# I How do you derive those basis vectors in GR?

#### space-time

You may be familiar with how you can express a vector field as a linear combination of basis vectors like so:

X = Xii

Now, I know that normally, the basis vectors ∂i can be derived by taking the derivatives of the position vector for the coordinate system with respect to all the axes like this:

i = ∂R/∂xi

where R is a position vector such as R = [rsin(θ)cos(∅), rsin(θ)sin(∅), rcos(θ)] for spherical coordinates.

This is how you normally derive such basis vectors.

However, how do you do it in GR where you don't have such R vectors (or at least they don't seem obvious to me)?

For instance, how would I derive the basis vectors for a metric like the Godel metric, which has the following line element?:

ds2 = (1/(2ω2))[-c2dt2 - 2cexdzdt + dx2 + dy2 - (1/2)e2xdz2]

I don't think that it would just be a matter of differentiating the position vector with respect to the axes, because in the case of this metric I'm about 95% sure that the position vector here is just [ ct, x, y, z ]. However, the metric tensor has terms of ω in it as well as terms of e. Obiviously, none of the derivatives of this position vector contain any such terms. This is problematic because you are supposed to be able to dot product two basis vectors together in order to get an element of the metric tensor (which I just noted to contain terms of ω and e).

This is not even to mention issues of sign convention and attaching a negative to at least one basis vector or metric tensor element?

Long story short:

How do I derive the basis vectors for spacetime metrics in GR where I don't have a clear cut R vector like I do with spherical coordinates in 3D?

• dsaun777
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#### Orodruin

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It is correct that there is no position vector in a general manifold. Instead, everything comes down to how a tangent vector on a manifold is defined. It turns out that the important property of the definition you mention was not the position vector, but the partial derivatives wrt the coordinates. The partial derivatives are the basis vectors.

This should be described in more detail in any text discussing calculus on manifolds.

• Martin Scholtz

#### dsaun777

It is correct that there is no position vector in a general manifold. Instead, everything comes down to how a tangent vector on a manifold is defined. It turns out that the important property of the definition you mention was not the position vector, but the partial derivatives wrt the coordinates. The partial derivatives are the basis vectors.

This should be described in more detail in any text discussing calculus on manifolds.
Is it possible to define a position on any manifold?

#### Ibix

Is it possible to define a position on any manifold?
Better to say you label a position with coordinates, and yes you can do that on any manifold. What you can't do in general is treat coordinates as if they were vectors and add and subtract them.

• Martin Scholtz

#### dsaun777

Better to say you label a position with coordinates, and yes you can do that on any manifold. What you can't do in general is treat coordinates as if they were vectors and add and subtract them.
I meant to say is it possible to have a position vector on any manifold or do you just throw away the concept of position vector away completely?

#### Orodruin

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A position vector requires a concept of having an identification of the points in your space with the tangent space at a particular arbitrary point (the origin). This is the basic concept of an affine space.

In a general manifold there is no such thing as a position vector.

#### Martin Scholtz

Gold Member
I meant to say is it possible to have a position vector on any manifold or do you just throw away the concept of position vector away completely?
On general manifold, vectors live in different mathematical spaces. Imagine a sphere and consider any vector tangent to the sphere. What is well defined is the starting point of the vector, but the end point (the end of the arrow) lies outside the sphere and hence does not belong to the manifold. The only reason why we can imagine this situation is that the sphere is 2d manifold which can be embedded into 3d Euclidian space which we can still imagine, and therefore also objects like vectors pointing off the sphere.

That's also the reason why you cannot define the position vector: position vector is an arrow connecting the origin with a point, but for the sphere it would mean that the connecting arrow lies in the interior of the sphere.

If you want to analyze the intrinsic properties of the manifold, there is know such thing as vector pointing off the sphere. Like in GR: spacetimes are not embedded into bigger manifolds.

So you need to generalize the concept of vector and there are more ways how to do it (they are equivalent).

First one is to consider curve lying on the sphere starting from the point to which the vector is attached and then going on the direction of the vector. More precisely, vector is a class of curves such that their coordinate expressions have the same derivatives wrt the parameter, i.e. they have the same direction up to first order. Each such class defines a vector and it can be checked that this definition is independent of coordinates and these equivalence classes form an #n# dimensional linear space called the tangent the tangent space.

That also means that derivative of a function #f# along any curve from such class is the same. It can be then shown that the tangent space is isomorphic to the space of derivatives of functions. That's why we can represent vectors as differential operators. Acting with vector on a function, you get its derivative in the direction of the vector.

• kent davidge

#### PeterDonis

Mentor
This is how you normally derive such basis vectors.
What makes you think that? Can you give a reference that "derives" basis vectors this way?

#### pervect

Staff Emeritus
You may be familiar with how you can express a vector field as a linear combination of basis vectors like so:

X = Xii

Now, I know that normally, the basis vectors ∂i can be derived by taking the derivatives of the position vector for the coordinate system with respect to all the axes like this:

i = ∂R/∂xi

where R is a position vector such as R = [rsin(θ)cos(∅), rsin(θ)sin(∅), rcos(θ)] for spherical coordinates.
There are some terminology issues here.

