Local Lorentz Frames: Definition & Context for GR

[Fredrik]

OMG Fredrik, this is absolutely false!

Orthonormality is e^μ₀ e_μ0 = 1, e^μ₀e_μ1 = 0, and so on, ten equations. Do you see, I hope, that each equation is summed on μ?? These are ten vector dot products, each one of them equal to either 0 or 1.

The condition you're confusing this with is e_μ⁰ e_ν0 + e_μ¹ e_ν1 + e_μ² e_ν2 + e_μ³ e_ν3 = η_μν (whatever η_μν is supposed to mean in a general coordinate system!) Do you see that this set of equations is summed on the Lorentz index a and is totally different from the other one??

You managed to make me think that I was wrong for a while, but it looks like I was right all along.

If we had been talking about orthogonality in $\mathbb R^4$ with the Euclidean inner product, the statement that $\{e_\mu\}$ is an orthonormal set would have been equivalent to the 16 equations
$$\langle e_\mu,e_\nu\rangle=\delta_{\mu\nu}.$$ When we're dealing with tangent spaces in GR, what we have instead of the Euclidean inner product is the metric tensor. And what corresponds to the Kronecker delta is the matrix
$$\eta=\begin{pmatrix}-1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{pmatrix}.$$ $\eta_{\mu\nu}$ denotes the entry on row $\mu$, column $\nu$ of this matrix. Now the statement that $\{e_\mu\}$ is orthonormal (with respect to g) is equivalent to the 16 equations
$$g(e_\mu,e_\nu)=\eta_{\mu\nu}.$$ Now, assuming that the index you put on the right is the one that labels what vector we're talking about, and the one on the left labels components of that vector (this seems like a weird convention), your orthogonality condition is the same as mine. The only thing you did different was to use a different (more complicated) way to write the left-hand side, and to write down two specific equations instead of an arbitrary one. Edit: I see now that this isn't necessarily what your notation meant. See my edit of post #39.

Note that I proved in the first paragraph of my previous post that the complicated way to write the left-hand side (but with the vector index to the left and the component index to the right) is equivalent to this way.

 

[dextercioby]

[...]
The condition $(e_\mu)^a (e_\nu)_a=\eta_{\mu\nu}$ on the other hand, is the definition of what it means for $\{e_\mu\}_{\mu=0}^3$ to be an orthonormal set. [...]

No, not really.

$\{e_{\mu}\}$ is orthonormal wrt the flat space time metric $\eta_{ab}$ iff

$e_{\mu a} e^{\mu}_{~b} = \eta_{ab}$, not really what you wrote and is exactly what Bill wrote.

EDIT: @Ben: tell that to Wald. :D

 

[Ben Niehoff]

The usual convention is that $a, b, c$ are "flat space" indices, and $\mu,\nu,\rho$ are "curved space" indices, such that

g^{\mu\nu} \, e_\mu{}^a e_\nu{}^b = \eta^{ab}, \qquad \text{and} \qquad \eta_{ab} \, e_\mu{}^a e_\nu{}^b = g_{\mu\nu}

 

[WannabeNewton]

Ok using Carroll's book I was able to make sense of the apparent "conflicts" here. Let the Latin index $a$ denote the element of the orthonormal basis $(e_{a})$ for $T_p M$. This basis satisfies $g(e_a,e_b) = \eta_{ab}$ as noted by Carroll's text. Now, say we have a chart containing $p$ and have an associated basis $\partial_{\mu}$ where the index $\mu$ now denotes the element of the coordinate basis. Then, we can write the elements of the coordinate basis as linear combinations of elements of the orthonormal basis i.e. $\partial_{\mu} = e^{a}_{\mu}e_{a}$. Then, $g_{\mu\nu} = g(\partial_{\mu},\partial_{\nu}) = e^{a}_{\mu}e^{b}_{\nu}g(e_{a},e_{b}) = e^{a}_{\mu}e^{b}_{\nu}\eta_{ab}$ and, after inverting the linear relationship between the two bases, we have $\eta_{ab} = g(e_{a},e_{b}) = e^{\mu}_{a}e^{\nu}_{b}g(\partial_{\mu},\partial_{\nu}) = e^{\mu}_{a}e^{\nu}_{b}g_{\mu\nu}$. But even in light of this, I still don't know what Padmanabhan means by $\mathbf{e}_{a}\cdot \mathbf{e}_{b} = g_{ab}$ right after saying that $(e_a)$ is an orthonormal basis.

