Clock rates and effective ptotential.

[yuiop]

Clock rates and effective potential.

This is subject that came up in another thread that was getting "off topic" so I decided to start a new thread.

I quoted some formulas that I had informally derived (so they might be wrong) for the relative rates of moving clocks in a gravitational field as:

The equation for the time dilation of a satellite with natural motion was given by me in the other thread as:

$$\frac{T '}{T} = \sqrt{\left(1-\frac{GM}{(Rc^2-2GM)}\right)\left(1-\frac{2GM}{Rc^2}\right)}$$

where T' is the proper time of the satellite clock and T is the time recoded by the Schwarzschild observer at infinity. The only variables are the mass of the massive gravitational body that is being orbited and the orbital radius because those factors determine the orbital velocity. As Jesse quite correctly observed, the contribution of the altitude related time dilation is greater than the contribution of the velocity related time dilation factor for any orbit that is greater than R=4Gm/c^2 which applies to any circular Earth orbit. It can also be noted that increased orbital radius causes reduced tangential velocity of the orbiting satellite and reduced gravitational time dilation and both effects speed up the satellite clock with increased altitude. The upshot is that the higher the satellite orbit, the faster the clock runs, relative to clocks in lower orbit, as Ich said.

If you need to calculate the time dilation of a clock with arbitrary motion, such as clocks in aircraft, helicopters or towers, then you need to use the more general equation that I gave in the same post which is:

$$T '/T = \sqrt{1-v^2/c^2}\sqrt{1-2GM/(Rc^2)}$$

If you wish to compare the clock rate T1 of a clock at R1 with velocity V1 with the clock rate T2 of a clock at R2 with velocity V2, then you can use:

$$\frac{T_1}{T_2} = \sqrt{\left(\frac{c^2-v_1^2}{c^2-v_2^2}\right) \, \left(\frac{1-\frac{2GM}{r_1c^2}}{1-\frac{2GM}{r_2c^2}\right) }$$

Passionflower gave this alternative equation for the relative clocks:

To make things easy let's consider the simplest case a satellite on a circular polar orbit and a clock at one of the poles on Earth. The time ratio is:

$$\frac{\tau_S}{\tau_E} = 1+\frac{GM}{Rc^2} - \frac{3GM}{2rc^2}$$

At first glance the formulas do not seem to be in agreement. Passionflower's equation looks a bit like an "effective potential", but that might be a red herring.

I am pretty sure that one difference between my equations and the others is that I am using "local velocity" while the others are using coordinate velocity as measured by an observer at infinity.

Starthaus gave another version of passionflowers equation as:

Yes, this is correct. Here is a more precise (but with same results) calculation:

Start with the Schwarzschild metric for $$dr=d \theta =0$$

$$(cd\tau)^2=\alpha (cdt)^2-(r d\phi)^2$$

valid for your satellite

and:

the Schwarzschild metric for $$dr=d \theta =d \phi=0$$

$$(cd\tau)^2=\alpha (cdt)^2$$

valid at the pole

where $$\alpha=1-2m/r$$

and $$m=\frac{GM}{c^2}$$

$$\frac{d \tau}{dt}=\sqrt{1-2m/r-(r \omega/c)^2}<1$$ for any r



The above gives:

$$\frac{d \tau_s}{d \tau_p}=\sqrt{\frac{1-2m/r-(r \omega/c)^2}{1-2m/R}}$$

So:

$$\frac{d \tau_s}{d \tau_p}>1$$ if

$$2m/r+(r \omega/c)^2<2m/R$$

But, from the Kepler's law, it can be shown that :

$$(r \omega/c)^2=m/r$$

so, the above becomes:

$$3m/r<2m/R$$ i.e., your condition $$r>\frac{3R}{2}$$

...
Now, GPS satellites are moving at about 20,000km above the Earth, so , if it weren't for the frequency precompensation at launch, their clocks would be faster than the ones left on Earth.

It would be nice to sort these equations out and see how they relate to each other and fix any errors (even if they are mine).

Passionflower also asked

Reasoning why we have the 3/2R ratio would be very interesting. I wonder if we can relate it to the fact that the photon sphere has the same ratio with respect to the Schwarzschild radius.

which was also getting off topic so I have put that question here too.

 

[starthaus]

I understand that two clocks at the same effective potential are running at the same rate, but if they are different potentials are the relative clock rates exactly proportional to the difference in effective potentials

You can find that out by Starting the derivation from scratch from the Schwarzschild metric, in the most general case, if there is radial and/or orbital motion, the relative rates are not proportional to the difference in potential. Even for the simplest case $$dr=d \phi = d \theta =0$$ you should get :

$$\frac{d \tau_1}{d \tau_2}=\sqrt{\frac{1-2m/r_1}{1-2m/r_2}}$$

where $$m=\frac{GM}{c^2}$$

The above is the basic theory behind the Pound-Rebka experiment.

So, the answer to your question is "no"

or does it just give an indication of faster/slower?

