Is There a Definitive Answer to the Existence of Action in Quantum Mechanics?

[smallphi]

That was discussed before but the thread died inconclusively.

Anybody knows of necessary or sufficient conditions the differential equations of motion of a given system have to satisfy to be derivable by variations of some action?

 

[lugita15]

That was discussed before but the thread died inconclusively.

Anybody knows of necessary or sufficient conditions the differential equations of motion of a given system have to satisfy to be derivable by variations of some action?

I am not sure whether this is a sufficient condition, but it is necessary that the net force acting on the system must be conservative.

 

[smallphi]

My question is more mathematical than physical. Someone writes some differential equations for some quantities that describe some imaginary system. I need the mathematical conditions these equations have to satisfy to be derivable from action principle.

 

[Klaus_Hoffmann]

I am not sure whether this is a sufficient condition, but it is necessary that the net force acting on the system must be conservative.

Non-conservative forces such as 'friction' can be incorporated imposing several constraints to the system in this case you have a Lagrangian

$$\mathcal L =T(\dot x) -V(x)+\lambda U(\dot x , x ,t)$$

from this you can construct the Hamiltonian, of course the H-J equation with Constraints goes in a similar way to QM with constraints by Dirac so i guess you should keep a look at.

http://en.wikipedia.org/wiki/Second_class_constraints

 

[olgranpappy]

...it is necessary that the net force acting on the system must be conservative.

here's another obvious counterexample: electromagnetism.

 

[Klaus_Hoffmann]

I am not sure whether this is a sufficient condition, but it is necessary that the net force acting on the system must be conservative.

Non-conservative forces such as 'friction' can be incorporated imposing several constraints to the system in this case you have a Lagrangian

$$\mathcal L =T(\dot x) -V(x)+\lambda U(\dot x , x ,t)$$

from this you can construct the Hamiltonian, of course the H-J equation with Constraints goes in a similar way to QM with constraints by Dirac so i guess you should keep a look at.

http://en.wikipedia.org/wiki/Second_class_constraints

 

[lugita15]

here's another obvious counterexample: electromagnetism.

When you're dealing with electromagnetism, the net force is obviously conservative. Here's why:
The definition of a conservative vector field in 3 dimensions is that the curl of the field must be zero, or equivalently, that the field be the gradient of some scalar field. The electric field is conservative because it is the gradient of electric potential. Therefore, the electric force is also conservative since it is proportional to the electric field. An alternate definition of conservative force is that the work around a closed path must be zero. This definition is obviously satisfied by the magnetic force because the work done by the magnetic force along any path, whether open or closed, is 0.
It is for this reason that the net force for electromagnetism is conservative.

 

[olgranpappy]

When you're dealing with electromagnetism, the net force is obviously conservative.

wrong. you are talking about electrostatics.

 

[lugita15]

wrong. you are talking about electrostatics.

You're right, I was only talking about the case in which electric and magnetic fields are constant. However, if you consider an arbitrary closed path, then the work done along that path by the changing electric field always cancels exactly the work done along the path by the changing magnetic force. Therefore, the net electromagnetic force is conservative even when the fields are changing.

 

[olgranpappy]

still wrong

the work done along that path by the changing electric field always cancels exactly the work done along the path by the changing magnetic force.

...didn't you just say [quite correctly] that the magnetic force does no work? Anyways, you should stop trying to show that the electromagentic force is conservative because it simply can not be derived from a potential... hence it is not conservative. hence, as you said, I am right and you are still wrong. Let me help you out a bit; in electromagnetism the force is given by:
$$\vec F = q \left(\vec E + \vec v \times \vec B\right)$$
where q is charge and B is mag field and most importantly
$$\vec E = -\nabla \phi - \frac{\partial}{\partial t}\vec A\;.$$
In the above equation you see the 2nd term there with the vector potential $$\vec A$$ that is not zero in general... That term keeps the force from being conservative because there is no way to write it as the gradient of a scalar... because if there were then $$\nabla\times E$$ would be zero, which it is not.

 

[lugita15]

...didn't you just say [quite correctly] that the magnetic force does no work?

The work done by the force associated with a static magnetic field is 0. However, the work done by the force associated with a changing magnetic field is not necessarily 0. If you consider a closed path, this nonzero work cancels with the nonzero work produced by the nonconservative electric force. So the net electromagnetic force is still conservative.
Here's a simple example of a changing magnetic field doing nonzero work:
Consider two parallel current-carrying wires. There is an attractive magnetic force between them. For this reason, the will move closer together, and the magnetic field produced by each wire will do nonzero work on the other wire. The reason that work can be done is that as the wires move, the magnetic field produced by each wire constantly changes.

