What is the meaning of the Equation of Geodesic Deviation?

Question

I've seen the Equation of Geodesic Deviation stated in several text as:$$\frac{D\xi^2}{dt^2}+Riemann(\textbf U,\xi,\textbf U)=0$$ but I haven't seen a real good explanation of why it works or the meaning of the right hand side of the equation. Why is it zero? Some texts and posts seem to suggest that this is Newton's First Law at work or an assumption of Minkowski space (objects at rest stay at rest, objects in motion stay in motion, acceleration is zero). Is the zero on the right hand side of the equation an expression of Newton's First Law of Motion?

EDIT: The Equation of Geodesic Deviation appears to be asking the question, what shape would allow these objects that appear to be accelerating (e.g. the Moon, the Earth, the Sun), to be in "free fall". The Riemann appears to be the answer to the 'what shape' part of the question, but the right hand side, zero in Newtonian Dynamics, appears to be the desired answer of the equation: a test particle in "free fall" with respect to a fudicial geodesic.

Answer 1

The geodesic deviation equation describes the behaviour of a one-parameter family of neighboring geodesics $\gamma_s(t)$, that is for each $s$, $\gamma_s$ is a geodesic parameterized by the affine parameter $t$. Defining the tangent vector $T^\mu = \partial x^\mu / \partial t$ and the deviation vector $S^\mu = \partial x^\mu / \partial s$, we have

$A^\mu = D^2 S^\mu / dt^2 = R^\mu_{\nu \rho \sigma} T^\nu T^\rho S^\sigma$

where:
$A^\mu$ relative acceleration of geodesics
$D / dt = (dx^\mu / dt) \nabla _\mu$ directional covariant derivative

Coming to your first question: "Why the R.H.S. of the equation is zero?", as you can read it is not. The explanation why in many texts all the terms are in the L.H.S. is because they use the antisymmetry of the last two indices of the Riemann tensor. They swap the indices $\rho$ and $\sigma$ getting a minus sign which is then offset by moving the Riemann term to the L.H.S. The swap of the indices attaches to the deviation vector the third index of the Riemann tensor, thus posting the vector in the second slot out of the three available in Riemann.

As for the second question, you have to explicit the Riemann tensor in that geometry.

Answer 2

What you call the equation of geodesic deviation is called the Jacobi equation and the vector field $\xi$ is called the Jacobi field. The idea is the following.

A. Assume that you are given a space (smooth manifold) $M$ that has some Riemannian or pseudo-Riemannian metric.

B. Let $\gamma(t)$ be one fixed geodesic of that metric.

Definition. A Jacobi field $\xi(t)$ along the fixed geodesic $\gamma(t)$ is a vector field defined as follows:

1. let us assume that $\gamma(t)$ is included in a smooth one-parameter family of geodesics $\gamma(t, \sigma)$ such that $\gamma(t, 0) = \gamma(t)$. Smooth one-parameter family of geodesics means that the map $$\gamma \, : \, (t_1, t_2) \times (- \,\epsilon,\, \epsilon) \, \to \, M $$ is a smooth map such that for any fixed $\sigma \in (- \,\epsilon,\, \epsilon)$ the curve $\gamma(t, \sigma)$ is a geodesic of the space $M$.

2. Then the Jacobi vector field $\xi$ along $\gamma$ for the family of geodesics is $$\xi(t) = \frac{\partial}{\partial \sigma} \, \gamma(t, \sigma)\Big|_{\sigma=0} \,\,\,\,\,\text{ (i.e. first differentiate with respect to $\sigma$ and then plug $\sigma = 0$ )}$$

In other words, Jacobi vector fields along some fixed geodesic $\gamma$ describe in what directions and how much one should push $\gamma$ to turn it into a near-by geodesic. The Jacobi vector fields are very special because if you take an arbitrary vector field along $\gamma$ and use it to push $\gamma$ in the direction of that arbitrary field, the result may not be another geodesic but an arbitrary curve.

So the converse question arises: How can I tell whether a vector field $\xi(t)$ defined along a geodesic $\gamma(t)$ is a Jacobi vector field -- i.e. if for each $t$ I push $\gamma(t)$ along $\xi(t)$ would the result be another geodesic? Well, the answer is the Jacobi differential equation. In other words,

Theorem. A vector field $\xi(t)$ along the geodesic $\gamma(t)$ satisfies the Jacobi equation $$\frac{D^2}{dt^2}\xi + R\left(\xi, \frac{d\gamma}{dt}\right)\,\frac{d\gamma}{dt} = 0 $$ if and only if there is a smooth one parameter family of geodesics $\gamma(t, \sigma)$ sich that $\gamma(t, 0) = \gamma(t)$ and $$\xi(t) = \frac{\partial}{\partial \sigma} \, \gamma(t, \sigma)\Big|_{\sigma=0} $$

To justify this theorem you have to prove it in both directions, but I can sketch the direction which states that if $\xi(t)$ is a Jacobi field, then it must satisfy the Jacobi equation, i.e. deriving the Jacobi equation.

