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I am trying to derive some equations which will let me simulate the motion of a spaceship in different geometries of spacetime. Suppose I know the metric and Christoffel symbols at the location of the spaceship, and I know the force applied on it (for example the spaceship is firing its thrusters). How do I calculate its coordinate acceleration from this information?

What I have tried so far:

As far as I understand the most natural thing to calculate from the force provided by the thrusters, is the proper acceleration of the spaceship. Then the wikipedia page for proper acceleration gives the following equation relating proper acceleration to coordinate acceleration:

$$ A^{\lambda} := \frac{DU^{\lambda}}{d\tau} = \frac{dU^{\lambda}}{d\tau} + \Gamma^{\lambda}\,_{\mu\nu}\,U^{\mu}U^{\nu} $$

One can rearrange the equation (following the Wikipedia page) to

$$ \frac{dU^{\lambda}}{d\tau} = A^{\lambda} - \Gamma^{\lambda}\,_{\mu\nu}\,U^{\mu}U^{\nu} $$

The rightmost term can be evaluated by just expanding the sum and putting in the Christoffel symbols for the spacetime around the black hole (which I can find in a reference). So far everything seems to make sense.

However, finding $A^{\lambda}$ from the force provided by the thrusters is something I can't figure out. Wikipedia says this about the proper acceleration:

... is the object's proper-acceleration 3-vector combined with a null time component as seen from the vantage point of a reference or book-keeper coordinate system in which the object is at rest.

From this I understand that

$$ (A^{x}, A^{y}, A^{z}) = \mathrm{\frac{(vector\; force\; provided\; by\; thrusters)}{(mass\; of\; spaceship)}} $$

in the either the reference of the spaceship, or a stationary reference frame which is instantaneously comoving with the spaceship at a given moment in time (I can't tell which one).

The part about the null time component completely confuses me.

From what I understand of the wikipedia article, the proper acceleration is also equal to the covariant derivative of the coordinate velocity. However, I don't know much differential geometry and the covariant derivative is completely meaningless to me.

Questions:

  • How to go from force provided by thrusters (as a classical 3-vector) to proper acceleration as a 4-vector?
  • Is it correct to say that the three spacelike components of the 4-vector proper acceleration are equal to the resultant acceleration in the frame of the spaceship? If not, how are they defined?
  • What is the timelike component $A^{t}$ equal to and how can one work it out?
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  • $\begingroup$ The key thing to remember is that the 4-velocity always has unit norm, so the time component of the acceleration is always determined by the spatial components. $\endgroup$
    – Zo the Relativist
    Apr 21, 2021 at 17:45
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    $\begingroup$ One point, "the proper acceleration is also equal to the covariant derivative of the coordinate velocity" is just your first equation put into words. $\endgroup$
    – m4r35n357
    Apr 21, 2021 at 17:48
  • $\begingroup$ @m4r35n357 Thank you, I didn't realise that! Maybe I'll edit the Wikipedia article to make it clearer for other readers like me... $\endgroup$
    – DarthVlader
    Apr 21, 2021 at 19:36

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How to go from force provided by thrusters (as a classical 3-vector) to proper acceleration as a 4-vector?

I will introduce the concept of a momentarily co-moving inertial frame (MCIF). This is the inertial reference frame where the rocket is momentarily at rest. Of course, since the rocket is accelerating and the frame is inertial, the rocket does not remain at rest for more than just a moment.

So the thrusters are usually described in terms of their force, $\vec F$ in the MCIF. So your 3-acceleration is just the usual Newtonian $\vec a = \vec F/m = (a_x,a_y,a_z)$. The four-acceleration in the MCIF is then simply $$\mathbf A = (0,\vec a) = (0, a_x, a_y, a_z)$$ To get the four-acceleration in another frame you simply use the Lorentz transform to transform to the required frame. Transforming to a frame which is moving relative to the MCIF at a velocity $v$ in the $x$ direction gives $$\mathbf A = \left(\gamma \frac{v}{c} a_x, \gamma a_x, a_y, a_z \right)$$

Is it correct to say that the three spacelike components of the 4-vector proper acceleration are equal to the resultant acceleration in the frame of the spaceship? If not, how are they defined?

Not in the frame of the spaceship, which is non-inertial, but in the MCIF, yes.

What is the timelike component At equal to and how can one work it out?

See above. The "null" that was confusing you is simply the 0 for the time component in the MCIF.

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  • $\begingroup$ Thank you for your answer, things are a lot clearer now. Just one more question, are Lorentz transforms valid in curved spacetime? So if my spaceship is around a black hole and I do my calculations relative to a faraway observer, do I just do a Lorentz transform to get the 4-acceleration in the observer's frame? $\endgroup$
    – DarthVlader
    Apr 21, 2021 at 19:31
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    $\begingroup$ The Lorentz transforms are valid, but only locally. They won’t help you with far away observers in curved spacetime. $\endgroup$
    – Dale
    Apr 21, 2021 at 20:54

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