Non-Coordinate Bases Explained

[Silviu]

Hello! I am a bit confused about non-coordinate basis. I understand the way they are defined (I think) and the main purpose is to get on a manifold a coordinate system that is orthonormal at any point on the manifold (right?). So if you have a coordinate basis $e_\alpha$, you get to a non-coordinate by doing a transformation $\hat{e_\mu}=a_\mu^\alpha e_\alpha$, such that $g_{\mu\nu}e_\alpha^\mu e_\beta^\nu = \delta_{\alpha\beta}$ (or $=\eta_{\alpha \beta}$). In polar coordinates in $R^2$, the way it is done is to divide the angular basis by r i.e. $\hat{e}_r = \partial_r$ and $\hat{e}_\phi = \frac{1}{r}\partial_\phi$. I am a bit confused why this new basis is non-coordinate. If initially a point had coordinates $(2,1)$, not it has coordinates $(2,1/2)$. Why is this not a coordinate basis? You can still identify any point in space by providing 2 numbers, so it seems to be a good coordinate system. Can someone explain this to me? Thank you!

 

[Orodruin]

I am a bit confused why this new basis is non-coordinate. If initially a point had coordinates (2,1)(2,1)(2,1), not it has coordinates (2,1/2)(2,1/2)(2,1/2).

No, it doesn't. The coordinates are fixed, they are polar coordinates. The polar coordinates of a point do not depend on the tangent vector basis you are using at some arbitrary point. The coordinate basis for the tangent space is the set of partial derivatives with respect to the coordinates.

 

[Silviu]

No, it doesn't. The coordinates are fixed, they are polar coordinates. The polar coordinates of a point do not depend on the tangent vector basis you are using at some arbitrary point. The coordinate basis for the tangent space is the set of partial derivatives with respect to the coordinates.

I am confused. If initially a point had coordinates $(2,1)$, how you represent it in non-coordinate basis?

 

[Orodruin]

I am confused. If initially a point had coordinates $(2,1)$, how you represent it in non-coordinate basis?

You do not represent points in bases. Points are not vectors.

 

[Silviu]

You do not represent points in bases. Points are not vectors.

OK... can you explain to me the idea of non-coordinate basis. I mean say we have a vector (2,1) how do you represent it in non-coordinate basis?

 

[Orodruin]

I mean say we have a vector (2,1) how do you represent it in non-coordinate basis?

That is not a vector. It is at best two components of a vector. Given two different bases (coordinate bases or not), there is always a linear transformation that takes the components of the vector in one of the bases to those in the second. A basis is just a collection of linearly independent vectors at a point such that any vector can be written as a linear combination of them.

 

[Silviu]

That is not a vector. It is at best two components of a vector. Given two different bases (coordinate bases or not), there is always a linear transformation that takes the components of the vector in one of the bases to those in the second. A basis is just a collection of linearly independent vectors at a point such that any vector can be written as a linear combination of them.

Why is it not a vector? You need just r and $\theta$ to specify it (in the example I gave)? And still I don't see the difference between coordinate and non-coordinate basis...

 

[Orodruin]

You need just r and θθ\theta to specify it (in the example I gave)?

I am sorry, it is unclear what you mean. A point is not a vector, nor is it defined by one. Just referencing components is not sufficient, you need to also specify the point in the manifold as well as the basis that you use. You may be confused by the fact that points in Euclidean spaces often are referred to using a position vector. There is no such thing in a general manifold.

 

[Silviu]

I am sorry, it is unclear what you mean. A point is not a vector, nor is it defined by one. Just referencing components is not sufficient, you need to also specify the point in the manifold as well as the basis that you use. You may be confused by the fact that points in Euclidean spaces often are referred to using a position vector. There is no such thing in a general manifold.

Oh, I apologize if I was not clear enough. Let me try again. Let's say that the manifold is the unit sphere. As you said, you need to specify a point on the manifold. The way I thought you do it is by giving 2 numbers like (2,1) (the same way you do on the surface of Earth, longitude and latitude). You said that my representation of the point like that is not correct, because points are not represented in basis. I am not sure I understand this. Could you please explain it a bit more and tell me how do you specify a point on the manifold? Then, after you go to a point and take the tangent plane to that point, you have a vector space of dim 2 so to specify a vector you again need 2 coordinates, (2,1) for example, in a given basis. Again you said this is not a vector, so could you please explain to me how do you represent a vector? Lastly, if you represent a vector in a coordinate basis, how do you transform it to a non-coordinate basis and what is the difference between the 2? Sorry for the long post but I got confused. Thank you!

