#Toroidal Coordinates - Scale factors
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Can you ask for a helper in your own channel
What's a scaling factor?
For a diagonal metric tensor g_(ij)=g_(ii)delta_(ij), where delta_(ij) is the Kronecker delta, the scale factor for a parametrization x_1=f_1(q_1,q_2,...,q_n), x_2=f_2(q_1,q_2,...,q_n), ..., is defined by h_i = sqrt(g_(ii)) (1) = sqrt(sum_(k=1)^(n)((partialx_k)/(partialq_i))^2). (2) The line element (first fundamental form) is then given by ...
I didn't know, how do I make my own channel?
This is your channel
Show how you did the derivatives directly
One of the ways I tried to do it was like this. but something is missing.
you're super close
hold on gimme 2 sec I made a mistake XD
ok I got it
so note that from the last line, if we combine the first and last terms, then we have $\sinh^2{\tau}\sin^2{\sigma}+1 = \cosh^2{\tau}\sin^2{\sigma$
lgkoo
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@proven echo see if that's enough to help you do the rest? Lmk if you need more hints
I will analyze it here.
wait no @proven echo I'm sorry this is absolute bullshit from me XD. It should be $\sinh^2{\tau}\sin^2{\sigma}+ \sin^2{\sigma} = \cosh^2{\tau}\sin^2{\sigma$.
lgkoo
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The right thing to do should be
from the last line, if we write the $\cos^2{\sigma} = 1 - \sin^2{\sigma}$ in the second term, then we have $\sinh^2{\tau}\sin^2{\sigma} + \cosh^2{\tau}(1-\sin^2{\sigma}) - 2\cosh{\tau}\cos{\sigma} +1 = \sinh^2{\tau}\sin^2{\sigma} - \cosh^2{\tau}\sin^2{\sigma} + \cosh^2{\tau} - 2\cosh{\tau}cos{\sigma} +1$
lgkoo
which should now correctly simplify down to $(\cosh{\tau} - \cos{\sigma})^2$ after a bit more trig manipulation and factorisation
lgkoo
and now?
in this picture, from second to third line, you forgot the $\cosh^2{\tau}$ when you expanded $\cosh^2{\tau}(1-\sin^2{\sigma})$
lgkoo