# Equation for time dilation.

1. May 25, 2008

Hey I have done a few google searches, and I do not have a relativity book to let me know what the equation for time dilation is. So I tried to figure it out on my own and come up with this for instantanious time dilation...

Td = cos$$^{-1}$$($$\frac{Ag' + Am'}{1-Ag'*Am'/c^2)}$$​

Where Ag' is the first dirivitive of the acceleration of gravity and Am' is the instionious velocity of the object.

If this is wrong can anyone help me find the actual equation for time dilation?

2. May 25, 2008

I don't know what I was thinking the equation I ment was...

Td = sqrt[1-($$\frac{Ag' + Am')^2}{(1-Ag'*Am'/c^2)^2]}$$​

3. May 25, 2008

### Mentz114

I think you need to specify clearly what time-dilation you are talking about. Usually we talk about two clocks running at different rates as time dilation.

4. May 25, 2008

Very good I used the velocity addition formula to derive the amount of time dilation someone will see when looking at another object. I should probably add that Ag' is the difference in your acceleration due to gravity and their's. Just as Am' is really just your difference in velocity.

5. May 25, 2008

### Mentz114

I'm still not clear. You need to explicitly show the form of Am' and Ag'.

The dimensions (units) don't seem to match up in any case.

Special relativity says that an observer moving at speed v wrt to a clock will measure that clock slowed down by a factor $$\gamma = \sqrt{1 - \frac{v^2}{c^2}}$$ and this effect depends only on the relative velocity.

Gravitational time dilation is a separate effect. The dilation is approximately

$$\sqrt{1 - \frac{2\phi}{rc^2}}$$ where $$\phi$$ is the gravitational potential.

6. May 25, 2008

Well it looks like I am close. Am's = v, and Ag' = 2θ/r. How does one calculate total time dilation for an object in a gravitational field and moving?

7. May 25, 2008

### shalayka

From what I can gather, it depends on whether the massive object is being accelerated transversely or not.

If a massive object is infalling radially (ex: the massive object's direction vector is equal to the direction vector formed by the gravitational time dilation gradient), then it seems that the kinematic time dilation is simply due to gravitational time dilation. If they were not equal in this instance, then the infalling massive object's velocity would become practically c at a distance of 1.5 Rs, and not Rs.

If a massive object is orbiting (ex: the massive object's direction vector is perpendicular to the direction vector formed by the gravitational time dilation gradient), then kinematic time dilation is in addition to gravitational time dilation. My reasoning behind this comes from the relativistic precession of orbit, and how two-thirds of it results from the difference in the rate of time due to Newtonian vs. Einsteinian "potential" (gravitational time dilation), and one-third results from the difference in the rate of time due Newtonian vs. Einsteinian kinematics (kinematic time dilation).

This is all considering that an approximate formula for relativistic precession of orbit (measured in orbits per orbit) is:

$$r$$ is semi-major axis of orbit
$$e$$ is eccentricity of orbit

$$\delta = \frac{3GM}{rc^2 (1 - e^2)}$$

(multiply by $$2\pi$$ to get radians per orbit)

and that:

$$\sqrt{1 - \frac{GM}{rc^2} } = \sqrt{1 - \frac{v^2}{c^2} }$$

Last edited: May 25, 2008
8. May 25, 2008

### Mentz114

Use general relativity. Unfortunately it's not trivial. I suppose you could choose two freely falling frames at different points on the same geodesic

Shalayka is using GR.

M

9. May 25, 2008

### shalayka

Mentz is being very generous. I might *look* like I'm using GR, but what I'm really doing is using a whole bunch of half-a**ed logic that I've pieced together over the last year during the five minutes a day that I've got to study GR. :)

I personally couldn't verify if my reasoning is correct, but so far no one's pointed out a major flaw in it, and boy oh boy -- when you're wrong regarding physics, EVERYONE lets you know (in a not so nice way, generally).

Thanks for the vote of confidence Mentz. It's pretty much the first I've ever gotten.

Last edited: May 25, 2008
10. May 25, 2008

### yuiop

Hi Shalayka,
a derivation given in this post https://www.physicsforums.com/showpost.php?p=1526798&postcount=25 that simply assumes the proper time of an orbiting clock (circular orbit) is the classic Kepler orbital period multiplied by the gravitational time dilation and the kinematic time dilation due to the orbiting clocks orbital velocity, agrees with a derivation obtained by Pervect using a different method. The derivations seem to agree with your conclusions for the time dilation experienced by an orbiting body.

11. May 25, 2008

### Mentz114

Hi shalayka,

whenever I see this

$$\sqrt{1 - \frac{GM}{rc^2} }$$

I think Schwarzschild metric and GR. But I suppose that expression can be got outside GR.

Five minutes a day is not a lot - try to get it up to ten ...

M

12. May 26, 2008

Since $$\sqrt{\frac{GM}{r}}$$ and v are both velocities if you want to find the total time dilation for an object in a gravitational field with a velocity would you need to add the two velocities using the relatavistic velocity addition like this?

$$\sqrt{1 - \frac{\frac{\frac{GM}{r}+V}{1-\frac{GMV}{rc^2}}}{c^2} }$$

Assuming no gravitational forces on the viewer of course.

Last edited: May 26, 2008
13. May 27, 2008

### shalayka

My crack at this would be:

- $$\vec{A}$$ is the unit 3-vector pointing from the center of the orbiting body to the centre of the gravitating body

- $$\vec{B}$$ is the unit 3-vector describing the direction of the orbiting body's travel

- $$\phi$$ is the angle between $$\vec{A}$$ and $$\vec{B}$$: $$\phi = \arccos(\vec{A}\cdot\vec{B}) = \arccos({A}_x{B}_x + {A}_y{B}_y + {A}_z{B}_z)$$

- $$\omega$$ is the time dilation factor: $$2 + (1 - \cos\phi)$$, or if you like: $$3 - \vec{A}\cdot\vec{B}$$

Note: This calculation does NOT take into consideration the scenario where the body is moving directly away from the gravitating body (ex: $$\cos\phi < 0$$). I don't think that counts as freefall, so I'm avoiding it.

$$\sqrt{1 - \frac{\omega GM}{rc^2}}$$

Where $$\vec{A} = \vec{B}, \vec{A}\cdot\vec{B} = 1, \phi = 0, \omega = 2$$, the body is in freefall directly toward the gravitating body.

Where $$\vec{A}\cdot\vec{B} = 0, \phi = \frac{\pi}{2}, \omega = 3$$, the body is in freefall transversely (circular orbit).

I've only had one cup of coffee yet, so take this calculation very cautiously. :)

Last edited: May 27, 2008