Timeline for Why do we do partial and not covariant differentiation with $x^{\nu}$?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Aug 4, 2014 at 13:10 | vote | accept | Yossarian | ||
| Aug 4, 2014 at 5:43 | answer | added | user10851 | timeline score: 1 | |
| Aug 2, 2014 at 22:27 | comment | added | ACuriousMind♦ | It is not legit to set $A^\mu = x^\mu$ because the $A^\mu$ is meant to be a vector field along the curve $x^\mu(\tau)$ (a section of the tangent bundle along the curve, in some dictions), but $x^\mu$ is not a vector field. (@Ben Crowell: You're right, in GR, the thing the covariant derivative acts upon are indeed vector fields (and, by extension, arbitary tensor fields). I tend to mix this up with the gauge covariant derivative, which acts more naturally upon n-forms.) | |
| Aug 2, 2014 at 22:21 | comment | added | Yossarian | With this question I meant this. The way in which the covariant derivative was introduced to me was using this formula $\frac{\nabla}{d\tau} A^{\lambda} = \frac{d}{d\tau} A^{\lambda} + \Gamma_{\mu\nu}^{\lambda} A^{\mu} \frac{dx^{\nu}}{d\tau}$. So, if we (maybe naively) set $A^{\lambda}=x^{\lambda}$ we might define a covariant derivative of $x^{\lambda}$. why is this not legit? | |
| Aug 2, 2014 at 22:17 | comment | added | user4552 | Here's another way to see that defining four-velocity in terms of a covariant derivative wouldn't make sense. Write out the definition of the covariant derivative. You're going to have a Christoffel symbol with an index that stands for $\tau$, but $\tau$ isn't a coordinate. Also, you'll have a vector $x^\nu$, but a 4-tuple of coordinates isn't a vector. | |
| Aug 2, 2014 at 22:14 | comment | added | user4552 | @ACuriousMind: Yes, except: The covariant derivative acts upon n-forms. This doesn't quite make sense to me. Do you really mean n-forms here? I would have just said tensor fields. | |
| Aug 2, 2014 at 21:14 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
added 4 characters in body; edited tags
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| Aug 2, 2014 at 21:07 | comment | added | ACuriousMind♦ | The covariant derivative acts upon n-forms. $x^\nu(\tau)$ is a curve in spacetime. How would you even define its covariant derivative? | |
| Aug 2, 2014 at 21:03 | history | asked | Yossarian | CC BY-SA 3.0 |