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Oct
29
comment Derivation of one-form/vector equation in Carroll confusion
Equation 2.17 on p60 of Gravitation is $\partial_{v}f=\left\langle \mathrm{d}f,v\right\rangle$ (equivalent, I pray, to $\frac{df}{d\lambda}=\mathrm{d}f\left(\frac{d}{d\lambda}\right)$). As they explain, Eqn 2.17 is itself derived (by applying $\frac{d}{d\lambda}$) from Eqn 2.15 (p59): $f\left(P\right)=f\left(P_{0}\right)+\left\langle \mathrm{d}f,P-P_{0}\right\rangle +\textrm{(non linear terms)}$.
Oct
29
comment Derivation of one-form/vector equation in Carroll confusion
Also, (take a deep breath, try to stay calm, do NOT kick the cat!), if the answers to my latest questions are all “yes”, I cannot see much in the way of difference between your definition of a one-form $d_{x}f(v)=\partial_{v}f(x)$ and my original equation $\mathrm{d}f\left(\frac{d}{d\lambda}\right)=\frac{df}{d\lambda}$. Apart from your definition applying to a particular position $x$.
Oct
29
comment Derivation of one-form/vector equation in Carroll confusion
So when you write $d_{x}f(v)=\partial_{v}f(x)$ you mean the one-form $d_{x}f$ acting on the vector $v$ equals the derivative of $f$ with respect to $\lambda$ at position $x$? So the two sets of brackets mean different things. In the case of $\left(v\right)$, they mean “acting on”, and in the case of $\left(x\right)$, they mean “at the position”?
Oct
28
comment Derivation of one-form/vector equation in Carroll confusion
Thanks for you patience, but I still don't see (hence the bounty) whether $\mathrm{d}f\left(\frac{d}{d\lambda}\right)=\frac{df}{d\lambda}$ can be derived without assuming $\mathrm{\mathrm{d}x^{i}}\left(\frac{\partial}{\partial x^{j}}\right)=\delta_{j}^{i}$. Is that a yes or no? I've just noticed that on page 60 of Gravitation by Misner, Thorne and Wheeler they appear to derive the equation (2.17) from some kind of Taylor series (equation 2.15 on the previous page), but I can't follow that derivation either.
Oct
26
comment Derivation of one-form/vector equation in Carroll confusion
But isn't $\partial_{v}f(x)=d_{x}f(v)$ the equivalent of $\frac{df}{d\lambda}\left(x\right)= \textrm{d}f\left(\frac{dx^{\mu}}{d\lambda}\right)$, which is equivalent to $\frac{df}{d\lambda}=\textrm{d}f\left(\frac{d}{d\lambda}\right)$, in other words, the very equation we are trying to derive?
Oct
26
comment Derivation of one-form/vector equation in Carroll confusion
So I need to think of everything in your derivation being evaluated at a point $x$? Are you saying $\partial_{\mu}f(x)=df$? I can't see why that is. Doesn't $df=\frac{\partial f}{\partial x^{\mu}}dx^{\mu}$?
Oct
26
comment Derivation of one-form/vector equation in Carroll confusion
Apologies for my limited (high school) maths, but what's the difference between my $\frac{df}{d\lambda}$ and your $\frac{df}{d\lambda}\left(x\right)$? Thanks.
Oct
25
comment Derivation of one-form/vector equation in Carroll confusion
If you can see it, here's the link to p56 of the Schutz book: books.google.co.uk/…
Oct
25
comment Derivation of one-form/vector equation in Carroll confusion
So to speak....
Oct
25
revised Derivation of one-form/vector equation in Carroll confusion
added 2 characters in body
Oct
25
comment Derivation of one-form/vector equation in Carroll confusion
I've edited my question to try to make my confusion clearer.
Oct
25
revised Derivation of one-form/vector equation in Carroll confusion
Reaction to comment by @ACuriousMind
Oct
25
asked Derivation of one-form/vector equation in Carroll confusion
Sep
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Sep
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reviewed Approve Getting started self-studying general relativity
Sep
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comment Can a pilot in a small spaceship feel G force in space?
@CuriousOne - Honing "physics intuition" needn't involve flip, superior comments. Not the best way to encourage learners to use this site.
Sep
4
comment Can a pilot in a small spaceship feel G force in space?
@CuriousOne - that is not really a helpful comment.
Aug
29
accepted Derivation of geodesic deviation equation from two neighbouring geodesics
Aug
29
comment Derivation of geodesic deviation equation from two neighbouring geodesics
Thanks. I would never have seen that.