Do Gravitating Bodies Warp the Fabric of Space?

[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:

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