Starthaus gave another version of passionflowers equation as:

The formalism, if done correctly, will tell you the differences in clock rates. Both I and Passionflower dealt with a specific case and our results agree. He started from the result of an exercise from a book, I started from scratch. Do you understand my derivation?

 

[Passionflower]

Indeed, and generally we can say that in the weak field and for slow moving particles:

$$g_{00} = \left 1+\frac{2\Phi}{c^2}\right$$

Since:

$$d\tau^2 = g_{00}dt^2$$

We get:

$$d\tau = \left( 1+\frac{2\Phi}{c^2}\right)^{1/2}dt$$

 

[bcrowell]

My question is "is time dilation and effective potential exactly the same thing?"

Sort of. First, the whole thing only works in a stationary field. Second, you need to separate the gravitational time dilation from the kinematic effect (which can be done in a uniquely well defined way in the case of a stationary field). Then you can simply define the potential in terms of the gravitational time dilation. See Rindler, Essential Relativity, 2nd ed., p. 120, or http://www.lightandmatter.com/html_books/genrel/ch07/ch07.html#Section7.3 .

I understand that two clocks at the same effective potential are running at the same rate, but if they are different potentials are the relative clock rates exactly proportional to the difference in effective potentials or does it just give an indication of faster/slower?

The time dilation factor equals $e^{-\phi}$ (with c=1). This can be taken as the definition of the potential $\phi$.

The gravitational time dilation factor in the Schwarzschild spacetime is simply the square root of the time-time component of the metric. Equating that to $e^{-\phi}$ gives $\phi=(1/2)\ln(1-2m/r)$ (Rindler, p. 148). This is an exact expression. The expressions you listed in the OP are probably all weak-field approximations to this, but to check that, you'd need to separate out the kinematic effects from the gravitational ones.

[Edit] Oops, I implied incorrectly in this post that the logarithmic expression was not exactly consistent with kev's expression. Actually I think it is exactly consistent -- see #15.

 

[yuiop]

The formalism, if done correctly, will tell you the differences in clock rates. Both I and Passionflower dealt with a specific case and our results agree. He started from the result of an exercise from a book, I started from scratch. Do you understand my derivation?

I understand it and it seems OK to me.

Continuing your derivation:

The above gives:

$$\frac{d \tau_s}{d \tau_p}=\sqrt{\frac{1-2m/r-(r \omega/c)^2}{1-2m/R}}$$

So:

$$\frac{d \tau_s}{d \tau_p}>1$$ if

$$2m/r+(r \omega/c)^2<2m/R$$

But, from the Kepler's law, it can be shown that :

$$(r \omega/c)^2=m/r$$

.. to simplify it a bit, substitute the Kepler equation into the equation at the top and you obtain:

$$\frac{d \tau_s}{d \tau_p}=\sqrt{\frac{1-2m/r-m/r}{1-2m/R}} = \sqrt{\frac{1-3m/r}{1-2m/R}}$$

Now if I take the equation given by Passionflower:

To make things easy let's consider the simplest case a satellite on a circular polar orbit and a clock at one of the poles on Earth. The time ratio is:

$$\frac{\tau_S}{\tau_E} = 1+\frac{GM}{Rc^2} - \frac{3GM}{2rc^2}$$

and alter it so that the symbols are the same as the ones you are using, then Passion's equation can be stated as:

$$\frac{d \tau_s}{d \tau_p}= 1+\frac{m}{R} - \frac{3m}{2r}$$

Now if your two equations agree, then the following should be true:

$$\sqrt{\frac{1-3m/r}{1-2m/R}} - \left(1+\frac{m}{R} - \frac{3m}{2r} \right) = 0$$

..but that does not seem to be the case. I suspect that is because Passion's equation is a weak field approximation.

 

[yuiop]

The expressions you listed in the OP are probably all weak-field approximations to this, but to check that, you'd need to separate out the kinematic effects from the gravitational ones.

The expressions given by me are intended to be exact and the expression given by Starthaus is also exact, coming directly from the Schwarzschild metric. As I mentioned in the last post I suspect that Passionflower's equation is (possibly?) a weak field approximation. Alomost certainly the equation given by Starthaus is exact and if Passionflower's expression is exact it should agree with it.

 

[Passionflower]

You got an extra c^2 in my formula after transition to m.

 

[yuiop]

You got an extra c^2 in my formula after transition to m.

Thanks Fixed the equation, but I do not think that solves the problem of your equation and Starthaus's not being equivalent.

 

[starthaus]

Now if your two equations agree, then the following should be true:

$$\sqrt{\frac{1-3m/r}{1-2m/R}} - \left(1+\frac{m}{R} - \frac{3m}{2r} \right) = 0$$

..but that does not seem to be the case. I suspect that is because Passion's equation is a weak field approximation.

