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Reply 60
Avatar for Zhen Lin
Zhen Lin
12
Original post by latentcorpse
Hmmm. So I get (ϕ∗(X))α∂∂yα(f)=Xμ∂∂xμ(ϕ∗(f))( \phi_* ( X) ) ^\alpha \frac{\partial}{\partial y^\alpha} (f) = X^\mu \frac{\partial}{\partial x^\mu} ( \phi^* (f) )

Now I feel like I should be able to cross multiply the y term over and that gives me something that looks similar to what I want. How do I get rid of the f's though?


Just drop them.

That's what I thought but we have
f′(x)=∂f(x)∂x=lim⁡h→0f(x+h)−f(x)hf'(x)=\frac{\partial f(x)}{\partial x} = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} so if h is the thing in the limit we take the derivative wrt x. In our case, t is the thing in the limit so surely we should take the derivative wrt s, no?


Read the notation carefully. That is what he's done.

Can you explain how our choice of coordinates causes this term to vanish? Isn't X=∂∂tX=\frac{\partial}{\partial t} from the paragraph at the top of the page? How does this cause it to vanish?


Write it out in components. If t is the first coordinate then X = (1, 0, 0, ...) - so all its partial derivatives are zero.

So if I have something like
Unparseable latex formula:

g_\mu \rho} \partial_\nu X^\rho

, that is NOT equal to ∂νXμ\partial_\nu X_\mu?


In general, yes. When the metric is especially nice (e.g. constant) then it will commute.

However the metric does commute with the covariant derivative doesn't it?


Yes, but this is not the covariant derivative.

So why do all the partial derivatives of the metric disappear except the ∂gμν∂z\frac{\partial g_{\mu \nu}}{\partial z} term?


They're annihilated by the components of X: Xρ∂ρgμν=∂gμν∂zX^\rho \partial_\rho g_{\mu \nu} = \frac{\partial g_{\mu \nu}}{\partial z}.

Where did we define V this way? I still don't see this step.


V is tangent to an affinely-parametrised geodesic.

And further down on p74:
What is an isometry group? He has introduced the idea of an isometry but nothing to do with an isometry group?


It's literally what it says: a group of isometries. If you don't know what a group is, you probably should go quickly learn some elementary group theory. (I'm sure you must have learned what a group is, whether you came from a physics background or a maths background.)

Then he says "Any 1 dimensional subgroup of SO(3) gives a 1 parameter family of isometries, and hence a Killing vector field" I don't understand this at all! Can you explain please? In particular, why does it give a Killing vector field?


SO(3) is the rotation group of a 2-sphere. SO(3) itself is a 3-dimensional manifold (basically a twisted 3-sphere) and it has subgroups of various topological dimensions. A 1-dimensional subgroup, by definition, is a 1-parameter family of isometries. And a 1-parameter family of isometries gives rise to a Lie derivative, which gives rise to a Killing vector field.

On the remarks on p75:
He says that r is not the distance from the origin. What is it then?


Just a coordinate.

And he asks us why this doesn't make sense, I assume that's just because if we went to r=0 (the origin) then we would get a coordinate singularity i.e. 2Mr\frac{2M}{r} would be undefined. Is this correct?


That's not the only reason. Firstly, note that inside the Schwarzschild radius, moving in the r direction is timelike, not spacelike. (Conversely, moving in the t direction is spacelike, not timelike.)

And what is he on about in the 3rd remark? Surely changing either r or M to -r or -M is physically unrealistic?


r is just a coordinate. He's pointing out that you can assume M and r have the same sign, i.e. positive.

At the bottom of p70, he says that since ∂∂t\frac{\partial}{\partial t} and ∂∂ϕ\frac{\partial}{\partial \phi} are Killing vector fields they give rise to conserved quantities. Yet as far as I can see in all teh stuff on Killing fields, he has not shown that they give rise to conserved quantities. How do we know Killing fields correspond to conservation laws?


I presume this is a reference to Noether's theorem.

How does he derive (250)?


A simple substitution, ua=∂xa∂τ\displaystyle u^a = \frac{\partial x^a}{\partial \tau}.

Do null geodesics have zero length?


Sure, if you want to define length that way.

Doesn't this violate causality?


No.

Because that would mean two points seperated by a null geodesic would be able to communicate with one another "instantaneously". Surely I must be interpreting this wrongly?


