I'm reading Quantum Theory for Mathematicians, by Brian C. Hall. Although the book is about Quantum Mechanics, it's chapter 2 is actually about Classical Mechanics, in which I encountered the following theorem:

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Rephrasing in English, suppose a force acting on a particle has two components. The first component comes from a potential function and the second component is orthogonal to the velocity. Then the energy of the particle is conserved.

What is the physical intuition behind this theorem?


2 Answers 2


This is a generalization of a result commonly discussed in Electrodynamics: magnetic forces do no work.

The idea is that for the energy of the particle to change, some force must do work on the particle due to the work-energy theorem. For the force Hall chose, we can split the analysis in two parts: the potential and the "magnetic" force (I'll give it this name in analogy with Electrodynamics).

The potential can do work, but it also goes in the very definition of energy. It is what we call a conservative force. If it accelerates the particle, the potential energy drops in exchange. Notice there is some sort of equilibrium in here: this force can make the particle go faster, at the expense of the potential getting lower. In general, this means there is a maximum possible speed the particle can get by being accelerated by the force, for eventually one will "run out" of potential energy and all energy will be kinetic. Furthermore, if the particle slows down, the loss of kinetic energy is just being traded for potential energy, which can once again be converted in kinetic and so on.

For the magnetic force, things are quite different. It does not enter the definition of the energy, so in principle it could spoil this idea and either accelerate the particle indefinitely (which would be non-physical) or "steal way energy" (which is the case for friction). What the theorem is showing is that forces that are always normal to the velocity do neither: they can only change direction. The fact they are always orthogonal to the velocity means that the infinitesimal amount of work they can do will always be $$ \begin{align} dW &= \mathbf{F \cdot} \text{d}\mathbf{l}, \\ &= \mathbf{F \cdot} \frac{\text{d}\mathbf{l}}{\text{d} t} \ \text{d} t, \\ &= \mathbf{F \cdot v} \ \text{d} t, \\ &= 0, \end{align} $$ hence, it can't do any work. As a consequence, it can't change the energy. (I think/hope this doesn't solve Exercise 8, since this argument is usually fair game for physicists, but too sloopy for mathematicians hahaha).

The fact it can't change the energy is a different way of saying it can never lead to the particle speeding up or down. If it increased the energy, it could be possible to turn this extra energy into kinetic and get the particle to move faster. If it decreased the energy, it could be possible to slow down the particle in a similar way.

In short, one possible physical intuition is that if a force is orthogonal to the velocity, it can only change the direction, it can never lead directly or indirectly to an increase or decrease in speed. As commonly stated in Electrodynamics, "magnetic forces do no work".


Not sure if this helps, but...

If you start off with Newtonian mechanics, energy conservation has to be added in as an additional axiom. (As in "... and only forces which conserve energy are found in the wild").

If you start off with Lagrangian mechanics it is simply not possible to write a Lagrangian which does not lead to a conservative force.

At first I thought Newtonian and Lagrangian formulations were totally equivalent, but once I grasped the above it suddenly made sense why people consider Lagrangian to be more fundamental.

So the physical intuition is simply that a formulation of mechanics based around conserved energy is more fundamental than one based around force, and forces are simply derived from that.

  • $\begingroup$ This is not true! Energy conservation follows from Newtons laws, for a certain class of force fields. Also you can write down Lagrangians that are not invariant under time translations and therefore describe systems which do not conserve energy. $\endgroup$ Dec 28, 2021 at 10:58
  • $\begingroup$ For a single particle 2D universe F(x,y) = yi -xj is not conservative. In fact almost anything you could write down would not be in any context. So the "certain class of force fields" is just those which come from a Lagrangian. I take your point though that this only applies to Lagrangians where L is not a function of time. But I think stating that L is not a function of time is a simpler constraint than saying "only conservative force fields appear in closed systems". $\endgroup$ Dec 29, 2021 at 11:48
  • $\begingroup$ I'm not sure why this has been marked down. The OP was asking for an intuitive explanation. The argument above is essentially the same as made in Susskind's Theoretical Minimum books. Of course it is possible to take examples of forces and show by calculation that they're conservative, but that doesn't really tell you why do they always turn out to be conservative, or provide any intuition. $\endgroup$ Dec 29, 2021 at 12:05
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    $\begingroup$ I think the work energy theorem is only a proper theorem, ie something that is derived, in the case of a specific force. Then you can calculate work and show it to be true. In the general case, it is a principle, ie a statement about what sort of forces can exist. Why does it work? An intuitive explanation might be that energy comes first and force is derived from it. It's up to you whether you consider this to be of intuitive help. If you don't you just need to say, well it just turns out that this principle holds about this particular quantity work, derived from force, and KE. $\endgroup$ Dec 30, 2021 at 17:34
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    $\begingroup$ @AlexZeffertt There is the following consideration: as we know there is no intrinsic zero point of potential energy; all calculations use difference of potential energy; choice of zero point of potential energy is arbitrary. Similarly, kinetic energy is a function of the relative velocity of two (or more) objects. In that sence attribution of kinetic energy is frame dependent. Magnitude of acceleration, on the other hand, does have an intrinsic zero point. What comes first? In physics you can often run derivations in both directions. $\endgroup$
    – Cleonis
    Dec 31, 2021 at 13:12

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