Why is the 'motionless solution' not always valid in classical mechanics?

[jmb]

TL;DR Summary

(Slightly philosophical). Conservation of energy would be satisfied by a totally stationary state, is there any principle, other than empiricism, that allows us to discard this solution?

I realize to many people this might seem a silly or at least pointlessly philosophical question, but I was wondering if there's a deeper answer I've missed...

After a lengthy break in physics I've been playing around with some classical mechanics to try and probe my understanding (and memory!). And I realized this...

Conservation of Energy is not violated if nothing moves: a ball hanging in mid-air is a perfectly valid solution if conservation of energy is the only constraint.

My first reaction was that: "No, conservation of energy doesn't require anything to move, but the Euler-Lagrange equations for a system do. (In fact you can derive conservation of energy as a first integral of the system specifically by multiplying both sides of the E-L equations by $\dot{x}_i$, which actually introduces this motionless solution: because $\dot{x}_i$ could be zero)." But then I realized this isn't true...

To derive the Euler-Lagrange equations you search for a path of the system between two points $x_i\left(t_0\right)$ and $x_i\left(t_1\right)$ for which the integral over the path of some quantity $L\left(x_i,\dot{x}_i\right)$ (the Lagrangian) is stationary w.r.t. changes in the path. However, typically in this derivation we implicitly assume that $x_i\left(t_0\right) \neq x_i\left(t_1\right)$. Clearly if $x_i\left(t_0\right) = x_i\left(t_1\right)$ then (at least for the subset of problems where the Lagrangian is strictly positive, such as movement in a gravitational potential well) a path that does not go anywhere does indeed represent a stationary value of this integral: as the integral will equate to zero.

Obviously, we don't live in a totally motionless universe (and one could also argue that such solutions are unstable to perturbation), so it is easy to discard this solution on empirical grounds, but I was wondering: is there any other principle of classical physics that requires us to throw out the the 'motionless solution'?

To put it another way:

Conservation of energy says we can exchange potential energy for kinetic energy, and the Euler-Lagrange equations tell us how we would do so, but is there any principle that says that if a system can exchange potential energy for kinetic energy then it will?

Appealing to Newton's Laws doesn't help, as you either have to consider them as empirical or as a consequence of the Euler-Lagrange equations.

 

[PeroK]

To put it another way:

Conservation of energy says we can exchange potential energy for kinetic energy, and the Euler-Lagrange equations tell us how we would do so, but is there any principle that says that if a system can exchange potential energy for kinetic energy then it will?

Appealing to Newton's Laws doesn't help, as you either have to consider them as empirical or as a consequence of the Euler-Lagrange equations.

Newton's law require that you interpret a gradient in the potential as a force: $$\vec F = - \vec \nabla U$$

 

[russ_watters]

The fact that a stationary system satisfies conservation of energy doesn't mean a moving system can't or require that there can only be stationary systems. Does 1+1=2 mean you can never have 3 of something?

 

[jmb]

The fact that a stationary system satisfies conservation of energy doesn't mean a moving system can't or require that there can only be stationary systems. Does 1+1=2 mean you can never have 3 of something?

I'm not saying there can't be dynamic systems. I'm saying that in many cases there is an apparent 'motionless solution' to the Euler-Lagrange equations, which is unphysical and I'm asking what principal (other than empiricism) we use to discard this solution.

Consider a point particle in a gravitational potential well at some position other than the centre of the well. Clearly this particle should move: there is a force on it. But (because the Lagrangian of the system is strictly positive) there is technically a solution that satisfies the principal of least action in which the particle remains totally stationary. What do we appeal to in order to discard this solution?

 

[jmb]

Newton's law require that you interpret a gradient in the potential as a force

Yes, but Newton's Law is empirical, or else a consequence of solving the Euler-Lagrange equations for the system. But in many cases the E-L equations would also admit a 'motionless solution'.

 

[PeroK]

Yes, but Newton's Law is empirical, or else a consequence of solving the Euler-Lagrange equations for the system. But in many cases the E-L equations would also admit a 'motionless solution'.

First, E-L are equivalent to Newton's laws, so that can't be right.

Second, if we imagine a ball hanging in midair in a gravitational field, then the action is not stationary. A variation on the path to have it move up and back down would reduce the action integral.

 

[PeroK]

Yes, but Newton's Law is empirical, or else a consequence of solving the Euler-Lagrange equations for the system. But in many cases the E-L equations would also admit a 'motionless solution'.