Yes, it's true that no proper time passes on a null geodesic, but I'll just point out that light rays travel on null geodesics. (The converse is also true; if something is moving along a null geodesic then it is moving at the speed of light.)
(edited 13 years ago)
Original post by Zhen Lin

Read the notation carefully. That is what he's done.

He has (∂∂tT(t,xi))(s,xi)=(∂∂tT(s,xi))\left( \frac{\partial}{\partial t} T ( t, x^i ) \right)_{(s,x^i)} = \left( \frac{\partial}{\partial t} T ( s, x^i ) \right) as his answer
I think that from the definition it should be
(∂∂sT(s,xi))\left( \frac{\partial}{\partial s} T ( s, x^i ) \right) though?
Why does he get ∂∂t\frac{\partial}{\partial t} instead?

Original post by Zhen Lin

V is tangent to an affinely-parametrised geodesic.

I can't find this in my notes. Can you explain why this means ddτ(XaVa)=V(XaVa)\frac{d}{d \tau} ( X_a V^a)=V(X_aV^a)?

Original post by Zhen Lin

That's not the only reason. Firstly, note that inside the Schwarzschild radius, moving in the r direction is timelike, not spacelike. (Conversely, moving in the t direction is spacelike, not timelike.)

How can you tell that moving in the r direction is timelike?


Thanks
(edited 13 years ago)
Reply 62
Avatar for Zhen Lin
Zhen Lin
12
Original post by latentcorpse
He has (∂∂tT(t,xi))(s,xi)=(∂∂tT(s,xi))\left( \frac{\partial}{\partial t} T ( t, x^i ) \right)_{(s,x^i)} = \left( \frac{\partial}{\partial t} T ( s, x^i ) \right) as his answer
I think that from the definition it should be
(∂∂sT(s,xi))\left( \frac{\partial}{\partial s} T ( s, x^i ) \right) though?
Why does he get ∂∂t\frac{\partial}{\partial t} instead?

Then you misunderstand the notation. ∂∂t[T(t,xi)]∣(s,xi)=∂∂ξ[T(ξ,xi)]∣(s,xi)\displaystyle \left. \frac{\partial}{\partial t} \left[ T(t, x^i) \right] \right|_{(s, x^i)} = \left. \frac{\partial}{\partial \xi} \left[ T(\xi, x^i) \right] \right|_{(s, x^i)}. The point is that the letter used doesn't matter; what matters is where it's evaluated.

I can't find this in my notes. Can you explain why this means ddτ(XaVa)=V(XaVa)\frac{d}{d \tau} ( X_a V^a)=V(X_aV^a)?


It may not be in your notes, but it's a straightforward consequence of the definitions. And we've discussed it before.

How can you tell that moving in the r direction is timelike?


The coefficient on the dr2dr^2 term is negative and the coefficient on the dt2dt^2 term is positive, when r<2Mr < 2M.
(edited 13 years ago)
Original post by Zhen Lin
Then you misunderstand the notation. ∂∂t[T(t,xi)]∣(s,xi)=∂∂ξ[T(ξ,xi)]∣(s,xi)\displaystyle \left. \frac{\partial}{\partial t} \left[ T(t, x^i) \right] \right|_{(s, x^i)} = \left. \frac{\partial}{\partial \xi} \left[ T(\xi, x^i) \right] \right|_{(s, x^i)}. The point is that the letter used doesn't matter; what matters is where it's evaluated.


Sorry to drag this out. But the basic defn is
f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}
so if we take the limit with respect to h then we use the OTHER variable in the ∂∂x\frac{\partial}{\partial x}
but in what he has written down we use t in the limit AND in the ∂∂t\frac{\partial}{\partial t}.
So shouldn't it be ∂∂s\frac{\partial}{\partial s} if we take the limit with respect to t?

Original post by Zhen Lin

It may not be in your notes, but it's a straightforward consequence of the definitions. And we've discussed it before.

I'm not sure we have discussed it before - I just had a look through all the stuff we talked about on geodesics and can't see anything to do with this.

Original post by Zhen Lin

The coefficient on the dr2dr^2 term is negative and the coefficient on the dt2dt^2 term is positive, when r<2Mr < 2M.


So I asusme that your point here is that in order for there not to be a problem (as discussed in remark 2 on p75), they should both be either timelike or both spacelike and that since moving one direction is timelike and the other is spacelike, tthis is unphysical or something?
I guess I now understand the maths but not why this causes a problem with defining r as the distance from the origin?

Do you have any ideas about how to do the exercise on p82?

