# If angular momentum corresponds to linear momentum, what corresponds to energy?

Angular momentum is defined from linear momentum via $\vec L = \vec r\times\vec p$, and is conserved in a closed system. Since energy is the time part of the linear four-momentum, is there a quantity defined from energy that's also conserved?

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Torque seems related but it strikes me as a bit of an apples-to-oranges comparison. –  Brandon Enright May 26 '13 at 23:24
Duplicate of physics.stackexchange.com/q/9864 –  Ben Crowell May 27 '13 at 3:11
For one possible interpretation of the question, this might be a nice resource: math.ucr.edu/home/baez/boosts.html –  Siva May 27 '13 at 8:32

In special relativity momentum is part of a 4-vector with energy as the time component as you correctly said. Angular momentum is not part of a 4-vector. The cross product you used in the question does not give a vector when generalised from 3 to 4 dimensions. Instead it gives an antisymmetric matrix. So angular momnetum is part of an antisymmetric rank-2 tensor (a matrix) which has six independent components. The angular momentum vector from 3d space forms three of those components and the other three components form another 3d vector, so this works differently from a 4-vector.

In 3d an antisymmetric tensor is a hodge dual to a vector which is why we think of angular momentum as a vector like momentum, but in 4d spacetime an antisymmetric tensor is hodge dual to another antisymmetric tensor and cannot be thought of as equivalent to a 4-vector. This is obvious because the number of components is different.

To understand what the other 3-components of this antisymmetric matrix are you have to look at the relationship between conserved quantities and symmetry as understood via Noether's theorem. Energy and momentum correspond to symmetry under translations in time and space, but angular momentum is symmetry under rotations. In special relativity rotations and combined with Lorentz boosts to form the six parameter group of Lorentz transformations. So your question is equivalent to asking what conserved quantity corresponds to Lorentz boosts. This has been answered here before See there for the details, but in short it is the centre of mass at a fixed time.

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It looks like the center of energy is conserved to me. –  Larry Harson May 27 '13 at 13:44
@Larry Hanson, Energy is equivalent to mass so you can call it centre of energy if you prefer. That makes it more connected to the energy in the momentum 4-vector. However, be careful what you state. If centre of energy were conserved then it would mean that the centre does not move. It is the projection back to where it was at a fixed time (call it t=0) that is fixed. This is a strange conservation law because it has an explicit time dependence but that is because the Lorentz boost is a transformation with a time dependence. –  Philip Gibbs - inactive May 27 '13 at 14:32
No it is $tp-xE$ where $x$ is the position and $p$ is the momentum of the centre of energy –  Philip Gibbs - inactive May 27 '13 at 15:56

First, be careful about assuming that a closed system with linear momentum necessarily includes angular momentum. To ask the question you are trying to ask, you need to be careful that you really do have a one-to-one analogy. Since it is easy to construct linear momentum systems that have vanishingly small angular momenta, the analogy you just suggested is not quite right.

Understanding the simplest possible closed system with angular momentum helps analyze your question, however. Consider for example two particles approaching each other. If at their point of closest approach the particles will collide -- that is, if their paths are precisely co-axial -- then the angular momentum of the closed system is zero. Conversely, if there will be any separation between the two particles at the moment of closet approach, the system has angular momentum. I'll explain below why this is important for your question.

As it happens, it is more accurate (and conventional) to compare rest-mass energy not to angular momentum, but to linear momentum. In fact, if you express rest mass as $mc$, you get the same units as momentum. You can even loosely interpret the quantity $mc$ as the momentum of an object of mass $m$ moving through time at velocity $v=c$.

So, using that, allow me to alter your question slightly to keep the analogies properly aligned:

If linear momentum (mass moving through space) corresponds to energy (mass moving through time), what does angular momentum (two masses moving in opposite directions with separation in space) correspond to in the time domain?

The problem of course is that in classical physics, everything moves the same direction in time, making it impossible to set up the two-particles-moving-oppositely scenario needed to create the simplest possible classical example of angular momentum. Does that mean that the analogy simply breaks, and that there simply is no equivalent of angular momentum in the time domain?

Not quite.

If you know Feynman diagrams, think of the case of a gamma ray breaking apart into two half loops, one moving forward in time (an electron), and the other moving backward in time (a positron). As best I can tell, this tiny loops represent virtual examples of the time equivalent of angular momentum, since they include two electrons moving opposite directions in time, with a tiny but finite separation distance between them. Those are the ingredients for angular momentum, but with the paths of the particles moving in opposite directions in time, rather than in space.

I don't know if there is a specific name for this angular-momentum-like quantity that is created very transiently by such loops. Perhaps someone else here does? I also don't know if this quantity exists in a more permanent form. For example, when a gamma ray passing near a heavy nucleus splits into a real electron and a real positron, does the resulting pair of real and persistent particles have associated with it a named quantity that is equivalent to angular momentum in the time domain? And would that quantity be related in some way to the electric field between the particles that comes into existence at the same time? Again, I don't know. Perhaps someone else here does.

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You may want to look at the analogy in a different manner. Energy is the time part of the 4-momentum while linear momentum is its space part. Energy conservation is a consequence of homegeneity of time while momentum conservation is a consequence of homogeneity of space.

If you were interested in knowing is the wedge product of 4-position and 4-momentum also conserved then you may want to look at the paper http://panda.unm.edu/Courses/Finley/P495/TermPapers/relangmom.pdf which explains how conservation of angular momentum implies a uniform motion of the centroid of a system of particles. The centroid is defined taking into account an equivalence of mass and energy.

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Yes. There is not just one "energy" associated with an object, there is translational energy and rotational energy. Consider an object whose center of mass is stationary, but it's rotating. It's not moving through space, but it clearly has kinetic energy. This is rotational kinetic energy, which is related to the angular velocity in much the same way as linear velocity:

$$K=\frac{1}{2}I\omega^2$$

In an elastic collision, rotational kinetic energy is conserved separately from translational kinetic energy.

Likewise there is also rotational potential energy (consider a spinning top or a wind up toy, for example).

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The question is a question about relativity. –  Ben Crowell May 27 '13 at 2:34