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variables in triangle Refer to the above figure. Assuming the length of the 3 edges of triangle are $r_0,z_0,\xi_0$. And we have $\xi=\sqrt{r^2+z^2}$ (Eqn.1)and $\xi_0=\sqrt{r_0^2+z_0^2}$. The normal vector on the hypotenuse is $\hat{n}$, and the components are $n_r=\frac{z_0}{\xi_0}$ and $n_z=-\frac{r_0}{\xi_0}$ Meanwile$\frac{\xi}{r}=\frac{\xi_0}{r_0}=\frac{1}{\cos{\theta}}$(Eqn.2) and $\frac{\xi}{z}=\frac{\xi_0}{z_0}=\frac{1}{\sin{\theta}}$(Eqn.3).

Now I have a funtion $T=T(r,z)$ on the hypotenuse, and I need to calculate $f=\frac{\partial{T}}{\partial{z}}n_r-\frac{\partial{T}}{\partial{r}}n_z=\frac{\partial{T}}{\partial{\xi}}\frac{\partial{\xi}}{\partial{z}}n_r-\frac{\partial{T}}{\partial{\xi}}\frac{\partial{\xi}}{\partial{r}}n_z$ (Eqn.4)

If using Eqn.1, then $\frac{\partial{\xi}}{\partial{z}}=\frac{z}{\sqrt{r^2+z^2}}$ (Eqn.1*a)and $\frac{\partial{\xi}}{\partial{r}}=\frac{r}{\sqrt{r^2+z^2}}$(Eqn.1*b) and

$f=\frac{\partial{T}}{\partial{\xi}}(\frac{z_0z}{\xi_0\sqrt{r^2+z^2}}+\frac{r_0r}{\xi_0\sqrt{r^2+z^2}})$. (Eqn.5)

If using Eqn.2 and Eqn.3, $\frac{\partial{\xi}}{\partial{z}}=\frac{\xi_0}{z_0}$ (Eqn.6)and$\frac{\partial{\xi}}{\partial{r}}=\frac{\xi_0}{r_0}$(Eqn.7) and

$f=\frac{\partial{T}}{\partial{\xi}}(\frac{\xi_0z_0}{z_0\xi_0}+\frac{\xi_0r_0}{r_0\xi_0})=2\frac{\partial{T}}{\partial{\xi}}$(Eqn.8)

It looks to me that Eqn.6 and Eqn.7 are right. but Eqn.8 is wrong. because when $\theta$ goes to 90degree,it should be $f=\frac{\partial{T}}{\partial{z}}=\frac{\partial{T}}{\partial{\xi}}$

So what is the right expression for $f$, and why?

Thanks in advance!

george andrew
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    Can you clarify the origin of your $f$ a bit? At first glance I thought it was the azimuthal component of the curl of $T$, but $T$ isn't a vector field. (I raise that question since the vector form of the calculation might allow you to check your work via the RH-rule...) – Semiclassical Sep 06 '14 at 01:38
  • Thank you Semiclassical. You are right. It is the azimuthal component of the curl of $T\hat{n}$ – george andrew Sep 06 '14 at 05:12
  • Ah, very good. That means you can actually look up the answer directly by writing your field in spherical coordinates. Proceeding that way looks to give an answer of $f=T+\xi \dfrac{\partial T}{\partial \xi}$; I'll see if I can work out an answer along the lines given in your question (perhaps posted tomorrow.) – Semiclassical Sep 06 '14 at 06:05
  • Cross-posted to http://physics.stackexchange.com/q/134181/2451 – Qmechanic Sep 06 '14 at 06:55
  • Thank you very much Semiclassical, but I don't understand why becomes $f=T+\xi\frac{\partial{T}}{\partial{\xi}}$... – george andrew Sep 06 '14 at 11:46
  • Meant to say $f=\frac{T}{\xi}+\frac{\partial T}{\partial \xi}$, which is me simply looking at the curl in spherical coordinates (see [1]. But that's nto a direct proof in any case. – Semiclassical Sep 06 '14 at 17:13
  • I think I see your issue: eqs. 2 and 3 are upside-down – Semiclassical Sep 06 '14 at 18:03
  • Thank you Semiclassical. Basically I now think Eqn.5 is correct, since $\frac{z_0}{\xi_0}=\frac{z}{\sqrt{r^2+z^2}}=\sin{\theta}$ and $\frac{r_0}{\xi_0}=\frac{r}{\sqrt{r^2+z^2}}=\cos{\theta}$ and Eqn.5 equals $\frac{\partial{T}}{\partial{\xi}}$. This makes sense, that no matter what $\theta$ is, the expression(the azimuthal component expression is unchanged) – george andrew Sep 07 '14 at 02:49
  • BTW, I did it in cylindrical coordinate system. – george andrew Sep 07 '14 at 02:56
  • Yes, Semiclassical, I am really confused, Eqn.2/3 compared to Eqn.1*a/b are just upside down. I don't understand which is right. – george andrew Sep 07 '14 at 03:00
  • Equation 5 is definitely correct, but you're right that something seems odd in that derivation. I'll see if I can work out why. – Semiclassical Sep 07 '14 at 03:55

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ok, Eqn.6/7 are wrong. they are not the partial derivative.

george andrew
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