Kinetics of circular orbits

World line of a circular orbit about the Earth depicted in two spatial dimensions X and Y (the plane of the orbit) and a time dimension, usually put as the vertical axis. Note that the orbit about the Earth is a circle in space, but its worldline is a helix in spacetime.

For definiteness consider a circular Earth orbit (helical world line) of a particle. The particle travels with speed v. An observer on Earth sees that length is contracted in the frame of the particle. A measuring stick traveling with the particle appears shorter to the Earth observer. Therefore, the circumference of the orbit, which is in the direction of motion appears longer than ${\displaystyle \pi }$ times the diameter of the orbit.^[1]

In special relativity the 4-proper-velocity of the particle in the inertial (non-accelerating) frame of the earth is

${\displaystyle u=\left(\gamma ,\gamma {\mathbf {v} \over c}\right)}$

where c is the speed of light, ${\displaystyle \mathbf {v} }$ is the 3-velocity, and ${\displaystyle \gamma }$ is

${\displaystyle \gamma ={1 \over {\sqrt {1-{{\mathbf {v} \cdot \mathbf {v} } \over c^{2}}}}}}$.

The magnitude of the 4-velocity vector is always constant

${\displaystyle u_{\alpha }u^{\alpha }=-1}$

where we are using a Minkowski metric

${\displaystyle \eta ^{\mu \nu }=\eta _{\mu \nu }={\begin{pmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}}}$.

The magnitude of the 4-velocity is therefore a Lorentz scalar.

The 4-acceleration in the Earth (non-accelerating) frame is

${\displaystyle a\equiv {{du} \over {d\tau }}={d \over {d\tau }}{\left(\gamma ,\gamma {\mathbf {v} \over c}\right)}={\left(0,\gamma ^{2}{\mathbf {a} \over c^{2}}\right)}={\left(0,-\gamma ^{2}{{\mathbf {v} \cdot \mathbf {v} } \over c^{2}}{{\mathbf {r} } \over r^{2}}\right)}}$

where ${\displaystyle d\tau }$ is c times the proper time interval measured in the frame of the particle. This is related to the time interval in the Earth's frame by

${\displaystyle cdt=\gamma d\tau }$.

Here, the 3-acceleration for a circular orbit is

${\displaystyle \mathbf {a} =-\omega ^{2}\mathbf {r} =-{\mathbf {v} \cdot \mathbf {v} }{{\mathbf {r} } \over r^{2}}}$

where ${\displaystyle \omega }$ is the angular velocity of the rotating particle and ${\displaystyle \mathbf {r} }$ is the 3-position of the particle.

The magnitude of the 4-velocity is constant. This implies that the 4-acceleration must be perpendicular to the 4-velocity. The inner product of the 4-acceleration and the 4-velocity is therefore always zero. The inner product is a Lorentz scalar.

Curvature of spacetime: Geodesic equation

The equation for the acceleration can be generalized, yielding the geodesic equation

${\displaystyle {{du^{\mu }} \over {d\tau }}-a^{\mu }=0}$

${\displaystyle {{du^{\mu }} \over {d\tau }}+{R^{\mu }}_{\alpha \nu \beta }u^{\alpha }x^{\nu }u^{\beta }=0}$

where ${\displaystyle x^{\mu }}$ is the 4-position of the particle and ${\displaystyle {R^{\mu }}_{\alpha \nu \beta }}$ is the curvature tensor given by

${\displaystyle {R^{\mu }}_{\alpha \nu \beta }={1 \over r^{2}}\eta _{\alpha \beta }{\delta ^{\mu }}_{\nu }}$

where ${\displaystyle {\delta ^{\mu }}_{\nu }}$ is the Kronecker delta function, and we have the constraints

${\displaystyle u_{\alpha }u^{\alpha }=-1}$

and

${\displaystyle a_{\alpha }u^{\alpha }=0}$.

It is easily verified that circular orbits satisfy the geodesic equation. The geodesic equation is actually more general. Circular orbits are a particular solution of the equation. Solutions other than circular orbits are permissible and valid.

