# Physical significance of divergence [closed]

In my textbook They considered a parallelopiped $$ABCDEFGH$$ with sides $$dx,dy,dz$$ parallel to $$x,y,z$$ axis respectively

• $$\vec V$$ represents the vector velocity of the fluid at the centre $$P$$ of f the volume (parallelopiped) with components $$V_x,V_y,V_z$$ along three axis
• the fluid flows in or out through the all six faces.
• let us consider the fluid flow through the two opposite faces 1 and 2 of the volume element each being normal to x-axis and has area=dydz
• The value of the x component of $$\vec V$$ at the centre of the face 1 and 2 will be different from $$V_x$$ at the at the centre $$P$$.

• $$\frac {\partial V_x}{\partial x}$$ is the rate of change of $$V_x$$ along x-axis, then the change in value of $$V_x$$ in going from $$P$$ to the centre of vertical faces 1 or 2 = $$\frac{\partial V_xdx}{\partial x2}$$

• the value of x-component of the velocity $$\vec V$$ at the centre of face 1 = $$V_x-\frac{1}{2}\frac {\partial V_x}{x}dx$$

Now my problem is

• "The value of the $$x$$ component of $$\vec V$$ at the centre of the face 1 and 2 will be different from $$V_x$$ at the at the centre $$P$$" Why?

• $$\frac {\partial V_x}{\partial x}$$ is the rate of change of $$V_x$$ along $$x$$-axis, then the change in value of $$V_x$$ in going from $$P$$ to the centre of vertical faces 1 or 2 = $$\frac{\partial V_xdx}{\partial x2}$$

• $$\frac {\partial V_x}{\partial x}$$ is the rate of change of $$V$$ along x-axis because here water is flowing in all directions and we are concerned with the $$x$$ axis only, okay. But, change in the value of $$V_x$$ in going from $$p$$ to the centre of vertical faces of 1 or 2 = $$\frac{\partial V_xdx}{\partial x2}$$ but how and why. Please explain

• the value of $$x$$-component of the velocity $$\vec V$$ at the centre of face 1 = $$V_x-\frac{1}{2}\frac {\partial V_x}{x}dx$$

• Possible duplicates: physics.stackexchange.com/q/264509/2451 and links therein. Oct 28 '18 at 6:40
• 1. Because $v_x$ changes with $x$. Oct 28 '18 at 8:01
• @RobJeffries what?? Oct 28 '18 at 8:02
• @Qmechanici checked out that...i didn't get my answer ..that is not a duplicate Oct 28 '18 at 8:10

“The value of the 𝑥 component of $$\vec{V}$$ at the centre of the face 1 and 2 will be different from $$𝑉_x$$ at the centre 𝑃.”

“The $$x$$ component of $$\vec{V}$$” And “$$V_x$$” are the same thing. This just says $$V_x$$ is a function, so it’s value changes because $$x$$ changed going from center to a face. (See note about their bad terminology at the bottom of this.)

Unless otherwise specified, $$\vec{V}$$, and $$V_x$$ (the $$x$$-component of $$V$$), are specified at the point being considered, point P. Let’s call that point (a,b,c).

The center of face I is at ($$a - \frac{dx}{2}$$, b, c), and center of II is at ($$a + \frac{dx}{2}$$, b, c). If y =b and z=c don’t change, then on that line from center of I... to P... to center of II, everything is only a function of x.

Because dx is small, we can assume the change in $$V_x$$ is just the change in $$x$$ times $$\frac{\partial V_x}{\partial x}$$. But notice that the change in $$x$$ is $$\frac{d x}{2}$$.

(That’s what $$\frac{\partial V_x}{\partial x}$$ means: how much $$V_x$$ changes per change in $$x$$)

The value of x-component of the velocity 𝑉⃗ at the centre of face 1 $$= V_x- \left( \frac{\partial V_x}{\partial x}\right)~\left( \frac{dx}{2}\right)$$ How? Why?

Change in $$V_x$$ going from P to face I

= Change in $$V_x$$ from $$x=a$$ to $$x=a - \frac{dx}{2}$$

= Change in $$V_x$$ from decreasing x by $$\frac{dx}{2}$$

$$=~ \left( \frac{\partial V_x}{\partial x}\right)~\left(\Delta x\right)=~ \left( \frac{\partial V_x}{\partial x}\right)~\left(- \frac{dx}{2}\right)$$ $$\implies V_{x~\text{(face-}I)}= V_{x~(P)}- \left( \frac{\partial V_x}{\partial x}\right)~\left( \frac{dx}{2}\right)$$

Note, they seem at times to be implying “When I say $$V_x$$, Im talking about the x-component of $$V$$, at point P, but when I say ‘the $$x$$ component of V’ that could be referring to anywhere”. Which is dumb.

• Use \left(content\right) instead of (content) in order to adjust their height to that of the content. Similarly for \left(content\right). Sep 3 at 9:34
• @Frobenius Was wondering exactly that. But Im missing something about what you wrote, if we could do an example “( \frac{\partial V_x}{\partial x} ) ” .. you are talking about parenthesis right? Sep 3 at 9:38
• Ok I see thanks Sep 3 at 9:40
• I edit some parentheses in your question. There are also bigger symbols for example \bigl(content\bigr),\biggl(content\biggr),\Bigl(content\Bigr). Sep 3 at 9:42
• @Frobenius Thanks much 🙏🏻👍🏻 Sep 3 at 9:43

The velocity $$V$$ is actually a vector field i.e it has different values of velocity at different points in space. That is why you get different components at different points.

Rate of change of $$x-component$$ of velocity in the x- direction is $$\frac{\partial V}{\partial x}$$. Now since the centre of parallelopiped is $$\frac{dx}{2}$$ units away from the centres of faces of parallelopiped, we multiply by $$\frac{dx}{2}$$ to obtain change in velocity.

Using the same logic, since the centre of face 1 is $$\frac{dx}{2}$$ away from the centre of parallelopiped, we subtract the change in velocity from the velocity at the centre.

• So,according to your 3rd paragraph, the velocity $V_x$ is changing at a constant rate as ths velocity at face 1 is find out by $V_x-\frac{1}{2}\frac {\partial V_x}{x}dx$. Oct 28 '18 at 23:52