Why are central force fields irrotational and conservative?

  • #1
sams
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In Mathematical Methods for Physicists, 6th Edition, page 44, Example 1.8.2, the curl of the central force field is zero.

1. Why are central force fields irrotational?
2. Why are central force fields conservative?

Any help is much appreciated...
 

Answers and Replies

  • #2
dRic2
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Because the central force field is a function of only ## \mathbf r## (position vector). Try to evaluate the curl of ## f(\phi, \theta, \mathbf r) = f(\mathbf r)##
 
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  • #3
sams
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In case f is in function of ϕ, θ, and r, do we consider f to be solenoidal and thus not a conservative force field?
 
  • #4
dRic2
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Why would you say so? If ##f## is a generic function then you can't say much about it.
 
  • #5
bobob
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Why would you say so? If ##f## is a generic function then you can't say much about it.
It is not a generic function. It's the gradient of a scalar. That and vector identities are all you need.
 
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ZapperZ
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In case f is in function of ϕ, θ, and r, do we consider f to be solenoidal and thus not a conservative force field?
Why can't you just take the curl of the function as @dRic2 has suggested? This is not something ambiguous. You should be able to tell mathematically if a function is conservative or not.

Zz.
 
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  • #7
sams
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Thank you all for your replies and I am sorry for this confusion. In fact, I am not asking from a mathematical perspective. The mathematics is straightforward. In my questions above, I just would like to know why rotational fields are non-conservative. Is there any physical meaning or physical interpretation? This leads to the questions raised above: why are central force fields irrotational and conservative? In other words, why also irrotational fields are conservative?

Thank you once again for your kind efforts...
 
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ZapperZ
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Thank you all for your replies and I am sorry for this confusion. In fact, I am not asking from a mathematical perspective. The mathematics is straightforward. In my questions above, I just would like to know why rotational fields are non-conservative. Is there any physical meaning or physical interpretation? This leads to the questions raised above: why are central force fields irrotational and conservative? In other words, why also irrotational fields are conservative?

Thank you once again for your kind efforts...
I'm a bit puzzled by this.

A field having a "rotational"component, by definition, will have a "rotational impulse" (I made up that phrase), and so, there is something "gained" along a rotational path. Something that has no rotational component, shouldn't have such an impulse and will not provide any type of gain along that path.

So it is conservative or non-conservative by definition. Your question is like asking why a straight line has no curvature.

Zz.
 
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  • #9
sams
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Something that has no rotational component, shouldn't have such an impulse and will not provide any type of gain along that path.
Can't a linear field be also non-conservative? For example, a time-varying magnetic field!

Your question is like asking why a straight line has no curvature.
So again, is it always the case of a straight line that results in a conservative field? Are there other factors that affects conservative fields?
 
  • #10
ZapperZ
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Can't a linear field be also non-conservative? For example, a time-varying magnetic field!
First of all, I didn't see any indication that we're talking about time-varying fields. Secondly, a magnetic field is already non-conservative in the static case.

So again, is it always the case of a straight line that results in a conservative field? Are there other factors that affects conservative fields?
Your question then is: is there any mathematical function, whose curl is zero, that ........ So this IS a mathematical issue, contrary to what you stated earlier.

Zz.
 
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  • #11
Charles Link
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If ## \nabla \times E \neq 0 ## anywhere in a region, that means by Stokes theorem, ## \\ ## (Note: Stokes theorem says## \int (\nabla \times E) \cdot \hat{n} \, dA=\oint E \cdot dl ##),## \\ ## ## \oint E \cdot dl \neq 0 ## for at least one path that this integral might take where, in integrating over the loop, the finish point is the starting point. This means that there is non-zero work over this loop. Thereby, the force is non-conservative if ## \nabla \times E \neq 0 ##.## \\ ## If ## \nabla \times E =0 ## everywhere, by similar arguments we see we have a conservative force.
 
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