Ricci curvature tensor and trace

The Ricci curvature tensor is a special curvature tensor given by the contraction

${\displaystyle R_{\alpha \beta }\equiv {R^{\nu }}_{\alpha \nu \beta }}$.

The trace of the Ricci tensor, called the scalar curvature, is

${\displaystyle R\equiv {R^{\alpha }}_{\alpha }}$.

The geodesic equation in a local coordinate system

Circular orbits at the same radius

Consider the situation in which there are now two particles in nearby circular polar orbits of the Earth at radius ${\displaystyle r}$ and speed ${\displaystyle v}$.

The particles execute simple harmonic motion about the Earth and with respect to each other. They are at their maximum distance from each other as they cross the equator. Their trajectories intersect at the poles.

Imagine a spacecraft co-moving with one of the particles. The ceiling of the craft, the ${\displaystyle {\acute {\mathbf {z} }}}$ direction, coincides with the ${\displaystyle \mathbf {r} }$ direction. The front of the craft is in the ${\displaystyle {\acute {\mathbf {x} }}}$ direction, and the ${\displaystyle {\acute {\mathbf {y} }}}$ direction is to the left of the craft. The spacecraft is small compared with the size of the orbit so that the local frame is a local Lorentz frame. The 4-separation of the two particles is given by ${\displaystyle {\acute {x}}^{\mu }}$. In the local frame of the spacecraft the geodesic equation is given by

${\displaystyle {{d^{2}{\acute {x}}^{\mu }} \over {d\tau ^{2}}}+{\acute {{R}^{\mu }}}_{\alpha \nu \beta }{\acute {u}}^{\alpha }{\acute {x}}^{\nu }{\acute {u}}^{\beta }=0}$

where

${\displaystyle {\acute {u}}^{\mu }={{d{\acute {x}}^{\mu }} \over {d\tau }}}$

and

${\displaystyle {\acute {{R}^{\mu }}}_{\alpha \nu \beta }}$

is the curvature tensor in the local frame.

Geodesic equation as a covariant derivative

The equation of motion for a particle in flat spacetime and in the absence of forces is

${\displaystyle {{d{u}^{\mu }} \over {d\tau }}=0}$.

If we require a particle to travel along a geodesic in curved spacetime, then the analogous expression in curved spacetime is

${\displaystyle {{D{\acute {u}}^{\mu }} \over {D\tau }}={{d{\acute {u}}^{\mu }} \over {d\tau }}+{\Gamma ^{\mu }}_{\alpha \beta }{\acute {u}}^{\alpha }{\acute {u}}^{\beta }=0}$

where the derivative on the left is the covariant derivative, which is the generalization of the normal derivative to a derivative in curved spacetime. Here

${\displaystyle {\Gamma ^{\mu }}_{\alpha \beta }}$

is a Christoffel symbol.

The curvature is related to the Christoffel symbol by

${\displaystyle {\acute {{R}^{\mu }}}_{\alpha \nu \beta }={{\partial {{\Gamma }^{\mu }}}_{\alpha \beta } \over {\partial x^{\nu }}}-{{\partial {{\Gamma }^{\mu }}}_{\alpha \nu } \over {\partial x^{\beta }}}+{{\Gamma }^{\mu }}_{\gamma \nu }{{\Gamma }^{\gamma }}_{\alpha \beta }-{{\Gamma }^{\mu }}_{\gamma \beta }{{\Gamma }^{\gamma }}_{\alpha \nu }}$.

Metric tensor in the local frame

The interval in the local frame is

${\displaystyle ds^{2}=dx^{2}+dy^{2}+dz^{2}-c^{2}dt^{2}\equiv g_{\mu \nu }d{\acute {x}}^{\mu }d{\acute {x}}^{\nu }}$

${\displaystyle =d{\acute {x}}^{2}+d{\acute {y}}^{2}+d{\acute {z}}^{2}-c^{2}d{\acute {t}}^{2}+2\gamma \cos(\theta )\cos(\phi )\,v\,d{\acute {t}}\,d{\acute {x}}+2\gamma \cos(\theta )\sin(\phi )v\,d{\acute {t}}\,d{\acute {y}}-2\gamma \sin(\theta )v\,d{\acute {t}}\,d{\acute {z}}}$

where

${\displaystyle \theta }$ is the angle with the ${\displaystyle z}$ axis (longitude) and

${\displaystyle \phi }$ is the angle with the ${\displaystyle x}$ axis (latitude).

