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Wald, for instance, has a more formal derivation in which we have a smooth one -parameter family of geodesics. So every ##\gamma_s## is a geodesic curve, where s is the selector paramter that determines which geodesic in the family is beling selected. Each geodesic has an affine paramater t, so the one-parameter family of geodesics is represented by ##\gamma_s(t)##.

Wald states that this one parameter family of geodesics spans a sub-manifold, but didn't offer any proof. Also, re-reading this, it's not clear to me how one goes from Wald's one parameter family of geodesics (which is essentially a 1-space, 1 time case) to a three-parameter family of geodesics (3 space, 1 time).

If we accept this so far, then the short version is that we use the notion of derivative operators as vectors to more formally define the separation vector. Derivative operators form a vector space, because they can be multiplied by scalars and added together (that's the basic properties a vector space needs to satisfy). Then in terms of derivative operators, ##\partial / \partial s## represents the separation vector, and ##\partial / \partial t## points along the curve. The affine paramterization t of each curve needs to be chosen specially though. For ease of physical interpretation, It can (and should) be be chosen so that so that ##\frac{\partial}{\partial s} \cdot \frac{\partial}{\partial t}## is zero, i.e so that the separation vector is orthogonal to the tangent vectors of the geodesics. Wald has a short proof that this is possible, but this is involved enough that you might want to read the original text if you want to use Wald's approach. There may be a better/simpler approach out there, perhaps someone will suggest one.

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Thank you for this. So in layman terms, you mean that you take many geodesics (1D) and "stick" them together to form a volume of geodesics, then you use the operator that moves you from a (1D) geodesic to a neighboring one (perpendicular to the direction along the geodesic) to define the separation vector. Is this correct?

Wald, for instance, has a more formal derivation in which we have a smooth one -parameter family of geodesics. So every ##\gamma_s## is a geodesic curve, where s is the selector paramter that determines which geodesic in the family is beling selected. Each geodesic has an affine paramater t, so the one-parameter family of geodesics is represented by ##\gamma_s(t)##.

Wald states that this one parameter family of geodesics spans a sub-manifold, but didn't offer any proof. Also, re-reading this, it's not clear to me how one goes from Wald's one parameter family of geodesics (which is essentially a 1-space, 1 time case) to a three-parameter family of geodesics (3 space, 1 time).

If we accept this so far, then the short version is that we use the notion of derivative operators as vectors to more formally define the separation vector. Derivative operators form a vector space, because they can be multiplied by scalars and added together (that's the basic properties a vector space needs to satisfy). Then in terms of derivative operators, ##\partial / \partial s## represents the separation vector, and ##\partial / \partial t## points along the curve. The affine paramterization t of each curve needs to be chosen specially though. For ease of physical interpretation, It can (and should) be be chosen so that so that ##\frac{\partial}{\partial s} \cdot \frac{\partial}{\partial t}## is zero, i.e so that the separation vector is orthogonal to the tangent vectors of the geodesics. Wald has a short proof that this is possible, but this is involved enough that you might want to read the original text if you want to use Wald's approach. There may be a better/simpler approach out there, perhaps someone will suggest one.

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Yes, that's a good (IMO) non-technical summary of what I said.Thank you for this. So in layman terms, you mean that you take many geodesics (1D) and "stick" them together to form a volume of geodesics, then you use the operator that moves you from a (1D) geodesic to a neighboring one (perpendicular to the direction along the geodesic) to define the separation vector. Is this correct?

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So if my (formal) understanding is correct, ##\partial/\partial s## seems to be the vector in the tangent space of a point on a given geodesic, perpendicular to ##\partial/\partial t##, in that tangent plane. So they define the separation vector similar to MTW, but they specify clearly that the vector they use is the one in the tangent plane, which in the limit of 2 infinitely closed geodesics (and presumably in a local inertial frame at that point) is an actual well defined vector? So it seems to me that they do the same thing as MTW (but they explain in more details what vector do they use). Is this right?Yes, that's a good (IMO) non-technical summary of what I said.

