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Say I have a function of three variables, $F=F(s_{12},s_{23},s_{13}) = F(s,t,-s-t)$, where $s_{12}=s,s_{23}=t$ and $s_{13}=u = -s-t$. I want to compute the differential operators $$\frac{\partial}{\partial s}, \frac{\partial}{\partial t}\,\,\text{and}\,\,\frac{\partial}{\partial u}.$$

I can write $$\frac{\partial}{\partial s} = \frac{\partial s_{12}}{\partial s} \frac{\partial}{\partial s_{12}} + \frac{\partial s_{23}}{\partial s} \frac{\partial}{\partial s_{23}} + \frac{\partial s_{13}}{\partial s} \frac{\partial}{\partial s_{13}} = \frac{\partial}{\partial s_{12}} - \frac{\partial}{\partial s_{13}}$$

Similarly, $$\frac{\partial}{\partial t} = \frac{\partial}{\partial s_{23}} - \frac{\partial}{\partial s_{13}}$$ How should I go about computing $\partial/\partial u$? I can also write $$\frac{\partial}{\partial u} = \frac{\partial s_{12}}{\partial u} \frac{\partial}{\partial s_{12}} + \frac{\partial s_{23}}{\partial u} \frac{\partial}{\partial s_{23}} + \frac{\partial s_{13}}{\partial u} \frac{\partial}{\partial s_{13}}$$ but I am not sure how to simplify the first two terms.

Thanks!

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$\partial\over\partial u$ is only defined in terms of a complete set of independent variables including $u$. You have not defined such a set, as $u$ is a function of $s$ and $t$, and is thus not independent. So $\partial\over\partial u$ is not well-defined.

You could consider either $s$ or $t$ to be a function of $u$ and the other. In this case, you could define $\partial\over\partial u$ in that sense, but you would find (in general - I have not checked here) that the meaning of $\partial\over\partial u$ would differ depending on whether $s$ or $t$ was chosen as the other independent variable. And also, $\partial\over\partial s$ or $\partial\over\partial t$ would also change from what it is when $s$ and $t$ are considered independent. (I would not be surprised if in this simple case, though, if these variant definitions actually coincided.)

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  • $\begingroup$ Ok many thanks, I see the issue. Basically, what I want to do is to rewrite the operators $$\frac{\partial}{\partial s_{12}}, \frac{\partial}{\partial s_{23}}, \frac{\partial}{\partial s_{13}}$$ in terms of operators involving s,t and u. Naively, I thought that I could write e.g $$\frac{\partial}{\partial s_{12}} = \frac{\partial}{\partial s}$$ but I was told this is not the case. $\endgroup$ – CAF Nov 4 '15 at 12:04
  • $\begingroup$ Since $$\frac{\partial}{\partial s_{12}} = \frac{\partial}{\partial s} \frac{\partial s}{\partial s_{12}} + \frac{\partial}{\partial t} \frac{\partial t}{\partial s_{12}} + \frac{\partial}{\partial u} \frac{\partial u}{\partial s_{12}} = \frac{\partial}{\partial s} - \frac{\partial}{\partial u} \neq \frac{\partial}{\partial s} $$ $\endgroup$ – CAF Nov 4 '15 at 12:06
  • $\begingroup$ What you have is a map $\phi\ :\ \Bbb R^2 \to \Bbb R^3 \ :\ (s, t) \mapsto (s, t, -s-t)$ with which you are composing $F$. $\phi$ is the coordinatization of a plane in $\Bbb R^3$. The differential operators ${\partial\over\partial s}, {\partial\over\partial t}$ (or any other set) on this plane can only measure changes in directions along the plane. They can never measure how a function changes as it moves away from that plane. So it is impossible to express all three differential operators in space in terms of the operators in the plane. $\endgroup$ – Paul Sinclair Nov 4 '15 at 14:33
  • $\begingroup$ Hmm what did you mean by differential operators in space? What I am trying to figure out is whether the equation I posted for $\partial/\partial s_{12}$ is correct or not. $\endgroup$ – CAF Nov 4 '15 at 19:24
  • $\begingroup$ Hmm, we are talking about $\Bbb R^3$ which is commonly called "space", and you have 3 coordinate variables $s_{12}, s_{23}, s_{13}$ whose partial differential operators you are trying to represent. So obviously, I am referring to ${\partial\over\partial s_{12}}, {\partial\over\partial s_{23}}, {\partial\over\partial s_{13}}$. Your expression for $\partial\over\partial s_12$ is not correct. All three of the $s_{ij}$ partial differential operators measure changes in the function in directions that point outside the $s\ t$ plane. None of them can be expressed in terms of the $s, t, u$ partials, $\endgroup$ – Paul Sinclair Nov 4 '15 at 23:22

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