This gives a metric of

${\displaystyle g_{\mu \nu }={\begin{pmatrix}-1&\gamma \cos(\theta )\cos(\phi ){\frac {v}{c}}&\gamma \cos(\theta )\sin(\phi ){\frac {v}{c}}&-\gamma \sin(\theta ){\frac {v}{c}}\\\gamma \cos(\theta )\cos(\phi ){\frac {v}{c}}&1&0&0\\\gamma \cos(\theta )\sin(\phi ){\frac {v}{c}}&0&1&0\\-\gamma \sin(\theta ){\frac {v}{c}}&0&0&1\end{pmatrix}}}$

in the local frame.

The inverse of the metric tensor ${\displaystyle g^{\mu \nu }}$ is defined such that

${\displaystyle g_{\mu \alpha }g^{\alpha \nu }=\delta _{\mu }^{\nu }}$

where the term on the right is the Kronecker delta.

The transformation of the infinitesimal 4-volume ${\displaystyle d\Omega }$ is

${\displaystyle d{\acute {\Omega }}={\sqrt {-g}}d{\Omega }}$

where g is the determinant of the metric tensor.

The differential of the determinant of the metric tensor is

${\displaystyle dg=gg^{\mu \nu }dg_{\mu \nu }=-gg_{\mu \nu }dg^{\mu \nu }}$.

The relationship between the Christoffel symbols and the metric tensor is

${\displaystyle {{\Gamma }^{\alpha }}_{\mu \nu }=g^{\alpha \beta }{\Gamma }_{\beta \mu \nu }}$

${\displaystyle {\Gamma }_{\beta \mu \nu }={\frac {1}{2}}\left({{\partial {g}}_{\beta \nu } \over {\partial x^{\mu }}}+{{\partial {g}}_{\beta \mu } \over {\partial x^{\nu }}}-{{\partial {g}}_{\mu \nu } \over {\partial x^{\beta }}}\right)}$.

Principle of least action in general relativity

The principle of least action states that the world line between two events in spacetime is that world line that minimizes the action between the two events. In classical mechanics the principle of least action is used to derive Newton's laws of motion and is the basis for Lagrangian dynamics. In relativity it is expressed as

${\displaystyle S=\int _{1}^{2}{\mathcal {L}}\,d\Omega }$

between events 1 and 2 is a minimum. Here S is a scalar and

${\displaystyle {\mathcal {L}}}$

is known as the Lagrangian density. The Lagrangian density is divided into two parts, the density for the orbiting particle ${\displaystyle {\mathcal {L}}_{p}}$ and the density ${\displaystyle {\mathcal {L}}_{e}}$ of the gravitational field generated by all other particles including those comprising the Earth,

${\displaystyle {\mathcal {L}}={\mathcal {L}}_{p}+{\mathcal {L}}_{e}}$.

In curved spacetime, the "shortest" world line is that geodesic that minimizes the curvature along the geodesic. The action then is proportional to the curvature of the world line. Since S is a scalar, the scalar curvature is the appropriate measure of curvature. The action for the particle is therefore

${\displaystyle S_{p}=C\int _{1}^{2}{\acute {R}}\,d{\acute {\Omega }}=C\int _{1}^{2}{\acute {R}}{\sqrt {-g}}\,d{\Omega }=C\int _{1}^{2}g^{\alpha \beta }{\acute {R}}_{\alpha \beta }{\sqrt {-g}}\,d{\Omega }}$

where ${\displaystyle C}$ is an unknown constant. This constant will be determined by requiring the theory to reduce to Newton's law of gravitation in the nonrelativistic limit.

The Lagrangian density for the particle is therefore

${\displaystyle {\mathcal {L}}_{p}=Cg^{\alpha \beta }{\acute {R}}_{\alpha \beta }{\sqrt {-g}}}$.

