# Can conservation of angular momentum be derived from energy consideration? [closed]

About the context of this question:
A dramatic illustration of back and forth conversion of kinetic energy and gravitational potential energy is when a celestial body has an highly eccentric orbit. Vivid example: Halley's comet.

As we know, at perihelion Halley's comet has far more velocity than at aphelion. As we know, the reason for that is that at perihelion Halley's comet has been falling towards the Sun for 38 years. For 38 years the gravity of the Sun has been doing work, increasing the kinetic energy of Halley's comet.

The conversion of gravitational potential energy to kinetic energy completely determines the shape and velocity profile of the comet's trajectory.

I have always wondered: is it possible to use the work-energy theorem to derive conservation of angular momentum? As I mentioned: in the case of a central force just the energy conversion completely determines the trajectory.

Of course, implicit in such an attempt is that the setup has rotational symmetry. The force is a central force, the force law is the same for every orientation.

Here is an attempt at such a derivation, but it falls short. The expression for kinetic energy that it uses is in terms of the angular velocity $$\omega$$, but in order for a celestial object to move from aphelion to perihelion there must be some radial velocity, that is inconsistent. The attempt does arrive at conservation of angular momentum, but well... maybe that's just a fluke.

So my question is: can the outline that is below be improved to the level of an actual derivation?

Rotational kinetic energy:

$$E_k = \frac{1}{2}mr^2\omega^2 \quad (1)$$

Required centripetal force at angular velocity $$\omega$$ and radial distance $$r$$:

$$F = -m\omega^2r \quad (2)$$

The Work-Energy theorem in differential form:

$$\frac{dE}{dr} = F \quad (3)$$

Substituting the expressions for $$E$$ and $$F$$ in the differential equation:

$$\frac{d(\frac{1}{2}r^2\omega^2)}{dr} = -\omega^2r \quad (4)$$

To show how this implies a relation between $$\omega$$ and $$r$$ the factor $$\omega^2$$ is substituted with $$\frac{C}{r^4}$$, ($$C$$ is a constant), and where applicable factors of $$r$$ are allowed to drop away against each other.

$$\frac{d(\frac{1}{2}\frac{C}{r^2})}{dr} = -\frac{C}{r^3} \quad (5)$$

So we have the following relation:

$$\omega^2 = \frac{C}{r^4} \quad (6)$$

Hence:

$$\omega^2r^4 = C \quad (7)$$

Which is of course a squared version of the expression that says that angular momentum $$\omega r^2$$ is a constant.

Pushing the attempt further:
I assume using polar coordinates is best in this case, because that aligns with the rotational symmetry of the setup.

Assumption: the total kinetic energy is the following sum:

$$E_{kin.total} = \tfrac{1}{2}m\left(\tfrac{dr}{dt}\right)^2 + \tfrac{1}{2}mr^2\omega^2 \qquad (8)$$

The component velocities of these contributions are perpendicular, so I expect these component velocities to sum as the sides of a right triangle.

Assumption:
The amount of centripetal force that is available for radial acceleration is the total centripetal force minus the required centripetal force.

The thing that is tripping me up here is the assignment of minus signs. What is needed is an expression with the total centripetal force on one side, and the radial acceleration and required centripetal force on the other side. Of course everything in the radial direction must be treated like a vector; the minus signs must be correct.

Assumption on how to apply the work-energy theorem:
The motion of a celestial body from aphelion to perihelion can be thought of as spiraling in. Compare that gradual motion to moving down a ramp. To apply the work-energy theorem it isn't necessary to know the shape of the ramp. The total change in energy is always equal to the total difference in energy potential from top to bottom of the ramp, independent of its shape.

So I think the derivation only needs to consider the change in potential energy as a function of the radial distance, no need for separate radial and tangential component there.

Pushing the attempt further:

The image illustrates the kind of case that this question is about. Image credit:
University of Toronto, Department of Physics, Angular Momentum

Of course, the case will be treated in an idealized form, so that exclusively the principle is examined.

The idealization:
All motion is treated as frictionless.

Initial condition: the radial distance of the circumnavigating mass is kept constant. As we know, in the case of circular motion the magnitude of the required centripetal force is $$m\omega^2r$$ From here on I will use the expression 'required centripetal force' for 'magnitude of centripetal force required to sustain circular motion'.

In the next phase the centripetal force is increased. So in this fase there is a surplus of centripetal force, and the circular motion contracts.

As long as the contraction continues the centripetal force is doing work, increasing the rotational kinetic energy.

In the final state the increase of centripetal force is leveled off.
As we know, in the final state the angular velocity is much higher than in the initial state. The angular velocity is higher because the centripetal force has been doing work.

What is interesting here is that even though the centripetal force acts purely in radial direction, the end result is an increase of angular velocity. That is the nature of circumnavigating motion: any change of radial velocity transforms to angular velocity.

So: initial state is unambiguious; circular motion at a particular angular velocity, and the end state is unambiguous too: circular motion at a decreased radial distance, with an increased angular velocity.

