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    (Original post by Zhen Lin)
    Yes, it is. But I'm explaining something different - namely the definition of \displaystyle \frac{\partial}{\partial x^\mu}, and how this leads to equation 27. Obviously, if I'm defining \displaystyle \frac{\partial}{\partial x^\mu}, \mu is a fixed index thoughout my calculation.



    So it is. Nevermind then.



    I have never, ever seen it used for function composition.
    ok. so there is just hte one thing left that i don't get: the same thign as earlier where i had

     \frac{\partial F}{\partial x^\mu} \frac{ d x^\mu \phi \circ \lambda(t)}{dt}

    now i realise that the 2nd term doesn't make sense for the reason you said, namely that x^\mu,\phi : M \rightarrow \mathbb{R}^n and so x^\mu \circ \phi 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 x^\mu to give us the  \frac{\partial F}{\partial x^\mu} term then we need to multiply by it to cancel this out - this is why i think there should be a x^\mu \circ \phi term. Can you explain where I am going wrong here?

    Cheers!
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    (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 x^\mu to give us the  \frac{\partial F}{\partial x^\mu} term then we need to multiply by it to cancel this out - this is why i think there should be a x^\mu \circ \phi term. Can you explain where I am going wrong here?
    This is a problem with the way our notation for multivariable calculus works. It's much clearer in the one-variable case: (f \circ g)' = f' \circ g \cdot g'. So, analogously, (F \circ \phi \circ \lambda)' = \partial_\mu F \circ \phi \circ \lambda \cdot (x^\mu \circ \lambda)'. Rewriting it with t, \displaystyle \frac{d}{dt} \left[F \circ \phi \circ \lambda(t) \right] = \partial_\mu F (\phi \circ \lambda(t)) \frac{d}{dt} \left[x^\mu \circ \lambda(t) \right]. (I don't write \displaystyle \frac{\partial F}{\partial x^\mu} here because (a) it conflicts with the notation \displaystyle \frac{\partial}{\partial x^\mu} for the operator on manifold functions and (b) it's bad notation which sometimes confuses people.)
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    (Original post by Zhen Lin)
    This is a problem with the way our notation for multivariable calculus works. It's much clearer in the one-variable case: (f \circ g)' = f' \circ g \cdot g'. So, analogously, (F \circ \phi \circ \lambda)' = \partial_\mu F \circ \phi \circ \lambda \cdot (x^\mu \circ \lambda)'. Rewriting it with t, \displaystyle \frac{d}{dt} \left[F \circ \phi \circ \lambda(t) \right] = \partial_\mu F (\phi \circ \lambda(t)) \frac{d}{dt} \left[x^\mu \circ \lambda(t) \right]. (I don't write \displaystyle \frac{\partial F}{\partial x^\mu} here because (a) it conflicts with the notation \displaystyle \frac{\partial}{\partial x^\mu} for the operator on manifold functions and (b) it's bad notation which sometimes confuses people.)
    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 x^\mu have a subscript s on it? is that important? if so, why does the x^\mu 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.
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    (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 x^\mu have a subscript s on it? is that important? if so, why does the x^\mu in the second term not have it aswell?
    I have no idea. Probably a typo.

    (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!
    I'm pretty sure that's obtained by differentiating G.

    (iii) And finally, do you have any ideas for the exercise at the bottom of p32?
    Just follow the hint? Take g_{ab} = \eta_{ab}, then e.g. (1, 0, 0, 0) is a timelike vector, (1, 1, 0, 0) is null and (0, 1, 0, 0) is spacelike.
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    (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 g_{ab} = \eta_{ab}, then e.g. (1, 0, 0, 0) is a timelike vector, (1, 1, 0, 0) is null and (0, 1, 0, 0) is spacelike.
    Thanks. As for the differentiating G stuff, I'm out by a factor of G^{\frac{1}{2}} in both cases.

    If G = \sqrt{ -g_{\mu \nu} \dot{x}^\mu \dot{x}^\nu }

    then \frac{\partial G}{\partial \dot{x}^\mu} = \frac{1}{2} G^{-\frac{1}{2}} \cdot -2 g_{\mu \nu} \dot{x}^\nu

    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 \frac{\partial G}{\partial \dot{x}^\mu} or \frac{\partial G}{\partial x^\mu} terms?
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    (Original post by latentcorpse)
    Thanks. As for the differentiating G stuff, I'm out by a factor of G^{\frac{1}{2}} in both cases.

    If G = \sqrt{ -g_{\mu \nu} \dot{x}^\mu \dot{x}^\nu }

    then \frac{\partial G}{\partial \dot{x}^\mu} = \frac{1}{2} G^{-\frac{1}{2}} \cdot -2 g_{\mu \nu} \dot{x}^\nu

    but they have it over G not root(G). Any ideas where I've gone wrong?
    I don't see how you can be getting \sqrt{G} anywhere. G = \sqrt{\cdots} - so if you get \displaystyle \frac{1}{\sqrt{\cdots}} that's just 1/G. (In other words: check your substitutions!)