The Jacobi equation is simply the first variation of the geodesic equation. That is why the right hand side is equal to zero. Because the right hand side of the geodesic equation is zero. Here is what I mean: for each $\sigma$ the curve $\gamma(t, \sigma)$ is a geodesic. Then the geodesic equation says $$\nabla_{\frac{\partial \gamma}{\partial t}(t, \sigma)} \,\frac{\partial \gamma}{\partial t}(t, \sigma) = 0 $$ Now, if you differentiate this geodesic equation with respect to $\sigma$ and set $\sigma = 0$ you get the Jacobi equation $$\frac{D^2}{dt^2}\xi + R\left(\xi, \frac{d\gamma}{dt}\right)\,\frac{d\gamma}{dt} = 0 $$

Indeed, differentiating with respect to $\sigma$ means taking the covariant derivative with respect to $\sigma$, i.e. $$\nabla_{\frac{\partial \gamma}{\partial \sigma}(t, \sigma)} \left( \nabla_{\frac{\partial \gamma}{\partial t}(t, \sigma)} \,\frac{\partial \gamma}{\partial t}(t, \sigma) \right) = 0$$ However, by the definition of the Riemann tensor $$\nabla_{\frac{\partial \gamma}{\partial \sigma}(t, \sigma)} \nabla_{\frac{\partial \gamma}{\partial t}(t, \sigma)} = \nabla_{\frac{\partial \gamma}{\partial t}(t, \sigma)} \nabla_{\frac{\partial \gamma}{\partial \sigma}(t, \sigma)} + R\left( \frac{\partial \gamma}{\partial \sigma}(t, \sigma), \, \frac{\partial \gamma}{\partial t}(t, \sigma) \right)$$ so \begin{align} 0 =& \nabla_{\frac{\partial \gamma}{\partial \sigma}(t, \sigma)} \nabla_{\frac{\partial \gamma}{\partial t}(t, \sigma)} \,\frac{\partial \gamma}{\partial t}(t, \sigma) = \\ =& \nabla_{\frac{\partial \gamma}{\partial t}(t, \sigma)} \nabla_{\frac{\partial \gamma}{\partial \sigma}(t, \sigma)}\, \frac{\partial \gamma}{\partial t}(t, \sigma) + R\left( \frac{\partial \gamma}{\partial \sigma}(t, \sigma), \, \frac{\partial \gamma}{\partial t}(t, \sigma) \right) \frac{\partial \gamma}{\partial t}(t, \sigma) = \\ =& \nabla_{\frac{\partial \gamma}{\partial t}}\left( \nabla_{\frac{\partial \gamma}{\partial \sigma}}\, \frac{\partial \gamma}{\partial t}\right) + R\left( \frac{\partial \gamma}{\partial \sigma}, \, \frac{\partial \gamma}{\partial t} \right) \frac{\partial \gamma}{\partial t} \end{align} However, if by we come back to the interpretation of $\gamma$ as a two-parameter smooth map, as given in point 1 above, then $$\frac{\partial \gamma}{\partial t}(t, \sigma) = \gamma_{*(t,\sigma)} \frac{\partial}{\partial t} \,\,\, \text{ and }\,\,\, \frac{\partial \gamma}{\partial \sigma}(t, \sigma) = \gamma_{*(t,\sigma)} \frac{\partial}{\partial \sigma}$$ where $\gamma_{*(t,\sigma)}$ is the derivative map (the tangent map) of the map $\gamma$ at the point $(t,\sigma)$. Based on the fact that there is no torsion of the covariant derivative, since it comes from a metric \begin{align} \nabla_{\frac{\partial \gamma}{\partial \sigma}}\, \frac{\partial \gamma}{\partial t} =& \nabla_{\gamma_{*} \frac{\partial}{\partial \sigma}}\, \gamma_{*} \frac{\partial}{\partial t} = \nabla_{\gamma_{*} \frac{\partial}{\partial t}}\, \gamma_{*} \frac{\partial}{\partial \sigma} + \left[\gamma_{*} \frac{\partial}{\partial t}, \gamma_{*} \frac{\partial}{\partial \sigma}\right] = \\ =& \nabla_{\gamma_{*} \frac{\partial}{\partial t}}\, \gamma_{*} \frac{\partial}{\partial \sigma} + \gamma_{*}\left[\frac{\partial}{\partial t}, \frac{\partial}{\partial \sigma}\right] = \nabla_{\gamma_{*} \frac{\partial}{\partial t}}\, \gamma_{*} \frac{\partial}{\partial \sigma} + 0 \\ =& \nabla_{\frac{\partial \gamma}{\partial t}}\, \frac{\partial \gamma}{\partial \sigma} \end{align}