 

[Orodruin]

The way I thought you do it is by giving 2 numbers like (2,1) (the same way you do on the surface of Earth, longitude and latitude). You said that my representation of the point like that is not correct, because points are not represented in basis. I am not sure I understand this. Could you please explain it a bit more and tell me how do you specify a point on the manifold?

You do, but those two numbers have no relation whatsoever to any vector. They will not change if you change the basis for the tangent space at that point.

you have a vector space of dim 2 so to specify a vector you again need 2 coordinates, (2,1) for example, in a given basis.

Those are vector components, not coordinates of the manifold. A tangent vector is a linear combination of basis vectors in the appropriate tangent space. The coordinate basis is a particular way of assigning such a set of basis vectors based on the coordinates of the manifold. A non-coordinate basis is just a different basis that is a different set of linearly independent basis vectors not directly given by the coordinates.

 

[Silviu]

You do, but those two numbers have no relation whatsoever to any vector. They will not change if you change the basis for the tangent space at that point.Those are vector components, not coordinates of the manifold. A tangent vector is a linear combination of basis vectors in the appropriate tangent space. The coordinate basis is a particular way of assigning such a set of basis vectors based on the coordinates of the manifold. A non-coordinate basis is just a different basis that is a different set of linearly independent basis vectors not directly given by the coordinates.

Ok, I am starting to get it I think. Now, isn't the tangent space $R^n$, for any point on the manifold? So, in our case, for a given point p on the manifold, you have $R^2$ tangent to it. And it is in this $R^2$ where you do the coordinate transformations. And for any p, you can make a different transformation. Is this correct?

 

[PeroK]

Hello! I am a bit confused about non-coordinate basis. I understand the way they are defined (I think) and the main purpose is to get on a manifold a coordinate system that is orthonormal at any point on the manifold (right?). So if you have a coordinate basis $e_\alpha$, you get to a non-coordinate by doing a transformation $\hat{e_\mu}=a_\mu^\alpha e_\alpha$, such that $g_{\mu\nu}e_\alpha^\mu e_\beta^\nu = \delta_{\alpha\beta}$ (or $=\eta_{\alpha \beta}$). In polar coordinates in $R^2$, the way it is done is to divide the angular basis by r i.e. $\hat{e}_r = \partial_r$ and $\hat{e}_\phi = \frac{1}{r}\partial_\phi$. I am a bit confused why this new basis is non-coordinate. If initially a point had coordinates $(2,1)$, not it has coordinates $(2,1/2)$. Why is this not a coordinate basis? You can still identify any point in space by providing 2 numbers, so it seems to be a good coordinate system. Can someone explain this to me? Thank you!

I'm not sure using polar coordinates in a plane is such a helpful example as you still have the concept of position vector.

In general relativity vectors are only a local concept. At each point you can define vectors and use vector operations such as addition or the scalar product.

At each point, therefore, you have a tangent space of vectors in which you work.

One way to define a basis for this tangent space is to use the coordinates to generate the basis. in general this basis is neither orthogonal nor of unit length. In your example, the coordinate basis vectors are orthogonal, but the angular basis vectors must be rescaled.

If you do this, then the basis is no longer exactly the coordinate basis, although it is not too different.

If, therefore, you rescale your global coordinates you get a new co ordinate system whose coordinate basis at that point coincides with the orthonormal basis at that point.

I think your idea is to move these basis vectors to the origin and use them as a pair of cartesian basis vectors for the whole plane. But that is not the idea at all.

 

[Silviu]

I'm not sure using polar coordinates in a plane is such a helpful example as you still have the concept of position vector.

In general relativity vectors are only a local concept. At each point you can define vectors and use vector operations such as addition or the scalar product.

At each point, therefore, you have a tangent space of vectors in which you work.

One way to define a basis for this tangent space is to use the coordinates to generate the basis. in general this basis is neither orthogonal nor of unit length. In your example, the coordinate basis vectors are orthogonal, but the angular basis vectors must be rescaled.

If you do this, then the basis is no longer exactly the coordinate basis, although it is not too different.

If, therefore, you rescale your global coordinates you get a new co ordinate system whose coordinate basis at that point coincides with the orthonormal basis at that point.

I think your idea is to move these basis vectors to the origin and use them as a pair of cartesian basis vectors for the whole plane. But that is not the idea at all.