Let's look at the spherical coordinates you wrote.

Then $r, \theta, \phi$ are spherical coordinates. x,y, and z are cartesian coordinates, given by some diffeomorphism. Wiki gives <<link>>:

$$x = r\sin \theta \cos \phi \quad y= r \sin \theta \sin \phi \quad z= r \cos \theta$$

You have a bit of a funny ordering in your R, but otherwise it looks OK (I could have missed something)

The coordinate basis vectors for spherical coordinates are just

$$\frac{\partial}{\partial r} \quad \frac{\partial}{\partial \theta} \quad \frac{\partial}{\partial \phi}$$

The coordinate basis vectors for the cartesian coordiantes are just

$$\frac{\partial}{\partial x} \quad \frac{\partial}{\partial y} \quad \frac{\partial}{\partial z}$$

Using the chain rule, and the diffeomorphism between coordinates, one can use the chain rule to go from one set of vectors to the other.

For instance, knowing x,y,z as functions of r,$\theta$, $\phi$, one can write.

$$\frac{\partial}{\partial r} = \frac{\partial x}{\partial r} \frac{\partial}{\partial x} + \frac{\partial y}{\partial r}\frac{\partial}{\partial y} + \frac{\partial z}{\partial r} \frac{\partial}{\partial z} = \sin \theta \cos \phi \frac{\partial}{\partial x} + \sin \theta \sin \phi \frac{\partial}{\partial y} + \cos \theta \frac{\partial}{\partial z}$$

Thus, the coordinate basis vectors in spherical coordinates can be expresssed as a linear sum of the basis vectors in cartesian coordinates. With the inverse mapping, one can do the reverse.

Now, so far, I've been talking about "coordinate basis vectors". These are more specific than "basis vectors". Vectors, by defintion can be scaled (multipled by a scalar), and added together. An example of this is given above, where we write the coordinate basis vectors in one set of coordinates as a weighted sum of coordinate basis vectors in another basis.

There is another sort of basis vectors that you may be intersted in, an orthonormal basis. To describe this, though, you need to know how to use the metric tensor to take the inner product of two vectors.

The inner product of a vector with itself gives the length of a vector, the inner product of two orthogonal vectors is zero. So, given a knowledge of how to take the inner product, one can determine how to weight and combine vectors to make coordinate basis vectors an orthonormal basis.

But I don't want to get into the details of how to take the inner product of two vectors. That's something you might want to work out on your own. A good goal would be "what is the length of $\frac{\partial}{\partial \theta}$. Another good exercise would be "Is $\frac{\partial}{\partial r}$ orthogonal to $\frac{\partial}{\partial \theta}$".

To get started, you'd want the metric tensor for spherical coordinates. I should probably give it, but I'd frankly want to look it up, my memory isn't what it used to be. So it's best if you dig it out on your own.

#### space-time

There are some terminology issues here.

Let's look at the spherical coordinates you wrote.

Then $r, \theta, \phi$ are spherical coordinates. x,y, and z are cartesian coordinates, given by some diffeomorphism. Wiki gives <<link>>:

$$x = r\sin \theta \cos \phi \quad y= r \sin \theta \sin \phi \quad z= r \cos \theta$$

You have a bit of a funny ordering in your R, but otherwise it looks OK (I could have missed something)

The coordinate basis vectors for spherical coordinates are just

$$\frac{\partial}{\partial r} \quad \frac{\partial}{\partial \theta} \quad \frac{\partial}{\partial \phi}$$

The coordinate basis vectors for the cartesian coordiantes are just

$$\frac{\partial}{\partial x} \quad \frac{\partial}{\partial y} \quad \frac{\partial}{\partial z}$$

Using the chain rule, and the diffeomorphism between coordinates, one can use the chain rule to go from one set of vectors to the other.

For instance, knowing x,y,z as functions of r,$\theta$, $\phi$, one can write.

$$\frac{\partial}{\partial r} = \frac{\partial x}{\partial r} \frac{\partial}{\partial x} + \frac{\partial y}{\partial r}\frac{\partial}{\partial y} + \frac{\partial z}{\partial r} \frac{\partial}{\partial z} = \sin \theta \cos \phi \frac{\partial}{\partial x} + \sin \theta \sin \phi \frac{\partial}{\partial y} + \cos \theta \frac{\partial}{\partial z}$$

Thus, the coordinate basis vectors in spherical coordinates can be expresssed as a linear sum of the basis vectors in cartesian coordinates. With the inverse mapping, one can do the reverse.

Now, so far, I've been talking about "coordinate basis vectors". These are more specific than "basis vectors". Vectors, by defintion can be scaled (multipled by a scalar), and added together. An example of this is given above, where we write the coordinate basis vectors in one set of coordinates as a weighted sum of coordinate basis vectors in another basis.

There is another sort of basis vectors that you may be intersted in, an orthonormal basis. To describe this, though, you need to know how to use the metric tensor to take the inner product of two vectors.

The inner product of a vector with itself gives the length of a vector, the inner product of two orthogonal vectors is zero. So, given a knowledge of how to take the inner product, one can determine how to weight and combine vectors to make coordinate basis vectors an orthonormal basis.

