Equality with curl and gradient

In summary, the conversation discusses the proof of $\nabla\times (f\nabla g)=\nabla f\times \nabla g$, showing that the second term in the last expression becomes the zero vector and that the remaining expression is $\nabla f\times \nabla g$. The conversation also mentions the use of Schwarz's Theorem in Symmetry of second derivatives.
  • #1
mathmari
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Hey! :eek:

I want to show that $\nabla\times (f\nabla g)=\nabla f\times \nabla g$.

We have that $f\nabla g=f\sum\frac{\partial g}{\partial x_i}\hat{x}_i$, therefore we get \begin{align*}&\nabla\times (f\nabla g)=\nabla\times \left (f\sum\frac{\partial g}{\partial x_i}\hat{x}_i \right )\\ & =\nabla\times \left (\sum f\frac{\partial g}{\partial x_i}\hat{x}_i \right ) \\ & =\nabla\times \left (f\frac{\partial g}{\partial x_1}, f\frac{\partial g}{\partial x_2} ,f\frac{\partial g}{\partial x_3} \right ) \\ & = \left [ \frac{\partial}{\partial x_2} \left (f\frac{\partial g}{\partial x_3}\right )-\frac{\partial}{\partial x_3} \left (f\frac{\partial g}{\partial x_2}\right ), \frac{\partial}{\partial x_3} \left (f\frac{\partial g}{\partial x_1}\right )-\frac{\partial}{\partial x_1} \left (f\frac{\partial g}{\partial x_3}\right ), \frac{\partial}{\partial x_1} \left (f\frac{\partial g}{\partial x_2}\right )-\frac{\partial}{\partial x_2} \left (f\frac{\partial g}{\partial x_1}\right )\right ]\\ & = \left [ \frac{\partial}{\partial x_2} \left (f\right )\frac{\partial g}{\partial x_3}+f\frac{\partial}{\partial x_2} \left (\frac{\partial g}{\partial x_3}\right )-\frac{\partial}{\partial x_3} \left (f\right )\frac{\partial g}{\partial x_2}-f\frac{\partial}{\partial x_3} \left (\frac{\partial g}{\partial x_2}\right ), \frac{\partial}{\partial x_3} \left (f\right )\frac{\partial g}{\partial x_1}+f\frac{\partial}{\partial x_3} \left (\frac{\partial g}{\partial x_1}\right )-\frac{\partial}{\partial x_1} \left (f\right )\frac{\partial g}{\partial x_3}-f\frac{\partial}{\partial x_1} \left (\frac{\partial g}{\partial x_3}\right ), \frac{\partial}{\partial x_1} \left (f\right )\frac{\partial g}{\partial x_2}+f\frac{\partial}{\partial x_1} \left (\frac{\partial g}{\partial x_2}\right )-\frac{\partial}{\partial x_2} \left (f\right )\frac{\partial g}{\partial x_1}-f\frac{\partial}{\partial x_2} \left (\frac{\partial g}{\partial x_1}\right )\right ]\\ & = \left [ \frac{\partial f}{\partial x_2} \frac{\partial g}{\partial x_3}+f\frac{\partial}{\partial x_2} \left (\frac{\partial g}{\partial x_3}\right )-\frac{\partial f}{\partial x_3} \frac{\partial g}{\partial x_2}-f\frac{\partial}{\partial x_3} \left (\frac{\partial g}{\partial x_2}\right ), \frac{\partial f}{\partial x_3} \frac{\partial g}{\partial x_1}+f\frac{\partial}{\partial x_3} \left (\frac{\partial g}{\partial x_1}\right )-\frac{\partial f}{\partial x_1} \frac{\partial g}{\partial x_3}-f\frac{\partial}{\partial x_1} \left (\frac{\partial g}{\partial x_3}\right ), \frac{\partial f}{\partial x_1} \frac{\partial g}{\partial x_2}+f\frac{\partial}{\partial x_1} \left (\frac{\partial g}{\partial x_2}\right )-\frac{\partial f}{\partial x_2} \frac{\partial g}{\partial x_1}-f\frac{\partial}{\partial x_2} \left (\frac{\partial g}{\partial x_1}\right )\right ]\\ & = \left [ \frac{\partial f}{\partial x_2} \frac{\partial g}{\partial x_3}-\frac{\partial f}{\partial x_3} \frac{\partial g}{\partial x_2}, \frac{\partial f}{\partial x_3} \frac{\partial g}{\partial x_1}-\frac{\partial f}{\partial x_1} \frac{\partial g}{\partial x_3}, \frac{\partial f}{\partial x_1} \frac{\partial g}{\partial x_2}-\frac{\partial f}{\partial x_2} \frac{\partial g}{\partial x_1}\right ]+\left [ f\frac{\partial}{\partial x_2} \left (\frac{\partial g}{\partial x_3}\right )-f\frac{\partial}{\partial x_3} \left (\frac{\partial g}{\partial x_2}\right ), f\frac{\partial}{\partial x_3} \left (\frac{\partial g}{\partial x_1}\right )-f\frac{\partial}{\partial x_1} \left (\frac{\partial g}{\partial x_3}\right ), f\frac{\partial}{\partial x_1} \left (\frac{\partial g}{\partial x_2}\right )-f\frac{\partial}{\partial x_2} \left (\frac{\partial g}{\partial x_1}\right )\right ]\end{align*}