So what is the point then? Why is it called non-coordinate basis. How is it different than the original one? Like what is the point of all this and how do you differentiate between coordinate and non-coordinate bases?

 

[Ibix]

In Euclidean space with Cartesian coordinates you can get away with being sloppy about the distinction between points and vectors, regarding a point as the end of a vector from the origin. That doesn't work well with curved spaces, since there often isn't a uniquely picked-out line from A to B. It has problems even with curved coordinates on flat space - what is 1m in the tangential direction at radial coordinate 1m?

The way things work in differential geometry is that vectors aren't things in space (or spacetime). They exist in a separate flat space called the tangent space, one of which is associated with every point. For example, the electric field at a given location is a vector, but it isn't a vector "from" that location "to" any other, it's just a direction and a magnitude. And that direction and magnitude are represented in the tangent space associated with their location.

But there isn't naturally an association between the tangent space and spacetime. They're separate things. So when I say "the vector (in the tangent space) points in such-and-such a direction", I need some way to associate what I mean by "direction in the tangent space" to "direction in spacetime". The way you do that is be specifying basis vectors in the tangent space that have some relation to something you can reference in the real world.

A standard way to do that is to use basis vectors that point along the directions in which only one coordinate is changing. Latitude and longitude on the surface of a sphere, for example. You can show that the differential operator $\partial_\mu$ points along the direction in the tangent space that corresponds to changing $x^\mu$ in spacetime, so the differentials make a good basis for the tangent space. But they aren't the only possible basis. Picking a coordinate basis means that the basis vectors in tangent spaces are related to how the coordinates change around the location - which may be different at different places. You may wish to adjust your basis vectors so that they are orthonormal at each location, as you say. Or something else if you're feeling masochistic, I guess.

Take the surface of the Earth as an example, and pick latitude and longitude as your coordinates. You can specify your location using the coordinates. And if you've got a scalar field on the surface of the Earth (e.g. temperature), that's all you need. The temperature at my location is 12C, full stop. But if you've got a vector field, such as the wind speed, you need some notion of "velocity". Wind speed, like the electric field I mentioned earlier, doesn't point from your location to another location. It's just a direction and a magnitude. What I need to do is specify direction somehow. One way to do it would be to express the wind speed in terms of arcseconds per second eastwards and arcseconds per second northwards. Or I could use mph eastwards and mph northwards. The former would be using a coordinate basis; the latter a non-coordinate basis. In either case, I could draw a little arrow at your location to represent your windspeed. Note that a wind velocity of (1,1) in the coordinate basis would always give the same arrow at every location on a Mercator projection map, but would mean radically different things in terms of blowing you off your feet. A windspeed of (1,1) in the non-coordinate basis would be the same zephyr everywhere, but the arrows drawn on my Mercator projection would be of very different sizes.

Hope that helps.

 

[PeroK]

So what is the point then? Why is it called non-coordinate basis. How is it different than the original one? Like what is the point of all this and how do you differentiate between coordinate and non-coordinate bases?

The point physically is that a local orthonormal basis is what a local observer would use for their measurements. Whereas the coordinates are used to study tge geometry globally.

Mathematically the point is simply to define an orthornormal basis as it us more convenient, especially if you are using the scalar product.

 

[vanhees71]

The problem is that physicists often call the components of vectors simply vectors. A vector (or a more general tensor), however, is an object invariant object, independent of the choice of basis.The components, of course depend on the choice of basis.

Take the example of polar coordinates in the Euclidean plane from above. The "coordinate basis" (holonomous basis) $\vec{b}_j$ is given in terms of a fixed Cartesian basis $\vec{c}_j$ as
$$\vec{b}_1=\vec{b}_r=\vec{c}_1 \cos \varphi + \vec{c}_2 \sin \varphi, \quad \vec{b}_2=\vec{b}_{\varphi}=-\vec{c}_1 r \sin \varphi + \vec{c}_2 r \cos \varphi.$$
Obviously $\vec{b}_1 \cdot \vec{b}_2=0$, and thus one usually uses orthonormalized basis vectors
$$\vec{e}_1=\frac{1}{|\vec{b}_1|}\vec{b}_1=\vec{b}_1, \quad \vec{e}_2=\frac{1}{|\vec{b}_2|}\vec{b}_2=\frac{1}{r} \vec{b}_2.$$
Now you can easily express one and the same vector in all three bases involved here. It stays always the same vector. E.g., take the
$$\vec{v}=\vec{c}_1+\vec{c}_2.$$
Since $\varphi=\pi/4$ in this case and $\cos(\pi/4)=\sin(\pi/4)=\sqrt{2}/2$ you simply have
$$\vec{v}=\sqrt{2} \vec{b}_1=\sqrt{2} \vec{e}_1.$$
It's still the same vector, but its components are obviously $(1,1)$ wrt. the Cartesian basis $\vec{c}_j$ and $(1,0)$ with respect to the bases $\vec{b}_j$ and $\vec{e}_j$.

