The contiunity equation and differentiation

[Cyrus]

The contiunity equation is given as:

$$ln(\rho) + ln(A) + ln(V)=constant$$

"Differentiating Eq. 11.43 we get:

$$\frac{d\rho}{\rho}+ \frac{dA}{A} +\frac{dV}{V}=0$$

My question is, what is this differentiated with resepct to?

Its clearly not $$\rho$$, A or V alone.

 

[Gokul43201]

It depends on what your "constant" is actually independent of. For instance, if the top equation is a constant in space or time, then a derivative with repsect to space or time co-ordinates will give you the bottom equation.

 

[Cyrus]

Then, shouldn't it be $$\frac{d\rho}{dx}\frac{1}{\rho}$$, (assuming d.w.r.t. x).

$$d\rho$$ alone, is meaningless.

 

[Gokul43201]

While in the above case, it may simply be laziness of notation, I don't believe it is, in general, meaningless. But I'm not qualified to talk about differential forms, so I'll let someone else do it.

 

[HallsofIvy]

It's not, strictly speaking, a derivative, it is a total differential. Just as, with f(x,y), both x and y are functions of t, we have (chain rule)
$$\frac{df}{dt}= \frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}$$
and the total differential is
$$df= \frac{df}{dx}dx+ \frac{df}{dy}dy$$
with no mention of a specific variable, t.

 

[arildno]

Go along the curve of constant value C, with an accretion $$(d\rho,dA,dV)$$
Then, we have two equations:
$$\ln(\rho)+\ln(A)+\ln(V)=C$$
$$\ln(\rho+d\rho)+\ln(A+dA)+\ln(V+dV)=C$$
Subtracting, we get:
$$(\ln(\rho+d\rho)-\ln(\rho))+(\ln(A+dA)-\ln(A))+(\ln(V+dV)-\ln(V))=0$$
an equation for the related accretions whose linear part is given by your second expression.

 

[Gib Z]

Go along the curve of constant value C, with an accretion $$(d\rho,dA,dV)$$
Then, we have two equations:
$$\ln(\rho)+\ln(A)+\ln(V)=C$$
$$\ln(\rho+d\rho)+\ln(A+dA)+\ln(V+dV)=C$$
Subtracting, we get:
$$(\ln(\rho+d\rho)-\ln(\rho))+(\ln(A+dA)-\ln(A))+(\ln(V+dV)-\ln(V))=0$$
an equation for the related accretions whose linear part is given by your second expression.

I know this completely goes against everything I know about mathematical rigor, especially since I haven't learned about differential geometry yet, but...

$$(\ln(\rho+d\rho)-\ln(\rho))+(\ln(A+dA)-\ln(A))+(\ln(V+dV)-\ln(V))=0$$

Dividing through by $$dA\cdot dV \cdot d\rho$$ we get some nice difference quotients giving:

$$\frac{1}{\rho \cdot dA \cdot dV} + \frac{1}{A\cdot d\rho \cdot dV}+\frac{1}{V\cdot dA \cdot d\rho} = 0$$

Then, multiplying by $$dA\cdot dV \cdot d\rho$$ yields the desired result. Would this be an acceptable thing to do, pretending I knew the rigorous theory of differentials?

 

[arildno]

You have, for finite accretions:
$$\ln(\rho+d\rho)-\ln(\rho)=\frac{d\rho}{\rho}+O(d\rho^{2})$$
and so on, by simple Taylor expansions.

The retained LINEAR equation is that relation which must hold between the differential accretions.

 

[Gib Z]

What is a finite accretion :( ? And doesn't that Taylor series only have a radius of convergence of 1?

 

[arildno]

A triple of real numbers (dp,dA,dV), not all zero.

 

[Gib Z]

Can those differentials really be considered real numbers? Even when they are considered numbers in non-standard analysis they had to construct the hyper reals to do so.

 

[arildno]

Well, I made the notational sin using d's rather than $\bigtriangleup$'s instead.

 

[OrderOfThings]

Differentiating is the calculation of tangent planes to a surface. There is no need for "with respect to" to do that. The "wrt" has the meaning of picking a preferred parameterisation (of curves on the surface).

 

[Cyrus]

Thanks HallyofIvy and arildno.

Also, the equation above it says:

$$dp + \frac{1}{2}d(V^2)+\gamma dz=0$$

Here, $$d(V^2)=2VdV$$. But in this case isn't it d.w.r.t. V? I.e. $$\frac{d}{dV}(V^2)$$

Or is that ok in terms of differential notation? I suppose its just another application of the total differential on $$V^2$$.

Edit: Ah, it can't be $$\frac{d}{dV}(V^2)$$ because that would just yield 2V, and no dV term.

