DanMP Posted March 3, 2017 Share Posted March 3, 2017 We know (measured) that a clock at sea level "ticks"slower than a clock on a mountain. As we get higher, the clock runs even faster. Now, if we go much higher, towards the Sun, the clock should begin, at some point, to run slower, influenced by Sun's gravitation. Where is that "point", at what distance from Earth? It was calculated and/or tested experimentally? Link to comment Share on other sites More sharing options...

Lord Antares Posted March 3, 2017 Share Posted March 3, 2017 (edited) There is no "point" where this happens. It is completely gradual. It is always influenced influenced by both masses of the sun and the earth, wherever it is in between them.There is no point where it stops being influenced by the earth and starts being influenced by the sun.There is, however, a point where it starts being influenced by the sun MORE than by the earth. I would guess that breaking point is the reverse center of mass between the sun and the earth. Edited March 3, 2017 by Lord Antares Link to comment Share on other sites More sharing options...

swansont Posted March 3, 2017 Share Posted March 3, 2017 You can calculate it. The dilation is the gravitational potential divided by c^2: GM/rc^2 A clock on the earth's surface ticks slower by 7 x 10^-10 relative to a clock far, far away from it. (Locally that change is about a part in 10^16 per meter) What you have to do is calculate the effect of the sun and the earth. You can probably assume you are far enough away from the earth that its contribution is negligible, for a good approximation. There is no "point" where this happens. It is completely gradual. It is always influenced influenced by both masses of the sun and the earth, wherever it is in between them. There is no point where it stops being influenced by the earth and starts being influenced by the sun. There is, however, a point where it starts being influenced by the sun MORE than by the earth. That breaking point is the center of mass between the sun and the earth. Since there is a 1/r dependence of the potential, it's not obvious to me that the center of mass is where that should occur. The center of mass in this system is within the sun itself. But I'm pretty sure the dilation on the sun's surface will slow the clock by a much greater amount than it will speed up by leaving the earth. 1 Link to comment Share on other sites More sharing options...

zztop Posted March 3, 2017 Share Posted March 3, 2017 I would guess that breaking point is the reverse center of mass between the sun and the earth. this is incorrect, guess is a bad thing in physics Link to comment Share on other sites More sharing options...

Lord Antares Posted March 3, 2017 Share Posted March 3, 2017 /cut I corrected my post to say ''reversed center of mass''. That's what I meant to say. Zztop says I'm wrong about that. I trust him, but the central point still stands, right? Link to comment Share on other sites More sharing options...

zztop Posted March 3, 2017 Share Posted March 3, 2017 (edited) We know (measured) that a clock at sea level "ticks"slower than a clock on a mountain. As we get higher, the clock runs even faster. Now, if we go much higher, towards the Sun, the clock should begin, at some point, to run slower, influenced by Sun's gravitation. Where is that "point", at what distance from Earth? It was calculated and/or tested experimentally? the distance between the Earth and the Sun is d the potential of the two bodies as calculated at a distance x from the Sun is: [latex]-\frac{GM}{x}+\frac{Gm}{d-x}[/latex] The effect is null for .....[latex]x=\frac{Md}{M+m} \approx d(1-\frac{m}{M})[/latex]. This works out to be about 279 mi from the Earth center or 275 mi above its surface. This falls approximately into the Earth exosphere Edited March 3, 2017 by zztop Link to comment Share on other sites More sharing options...

swansont Posted March 3, 2017 Share Posted March 3, 2017 I corrected my post to say ''reversed center of mass''. That's what I meant to say. Zztop says I'm wrong about that. I trust him, but the central point still stands, right? What is a reversed center of mass? the distance between the Earth and the Sun is d the potential of the two bodies as calculated at a distance x from the Sun is: [latex]\frac{GMm}{x}-\frac{GMm}{d-x}[/latex] The effect is null for .....[latex]x=\frac{d}{2}[/latex]. Surprising but true. You have not used the equation for the gravitational potential. You have used the equation for the gravitational potential energy. 1 Link to comment Share on other sites More sharing options...

zztop Posted March 3, 2017 Share Posted March 3, 2017 What is a reversed center of mass? You have not used the equation for the gravitational potential. You have used the equation for the gravitational potential energy. Yep, you are correct, I fixed it. Thank you for the catch, this is what happens when you cite formulas from memory, serves me right. Link to comment Share on other sites More sharing options...