 

[Silviu]

The problem is that physicists often call the components of vectors simply vectors. A vector (or a more general tensor), however, is an object invariant object, independent of the choice of basis.The components, of course depend on the choice of basis.

Take the example of polar coordinates in the Euclidean plane from above. The "coordinate basis" (holonomous basis) $\vec{b}_j$ is given in terms of a fixed Cartesian basis $\vec{c}_j$ as
$$\vec{b}_1=\vec{b}_r=\vec{c}_1 \cos \varphi + \vec{c}_2 \sin \varphi, \quad \vec{b}_2=\vec{b}_{\varphi}=-\vec{c}_1 r \sin \varphi + \vec{c}_2 r \cos \varphi.$$
Obviously $\vec{b}_1 \cdot \vec{b}_2=0$, and thus one usually uses orthonormalized basis vectors
$$\vec{e}_1=\frac{1}{|\vec{b}_1|}\vec{b}_1=\vec{b}_1, \quad \vec{e}_2=\frac{1}{|\vec{b}_2|}\vec{b}_2=\frac{1}{r} \vec{b}_2.$$
Now you can easily express one and the same vector in all three bases involved here. It stays always the same vector. E.g., take the
$$\vec{v}=\vec{c}_1+\vec{c}_2.$$
Since $\varphi=\pi/4$ in this case and $\cos(\pi/4)=\sin(\pi/4)=\sqrt{2}/2$ you simply have
$$\vec{v}=\sqrt{2} \vec{b}_1=\sqrt{2} \vec{e}_1.$$
It's still the same vector, but its components are obviously $(1,1)$ wrt. the Cartesian basis $\vec{c}_j$ and $(1,0)$ with respect to the bases $\vec{b}_j$ and $\vec{e}_j$.

But as far as I understand, $(b_1,b_2)$ is coordinate (holonomous) basis while $(e_1,e_2)$ is non-coordinate (non-holonomous). I am confused about nomenclature. The second one is just a normalization of the first one. Why do they have different names? Thank you!

 

[vanhees71]

Yes, $\vec{b}_j$ is the holonomous basis, while $\vec{e}_j$ are not holonomous, although in this case it's just normalizing the non-holonomous basis vectors. More generally you can introduce arbitrary bases using the holonomous basis by using any invertible transformation matrix via (Einstein summation convention implied)
$$\vec{g}_j=\vec{b}_i {T^i}_j.$$

 

[Silviu]

Yes, $\vec{b}_j$ is the holonomous basis, while $\vec{e}_j$ are not holonomous, although in this case it's just normalizing the non-holonomous basis vectors. More generally you can introduce arbitrary bases using the holonomous basis by using any invertible transformation matrix via (Einstein summation convention implied)
$$\vec{g}_j=\vec{b}_i {T^i}_j.$$

Yes, yes, my question is why one of them is holonomous and the other one is not. What is the difference between them? What is the general definition of each of them?

 

[Orodruin]

Yes, yes, my question is why one of them is holonomous and the other one is not. What is the difference between them? What is the general definition of each of them?

A holonomic basis consists of the partial derivatives with respect to some set of local coordinates. A non-holonomic basis does not.

 

[Silviu]

A holonomic basis consists of the partial derivatives with respect to some set of local coordinates. A non-holonomic basis does not.

Ah, I see. Makes sense now. Thank you! One more question. The way you play with the coordinates is independent from a point to another? I.e. can you at a point keep the coordinates still while at a nearby point on the manifold you rotate the coordinate system by 90 degrees? Or the transformation should be smooth along the manifold? So if the basis vectors at a point transform using a matrix M, is this matrix a smooth function M(p), or it can take any value at any point on the manifold?

