ok. so there is just hte one thing left that i don't get: the same thign as earlier where i had(Original post by Zhen Lin)
Yes, it is. But I'm explaining something different  namely the definition of , and how this leads to equation 27. Obviously, if I'm defining , is a fixed index thoughout my calculation.
So it is. Nevermind then.
I have never, ever seen it used for function composition.
now i realise that the 2nd term doesn't make sense for the reason you said, namely that and so doesn't make sense. I understand this aspect of why it is wrong.
What I don't get is how we use the chain rule properly here:
on a very basic level (lol!) if we "divide" by to give us the term then we need to multiply by it to cancel this out  this is why i think there should be a term. Can you explain where I am going wrong here?
Cheers!

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 21122010 20:32

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 22122010 02:27
(Original post by latentcorpse)
What I don't get is how we use the chain rule properly here:
on a very basic level (lol!) if we "divide" by to give us the term then we need to multiply by it to cancel this out  this is why i think there should be a term. Can you explain where I am going wrong here? 
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 22122010 12:43
(Original post by Zhen Lin)
This is a problem with the way our notation for multivariable calculus works. It's much clearer in the onevariable case: . So, analogously, . Rewriting it with t, . (I don't write here because (a) it conflicts with the notation for the operator on manifold functions and (b) it's bad notation which sometimes confuses people.)
(i) In eqn 86, why does that one have a subscript s on it? is that important? if so, why does the in the second term not have it aswell?
(ii)Given the Euler Lagrange equation 88, how do we get eqns 89 and 90. Presumably by integrating eqn 84 but I'm struggling to do this!
(iii) And finally, do you have any ideas for the exercise at the bottom of p32?
Thank you.Last edited by latentcorpse; 22122010 at 13:13. 
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 22122010 13:50
(Original post by latentcorpse)
Awesome! Think I'm finally starting to get the hang of this stuff! I do have a few more questions about some of the material alter on if you're able to help with that?
(i) In eqn 86, why does that one have a subscript s on it? is that important? if so, why does the in the second term not have it aswell?(ii)Given the Euler Lagrange equation 88, how do we get eqns 89 and 90. Presumably by integrating eqn 84 but I'm struggling to do this!(iii) And finally, do you have any ideas for the exercise at the bottom of p32? 
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 22122010 15:20
(Original post by Zhen Lin)
I have no idea. Probably a typo.
I'm pretty sure that's obtained by differentiating G.
Just follow the hint? Take , then e.g. (1, 0, 0, 0) is a timelike vector, (1, 1, 0, 0) is null and (0, 1, 0, 0) is spacelike.
If
then
but they have it over G not root(G). Any ideas where I've gone wrong?
I'm also struggling to get eqn 93 from eqn 88. I don't know how to deal with the or terms? 
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 22122010 15:50
(Original post by latentcorpse)
Thanks. As for the differentiating G stuff, I'm out by a factor of in both cases.
If
then
but they have it over G not root(G). Any ideas where I've gone wrong?
Equation 93 doesn't follow from 88. (Well, actually, it might. But that's irrelevant.) It's the EulerLagrange equation for the action . 
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 22122010 16:10
(Original post by Zhen Lin)
I don't see how you can be getting anywhere.  so if you get that's just 1/G. (In other words: check your substitutions!)
Equation 93 doesn't follow from 88. (Well, actually, it might. But that's irrelevant.) It's the EulerLagrange equation for the action .
So I'm now trying to get (103):
EL for Lagrangian as given in (97) is
But is it correct to say that as there is no explicit dependence, ?
in which case we get to
But I'm fairly sure I've messed up because this doesn't give me any of the delta terms that have appeared?
And secondly, why would (106) follow from (80)? Is it because (80) tells us that
But in an orthonormal basis such as ,
and so (80) tells us that and if then . Is this correct? 
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 22122010 17:26

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 22122010 17:50
(Original post by Zhen Lin)
No. The metric depends on the point!(Original post by Zhen Lin)
No. You can't do that. Those are the components in the coordinate basis, so you must use the given metric: so , where the dot denotes differentiation w.r.t. proper time. Fortunately, we're given that is negligible, so . Then we're also given that is negligible, so it immediately follows that . (Note that is a valid conclusion, but conventionally we choose so that .)
And in (116), why does this follow from (115) only in a coordinate basis?
Surely, is just like from (111) and then since for scalar functions, we would have in any basis, no?
Thanks!Last edited by latentcorpse; 22122010 at 18:03. 
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 23122010 02:07
(Original post by latentcorpse)
But in exercise 2 at the bottom of p36, they say that since L has no explicit tau dependence but it looks to me, like what you were saying, the metric does depend on tau (), no? 