You are trying to equate my exact formula with passionflower Taylor expansion of the exact formula, this is why it isn't working.

$$\sqrt{\frac{1-3m/r}{1-2m/R}}$$

can be approximated as:$$\frac{1-3m/2r}{1-m/R}=(1-3m/2r)(1+m/R)=1+m/R-3m/2r$$

 

[yuiop]

If I take this equation I gave earlier:

$$\frac{T_1}{T_2} = \sqrt{\left(\frac{c^2-v_1^2}{c^2-v_2^2}\right) \, \left(\frac{1-\frac{2GM}{r_1c^2}}{1-\frac{2GM}{r_2c^2}\right) }$$


and subsitute R2 = R and V2=0 for the clock at the pole and R1=r and the local version of the Kepler orbital velocity V1 = rw = $\sqrt{(mc^2/(r-2m))}$ for the the clock in orbit (See #10 of https://www.physicsforums.com/showthread.php?p=2690217#post2690217) then after simplification I recover:

$$\frac{T_1}{T_2} = \sqrt{\frac{1-3m/r}{1-2m/R}}$$

which means my equation is in agreement with the equation given by Starthaus.

 

[yuiop]

You are trying to equate my exact formula with passionflower Taylor expansion of the exact formula, this is why it isn't working.

$$\sqrt{\frac{1-3m/r}{1-2m/R}}$$

can be approximated as:


$$\frac{1-3m/2r}{1-m/R}=(1-3m/2r)(1+m/R)=1+m/R-3m/2r$$

Yep. I suspected that Passion's equation was an approximation. Nice demonstration

 

[starthaus]

If I take this equation I gave earlier:

$$\frac{T_1}{T_2} = \sqrt{\left(\frac{c^2-v_1^2}{c^2-v_2^2}\right) \, \left(\frac{1-\frac{2GM}{r_1c^2}}{1-\frac{2GM}{r_2c^2}\right) }$$

Yes, this is the equation derived by DrGreg some time ago. It is correct.

which means my equation is in agreement with the equation given by Starthaus.

Both DrGreg and I use the same exact derivation, so the formulas are bound to agree.

 

[yuiop]

Reasoning why we have the 3/2R ratio would be very interesting. I wonder if we can relate it to the fact that the photon sphere has the same ratio with respect to the Schwarzschild radius.

Perhaps there is some merit in considering that a stationary observer hovering just above the event horizon has the maximum (coordinate) time dilation due to gravitational time dilation and zero 'SR time dilation' and an observer in a circular orbit slightly outside the photon sphere has the maximum (coordinate and proper) velocity approaching c which implies the maximum 'SR time dilation'. These 'maximums' are 3/2 away from each other.

Using the exact equation for circular orbits:

$$\frac{d \tau_s}{d \tau_p} = \sqrt{\frac{1-3m/r}{1-2m/R}}$$

The ratio becomes imaginary for r<3m and R>2M. This is because it is not possible to have a circular orbit below r=3m (which would require the particle to orbit at greater than the local speed of light).

 

[starthaus]

Using the exact equation for circular orbits:

$$\frac{d \tau_s}{d \tau_p} = \sqrt{\frac{1-3m/r}{1-2m/R}}$$

The ratio becomes imaginary for r<3m and R>2M. This is because it is not possible to have a circular orbit below r=3m (which would require the particle to orbit at greater than the local speed of light).

R is always larger than 2m since 2m is the Schwarzshild radius.
r<3m makes little sense since it would mean orbiting inside the nassive body, very close to its core. For example, for the Earth $$2m=r_s=9mm$$. This makes no physical sense.

 

[bcrowell]

The expressions given by me are intended to be exact and the expression given by Starthaus is also exact, coming directly from the Schwarzschild metric. As I mentioned in the last post I suspect the Passionflower's equation is (possibly?) a weak field approximation. Alomost certainly the equation given by Starthaus is exact and if Passionflower's expression is exact it should agree with it.

Oops, sorry, I wasn't thinking straight in my #4 when I implied that the logarithmic expression for the potential was inconsistent with your expression. Actually if you take the exponential of Rindler's potential, you get your expression (without the kinematic factor). So I think this confirms that yours is correct, and also exact, not a weak-field approximation, when r is taken to be the Schwarzschild radius.

 

[yuiop]

The equation for the time dilation of a satellite with natural motion was given by me in the other thread as:

$$\frac{T '}{T} = \sqrt{\left(1-\frac{GM}{(Rc^2-2GM)}\right)\left(1-\frac{2GM}{Rc^2}\right)}$$

where T' is the proper time of the satellite clock and T is the time recoded by the Schwarzschild observer at infinity.

Unfortunately the above equation is wrong.

The above equation simplifies to:

$$\frac{T '}{T} = \sqrt{\left(1-\frac{3GM}{Rc^2}\right)}$$

which is in agreement with the equation given by Starthaus, so I do not see his grounds for claiming it is incorrect.

 

[starthaus]

The above equation simplifies to:

$$\frac{T '}{T} = \sqrt{\left(1-\frac{3GM}{Rc^2}\right)}$$

which is in agreement with the equation given by Starthaus, so I do not see his grounds for claiming it is incorrect.

This is nowhere close to the formula I derived, nor is it correct, do you want to try again?