We can calculate the increase of angular velocity in two ways:

• Apply the principle of conservation of angular momentum

• Solve analytically for the amount of work that the centripetal force has done when the system has contracted. The change in energy from initial state to end state is equal to the amount of work done, hence the end state is uniquely determined.

Those two independent calculations must end up with the same answer.

So: how to arrive at an analytical solution?

As noted: when the total force is larger than the required centripetal force there will be an acceleration in the radial direction. For the magnitudes of the total force, the radial acceleration and the required centripetal force respectively:

$$F_{total} = m\tfrac{d^2r}{dt^2} + m\omega^2r \qquad (9)$$

Repeating equation (3):

$$\frac{dE}{dr} = F \qquad (10)$$

Equation (10) expresses that in this case the force is at all times acting in radial direction, which means that all the work that is done is in the radial direction. Therefore the derivative of the total energy with respect to $$r$$ should be equal to the force.

The glaring problem with this approach: In equation (8) a decomposition in two perpendicular components is proposed, but in the case of the force the two contributions on the right side of equation (9) are not perpendicular components, those two contributions are aligned. This way of approaching the problem is neither purely scalar, nor purely vectorial, but something in between, which may well invalidate the whole approach.

Subsitituting (8) and (9) into (10) $$\frac{d(\tfrac{1}{2}m\left(\tfrac{dr}{dt}\right)^2 + \tfrac{1}{2}mr^2\omega^2)}{dr} = m\tfrac{d^2r}{dt^2} + m\omega^2r \qquad (11)$$

the following two equations are a separation: equation (11) is separated into two statements:

$$\frac{d(\tfrac{1}{2}m\left(\tfrac{dr}{dt}\right)^2)}{dr} = m\tfrac{d^2r}{dt^2} \qquad (12)$$

$$\frac{d(\tfrac{1}{2}mr^2\omega^2)}{dr} = m\omega^2r \qquad (13)$$

Equation (12) has on the left the derivative with respect to $$r$$ of the kinetic energy in radial direction, and on the right the acceleration in radial direction.
Assertion: the term on the left side and the term on the right side will be the same at every point in time; it's the work-energy theorem.
Assertion: equation (12) is satisfied at every point in time, that implies that equation (13) is constrained to be satisfied at every point in time too.

Equation (13) is very similar in form to equation (4), except in equation (4) the $$m\omega^2r$$ term has a minus sign.

Therefore in the case of equation (13) to accommodate the non-presence of a minus sign you would have to relax the constraint to a statement about magnitude; asserting that the magnitude of the term on the left side must be equal to the magnitude of the term on the right side. Presumably this minus-sign-problem arises because the whole approach is neither purely scalar, nor purely vectorial

Now to what would count as a useful stackexchange answer:
In this form this is a problem in mathematical physics. When a rotating system contracts, find a way to calculate the increase of rotational kinetic energy. The initial state and the end state are both unambiguous, therefore an unambiguous calculation must be possible. I have outlined my own attempt, and I have outlined the glaring problem with it: neither purely scalar nor purely vectorial. Can the outline be improved?

More generally:
Discussion of necessary and sufficient condition.
First the case of conservation of linear momentum, in the case of two masses that exert a force upon each other, which of course causes mutual acceleration. In that case a necessary condition for conservation of momentum is that the force law supports it. $$F=ma$$ supports conservation of linear momentum. By contrast: if instead of $$F=ma$$ the force law would be this:

$$a = \frac{F}{m^2}$$

Then you don't have conservation of linear momentum.

In the case of angular momentum it is of course a necessary condition is that the force that the masses exert upon each other is a central force, but that is not a sufficient condition. In addition the force law must support it, just as in the case of linear momentum.

## closed as unclear what you're asking by Aaron Stevens, GiorgioP, Jon Custer, Buzz, ZeroTheHeroApr 15 at 3:57

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• The work-kinetic energy theorem is basically a slightly different form of $F=ma$, so you're asking if you can derive conservation laws given the equation of motion. And the answer is, of course you can! – knzhou Apr 14 at 0:04
• @knzhou Yeah, it's about $F = ma$. So, in the case of conservation of linear momentum when two objects collide in a perfectly elastic collision: then conservation of linear momentum in that forceful interaction can be regarded as a consequence of $F = ma$. If the two objects have unequal mass then the least massive object will undergo the largest acceleration. Like collision, orbital mechanics is two objects, both accelerating each other. – Cleonis Apr 14 at 0:30
• Perhaps another way to say this is: conservation laws are just intuitive and useful ways of summarizing some of the information that comes out of $F=ma$. If you have all the $F=ma$ (and similar) equations, then it's not surprising you get everything. – knzhou Apr 14 at 1:18
• Your talk about total, required, and available centripetal force doesn't make sense to me. Honestly I'm trying hard to put something together here, but anything I try just ends up being circular or conservation of angular momentum pops up without energy being directly tied in. – Aaron Stevens Apr 14 at 4:36
• The question has been put on hold. Independent of that I had already come to the conclusion that this question is not suitable for stackexchange. I still think the question is an interesting one to raise, but stackexchange isn't a good place for it. So I intend to delete it, in a day or so. My thanks to knzhou, for pointing out that it is about $F=ma$, the work-energy theorem coming in because the work-energy theorem is equivalent with $F=ma$ – Cleonis Apr 15 at 16:26

## 3 Answers

I did not read the whole of the derivation, but

The force is a central force

implies conservation of angular momentum with respect to the center the force is directed to. So a derivation that starts from that assumption is not a derivation of conservation of angular momentum from the work-energy theorem.

