Tensors

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?

(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?

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. $\gamma_p(0,X_p)=p , \gamma_p'(0,X_p)=X_p$

(iv) Given this geodesic, you tell me that $\gamma_p(t,X_p)=\gamma_p(1,tX_p)$ Why?

(v) And how does that then equal $exp(t X_p)$?

(vi) Ok so finally we prove that $\phi \circ \gamma(t,X_p) = t d \phi_p (X_p)$. Can you explain in words how this proves $X^\mu(t)=tX^\mu_p$?

Ok. So I would like this method to work out seeing as I managed it myself for a change. However, if you look at the exercise on p46, it is considering the case of the Levi Civita connection. However, I don't think I have made use of the Levi Civita proiperty anywhere in my answer. I guess I used the fact that $\nabla_XX=0$ but this is a consequence of affine parameterisation, not of the Levi Civita connection. Have I used the property somewhere without realising it?

I'm also confusing myself on (150)

Yes, but your equation is wrong. Check your differentiation!

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)?

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?)

I imagine the first coordinate is indexed by 1, so this is effectively automatic. I went through this with you on the first page!

That's what it means to have a local inertial frame.

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.

This isn't strictly speaking index notation. (Unfortunately, this sort of confusion frequently arises when trying to calculate components using the coordinate-free definitions...) $\Gamma^tau_{\phantom{\tau}\nu\sigma} e_\tau$ is a family of (1,0) vector fields indexed by $\nu$ and $\sigma$.

Just calculate it from the coordinate-free definitions, I guess — show that $\nabla_\rho \nabla_\sigma [Z^\mu e_\mu] - \nabla_\sigma \nabla_\rho [Z^\mu e_\mu] = R^\mu_{\phantom{\mu}\nu\rho\sigma} Z^\nu e_\mu$. 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.)

So we are considering some curve whose tangent vector is $X^\mu=(t,0, \dots , 0)$

And then, how can $e_1$ be a tangent?

I mean even the first line of this paragraph "Now consider $\frac{\partial}{\partial X^1}$ in normal coordinates" - what does this even mean?

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 $g_{\mu \nu , \rho}=0$.

Where did we show this though? What we showed on p48, wasn't for a local inertial frame, was it?

Ok. Got it. Out of interest, what is this isomorphism?

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?

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 $Z^\mu e_\mu$?

Now I think this might work out if I am allowed to write $e_\rho ( e_\sigma (Z^\mu)) = \nabla_{\sigma \rho}$

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?

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 $\displaystyle \frac{d}{d \tau} \left[ 2 g_{\rho \nu} \dot{x}^\nu \right]$.

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.

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.

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 $\nu$.

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.

Notation again. Here the curve parameter is t, not $\tau$, and it is certainly the case that $\frac{d^2}{dt^2} \left[ t X^\mu_p \right] = 0$.

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?

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?

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?

Hmmm. Well isn't $X^\mu(t) \in T_p(M)$ though?

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?

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.

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.)

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.

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 $X^\mu(t) \in M$.

No. $X^\mu (t)$ is the coordinates, not the tangent vectors.

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?

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
$\nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{X^a Y^b{}_{;a}-Y^aX^b{}_{;a}}$

I'm not sure I follow...surely it owuld be $exp(X^\mu(t))$ that is in the manifold, no?

But you just said above that it's in both spaces...so how do we distinguish between the two possibilities?

Yes, exactly.

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.

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.

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.)

So are you wanting me to calculate the Christoffel symbols explicitly using the formula involving partial derivatives of the metric?

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.

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...

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?
Thanks