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 23122010 12:46
(Original post by Zhen Lin)
Yes, the metric does depend on when you write it in that form. But the correct way to write the Lagrangian is like this: . In this form it's clear that there's no dependence. L is to be understood as a function of 9 independent variables here. Again, this is difficult to explain due to poor notation  the point is, at the stage where you're partialdifferentiating the Lagrangian, you haven't yet put in the relation between and yet, but when you do the totaldifferentiation, you have set to be the coordinate along some path.
Yes. This is a physics question, not a geometry question.
is defined as differentiating a function w.r.t. to the coordinates. There are bases which may not correspond to any coordinate chart at all. For instance, I might work in an orthonormalised polar coordinate system where . Then, but . Perhaps this is indicative of bad notation...
then
so
=2(1+2\Phi) \dot{t}  2(12 \Phi)(\dot{x}+\dot{y}+\dot{z}) [/latex]
= 2 \partial_i \Phi \dot{t}^2 + 2 \partial_i \Phi ( \dot{x}^2 + \dot{y}^2 + \dot{z}^2)[/latex]
I get the feeling this has gone wrong though! How is it looking?
Then, in the second line of chapter 15, he claims that are the components of in a coordinate basis. However, in equation 36, it would appear that the components are in fact . Which is correct?
And how do we derive (118)?
Also, given that (118) tells us that the Christoffel symbols don't transform as tensors (due to presence of 2nd term in (118)), we also need to show that the 1st term in (116) doesn't transform as a tensor according to the paragraph after (118). However, I find that it does transform as a tensor:
which shows it transforms as a tensor, no?
And in (128), this is the requirement for a torsion free connection apparently. However, I thought that because the Christoffel symbols are symmetric in the two lower indices, all the time. Therefore, how is it possible to ever have a connectino that isn't torsion free?
And in (131), what the hell has happened here lol?
Firstly, how does ?
And secondly, I don't see how the Leibniz rule has been used to get the second equality. The Leibniz rule is for fY NOT f(Y) and here we have g(Y,Z). Do we assume f(Y) and fY are the same thing? If so, I don't understand how to apply the Leibniz rule to a function of 2 vectors?
And finally, do you have any ideas about the exercise on p40? I had a look through Wald's textbook and couldn't find anything useful to help with this either.
Thanks again!Last edited by latentcorpse; 23122010 at 13:40. 
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 23122010 14:22
(Original post by latentcorpse)
Ok. So if we have
then
so
=2(1+2\Phi) \dot{t}  2(12 \Phi)(\dot{x}+\dot{y}+\dot{z}) [/latex]
= 2 \partial_i \Phi \dot{t}^2 + 2 \partial_i \Phi ( \dot{x}^2 + \dot{y}^2 + \dot{z}^2)[/latex]
I get the feeling this has gone wrong though! How is it looking?And how do we derive (118)?Also, given that (118) tells us that the Christoffel symbols don't transform as tensors (due to presence of 2nd term in (118)), we also need to show that the 1st term in (116) doesn't transform as a tensor according to the paragraph after (118). However, I find that it does transform as a tensor:
which shows it transforms as a tensor, no?And in (131), what the hell has happened here lol?
Firstly, how does ?
And secondly, I don't see how the Leibniz rule has been used to get the second equality. The Leibniz rule is for fY NOT f(Y) and here we have g(Y,Z). Do we assume f(Y) and fY are the same thing? If so, I don't understand how to apply the Leibniz rule to a function of 2 vectors?And finally, do you have any ideas about the exercise on p40? I had a look through Wald's textbook and couldn't find anything useful to help with this either.Thanks again!Last edited by Zhen Lin; 23122010 at 14:29. 
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 23122010 14:57
(Original post by Zhen Lin)
That's gone horribly wrong. Just work with the general case and you should be fine. (For what it's worth, and , for example.)(Original post by Zhen Lin)
Well, we have , so yes, the components are what he claims. Equation 36 is irrelevant — the subscripts there indicate evaluation at a point, not components. (Again, indicative of bad notation.)(Original post by Zhen Lin)
Your calculation assumes the tensoriality. The definition is , but is in general not a constant and doesn't commute with derivatives!(Original post by Zhen Lin)
Yes, the Christoffel symbol, in a coordinate basis, is symmetric in the lower two indices. But that's because the Christoffel symbol is the coordinate expression for the Levi–Civita conection; there are other connections which may not be torsionfree.(Original post by Zhen Lin)
If I write it out in index notation, is it clearer? It's just the statement . The Leibniz rule applies because the metric is bilinear.
now I definitely think I've messeg up here  I have the two terms I want plus these two additional terms. Surely they don't make sense as I just have the pair (Y,Z) sitting on their own without anything acting on them!(Original post by Zhen Lin)
Have you considered asking on a different forum? I feel like I'm the only one looking at this thread, and I haven't even taken this course (or differential geometry, for that matter).
