User:Maschen/Relativistic angular momentum


 * "Relativistic angular momentum" redirects to here.

In physics, relativistic angular momentum refers to the mathematical formalisms and physical concepts that define angular momentum in special relativity (SR) and general relativity (GR). The relativistic quantity is subtly different from the three dimensional quantity in classical mechanics. Angular momentum is a dynamical quantity derived position and momentum, and is important; in the same way momentum conservation corresponds to translational symmetry, angular momentum conservation corresponds to rotational symmetry - the connection between symmetries and conservation laws is made by Noether's theorem. More abstractly, invariance of angular momentum, four momentum, and other symmetries, in spacetime, are described by the Poincaré group and Lorentz group.

Physical quantities which remain separate in classical physics are naturally combined in SR and GR by enforcing the postulates of relativity, an appealing characteristic. Most notably; space and time coordinates combine into the four position, and energy and momentum combine into the four momentum. These four vectors depend on the frame of reference used, and change under Lorentz transformations to other inertial frames or accelerated frames.

Relativistic angular momentum is less obvious. The classical definition of angular momentum is the cross product of position x with momentum p to obtain a pseudovector x×p, or alternatively as the exterior product to obtain a 3d bivector x&and;p. What does this combine with, if anything? There is another vector quantity not often discussed - it is the time-varying moment of mass (not the moment of inertia) related to the boost of the centre of mass of the system, and this combines with the classical angular momentum to form an antisymmetric tensor of second order.

In general relativity, angular momentum becomes complicated, since even for an isolated system, the total angular momentum is not conserved in curved spacetime.

In relativistic quantum mechanics it is important to make it clear that the angular momentum tensor is for orbital motion, since in quantum theory elementary particles have spin and this is an additional contribution to the total angular momentum tensor operator of relativistic quantum mechanics.

Cross product definition
In classical mechanics, the orbital angular momentum of a particle with instantaneous three dimensional position vector x and momentum vector p, is defined as the axial vector


 * $$\mathbf{L} = \mathbf{r}\times \mathbf{p}$$

which has three components:


 * $$L_3 = x_1 p_2 - x_2 p_1$$
 * $$L_1 = x_2 p_3 - x_3 p_2$$
 * $$L_2 = x_3 p_1 - x_1 p_3\,.$$

This quantity is additive, and for an isolated system, the total angular momentum of a system is conserved. However, this definition can be used in three dimensions only - considering that the cross product in the definition defines an axial vector perpendicular to the plane spanned by x and p. In four dimensions, there is not one axis uniquely perpendicular to a two-dimensional plane, but two such axes, allowed by the additional dimension.

Exterior product definition
An alternative definition, which avoids any axes about which objects rotate, is to conceive orbital angular momentum as a plane element. This can be achieved in the language of exterior algebra or geometric algebra as a contravariant second order antisymmetric tensor:


 * $$\mathbf{L}=2\mathbf{x}\wedge\mathbf{p}$$

with components


 * $$L^{ij} = x^i p^j - x^j p^i = 2 x^{[i} p^{j]}$$

where the indices i and j take the values 1, 2, 3. The components can be systematically collected into a 3 × 3 antisymmetric matrix:


 * $$\mathbf{L} = \begin{pmatrix}

L^{11} & L^{12} & L^{13} \\ L^{21} & L^{22} & L^{23} \\ L^{31} & L^{32} & L^{33} \\ \end{pmatrix} = \begin{pmatrix} 0 & L_{xy} & L_{xz} \\ L_{yx} & 0 & L_{yz} \\ L_{zx} & L_{zy} & 0 \end{pmatrix} = \begin{pmatrix} 0 & L_{xy} & -L_{zx} \\ -L_{xy} & 0 & L_{yz} \\ L_{zx} & -L_{yz} & 0 \end{pmatrix}$$

This is generalized below.

