Defining the components of a metric

  • #1
184
42

Main Question or Discussion Point

This has been bugging me for a while, and I feel like I’m missing or misunderstanding some crucial piece of information, so please advise me: the scalar product of two vectors (say ##\mathbf v## and ##\mathbf w##) is given using the metric: ##g_{\alpha \beta} \mathrm v^{\alpha} \mathrm w^{\beta}##. So how can we define the components of the metric, ##g_{\alpha \beta}##, as the scalar product of the basis vectors? Wouldn’t calculating the scalar product of the basis vectors require knowledge of the components of the metric to begin with? Or is "calculating" the scalar product of the basis vectors just nonsense- i.e. you don't calculate them, instead, you choose them, and in so doing, simultaneously choose the components of the metric?
 

Answers and Replies

  • #2
PeterDonis
Mentor
Insights Author
2019 Award
30,073
9,256
the scalar product of two vectors (say ##\mathbf v## and ##\mathbf w##) is given using the metric: ##g_{\alpha \beta} \mathrm v^{\alpha} \mathrm w^{\beta}##.
That is a formula you can use if you know the components of the metric and the vectors in a particular coordinate chart, yes.

So how can we define the components of the metric, ##g_{\alpha \beta}##, as the scalar product of the basis vectors?
Because the scalar product is not defined in terms of the components of the metric. The components of the metric (and of the vectors themselves) require you to have chosen a coordinate chart; but the scalar product of two vectors is an invariant, independent of any choice of coordinates.

Wouldn’t calculating the scalar product of the basis vectors require knowledge of the components of the metric to begin with?
No. You pick the scalar products of the basis vectors when you choose coordinates, which, as above, you have to do before you can even know the components of the metric (or of vectors).

More generally, you can have information that tells you the scalar product of two vectors without having to choose coordinates at all. For example, if you know the Doppler shift of light emitted by one observer as seen by a second observer, then you know the scalar product of their 4-velocity vectors, without ever having to choose coordinates.

Or is "calculating" the scalar product of the basis vectors just nonsense- i.e. you don't calculate them, instead, you choose them, and in so doing, simultaneously choose the components of the metric?
Yes. See above.
 
  • Like
Likes sweet springs and Pencilvester
  • #3
andrewkirk
Science Advisor
Homework Helper
Insights Author
Gold Member
3,836
1,418
The response is easier to see writing in coordinate-free form. Recall that the metric is a 2-tensor, which is a function that takes two vectors as inputs and returns a scalar as output. Then the inner product of ##\mathbf v## with ##\mathbf w## is just ##\mathbf g(\mathbf v,\mathbf w)##, the result of applying tensor ##\mathbf g## to ##\mathbf v## and ##\mathbf w##.

The ##\alpha,\beta## component of ##\mathbf g## in an orthonormal basis ##(\mathbf e_1,...,\mathbf e_n)## is the result of applying tensor ##\mathbf g## to ##\mathbf e_\alpha## and ##\mathbf e_\beta##, which is:
\begin{align*}
g_{\alpha\beta}&=\mathbf g(\mathbf e_\alpha,\mathbf e_\beta)\\
&=g_{\mu\nu}(\mathbf e_\alpha)^\mu(\mathbf e_\beta)^\nu\\
&=g_{\mu\nu}\,\delta_\alpha^\mu \,\delta_\beta^\nu \quad \textrm{where $\delta$ is the Kronecker Delta}\\
&=g_{\alpha\beta}
\end{align*}

Note how all items in these formulas are coordinate/component-free until the third-last line.
 
  • #4
184
42
The ##\alpha,\beta## component of ##\mathbf g## in an orthonormal basis ##(\mathbf e_1,...,\mathbf e_n)## is the result of applying tensor ##\mathbf g## to ##\mathbf e_\alpha## and ##\mathbf e_\beta##
Won't the components of ##\mathbf g## in an orthonormal basis always be the same as those of ##\boldsymbol {\delta}## (or ##\boldsymbol {\eta}## in the case of relativistic spacetime)?
The ##\alpha,\beta## component of ##\mathbf g## in an orthonormal basis ##(\mathbf e_1,...,\mathbf e_n)## is the result of applying tensor ##\mathbf g## to ##\mathbf e_\alpha## and ##\mathbf e_\beta##, which is:
\begin{align*}
g_{\alpha\beta}&=\mathbf g(\mathbf e_\alpha,\mathbf e_\beta)\\
&=g_{\mu\nu}(\mathbf e_\alpha)^\mu(\mathbf e_\beta)^\nu\\
&=g_{\mu\nu}\,\delta_\alpha^\mu \,\delta_\beta^\nu \quad \textrm{where $\delta$ is the Kronecker Delta}\\
&=g_{\alpha\beta}
\end{align*}
Also, the step from line 2 to line 3 here won't applicable in general, right? (e.g. when the basis vectors don't form an orthonormal basis?)
Yes. See above.
Thanks!
 