Can you post the solution you are talking about. There's not enough detail to see what you are saying.

 

[jmb]

First, E-L are equivalent to Newton's laws, so that can't be right.

Newton's laws do indeed follow from solving the E-L equations, but what I'm trying (unfortunately not very well!) to say is that when we set up the E-L equations there is also another solution that we typically ignore. Let me take the concrete example you gave...

if we imagine a ball hanging in midair in a gravitational field

I'm going to make this slightly more specific and say "in a potential well", because the case of a uniform gravitational field (like the surface of the Earth) isn't one of the cases where this extra solution rears its head.

So we have a ball of point mass $m$ sitting in a potential $V\left(r\right) = - \frac{km}{r}$ at some position $r \neq 0$. If we were going to derive the Euler-Lagrange equations from first principles we would write
$$I= \int_{t_0}^{t_1}L\left(x_i,\dot{x}_i\right)\; dt$$ and try to find a path $x_i\left(t\right)$ between $x\left(t_0\right)$ and $x\left(t_1\right)$ for which $I$ is stationary w.r.t. further variations in the path.

But it is possible (though often disregarded) that $$x_i\left(t_0\right) = x_i\left(t_1\right)$$ (for instance the particle could be in a closed orbit). If we choose to look at that case, then because $L = T - V = \frac{1}{2}m\dot{x}_i\dot{x}_i + \frac{km}{r} \geq 0$ we can conclude that a path of zero length (i.e. $x_i\left(t\right) = \rm{const}$) must be a stationary point of $I$ because it gives $I=0$ (the integral along a path of zero length is zero) and any non-zero length path would necessarily result in an action larger than zero.

So my question is, is this just a mathematical curiosity we have learned to disregard as 'unphysical', or is there a physical principal we can appeal to in order to discard this solution?

 

[Frabjous]

There is a position dependence in km/r that you are ignoring which generates the force.

edit: You are integrating over time. If the particle is initially stationary, there is a force on it which causes a velocity at later times.

 

[jmb]

There is a position dependence in km/r that you are ignoring which generates the force.

Sorry, I'm being slightly loose in my notation for brevity: $r^2=x_i x_i$, so yes $V$ is a function of position. But I'm not ignoring it: finding a stationary point of the action can indeed produce a solution which results in the particle moving in a way corresponding to what a non-Lagrangian treatment would label as a force due to a gradient in the potential, but what I'm grappling with is that when starting from first principles a 'motionless' solution would also seem to be a stationary point of the action integral in this example.

 

[Frabjous]

Sorry, I'm being slightly loose in my notation for brevity: $r^2=x_i x_i$, so yes $V$ is a function of position. But I'm not ignoring it: finding a stationary point of the action can indeed produce a solution which results in the particle moving in a way corresponding to what a non-Lagrangian treatment would label as a force due to a gradient in the potential, but what I'm grappling with is that when starting from first principles a 'motionless' solution would also seem to be a stationary point of the action integral in this example.

You need to do the math. Do your solution and then calculate with the “standard solution”. The path is through time so your solution is non-zero.

 

[PeroK]

But it is possible (though often disregarded) that $$x_i\left(t_0\right) = x_i\left(t_1\right)$$ (for instance the particle could be in a closed orbit). If we choose to look at that case, then because $L = T - V = \frac{1}{2}m\dot{x}_i\dot{x}_i + \frac{km}{r} \geq 0$ we can conclude that a path of zero length (i.e. $x_i\left(t\right) = \rm{const}$) must be a stationary point of $I$ because it gives $I=0$ (the integral along a path of zero length is zero) and any non-zero length path would necessarily result in an action larger than zero.

I don't see this at all. Having the $x_i$ constant is not a local minimum of the action integral. You can see this from the E-L equations: $$m\ddot x_i = -\frac{kmx_i}{r^3}$$
If you have the patience, you should be able to construct a variation of the stationary trajectory that reduces $I$. Alternatively, you can trust the mathematics that produces the E-L equations.

 

[jmb]

The path is through time so your solution is non-zero.

After thinking on it again, it suddenly came to me, I was about to post my mistake but it seems you beat me to it!

I was so fixated on the path taken by the particle as a physical trajectory that I missed/forgot the fact that the action integral is over time: viewed over space and time the path length of the stationary case is as you point out non-zero, and my argument is thus bogus.

Thankyou everyone for your patience and help in rooting out the mistake in my thinking!