Lord Antares Posted March 3, 2017 Share Posted March 3, 2017 What is a reversed center of mass? Well, if you imagine that the center of mass between two masses is 1/10 d, then I meant 9/10 d to be the reversed center of mass. I would intuitively think that this would the point where the gravity of the two objects would be nullified. Link to comment Share on other sites More sharing options...

zztop Posted March 3, 2017 Share Posted March 3, 2017 Well, if you imagine that the center of mass between two masses is 1/10 d, then I meant 9/10 d to be the reversed center of mass. I would intuitively think that this would the point where the gravity of the two objects would be nullified. Yes, with this definition, you are correct, see the exact calculations in post 6 Link to comment Share on other sites More sharing options...

Janus Posted March 4, 2017 Share Posted March 4, 2017 We know (measured) that a clock at sea level "ticks"slower than a clock on a mountain. As we get higher, the clock runs even faster. Now, if we go much higher, towards the Sun, the clock should begin, at some point, to run slower, influenced by Sun's gravitation. Where is that "point", at what distance from Earth? It was calculated and/or tested experimentally? Okay, are you looking for the point where the clock would stop increasing its tick rate and then start to decrease again, or the point where it goes from running faster than the Earth surface clock to running slower than it? Also, when you say "towards the sun" do you mean remaining on the Earth-Sun line? The reason for the second question is that you can't just consider the clock's distance from the Sun but also its motion with respect to it. If you are looking for the point where the clock rate quits increasing, then you would have to work out the equation for time dilation with respect to h (the height above the Earth's surface) including how h effects time dilation due to the increasing distance from the Earth, the distance from the Sun, and how this changes speed of the clock with respect to the Sun. Then you can differentiate this equation and solve for h when the derivative equals zero. Link to comment Share on other sites More sharing options...

Tim88 Posted March 6, 2017 Share Posted March 6, 2017 the distance between the Earth and the Sun is d the potential of the two bodies as calculated at a distance x from the Sun is: [latex]-\frac{GM}{x}+\frac{Gm}{d-x}[/latex] The effect is null for .....[latex]x=\frac{Md}{M+m} \approx d(1-\frac{m}{M})[/latex]. This works out to be about 279 mi from the Earth center or 275 mi above its surface. This falls approximately into the Earth exosphere That still looks not quite ok, I think that something went wrong there: for the Earth's radius isn't 279-275=4 miles... 1 Link to comment Share on other sites More sharing options...

zztop Posted March 6, 2017 Share Posted March 6, 2017 (edited) That still looks not quite ok, I think that something went wrong there: for the Earth's radius isn't 279-275=4 miles... Yes, you are correct. x would fall inside the Earth, meaning that , in fact, the potential doesn't go to 0 anywhere between the Earth and the Sun. Edited March 6, 2017 by zztop Link to comment Share on other sites More sharing options...

