Re-writing the geodesic deviation eqn in matrix notation (3d only)

In summary, the geodesic equation in 3 dimensions and +++ signature can be expressed in matrix notation by assuming an orthonormal basis and using the Riemann tensor to map two bivectors to a scalar. The resulting relative acceleration between objects following geodesics can be written as the triple product of the velocity vector, the symmetric matrix C, and the displacement vector. This can be further simplified by considering the case when the vector C times D points in the same direction as D, making the calculation of the acceleration vector simpler. Ultimately, this leads to the desired result for the geodesic equation expressed in matrix notation, with the magnitude of the acceleration being proportional to the magnitude of the displacement and the square of the velocity,
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This is my attempt to re-write the geodesic deviation equation in the special case of 3 dimensions and +++ signature in matrix notation.

We start with assuming an orthonormal basis. Matrix notation allows one to express vectors as column vectors, and dual vectors as row vectors, but by assuming an orthonormal basis, we insure that the row vectors are simply the transpose of the column vectors. And we can essentially conflate the vectors and their duals as a consequence, aided by the fact that with the +++ signature there are no sign issues to get in the way.

Next we observe that the Riemann tensor maps two bivectors to a scalar. Bivectors use the tensor language that I'm trying to eliminate, but because we are working in three dimensions, we can replace the bivectors with a cross product of two vectors.

Essentially then, the Riemann tensor is taking two directed plane surfaces represented by the cross-product and mapping it to a scalar, which represents a component of the curvature. At this point we also want to note that the symmetries of the Riemann make C a symmetric matrix, swapping the order of the two vector arguments produces the same result.

Finally, our three dimensional space obviously has no time. We will still talk about "velocities" and "accelerations", though. We do this by essentially adopting a Newtonian time parameter to replace the less familiar idea of affine parameterization of geodesics. There is no time in the 3d manifold, but it aids the intuition (at least mine) to imagine objects actually "moving" along the geodesics with some velocity, and some relative acceleration. But we are using an absolute Newtonian time here, as an aid to intuition, not time as it is understood in special relativity.

With these preliminaries out of the way, we write down the geodesic equation in tensor notation and then manipulatge it into matrix notation. We will need one further step after this.

What we have are two parallel velocity vectors ##\vec{u}## separated by some infintesimal displacement vector ##\vec{d}##, where the displacement vector is by defintion orthogonal to the velocity vectores. Then the geodesic equation gives the resulting relative acceleration vector ##D^2 \vec{d} / dt^2## which we write as ##\vec{a}##. The geodesic equation gives us the relative acceleration between objects following geodesics, such as the relative acceleration between two objects moving along great circles on a sphere in the 2D case that I've used as a motivational example in other posts.

In tensor notation, we'd write ##a_a =- R_{abcd} u^b d^c u^d## or, equivalently ##R = -R_{abcd} a^a u^b d^c u^d##, where R is a scalar constant. But we can replace the rank 4 tensor operation on two bivectors as previously discussed with a rank 2 matrix operating on two vectors in order to replace the rank 4 tensor R with the rank 2 matrix C. Then the second form of the geodesic equation becomes ##R = \left( \vec{a} \times \vec{u} \right)^T C \left(\vec{d} \times \vec{u} \right)##.

But what we'd really like is an expression for ##\vec{a}##. We can think of the vector ##\vec{a}## as having some magnitude and direction ##\vec{aa}##, where ##\vec{aa}## is a unit vector. We note that ##C \left( \vec{d} \times \vec{u} \right)## is just some vector which we will call ##\vec{D}##. Then the problem of finding the direction ##\vec{aa}## of the accleration is the problem of maximizing the triple product ##\vec{aa} \times \vec{u} \cdot \vec{D}##. Modulo sign issues, the triple product will be maximized when we have the maximum 3D volume of the triple product, which occurs when ##\vec{aa}## is perpendicular to both ##\vec{u}## and ##\vec{D}##. Putting this all together allows us to write our desired result:

$$\vec{a} = \vec{u} \times C \vec{D} = \vec{u} \times C \left( \vec{d} \times \vec{u} \right)$$

once we confirm that ##\vec{a}## has the proper magnitude and sign are correct.

Considering the case when ##C \vec{D}## points in the same direction as ##\vec{D}## makes this simpler. This happens when ##\vec{D}## is an eigenvector of the symmetric matrix C. Letting K be the associated eigenvalue, we then have ##C \vec{D} = K \vec{D}##. Then we observe that ##\vec{aa}## is parallel to ##\vec{d}## and perpendicular to ##\vec{D}## and ##\vec{u}## as required, and that the magnitude and direction of ##\vec{a}## is also correct. The magnitude of ##\vec{a}## is proportional to the mangitude of the displacement and the square of the magnitude of the velocity, and for a positive curvature, the relative acceleration is negative, i.e. the geodesics approach each other.
 
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  • #2
This then is our desired result for the geodesic equation in 3 dimensions and +++ signature expressed in matrix notation.
 

FAQ: Re-writing the geodesic deviation eqn in matrix notation (3d only)

What is the purpose of re-writing the geodesic deviation equation in matrix notation?

Re-writing the geodesic deviation equation in matrix notation allows for a more concise and efficient representation of the equation, making it easier to manipulate and solve. It also allows for a better understanding of the underlying mathematical concepts.

How does matrix notation differ from the standard notation of the geodesic deviation equation?

In standard notation, the geodesic deviation equation is written as a set of differential equations. In matrix notation, it is represented as a single matrix equation, which is a more compact and elegant form.

Can the geodesic deviation equation be re-written in matrix notation for any dimension?

Yes, the geodesic deviation equation can be re-written in matrix notation for any dimension. However, the specific form of the matrix may vary depending on the dimensionality of the space.

What are the advantages of using matrix notation for the geodesic deviation equation?

Using matrix notation for the geodesic deviation equation allows for easier manipulation and analysis, as well as a more intuitive understanding of the relationship between geodesics and deviation vectors. It also allows for a more efficient representation of the equation, which can be useful in computational applications.

Are there any limitations or drawbacks to using matrix notation for the geodesic deviation equation?

One potential limitation of using matrix notation for the geodesic deviation equation is that it may be more difficult to visualize and interpret compared to the standard notation. Additionally, it may require a deeper understanding of linear algebra concepts to fully grasp the implications of the matrix representation.

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