Is the Lorentz force conservative?

[Dash-IQ]

When a wire has current I in a magnetic field B, there is the Lorentz force is it considered a conservative force or not? Please do explain as to why it is.

 

[UltrafastPED]

The force on the wire is given here: http://en.wikipedia.org/wiki/Lorentz_force#Force_on_a_current-carrying_wire

If the electric field is static then it's curl is zero - and the electric field is conservative.

But when we consider the magnetic field Maxwell's equations tell us that it's divergence is always zero -so if it's curl were also zero we would have B=0. Thus the magnetic field is not conservative, and the exact situation shouldn't matter.

 

[Dash-IQ]

The force on the wire is given here: http://en.wikipedia.org/wiki/Lorentz_force#Force_on_a_current-carrying_wire

If the electric field is static then it's curl is zero - and the electric field is conservative.

But when we consider the magnetic field Maxwell's equations tell us that it's divergence is always zero -so if it's curl were also zero we would have B=0. Thus the magnetic field is not conservative, and the exact situation shouldn't matter.

If the magnetic field is not conservative, yet the electric field is... the Lorentz force shouldn't be conservative? Why does it not matter?

 

[UltrafastPED]

It doesn't matter what the situation is: the magnetic field is not conservative is what I meant to say; the electric force is conservative only when static.

 

[Dash-IQ]

Ah, so all in examples the demonstrate the Lorentz force are always nonconservative?

 

[Robert_G]

No, It's not, there is an easy way to see if a force is conservative, or not. If the force is only depends on the position, or in other word.
$\mathbf{F}=\mathbf{f(\mathbf{r})}$,
then, is conservative. because, only if the force only depends on $\mathbf{r}$, A plane with the equal potential energy can be introduced.

now for the Lorentz force, this force actually has something to do with the velocity of the charged particles. Different velocity means Different force, that plane will never be introduced. So it's not conservative.

 

[DrStupid]

When a wire has current I in a magnetic field B, there is the Lorentz force is it considered a conservative force or not? Please do explain as to why it is.

Lorentz force is conservative because the work done between two points is independent from the path.

 

[Dash-IQ]

now for the Lorentz force, this force actually has something to do with the velocity of the charged particles. Different velocity means Different force, that plane will never be introduced. So it's not conservative.

What about the case of the wire? F = IL x B?
Different current values ?

 

[Dash-IQ]

Lorentz force is conservative because the work done between two points is independent from the path.

Hm, what about the statements of the rest?

 

[DrStupid]

Hm, what about the statements of the rest?

I can't speak for the rest but for myself only. I just checked if Lorentz force meets the condition for a conservative force and due to
$d\vec E = \vec F \cdot d\vec s = q \cdot \left( {\vec v \times \vec B} \right) \cdot \vec v \cdot dt = 0$
it does.

 

[UltrafastPED]

Read #2 again. What is your conclusion?

 

[DrStupid]

Read #2 again. What is your conclusion?

I conclude that we need to distinguish between Lorentz force and magnetic field. As Lorentz force has no force field the corresponding formalisms does not apply. Thus there is only one condition left: conservative forces conserve mechanical energy.

 

[BruceW]

yeah... I think there are several possible definitions of a conservative force. 1) does work done by the force on a particle depend on the path, or just the end points? 2) Does the force depend only on the position of the particle?

These two ways to define conservative force are similar, but not the same, so the answer will depend on which definition you choose. Also, why can't the Lorentz force be a force field?

edit: ah, OK, definition 2) defines a force field. I just looked this up, since I didn't know the standard definition of a force field.

 

[Dash-IQ]

I conclude that we need to distinguish between Lorentz force and magnetic field. As Lorentz force has no force field the corresponding formalisms does not apply. Thus there is only one condition left: conservative forces conserve mechanical energy.

Im confused, potential energy can't be predicted here? How can it still be conservative?

 

[BruceW]

It looks like there's more than one possible definition for 'conservative force'. One possible definition is that there must be an associated potential. But another definition (as DrStupid is saying) would be that the work done on the particle does not depend on the path, but only on the endpoints. Anyway, look at the equation for Lorentz force:
$$\vec{F} = q \vec{E} + q \vec{v} \wedge \vec{B}$$
The bit due to the electric field is 'nice', and the bit due to the magnetic field is not so nice. But what happens when we integrate this force, over the path of the particle? What happens to the term due to the magnetic field?

 

[Dash-IQ]

But what happens when we integrate this force, over the path of the particle? What happens to the term due to the magnetic field?

No idea...

 

[DrStupid]

But what happens when we integrate this force, over the path of the particle? What happens to the term due to the magnetic field?

See my equation above. The integral of the magnetic term is always zero.

 

[Meir Achuz]

The physical meaning of a force being conservative is that any work done by the force can be retrieved.
The Lorentz force on the current in a wire does no work. Since energy conservation is a statement about the work done by a force, the concept of conservative force is not relevant to the Lorentz force, except in the trivial sense that 0 = 0.

 

[Meir Achuz]

If the force is only depends on the position, or in other words.
$\mathbf{F}=\mathbf{f(\mathbf{r})}$,
then, is conservative. because, only if the force only depends on $\mathbf{r}$, A plane with the equal potential energy can be introduced.

That is wrong.

 

[Meir Achuz]

The force on the wire is given here: http://en.wikipedia.org/wiki/Lorentz_force#Force_on_a_current-carrying_wire

If the electric field is static then it's curl is zero - and the electric field is conservative.

But when we consider the magnetic field Maxwell's equations tell us that it's divergence is always zero -so if it's curl were also zero we would have B=0. Thus the magnetic field is not conservative, and the exact situation shouldn't matter.

