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    (Original post by latentcorpse)
    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?
    Yes, but your equation is wrong. Check your differentiation!

    (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?
    Every diffeomorphism \phi: M \to N of manifolds induces an isomorphism d\phi_p : T_p M \to T_{\phi(p)} N. This map is called the differential or pushforward. Setting N = \mathbb{R}^n we get the special case for charts, and here T_{\phi(p)} N = \mathbb{R}^n. Just as \phi puts a coordinate system on M, d\phi_p puts a basis on T_p M (the coordinate basis, of course).

    Geodesic normal coordinates are defined so that we have the property \phi \circ \exp \circ (d\phi_p)^{-1} = \mathrm{id}. Note they're not unique — because each basis of T_p M induces a different chart. Formally: Fix a point p \in M and an isomorphism \psi: \mathbb{R}^n \to T_p M (i.e. choose a basis). Then, the claim is that there is an open subset E \subseteq \mathbb{R}^n such that the restriction \exp \circ \psi: E \to M is a diffeomorphism onto U \subseteq M. Then the corresponding geodesic polar coordinates are given by the chart \phi = (\exp \circ \psi)^{-1}: U \to \mathbb{R}^n.

    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
    Hmmm. That's an interesting point — we want I to be open so that differentiation works nicely. On the other hand, everywhere else in geometry we want I to be closed so that we can talk about its endpoints. Anyway, it doesn't really matter - as long as [0, 1] \subseteq I that's good enough. (Note that in general, it may not be possible to arbitrarily extend a given affinely-parametrised geodesic. Recall that such a geodesic has constant velocity — now imagine a manifold which has an edge. The open ball in \mathbb{R}^n is a perfectly good smooth manifold, for instance, and every geodesic there has finite length, so you can't arbitrarily extend them!)

    (iv) Given this geodesic, you tell me that \gamma_p(t,X_p)=\gamma_p(1,tX_p) Why?
    Go stand at p, facing the direction X_p. If I tell you to walk for t hours at 1 km/h, that's the same as telling you to walk for 1 hour at t km/h, no?

    Remember, affinely parametrised geodesics have constant velocity.

    (v) And how does that then equal exp(t X_p)?
    By definition of the exponential map.

    (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?
    It's just abuse of notation. If we let X^\mu_p be the \mu-th component of the initial velocity vector, and X^\mu(t) to be the \mu-th coordinate at time t, then it's just a (confusing) restatement of my equation. Admittedly it captures the intuitive idea better.

    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?
    If you ever wrote \nabla_X g = 0 then you have used the defining property of the Levi–Civita connection, and you almost surely have, because you're calculating (basically) \nabla_X [g(X, X)].

    I'm also confusing myself on (150)
    Writing it out in full, we get 0 = g_{\mu \nu, \rho} - g_{\nu \mu, \rho} + g_{\mu \rho, \nu} - g_{\nu \rho, \mu} - g_{\nu \rho, \mu} + g_{\mu \rho, \nu}. So the first two terms cancel and 0 = g_{\mu \rho, \nu} - g_{\nu \rho, \mu}, as claimed. Substitute back to the original equation 0 = g_{\mu \nu, rho} + g_{\mu \rho, \nu} - g_{\nu \rho, \mu} and conclude 0 = g_{\mu \nu, \rho}.
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    (Original post by Zhen Lin)
    Yes, but your equation is wrong. Check your differentiation!
    Ok.
    L=-g_{\mu \nu}(x) \dot{x}^\mu \dot{x}^\nu
    \frac{d}{d \tau} ( \frac{\partial L}{\partial \dot{x}^\rho})  - ( \frac{\partial L}{\partial x^\rho})=0
    \frac{d}{d \tau} (-2g_{\mu \nu} \dot{x}^\rho) + g_{\mu \nu , \rho} \dot{x}^\mu \dot{x}^\nu=0
    \frac{d}{d \tau} (2g_{\mu \nu} \dot{x}^\rho) = g_{\mu \nu , \rho} \dot{x}^\mu \dot{x}^\nu=0
    Is this better?

    Now I think I understand your argument with the exp map and geodesics. Thanks!

    Now if you take a look at the second paragraph on p48, we are asked to consider \frac{\partial}{\partial X^1} in normal coordinates at p...
    (i) How is it the tangent to X^\mu=(t,0, \dots , 0).

    (ii) It says that "from above, this is the same as the geodesic through p with tangent vector e_1 at p. How is that true?

    (iii) How was our choice of basis \{ e_\mu \} arbitrary?

    (iv)Why does having an orthonormal basis \frac{\partial}{\partial X^\mu} allow us to choose coordinates s.t. g_{\mu \nu , \rho}(p)=0 and g_{\mu \nu}(p) = \begin{cases} \eta_{\mu \nu} &\text{(Lorentzian)} \\ \delta_{\mu \nu} &\text{(Riemannian)} \end{cases}

    And in the paragraph on p50, he talks about how in a local inertial frame, we have \Gamma^\mu{}_{\nu \rho}(p)=0. How is this?
    We only proved this on p47 and that was proved for a manifold with a torsion free connection on which we have used normal coordinates.

    Still in this same paragraph, why does knowing that covariant derivatives reduce to partial derivatives mean that in any chart \nabla^\mu \nabla_\mu \Phi = g^{\mu \nu} \nabla_\mu \partial_\nu \Phi? i.e. why do we still have a covariant derivatice left until we consider it only at the point p \in M where it becomes \eta^{\mu \nu} \partial_\mu \partial_\nu \Phi?

    And in the remark on page 53, he talks about how T_{ab}(p) u^au^b in a local inertial frame is the energy density of matter - I don't see how we are able to recover eqn (165) from this though?

    And still in this remark, he talks about the component of j^a alon u^a - how do we write that? Is it j^au_a?

    Additionally, one or two small things concerning chapter 21 on parallel transport:
    a) In the 2ns remark on p54, he writes that "a geodesic is a curve whose tangent vector is parallelly propagated along the curve". Well, this doesn't make sense to me. For starters in that stuff we were debating a few posts ago, we ahd X^\mu(t)=t X^\mu_p: surely the mere presence of t in that equation suggest that the vector will change as we go along the curve.
    Imagine a U shaped curve on a manifold: clearly the tangent vector will change direction as you go round the curve. But here he seems to suggest that the tangent vector is always parallel to the initial tangent vector, no?
    b)How do we get equation (177).
    It comes from \nabla_XT=0
    Do we write X^\rho T^\mu{}_{\nu ; \rho}=0
    I don't see how that will work out though? In particular, how are we goign to get a \frac{d T^\mu{}_\nu}{dt} term?

    And then in the last paragraph of chapter 21 (on page 55), he says "if we use the Levi Civita connection in Minkowski spacetime and use the inertial frame coordinates then the Christoffel symbols vanish". How is this? I have a feeling it relates to eqn (149) and the paragraph under (152).

    Does the next sentence "A tensor is parallelly transported along a curve if its components are constant along the curve. Hence if we have two different curves from p to q then the result of parallelly transporting T from p to q is independent of which curve we choose". mean that because the components are constant when we parallel transport the tensor, the tensor will not change under parallel transport and so regardless of which path we go along, we will always end up at the same vector (and this is becuase the spacetime is flat)?

    If you look at the calculation in (181), how do we go from the 2nd line to the 3rd? I tried to use (125) but this didn't work since both of these are (0,2) tensors, we would get ONLY - \Gamma terms (and we would also get two \Gamma terms from each covariant derivative).

    Finally, how do we show (187)?

    Thanks!
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    (Original post by latentcorpse)
    Ok.
    L=-g_{\mu \nu}(x) \dot{x}^\mu \dot{x}^\nu
    \frac{d}{d \tau} ( \frac{\partial L}{\partial \dot{x}^\rho})  - ( \frac{\partial L}{\partial x^\rho})=0
    \frac{d}{d \tau} (-2g_{\mu \nu} \dot{x}^\rho) + g_{\mu \nu , \rho} \dot{x}^\mu \dot{x}^\nu=0
    \frac{d}{d \tau} (2g_{\mu \nu} \dot{x}^\rho) = g_{\mu \nu , \rho} \dot{x}^\mu \dot{x}^\nu=0
    Is this better?
    A little bit. But the indices are wrong — for a start, the position of \rho doesn't match up. And then the free and bound indices don't match either. So it's still wrong. Please be more careful! (Have you never used index notation before?)

