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    Hi, in the attachment, in the calculation at teh top of page 27,
    (i) how do we go from the 2nd to 3rd line? surely this assumes X^\mu and Y^\nu commute?
    (ii)how do we go from the 4th to the 5th line?

    Thanks you.
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  1. File Type: pdfGR Course Notes.pdf (527.9 KB, 188 views)
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    A bit off topic but where is part 1 to this if you don't mind me asking?
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    (Original post by boromir9111)
    A bit off topic but where is part 1 to this if you don't mind me asking?
    part i of what sorry?
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    (Original post by latentcorpse)
    part i of what sorry?
    It says part 3 General relativity, I assumed there were other 2 parts to it? or just a continuation on from another topic?
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    oh i don't know. presumably they're available online. the part 3 bit is like a masters course which i am doing so i did my undergrad elsewhere.
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    Oh, ok.....thanks for that and sorry for not being of no help at all!
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    i) They're just functions. Why shouldn't they commute?
    ii) Switch the indices summed over in the first. They're just dummy variables after all.
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    (Original post by SimonM)
    i) They're just functions. Why shouldn't they commute?
    To clarify, X^\mu are just functions M \to \mathbb{R} with the property that \displaystyle X^\mu \frac{\partial}{\partial x^\mu} = X (summation convention). So when we write X^\mu Y^\nu it's just pointwise multiplication of two real-valued functions.

    In a less-coordinateful way: X^\mu Y^\nu can be understood as the components of the tensor product X \otimes Y. The ordering of the tensor product is fixed by the indices, not the order of the factors - in the sense that (X \otimes Y)^{\mu \nu} = X^\mu Y^\nu = Y^\nu X^\mu. (And yes, in a very real way (X \otimes Y)^{\mu \nu} = (Y \otimes X)^{\nu \mu}. In abstract index notation this is just saying that I can twist the indices of a tensor around and it's the same as reversing the order of multiplication to begin with.)
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    (Original post by SimonM)
    i) They're just functions. Why shouldn't they commute?
    ii) Switch the indices summed over in the first. They're just dummy variables after all.
    ok. thanks. i understand (i) now but for (ii), isn't it just going to be identifying eqn 64 in the 4th line and then realising that
    \frac{\partial F}{\partial x^\mu} = (\frac{\partial}{\partial x^\mu}) (f) from eqn 27?

    Also, could you perhaps explain to me where eqn 27 comes from? It gets used loads in these notes and it's bugging me that I don't seem to understand it at all!
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    (Original post by latentcorpse)
    Also, could you perhaps explain to me where eqn 27 comes from? It gets used loads in these notes and it's bugging me that I don't seem to understand it at all!
    That's practically a definition of what it means to partial-differentiate a function on a manifold. (f is defined on a manifold, F is defined on \mathbb{R}^n.) To obtain the derivation, note that you can obtain \displaystyle \frac{\partial}{\partial x^\mu} as a tangent vector by considering the curve which is constant in all coordinates except one and constant velocity in x^\mu.
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    (Original post by Zhen Lin)
    That's practically a definition of what it means to partial-differentiate a function on a manifold. (f is defined on a manifold, F is defined on \mathbb{R}^n.) To obtain the derivation, note that you can obtain \displaystyle \frac{\partial}{\partial x^\mu} as a tangent vector by considering the curve which is constant in all coordinates except one and constant velocity in x^\mu.
    I don't understand. Sorry!
    So this curve you are talking about is given by eqn 26 but then he just says "the tangent to this curve is \frac{\partial}{\partial x^\mu}...
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    (Original post by latentcorpse)
    I don't understand. Sorry!
    So this curve you are talking about is given by eqn 26 but then he just says "the tangent to this curve is \frac{\partial}{\partial x^\mu}...
    I don't know if I'm misremembering, but I thought you did differential geometry before? Anyway. It's hard to talk about this formally because you'll run into notational difficulty quite fast. But it basically follows by definition - for example, the tangent to the curve x = 0 in Euclidean space, parametrised appropriately, is the standard unit vector \mathbf{e}_x.
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    (Original post by Zhen Lin)
    I don't know if I'm misremembering, but I thought you did differential geometry before? Anyway. It's hard to talk about this formally because you'll run into notational difficulty quite fast. But it basically follows by definition - for example, the tangent to the curve x = 0 in Euclidean space, parametrised appropriately, is the standard unit vector \mathbf{e}_x.
    Hmmmm. Yeah I studied it a couple fo years ago but it was poorly taught by a man who spoke very little English and clearly I remember very little of it.

