How to transform this into partial derivatives? (Arfken)

[physicophysiology]

Hello. Glad to meet you, everyone
I am studying the [Mathematical Methods for Physicists; A Comprehensive Guide (7th ed.) - George B. Arfken, Hans J. Weber, Frank E. Harris]
In Divergence of Vector Field,
I do not understand that

How to transform the equation in left side into that in right side?

Besides, the other question is

How to transform the equation in left side into that in right side?

 

[RPinPA]

In physics derivations, we often tend to be a little careless about distinguishing between small finite things and the limit as those things go to 0. But essentially the first one is using the definition of derivative.

Suppose instead of $dx$ you use $h$.
Define $f(x) = \rho v_x$.
Then $\left. -(\rho v_x) \right |_{x-dx/2} = f(x - (h/2))$ and $\left. -(\rho v_x) \right |_{x+dx/2} = f(x + (h/2))$ and the left hand side of that equation becomes $f(x + (h/2)) - f(x - (h/2))$.
We know that in the limit as $h \rightarrow 0$, $\frac {f(x + (h/2)) - f(x - (h/2))} {h} \rightarrow \frac {\partial f}{\partial x}$ which is being rearranged as
$f(x + (h/2)) - f(x - (h/2)) = h \frac {\partial f}{\partial x}$
where $h$ is the infinitesimal $dx$.

It is as I said not exactly rigorous mathematics.

As for your second question, you can not transform the left into the right. That's not a statement that the two quantities are equal in all cases. It's a restatement of some earlier differential equation. The left is the divergence of $\rho \vec v$ and the right is the time derivative of $\rho$. There must be some other equation that says that those two things are equal for $\rho$.

 

[tnich]

In physics derivations, we often tend to be a little careless about distinguishing between small finite things and the limit as those things go to 0. But essentially the first one is using the definition of derivative.

Suppose instead of $dx$ you use $h$.
Define $f(x) = \rho v_x$.
Then $\left. -(\rho v_x) \right |_{x-dx/2} = f(x - (h/2))$ and $\left. -(\rho v_x) \right |_{x+dx/2} = f(x + (h/2))$ and the left hand side of that equation becomes $f(x + (h/2)) - f(x - (h/2))$.
We know that in the limit as $h \rightarrow 0$, $\frac {f(x + (h/2)) - f(x - (h/2))} {h} \rightarrow \frac {\partial f}{\partial x}$ which is being rearranged as
$f(x + (h/2)) - f(x - (h/2)) = h \frac {\partial f}{\partial x}$
where $h$ is the infinitesimal $dx$.

It is as I said not exactly rigorous mathematics.

As for your second question, you can not transform the left into the right. That's not a statement that the two quantities are equal in all cases. It's a restatement of some earlier differential equation. The left is the divergence of $\rho \vec v$ and the right is the time derivative of $\rho$. There must be some other equation that says that those two things are equal for $\rho$.

I think the second equation is just conservation of mass.

 

[Chestermiller]

The second equation is required by conservation of mass. It is equivalent to saying that the rate of increase of mass within a fixed control volume is equal to the net rate of flow of mass into the control volume (in minus out).

 

[physicophysiology]

In physics derivations, we often tend to be a little careless about distinguishing between small finite things and the limit as those things go to 0. But essentially the first one is using the definition of derivative.

Suppose instead of $dx$ you use $h$.
Define $f(x) = \rho v_x$.
Then $\left. -(\rho v_x) \right |_{x-dx/2} = f(x - (h/2))$ and $\left. -(\rho v_x) \right |_{x+dx/2} = f(x + (h/2))$ and the left hand side of that equation becomes $f(x + (h/2)) - f(x - (h/2))$.
We know that in the limit as $h \rightarrow 0$, $\frac {f(x + (h/2)) - f(x - (h/2))} {h} \rightarrow \frac {\partial f}{\partial x}$ which is being rearranged as
$f(x + (h/2)) - f(x - (h/2)) = h \frac {\partial f}{\partial x}$
where $h$ is the infinitesimal $dx$.

It is as I said not exactly rigorous mathematics.

As for your second question, you can not transform the left into the right. That's not a statement that the two quantities are equal in all cases. It's a restatement of some earlier differential equation. The left is the divergence of $\rho \vec v$ and the right is the time derivative of $\rho$. There must be some other equation that says that those two things are equal for $\rho$.

Thank you very much
And I found the derivation of continuity equation so I understood the second equation

 

[RPinPA]

Thank you very much
And I found the derivation of continuity equation so I understood the second equation

There was a typo in my response, a minus sign that doesn't belong there. But glad you got the sense of my argument anyway.