- #31
JDoolin
Gold Member
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WannabeNewton said:
You may have noticed that I have posted and deleted twice now. (and now I have edited again, for emphasis) I couldn't seem to get the notation right. But my main point was that if we define a parametric path on the surface of a sphere, (\theta(\lambda),\phi(\lambda))
and you are asked to calculate the length of a differential section of that path, \left \| \left (\frac{\mathrm{d \theta} }{\mathrm{d} \lambda},\frac{\mathrm{d} \phi }{\mathrm{d} \lambda} \right )| \right \| =\sqrt {r^2 \frac{d \theta}{d\lambda}+ r^2 \sin^2(\theta)\frac{d \phi}{d\lambda}}
you need to take into account the unit vectors; (i.e. the scale factors) how long a differential change in theta or phi actually is in the three-dimensional space.
However, if you are just asked for the differential of that path
\frac{\mathrm{d} \vec s }{\mathrm{d} \lambda}=\left (\frac{\mathrm{d \theta} }{\mathrm{d} \lambda},\frac{\mathrm{d} \phi }{\mathrm{d} \lambda} \right )
you don't need to take into account how long the unit vectors are in the theta direction or phi direction, because the coordinates of the path already take that into account.
In particular, \frac{\mathrm{d} \vec s}{\mathrm{d} r},\frac{\mathrm{d} \vec s}{\mathrm{d} \theta}, \, \mathrm{and}\, \frac{\mathrm{d} \vec s}{\mathrm{d} \phi}
should be identical to \frac{\partial \vec s}{\partial r},\frac{\partial \vec s}{\partial \theta}, \, \mathrm{and}\, \frac{\partial \vec s}{\partial \phi}.
If there is no flaw in my reasoning, this seems to directly conflict with:
d _{\alpha }\mathbf{\overrightarrow{V}} = \left (\partial _{\alpha }V^{\beta } \right )\overrightarrow{e_{\beta }} + V^{\beta }\left (\partial _{\alpha }\overrightarrow{e_{\beta }} \right )
This equation also appears in various forms as equations 1a, 1b, 2a, 2b, here: http://www.mathpages.com/rr/appendix/appendix.htm
surely there must be a simply explained flaw in either my reasoning, or my understanding of what is meant.
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