  • #5
PeterDonis
Mentor
Insights Author
2019 Award
30,073
9,256
The ##\alpha,\beta## component of ##\mathbf g## in an orthonormal basis
I don't think the basis needs to be orthonormal; any coordinate basis will do. The key fact is that ##(\mathbf{e}_\alpha)^\mu = \delta^\mu_\alpha##, which is what defines a coordinate basis.
 
  • #6
PeterDonis
Mentor
Insights Author
2019 Award
30,073
9,256
Won't the components of ##\mathbf g## in an orthonormal basis always be the same as those of ##\boldsymbol {\delta}## (or ##\boldsymbol {\eta}## in the case of relativistic spacetime)?
No. You are confusing an orthonormal basis with a (local) inertial frame. For example: in Schwarzschild spacetime, the following set of four vectors is an orthormal basis, expressed in Schwarzschild coordinates:

$$
\hat{\mathbf{e}}_0 = \left( \frac{1}{\sqrt{1 - \frac{2M}{r}}}, 0, 0, 0 \right)
$$

$$
\hat{\mathbf{e}}_1 = \left( 0, \sqrt{1 - \frac{2M}{r}}, 0, 0 \right)
$$

$$
\hat{\mathbf{e}}_2 = \left( 0, 0, \frac{1}{r}, 0 \right)
$$

$$
\hat{\mathbf{e}}_3 = \left( 0, 0, 0, \frac{1}{r \sin \theta} \right)
$$

Note, also, that this basis is not a coordinate basis; that would be

$$
\mathbf{e}_0 = \left( 1, 0, 0, 0 \right)
$$

$$
\mathbf{e}_1 = \left( 0, 1, 0, 0 \right)
$$

$$
\mathbf{e}_2 = \left( 0, 0, 1, 0 \right)
$$

$$
\mathbf{e}_3 = \left( 0, 0, 0, 1 \right)
$$
And the latter is the basis we would use if we wanted to apply @andrewkirk 's formula.
 
  • #7
andrewkirk
Science Advisor
Homework Helper
Insights Author
Gold Member
3,836
1,418
I don't think the basis needs to be orthonormal; any coordinate basis will do. The key fact is that ##(\mathbf{e}_\alpha)^\mu = \delta^\mu_\alpha##, which is what defines a coordinate basis.
Oh yes. I see I was being excessively cautious. That also answers the OP's second concern expressed in post 4.
 
  • #8
184
42
No. You are confusing an orthonormal basis with a (local) inertial frame. For example: in Schwarzschild spacetime, the following set of four vectors is an orthormal basis, expressed in Schwarzschild coordinates:

$$
\hat{\mathbf{e}}_0 = \left( \frac{1}{\sqrt{1 - \frac{2M}{r}}}, 0, 0, 0 \right)
$$

$$
\hat{\mathbf{e}}_1 = \left( 0, \sqrt{1 - \frac{2M}{r}}, 0, 0 \right)
$$

$$
\hat{\mathbf{e}}_2 = \left( 0, 0, \frac{1}{r}, 0 \right)
$$

$$
\hat{\mathbf{e}}_3 = \left( 0, 0, 0, \frac{1}{r \sin \theta} \right)
$$
Okay, I think we're getting to the root of some of my misunderstandings. I thought if we applied the Schwarzchild metric to get the scalar product of basis vectors ##\hat{\mathbf e}_{\alpha}## and ##\hat{\mathbf e}_{\beta}##, we would end up with ##\eta_{\alpha \beta}##. And we do, right? But you're saying that the scalar product of basis vectors yielding components of the metric only applies to coordinate basis vectors, where, like you said, ##(\mathbf e_{\alpha})^{\mu} = \delta^{\mu}_{\alpha}##?
 
