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Suppose $B$ is a covariant vector in $\mathbb{R}^{2}$ and that $B = <x^{2},x^{1}+3x^{2}>$ in the $x^{1},x^{2}$ coordinate system. Find $B$ in the new coordinate system $\bar{x}^{1},\bar{x}^{2}$ if $$ \bar{x}^{1} = (x^{2})^{3}$$ $$ \bar{x}^{2} = x^{1} + 2x^{2}$$

How can I do this in terms of $\bar{x}^{1},\bar{x}^{2}$?

user8290579
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1 Answers1

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When you say that $B$ is a covariant tensor field, you are considering the $1$-form $\omega = x^2\,dx^1 + (x^1+3x^2)dx^2$. Note that $x^2 = (\bar x^1)^{1/3}$ and $x^1=\bar x^2 - 2(\bar x^1)^{1/3}$, so $$dx^1 = d\bar x^2 - \tfrac23(\bar x^1)^{-2/3}d\bar x^1 \quad\text{and}\quad dx^2 = \tfrac13(\bar x^1)^{-2/3}d\bar x^1.$$ Therefore $$\omega = (\bar x^1)^{1/3}\left(d\bar x^2 - \tfrac23(\bar x^1)^{-2/3}d\bar x^1\right) + \left(\bar x^2 - 2(\bar x^1)^{1/3}+3(\bar x^1)^{1/3}\right)\left(\tfrac13(\bar x^1)^{-2/3}d\bar x^1\right).$$ Now collect terms and find the coefficients of $d\bar x^1$ and $d\bar x^2$. These will be your new coordinates $\bar B$ of your covariant tensor field. (If I may be permitted to say so, yuck.)

If I haven't messed up, the answer is $\bar B = \langle \frac13(\bar x^1)^{-2/3}\big(\bar x^2 - (\bar x^1)^{1/3}\big), (\bar x^1)^{1/3}\rangle$.

Ted Shifrin
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  • Thank you for the answer, I understand now. If however, B was a contravariant field instead, what would you consider $\omega$? – user8290579 Sep 12 '17 at 20:10
  • Then you'd start with the vector field $X = x^2 \dfrac{\partial}{\partial x^1} + (x^1+3x^2)\dfrac{\partial}{\partial x^2}$ and proceed analogously. – Ted Shifrin Sep 12 '17 at 20:36
  • Do you mean $\frac{\partial}{\partial\bar{x}^1}$ and $\frac{\partial}{\partial\bar{x}^1}$? How would I take the partial with respect to $x^1$ and $x^2$ if the variables are in terms of $\bar{x}^1$ and $\bar{x}^2$? – user8290579 Sep 13 '17 at 16:23
  • and maybe a more trivial question but is $\frac{\partial}{\partial x^1}$ the same thing as $dx^1$ and $dx^2$? – user8290579 Sep 13 '17 at 16:24
  • No, the vector fields $\partial/\partial x^1,\partial/\partial x^2$ give the dual basis to the $1$-forms $dx^1,dx^2$. You can use linear algebra to say how they change or you can write down the chain rule. – Ted Shifrin Sep 13 '17 at 16:28
  • So does that mean $\frac{\partial}{\partial x^1} = \frac{\partial}{\partial\bar{x}^1} + \frac{\partial\bar{x}^1}{\partial x^1}$? and similarly for $x^{2}$? and I would compute that? – user8290579 Sep 13 '17 at 16:31
  • Yikes, no. Write down the chain rule computing $\partial f/\partial x^1$ in terms of $\partial f/\partial \bar x^1$ and $\partial f/\partial\bar x^2$ (I'm using the standard abuse of notation here to write the composed function still as $f$). – Ted Shifrin Sep 13 '17 at 16:33
  • Sorry! I'm just a little confused. So I would need to use the chain rule to for $\partial f / \partial x^1$ in terms of two other partials? Why would that me different from what I said? – user8290579 Sep 13 '17 at 16:45