DanMP Posted March 6, 2017 Author Share Posted March 6, 2017 the distance between the Earth and the Sun is d the potential of the two bodies as calculated at a distance x from the Sun is: [math]-\frac{GM}{x}+\frac{Gm}{d-x}[/math] The effect is null for .....[latex]x=\frac{Md}{M+m} \approx d(1-\frac{m}{M})[/latex]. This works out to be about 279 mi from the Earth center or 275 mi above its surface. This falls approximately into the Earth exosphere I don't think that your result is corect, because gravitational potential is defined: The potential V of a unit mass m at a distance x from a point mass of mass M can be defined as the work W that needs to be done by an external agent to bring the unit mass in from infinity to that point and on the Earth - Sun line the work that needs to be done by an external agent to move the unit mass is affected by both the Sun and the Earth ... and by the centrifugal force ... Also, it's not clear (at least for me) why/how did you reach the conclusion that the point I'm looking for is exactly there. The equation for gravitational time dilation is: A common equation used to determine gravitational time dilation is derived from the Schwarzschild metric, which describes spacetime in the vicinity of a non-rotating massive spherically symmetric object. The equation is [latex]t_0 = t_f \sqrt{1-\frac{2GM}{rc^2}}[/latex] where [latex]t_0[/latex] is the proper time between events A and B for a slow-ticking observer within the gravitational field, [latex]t_f[/latex] is the coordinate time between events A and B for a fast-ticking observer at an arbitrarily large distance from the massive object (this assumes the fast-ticking observer is using Schwarzschild coordinates, a coordinate system where a clock at infinite distance from the massive sphere would tick at one second per second of coordinate time, while closer clocks would tick at less than that rate), Maybe we should use it ... Okay, are you looking for the point where the clock would stop increasing its tick rate and then start to decrease again, or the point where it goes from running faster than the Earth surface clock to running slower than it? Also, when you say "towards the sun" do you mean remaining on the Earth-Sun line? The reason for the second question is that you can't just consider the clock's distance from the Sun but also its motion with respect to it. If you are looking for the point where the clock rate quits increasing, then you would have to work out the equation for time dilation with respect to h (the height above the Earth's surface) including how h effects time dilation due to the increasing distance from the Earth, the distance from the Sun, and how this changes speed of the clock with respect to the Sun. Then you can differentiate this equation and solve for h when the derivative equals zero. The point where the clock would stop increasing its tick rate and then start to decrease again. Yes, remaining on the Earth-Sun line. I don't know how to solve the problem. That's why I asked here if there is a proper calculation/solution. Link to comment Share on other sites More sharing options...

zztop Posted March 6, 2017 Share Posted March 6, 2017 (edited) I don't think that your result is corect, because gravitational potential is defined: and on the Earth - Sun line the work that needs to be done by an external agent to move the unit mass is affected by both the Sun and the Earth ... and by the centrifugal force .. Actually, it is correct. The Force acting on the test probe of mass [latex]m_p[/latex] is: [latex]\frac{Gmm_p}{(d-x)^2}-\frac{GMm_p}{x^2}[/latex] Edited March 6, 2017 by zztop Link to comment Share on other sites More sharing options...

swansont Posted March 6, 2017 Share Posted March 6, 2017 Actually, it is 4 ≠ 4000 1 Link to comment Share on other sites More sharing options...

DanMP Posted March 6, 2017 Author Share Posted March 6, 2017 (edited) Actually, it is correct. I don't think that there is covered the case of a unit mass located between 2 massive objects ... Also there is no mention of centrifugal forces. More of it, you did't use the gravitational time dilation equation + speed related time dilation (due to orbital speeds) ... Without them one can propose the solution as being on the Earth's Hill sphere, between Sun and Earth ... Edited March 6, 2017 by DanMP Link to comment Share on other sites More sharing options...