 

[yuiop]

The above equation simplifies to:

$$\frac{T '}{T} = \sqrt{\left(1-\frac{3GM}{Rc^2}\right)}$$

which is in agreement with the equation given by Starthaus, so I do not see his grounds for claiming it is incorrect.

This is nowhere close to the formula I derived, nor is it correct, do you want to try again?

Do you read the accompanying text? It is the equation for the satellite's proper time relative to Schwarzschild coordinate time. You derivation of the satellite's proper time was:

Yes, this is correct. Here is a more precise (but with same results) calculation:

Start with the Schwarzschild metric for $$dr=d \theta =0$$

$$(cd\tau)^2=\alpha (cdt)^2-(r d\phi)^2$$

valid for your satellite

So for the satellite:

$$\frac{d \tau_s}{dt}=\sqrt{1-2m/r-(r \omega/c)^2}$$

You also gave an equation for Kepler's third law as $$(r \omega/c)^2=m/r$$ and after substituting this equation directly into the equation above you get:

$$\Implies \frac{d \tau_s}{dt}=\sqrt{1-3m/r} \qquad (Eq1)$$

which is in agreement with the equation I gave above, so you are are wrong. The only difference is that I was using the symbol R rather than r for the radius of the satellite's orbit, but I was quoting the equation from an old post that was written way before your post and you are using the symbol m to represent $$GM/c^2$$. So what is your problem?

For the clock at the pole at radius R with no orbital motion, $$\omega=0$$ so:

$$\frac{d \tau_p}{dt}=\sqrt{1-2m/R-(R \omega/c)^2}$$

$$\Implies \frac{d \tau_p}{dt}=\sqrt{1-2m/R} \qquad (Eq2)$$

For the ratio of the proper time of the satellite relative to the proper time of the clock at the pole, divide (Eq1) by (Eq2) to get:

$$\frac{d \tau_s}{d \tau_p}=\sqrt{\frac{1-3m/r}{1-2m/R}}$$

Surely you did not really need me to explain that to you?

 

[yuiop]

Yes, I read, it is total nonsense. Try showing your derivation.

I did show the derivation here:

It is the equation for the satellite's proper time relative to Schwarzschild coordinate time. You derivation of the satellite's proper time was:

So for the satellite:

$$\frac{d \tau_s}{dt}=\sqrt{1-2m/r-(r \omega/c)^2}$$

You also gave an equation for Kepler's third law as $$(r \omega/c)^2=m/r$$ and after substituting this equation directly into the equation above you get:

$$\Implies \frac{d \tau_s}{dt}=\sqrt{1-3m/r} \qquad (Eq1)$$

Which is the same as my equation because for the satellite's proper time relative to Schwarzschild coordinate time, $$T' = d\tau_s$$ and $$T=dt$$ and you have defined $$m = GM/c^2$$

No, it isn't you are missing terms, it is quite obvious. How is your $$\frac{T '}{T} = \sqrt{\left(1-\frac{3GM}{Rc^2}\right)}$$ consistent with my derivation? Don't you see that you are missing terms?

What missing terms when $$T' = d\tau_s$$ and $$T=dt$$ and you have defined $m = GM/c^2$ ?

I think that you did not understand the exchange with Passionflower, the problem we were working on compares the clock rate of the satellite orbiting at r with the clock rate at the pole (at R).

I did understand your exchange with Passionflower. Comparing the clock rate of the satellite orbiting at r with the clock rate at the pole (at R), involves calculating the clock rate of the satellite as an intermediate step. My equation was for that intermediate step, i.e. It is the equation for the satellite's proper time relative to Schwarzschild coordinate time.

In your intermediate step, you gave the equation for the satellite as:

$$\frac{d \tau_s}{dt}=\sqrt{1-2m/r-(r \omega/c)^2}$$

After substitution of the equation obtained from Kepler's 3rd law, this becomes:

$$\Implies \frac{d \tau_s}{dt}=\sqrt{1-3m/r} \qquad (Eq1)$$

This equation (Eq1) is the same as my equation for the satellite's proper time relative to Schwarzschild coordinate time.

To obtain it substitute $$T' = d\tau_s$$, $$T=dt$$ and $GM/c^2 = m$ into (Eq1).

Really, I can only assume that a person of your mathematical ability is pretending not understand, for reasons that are beyond me.

 

[starthaus]

You also gave an equation for Kepler's third law as $$(r \omega/c)^2=m/r$$ and after substituting this equation directly into the equation above you get:

$$\Implies \frac{d \tau_s}{dt}=\sqrt{1-3m/r} \qquad (Eq1)$$

...which, as I explained a few times to you already , is valid only for the class of satellites that satisfy $$(r \omega/c)^2=m/r$$. It isn't valid for the planes in the Haefele-Keating experiment, it isn't valid for GPS. For those cases you need to use the general formula I derived:

$$\frac{d \tau_s}{dt}=\sqrt{1-2m/r-(r \omega/c)^2}$$

The terms in $$r \omega$$ are the obviously missing terms in your formula.

