4
$\begingroup$

Consider the standard transformation equations between Cartesian and polar coordinates:

\begin{align*} x&=r \cos \theta\\ y&=r \sin \theta \end{align*}

and the inverse: $r=\sqrt{x^2+y^2}, \theta=\arctan\frac{y}{x}$.

Now consider the following product of derivatives: ${\displaystyle f=\frac{\partial r(x,y)}{\partial y}\frac{\partial y(r,\theta)}{\partial r}}.$ By the chain rule ${\displaystyle f=\frac{\partial r(x,y)}{\partial y}\frac{\partial y(r,\theta)}{\partial r} = \frac{\partial r}{\partial r} = 1}.$ However, if we calculate each multiplicand in isolation, then transform the mixed-coordinate result into a single coordinate system, we get:

\begin{align*} \frac{\partial r(x,y)}{\partial y}& =\frac{y}{\sqrt{x^2+y^2}}=\sin\theta\\ \frac{\partial y(r,\theta)}{\partial r}& = \sin\theta \end{align*}

and therefore, ${\displaystyle f=\frac{\partial r(x,y)}{\partial y}\frac{\partial y(r,\theta)}{\partial r} = \sin^2\theta}$

But we've shown by the chain rule that $f=1$!

:giantfireball:

I must be abusing the chain rule in some way (in the original context in which I stumbled on this, the correct result is $\sin^2\theta$), but I can't see what I did wrong. What's going on?

$\endgroup$
  • 6
    $\begingroup$ That is not quite how the multivariate chain rule works. IIRC, you need to calculate the Jacobian: en.wikipedia.org/wiki/Chain_rule#Higher_dimensions $\endgroup$ – InterstellarProbe May 22 '18 at 19:24
  • 10
    $\begingroup$ Karma is a funny thing: the harmless misdemeanors committed in calc 1 by cancelling differentials as if they were numbers turn into felonies in higher dimensions. $\endgroup$ – zhw. May 22 '18 at 19:35
  • $\begingroup$ The notation warns us against doing this. $\partial_x $ is not $dx$. The chain rule does say it's o.k. to "formally" cancel differentials with single-variable functions. $\endgroup$ – Jane Kamden May 26 '18 at 0:54
  • $\begingroup$ Somewhat related is saulspatz' confusion raised in the comments of this question. $\endgroup$ – Arnaud Mortier Jun 15 '18 at 21:18
5
$\begingroup$

This is what Penrose calls "the second fundamental confusion of calculus". The problem is that with partial derivatives, what is held constant is just as important as what is varied. A simple example is if we change coordinates by $$ x = u+v \\ y = v, $$ or $$ u = x-y \\ v = y $$ Then $$ \frac{\partial y}{\partial u} = 0, \quad \text{but} \quad \frac{\partial u}{\partial y} = -1, $$ which is quite different from in one dimension. Even weirder, $$ \frac{\partial u}{\partial v} = 0, \quad \text{ but } \quad \frac{\partial u}{\partial y} = -1, $$ even though $v=y$!

The reason is that in the first expression, $u$ is implicitly held constant, whereas in the second, $x$ is held constant, and although $u=y$, clearly $u \neq x$. This is why partial derivatives are such a nuisance compared to differentials: a partial derivative is moving along a line, and a line requires specifying everything that is constant along it, whereas differentials act more like hyperplanes, which only need a normal to be specified.

$\endgroup$
4
$\begingroup$

Thanks to @InterstellarProbe for the memory jog.

Recall the multivariable chain rule:

$\displaystyle \frac{\partial u(r,\theta)}{\partial r}=\frac{\partial u(x(r,\theta),y(r,\theta))}{\partial r} = \frac{\partial u}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial u}{\partial r}\frac{\partial r}{\partial y}$

Clearly,

$\displaystyle \frac{\partial u(x(r,\theta),y(r,\theta))}{\partial r} \ne \frac{\partial u}{\partial r}\frac{\partial r}{\partial y}$

which is what the OP assumes, for the case $u=r$. You could calculate each term seperately as in OP, or, just to show the multivariable chain rule hasn't failed us yet ( using the chain rule identity above with $u=r$):

${\displaystyle f=\frac{\partial r(x,y)}{\partial y}\frac{\partial y(r,\theta)}{\partial r} = \frac{\partial r}{\partial r} - \frac{\partial r}{\partial r}\frac{\partial r}{\partial y} = 1 - \cos(\theta)^2 = \sin(\theta)^2 }$

Which agrees with result in OP.

In summary, with multivariate functions you cannot simply "cancel" differentials. Any muscle memory from single variable calculus needs to be corrected.

Update: There is a case where partial derivatives can be cancelled, though it's a matter of notation. If we use Einstein Summation, then it is true that:

$$ \frac{\partial u(x^1,x^2,\ldots,x^n)}{\partial \bar{x}^i}\frac{\partial \bar{x}^i}{\partial x^j}=\frac{\partial u}{\partial x^j} $$ Assuming $x^i=f_i(\bar{x}^1,\bar{x}^2,\ldots,\bar{x}^n)$.

Of course this works only because the repeated $i$ index on the LHS indicates a summation, so that the LHS is actually the correct application of the chain rule for the partial derivative on the RHS.

$\endgroup$
  • $\begingroup$ There are typos all over this, I think - in particular, in the first line (and several other spots), shouldn't that be $\frac{\partial u}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial r}$? $\endgroup$ – Steven Stadnicki Jun 15 '18 at 21:15
3
$\begingroup$

One of my favorites in the hall of shame of differential cancellation: Consider the plane $x+y+z=0.$ At every point on this plane, each of the variables can be solved explicitly(!) as a function of the other two. We easily find

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

Now consider

$$\frac{\partial x}{\partial y}\cdot\frac{\partial y}{\partial z} \cdot \frac{\partial z}{\partial x}$$

This is a merry canceler's delight: All differentials cancel and so the product is $1$! But as we've seen, each factor is $-1,$ so the product is actually $-1.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.