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I have just been introduced to the concept of central forces, and to the fact that they are per definition conservative forces. I have looked up several examples of central forces (gravity, electric, and spring), but they cover just about all the conservative forces I have ever heard about. Are there any conservative forces that are not central?

There must be, because otherwise there would not be any point in having a subcategory for central forces, yet I cannot find any examples anywhere.

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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/213061/2451 , physics.stackexchange.com/q/38874/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Aug 22, 2018 at 17:29
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    $\begingroup$ You need to be careful about the phrase "per/by definition". In colloquial English we usually use this phrase to mean "it logically follows from the definition", but in math this usage isn't useful because literally everything "logically follows from the definition(s)". So in math and physics we usually use that phrase to mean "it is a necessary part of the definition." Under this usage, (spherically symmetric) central forces are conservative, but they are not conservative "per definition". $\endgroup$
    – tparker
    Commented Aug 22, 2018 at 17:38
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    $\begingroup$ @tparker, in mathematics one should distinguish between "by definition/axiom" and "by theorem". Although everything follows from the basic set of axioms theorems are not set to be true intentionally. In natural sciences, however, the phrase has even greater importance and different sense, as there are postulates and observations, to be distinguished. $\endgroup$
    – user168013
    Commented Aug 22, 2018 at 23:58

4 Answers 4

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If $\phi=-xy$, and ${F}=-\nabla \phi=y \hat{i}+x \hat{j}$ is conservative but not central

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A constant force is conservative but not central.

For example: $\vec F=F \hat x$

You can check that the curl of this force is $0$, hence it is conservative. Its potential energy function in 3D space would just be $V(x,y,z)=-Fx+V_0$, where $V_0$ is some constant value.

An example of this is the approximation of gravity near the Earth's surface. In this regime the force is assumed to be constant, and we get the same form as above for the potential energy. Of course gravity on larger scales is a central force for, say, planets in orbit around a central star, which is why I gave the general form first.

Another example of a conservative, non-central force is one that is a superposition (sum) of two central forces.

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    $\begingroup$ This is a limit of central forces though where the center goes to infinity, so it's in the closure of the set of nonexamples :) $\endgroup$ Commented Aug 22, 2018 at 17:30
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    $\begingroup$ If the OP has "just been introduced to the concept of Central Forces" then they probably don't know what the curl is $\endgroup$
    – tparker
    Commented Aug 22, 2018 at 17:30
  • $\begingroup$ @RyanThorngren just because a constant force is the limit of a central force where the center is at infinity doesn't mean a constant force is a non-example. You can have an actual constant force that is not this limit. $\endgroup$ Commented Aug 22, 2018 at 17:52
  • $\begingroup$ @tparker I have posted answers at simpler levels when the OP is at a simpler level and gotten strong responses from others here that I should go into more detail when possible. I also show that there is a potential energy function for someone who doesn't know about the curl of a vector field. $\endgroup$ Commented Aug 22, 2018 at 17:54
  • $\begingroup$ @tparker example in comments of the question physics.stackexchange.com/questions/419626/… $\endgroup$ Commented Aug 22, 2018 at 17:58
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Paul's answer is great. But I just found out an error in the background information you mentioned: central forces aren't necessarily conservative forces. I'm writing it down so you may have a clearer understanding of the logic relationship between a 'central' force and a 'conservative' force.
For example we may take $$\vec F = x \cdot \hat r$$ and with a bit of calculation we have$$\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}= \frac{y^3+yz^2}{(x^2+y^2+z^2)^{3/2}}-\frac{-yx^2}{(x^2+y^2+z^2)^{3/2}}=\frac y r \neq 0$$ Since that's the $\hat z$ term in $\nabla \times \vec F$, we can tell that the curl is not zero, hence the force being nonconservative.
For a central force to be conservative, it must also be spherically symmetric, namely its magnitude must be a function of distance $r$ only. With that we can express $F$ as the gradient of some scalar $T$ $$\vec F = f(r)\cdot \hat r = \nabla T$$ with T being the indefinite integrationof $f(r)$ $$T = \int f(r)$$ Since the curl of a gradient is always zero,that gives us $\nabla \times \vec F = 0$, the proof of $F$ being conservative we're looking for.

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The total potential of any distribution of sources of a conservative potential is itself conservative, but not necessarily central.

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