 

[olgranpappy]

no. that's just wrong.

 

[lugita15]

no. that's just wrong.

Why is it wrong? My reasoning seems to be perfectly valid.

 

[jackiefrost]

... Here's a simple example of a changing magnetic field doing nonzero work:
Consider two parallel current-carrying wires. There is an attractive magnetic force between them. For this reason, the will move closer together, and the magnetic field produced by each wire will do nonzero work on the other wire. The reason that work can be done is that as the wires move, the magnetic field produced by each wire constantly changes.

Consider a source of the emf causing the current in the conductors. Just for fun, let's say it's an ideal DC generator being driven from by some perfectly constant mechanical driver. Let the mechanical driver be coupled to the generator by an infinitely precise torque meter. Imagine we somehow keep the wires perfectly stationary in the presence of the attractive force, do you think that we should read some constant value on the torque meter (again, assume ideal conditions so we don't have to get crazy over irrelevant details). What would happen to the meter reading if the conductors were suddenly allowed to respond to the attraction?

Next, let's assume we use a high frequency generator for the source emf. Now we have some highly varying [electro]magnetic fields. How would the torque meter respond if we repeated the above experiment?

 

[lugita15]

Consider a source of the emf causing the current in the conductors. Just for fun, let's say it's an ideal DC generator being driven from by some perfectly constant mechanical driver. Let the mechanical driver be coupled to the generator by an infinitely precise torque meter. Imagine we somehow keep the wires perfectly stationary in the presence of the attractive force, do you think that we should read some constant value on the torque meter (again, assume ideal conditions so we don't have to get crazy over irrelevant details). What would happen to the meter reading if the conductors were suddenly allowed to respond to the attraction?

Next, let's assume we use a high frequency generator for the source emf. Now we have some highly varying [electro]magnetic fields. How would the torque meter respond if we repeated the above experiment?

I'm not sure what kind of situation you're imagining. I'm talking about two straight, infinitely long, parallel wires with currents flowing in the same direction.

 

[jackiefrost]

Please forgive me for being unclear. I'll take a different approach. It is my understanding that as the wires move toward each other (conductors' currents in same direction), each will be moving in an area of increasing flux density that surrounds the other wire. Then, each would experience an increasing change in flux which would induce an increasing emf in the direction that would oppose the wires' motion (Lenz's Law). Therefore, I don't see any connection between the changing magnetic field and any work done in moving the wires. The magnetic field deflects the moving charges' direction of motion in the conductor but it is the prime mover of the charges (the source of the potential driving the charges in the conductors) that ends up doing the work in moving the wires. The induced emf from the changing flux due to the wires motion ends up opposing that motion. As a result, I believe that even for changing magnetic fields, their influence will still always be at right angles to the motion of any particular charge and, therefore, will not do any work throughout the conductor. If I'm incorrect in this, I hope someone will show me.

 

[lugita15]

Please forgive me for being unclear. I'll take a different approach. It is my understanding that as the wires move toward each other (conductors' currents in same direction), each will be moving in an area of increasing flux density that surrounds the other wire. Then, each would experience an increasing change in flux which would induce an increasing emf in the direction that would oppose the wires' motion (Lenz's Law). Therefore, I don't see any connection between the changing magnetic field and any work done in moving the wires. The magnetic field deflects the moving charges' direction of motion in the conductor but it is the prime mover of the charges (the source of the potential driving the charges in the conductors) that ends up doing the work in moving the wires. The induced emf from the changing flux due to the wires motion ends up opposing that motion. As a result, I believe that even for changing magnetic fields, their influence will still always be at right angles to the motion of any particular charge and, therefore, will not do any work throughout the conductor. If I'm incorrect in this, I hope someone will show me.

Unfortunately, I'm still having trouble with your explanation. Faraday's Law (in integral form) states that the magnitude of the induced emf ALONG A CLOSED PATH is equal to the rate of change of magnetic flux through that CLOSED PATH. When you say that each wire "would experience an increasing change in flux," what CLOSED PATH are you considering?

 

[jackiefrost]

I keep overlooking the fact that your parallel conductors are infinite and never loop back to a potential source. I was picturing something like shown in the applet at http://www.magnet.fsu.edu/education/tutorials/java/parallelwires/index.html"

My bad...