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Yes, once you've restricted yourself to two dimensons (1 space + 1 time), the 2d space-time is spanned by two vectors. One of these is defined by the tangent vector to the geodesics, which I'd call "time". The remaining vector needed to span the space, which we can take as perpendicular, describes the separation of the geodesics, which I'd call space.So if my (formal) understanding is correct, ##\partial/\partial s## seems to be the vector in the tangent space of a point on a given geodesic, perpendicular to ##\partial/\partial t##, in that tangent plane. So they define the separation vector similar to MTW, but they specify clearly that the vector they use is the one in the tangent plane, which in the limit of 2 infinitely closed geodesics (and presumably in a local inertial frame at that point) is an actual well defined vector? So it seems to me that they do the same thing as MTW (but they explain in more details what vector do they use). Is this right?

To compute the relative acceleration of geodesics, we do need to parallel transport the velocities (which are just the tangent vectors of the geodesics curves) to the same point to properly compare them and compute the covariant derivative.

What curve do we parallel transport the velocities along? I'd say that we parallel transport them "through space", i.e. in a curve whose tangent vector is everywhere orthogonal to the tangent vector of the geodesic curves. Be warned that this isn't straight from a textbook, so it may contain a personal quirks.

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strangerep

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Geodesic Deviation

We adapt the treatments of Carroll [sect 3.10]and Callahan [ch7].

To generalize the previous concept of Newtonian tidal acceleration to a general relativistic situation, we consider a family of geodesic curves ##\gamma^\mu(s,q)## in 2 parameters ##s,q##, where now ##s## is an

[Edit: so the idea is that we have a 2D manifold of geodesic curves -- i.e., a geodesic congruence. My ##s## is a coordinate along the geodesics, and my ##q## is a coordinate on the 2D manifold

The 4D-generalized notion of

V^\mu ~:=~ \Pdrv{\gamma^\mu(s,q)}{s} ~,

$$and once again ##V^\mu## has dimensions of

X^\mu ~:=~ \Pdrv{\gamma^\mu(s,q)}{q} ~,

$$and ##X^\mu## thus has dimensions of

U^\mu ~:=~ \Pdrv{X^\mu}{s} ~=~ V^\rho \nabla_\rho X^\mu

~\equiv~ \nabla_V X^\mu ~,

$$where ##\nabla## now denotes the usual covariant derivative. ##U^\mu## still has dimensions of

Similarly, the 4D

A^\mu ~:=~ \Pdrv{U^\mu}{s} ~=~ V^\rho \nabla_\rho U^\mu

~\equiv~ \nabla_V^2 X^\mu ~,

$$and we see easily that ##A^\mu## does indeed have dimensions of

Note that the

$$##a^\mu## is zero on a geodesic, since motion along a geodesic corresponds to an observer in free-fall, i.e., inertial motion. In contrast, an observer for whom ##a^\mu## is nonzero can detect her acceleration via an accelerometer. Two inertial observers moving along the neighbouring geodesics each perceive the other as accelerating relative to themselves, even though their own respective accelerometers both read 0.

We now compute the relative acceleration ##A^\mu## under the assumption that the ##X## and ##V## are basis vectors adapted to a coordinate system. This means they satisfy ##[X,V]=0##, which can be re-expressed as $$

0 ~=~ X^\sigma \nabla_\sigma V^\mu ~-~ V^\sigma \nabla_\sigma X^\mu

~+~ X^\alpha V^\beta T^\mu_{~\alpha\beta} ~,

$$where ##T## is the torsion (usually 0 in ordinary GR).

Several tedious lines of index manipulation eventually give $$

A^\mu ~=~

R^\mu_{~\nu\rho\sigma} V^\nu V^\rho X^\sigma - (\nabla_\lambda V^\mu) T^\lambda_{~\rho\sigma} V^\rho X^\sigma

~+~ (V^\nu \nabla_\nu X^\rho) V^\sigma T^\mu_{~\rho\sigma}

~-~ (V^\nu \nabla_\nu T^\mu_{~\rho\sigma}) V^\rho X^\sigma ~.

$$In the torsionless case, only the first term on the right hand side survives. I.e.,$$

A^\mu ~=~ R^\mu_{~\nu\rho\sigma} V^\nu V^\rho X^\sigma ~.

$$

HTH.

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