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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.
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.