Torsion tensor derivation

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  • #26
stevendaryl
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Let me first show how it works in the good-old 2D Euclidean plane. Since that space is flat, you can choose Cartesian coordinates, and you don't have to worry about connection coefficients.

Now, suppose we have two different vector fields, [itex]\vec{u}[/itex] and [itex]\vec{v}[/itex]. Then consider two different paths starting from the point [itex]p_0[/itex]
  1. Move along displacement vector [itex]\alpha \vec{u}[/itex], and then move along displacement vector [itex]\beta \vec{v}[/itex].
  2. Move along displacement vector [itex]\beta \vec{v}[/itex], and then move along displacement vector [itex]\alpha \vec{u}[/itex].
As shown in the attachment, path 1 consists of going from [itex]p_0[/itex] to [itex]p_1[/itex] to [itex]p_2[/itex]. Path 2 consists of going from [itex]p_0[/itex] to [itex]p_3[/itex] to [itex]p_4[/itex]

In general, [itex]p_2 \neq p_4[/itex]. We can compute, to lowest order, these two locations, as follows:

[itex]p_1 = p_0 + \beta\vec{v}(p_0)[/itex]
[itex]p_2 = p_1 + \alpha \vec{u}(p_1) \approx p_0 + \beta\vec{v}(p_0) + \alpha\vec{u}(p_0) + \alpha \beta ((\vec{v} \cdot \nabla) \vec{u})|_{p_0}[/itex]
[itex]p_3 = p_0 + \alpha \vec{u}(p_0)[/itex]
[itex]p_4 = p_3 + \beta \vec{v}(p_3) \approx p_0 + \beta\vec{v}(p_0) + \alpha \vec{u}(p_0) + \alpha \beta ((\vec{u} \cdot \nabla) \vec{v})|_{p_0}[/itex]

So the displacement vector from [itex]p_2[/itex] to [itex]p_4[/itex] is given by:

[itex]\vec{\delta_{24}} = p_4 - p_2 = \alpha \beta ((\vec{u} \cdot \nabla) \vec{v} - (\vec{v} \cdot \nabla) \vec{u})|_{p_0}[/itex]
[itex] \equiv [u,v]|_{p_0}[/itex]
 

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  • #27
stevendaryl
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Let me first show how it works in the good-old 2D Euclidean plane. Since that space is flat, you can choose Cartesian coordinates, and you don't have to worry about connection coefficients.

Now, suppose we have two different vector fields, [itex]\vec{u}[/itex] and [itex]\vec{v}[/itex]. Then consider two different paths starting from the point [itex]p_0[/itex]
  1. Move along displacement vector [itex]\alpha \vec{u}[/itex], and then move along displacement vector [itex]\beta \vec{v}[/itex].
  2. Move along displacement vector [itex]\beta \vec{v}[/itex], and then move along displacement vector [itex]\alpha \vec{u}[/itex].
As shown in the attachment, path 1 consists of going from [itex]p_0[/itex] to [itex]p_1[/itex] to [itex]p_2[/itex]. Path 2 consists of going from [itex]p_0[/itex] to [itex]p_3[/itex] to [itex]p_4[/itex]

In general, [itex]p_2 \neq p_4[/itex]. We can compute, to lowest order, these two locations, as follows:

[itex]p_1 = p_0 + \beta\vec{v}(p_0)[/itex]
[itex]p_2 = p_1 + \alpha \vec{u}(p_1) \approx p_0 + \beta\vec{v}(p_0) + \alpha\vec{u}(p_0) + \alpha \beta ((\vec{v} \cdot \nabla) \vec{u})|_{p_0}[/itex]
[itex]p_3 = p_0 + \alpha \vec{u}(p_0)[/itex]
[itex]p_4 = p_3 + \beta \vec{v}(p_3) \approx p_0 + \beta\vec{v}(p_0) + \alpha \vec{u}(p_0) + \alpha \beta ((\vec{u} \cdot \nabla) \vec{v})|_{p_0}[/itex]

So the displacement vector from [itex]p_2[/itex] to [itex]p_4[/itex] is given by:

[itex]\vec{\delta_{24}} = p_4 - p_2 = \alpha \beta ((\vec{u} \cdot \nabla) \vec{v} - (\vec{v} \cdot \nabla) \vec{u})|_{p_0}[/itex]
[itex] \equiv [u,v]|_{p_0}[/itex]
If you do the same thing with non-Cartesian coordinates in curved space, you get additional terms in [itex]p_4 - p_2[/itex] that are due to connection coefficients, in addition to [itex][u,v][/itex].