    I'm also struggling to get eqn 93 from eqn 88. I don't know how to deal with the \frac{\partial G}{\partial \dot{x}^\mu} or \frac{\partial G}{\partial x^\mu} terms?
    Equation 93 doesn't follow from 88. (Well, actually, it might. But that's irrelevant.) It's the Euler-Lagrange equation for the action \displaystyle \int G^2 \, du.
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    (Original post by Zhen Lin)
    I don't see how you can be getting \sqrt{G} anywhere. G = \sqrt{\cdots} - so if you get \displaystyle \frac{1}{\sqrt{\cdots}} 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 Euler-Lagrange equation for the action \displaystyle \int G^2 \, du.
    OK. That was a bad mistake on the \sqrt{G} stuff!

    So I'm now trying to get (103):

    E-L for Lagrangian as given in (97) is

    \frac{d}{d \tau} \left( \frac{\partial}{\partial \dot{x}^\mu} \left( - g_{\mu \nu} \dot{x}^\mu \dot{x}{^\nu} \right) \right)= \frac{ \partial L}{\partial x^\mu}

    But is it correct to say that as there is no explicit x^\mu dependence, \frac{\partial L}{\partial x^\mu}=0?

    in which case we get to

    \frac{d}{d \tau} \left( g_{\mu \nu} \dot{x}^\nu \right)=0 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
    g_{\mu \nu} \dot{x}^\mu \dot{x}^\nu =-1
    But in an orthonormal basis such as (t,x,y,z), g_{\mu \nu}=\eta_{\mu \nu} = \text{diag}(-1,1,1,1)
    and so (80) tells us that -(\frac{dt}{d \tau})^2+(\frac{dx}{d \tau})^2+(\frac{dy}{d \tau})^2+(\frac{dz}{d \tau})^2=-1 and if (\frac{d x^i}{d \tau})^2 \simeq 0 then \frac{dt}{d \tau} \simeq 1. Is this correct?
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    (Original post by latentcorpse)
    But is it correct to say that as there is no explicit x^\mu dependence, \frac{\partial L}{\partial x^\mu}=0?
    No. The metric depends on the point!

    And secondly, why would (106) follow from (80)? Is it because (80) tells us that
    g_{\mu \nu} \dot{x}^\mu \dot{x}^\nu =-1
    But in an orthonormal basis such as (t,x,y,z), g_{\mu \nu}=\eta_{\mu \nu} = \text{diag}(-1,1,1,1)
    and so (80) tells us that -(\frac{dt}{d \tau})^2+(\frac{dx}{d \tau})^2+(\frac{dy}{d \tau})^2+(\frac{dz}{d \tau})^2=-1 and if (\frac{d x^i}{d \tau})^2 \simeq 0 then \frac{dt}{d \tau} \simeq 1. Is this correct?
    No. You can't do that. Those are the components in the coordinate basis, so you must use the given metric: so -(1 + 2 \Phi) \dot{t}^2 + (1 - 2 \Phi) (\dot{x}^2 + \dot{y}^2 + \dot{z}^2) = -1, where the dot denotes differentiation w.r.t. proper time. Fortunately, we're given that \Phi is negligible, so -(1 + 2 \Phi) \dot{t}^2 + (1 - 2 \Phi) (\dot{x}^2 + \dot{y}^2 + \dot{z}^2) \approx -\dot{t}^2 + \dot{x}^2 + \dot{y}^2 + \dot{z}^2. Then we're also given that \dot{x}^2 + \dot{y}^2 + \dot{z}^2 is negligible, so it immediately follows that -\dot{t}^2 \approx -1. (Note that \dot{t} \approx -1 is a valid conclusion, but conventionally we choose \tau so that \dot{t} > 0.)
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    (Original post by Zhen Lin)
    No. The metric depends on the point!
    But in exercise 2 at the bottom of p36, they say that \frac{\partial L}{\partial \tau}=0 since L has no explicit tau dependence but it looks to me, like what you were saying, the metric does depend on tau (g_{\mu \nu} ( x ( \tau))), no?




    (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 -(1 + 2 \Phi) \dot{t}^2 + (1 - 2 \Phi) (\dot{x}^2 + \dot{y}^2 + \dot{z}^2) = -1, where the dot denotes differentiation w.r.t. proper time. Fortunately, we're given that \Phi is negligible, so -(1 + 2 \Phi) \dot{t}^2 + (1 - 2 \Phi) (\dot{x}^2 + \dot{y}^2 + \dot{z}^2) \approx -\dot{t}^2 + \dot{x}^2 + \dot{y}^2 + \dot{z}^2. Then we're also given that \dot{x}^2 + \dot{y}^2 + \dot{z}^2 is negligible, so it immediately follows that -\dot{t}^2 \approx -1. (Note that \dot{t} \approx -1 is a valid conclusion, but conventionally we choose \tau so that \dot{t} > 0.)
    Ok. But how do we know to pick \frac{ds^2}{d \tau^2} = -1? I assume this is because we are treating it as a massive particle that will follow a timelike curve? If this is right, how did you know this in the first place?

    And in (116), why does this follow from (115) only in a coordinate basis?
    Surely, e_\nu(Y^\mu) is just like X(f) = \nabla_X(f) from (111) and then since for scalar functions, f_{;a}=f_{,a} we would have e_\nu(Y^\mu)=y^\mu{}_{,\nu} in any basis, no?

