Gravitational time dilation near Earth

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No, the gravitational forces do not cancel in the above point ...

Yes, they do cancel.

Anyway, thank you very much for your replies.

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And if we include the orbital rotation the cancellation of forces takes place much higher, on the Hill sphere.

Not necessarily, you will need to calculate instead the extremum of the function:

$\frac{f®}{f(x)}=\sqrt{\frac{1+2\Phi(x)/c^2-((d-x) \omega/c)^2}{1+2\Phi®/c^2}}$

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Not necessarily, you will need to calculate instead the extremum of the function:

$\frac{f®}{f(x)}=\sqrt{\frac{1+2\Phi(x)/c^2-((d-x) \omega/c)^2}{1+2\Phi®/c^2}}$

Why? And what is R?

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Why? And what is R?

R is the Earth radius. The formula (derived per your earlier request) gives the ratio of frequencies of a clock placed on the Earth surface vs a clock situated at distance d-x from the Earth center moving with angular speed $\omega$

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R is the Earth radius. The formula (derived per your earlier request) gives the ratio of frequencies of a clock placed on the Earth surface vs a clock situated at distance d-x from the Earth center moving with angular speed $\omega$

Then the equation should be:

$\frac{f®}{f(x)}=\sqrt{\frac{1+2\Phi(x)/c^2-((d-x) \omega/c)^2}{1+2\Phi®/c^2-((d-R) \omega/c)^2}}$

where:

- d is the distance between the Sun and the Earth

- x is the distance between the center of the Earth and the test clock on the Earth-Sun line

- $\omega$ is the angular speed of the Earth around the Sun

Edited by DanMP
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Then the equation should be:

$\frac{f®}{f(x)}=\sqrt{\frac{1+2\Phi(x)/c^2-((d-x) \omega/c)^2}{1+2\Phi®/c^2-((d-R) \omega/c)^2}}$

where:

- d is the distance between the Sun and the Earth

- x is the distance between the center of the Earth and the test clock on the Earth-Sun line

- $\omega$ is the angular speed of the Earth around the Sun

Yes, the standard is given in post 23. Doesn't change the answer since you need to take the derivative wrt x.

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Doesn't change the answer since you need to take the derivative wrt x.

True, but in your equation (#27), the angular speed, $\omega$, seemed to be of a test clock orbiting the Earth, which is not the case.

Anyway, I finally got the time to do & now to write the math:

$\frac{f®}{f(x)}=\sqrt{\frac{1+2\Phi(x)/c^2-((d-x) \omega/c)^2}{1+2\Phi®/c^2-((d-R) \omega/c)^2}}$

where:

- d is the distance between the Sun and the Earth

- x is the distance between the center of the Earth and the test clock on the Earth-Sun line

- $\omega$ is the angular speed of the Earth around the Sun

- R is the Earth radius

so $f(x)=f®\sqrt{\frac{1+2\Phi®/c^2-((d-R) \omega/c)^2}{1+2\Phi(x)/c^2-((d-x) \omega/c)^2}}$

and we need to find the point where $f'(x)=0$

This leads to: $\frac {2\Phi'(x)}{c^2}+\frac {2(d-x)\omega^2}{c^2}=0$ (1)

As in #24:

$\Phi(x)=-\frac{GM}{d-x}-\frac {Gm}x$ and $\Phi'(x)=-\frac{GM}{(d-x)^2}+\frac {Gm}{x^2}$

So (1) become: $-\frac{GM}{(d-x)^2}+\frac {Gm}{x^2}+(d-x)\omega^2=0$ or

$\frac{GM}{(d-x)^2}=\frac {Gm}{x^2}+(d-x)\omega^2$

Meaning that the point where a test clock going from the Earth's surface towards the Sun would switch from increasing its tick rate to decrease it again is exactly where the gravitational pull of the Sun on the clock is cancelled by the gravitational pull of the Earth and the centrifugal force (in the above equation the mass of the clock is 1 kg). This happens at x=1482493941.5 m (approx. 1,5 million km), probably on the Hill sphere, as I kind of suggested in #24. Interesting.

This can and should be tested using 2 atomic clocks linked with fibre optic cable, the first pulling the other with a few km long cable, spiraling upwards from the Earth.

Edited by DanMP
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True, but in your equation (#27), the angular speed, $\omega$, seemed to be of a test clock orbiting the Earth, which is not the case.

Anyway, I finally got the time to do & now to write the math:

$\frac{f®}{f(x)}=\sqrt{\frac{1+2\Phi(x)/c^2-((d-x) \omega/c)^2}{1+2\Phi®/c^2-((d-R) \omega/c)^2}}$

where:

- d is the distance between the Sun and the Earth

- x is the distance between the center of the Earth and the test clock on the Earth-Sun line

- $\omega$ is the angular speed of the Earth around the Sun

- R is the Earth radius

so $f(x)=f®\sqrt{\frac{1+2\Phi®/c^2-((d-R) \omega/c)^2}{1+2\Phi(x)/c^2-((d-x) \omega/c)^2}}$

and we need to find the point where $f'(x)=0$

This leads to: $\frac {2\Phi'(x)}{c^2}+\frac {2(d-x)\omega^2}{c^2}=0$ (1)

As in #24:

$\Phi(x)=-\frac{GM}{d-x}-\frac {Gm}x$ and $\Phi'(x)=-\frac{GM}{(d-x)^2}+\frac {Gm}{x^2}$

So (1) become: $-\frac{GM}{(d-x)^2}+\frac {Gm}{x^2}+(d-x)\omega^2=0$ or

$\frac{GM}{(d-x)^2}=\frac {Gm}{x^2}+(d-x)\omega^2$

Meaning that the point where a test clock going from the Earth's surface towards the Sun would switch from increasing its tick rate to decrease it again is exactly where the gravitational pull of the Sun on the clock is cancelled by the gravitational pull of the Earth and the centrifugal force (in the above equation the mass of the clock is 1 kg). This happens at x=1482493941.5 m (approx. 1,5 million km), probably on the Hill sphere, as I kind of suggested in #24. Interesting.

