Causal structure of metric

• A
Proposition: Consider an ##n + 1##-dimensional metric with the following product structure:

$$g=\underbrace{g_{rr}(t,r)\mathrm{d}r^2+2g_{rt}(t,r)\mathrm{d}t\mathrm{d}r+g_{tt}(t,r)\mathrm{d}t^2}_{:=^2g}+\underbrace{h_{AB}(t,r,x^A)\mathrm{d}x^A\mathrm{d}x^B}_{:=h}$$

where ##h## is a Riemannian metric in dimension ##n-1##. Then any causal vector for ##g## is also a causal vector for ##^2g##, and drawing light-cones for ##^2g## gives a good idea of the causal structure of ##(\mathcal{M},g)##.

I really don't understand this proposition. The metric tensor ##^2g## can be represented in a ##2\times 2## matrix, and ##g## corresponds to a ##(n+1) \times (n+1)## matrix. How can I then check ##\bigg(g(x^\mu, x^\nu)<0\bigg) \implies \bigg(\; ^2g(x^\mu, x^\nu)<0\bigg) ## for the same vector ##x=x^\mu \partial_\mu## if the dimension of the corresponding matrices is different?

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Orodruin
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Their dimensions are not different. ##^2g## and ##h## are metrics only on submanifolds of your space-time and it is here more relevant to consider them simply as tensors in ##n+1## dimensions. Obviously, neither ##^2g## nor ##h## are metric tensors on their own in the full space-time, but their sum ##g = {}^2g + h## is. However, ##h## is still positive semi-definite and so ##h(X,X) \geq 0## for any vector ##X## (you really should not use coordinates ##x^\mu## as arguments for your metric, the coordinates are not vectors). It follows that
$$g(X,X) = {}^2g(X,X) + h(X,X) \geq {}^2g(X,X).$$
Thus, if ##g(X,X) \leq 0##, then
$$^2g(X,X) \leq g(X,X) \leq 0.$$

Thanks!

Yeah, I forgot the basis vectors in the metric, that was a typo.

Orodruin
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Yeah, I forgot the basis vectors in the metric, that was a typo.
This was not my main issue. My main issue was that if ##x^\mu## are the coordinates, which is rather conventional, then ##x^\mu\partial_\mu## is not a vector field. You should therefore typically use some other notation for a general vector.

This was not my main issue. My main issue was that if ##x^\mu## are the coordinates, which is rather conventional, then ##x^\mu\partial_\mu## is not a vector field. You should therefore typically use some other notation for a general vector.
Why is ##x^\mu \partial_\mu## not a vector field?

Orodruin
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Why is ##x^\mu \partial_\mu## not a vector field?
Because it does not transform correctly under coordinate transformations.

If you just look at a particular coordinate system, it might define a vector field (at least on that coordinate patch), but that vector field will generally not take the form ##x'^{\mu'}\partial'_{\mu'}## in a different coordinate system.

Edit: Furthermore, it is not an arbitrary vector field, which is really what you want in this case.

Edit 2: To illustrate. Consider the position vector in 2D Euclidean space. It can be written ##\vec x = x^i \vec e_i## in Cartesian coordinates. However, it is not equal to ##r\vec e_r + \theta \vec e_\theta## in polar coordinates.

$$x'=x'^\nu \frac{\partial}{\partial x'^\nu}$$

Then
$$\frac{\partial}{\partial x'^\nu}=\frac{\partial x^\mu}{\partial x'^\nu}\frac{\partial}{\partial x^\mu}\\ x'^\alpha=\frac{\partial x'^\alpha}{\partial x^\beta}x^\beta$$
Then
$$x'=x'^\nu \frac{\partial}{\partial x'^\nu}=\frac{\partial x'^\alpha}{\partial x^\beta}x^\beta \frac{\partial x^\mu}{\partial x'^\alpha}\frac{\partial}{\partial x^\mu}=\delta^\mu_{\:\:\beta} x^\beta \frac{\partial}{\partial x^\mu}=x^\mu \frac{\partial}{\partial x^\mu}$$

What am I missing here?

Orodruin
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$$x'^\alpha=\frac{\partial x'^\alpha}{\partial x^\beta}x^\beta$$
...
What am I missing here?
This is not correct unless ##x'## is a linear function of ##x##.

strangerep and dextercioby
This is not correct unless ##x'## is a linear function of ##x##.
Are you sure? I see no assumption that the transformation need be linear in the derivation of this transformation.

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Orodruin
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Are you sure?
I only wrote a textbook with a chapter on the subject. I am pretty sure.

I see no assumption that the transformation need be linear in the derivation of this transformation.

View attachment 230082
What you attached is the transformation rule for vector components. The coordinates are not components of a vector.

Edit: As a counter example, consider the one-dimensional manifold ##\mathbb R## with coordinates ##x## and ##y = x^3##. Then
$$y = x^3 \neq \frac{dy}{dx} x = 3x^3.$$

strangerep
As a counter example, consider the one-dimensional manifold ##\mathbb R## with coordinates ##x## and ##y = x^3##. Then
$$y = x^3 \neq \frac{dy}{dx} x = 3x^3.$$
Those are just the coefficients, though, aren't they? ##\big( x^i \partial_i = x'^j \partial'_j \big)## does not imply ##x^i=x'^i##.

Orodruin
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Those are just the coefficients, though, aren't they? ##\big( x^i \partial_i = x'^j \partial'_j \big)## does not imply ##x^i=x'^i##.
I literally took your transformation law and showed you that, in the simplest case, it would mean that 3=1. You don't have a problem with that?

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Orodruin
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Also consider the case of polar coordinates in two-dimensional Euclidean space. It is not true that ##x\partial_x + y\partial_y = r \partial_r + \theta \partial_\theta##, which would be the case if the coordinates were actually the components of a vector. For any coordinate system, there exists a vector field that has this property, but those vector fields will generally not be the same vector fields. Indeed, it is true that ##x\partial_x + y\partial_y = r \partial_r##. This has the form ##x^i\partial_i## in Cartesian coordinates, but certainly not in polar coordinates.

strangerep and vanhees71