    Now if you take a look at the second paragraph on p48, we are asked to consider \frac{\partial}{\partial X^1} in normal coordinates at p...
    (i) How is it the tangent to X^\mu=(t,0, \dots , 0).
    I imagine the first coordinate is indexed by 1, so this is effectively automatic. I went through this with you on the first page!

    (ii) It says that "from above, this is the same as the geodesic through p with tangent vector e_1 at p. How is that true?
    By definition, essentially.

    (iii) How was our choice of basis \{ e_\mu \} arbitrary?
    By the argument in my previous post.

    (iv)Why does having an orthonormal basis \frac{\partial}{\partial X^\mu} allow us to choose coordinates s.t. g_{\mu \nu , \rho}(p)=0 and g_{\mu \nu}(p) = \begin{cases} \eta_{\mu \nu} &\text{(Lorentzian)} \\ \delta_{\mu \nu} &\text{(Riemannian)} \end{cases}
    Read my previous post.

    And in the paragraph on p50, he talks about how in a local inertial frame, we have \Gamma^\mu{}_{\nu \rho}(p)=0. How is this?

    We only proved this on p47 and that was proved for a manifold with a torsion free connection on which we have used normal coordinates.
    That's what it means to have a local inertial frame.

    Still in this same paragraph, why does knowing that covariant derivatives reduce to partial derivatives mean that in any chart \nabla^\mu \nabla_\mu \Phi = g^{\mu \nu} \nabla_\mu \partial_\nu \Phi? i.e. why do we still have a covariant derivatice left until we consider it only at the point p \in M where it becomes \eta^{\mu \nu} \partial_\mu \partial_\nu \Phi?
    \Phi is a scalar field, so \nabla_\mu \Phi = \partial_\mu \Phi automatically.

    The Christoffel symbols only vanish at the chosen point p and not in general; so at p (but not necessarily anywhere else!) \nabla_\mu \partial_\nu \Phi = \partial_\mu \partial_\nu \Phi.

    And in the remark on page 53, he talks about how T_{ab}(p) u^au^b in a local inertial frame is the energy density of matter - I don't see how we are able to recover eqn (165) from this though?
    This is going into the realm of physics, and electromagnetism in particular I know nothing about.

    And still in this remark, he talks about the component of j^a alon u^a - how do we write that? Is it j^au_a?
    I guess so.

    Additionally, one or two small things concerning chapter 21 on parallel transport:
    a) In the 2ns remark on p54, he writes that "a geodesic is a curve whose tangent vector is parallelly propagated along the curve". Well, this doesn't make sense to me. For starters in that stuff we were debating a few posts ago, we ahd X^\mu(t)=t X^\mu_p: surely the mere presence of t in that equation suggest that the vector will change as we go along the curve.
    Imagine a U shaped curve on a manifold: clearly the tangent vector will change direction as you go round the curve. But here he seems to suggest that the tangent vector is always parallel to the initial tangent vector, no?
    Ah, but you see, what does it mean for two vectors to be parallel on a curved surface? We're defining what it means to be parallel. Here's a subtle point: the tangent spaces at each point are all isomorphic to each other, but there's no natural way to identify vectors in one tangent space with vectors in another tangent space. Parallel transport gives us the means to do so, but this requires a choice of connection.

    b)How do we get equation (177).
    It comes from \nabla_XT=0
    Do we write X^\rho T^\mu{}_{\nu ; \rho}=0
    I don't see how that will work out though? In particular, how are we goign to get a \frac{d T^\mu{}_\nu}{dt} term?
    Reverse chain rule. \displaystyle X^rho T^\mu_{\phantom{\mu}\nu,\rho} = \frac{dx^\rho}{d\tau} \frac{\partial T^\mu_{\phantom{\mu}\nu}}{\parti  al x^\rho} = \frac{dT^\mu_{\phantom{\mu}\nu}}  {d\tau}.

    And then in the last paragraph of chapter 21 (on page 55), he says "if we use the Levi Civita connection in Minkowski spacetime and use the inertial frame coordinates then the Christoffel symbols vanish". How is this? I have a feeling it relates to eqn (149) and the paragraph under (152).
    The Levi–Civita connection gives the Christoffel symbols in terms of partial derivatives of the metric. The metric is constant in Minkowski spacetime under inertial coordinates, so of course the Christoffel symbols disappear.

    Does the next sentence "A tensor is parallelly transported along a curve if its components are constant along the curve. Hence if we have two different curves from p to q then the result of parallelly transporting T from p to q is independent of which curve we choose". mean that because the components are constant when we parallel transport the tensor, the tensor will not change under parallel transport and so regardless of which path we go along, we will always end up at the same vector (and this is becuase the spacetime is flat)?
    Yes.

    If you look at the calculation in (181), how do we go from the 2nd line to the 3rd? I tried to use (125) but this didn't work since both of these are (0,2) tensors, we would get ONLY - \Gamma terms (and we would also get two \Gamma terms from each covariant derivative).
    This isn't strictly speaking index notation. (Unfortunately, this sort of confusion frequently arises when trying to calculate components using the coordinate-free definitions...) \Gamma^tau_{\phantom{\tau}\nu\si  gma} e_\tau is a family of (1,0) vector fields indexed by \nu and \sigma.

    Finally, how do we show (187)?
    Just calculate it from the coordinate-free definitions, I guess — show that \nabla_\rho \nabla_\sigma [Z^\mu e_\mu] - \nabla_\sigma \nabla_\rho [Z^\mu e_\mu] = R^\mu_{\phantom{\mu}\nu\rho\sigm  a} Z^\nu e_\mu. It's essentially tautological in index notation as well. (I think the point of the question is for you to show that Z \mapsto R(X, Y)Z is a linear map of vector fields.)

    You seem to be trying to go through these notes quickly. I suggest you slow down and try to understand each section before moving on to the next. I learned this differential geometry stuff on my own over several years and I can tell you, it's just a matter of getting used to the notation and the language — and you can't get used to something by rushing through it!
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    (Original post by Zhen Lin)
    A little bit. But the indices are wrong — for a start, the position of \rho doesn't match up. And then the free and bound indices don't match either. So it's still wrong. Please be more careful! (Have you never used index notation before?)
    Ohh dear. This is rather embarassing - I have been using it for a gd couple of years now! Anyway checking it all twice I get

    \frac{d}{d \tau} ( 2 g_{\rho \nu} \dot{x}^\nu ) - g_{\mu \nu , \rho} \dot{x}^\mu \dot{x}^\nu=0

    So then in the first term I want to set \nu=i and I would ahve thought that in the 2nd I would want to set \mu=\nu=i however I guess that's not quite right because in the answer in (103) he has \dot{x}^j , \dot{x}^k and \delta terms... so should I set \mu=j,\nu=k in the 2nd term? How do we know to do that? Surely it would make more sense just to acknowledge the metric is diagonal and just set everything equal to i, no?

    (Original post by Zhen Lin)
    I imagine the first coordinate is indexed by 1, so this is effectively automatic. I went through this with you on the first page!
    So we are considering some curve whose tangent vector is X^\mu=(t,0, \dots , 0)
    I'm sorry I don't see how \frac{\partial}{\partial X^1} can be a tangent vector to (t,0,...,)? We know that it will be the tangent vector to a curve which under a chart will be constant in all components except the first where it has unit velocity. Ok well that explains why all the other components are 0. But why is the 1st component t and not 1 (you said it would have unit velocity in this component after all?


    And then, how can e_1 be a tangent?
    Is it because we have just shown that X^\mu(t)=tX^\mu_p and so that tells us that X^\mu_p=(1,0, \dots , 0) and since e_1=(1,0, \dots , 0) by definition (not sure if this is true because I may have assumed that we are in some basis that we are not - can you confirm?) we get that \frac{\partial}{\partial X^1}=e_1 at p \in M


    To be honest, I think I'm missing something fundamental to do with normal coordinates here - he writes as though once we know \frac{\partial}{\partialX^1} is a tangent vector then it follows that e_1 is also.
    I mean even the first line of this paragraph "Now consider \frac{\partial}{\partial X^1} in normal coordinates" - what does this even mean?
    I'm also sorry but I don't get either of how \{e_\mu was arbitrary - it seems to me like if we are working normal coordinates then it will be determined by the exponential map as the element of T_p(M) that gets mapped to \frac{\partial}{\partial X^1}, no?
    And also, could you please go over how having an orthonormal \frac{\partial}{\partial X^\mu} allows us to choose coordinates such that g_{\mu \nu}(p) = \begin{cases} \eta_{\mu \nu} &\text{(Lorentzian)} \\ \delta_{\mu \nu} &\text{(Riemannian)} \end{cases} and g_{\mu \nu , \rho}=0.




