Gravitational time dilation near Earth

[DanMP]

  On 3/8/2017 at 3:16 PM, zztop said:

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

Yes, they do cancel.

 

Anyway, thank you very much for your replies.

[zztop]

  On 3/8/2017 at 2:26 PM, DanMP said:

 

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:

 

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

[DanMP]

  On 3/8/2017 at 5:12 PM, zztop said:

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?

[zztop]

  On 3/10/2017 at 3:15 PM, DanMP said:

 

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 $\omega$

[DanMP]

  On 3/10/2017 at 6:36 PM, zztop said:

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 March 17, 2017 by DanMP

[zztop]

  On 3/17/2017 at 2:32 PM, DanMP said:

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.

[DanMP]

  On 3/17/2017 at 3:05 PM, zztop said:

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:

 

Your (corrected) equation is:

 

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

 

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 March 21, 2017 by DanMP

[zztop]

  On 3/21/2017 at 1:53 PM, DanMP said:

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:

 

Your (corrected) equation is:

 

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

 

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.

[DanMP]

  On 3/21/2017 at 2:14 PM, zztop said:

...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 March 21, 2017 by DanMP

[zztop]

  On 3/21/2017 at 2:24 PM, DanMP said:

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 March 21, 2017 by zztop

[DanMP]

  On 3/21/2017 at 2:26 PM, zztop said:

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?

[zztop]

  On 3/22/2017 at 12:08 PM, DanMP said:

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.

[DanMP]

  On 3/22/2017 at 2:11 PM, zztop said:

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 March 23, 2017 by DanMP

[zztop]

  On 3/23/2017 at 1:19 PM, DanMP said:

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 March 23, 2017 by zztop

[imatfaal]

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

 

!

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

 

[zztop]

  On 3/23/2017 at 2:36 PM, imatfaal said:

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

 

!

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.

[DanMP]

  On 3/23/2017 at 1:52 PM, zztop said:

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.

[zztop]

  On 3/24/2017 at 12:26 PM, DanMP said:

 

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)

 

 

 

  Quote

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 March 24, 2017 by zztop

[DanMP]

  On 3/24/2017 at 1:52 PM, zztop said:

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.

[zztop]

  On 3/27/2017 at 1:25 PM, DanMP said:

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.