#physics

Second derivatives with $\overrightarrow{\mathrm{\nabla }}$

1. Div of grad $\mathrm{\nabla }\cdot (\mathrm{\nabla }T)$
2. Curl of grad $\mathrm{\nabla }\times (\mathrm{\nabla }T)==0$
   It is always $0$.
3. Div of curl $\mathrm{\nabla }\cdot (\mathrm{\nabla }\times \mathbf{v})==0$
4. Curl of curl $\mathrm{\nabla }\times (\mathrm{\nabla }\times \mathbf{v}\mathbf{)}$

Fundamental Theorems $$\begin{align} \boxed{\int_{\mathbf{a}}^\mathbf{b} (\nabla T)d\mathbf{l} = T(\mathbf{b} ) - T(\mathbf{a})} \ \ \boxed{\int_{\mathcal{V}} (\mathbf{\nabla \cdot v})dV = \oint_{\mathcal{S} }\mathbf{v}\cdot d\mathbf{a}} \ \ \boxed{\int_{\mathcal{S}}(\nabla \times \mathbf{v})\cdot d\mathbf{a} = \oint_{\mathcal{P}}\mathbf{v}\cdot d\mathbf{l}}\end{align}$$

Fundamental Theorem of Gradient

Their line integrals are path independent.
Integral around closed loop is always 0.

The above statement doesn't obviously hold for non-gradient vector fields.
This gives rise to a special class of vector fields, ones which are gradients of some scalar fields, they are called conservative fields.

• Their line integrals are path independent
  What remains is called non-conservative. Their line integrals are path dependent

We already know about this.

If a field is conservative, ie it was gradient of some scalar field, then it's curl must always be 0, since $\mathrm{\nabla }\times (\mathrm{\nabla }T)==0$ for any case.

Fundamental Theorem of divergence $$\boxed{\int_{\mathcal{V}} (\mathbf{\vec{\nabla} \cdot v})dV = \oint_{\mathcal{S} }\mathbf{v}\cdot d\mathbf{a}}$$

The volume integral of the divergence is equal to the flux of that vector field through the surface that bounds the volume

Since the flux represents outwards flow anyways, the sum of all sources/faucets inside a volume is equal to the flux going out the surface of that volume.

Fundamental Theorem of curl $$\boxed{\int_{\mathcal{S}}(\vec{\nabla} \times \mathbf{v})\cdot d\mathbf{a} = \oint_{\mathcal{P}}\mathbf{v}\cdot d\mathbf{l}}$$

The line integral of a vector field is equal to the flux of it's curl through any surface that borders the line

The ambiguity here is with sign convention. Which side is the correct, positive direction of $d\mathbf{l}$?
The answer lies in consistency.
The area and line vectors must agree.
If we wrap our right hand fingers around the direction of $d\mathbf{l}$, then our thumb must point in the direction of $d\mathbf{a}$

For any closed surface, since the boundary vanishes, the integral must evaluate to zero.