Can a Force of Gravitation Be Considered Conservative?

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Discussion Overview

The discussion revolves around whether the force of gravitation can be considered a conservative force. Participants explore the conditions required for a force to be classified as conservative, particularly in relation to the gravitational field and its definition at certain points.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant notes that for a force to be conservative, it must have a curl of zero and be defined and differentiable everywhere, citing an example where a force is not conservative due to being undefined at a specific point.
  • Another participant suggests that a vector field can still be conservative if it is defined on a proper domain, implying that the conditions for conservativeness can be relaxed under certain circumstances.
  • There is a clarification that having curl zero and being defined everywhere is sufficient for a vector field to be conservative, but not necessary, as a field may still be conservative under different conditions.
  • A participant questions how to verify if a force is conservative, suggesting that expressing it as the gradient of a function might be a method of verification.
  • One participant indicates that they found clarification in later lectures, suggesting that further exploration of the topic may resolve their doubts.

Areas of Agreement / Disagreement

Participants express differing views on the conditions necessary for a force to be considered conservative, with some arguing that the force of gravitation can be treated as conservative despite being undefined at certain points, while others emphasize the strict conditions required for a force to be classified as conservative.

Contextual Notes

The discussion highlights the complexity of defining conservative forces, particularly in relation to singularities and the mathematical properties of vector fields. There is an acknowledgment of the need for proper domains in the context of defining forces.

WiFO215
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I was watching 18.02 now and Denis Aurox, the prof. who is lecturing, mentioned that for a force to be conservative, not only was it necessary for its curl = 0 but also that F is defined and differentiable everywhere. He then went on to give the example y<i> + x<j>/x^2 + y ^2 will not be a conservative force for it is not defined at the origin.

However, I have a doubt. The first example that hit me was the Force of Gravitation. It's given by Const./ r^2 where r is the displacement between the masses. When we look at it in terms of a field, the field is not defined on the mass itself (i.e. on the point mass) and shoots to infinity as r shoots to zero. How come we define a potential in this case?

Anirudh
 
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Curl is just a consequence of a conservative vector field. For some smooth function f which is defined on some domain D, then we can define a conservative vector such that F = grad(f).

So in your first example it is conservative, you just have to give a proper domain.
 
You are confusing "if" with "only if". If a vector field has curl 0 and is defined and differentiable everywhere, it is conservative. If a vector field does not satisfy those conditions, it may or may not be conservative.
 
HallsofIvy said:
You are confusing "if" with "only if". If a vector field has curl 0 and is defined and differentiable everywhere, it is conservative. If a vector field does not satisfy those conditions, it may or may not be conservative.

I see. So how do you verify if it is or isn't conservative? By finding out if it can be expressed as a gradient of a function?

So in the case of Gravitation, even though the field is not defined at r = 0, we get a gradient field V satisfying grad V = G field because ...?
 
Okay. I got it. I got it. I just went ahead a few lectures where he talks about this in more detail and that clarified things.

<<SO THIS IS DONE>>

Could someone finish up my other thread lurking below this one?
 

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