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Is it possible to deduce the law of conservation of angular momentum from the law of conservation of energy? If possible, by what sense the conservation of angular momentum has the status of law, if it can be deduced from other law (and therefore, strictly speaking, would be a theorem)?

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Note that conservation of momentum (not angular momentum) can be deduced from conservation of energy plus Galilean or Lorentz invariance. –  Ben Crowell Aug 24 '13 at 20:10
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2 Answers

No. Angular momentum and energy are two different quantities. In fact, angular momentum can be conserved when energy is not. See, for instance, the famous case of the figure skater pulling their arms in and spinning more quickly.

Let's say our figure skater reduces their moment of inertia to some fraction $n$ of their original moment of inertia. Then:

$$\begin{align} I_{0}\omega_{0} &= I_{f}\omega_{f}\\ I_{0}\omega_{0} &= nI_{0}\omega_{f}\\ \omega_{f} &= \frac{1}{n}\omega_{0} \end{align}$$

However, we have:

$$KE_{i} = \frac{1}{2}I_{0}\omega_{0}^{2}$$

And

$$\begin{align} KE_{f} &= \frac{1}{2}I_{f}\omega_{f}^{2}\\ &= \frac{1}{2}nI_{0}\left(\frac{1}{n}\omega_{0}\right)^{2}\\ &= \frac{1}{2n}I_{0}\omega_{0}^{2}\\ &= \frac{1}{n}KE_{i} \end{align}$$

So, the final kinetic energy of our figure skater is greater than their initial kinetic energy! Angular momentum is conserved, but energy is not.

(of course, we know that overall energy is conserved, so you may ask where this extra energy came from--the answer is that it requires net work in order to pull your hands in while you're spinning. So no violation of the rules of physics here.)

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The conservation of angular momentum and energy in a mechanical system are distinct, independent conservation laws. The deep reason for this is that, because of Noether's theorem, conservation laws are in correspondence with the symmetries of the system. Thus if a system is time-translation invariant - if experiments behave the same no matter what the starting time is - then energy is conserved. Analogously, if a system is rotation invariant - if the outcome of experiments is independent of the system's orientation - then angular momentum is conserved.

Because of that, angular momentum conservation can no more be derived from energy conservation than it is possible to derive rotation invariance from time-translation invariance.


To be clear: This answer explains the fundamental reason why the two conservation laws are independent, and why it would be useless to even try. This argument can be strengthened using this converse to Noether's theorem: conservation of energy and of angular momentum are equivalent to time-translation and rotation invariance. These symmetries are independent, and contribute independent dimensions to the Lie symmetry group of the system's evolution, so they cannot be derived from each other.

As Jerry Schirmer said, it is trivial to construct a counterexample, which is what his answer contains.

Further, this independence argument does also extend to linear momentum. (For a simple counterexample, try a particle in a box - conserved energy but not momentum - or a free particle in a time-dependent but spatially homogeneous electrostatic potential - conserved energy but not momentum.) Of course, neither of these systems obeys boost invariance.

The underlying theme is this. If I have two different conserved quantities, $A$ and $B$, then I can sometimes find a third conserved quantity by taking their Poisson bracket (or their commutator in QM), so long as that is not zero. For example, conservation of $p_x$ and $L_z$ implies conservation of $p_y$. The key to this is that the associated symmetries are also related: translation invariance along $x$ and rotation invariance about the $z$ axis, together, imply translation invariance along the $y$ axis. Similarly, space translation invariance and boost invariance imply time translation invariance.

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The Noether's theorem says there is a conserved quantity, but does not prevent such is consequence of other conservation law. –  Eli MC Aug 24 '13 at 18:00
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@EliMC: the two symmetries are independent of each other. It is trivial to construct a model that is not rotationally invariant but is time-translation invariant, or vice versa. –  Jerry Schirmer Aug 24 '13 at 20:48
    
This is not quite right, because if I take your argument and replace "angular momentum" with "momentum" everywhere, the conclusion is false, and yet you haven't invoked any properties of angular momentum that differ from those of momentum. The difference is that time translation invariance plus boost-invariance (Galilean or Lorentz) implies spatial translation invariance but not rotational invariance. –  Ben Crowell Aug 24 '13 at 21:23
    
@BenCrowell See response in edit. –  Emilio Pisanty Aug 24 '13 at 21:46
    
OK, that works better. –  Ben Crowell Aug 25 '13 at 0:06
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