 

[dextercioby]

Ok using Carroll's book I was able to make sense of the apparent "conflicts" here. Let the Latin index $a$ denote the element of the orthonormal basis $(e_{a})$ for $T_p M$. This basis satisfies $g(e_a,e_b) = \eta_{ab}$ as noted by Carroll's text. Now, say we have a chart containing $p$ and have an associated basis $\partial_{\mu}$ where the index $\mu$ now denotes the element of the coordinate basis. Then, we can write the elements of the coordinate basis as linear combinations of elements of the orthonormal basis i.e. $\partial_{\mu} = e^{a}_{\mu}e_{a}$. Then, $g_{\mu\nu} = g(\partial_{\mu},\partial_{\nu}) = e^{a}_{\mu}e^{b}_{\nu}g(e_{a},e_{b}) = e^{a}_{\mu}e^{b}_{\nu}\eta_{ab}$ and, after inverting the linear relationship between the two bases, we have $\eta_{ab} = g(e_{a},e_{b}) = e^{\mu}_{a}e^{\nu}_{b}g(\partial_{\mu},\partial_{\nu}) = e^{\mu}_{a}e^{\nu}_{b}g_{\mu\nu}$.[...]

Textbook stuff. If you ever want to go from GR to SUGRA, this should be too familiar to you.

But even in light of this, I still don't know what Padmanabhan means by $\mathbf{e}_{a}\cdot \mathbf{e}_{b} = g_{ab}$ after noting that $(e_a)$ is an orthonormal basis.

I would argue that there's only one metric, hence one <dot product> (a term highly unreccomendable), so that if the $e_a$ form a Lorentzian base in the tangent space of the point p, then $e^a$ would be a Lorentzian base in the cotangent space of the same point. Relating the tangent space and the cotangent space is done through g only, hence

$g(e_a , e_b) = \eta_{ab}$

 

[WannabeNewton]

So how am I supposed to make sense of $\mathbf{e}_a\cdot \mathbf{e}_b = g_{ab}$? I have no idea what the dot notation means in this context. $g(e_a,e_b) = \eta_{ab}$ I'm perfectly fine with however. Thanks dexter (p.s. what's SUGRA?).

 

[Ben Niehoff]

I don't have your source in front of me, so I can't guess what it means. I would avoid using "dot" notation for inner products in this context. As you can see, it can be confusing.

 

[WannabeNewton]

I don't have your source in front of me, so I can't guess what it means. I would avoid using "dot" notation for inner products in this context. As you can see, it can be confusing.

Here's the relevant passage: http://s12.postimg.org/84fkdrvhp/padma_tetrad_stuffs.png

 

[Fredrik]

No, not really.

$\{e_{\mu}\}$ is orthonormal wrt the flat space time metric $\eta_{ab}$ iff

$e_{\mu a} e^{\mu}_{~b} = \eta_{ab}$, not really what you wrote and is exactly what Bill wrote.

I don't understand how you got that left-hand side, but all I'm saying is that $\{e_\mu\}$ is orthonormal with respect to g (not $\eta$) if and only if $g(e_\mu,e_\nu)=\eta_{\mu\nu}$. I don't see how you guys can say that this is wrong.

It seems strange and unnecessary to involve the dual basis in the statement of the orthogonality condition.Edit: I see a way to get that left-hand side. The definition of the dual basis implies that
$$(e_\nu)^\mu =(e^\mu)_\nu =\delta^\mu_\nu.$$ This implies that
$$(e_\mu)_\nu=g_{\mu\nu} =g_{\nu\mu} =(e_\nu)_\mu.$$ So
$$g(e_\mu,e_\nu)= (e_\mu)^\rho (e_\nu)_\rho = (e^\rho)_\mu (e_\rho)_\nu.$$ So your left-hand side is equal to mine, and this means that you and Bill_K were wrong to say that I was wrong.

I don't get why anyone would write the orthonormality condition the way you guys did. You have transformed it so that it looks very different from orthonormality.

 

[dextercioby]

So how am I supposed to make sense of $\mathbf{e}_a\cdot \mathbf{e}_b = g_{ab}$? I have no idea what the dot notation means in this context. $g(e_a,e_b) = \eta_{ab}$ I'm perfectly fine with however. Thanks dexter (p.s. what's SUGRA?).