Cheers 
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 23122010 17:07
(Original post by latentcorpse)
What do you mean the general case, sorry?I see that eqn 36 is evaluated at a point but underneath he writes that it gives the components? Is it because eqn 36 gives the components of the tensor and here in chapter 15, he is talking about the components of the tensor FIELD?I don't understand what you're on about here. Could you elaborate please?I think I "kind of" understand this. I can't really see how it would ever be nonzero though? Or do you mean that for a different connection (or for the same connection in a different i.e. non coordinate basis), eqn 125 would change and then 126 would change and so we would end up with a different condition for torsion free connections?
Because the action of the covariant derivative on a scalar field is exactly the same as forming the covector of partial derivatives with respect to the coordinates (in a coordinate basis, of course). 
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 23122010 18:20
(Original post by latentcorpse)
And how do we derive (118)? 
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 23122010 18:58
(Original post by Zhen Lin)
Let and compute the derivatives. You should get and . Then write out the components explicitly for your given metric.
Take
How is that looking? Any tips on how to proceed?
I'd also like to ask about (138):
(i) In the first equality, we go through and stick an f on the left of every X (obviously!) but then take for example the 2nd term on the 1st line . How are we able to move the f out from to give this?
(ii) In the second equality of (138), he rewrites . How does this work? I can see it is some sort of product rule but can we write it in indices to make it clearer?
And just under (137), he writes "This determines uniquely because the metric is nondegenerate. How does this work?
And I'm a bit confused by the example on p39, he says if we take to be partial differentiation, it satisfies all the above conditions for a covariant derivative  how can this be? He explicitly states at the start of chapter 15 that partial derivatives are no good as covariant derivatives as they don't transform as tensors!
Also, in (148), where does this come from? I don't really understand the bit above it where he says that affinely parameterised geodesics through p are given in normal coordinates by .
Then a few small things about the next few lines:
(i) in (149), why do we symmetrise the lower indices?
(ii) He then says "hence, for a torsion free connection......" but I don't see where he has made use of the connection being torsion free at any point in this argument?
(iii) And in (150) he says he has applied the above to the Levi Civita connection  how has this given us (150) though?
And finally, any thoughts for the exercise on p46?
Edit: As for your post on 118, I'm fairly sure there should be a minus sign because of what you showed me earlier when we worked out how transforms  its unwanted term had a + sign so this way they would cancel and the covariant derivative would (as expected) transform as a tensor!
However, in your post above, why have you equated the primed version with the coordinate basis version? Just because we know what this looks like? And when I try and prove it, should I expand on the RHS or the LHS of that equation?
Thanks again! And apologies for bombarding you with so many questions!Last edited by latentcorpse; 23122010 at 23:57. 
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 24122010 04:35
(Original post by latentcorpse)
So I get the t equation fine. Now for the i equation
Take
How is that looking? Any tips on how to proceed?And I'm a bit confused by the example on p39, he says if we take to be partial differentiation, it satisfies all the above conditions for a covariant derivative  how can this be? He explicitly states at the start of chapter 15 that partial derivatives are no good as covariant derivatives as they don't transform as tensors!
Equation 148 follows trivially once we have this — just differentiate!
The intuitive picture is much clearer, so I suggest you think about the definition of geodesic normal coordinates more thoroughly before proceeding.Then a few small things about the next few lines:
(i) in (149), why do we symmetrise the lower indices?(ii) He then says "hence, for a torsion free connection......" but I don't see where he has made use of the connection being torsion free at any point in this argument?(iii) And in (150) he says he has applied the above to the Levi Civita connection  how has this given us (150) though?And finally, any thoughts for the exercise on p46?Edit: As for your post on 118, I'm fairly sure there should be a minus sign because of what you showed me earlier when we worked out how transforms  its unwanted term had a + sign so this way they would cancel and the covariant derivative would (as expected) transform as a tensor!
However, in your post above, why have you equated the primed version with the coordinate basis version? Just because we know what this looks like?And when I try and prove it, should I expand on the RHS or the LHS of that equation? 
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 24122010 13:15
(Original post by Zhen Lin)
That doesn't look right — you seem to be missing a term where the index the metric is differentiated w.r.t. is contracted with a velocity vector. I can't be bothered to check your calculations — this is nothing more than a simple exercise in differentiation — so I'll just tell you what the general result is: . Raising indices, it's . You should check that this is the same as .(Original post by Zhen Lin)
Basically, by the existence and uniqueness theorem for ODEs, at every point , and every tangent vector , there is a geodesic which passes through p with tangent vector (at p). The equation is basically justifying the abuse of notation (but I still think it's bad notation) — let's try to make this clearer. Let be the unique geodesic s.t. and . Let be the normal coordinate chart, i.e. the chart such that . Then, , where gives the components of a vector w.r.t the coordinate basis induced by . So, if we write for the th component of and for the th component of we get the claimed equation.
