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poynting divergence doesn't use dt, and can have default dxes

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This commit is contained in:
Jan Petykiewicz 2019-07-17 00:50:49 -07:00
parent dd4e6f294f
commit 79e14af4db
1 changed files with 5 additions and 1 deletions
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6
fdfd_tools/fdtd.py
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@ -263,9 +263,13 @@ def poynting(e, h):
return numpy.array(s) return numpy.array(s)
def poynting_divergence(dt, dxes, s=None, *, e=None, h=None): # TODO dxes def poynting_divergence(s=None, *, e=None, h=None, dxes=None): # TODO dxes
if dxes is None:
dxes = tuple(tuple(numpy.ones(1) for _ in range(3)) for _ in range(2))
if s is None: if s is None:
s = poynting(e, h) s = poynting(e, h)
ds = ((s[0] - numpy.roll(s[0], 1, axis=0)) / numpy.sqrt(dxes[0][0] * dxes[1][0])[:, None, None] + ds = ((s[0] - numpy.roll(s[0], 1, axis=0)) / numpy.sqrt(dxes[0][0] * dxes[1][0])[:, None, None] +
(s[1] - numpy.roll(s[1], 1, axis=1)) / numpy.sqrt(dxes[0][1] * dxes[1][1])[None, :, None] + (s[1] - numpy.roll(s[1], 1, axis=1)) / numpy.sqrt(dxes[0][1] * dxes[1][1])[None, :, None] +
(s[2] - numpy.roll(s[2], 1, axis=2)) / numpy.sqrt(dxes[0][2] * dxes[1][2])[None, None, :] ) (s[2] - numpy.roll(s[2], 1, axis=2)) / numpy.sqrt(dxes[0][2] * dxes[1][2])[None, None, :] )