}+}\frac{2\Phi_2}{c^2}}}

At the Earth surface :

\Phi_1=-\frac{GM}{R}

At the Earth center:

\Phi_2=-3/2\frac{GM}{R}

Now, due to the fact that \frac{\Phi}{c^2}<<1 you can obtain the approximation:

\frac{d\tau_1}{d\tau_2}=1{\color{red}+}\frac{\Phi_1-\Phi_2}{c^2}=1{\color{red}+}\frac{GM}{2Rc^2}{\color{red}>}1

So, f_1>f_2 where f_1 is the clock frequency on the Earth crust and f_2 is the frequency of the clock at the center of the Earth.

When corrected, your aproximation is in close agreement with the equations I gave and is on the right side of unity.
...

Strange that when you found our calculations did not agree you did not check your own calculations and assumed mine were wrong. You then compounded this error by using a misconception to justify your erronous calculation:

Nope, you got this backwards:

\frac{d\tau_s}{d\tau_c}>1

implies:

\frac{f_1}{f_2} < 1

which is clearly wrong. Remember, clock frequency is the inverse of clock period.

Yes, clock frequency is the inverse of clock periods, but d\tau_1/d\tau_2 is not a ratio of clock periods as I demonstrate below.

Lets compare two clocks in SR:

\frac{d\tau_1}{d\tau_2} = \frac{\sqrt{1-v_1^2/c^2}}{\sqrt{1-v_2^2/c^2}}

When v_2>v_1 the clock rate of clock 2 is slower than that of clock 1 and the ratio of clock rates is:

\frac{d\tau_1}{d\tau_2}>1

clearly indicating that d\tau_1/d\tau_2 is a ratio of clock frequencies and not periods as you claim.

Now let's compare two clocks in exterior Schwarzschild coordinates:

\frac{d\tau_1}{d\tau_2} = \frac{\sqrt{1-2GM/(r_1 c^2)}}{\sqrt{1-2GM/(r_2 c^2)}}

When r_2<r_1 the clock rate of clock 2 is slower than that of clock 1 and the ratio of clock rates is:

\frac{d\tau_1}{d\tau_2}>1

clearly indicating that d\tau_1/d\tau_2 is a ratio of clock frequencies in GR as well as in SR and not periods as you claim.

In Schwarzschild coordinates the time dilation of a stationary clock at r compared to a clock at infinity is given by:

\frac{d\tau}{dt} = \sqrt{1-2GM/(r c^2 )}

The term on the LHS of the equation can be read as ticks of the test clock (d\tau) per second of the reference clock at infinity (dt). Expressed like this it is easy to see that d\tau/dt is in fact a frequency, which often expressed in terms of "per second".

So:

\frac{d\tau_1/dt}{d\tau_2/dt} = \frac{f_1}{f_2} = \frac{d\tau_1}{d\tau_2} = \frac{f_1}{f_2}

Hopefully, I have clearly established that d\tau_1/d\tau_2 is a ratio of frequencies and not a ratio of clock periods as you claim.

 

[yuiop]

Nah, why don't you invest in buying a good book on relativity, I recommended Rindler to you several times:

ds^2=(1-\frac{2m}{r})dt^2-(1-\frac{2m}{r})^{-1}dr^2-... (see Rindler (11.13))

with

m=\frac{GM}{c^2}

Yep, but when YOU define \Phi as -\frac{GM}{R} the equation becomes:

(cd\tau)^2=(1+\frac{2\Phi}{c^2})(cdt)^2-(1+\frac{2\Phi}{c^2})^{-1}(dr)^2-...

and not

(cd\tau)^2=(1-\frac{2\Phi}{c^2})(cdt)^2+(1-\frac{2\Phi}{c^2})^{-1}(dr)^2+...

as you posted. (You got the signs between the consecutive terms wrong too. You can have a signiture of +,-,-,- or -,+,+,+ but not +,+,+,+.

Its OK. You are allowed to admit you are wrong sometimes.

Second of all, your equation in this thread contradicts the equation you gave in this other thread https://www.physicsforums.com/showthread.php?t=397403&page=12 so one of them has to be wrong.

 

[starthaus]

Yep, but when YOU define \Phi as -\frac{GM}{R} the equation becomes:

(cd\tau)^2=(1+\frac{2\Phi}{c^2})(cdt)^2-(1-\frac{2\Phi}{c^2})^{-1}(dr)^2-...

I never wrote this meaningless expression.

and not

(cd\tau)^2=(1-\frac{2\Phi}{c^2})(cdt)^2+(1-\frac{2\Phi}{c^2})^{-1}(dr)^2+...

as you posted.

Its OK. You are allowed to admit you are wrong sometimes.

