Tensors

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.

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.)

(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?

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?

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 $\displaystyle \int G^2 \, du$.

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

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. The metric depends on the point!

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 $\dot{t} > 0$.)

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?

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?

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})

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!

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.)

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.)

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!

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.

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.

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).

I see that eqn 36 is evaluated at a point but underneath he writes that it gives the components? Is it because eqn 36 gives the components of the tensor and here in chapter 15, he is talking about the components of the tensor FIELD?

I don't understand what you're on about here. Could you elaborate please?

I think I "kind of" understand this. I can't really see how it would ever be 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?

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!

And how do we derive (118)?

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'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?

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 $\displaystyle \ddot{x}^\sigma + \Gamma^{\sigma}_{\phantom{\sigma}\mu \nu} \dot{x}^\mu \dot{x}^\nu = 0$.

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!