Local Lorentz Frames: Definition & Context for GR

[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?