    Thanks!
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    (Original post by latentcorpse)
    But in exercise 2 at the bottom of p36, they say that \frac{\partial L}{\partial \tau}=0 since L has no explicit tau dependence but it looks to me, like what you were saying, the metric does depend on tau (g_{\mu \nu} ( x ( \tau))), no?
    Yes, the metric does depend on \tau when you write it in that form. But the correct way to write the Lagrangian is like this: L(\tau, x^\sigma, \dot{x}^\sigma) = g_{\mu \nu}(x^\sigma) \dot{x}^\mu \dot{x}^\nu. In this form it's clear that there's no \tau 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 partial-differentiating the Lagrangian, you haven't yet put in the relation between x^\sigma and \tau yet, but when you do the total-differentiation, you have set x^\sigma to be the coordinate along some path.

    Ok. But how do we know to pick \frac{ds^2}{d \tau^2} = -1? I assume this is because we are treating it as a massive particle that will follow a timelike curve? If this is right, how did you know this in the first place?
    Yes. This is a physics question, not a geometry question.

    And in (116), why does this follow from (115) only in a coordinate basis?
    Surely, e_\nu(Y^\mu) is just like X(f) = \nabla_X(f) from (111) and then since for scalar functions, f_{;a}=f_{,a} we would have e_\nu(Y^\mu)=y^\mu{}_{,\nu} in any basis, no?
    f_{,\nu} 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 \displaystyle e_1= \frac{\partial}{\partial r}, e_2 = \frac{1}{r} \frac{\partial}{\partial \theta}, e_3 = \frac{1}{r \sin \theta} \frac{\partial}{\partial \phi} . Then, \displaystyle e_2 (f) = \frac{1}{r} \frac{\partial f}{\partial \theta} but \displaystyle f_{,2} = \frac{\partial f}{\partial \theta}. Perhaps this is indicative of bad notation...
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    (Original post by Zhen Lin)
    Yes, the metric does depend on \tau when you write it in that form. But the correct way to write the Lagrangian is like this: L(\tau, x^\sigma, \dot{x}^\sigma) = g_{\mu \nu}(x^\sigma) \dot{x}^\mu \dot{x}^\nu. In this form it's clear that there's no \tau 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 partial-differentiating the Lagrangian, you haven't yet put in the relation between x^\sigma and \tau yet, but when you do the total-differentiation, you have set x^\sigma to be the coordinate along some path.



    Yes. This is a physics question, not a geometry question.



    f_{,\nu} 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 \displaystyle e_1= \frac{\partial}{\partial r}, e_2 = \frac{1}{r} \frac{\partial}{\partial \theta}, e_3 = \frac{1}{r \sin \theta} \frac{\partial}{\partial \phi} . Then, \displaystyle e_2 (f) = \frac{1}{r} \frac{\partial f}{\partial \theta} but \displaystyle f_{,2} = \frac{\partial f}{\partial \theta}. Perhaps this is indicative of bad notation...
    Ok. So if we have

    ds^2=-(1+2 \Phi) dt^2 + (1-2 \Phi)(dx^2+dy^2+dz^2)
    then
    L=-g_{\mu \nu} \dot{x}^\mu \dot{x}^\nu = (1+2 \Phi) \dot{t}^2 - (1-2 \Phi)(\dot{x}^2+\dot{y}^2+\dot{z  }^2)

    so

    \frac{\partial L}{\partial \dot{x}^\mu} =2(1+2\Phi) \dot{t} - 2(1-2 \Phi)(\dot{x}+\dot{y}+\dot{z}) [/latex]

    \frac{\partial L}{\partial x^\mu} = 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 \frac{\partial f}{\partial x^\mu} are the components of (df)_a in a coordinate basis. However, in equation 36, it would appear that the components are in fact (\frac{\partial F}{\partial x^\mu})_{\phi(p)}. 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:

    Y'^\mu{}_{, \mu} = \frac{\partial Y'^\mu}{\partial x'^\nu} = \frac{\partial x'^\mu}{\partial x^\rho} \frac{\partial Y^\rho}{\partial x'^\nu} = \frac{\partial x'^\mu}{\partial x^\rho} \frac{\partial x^\lambda}{\partial x'^\nu} \frac{\partial Y^\rho}{\partial x^\lambda} = \frac{\partial x'^\mu}{\partial x^\rho} \frac{\partial x^\lambda}{\partial x'^\nu} Y^\rho_{, \lambda}
    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, \Gamma^\rho{}_{[\mu \nu]}=0 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 X(g(Y,Z))=\nabla_X(g(Y,Z))?
    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!
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    (Original post by latentcorpse)
    Ok. So if we have

    ds^2=-(1+2 \Phi) dt^2 + (1-2 \Phi)(dx^2+dy^2+dz^2)
    then
    L=-g_{\mu \nu} \dot{x}^\mu \dot{x}^\nu = (1+2 \Phi) \dot{t}^2 - (1-2 \Phi)(\dot{x}^2+\dot{y}^2+\dot{z  }^2)

    so

    \frac{\partial L}{\partial \dot{x}^\mu} =2(1+2\Phi) \dot{t} - 2(1-2 \Phi)(\dot{x}+\dot{y}+\dot{z}) [/latex]

    \frac{\partial L}{\partial x^\mu} = 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?
    That's gone horribly wrong. Just work with the general case and you should be fine. (For what it's worth, \displaystyle \frac{\partial T}{\partial \dot{x}} = -2(1 - 2 \Phi) \dot{x} and \displaystyle \frac{\partial L}{\partial x} = 2 \frac{\partial \Phi}{\partial x} ( \dot{t}^2 + \dot{x}^2 + \dot{y}^2 + \dot{z}^2 ), for example.)

