Actually, time travel is science friction, but that isn't important right now.

Time travel in relativity theory is traveling into the future of a different proper frame of reference. For example: For the relativistic rocket, constant acceleration g_earth, the prediction for proper time in the rocket frame is 28 earth years to Andromeda. The prediction for the proper frame of the earth is 2,000,000 earth year to Andromeda. The distance in the rocket proper frame is 28 light years and 2,000,000 light years in earth's proper frame.

So the tick ratio

dTau_rocket/dTau_earth = 28_earth year/2,000,000_earth year = 1.4E-15.

Theoretically the rocket could return to earth 4,000,000 years later than when they left. It's happening 'every tick' between the GPS satellite and earth based clocks. On an infinitesimal scale. Time travel in your own proper frame is what's impossible.

This is a cool formula you can derive from the Schwarzschild metric which I like to call a relativistic time travel equation. Just for the fun of it. This is for an orbit and both the SR and GR relativistic time dilation are included in the derivation.

Problem 7, page 4-32, Exploring Black Holes.\

The derivation

Put the derivative of the effective potential term (from the equation of motion) into quadratic form (to find critical values)

r*^2 - L*^2r + 3L*^2 = 0

Where

r* = r/M, and L* = L/mM

Then divide through by L*^2 and manipulate to get

r*^2/L*^2 = r* - 3 [saving this for a later

substitution]

Setting dr = 0 in the Schwarzchild metric and

substituting dphi = (L*/r*^2)dTau the metric becomes

dTau^2 = (1 - 2/r*)dt^2 - (L*^2/r*^2)dTau^2

To find the ratio dTau^2/dt^2 divide through by the

bookkeeper time dt^2 and simplify to

(dTau/dt)^2 = (1 - 2/r*) / (1 + L*^2/r*^2)

Now substitute 1/(r*-3) for L*^2/r*^2 and simplify to

dTau/dt = (1 - 3M/r)^1/2

This following example uses the time travel equation in a way which might be interesting to you. Note that the time intervals dTau and dt are different but that they both measure time intervals as recorded by a clock in a specific coordinate system. dt for Earth and dTau for the spaceship orbiting the black hole.

Abe signs on with the crew of 'Warp Drive 1' while Bill remains on Earth. Warp Drive 1's maiden voyage is to visit a solar mass (M = 1477 meters) black hole, free fall to a knife edge orbit just outside the photon sphere at r = 3.000001M, remain in orbit for 172,800 seconds (Two Earth days) wristwatch time (dTau), and then return to Earth arriving ~ 9.5 years in Earths future. During the warp phase of the journey the ship remains in an inertial rest frame so the difference in wristwatch rate for Abe and Bill is minimal during this phase of the journey.

What will be the difference in elapsed wristwatch time, dTau, for Abe with respect to the elapsed coordinate time, dt, for

Bill when they meet upon Abe's return? Keeping in mind that dTau is measured with Abe's wristwatch (shipframe) and the wristwatch time for

Bill is measured with Bill's wristwatch (Earthframe).

From the Schwarzchild geometry a time travel equation

dTau/dt = (1-3M/r)^1/2

Substituting 172,800 seconds for dTau, 3.000001M for r, and 1477 meter for M, then solving for dt

dt = 172,800s / 5.7735x10-4 = 299298529.2s

Since there are 3.156x107 seconds / Earth year

299298529.2 seconds / 3.156x107 seconds / Earth year

= ~ 9.48 Earth year

When Abe returns the twins can determine whether the prediction was correct by just comparing clocks.

So you might play with this in the weak field. Set r for an orbit around the Sun. This equation includes the gravitational and SR component of time dilation.

dTau/dt = (1-3M/r)^1/2