  • #9
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,817
6,626
For example, if you know the Doppler shift of light emitted by one observer as seen by a second observer, then you know the scalar product of their 4-velocity vectors, without ever having to choose coordinates.
As an aside, this is not true unless you also know some additional information, such as their separation being colinear with the relative velocity. All you know is the scalar product of their 4-velocities with the 4-frequency of the wave.
 
  • #10
PeterDonis
Mentor
Insights Author
2019 Award
30,073
9,256
As an aside, this is not true unless you also know some additional information, such as their separation being colinear with the relative velocity.
Yes, you're right. "Separation colinear with relative velocity" is also a coordinate-independent fact, so with your correction the example still works. :wink:
 
  • #11
PeterDonis
Mentor
Insights Author
2019 Award
30,073
9,256
I thought if we applied the Schwarzchild metric to get the scalar product of basis vectors ##\hat{\mathbf e}_{\alpha}## and ##\hat{\mathbf e}_{\beta}##, we would end up with ##\eta_{\alpha \beta}##. And we do, right?
Have you done the calculation? If not, I suggest that you do it and see for yourself.

But you're saying that the scalar product of basis vectors yielding components of the metric only applies to coordinate basis vectors, where, like you said, ##(\mathbf e_{\alpha})^{\mu} = \delta^{\mu}_{\alpha}##?
Yes. Again, if you haven't explicitly done the calculation, do it. And compare it with the calculation using the "hatted" vectors. It will be instructive.
 
  • #12
PeterDonis
Mentor
Insights Author
2019 Award
30,073
9,256
Won't the components of ##\mathbf g## in an orthonormal basis always be the same as those of ##\boldsymbol {\delta}## (or ##\boldsymbol {\eta}## in the case of relativistic spacetime)?
As another response to this, note that "the components of ##\mathbf g## in a basis" does not make sense. Components are components in a coordinate chart. That's why it's the scalar product of coordinate basis vectors that gives the components of the metric tensor in the coordinate chart.

However, there is another concept that's often used in these computations that you are sort of getting at here. It's called a "vierbein" (in the context of 4-d spacetime--or a "vielbein" in a more general context). The term "tetrad" is also sometimes used. See here for a quick overview:

https://en.wikipedia.org/wiki/Cartan_formalism_(physics)#Example:_general_relativity

The terminology here is a bit abstract, but if you consider the Schwarzschild spacetime example, "tangent space coordinates" are the usual Schwarzschild coordinates (the ones I wrote the vectors in) and "Minkowski coordinates" are the coordinates of a local inertial frame centered on a chosen event. The "hatted" vectors I wrote down are the vierbein vectors.
 
  • Like
Likes Pencilvester
  • #13
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,817
6,626
As another response to this, note that "the components of g in a basis" does not make sense. Components are components in a coordinate chart. That's why it's the scalar product of coordinate basis vectors that gives the components of the metric tensor in the coordinate chart.
Sorry to just complain, but I do not agree with this. You can always write the components of any tensor in any basis, the basis does not need to be holonomic, just that it needs to span the tangent space at each point. As long as it does that I would certainly call the scalar fields ##V^\alpha## in ##V = V^\alpha E_\alpha## the components of ##V## in the basis ##E_\alpha##. The issue is just that what is normally meant when you say "components of metric tensor in so and so coordinates" is an implicit reference to the corresponding coordinate basis, but this does not stop you from choosing to write the components down in any basis of your choice.
 
  • #14
PeterDonis
Mentor
Insights Author
2019 Award
30,073
9,256
the basis does not need to be holonomic, just that it needs to span the tangent space at each point. As long as it does that I would certainly call the scalar fields ##V^\alpha## in ##V = V^\alpha E_\alpha## the components of ##V## in the basis ##E_\alpha##.
I can see that this usage is reasonable, yes, it's just not usage I'm familiar with.
 