    (Original post by Zhen Lin)
    That's what it means to have a local inertial frame.
    Where did we show this though? What we showed on p48, wasn't for a local inertial frame, was it?


    (Original post by Zhen Lin)
    Ah, but you see, what does it mean for two vectors to be parallel on a curved surface? We're defining what it means to be parallel. Here's a subtle point: the tangent spaces at each point are all isomorphic to each other, but there's no natural way to identify vectors in one tangent space with vectors in another tangent space. Parallel transport gives us the means to do so, but this requires a choice of connection.
    Ok. Got it. Out of interest, what is this isomorphism?


    (Original post by Zhen Lin)
    This isn't strictly speaking index notation. (Unfortunately, this sort of confusion frequently arises when trying to calculate components using the coordinate-free definitions...) \Gamma^tau_{\phantom{\tau}\nu\si  gma} e_\tau is a family of (1,0) vector fields indexed by \nu and \sigma.
    I don't get this at all. How do we know we are working with coordinate free definitions? The definition in (125) is in a coordinate basis, is it not? And how on earth do we tell whether it's a (0,2) tensor field or a family of (1,0) ones? And how can it be a family of (1,0) ones - does one index label the element in the family and the other label the components of that element?

    (Original post by Zhen Lin)
    Just calculate it from the coordinate-free definitions, I guess — show that \nabla_\rho \nabla_\sigma [Z^\mu e_\mu] - \nabla_\sigma \nabla_\rho [Z^\mu e_\mu] = R^\mu_{\phantom{\mu}\nu\rho\sigm  a} Z^\nu e_\mu. It's essentially tautological in index notation as well. (I think the point of the question is for you to show that Z \mapsto R(X, Y)Z is a linear map of vector fields.)
    Two things here:
    (i) when presented with a problem like this, how do we know to approach it by evaluating it in a basis i.e. how do you know to change Z to Z^\mu e_\mu?
    (ii) I have reduced the problem to
    [Z^\mu \nabla_\rho \nabla_\sigma - Z^\mu \nabla_\sigma \nabla_\rho + e_\rho ( e_\sigma * Z^\mu)) - e_\sigma ( e_\rho ( Z^\mu)) ] e_\mu

    Now I think this might work out if I am allowed to write e_\rho ( e_\sigma (Z^\mu)) = \nabla_{\sigma \rho}
    Is this allowed? And where does it come from? It's not one of the rules for connections given at the start of chapter 15?

    And a very quick question about (148), he says the geodesic equation reduces to this i.e. the 1st term vanishes. But the 1st term is:
    \frac{d^2 x^\mu(t)}{d \tau^2} = \frac{d X^\mu(t)}{d \tau} = \frac{dt}{d \tau} X^\mu_p. Why does this vanish? t could be a function of tau surely?

    Thanks! And Merry Christmas!
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    (Original post by latentcorpse)
    Ohh dear. This is rather embarassing - I have been using it for a gd couple of years now! Anyway checking it all twice I get

    \frac{d}{d \tau} ( 2 g_{\rho \nu} \dot{x}^\nu ) - g_{\mu \nu , \rho} \dot{x}^\mu \dot{x}^\nu=0

    So then in the first term I want to set \nu=i and I would ahve thought that in the 2nd I would want to set \mu=\nu=i however I guess that's not quite right because in the answer in (103) he has \dot{x}^j , \dot{x}^k and \delta terms... so should I set \mu=j,\nu=k in the 2nd term? How do we know to do that? Surely it would make more sense just to acknowledge the metric is diagonal and just set everything equal to i, no?
    Now I'm lost. You can't only consider the spatial components of a summed index, unless you know that the time components are zero or cancel out or something. The only free index here is \rho.

    Before proceeding, it might be useful if you expand \displaystyle \frac{d}{d \tau} \left[ 2 g_{\rho \nu} \dot{x}^\nu \right].

    So we are considering some curve whose tangent vector is X^\mu=(t,0, \dots , 0)
    No, we're considering a curve whose coordinates are those. (A sign of poor notation? Sometimes X is a tangent vector, other times it's just the coordinates.)

    And then, how can e_1 be a tangent?
    I would imagine this is because \displaystyle e_1 = \frac{\partial}{\partial X^1}.

    To be quite honest, I don't have the patience to try to read through these notes and clarify them. I can explain things my own way. You might be better off finding a different book — perhaps one written with an emphasis on explaining the geometry clearly.

    I mean even the first line of this paragraph "Now consider \frac{\partial}{\partial X^1} in normal coordinates" - what does this even mean?
    \displaystyle \frac{\partial}{\partial X^1} refers to the first coordinate basis vector, so it depends directly on the choice of coordinates.

    And also, could you please go over how having an orthonormal \frac{\partial}{\partial X^\mu} allows us to choose coordinates such that g_{\mu \nu}(p) = \begin{cases} \eta_{\mu \nu} &\text{(Lorentzian)} \\ \delta_{\mu \nu} &\text{(Riemannian)} \end{cases} and g_{\mu \nu , \rho}=0.
    You have it backwards. Choose an orthonormal basis for T_p M first. Then construct the geodesic normal coordinates associated with that basis. The coordinate basis at p will then coincide with the orthonormal basis you chose.

    Where did we show this though? What we showed on p48, wasn't for a local inertial frame, was it?
    A local inertial frame is just a geodesic normal coordinate system constructed so that the coordinate basis is orthonormal at your chosen point. So what we said before about geodesic normal coordinates applies.

    Ok. Got it. Out of interest, what is this isomorphism?
    What isomorphism? I claimed the tangent spaces are isomorphic — this simply means an isomorphism exists. (This is because the tangent space at every point has the same dimension.) In fact, there are (uncountably) infinitely many isomorphisms, and there's no natural choice.

    When you fix a geodesic and a connection, this induces an isomorphism of tangent spaces of the points on the geodesic. Note that the isomorphism depends on the path and is not globally defined!

    I don't get this at all. How do we know we are working with coordinate free definitions? The definition in (125) is in a coordinate basis, is it not? And how on earth do we tell whether it's a (0,2) tensor field or a family of (1,0) ones? And how can it be a family of (1,0) ones - does one index label the element in the family and the other label the components of that element?
    Equation 125 is only valid in the coordinate basis, yes. With a small modification it's valid in any basis, with the correct (re)definition of various symbols.

    If I just write down a symbol like T^{\rho}_{\phantom{\rho}\mu\nu} there's no way to tell what the intended interpretation is. Following the convention that Greek indices indicate components and Latin indices are abstract, then these are the components of some object T. The important thing is that they're just scalars. However, if I write T^a_{\phantom{a}bc} following this convention, then this means a tensor of type (1, 2). On the other hand, if I say e_\rho is a family of vector fields indexed by \rho, then T^{\rho}_{\phantom{\rho}\mu\nu} e_\rho is a family of vector fields indexed by \mu and \nu.

    Here we really start running into terminological issues. As emphasised before, \Gamma^\rho_{\phantom{\rho}\mu\n  u} are not the components of a tensor because they don't transform the right way under a coordinate change. Nonetheless, I could define a (1, 2) tensor T which happens to have the same components — under some fixed coordinate basis. Its components in other coordinate bases are forced by tensoriality. Similarly, I could define a family of scalar fields indexed by \rho, \mu, \nu whose values happen to coincide with the components of some tensor in some fixed coordinate basis. There's nothing in the definitions which prevents me from doing this, because — as it tends to turn out — it's very difficult to mathematically capture the physicist's intuitive notion of what it means to be coordinate-independent. It may well be impossible, but I'll stay clear of claiming that — after all, category theorists managed to invent a mathematical definition of naturality which seems to agree with intuition!

    Two things here:
    (i) when presented with a problem like this, how do we know to approach it by evaluating it in a basis i.e. how do you know to change Z to Z^\mu e_\mu?
    Actually, I've just realised the indices of equation 187 are abstract. The point of the exercise is to show that the third term of the coordinate-free definition, i.e. the one with the commutator, vanishes when your connection is torsion-free. This is the case when e.g. the connection is the Levi–Civita connection, which is why it is absent in equation 181.