Get over the dot. It's shorthand from g(,). And g has one type of indices, the (traditionally small case Greek ) ones coming from using coordinate (holonomic) basis in each tangent space of each spacetime point. So no wonder the notation is confusing in the RHS as well, since the letter g shouldn't be there.

 

[WannabeNewton]

Get over the dot. It's shorthand from g(,). And g has one type of indices, the (traditionally small case Greek ) ones coming from using cordinate (holonomic) basis in each tangent space of each spacetime point. So no wonder the notation is confusing in the RHS as well, since the letter g shouldn't be there.

Thanks, so by his notation it should really be $e_a \cdot e_b = \eta_{ab}$? Not sure why he choose to use the dot, haven't seen that notation in a GR book before.

 

[dextercioby]

I don't understand how you got that left-hand side, but all I'm saying is that $\{e_\mu\}$ is orthonormal with respect to g (not $\eta$) if and only if $g(e_\mu,e_\nu)=\eta_{\mu\nu}$. I don't see how you guys can say that this. [...]

It seems we're speaking of different things. We (i.e. both Bill and I) were arguing about the orthonormality of the 4 tetrads at a spacetime point and what does that mean for the transition coefficients, while your last statement doesn't seem to address tetrads at all.

 

[dextercioby]

Thanks, so by his notation it should really be $e_a \cdot e_b = \eta_{ab}$? Not sure why he choose to use the dot, haven't seen that notation in a GR book before.

No, from a geometric perspective it's not necessarily ita, i.e. special relativity metric. It can generally be any 4x4 matrix, but if I'm not mistaking the physics (principle of equivalence) forces the matrix to be really ita.

If you like GR with a strong flavor of maths, try Hawking and Ellis.

 

[WannabeNewton]

No, from a geometric perspective it's not necessarily ita, i.e. special relativity metric. It can generally be any 4x4 matrix, but if I'm not mistaking the physics (principle of equivalence) forces the matrix to be really ita.

I mean if the $(e_a)$ are an orthonormal basis for $T_p M$ with respect to $g_p ( , )$ then shouldn't $e_0 \cdot e_0 = -1, e_1 \cdot e_1 = 1$ etc. if the dot is to be interpreted as $g_p ( , )$?

 

[dextercioby]

Yes. That's what Bill and I were arguing all along.

 

[Fredrik]

It seems we're speaking of different things. We (i.e. both Bill and I) were arguing about the orthonormality of the 4 tetrads at a spacetime point and what does that mean for the transition coefficients, while your last statement doesn't seem to address tetrads at all.

I'm not very familiar with the "tetrad" terminology, but posts #1 (WannabeNewton) and #11 (Ben Niehoff) define a tetrad as an orthonormal frame field, i.e. a collection of four vector fields that form an orthonormal basis for the tangent space at each point. So I wrote down the condition for orthnormality in a straightforward way $g(e_\mu,e_\nu)=\eta_{\mu\nu}$. OK, I wrote the left-hand side as $(e_\mu)^a (e_\nu)_a$, but I had just proved that this is equal to $g(e_\mu,e_\nu)$. So it seems to me that this is precisely what needed to be said to address tetrads.

 

[Fredrik]

I mean if the $(e_a)$ are an orthonormal basis for $T_p M$ with respect to $g_p ( , )$ then shouldn't $e_0 \cdot e_0 = -1, e_1 \cdot e_1 = 1$ etc. if the dot is to be interpreted as $g_p ( , )$?

Yes. That's what Bill and I were arguing all along.

Me too. I didn't comment specifically on the dot notation, but I've been saying that $g(e_\mu,e_\nu)=\eta_{\mu\nu}$ holds if and only if $\{e_\mu\}$ is an orthonormal set.

 

[WannabeNewton]

Me too. I didn't comment specifically on the dot notation, but I've been saying that $g(e_\mu,e_\nu)=\eta_{\mu\nu}$ holds if and only if $\{e_\mu\}$ is an orthonormal set.

Alrighty then, screw the dot notation the author uses lol. It's about as confusing as Wald's abstract index notation in the context of frame fields. So $g(e_a,e_b) = \eta_{ab}$ at each event, for a tetrad. Is that fine?