Proving the Irrotational Property of Vector Fields with an Example Solution

Thread starter neelakash
Start date May 8, 2007
Tags

Field Vector Vector field

AI Thread Summary

The discussion focuses on demonstrating that a non-irrotational vector field V can be modified by a scalar function f to create an irrotational field fV. The user attempts to apply the operator \nabla to the product fV but encounters difficulties in isolating f from the resulting equation. They express confusion about the application of the \nabla operator and seek assistance in resolving this issue. The conversation highlights the mathematical challenge of manipulating vector fields and scalar functions in the context of irrotational properties. Clarifying these concepts is essential for successfully proving the irrotational property.

May 8, 2007

neelakash

Homework Statement

A vector field V is not irrotational.Show that it is always possible to find f such that fV is irrotational.

Homework Equations

The Attempt at a Solution

\nablax[fV]=f\nablaxV-Vx\nablaf
I have to equate the LHS to zero.But then,how can I extract f out of the resulting equation?
Please help

Physics news on Phys.org

May 8, 2007

neelakash

I don't know why that \nabla didn't work.Please bear with this problem and help me.

Thread 'Collision of a bullet on a rod-string system: query'

In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...

View full post »

Thread 'Explain this asphalt sheen'

This problem is two parts. The first is to determine what effects are being provided by each of the elements - 1) the window panes; 2) the asphalt surface. My answer to that is The second part of the problem is exactly why you get this affect. And one more spoiler:

View full post »

Thread 'A cylinder connected to a hanged mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

View full post »