Second of al,l your equation in this thread contradicts the equation you gave in this other thread https://www.physicsforums.com/showthread.php?t=397403&page=12 so one of them has to be wrong.

Yes, I cited the metric from memory. I am quite sure that you will be happy to have scored a victory :-)
It's indeed the expression I used in the thread where I was trying to teach you how to use metrics to derive the Lagrangians and how to use the Lagrangian for deriving the equations of motion.

(cd\tau)^2=(1+\frac{2\Phi}{c^2})(cdt)^2-(1+\frac{2\Phi}{c^2})^{-1}(dr)^2-...

Yet, you STILL have the physical interpretation wrong:

\frac{d\tau_1}{d\tau_2}>1

means

\frac{f_1}{f_2}<1

where f_1 is the frequency measured at the Earth surface for a wave that had the frequency f_2 when emitted from the center of the Earth. In other words, the observer at the top of the well sees the frequency emitted from the bottom redshifted.

 

[yuiop]

Yes, I cited the metric from memory. I am quite sure that you will be happy to have scored a victory :-)

You have gone up considerably in my estimation in being able to admit a mistake, even it did take 7 exchanges to convince you of it.

{EDIT} Damn.. just noticed you edited your post and are still going on about the frequency/ period thing. I can't add any more to the extensive arguments I have already made. Maybe someone else can give a independent point of view.

 

[starthaus]

You have gone up considerably in my estimation in being able to admit a mistake, even it did take 7 exchanges to convince you of it.

{EDIT} Damn.. just noticed you edited your post and are still going on about the frequency/ period thing. I can't add any more to the extensive arguments I have already made. Maybe someone else can give a independent point of view.

well, you need to learn the basics. :-)

 

[yuiop]

I never wrote this meaningless expression.

Yes, I made a obvious typo that I have corrected. You wrote a different meaningless expression.

Yet, you STILL have the physical interpretation wrong:

\frac{d\tau_1}{d\tau_2}>1

means

\frac{f_1}{f_2}<1

where f_1 is the frequency measured at the Earth surface for a wave that had the frequency f_2 when emitted from the center of the Earth. In other words, the observer at the top of the well sees the frequency emitted from the bottom redshifted.

OK, let's establish where the goal posts are now.

Do you disagree with the conclusion that

\frac{d\tau_1}{d\tau_2}>1

when clock 2 is lower down, even though your calculations (after my corrections) show that is true?

When I said

\frac{f_1}{f_2}>1

I was talking about the relative clock frequencies of two clocks with clock 2 lower down.

You seem to be talking about \frac{f_1}{f_2} in relation to frequencies of lightwaves which is a different thing (though related) so what exactly are you claiming?

Do you disagree with my claim that the frequency or clock rate of a clock higher up, is relatively higher than the frequency of clock lower down?

 

[starthaus]

Lets compare two clocks in SR:

\frac{d\tau_1}{d\tau_2} = \frac{\sqrt{1-v_1^2/c^2}}{\sqrt{1-v_2^2/c^2}}

When v_2>v_1 the clock rate of clock 2 is slower than that of clock 1 and the ratio of clock rates is:

\frac{d\tau_1}{d\tau_2}>1

clearly indicating that d\tau_1/d\tau_2 is a ratio of clock frequencies and not periods as you claim.

d\tau is the proper time difference that one derives from the SR metric:

(c\dtau)^2=(cdt)^2-dr^2=(cdt)^2(1-\frac{dr^2}{c^2dt^2})=(cdt)^2(1-\frac{v^2}{c^2})

so:

d\tau=dt\sqrt{1-(v/c)^2}

If you write the above for two different speeds you get:

d\tau=dt\sqrt{1-(v_1/c)^2}

d\tau=dt\sqrt{1-(v_2/c)^2}

Contrary to your beliefs, both expressions have dimensions of time, not frequency.

Divide the expressions and you get rid of dt. It is all very simple.

 

[yuiop]

d\tau is the proper time difference that one derives from the SR metric:

(c\dtau)^2=(cdt)^2-dr^2=(cdt)^2(1-\frac{dr^2}{c^2dt^2})=(cdt)^2(1-\frac{v^2}{c^2})

so:

d\tau=dt\sqrt{1-(v/c)^2}

If you write the above for two different speeds you get:

d\tau=dt\sqrt{1-(v_1/c)^2}

d\tau=dt\sqrt{1-(v_2/c)^2}

Contrary to your beliefs, both expressions have dimensions of time, not frequency.

Divide the expressions and you get rid of dt. It is all very simple.

and you are claiming d\tau_1/d\tau_2 is the ratio of the periods between consecutive clock ticks for clocks 1 and 2?

In #78 I explicitly asked for clarification of your statements and as always you are being evasive and clarified nothing.

 

[starthaus]

Do you disagree with my claim that the frequency or clock rate of a clock higher up, is relatively higher than the frequency of clock lower down?