 

[Cyrus]

Another thing I was not sure about:

If you follow the derivation of the product rule it says:

$$\Delta (uv)=u \Delta v +v\Delta u + \Delta u \Delta v$$

Am I not allowed to change the $$\Delta$$'s into d's if I consider $$\Delta$$ very small? In which case this could also be viewed as an application of the product rule?

I.e, use the product rule on $$\Delta (\rho VA)=0$$ and divide by the accretion $$(d\rho,dA,dV)$$
to get the same result.

 

[Hurkyl]

Then, shouldn't it be $$\frac{d\rho}{dx}\frac{1}{\rho}$$, (assuming d.w.r.t. x).

$$d\rho$$ alone, is meaningless.

It's meaningful, you just haven't been taught the meaning.


You are already encouraged to think of things like $\rho$, P, and V as 'numbers'. You know that they're really functions of your state, but you manipulate them as if they were ordinary numbers. For example, you write $\ln \rho$ when you really mean function composition; it is another function of your state, and evaluating it on a state $\xi$ gives you the number $(\ln \rho)(\xi) := \ln(\rho(\xi))$. You could view this as sloppiness, but it also turns out that there is a deep reason that suggests this really is the right way to be treating these things.


Expressions like $d\rho$ are differential forms. But to keep in spirit, you simply think of a new kind of object called a 'differential', and it turns out that 'numbers' and 'differentials' relate in all the ways you would like them to relate, except for one -- you you can't divide 'differentials' to get a derivative¹. In order to get derivatives, you have to pair a 'differential' with a 'vector'. (really, a vector field. But we call it a 'vector' to keep in the right spirit!) e.g. if you've parametrized your state space with the variables (s, t), then you have the 'vectors' $\partial/\partial s$ and $\partial / \partial t$... and now everything works out:

$$\left( d\rho \cdot \frac{\partial}{\partial s} \right) (\xi(s, t)) = \frac{\partial}{\partial s} \left( \rho(\xi(s, t)) \right)$$

where the right hand side is an ordinary partial derivative that you learned in your calculus courses. As you might imagine, the left hand side is usually given the same notation: $\partial \rho / \partial s := d\rho \cdot \partial/\partial s$.


1: Well, you probably can make some sense of division, but I won't try to do it here

 

[Cyrus]

-- you you can't divide 'differentials' to get a derivative1

Why not? $$\frac{dy}{dx} = _{lim \Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}$$

This is the differential, dy, divided by the differential dx.

I.e. $$dy=f'(x)dx$$

Its exactly the division of dy by dx.

 

[Hurkyl]

Why not? $$\frac{dy}{dx} = _{lim \Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}$$

This is the differential, dy, divided by the differential dx.

I.e. $$dy=f'(x)dx$$

Its exactly the division of dy by dx.

Division does make sense in some cases, but it doesn't work out so nicely in others. One of matt grime's favorite counterexamples is the identity

$$\frac{\partial x}{\partial y} \frac{\partial y}{\partial z} \frac{\partial z}{\partial x} = -1$$




For example, if our state space is the plane (parametrized by (x, y)), and we define the scalar z by

z = x + y

then we also have

x = z - y
y = z - x

and we compute

$$\frac{\partial x}{\partial y} \frac{\partial y}{\partial z} \frac{\partial z}{\partial x} = (-1) \cdot 1 \cdot 1 = -1$$


The tricky point here is that the vector $\partial / \partial y$ is not well-defined on its own -- here, it was used to denote the directional derivative in the direction in which "y increases but z remains fixed". Similarly for $\partial / \partial z$ and $\partial / \partial x$.

In other words, you have to have chosen a parametrization before these partial differentiation operators make sense, and $\partial / \partial y$ can mean different things depending on the parametrization. (e.g. the direction in which "y increases but z remains fixed" is different from the direction in which "y increases but x remains fixed", but alas, in the simplest notation, both would be notated $\partial / \partial y$)

On the other hand, you need no such thing to make sense of differentials.


Roughly speaking, differentials are to functions as vectors are to parametrizations. And to compute a derivative, you need both a differential and a vector.

 

[arildno]

Another thing I was not sure about:

If you follow the derivation of the product rule it says:

$$\Delta (uv)=u \Delta v +v\Delta u + \Delta u \Delta v$$

Am I not allowed to change the $$\Delta$$'s into d's if I consider $$\Delta$$ very small? In which case this could also be viewed as an application of the product rule?

I.e, use the product rule on $$\Delta (\rho VA)=0$$ and divide by the accretion $$(d\rho,dA,dV)$$
to get the same result.