    Then, in the second line of chapter 15, he claims that \frac{\partial f}{\partial x^\mu} are the components of (df)_a in a coordinate basis. However, in equation 36, it would appear that the components are in fact (\frac{\partial F}{\partial x^\mu})_{\phi(p)}. Which is correct?
    Well, we have \displaystyle df = \frac{\partial f}{\partial x^\mu} dx^\mu, 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.)

    And how do we derive (118)?
    I have no idea — I don't know what the definition of the LHS is.

    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:

    Y'^\mu{}_{, \mu} = \frac{\partial Y'^\mu}{\partial x'^\nu} = \frac{\partial x'^\mu}{\partial x^\rho} \frac{\partial Y^\rho}{\partial x'^\nu} = \frac{\partial x'^\mu}{\partial x^\rho} \frac{\partial x^\lambda}{\partial x'^\nu} \frac{\partial Y^\rho}{\partial x^\lambda} = \frac{\partial x'^\mu}{\partial x^\rho} \frac{\partial x^\lambda}{\partial x'^\nu} Y^\rho_{, \lambda}
    which shows it transforms as a tensor, no?
    Your calculation assumes the tensoriality. The definition is \displaystyle Y'^\mu = \frac{\partial x'^\mu}{\partial x^\rho} Y^\rho, but \displaystyle \frac{\partial x'^\mu}{\partial x^\rho} is in general not a constant and doesn't commute with derivatives!

    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, \Gamma^\rho{}_{[\mu \nu]}=0 all the time. Therefore, how is it possible to ever have a connectino that isn't torsion free?
    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 torsion-free.

    And in (131), what the hell has happened here lol?
    Firstly, how does X(g(Y,Z))=\nabla_X(g(Y,Z))?
    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?
    If I write it out in index notation, is it clearer? It's just the statement \displaystyle X^\rho \frac{\partial}{\partial x^\rho} \left[ g_{\mu \nu} Y^\mu Z^\nu \right] = X^\rho \left( g_{\mu \nu} Y^\mu Z^\nu \right)_{; \rho} = g_{\mu \nu} Y^{\mu}_{;\rho} Z^{\nu} + g_{\mu \nu} Y^\mu Z^{\nu}_{;\rho}. The Leibniz rule applies because the metric is bilinear.

    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.
    I'm not certain, but I believe the calculation has to do with non-holonomic bases. Try looking for references to do with those. If you don't find anything, try looking up tetrads/vierbeins/vielbeins or Einstein—Cartan theory.

    Thanks again!
    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).
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    (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, \displaystyle \frac{\partial T}{\partial \dot{x}} = -2(1 - 2 \Phi) \dot{x} and \displaystyle \frac{\partial L}{\partial x} = 2 \frac{\partial \Phi}{\partial x} ( \dot{t}^2 + \dot{x}^2 + \dot{y}^2 + \dot{z}^2 ), for example.)
    What do you mean the general case, sorry?

    (Original post by Zhen Lin)
    Well, we have \displaystyle df = \frac{\partial f}{\partial x^\mu} dx^\mu, 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.)
    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?

    (Original post by Zhen Lin)
    Your calculation assumes the tensoriality. The definition is \displaystyle Y'^\mu = \frac{\partial x'^\mu}{\partial x^\rho} Y^\rho, but \displaystyle \frac{\partial x'^\mu}{\partial x^\rho} is in general not a constant and doesn't commute with derivatives!
    I don't understand what you're on about here. Could you elaborate please?

    (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 torsion-free.
    I think I "kind of" understand this. I can't really see how it would ever be non-zero 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?

    (Original post by Zhen Lin)
    If I write it out in index notation, is it clearer? It's just the statement \displaystyle X^\rho \frac{\partial}{\partial x^\rho} \left[ g_{\mu \nu} Y^\mu Z^\nu \right] = X^\rho \left( g_{\mu \nu} Y^\mu Z^\nu \right)_{; \rho} = g_{\mu \nu} Y^{\mu}_{;\rho} Z^{\nu} + g_{\mu \nu} Y^\mu Z^{\nu}_{;\rho}. The Leibniz rule applies because the metric is bilinear.
    Why is \frac{\partial}{\partial x^\rho} the same as covariant derivative here? And if we write out hte Leibniz rule stuff we get
    \nabla_X(g(Y,Z))=X(g)(Y,Z) + g(\nabla_X Y, Z) + X(g)(Y,Z) + g(X, \nabla_X Z)
    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).
    Well, you're certainly doing an excellent job of answering my questions lol! The differential geometry stuff ends about p60ish I think so hopefully once I get into the more physics-y side of things I won't be pestering you as much!

    Cheers
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    (Original post by latentcorpse)
    What do you mean the general case, sorry?
    Let \displaystyle L = -g_{\mu \nu} \dot{x}^\mu \dot{x}^\nu and compute the derivatives. You should get \displaystyle \frac{\partial L}{\partial x^\rho} = -g_{\mu \nu, \rho} \dot{x}^\mu \dot{x}^\nu and \displaystyle \frac{\partial L}{\partial \dot{x}^\rho} = -g_{\mu \rho} \dot{x}^\mu - g_{\rho \nu} \dot{x}^\nu. Then write out the components explicitly for your given metric.