 

[yuiop]

...which, as I explained a few times to you already , is valid only for the class of satellites that satisfy $$(r \omega/c)^2=m/r$$. It isn't valid for the planes in the Haefele-Keating experiment, it isn't valid for GPS. For those cases you need to use the general formula I derived:

$$\frac{d \tau_s}{dt}=\sqrt{1-2m/r-(r \omega/c)^2}$$

Yes, my equation is specifically for a satellite in circular orbit.

In your exchange with Passionflower, he specified:

To make things easy let's consider the simplest case a satellite on a circular polar orbit and a clock at one of the poles on Earth.

Satellites in circular orbit, excludes planes in the Haefele-Keating experiment.

In Newtonian physics, Kepler's 3rd law applies to circular and elliptical orbits, but in GR things are more complicated and for GPS satellites with highly elliptical orbits the equation for a satellite with a circular orbit is only an approximation.

 

[starthaus]

Yes, my equation is specifically for a satellite in circular orbit.

In your exchange with Passionflower, he specified:
Satellites in circular orbit, excludes planes in the Haefele-Keating experiment.

I explained earlier that the laws of physics apply equally to all vehicles, so, you need to derive formulas that don't differentiate between rockets and planes.

In Newtonian physics, Kepler's 3rd law applies to circular and elliptical orbits, but in GR things are more complicated and for GPS satellites with highly elliptical orbits the equation for a satellite with a circular orbit is only an approximation.

I know all that, I have been pointing out to you that your formula is incorrect for the general case since it was obviously missing the angular speed terms. As a matter of fact, your original derivation is highly dubious, even for the circular orbit case. You tack in a "time dilation" that is in reality a time "contraction".
Anyways, you know now how to derive the general case formula.

 

[JesseM]

Responding to starthaus' post #26 on the 'Laws of physics are the same' thread:

The above cannot possibly be right since it is falsified by experiment. In the Hafele-Keating experiment one of the atomic clocks lagged behind the ground clock while the other was ahead. So, something is wrong in kev's computations.

kev's calculation was only for comparing proper time on a clock moving in a circular orbit with coordinate time in Schwarzschild coordinates (which matches up to proper time for a clock at infinity).

Even that seems wrong since $$\frac{d\tau}{dt}<1$$ for any value of the Schwarzschild coordinate $$r$$ for the case of objects in circular orbits around a massive object like the Earth.
If you point to the exact post, I will be able to point out the error in kev's calculations. Did you check kev's calculations for correctness?

Why do you think it seems wrong? Do you understand that the clock on Earth is moving in Schwarzschild coordinates due to the Earth's rotation, and that the plane that flies in the direction opposite to the rotation of the Earth would have a smaller speed in Schwarzschild coordinates than the clock on the ground?

Look, I am not going to engage in this Q&A game with you again. I know GR quite well and it takes one line of calculations to show that $$\frac{d\tau}{dt}<1$$ for any r.

Of course, I don't dispute that and never did. The point is that this appears to be a complete non sequitur argument, you have never explained why you think that this trivial fact somehow conflicts with kev and pervect's claim about the total time dilation for a clock moving in a circular path of constant radius (which according to their equations is indeed always running slower than coordinate time, i.e. their equations imply that $$\frac{d\tau}{dt} < 1$$ for any clock orbiting at a finite radius). Likewise you never explained why you think the Hafele-Keating experiment conflicts with pervect and kevs' results--it seems to me it matches them just fine, since the clock on the plane that went against the Earth's rotation would have a smaller coordinate velocity than the ground clock in Schwarzschild coordinates (assuming we are approximating the Earth's metric as a Schwarzschild metric, which isn't exactly correct but is probably close enough to deal with the HK experiment), so pervect's formula predicts that if they are both moving in circular paths at about the same radius, the ground clock should elapse less time (be more time dilated relative to Schwarzschild coordinate time) than the clock on that plane. Meanwhile the clock on the plane that was traveling with the Earth's rotation should have a greater coordinate velocity than the ground clock, so according to pervect's formula it should elapse less time than the ground clock. So, one plane clock should be ahead of the ground clock and the other should be behind, just as was seen in the experiment.

I looked, pervect is using an incorrect metric, his formula for $$g_{tt}$$ is incorrect.
If you want, I can do the correct one-line calculation.

Here you are simply repeating an argument I already disposed of on the other thread in post #119 :

Yes, obviously. pervect not only truncated the metric, he also got the wrong. Do you disagree?

He made a minor error with $$g_{tt}$$ which kev corrected in his response, but other than that I still don't know what you mean by "truncated", in his post he started from the abstract form of the metric which included all the terms:
$$c^2 d\tau^2 = g_{tt} dt^2 + g_{rr} dr^2 + g_{\theta\theta} d\theta^2 + g_{\phi\phi} d\phi^2$$
Are you calling this "truncated" just because he didn't actually write out the equations for each "g"? There's nothing non-rigorous about this, anyone can look up what they'd all be in the Schwarzschild metric. Or do you think there was something wrong with pervect's next step of specifying that he was talking about a circular orbit where $$dr/dt$$ and $$d\theta /dt$$ would be zero, and eliminating the appropriate terms given these conditions?