And if you look at (258), we are asked just above to prove this as an exercise. I tried to do this using E2=2V(r)E^2=2V(r) where V(r) is given by (254) with σ=1\sigma=1 for timelike geodesics. However, he says that we get (258) by evaluating at r=r±r=r_\pm. Well r±r_\pm is given in (257) so how the hell are there any rr terms left in the final answer (258)?

And underneath (258) he says that an orbit with large r has E≃1−M2rE \simeq 1 - \frac{M}{2r}. How did he get that factor of 2 on the bottom?

And further down this paragraph on p83, he says that "eventually the particle will reach the ISCO" ( this is defined at the bottom of p82 as having r=6M) - anyway he says it will have E=89E=\sqrt{\frac{8}{9}}.
I find
E=6M−2M6M6M−3M=4M18M2=418=432≠89E=\frac{6M-2M}{\sqrt{6M}\sqrt{6M-3M}}=\frac{4M}{\sqrt{18M^2}}= \frac{4}{\sqrt{18}}=\frac{4}{3 \sqrt{2}} \neq \sqrt{\frac{8}{9}}
How on earth does he get this answer then?

On p84, above (260), he says it's clear that an incoming geodesic will reach r=2M in finite affine parameter. How do we know that it will be finite?

An just below (260) he says that since dtdr\frac{dt}{dr} has a simple pole at r=2Mr=2M, t will diverge logarithmically as r→2Mr \rightarrow 2M. Why logarithmically?

And in (261), I'm trying to find dr∗dr\frac{dr_*}{dr}:
dr∗dr=1+2Mr2M−112M\frac{dr_*}{dr}=1+\frac{2M}{\frac{r}{2M}-1} \frac{1}{2M} where we used the chain rule
=1+1r2M−1=1+\frac{1}{\frac{r}{2M}-1}
and then I cannot rearrange it to what we want....

At the top of p85, is v constant along ingoing radial null geodesics because
dvdt=dtdt+dr∗dt=1+(−1)=0\frac{dv}{dt}=\frac{dt}{dt}+ \frac{dr_*}{dt}=1+(-1)=0 where dr∗dt=−1\frac{dr_*}{dt}=-1 since dtdr∗=−1\frac{dt}{dr_*}=-1 for incoming radial null geodesics?
(edited 13 years ago)
Reply 64
Avatar for Zhen Lin
Zhen Lin
12
Original post by latentcorpse
Sorry to drag this out. But the basic defn is
f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}
so if we take the limit with respect to h then we use the OTHER variable in the ∂∂x\frac{\partial}{\partial x}
but in what he has written down we use t in the limit AND in the ∂∂t\frac{\partial}{\partial t}.
So shouldn't it be ∂∂s\frac{\partial}{\partial s} if we take the limit with respect to t?

*sigh* Look, all I'm saying is that f′(x)=ddt[f(t)]∣xf'(x) = \left. \frac{d}{dt} \left[ f(t) \right] \right|_x. It's really just a notation thing. To make another analogy, AabBbc=AadBdcA^{ab} B_{bc} = A^{ad} B_{dc}. It doesn't matter whether I write "b" or "d" or any letter, what matters is that I use the same letter twice.

I'm not sure we have discussed it before - I just had a look through all the stuff we talked about on geodesics and can't see anything to do with this.


Let γ:I→M\gamma : I \to M be an affinely-parametrised geodesic. Then, writing xμx^\mu for the coordinates of γ(τ)\gamma(\tau), let Vμ=dxμdτ\displaystyle V^\mu = \frac{d x^\mu}{d \tau}. Then, the geodesic equation tells us dVμdτ+ΓμνρμVνVρ=0\displaystyle \frac{d V^\mu}{d \tau} + \Gamma^{\mu}_{\phantom{\mu} \nu \rho} V^\nu V^\rho = 0. So, in particular, ∇Vf=Vρ∂ρf\nabla_V f = V^\rho \partial_\rho f when f:M→Rf: M \to \mathbb{R} is a scalar field. But by the chain rule, ddτ=dxμdτ∂∂xμ=Vμ∂μ\displaystyle \frac{d}{d\tau} = \frac{d x^\mu}{d \tau} \frac{\partial}{\partial x^\mu} = V^\mu \partial_\mu, so ∇Vf=V(f)\nabla_V f = V(f), as claimed.

So I asusme that your point here is that in order for there not to be a problem (as discussed in remark 2 on p75), they should both be either timelike or both spacelike and that since moving one direction is timelike and the other is spacelike, tthis is unphysical or something?


