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by sfpotter
441 days ago
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I'm not a big Riemannian geometry buff, but I took a look at the definition in Do Carmo's book and it appears that "grad f" actually lies in TM, consistent with what I said above. Would love to learn more if I've got this mixed up. This would be nice, because it would generalize the "gradient" from vector calculus, which is clearly and unambiguously a vector. |
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I'm a simple-minded physicist. I just know if you apply the same coordinate transformation to the gradient and to the displacement vector, you get the wrong answer.
My usual reference is Schutz's Geometrical Methods of Mathematical Physics, and he defines the gradient as df, but other sources call that the "differential" and say the gradient is what you get if you use the metric to raise the indices of df.
But that raised-index gradient (i.e. g(df)), is weird and non-physical. It doesn't behave properly under coordinate transformations. So I'm not sure why folks use that definition.
You can see difference by looking at the differential in polar coordinates. If you have f=x+y, then df=dx+dy=(cos th + sin th)dr + r(cos th - sin th)d th. If you pretend this is instead a vector and transform it, you'd get "df"=(cos th + sin th)dr + (1/r)(cos th - sin th)d th, which just gives the wrong answer.
To be specific, if v=(1,1) in cartesian (ex,ey), then df(v)=2. But (1,1) in cartesian is (1,1/r) in polar (er, etheta). The "proper" df still gives 2, but the "weird metric one" gives 1+1/r^2, since you get the 1/r factor twice, instead of a 1/r and a balancing r.