# Clock rates and effective ptotential.

1. Aug 27, 2010

### 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:

Passionflower gave this alternative equation for the relative clocks:

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:

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).

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

Last edited: Aug 27, 2010
2. Aug 27, 2010

### starthaus

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.

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?

3. Aug 27, 2010

### 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:

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

4. Aug 27, 2010

### bcrowell

Staff Emeritus
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 [Broken] .

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.

 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.

Last edited by a moderator: May 4, 2017
5. Aug 27, 2010

### yuiop

I understand it and it seems OK to me.

.. 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:

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.

Last edited: Aug 27, 2010
6. Aug 27, 2010

### yuiop

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.

Last edited: Aug 27, 2010
7. Aug 27, 2010

### Passionflower

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

8. Aug 27, 2010

### yuiop

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

9. Aug 27, 2010

### starthaus

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:

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

Last edited: Aug 27, 2010
10. Aug 27, 2010

### 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.

11. Aug 27, 2010

### yuiop

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

12. Aug 27, 2010

### starthaus

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

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

13. Aug 27, 2010

### yuiop

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).

14. Aug 27, 2010

### starthaus

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.

Last edited: Aug 27, 2010
15. Aug 27, 2010

### bcrowell

Staff Emeritus
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.

16. Aug 27, 2010

### 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.

17. Aug 27, 2010

### starthaus

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

Last edited: Aug 27, 2010
18. Aug 27, 2010

### yuiop

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:

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?

19. Aug 27, 2010

### yuiop

I did show the derivation here:

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 $$m = GM/c^2$$

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

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.

20. Aug 27, 2010

### starthaus

...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.

Last edited: Aug 27, 2010