If you want proof that pervect's error makes no difference to his final result, here is a modified version of his derivation which corrects the error (edit: I also noticed another error where pervect treated g_tt as equal to (1 - r:

It might be instructive to work out the problem using GR in the Earth centered frame, using the Schwarzschild metric, for a circular orbit at the equator. This will also show that the usual SR velocity time dilation formula is only an approximation.

It might be helpful to recap how we get the equation for time dilation in SR, first.


The time dilation that we wish to solve for is just
$$\frac{dt}{d\tau}$$

In SR, we know the flat space-time metric is

$$c^2 d\tau^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2$$

dividing both sides by dt^2, we get

$$c^2 \left( \frac{d\tau}{dt} \right) ^2 = c^2 - \left( \frac{dx}{dt} \right)^2 - \left( \frac{dy}{dt} \right)^2 - \left( \frac{dz}{dt} \right)^2$$

Solving for $d\tau / dt$ we get

$$\frac{d\tau}{dt} = \sqrt{1 - \frac{ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2 }{c^2} }$$

We recognize this as
$$\frac{d\tau}{dt} = \sqrt{1 - \frac{v^2}{c^2}}$$

but since we want the reciprocal, $dt / d\tau$, we get the usual relation

$$\frac{dt}{d\tau} = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$

Now, we just need to repeat this for the case of a gravitating body. We'll use Schwarzschild coordinates to get the curved space-time metric, rather than the flat metric we used in SR.

Note that in Schwarzschild coordinates, $\theta$ is Pi at the equator, and $\phi$ varies from 0 to 2Pi as it sweeps out the orbit. r is constant for a circular orbit.

So we write

$$c^2 d\tau^2 = g_{tt} dt^2 + g_{rr} dr^2 + g_{\theta\theta} d\theta^2 + g_{\phi\phi} d\phi^2$$

dividing both sides of the equation by dt^2, and dropping some terms that we know to be zero, such as $\frac{dr}{dt}$ and $\frac{d\theta}{dt}$, we get

$$c^2 \left( \frac{d\tau}{dt} \right)^2 = g_{tt} + g_{\phi\phi}\left(\frac{d\phi}{dt}\right)^2$$

We know that for the schwarzschild metric

http://en.wikipedia.org/w/index.php?title=Schwarzschild_metric&oldid=325313722

$$g_{tt} = \left(1-\frac{r_s}{r}\right)\,c^2$$

and we know that

$$g_{\phi\phi} = -r^2$$

as sin$\phi$ is one.

Putting this together we get
$$c^2 \left( \frac{d\tau}{dt} \right)^2 = c^2 \, \left(1-\frac{r_s}{r}\right)\ - r^2 \left(\frac{d\phi}{dt}\right)^2$$

This can be rewritten as

$$\frac{d\tau}{dt}= \sqrt{ (1-\frac{r_s}{r}) } \sqrt{1 - \frac{1}{(1-\frac{r_s}{r})} \left( \frac{r \frac{d\phi}{dt} }{c} \right)^2$$


This is almost in the form of the product of the GR and SR time dilation but not quite exactly.

Note that if we set $d\phi / dt$ to zero, we see that the time dilation is just the gravitational time dilation

$$\frac{dt}{d\tau}= \frac{1}{\sqrt{(1-\frac{r_s}{r})}}$$

But because the gravitational time dilation is so nearly unity, it provides only a tiny correction to the velocity in the SR formula to multiply it by g_tt so it's approximately correct to multiply the SR time dilation by the gravitational time dilation for a non-moving object to get the total time dilation.

Do you see any error in this slightly edited version?

kev's calculation is just a slight modification where he substitutes the local velocity (velocity in a locally inertial frame instantaneously at rest in Schwarzschild coordinates) for the Schwarzschild coordinate velocity, see this post or this one.

I looked. kev's calculation has two mistakes:

1. He uses the wrong metric , i.e. wrong $$g_{tt}$$

Oh really? He writes $$g_{tt}$$ as $$c^2 (1 - r_s /r )$$ (correcting pervect's typo of accidentally squaring the (1 - r_s/r) ), exactly the same as it's written here for example, do you claim this is wrong?

2. even worse, he does not calculate the time dilation for orbital motion

Looking at the derivation again, it seems to me it is not specific to orbital motion, it applies to any circular motion of constant radius and constant angular velocity (circular orbits are a special case, where we can determine the coordinate orbital velocity using Kepler's law according to kev's post #10 here). Do you disagree with his claim that the Schwarzschild coordinate velocity u for an object at constant radius with constant angular velocity would be $$u = r \left(\frac{d\phi}{dt}\right)$$? Do you disagree that for an orbiting clock, the local velocity is related to the coordinate velocity by the equation $$u = v \sqrt{1-\frac{r_s}{r}}$$? These are the only aspects of kev's derivation which differ from the corrected version of pervect's derivation (and note that DrGreg agreed with kev's result, see post #8 here)

I remember now that we already discussed pervect's calculations on this thread where you raised a lot of spurious objections and wouldn't give straight answers to the questions I and others asked you, let's not have a repeat of that please (I will only agree to discuss this again if you agree in advance to give definite answers to any questions asked of you).

