Question

ques

Answer 1

Let
\[\phi(x,y)\]
define a scalar field
then will
\[\frac{\partial^2\phi}{\partial x\partial y}\]
and
\[\frac{\partial^2 \phi}{\partial y \partial x}\]
always be continuous?
Also if
\[\frac{\partial \phi}{\partial x}=0\]
then
\[\implies \frac{\partial^2 \phi}{\partial x \partial y}=\frac{\partial}{\partial x}(\frac{\partial \phi}{\partial y})=0?\]

Answer 2

.

Answer 3

Essentially, you're asking if the mixed partials are always equal ?

Answer 4

I know they are not always, they must be continuous, but is it true for all cases when we talk about a scalar field?

Answer 5

I was thinking about the different ambiguities that arise when we say the following determinant is 0
\[\left[\begin{vmatrix}i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y}& \frac{\partial}{\partial z}\\ \frac{\partial \phi}{\partial x} & \frac{\partial \phi}{\partial y} & \frac{\partial \phi}{\partial z}\end{vmatrix}\right]\]

Answer 6

and it IS 0 that's an identity

Answer 7

Curl of gradient is 0, yeah

Answer 8

How can it be 0 if we are ambigious about for example the expression
\[\frac{\partial^2 \phi}{\partial y \partial z}-\frac{\partial^2 \phi}{\partial z \partial y}\]
This is not always 0 yet the curl of gradient is 0 vector

Answer 9

Clearly \(\phi \) is the potential function of vector field \(F=\nabla \phi\)
Since a potential function exists, the field \(F\) is conservative by definition.
Equivalently \(\nabla \times F\) is 0.

Answer 10

wait a second, i think i got your question
you're saying the mixed partials are "required" to be equal for the curl to be 0 ?

Answer 11

yep that's what im saying

Answer 12

at least when we talk about a scalar field, not talking about other multivariable functions

Answer 13

The individual components need to be 0 so that curl(grad phi) is a null vector

Answer 14

Rigt, i get it... I don't seem to have a convincing explanation for myself.. let me think

Answer 15

ok

Answer 16

I guess I'll doze off for a few hours, really tired. I will check this out later if you have found an explanation

Answer 17

let me tag few others
@eliassaab @zzr0ck3r @SithsAndGiggles @Michele_Laino @Empty @UnkleRhaukus @oldrin.bataku @Astrophysics

Answer 18

@mukushla @adxpoi @jtvatsim

Answer 19

Ok, so I'm back, not much luck huh lol

Answer 20

O-O i jst know that we can use curl method to check if a force is conservative or not.

Answer 21

@ganeshie8 Looks like I've the found the solution to this,
apparently in physics, we assume our functions to be "nice"

http://mathworld.wolfram.com/PartialDerivative.html

Answer 22

I had some problems when I asked a similar question, you might enjoy because it wasn't given a satisfactory answer I feel.

http://math.stackexchange.com/questions/956095/when-does-order-of-partial-derivatives-matter