The Coulomb force has zero divergence and curl.

 

[BruceW]

That is wrong.

why is that wrong? It is a different definition of conservative force. But I've seen more than one definition used. For example, on the wikipedia page, they seem to use at least two different definitions.

 

[Khashishi]

All fundamental forces are conservative. The Lorentz force is just the electromagnetic force, which is conservative.

 

[DrStupid]

why is that wrong?

Counterexample: F = [x-z,y,0]

 

[Meir Achuz]

To clarify my post #18:

I was referring to only the magnetic part of the Lorentz force. Of course the electric part is conservative.

Another case is the magnetic force on a contained current distribution, such as a current loop. That can be described in terms of the magnetic moment of the current distribution as
$${\bf F}=\nabla(\mu\cdot{\bf B})$$. This is a conservative force, but I would not call it the 'Lorentz force'.

 

[Meir Achuz]

No, It's not, there is an easy way to see if a force is conservative, or not. If the force is only depends on the position, or in other word.
$\mathbf{F}=\mathbf{f(\mathbf{r})}$,
then, is conservative. because, only if the force only depends on $\mathbf{r}$, A plane with the equal potential energy can be introduced.

now for the Lorentz force, this force actually has something to do with the velocity of the charged particles. Different velocity means Different force, that plane will never be introduced. So it's not conservative.

It is the two onlys that make that statement wrong. The equation given is for a conservative force, but there are many other examples of conservative forces. One example, among many, is the force in my previous post.

 

[Meir Achuz]

All fundamental forces are conservative. The Lorentz force is just the electromagnetic force, which is conservative.

Dissipative forces like ${\bf F}=-k({\bf v\cdot r)r}$ are not conservative.

 

[BruceW]

Counterexample: F = [x-z,y,0]

ah right. The force field would also need to have zero curl, to be able to write it as a gradient of a potential energy.

 

[Dash-IQ]

@Meir , the force moving the wire is due to the electric field? If so, is it due to the change in magnetic field's flux? And if so... Why would it be conservative?

 

[Meir Achuz]

The force moving a closed current loop is due to an inhomogeneous magnetic field B. there need not be a change in flux through the loop, although the magnetic flux will change if the loop starts to move. Any force that can be written as the gradient of the scalar is conservative

 

[mattt]

This mathematical theorem is useful:

Let $$U\subset\mathbb{R}^3$$ be an open connected set, $$\vec{F}: U\subset\mathbb{R}^3\to\mathbb{R}^3$$ (a continuous vector field), then the following three statements are equivalents:

i) For any two points A,B in U, and any (continuous) curve C connecting A to B, $$\int_C\vec{F}\cdot\vec{dr}$$ does not depend on the concrete (continuous) path (curve) connecting A to B.ii) For any close path C (included in U), $$\int_C\vec{F}\cdot\vec{dr} = 0$$iii) There exists a $$C^1(U)$$ function $$V:U\subset\mathbb{R}^3\to\mathbb{R}$$ (a scalar field) such that:

$$\vec{F}(x,y,z) = (\frac{\partial V}{\partial x}(x,y,z), \frac{\partial V}{\partial y}(x,y,z), \frac{\partial V}{\partial z}(x,y,z))$$

(i.e. $$\vec{F}=\nabla V$$)If a vector field satisfy one (and hence all three) of the previous conditions, it is called "conservative".

If a vector field is conservative (and C^1), then rot F = 0 in region U.

If rot F = 0 ( in the region U ) AND U is not only an open connected region, but also simply-connected, then F is conservative.

 

[vanhees71]

This is a bit subtle, and I think there are already many answers to your question. So here are my 2cts in addition.

Usually in classical mechanics an external force is called "conservative", if it only depends on the position of the particle and if it has a scalar potential, i.e., if it is of the form
$$\vec{F}(\vec{x})=-\vec{\nabla} U(\vec{x}).$$

An interaction force between two particles is called conservative, if there exists a potential $U(\vec{x}_1-\vec{x}_2)$ such that
$$\vec{F}_{12}=-\vec{\nabla}_1 U=-\vec{\nabla}_2 U=-F_{21}.$$
The specific dependence on the difference of the position vectors of the two particles is demanded by Newton's 3rd Law ("actio=reactio").

Other posters have already stated the conditions on the force/interaction field(s) that guarantee the existence of a potential and thus the criteria for a force/interaction being conservative.

Examples are the motion of a charged particle in an electrostatic field or the mutual gravitative attraction between bodies (in Newtonian approximation).

This is a pretty limited view on forces, however. E.g., there are nearly no relativistically covariant forces. An exception is the motion in an electrostatic field. Here, the equation of motion is governed by the more general Lorentz force, including the interaction with the magnetic field, but this force is not conservative in the above given sense, because the force depends on the velocity of the particle and not only on its position.

Nevertheless the conservation of energy holds for static (and only static!) electromagnetic fields, as can be easily proven by taking the line integral of the force along the particle's trajectory under the influence of the Lorentz force. It turns out that only the electric field does work on the particle while the magnetic field only changes the direction of the particle's trajectory. Indeed the power is
$$P=\vec{v} \cdot \vec{F}=\vec{v} \cdot q \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right )=q \vec{v} \cdot \vec{E}.$$
The total energy of the particle is given by
$$\mathcal{E}=\frac{m c^2}{\sqrt{1-\vec{v}^2/c^2}}+q \Phi,$$
where $\Phi$ is the potential of the electrostatic field,
$$\vec{E}=-\vec{\nabla} \Phi.$$