Dynamic mass moment
Additionally in classical mechanics, the three dimensional quantity for a particle of mass m moving with velocity u:


 * $$\mathbf{N} = m \left( \mathbf{x} - t \mathbf{u} \right) = m \mathbf{x} - t \mathbf{p} $$

has the dimensions of mass moment - length multiplied by mass. It is related to the boost (relative velocity) of the centre of mass (COM) of the particle or system of particles, as measured in the lab frame. There is no universal symbol, nor even a universal name, for this quantity - different authors denote it by various other symbols (including e.g. μ), may designate other names, and may define N to be the negative of what is used here - the above form has the advantage that it resembles the familiar Galilean transformation for position, which in turn is the classical boost transformation between classical inertial frames. This vector is also additive: for a system of particles, the vector sum is the resultant:


 * $$\sum_n \mathbf{N}_n = \sum_n m_n \left(\mathbf{x}_n - t \mathbf{u}_n \right) = \left(\mathbf{x}_\mathrm{com}\sum_n m_n - t \sum_n m_n \mathbf{u}_n \right) $$

where the system's centre of mass is:


 * $$\mathbf{x}_\mathrm{com} = \frac{\sum_n m_n\mathbf{x}_n}{\sum_n m_n}$$

For an isolated system, N is conserved in time, apparent by differentiating with respect to time. Unlike L, N is a (polar) vector, not a pseudovector, and is therefore invariant under rotations.

The resultant Ntotal for a multiparticle system has the physical visualization that, whatever the complicated motion of all the particles are, they move in such a way that the system's COM moves in a straight line. This doesn't necessarily mean all particles "follow" the COM, nor that all particles all move in almost the same direction simultaneously, just that the motion of each particle is coupled with respect to the COM.

In special relativity, if the particle moves with velocity u relative to the lab frame, then


 * $$E = \gamma(\mathbf{u})m_0c^2, \quad \mathbf{p}=\gamma(\mathbf{u})m_0\mathbf{u} $$

where γ is the Lorentz factor and m0 the rest mass of the particle. Some authors use relativistic mass:


 * $$m = \gamma(\mathbf{u})m_0$$

The corresponding relativistic mass moment in terms of m0, m, u, p, E, in the same lab frame is:


 * $$\gamma(\mathbf{u})\mathbf{N} = m\mathbf{x} - \mathbf{p}t = \frac{E}{c^2}\mathbf{x} - \mathbf{p}t = \gamma(\mathbf{u})m_0(\mathbf{x} - \mathbf{u}t) $$

defined here so that the relativistic equation in terms of the rest mass, and classical definition, have the same form.

The relativistic mass simplifies the expressions in this context as it removes extra Lorentz factors. However rest mass is discouraged by some authors since it can be a misleading quantity to apply in certain equations. In the following, N is given in terms of the rest and relativistic masses.

<!---LEAVE FOR NOW: ===Intertwine of L and N: Lorentz transformations===

Consider the Lorentz boost in standard setup with velocity V = (V, 0, 0) in the direction of the coincident xx&prime; axes. The mass-energy E = mc2 and momentum components p = (px, py, pz) of an object, as well as position coordinates x = (x, y, z) and time t in frame F are transformed to E&prime; = m&prime;c2, p&prime; = (px&prime;, py&prime;, pz&prime;), x&prime; = (x&prime;, y&prime;, z&prime;), and t&prime; in F&prime; according to:


 * $$t' = \gamma(V) \left(t -\frac{Vx}{c^2}\right)\,,\quad \frac{E'}{c} = \gamma(V) \left(\frac{E}{c} - \frac{Vp_x}{c}\right) $$
 * $$x' = \gamma(V) (x - vt)\,,\quad p_x' = \gamma(V) \left(p_x - \frac{VE}{c^2}\right)$$
 * $$y' = y' \,,\quad p_y' = p_y $$
 * $$z' = z' \,,\quad p_z' = p_z $$

additionally for massive objects:


 * $$E' = m'c^2 \,\quad E = mc^2 $$

where m and m&prime; are the relativistic masses of the object as measured in frames F and F&prime; corresponding to its energies E and E&prime; in the respective frames. The velocity V here is the relative velocity between the frames, not necessarily of the object, neither F nor F&prime; is the rest frame of the object. If the object moves with velocity u relative to F (not F&prime;) then


 * $$E = \gamma(\mathbf{u})m_0c^2, \quad \mathbf{p}=\gamma(\mathbf{u})m_0\mathbf{u} $$

where m0 is the rest mass of the object.

