Do Gravitating Bodies Warp the Fabric of Space?

[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:

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