    Now I think this might work out if I am allowed to write e_\rho ( e_\sigma (Z^\mu)) = \nabla_{\sigma \rho}
    That's not allowed and is not true in general. What is true, though, is if you work in a coordinate basis, then e_\rho (e_\sigma (Z^\mu)) = Z^\mu_{\phantom{\mu},\sigma\rho}. Note these are partial derivatives rather than covariant derivatives.

    And a very quick question about (148), he says the geodesic equation reduces to this i.e. the 1st term vanishes. But the 1st term is:
    \frac{d^2 x^\mu(t)}{d \tau^2} = \frac{d X^\mu(t)}{d \tau} = \frac{dt}{d \tau} X^\mu_p. Why does this vanish? t could be a function of tau surely?
    Notation again. Here the curve parameter is t, not \tau, and it is certainly the case that \frac{d^2}{dt^2} \left[ t X^\mu_p \right] = 0.
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    (Original post by Zhen Lin)
    Now I'm lost. You can't only consider the spatial components of a summed index, unless you know that the time components are zero or cancel out or something. The only free index here is \rho.

    Before proceeding, it might be useful if you expand \displaystyle \frac{d}{d \tau} \left[ 2 g_{\rho \nu} \dot{x}^\nu \right].
    Well g has tau dependence as well so

    2 \frac{dg_{\rho \nu}}{d \tau} \dot{x}^\nu + 2 g_{\rho \nu} \ddot{x}^\nu} - g_{\mu \nu , \rho} \dot{x}^\mu \dot{x}^\nu=0
    How's that?

    (Original post by Zhen Lin)
    No, we're considering a curve whose coordinates are those. (A sign of poor notation? Sometimes X is a tangent vector, other times it's just the coordinates.)
    I would imagine this is because \displaystyle e_1 = \frac{\partial}{\partial X^1}.
    To be quite honest, I don't have the patience to try to read through these notes and clarify them. I can explain things my own way. You might be better off finding a different book — perhaps one written with an emphasis on explaining the geometry clearly.
    I think I've got the hang of what's going on at the beginning of this paragraph now.
    I was going to say that since X^\mu(t)=tX^\mu_p, if X^\mu(t)=(t, 0 , \dots, 0) then X^\mu_p=(1,0, \dots , 0)=e_1 and this is why \frac{\partial}{\partial X^1}=e_1 at p. Is this true? I'm not sure if it works because I just treated X^\mu(t) as the tangent vector to the geodesic at time t whereas you said it was the coordinates of the geodesic! But I'm concerned about how he gets these two to be equal especially when he writes that it follows "from above"!

    (Original post by Zhen Lin)
    You have it backwards. Choose an orthonormal basis for T_p M first. Then construct the geodesic normal coordinates associated with that basis. The coordinate basis at p will then coincide with the orthonormal basis you chose.
    Sorry but I still don't get this. He writes "Hence, we can choose coordinates ..." as if it is a consequence of the orthonormal basis, no? And I still don't follow how that makes the metric either eta or delta?

    (Original post by Zhen Lin)
    Equation 125 is only valid in the coordinate basis, yes. With a small modification it's valid in any basis, with the correct (re)definition of various symbols.

    If I just write down a symbol like T^{\rho}_{\phantom{\rho}\mu\nu} there's no way to tell what the intended interpretation is. Following the convention that Greek indices indicate components and Latin indices are abstract, then these are the components of some object T. The important thing is that they're just scalars. However, if I write T^a_{\phantom{a}bc} following this convention, then this means a tensor of type (1, 2). On the other hand, if I say e_\rho is a family of vector fields indexed by \rho, then T^{\rho}_{\phantom{\rho}\mu\nu} e_\rho is a family of vector fields indexed by \mu and \nu.
    Hmmm. But supposing I presented you just with the line
    \nabla_\rho ( \Gamma^\tau{}_{\nu \sigma} e_\tau)-\nabla_\sigma ( \Gamma^\tau{}_{\nu\ rho} e_\tau)
    How would you know whether they are to be treated as a family of (1,0) tensor fields? Is this from what you were saying that they don't transform as tensors so the next things we can treat them as is a family of (1,0) tensors?


    (Original post by Zhen Lin)
    Actually, I've just realised the indices of equation 187 are abstract. The point of the exercise is to show that the third term of the coordinate-free definition, i.e. the one with the commutator, vanishes when your connection is torsion-free. This is the case when e.g. the connection is the Levi–Civita connection, which is why it is absent in equation 181.
    So does coordinate free just mean before we put any indices on anything?

    So for this we can just write
    R^a{}_{bcd} Z^b X^c Y^d = (R(X,Y)Z)^a =\nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]}Z=\nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z-\nabla_{XY}Z+\nabla_{YX}Z
    I am not sure about the 2nd equality I used here. If you look at p55 you'll see what I was trying to make it equal to!
    And also, surely in order to use the torsion free condition I would need that
    -\nabla_{XY}Z +\nabla_{YX}Z = \nabla_Y \nabla_X Z - \nabla_X \nabla_Y Z=0 but you are saying that I cannot write \nabla_{XY}=\nabla_X \nabla_Y. How should I proceed then?

    (Original post by Zhen Lin)
    Notation again. Here the curve parameter is t, not \tau, and it is certainly the case that \frac{d^2}{dt^2} \left[ t X^\mu_p \right] = 0.
    Hmmm. Well isn't X^\mu(t) \in T_p(M) though?
    That would mean \frac{dx^\mu}{dt}=X^\mu(t) and then [latex]\frac{d^2 x^\mu}{dt}=\frac{d X^\mu}{dt}=X^\mu_p
    And where does it say this curve is now parameterised by t and not \tau? Is this just from the definitions above which have all been using t?

    Cheers.
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    (Original post by latentcorpse)
    Well g has tau dependence as well so

    2 \frac{dg_{\rho \nu}}{d \tau} \dot{x}^\nu + 2 g_{\rho \nu} \ddot{x}^\nu} - g_{\mu \nu , \rho} \dot{x}^\mu \dot{x}^\nu=0
    How's that?
    g doesn't explicitly depend on \tau though — it's defined as a function of spacetime coordinates. So you need to rewrite \displaystyle \frac{dg_{\rho \nu}}{d \tau} in terms of derivatives you do know about.

    Sorry but I still don't get this. He writes "Hence, we can choose coordinates ..." as if it is a consequence of the orthonormal basis, no? And I still don't follow how that makes the metric either eta or delta?
    How do you know that you have an orthonormal basis? It's precisely when your metric is either \eta (pseudo-Riemannian / Minkowskian) or \delta (Riemannian / Euclidean).

    Hmmm. But supposing I presented you just with the line
    \nabla_\rho ( \Gamma^\tau{}_{\nu \sigma} e_\tau)-\nabla_\sigma ( \Gamma^\tau{}_{\nu\ rho} e_\tau)
    How would you know whether they are to be treated as a family of (1,0) tensor fields? Is this from what you were saying that they don't transform as tensors so the next things we can treat them as is a family of (1,0) tensors?
    Would it help if I start writing vectors in bold? I'm saying that \mathbf{e}_\tau is a vector and \Gamma^\tau_{\phantom{\tau}\nu \sigma} is a scalar, so when you put them together you get a vector \mathbf{v}_{\nu \sigma}. These are not abstract indices! (In abstract index notation, I'd have to write e_\tau^a, where the lower index identifies the member of the family and the upper index indicates it's a (1, 0) tensor.)

    So for this we can just write
    R^a{}_{bcd} Z^b X^c Y^d = (R(X,Y)Z)^a =\nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]}Z=\nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z-\nabla_{XY}Z+\nabla_{YX}Z
    I am not sure about the 2nd equality I used here. If you look at p55 you'll see what I was trying to make it equal to!
    You can't do that. What's XY? That's not a vector field. [X, Y] = XY - YX is a vector field, however. In index notation, under a coordinate basis, the components of the commutator are X^\mu Y^\nu_{\phantom{\nu},\mu} - Y^\mu X^\nu_{\phantom{\nu},\mu}; and in abstract index notation, for a torsion-free connection, X^a Y^b_{\phantom{b};a} - Y^a X^b_{\phantom{a};b}.

