Just drop them.(Original post by latentcorpse)
Hmmm. So I get
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?Read the notation carefully. That is what he's done.Write it out in components. If t is the first coordinate then X = (1, 0, 0, ...)  so all its partial derivatives are zero.
In general, yes. When the metric is especially nice (e.g. constant) then it will commute.Yes, but this is not the covariant derivative.However the metric does commute with the covariant derivative doesn't it?
They're annihilated by the components of X: .V is tangent to an affinelyparametrised geodesic.Where did we define V this way? I still don't see this step.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.)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?SO(3) is the rotation group of a 2sphere. SO(3) itself is a 3dimensional manifold (basically a twisted 3sphere) and it has subgroups of various topological dimensions. A 1dimensional subgroup, by definition, is a 1parameter family of isometries. And a 1parameter family of isometries gives rise to a Lie derivative, which gives rise to a Killing vector field.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?Just a coordinate.On the remarks on p75:
He says that r is not the distance from the origin. What is it then?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.)r is just a coordinate. He's pointing out that you can assume M and r have the same sign, i.e. positive.And what is he on about in the 3rd remark? Surely changing either r or M to r or M is physically unrealistic?I presume this is a reference to Noether's theorem.A simple substitution, .How does he derive (250)?Sure, if you want to define length that way.Do null geodesics have zero length?No.Doesn't this violate causality?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.)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?

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 02012011 01:46
Last edited by Zhen Lin; 02012011 at 01:48. 
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 02012011 12:38
(Original post by Zhen Lin)
Read the notation carefully. That is what he's done.
I think that from the definition it should be
though?
Why does he get instead?(Original post by Zhen Lin)
V is tangent to an affinelyparametrised geodesic.(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.)
ThanksLast edited by latentcorpse; 02012011 at 13:20. 
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 02012011 13:47
(Original post by latentcorpse)
He has as his answer
I think that from the definition it should be
though?
Why does he get instead?
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?Last edited by Zhen Lin; 02012011 at 13:56. 
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 02012011 17:43
(Original post by Zhen Lin)
Then you misunderstand the notation. . The point is that the letter used doesn't matter; what matters is where it's evaluated.
so if we take the limit with respect to h then we use the OTHER variable in the
but in what he has written down we use t in the limit AND in the .
So shouldn't it be 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.(Original post by Zhen Lin)
The coefficient on the term is negative and the coefficient on the term is positive, when .
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 where V(r) is given by (254) with for timelike geodesics. However, he says that we get (258) by evaluating at . Well is given in (257) so how the hell are there any terms left in the final answer (258)?
And underneath (258) he says that an orbit with large r has . 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 .
I find
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 has a simple pole at , t will diverge logarithmically as . Why logarithmically?
And in (261), I'm trying to find :
where we used the chain rule
and then I cannot rearrange it to what we want....
At the top of p85, is v constant along ingoing radial null geodesics because
where since for incoming radial null geodesics?Last edited by latentcorpse; 02012011 at 18:40. 
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 02012011 18:43
(Original post by latentcorpse)
Sorry to drag this out. But the basic defn is
so if we take the limit with respect to h then we use the OTHER variable in the
but in what he has written down we use t in the limit AND in the .
So shouldn't it be if we take the limit with respect to t?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.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?
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?Last edited by Zhen Lin; 02012011 at 18:46. 
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 02012011 18:59
(Original post by Zhen Lin)
No. I can't answer any of your other questions either, since I haven't studied general relativity itself.
Also, if we had in a coordinate chart
would the metric be
or
How does the metric in (266) have a smooth inverse metric? Surely which is definitely not smooth at ?
Thank you!Last edited by latentcorpse; 02012011 at 19:07. 
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 03012011 01:53
(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.
Yes. You could have easily checked this yourself by writing it all out. (In matrix form: compute .)(Original post by latentcorpse)
How does the metric in (266) have a smooth inverse metric? Surely which is definitely not smooth at ?Last edited by Zhen Lin; 03012011 at 01:55. 
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 03012011 22:26
(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 .)
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. . 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, ,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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 04012011 03:04
(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. . 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?
This is approximation stuff, and really not something I know about. (I don't do much applied mathematics in general.) 
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 04012011 15:52
(Original post by Zhen Lin)
Your derivation is incorrect. In general .
This is approximation stuff, and really not something I know about. (I don't do much applied mathematics in general.)
Just above (288) though, he says we can use (239) to write
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 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 term in (291)? i.e. why is there no 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 and in (310)? This seems quite important and I don't understand!
Thanks!Last edited by latentcorpse; 04012011 at 16:41. 
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 04012011 16:19
(Original post by latentcorpse)
Ok. Thanks.
Just above (288) though, he says we can use (239) to write
I have no idea.Why does (300) imply the waves are transverse?
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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 04012011 16:45
(Original post by Zhen Lin)
Well, that is true, but it follows from (238) rather than (239). Basically it's because is constant. (I presume 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.
A couple of last (mathematical) things:
at the top of p97, why is ?
and how do you go from (313) to (314)?
Say you have as in (311),
we can write this as
why is this equal to ?
i don't see why the should become a  it's just the components (i.e. a function) not a vector, isn't it?Last edited by latentcorpse; 04012011 at 16:57. 
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 04012011 16:57
and how do you go from (313) to (314)? 
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 04012011 17:22
(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" 4velocity, whatever that actually means.
I think this is the third time I've had to explain this, or something like this: , by the chain rule. So the action of on a scalar is just . 
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 04012011 17:36
(Original post by latentcorpse)
but you've use 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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 05012011 01:35
Why in (240), does he say the solutions of the equation are Killing vector fields? Surely they are covectors? 
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 05012011 03:31
(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? 
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 05012011 11:38
(Original post by Zhen Lin)
Just dualise with the metric and you'll have a vector field.
I was trying to do this question on covectors and got stuck:
If is a covector orthogonal to the surface . Show that
.
Any ideas? 
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 05012011 11:51
(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 is a covector orthogonal to the surface . Show that
.
Any ideas? 
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 05012011 12:25
(Original post by Zhen Lin)
Wha? That's just an expression. Where's the RHS?
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