That thread dealt with yet different errors in calculating time dilation for radial (not orbital) motion.

No it didn't, the original post by espen180 said nothing about radial vs. orbital motion (you had some kind of weird memory conflation where you imagined that the thread was actually a split from another thread started by Dmitry67, but I pointed out in post #40 that the timestamps didn't fit with your memory, and espen180 confirmed in post #44 that the thread was started by him and was not a split from another thread). And my post #3 on that thread said:

For an object in a circular orbit, the total time dilation is a product of gravitational and velocity-based time dilation--see kev's post #8 on this thread and post #10 here. But cases other than a circular orbit would probably be more complicated.

To which you replied in post #6:

Hi Jesse,

I don't think the expressions put down by kev in that post are correct.

So, clearly you were disputing that kev's results were correct for the case of a circular orbit, and that's what led to the long protracted debate which took up most of the remainder of that thread.

 

[yuiop]

I explained earlier that the laws of physics apply equally to all vehicles, so, you need to derive formulas that don't differentiate between rockets and planes.

If you insists all equations are fully general, then your equation:

$$\frac{d \tau_s}{d \tau_p}=\sqrt{\frac{1-2m/r-(r \omega/c)^2}{1-2m/R}}$$

is not sufficiently general enough to analyse the Hafele-Keating experiment.

If we define the proper time of the aircraft as $d\tau_a$ and the proper time of the ground clock as $d\tau_g$ then the equation you need to analyse the HK experiment is:

$$\frac{d \tau_a}{d \tau_g}=\sqrt{\frac{1-2m/r-(r \omega_a/c)^2}{1-2m/R - (R \omega_g/c)^2}}$$

where $$\omega_a$$ is the angular velocity of the aircraft and $$\omega_g$$ is the angular velocity of the ground clock. In the real HK experiment the ground clock was not at the pole and has angular velocity by virtue of the Earth's rotation.

Insisting on fully general formulas that are applicable to every conceivable situation when analysing simple situations is pedantic and over complicates things unnecessarily. Using simplified equations where the simplifying assumptions are stated should be sufficient and not automatically "wrong" as you keep insisting. If non fully general equations are wrong, then your equation is wrong because it does not consider the case where the ground clock is moving, or where the vehicles altitude is changing, etc, etc.

 

[starthaus]

If you insists all equations are fully general, then your equation:

$$\frac{d \tau_s}{d \tau_p}=\sqrt{\frac{1-2m/r-(r \omega/c)^2}{1-2m/R}}$$

is not sufficiently general enough to analyse the Hafele-Keating experiment.

Obviously not. The clock left on Earth in Passionflower's exercise is left at the pole, so it undergoes no rotation.

If we define the proper time of the aircraft as $d\tau_a$ and the proper time of the ground clock as $d\tau_g$ then the equation you need to analyse the HK experiment is:

$$\frac{d \tau_a}{d \tau_g}=\sqrt{\frac{1-2m/r-(r \omega_a/c)^2}{1-2m/R - (R \omega_g/c)^2}}$$

True, I have shown you the general formula long ago in another thread,. It is good to see that you are starting to understand that your formulas are insufficient in representing the physical reality.

Now, if you want to be absolutely correct, you should write:

$$\frac{d \tau_a}{d \tau_g}=\sqrt{\frac{1-2m/r-(dr/cdt)^2(1-2m/r)^{-1}-(r \omega_a/c)^2}{1-2m/R - (R \omega_g/c)^2}}$$

iin order to take into consideration the fact that the planes climbed to their respective flight altitudes.

Insisting on fully general formulas that are applicable to every conceivable situation when analysing simple situations is pedantic and over complicates things unnecessarily.

The issue came up in my pointing out to JesseM that he misunderstood the GPS time dilation, he brought in your formula and I pointed out to him that your formula is not applicable due to its obvious limitations. I have also pointed out to him that you gave a wrong derivation initially (I missed the fact that you went back and you gave a correct derivation for a restricted case).

If non fully general equations are wrong, then your equation is wrong because it does not consider the case where the ground clock is moving, or where the vehicles altitude is changing, etc, etc.

I have given you the complete formula in a different thread. I have used the appropriate formula for solving Passionflower's exercise. Anything else I can do for you?

 

[JesseM]

The issue came up in my pointing out to JesseM that he misunderstood the GPS time dilation, he brought in your formula and I pointed out to him that your formula is not applicable due to its obvious limitations.

I didn't say anything about GPS, I was just talking about objects in circular orbits.

 

[yuiop]

Obviously not. The clock left on Earth in Passionflower's exercise is left at the pole, so it undergoes no rotation.