    And also, surely in order to use the torsion free condition I would need that
    -\nabla_{XY}Z +\nabla_{YX}Z = \nabla_Y \nabla_X Z - \nabla_X \nabla_Y Z=0 but you are saying that I cannot write \nabla_{XY}=\nabla_X \nabla_Y. How should I proceed then?
    I have no idea. I've never tried to do this calculation before. Presumably a first step would be to show that it is the case that you can replace the partial derivatives in the component definition of the commutator by covariant derivatives under the torsion-free definition.

    Hmmm. Well isn't X^\mu(t) \in T_p(M) though?
    Hmmm. Well, this is a bit complicated. Remember that the exponential map gives local coordinates in terms of tangent vectors: \exp_p: T_p M \to M. So, with some abuse of notation, and some carelessness about distinguishing between the manifold \mathbb{R}^n and the vector space \mathbb{R}^n, it is both the case that X^\mu(t) \in T_p M and that X^\mu(t) \in M.

    That would mean \frac{dx^\mu}{dt}=X^\mu(t) and then [latex]\frac{d^2 x^\mu}{dt}=\frac{d X^\mu}{dt}=X^\mu_p
    No. X^\mu (t) is the coordinates, not the tangent vectors.

    And where does it say this curve is now parameterised by t and not \tau? Is this just from the definitions above which have all been using t?
    Where does it say that every curve must be parametrised by \tau? It's just an arbitrary symbol. (Yes, it usually means proper time. But that's not what it means here. And you can only parametrise timelike curves with proper time; proper time is meaningless along null curves and spacelike curves.)
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    (Original post by Zhen Lin)
    g doesn't explicitly depend on \tau though — it's defined as a function of spacetime coordinates. So you need to rewrite \displaystyle \frac{dg_{\rho \nu}}{d \tau} in terms of derivatives you do know about.
    ok. yep so 2 \frac{d g_{\rho \nu}}{d \tau} = 2 \frac{\partial g_{\rho \nu}}{\partial x^\mu} \frac{d x^\mu}{d \tau} = 2 g_{\rho \nu, \mu} \dot{x}^\mu

    but we can use the symmetry of the metric to break the above into two seperate terms. then divide by 2 and use an inverse metric and we reproduce the geodesic equation
    \ddot{x}^\mu + \Gamma^\mu{}_{\nu \rho} \dot{x}^\nu \dot{x}^\rho=0
    where \Gamma^\mu{}_{\nu \rho} = \frac{1}{2} g^{\mu \sigma} \left( g_{\nu \sigma, \rho} + g_{\rho \sigma, \rho} - g_{\nu \rho, \sigma}
    Now I should set \mu=i yes?


    (Original post by Zhen Lin)
    Would it help if I start writing vectors in bold? I'm saying that \mathbf{e}_\tau is a vector and \Gamma^\tau_{\phantom{\tau}\nu \sigma} is a scalar, so when you put them together you get a vector \mathbf{v}_{\nu \sigma}. These are not abstract indices! (In abstract index notation, I'd have to write e_\tau^a, where the lower index identifies the member of the family and the upper index indicates it's a (1, 0) tensor.)
    But \Gamma isn't a scalar though is it? If we describe it as a scalar then it would surely transform as a (0,0) tensor butw e know that it does not transform tensorially?

    (Original post by Zhen Lin)
    You can't do that. What's XY? That's not a vector field. [X, Y] = XY - YX is a vector field, however. In index notation, under a coordinate basis, the components of the commutator are X^\mu Y^\nu_{\phantom{\nu},\mu} - Y^\mu X^\nu_{\phantom{\nu},\mu}; and in abstract index notation, for a torsion-free connection, X^a Y^b_{\phantom{b};a} - Y^a X^b_{\phantom{a};b}.

    I have no idea. I've never tried to do this calculation before. Presumably a first step would be to show that it is the case that you can replace the partial derivatives in the component definition of the commutator by covariant derivatives under the torsion-free definition.
    So is it ok to write (R(X,Y)Z)^a = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]}?
    I wasn't sure because I have an abstract index on one side but not on the other? If not, how are these two objects related?

    Anyway so we get it to the point of
    \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{X^a Y^b{}_{;a}-Y^aX^b{}_{;a}}
    Now if we expand into a coordinate basis, we can rewrite that last term as
    X^\nu \nabla_{Y^\mu{}_{, \nu}} Z - Y^\nu \nabla_{X^\mu{}_{,v}}Z but this doesn't seem to be helping much?



    (Original post by Zhen Lin)
    Hmmm. Well, this is a bit complicated. Remember that the exponential map gives local coordinates in terms of tangent vectors: \exp_p: T_p M \to M. So, with some abuse of notation, and some carelessness about distinguishing between the manifold \mathbb{R}^n and the vector space \mathbb{R}^n, it is both the case that X^\mu(t) \in T_p M and that X^\mu(t) \in M.
    I'm not sure I follow...surely it owuld be exp(X^\mu(t)) that is in the manifold, no?


    (Original post by Zhen Lin)
    No. X^\mu (t) is the coordinates, not the tangent vectors.
    But you just said above that it's in both spaces...so how do we distinguish between the two possibilities?

    Thanks!
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    (Original post by latentcorpse)
    ok. yep so 2 \frac{d g_{\rho \nu}}{d \tau} = 2 \frac{\partial g_{\rho \nu}}{\partial x^\mu} \frac{d x^\mu}{d \tau} = 2 g_{\rho \nu, \mu} \dot{x}^\mu

    but we can use the symmetry of the metric to break the above into two seperate terms. then divide by 2 and use an inverse metric and we reproduce the geodesic equation
    \ddot{x}^\mu + \Gamma^\mu{}_{\nu \rho} \dot{x}^\nu \dot{x}^\rho=0
    where \Gamma^\mu{}_{\nu \rho} = \frac{1}{2} g^{\mu \sigma} \left( g_{\nu \sigma, \rho} + g_{\rho \sigma, \rho} - g_{\nu \rho, \sigma}
    Now I should set \mu=i yes?
    Yes, exactly.

    But \Gamma isn't a scalar though is it? If we describe it as a scalar then it would surely transform as a (0,0) tensor butw e know that it does not transform tensorially?
    No, \Gamma is not a scalar. And don't think about transformation laws here — my point is that there's no way to talk about them in a coordinate-free picture. The idea is that when all is said and done, the components of a tensor are just numbers and we can treat them as scalars if we want. If I define a scalar field that coincides with the components of \Gamma in a particular coordinate basis, that doesn't mean it will coincide with the components of \Gamma in every coordinate basis. In fact it won't, precisely because the transformation law for scalars says they're invariant under coordinate change.

    This is something you need to think long and hard about. It's not essential but it's certainly interesting.

    So is it ok to write (R(X,Y)Z)^a = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]}?
    I wasn't sure because I have an abstract index on one side but not on the other? If not, how are these two objects related?
    You can't write that as an equation — that's mixing notation. The LHS should be just R(X, Y)Z.

    Anyway so we get it to the point of
    \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{X^a Y^b{}_{;a}-Y^aX^b{}_{;a}}
    You can't really write that. But you can write X^b \nabla_b Z - Y^b \nabla_b Z - (X^a Y^b_{;a} - Y^a X^b_{;a}) \nabla_b Z though.

    I'm not sure I follow...surely it owuld be exp(X^\mu(t)) that is in the manifold, no?
    Yes, that's true too, because we've identified part of the manifold with (part of the) tangent space at p. Note though that \exp_p (X^\mu(t)) = X^\mu (t). (However, if T^\mu (t) is the tangent vector at X^\mu (t), then \exp_p of it will in general be different.)

    But you just said above that it's in both spaces...so how do we distinguish between the two possibilities?
    As usual, context! It's obvious to me which one is meant based on the discussion. If it's not obvious to you, then you need to check that you really understand the preceding material. Also, the paragraph above equation 148 explicitly says that they're coordinates.
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    (Original post by Zhen Lin)
    Yes, exactly.
    So are you wanting me to calculate the Christoffel symbols explicitly using the formula involving partial derivatives of the metric?