Yep, I disagree.
The correct statement is : the rate of the clock "higher up" in a gravitational field appears blueshifted when measured by an observer "lower down".
Conversely: the observer "higher up" measures the frquency of a clock "lower down" as redshifted.
See the Pound-Rebka experiment.

 

[starthaus]

and you are claiming d\tau_1/d\tau_2 is the ratio of the periods between consecutive clock ticks for clocks 1 and 2?

What in "time difference" did you not understand in the derivation I just shoowed you?

 

[starthaus]

Start with the Schwarzschild solution in the weak field approximation:

(cd\tau)^2=(1-\frac{2\Phi}{c^2})(cdt)^2+(1-\frac{2\Phi}{c^2})^{-1}(dr)^2+...

For the case dr=d\theta=d\phi=0 you get the well known relationship:

d\tau=\sqrt{1-\frac{2\Phi}{c^2}}dt

Writing the above for two different gravitational potentials \Phi_1 and \Phi_2 you obtain the well-known time dilation relationship:

\frac{d\tau_1}{d\tau_2}=\sqrt{\frac{1-\frac{2\Phi_1}{c^2}}{1-\frac{2\Phi_2}{c^2}}}

At the Earth surface :

\Phi_1=-\frac{GM}{R}

At the Earth center:

\Phi_2=-3/2\frac{GM}{R}

Now, due to the fact that \frac{\Phi}{c^2}<<1 you can obtain the approximation:

\frac{d\tau_1}{d\tau_2}=1-\frac{\Phi_1-\Phi_2}{c^2}=1-\frac{GM}{2Rc^2}<1

So, f_1>f_2 where f_1 is the clock frequency on the Earth crust and f_2 is the frequency of the clock at the center of the Earth.

In addition, the time dilation depends on the difference in the gravitational field \Phi_1-\Phi_2.

Generalization:

At a distance r<R from the center of the sphere, inside the sphere, the gravitational potential is:

\Phi_2(r)=-\frac{GM}{R}(\frac{3}{2}-\frac{r^2}{2R^2})

The above gives:

\frac{d\tau_1}{d\tau_2}=1-\frac{\Phi_1-\Phi_2}{c^2}=1-\frac{GM}{2Rc^2}(1-\frac{r^2}{R^2})<1

For r=0 (clock2 at the center of the Earth) you recover the results from above.

For r=R you get the expected:

\frac{d\tau_1}{d\tau_2}=1

There was an unfortunate error in the above due to my citing the metric from memory.The derivation is correct modulo the sign of the potential in the metri. Here is the corrected version:

Start with the Schwarzschild solution in the weak field approximation:

(cd\tau)^2=(1+\frac{2\Phi}{c^2})(cdt)^2+(1+\frac{2\Phi}{c^2})^{-1}(dr)^2+...

For the case dr=d\theta=d\phi=0 you get the well known relationship:

d\tau=\sqrt{1+\frac{2\Phi}{c^2}}dt

Writing the above for two different gravitational potentials \Phi_1 and \Phi_2 you obtain the well-known time dilation relationship:

\frac{d\tau_1}{d\tau_2}=\sqrt{\frac{1+\frac{2\Phi_1}{c^2}}{1+\frac{2\Phi_2}{c^2}}}

At the Earth surface :

\Phi_1=-\frac{GM}{R}

At the Earth center:

\Phi_2=-3/2\frac{GM}{R}

Now, due to the fact that \frac{\Phi}{c^2}<<1 you can obtain the approximation:

\frac{d\tau_1}{d\tau_2}=1+\frac{\Phi_1-\Phi_2}{c^2}=1+\frac{GM}{2Rc^2}>1

So, f_1<f_2 where f_1 is the frequency measured by an observer on the Earth crust and f_2 is the frequency emitted by a source at the center of the Earth. In other words, the observer at the top of the "gravitational well" measures the frequency emitted from the bottom of the gravitational well as redshifted. Conversely, an observer at the bottom of a gravity well will measure the frequency emitted from the top as blueshifted. See the Pound-Rebka experiment.

In addition, the time dilation depends on the difference in the gravitational field \Phi_1-\Phi_2.

Generalization:

At a distance r<R from the center of the sphere, inside the sphere, the gravitational potential is:

\Phi_2(r)=-\frac{GM}{R}(\frac{3}{2}-\frac{r^2}{2R^2})

The above gives:

\frac{d\tau_1}{d\tau_2}=1+\frac{\Phi_1-\Phi_2}{c^2}=1+\frac{GM}{2Rc^2}(1-\frac{r^2}{R^2})

For r=0 (clock2 at the center of the Earth) you recover the results from above.

For r=R you get the expected:

\frac{d\tau_1}{d\tau_2}=1