- #1
scope
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hey,
does anyone there know how the angular momentum (L=r x p) is transformed under Lorentz transformations?
does anyone there know how the angular momentum (L=r x p) is transformed under Lorentz transformations?
bcrowell said:There is no cross product in four dimensions, so the generalization of rxp is a rank-2 tensor, not a vector: [itex]J^{jk}=r^jp^k[/itex]. 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 Jl according to [itex]J^{jk}=\epsilon^{jkl}J^l[/itex], where [itex]\epsilon[/itex] is the Levi-Civita symbol. The advantages of Jl are that it's less unwieldy than [itex]J^{jk}[/itex], and it matches up with the Newtonian angular momentum in the appropriate limit. The advantage of [itex]J^{jk}[/itex] is that it transforms in a simple way, as a rank-2 tensor.
Dickfore said:[tex]
M^{\mu \nu} \equiv X^{\mu} P^{\nu} - X^{\nu} P^{\mu}, \; M^{\nu \mu} = -M^{\mu \nu}
[/tex]
[tex]
M'^{\mu \nu} = \Lambda^{\mu}{}_{\rho} \, \Lambda^{\nu}{}_{\pi} \, M^{\rho \pi}
[/tex]
scope said: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?
Dickfore said:So, for example, to find how [itex]L_{z}[/itex] transforms under a conventional Lorentz transformation, you should look how [itex]M^{1 2} = L_{z}[/itex] transforms according to the general rule:
[tex]
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)
[/tex]
scope said:thank you, but then how is the "contraction factor" of the angular momentum for Lz or any other coordinate, (simply) calculated?
arkajad said: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 said: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.
Lorentz transformations of angular momentum refer to the mathematical equations that describe how the angular momentum of an object changes as it moves through space and time in accordance with Einstein's theory of special relativity.
Lorentz transformations are important in physics because they allow us to accurately describe the behavior of objects that are moving at high speeds, near the speed of light. They also play a crucial role in understanding the fundamental principles of relativity and the nature of space and time.
Lorentz transformations do not affect the conservation of angular momentum. This principle still holds true, meaning that the total angular momentum of a system remains constant in all reference frames, even when using Lorentz transformations.
Yes, Lorentz transformations can be applied to all types of angular momentum, including orbital angular momentum (the motion of an object around an axis) and spin angular momentum (the intrinsic angular momentum of particles).
Yes, Lorentz transformations of angular momentum have many real-world applications, particularly in the fields of particle physics and astrophysics. They are used to accurately describe the behavior of subatomic particles, as well as the motion of objects in space, such as planets and stars.