B What is the Significance of the Laplace Operator in Vector Calculus?

[erocored]

$\frac {\partial \vec F} {\partial x}$ + $\frac{\partial \vec F} {\partial y}$ = vector which gives me a direction of the greatest increase of the greatest increase of the function, where $\vec F$ = gradient of the function. If I multiple the first by $\hat i$ and the second by $\hat j$ then I will get the length of the x-component of vector $\frac {\partial \vec F} {\partial x}$ plus the length of the y-component of vector $\frac{\partial \vec F} {\partial y}$. What does this sum mean?

 

[BvU]

Hi,

Your thread name is 'Laplace operator' ...

If I go from your $$F = \nabla f = {\partial f\over\partial x} {\bf \hat\imath} + {\partial f\over\partial y} {\bf \hat\jmath} $$ in 2 dimensions to your vector $$\frac{\partial \vec F} {\partial x} +\frac{\partial \vec F} {\partial y} = \Biggl({\partial^2 f\over\partial x^2}+{\partial^2 f\over\partial y\partial x} \Biggr) {\bf \hat\imath} +
\Biggl( {\partial^2 f\over\partial y^2} + {\partial^2 f\over\partial x\partial y} \Biggr) {\bf \hat\jmath} $$ Then I do not understand what you mean with

​

If I multiple the first by $\hat i$ and the second by $\hat j$​

​

but let's assume you mean 'take the inner product', then I get
[edit] wrong ! Instead of first inner product then adding up I made the mistake of first adding up !
$$\frac{\partial \vec F} {\partial x} \cdot\hat\imath +\frac{\partial \vec F} {\partial y} \cdot\hat\jmath = {\partial^2 f\over\partial x^2}+{\partial^2 f\over\partial y\partial x} +
{\partial^2 f\over\partial y^2} + {\partial^2 f\over\partial x\partial y} $$And that is NOT the 2 D Laplacian ...
[edit] first doing the inner product and then the addition does yield the Laplacian:
$$\Delta \vec F = \nabla^2 \vec F = \nabla \cdot (\nabla\vec F ) =
{\partial^2 f\over\partial x^2}+{\partial^2 f\over\partial y^2}$$which you probably meant, and which does have a physical meaning .

$\$

 

[erocored]

Hello, I am sorry for my mistake. Here is what I meant exactly:

I have vector $\vec v_1$ at point $\left(x, y \right)$ , vector $\vec v_2$ at point $\left(x+dx, y \right)$ and vector $\vec v_3$ at point $\left(x, y+dy \right)$ ($\vec v_1$ = $\vec F\left(x, y \right)$, $\vec v_2$ = $\vec F \left(x+dx, y \right)$, $\vec v_3$ = $\vec F \left(x, y+dy \right)$).

Then difference between $\vec v_2$ and $\vec v_1$ will be $d\vec v_{21}$, difference between $\vec v_3$ and $\vec v_1$ will be $d\vec v_{31}$ ($d\vec v_{21}$ = $\vec F \left(x+dx, y \right)$ - $\vec F\left(x, y \right)$, $d\vec v_{31}$ = $\vec F \left(x, y+dy \right)$ - $\vec F\left(x, y \right)$).

Next I will get vector $\frac {d\vec v_{21}} {dx}$ which is collinear with $d\vec v_{21}$, and so for $\frac {d\vec v_{31}} {dy}$.

Now if I multiple $\frac {d\vec v_{21}} {dx}$ by $\hat i$ I will get a length of the x-component of that vector and the y-component of vector $\frac {d\vec v_{31}} {dy}$ if multiple by $\hat j$.