    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?
    Well, if you try to evaluate \displaystyle \frac{\partial f}{\partial x^\mu} at a point p \in M under the chart \phi: M \to \mathbb{R}^n you'll get the expression under equation 36.

    I don't understand what you're on about here. Could you elaborate please?
    You're skipping steps. Write it out properly: \displaystyle \frac{\partial Y'^\mu}{\partial x'^\nu} = \frac{\partial}{\partial x'^\nu} \left[ \frac{\partial x'^\mu}{\partial x^\rho} Y^\rho \right]. Apply the product rule and you'll get the expression you want, \displaystyle \frac{\partial x'^\mu}{\partial x^\rho} \frac{\partial Y^\rho}{\partial x'^\nu} plus an unwanted term, \displaystyle \frac{\partial^2 x'^\mu}{\partial x'^\nu \partial x^\rho} Y^\rho. It's the latter unwanted term that makes this non-tensorial. It's also this term which (the transformation law of) the Christoffel symbol kills off, which makes the covariant derivative tensorial.

    I think I "kind of" understand this. I can't really see how it would ever be non-zero 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?
    No. A connection, in coordinate terms, is expressed as an object \tilde{\Gamma}^{\rho}_{\phantom{  \rho}\mu\nu} which satisfies the same transformation law as the Christoffel symbol. It is therefore something which makes the expression T^{\rho}_{\phantom{\rho},\nu} + \tilde{\Gamma}^{\rho}_{\phantom{  \rho}\mu\nu} T^\mu tensorial. (Strictly speaking, actually, a connection is defined as something which makes the covariant derivative expression tensorial, but a necessary and sufficient condition is the transformation law, if I'm not mistaken.)

    Why is \frac{\partial}{\partial x^\rho} the same as covariant derivative here?
    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).

    And if we write out hte Leibniz rule stuff we get
    \nabla_X(g(Y,Z))=X(g)(Y,Z) + g(\nabla_X Y, Z) + X(g)(Y,Z) + g(X, \nabla_X Z)
    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!
    Well, applying the Leibniz rule yields \nabla_X [g(Y,Z)] = \nabla_X [g](Y, Z) + g(\nabla_X Y, Z) + g(Y, \nabla_X Z). \nabla_X[g] is simply another bilinear form, so (Y, Z) isn't just sitting there.
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    (Original post by latentcorpse)
    And how do we derive (118)?
    I think I figured it out. I get a minus sign in front of the second term in the equation though. We define \Gamma'^\mu_{\phantom{\mu}\rho\n  u} as the thing which makes this equation true: \nabla_X Y = X'^\nu ( e'_\nu [Y'^\mu] + \Gamma'^\mu_{\phantom{\mu}\rho\n  u} Y'^\rho ) e'_\mu. So, equate it with the usual definition to get \displaystyle X'^\nu ( e'_\nu [Y'^\mu] + \Gamma'^\mu_{\phantom{\mu}\rho\n  u} Y'^\rho ) e'_\mu = X^\nu (Y^\mu_{\phantom{\mu},\nu} + \Gamma^\mu_{\phantom{\mu}\rho\nu  } Y^\rho) \frac{\partial}{\partial x^\mu} and then do lots of algebra. It's useful to note that if \displaystyle e'_\mu = (A^{-1})^{\nu}_{\phantom{\nu}\mu} \frac{\partial}{\partial x^\nu} then \displaystyle \frac{\partial}{\partial x^\nu} = A^{\mu}_{\phantom{\mu}\nu} e'_\mu.
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    (Original post by Zhen Lin)
    Let \displaystyle L = -g_{\mu \nu} \dot{x}^\mu \dot{x}^\nu and compute the derivatives. You should get \displaystyle \frac{\partial L}{\partial x^\rho} = -g_{\mu \nu, \rho} \dot{x}^\mu \dot{x}^\nu and \displaystyle \frac{\partial L}{\partial \dot{x}^\rho} = -g_{\mu \rho} \dot{x}^\mu - g_{\rho \nu} \dot{x}^\nu. Then write out the components explicitly for your given metric.
    So I get the t equation fine. Now for the i equation

    \frac{d}{d \tau} \left( \frac{\partial L}{\partial \dot{x}^\rho} \right) - \frac{\partial L}{\partial x^\rho}=0
    Take \rho=i

    \frac{d}{d \tau} ( -2g_{ii} \dot{x}^i) - g_{\mu \nu , i} \dot{x}^\mu \dot{x}^\nu =0
    -2(1-2\Phi) \ddot{x}^i - \partial_i g_{00} (\dot{x}^0)^2 - \partial_i g_{jk} \dot{x}^j \dot{x}^k=0

    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 Y(fg(Z,X)). How are we able to move the f out from Y(g(Z,fX)) to give this?

    (ii) In the second equality of (138), he rewrites Y(fg(Z,X))=fY(g(Z,X)) + Y(f)g(Z,X). 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 \nabla_XY 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 \nabla 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 X^\mu(t)=t X^\mu_p.

