I Covariant derivative of tangent vector for geodesic

  • Thread starter Apashanka
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For the simple case of a 2-D curve in polar coordinated (r,θ) parametrised by λ (length along the curve).
At any λ the tangent vector components are V1=dr(λ)/dλ along ##\hat r## and V2=dθ(λ)/dλ along ##\hat θ##.
The non-zero christoffel symbol are Γ122 and Γ212.
From covariant derivative
(1)∇1V1=∂rV1
(2)∇2V2=∂θV2212V1.
(3)∇2V1=∂θV1122V2
(4)∇1V2=∂rV2212V2
For this curve to be geodesic V1=1 and V2=0
And (1),(3) and (4) becomes 0 and (2)≠0.
But for a geodesic the covariant derivatives of tangent vectors are 0.
Am I missing something??
I am trying to interpret physically by taking these examples.
Thank you
 

Orodruin

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But for a geodesic the covariant derivatives of tangent vectors are 0.
Am I missing something??
No, for a geodesic the covariant derivative of the tangent vector in the direction of the tangent vector is zero. In your case, the direction of the tangent vector is the 1-direction.
 
392
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No, for a geodesic the covariant derivative of the tangent vector in the direction of the tangent vector is zero. In your case, the direction of the tangent vector is the 1-direction.
Ohh that means sir here in this case ∇1V1 and ∇1V2 has to be necessarily 0.(for geodesic)
Sir what could be the physical meaning behind these zero and non-zero terms of these covariant derivative ....as I am taking these simple examples to learn.
Thank you
 
392
13
For the simple case of a 2-D curve in polar coordinated (r,θ) parametrised by λ (length along the curve).
At any λ the tangent vector components are V1=dr(λ)/dλ along ##\hat r## and V2=dθ(λ)/dλ along ##\hat θ##.
The non-zero christoffel symbol are Γ122 and Γ212.
From covariant derivative
(1)∇1V1=∂rV1
(2)∇2V2=∂θV2212V1.
(3)∇2V1=∂θV1122V2
(4)∇1V2=∂rV2212V2
For this curve to be geodesic V1=1 and V2=0
And (1),(3) and (4) becomes 0 and (2)≠0.
But for a geodesic the covariant derivatives of tangent vectors are 0.
Am I missing something??
I am trying to interpret physically by taking these examples.
Thank you
The same curve parametrised by λ in 2-D cartesian coordinate system and if this curve is geodesic.(straight line with slope m).
x(λ)=λ/√(1+m2) and y(λ)=mλ/√(1+m2) .therefore the tangent vector components at any λ becomes V1≠f(λ) along ##\hat x## and V2≠f(λ) along ##\hat y##.
Which is consistent that the tangent vector is fixed in direction throughout the curve (as taken geodesic).
Now the tangent vector direction is ##\hat n=\frac{\hat x+m\hat y}{\sqrt(1+m^2)}## and the tangent vector is given as V=V1(≠λ)##\hat x+##V2(≠λ)##\hat y##
From the covariant derivative ##\nabla_{\hat n}V=0##,
My question is sir how to prove it ??
For polar coordinates it is easy
 
799
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The same curve parametrised by λ in 2-D cartesian coordinate system and if this curve is geodesic.(straight line with slope m).
x(λ)=λ/√(1+m2) and y(λ)=mλ/√(1+m2) .therefore the tangent vector components at any λ becomes V1≠f(λ) along ##\hat x## and V2≠f(λ) along ##\hat y##.
Which is consistent that the tangent vector is fixed in direction throughout the curve (as taken geodesic).
Now the tangent vector direction is ##\hat n=\frac{\hat x+m\hat y}{\sqrt(1+m^2)}## and the tangent vector is given as V=V1(≠λ)##\hat x+##V2(≠λ)##\hat y##
From the covariant derivative ##\nabla_{\hat n}V=0##,
My question is sir how to prove it ??
For polar coordinates it is easy
In 2D, Cartesian Coordiantes, the parameters are given by ##(x(\lambda), y(\lambda)) = (a_1 \lambda + b_1, a_2 \lambda + b_2)## and the tangent vector components are ##(a_1,a_2)## at any point with coordinates ##(x,y)##. This tangent vector can also be expressed as ##a_1 e_x + a_2 e_y \equiv V##. Then, $$\nabla_V V = V^\mu \nabla_\mu (V^\nu e_\nu) = V^\mu (\partial_\mu V^\nu) e_\nu = 0$$
 
392
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In 2D, Cartesian Coordiantes, the parameters are given by ##(x(\lambda), y(\lambda)) = (a_1 \lambda + b_1, a_2 \lambda + b_2)## and the tangent vector components are ##(a_1,a_2)## at any point with coordinates ##(x,y)##. This tangent vector can also be expressed as ##a_1 e_x + a_2 e_y \equiv V##. Then, $$\nabla_V V = V^\mu \nabla_\mu (V^\nu e_\nu) = V^\mu (\partial_\mu V^\nu) e_\nu = 0$$
Will you please elaborate your last step I didn't get it....
 