That's part of the problem, yes. A curve which is constant in every coordinate except r cannot be a geodesic, so you can't integrate along that to get a distance from the origin.

I guess I now understand the maths but not why this causes a problem with defining r as the distance from the origin?


Well, sure, you could define it as that if you want. But really you should be defining distances as the integral of some function over an extremising curve of some kind. The problem is this: what function should you use? If you think about it, proper time isn't an exact analogue of spatial distance, since it depends on the exact path travelled. But there's no physically meaningful way of defining what it means for two points to be "the same point in space but separated in time", so there's no good answer anyway.

Have you done a course in special relativity before? It may be worth revising some of the philosophical and physical points from that theory.

Do you have any ideas about how to do the exercise on p82?


No. I can't answer any of your other questions either, since I haven't studied general relativity itself.
(edited 13 years ago)
Original post by Zhen Lin

No. I can't answer any of your other questions either, since I haven't studied general relativity itself.


I think I must have wrote a few other questions in that last post whilst you were writing your last post. Could you maybe take a look at some of the last few which I think are all just algebra.

Also, if we had ds2=−adv2+2bdvdr+cdr2+ddθ2+edϕ2ds^2=-a dv^2 + 2 b dv dr + c dr^2 + d d \theta^2 + e d \phi^2 in a coordinate chart (v,r,θ,ϕ)(v,r, \theta, \phi)
would the metric be

gμν=(−ab00bc0000d0000e)g_{\mu \nu} = \begin{pmatrix} -a & b & 0 & 0 \\ b & c & 0 & 0 \\ 0 & 0 & d & 0 \\ 0 & 0 & 0 & e \end{pmatrix}

or
gμν=(−a2b002bc0000d0000e)g_{\mu \nu} = \begin{pmatrix} -a & 2b & 0 & 0 \\ 2b & c & 0 & 0 \\ 0 & 0 & d & 0 \\ 0 & 0 & 0 & e \end{pmatrix}

How does the metric in (266) have a smooth inverse metric? Surely gvv=−11−2Mrg^{vv}=-\frac{1}{1-\frac{2M}{r}} which is definitely not smooth at r=2Mr=2M?

Thank you!
(edited 13 years ago)
Reply 66
Avatar for Zhen Lin
Zhen Lin
12
Original post by latentcorpse
I think I must have wrote a few other questions in that last post whilst you were writing your last post. Could you maybe take a look at some of the last few which I think are all just algebra.

I don't have the patience to do the algebraic manipulations in my own work, let alone other people's!

Also, if we had ds2=−adv2+2bdvdr+cdr2+ddθ2+edϕ2ds^2=-a dv^2 + 2 b dv dr + c dr^2 + d d \theta^2 + e d \phi^2 in a coordinate chart (v,r,θ,ϕ)(v,r, \theta, \phi)
would the metric be

gμν=(−ab00bc0000d0000e)g_{\mu \nu} = \begin{pmatrix} -a & b & 0 & 0 \\ b & c & 0 & 0 \\ 0 & 0 & d & 0 \\ 0 & 0 & 0 & e \end{pmatrix}


Yes. You could have easily checked this yourself by writing it all out. (In matrix form: compute (dvdrdθdϕ)(−ab00bc0000d0000e)(dvdrdθdϕ)\begin{pmatrix} dv & dr & d\theta & d\phi \end{pmatrix} \begin{pmatrix} -a & b & 0 & 0 \\ b & c & 0 & 0 \\ 0 & 0 & d & 0 \\ 0 & 0 & 0 & e \end{pmatrix} \begin{pmatrix} dv \\ dr \\ d\theta \\ d\phi \end{pmatrix}.)

Original post by latentcorpse
How does the metric in (266) have a smooth inverse metric? Surely gvv=−11−2Mrg^{vv}=-\frac{1}{1-\frac{2M}{r}} which is definitely not smooth at r=2Mr=2M?

I think you need to compute the inverse matrix there, since the metric is not diagonal.
(edited 13 years ago)
Original post by Zhen Lin
I don't have the patience to do the algebraic manipulations in my own work, let alone other people's!



