Understanding Scalar Fields: Div, Curl, RotGrad & DivGrad

Engels
Messages
12
Reaction score
0
Curl div...

Homework Statement



f is a scalar field. What does div(f) curl(f) rotgrad(f) divgrad(f) stand for?

I need to know if a scalar field can have the meanings of roration and diverge like a vector field
 
Physics news on Phys.org


Engels said:

Homework Statement



f is a scalar field. What does div(f) curl(f) rotgrad(f) divgrad(f) stand for?

I need to know if a scalar field can have the meanings of roration and diverge like a vector field
if f is a scalar field, then grad(f) is a vector fields

div(f) makes no sense as f is a scalar, and div operates on vectors , curl(f) doesn't make sense for the same reasons

i'm guess rotgrad(f) = curl(grad(f)) which is ok, though i remember correctly its zero

and divgrad(f) = div(grad(f)) which is ok as well

have a look at this
http://en.wikipedia.org/wiki/Vector_calculus_identities
 
Last edited:


div and curl are only defined for vector fields. grad is only defined for scalar fields.

The result of div is a scalar and the result of grad and curl is a vector. Therefore, these are the second spatial derivatives that you can construct:

<br /> \mathrm{div} (\mathbf{grad} \, \phi) = \nabla^{2} \, \phi<br />

<br /> \mathrm{div} (\mathbf{curl} \, \mathbf{A}) = 0<br />

<br /> \mathbf{grad} (\mathrm{div} \, \mathbf{A})<br />

<br /> \mathbf{curl}(\mathbf{grad} \, \phi) = \mathbf{0}<br />
<br /> \mathbf{curl} (\mathbf{curl} \, \mathbf{A}) = \mathbf{grad} (\mathrm{div} \, \mathbf{A}) - \nabla^{2} \, \mathbf{A}<br />

where \nabla^{2} stands for the Laplace differential operator (Laplacian).
 


fixed up above - missed the curl(f) bit
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top