  • #15
184
42
Have you done the calculation? If not, I suggest that you do it and see for yourself.
Sorry, I can't tell if you're saying I'm mistaken or if you're encouraging me to exercise my algebraic brain muscles. In either case, the non-zero component of ##\hat {\mathbf e}_{\alpha}## is ##(\hat {\mathbf e}_{\alpha})^{\alpha}## and is simply the square root of the multiplicative inverse of ##g_{\alpha \alpha}## (for Schwarzschild metric in Schwarzschild coordinates), and since said metric is diagonal, ##g_{\mu \nu} (\hat{\mathbf e}_{\alpha})^{\mu} (\hat{\mathbf e}_{\beta})^{\nu} = 0## whenever any index differs from any other, and ##g_{\alpha \alpha} (\hat{\mathbf e}_{\alpha})^{\alpha} (\hat{\mathbf e}_{\alpha})^{\alpha} = \pm 1## depending on ##\alpha## and the metric's signature. That description also fits ##\pm \eta_{\alpha \beta}##. And for the coordinate basis, the simplicity of the basis vectors means that the result ##g_{\mu \nu} (\mathbf e_{\alpha})^{\mu} (\mathbf e_{\beta})^{\nu} = g_{\alpha \beta}## is trivial. I think I understand now that my mistake was in thinking that the scalar products of the orthonormal basis vectors yield the components of the metric in that basis. Am I getting that right?
 
  • #16
184
42
I think I understand now that my mistake was in thinking that the scalar products of the orthonormal basis vectors yield the components of the metric in that basis. Am I getting that right?
As another response to this, note that "the components of ##\mathbf g## in a basis" does not make sense. Components are components in a coordinate chart. That's why it's the scalar product of coordinate basis vectors that gives the components of the metric tensor in the coordinate chart.
Okay, I see what you’re saying now. This makes much more sense. Thanks!
 
  • #17
PeterDonis
Mentor
Insights Author
2019 Award
30,073
9,256
I think I understand now that my mistake was in thinking that the scalar products of the orthonormal basis vectors yield the components of the metric in that basis.
Yes, I agree.
 
  • #18
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,401
2,580
This is a terminological confusion here. I would have said that the scalar product of two basis vectors ##e_a## and ##e_b## is always equal to the corresponding component of the metric tensor in that basis:

##e_a \cdot e_b = g_{ab}##

It doesn't matter whether the basis is a coordinate basis, or not.

So if I'm right about that, strictly speaking, the components of the metric tensor are not determined by the coordinate system, but by the basis, and the basis is not uniquely determined by the coordinate system.

To give a simple example that came up in another thread, look at polar coordinates for flat 2-D space. You have two coordinates ##r## and ##\theta## related to the usual Cartesian coordinates ##x## and ##y## through: ##x = r\ cos(\theta)##, ##y = r\ sin(\theta)##. You can define a coordinate basis ##e_r## and ##e_\theta##, leading to the nonzero metric components ##g_{rr} = 1##, ##g_{\theta \theta} = r^2##. You can also define an orthonormal basis ##\hat{r}##, ##\hat{\theta}## related to the first basis through ##\hat{r} = e_r##, ##\hat{\theta} = \frac{1}{r} e_\theta##. I would say that in this basis, ##g_{\hat{r} \hat{r}} = g_{\hat{\theta} \hat{\theta}} = 1## (or maybe the ##\hat{}## should be over ##g##, rather than ##r## and ##\theta##?) You haven't changed coordinate systems---you're still using ##\theta## and ##r##, but you've changed your basis, and so your metric has different components.
 
  • #19
184
42
This is a terminological confusion here. I would have said that the scalar product of two basis vectors ##e_a## and ##e_b## is always equal to the corresponding component of the metric tensor in that basis:

##e_a \cdot e_b = g_{ab}##

It doesn't matter whether the basis is a coordinate basis, or not... I would say that in this basis, ##g_{\hat{r} \hat{r}} = g_{\hat{\theta} \hat{\theta}} = 1## You haven't changed coordinate systems---you're still using ##\theta## and ##r##, but you've changed your basis, and so your metric has different components.
Am I to understand that this is more a matter of semantics, i.e. how you choose to define metric components? If we used this definition (which is the one I’ve been laboring under), wouldn’t my statement in post 4 be correct:
Won't the components of ##\mathbf g## in an orthonormal basis always be the same as those of ##\boldsymbol {\delta}## (or ##\boldsymbol {\eta}## in the case of relativistic spacetime)?
Wouldn’t this follow from the very definition of an orthonormal basis?
 
  • #20
PeterDonis
Mentor
Insights Author
2019 Award
30,073
9,256
You haven't changed coordinate systems---you're still using ##\theta## and ##r##, but you've changed your basis, and so your metric has different components.
If one adopts the usage that @Orodruin described, then yes, this would follow--in fact, with this usage, in any orthonormal basis, the metric components would, as @Pencilvester said, be the Kronecker delta (in a Riemannian manifold) or the Minkowski eta (in a Lorentzian manifold).