Is everything correct so far? How could we continue? (Wondering)

Or is there an other way to show this? (Wondering)
 
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  • #2
Hey mathmari! (Smile)

It looks correct to me, and I'm not aware of a different way to prove it.

To continue we need to realize that:
$$\pd {}{x_2}\left(\pd{g}{x_3}\right)=\pd {}{x_3}\left(\pd{g}{x_2}\right)=\frac {\partial^2g}{\partial x_2 \partial x_3}
$$
See Schwarz's Theorem in Symmetry of second derivatives. (Thinking)
 
  • #3
I like Serena said:
To continue we need to realize that:
$$\pd {}{x_2}\left(\pd{g}{x_3}\right)=\pd {}{x_3}\left(\pd{g}{x_2}\right)=\frac {\partial^2g}{\partial x_2 \partial x_3}
$$
See Schwarz's Theorem in Symmetry of second derivatives. (Thinking)

How could we continue using this property? I got stuck right now... (Wondering)
 
  • #4
mathmari said:
How could we continue using this property? I got stuck right now... (Wondering)

Doesn't it make the second term in your last expression become the zero vector?

And which expression are we aiming for? That is, what is $\nabla f \times \nabla g$? (Wondering)
 
  • #5
I like Serena said:
Doesn't it make the second term in your last expression become the zero vector?

Ah yes! (Blush)

I like Serena said:
And which expression are we aiming for? That is, what is $\nabla f \times \nabla g$? (Wondering)

The expression that is remained is $\nabla f \times \nabla g$, so we are done! (Yes)

Thank you! (Smile)
 

Related to Equality with curl and gradient

1. What is equality with curl and gradient?

Equality with curl and gradient is a mathematical concept in vector calculus that describes the relationship between two vector fields. It states that if the curl of one vector field is equal to the gradient of another vector field, then the two vector fields are equal up to a constant.

2. How is equality with curl and gradient used in physics?

In physics, equality with curl and gradient is used to describe the behavior of vector fields in three-dimensional space. It is commonly applied in fluid dynamics, electromagnetism, and other areas of physics to analyze the flow of fluids, the motion of charged particles, and other physical phenomena.

3. What is the significance of equality with curl and gradient?

The significance of equality with curl and gradient lies in its ability to simplify complex vector fields and provide a deeper understanding of their behavior. It also allows for the application of powerful mathematical tools, such as Green's theorem and Stokes' theorem, to solve problems in physics and other fields.

4. How is equality with curl and gradient related to conservative fields?

Equality with curl and gradient is closely related to conservative fields, which are vector fields that satisfy the condition of being equal to the gradient of a scalar function. In fact, a vector field is conservative if and only if its curl is equal to the gradient of another vector field.

5. Can equality with curl and gradient be extended to higher dimensions?

Yes, equality with curl and gradient can be extended to higher dimensions. In three dimensions, it is known as the fundamental theorem of vector calculus. In higher dimensions, it is known as the generalized Stokes' theorem, which includes the curl and gradient as special cases.

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