What is the angular-momentum 4-vector?

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In summary: The center of mass (i.e. the time-space part of the tensor) is the part that gets transformed by the Lorentz transformation when you go to a frame with a different velocity. This is the same way that the energy-momentum vector transforms.
  • #1
Zarathustra0
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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 [tex]\times[/tex] p

still valid for the 3-angular momentum in special relativity?
 
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  • #2
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.
 
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  • #3
Thanks for the help--I assume then that this tensor transforms by applying the Lorentz-transformation matrix twice?
 
  • #4
Welcome to PF!

Hi Zarathustra0! Welcome to PF! :wink:

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) :smile:
 
  • #5
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.
 
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  • #6
Hi Zarathustra0! :smile:
Zarathustra0 said:
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

perform the Lorentz transformation on each by two applications of the transformation matrix, and then add up the resulting six matrices?

I don't recommend trying to mix up tensors and 2-forms. :redface:

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! :wink:

(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 :smile:)​
 
  • #7
OK, in that case to apply the transformation twice to each basis element, should I simply make the substitutions

ct' --> [tex]\gamma[/tex]ct - [tex]\beta\gamma[/tex]x,

x' --> -[tex]\beta\gamma[/tex]ct + [tex]\gamma[/tex]x,

y' --> y, and

z' --> z

twice to the six basis elements?
 
  • #8
yes! :smile:

so for example x'Λt' = … ? :wink:

(btw, you may find it easier to use the rapidity α, where tanhα = v = β, and coshα = γ)
 
  • #9
OK, cool--thanks for all the help!

Actually I think I have one last (perhaps trivial) question. In the position 4-vector (in r [tex]\wedge[/tex] 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
I can't remember. :redface:

I can't remember ever seeing a useful application of rΛp !
 
  • #11
Zarathustra0 said:
In the position 4-vector (in [itex]r \wedge p[/itex]), should the temporal component ct be 0?
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
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.
 
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1. What is the definition of the angular-momentum 4-vector?

The angular-momentum 4-vector is a mathematical representation of the angular momentum of a particle in four-dimensional spacetime. It combines the three-dimensional angular momentum vector with the energy and momentum components of the particle, providing a comprehensive description of its angular motion.

2. How is the angular-momentum 4-vector calculated?

The angular-momentum 4-vector is calculated by multiplying the energy and momentum components of a particle by its three-dimensional angular momentum vector and combining them using the principles of special relativity. This results in a four-dimensional vector with components that describe the magnitude and direction of the particle's angular momentum.

3. What is the significance of the angular-momentum 4-vector in physics?

The angular-momentum 4-vector is a fundamental concept in physics, particularly in the field of relativity. It is used to describe the angular momentum of particles in motion and is essential in understanding the principles of conservation of angular momentum in relativistic systems.

4. How does the angular-momentum 4-vector differ from the classical angular momentum?

The angular-momentum 4-vector differs from classical angular momentum in that it takes into account the relativistic effects of energy and momentum on the angular motion of a particle. This means that the magnitude and direction of the angular-momentum 4-vector can change as the particle's energy and momentum change, whereas classical angular momentum remains constant.

5. Can the angular-momentum 4-vector be used for particles at rest?

Yes, the angular-momentum 4-vector can be used for particles at rest. In this case, the energy and momentum components of the particle are both zero, and the angular-momentum 4-vector represents the intrinsic angular momentum, or spin, of the particle. This is an important concept in quantum mechanics and is used to describe the properties of subatomic particles.

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