# Covariant derivative

#### fab13

Hello,

I try to apprehend the notion of covariant derivative. In order to undertsand better, here is a figure on which we are searching for express the difference $\vec{V} = \vec{V}(M') - \vec{V}(M)$ :

In order to evaluate this difference, we do a parallel transport of $\vec{V}(M')$ at point $M$.

When one writes what we call "the absolute differential" $\text{D}\,v_{k}$ :

$\text{D}\,v_{k} = (\partial_{j}\,v_{k} - v_{i}\Gamma_{jk}^{i})\,dy^{j}$

does $\text{D}\,v_{k}$ represent the difference that we could get if the 2 vectors $\vec{V}(M')$ and $\vec{V}(M)$ were expressed in the Cartesian referential ?

So that we could write also from the above figure (with $\theta=\theta(M)$ :

$\text{D}\,v_{x}=(\vec{V}_{\rho}(M')-\vec{V}_{\rho}(M))cos(\theta)$

and

$\text{D}\,v_{y}=(\vec{V}_{\rho}(M')-\vec{V}_{\rho}(M))sin(\theta)$

??
Then, if this covariant derivative is zero, one can write :

$\text{d}\,v_{k} = v_{j}\,\Gamma_{ik}^{j}\,dy^{i}$

does it significate curvilinear coordinates changes in the way that they are equal to $v_{j}\,\Gamma_{ik}^{j}\,dy^{i}$ ?

In the case of a parallel transport of vector $\vec{V}(M')$ from $M'$ point to $M$ point, why absolute differential between the
two vectors is equal to zero whereas the two vectors are not equal in the Cartesian coordinates system (O,x,y) ?

And if I take the same curvilinear coordinates for the two vectors, I get for the absolute differential :

$\text{D}\,v_{k} = v_{j}\,\Gamma_{ik}^{j}\,dy^{i}$
What does this case represent on the above figure ?
If anyone could help me to clarify this notion of covariant derivative and parallel transport.

Thanks.

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#### Matterwave

Gold Member
Which resource are you using to learn this? It would be good to know.

Sometimes an emphasis is placed on coordinate systems, and sometimes an emphasis is placed on a more geometrical understanding of the problem. It depends on the approach of the author. The notation is a little weird in your presentation, I'm having a little trouble making heads or tails of it. For example, you talk about covariantly differentiating vectors, but in your mathematical notation, you are working with components $v_k$ which usually would denote the components of a one-form.

#### fab13

Sorry for my notation, my ressource comes from a french book.

to be clairly, here are the notations :

$v_{k}$ represents the "k" covariant composant of vector $\vec{V}$

$\text{D}\,V_{k}$ represents the absolute derivate of the "k" covariant composant of vector $\vec{V}$

$dy^{i}$ represents the differential of "k" curvilinear coordinate ( curvilinear coordinate system is represented by all ($y^{i}$) composants).

From all these notations, the covariant and absolute differential are defined by :

$\text{D}\,v_{k} = \nabla_{j}\,v_{k}\,dy^{j}=(\partial_{j}\,v_{k} - v_{i}\Gamma_{jk}^{i})\,dy^{j}$

One other precision for my above post, I wanted to write :

$\text{D}\,v_{x}=(V_{\rho}(M')-V_{\rho}(M))cos(\theta)$

and

$\text{D}\,v_{y}=(V_{\rho}(M')-V_{\rho}(M))sin(\theta)$

Thanks for you help

#### Matterwave

Gold Member
Hello,

I try to apprehend the notion of covariant derivative. In order to undertsand better, here is a figure on which we are searching for express the difference $\vec{V} = \vec{V}(M') - \vec{V}(M)$ :

In order to evaluate this difference, we do a parallel transport of $\vec{V}(M')$ at point $M$.

When one writes what we call "the absolute differential" $\text{D}\,v_{k}$ :

$\text{D}\,v_{k} = (\partial_{j}\,v_{k} - v_{i}\Gamma_{jk}^{i})\,dy^{j}$

does $\text{D}\,v_{k}$ represent the difference that we could get if the 2 vectors $\vec{V}(M')$ and $\vec{V}(M)$ were expressed in the Cartesian referential ?
I'll try to answer these questions, but bear in mind that sometimes the notation is ambiguous in meaning. There is a lot of abuse of notation used in physics texts especially, so it's not always possible to know exactly what the author means without reading the full text.

It appears that $\text{D} v_k$ represents either an infinitesimal change in the co-vector $v_k$ itself or in the components of the co-vector in any coordinate system. The quantity $\text{D} v_k$ should not be coordinate system dependent, since you have corrected for that via the Christoffel symbols.