The action for the particle and the Earth is

${\displaystyle S=\int _{1}^{2}Cg^{\alpha \beta }{\acute {R}}_{\alpha \beta }{\sqrt {-g}}\,d\Omega +\int _{1}^{2}{\mathcal {L}}_{e}\,d\Omega }$.

Thus the world line that lies on the surface of the sphere of radius r by varying the metric tensor. Minimization and neglect of terms that disappear on the boundaries, including terms second order in the derivative of g, yields

${\displaystyle 0=\delta S=\int _{1}^{2}C\left({\acute {R}}_{\alpha \beta }-{1 \over 2}{\acute {R}}g^{\alpha \beta }\right)\delta g^{\alpha \beta }{\sqrt {-g}}\,d\Omega -\int _{1}^{2}{\acute {T}}_{\alpha \beta }\delta g^{\alpha \beta }{\sqrt {-g}}\,d\Omega }$

where^[2]

${\displaystyle {\acute {T}}_{\alpha \beta }={1 \over {\sqrt {-g}}}\left({d \over {dx^{\nu }}}{{\partial {\mathcal {L}}_{e}} \over {\partial \left({{dg^{\alpha \beta }} \over {dx^{\nu }}}\right)}}-{{\partial {\mathcal {L}}_{e}} \over {\partial g^{\alpha \beta }}}\right)}$

is the Hilbert stress–energy tensor of the field generated by the Earth.

The relationship, to within an unknown constant factor, between the stress-energy and the curvature is

${\displaystyle {\acute {T}}_{\alpha \beta }=C\left({\acute {R}}_{\alpha \beta }-{1 \over 2}{\acute {R}}\,g_{\alpha \beta }\right)}$.

Newton's law of gravitation

Diagram 1. Changing views of spacetime along the world line of a rapidly accelerating observer. In this animation, the dashed line is the spacetime trajectory ("world line") of a particle. The balls are placed at regular intervals of proper time along the world line. The solid diagonal lines are the light cones for the observer's current event, and intersect at that event. The small dots are other arbitrary events in the spacetime. For the observer's current instantaneous inertial frame of reference, the vertical direction indicates the time and the horizontal direction indicates distance. The slope of the world line (deviation from being vertical) is the velocity of the particle on that section of the world line. So at a bend in the world line the particle is being accelerated. Note how the view of spacetime changes when the observer accelerates, changing the instantaneous inertial frame of reference. These changes are governed by the Lorentz transformations. Also note that: * the balls on the world line before/after future/past accelerations are more spaced out due to time dilation. * events which were simultaneous before an acceleration are at different times afterwards (due to the relativity of simultaneity), * events pass through the light cone lines due to the progression of proper time, but not due to the change of views caused by the accelerations, and * the world line always remains within the future and past light cones of the current event.

Newton's Law of Gravitation in non-relativistic mechanics states that the acceleration on an object of mass ${\displaystyle m}$ due to another object of mass ${\displaystyle M}$ is equal to

${\displaystyle \mathbf {f} ={d^{2}\mathbf {r} \over d\tau ^{2}}=-{GM \over {c^{2}r^{3}}}\mathbf {r} }$

where ${\displaystyle G}$ is the gravitational constant, ${\displaystyle \mathbf {r} }$ is a vector from mass ${\displaystyle M}$ to mass ${\displaystyle m}$ and ${\displaystyle r}$ is the magnitude of that vector. The time t is scaled with the speed of light c

${\displaystyle \tau \equiv ct}$.

The acceleration ${\displaystyle \mathbf {f} }$ is independent of ${\displaystyle m}$.

For definiteness. consider a particle of mass ${\displaystyle m}$ orbiting in the gravitational field of the Earth with mass ${\displaystyle M}$. The law of gravitation can be written

${\displaystyle \mathbf {f} =-{4\pi G \over {3c^{2}}}\rho (r)\mathbf {r} }$

where ${\displaystyle \rho (r)}$ is the average mass density inside a sphere of radius ${\displaystyle r}$.

