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Posted: Mon Feb 19, 2007 9:59 am
When we say condition of a vector field F being conservative is curl F=0,does it mean that F=F(r)?.I know normally it does not look so.Please,then site an example where F is not a function of r,but still curl F=0.
Posted: Mon Feb 19, 2007 10:20 am
No, a vector field does not have to be radially/spherically symmetric to be conservative. As a simple example, take F(x,,y,z) = x*i
Posted: Mon Feb 19, 2007 10:45 am
you did not understand my question.F=F(x) is a particular case of F=F(r).My question is if F=F(t) or,F=F(v) or F=F(r,v,t) may be conservative.
Posted: Mon Feb 19, 2007 11:54 am
You're right, I didn't understand, I read F(r) as r being scalar r rather than vector r. The definition of Curl does not include any time-derivatives, thus a field is considered conservative at any given instant. So you can have F(r,t) be conservative for all t. If you want to think of a conservative field as have a zero integral of F.dr for any path, and this definition applies for instantaneous paths.
If you want to consider a finite-time path r(t) with F(r,t), this doesn't apply to the definition of conservative, and in fact you can always get a non-zero integral of F.dr. For example, start at a point at time t, go in dx direction while F has one value, then when F has a different value at another time t, go backwards on the same path. You have a closed path but non-zero integral of F.dr.
I hope I understood your question this time, and that my answer makes sense...[/b]