Lorentz transformations of the angular momentum

[scope]

hey,

does anyone there know how the angular momentum (L=r x p) is transformed under Lorentz transformations?

 

[bcrowell]

There is no cross product in four dimensions, so the generalization of rxp is a rank-2 tensor, not a vector: $J^{jk}=r^jp^k$. In the center of mass frame of a system, with the axis taken to be the center of mass, you can define a three-vector J^l according to $J^{jk}=\epsilon^{jkl}J^l$, where $\epsilon$ is the Levi-Civita symbol. The advantages of J^l are that it's less unwieldy than $J^{jk}$, and it matches up with the Newtonian angular momentum in the appropriate limit. The advantage of $J^{jk}$ is that it transforms in a simple way, as a rank-2 tensor.

 

[scope]

There is no cross product in four dimensions, so the generalization of rxp is a rank-2 tensor, not a vector: $J^{jk}=r^jp^k$. In the center of mass frame of a system, with the axis taken to be the center of mass, you can define a three-vector J^l according to $J^{jk}=\epsilon^{jkl}J^l$, where $\epsilon$ is the Levi-Civita symbol. The advantages of J^l are that it's less unwieldy than $J^{jk}$, and it matches up with the Newtonian angular momentum in the appropriate limit. The advantage of $J^{jk}$ is that it transforms in a simple way, as a rank-2 tensor.

thank you so angular momentum becomes a tensor that is 2 times countervariant? but how are these tensors transformed under Lorentz transformations?

 

[Dickfore]

$$M^{\mu \nu} \equiv X^{\mu} P^{\nu} - X^{\nu} P^{\mu}, \; M^{\nu \mu} = -M^{\mu \nu}$$$$M'^{\mu \nu} = \Lambda^{\mu}{}_{\rho} \, \Lambda^{\nu}{}_{\pi} \, M^{\rho \pi}$$

 

[scope]

$$M^{\mu \nu} \equiv X^{\mu} P^{\nu} - X^{\nu} P^{\mu}, \; M^{\nu \mu} = -M^{\mu \nu}$$


$$M'^{\mu \nu} = \Lambda^{\mu}{}_{\rho} \, \Lambda^{\nu}{}_{\pi} \, M^{\rho \pi}$$

thank you, does that mean that if the length is contracted by k, the momentum is also contracted by k, and the angular momentum by k^2?

 

[bcrowell]

thank you, does that mean that if the length is contracted by k, the momentum is also contracted by k, and the angular momentum by k^2?

In general a Lorentz transformation doesn't reduce to just length contraction and time dilation. I believe the way you've stated it would work if the angular momentum tensor was diagonal, but actually it's the off-diagonal components that are interpreted as the angular momentum.

 

[Dickfore]

The matrix corresponding to $\Lambda^{\mu}{}_{\nu}$ uses $\mu$ as a row index and $\nu$ as a column index. For example, for the common relative motion along the $Ox_{1}$-axis, the matrix is:

$$\hat{\Lambda} \rightarrow \left[\begin{array}{cccc}<br /> \gamma & -\beta \, \gamma & 0 & 0 \\<br /> <br /> -\beta \, \gamma & \gamma & 0 & 0 \\<br /> <br /> 0 & 0 & 1 & 0 \\<br /> <br /> 0 & 0 & 0 & 1<br /> \end{array}\right]$$

where:

$$\gamma = (1 - \beta)^{-1/2}, \; \beta = V/c$$

For the angular momentum tensor, one needs $X^{\mu} = (c t, \mathbf{r})$ and $P^{\mu} = (E/c, \mathbf{p})$. Therefore, we can write:

$$M^{i 0} = X^{i} \, P^{0} - X^{0} \, P^{i} = x_{i} \, \frac{E}{c} - c t \, p_{i}$$

i.e. they are the Cartesian components of the vector:

$$\frac{E}{c} \, \mathbf{r} - c t \, \mathbf{p}$$

$$M^{i j} = X^{i} \, P^{j} - X^{j} \, P^{i} = x_{i} \, p_{j} - x_{j} \, p_{i} = \epsilon_{i j k} \, L_{k}$$

where $\epsilon_{i j k}$ is the Levi-Civita completely antisymmetric symbol ($\epsilon_{1 2 3} = +1$) and $L_{k}$ are the Cartesian components of the angular momentum pseudo-vector $\mathbf{L} = \mathbf{r} \times \mathbf{p}$.

 

[Dickfore]

So, for example, to find how $L_{z}$ transforms under a conventional Lorentz transformation, you should look how $M^{1 2} = L_{z}$ transforms according to the general rule:

$$L'_{z} \equiv M'^{1 2} = \Lambda^{1}{}_{0} \, \Lambda^{2}{}_{2} \, M^{0 2} + \Lambda^{1}{}_{1} \, \Lambda^{2}{}_{2} \, M^{1 2} = \gamma \left(L_{z} - \beta \, c (t p_{y} - \frac{E}{c^{2}} \, y) \right)$$

 

[scope]

So, for example, to find how $L_{z}$ transforms under a conventional Lorentz transformation, you should look how $M^{1 2} = L_{z}$ transforms according to the general rule:

$$L'_{z} \equiv M'^{1 2} = \Lambda^{1}{}_{0} \, \Lambda^{2}{}_{2} \, M^{0 2} + \Lambda^{1}{}_{1} \, \Lambda^{2}{}_{2} \, M^{1 2} = \gamma \left(L_{z} - \beta \, c (t p_{y} - \frac{E}{c^{2}} \, y) \right)$$

thank you, but then how is the "contraction factor" of the angular momentum for Lz or any other coordinate, (simply) calculated?

 

[Dickfore]

I don't have a clue what a 'contraction factor for angular momentum' means. This is how the components of angular momentum are transformed when we go from one frame of reference to another.

 

[bcrowell]

thank you, but then how is the "contraction factor" of the angular momentum for Lz or any other coordinate, (simply) calculated?

It doesn't reduce to a simple contraction factor. That's the point of #6.

 

[arkajad]

Each Lorenz transformation can be composed of space rotations and a simple boost. It is clear how angular momentum transforms under simple space rotations. So, it remains to look for a simple boost affecting only, say, x and t coordinates. Then the contraction factor is the standard one.

 

[Dickfore]

Each Lorenz transformation can be composed of space rotations and a simple boost. It is clear how angular momentum transforms under simple space rotations. So, it remains to look for a simple boost affecting only, say, x and t coordinates. Then the contraction factor is the standard one.

The only problem is that the angular momentum is a 2-fold (antisymmetric) tensor, so it carries two Lorentz indices and you might perform a rotation with respect to one of them and a boost with respect to the other.

 

[arkajad]

That is not a problem, because even when it has two indices, the same Lorentz transformation acts on both of them, not two different.

 

[bcrowell]

Each Lorenz transformation can be composed of space rotations and a simple boost. It is clear how angular momentum transforms under simple space rotations. So, it remains to look for a simple boost affecting only, say, x and t coordinates. Then the contraction factor is the standard one.

No, it's a rank-2 tensor. It doesn't transform like a rank-1 tensor. Write out the tensor transformation law for a simple case, and you'll see that the result looks nothing like what you'd have for a 4-vector.

 

[arkajad]

A tensor transforms under the tensor product of the representation. There is nothing mysterious about it. 2-tensor transforms with $$R\otimes R$$ where $$R$$ is the vector (or -co-vector) representation. Then there can be covariant and contravariant tensors. For a covariant tensor you need to take the contragradient representation - that's all.

Or, in simple terms, each index transforms with the same transformation matrix (perhaps a contragradient one).