What is meant by rate of change with respect to volume?

[oliverkahn]

In physics we often come across $$\rho=\dfrac{dq}{dV}$$ Does it mean:

$(i)$ $\displaystyle \lim_{\Delta V \to 0} \dfrac{\Delta q}{\Delta V}$

OR

$(ii)$ $\dfrac{\partial}{\partial z} \left( \dfrac{\partial}{\partial y} \left( \dfrac{\partial q}{\partial x} \right) \right)$

What does the first one mean?

The second one is a mixed partial derivative. Can it be found if $q, \dfrac{\partial q}{\partial x}, \dfrac{\partial}{\partial y} \left( \dfrac{\partial q}{\partial x} \right)$ are differentiable?

 

[PeroK]

In physics we often come across $$\rho=\dfrac{dq}{dV}$$ Does it mean:

$(i)$ $\displaystyle \lim_{\Delta V \to 0} \dfrac{\Delta q}{\Delta V}$

What does the first one mean?

It means this. The derivative of $q$ with respect to $V$. If you can express $q$, explicitly or implicity, as a function of $V$ then you can differentiate it.

 

[troglodyte]

In this case you argue with a density which could be constant .If it is not constant then you get a space dependence in your density
So,this notion leads to an Integration of dq over the whole space instead of a differentiation process.

troglodyte

 

[oliverkahn]

It means this. The derivative of $q$ with respect to $V$. If you can express $q$, explicitly or implicity, as a function of $V$ then you can differentiate it.

But this doesn't look analogous to the definition of one dimensional derivative. Consider:

$\dfrac{dy}{dx}=\displaystyle \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}=\lim_{\Delta x \to 0} \frac{y(x+\Delta x)-y(x)}{\Delta x}$

There exists only one line segment from origin to each point $P \in \mathbb{R}$. Thus we know what $x$ is at the point where we are taking the derivative (w.r.t. $x$)

Now consider:

$\displaystyle \lim_{\Delta V \to 0} \frac{\Delta q}{\Delta V}=\lim_{\Delta V \to 0} \frac{q(V+\Delta V)-q(V)}{\Delta V}$

Infinitely many volumes can be constructed from origin to each point $P\in \mathbb{R^3}$. Thus $V$ is ambiguous at the point where we are taking the derivative (w.r.t. $V$). How shall we deal with this ambiguity?

 

[PeroK]

But this doesn't look analogous to the definition of one dimensional derivative. Consider:

$\dfrac{dy}{dx}=\displaystyle \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}=\lim_{\Delta x \to 0} \frac{y(x+\Delta x)-y(x)}{\Delta x}$

There exists only one line segment from origin to each point $P \in \mathbb{R}$. Thus we know what $x$ is at the point where we are taking the derivative (w.r.t. $x$)

Now consider:

$\displaystyle \lim_{\Delta V \to 0} \frac{\Delta q}{\Delta V}=\lim_{\Delta V \to 0} \frac{q(V+\Delta V)-q(V)}{\Delta V}$

Infinitely many volumes can be constructed from origin to each point $P\in \mathbb{R^3}$. Thus $V$ is ambiguous at the point where we are taking the derivative (w.r.t. $V$). How shall we deal with this ambiguity?

There's no ambiguity. There is nothing in the definition of a derivative that identifies $x$ as a spatial variable. It is a mathematical construction, which can be applied to any function of any variable.

For example, you could have derivatives involving economics data; or, topically, a derivative of variables determining the spread of an infectious disease.

 

[PeroK]

But this doesn't look analogous to the definition of one dimensional derivative. Consider:

$\dfrac{dy}{dx}=\displaystyle \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}=\lim_{\Delta x \to 0} \frac{y(x+\Delta x)-y(x)}{\Delta x}$

There exists only one line segment from origin to each point $P \in \mathbb{R}$. Thus we know what $x$ is at the point where we are taking the derivative (w.r.t. $x$)

Now consider:

$\displaystyle \lim_{\Delta V \to 0} \frac{\Delta q}{\Delta V}=\lim_{\Delta V \to 0} \frac{q(V+\Delta V)-q(V)}{\Delta V}$

Infinitely many volumes can be constructed from origin to each point $P\in \mathbb{R^3}$. Thus $V$ is ambiguous at the point where we are taking the derivative (w.r.t. $V$). How shall we deal with this ambiguity?

To give an example. Let's look at the volume $V$ of a cube of side length $l$. We have:
$$V = l^3 \ \ \text{and} \ \ \frac{dV}{dl} = 3l^2$$
That means that, for example, if the cube is expanding then the rate of change of volume is related to the rate of change of length by that equation.

But, we can equally say that:
$$l = V^{1/3} \ \ \text{and} \ \ \frac{dl}{dV} = \frac 1 3 V^{-2/3}$$
If we know the rate of change of volume we can calculate the rate of change of the side length from that.

Note that in both cases $l$ and $V$ are simple real-valued functions of each other. There is no mathematical difference between $l = V^{1/3}$ and $y = x^{1/3}$.