But I don't want to get into the details of how to take the inner product of two vectors. That's something you might want to work out on your own. A good goal would be "what is the length of $\frac{\partial}{\partial \theta}$. Another good exercise would be "Is $\frac{\partial}{\partial r}$ orthogonal to $\frac{\partial}{\partial \theta}$".

To get started, you'd want the metric tensor for spherical coordinates. I should probably give it, but I'd frankly want to look it up, my memory isn't what it used to be. So it's best if you dig it out on your own.
I already know how to take the inner product of two vectors using the metric tensor. That's just:

gabUaVb

where U and V are vectors.

#### space-time

What makes you think that? Can you give a reference that "derives" basis vectors this way?
It's just that every video playlist on tensor calculus/differential geometry that I watch starts with the R vector using either spherical or cylindrical coordinates and then calculates the basis vectors and then the metric tensor this way.

Nobody I've seen ever goes on to point out some different way to derive basis vectors.

#### PeterDonis

Mentor
every video playlist on tensor calculus/differential geometry that I watch starts with the R vector using either spherical or cylindrical coordinates and then calculates the basis vectors and then the metric tensor this way
First, the words "video playlist" already raises a red flag. Have you looked at textbooks?

Second, can you give a specific link to one of these?

#### Orodruin

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First, the words "video playlist" already raises a red flag. Have you looked at textbooks?

Second, can you give a specific link to one of these?
It is quite standard practice to start a discussion on curvilinear coordinate bases in Euclidean space (or affine space, but at that point it is usually just Euclidesn space) by defining a tangent vector basis as the partial derivatives of the position vector wrt the coordinates. I do this in my book for example. Of course, this no longer works when you leave affine space behind, which should be pointed out in any text discussing manifolds.

#### pervect

Staff Emeritus
I already know how to take the inner product of two vectors using the metric tensor. That's just:

gabUaVb

where U and V are vectors.
OK, then, you should be able to determine that the length squared of the vector $\frac{\partial}{\partial \theta}$ is $g_{\theta \theta}$. As I recall this equal to $r^2$. This makes the length of $\partial_\theta$ equal to r, since the length squared is r^2. And thus $\frac{1}{r} \frac{\partial}{\partial \theta}$ is a unit length vector in the same direction as the coordinate basis. Here I've used some more compact notation

$$\partial_\theta = \frac{\partial}{\partial \theta}$$

The answer to the question you actually asked in your first post is so trivial that I suspect you wanted to ask something different. For the line element you wrote, the coordinate basis vectors for the coordinates you specified, namely t,x,y,z are simply, by definition, the partial derivatives with respect to the coordinates, i.e.:

$$\partial_t \quad \partial_x \quad \partial_y \quad \partial_z$$

where I've used the same abbreviated notation for convenience.

Like the case for spherical coordinates, these vectors are not all of unit length. Furthermore $\partial_t$ is not orthogonal to $\partial_z$.

I am guessing that perhaps you wanted, not a coordinate basis, but an orthonormal basis? Which is why I mentioned how to find lengths of vectors, and whether or not they are orthogonal. Which you apparently already know, which is good.

But I may have gotten off track. If you are just happy with any old basis, the coordinate basis above is fine. But I suspect there's more to your question than that. I am suggesting that you wanted to find an orthonormal basis, but possibly that's not what you want. If you want something else, you'll need to clarify. If it is what you want, I can go into how you might do it at more length, if you ask.

#### vanhees71

Gold Member
In this case the trick is to read about "differentiable manifolds" in a math book rather than a physics book. The trick is that a differentiable manifold is a manifold consisting of points, whose topology is locally equivalent to an affine $d$-dimensional manifold and thus $\mathbb{R}^d$ as a metric space (for convenience you can use the Euclidean metric for that matter). That means for each point $X$ of the manifold there's an open subset that can be mapped one-to-one to $\mathbb{R}^d$. Now you can do calculus as usual.

Particularly if you have a curve in the manifold through the point $X$ there's through the one-to-one mapping to $\mathbb{R}^d$ a curve in $\mathbb{R}^d$ and you can define the tangent vector to this curve. Using all curves through $X$ this defines a vector space of tangent vectors at $X$. Now you can do the usual vector algebra on this tangent space. Particularly there's also a dual vector space and co-vectors, which are now naturally called co-tangent vectors.

The tricky point is that usually you need many local maps of the manifold to $\mathbb{R}^d$ and you must make sure that in the regions where two maps overlap nothing principally changes, i.e., that the properties of the tangent vectors and co-tangent vectors etc. are the same, no matter through which map you define them. of course having one-to-one maps in the overlap region you can define a mapping $\mathbb{R}^d \rightarrow \mathbb{R}^d$, and to make sure that everything works out fine in the just said sense you must make sure that this mapping is a diffeomorphism, i.e., that it is one-to-one, differentiable and the inverse is differentiable too. Covering the entire manifold with such compatible maps defines an atlas, and given an atlas you can do all the manipulations of vector calculus known from the vector calulus on $\mathbb{R}^d$ with the tangent vectors on every point of the manifold.

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