For the orbital 3-angular momentum L as a pseudovector, we have:


 * $$L_x' = y' p_z' - z' p_y' = y p_z - z p_y = L_x $$
 * $$L_y' = z' p_x' - x' p_z' = \gamma(V)[ (zp_x - x p_z) + V(p_z t - z E/c^2) ] = \gamma(V) [ L_y - V (m z - p_z t) ]$$
 * $$L_z' = x' p_y' - y' p_x' = \gamma(V) [ (xp_y - y p_x) + V(y E/c^2 - p_y t) ] = \gamma(V) [L_z + V (m y - p_y t) ] $$

In the second terms of Ly&prime; and Lz&prime;, there are cyclic permutations in the components of V and N', that are entirely in the y and z directions perpendicular to v in the x direction, and hence no components actually in the x direction. The cross product of the vectors V and N can be inferred:


 * $$ V (m y - p_y t) = V_x N_y - V_y N_x = \left(\mathbf{V}\times\mathbf{N}\right)_z $$
 * $$ - V (p_z t - m z) = V_z N_x - V_x N_z = \left(\mathbf{V}\times\mathbf{N}\right)_y $$

Since Lx is parallel to the relative velocity V, and the other components Ly and Lz are perpendicular to V, we can collect the components into the pseudovector equations:


 * $$\mathbf{L}_\perp ' = \mathbf{L}_\perp $$
 * $$\mathbf{L}_\parallel ' = \gamma(\mathbf{V})\left(\mathbf{L}_\parallel + \mathbf{V} \times \mathbf{N} \right) $$

using the decomposition of 3-angular momentum in each frame into components parallel and perpendicular to V, respectively subscripted by &#8741; and &#8869;:


 * $$\mathbf{L} = \mathbf{L}_\parallel + \mathbf{L}_\perp \,,\quad \mathbf{L}' = \mathbf{L}_\parallel' + \mathbf{L}_\perp'\,.$$

These transformations are true for all V, not just for motion along the xx&prime; axes.

Considering L as a tensor, we get a similar result:


 * $$\mathbf{L}_\parallel' = \gamma(\mathbf{V})\left(\mathbf{L}_\parallel + 2 \mathbf{V} \wedge \mathbf{N} \right) $$

where the exterior product term has the factor of two for the antisymmetrization of position and momentum components:


 * $$ V (m y - p_y t) = V_x N_y - V_y N_x = 2\left(\mathbf{V}\wedge\mathbf{N}\right)_{xy} $$
 * $$ V (p_z t - m z) = V_z N_x - V_x N_z = 2\left(\mathbf{V}\wedge\mathbf{N}\right)_{zx} $$

For the dynamic mass moment:


 * $$N_x' = m' x' - p_x' t' = \gamma(V)\left(m-\frac{V p_x}{c^2}\right)\gamma(V)(x - vt) - \gamma(V)\left(p_x - \frac{VE}{c^2}\right) \gamma(V)\left(t - \frac{Vx}{c^2}\right) = N_x $$
 * $$N_y' = m' y' - p_y' t' = \gamma(V)\left(m-\frac{V p_x}{c^2}\right)y - p_y \gamma(V)\left(t-\frac{Vx}{c^2}\right) = \gamma(V)\left(N_y - \frac{V L_z}{c^2}\right) $$
 * $$N_z' = m' z' - p_z' t' = \gamma(V)\left(m-\frac{V p_x}{c^2}\right)z - p_z \gamma(V)\left(t-\frac{Vx}{c^2}\right) = \gamma(V)\left(N_z + \frac{V L_y}{c^2}\right) $$

where:


 * $$\gamma(u) N_x = mx - p_xt = \gamma(u)m_0(x - u_x t) $$
 * $$\gamma(u) N_y = my - p_yt = \gamma(u)m_0(y - u_y t) $$
 * $$\gamma(u) N_z = mz - p_zt = \gamma(u)m_0(z - u_z t) $$

and collecting parallel and perpendicular components as before:


 * $$\mathbf{N}_\parallel' = \mathbf{N}_\parallel $$
 * $$\mathbf{N}_\perp' = \gamma(\mathbf{V})\left(\mathbf{N}_\perp + \frac{1}{c^2}\mathbf{V}\times\mathbf{L}\right)$$

which has the implication that orbital motion has a contribution to the motion of the centre of mass under a Lorentz boost.--->

