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Mortimer

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- Thread starter Mortimer
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Mortimer

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Stingray

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But angular momentum is also defined with respect to a particular base point (more precisely, it is defined along a chosen worldline, and the hypersurfaces orthogonal to that line also enter the definition). This is true even in Newtonian physics (e.g. a ball moving in a straight line has an angular momentum unless it is moving radially away from the chosen base point). Transforming from one point to another is considerably more complicated. The ideas in general are rather intricate to describe in any detail here.

I'll just say that if the base point is chosen to be the center-of-mass, then it happens that [tex]u_{a} S^{ab} =0[/tex] (this condition is actually used to

[tex]

S^{a} = \epsilon^{abcd} u_{b} S_{cd}

[/tex]

Now, [tex]S^{a}u_{a}=0[/tex], so this is basically a 3-vector (in the center-of-mass frame). It is called the 'spin' angular momentum, and is closely analogous to what you're used to in Newtonian physics. If z is not at the center-of-mass, there is also an 'orbital' component which which basically just takes into account things like the offset ball example above.

By far the best treatment I know of these things is unfortunately in a hard-to-find book by WG Dixon written about 25 years ago (titled "Special Relativity: The Foundation of Macroscopic Physics"). I don't know what level you want to learn things at, though. That book is nowhere near as introductory as its title suggests!

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Mortimer

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If [itex]u_aS^{ab}=0[/itex] for all [itex]u_a[/itex] with the center of mass as base point than it seems to me that this tensor cannot include the quantum spin, am I right? After all that spin should be invariant.

- #4

Stingray

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There is no lack of invariant quantities either. [tex]S^{a} S_{a}[/tex] is a scalar invariant, for example. More generally, [tex]S^{ab} S_{ab}[/tex] is always an invariant no matter whether we're using center-of-mass definitions or not. These quantities are related to the square of the angular momentum operator in quantum mechanics. In any case, here are some details of the classical theory:

As mentioned, the angular momentum depends on both a choice of base point and a preferred time direction. This would usually (though not always) be dealt with by defining a 'reference' worldline [tex]z^{a}(s)[/tex]. Of course, this has a 4-velocity [tex]u^{a}(s) = \dot{z}^{a}(s)[/tex] (assume it is normalized, so s is a proper time for someone moving along z). Now, the direction of the 4-velocity is naturally interpreted as the time direction of an observer on z, so define a set of spacelike hyperplanes [tex]\Sigma(s)[/tex] that intersect [tex]z^{a}(s)[/tex], and are orthogonal to [tex]u^{a}(s)[/tex]. These can be taken as naturally-defined 'moments of time.' Note, however, that these surfaces will intersect each other if they are extended too far (unless [tex]\dot{u}^{a} =0[/tex]).

Now, both the linear and angular momenta are naturally defined as objects on z. In terms of 'normal' variables, though, they are integrals over the hyperplanes. The most common definitions (again, these are not universal) look like

[tex]

p^{a}(s) = \int_{\Sigma(s)} T^{ab}(x) d\Sigma_{b} ,

[/tex]

[tex]

S^{ab}(s) = \int_{\Sigma(s)} \left(x-z(s) \right)^{[a} T^{b]c}(x) d\Sigma_{c} .

[/tex]

In the quantum theory, particles have intrinsic spins that can add extra terms here.

Anyway, you can see that the angular momentum depends on the base point z, but also on the surface of integration (defined here in terms of z and u). It is therefore instructive to rewrite these quantities more abstractly as [tex]p^{a}\left(\Sigma(u,z) \right) [/tex] and [tex]S^{ab}\left(z, \Sigma(u,z) \right) [/tex].

Let's now define a center-of-mass. This fixes the reference line, so let's forget about the one we've assumed to exist for now. Instead, choose a single point [tex]z(s_{0})[/tex]. Then it can be shown that under reasonable conditions, there exists a unique vector [tex]u^{a}[/tex] such that

[tex]

p^{a} \left( \Sigma(z,u) \right) \propto u^{a}(z)

[/tex]

From looking at the integral form of p, you can see that this is actually a highly implicit definition. In any case, given any z, it fixes a u.

We now need to find a way to pick out a unique z. This is done with

[tex]

S^{ab} \left( z, \Sigma(z,u) \right) p_{b} \left( \Sigma(z,u) \right) = S^{ab} u_{b} =0

[/tex]

Again, this is extremely implicit, but it works. Using the integral expression for the angular momentum, you can show that this is equivalent to the usual expression defining a center-of-mass expression:

[tex]

\int_{\Sigma} (x-z)^{a} \left( u_{b} u_{c} T^{bc} \right) d\Sigma =0

[/tex]

where the quantity in parentheses is usually interpreted as the mass density seen by on observer with velocity u.