Yes. You could have easily checked this yourself by writing it all out. (In matrix form: compute (dvdrdθdϕ)(−ab00bc0000d0000e)(dvdrdθdϕ)\begin{pmatrix} dv & dr & d\theta & d\phi \end{pmatrix} \begin{pmatrix} -a & b & 0 & 0 \\ b & c & 0 & 0 \\ 0 & 0 & d & 0 \\ 0 & 0 & 0 & e \end{pmatrix} \begin{pmatrix} dv \\ dr \\ d\theta \\ d\phi \end{pmatrix}.)



I think you need to compute the inverse matrix there, since the metric is not diagonal.



Okay what about this dillema:
(238) is in basis indices. we convert it to a generally covariant (239) using our rules. Now (239) is true in any basis as it's written in abstract indices so by simply replacing these with Greek basis indices we should get an equation true in all bases i.e. (LXg)μν=∂μXν+∂νXμ(L_Xg)_{\mu \nu} = \partial_\mu X^\nu + \partial_\nu X_\mu. So it would appear (238) is wrong since it has this additional first term. Has he just left that in there for the derivation or what?

And if you look at the calculation (286) at the bottom of p92, he then goes on to say at the top of p93 that the first term T is small by assumptiona nd so we can ignore that. That's fine. However he then says the second term, LξTL_\xi T,is higher order and can be neglected - I don't get this?

And also in that first paragraph on p93, he says the same is true for any tensor that vanishes in unperturbed spacetime - why is that?

Cheers!
Reply 68
Avatar for Zhen Lin
Zhen Lin
12
Original post by latentcorpse
Okay what about this dillema:
(238) is in basis indices. we convert it to a generally covariant (239) using our rules. Now (239) is true in any basis as it's written in abstract indices so by simply replacing these with Greek basis indices we should get an equation true in all bases i.e. (LXg)μν=∂μXν+∂νXμ(L_Xg)_{\mu \nu} = \partial_\mu X^\nu + \partial_\nu X_\mu. So it would appear (238) is wrong since it has this additional first term. Has he just left that in there for the derivation or what?

Your derivation is incorrect. In general ∂μ≠∇μ\partial_\mu \ne \nabla_\mu.

And if you look at the calculation (286) at the bottom of p92, he then goes on to say at the top of p93 that the first term T is small by assumptiona nd so we can ignore that. That's fine. However he then says the second term, LξTL_\xi T,is higher order and can be neglected - I don't get this?


This is approximation stuff, and really not something I know about. (I don't do much applied mathematics in general.)
Original post by Zhen Lin
Your derivation is incorrect. In general ∂μ≠∇μ\partial_\mu \ne \nabla_\mu.



This is approximation stuff, and really not something I know about. (I don't do much applied mathematics in general.)


Ok. Thanks.

Just above (288) though, he says we can use (239) to write Lξη)μν=∂μξν+∂νξμL_\xi \eta)_{\mu \nu} = \partial_\mu \xi_\nu + \partial_\nu \xi_\mu

But if you look at (235) it uses covariant not partial derivatives - why does this coordinate system allow us to change the covariants to partials? I thought this was only allowed when we were working at a particular point p∈Mp \in M and had defined normal coordinates at that point (since then the Christoffel symbols would vanish at that point).

And what has happened to the second ξ\xi term in (291)? i.e. why is there no ∂ν∂μξν\partial^\nu \partial_\mu \xi_\nu term also?

Why does (300) imply the waves are transverse?

How do we obtain (304)?

I also can't figure out how to get to (305)?

Why do uaeia=0u_ae^a_i=0 and gabeiaejb=δijg_{ab}e^a_ie^b_j=\delta_{ij} in (310)? This seems quite important and I don't understand!

Thanks!
(edited 13 years ago)
Reply 70
Avatar for Zhen Lin
Zhen Lin
12
Original post by latentcorpse
Ok. Thanks.

Just above (288) though, he says we can use (239) to write Lξη)μν=∂μξν+∂νξμL_\xi \eta)_{\mu \nu} = \partial_\mu \xi_\nu + \partial_\nu \xi_\mu

Well, that is true, but it follows from (238) rather than (239). Basically it's because ημν\eta_{\mu \nu} is constant. (I presume gμν=ημνg_{\mu \nu} = \eta_{\mu \nu} here. I can't be bothered to read the whole section.)

And what has happened to the second ξ\xi term in (291)? i.e. why is there no ∂ν∂μξν\partial^\nu \partial_\mu \xi_\nu term also?


I have no idea.

Why does (300) imply the waves are transverse?


This is physics. I don't even know what a transverse wave is!

