Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Chain rule and multivariable calculus

  1. Nov 3, 2015 #1

    CAF123

    User Avatar
    Gold Member

    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!
     
  2. jcsd
  3. Nov 3, 2015 #2

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Strictly speaking, partial derivatives only make sense applied to functions, not to amounts. So it's important to keep a clear view of what is a function and what is an amount. Being loose with that distinction can easily lead to confusion, and I think that's what's happening here.

    We have a function ##F:\mathbb{R}^3\to\mathbb{R}##, and have labelled the three arguments to the function as ##s_{12},s_{23},s_{13}##.

    But we can't then go and write ##F=F(s_{12},s_{23},s_{13}) = F(s,t,-s-t)##. The second and third items are amounts (values), while the first is a function.

    We can define a function ##G:\mathbb{R}^2\to\mathbb{R}^3## by ##G(s,t)=(s,t,-s-t)## and then the composed function ##H=F\circ G## gives ##H(s,t)=F(s,t,-s-t)## and we have well-defined partial derivatives ##\frac{\partial H}{\partial s},\ \frac{\partial H}{\partial t}## which will be equal to the first two things you calculated. But ##\frac{\partial H}{\partial u}## is meaningless, as we have not labelled either of the two arguments to ##H## as ##u##.

    Or we can define a function ##P:\mathbb{R}^3\to\mathbb{R}^3## by ##P(s,t,u)=(s,t,u)## and then the composed function ##N=F\circ P## gives ##H(s,u)=F(s,t,u)## and we have well-defined partial derivatives ##\frac{\partial N}{\partial s},\ \frac{\partial N}{\partial t},\ \frac{\partial N}{\partial u}##. But the first two of these will not equal the first two things you calculated because the second terms of each simplified expression will disappear. And we will have ##\frac{\partial N}{\partial u}=\frac{\partial F}{\partial s_{13}}|_{s_{13}=u}##.

    Or we can define a function ##Q:\mathbb{R}^3\to\mathbb{R}^3## by ##Q(s,t,u)=(s,t,-s-t)## and then the composed function ##M=F\circ Q## gives ##M(s,u)=F(s,t,-s-t)## and we have well-defined partial derivatives ##\frac{\partial M}{\partial s},\ \frac{\partial M}{\partial t},\ \frac{\partial M}{\partial u}##. The first two of these will equal the first two things you calculated, and the third will be zero.

    In summary, it is necessary to clearly set out what the functions being differentiated are, and then a clear solution emerges.
     
  4. Nov 4, 2015 #3

    CAF123

    User Avatar
    Gold Member

    Hi andrewkirk,
    Thanks for your answer, although I think I was not careful enough to say what I mean by these variables. The ##s_{ij}## is just a shorthand notation for the s,t and u (so ##s_{12}## is identically s, ##s_{23}## is identically t and ##s_{13}## is identically u). So the equality in ##F(s_{12}, s_{23}, s_{13}) = F(s,t,-s-t)## means to say I have a function of s,t,u and not all three of these are independent since u=-s-t. (These are the Mandelstam variables from Physics, but this is not relevant to the discussion, although perhaps I should have pointed this out from the outset)

    Basically, what I am doing is I have some scattering amplitude (i.e some function ##F##) that depends on s and t. I want to compute ##\partial_{s_{ij}} F## but to do this I need to express the differential operators in ##s_{ij}## in terms of operators in s and t, since s and t appear in my function.

    Naively, my first thought was that ##\partial_{s_{12}} = \partial_s##, however I was told this is not the case because of the fact that $$\frac{\partial}{\partial s_{12}} = \frac{\partial}{\partial s} \frac{\partial s}{\partial s_{12}} + \frac{\partial}{\partial t} \frac{\partial t}{\partial s_{12}} + \frac{\partial}{\partial u} \frac{\partial u}{\partial s_{12}} = \frac{\partial}{\partial s} - \frac{\partial}{\partial u} \neq \frac{\partial}{\partial s},$$ which I still think is non intuitive but that is what I have been given.

    Do you agree with this?
     
  5. Nov 5, 2015 #4

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No, I don't agree with the person that corrected you, but the whole problem is so ill-defined that there's no clear right or wrong answer until some fundamental ground rules are established.

    The trouble is that symbols like ##\frac{\partial}{\partial s_{12}}## and ##\frac{\partial}{\partial s}## have no intrinsic, stand-alone meaning. Their meaning depends on context, because they are only meaningful as an operator on a particular function, or at most on a class of functions that share the same domain and range. That is a key difference from total differentials such as ##\frac{d}{dt}##, which can have a stand-alone meaning.

    Perhaps the best way to get to the heart of this is to ask what you think the differential operator ##\partial_{s_{12}}## means (or what you want it to mean) in the context of the problem you are working on. If you can find a clear answer to that question, a clear answer to the original question will not be far behind.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Chain rule and multivariable calculus
Loading...