zztop Posted March 6, 2017 Share Posted March 6, 2017 (edited) I don't think that there is covered the case of a unit mass located between 2 massive objects ... Also there is no mention of centrifugal forces. More of it, you did't use the gravitational time dilation equation ... Without it one can propose the solution as being on the Earth's Hill sphere, between Sun and Earth ... actually , is is exactly the case of mass probe between two other bodies. there is no centrifugal force the gravitational time dilation calculation REQUIRES the potential above [latex]\Phi(x)=\frac{Gm}{d-x}+\frac{GM}{x}[/latex] The ratio of clock frequencies is: [latex]\frac{f®}{f(x)}=\sqrt{\frac{\Phi(x)}{\Phi®}}[/latex] The extremum occurs where [latex]\frac{d \Phi}{dx}=0[/latex] This happens for [latex]x=d(1-\sqrt{\frac{m}{M}})[/latex] This works out to be [latex]93 \sqrt{3}*10^3[/latex] miles from the center of the Earth. If you want to take into consideration the rotation of the Earth, the problem becomes a lot more complicated since the above formula for ratio of clock frequencies no longer applies, you will need to redo the derivation starting from the Schwarzschild metric in order to account for any kinematic effects. 4 ≠ 4000 yep, my bad Edited March 6, 2017 by zztop Link to comment Share on other sites More sharing options...

Pugdaddy Posted March 6, 2017 Share Posted March 6, 2017 Would the point be the same as Lagrange points? Link to comment Share on other sites More sharing options...

zztop Posted March 6, 2017 Share Posted March 6, 2017 (edited) Would the point be the same as Lagrange points? Hard to tell, Lagrange points are determined by forces , the clock frequencies are a function of potentials.... ETA: the Lagrange points are much farther than the point calculated above Edited March 6, 2017 by zztop Link to comment Share on other sites More sharing options...

DanMP Posted March 7, 2017 Author Share Posted March 7, 2017 The ratio of clock frequencies is: [latex]\frac{f®}{f(x)}=\sqrt{\frac{\Phi(x)}{\Phi®}}[/latex] The extremum occurs where [latex]\frac{d \Phi}{dx}=0[/latex] This happens for [latex]x=d(1-\sqrt{\frac{m}{M}})[/latex] This works out to be [latex]93 \sqrt{3}*10^3[/latex] miles from the center of the Earth. If you want to take into consideration the rotation of the Earth, the problem becomes a lot more complicated since the above formula for ratio of clock frequencies no longer applies, you will need to redo the derivation starting from the Schwarzschild metric in order to account for any kinematic effects. yep, my bad This value (about 252,850 km above Earth's surface) looks much better Tank you! Still, I don't quite understand your calculation. Please be kind and explain it again, beginning with the origin of this: [latex]\frac{f®}{f(x)}=\sqrt{\frac{\Phi(x)}{\Phi®}}[/latex] Link to comment Share on other sites More sharing options...

zztop Posted March 7, 2017 Share Posted March 7, 2017 (edited) This value (about 252,850 km above Earth's surface) looks much better Tank you! Still, I don't quite understand your calculation. Please be kind and explain it again, beginning with the origin of this: [latex]\frac{f®}{f(x)}=\sqrt{\frac{\Phi(x)}{\Phi®}}[/latex] It comes from the following: The Schwarzschild solution to the EFE's is (see Rindler "Relativity, Special, General and Cosmological"): [latex](c d \tau)^2=(1+2 \frac{\Phi}{c^2})(cdt)^2-dr^2-(rd \theta)^2[/latex] If [latex]dr=0[/latex] (no radial motion) and [latex]d \theta=0[/latex] (no angular motion) then: [latex]\frac{d \tau}{dt}=\sqrt{1+2 \frac{\Phi}{c^2}}[/latex] For the case of two clocks A and B situated in gravitational potentials [latex]\Phi_A, \Phi_B[/latex] : [latex]\frac{d \tau_A}{dt}=\sqrt{1+2 \frac{\Phi_A}{c^2}}[/latex] [latex]\frac{d \tau_B}{dt}=\sqrt{1+2 \frac{\Phi_B}{c^2}}[/latex] So, the clock periods are in the relation: [latex]\frac{d \tau_A}{d \tau_B}=\sqrt{\frac{1+2 \frac{\Phi_A}{c^2}}{1+2 \frac{\Phi_B}{c^2}}}[/latex] The frequencies are in the inverse relation: [latex]\frac{f_B}{f_A}=\sqrt{\frac{1+2 \frac{\Phi_A}{c^2}}{1+2 \frac{\Phi_B}{c^2}}}[/latex] So, what I should have written is: [latex]\frac{f®}{f(x)}=\sqrt{\frac{1+2\Phi(x)/c^2}{1+2\Phi®/c^2}}[/latex] The extremum calculation shown earlier is not affected. If there is angular motion, things get more complicated: [latex]\frac{d \tau_A}{d \tau_B}=\sqrt{\frac{1+2 \frac{\Phi_A}{c^2}-(r_A \omega_A)^2}{1+2 \frac{\Phi_B}{c^2}-(r_B \omega_B)^2}}[/latex] Edited March 7, 2017 by zztop Link to comment Share on other sites More sharing options...