    Anyway, I would quite like to try and get my head around this. So, looking at the notes, it tells us (at the bottom of p15) that we can derive eqn27 using eqn 24 and additionally that we can denote the tangent vector by ( \frac{\partial}{\partial x^\mu})_p.

    This means that we can look at eqn 24 and replace X_p(f) by ( \frac{\partial}{\partial x^\mu})_p (f) .
    Now we just need to sort out the RHS.
    We have

    ( \frac{\partial}{\partial x^\mu})_p (f) = (\frac{ \partial F}{ \partial x^\mu})_{\phi(p)} ( \frac{d x^\mu ( \lambda (t))}{dt} )_{t=0}

    So we need the second term on the RHS to equal 1 and we'll be done.
    I'm not 100% sure why this is - is it because from the defn of \lambda at the bottom of p15, we see that the derivative of the \muth term wrt t is just 1? I think I may be getting confused here....

    Thanks again for your help!
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    Actually I just realised I wrote down a contradiction. Rather, it's the curve (x, y, z) = (t, 0, 0) that has tangent vector \mathbf{e}_x. But anyway, let me try to give a full derivation. (No pun, haha.)

    Firstly, we need to define what we mean by \displaystyle \frac{\partial}{\partial x^\mu}. It's the tangent vector to the curve \lambda: [-1, 1] \to M such that under the chart \phi: M \to \mathbb{R}^n, \phi \circ \lambda is constant in all components except the \mu-th component, where it has unit velocity, i.e. the \mu-th component of \displaystyle \frac{d}{dt} \left[ \phi \circ \lambda(t) \right] is 1. Now, let p = \lambda(0) be a point on the manifold. Then, by definition, (pardon the difference in notation) \displaystyle \left. \frac{\partial}{\partial x^\mu} \right| _p is the differential operator such that \displaystyle \left. \frac{\partial}{\partial x^\mu} \right| _p f = \left. \frac{d}{dt} \right|_0 \left[ f \circ \lambda(t) \right].

    We can further expand the RHS using the charts and the chain rule: write F = f \circ \phi^{-1} : \mathbb{R}^n \to \mathbb{R}, and then f \circ \lambda = F \circ (\phi \circ \lambda). So, writing \partial_\mu F for the \mu-th component of \nabla F, we have, by the chain rule, \displaystyle  \left. \frac{d}{dt} \right| _0 \left[ f \circ \lambda(t) \right] = \partial_\nu F(\phi(p)) \left. \frac{d}{dt} \right| _0 \left[ x^\nu \circ \lambda(t) \right] (summation convention; x^\nu is the \nu-th component of \phi). But \displaystyle \frac{d}{dt} \left[ x^\nu \circ \lambda(t) \right] = \begin{cases} 0 & \nu \ne \mu \\ 1 & \nu = \mu \end{cases}, so \displaystyle \partial_\nu F(\phi(p)) \left. \frac{d}{dt} \right| _0 \left[ x^\nu \circ \lambda(t) \right] = \partial_\mu F(\phi(p)). So  \displaystyle \left. \frac{\partial}{\partial x^\mu} \right| _p f = \partial_\mu F(\phi(p)).

    Of course, there are certain details being swept under the rug here. For instance, there are many curves \lambda: [-1, 1] \to M where the tangent vector at 0 acts on functions exactly the same way as \displaystyle \left. \frac{\partial}{\partial x^\mu} \right| _{\lambda(0)}. When you do the construction of the tangent space, you quotient out all these equivalent operators so that the remaining tangent vectors act on functions in a unique way. And then this only defines the tangent space (and its action on functions) at one point - so you have to repeat this construction everywhere on the manifold, and then define a vector field in order to get \displaystyle \left. \frac{\partial}{\partial x^\mu}. And even then, in the general case this vector field is only defined on one patch on the manifold, because in general a chart is only defined on one patch on the manifold.
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    (Original post by Zhen Lin)
    Actually I just realised I wrote down a contradiction. Rather, it's the curve (x, y, z) = (t, 0, 0) that has tangent vector \mathbf{e}_x. But anyway, let me try to give a full derivation. (No pun, haha.)