4d Angular momentum as a bivector
In relativistic mechanics, the COM boost and orbital 3-angular momentum of a rotating object are combined into a four dimensional bivector in terms of the 4-position X and the 4-momentum P of the object:


 * $$\mathbf{M} = 2\mathbf{X}\wedge\mathbf{P}$$

In components:


 * $$M^{\alpha\beta} = X^\alpha P^\beta - X^\beta P^\alpha = 2 X^{[\alpha} P^{\beta]} $$

which are six independent quantities altogether. Three components:


 * $$M^{ij} = x^i p^j - x^j p^i = L^{ij} $$

are the components of the familiar classical 3-orbital angular momentum, and the other three:


 * $$M^{0i} = x^0 p^i - x^i p^0 = c\,\left(t p^i - x^i \frac{E}{c^2} \right) = c\,\left(t p^i - m x^i \right) = \gamma(u)m_0 c\,\left(t u^i - x^i \right) = - \gamma(u)c N^i $$

correspond to the relativistic mass moment given above, multiplied by c. The quantities Lij and N0i are all frame-dependent as they are derived from the frame-dependent quantities t, x, E, and p. Consequently, the mass moment is also frame-dependent.

The components of the tensor can be systematically displayed as a matrix:


 * $$\mathbf{M} = \begin{pmatrix}

M^{00} & M^{01} & M^{02} & M^{03} \\ M^{10} & M^{11} & M^{12} & M^{13} \\ M^{20} & M^{21} & M^{22} & M^{23} \\ M^{30} & M^{31} & M^{32} & M^{33} \end{pmatrix} = \left(\begin{array}{c|ccc} 0 & -\gamma(u) N^1 c & -\gamma(u) N^2 c & -\gamma(u) N^3 c \\ \hline \gamma(u) N^1 c & 0 & L^{12} & -L^{31} \\ \gamma(u) N^2 c & -L^{12} & 0 & L^{23} \\ \gamma(u) N^3 c & L^{31} & -L^{23} & 0 \end{array}\right) = \left(\begin{array}{c|c} 0 & - \gamma(\mathbf{u}) \mathbf{N} c \\ \hline \gamma(\mathbf{u}) \mathbf{N}^\mathrm{T} c & 2\mathbf{x}\wedge\mathbf{p} \\ \end{array}\right)$$

in which the last array is a block matrix formed by treating N as a row vector which matrix transposes to the column vector NT, and 2x&and;p as a 3 × 3 antisymmetric matrix.

The components of the angular momentum pseudovector enter the angular momentum tensor in the same way as if it were a 3d bivector.

Again, this tensor is additive: the total angular momentum of a system is the sum of the angular momentum tensors for each constituent of the system:


 * $$\mathbf{M}_\mathrm{total} = \sum_n \mathbf{M}_n = 2 \sum_n \mathbf{X}_n \wedge \mathbf{P}_n \,.$$

Each of the six components forms a conserved quantity when aggregated with the corresponding components for other objects and fields.

Lorentz transformation
The angular momentum tensor M is indeed a tensor which changes according to a Lorentz transformation matrix Λ, as illustrated in the usual way by tensor index notation:


 * $$\begin{align}

{\bar{M}}^{\alpha\beta} & = {\bar{X}}^\alpha {\bar{P}}^\beta - {\bar{X}}^\beta {\bar{P}}^\alpha \\ & = \Lambda^\alpha {}_\gamma X^\gamma \Lambda^\beta {}_\delta P^\delta - \Lambda^\beta {}_\delta X^\delta \Lambda^\alpha {}_\gamma P^\gamma \\ & = \Lambda^\alpha {}_\gamma \Lambda^\beta {}_\delta \left( X^\gamma P^\delta - X^\delta P^\gamma \right) \\ & = \Lambda^\alpha {}_\gamma \Lambda^\beta {}_\delta M^{\gamma \delta}  \\ \end{align}$$

In fact, one can Lorentz-transform the four position and four momentum separately, and then antisymmetrize those newly-found components to obtain the angular momentum tensor in the new frame.

Angular momentum in general relativity
Angular momentum is more complicated in GR, and the stress-energy tensor and killing vectors from general relativity become involved with calculating the angular momentum tensor. See the references.