Equation 148 follows trivially once we have this — just differentiate!
So you've definde as the geodesic through p with tangent vector at
is taken to be a chart....fair enough but then why is ?
Moreover, how does this map even make sense, the domain of exp is but the codomain of is ?
And where does come from?(Original post by Zhen Lin)
I don't know how to do it in that formalism. It's fairly straightforward in index notation: An affinely parametrised geodesic has velocity vector satisfying . Differentiate using the product rule and you should get terms which correspond to the previous equation contracted with .
"An affinely parameterised geodesic is an integral curve of a vector field where satisfies .
Now my interpretation is that given , this is the same as eqn (144) beign true which is the same as (142) being true and so it reduces to your definition i.e. an affinely parameterised geodesic has velocity vector . Is this correct?
Secondly, in doing this calculation, I don't understand what you are asking me to do? What do I differentiate wrt to?
Surely I want to write
where in the last equality we used the fact that for the Levi Civita connection, the metric is covariantly constant.
but then the symmetry of the metric can reduce this to
But this is just
and for affinely parameterised geodesics so we get
Now I think there should be some way of saying this is zero but I don't know how? Something to do with nondegeneracy perhaps?
I'm also confusing myself on (150),
we have
now if we antisymmetrise on we get:
and this appears to reduce to 0=0
Aargh!
What am I doing wrong here? When he says antisymmetrise I'm sure he expects us to jump to
But I don't accept this. To me it appears he has antisymmetrised the 1st term and just combined ( i.e. not bothered to antisymmetrise) the 2nd and 3rd.
Surely you can't do different operations on different terms?
Cheers!Last edited by latentcorpse; 24122010 at 13:38. 
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 24122010 13:43
(Original post by latentcorpse)
Yeah but the whole point fo the EL equations is that we produce a new equation that we can then compare to the geodesic equation in order to read offf the Christoffel symbols. At what point in the calculation should I substitute stuff in in order to get my equation?
Linearity, basically. , and so .Ok. As far as these notes go, affine parameterisation is defined on p45 as follows:
"An affinely parameterised geodesic is an integral curve of a vector field where satisfies .
Now my interpretation is that given , this is the same as eqn (144) beign true which is the same as (142) being true and so it reduces to your definition i.e. an affinely parameterised geodesic has velocity vector . Is this correct?Secondly, in doing this calculation, I don't understand what you are asking me to do? What do I differentiate wrt to?Last edited by Zhen Lin; 24122010 at 13:45. 
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 24122010 15:38
(Original post by Zhen Lin)
You can do it at any point, as long as you do it correctly!
To get the equations for what should i be setting my indices equal to?
I reckon it should be , no?(Original post by Zhen Lin)
I miswrote. Fix a point , then the exponential map at that point is . Then, the geodesic normal coordinate chart has the property that , essentially by definition. This is just formalising the intuitive idea that the exponential map identifies points on the manifold with vectors in the tangent space.
Linearity, basically. , and so .
So, can we talk through this slowly:
(i) We're trying to show that affinely parameterised geodesics through p, in normal coordinates, satisfy .
(ii) So we set up exp as described at teh top of p47 and our chart as we've been using previously. Now you go and introduce this . How did you know to introduce this and how did you know what form it would have to take?
(iii) Ok. so given the above, we define these maps s.t.
(iv)Now we consider the geodesic
First of all, back on p14, we described a smooth curve as a smooth function where I is an open interval. Your interval is closed though?
Also, why does your interval not just go from 0 to 1. Our initial conditions are at 0 i.e.
And just now that I've raised the point, we discussed smooth curves as going from open intervals into the manifolds but when we were dealing with geodesics earlier in chapter 14 we were talking about curves from p to q and how and . But how can this make sense as so neither 0 nor 1 are in the domain?
(iv) Given this geodesic, you tell me that Why?
(v) And how does that then equal ?
(vi) Ok so finally we prove that . Can you explain in words how this proves ?(Original post by Zhen Lin)
g is bilinear, so automatically. This should be familiar from linear algebra, I hope. There's no need to invoke nondegeneracy or anything sophisticated.
I'm also confusing myself on (150),
we have
now if we antisymmetrise on we get:
and this appears to reduce to 0=0
Aargh!
What am I doing wrong here? When he says antisymmetrise I'm sure he expects us to jump to
But I don't accept this. To me it appears he has antisymmetrised the 1st term and just combined ( i.e. not bothered to antisymmetrise) the 2nd and 3rd.
Surely you can't do different operations on different terms?
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