 

[jackiefrost]

I have a couple questions maybe someone (anyone) could help me with. I'm kinda confused about the possibility of induction in this infinite parallel straight conductor scenario. Since we are talking conductors where there's no loop by which to perform the normal closed path integrals related to induction, as the wires move due to the Lorrentz force, each conductor is still moving through a region of changing B field density. So here's my questions:

1. This question is strictly about the Lorrentz force. Assume the wires are being attracted (conductor currents in the same direction). Does each conductor experience an increasing Lorrentz force as it moves closer to the other wire and thus their motion is accelerated? It seems to me that it would since at a smaller distance the other wire's B field magnitude is stronger than it is at a greater distance.

2. Even though we have no actual conductor loop in this hypothetical scenario, would there still be induction in each conductor as it moves through the other conductor's magnetic field - especially since that field strength varies as a function of the distance from its source? Or, owing to the geometry of the two wire's spatial relationship, there can be no difference in B field magnitude anywhere inside the conductor (except across its radial plane) at any instant in time, so no induction can take place?

3. If induction is possible, would the induced emf be in the direction that would tend to oppose the wire's motion? i.e. buck the other wire's current?

 

[lugita15]

I have a couple questions maybe someone (anyone) could help me with. I'm kinda confused about the possibility of induction in this infinite parallel straight conductor scenario. Since we are talking conductors where there's no loop by which to perform the normal closed path integrals related to induction, as the wires move due to the Lorrentz force, each conductor is still moving through a region of changing B field density. So here's my questions:

1. This question is strictly about the Lorrentz force. Assume the wires are being attracted (conductor currents in the same direction). Does each conductor experience an increasing Lorrentz force as it moves closer to the other wire and thus their motion is accelerated? It seems to me that it would since at a smaller distance the other wire's B field magnitude is stronger than it is at a greater distance.

I could be mistaken, but this is my reasoning:
F is proportional to Bd, where B is the magnetic force and d is the distance between the wires. However, B is proportional to 1/d. Therefore, F should be constant as the wires move closer together, however counterintuitive this may seem. And it is highly counterintuitive; how can two tiny wires a thousand miles away from each other exert the same force on one another as if they were two inches from each other?

 

[lugita15]

2. Even though we have no actual conductor loop in this hypothetical scenario, would there still be induction in each conductor as it moves through the other conductor's magnetic field - especially since that field strength varies as a function of the distance from its source? Or, owing to the geometry of the two wire's spatial relationship, there can be no difference in B field magnitude anywhere inside the conductor (except across its radial plane) at any instant in time, so no induction can take place?

I don't believe there'd be any induced current in the wires.

 

[jdstokes]

To make things even more confusing, it turns out that even though the Lorentz force is in general nonconservative (it cannot be written as the gradient of a potential), the Lorentz force can still be derived from a Lagrangian, see my thread

https://www.physicsforums.com/showthread.php?t=176428

 

[lugita15]

To make things even more confusing, it turns out that even though the Lorentz force is in general nonconservative (it cannot be written as the gradient of a potential), the Lorentz force can still be derived from a Lagrangian, see my thread

https://www.physicsforums.com/showthread.php?t=176428

In that thread, you claimed that the lagrangian for the electromagnetic force was $L = (1/2)m \dot{\mathbf{r}} \cdot \dot{\mathbf{r}} + e V(\mathbf{r},t) + e \dot{\mathbf{r}}\cdot \mathbf{A}(\mathbf{r},t)$
If this is correct (I haven't verified it myself), then isn't the electromagnetic force conservative, being the gradient of the difference between the lagrangian and the kinetic energy?
$F=\nabla(e V(\mathbf{r},t) + e \dot{\mathbf{r}}\cdot \mathbf{A}(\mathbf{r},t))$

 

[olgranpappy]

In that thread, you claimed that the lagrangian for the electromagnetic force was $L = (1/2)m \dot{\mathbf{r}} \cdot \dot{\mathbf{r}} + e V(\mathbf{r},t) + e \dot{\mathbf{r}}\cdot \mathbf{A}(\mathbf{r},t)$
If this is correct (I haven't verified it myself), then isn't the electromagnetic force conservative, being the gradient of the difference between the lagrangian and the kinetic energy?
$F=\nabla(e V(\mathbf{r},t) + e \dot{\mathbf{r}}\cdot \mathbf{A}(\mathbf{r},t))$

Jesus-christ-on-the-f-ing-cross, how many times do I have to tell you that you are wrong. That gradient does *not* give you the Lorentz force and electromagnetism is *not* conservative. You obviously do not understand the Lagrangian formulation of dynamics.