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$$\nabla_V V = V^\mu \nabla_\mu (V^\nu e_\nu) = V^\mu (\partial_\mu V^\nu) e_\nu = 0$$
Will you please elaborate your last step I didn't get it....
In flat spaces, using cartesian coordinates, ##\nabla_\mu = \partial_\mu## and ##\partial_\mu e_\nu = 0##, and since ##V^\nu## are just constants, ##\partial_\mu V^\nu = 0##. Does that clear up your confusion?
 
392
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In flat spaces, using cartesian coordinates, ##\nabla_\mu = \partial_\mu## and ##\partial_\mu e_\nu = 0##, and since ##V^\nu## are just constants, ##\partial_\mu V^\nu = 0##. Does that clear up your confusion?
It's ok ,my question is why ##\nabla_V## is written as ##V^\mu \nabla_\mu##
 
160
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It's ok ,my question is why ##\nabla_V## is written as ##V^\mu \nabla_\mu##
That’s how you take derivatives (covariant or otherwise) along an arbitrary curve.
 
392
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That’s how you take derivatives (covariant or otherwise) along an arbitrary curve.
Okk for this it is the covariant vector along the tangent vector ,but for any arbitrary direction will it be same??
 

Orodruin

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It's ok ,my question is why ##\nabla_V## is written as ##V^\mu \nabla_\mu##
The definition of ##\nabla_\mu## is ##\nabla_{\partial_\mu}## and the connection is linear in the direction (note that ##\partial_\mu## is a basis vector of the tangent space). Hence, for an arbitrary vector ##V = V^\mu \partial_\mu##, you find
$$
\nabla_V = \nabla_{V^\mu \partial_\mu} = \{\mbox{linearity}\} = V^\mu \nabla_{\partial_\mu} \equiv V^\mu \nabla_\mu.
$$
 
392
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The definition of ##\nabla_\mu## is ##\nabla_{\partial_\mu}## and the connection is linear in the direction (note that ##\partial_\mu## is a basis vector of the tangent space). Hence, for an arbitrary vector ##V = V^\mu \partial_\mu##, you find
$$
\nabla_V = \nabla_{V^\mu \partial_\mu} = \{\mbox{linearity}\} = V^\mu \nabla_{\partial_\mu} \equiv V^\mu \nabla_\mu.
$$
For any vector ##V=V^\mu e_\mu## .
Therefore ##e_\mu=∂_\mu##??
 

Orodruin

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For any vector ##V=V^\mu e_\mu## .
Therefore ##e_\mu=∂_\mu##??
You can choose whatever basis you like, but if you want to express the covariant derivative in terms of ##\nabla_\mu \equiv \nabla_{\partial_\mu}##, then you must use the coordinate basis.
 

stevendaryl

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For any vector ##V=V^\mu e_\mu## .
Therefore ##e_\mu=∂_\mu##??
I'm not sure that @Orodruin's notation is familiar to you. I know, I never completely got used to it.

A vector ##V## uniquely determines an operator on scalar fields: ##V(\phi) \equiv V^\mu \partial_\mu \phi##. Because of this, the modern way of teaching differential geometry simply defines a vector to be a differential operator on scalar fields. So we let it be an equality:

##V = V^\mu \partial_\mu##

Once you've identified a vector with the corresponding differential operator, it follows that for the basis vectors ##e_\mu##:

##e_\mu = \partial_\mu##

(That is, when ##e_\mu## is the basis vector corresponding to the coordinate ##x^\mu##).
 

Orodruin

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I would add that this is actually very natural. In curvilinear coordinates in Euclidean space you typically introduce the tangent vector basis as the partial derivatives of the position vector with respect to the coordinates. However, the position vector never actually does anything there, it kind of just comes along. Once in a general manifold there is anyway no longer any notion of a position vector so the most elegant solution is to realise that it was never needed and you can just get rid of it.
 

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