So that we could write also from the above figure (with $\theta=\theta(M)$ :

$\text{D}\,v_{x}=(\vec{V}_{\rho}(M')-\vec{V}_{\rho}(M))cos(\theta)$

and

$\text{D}\,v_{y}=(\vec{V}_{\rho}(M')-\vec{V}_{\rho}(M))sin(\theta)$

??
I believe in the notation in the figure, the quantity $\text{D} v_k$ should be basically the infinitesimal version of $\text{D} v_k = (V(M)-V_M(M'))_k$. I don't understand your notation here, the indices do not match on the left and right side, I don't know which vector you are denoting. In the figure $V(M)$ is the vector field's value at $M$ while $V_M(M')$ is the parallel transported vector from $V(M')$ to the point $M$.

Then, if this covariant derivative is zero, one can write :

$\text{d}\,v_{k} = v_{j}\,\Gamma_{ik}^{j}\,dy^{i}$

does it significate curvilinear coordinates changes in the way that they are equal to $v_{j}\,\Gamma_{ik}^{j}\,dy^{i}$ ?
Yes, if I interpreted your question correctly, I think this is correct. If the covariant derivative is 0, it means that the vector field is parallel transported along the curve. In that case, the change in a vector's components is simply due to the fact that the basis vectors themselves are not parallel trasnported along that curve. In that case, this above equation tells you, in some sense, how you have to compensate for that effect. Curvilinear coordinates in flat-space will not have parallel transported basis vectors (along some curves), unlike Cartesian coordinate basis vectors (which are parallel transported along every curve).

In the case of a parallel transport of vector $\vec{V}(M')$ from $M'$ point to $M$ point, why absolute differential between the
two vectors is equal to zero whereas the two vectors are not equal in the Cartesian coordinates system (O,x,y) ?
Looking at the figure, the absolute differential should not be equal to 0 for this vector field. We just wrote out what it was above.
And if I take the same curvilinear coordinates for the two vectors, I get for the absolute differential :

$\text{D}\,v_{k} = v_{j}\,\Gamma_{ik}^{j}\,dy^{i}$
What does this case represent on the above figure ?
If anyone could help me to clarify this notion of covariant derivative and parallel transport.

Thanks.
I don't understand your "if I take the same curvilinear coordinates..." statement. But this equation would mean that the components of the vector did not change along the curve, i.e. $dy^i \partial_i v_k=0$. In other words, the vector field rotated in the exactly the same way as the basis vector fields so that its components did not change. But because the basis vector fields are themselves not parallel transported along the curve, you still have an "absolute differential" present.

#### fab13

Thank you very much, I understand better this notion, especially for the two cases :

1) when $\text{D}\,v_{k} = 0$, it implies that $\text{d}\,v_{k} = v_{j}\,\Gamma_{ik}^{j}\,dy^{i}$, so the curvilinear components change to compensate the rotation of vector basis.

2) when $dy^i \partial_i v_k=0$, it implies that $\text{D}\,v_{k} = v_{j}\,\Gamma_{ik}^{j}\,dy^{i}$ : so I have an the same expression for two situations (above 1) and this case 2)), i.e :

$\text{d}\,v_{k} = v_{j}\,\Gamma_{ik}^{j}\,dy^{i}$ is the differential of covariant components between the vector $\vec{V}(M')$ and $\vec{V}(M)$.

and for the case 2), we have the absolute differential equal to : $\text{D}\,v_{k} = v_{j}\,\Gamma_{ik}^{j}\,dy^{i}$, i.e the same quantity above for 1), which simply means that the absolute differential (what you call the infinitesimal version of $\text{D}\,v_{k}$) is also equal to $v_{j}\,\Gamma_{ik}^{j}\,dy^{i}$.

Now, you make me think to another problem :

on my figure, If I want for example study the absolute differential of the tangential vector $\vec{e}_2$ ( $\vec{e}_{\theta}$ in polar coordinates ) :

Why does one say generally that a tangential vector is parallely transported along the geodesics curve (and have $\text{D}\,e_{\theta} = 0$) and saying that the angle between the curve and the tangential vector is a constant ?

I think that it can't be called "parallely transported" because the difference of two vectors ( like $\vec{e}_{\theta}(M')$ and $\vec{e}_{\theta}(M)$ ) in an absolute referential is not equal to zero.

Regards

#### Matterwave

Gold Member
Thank you very much, I understand better this notion, especially for the two cases :

1) when $\text{D}\,v_{k} = 0$, it implies that $\text{d}\,v_{k} = v_{j}\,\Gamma_{ik}^{j}\,dy^{i}$, so the curvilinear components change to compensate the rotation of vector basis.

