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  • 92 votes cast
Feb
24
comment How were the solar masses and distance of the GW150914 merger event calculated from the signal?
Shouldn't the term in square brackets be raised to the power of 3/5?
Feb
20
comment Physical and non-physical solutions to Einstein's field equations
Thanks for that.
Feb
5
comment How far do we need to be removed from the earth to show the curvature with a viewing angle between 42 and 48 degrees?
If I'm standing on a beach and hold up a straight edge to the horizon, surely I'll see the horizon falling away from the straight edge. Doesn't that demonstrate the curvature of the Earth?
Jan
15
comment Two ways of writing coordinate basis vectors confusion
Expressing basis vectors as partial derivative operators still looks weird to me, but you've answered my question. Thanks.
Jan
15
comment Two ways of writing coordinate basis vectors confusion
Sorry to be so slow, but do you mean my first equation can be written as $$\vec{e_{r}}=\frac{\partial}{\partial r}=\frac{\partial x}{\partial r}\frac{\partial}{\partial x}+\frac{\partial y}{\partial r}\frac{\partial}{\partial y}.$$
Jan
10
comment why does the higgs mechanism need to exist?
"Pleb" is a term senior Tory politicians in the UK may or may not use when asking police officers permission to cycle through the main gates at Downing Street. Possibly.
Nov
26
comment Galileo 5 and 6 satellites testing gravitational time dilation
Excellent link. Thanks.
Nov
11
comment Is partial derivative a vector or dual vector?
Where does $\partial_{\mathbf v}f = \langle\mathbf df, \mathbf v \rangle$ come from? Is it a definition or can you derive it? In "Gravitation" it (Eqn 2.17) seems to be derived from some sort of Taylor Series (Eqn 2.15), but I've never seen this derivation in any other textbook. Usually $\partial_{\mathbf v}f = \langle\mathbf df, \mathbf v \rangle$ is just stated. Apologies if I'm missing the obvious.
Nov
10
comment Time taken to cook frozen peas
Must admit, I've never heard of cooking frozen peas without water. It works - I've just tried it - but you need to be careful, the saucepan gets very hot (my stainless steel saucepan began to discolour with the heat). Don't burn down the kitchen!
Nov
10
comment Time taken to cook frozen peas
Thanks. I thought that would be the case but didn't have the physics to back it up.
Oct
30
comment Derivation of one-form/vector equation in Carroll confusion
Hmmm, there's a lot to take in here for a plodder like me. You've been very patient. Thanks again for your help.
Oct
29
comment Derivation of one-form/vector equation in Carroll confusion
Could this be a more straightforward answer (at my level) to my question of whether $\frac{df}{d\lambda}=\mathrm{d}f\left(\frac{d}{d\lambda}\right)$ can be derived without assuming $\mathrm{\mathrm{d}x^{i}}\left(\frac{\partial}{\partial x^{j}}\right)=\delta_{j}^{i}$? Much further on (p203) they actually use Eqn 2.17 to derive $\mathrm{\mathrm{d}x^{i}}\left(\frac{\partial}{\partial x^{j}}\right)=\delta_{j}^{i}$.
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/…