    (Original post by Zhen Lin)
    No, \Gamma is not a scalar. And don't think about transformation laws here — my point is that there's no way to talk about them in a coordinate-free picture. The idea is that when all is said and done, the components of a tensor are just numbers and we can treat them as scalars if we want. If I define a scalar field that coincides with the components of \Gamma in a particular coordinate basis, that doesn't mean it will coincide with the components of \Gamma in every coordinate basis. In fact it won't, precisely because the transformation law for scalars says they're invariant under coordinate change.
    This is something you need to think long and hard about. It's not essential but it's certainly interesting.
    So if I do treat them as a family of (0,1) tensors, I now want to apply (125) to find their covariant derivative. Consider \nabla_\rho(\Gamma^\tau{}_{\nu \sigma} e_\tau)=\partial_\rho \Gamma^\tau{}_{\nu \sigma} e_\tau - \Gamma^\tau{}_{\nu \rho} \Gamma^\mu{}_{\tau \sigma} e_\mu
    So my question is how do we know which of \nu or \sigma to treat as the free index when applying (125). I checked both cases and clearly here he is treating \nu as the free index.

    (Original post by Zhen Lin)
    You can't really write that. But you can write X^b \nabla_b Z - Y^b \nabla_b Z - (X^a Y^b_{;a} - Y^a X^b_{;a}) \nabla_b Z though.
    I think you might have missed some stuff in the first two terms, no? Aren't there two covariant derivatives?

    And for expanding the last term, is the following correct:
    \nabla_{[X,Y]}Z=[X,Y]^b \nabla_b Z = (X^a Y^b_{;a} - Y^a X^b_{;a}) \nabla_b Z????

    So I haven't really got any ideas where to go with this now...

    (Original post by Zhen Lin)
    Yes, that's true too, because we've identified part of the manifold with (part of the) tangent space at p. Note though that \exp_p (X^\mu(t)) = X^\mu (t). (However, if T^\mu (t) is the tangent vector at X^\mu (t), then \exp_p of it will in general be different.)
    Sorry, I just don't see how exp(X^\mu(t))=X^\mu(t)?


    And in (188):
    Just in the paragraph above (188), he calculates that \frac{d^2Z^\mu}{ds^2}=-( \Gamma^\mu{}_{\nu \rho} Z^\nu X^\rho)_{, \sigma} X^\sigma but when he subs this into the Taylor expansion in (188), we have \Gamma^\mu{}_{\nu\ rho , \sigma} Z^\nu X^\rho X^\sigma)_p
    So why does evaluating at p \in M mean the derivative only acts on the Christoffel symbol and not the tangent vectors X^\mu or Z^\mu?
    And in (189), how do you go from the 2nd to the 3rd line?

    I've also managed to get myself confused about the first remarks on p39. If you could clarify this it would help a lot. I understand everything except the bit "or \partial_a f or f_{, a}. Why does the covariant derivative reduce to the partial derivative on the scalar functions?

    Also, to me (119) appears to be "pulled out of thin air" but he claims it comes from the Leibniz rule. I tried to expand (\nabla_X \eta)(Y) in a coordinate basis but got nowhere. Is this doable?

    And finally, I can't understand where (196) comes from. What is he on about when he says that \partial R = \partial \partial \Gamma - \Gamma \partial \Gamma? How will taking the partial derivative of (128) in normal coordinates give that?

    Thanks
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    (Original post by latentcorpse)
    So are you wanting me to calculate the Christoffel symbols explicitly using the formula involving partial derivatives of the metric?
    Well, that is basically what it boils down to, yes. Or you can try to calculate the Euler—Lagrange equations directly, but that's probably not such a good idea unless you're confident with your manipulations.

    So if I do treat them as a family of (0,1) tensors, I now want to apply (125) to find their covariant derivative. Consider \nabla_\rho(\Gamma^\tau{}_{\nu \sigma} e_\tau)=\partial_\rho \Gamma^\tau{}_{\nu \sigma} e_\tau - \Gamma^\tau{}_{\nu \rho} \Gamma^\mu{}_{\tau \sigma} e_\mu
    So my question is how do we know which of \nu or \sigma to treat as the free index when applying (125). I checked both cases and clearly here he is treating \nu as the free index.
    Neither. The indices which indicate membership in the family do not ever come into the calculation — ignore them. The free index is actually \tau — it looks bound in this notation because \Gamma^\tau_{\phantom{\tau} \nu \sigma} is a scalar and you need to contract it with the basis vectors in order to produce a vector. But remember this notation is index-free (so any "free" indices which appear therefore indicate membership in a family).

    I think you might have missed some stuff in the first two terms, no? Aren't there two covariant derivatives?
    Oops, yes. Uh, something more like X^a \nabla_a Y^b \nabla_b Z^c - Y^a \nabla_a X^b \nabla_b Z^c - \cdots.

    And for expanding the last term, is the following correct:
    \nabla_{[X,Y]}Z=[X,Y]^b \nabla_b Z = (X^a Y^b_{;a} - Y^a X^b_{;a}) \nabla_b Z????
    Yes. But have you shown that you can replace the commas by semicolons?

    So I haven't really got any ideas where to go with this now...
    Nor I.

    Sorry, I just don't see how exp(X^\mu(t))=X^\mu(t)?
    Then you need to go back and check that you understand geodesic normal coordinates! This is basically by definition.

    And in (188):
    Just in the paragraph above (188), he calculates that \frac{d^2Z^\mu}{ds^2}=-( \Gamma^\mu{}_{\nu \rho} Z^\nu X^\rho)_{, \sigma} X^\sigma but when he subs this into the Taylor expansion in (188), we have \Gamma^\mu{}_{\nu\ rho , \sigma} Z^\nu X^\rho X^\sigma)_p
    So why does evaluating at p \in M mean the derivative only acts on the Christoffel symbol and not the tangent vectors X^\mu or Z^\mu?
    And in (189), how do you go from the 2nd to the 3rd line?
    I can't be bothered to follow all those calculations now. Personally I'd just ignore the whole section and take it on faith that the Riemann tensor has something to do with the path-dependence of parallel transport. Personally I regard it (in the torsion-free case) as a measure of the non-commutativity of the covariant derivatives.

    I've also managed to get myself confused about the first remarks on p39. If you could clarify this it would help a lot. I understand everything except the bit "or \partial_a f or f_{, a}. Why does the covariant derivative reduce to the partial derivative on the scalar functions?
    By definition.

    Also, to me (119) appears to be "pulled out of thin air" but he claims it comes from the Leibniz rule. I tried to expand (\nabla_X \eta)(Y) in a coordinate basis but got nowhere. Is this doable?
    Thanks
    This is basically a calculation along the lines of "we want the Leibniz rule to work, and we haven't yet defined how the covariant derivative acts on general tensors, so let's define it in a way that makes the Leibniz rule true." So, for example, if Y is a vector field and \eta is a covector field, then \eta (Y) is a scalar field. So \nabla_X (\eta(Y)) = \nabla_X (\eta) Y + \eta(\nabla_X Y) by the Leibniz rule; we know how what the covariant derivative of a scalar field and a vector field is, so that's enough to define what the covariant derivative of a covector field is.
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    (Original post by Zhen Lin)
    Neither. The indices which indicate membership in the family do not ever come into the calculation — ignore them. The free index is actually \tau — it looks bound in this notation because \Gamma^\tau_{\phantom{\tau} \nu \sigma} is a scalar and you need to contract it with the basis vectors in order to produce a vector. But remember this notation is index-free (so any "free" indices which appear therefore indicate membership in a family).
    Can you explain what you mean by index free?
    So I'm struggling to get this covariant derivative to work out. Am I still able to use (125)? What do I do about the e_\tau?

    (Original post by Zhen Lin)
    Oops, yes. Uh, something more like X^a \nabla_a Y^b \nabla_b Z^c - Y^a \nabla_a X^b \nabla_b Z^c - \cdots.
    Yes. But have you shown that you can replace the commas by semicolons?
    Ermmm.....no.

    So we have
    \nabla_{[X,Y]^b e_b} Z
    How can we bring the [X,Y]^b up to the front? We are only allowed to do that with functions and seeing as we have an upstairs abstract index b on it, that suggests it is a vector, no?
    Secondly, once we do bring it up we get
    (X^a Y^b_{,a} - Y^a X^b_{,a}) \nabla_b Z
    Is it true that we are able to change the commas to semi colons because we are workign with a torsion free connection and therefore the Christoffel symbols with the lower indices antisymmetrised will vanish? Exactly as in (130)?