    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 Y^\mu{}_{, \nu} 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!
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    (Original post by latentcorpse)
    So I get the t equation fine. Now for the i equation

    \frac{d}{d \tau} \left( \frac{\partial L}{\partial \dot{x}^\rho} \right) - \frac{\partial L}{\partial x^\rho}=0
    Take \rho=i

    \frac{d}{d \tau} ( -2g_{ii} \dot{x}^i) - g_{\mu \nu , i} \dot{x}^\mu \dot{x}^\nu =0
    -2(1-2\Phi) \ddot{x}^i - \partial_i g_{00} (\dot{x}^0)^2 - \partial_i g_{jk} \dot{x}^j \dot{x}^k=0

    How is that looking? Any tips on how to proceed?
    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: \displaystyle g_{\rho \nu} \ddot{x}^\nu + g_{\rho \nu, \mu} \dot{x}^\mu \dot{x}^\nu - \frac{1}{2}  g_{\mu \nu, \rho} \dot{x}^\mu \dot{x}^\nu = 0. Raising indices, it's \displaystyle \ddot{x}^\sigma + g^{\rho \sigma} \left( g_{\rho \nu, \mu} - \frac{1}{2} g_{\mu \nu, \rho} \right) \dot{x}^\mu \dot{x}^\nu = 0. You should check that this is the same as \displaystyle \ddot{x}^\sigma + \Gamma^{\sigma}_{\phantom{\sigma  }\mu \nu} \dot{x}^\mu \dot{x}^\nu = 0.


    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 Y(fg(Z,X)). How are we able to move the f out from Y(g(Z,fX)) to give this?
    f is a scalar field and g is bilinear. In index notation, Y^\rho (g_{\mu \nu} Z^\mu (f X)^\nu)_{, \rho} = Y^\rho (f g_{\mu \nu} Z^\mu X^\nu)_{,\rho}.

    (ii) In the second equality of (138), he rewrites Y(fg(Z,X))=fY(g(Z,X)) + Y(f)g(Z,X). 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?
    It's precisely the product rule. In index notation, Y^\rho (f g_{\mu \nu} Z^\mu X^\nu)_{,\rho} = f Y^\rho (g_{\mu \nu} Z^\mu X^\nu)_{,\rho} + Y^\rho f_{, \rho} g_{\mu \nu} Z^\mu X^\nu.

    And just under (137), he writes "This determines \nabla_XY uniquely because the metric is nondegenerate. How does this work?
    I have no idea — this is something I've always taken on trust. I guess it's an appeal to the notion that you could, in principle, drop various vectors and invert various matrices to obtain an equation for \nabla_X Y directly.

    And I'm a bit confused by the example on p39, he says if we take \nabla 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!
    Well, this is difficult to explain. I think it's true and basically boils down to fact that the coordinateful definition of tensor and the coordinate-free definition of tensor don't quite match up. In the coordinate-free definition, there is no condition on transformation laws — because transformation laws are inherently coordinateful! To make it precise — in the coordinate-free picture, a tensor is just a multilinear map which takes vectors and returns vectors. A tensor field is a tensor-valued function on the manifold. The partial derivative operator acts on a vector field to produce a function which, at each point on the manifold, defines a linear map taking a vector to a vector — so the partial derivative of a vector field is certainly a tensor field by this definition.

    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 X^\mu(t)=t X^\mu_p.
    Basically, by the existence and uniqueness theorem for ODEs, at every point p \in M, and every tangent vector X_p \in T_p M, there is a geodesic which passes through p with tangent vector X_p (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 \gamma_p( - ; X_p) : [-1, 1] \to M be the unique geodesic s.t. \gamma_p(0; X_p) = p and \gamma'_p(0; X_p) = X_p. Let \phi: M \to \mathbb{R}^n be the normal coordinate chart, i.e. the chart such that \exp \circ \phi = \mathrm{id}. Then, \phi \circ \gamma_p(t; X_p) = t d\phi_p (X_p), where d\phi_p : T_p M \to \mathbb{R}^n gives the components of a vector w.r.t the coordinate basis induced by \phi. So, if we write X^\mu (t) for the \mu-th component of \phi \circ \gamma_p(t; X_p) and X_p^\mu for the \mu-th component of d\phi_p (X_p) we get the claimed equation.

    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?
    Because X_p^\nu X_p^\rho is a symmetric tensor. If we had checked that \Gamma^{\mu}_{\phantom{\mu}\nu \rho} U^\nu V^\rho = 0 for arbitrary vectors U^\nu, V^\rho then we wouldn't need to symmetrise.

    (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?
    A torsion-free connection is one such that \Gamma^{\mu}_{\phantom{\mu}[\nu\rho]} = 0. Combine this with the previous equation to obtain that \Gamma^{\mu}_{\phantom{\mu} \nu \rho} = 0.

    (iii) And in (150) he says he has applied the above to the Levi Civita connection - how has this given us (150) though?
    The Levi–Civita connection is the one such that \displaystyle \Gamma^{\mu}_{\phantom{\mu}\nu \rho} = \frac{1}{2} g^{\mu \sigma} ( g_{\sigma \nu, \rho} + g_{\sigma \rho, \nu} - g_{\nu \rho, \sigma} ).