(Note: in curved space, you can't interpret a tangent vector as a "displacement", but you can sloppily think of it that way as long as the size of the parallelogram is small enough.)
 
  • #28
stevendaryl
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If you do the same thing with non-Cartesian coordinates in curved space, you get additional terms in [itex]p_4 - p_2[/itex] that are due to connection coefficients, in addition to [itex][u,v][/itex].

(Note: in curved space, you can't interpret a tangent vector as a "displacement", but you can sloppily think of it that way as long as the size of the parallelogram is small enough.)
In curved space,

[itex]\nabla_u v = \nabla_u (\sum_j v^j e_j)[/itex] where [itex]e_j[/itex] is a set of basis vectors. We can use the properties of derivatives to write:

[itex]\nabla_u v = (\sum_j \nabla_u v^j) e_j + \sum_j v^j (\nabla_u e_j)[/itex]

Similarly, [itex]\nabla_v u = (\sum_j \nabla_v u^j) e_j + \sum_j u^j (\nabla_v e_j)[/itex]

So the difference is:

[itex]\nabla_u v - \nabla_v u = \sum_j (\nabla_u v^j - \nabla_v u^j) e_j + \sum_j (v^j \nabla_u e_j) - \sum_i (u^i \nabla_v e_i)[/itex]

We can identify the terms on the left:
[itex] \sum_j (\nabla_u v^j - \nabla_v u^j) e_j \equiv [u,v][/itex]

[itex]\sum_j (v^j \nabla_u e_j) \equiv \sum_{ijk} u^i v^j \Gamma^k_{ij} e_k[/itex]
[itex]\sum_i (u^i \nabla_v e_i) \equiv \sum_{ijk} u^i v^j \Gamma^k_{ji} e_k[/itex]

So:
[itex]\nabla_u v - \nabla_v u = [u,v] + \sum_{ijk} (u^i v^j (\Gamma^k_{ij} - \Gamma^k_{ji}))[/itex]

In a coordinate basis (which is what I'm assuming), [itex]T(u,v) = \sum_{ijk} (u^i v^j (\Gamma^k_{ij} - \Gamma^k_{ji}))[/itex], so we have:
[itex]\nabla_u v - \nabla_v u = [u,v] + T(u,v)[/itex]

[edit] That should be: [itex]T(u,v) = \sum_{ijk} (u^i v^j (\Gamma^k_{ij} - \Gamma^k_{ji})e_k)[/itex],

which is what is being shown pictorially in your diagram.
 
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  • #29
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In curved space,

[itex]\nabla_u v = \nabla_u (\sum_j v^j e_j)[/itex] where [itex]e_j[/itex] is a set of basis vectors. We can use the properties of derivatives to write:

[itex]\nabla_u v = (\sum_j \nabla_u v^j) e_j + \sum_j v^j (\nabla_u e_j)[/itex]

Similarly, [itex]\nabla_v u = (\sum_j \nabla_v u^j) e_j + \sum_j u^j (\nabla_v e_j)[/itex]

So the difference is:

[itex]\nabla_u v - \nabla_v u = \sum_j (\nabla_u v^j - \nabla_v u^j) e_j + \sum_j (v^j \nabla_u e_j) - \sum_i (u^i \nabla_v e_i)[/itex]

We can identify the terms on the left:
[itex] \sum_j (\nabla_u v^j - \nabla_v u^j) e_j \equiv [u,v][/itex]

[itex]\sum_j (v^j \nabla_u e_j) \equiv \sum_{ijk} u^i v^j \Gamma^k_{ij} e_k[/itex]
[itex]\sum_i (u^i \nabla_v e_i) \equiv \sum_{ijk} u^i v^j \Gamma^k_{ji} e_k[/itex]

So:
[itex]\nabla_u v - \nabla_v u = [u,v] + \sum_{ijk} (u^i v^j (\Gamma^k_{ij} - \Gamma^k_{ji}))[/itex]

In a coordinate basis (which is what I'm assuming), [itex]T(u,v) = \sum_{ijk} (u^i v^j (\Gamma^k_{ij} - \Gamma^k_{ji}))[/itex], so we have:
[itex]\nabla_u v - \nabla_v u = [u,v] + T(u,v)[/itex]

which is what is being shown pictorially in your diagram.
Initially, I really appreciate your effort, your endeavour "stevendarly" :))Your sharing is so valuable, remarkable, and significant in my opinion. To sum up, I would like to ask one more final question about the terminology of your sharing in Torsion tensor derivation: I can not catch the meaning of
upload_2016-5-24_20-57-48.png
entirely in
upload_2016-5-24_20-58-33.png
( post#26). I have some thoughts as well as doubts about this situation, hence want to be sure: for what does
upload_2016-5-24_20-58-6.png
stand??? What is the aim and meaning of it??
 
