Calculating gravitational time dilation

ssope

I'm trying to calculate the time difference for something orbiting very high above the moon in comparison to the surface of the Earth.

DaleSpam

Here is the Hyperphysics page on the topic. Unfortunately it does not have a calculator, but it does have the formula.
http://hyperphysics.phy-astr.gsu.edu...gratim.html#c4

JesseM

Like I asked before:
And if you want to include velocity, kev calculated in post #8 of this thread that for a clock in a circular orbit, its time dilation (relative to a clock at infinity I think) would just be a product of velocity-based time dilation and gravitational time dilation.

ssope

I must also be a product of time-dilation based on gravity because if on the surface of the moon its 1/6th earth gravity, it must be far more orbiting high above the moon. I do understand that velocity based time-dilation would indeed come into play, and yes I would like to accurately calculate for both.

JesseM

What must be far more? Gravitational time dilation is less when you're farther from a source of gravity, not more.
As kev showed on that thread I linked to, the total time dilation for an object in a circular orbit is just the velocity-based time dilation (velocity measured relative to the center of the body it's orbiting) multiplied by the gravitational time dilation.

yuiop

If you are interested an equation given by Dr Greg in this thread http://www.physicsforums.com/showthr...=383884&page=3 arrives at the same conclusion from a different direction. (See posts#9 and #35 of that thread).

ssope

Does anyone here want, better, can anyone here to calculate the time difference for two clocks. One clocking orbiting the moon, the other clock on the surface? I need to double check my math. It doesn't matter what altitude you use, as long as you show your work I'll be able to cross check it with mine.

yuiop

Can we assume a moon in otherwise empty space, because if we have to take the gravity of the Earth and Sun into account then things get really complicated.

Do you want answers for a rotating moon moon or a static moon?

[EDIT] I have re-read your original post and I now think you want to compare the rate of a clock orbiting the moon, with the rate of a clock on the surface of the Earth? Is that right?

JesseM

What about the second question I asked?
Also, when you say you want to double check your math, does that mean you've already calculated the gravitational and velocity-based time dilation for each clock? If so what'd you get?

ssope

Yes you are correct

ssope

No it means I'm struggling to arrive at the correct answer and that my pride wouldn't let me post that in my previous post. If I had the math I'd show my work.

ssope

I wonder if putting one of the clocks in a Geosynchronous orbit above the moon to cut back on velocity based time-dilation when compared to a clock on the surface of the earth.

[Sorry about the triple post, but the forum won't let me delete posts, I thought they'd be automerged]

Jonathan Scott

Time dilation is a tiny effect. It is close to zero everywhere except close to massive objects. At distance r from spherical mass m the fraction by which clocks are slowed compared with their rate a long way away is simply Gm/rc², equal to the Newtonian potential difference divided by c². (This is accurate provided that the resulting fraction is small compared with 1, which applies for example to all gravitational potentials within the solar system). Another way of putting this is that a clock at that location will run (1-Gm/rc²) times more slowly than a distant one in empty space.

You can plug in the mass and radius of the earth (for example using the calculator built into Google search) to find out what this is at the surface of the earth.

If you want to also take account of the time dilation due to velocity in a circular orbit, you can make use of the fact that for such an orbit v²/r = Gm/r² where r is the radius of the orbit and m is the mass about which it is orbiting. If you multiply both sides by r/c² you find that v²/c² is equal to Gm/rc², the time dilation fraction.

From the usual Special Relativity formula, the time dilation due to the speed is given by sqrt(1-v²/c²), which for non-relativistic speeds is equal to 1-1/2(v²/c²), where the last part corresponds to the kinetic energy per unit energy. That's half as much as the gravitational time dilation.

If you want an accurate time dilation for something in orbit around the moon compared with the surface of the earth, you also have to take into account the time dilation in the vicinity of the moon due to the earth, which is the value of Gm/rc² when m is the mass of the earth and r is the radius of the moon's orbit around the earth. That is, time dilation at orbiting object is made up of time dilation relative to empty space caused by the moon plus time dilation caused by the orbital speed plus time dilation of the zone around the moon caused by the earth plus the time dilation due to the speed at which the moon moves around the earth. In general, you'd also need to think about how the orbital speeds add and subtract depending on the orientation of the orbits.

However, the difference in time dilation is dominated by the time dilation at the surface of the earth, with the other factors making only a little difference, so as an approximate estimate you can ignore everything else!

yuiop

I would not worry about your pride too much. It would take a rocket scientist to come up with an exact answer. For our purposes we are are going to have to make a lot of simplifications to come to an approximate answer. Initially I propose we assume the following simplifications with maybe more to be added later:

1) The Earth, Moon and Sun are perfect spheres.
2) The Earth's orbit is exactly on the equatorial plane of the sun.
3) The Earth's axis is exactly orthogonal to the Sun's equatorial plane. (i.e. the equator of the Earth always lies on the equatorial plane of the Sun.
4) The Moon's orbit is exactly on the equatorial plane of the Earth and Sun.
5) All orbits are perfectly circular.
6) We tidy up the Solar system by removing all the other planets and orbiting junk.
7) The mass of the satellite is insignificant compared to the mass of the moon.

