# Talk:Conservative force

Jump to navigation Jump to search

## Force field vs. Force

Isn't it the force field, rather than the force itself, that is conservative? Michael Hardy 20:55, 29 May 2004 (UTC)

Certain forces (in the sense of phenomena, not vectors; e.g. gravity) always generate conservative force fields, and others (like the obvious friction) never do (and in fact could be argued not to generate force fields at all). --Tardis 04:52, 31 October 2006 (UTC)

I removed magnetism as a nonconservative force; the cyclotron frequency (neglecting cyclotron radiation) seems to counterindicate it. Just because ${\displaystyle \nabla \times \mathbf {B} \not =0}$ doesn't mean that the force is nonconservative! (Of course, time-changing magnetic fields can impart momentum, but that's separate.) Just drawing attention to this edit in case I'm crazy. --Tardis 04:52, 31 October 2006 (UTC)

## My edit

I tried to make the page a bit more accessible to the general public, to start with. I'll try to create an image, but I'm not good at that so I'll hope someone improve on me :) Considering the above discussion, I inserted the magnetic force again, but added a remark about time-independency of the electric field (Maxwell says: rot B = - dE/dt). Hope this will give people a nudge to start editing. --CompuChip 16:39, 5 December 2006 (UTC)

And I apologize for the apostrophe abuse... Should have known better --CompuChip 18:22, 15 December 2006 (UTC)

## Edit

CompuChip, you are mistaken here : the magnetic force conserves energy even in time-dependent electric fields, not because of Maxwell but because of the Lorentz Force, which is what matters here (we are talking of forces, not fields). F = q V^B, thus P = F.V = q(V^B).V = 0, and the energy is conserved in every path. Besides, the Maxwell equation you cite is false, it's rotE = -dB/dt : the electric force is conservative in time-independant magnetic fields. I'm editing this, as well as other examples ("it is known from experiment" : the experiment has no place in this, it follows from the law of gravity). I'm also removing the proof of path independence, which is elementary and can be simply described instead of a "heavy" mathematical formulation. It would be nice to speak of potential energy in this article, i dunno how to do it without being too specific. Also, english is not my primary language, so feel free to correct me if i have done any mistake. Smeuuh 21:44, 5 August 2007 (UTC)

Any given force doesn't necessarily conserve energy in itself in time dependent fields- however conservation of energy still applies- the whole system has to conserve energy.WolfKeeper 20:09, 19 September 2007 (UTC)

## Magnetic field

I see there's been discussion of whether the magnetic force counts as conservative or not. I put in a paragraph describing which definitions of conservative force do and do not admit the magnetic force, and also found a reference for and reference against (although neither is particularly notable...these were the first two I found on Google Book search). It would be nice if more and better references could be found; it's possible that my phrasing "some authors classify the magnetic force as conservative, while others do not" should be modified to "most...a few..." or "a few...most...". :-) --Steve (talk) 16:14, 30 June 2008 (UTC)

## Magnetic field 2

If the three conditions for a conservative field are sufficient and equivalent, the magnetic field cannot satisfy condition 2 but not conditions 1 and 3. This is stated in the Mathematical Description section. —Preceding unsigned comment added by 68.34.215.33 (talk) 04:30, 25 July 2008 (UTC)

The Mathematical Description section also explains that they're only equivalent for force fields, and that the magnetic force is not a force field. Did you miss that part, or are you disputing its accuracy? If the former, I guess it could be made clearer. If the latter, exactly what aspect are you disputing? --Steve (talk) 05:57, 25 July 2008 (UTC)

## a problem with zero curl

The page says that a force is conservative iff ${\displaystyle \nabla \times \mathbf {F} =0}$, but I think there's a problem with that. This condition is equivalent to condition 3 only if the curl is defined everywhere. But requiring that F be differentiable everywhere excludes a lot of conservative fields we would like to include. Consider the following two examples:

${\displaystyle \mathbf {F} (\mathbf {r} )={\frac {x\mathbf {i} +y\mathbf {j} }{x^{2}+y^{2}}},\quad \mathbf {G} (\mathbf {r} )={\frac {x\mathbf {j} -y\mathbf {i} }{x^{2}+y^{2}}}}$

with F=G=0 along the line x=y=0. Both F and G have zero curl outside the line x=y=0, and both F and G have undefined curl on the line since they aren't differentiable there. F represents the electric field of a infinite wire with a constant linear charge density, which is a conservative force. G is clearly not conservative. Rckrone (talk) 22:30, 30 July 2009 (UTC)

Actually, Conservative_vector_field#Irrotational_vector_fields does a better job explaining the issue. Rckrone (talk) 22:37, 30 July 2009 (UTC)

A more physics-y and less mathematical person would say that you can compute the curl of F and G everywhere...just because something has a singularity doesn't mean you can't compute its curl, as long as you know how to deal with delta functions and related things. Anyway, F truly does have zero curl everywhere, including the z-axis. The curl of G is some sort of two-dimensional delta-function times k (I haven't bothered to compute the prefactors), so that it's infinite along the z-axis. :-) --Steve (talk) 23:10, 30 July 2009 (UTC)
I guess it would have to be 2π by Stoke's Theorem. Thanks for the explanation. That answer makes me a little sad though. :-( Rckrone (talk) 23:33, 30 July 2009 (UTC)

### again

I was about to point out the same thing. There is an example at Conservative_vector_field#Irrotational_vector_fields of a vector field defined on the plane minus the origin that has zero curl everywhere it is defined but which does not have path-independence and is not the gradient field of any function on its domain. This article should have some stipulation that the domain of the vector field is simply connected, or a similar condition, in order to claim that having curl 0 makes a field conservative. — Carl (CBM · talk) 19:24, 2 March 2010 (UTC)

I added this stipulation. Is it OK now? --Steve (talk) 23:13, 2 March 2010 (UTC)
Adding the hypothesis makes the proof work. I am not sure how this is usually handled in physics; maybe there is usually an assumption that the field is defined on all of R^3. — Carl (CBM · talk) 02:33, 3 March 2010 (UTC)
I wasn't the one who wrote the proof originally, but I've edited it a few times. Yes, I thought it went without saying that a vector field meant a vector field in R3. Anyway it sure never hurts to be clear and explicit, so thanks for the suggestion! :-) --Steve (talk) 05:03, 3 March 2010 (UTC)
I can believe that would be a reasonable assumption. Although, I vaguely remember hearing someone claim that there models of relativity in which the universe is not simply connected. But I am a mathematician, not a physicist, and so I often run into things that physicists take for granted that a mathematician would note explicitly. — Carl (CBM · talk) 14:06, 3 March 2010 (UTC)

## Conservative forces can cause rotation

It's misleading or at least confusing to say "a conservative force is one that does not cause rotation (i.e. it is irrotational) in a object within that force field. This does not mean the object does not or can not rotate, merely that no rotation is caused by the force." Here are some examples:

• A stationary pendulum that's oriented horizontally will start rotating (in an oscillatory way) because of gravity (a conservative force).
• Magnetic forces, which many people define to be conservative, are the basis for motors--making the wheels in an electric car spin. If that's not "causing rotation" I don't know what is!!
• An electric force exerts a torque that causes electric dipoles to start rotating.
• The force of one toothed gear on another, causing the second to rotate, is conservative. (Of course there is dissipation in the real world, but the gears would work just fine if there was no friction losses, just normal force).
• An object in an inhomogeneous gravity field can start rotating as it falls.

Also, whether a force is applied homogeneously to all atoms or inhomogeneously to surface atoms is not really related to whether it is conservative. For example, an electric force on a charged metal object is only applied to less than a nanometer-thick layer on the outside surface, which is where all the excess charge resides. But electric force is still conservative.

So I deleted that addition, and replaced it with something more about where the energy goes. --Steve (talk) 14:30, 2 October 2011 (UTC)