• I misread it too, but the OP isn't asking about energy conservation. – Aaron Stevens Apr 13 at 22:26
• It seems to me that the condition of rotational symmetry in itself only constrains that something angular must be conserved. Additional steps are required to arrive at conservation of $\omega r^2$. It seems to me there are multiple ways to arrive at conservation of $\omega r^2$, that is why I think exploring the a route via energy conversion can be interesting. If possible then that would be an interesting illustration of the interconnectedness of energy considerations and momentum considerations – Cleonis Apr 13 at 22:31
• As far as I understand, he is asking whether it is possible to use energy conservation to derive the conservation of angular momentum: starts with "is it possible to use the work-energy theorem to derive conservation of angular momentum" and ends with "Which is of course a squared version of the expression that says that angular momentum $ωr^2$ is a constant." and asks "can the outline that is below be improved to the level of an actual derivation?". The answer is no, with a central force angular momentum is conserved, independently of anything else. – JTS Apr 13 at 22:32
• @Cleonis: here you are a case in which energy is not conserved and angular momentum is conserved: the force has a component proportional to the component of velocity directed towards the center. – JTS Apr 13 at 22:33
• @JTS The work-energy theorem isn't energy conservation – Aaron Stevens Apr 13 at 22:35

In general this is not possible: Conservation of energy and conservation of angular momentum pertain to two distinctly different symmetries in the Lagrangians describing physical systems. And thanks to Noether's theorem, we know that there is a 1-to-1 correspondence between symmetries and conserved quantities.

• There seems to be a confusion here. My question hinges on energy conversion (in this case between potential energy and kinetic energy). This question does not necessarily involve energy conservation – Cleonis Apr 14 at 1:29
• Your question title speaks about conservation though. – AtmosphericPrisonEscape Apr 14 at 1:30
• Also you get an important thing wrong in your question: The energy landscape does not determine the trajectory in the Kepler problem fully. The other free parameter is the angular momentum. – AtmosphericPrisonEscape Apr 14 at 1:32
• The title says 'energy consideration', which unfortunately is easily misread as energy conservation. Too easy to misread; in retrospect not a good choice of word, in the title. – Cleonis Apr 14 at 1:43
• About how much is determined by the energy landscape: given a particular radial distance at aphelion, and a particular amount of kinetic energy at aphelion, it seems to me the subsequent motion is completely determined. Of course, if the starting specification is only a total amount of energy then additional information is needed to narrow down to one particular trajectory. Anyway, these details do not materially affect the purpose of the question – Cleonis Apr 14 at 2:24

Your "proof" is circular. You substitute in $$\omega^2=C/r^4$$ and show that $$\omega^2r^4=C$$

You assume angular momentum is conserved in your derivation then.

• I have edited the question to add enumeration to the expressions. About the subsitution from exp. (4) to exp. (5): it is not the case that any substitution will result in a satisfied equation. The point is that eq. (4) imposes a constraint. There is only one subsitution that results in a satisfied equation; in that sense that particular substitution is enforced. So the suggestion of circular reasoning is incorrect. (Of course, the outline does fall short, but for the reason given in the original question.) – Cleonis Apr 13 at 22:58
• @Cleonis No it is circular. You will always get a true expression no matter what you replace $\omega$ with. This is because the left and right sides of equation 4 are exactly the same thing. Just take the derivative and you will see this. Therefore you are in fact assuming angular momentum conservation and then getting angular momentum conservation. The angular momentum conservation doesn't arise from something else. I could argue $\omega^2r^5$ is also conserved using your arguments – Aaron Stevens Apr 13 at 23:06
• Well, expression (4) is not one where a partial derivative is taken. When you don't do a subsitution then writing out the derivative means also writing out the derivative of the $\omega$ term with respect to $r$ – Cleonis Apr 13 at 23:16
• @Cleonis Oh. By saying $F_{rad}=-m\omega^2r$ you are assuming a constant $r$ and therefore constant $\omega$. In general the radial component of the force is given by $F_{rad}=m\ddot r-m\omega^2r$ where $\ddot r$ is the second rate of change of $r$. I thought you were assuming uniform circular motion by using this expression. – Aaron Stevens Apr 13 at 23:39
• @Cleonis And of course you have to add in the energy the radial part of the motion too. I think it will not be possible to deduce how to split the energy into the parts that are due to radial and tangential velocity using the work-energy theorem alone, as in the work-energy theorem (scalar) you lose the information about the vectorial character of motion. The correct splitting is what will give you the conservation of angular momentum. This is for the moment in part guessing but it might be correct and also the thing that I could add to my answer. Need to think about it a bit more. – JTS Apr 13 at 23:49