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    (Original post by latentcorpse)
    Hmmm. So I get ( \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)=\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=\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 g_\mu \rho} \partial_\nu X^\rho, that is NOT equal to \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 \frac{\partial g_{\mu \nu}}{\partial z} term?
    They're annihilated by the components of X: X^\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. \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 \frac{\partial}{\partial t} and \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, \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.)
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    (Original post by Zhen Lin)
    Read the notation carefully. That is what he's done.
    He has \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
    \left( \frac{\partial}{\partial s} T ( s, x^i ) \right) though?
    Why does he get \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 \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
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    (Original post by latentcorpse)
    He has \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
    \left( \frac{\partial}{\partial s} T ( s, x^i ) \right) though?
    Why does he get \frac{\partial}{\partial t} instead?
    Then you misunderstand the notation. \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 \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 dr^2 term is negative and the coefficient on the dt^2 term is positive, when r < 2M.
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    (Original post by Zhen Lin)
    Then you misunderstand the notation. \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 \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 \frac{\partial}{\partial x}
    but in what he has written down we use t in the limit AND in the \frac{\partial}{\partial t}.
    So shouldn't it be \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 dr^2 term is negative and the coefficient on the dt^2 term is positive, when r < 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 E^2=2V(r) where V(r) is given by (254) with \sigma=1 for timelike geodesics. However, he says that we get (258) by evaluating at r=r_\pm. Well r_\pm is given in (257) so how the hell are there any r terms left in the final answer (258)?

    And underneath (258) he says that an orbit with large r has E \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=\sqrt{\frac{8}{9}}.
    I find
    E=\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 \frac{dt}{dr} has a simple pole at r=2M, t will diverge logarithmically as r \rightarrow 2M. Why logarithmically?

    And in (261), I'm trying to find \frac{dr_*}{dr}:
    \frac{dr_*}{dr}=1+\frac{2M}{\fra  c{r}{2M}-1} \frac{1}{2M} where we used the chain rule
    =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
    \frac{dv}{dt}=\frac{dt}{dt}+ \frac{dr_*}{dt}=1+(-1)=0 where \frac{dr_*}{dt}=-1 since \frac{dt}{dr_*}=-1 for incoming radial null geodesics?
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    (Original post by latentcorpse)
    Sorry to drag this out. But the basic defn is
    f'(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 \frac{\partial}{\partial x}
    but in what he has written down we use t in the limit AND in the \frac{\partial}{\partial t}.
    So shouldn't it be \frac{\partial}{\partial s} if we take the limit with respect to t?
    *sigh* Look, all I'm saying is that f'(x) = \left. \frac{d}{dt} \left[ f(t) \right] \right|_x. It's really just a notation thing. To make another analogy, A^{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 \gamma : I \to M be an affinely-parametrised geodesic. Then, writing x^\mu for the coordinates of \gamma(\tau), let \displaystyle V^\mu = \frac{d x^\mu}{d \tau}. Then, the geodesic equation tells us \displaystyle \frac{d V^\mu}{d \tau} + \Gamma^{\mu}_{\phantom{\mu} \nu \rho} V^\nu V^\rho = 0. So, in particular, \nabla_V f = V^\rho \partial_\rho f when f: M \to \mathbb{R} is a scalar field. But by the chain rule, \displaystyle \frac{d}{d\tau} = \frac{d x^\mu}{d \tau} \frac{\partial}{\partial x^\mu} = V^\mu \partial_\mu, so \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.
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    (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 ds^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, \theta, \phi)
    would the metric be

    g_{\mu \nu} = \begin{pmatrix} -a & b & 0 & 0 \\ b & c & 0 & 0 \\ 0 & 0 & d & 0 \\ 0 & 0 & 0 & e \end{pmatrix}

    or
    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 g^{vv}=-\frac{1}{1-\frac{2M}{r}} which is definitely not smooth at r=2M?

    Thank you!
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    (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 ds^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, \theta, \phi)
    would the metric be

    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 \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 g^{vv}=-\frac{1}{1-\frac{2M}{r}} which is definitely not smooth at r=2M?
    I think you need to compute the inverse matrix there, since the metric is not diagonal.
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    (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 \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. (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, L_\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!
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    (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. (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 \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, L_\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.)
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    (Original post by Zhen Lin)
    Your derivation is incorrect. In general \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 L_\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 \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 \xi term in (291)? i.e. why is there no \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 u_ae^a_i=0 and g_{ab}e^a_ie^b_j=\delta_{ij} in (310)? This seems quite important and I don't understand!

    Thanks!
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    (Original post by latentcorpse)
    Ok. Thanks.

    Just above (288) though, he says we can use (239) to write L_\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 \eta_{\mu \nu} is constant. (I presume g_{\mu \nu} = \eta_{\mu \nu} here. I can't be bothered to read the whole section.)

    And what has happened to the second \xi term in (291)? i.e. why is there no \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.
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    (Original post by Zhen Lin)
    Well, that is true, but it follows from (238) rather than (239). Basically it's because \eta_{\mu \nu} is constant. (I presume g_{\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 e^a_0=u^a?

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

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

    i don't see why the u^\mu should become a u^b - it's just the components (i.e. a function) not a vector, isn't it?
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    (Original post by latentcorpse)
    at the top of p97, why is e^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: \displaystyle u^a \partial_a = \frac{d}{d\tau}, by the chain rule. So the action of u^a \nabla_a on a scalar is just \displaystyle \frac{d}{d\tau}.
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    (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: \displaystyle u^a \partial_a = \frac{d}{d\tau}, by the chain rule. So the action of u^a \nabla_a on a scalar is just \displaystyle \frac{d}{d\tau}.
    but you've use \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?
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    (Original post by latentcorpse)
    but you've use \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. \nabla_a f = \partial_a f for scalar fields f, by definition.
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    (Original post by Zhen Lin)
    No. \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?
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    (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.
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    (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 n_a is a covector orthogonal to the surface \beta(x)=0. Show that
    n_{[a;b}n_{c]}.
    Any ideas?
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    (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 n_a is a covector orthogonal to the surface \beta(x)=0. Show that
    n_{[a;b}n_{c]}.
    Any ideas?
    Wha? That's just an expression. Where's the RHS?
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    (Original post by Zhen Lin)
    Wha? That's just an expression. Where's the RHS?
    oops. sorry. we want to show n_{[a;b}n_{c]}=0 given that n_a is orthogonal to the surface \beta(x)=0.
 
 
 
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