zztop Posted March 8, 2017 Share Posted March 8, 2017 If there is angular motion, things get more complicated: [latex]\frac{d \tau_A}{d \tau_B}=\sqrt{\frac{1+2 \frac{\Phi_A}{c^2}-(r_A \omega_A)^2}{1+2 \frac{\Phi_B}{c^2}-(r_B \omega_B)^2}}[/latex] Correction: [latex]\frac{d \tau_A}{d \tau_B}=\sqrt{\frac{1+2 \frac{\Phi_A}{c^2}-(r_A \omega_A/c)^2}{1+2 \frac{\Phi_B}{c^2}-(r_B \omega_B/c)^2}}[/latex] Link to comment Share on other sites More sharing options...

DanMP Posted March 8, 2017 Author Share Posted March 8, 2017 (edited) Ok, thank you very much. I think I got it. Like here we can express the gravitational potential at the point of interest as: [math]V(x)=-\frac{GM}{R-x}-\frac {Gm}x[/math] where x is the distance from the Earth to the point of interest and R is the Sun-Earth distance. [math]\frac {d V(x)}{dx}=-\frac{GM}{(R-x)^2}+\frac {Gm}{x^2}[/math] The point of interest is where the above is 0, so: [math]\frac{M}{(R-x)^2}=\frac {m}{x^2}[/math] and [math]x=\frac{R\sqrt{\frac m M}}{1+\sqrt{\frac m M}}= 258,775 km[/math] This is the point where Sun/Earth gravitational forces cancel each other. And if we include the orbital rotation the cancellation of forces takes place much higher, on the Hill sphere. So where is the real point in which a clock going towards the Sun would reverse its ticking rate? Do we have some experimental data? Edited March 8, 2017 by DanMP Link to comment Share on other sites More sharing options...

zztop Posted March 8, 2017 Share Posted March 8, 2017 (edited) Ok, thank you very much. I think I got it. Like here we can express the gravitational potential at the point of interest as: [math]V(x)=-\frac{GM}{R-x}-\frac {Gm}x[/math] where x is the distance from the Earth to the point of interest and R is the Sun-Earth distance. [math]\frac {d V(x)}{dx}=-\frac{GM}{(R-x)^2}+\frac {Gm}{x^2}[/math] The point of interest is where the above is 0, so: [math]\frac{M}{(R-x)^2}=\frac {m}{x^2}[/math] and [math]x=\frac{R\sqrt{\frac m M}}{1+\sqrt{\frac m M}}= 258,775 km[/math] This is the point where Sun/Earth gravitational forces cancel each other. No, the gravitational forces do not cancel in the above point, the gravitational potential has an extremum. This is different. The gravitational forces cancel each other for : [latex]\frac{GMm_p}{(R-x)^2}=\frac{Gmm_p}{x^2}[/latex] It so happens that the two equations have the same solution.This is because the force is the gradient (derivative wrt x) of the potential. So where is the real point in which a clock going towards the Sun would reverse its ticking rate? At the distance x calculated above. I do not think that there is any experiment confirming it but it would be fairly straightforward. Edited March 8, 2017 by zztop Link to comment Share on other sites More sharing options...

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