Anyway, taking these two conditions together defines a unique reference line usually called the center of mass. Rewriting the integral for [tex]S^{ab}[/tex] so that it gives [tex]S^{a}[/tex] shows that this quantity is basically an integral over the cross products [tex] \bf{r} \times \bf{\mathcal{P}}[/tex], where [tex]\bf{\mathcal{P}}[/tex] is a momentum density. This is just as it is in Newton's theory.

We also have (at least) two natural scalar invariants in this formalism:

[tex] m^{2} := -p^{a} p_{a} [/tex]

[tex] |S|^{2} := S^{a} S_{a} [/tex]

Neither of these quantities are necessarily constants (ie independent of s) unless there are no forces acting on the system.

Continuing in the next post...

- #5

Stingray

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[tex]

S^{a} = \int \epsilon^{abcd} u_{b} (x-z)_{c} T_{df} u^{f} d\Sigma

\\

= \int \epsilon^{abcd} u_{b} (x-z)_{c} \mathcal{P}_{d} d\Sigma

[/tex]

where again,

[tex]

\mathcal{P}^{a} = T^{bc} \left( \delta_{b}^{a} + u_{b} u^{a} \right) u_{c}

[/tex]

The quantity in parentheses is a projection operator if you're not familiar with it.

Anyway,

[tex]

S^{ab} = \epsilon^{abcd} u_{c} S_{d} + S^{ab}_{\rm{orb}},

[/tex]

where the second term vanishes in the center of mass frame. I might have a sign or factor of two wrong on the first term, but let's assume that its correct (I always forget how to contract [tex]\epsilon^{abcd}[/tex]'s).

Combining these expressions gives

[tex]

S^{ab}_{\rm{orb}} = - \int_{\Sigma} (x-z)^{[a} u^{b]} \left( u_{c} u_{d} T^{cd} \right) d\Sigma

[/tex]

In this context, we now have more scalar quantities, but they are not as fundamental as the previous ones.

To back up a bit, all of this stuff about defining reference lines and so on is actually quite standard whenever one wants to decompose something in terms of multipole moments. The linear and angular momenta are basically the monopole and dipole moments of the stress-energy tensor.

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Mortimer

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Rob

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I'm not sure much the following will help - it's an old link I found. It would probably help to know about p-forms and the wedge product, and/or Clifford algebra

http://panda.unm.edu/Courses/finley/P495/TermPapers/relangmom.pdf [Broken]

The thing you need to know is that the cross product in three dimensions is actually the dual of a geometric entity known as a "bi-vector". You can think of a bi-vector as an oriented surface

http://www.mrao.cam.ac.uk/~clifford/introduction/intro/node4.html

In three dimensions, there is only one vector perpendicular to a specified bi-vector (a specified oriented surface), the operation of going from a bi-vector to a vector is called "taking the dual".

The root link to the above is Imaginary numbers are not real which goes into Clifford algebras in great detail.

http://panda.unm.edu/Courses/finley/P495/TermPapers/relangmom.pdf [Broken]

The thing you need to know is that the cross product in three dimensions is actually the dual of a geometric entity known as a "bi-vector". You can think of a bi-vector as an oriented surface

http://www.mrao.cam.ac.uk/~clifford/introduction/intro/node4.html

In three dimensions, there is only one vector perpendicular to a specified bi-vector (a specified oriented surface), the operation of going from a bi-vector to a vector is called "taking the dual".

The root link to the above is Imaginary numbers are not real which goes into Clifford algebras in great detail.

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Mortimer

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selfAdjoint

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Mortimer said:

No I don't think so. That's a canard anyway; SR can perfectly well treat accelerating frames, even spinning ones. But the product of two boosts is a bost plus a rotation, which means that anything with a rotation in it is going to be more complicated math. See Thomas precession in almost any good text.

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Stingray

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It's a little difficult to say where to find things because angular momentum can be discussed from many different points of view. Depending on what you want to do with it, it might even be convenient to adopt completely different (and incompatible) definitions of the angular momentum tensor.

Anyway, I think that these things are not often discussed at length in SR books because almost all of them are very elementary. SR is usually only taught properly in GR courses (as a special case).

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