2) when $dy^i \partial_i v_k=0$, it implies that $\text{D}\,v_{k} = v_{j}\,\Gamma_{ik}^{j}\,dy^{i}$ : so I have an the same expression for two situations (above 1) and this case 2)), i.e :

$\text{d}\,v_{k} = v_{j}\,\Gamma_{ik}^{j}\,dy^{i}$ is the differential of covariant components between the vector $\vec{V}(M')$ and $\vec{V}(M)$.

and for the case 2), we have the absolute differential equal to : $\text{D}\,v_{k} = v_{j}\,\Gamma_{ik}^{j}\,dy^{i}$, i.e the same quantity above for 1), which simply means that the absolute differential (what you call the infinitesimal version of $\text{D}\,v_{k}$) is also equal to $v_{j}\,\Gamma_{ik}^{j}\,dy^{i}$.
You are putting a bit too much emphasis on coordinates and coordinate bases. The point of parallel transport is to define a way of taking the difference of two vectors (along some path) in a curved manifold. Right now we are working in a flat manifold, where parallel transport is simple. We can still define a notion of a co variant derivative in this case, but in a flat space-time there is a preferred Cartesian basis in which the co variant derivative is simply the Cartesian derivative. In a curved manifold, no such preferred coordinate system may exist, and it is there that we actually need to define this notion of parallel transport.

Also, do not forget the minus sign. If $\text{d}v_k=0$, you have $\text{D}v_k=-dy^j \Gamma^i_{jk} v_i$.

Now, you make me think to another problem :

on my figure, If I want for example study the absolute differential of the tangential vector $\vec{e}_2$ ( $\vec{e}_{\theta}$ in polar coordinates ) :

Why does one say generally that a tangential vector is parallely transported along the geodesics curve (and have $\text{D}\,e_{\theta} = 0$) and saying that the angle between the curve and the tangential vector is a constant ?

I think that it can't be called "parallely transported" because the difference of two vectors ( like $\vec{e}_{\theta}(M')$ and $\vec{e}_{\theta}(M)$ ) in an absolute referential is not equal to zero.

Regards
The vector $\vec{e}_\theta$ is NOT parallel transported as you say along the curve in the figure, as such $\text{D}\vec{e}_\theta \neq 0$. The curve in the figure is NOT a geodesic. Geodesics in the figure are straight lines. Tangent vectors to straight lines will be parallel transported along their respective straight lines. You can check this. (But it's rather trivial).

But the fact that a geodesic's tangent vector is parallel transported along the geodesic is in fact the actual definition of a (affine) geodesic.

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#### fab13

The curve in the figure is NOT a geodesic. Geodesics in the figure are straight lines. Tangent vectors to straight lines will be parallel transported along their respective straight lines
Could you give please a simple example of metric where geodesics are not straight lines ? is it possible in 2D dimension ? I saw a basic example with sphere surface (with a constant radius) where the tangent vector is transported ($$\vec{e_{\theta}}$$) but this tangent vector is expressed in 3D dimensions ( $$r, \theta, \phi$$ )

Thanks

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#### Matterwave

Gold Member
The surface of a sphere is a perfect example of a 2-D surface which has curvature. The geodesics are great circles around the sphere. There should only be 2 basis tangent vectors $e_\theta,~e_\phi$. The $e_r$ vector is not a tangent vector as that vector points normal to the sphere.

#### fab13

If I take the example of the surface of a sphere, how can I define the 2 basis tangent vector $$\vec{e_{\theta}}$$ and $$\vec{e_{\phi}}$$.

For example, with polar coordinates, I can define $$\vec{e_{\rho}}$$ and $$\vec{e_{\theta}}$$ by :

$$\vec{e_{\rho}}=\dfrac{\partial \vec{OM}}{\partial\rho} = \dfrac{\partial (\rho cos(\theta) \vec{e_{x}} + \rho sin(\theta)\vec{e_{y}})}{\partial \rho} = cos(\theta) \vec{e_{x}} + sin(\theta) \vec{e_{y}}$$

and

$$\vec{e_{\theta}}=\dfrac{\partial \vec{OM}}{\partial \theta} = \dfrac{\partial (\rho cos(\theta) \vec{e_{x}} + \rho sin(\theta)\vec{e_{y}})}{\partial \theta} = -\rho sin(\theta) \vec{e_{x}} + \rho cos(\theta) \vec{e_{y}}$$

So, on the surface of a sphere, How can I express the 2 basis tangent vector $$\vec{e_{\theta}}$$ and $$\vec{e_{\phi}}$$ like I did for polar coordinates above ?

Can I take the same definition ? , i.e :

$$\vec{e_{\theta}}=\dfrac{\partial \vec{OM}}{\partial\theta}$$

$$\vec{e_{\phi}}=\dfrac{\partial \vec{OM}}{\partial \phi}$$

but in this case, what have I to take for the vector $$\vec{OM}$$ for the surface of a sphere ?