So we continue the derivation of the Jacobi equation \begin{align} 0 =& \nabla_{\frac{\partial \gamma}{\partial \sigma}(t, \sigma)} \nabla_{\frac{\partial \gamma}{\partial t}(t, \sigma)} \,\frac{\partial \gamma}{\partial t}(t, \sigma) = \\ =& \nabla_{\frac{\partial \gamma}{\partial t}(t, \sigma)} \nabla_{\frac{\partial \gamma}{\partial \sigma}(t, \sigma)}\, \frac{\partial \gamma}{\partial t}(t, \sigma) + R\left( \frac{\partial \gamma}{\partial \sigma}(t, \sigma), \, \frac{\partial \gamma}{\partial t}(t, \sigma) \right) \frac{\partial \gamma}{\partial t}(t, \sigma) = \\ =& \nabla_{\frac{\partial \gamma}{\partial t}}\left( \nabla_{\frac{\partial \gamma}{\partial \sigma}}\, \frac{\partial \gamma}{\partial t}\right) + R\left( \frac{\partial \gamma}{\partial \sigma}, \, \frac{\partial \gamma}{\partial t} \right) \frac{\partial \gamma}{\partial t} = \\ =& \nabla_{\frac{\partial \gamma}{\partial t}}\left( \nabla_{\frac{\partial \gamma}{\partial t}}\, \frac{\partial \gamma}{\partial \sigma}\right) + R\left( \frac{\partial \gamma}{\partial \sigma}, \, \frac{\partial \gamma}{\partial t} \right) \frac{\partial \gamma}{\partial t} = \\ =& \nabla_{\frac{\partial \gamma}{\partial t}(t, \sigma)}\left( \nabla_{\frac{\partial \gamma}{\partial t}(t, \sigma)}\, \frac{\partial \gamma}{\partial \sigma}(t, \sigma)\right) + R\left( \frac{\partial \gamma}{\partial \sigma}(t, \sigma), \, \frac{\partial \gamma}{\partial t}(t, \sigma) \right) \frac{\partial \gamma}{\partial t}(t, \sigma) \end{align} Now, set $\sigma = 0$ in which case $$\frac{\partial \gamma}{\partial t}(t, 0) = \frac{d\gamma}{dt}(t) \,\,\, \text{ and } \,\,\, \frac{\partial \gamma}{\partial \sigma}(t, 0) = \xi(t)$$ Consequently, the equation above turns into \begin{align} 0 =& \nabla_{\frac{d \gamma}{d t}(t)}\left( \nabla_{\frac{d \gamma}{d t}(t)}\, \xi(t)\right) + R\left( \xi(t), \,\frac{d \gamma}{d t}(t) \right) \frac{d \gamma}{d t}(t) \end{align} Recall that $$\nabla_{\frac{d \gamma}{d t}(t)} \nabla_{\frac{d \gamma}{d t}(t)} = \frac{D^2}{dt^2}$$ as a notation and we arrive at the Jacobi equation $$\, \frac{D^2 \xi}{dt^2} + R\left( \xi, \,\frac{d \gamma}{d t}(t) \right) \frac{d \gamma}{d t}(t) = 0$$

I assume your last question about the sphere with changing radius is asked out of confusion. The topic of Jacobi equations and Jacobi fields does not require for the metric and the whole geometry to be changing. The geometry usually stays fixed. If the geometry changes with respect to some parameters, simply all the quantities involved in the Jacobi equation will additionally depend on these parameters.

Answer 3

What I'll do here is lay out the "geodesic" deviation equation in its most general form - with torsion - explain what it means, and provide a simple derivation of it; which will require a little exposition on a (very liberating) notational convention that will make everything transparent.

The geodesic deviation equation is generic to affine manifolds, independent of what metric, if any, is defined on it. In its most general form, it involves the curvature tensor and covariant derivative of the torsion. Strictly speaking, it is not "geodesic" deviation, but auto-parallel deviation, since the term "geodesic" is reserved only for metric-affine manifolds, where the connection is a Levi-Civita connection. And, the curves in it are not geodesics, in general, but auto-parallels.

