In Schutz's A First Course in General Relativity (p122) he derives the polar coordinate basis vector$$\vec{e_{r}}=\frac{\partial x}{\partial r}\vec{e_{x}}+\frac{\partial y}{\partial r}\vec{e_{y}.}$$ But in other relativity texts I've seen coordinate basis vectors given simply as the partial derivative operator:$$\frac{\partial}{\partial x^{\mu}}.$$ These two equations look different to me. How do they relate to each other?
2 Answers
Writing $\vec{e}_r = \partial_r$, $\vec{e}_x = \partial_x$ transforms the "$\vec{e}$-notation" into the partial-derivative-notation, so the "relation" is just that $\vec{e}_{x^\mu} = \frac{\partial}{\partial x^\mu}$.
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$\begingroup$ Sorry to be so slow, but do you mean my first equation can be written as $$\vec{e_{r}}=\frac{\partial}{\partial r}=\frac{\partial x}{\partial r}\frac{\partial}{\partial x}+\frac{\partial y}{\partial r}\frac{\partial}{\partial y}.$$ $\endgroup$ Commented Jan 15, 2016 at 16:26
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$\begingroup$ @Peter4075: Yes - that's just the chain rule, if you look closely. $\endgroup$– ACuriousMind ♦Commented Jan 15, 2016 at 16:49
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$\begingroup$ Expressing basis vectors as partial derivative operators still looks weird to me, but you've answered my question. Thanks. $\endgroup$ Commented Jan 15, 2016 at 17:02
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1$\begingroup$ @Peter4075: You have to realize those are not arbitrary vectors we express as derivatives, but tangent vectors, and tangent vectors are rather naturally associated to derivatives $\endgroup$– ACuriousMind ♦Commented Jan 15, 2016 at 17:08
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1$\begingroup$ @ArturodonJuan the link isn’t working. Can you provide the correct link again please $\endgroup$ Commented Feb 21, 2021 at 19:25
The first expression is a single equation expressing equality between two vectors. The second expression is in Einstein notation or rather index notation. It is a set of 4 equations (for mu=0..3) expressing equality between components of vectors. Both expressions are equivalent ways of working with vectors. Einstein notation can be much faster to manipulate.