 

[oliverkahn]

In this case you argue with a density which could be constant .If it is not constant then you get a space dependence in your density
So,this notion leads to an Integration of dq over the whole space instead of a differentiation process.

troglodyte

Do you mean if density is not constant, we cannot write:

$\displaystyle\rho=\dfrac{dq}{dV}=\lim_{\Delta V \to 0} \frac{\Delta q}{\Delta V}=\lim_{\Delta V \to 0} \frac{q(V+\Delta V)-q(V)}{\Delta V}$

 

[troglodyte]

Do you mean if density is not constant, we cannot write:

$\displaystyle\rho=\dfrac{dq}{dV}=\lim_{\Delta V \to 0} \frac{\Delta q}{\Delta V}=\lim_{\Delta V \to 0} \frac{q(V+\Delta V)-q(V)}{\Delta V}$

Of course you can write it in such a way.But in Physics you are not mainly interested in the density itself.You are more interested in the charge density which means you are more interested in how much charge includes an infinitesimal volume elment.This volume element include all informations about the whole system.In your case the whole informations are encoded in differential dq.Maybe you want to measure how much charge a sphere volume inlvole then you have to integrate over the whole sphere volume.

Hopefully,this informations leads to more clarity than to more confusion.
In some cases one have to switch his views in Physics which means the truth lies in both representations.troglodyte

 

[PeroK]

Do you mean if density is not constant, we cannot write:

$\displaystyle\rho=\dfrac{dq}{dV}=\lim_{\Delta V \to 0} \frac{\Delta q}{\Delta V}=\lim_{\Delta V \to 0} \frac{q(V+\Delta V)-q(V)}{\Delta V}$

The other relevant thing here is what do you mean by $\rho=\dfrac{dq}{dV}$? If $q$ is electric charge, $\rho$ is charge density and $V$ is volume, then simply $\rho = \dfrac q V$.

 

[PeroK]

In physics we often come across $$\rho=\dfrac{dq}{dV}$$ Does it mean:

$(i)$ $\displaystyle \lim_{\Delta V \to 0} \dfrac{\Delta q}{\Delta V}$

Okay, I need to add to what I said above. There is another interpretation of this, which is where $dV$ is a differential volume element. And, this might be what you mean. In this case $dV$ is shorthand for $dxdydz$.

Technically, what you have then is:
$$\rho dV = dq$$
And then you do have to be careful if you "divide by $dV$". I think is what @troglodyte has been saying. That this implies that you are going to integrate. And you get:
$$\int \rho dV = \int dq = Q$$
There's a good section here on the difference between "differentials" like $dq, dV$ and the derivative $\frac{dq}{dV}$:

http://tutorial.math.lamar.edu/Classes/CalcI/Differentials.aspx

Apologies. I should have worked out that you were dealing with differentials.

PS and something on the volume element $dV$:

https://en.wikipedia.org/wiki/Volume_element#Volume_element_in_Euclidean_space

 

[Stephen Tashi]

But this doesn't look analogous to the definition of one dimensional derivative.

If the position of a particle is given by $x = f(t)$, an unconventional way to think of velocity is to think of it as a density.

The "average amount of position" of the particle in the interval from $t$ to $t + \triangle t$ can be estimated by $f(t + \triangle t) - f(t)$. So the average density of position per unit time is $\frac{f(t + \triangle t) - f(t)}{\triangle t}$ and the position density per unit time at time $t$ is $\lim_{\triangle t \rightarrow 0} \frac{f(t + \triangle t) - f(t)}{\triangle t}$

We could express the concept of position density per unit time by writing it as $\lim_{\triangle t \rightarrow 0} \frac{ P(t)_{av}}{\triangle t}$ where $P(t)_{av}$ is the average amount of position in the interval from $t$ to $t + \triangle t$.

Thinking of what you call a "dimensional derivative" in this way makes it clear that there is a general concept for derivatives that follows the pattern: density of such-and-such per unit so-and-so. So we can have derivatives that represent heat per unit volume, or force per unit area, etc.

We can write densities using notation like like $\frac{ \triangle g}{\triangle A}$. These expressions show concepts, but they are not of practical use unless we know where $g$ and $A$ are - with respect to space or time or whatever coordinates are relevant. To include the information about the location of $g$ we need express $g$ as a function of those coordinates.

$\displaystyle \lim_{\Delta V \to 0} \frac{\Delta q}{\Delta V}=\lim_{\Delta V \to 0} \frac{q(V+\Delta V)-q(V)}{\Delta V}$

Infinitely many volumes can be constructed from origin to each point $P\in \mathbb{R^3}$.

And the limit you mention doesn't exist unless all reasonable ways of constructing the finite volumes give the same result as the volumes approach zero. By "reasonable", I mean that the method of computing $\triangle q$ must give a good approximation of the amount of $q$ in the volumes that are used.

Suppose $A$ is an area. To relate the area density of $g$ to the partial derivatives of $g$ , we need to convince ourselves that the expression $\frac{ (g(x + \triangle x,y) - g(x,y)}{\triangle x} \frac{ (g(x,y+\triangle y) - g(x,y))}{\triangle y}$ is a plausible approximation (in cartesian coordinates) for the for the average density of $g$ per unit area near the point $(x,y)$. Is it reasonable that an area density can be approximated by the product of two linear densities? It sounds reasonable, but we need to think of a convincing example or a more detailed intuitive argument.