How Does Acceleration Affect the Geodesic Equation in General Relativity?

  • Thread starter Karlisbad
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In summary: M. Math., PhD, PEngIn summary, acceleration in a geodesic equation can be caused by non-gravitational physical interactions or processes, such as hydrostatic equilibrium, the Lorentz force, or spin-spin forces. In these scenarios, the geodesic equation becomes \nabla _{u} u -a^{\mu}=0, and the 4-dimensional force is derived from the weak field approximation and potential of the particle. This information can then be used to derive the Einstein field equation, R_{ab}=0.
  • #1
Karlisbad
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Whenever there's no acceleration then: (Geodesic equation)

[tex] \nabla _{u} u =0[/tex]

where the covariant derivative includes "Christoffel symbols" so [tex] \Gamma_{kl}^{i} =0 [/tex] for an Euclidean space-time..however when generalizing to a system under an acceleration then Gedesic equation becomes:

[tex] \nabla _{u} u -a^{\mu}=0 [/tex] (foruth dimensional acceleration)

for a test particle [tex] \acute{{R}^{\mu}}_{\alpha \nu \beta} \acute{u}^{\alpha} \acute{x}^{\nu} \acute{u}^{\beta} = - \acute{f}^{\mu} [/tex]

where does this 4-dimensional force come from?? :confused: :confused: but if you use "Weak field approximation" then the potential of the particle (and force) is: [tex] \Gamma^{i}_{00} [/tex]

and using all that how you derive Einstein field equation so [tex] R_{ab}=0 [/tex]
 
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  • #2
What causes acceleration?

Karlisbad said:
where does this 4-dimensional force come from??

From some non-gravitational physical interaction or process.

Some typical examples:

1. In a static spherically symmetric perfect fluid solution (often used as an idealized stellar model), in the interior of the fluid ball, bits of fluid are accelerated radially outward by the differential pressure of the fluid, so that their world lines are timelike but nongeodesic curves. This is simply "hydrostatic equilibrium".

2. In the vacuum region outside such a fluid ball (modeled by a portion of the famous Scharzschild vacuum solution), we might consider "static test particles". These would also have world lines with outward pointing radial acceleration vector. We can think of such test particles as tiny rocket ships firing idealized rocket engines directly inward with just the right thrust to oppose the gravitational attraction of the "star". We would not of course generally attempt to model such rocketry using gtr! Fortunately, in the test model approximation, we don't have to.

3. In an "electrovacuum solution" (say the region outside an isolated charged object), a charged particle will in general be accelerated by the Lorentz force, just as in ordinary electromagnetism, but fixed up to work in a curved spacetime.

A more exotic example: according to gtr, nonspinning uncharged test particles (in an electrovacuum region outside a rotating magnetized neutron star, say) will have world lines which are timelike geodesics. However, if the test particle is spinning, in principle it will experience tiny spin-spin forces which will force its world line off a geodesic. In terms of the Bel decomposition of the Riemann curvature tensor, these spin-spin forces can be computed from the magnetogravitic tensor in a manner similar to how one can compute tidal forces from the electrogravitic tensor. They work a bit like the electromagnetic interaction of two dipole magnets, although such analogies are potentially misleading.

I should also stress that when you crunch the numbers, in most scenarios spin-spin forces turn out be very weak. Nonethelss, you might be wondering: could two spinning black holes, with coaxial and opposite spins, perhaps, in principle, be held in static equilibrium by the spin-spin force counteracting their mutual "mass monopole" gravitational attraction? This question was the focus of a protracted debate in the research literature some years ago, but this appears to have been settled in the negative.

Chris Hillman
 
  • #3


The first equation, \nabla _{u} u =0, is known as the geodesic equation and it describes the motion of a test particle in a space-time with no acceleration. This means that the particle follows the shortest path, known as a geodesic, in this space-time. This equation is based on the concept of covariant derivative, which takes into account the curvature of the space-time.

In an Euclidean space-time, where there is no curvature, the Christoffel symbols, denoted by \Gamma_{kl}^{i}, are equal to zero. However, when we consider a system under acceleration, the geodesic equation becomes \nabla _{u} u -a^{\mu}=0, where a^{\mu} represents the fourth-dimensional acceleration. This means that in a non-inertial frame, the particle will experience an acceleration due to the curvature of space-time.

The next equation, \acute{{R}^{\mu}}_{\alpha \nu \beta} \acute{u}^{\alpha} \acute{x}^{\nu} \acute{u}^{\beta} = - \acute{f}^{\mu}, introduces a four-dimensional force, denoted by \acute{f}^{\mu}. This force is a result of the curvature of space-time and it is related to the Riemann curvature tensor, represented by \acute{{R}^{\mu}}_{\alpha \nu \beta}. This equation is used to describe the motion of a test particle in a curved space-time.

In the weak field approximation, the potential of the particle and the force can be described by \Gamma^{i}_{00}. This approximation is used when the gravitational field is weak and the space-time is nearly flat. In this case, the Einstein field equations can be derived from the geodesic equation and the weak field approximation.

The Einstein field equations, represented by R_{ab}=0, describe the relationship between the curvature of space-time and the distribution of matter and energy. These equations are derived from the geodesic equation and the weak field approximation, taking into account the presence of matter and energy in the space-time. They are fundamental equations in general relativity and play a crucial role in understanding the behavior of gravity.
 

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