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cristina89
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Be f and g two differentiable scalar field. Proof that ([itex]\nabla[/itex]f) x ([itex]\nabla[/itex]g) is solenoidal.
Proof: $(\nabla f \times \nabla g)$ is Solenoidal
- Thread starter cristina89
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In summary, to show that a vector field is solenoidal, we can use the identity that ##\nabla \times \nabla f = 0## for any differential scalar field f. This means that for the given problem, (\nablaf) x (\nablag) is solenoidal.
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Dick
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cristina89 said:
Show what you've done so far. What would you do to show a vector field is solenoidal?
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cristina89
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Dick said:
Well, to be solenoidal, I know that [itex]\nabla[/itex] [itex]\cdot[/itex] ([itex]\nabla[/itex]f x [itex]\nabla[/itex]g) needs to be 0.
So,
[itex]\nabla[/itex] [itex]\cdot[/itex] ([itex]\nabla[/itex]f x [itex]\nabla[/itex]g) = [itex]\nabla[/itex]g [itex]\cdot[/itex] ([itex]\nabla[/itex] x [itex]\nabla[/itex]f) - [itex]\nabla[/itex]f [itex]\cdot[/itex] ([itex]\nabla[/itex] x [itex]\nabla[/itex]g)
Right? But why is this equal to zero?
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I like Serena
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Welcome to PF, cristina89! 
Did you know that ##\nabla \times \nabla f = 0## for any differential scalar field f?
It is one of the non-trivial identities of the del operator.
See for instance: http://en.wikipedia.org/wiki/Curl_(mathematics)#Identities
Did you know that ##\nabla \times \nabla f = 0## for any differential scalar field f?
It is one of the non-trivial identities of the del operator.
See for instance: http://en.wikipedia.org/wiki/Curl_(mathematics)#Identities
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cristina89
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I like Serena said:Welcome to PF, cristina89!
Did you know that ##\nabla \times \nabla f = 0## for any differential scalar field f?
It is one of the non-trivial identities of the del operator.
See for instance: http://en.wikipedia.org/wiki/Curl_(mathematics)#Identities
Thank you so much!
FAQ: Proof: $(\nabla f \times \nabla g)$ is Solenoidal
What does it mean for $(\nabla f \times \nabla g)$ to be Solenoidal?
In vector calculus, a vector field is considered solenoidal if its divergence is equal to zero. In the case of $(\nabla f \times \nabla g)$, this means that the vector field formed by taking the cross product of the gradient of two scalar fields has a divergence of zero.
Why is it important for $(\nabla f \times \nabla g)$ to be Solenoidal?
When a vector field is solenoidal, it has certain properties that make it useful in many applications. For example, solenoidal vector fields are often used to model fluid flow, electromagnetism, and other physical phenomena.
What is the physical interpretation of $(\nabla f \times \nabla g)$ being Solenoidal?
The physical interpretation of a solenoidal vector field is that it represents a flow of a quantity that is conserved. In the case of $(\nabla f \times \nabla g)$, this quantity is the product of the gradients of two scalar fields, which could represent things like temperature, pressure, or concentration.
How is the Solenoidal property of $(\nabla f \times \nabla g)$ used in real-world applications?
The solenoidal property of $(\nabla f \times \nabla g)$ is used in a variety of real-world applications, such as fluid mechanics, electromagnetism, and other areas of physics and engineering. It allows for the accurate modeling and prediction of physical phenomena and is essential in many fields of science and technology.
Can $(\nabla f \times \nabla g)$ still be Solenoidal if one of the scalar fields is constant?
Yes, $(\nabla f \times \nabla g)$ can still be solenoidal if one of the scalar fields is constant. This is because the cross product of a constant vector with any other vector is always equal to zero, which means that the divergence of the resulting vector field will also be zero.