- #1
EngageEngage
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I'm kind of confused as to how to determine when a vector field is conservative. For example, if we consider the following scalar field:
[tex]\varphi[/tex] = arctan([tex]\frac{y}{x}[/tex])
We see that the gradient is:
F = [tex]\nabla[/tex][tex]\varphi[/tex] = [tex]\frac{-y i + x j}{x^{2} + y^{2}}[/tex]
However, F is not a conservative vector field. Can someone please tell me why this is so? In class we've been discussing topology topics including simply connected domains, and other relevant topics.
I don't think its because F is undefined at the origin because when i look at the problem in the context of electrodynamics, we know that the scalar potential associated with the Electric Vector Field goes off as 1/r. This would therefore make the Electric field non-conservative, which isn't true since we can simply take the end points when computing potential differences.
So, is it because the field is not simply connected (i.e. the domain has a hole which is the z axis)?
If someone could please help me understand this or point me to a place which could help that would be great.
[tex]\varphi[/tex] = arctan([tex]\frac{y}{x}[/tex])
We see that the gradient is:
F = [tex]\nabla[/tex][tex]\varphi[/tex] = [tex]\frac{-y i + x j}{x^{2} + y^{2}}[/tex]
However, F is not a conservative vector field. Can someone please tell me why this is so? In class we've been discussing topology topics including simply connected domains, and other relevant topics.
I don't think its because F is undefined at the origin because when i look at the problem in the context of electrodynamics, we know that the scalar potential associated with the Electric Vector Field goes off as 1/r. This would therefore make the Electric field non-conservative, which isn't true since we can simply take the end points when computing potential differences.
So, is it because the field is not simply connected (i.e. the domain has a hole which is the z axis)?
If someone could please help me understand this or point me to a place which could help that would be great.