This can and should be tested using 2 atomic clocks linked with fibre optic cable, the first pulling the other with a few km long cable, spiraling upwards from the Earth.

...except that the above is just a simplification of the actual solution. It uses the Schwarzschild solution applicable to a single gravitating body. What is needed is the two-body EFE solution.

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...except that the above is just a simplification of the actual solution.

This is good, because I expected a smaller x

Can you calculate the correct value?

Edited by DanMP
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This is good, because I expected a smaller x

Can you calculate the correct value?

That would be difficult for the general case. See here the metric for particular cases. You already have the approach to the solution , in post 22. You know what to do.

Edited by zztop
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That would be difficult for the general case.

... You know what to do.

Well, I don't know what to do. You expect major differences? Can you calculate it? Anyone else?

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Well, I don't know what to do. You expect major differences? Can you calculate it? Anyone else?

Yes, I can. You should be able to , after I showed you.

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Yes, I can. You should be able to , after I showed you.

If you can calculate the "real" x, please do, because I can't.

Edited by DanMP
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If you can calculate the "real" x, please do, because I can't.

Damour, in a series of famous papers, has reduced the two-body problem to an equivalent one-body problem. The solution to the EFE's is the metric:

$(cd \tau)^2=-A[r](cdt)^2+B[r]dr^2+(r d \theta)^2$ (see page 4 for definitions of terms and coefficients).

PS: Latex doesn't seem to want to work this morning.

Edited by zztop
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$(cd \tau)^2=-A\left(r \right)(cdt)^2+B\left(r \right) dr^2+(r d \theta)^2$

!

Moderator Note

OK - So the Latex thing. ® (r ) was breaking the encoding. I have repeated the outside brackets with \left( and \right) - which allows correct interpretation by the renderer.

® is one of those annoying autotexts (registered trade box) so I wonder if something is happening before the text reaches the renderer. If I have got your formula wrong please pm me with the correction

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$(cd \tau)^2=-A\left(r \right)(cdt)^2+B\left(r \right) dr^2+(r d \theta)^2$

!

Moderator Note

OK - So the Latex thing. ® (r ) was breaking the encoding. I have repeated the outside brackets with \left( and \right) - which allows correct interpretation by the renderer.

® is one of those annoying autotexts (registered trade box) so I wonder if something is happening before the text reaches the renderer. If I have got your formula wrong please pm me with the correction

Thank you, it is correct. This is a very famous solution to a very difficult problem.

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Damour, in a series of famous papers, has reduced the two-body problem to an equivalent one-body problem. The solution to the EFE's is the metric ...

(see page 4 for definitions of terms and coefficients).

Sorry, but I didn't find the definitions of terms and coefficients there, nor the equation you posted ...

Anyway, the math I saw there is too much for me, so clearly I can't calculate the distance in question ...

Moreover, after, say, I manage to do it, you will say that it's not good, because it didn't include the spin, the Moon, the orbital speed around the center of our galaxy and so on ...

If you or someone else can understand and do the math you proposed, be my guest. I give up.

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Sorry, but I didn't find the definitions of terms and coefficients there, nor the equation you posted ...

Anyway, the math I saw there is too much for me, so clearly I can't calculate the distance in question ...

If you or someone else can understand and do the math you proposed, be my guest. I give up.

Actually, it DOES. It gives you the formulas for $A[R],B[R]$. It teaches you how it replaces the two bodies of masses $m_1,m_2$ with a single body of mass $\frac{m_1m_2}{m_1+m_2}$.....

You will need to find the extremum of the function $-A[R]+\frac{R^2 \omega^2}{c^2}$.

where $A[R]=1+\frac{a_1}{c^2R}$ (see page 3)

Moreover, after, say, I manage to do it, you will say that it's not good, because it didn't include the spin, the Moon, the orbital speed around the center of our galaxy and so on ...

This is beyond pale, after teaching you how this type of problem is being solved.

Edited by zztop
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Actually, it DOES. It gives you the formulas for $A[R],B[R]$. It teaches you how it replaces the two bodies of masses $m_1,m_2$ with a single body of mass ...

Sorry, I really can't understand your solution. What are $A[R],B[R]$? What is R? Where is x, with a single body replacing the two?

For the last time: if you think you can calculate the distance in question, please do. I can't and I give up.

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Sorry, I really can't understand your solution. What are $A[R],B[R]$? What is R? Where is x, with a single body replacing the two?

For the last time: if you think you can calculate the distance in question, please do. I can't and I give up.

You need to read the paper I linked, all the notions are explained on pages 3 and 4.

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