Sum of $\frac {d\vec v_{21}} {dx} \cdot \hat i$ and $\frac {d\vec v_{31}} {dy} \cdot \hat j$ is equal to $\frac {\partial \vec F} {\partial x} \cdot \hat i$ + $\frac {\partial \vec F} {\partial y}\cdot \hat j$. This is also equal to
$\frac \partial {\partial x} \cdot \left(\frac {\partial f} {\partial x}\cdot \hat i + \frac {\partial f} {\partial y}\cdot \hat j \right) \cdot \hat i$ + $\frac \partial {\partial y} \cdot \left(\frac {\partial f} {\partial x}\cdot \hat i + \frac {\partial f} {\partial y}\cdot \hat j \right) \cdot \hat j$ =
$\frac {\partial^2 f} {\partial^2 x} \cdot \hat i \cdot \hat i$ + $\frac {\partial^2 f} {\partial x \cdot \partial y} \cdot \hat i \cdot \hat j$ + $\frac {\partial^2 f} {\partial x \cdot \partial y} \cdot \hat i \cdot \hat j$ + $\frac {\partial^2 f} {\partial^2 y} \cdot \hat j \cdot \hat j$ =
$\frac {\partial^2 f} {\partial^2 x}$+$\frac {\partial^2 f} {\partial^2 y}$ = sum of the lengths of those dashed lines. I can not understand what does this sum mean.

 

[erocored]

[edit] first doing the inner product and then the addition does yield the Laplacian:
$$\Delta \vec F = \nabla^2 \vec F = \nabla \cdot (\nabla\vec F ) =
{\partial^2 f\over\partial x^2}+{\partial^2 f\over\partial y^2}$$which you probably meant, and which does have a physical meaning .

$\$

Thank you! But is it correct that the Laplace operator is equal to the sum of these dashed lines, and if so, what does this sum mean?

 

[BvU]

I am sorry for my mistake

I don't think you made one. In contrast I screwed up grossly and edited my reply to correct it.

I am afraid I let you make you drawing without that helping to shed light on your original question: it's imply too complicated (at least for me). In my case, what helps here to see and fix the mistake is a 'numerical' example (below). But it still doesn't clearly show what the result signifies $$
\begin{align*}
f & = x^3y^2 + x^2y^3 \\ \ &\ \\

\vec F =\vec \nabla f & =
\Bigl (3x^2y^2 + 2xy^3 \Bigr )\; {\bf \hat\imath} + \Bigl (2x^3y + 3x^2y^2 \Bigr )\;\hat\jmath \\ \ &\ \\
{\partial \vec F \over\partial x } & =
\Bigl (6xy^2 + 2y^3 \Bigr )\; {\bf \hat\imath} + \Bigl (6x^2y + 6xy^2 \Bigr )\;\hat\jmath \\ \ &\ \\
{\partial \vec F \over\partial y } & =
\Bigl (6x^2y + 6xy^2 \Bigr )\; {\bf \hat\imath} + \Bigl (2x^3 + 6x^2y \Bigr )\;\hat\jmath \\ \ &\ \\

\frac{\partial \vec F} {\partial x} \cdot\hat\imath +\frac{\partial \vec F} {\partial y} \cdot\hat\jmath
& =\Bigl (6xy^2 + 2y^3 \Bigr ) + \Bigl (2x^3 + 6x^2y \Bigr ) \\ \ \\

& ={\partial^2 f\over\partial x^2}+{\partial^2 f\over\partial y^2} = \Delta f

\end{align*}
$$

So after this error of mine we can finally come to the Laplacian

which does have a physical meaning .

The Laplacian is the divergence of a gradient: $$\Delta f = \vec \nabla\cdot(\vec\nabla f) $$ and it plays an important role in physics in the Laplace equation and in the (more general) Poisson equation.

It can be shown that a conservative force field is the gradient of some potential. And the divergence indicates the presence (or absence) of sources.

I wouldn't know how to come to that insight directly from the definitions. Only in steps in vector calculus.

Google things like Conservative vector field , Laplacian vector field , Divergence theorem ,
physical interpretation of ... (any of: gradient, curl, divergence, laplacian ),

$\$

 

[erocored]

Thank you!

 

[BvU]

The last smiley was a realization that my reply is rather circular ...