    (Original post by Zhen Lin)
    Then you need to go back and check that you understand geodesic normal coordinates! This is basically by definition.
    I thought I'd more or less got the just of this.
    Ok well if X^\mu(t) \in M is in the neighbourhood of p \in M then in the coordinate chart given by normal coordinates at p \in M, the coordinates of X^\mu(t) is just the preimage of X^\mu(t)under the exponential map (and will lie in T_p(M)). Then has it got something to do with X^\mu(t)=tX^\mu_p?

    And finally, I can't understand where (196) comes from. What is he on about when he says that \partial R = \partial \partial \Gamma - \Gamma \partial \Gamma? How will taking the partial derivative of (128) in normal coordinates give that?

    Cheers!
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    (Original post by latentcorpse)
    Can you explain what you mean by index free?
    It means exactly what it says. You don't use indices. There are basically three sorts of notation in use — concrete index notation, where objects with indices are interpreted as the components of tensors; abstract index notation, where the objects with indices actually represent the entire tensor; and index-free notation, where we don't use indices at all.

    So I'm struggling to get this covariant derivative to work out. Am I still able to use (125)? What do I do about the e_\tau?
    (125) is concrete index notation. You can use it if you want, but then you have to be really careful about which indices represent tensor components and which don't.

    So we have
    \nabla_{[X,Y]^b e_b} Z
    How can we bring the [X,Y]^b up to the front? We are only allowed to do that with functions and seeing as we have an upstairs abstract index b on it, that suggests it is a vector, no?
    No. An expression like [X, Y]^b e_b can only be understood in index-free notation (becase e_b is a family of vector fields indexed by b), and there it's usually just written [X, Y]. In abstract index notation we would just write [X, Y]^b. In index-free notation we write \nabla_X, in abstract index notation we write X^a \nabla_a. (This is justified by showing that \nabla Y: X \mapsto \nabla_X Y is a linear map.)

    Is it true that we are able to change the commas to semi colons because we are workign with a torsion free connection and therefore the Christoffel symbols with the lower indices antisymmetrised will vanish? Exactly as in (130)?
    Something like that, I think, yes. I haven't checked.

    Ok well if X^\mu(t) \in M is in the neighbourhood of p \in M then in the coordinate chart given by normal coordinates at p \in M, the coordinates of X^\mu(t) is just the preimage of X^\mu(t)under the exponential map (and will lie in T_p(M)). Then has it got something to do with X^\mu(t)=tX^\mu_p?
    If that's how you understand it, sure. I've given you my explanation; if you don't understand it, you'll have to ask someone else.

    And finally, I can't understand where (196) comes from. What is he on about when he says that \partial R = \partial \partial \Gamma - \Gamma \partial \Gamma? How will taking the partial derivative of (128) in normal coordinates give that?
    It doesn't. It's just a schematic expression which abbreviates the real equation. It looks plausible, at any rate. R = \partial \Gamma - \partial \Gamma + \Gamma \Gamma - \Gamma \Gamma, so \partial R = \partial \partial \Gamma - \partial \partial \Gamma + \partial \Gamma \Gamma + \Gamma \partial \Gamma - \partial \Gamma \Gamma - \Gamma \partial \Gamma and it's conceivable most of the terms cancel out by some symmetry properties — but you'd have to work it out properly with indices to be sure. At any rate it doesn't really matter — in geodesic normal coordinates based at p, \Gamma = 0 at p, so we're just left with \partial R = \partial \partial \Gamma - \partial \partial \Gamma, as claimed in equation 196.
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    (Original post by Zhen Lin)
    (125) is concrete index notation. You can use it if you want, but then you have to be really careful about which indices represent tensor components and which don't.
    Yeah so for example,
    \nabla_\rho ( \Gamma^\tau}{}_{\nu \sigma}e_\tau )
    What should i treat as the index representing the tensor components?

    (Original post by Zhen Lin)
    If that's how you understand it, sure. I've given you my explanation; if you don't understand it, you'll have to ask someone else.
    Could you explain how you would understand this please?
    And in this chapter on normal coordinates we said that X^\mu(t) was now the coordinates of the curve in this context. When he writes X^\mu(t)=X^\mu_p in normal coordinates, is X^\mu_p the coordinates of the curve at p \in M or is it the tangent vector to the curve at p \in M?

    And when we are dealing with the Levi Civita connection, we can write the Christoffel symbols in terms of partial derivatives of the metric. This is why they vanish for the Euclidean metric in Cartesion coordinates/ Minkowski metric in inertial coordinates (as the metric is constant). However, reading the remark on p29, they make it out as if, given any metric, we can use the fact that it is symmetric and non degenerate, to write it as a diagonal matrix where all the entries are \pm 1. Doesn't this mean that the Christoffel symbols should vanish for every metric. Clearly this is not true!! But I've confused myself as to why?

    And at the end of the remark above (197) he says that the acceleration of a curve with tangent vector T is \nabla_T T but by comparing to the blurb at the beginning of this remark, wouldn't \nabla_T T be the velocity of the curve? This doesn't make sense though because we are talking about geodesics here so \nabla_T T=0 and we expect them to have no acceleration but if my interpretation os right then they would have no velocity? What's going on?
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    (Original post by latentcorpse)
    Yeah so for example,
    \nabla_\rho ( \Gamma^\tau}{}_{\nu \sigma}e_\tau )
    What should i treat as the index representing the tensor components?
    The concrete index notation for this expression is \nabla_\rho ( \Gamma^\tau}_{\phantom{\tau} \nu \sigma}). Ignore the \nu and \sigma.

    X^\mu_p the coordinates of the curve at p \in M or is it the tangent vector to the curve at p \in M?
    The tangent vector.

    And when we are dealing with the Levi Civita connection, we can write the Christoffel symbols in terms of partial derivatives of the metric. This is why they vanish for the Euclidean metric in Cartesion coordinates/ Minkowski metric in inertial coordinates (as the metric is constant). However, reading the remark on p29, they make it out as if, given any metric, we can use the fact that it is symmetric and non degenerate, to write it as a diagonal matrix where all the entries are \pm 1. Doesn't this mean that the Christoffel symbols should vanish for every metric. Clearly this is not true!! But I've confused myself as to why?
    The expression as given is for a coordinate basis. It is true that every metric admits an orthonormal basis, but these orthonormal bases are in general non-holonomic. (So you need to use the general expression for the Christoffel symbol, which involves commutation coefficients.)

    And at the end of the remark above (197) he says that the acceleration of a curve with tangent vector T is \nabla_T T but by comparing to the blurb at the beginning of this remark, wouldn't \nabla_T T be the velocity of the curve? This doesn't make sense though because we are talking about geodesics here so \nabla_T T=0 and we expect them to have no acceleration but if my interpretation os right then they would have no velocity? What's going on?
    No, if T is a tangent vector then it is already a velocity vector. So \nabla_T T is the acceleration, and yes, it is 0 for affinely-parametrised geodesics.
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    (Original post by Zhen Lin)
    The concrete index notation for this expression is \nabla_\rho ( \Gamma^\tau}_{\phantom{\tau} \nu \sigma}). Ignore the \nu and \sigma.



    The tangent vector.



    The expression as given is for a coordinate basis. It is true that every metric admits an orthonormal basis, but these orthonormal bases are in general non-holonomic. (So you need to use the general expression for the Christoffel symbol, which involves commutation coefficients.)



    No, if T is a tangent vector then it is already a velocity vector. So \nabla_T T is the acceleration, and yes, it is 0 for affinely-parametrised geodesics.
    How do we go about the exercise at the bottom of p67/top of p68?

    And secondly, at the top of p69, he giveas the example and invites us to check that (\phi^* g)_{\mu \nu} = diag ( \sin^2{\theta})

    However I find that
    (\phi^* g)_{\mu \nu}= \left( \frac{ \partial y^\alpha }{ \partial \theta} \right) \left( \frac{\partial y^\beta}{ \partial \phi} \right) \delta_{\alpha \beta} =  \left( \frac{ \partial y^\alpha }{ \partial \theta} \right) \left( \frac{\partial y^\alpha}{ \partial \phi} \right) = \begin{pmatrix} \cos{\theta} \cos{\phi} & \cos{\theta} \sin{\phi} & -\sin{\theta} \end{pmatrix} \begin{pmatrix} -\sin{\theta} \sin{\phi} \\ \sin{\theta}\cos{\phi} \\ 0 \end{pmatrix}
    =-\cos{\theta} \cos{\phi} \sin{\theta} \sin{\phi}+\cos{\theta} \cos{\phi} \sin{\theta} \sin{\phi}=0

    Where have I gone wrong?