(We might have to remove the Sun later.)

To start you off here are some equations you might need.

The basic time dilation factor is:

[tex] \tau = t \sqrt{(1-2GM/(Rc^2)}\sqrt{(1-{v_L}^2/c^2)}[/tex]

where:

[itex]\tau[/itex] is the proper time interval measured by a clock on the satellite.
t is the time interval according to a clock at infinity.
R is the distance the satellite is from the gravitational body.
[itex]v_L[/itex] is the "local velocity" according to an observer at R that is stationary with respect to the distant stars.

Note that the above time dilation equation is based on the Schwarzschild metric which assumes a non rotating gravitational body which is not true for the Earth and Moon and really we should be using the Kerr metric but that is more complicated and for the rotation velocities of the massive bodies we are studying the Schwarzschild metric is a good enough approximation.

Note that all commonly know GR metrics assume a single gravitational body in otherwise empty space so they can not be directly applied here. However time dilation is a function of gravitational potential and I think it is safe to assume that gravitational potentials are simply additive.

Now we come to the problem of determining the velocity of the satellite. Calculating the orbital velocity for a given radius is not too difficult. However you should bear in mind that this orbital velocity in an equatorial orbit around the Moon (orbit type 1) subtracts from the velocity of the Moon around the Earth when the satellite is nearest the Earth in a prograde (anticlockwise looking from the North pole) orbit and adds to the velocity in a retrograde orbit. The gravitational potential will also vary periodically as the satellites distance from the Earth varies in an equatorial orbit. We can remove the periodic velocity variation by considering an orbit around the Moon that remains on a plane than intersects the centres of the both the Moon and the Earth which is at right angles to the equatorial plane (orbit type 2). Strictly speaking the direction is constantly changing but the magnitude of the velocity is constant and for time dilation we only need the magnitude (or speed). Unfortunately the variation in gravitational potential remains in this orbit. We could also consider an orbit around the moon that is orthogonal to a line joining the Moon and the Earth (orbit type 3) such that the satellite is always visible from the Earth and appears to roll like wheel. This removes the periodic variation of the gravitational potential but the periodic variation in velocity remains. Without doing the calculations I imagine the periodic variations in the time dilation are much smaller than the magnitude of the average time dilation. Anyway, you should specify what orbit type you have in mind.

We can also note that calculations for the time dilation of GPS satellite clocks can be calculated to a very good accuracy by ignoring the gravitational influence of the Sun, Moon etc and the velocity variations due to the orbit velocity of the GPS satellite adding or subtracting from the Earth's orbital velocity around the Sun.

You might want to consider changing the question to "How does the clock rate of clock on the surface of the Moon compare to the clock rate on the surface of the Earth?" as that might be slightly simpler.

You also need to specify where on the surface of the Earth the clock is situated. The rate of a clock at the equator of the Earth is different form one situated at one of the poles.

To calculate orbital velocities you need the following equation:

[tex] {v_{L}} = \sqrt{\frac{GM}{R(1-2GM/(R_{o}c^2)}}[/tex]

Where:

R is the orbital radius of the satellite.
Ro is the location of the "stationary observer".

For an observer at infinity (Ro = infinity) the equation reduces to the familiar Newtonian equation:

[tex] {v_{L}} = \sqrt{\frac{GM}{R}}[/tex]

For R = Ro = 3GM/c^2 the result is:

[tex] {v_{L}} = \sqrt{\frac{GMc^2}{(3GM(1-2GM/(3GM)}} = \sqrt{\frac{GMc^2}{(3GM-2GM)}} = c [/tex]

which is the photon orbit where particles need to travel at the speed of light to orbit.

That should give you a good start. For now it would be useful if you would specify the type of orbit (1,2 or 3) and whether you mean clockwise or anticlockwise as viewed from high above the North pole or from the surface of the Earth's equator.

Slinkey

I have a question, but if it has already been covered above then please forgive me for being lazy and just scanning rather than reading.

I am standing at sea level next to a 1000m tall skyscraper. By my foot is a clock.

At the top of the skyscraper is another clock.

I am not moving relative to either clock.

As I understand it these two clocks do not tick at the same rate. The one at the top of the skyscraper ticks faster than the clock by my foot.

I watch the clock by my foot for 1 day. How much time passes on the clock at the top of the skyscraper?

JesseM

Do you want to calculate the gravitational time dilation alone (idealizing the Earth as a nonrotating sphere) or do you want to include the time dilation due to different velocities in an Earth-centered coordinate system?

Slinkey

I want to calculate the gravitational time dilation from my perspective of standing at the foot of the skyscraper. Neither clock is moving relative to me so I would assume that there is no velocity component and wondering why I would need to idealise the Earth as a non-rotating sphere?