Let's say same point in two co-ordinate system has the following relation from partial derivatives,

$$dx'=\frac{\partial x'}{\partial x} dx + \frac{\partial x'}{\partial y} dy$$ and $$dy'=\frac{\partial y'}{\partial x} dx + \frac{\partial y'}{\partial y} dy.$$

I have to derive the conditions for $$ dx'^2 + dy'^2 \propto dx^2 + dy^2. $$

Now my question is how do define $dx'^2$? My memory and hunch says, it should be $$ dx'^2 = \left( \frac{\partial x'}{\partial x} \right)^2 dx^2 + 2 \frac{\partial x'^2}{\partial x \partial y} dxdy + \left( \frac{\partial x'}{\partial y} \right)^2 dy^2$$

However, I'm not sure. Is it correct?

Also, could anyone tell me a good text to revise these stuffs from. I did them a long time ago and forgot.

  • $\begingroup$ The expression you wrote is not $dx'^2$ but $d^2x'$, the second differential of $x'$ as a function of $x$ and $y$. $\endgroup$ – GReyes Feb 24 at 7:26
  • $\begingroup$ What does $\alpha$ mean? $\endgroup$ – William Elliot Feb 24 at 9:44
  • $\begingroup$ It is not $\alpha$, it is $\propto$ = proportional to. $\endgroup$ – ponir Feb 24 at 18:18
  • $\begingroup$ @GReyes, In that case what is the expression for $dx'^2$? $\endgroup$ – ponir Feb 24 at 18:19
  • $\begingroup$ Actually, what you have is neither of them (just did not notice that you have the squares, not the pure second derivatives in your expression). $dx'^2$ would simply be what you have, replacing the mixed second derivative by $ (dx'/dx)(dx'/dy)$ (you just square your expression for $dx'$). $\endgroup$ – GReyes Feb 24 at 23:31

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