    And what about the exercises at the bottom of p69?

    Also, could you explain the puzzle he describes in the 3rd remark on p70. Why should the Einstein eqn not determine the components of g uniquely?

    Thanks
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    (Original post by latentcorpse)
    How do we go about the exercise at the bottom of p67/top of p68?
    Just follow the definitions?

    Where have I gone wrong?
    Nothing's wrong. Although the first g_{\mu \nu} should really be g_{\theta \phi}. You've only calculated one component of the metric tensor, namely the off-diagonal component... which is 0, exactly as claimed. Now calculate g_{\theta \theta} and g_{\phi \phi}.

    And what about the exercises at the bottom of p69?
    Again, just follow the definitions? In particular you'll end up writing something like \displaystyle \frac{\partial y^\alpha}{\partial x^\mu} \frac{\partial x^\nu}{\partial y^\alpha} for the contraction case. This simplifies to \delta^\nu_\mu, since the matrices are mutually inverse.

    It might be helpful to do the second exercise first.

    Also, could you explain the puzzle he describes in the 3rd remark on p70. Why should the Einstein eqn not determine the components of g uniquely?
    The gauge symmetry tells us that if g is a solution, then the pullback of g by some arbitrary diffeomorphism (respecting the boundary conditions) is also a solution. A more detailed physical argument is given by Einstein himself — look up Einstein's hole argument.

    Of course, it might be that there are no non-trivial diffeomorphisms which respect the boundary conditions, but there's another argument: Suppose we regard the Einstein field equations (EFE) as a system of PDEs with both the metric and the stress-energy tensor as unknowns. Then we have 20 unknowns, since each tensor has 10 independent components. The EFE gives us 10 equations, and even when we add in the divergence-free condition on the stress-energy tensor, this only gives us 4 more equations, so it's still underdetermined.
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    (Original post by Zhen Lin)
    Just follow the definitions?
    What definitions?

    (Original post by Zhen Lin)
    Again, just follow the definitions? In particular you'll end up writing something like \displaystyle \frac{\partial y^\alpha}{\partial x^\mu} \frac{\partial x^\nu}{\partial y^\alpha} for the contraction case. This simplifies to \delta^\nu_\mu, since the matrices are mutually inverse.

    It might be helpful to do the second exercise first.
    Probably the same reason as above why I still can't do this...

    In the definition (228), why do we use \phi_{-t} and not \phi_t - I guess this is a fairly minor point (after all it is a definition so I should just accept it!) but I was just curious if you knew....

    In (231), how do we go from the 1st to the 2nd line?

    In (234), how does this work?
    I find that [ X,Y]^\mu = X^\nu Y^\mu_{, \nu} - Y^\nu X^\mu_{, \nu}. Why does the last term vanish?

    Why is (238) written the way it is? Why can we not use the metric to lower those indices on the X's i.e. why can't we write g_{\mu \rho} \partial_\nu X^\rho? Or can we and he just hasn't bothered to do so yet?

    Then at the bottom of p73, he talks about how if we had a metric that was independent of z, (238) would tell us that \frac{\partial}{\partial z} is a Killing vector field. I don't see how it tells us this at all!

    Or the converse at the top of p74, how does having a Killing vector field allow us to introduce coordinates so that the metric is independent of one of the coordinates?

    And in the lemma on p74, why does \frac{d}{d \tau} ( X_aV^a ) = V ( X_a V^a)?
    Why does V ( X_a V^a)=\nabla_V ( X_a V^a )?

    And finally, if we were trying to show a basis independent version of a formula, why would we go to normal coordinates and then apply the rules of general covariance? (This is what the answers to one of my tutorials says). I don' see why we can't just apply general covariance rules straight away? Is it because minimal coupling i.e. replacing partial derivatives with covariant ones requires the Christoffel symbols to vanish and that only happens in normal coordinates?

    Thanks again and happy new year!
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    (Original post by latentcorpse)
    What definitions?
    Pushforward is defined on page 67. This is just a calculation of the pushforward in a coordinate basis.

    In the definition (228), why do we use \phi_{-t} and not \phi_t - I guess this is a fairly minor point (after all it is a definition so I should just accept it!) but I was just curious if you knew....
    I have no idea. Presumably it's what's needed to make \mathcal{L}_X Y = [X, Y] in the case of vector fields. (I've never attempted this calculation before.)

    In (231), how do we go from the 1st to the 2nd line?
    Definition of partial derivative.

    In (234), how does this work?
    I find that [ X,Y]^\mu = X^\nu Y^\mu_{, \nu} - Y^\nu X^\mu_{, \nu}. Why does the last term vanish?
    Choice of coordinates.

    Why is (238) written the way it is? Why can we not use the metric to lower those indices on the X's i.e. why can't we write g_{\mu \rho} \partial_\nu X^\rho? Or can we and he just hasn't bothered to do so yet?
    Huh? I don't understand your question. I imagine it has something to do with the fact that the metric doesn't commute with partial differentiation.

    Then at the bottom of p73, he talks about how if we had a metric that was independent of z, (238) would tell us that \frac{\partial}{\partial z} is a Killing vector field. I don't see how it tells us this at all!
    Suppose the 1st coordinate is z. Then if \displaystyle X = \frac{\partial}{\partial z}, then the components of X are (1, 0, 0, ...) - so the partial derivatives of components of X disappear, and the partial derivatives of the metric disappear except for the \displaystyle \frac{\partial g_{\mu \nu}}{\partial z} term. But that term is zero anyway, so the whole thing is zero, and X is a Killing vector field.

    This is really just a straightforward exercise in substituting things into equations. Are you sure you understand the definitions?

    Or the converse at the top of p74, how does having a Killing vector field allow us to introduce coordinates so that the metric is independent of one of the coordinates?
    This is not clear to me either.

    And in the lemma on p74, why does \frac{d}{d \tau} ( X_aV^a ) = V ( X_a V^a)?
    By definition of V. Remember vectors are differential operators!

    Why does V ( X_a V^a)=\nabla_V ( X_a V^a )?
    By definition of covariant derivative on scalar fields.

    And finally, if we were trying to show a basis independent version of a formula, why would we go to normal coordinates and then apply the rules of general covariance? (This is what the answers to one of my tutorials says). I don' see why we can't just apply general covariance rules straight away? Is it because minimal coupling i.e. replacing partial derivatives with covariant ones requires the Christoffel symbols to vanish and that only happens in normal coordinates?
    I imagine it's something like that. The principle of minimal coupling is a postulate of physics, and there are some instances where it leads to the wrong answer, so don't worry too much about it.
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    (Original post by Zhen Lin)
    Pushforward is defined on page 67. This is just a calculation of the pushforward in a coordinate basis.
    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?

    (Original post by Zhen Lin)
    Definition of partial derivative.
    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?

    (Original post by Zhen Lin)
    Choice of coordinates.
    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?

    (Original post by Zhen Lin)
    Huh? I don't understand your question. I imagine it has something to do with the fact that the metric doesn't commute with partial differentiation.
    So if I have something like g_\mu \rho} \partial_\nu X^\rho, that is NOT equal to \partial_\nu X_\mu? However the metric does commute with the covariant derivative doesn't it?

    (Original post by Zhen Lin)
    Suppose the 1st coordinate is z. Then if \displaystyle X = \frac{\partial}{\partial z}, then the components of X are (1, 0, 0, ...) - so the partial derivatives of components of X disappear, and the partial derivatives of the metric disappear except for the \displaystyle \frac{\partial g_{\mu \nu}}{\partial z} term. But that term is zero anyway, so the whole thing is zero, and X is a Killing vector field.
    So why do all the partial derivatives of the metric disappear except the \frac{\partial g_{\mu \nu}}{\partial z} term?

    (Original post by Zhen Lin)
    By definition of V. Remember vectors are differential operators!
    Where did we define V this way? I still don't see this step.

    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?
    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?

    On the remarks on p75:
    He says that r is not the distance from the origin. What is it then?
    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?
    And what is he on about in the 3rd remark? Surely changing either r or M to -r or -M is physically unrealistic?

    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?

    How does he derive (250)?

    Do null geodesics have zero length? Doesn't this violate causality? 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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