Derive the curl of a conservative force..
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A force field is called conservative if its work between any points AA and BB does not depend on the path. This implies that the work over any closed path (circulation) is zero. This also implies that the force cannot depend explicitly on time. Consider for instance a time decaying force on a straight line. Choose a long closed path. The magnitude of the work from AA to BB will be greater than from BB to AA leading a non vanishing circulation. This is true even if you are able to write this force as F⃗ (r⃗ ,t)=−∇⃗ U(r⃗ ,t)F→(r→,t)=−∇→U(r→,t) or ∇⃗ ×F⃗ (r⃗ ,t)=0∇→×F→(r→,t)=0. Moreover one can show that this would imply dEdt=∂Udt≠0dEdt=∂Udt≠0.
Regarding the curl, I like to visualize it as an infinitesimal circulation. Let me try to elaborate the answer given by zeldredge. Imagine a two dimensional fluid flow. You can think of the curl as an infinitesimal paddle wheel put with its axis perpendicular to the fluid. Whenever the the fluid makes it rotate, the curl is different from zero. The angular velocity of the paddle wheel gives the magnitude of the curl. It is not hard to visualize that this paddle will rotate when put in a whirlpool and remain static when put in a laminar flow of a river. When you go to more abstract vector fields, like the electric or magnetic ones, you just to think about an abstract paddle wheel.
To finish let me mention that a vanishing curl (even for a position only dependent force) does not implies, in general, that the force is conservative. Those things are equivalent only when the space is simply connected. When this is not the case, the Stoke and Green theorems fail and zero curl does not imply zero circulation. The classic example is the two dimensional force F⃗ (x,y)=−yi^+xj^x2+y2F→(x,y)=−yi^+xj^x2+y2, which has vanishing curl and circulation 2π2π around a unit circle centerd at origin. If this vector field is meant to be a flow velocity field it clearly means the fluid is rotating around the origin. However it gets slower as we go away from the origin. Imagine an infinitesimal closed path, l1+l2+l3+l4l1+l2+l3+l4, in polar coordinates do not containing the origin, as shown in the figure. We have zero "work" over l3l3and l4l4 and mutually canceling "work" over l1l1 and l2l2. The longer path over l2l2compensate
hope you like this answer
A force field is called conservative if its work between any points AA and BB does not depend on the path. This implies that the work over any closed path (circulation) is zero. This also implies that the force cannot depend explicitly on time. Consider for instance a time decaying force on a straight line. Choose a long closed path. The magnitude of the work from AA to BB will be greater than from BB to AA leading a non vanishing circulation. This is true even if you are able to write this force as F⃗ (r⃗ ,t)=−∇⃗ U(r⃗ ,t)F→(r→,t)=−∇→U(r→,t) or ∇⃗ ×F⃗ (r⃗ ,t)=0∇→×F→(r→,t)=0. Moreover one can show that this would imply dEdt=∂Udt≠0dEdt=∂Udt≠0.
Regarding the curl, I like to visualize it as an infinitesimal circulation. Let me try to elaborate the answer given by zeldredge. Imagine a two dimensional fluid flow. You can think of the curl as an infinitesimal paddle wheel put with its axis perpendicular to the fluid. Whenever the the fluid makes it rotate, the curl is different from zero. The angular velocity of the paddle wheel gives the magnitude of the curl. It is not hard to visualize that this paddle will rotate when put in a whirlpool and remain static when put in a laminar flow of a river. When you go to more abstract vector fields, like the electric or magnetic ones, you just to think about an abstract paddle wheel.
To finish let me mention that a vanishing curl (even for a position only dependent force) does not implies, in general, that the force is conservative. Those things are equivalent only when the space is simply connected. When this is not the case, the Stoke and Green theorems fail and zero curl does not imply zero circulation. The classic example is the two dimensional force F⃗ (x,y)=−yi^+xj^x2+y2F→(x,y)=−yi^+xj^x2+y2, which has vanishing curl and circulation 2π2π around a unit circle centerd at origin. If this vector field is meant to be a flow velocity field it clearly means the fluid is rotating around the origin. However it gets slower as we go away from the origin. Imagine an infinitesimal closed path, l1+l2+l3+l4l1+l2+l3+l4, in polar coordinates do not containing the origin, as shown in the figure. We have zero "work" over l3l3and l4l4 and mutually canceling "work" over l1l1 and l2l2. The longer path over l2l2compensate
hope you like this answer
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