    And finally, any thoughts for the exercise on p46?
    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 U^\mu = \dot{x}^\mu satisfying \dot{U}^\mu + \Gamma^{\mu}_{\phantom{\mu}\nu \rho} U^\nu U^\rho = 0. Differentiate g_{\mu \nu} U^\mu U^\nu using the product rule and you should get terms which correspond to the previous equation contracted with g_{\mu \sigma} U^\sigma.

    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 Y^\mu{}_{, \nu} 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?
    Yeah. It should work for arbitrary bases though, by transitivity.

    And when I try and prove it, should I expand on the RHS or the LHS of that equation?
    Does it matter? You'll probably have to do both.
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    (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: \displaystyle g_{\rho \nu} \ddot{x}^\nu + g_{\rho \nu, \mu} \dot{x}^\mu \dot{x}^\nu - \frac{1}{2}  g_{\mu \nu, \rho} \dot{x}^\mu \dot{x}^\nu = 0. Raising indices, it's \displaystyle \ddot{x}^\sigma + g^{\rho \sigma} \left( g_{\rho \nu, \mu} - \frac{1}{2} g_{\mu \nu, \rho} \right) \dot{x}^\mu \dot{x}^\nu = 0. You should check that this is the same as \displaystyle \ddot{x}^\sigma + \Gamma^{\sigma}_{\phantom{\sigma  }\mu \nu} \dot{x}^\mu \dot{x}^\nu = 0.
    Yeah but the whole point fo the E-L 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 x^i equation?

    (Original post by Zhen Lin)
    Basically, by the existence and uniqueness theorem for ODEs, at every point p \in M, and every tangent vector X_p \in T_p M, there is a geodesic which passes through p with tangent vector X_p (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 \gamma_p( - ; X_p) : [-1, 1] \to M be the unique geodesic s.t. \gamma_p(0; X_p) = p and \gamma'_p(0; X_p) = X_p. Let \phi: M \to \mathbb{R}^n be the normal coordinate chart, i.e. the chart such that \exp \circ \phi = \mathrm{id}. Then, \phi \circ \gamma_p(t; X_p) = t d\phi_p (X_p), where d\phi_p : T_p M \to \mathbb{R}^n gives the components of a vector w.r.t the coordinate basis induced by \phi. So, if we write X^\mu (t) for the \mu-th component of \phi \circ \gamma_p(t; X_p) and X_p^\mu for the \mu-th component of d\phi_p (X_p) we get the claimed equation.

    Equation 148 follows trivially once we have this — just differentiate!
    I don't understand this calculation:
    So you've definde \gamma_p as the geodesic through p with tangent vector X_p at p \in M
    \phi is taken to be a chart....fair enough but then why is \exp \circ \phi = id?
    Moreover, how does this map even make sense, the domain of exp is T_p(M) but the codomain of \phi is \mathbb{R}^n?
    And where does \phi \circ \gamma_p(t; X_p) = t d\phi_p (X_p) 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 U^\mu = \dot{x}^\mu satisfying \dot{U}^\mu + \Gamma^{\mu}_{\phantom{\mu}\nu \rho} U^\nu U^\rho = 0. Differentiate g_{\mu \nu} U^\mu U^\nu using the product rule and you should get terms which correspond to the previous equation contracted with g_{\mu \sigma} U^\sigma.
    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 X where X satisfies \nabla_XX=0.
    Now my interpretation is that given \nabla_XX=0, 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 U^\mu=\frac{dx^\mu}{d \tau}. 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 X^\rho (g_{\mu \nu} X^\mu X^\nu)_{; \rho}=X^\rho  g_{\mu \nu; \rho} X^\mu X^\nu + X^\rho g_{\mu \nu} X^\mu{}_{; \rho} X^\nu + X^\rho g_\mu \nu} X^\my X^\nu{}_{; \rho}=X^\rho g_{\mu \nu} X^\mu{}_{; \rho} X^\nu + X^\rho g_\mu \nu} X^\my X^\nu{}_{; \rho}
    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
    2X^\rho g_{\mu \nu} X^\mu{}_{; \rho} X^\nu
    But this is just 2g(\nabla_XX,X)
    and \nabla_XX=0 for affinely parameterised geodesics so we get
    2g(0,X)
    Now I think there should be some way of saying this is zero but I don't know how? Something to do with non-degeneracy perhaps?

    I'm also confusing myself on (150),
    we have g_{\mu \nu , \rho} + g_{\mu \rho , \nu}+ g_{\nu \rho, \mu}=0
    now if we antisymmetrise on \mu.\nu we get:
    g_{\mu \nu , \rho}-g{\nu \mu , \rho} + g_{\mu \rho , \nu} - g_{\nu \rho , \mu} + g_{\nu \rho, \mu} - g_{\mu \rho , \nu}=0
    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
    g_{\mu \nu , \rho}-g{\nu \mu , \rho} +g_{[ \mu | \rho | , \nu]}=0 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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    (Original post by latentcorpse)
    Yeah but the whole point fo the E-L 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 x^i equation?
    You can do it at any point, as long as you do it correctly!

    I don't understand this calculation:
    So you've definde \gamma_p as the geodesic through p with tangent vector X_p at p \in M
    \phi is taken to be a chart....fair enough but then why is \exp \circ \phi = id?
    I miswrote. Fix a point p \in M, then the exponential map at that point is \exp_p : T_p M \to M. Then, the geodesic normal coordinate chart \phi : M \to \mathbb{R}^n has the property that \phi \circ \exp \circ (d\phi_p)^{-1} = \mathrm{id}, 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.

