# I A question about the definition of vector field

1. Nov 26, 2017

### Jianbing_Shao

In general relativity we demand that the physical law can be stated as a form which does not depend on the choose of particular coordinate system, So the vector field is defined as a changing object following a regular pattern under the transformation of coordinates. For example, we can define the covariant vector field as:

$v^\mu (x) \rightarrow v'^{\mu} (x)=J^\mu_\nu(x) v^\nu (x)$ . $J^\mu_\nu(x)\equiv \frac{\partial x'^{\mu}}{\partial x^\nu}$

In fact, this transformation is a global transformation. What interested me is the infinitesimal transformation induced by the global transformation between two neighboring point:$x$ and $x+dx$.

If we require that $v(x)=G(x)v(x_0)$, or $dv(x)=(\partial_iG(x))G^{-1}(x)dx^iv(x)$

Then $v'(x)=J(x)G(x)v(x_0)$. So the infinitesimal change is:

$dv'(x)=\partial_i(J(x)G(x)) (J(x)G(x))^{-1}dx^iv'(x)$,

And we can get:

$dv'(x)=\left((\partial_iJ)J^{-1} +(\partial_i G)G^{-1} +J[(\partial_i G)G^{-1}, J^{-1}]\right)dx^i v'(x)$

To us, the first two terms $(\partial_i J)J^{-1}dx^iv'(x)$ and$(\partial_i G)G^{-1}dx^i v'(x)$ is easily to be explained, They can be seemed to result from the original vector field and coordinate transformation, but the third term is hard to be explained, Does it has any exact physical interpretations?

2. Nov 27, 2017

### Orodruin

Staff Emeritus
This is not accurate. What changes when you change basis are the components of the vector field. The vector field itself stays the same.

Later you seem to want to compare vectors at different points in the manifold. Generally you cannot do this without introducing some form of connection. It is not clear to me exactly what you are trying to do.
It is not a global transformation, it is a local transformation. If it was a global transformation all components would transform in the same way but the Jacobian depends on the point.

3. Nov 27, 2017

### Jianbing_Shao

Last edited: Nov 27, 2017
4. Nov 28, 2017

### Orodruin

Staff Emeritus
The difference between the vectors at two different points is meaningless without a connection and (in general) a path between them. They belong to different tangent spaces.

5. Nov 28, 2017

### Jianbing_Shao

Of course we can define the connection, because we have basis field $e'_\mu$. so:
$de'_\mu=\omega^\nu_\mu e'_\nu$.
As to choosing a particular path, because we only discussing the change between two neighboring points, so we can require that the path should be a geodesic.

Last edited: Nov 29, 2017
6. Nov 29, 2017

### Jianbing_Shao

In fact, I am not interested in comparing two vector between two neighboring points, I only focus on the question that how $v^\mu (x)$ and Jacobian matrix $J^\mu_\nu(x)$ determine the local property of $v'^\mu (x)$ . and in fact. this problem result from Thomas precession.
We can consider a particle moving with velocity $v(t)=\beta/c$, The connection between the coordinates in the particle's rest frame at time $t$ and the coordinates in the laboratory frame always are described with boost matrix $A(\beta)$,
$A(\beta)= exp(-\zeta^i(t) K_i) , \zeta^i(t)=\tanh^{-1}{\beta^i}$
So the infinitesimal transformation between time $t$ and $t+\delta t$ is:
$A(\beta+\delta\beta)A^{-1}(\beta)=\exp\left(-\delta\zeta^iK_i+\frac{1}{2}[(\zeta^i+\delta\zeta^i)K_i,\zeta^jK_j]\right)$
Then we can get:
$x^{(t+\delta t)}= \exp\left(-\delta\zeta^iK_i+\frac{1}{2}[(\zeta^i+\delta\zeta^i)K_i,\zeta^jK_j]\right)x^{(t)}$
$\exp\left(\frac{1}{2}[\delta\zeta^iK_i,\zeta^jK_j]\right)$ describes a rotation called Thomas precession.
But in fact it is not hard to find the actual transformation formula is:
$x^{(t+\delta t)}= \exp\left(-\delta\zeta^iK_i\right)x^{(t)}$,
Then the actual coordinate transformation can be written as:
$x^{(t)}=T {\exp\left(-\int d\zeta^iK_i\right)}x^{(t_0) }$, $T$ represents time-ordered product.
So we can draw a conclusion that the appearance of Thomas precession only because we try to use $A(\beta)$ to describe the coordinate transformation. If we use the formula above to describe the coordinate transformation, then there is no need to introducing Thomas precession, Rotation appears in a more natural way.