Of particular relevance is this: if non-zero torsion is present, and if the covariant continuity equation is true for the stress tensor, then test bodies do not follow geodesics, but auto-parallels.

Test Bodies: Geodesic Or Auto-Parallel?

I'll still use the terms "geodesic" and "geodesic" deviation with that understanding and qualification.

What Does Geodesic Deviation Convey?

In reality, the geodesic deviation equation describes the warping of a coordinate grid over time, due to the differential effects of free-fall on neighboring points (i.e. due to the tidal force), where the points in the grid are geodesics. So, in place of a function $x(τ,σ)$ of time $τ$ and separation $σ$, you actually have a function $x\left(τ,σ^1,σ^2,σ^3\right)$ (if in $3+1$ dimensions), such that for each $ = \left(σ^1,σ^2,σ^3\right)$, $τ ↦ x(τ,)$ is a geodesic curve. Seen in this light (an equation describing the warping of the coordinate grid), it is closely connected to the Raychaudhuri Equation (which can also be generalized to cover the case of non-zero torsion).

For such a family of geodesics, one can write $$T = \frac{∂x^ρ}{∂τ} ∂_ρ,\quad S_i = \frac{∂x^ρ}{∂σ^i} ∂_ρ,$$ or, in component form, as $$T^ρ = \frac{∂x^ρ}{∂τ},\quad {S_i}^ρ = \frac{∂x^ρ}{∂σ^i}.$$ As differential operators applied to functions $Φ(x(τ,))$, they can be written, through the chain rule, as $$T Φ = T^ρ ∂_ρ Φ = \frac{∂Φ}{∂τ},\quad S_i Φ = {S_i}^ρ ∂_ρ Φ = \frac{∂Φ}{∂σ^i}\quad (0 < i < 4).$$

Notational Convention

The best way to approach this (and many other applications) is to adopt the following conventions:

• Tensors, and other objects (like connections) are functionals of vectors and co-vectors.
• Co-vector arguments are written as upstairs indices, vector arguments a downstairs indices.
• The Physicists' convention is subsumed by treating $$()^μ = ()^{dx^μ},\quad ()_μ = ()_{∂_μ}.$$
• It can even be applied recursively, e.g. $$dx^μ = dx^{dx^μ},\quad ∂_μ = ∂_{∂_μ}.$$
• An object is "tensorial" in an index, if it is linear in the index; and it is a tensor if it is linear in all of its indices.

This is a resume of the most common objects encountered:

• The contraction operator / Kronecker delta: $$δ^α_u = α·u = δ^ρ_μ α_ρ u^μ = α_u = u^α = α_μ u^μ.$$
• The differential operator $∂_u = u^μ ∂_μ$ and "coordinate differential" operator $dx^α = dx^μ α_μ$; where, by convention $α = dx^α$, $u = ∂_u$.
• The commutator / "structure coefficients": $$[u,v] = f^ρ_{uv} ∂_ρ, \quad f^ρ_{uv} = ∂_u v^ρ - ∂_v u^ρ,\quad f^α_{uv} = α·[u,v].$$
• The covariant derivative / connection: $$∇_u v = Γ^ρ_{uv} ∂_ρ,\quad Γ^α_{uv} = α·∇_u v = α_ρ u^μ \left(δ^ρ_ν ∂_μ + Γ^ρ_{μν}\right) v^ν.$$
• The torsion operator and torsion: $$T_{uv} = ∇_u v - ∇_v u - [u,v],\quad T^α_{uv} = α·T_{uv} = Γ^α_{uv} - Γ^α_{vu} - f^α_{uv}.$$
• The curvature operator and tensor: $$\begin{align} Ω_{uv} &= ∇_u ∇_v - ∇_v ∇_u - ∇_{[u,v]},\\ R^α_{wuv} &= α·Ω_{uv} w\\ &= α_ρ \left(∂_u Γ^ρ_{vw} - ∂_v Γ^ρ_{uw} + Γ^ρ_{uσ} Γ^σ_{vw} - Γ^ρ_{vσ} Γ^σ_{uw} - f^σ_{uv} Γ^ρ_{σw}\right). \end{align}$$

All the objects are tensors, except the connection $Γ^α_{uv}$ and structure coefficients $f^α_{uv}$: $Γ^α_{uv}$ is tensorial only in $α$ and $u$, while $f^α_{uv}$ is tensorial only in $α$ and reduces to 0 for coordinate frames: $f^α_{μν} = α·\left[∂_μ,∂_ν\right] = 0$.