    Firstly, we need to define what we mean by \displaystyle \frac{\partial}{\partial x^\mu}. It's the tangent vector to the curve \lambda: [-1, 1] \to M such that under the chart \phi: M \to \mathbb{R}^n, \phi \circ \lambda is constant in all components except the \mu-th component, where it has unit velocity, i.e. the \mu-th component of \displaystyle \frac{d}{dt} \left[ \phi \circ \lambda(t) \right] is 1. Now, let p = \lambda(0) be a point on the manifold. Then, by definition, (pardon the difference in notation) \displaystyle \left. \frac{\partial}{\partial x^\mu} \right| _p is the differential operator such that \displaystyle \left. \frac{\partial}{\partial x^\mu} \right| _p f = \left. \frac{d}{dt} \right|_0 \left[ f \circ \lambda(t) \right].

    We can further expand the RHS using the charts and the chain rule: write F = f \circ \phi^{-1} : \mathbb{R}^n \to \mathbb{R}, and then f \circ \lambda = F \circ (\phi \circ \lambda). So, writing \partial_\mu F for the \mu-th component of \nabla F, we have, by the chain rule, \displaystyle  \left. \frac{d}{dt} \right| _0 \left[ f \circ \lambda(t) \right] = \partial_\nu F(\phi(p)) \left. \frac{d}{dt} \right| _0 \left[ x^\nu \circ \lambda(t) \right] (summation convention; x^\nu is the \nu-th component of \phi). But \displaystyle \frac{d}{dt} \left[ x^\nu \circ \lambda(t) \right] = \begin{cases} 0 & \nu \ne \mu \\ 1 & \nu = \mu \end{cases}, so \displaystyle \partial_\nu F(\phi(p)) \left. \frac{d}{dt} \right| _0 \left[ x^\nu \circ \lambda(t) \right] = \partial_\mu F(\phi(p)). So  \displaystyle \left. \frac{\partial}{\partial x^\mu} \right| _p f = \partial_\mu F(\phi(p)).

    Of course, there are certain details being swept under the rug here. For instance, there are many curves \lambda: [-1, 1] \to M where the tangent vector at 0 acts on functions exactly the same way as \displaystyle \left. \frac{\partial}{\partial x^\mu} \right| _{\lambda(0)}. When you do the construction of the tangent space, you quotient out all these equivalent operators so that the remaining tangent vectors act on functions in a unique way. And then this only defines the tangent space (and its action on functions) at one point - so you have to repeat this construction everywhere on the manifold, and then define a vector field in order to get \displaystyle \left. \frac{\partial}{\partial x^\mu}. And even then, in the general case this vector field is only defined on one patch on the manifold, because in general a chart is only defined on one patch on the manifold.
    Hi. That was a really useful post! Thanks a lot!

    A couple of things:

    (i) why do we evalutate \frac{\partial F(x)}{\partial x^\mu} at \phi(p)?

    (ii) Just trying to get this chain rule thing to work out and having some trouble:

     \frac{d}{dt} ( F \cdot \phi \cdot \lambda)_{t=0}
    Now by taking the \muth component of \phi, we can get the x^\mu(\lambda(t)) that we want but I don't see how we get the \frac{\partial}{\partial x^\mu} in the first term?
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    (Original post by latentcorpse)
    (i) why do we evalutate \frac{\partial F(x)}{\partial x^\mu} at \phi(p)?
    As opposed to...? p \in M, so it's not in the domain of F: \mathrm{R}^n \to \mathrm{R}. On the other hand \phi(p) \in \mathbb{R}^n.

    (ii) Just trying to get this chain rule thing to work out and having some trouble:

     \frac{d}{dt} ( F \cdot \phi \cdot \lambda)_{t=0}
    Now by taking the \muth component of \phi, we can get the x^\mu(\lambda(t)) that we want but I don't see how we get the \frac{\partial}{\partial x^\mu} in the first term?
    \mu is fixed, so you can't use it as a summation index. Recall the chain rule from multivariable calculus: \displaystyle \frac{d}{dt} = \frac{dx^\nu}{dt} \frac{\partial}{\partial x^\nu}.

    Moreover, you can't extract components like that, since the multivariate part of the expression is embedded inside. The key point is that \displaystyle \left. \frac{d}{dt} \right| _0 \left[ \phi \circ \lambda(t) \right] is just the vector with 1 in the \mu-th component and 0 everywhere else. This is by construction of the curve.

