Originally posted by Metal Brain
So you think gravity causes time dilation? I suppose you think the cart pushes the horse and water skiers push the boat.
Why don't you explain how gravity causes time dilation? You are intelligent enough to do that, right? What is your source of information?
I have two degrees in Physics, Kazet is actively engaged in academic research as a physicist, googlefudge has a physics background, and Adam Warlock is a physicist. I don't know about humy, I can't remember. Basically you can't move in this forum for physicists. I had an undergraduate course in General Relativity, a course in differential geometry and my thesis title was Polymerisation of 2D Quantum Gravity models. I am to all intents and purposes infallible in this field as far as discussions in this forum are concerned.
Earlier I was trying to ask a series of questions to get at the root of your misunderstanding, but gave up when you tried to bring weather maps into it. Having looked back through the thread I think you have the concepts of proper time and time dilation confused.
Proper time is the length of a line interval in spacetime. It is the generalisation of Pythagoras' theorem for 4 dimensional pseudo-Riemannian manifolds. In Special Relativity the proper time along the straight path of a moving body is:
c²dτ² = c²dt² - dx² - dy² - dz²
where dτ is the interval of proper time. dt is the time elapsed for some observer, and dx is the distance travelled along the x-axis. c is the speed of light. Suppose the motion is along the x-axis so that dy = dz = 0 and v = dx/dt. We get:
dτ = dt√(1 - v²/c² )
so:
dt = dτ/√(1 - v²/c² )
Or that a co-moving observer, one keeping pace with the moving body will have a clock running slower by this factor. This is time dilation in special relativity.
In General Relativity the space-time is curved. The Einstein Field equations relate curvature to energy density, when they are solved on gets a formula for a line element. For a spherically symmetric, massive body the surrounding space is described by the Schwarzschild metric which is:
c²dτ² = c²A(r)dt² - B(r)dr² - r²(dθ² + sin²θ dφ² )
where
A(r) = 1 - rₛ/r
B(r) = 1/A(r)
and rₛ = 2GM/c², the famous Schwarzschild radius.
These coordinates correspond to what an asymptotic observer would see. For an observer stood at the north pole (so we can ignore the rotation of the earth), dr = 0 as their radial distance doesn't change. dθ doesn't change, that's the lattitude, in polar coordinates at the pole sin(θ ) = 0, so we are left with:
using g = GM/r², we have:
dτ = dt√A(r) = dt√(1 - 2GM/rc² ) = dt√(1 - 2gr/c² )
g = 9.81 m/s², r = 6.37 * 10⁶m, c = 3 * 10⁸m/s
so let x = gr/c² ~ 6.94*10⁻¹¹
Using the small x expansion for the square root:
dτ ≈ dt (1 - gr/c² ) = dt(1 - 6.94*10⁻¹¹ )
So a clock at the North Pole will lose a second relative to a co-moving clock a large distance from Earth every 456 or so years.
I've presented two calculations of time dilation. One due to relative motion in the absence of a gravitational field and one for stationary observers in the presence of a gravitational field. If all you knew was the time dilation you couldn't tell if it was due to gravity or relative motion. So this should be enough to show it is an effect and not a cause.