 

[Orodruin]

Take two sets of coordinates $x^a$ and $y^i$. Based on the chain rule for derivatives
$$
\newcommand{\dd}[2]{\frac{\partial #1}{\partial #2}}
\dd{}{y^i} = \dd{x^a}{y^i} \dd{}{x^a}
$$
so the matrix you are looking for just consists of partial derivatives of the smooth coordinate functions.

 

[Silviu]

Take two sets of coordinates $x^a$ and $y^i$. Based on the chain rule for derivatives
$$
\newcommand{\dd}[2]{\frac{\partial #1}{\partial #2}}
\dd{}{y^i} = \dd{x^a}{y^i} \dd{}{x^a}
$$
so the matrix you are looking for just consists of partial derivatives of the smooth coordinate functions.

But this is only for coordinate basis (where the transformation matrix is the Jacobian). However you said that non-coordinate basis need not be partial derivatives. So in that case is the transformation matrix still smooth?

 

[Orodruin]

If the basis is a basis of smooth vector fields.

 

[Silviu]

If the basis is a basis of smooth vector fields.

Thank you! One more question (sorry for taking so long...). So in the book it says that we pick the non-coordinate basis in such a way that $g(\hat{e}_\alpha,\hat{e}_\beta) = e_\alpha^\mu e_\beta^\nu g_{\mu \nu}=\delta_{\alpha \beta}$. So, based on what I understood this is defined for just a point on a manifold, i.e. you can't find a smooth transformation such that the metric at that point is the flat metric. By this I mean that you can apply a transformation such that $Mg=\delta$, but this can't hold over the whole manifold. Is this correct? So the equation above is local.

 

[pervect]

Sorry to come in late, but I thought I'd mention that one of the important properties of vectors is that they can be added together, and that the addition is commutative. So if A and B are vectors, A+B = B+A.

On a general manifold, displacements do not add like vectors, so for example if you go 100 miles north and 200 miles east (in that order) on the curved manifold representing the Earth's surface, you do not wind up in exactly the same spot as if you went 200 miles east first, then went 100 miles north.

This means the operation (which I haven't formally defined) of going "100 miles east", the operator I am calling a displacement cannot be a vector, because it doesn't add with other displacements. One of the things we study in GR is the failure of displacements to commute.

In this specific example of polar coordinates, it turns out that the displacements do commute because the plane is flat (the curvature tensor vanishes), but relying on this specific feature of the specific problem will cause problems if you assume it's always true.

I'm not sure if this will help, there are a lot of semantic problems with the meaning of words. I hope that this informal note helps, but I'm not sure it will.

 

[PeroK]

Thank you! One more question (sorry for taking so long...). So in the book it says that we pick the non-coordinate basis in such a way that $g(\hat{e}_\alpha,\hat{e}_\beta) = e_\alpha^\mu e_\beta^\nu g_{\mu \nu}=\delta_{\alpha \beta}$. So, based on what I understood this is defined for just a point on a manifold, i.e. you can't find a smooth transformation such that the metric at that point is the flat metric. By this I mean that you can apply a transformation such that $Mg=\delta$, but this can't hold over the whole manifold. Is this correct? So the equation above is local.

If you can find a coordinate transformation that transforms the metric to $\eta_{\alpha \beta}$ globally then what you started with was flat spacetime in some or other coordinate system.

In general, for each point, you can find a coordinate transformation that transforms the metric to $\eta_{\alpha \beta}$ at that one point.

In fact, you can also ensure that in these coordinates $\frac{\partial g_{\alpha \beta}}{\partial x^{\gamma}} = 0$. Again only at that one point.

This means that in these coordinates the Christoffel symbols vanish at that point and the equations of motion reduce to those of flat spacetime (at that point). Hence the term "Local Inertial Frame".

 

[Orodruin]

If you can find a coordinate transformation that transforms the metric to $\eta_{\alpha \beta}$ globally then what you started with was flat spacetime in some or other coordinate system.

In general, for each point, you can find a coordinate transformation that transforms the metric to $\eta_{\alpha \beta}$ at that one point.

In fact, you can also ensure that in these coordinates $\frac{\partial g_{\alpha \beta}}{\partial x^{\gamma}} = 0$. Again only at that one point.

This means that in these coordinates the Christoffel symbols vanish at that point and the equations of motion reduce to those of flat spacetime (at that point). Hence the term "Local Inertial Frame".

Just to be clear, this refers to a holonomic basis. It is often possible to find a non-holonomic basis where the metric is $\eta_{\alpha\beta}$ in an extended region.