    I'm afraid this is very difficult to explain clearly. Either you get bogged down in notation and lose sight of the essential intuition, or you focus on the intuitive ideas and don't get a good idea of what is formally happening...
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    (Original post by Zhen Lin)
    As opposed to...? p \in M, so it's not in the domain of F: \mathrm{R}^n \to \mathrm{R}. On the other hand \phi(p) \in \mathbb{R}^n.



    \mu is fixed, so you can't use it as a summation index. Recall the chain rule from multivariable calculus: \displaystyle \frac{d}{dt} = \frac{dx^\nu}{dt} \frac{\partial}{\partial x^\nu}.

    Moreover, you can't extract components like that, since the multivariate part of the expression is embedded inside. The key point is that \displaystyle \left. \frac{d}{dt} \right| _0 \left[ \phi \circ \lambda(t) \right] is just the vector with 1 in the \mu-th component and 0 everywhere else. This is by construction of the curve.

    I'm afraid this is very difficult to explain clearly. Either you get bogged down in notation and lose sight of the essential intuition, or you focus on the intuitive ideas and don't get a good idea of what is formally happening...
    ok. i get the 1st point now but as for the 2nd perhaps this will make my problem a bit clearer:

    we have

    \frac{d}{dt} [ F \cdot \phi \cdot \lambda ]

    and then by the chain rule we can write

    \frac{\partial F}{\partial x^\mu} \frac{ d x^\mu \phi \cdot \lambda}{dt}

    now i realise that x^\mu is just the \mu-th component of \phi but to me it appears that there should be an x^\mu^2 term, no?

    how do we combine the \phi and the x^\mu to give just a single x^\mu?
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    (Original post by latentcorpse)
    ok. i get the 1st point now but as for the 2nd perhaps this will make my problem a bit clearer:

    we have

    \frac{d}{dt} [ F \cdot \phi \cdot \lambda ]

    and then by the chain rule we can write

    \frac{\partial F}{\partial x^\mu} \frac{ d x^\mu \phi \cdot \lambda}{dt}

    now i realise that x^\mu is just the \mu-th component of \phi but to me it appears that there should be an x^\mu^2 term, no?

    how do we combine the \phi and the x^\mu to give just a single x^\mu?
    Your expression is incorrect in two ways: firstly, \mu is fixed and cannot appear as a summation index; secondly, it's just x^\nu \circ \lambda, not x^\nu \circ \phi \circ \lambda. The latter is an invalid expression - \phi and x^\nu are both functions of the type M \to \mathbb{R}^n so cannot be composed. (I realise my definition of x^\nu conflicts with the way it's used in your lecture notes. In the lecture notes, x^\nu is the \nu-th component of \phi \circ \lambda.)

    (Also, \circ and \cdot mean different things. Please distinguish between them.)
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    (Original post by Zhen Lin)
    Your expression is incorrect in two ways: firstly, \mu is fixed and cannot appear as a summation index; secondly, it's just x^\nu \circ \lambda, not x^\nu \circ \phi \circ \lambda. The latter is an invalid expression - \phi and x^\nu are both functions of the type M \to \mathbb{R}^n so cannot be composed. (I realise my definition of x^\nu conflicts with the way it's used in your lecture notes. In the lecture notes, x^\nu is the \nu-th component of \phi \circ \lambda.)

    (Also, \circ and \cdot mean different things. Please distinguish between them.)
    ok. so i don't understand why \mu can't appear as a summation index - surely it's summed over in eqn 24?

    also, surely x^\nu is the \nu-th component of \phi and not the \nu-th component of \phi \circ \lambda since 2 lines before eqn 24, it says \phi = ( x^1, \dots , x^n)?

    And finally, what do you take \cdot to mean if \circ is composition? Do you use it just for multiplication?

    Thanks and sorry for draggin this out!
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    (Original post by latentcorpse)
    ok. so i don't understand why \mu can't appear as a summation index - surely it's summed over in eqn 24?
    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.

    also, surely x^\nu is the \nu-th component of \phi and not the \nu-th component of \phi \circ \lambda since 2 lines before eqn 24, it says \phi = ( x^1, \dots , x^n)?
    So it is. Nevermind then.

    And finally, what do you take \cdot to mean if \circ is composition? Do you use it just for multiplication?
    I have never, ever seen it used for function composition.
 
 
 
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