## anonymous one year ago ques

1. anonymous

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 $\implies \frac{\partial^2 \phi}{\partial x \partial y}=\frac{\partial}{\partial x}(\frac{\partial \phi}{\partial y})=0?$

2. IrishBoy123

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3. ganeshie8

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

4. anonymous

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

5. anonymous

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

6. anonymous

and it IS 0 that's an identity

7. ganeshie8

Curl of gradient is 0, yeah

8. anonymous

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

9. ganeshie8

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.

10. ganeshie8

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 ?

11. anonymous

yep that's what im saying

12. anonymous

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

13. anonymous

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

14. ganeshie8

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

15. anonymous

ok

16. anonymous

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

17. ganeshie8

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

18. ganeshie8

19. anonymous

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

20. imqwerty

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

21. anonymous

@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

22. Empty

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