Gravitational force in terms of the 00 component of the stress–energy tensor

Newton's law can be written

${\displaystyle \mathbf {f} =-{4\pi G \over {3c^{4}}}\left({Mc^{2} \over V}\right)\mathbf {r} }$.

where ${\displaystyle V}$ is the volume of a sphere of radius ${\displaystyle r}$. The quantity ${\displaystyle Mc^{2}}$ will be recognized from special relativity as the rest energy of the large body, the Earth. This is the sum of the rest energies of all the particles that compose Earth. The quantity in the parentheses is then the average rest energy density of a sphere of radius ${\displaystyle r}$ about the Earth. The gravitational field is proportional to the average energy density within a radius r. This is the 00 component of the stress–energy tensor in relativity for the special case in which all the energy is rest energy. More generally

${\displaystyle T_{00}=-{T^{0}}_{0}=\sum _{i=1}^{N}\left({\gamma _{i}m_{i}c^{2} \over V}\right)}$

where

${\displaystyle \gamma _{i}\equiv {1 \over {\sqrt {1-{{\mathbf {v} _{i}\cdot \mathbf {v} _{i}} \over c^{2}}}}}}$

and ${\displaystyle \mathbf {v_{i}} }$ is the velocity of particle i making up the Earth and ${\displaystyle m_{i}}$ in the rest mass of particle i. There are N particles altogether making up the Earth.

Relativistic generalization of the energy density

The components of the stress–energy tensor

There are two simple relativistic entities that reduce to the 00 component of the stress–energy tensor in the nonrelativistic limit

${\displaystyle u^{\alpha }T_{\alpha \beta }u^{\beta }\rightarrow T_{00}}$

and the trace

${\displaystyle T\equiv {T^{\alpha }}_{\alpha }=-u_{\alpha }u^{\alpha }T=-u^{\alpha }T\eta _{\alpha \beta }u^{\beta }\rightarrow -T_{00}}$

where ${\displaystyle u^{\alpha }}$ is the 4-velocity.

The 00 component of the stress–energy tensor can be generalized to the relativistic case as a linear combination of the two terms

${\displaystyle T_{00}\rightarrow u^{\alpha }\left(AT_{\alpha \beta }+BT\eta _{\alpha \beta }\right)u^{\beta }}$

where

${\displaystyle A+B=1}$

4-acceleration due to gravity

The 4-acceleration due to gravity can be written

${\displaystyle f^{\mu }=-8\pi {G \over {3c^{4}}}\left({A \over 2}T_{\alpha \beta }+{B \over 2}T\eta _{\alpha \beta }\right)\delta _{\nu }^{\mu }u^{\alpha }x^{\nu }u^{\beta }}$.

Unfortunately, this acceleration is nonzero for ${\displaystyle \mu =0}$ as is required for circular orbits. Since the magnitude of the 4-velocity is constant, it is only the component of the force perpendicular to the 4-velocity that contributes to the acceleration. We must therefore subtract off the component of force parallel to the 4-velocity. This is known as Fermi–Walker transport.^[3] In other words,

${\displaystyle f^{\mu }\rightarrow f^{\mu }+u^{\mu }u_{\nu }f^{\nu }}$.

This yields

${\displaystyle f^{\mu }=-8\pi {G \over {3c^{4}}}\left({A \over 2}T_{\alpha \beta }+{B \over 2}T\eta _{\alpha \beta }\right)\left(\delta _{\nu }^{\mu }+u^{\mu }u_{\nu }\right)u^{\alpha }x^{\nu }u^{\beta }}$.

The force in the local frame is

${\displaystyle {\acute {f}}^{\mu }=-8\pi {G \over {3c^{4}}}\left({A \over 2}{\acute {T}}_{\alpha \beta }+{B \over 2}{\acute {T}}g_{\alpha \beta }\right)\left(\delta _{\nu }^{\mu }+{\acute {u}}^{\mu }{\acute {u}}_{\nu }\right){\acute {u}}^{\alpha }{\acute {x}}^{\nu }{\acute {u}}^{\beta }}$.