All this means is that you've put the information about how the basis vectors change as you move around in the manifold into the basis vector expressions in terms of the coordinates, instead of into the metric components in terms of the coordinates. This is basically the vierbein approach.
 
  • Like
Likes Pencilvester
  • #21
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,817
6,626
If one adopts the usage that @Orodruin described, then yes, this would follow--in fact, with this usage, in any orthonormal basis, the metric components would, as @Pencilvester said, be the Kronecker delta (in a Riemannian manifold) or the Minkowski eta (in a Lorentzian manifold).

All this means is that you've put the information about how the basis vectors change as you move around in the manifold into the basis vector expressions in terms of the coordinates, instead of into the metric components in terms of the coordinates. This is basically the vierbein approach.
I think it should also be mentioned that the vierbein a priori does not need to be tied to a coordinate system, however, given a coordinate system it can (of course) be related to the coordinate basis as this is just a change of basis in the tangent space. When you have an orthogonal coordinate system, such as spherical coordinates on ##\mathbb R^3##, it can however be convenient to define it such that the basis vectors are parallel to the coordinate basis vectors, but you could just as well use the Cartesian basis vectors and express them in terms of the spherical coordinates, e.g.,
$$
\newcommand{\dd}[2]{\frac{\partial #1}{\partial #2}}
E_x = \partial_x = \dd{r}{x} \partial_r + \dd{\theta}{x}\partial_\theta + \dd{\varphi}{x}\partial_\varphi
$$
etc.
 
  • #22
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,401
2,580
This has been bugging me for a while, and I feel like I’m missing or misunderstanding some crucial piece of information, so please advise me: the scalar product of two vectors (say ##\mathbf v## and ##\mathbf w##) is given using the metric: ##g_{\alpha \beta} \mathrm v^{\alpha} \mathrm w^{\beta}##. So how can we define the components of the metric, ##g_{\alpha \beta}##, as the scalar product of the basis vectors? Wouldn’t calculating the scalar product of the basis vectors require knowledge of the components of the metric to begin with? Or is "calculating" the scalar product of the basis vectors just nonsense- i.e. you don't calculate them, instead, you choose them, and in so doing, simultaneously choose the components of the metric?
In General Relativity, the metric is something you calculate, it's something you determine empirically.

The distinction you're making, between the metric and the scalar product, is non-existent. The metric simply is the operator that takes two vectors and returns a scalar. So writing ##g(U,V)## and writing ##U \cdot V## means the same thing.

Being a bilinear function (meaning that ##g(aX+Y, Z) = a g(X,Z) + g(Y,Z)## and ##g(U, bV + W) = b g(U,V) + g(U,W)##, where ##X,Y,Z,U,V,W## are all vectors, and ##a,b## are reals), it means that if you know any 4 independent vectors, ##V^1, V^2, V^3, V^4##, then you know everything there is to know about ##g## if you know the 10 values: ##g(V^i, V^j)## for all possible combinations for ##i## and ##j##. (There would be 16 combinations, but 6 of them are repetitions, since ##g(V^i, V^j) = g(V^j, V^i)##.
 
Last edited:
  • #23
1,225
75
Why do not you decide metric by observation of time and length in your coordinate?

As for two events between of which ##dx^1=dx^2=dx^2=0##, you measure time interval ##d\tau## by clock corresponding to ##dx^0## and get value of ##g_{00}## by the relation
[tex]d\tau=\frac{1}{c}\sqrt{g_{00}}dx^0[/tex]

Same shall be done for length measurement of the events simultaneous in ##dx^0=0## using light signals and clock.
Six times different measurement of length and the relation [tex]dl^2=(-g_{ab}+\frac{g_{0a}g_{0b}}{g_{00}})dx^a dx^b[/tex] give you value of other components of g.

You can choose coordinate as you like, then observed g pays you back.
 

Related Threads on Defining the components of a metric

  • Last Post
Replies
15
Views
985
  • Last Post
Replies
1
Views
959
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
922
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
1K
Replies
2
Views
594
Replies
9
Views
2K
Replies
28
Views
547
  • Last Post
Replies
5
Views
3K
Top