# Time dilation where there is no net gravity?

1. May 22, 2010

### jaketodd

Imagine a low-mass object between two massive objects. Their configuration is such that the low-mass object does not move due to the gravity of the two massive objects. Since the gravity is canceled for the low-mass object, is the time dilation canceled as well for the low-mass object?

Thanks,

Jake

2. May 22, 2010

### Jonathan Scott

No, time dilation relates to the potential, but the gravitational field relates to the gradient of the potential.

As another example, time runs a little slower at the middle of a massive body than it does outside it, but there is no field at the middle.

(Last time I saw the Wikipedia article on gravitational time dilation, it was wrong about that, so there's an example of why you shouldn't believe everything you read on Wikipedia).

3. May 22, 2010

### jaketodd

So, in my example, there would be the same amount of time dilation on the low-mass object as if you took one of the massive objects and combined it together with the other one?

Thanks,

Jake

4. May 22, 2010

### jaketodd

I'll try to clarify my question:

In the original example I was talking about a low-mass object mid way in between two equally massive objects and whether there would be time dilation for the low-mass object. That was answered with a Yes.

I am now asking, if you took one of the massive objects and combined it with the other massive object, making a super massive object, but the same size as before you combined them, would the time dilation on the low-mass object be the same as in the original example?

Thanks,

Jake

5. May 22, 2010

### A.T.

1) To determine time dilation you need two clocks to compare. So let's assume the reference clock is very far away from the massive bodies and your test clock?

2) The amount of time dilation in your first example depends on how far away the two masses are from each other. So how to compare it to your second scenario?

3) You don't specify the shape of the masses. Spheres in both scenarios?

4) Where is the test-clock relative to the big mass in the second scenario?

6. May 22, 2010

### Jonathan Scott

Provided that we are only talking about the weak gravity approximation case (where nothing as dense as a neutron star is involved) then the change in the clock rate relative to infinity, expressed as a fraction, is the same as the Newtonian potential expressed in units of energy per energy, the sum of -Gm/rc2 for each source mass and distance involved.

If the original objects were approximately spherical and the low mass object is now the same distance from the centre of the combined object as it was from the centres of both original objects, then the time dilation would be the same as before, but now there would also be a gravitational field. The size of the combined object does not matter as long as it is spherically symmetrical.

This all applies to the static case. For dynamic motion such as orbits, time dilation due to velocity is involved as well.

7. May 22, 2010

### jaketodd

1) Yes, the reference clock is very far away from the objects. The test clock is the low-mass object in both scenarios.

2) In the second scenario, the low-mass object has not moved. There is only one other object, the super massive object, which is twice the mass it was in the first scenario but the same size and in the same location.

3) All objects are spheres.

4) The test clock is still the low-mass object, in the second scenario.

Thanks!

Jake

8. May 22, 2010

### starthaus

You are absolutely correct. Start with the Schwarzschild solution in the weak field approximation:

$$(cd\tau)^2=(1-\frac{2\Phi}{c^2})(cdt)^2+(1-\frac{2\Phi}{c^2})^{-1}(dr)^2+...$$

For the case $$dr=d\theta=d\phi=0$$ you get the well known relationship:

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

Writing the above for two different gravitational potentials $$\Phi_1$$ and $$\Phi_2$$ you obtain the well-known time dilation relationship:

$$\frac{d\tau_1}{d\tau_2}=\sqrt{\frac{1-\frac{2\Phi_1}{c^2}}{1-\frac{2\Phi_2}{c^2}}}$$

At the Earth surface :

$$\Phi_1=-\frac{GM}{R}$$

At the Earth center:

$$\Phi_2=-3/2\frac{GM}{R}$$

Now, due to the fact that $$\frac{\Phi}{c^2}<<1$$ you can obtain the approximation:

$$\frac{d\tau_1}{d\tau_2}=1-\frac{\Phi_1-\Phi_2}{c^2}=1-1/2\frac{GM}{Rc^2}<1$$

So, $$f_1>f_2$$ where $$f_1$$ is the clock frequency on the Earth crust and $$f_2$$ is the frequency of the clock at the center of the Earth.

Last edited: May 22, 2010
9. May 22, 2010

### Jonathan Scott

Why did you have to spoil an otherwise excellent post with this incorrect assertion?

It is clear from what you have posted that the time dilation only depends on the difference in potential and not on the gradient.

10. May 22, 2010

### starthaus

Picky, picky :-)

Generalization:

At a distance $$r<R$$ from the center of the sphere, the gravitational potential is:

$$\Phi_2(r)=-\frac{GM}{R}(\frac{3}{2}-\frac{r^2}{2R^2})$$

The above gives:

$$\frac{d\tau_1}{d\tau_2}=1-\frac{\Phi_1-\Phi_2}{c^2}=1-\frac{GM}{2Rc^2}(1-\frac{r^2}{R^2})<1$$

For $$r=0$$ (clock2 at the center of the Earth) you recover the results from post 8.

For $$r=R$$ you get the expected:

$$\frac{d\tau_1}{d\tau_2}=1$$

Last edited: May 22, 2010