Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$\newcommand{\bs}[1]{\boldsymbol{#1}}$ $\newcommand{\xx}[0]{\boldsymbol{x}}$ $\newcommand{\XX}[0]{\boldsymbol{X}}$ $\newcommand{\pderiv}[2]{\frac{\partial{#1}}{\partial{#2}}}$

Hi! I've come across a mathematical contradiction which seems to imply that there is something lacking in my understanding of interchangeability of derivatives when it comes to tensors.

Say we have a 1st order tensor $\xx$ that depends on $\XX$ (also 1st order) and $t$:

$$\xx = \xx(\XX,t)$$

Are derivatives of $\xx$ with respect to $\xx$ itself and $t$ then interchangeable? That is, is the following true?

$$\pderiv{}{\xx}\left(\pderiv{\xx}{t}\right) = \pderiv{}{t}\left(\pderiv{\xx}{\xx}\right)$$

If they aren't, can someone please explain why not? If they are, then how can the following contradiction be explained?

Choose $\xx = t^2\XX$

$$\pderiv{\xx}{t} = 2t\XX = \frac{2}{t}\xx \quad \Rightarrow \quad \pderiv{}{\xx}\left(\pderiv{\xx}{t}\right) = \frac{2}{t}\bs{I}$$


$$\pderiv{}{t}\left(\pderiv{\xx}{\xx}\right) = \pderiv{}{t}\left(\bs{I}\right) = \bs{0}$$

(EDIT: The same problem obviously occurs with scalars)

share|cite|improve this question
Minus pathological cases, I think partial derivatives only commute when they are with respect to independent variables... – anon Jul 21 '11 at 13:27
Thanks, that would explain it, but what theorem or other can one reference here? – andreasdr Jul 21 '11 at 13:37
Possibly the partial derivative paradox I posted about in the following sci.math post could be of use: <>; – Dave L. Renfro Jul 21 '11 at 14:11
Notice there is another poster called drandreas, who is different from you--so dr does not commute either. – gary Jul 21 '11 at 21:02
Fascinating! :) – andreasdr Jul 23 '11 at 10:46
up vote 6 down vote accepted

Your expectation that these should commute may or may not be due to being confused by the ambiguous notation we use for partial derivatives, which doesn't mark what's being held fixed, only what's being varied. Note that you're taking the derivative w.r.t. $t$ while holding $X$ fixed and the derivative w.r.t. $x$ while holding $t$ fixed, so these are two unrelated partial derivatives in different coordinate systems $(X,t)$ and $(x,t)$ that we have no reason to expect to commute. See also Calculus of variations question from Darcogona.

P.S.: The reason that partial derivatives within one and the same coordinate system usually commute is basically that

$$ \begin{eqnarray} && \left(\frac{\left(f(x+\Delta x,y+\Delta y)-f(x+\Delta x,y)\right)}{\Delta y}-\frac{\left(f(x,y+\Delta y)-f(x,y)\right)}{\Delta y}\right)/\Delta x\\ &=& \left(\frac{\left(f(x+\Delta x,y+\Delta y)-f(x,y+\Delta y)\right)}{\Delta x}-\frac{\left(f(x+\Delta x,y)-f(x,y)\right)}{\Delta x}\right)/\Delta y\;. \end{eqnarray} $$

This works because you reach the same point whether you first move by $\Delta x$ keeping $y$ fixed and then move by $\Delta y$ keeping $x$ fixed or vice versa, whereas this is usually not the case if the two steps are along axes in different coordinate systems.

For $x=x(X,t)$ and $X=X(x,t)$, the general formula for transforming partial derivatives yields, with a vertical bar marking the variable held fixed for first derivatives, and all second derivatives referring to the coordinates $(X,t)$,

$$ \begin{eqnarray} \left.\frac{\partial}{\partial t}\right|_X \left.\frac{\partial}{\partial x}\right|_t &=& \left.\frac{\partial}{\partial t}\right|_X\left(\left.\frac{\partial t}{\partial x}\right|_t\left.\frac{\partial}{\partial t}\right|_X+\left.\frac{\partial X}{\partial x}\right|_t\left.\frac{\partial}{\partial X}\right|_t\right) \\ &=& \left.\frac{\partial}{\partial t}\right|_X\left(\left.\frac{\partial X}{\partial x}\right|_t\left.\frac{\partial}{\partial X}\right|_t\right) \\ &=& \left(\left.\frac{\partial}{\partial t}\right|_X\left.\frac{\partial X}{\partial x}\right|_t\right) \left.\frac{\partial}{\partial X}\right|_t + \left.\frac{\partial X}{\partial x}\right|_t \frac{\partial^2}{\partial t\partial X}\; \end{eqnarray} $$

