# What exactly is L, Lz and E in orbital mechanics?

## Main Question or Discussion Point

I often stumbled across the variables for angular momentum L and axial angular momentum Lz, which would be no problem if working in cartesian coordinates, then it would be Lz = px y - py x. Unfortunately I have no idea what to make of an Lz in spherical coordinates:

For example, in equations of motion given in r, θ and φ (radius, latitude and longitude) I have pr, pθ and pφ for the momentum components. But what do I make of an Lz then? Is this the vertical (radial) momentum, or is it the horizontal momentum, or the momentum only in the θ or φ direction? Why does an Lz appear when everything else is given in terms of r, θ and φ? Is it just pθ r, or something else?

And E for the conserved energy is m/2 v² in classical mechanics and mc²γ-mc² or just mγ in relativity?

If somebody more experienced could help me I will give a thumb up!

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stevendaryl
Staff Emeritus
I often stumbled across the variables for angular momentum L and axial angular momentum Lz, which would be no problem if working in cartesian coordinates, then it would be Lz = px y - py x. Unfortunately I have no idea what to make of an Lz in spherical coordinates:

For example, in equations of motion given in r, θ and φ (radius, latitude and longitude) I have pr, pθ and pφ for the momentum components. But what do I make of an Lz then? Is this the vertical (radial) momentum, or is it the horizontal momentum, or the momentum only in the θ or φ direction? Why does an Lz appear when everything else is given in terms of r, θ and φ? Is it just pθ r, or something else?

And E for the conserved energy is m/2 v² in classical mechanics and mc²γ-mc² or just mγ in relativity?

If somebody more experienced could help me I will give a thumb up!
I'll just answer the question about angular momentum, and let somebody else tangle the energy question.

Angular momentum is not actually a vector; it's a tensor. What we call $L_z$ doesn't actually have anything to do with the z-direction; it's really associated with rotations in the x-y plane. $L_x$ is associated with rotations in the y-z plane. $L_y$ is associated with rotations in the z-x plane. It's just a coincidence of 3 dimensions of space that we have exactly three orthogonal directions, and we have three orthogonal planes, so we just arbitrarily associate each plane with a direction in space. But that doesn't work in other numbers of dimensions. In 2 dimensional space, there is only one plane, so there is only one component of angular momentum. In 4 dimensional space, there are 6 planes, so there are 6 components of angular momentum.

Getting back to our 3-D world, $L_z$ represents rotations in the x-y plane. It doesn't have anything to do with the z-axis (other than the fact that that axis is left unchanged by rotations in the x-y plane). It doesn't matter what coordinate system you are using, there is still going to be an x-y plane, it's just that it won't be called that in other coordinates.

On the surface of a sphere, motion in the x-y plane corresponds to motion in which $\phi$ is allowed to vary, but $\theta$ is kept fixed. So $L_z$ corresponds to motion where $\phi$ changes. $L_y$ and $L_x$ are more complicated to describe in spherical coordinates, so they aren't actually used.

D H
Staff Emeritus
I often stumbled across the variables for angular momentum L and axial angular momentum Lz, which would be no problem if working in cartesian coordinates, then it would be Lz = px y - py x.
You have that backwards. Angular momentum is $\mathbf L = \mathbf r \times \mathbf p$, so the z component of angular momentum is given by $L_z = r_x p_y - r_y p_x$.

Unfortunately I have no idea what to make of an Lz in spherical coordinates:

For example, in equations of motion given in r, θ and φ (radius, latitude and longitude) I have pr, pθ and pφ for the momentum components. But what do I make of an Lz then? Is this the vertical (radial) momentum, or is it the horizontal momentum, or the momentum only in the θ or φ direction? Why does an Lz appear when everything else is given in terms of r, θ and φ? Is it just pθ r, or something else?
Spherical coordinates turn out to be not particularly useful (rather useless, in fact) in orbital mechanics. The solution is simple: Don't do that then!

Polar coordinates on the other hand are very useful because of the planar nature of Keplerian orbits. It is handy to have the polar axis pointing to periapsis. The angular momentum vector as expressed in arbitrarily oriented Cartesian coordinates that aren't aligned with the orbital plane provides a mechanism to find the first two elements of a z-x-z Euler rotation sequence that rotate the Cartesian x-y-z axes to those of the orbital plane, the right ascension of the ascending node (itex]\Omega[/itex]) and the the argument of periapsis ($\omega$).

And E for the conserved energy is m/2 v² in classical mechanics and mc²γ-mc² or just mγ in relativity?
Neither. Special relativity does not pertain to orbital mechanics. Special relativity pertains to regions of space where gravitation is so negligibly small as to be essentially non-existent. Generalizing special relativity to areas where gravitation is present is the subject of general relativity.

In classical mechanics, it's the total energy (kinetic energy plus potential energy) that is conserved.

Unfortunately I have no idea what to make of an Lz in spherical coordinates
It is the component along the axis of symmetry, which is im most cases aligned in a direct straight line from the northpole to the south pole.

Spherical coordinates turn out to be not particularly useful (rather useless, in fact) in orbital mechanics. The solution is simple: Don't do that then!
The Kerr Metric is given in spherical coordinates!

Special relativity does not pertain to orbital mechanics.
The gamma factor is also present in the Kerr and Schwarzschild equations, it just gets combined with other terms but you can still split them up into gravitational and local velocity dependend effects which in the end get multiplied again (for example see this thread)

In classical mechanics, it's the total energy (kinetic energy plus potential energy) that is conserved.
You also have the conserved energy (E0) in relativity, see this Wikipedia-article.