 

[jdstokes]

In that thread, you claimed that the lagrangian for the electromagnetic force was $L = (1/2)m \dot{\mathbf{r}} \cdot \dot{\mathbf{r}} + e V(\mathbf{r},t) + e \dot{\mathbf{r}}\cdot \mathbf{A}(\mathbf{r},t)$
If this is correct (I haven't verified it myself), then isn't the electromagnetic force conservative, being the gradient of the difference between the lagrangian and the kinetic energy?
$F=\nabla(e V(\mathbf{r},t) + e \dot{\mathbf{r}}\cdot \mathbf{A}(\mathbf{r},t))$

This Lagrangian is indeed correct, you can find it in Goldstein if you don't believe me :)

No, the electromagnetic (Lorentz) force is not always conservative. It is given by $m\ddot{\mathbf{r}} = e[\mathbf{E} + \dot{\mathbf{r}}\times \mathbf{B}]$ which you can verify by plugging the Lagrangian into the E-L equations.

If you take a closed line integral along some current loop, you get $e\left(\oint \mathbf{E} \cdot d\mathbf{l} + \oint \dot{\mathbf{r}}\times \mathbf{B} \cdot d\mathbf{l} \right)$. The second integral inside the brackets must vanish by perpendicularity of the magnetic force to the velocity field (which is tangential to the path of integration). The first term is nonzero whenever there is a changing magnetic field (nonzero curl). In either case the Lagrangian works fine.

James

 

[jdstokes]

That was discussed before but the thread died inconclusively.

Anybody knows of necessary or sufficient conditions the differential equations of motion of a given system have to satisfy to be derivable by variations of some action?

To answer the OPs question. The necessary and sufficient condition that the equation of motion be obtainable from a Lagrangian is given in Goldstein p. 19 as pointed out to me by pervect on my other thread. It holds as long as there exists a function U such that the generalized forces satisfy

$Q_j = - \frac{\partial U}{\partial q_j} + \frac{d}{dt}\frac{\partial U}{\partial \dot{q}_j}$

 

[smallphi]

That is not going to work. You are given only a bunch of differential equations, not Lagrangian nor generalized forces. What I am asking for is a mathematical test on those equations showing if they are derivable from action/Lagrangian without knowing the Lagrangian.

 

[jdstokes]

By ``bunch of differential equations'' I assume you mean ordinary differential equations of at most second order, otherwise the Lagrangian is meaningless. The necessary and sufficient conditions you seek can be found in P. S. Bauer, Proc. Natl. Acad. Sci. USA 17, 311 (1931).

Basically what Bauer does is to consider conservative and nonconservative systems. For the former, the necessary and sufficient condition is that all of the coefficients of the second order terms (physically the masses) are constant. For the latter case, the condition is that dissipation coefficients (multiplying the first order term) are identically equal to the rates of change of the corresponding masses.

 

[jdstokes]

By the way, the condition I gave you before also works fine. Since you are only dealing with second order ODEs in order for the Lagrangian formalism to make sense, you simple rearrange your system of ODEs so that the second order term is isolated onto one side, then treat the opposite side as your effective force. If the condition is satisfied then the variational approach is valid.

 

[smallphi]

Now we have some progress :) The Bauer article discusses systems of point particles, I was more thinking of fields with PDE's instead of ODE's. The article helped me though find the name of the problem:

inverse problem of calculus of variations
inverse problem of variational calculus
inverse variational problem

Google returns many hits with those. In one paper I found reference to 'Helmholtz theorem for inverse variational problem' that states something about self-adjointness of the differential operator of the equations in order for the Lagrangian to exist. I'll check into that later.

 

[lugita15]

Jesus-christ-on-the-f-ing-cross, how many times do I have to tell you that you are wrong. That gradient does *not* give you the Lorentz force and electromagnetism is *not* conservative. You obviously do not understand the Lagrangian formulation of dynamics.

You keep saying that I'm wrong, but you're not giving me any actualproof that the net electromagnetic force is nonconservative.

 

[olgranpappy]

You keep saying that I'm wrong, but you're not giving me any actualproof [sic] that the net electromagnetic force is nonconservative.

Hmm... Miss Manners told me that if I don't have anything nice to say then I shouldn't say anything--I'm just going to quote jdstokes (post 25, I believe) in which he has already answered your question.