I think you're very close to exhausting what little I know about differential geometry and general relativity. I really don't know very much about this subject at all, honest! I recommend you take your questions to another forum — I know there's at least one forum which (ostensibly) specialises in physics questions.
Original post by Zhen Lin
Well, that is true, but it follows from (238) rather than (239). Basically it's because ημν\eta_{\mu \nu} is constant. (I presume gμν=ημνg_{\mu \nu} = \eta_{\mu \nu} here. I can't be bothered to read the whole section.)



I have no idea.



This is physics. I don't even know what a transverse wave is!

I think you're very close to exhausting what little I know about differential geometry and general relativity. I really don't know very much about this subject at all, honest! I recommend you take your questions to another forum — I know there's at least one forum which (ostensibly) specialises in physics questions.


Ok. Thanks a lot for all your help! I really appreciate it!

A couple of last (mathematical) things:

at the top of p97, why is e0a=uae^a_0=u^a?

and how do you go from (313) to (314)?

Say you have as in (311), ∇ueiα\nabla_u e^\alpha_i
we can write this as uμ∇μeiαu^\mu \nabla_\mu e^\alpha_i
why is this equal to ub∇beαiu^b \nabla_b e^i_\alpha?

i don't see why the uμu^\mu should become a ubu^b - it's just the components (i.e. a function) not a vector, isn't it?
(edited 13 years ago)
Reply 72
Avatar for Zhen Lin
Zhen Lin
12
Original post by latentcorpse
at the top of p97, why is e0a=uae^a_0=u^a?


Essentially by definition. If I believe I'm stationary, then the direction my time flows in is precisely my "real" 4-velocity, whatever that actually means.

and how do you go from (313) to (314)?


I think this is the third time I've had to explain this, or something like this: ua∂a=ddτ\displaystyle u^a \partial_a = \frac{d}{d\tau}, by the chain rule. So the action of ua∇au^a \nabla_a on a scalar is just ddτ\displaystyle \frac{d}{d\tau}.
Original post by Zhen Lin
Essentially by definition. If I believe I'm stationary, then the direction my time flows in is precisely my "real" 4-velocity, whatever that actually means.



I think this is the third time I've had to explain this, or something like this: ua∂a=ddτ\displaystyle u^a \partial_a = \frac{d}{d\tau}, by the chain rule. So the action of ua∇au^a \nabla_a on a scalar is just ddτ\displaystyle \frac{d}{d\tau}.


but you've use
Unparseable latex formula:

\nbla_a=\partial_a

here. Is that because we are in a local inertial frame i.e. normal coordinates and so the christoffel symbols vanish and covariant reduces to partial?
Reply 74
Avatar for Zhen Lin
Zhen Lin
12
Original post by latentcorpse
but you've use
Unparseable latex formula:

\nbla_a=\partial_a

here. Is that because we are in a local inertial frame i.e. normal coordinates and so the christoffel symbols vanish and covariant reduces to partial?


No. ∇af=∂af\nabla_a f = \partial_a f for scalar fields f, by definition.
Original post by Zhen Lin
No. ∇af=∂af\nabla_a f = \partial_a f for scalar fields f, by definition.


Ah, yes.

Why in (240), does he say the solutions of the equation are Killing vector fields? Surely they are covectors?
Reply 76
Avatar for Zhen Lin
Zhen Lin
12
Original post by latentcorpse
Ah, yes.

Why in (240), does he say the solutions of the equation are Killing vector fields? Surely they are covectors?


Just dualise with the metric and you'll have a vector field.
Original post by Zhen Lin
Just dualise with the metric and you'll have a vector field.


Yeah, I was guessing that's what he meant.

I was trying to do this question on covectors and got stuck:
If nan_a is a covector orthogonal to the surface β(x)=0\beta(x)=0. Show that
n[a;bnc]n_{[a;b}n_{c]}.
Any ideas?
Reply 78
Avatar for Zhen Lin
Zhen Lin
12
Original post by latentcorpse
Yeah, I was guessing that's what he meant.

I was trying to do this question on covectors and got stuck:
If nan_a is a covector orthogonal to the surface β(x)=0\beta(x)=0. Show that
n[a;bnc]n_{[a;b}n_{c]}.
Any ideas?


Wha? That's just an expression. Where's the RHS?
Original post by Zhen Lin
Wha? That's just an expression. Where's the RHS?


oops. sorry. we want to show n[a;bnc]=0n_{[a;b}n_{c]}=0 given that nan_a is orthogonal to the surface β(x)=0\beta(x)=0.

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