Not sure what you are saying here.
By, definition, we have:
$$\bigtriangleup(uv)=(u+\bigtriangleup{u})(v+\bigtriangleup{v})-uv$$
By merely using the arithemical law of distributivity, you gain your result.

 

[Cyrus]

Im saying is it incorrect to replace the deltas with d's?

 

[Rainbow Child]

...

If you follow the derivation of the product rule it says:

$$\Delta (uv)=u \Delta v +v\Delta u + \Delta u \Delta v$$

Am I not allowed to change the $$\Delta$$'s into d's if I consider $$\Delta$$ very small? In which case this could also be viewed as an application of the product rule? ...

The answer is yes, but the justification depends from the ...way you thinking!

• In a world filled up with physicists the quantity $$\delta\,u\,\delta\,v$$ is too small and you can neglect it, thus if you divide, say with $$\delta\,t$$ you get the product rule
  $$\frac{d}{d\,t}(u\cdot v)=\frac{d\,u}{d\,t}\cdot v+u\cdot\frac{d\,v}{d\,t}$$
  no need for reference to limits, the word small makes it all work.
• In a world filled up with mathematicians "$\Delta t$ small" means to divide by $\Delta t$ and take the limit $\Delta t\rightarrow 0$ in order to formulate the derivatives of $u,\,v$ with respect to $t$. Of course you have to known that $u(t),\,v(t)$ are differentiable, therefor continuous.
  On the other hand
  
  $$\Delta (uv)=u \Delta v +v\Delta u + \Delta u \Delta v$$
  
  means nothing! What is the argument of $u(t),\,v(t)$ for the above equation? Thus the correct way to evaluate $$\Delta (uv)$$ is to write
  $$\begin{array}{lll}\Delta(u(t)\,v(t)) &=&u(t)\,v(t) -u(t_o)\,v(t_o)\\ &=& u(t)\,v(t) +u(t_o)\,v(t)-u(t_o)\,v(t)-u(t_o)\,v(t_o) \\ &=& \Delta(u(t))\,v(t) +u(t_o)\,\Delta(v(t))\end{array}$$
  Divide now by $$\Delta t$$ and take the limit $$\Delta t\rightarrow 0$$ you have the product rule again (using of course the continuity o the $u(t),\,v(t)$).

Bottom line: you can apply the product rule to the $$\Delta (uv)$$ and divide by $$\Delta t$$ in order to produce derivates!

P.S.1 The whole analysis is based on the fact that we are dealing with functions on the real line (one independent variable). For all the other cases ($$\mathbb{R}^n$$, manifolds) we must invite for help, the differential forms, as Hurkyl posted.

P.S.2 Physicist' s way is quicker (I prefer it! ), but in order to be sure that you applied it correctly you have to know the mathematician' s way!

 

[Gib Z]

Though I'm sure most Physicists or physics students actually haven't learned non-standard analysis, it is their barrier of criticism to mathematicians because that proof of the product rule is completely justified by the theory.

For non-standard analysis, in the 1960's Abraham Robinson constructed the set of the hyperreals, an extension of the real numbers to include infinitesimals and infinities (one was defined as the reciprocal of the other, just a matter of preference). In the hyperreals, the infinitesimals posses the property that they are smaller in modulus than any real number, and that zero isn't the only infinitesimal. It also had the property (well, definition from construction I believe) that its square was equal to (exactly) zero.

Edit: I should mention, I haven't offically taken a course in NSA either, so I shouldn't be talking at all.

 

[Hurkyl]

In the hyperreals, the infinitesimals posses the property that they are smaller in modulus than any real number, and that zero isn't the only infinitesimal. It also had the property (well, definition from construction I believe) that its square was equal to (exactly) zero.

The red part is not right.

The revolutionary property of the hyperreals is that they satisfy exactly the same set of 'internal' statements that the real numbers do. (this is called elementary equivalence, and in this context it's also called the transfer principle) For example, the real numbers satisfy the internal property that the only solution to x*x=0 is x=0. Therefore, the hyperreals satisfy the same property.

(The hyperreals are not useful on their own -- the power of nonstandard analysis comes from comparing the standard reals with the hyperreals)



Number systems that have the property you stated are actually easy to construct. For example, you can construct one in exactly the same way you would the complexes; except rather than adjoining a new number i satisfying $i^2 = -1$, you adjoin a new number $\epsilon$ satisfying $\epsilon^2 = 0$. In fact, this particular number system even has a name: the dual numbers (see wikipedia). I have them seen used fruitfully in purely algebraic contexts, and I think also in differential geometry.

 

[Gib Z]

Ahh that's odd, I could have sworn I read that, and on wikipedia too. But your logic is obviously correct, I must brush up :(