The latter point is the one of interest here. For a family of geodesics, one can write $T$, and $S_i$, for $0 < i < 4$, as differential operators: $$∂_T = \frac{∂x^μ}{∂τ} ∂_μ = \frac{∂}{∂τ},\quad ∂_{S_i} = \frac{∂x^μ}{∂σ^i} ∂_μ = \frac{∂}{∂σ^i}\quad (0 < i < 4).$$ Consequently, as vector fields, they commute: $$\left[T,S_i\right] = 0,\quad \left[S_i,S_j\right] = 0\quad (0 < i,j < 4),$$ which is what makes this a coordinate grid.

The geodesic (actually: auto-parallel) equation can then be written as $$0 = \frac{∂^2x^ρ}{∂τ^2} + Γ^ρ_{μν} \frac{∂x^μ}{∂τ} \frac{∂x^ν}{∂τ} = ∂_T T^ρ + Γ^ρ_{Tν} T^ν = Γ^ρ_{TT} = (∇_T T)^ρ.$$ Thus, $∇_T T = 0$. The "velocity" components for the coordinate grid are $V_i = ∇_T S_i$ and the "acceleration" components are $A_i = ∇_T V_i$ which is what the "geodesic" deviation equation apply to.

Derivation Of The "Geodesic" Deviation Equation

With the background provided and conventions adopted, the derivation is quite elementary. Let $S$ denote any of the $S_i$, for $0 < i < 4$. Then $$Ω_{TS}T = R^ρ_{TTS} ∂_ρ = \left(∇_T ∇_S - ∇_S ∇_T - ∇_{[T,S]}\right) T = ∇_T ∇_S T,$$ since $∇_T T = 0$ and $[T,S] = 0$. In addition, $$T_{TS} = ∇_T S - ∇_S T - [T,S] = ∇_T S - ∇_S T.$$ Thus $$R^ρ_{TTS} ∂_ρ + ∇_T T_{TS} = ∇_T ∇_T S = A.$$ In component form, since we're dealing with tensors, recalling $T^ρ = ∂x^ρ/∂τ$ and ${S_i}^ρ = ∂x^ρ/∂σ^i$, and using semi-colons with indexes to denote covariant derivative components. So, we can reduce this directly to component form and write $$\left(R^ρ_{λμν} + T^ρ_{μν;λ}\right) \frac{∂x^λ}{∂τ} \frac{∂x^μ}{∂τ} \frac{∂x^ν}{∂σ^i} = {∇_T}^2 \frac{∂x^ρ}{∂σ^i}\quad (0 < i < 4).$$

That's the "geodesic" deviation equation with torsion - or strictly speaking - the auto-parallel deviation equation.

The usual form it is written in is with just one parameter $σ$, instead of an entire grid's worth of parameters $ = \left(σ^1,σ^2,σ^3\right)$ as $$\left(R^ρ_{λμν} + T^ρ_{μν;λ}\right) \frac{∂x^λ}{∂τ} \frac{∂x^μ}{∂τ} \frac{∂x^ν}{∂σ} = {∇_T}² \frac{∂x^ρ}{∂σ},$$ or as just $$\left(R^ρ_{λμν} + T^ρ_{μν;λ}\right) T^λ T^μ S^ν = {∇_T}² S^ρ,$$ but the form it is presented, here, (with an entire grid's worth of parameters) better conveys the intent behind it.

This is not just for the chrono-geometries of space-time, but for purely spatial geometries as well. Thus, for instance, the "township" grid laid out on the old Northwest Territory, and beyond, in the United States has defects arising from the non-Euclidean'ness of the geometry of the Earth's surface. Here's one of them at the southern border of the old Granville township, Hampton Avenue Defect. The $20-30$ meter warp is about right, as an adjustment for the four township north-south stack comprising the county. The surveyors who laid out the township grid throughout the United States, to the degree that they failed to account for the Earth being curved, may have lost sight of the need to make appropriate adjustments for the north-south boundary lines and may have just treated the adjustments as "fixes for surveying errors". The north-south boundary lines are, for the most part, supposed to be geodesics, since they follow lines of equal longitude.

Answer 4

The Equation of Geodesic Deviation describes how nearby free-falling objects in a curved spacetime move relative to each other. It shows how gravity, as described by General Relativity, causes the separation between two nearby geodesics (paths followed by free-falling objects) to change over time. This is essential for understanding tidal forces, where different parts of an object experience different gravitational pulls, leading to stretching or compression.