Reputation
228
Next privilege 250 Rep.
View close votes
Badges
2 14
Newest
 Tumbleweed
Impact
~7k people reached

  • 0 posts edited
  • 0 helpful flags
  • 29 votes cast
Feb
3
awarded  Tumbleweed
Jan
27
revised How can two parameters define a point?
added 44 characters in body
Jan
27
asked How can two parameters define a point?
Jan
1
comment Total confusion about differential one-forms and non-coordinate bases
Thanks very much.
Jan
1
accepted Total confusion about differential one-forms and non-coordinate bases
Dec
31
comment Total confusion about differential one-forms and non-coordinate bases
But my question was asking for examples of non-coordinate bases.
Dec
31
comment Total confusion about differential one-forms and non-coordinate bases
Thanks for that. I was aware of the summation convention. I've come thus far via self-studying general relativity, where the usual bases are the coordinate basis vectors $e_{\mu}=\frac{\partial}{\partial x^{\mu}}$ and basis one-forms $\omega^{\nu}=dx^{\nu}$. That's why I was confused by janmarqz's answer, which seemed to me to imply $e_{k}=\frac{\partial}{\partial x^{k}}$, which is a coordinate basis? Also, as I asked in my previous comment, does $df=\frac{\partial f}{\partial x^{a}}dx^{a}$ for both coordinate and non-coordinate bases? If yes, what are the basis one-forms for $df$?
Dec
16
comment Total confusion about differential one-forms and non-coordinate bases
@janmarqz - But doesn't $e_{k}(f)={\rm grad}f\cdot e_{k}=\frac{\partial f}{\partial x^{k}}$ imply $e_{k}=\frac{\partial}{\partial x^{k}}$, which is a coordinate vector basis?
Dec
15
comment Total confusion about differential one-forms and non-coordinate bases
Afraid I'm still confused as to how there are any basis vectors on the rhs. Given the definition $\omega^{a}e_{k}=\delta_{k}^{a}$ I would have thought both the basis one-forms and basis vectors on the lhs would have all disappeared in a puff of 1's and 0's.
Dec
14
comment Total confusion about differential one-forms and non-coordinate bases
So is there a relationship between a coordinate vector basis $\frac{\partial}{\partial x^{j}}$ and a non-coordinate vector basis $e_{k}$?
Dec
14
asked Total confusion about differential one-forms and non-coordinate bases
Dec
10
comment Basis of differential one-form confusion
@Phoenix87 - thanks very much. That's much clearer.
Dec
10
comment Basis of differential one-form confusion
Profuse apologies. I stand corrected.
Dec
10
comment Basis of differential one-form confusion
I didn't know that (obviously, otherwise I wouldn't have asked the question - seems a tad harsh to downvote after six minutes). How would I write the first equation, with indices, to show it referred, for example, only to coordinate bases?
Dec
10
asked Basis of differential one-form confusion
Oct
18
awarded  Benefactor
Oct
18
accepted Basis one-form and basis vector confusion
Oct
17
comment Basis one-form and basis vector confusion
So are you saying, using my example of $z=r\cos\phi$, that $v\left(f\right)=dz$ where $v=\frac{\partial}{\partial x^{a}}$ and $f=z$?
Oct
17
comment Basis one-form and basis vector confusion
I was taken aback to read in your answer that $\partial_{\nu}\mathrm{d}x^{\mu}$ “definitely doesn't mean differentiate $\mathrm{d}x^{\mu}$.” That was the one thing I thought I understood in the whole problem, ie differentiating $dx^{1}$ wrt $dx^{1}$ gives 1, wrt to $dx^{a}$ where $a\neq1$ gives zero (because the coordinates are independent of each other). So are are saying $e_{\nu}\omega^{\mu}=\partial_{\nu}dx^{\mu}=\delta_{\nu}^{\mu}$ only because it is defined to do so and that I'm going to have to forget about my explanation in terms of coordinate independence?
Oct
17
comment Basis one-form and basis vector confusion
Sorry, I meant to use the total differential of $z$ $(\mathrm{d}z=\cos\phi\mathrm{d}r-r\sin\phi\mathrm{d}\phi)$. But at least now that I'm using a simple example, the language of coordinate functions is becoming a little easier to visualise. I used $a=4$ in $\mathrm{d}x^{4}(\partial_{a})=\delta_{a}^{4}$ because any other value would give zero. Does that mean time component of $v$ in this case equals 1?