- #1

- 2

- 0

Trying to prove that the gradient of a scalar field is symmetric(?) Struggling with the formatting here. Please see the linked image. Thanks.

http://i.imgur.com/9ZelT.png

http://i.imgur.com/9ZelT.png

Last edited:

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter tewaris
- Start date

- #1

- 2

- 0

http://i.imgur.com/9ZelT.png

Last edited:

- #2

lanedance

Homework Helper

- 3,304

- 2

[tex] \frac{\partial g_i}{\partial x_j} = \frac{\partial g_j}{\partial x_i} [/tex]

this is because the partial derivatives coummte, ie

[tex] \frac{\partial^2 V}{\partial x_j x_i} = \frac{\partial^2 V}{\partial x_i x_j} [/tex]

now you need to conisder the other direction of the proof

- #3

- 2

- 0

- #4

lanedance

Homework Helper

- 3,304

- 2

- #5

Office_Shredder

Staff Emeritus

Science Advisor

Gold Member

- 4,639

- 651

In two dimensions for example, we want to find a function V(x,y) such that

[tex] \frac{\partial V}{\partial x} = g_1(x,y) [/tex]

and

[tex] \frac{\partial V}{\partial y} = g_2(x,y)[/tex]

We can integrate the first equation w.r.t. x and the second w.r.t. y and get by the fundamental theorem of calculus

[tex] V(x,y) = \int g_1(x,y)dx + F(y)[/tex]

[tex] V(x,y) = \int g_2(x,y)dy + G(x)[/tex]

The constant of integration when you integrate w.r.t. x is really an arbitrary function of y, and vice versa. (here the integration is really just any choice of antiderivative that you want, because we're writing out the constant of integration explicitly). So we're set as long as we can find functions G(x) and F(y) such that

[tex]\int g_1(x,y)dx + F(y)=\int g_2(x,y)dy + G(x)[/tex]

So the question boils down to why do these functions F and G exist given the condition on the partial derivatives of g

Share:

- Replies
- 1

- Views
- 2K