  • #30
stevendaryl
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Initially, I really appreciate your effort, your endeavour "stevendarly" :))Your sharing is so valuable, remarkable, and significant in my opinion. To sum up, I would like to ask one more final question about the terminology of your sharing in Torsion tensor derivation: I can not catch the meaning of [deleted]
It's better if you used LaTex for formulas, instead of trying to get screenshots.

There is no point to [itex]\alpha[/itex] and [itex]\beta[/itex] other than that they are small positive real numbers, so that we can expand things in a power series in [itex]\alpha[/itex] and [itex]\beta[/itex] and just keep the lowest-order nonzero terms.

If [itex]\vec{u}[/itex] is a vector field, then that means that for each point [itex]p[/itex] in your space, then [itex]\vec{u}(p)[/itex] is a vector. In terms of coordinates (I'll assume [itex]x,y, z[/itex], then we have [itex]\vec{u}(x,y,z)[/itex].
  • Let [itex]p_0[/itex] be some point with coordinates [itex]x, y, z[/itex].
  • Let [itex]v_0[/itex] be the value of [itex]v[/itex] at that point. Let it have coordinates [itex]v_0^x, v_0^y, v_0^z[/itex]
  • Let [itex]\beta[/itex] be a small positive real number.
  • Let [itex]p_1[/itex] be the point [itex]p_0 + \beta \vec{v}[/itex], that is, the point with coordinates [itex]x + \beta v_0^x, y + \beta v_0^y, z + \beta v_0^z[/itex]
  • Let [itex]u_0[/itex] be the value of [itex]u[/itex] at point [itex]p_0[/itex]. Let it's coordinates be [itex]u_0^x, u_0^y, u_0^z[/itex]
  • Let [itex]u_1[/itex] be the value of [itex]u[/itex] at point [itex]p_1[/itex].
  • Let [itex]p_2[/itex] be the point [itex]p_1 + \alpha u_1[/itex].
How do we calculate the components of [itex]u_1[/itex]? Well, any smooth function of position can be expanded in a Taylor series:

[itex]f(p_1) \approx f(p_0) + \frac{\partial f}{\partial x}|_{p_0} \delta x + \frac{\partial f}{\partial y}|_{p_0} \delta y + \frac{\partial f}{\partial z}|_{p_0} \delta z[/itex], where [itex]|_{p_0}[/itex] means to evaluate the partial derivative at the point [itex]p_0[/itex] and where [itex]\delta x[/itex] means the change in [itex]x[/itex] in going from [itex]p_0[/itex] to [itex]p_1[/itex], and similarly for [itex]\delta y[/itex] and [itex]\delta z[/itex].

We can write that more compactly as [itex]f(p_1) \approx f(p_0) + \sum_j \delta x^j \frac{\partial f}{\partial x^j}[/itex], or even more compactly as:

[itex]f(p_1) \approx f(p_0) + (\vec{\delta p} \cdot \nabla) f|_{p_0}[/itex]

where [itex]\vec{\delta p} \cdot \nabla[/itex] is the operator [itex]\sum_j \delta x^j \frac{\partial}{\partial x^j}[/itex] , and where [itex]\delta p[/itex] is the displacement vector from [itex]p_0[/itex] to [itex]p_1[/itex].

So in our case, we are interested in [itex]u_1 \equiv u(p_1)[/itex]. So [itex]u_1 \equiv u(p_1) = u(p_0) + (\vec{\delta p} \cdot \nabla) u|_{p_0}[/itex]. The displacement vector from [itex]p_0[/itex] to [itex]p_1[/itex] is just [itex]\beta\vec{v}[/itex], so we have:

[itex]u_1 \equiv u(p_1) \approx u(p_0) + \beta (\vec{v} \cdot \nabla) u|_{p_0}[/itex]

Now, we define [itex]p_2[/itex] to be the point [itex]p_1 + \alpha \vec{u_1} = p_0 + \beta \vec{v_0} + \alpha \vec{u_0} + \alpha \beta (\vec{v} \cdot \nabla) u|_{p_0}[/itex]
 
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