    And where does \phi \circ \gamma_p(t; X_p) = t d\phi_p (X_p) come from?
    Linearity, basically. \gamma_p(t; X_p) = \gamma_p(1; t X_p) = \exp (t X_p) , and so \phi \circ \gamma_p(t; X_p) = \phi \circ \exp (t X_p) = d\phi_p (t X_p) = t \, d\phi_p(X_p).

    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 X where X satisfies \nabla_XX=0.
    Now my interpretation is that given \nabla_XX=0, 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 U^\mu=\frac{dx^\mu}{d \tau}. Is this correct?
    Yes, I suppose so.

    Secondly, in doing this calculation, I don't understand what you are asking me to do? What do I differentiate wrt to?
    The parameter of the curve, of course. You'll get g_{\mu \nu, \rho} U^\mu U^\nu U^\rho + 2 g_{\mu \nu} U^\mu \dot{U}^\nu. Write out the Christoffel symbol in terms of partial derivatives of the metric, and you should see that \displaystyle g_{\mu \sigma} \Gamma^\sigma_{\phantom{\sigma} \nu \rho} U^\mu U^\nu U^\rho = \frac{1}{2} g_{\mu \nu, \rho} U^\mu U^\nu U^\rho. It should be clear how to obtain the result you want from here.

    2g(0,X)
    Now I think there should be some way of saying this is zero but I don't know how? Something to do with non-degeneracy perhaps?
    g is bilinear, so g(0, X) = 0 automatically. This should be familiar from linear algebra, I hope. There's no need to invoke non-degeneracy or anything sophisticated.
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    (Original post by Zhen Lin)
    You can do it at any point, as long as you do it correctly!
    My attempts keep goin off course. Let's say I want to substitute in at:
    \frac{d}{d \tau} g_{\rho \nu} \dot{x}^\nu + g_{\mu \rho} \dot{x}^\mu + g_{\mu \nu, \rho} \dot{x}^\mu \dot{x}^\nu=0
    To get the equations for x^i what should i be setting my indices equal to?
    I reckon it should be \nu=\mu=i, no?

    (Original post by Zhen Lin)
    I miswrote. Fix a point p \in M, then the exponential map at that point is \exp_p : T_p M \to M. Then, the geodesic normal coordinate chart \phi : M \to \mathbb{R}^n has the property that \phi \circ \exp \circ (d\phi_p)^{-1} = \mathrm{id}, 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. \gamma_p(t; X_p) = \gamma_p(1; t X_p) = \exp (t X_p) , and so \phi \circ \gamma_p(t; X_p) = \phi \circ \exp (t X_p) = d\phi_p (t X_p) = t \, d\phi_p(X_p).
    Ok. I'm going to apologise because this kind fo stuff in particular just doesn't come very easily to me!
    So, can we talk through this slowly:
    (i) We're trying to show that affinely parameterised geodesics through p, in normal coordinates, satisfy X^\mu(t)=tX^\mu_p.
    (ii) So we set up exp as described at teh top of p47 and our chart \phi as we've been using previously. Now you go and introduce this d \phi. 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. \phi \circ exp \circ (d \phi)^{-1} = id
    (iv)Now we consider the geodesic \gamma_p(-,X_p) : [-1,1] \rightarrow M
    First of all, back on p14, we described a smooth curve as a smooth function I \rightarrow M 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. \gamma_p(0,X_p)=p , \gamma_p'(0,X_p)=X_p
    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 \lambda(0)=p and \lambda(1)=q. But how can this make sense as  \lambda : (0,1) \rightarrow M so neither 0 nor 1 are in the domain?
    (iv) Given this geodesic, you tell me that \gamma_p(t,X_p)=\gamma_p(1,tX_p) Why?
    (v) And how does that then equal exp(t X_p)?
    (vi) Ok so finally we prove that \phi \circ \gamma(t,X_p) = t d \phi_p (X_p). Can you explain in words how this proves X^\mu(t)=tX^\mu_p?


    (Original post by Zhen Lin)
    g is bilinear, so g(0, X) = 0 automatically. This should be familiar from linear algebra, I hope. There's no need to invoke non-degeneracy or anything sophisticated.
    Ok. So I would like this method to work out seeing as I managed it myself for a change. However, if you look at the exercise on p46, it is considering the case of the Levi Civita connection. However, I don't think I have made use of the Levi Civita proiperty anywhere in my answer. I guess I used the fact that \nabla_XX=0 but this is a consequence of affine parameterisation, not of the Levi Civita connection. Have I used the property somewhere without realising it?

    I'm also confusing myself on (150),
    we have g_{\mu \nu , \rho} + g_{\mu \rho , \nu}+ g_{\nu \rho, \mu}=0
    now if we antisymmetrise on \mu.\nu we get:
    g_{\mu \nu , \rho}-g{\nu \mu , \rho} + g_{\mu \rho , \nu} - g_{\nu \rho , \mu} + g_{\nu \rho, \mu} - g_{\mu \rho , \nu}=0
    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
    g_{\mu \nu , \rho}-g{\nu \mu , \rho} +g_{[ \mu | \rho | , \nu]}=0 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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    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

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