(since $\partial t/\partial x|_t=0$), whereas

$$ \begin{eqnarray} \left.\frac{\partial}{\partial x}\right|_t \left.\frac{\partial}{\partial t}\right|_X &=& \left(\left.\frac{\partial t}{\partial x}\right|_t\left.\frac{\partial}{\partial t}\right|_X+\left.\frac{\partial X}{\partial x}\right|_t\left.\frac{\partial}{\partial X}\right|_t\right)\left.\frac{\partial}{\partial t}\right|_X \\ &=& \left(\left.\frac{\partial X}{\partial x}\right|_t\left.\frac{\partial}{\partial X}\right|_t\right)\left.\frac{\partial}{\partial t}\right|_X \\ &=& \left.\frac{\partial X}{\partial x}\right|_t \frac{\partial^2}{\partial X\partial t} \\ &=& \left.\frac{\partial X}{\partial x}\right|_t \frac{\partial^2}{\partial t\partial X}\;. \end{eqnarray} $$

So commuting these two partial derivatives from different coordinate systems leads to an additional term

$$ \left.\frac{\partial}{\partial t}\right|_X \left.\frac{\partial}{\partial x}\right|_t - \left.\frac{\partial}{\partial x}\right|_t \left.\frac{\partial}{\partial t}\right|_X = \left(\left.\frac{\partial}{\partial t}\right|_X\left.\frac{\partial X}{\partial x}\right|_t\right) \left.\frac{\partial}{\partial X}\right|_t $$

proportional to a first derivative, which arises because the rate of change of $X$ with $x$ changes with $t$ (which is of course not the case in a single coordinate system, where the rate of change of $x$ with $x$ is constant, namely $1$). Indeed, in your case, where you apply the mixed derivative to $x$ itself, we have

$$ \begin{eqnarray} \left(\left.\frac{\partial}{\partial t}\right|_X\left.\frac{\partial X}{\partial x}\right|_t\right) \left.\frac{\partial}{\partial X}\right|_t x &=& \left(\left.\frac{\partial}{\partial t}\right|_Xt^{-2}\right) \left.\frac{\partial}{\partial X}\right|_t t^2X \\ &=& -2t^{-3}\cdot t^2 \\ &=& -\frac2t\;, \end{eqnarray} $$

in agreement with your calculation.

share|cite|improve this answer
Thanks! That shed some light on things. – andreasdr Jul 21 '11 at 15:08
...and your edit even more so, thanks! I'm still wondering about the following however, where does it come from? $\left.\frac{\partial}{\partial x}\right|_t = \left.\frac{\partial t}{\partial x}\right|_t\left.\frac{\partial}{\partial t}\right|_X+\left.\frac{\partial X}{\partial x}\right|_t\left.\frac{\partial}{\partial X}\right|_t$ – andreasdr Jul 23 '11 at 10:46
@andreasdr: This is just the standard chain rule for a change of variables from $(x,t)$ to $(X,t)$ (see e.g.…, especially the last formula before the example); it may look unfamiliar because $t$ occurs in two different roles; if you write it out for $(x,t)$ and $(X,T)$ instead, it might ring a bell. – joriki Jul 23 '11 at 11:14
Got it, thanks! As you suspected, it was the term involving $\partial t/\partial x|_t$ that confused me, but now I see why it would appear in the case of four independent variables. – andreasdr Jul 23 '11 at 12:40
Lemme just write it out for future reference :) $\left.\frac{\partial}{\partial T}\right|_X \left.\frac{\partial}{\partial x}\right|_t = \left.\frac{\partial}{\partial T}\right|_X\left(\left.\frac{\partial T}{\partial x}\right|_t\left.\frac{\partial}{\partial T}\right|_X+\left.\frac{\partial X}{\partial x}\right|_t\left.\frac{\partial}{\partial X}\right|_T\right)$ – andreasdr Jul 25 '11 at 12:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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