It is good to see that you are finally coming to understand that it is OK to use non-fully generalised equations for limited cases. in future could you stop announcing my equations are wrong simply because they are specialised equations for a limited case, especially when I have made that clear. It is extremely irritating and confusing to newcomers. Surely you must be picking up on the irritation and frustration expressed by other members of this forum, about your behavior?

For example when analysing Passionflowers exercise you used Kepler's 3rd law:

But, from the Kepler's law, it can be shown that :

$$(r \omega/c)^2=m/r$$

Now if I wanted to be as pedantic, irritating, petty and small minded as you, I would say your Kepler based equation is wrong, without any explanation.

It is wrong because the Kepler's 3rd law is actually:

$$T^2 = \frac{4 \pi^2 r^3}{G(M_1 + M_2)}$$

$$\Rightarrow \frac{(2\pi r)^2}{T^2} = (r\omega/c)^2 = \frac{GM_1+GM_2}{rc^2}$$

where M1 is the mass of the massive body and Massive body and M2 is the mass of the satellite. You have assumed the mass of the satellite to be zero (which is physically incorrect) and not stated that you have made this assumption.

Laws of physics should apply to all vehicles, not just vehicles with zero mass.

Second of all, you have not proven that Kepler's 3rd law which is a Newtonian equation, is still valid in GR without modification, but simply assumed that. I was the first to use Kepler's 3rd law in the satellite derivation and I did not prove it, I simply assumed it and showed that the assumption was consistent with other facts we know, such as the local orbital speed at the r=3M being the speed of light. You on the other hand expect everyone to prove all assumptions and so you are not living up to the standards you set everyone else.

See how petty this sort of behavior is? Could you please stop behaving like this. I do not suppose for a moment you are enough of a gentleman to go and withdraw your unjustified claims in all the multiple threads that my equations are "incorrect". I started this thread to stop you derailing other threads and yet you have still gone back to those old threads and continued stating "kev's equations are incorrect" even though they are mathematically correct and physically correct for the limited cases I was considering, where I stated my simplifying assumptions. By your standards all the Lorentz transformations are wrong, because they are not valid over extended regions in a gravitational field. Do you see how silly this attitude is? What you are doing amounts to a sustained smear campaign and it is about time it stopped. I am seriously considering making an official complaint.

Relativity is complicated enough without you introducing petty and imaginary complications all the time.

On the positive side you mathematical and analytical ability is pretty good (with the exception of a couple of major misconceptions even in this area that you have not conceded yet), but your physical intuition sucks big time and you would do well to listen to the likes of JesseM and Dalespam on these matters. If you stop being petty and put some of your genuine abilities to good use, then we all might benefit and learn and do some good analytical work together and this forum would be a more pleasant and constructive place for everybody.

 

[starthaus]

It is good to see that you are finally coming to understand that it is OK to use non-fully generalised equations for limited cases.

This is wrong. What I told you is that I used the exact solution for the problem posed by Passionflower. No hacks, kev.

in future could you stop announcing my equations are wrong simply because they are specialised equations

They aren't "specialised", they are plain wrong or ,at best, hacks.
I would recommend that you invested into buying the Rindler book (I know, I recommended that to you before).

Now if I wanted to be as pedantic, irritating, petty and small minded as you, I would say your Kepler based equation is wrong, without any explanation.

It is wrong because the Kepler's 3rd law is actually:

$$T^2 = \frac{4 \pi^2 r^3}{G(M_1 + M_2)}$$

$$\Rightarrow \frac{(2\pi r)^2}{T^2} = (r\omega/c)^2 = \frac{GM_1+GM_2}{rc^2}$$

If you bought the Rindler book, on page 238 you would find that , in GR:

$$\omega^2=\frac{m}{r^3}$$

which is precisely what I used in my derivation. Like I said, I don't use any hacks.

where M1 is the mass of the massive body and Massive body and M2 is the mass of the satellite. You have assumed the mass of the satellite to be zero (which is physically incorrect) and not stated that you have made this assumption.

Invest in a good book and study it.

Second of all, you have not proven that Kepler's 3rd law which is a Newtonian equation, is still valid in GR without modification, but simply assumed that.

Err, no. I have showed you that before. To refresh your memory, see https://www.physicsforums.com/blog.php?b=1957 , for the complete derivation. It falls straight from the Euler-Lagrange equations.

On the positive side you mathematical and analytical ability is pretty good (with the exception of a couple of major misconceptions even in this area that you have not conceded yet), but your physical intuition sucks big time and you would do well to listen to the likes of JesseM and Dalespam on these matters. If you stop being petty and put some of your genuine abilities to good use, then we all might benefit and learn and do some good analytical work together and this forum would be a more pleasant and constructive place for everybody.

I understand your frustrations but you should look at your errors rather than attacking me. I only point out your misconceptions and hacks. Only a very small percentage of my posts are concerned with correcting your errors. Unfortunately the threads become very long because you refuse to admit to error or, even worse, you repeat the same errors in various threads.