No, the electromagnetic (Lorentz) force is not always conservative. It is given by $m\ddot{\mathbf{r}} = e[\mathbf{E} + \dot{\mathbf{r}}\times \mathbf{B}]$ which you can verify by plugging the Lagrangian into the E-L equations.

If you take a closed line integral along some current loop, you get $e\left(\oint \mathbf{E} \cdot d\mathbf{l} + \oint \dot{\mathbf{r}}\times \mathbf{B} \cdot d\mathbf{l} \right)$. The second integral inside the brackets must vanish by perpendicularity of the magnetic force to the velocity field (which is tangential to the path of integration). The first term is nonzero whenever there is a changing magnetic field (nonzero curl). In either case the Lagrangian works fine.

James

 

[lugita15]

This Lagrangian is indeed correct, you can find it in Goldstein if you don't believe me :)

No, the electromagnetic (Lorentz) force is not always conservative. It is given by $m\ddot{\mathbf{r}} = e[\mathbf{E} + \dot{\mathbf{r}}\times \mathbf{B}]$ which you can verify by plugging the Lagrangian into the E-L equations.

If you take a closed line integral along some current loop, you get $e\left(\oint \mathbf{E} \cdot d\mathbf{l} + \oint \dot{\mathbf{r}}\times \mathbf{B} \cdot d\mathbf{l} \right)$. The second integral inside the brackets must vanish by perpendicularity of the magnetic force to the velocity field (which is tangential to the path of integration). The first term is nonzero whenever there is a changing magnetic field (nonzero curl). In either case the Lagrangian works fine.

James

Thank you for your proof, now I understand why the Lorentz force is nonconservative. However, this puzzles me: I had always that that at a fundamental level, all forces were conservative, and they only appeared nonconservative when degrees of freedom were ignored. For instance, friction seems to be nonconservative, but that is only because the motion of the individual molecules is not taken into account. So how can the electromagnetic force be truly nonconservative?

 

[olgranpappy]

Thank you for your proof, now I understand why the Lorentz force is nonconservative. However, this puzzles me: I had always that that at a fundamental level, all forces were conservative, and they only appeared nonconservative when degrees of freedom were ignored. For instance, friction seems to be nonconservative, but that is only because the motion of the individual molecules is not taken into account. So how can the electromagnetic force be truly nonconservative?

The integrals that jdstokes presented are equal to the work done as a charge traverses a closed path in space... but why would a particle ever do such a thing? In general such a closed path is not the same as the path a particle would travel along if it moved only under the forces of the electric fields. Some external agent has to grab the particle and move it along the path! If the net work done by the field is positive then the net work done by the external agent is negative; if the net work done by the field is negative then the net work done by the external agent is positive.

The total mechanical and electromagnetic energy of a closed system will remain constant even though the lorentz force (as we have just seen) is not a conservative force. Just like how if you take into account the heat generated by a friction force you can retain conservation of energy.

...if there is radiation it is hard to "close" the system, but the energy that departs from the system through its boundries can be taken into account as well via the Poynting vector.

Cheers.

 

[lugita15]

The total mechanical and electromagnetic energy of a closed system will remain constant even though the lorentz force (as we have just seen) is not a conservative force. Just like how if you take into account the heat generated by a friction force you can retain conservation of energy.

Of course there is always conservation of energy. I wasn't talking about conservation of energy. I was talking about the fact that fundamentally, there is no such thing as a nonconservative force, and forces seem nonconservative only when degrees of freedom are neglected.
As it says in the Feynman Lectures on Physics:
"We have spent a considerable time discussing conservative forces; what about nonconservative forces? We shall take a deeper view of this than is usual, and state that there are no nonconservative forces! As a matter of fact, all the fundamental forces in nature appear to be conservative. This is not a consequence of Newton's laws. In fact, so far as Newton himself knew, the forces could be nonconservative, as friction apparently is. When we say friction apparently is, we are taking a modern view, in which it has been discovered that all the deep forces, the forces between particles at the most fundamental level, are conservative."
Also, it says in a Wikipedia article on force, "However, for any sufficiently detailed description, all forces are conservative."
In a Wikipedia article on conservative forces, it says:
"Nonconservative forces arise due to neglected degrees of freedom. For instance, friction may be treated without resorting to the use of nonconservative forces by considering the motion of individual molecules; however that means every molecule's motion must be considered rather than handling it through statistical methods. For macroscopic systems the nonconservative approximation is far easier to deal with than millions of degrees of freedom."
This is why I am so surprised that the electromagnetic force is nonconservative, as it seems to contradict these quotes.