# What is the angular-momentum 4-vector?

1. May 10, 2011

### Zarathustra0

Uh, the title pretty much says it: I'm wondering what the 4-vector analog to the classical 3-angular momentum is. Also, is the definition

L = r $$\times$$ p

still valid for the 3-angular momentum in special relativity?

2. May 10, 2011

### Bill_K

In four dimensions, angular momentum is no longer represented by a vector. The position is a vector, xμ. The momentum is a vector, pν. But when you take the cross product you get a rank two tensor, Jμν = xμ pν - pμ xν.

Why does this happen in four dimensions? Well actually the cross product is supposed to do that. It's only in three dimensions that we can replace it with a vector.

Jμν has six components. If you look at three of them in which μ and ν are both space indices, Jij = xi pj - pi xj, then these values correspond to the three components of what we call the angular momentum vector: Lx = Jyz, Ly = Jzx, and Lz = Jxy.

3. May 10, 2011

### Zarathustra0

Thanks for the help--I assume then that this tensor transforms by applying the Lorentz-transformation matrix twice?

4. May 10, 2011

### tiny-tim

Welcome to PF!

Hi Zarathustra0! Welcome to PF!

to transform it, it's easiest to write it as a 2-form rΛp, with the six (antisymmetric) basis elements tΛx tΛy tΛz yΛz zΛx and xΛy

so yes you apply the Lorentz transformation matrix twice to each basis element (of course, it only makes a difference to t and x)

5. May 10, 2011

### Zarathustra0

To make sure I understand (and I may not because the notions of forms and wedge products are still a bit new to me), does this mean I should express the tensor as a linear combination of the tensors

[0 1 0 0]
[-1 0 0 0]
[0 0 0 0]
[0 0 0 0],

[0 0 1 0]
[0 0 0 0]
[-1 0 0 0]
[0 0 0 0],

[0 0 0 1]
[0 0 0 0]
[0 0 0 0]
[-1 0 0 0],

[0 0 0 0]
[0 0 1 0]
[0 -1 0 0]
[0 0 0 0],

[0 0 0 0]
[0 0 0 1]
[0 0 0 0]
[0 -1 0 0], and

[0 0 0 0]
[0 0 0 0]
[0 0 0 1]
[0 0 -1 0],

perform the Lorentz transformation on each by two applications of the transformation matrix, and then add up the resulting six matrices? In other words, do the above tensors form the basis in question? Earlier I naively tried to just apply the transformation matrix twice to the J-tensor itself, but the result wasn't antisymmetric, so I don't think it worked.

Then again, given the nature of matrix multiplication, it seems like writing the matrix as a linear combination of other matrices and multiplying each of them by the square of the transformation matrix would have the same effect as just multiplying the square of the transformation matrix directly with the original angular-momentum tensor itself, and I already tried that, so maybe I'm still not understanding.

Last edited: May 10, 2011
6. May 11, 2011

### tiny-tim

Hi Zarathustra0!
I don't recommend trying to mix up tensors and 2-forms.

Use one or the other.

2-forms, in my opinion, are easier.

There isn't much call for writing out the Minkowski angular momentum but the similar electromagnetic field is regularly used …

we can write it either as the antisymmetric tensor (matrix) 0 By -Bz Ex etc (which takes up four lines, and doesn't say anything intuitive), or as a neat 2-form …

ExtΛx + EytΛy + EztΛz + BxyΛz + ByzΛx + BzxΛy​

but once you get used to the idea, you can just write that 2-form as
(Ex,Ey,Ez,Bx,By,Bz),​
and you'll know what that means just as you know what (E,px,py,pz) means without having to write the basis vectors!

(try writing out the Lorentz force expression F* = q(t,x,y,z)Λ(Ex,Ey,Ez,Bx,By,Bz), to get a feel for how it all works ) ​

7. May 11, 2011

### Zarathustra0

OK, in that case to apply the transformation twice to each basis element, should I simply make the substitutions

ct' --> $$\gamma$$ct - $$\beta\gamma$$x,

x' --> -$$\beta\gamma$$ct + $$\gamma$$x,

y' --> y, and

z' --> z

twice to the six basis elements?

8. May 11, 2011

### tiny-tim

yes!

so for example x'Λt' = … ?

(btw, you may find it easier to use the rapidity α, where tanhα = v = β, and coshα = γ)

9. May 11, 2011

### Zarathustra0

OK, cool--thanks for all the help!

Actually I think I have one last (perhaps trivial) question. In the position 4-vector (in r $$\wedge$$ p), should the temporal component ct be 0? This seems intuitive since we're just talking about two points in space (the position of the orbiting particle and the center of rotation) and not events at different times.

10. May 11, 2011

### tiny-tim

I can't remember.

I can't remember ever seeing a useful application of rΛp !

11. May 11, 2011

### DrGreg

No, it needs to be ct so that r is a genuine 4-vector (= rank-1 tensor) and hence the angular-momentum is a genuine rank-2 tensor. Note that t contributes to the other 3 components of the tensor that don't correspond to angular momentum (see post 2).

12. May 12, 2011

### Sam Gralla

The time-space components of the angular momentum tensor give the center of mass of the system (minus t times the momentum, if you are in a frame with non-zero momentum). You can think of the angular momentum tensor as a way of combining the ordinary angular momentum and the center of mass into a single 4D object, much as the relativistic energy-momentum vector combines energy and momentum into a single 4D object.