Geodesics under coordinate transformation

[VladZH]

I will start with an example.
Consider components of metric tensor $g'$ in a coordinate system
$$ g'=
\begin{pmatrix}
xy & 1 \\
1 & xy \\
\end{pmatrix}
$$
We can find a transformation rule which brings $g'$ to euclidean metric $g=\begin{pmatrix} 1 & 0 \\ 0 & 1\\ \end{pmatrix}$, namely
$$A^T*g'*A=g$$
where $A=\begin{pmatrix} -\frac{1}{\sqrt{xy}} & 1 \\ 1 & -\frac{1}{\sqrt{xy}}\\ \end{pmatrix}$ .

Levi-Civita connection for $g$ has all components as zero but not all components are vanishing for $g'$.
So if I want to find geodseics given $g'$ I could find appropriate transformation where components of $g'$ looks like $g$ but in this case geodesics are going to be straight lines given ANY $g'$.

Is this a wrong statement?

 

[andrewkirk]

This question demonstrates the importance of distinguishing between coordinate-dependent and coordinate-independent properties of objects, when working with tensors.

A metric is a tensor, which is a coordinate-independent object. Strictly speaking, it is a function that takes two vectors as input, and gives a real number as output. It has a matrix representation, which is coordinate-dependent, but the function itself is not coordinate-dependent. The two matrices you wrote for $g$ and $g'$ are coordinate-dependent representations of the tensor.

Next consider what is meant by 'straight line'. In differential geometry this is usually intended to mean 'geodesic'. Being a geodesic is a coordinate-independent property, so a geodesic is a geodesic in any coordinate system.

However I suspect that by 'straight line' you mean "has equation of the form y = ax + b". That property is coordinate-dependent, and so may apply in some coordinate systems and not others.

A geodesic in this system will have equation of form y = ax+b in the second coordinate system, but not in the first. So if by 'straight line' you mean 'has the form y = ax + b' the statement is wrong. Alternatively, if 'straight line' means geodesic, the statement just becomes a restatement of the fact that the property of being a geodesic is coordinate-independent.

 

[VladZH]

Thanks for reply!

However I suspect that by 'straight line' you mean "has equation of the form y = ax + b". That property is coordinate-dependent, and so may apply in some coordinate systems and not others.

Yes, I mean "equation of the form y = ax + b" . The solution of geodesic equation in coordinate system with component of metric as identity matrix gives y = ax + b. So I can take any metric in the coordinate system where its components are identity matrix and solving geodesic equation I will get y = ax + b. if I draw the equation I will see a straight line given any starting point and direction. Is this right?

 

[Orodruin]

Thanks for reply!

Yes, I mean "equation of the form y = ax + b" . The solution of geodesic equation in coordinate system with component of metric as identity matrix gives y = ax + b. So I can take any metric in the coordinate system where its components are identity matrix and solving geodesic equation I will get y = ax + b. if I draw the equation I will see a straight line given any starting point and direction. Is this right?

If you find a coordinate system where the Christoffel symbols are all zero, yes. However, it is not guaranteed that such a coordinate system exists. Note that what you have written down is a basis transformation on the tangent space - not a coordinate transformation!

 

[VladZH]

If you find a coordinate system where the Christoffel symbols are all zero, yes. However, it is not guaranteed that such a coordinate system exists.

So in this specific example can I say that g' is equivalent to euclidean metric?

Note that what you have written down is a basis transformation on the tangent space - not a coordinate transformation!

For coordinate transformation I can use A^-1, right?

 

[Orodruin]

For coordinate transformation I can use A-1, right?

No. That is not a coordinate transformation. That is a basis transformation. A coordinate transformation is introducing new coordinates $x'$ and $y'$ that are functions of $x$ and $y$. Not changing basis in the tangent space.

 

[VladZH]

No. That is not a coordinate transformation. That is a basis transformation. A coordinate transformation is introducing new coordinates $x'$ and $y'$ that are functions of $x$ and $y$. Not changing basis in the tangent space.

I cannot fully understand what you mean.
If we want to know vector components $x^i$ in another coordinate system we can use equation
$$x'^j=B^j_i x^i$$
And for components of metric tensor respectively
$$g'_{ij}=A^l_iA^k_j g_{kl}$$
where $A = J$ and $B = J^{-1}$.
Why it is not coordinate transformation if we are literally getting $x'^j=f(x^i)$?

 

[Orodruin]

You are confusing coordinates with vector components. They are not the same thing.

 

[VladZH]

You are confusing coordinates with vector components. They are not the same thing.

What is the difference? What are vector components then?

 

[Orodruin]

Coordinates are a set of functions on a manifold whose values together uniquely specify a point on the manifold. A (tangent) vector is in the tangent space of a manifold and its components are the expansion coefficients of that vector in some given basis - often taken to be a holonomic basis.

The likely reason that you, and many others, confuse the two is that the space where people are typically used to working, i.e., Euclidean space is an affine space where there is a bijection from the tangent space of any given point and the space itself based on the translation map.

 

[stevendaryl]

Yes, I mean "equation of the form y = ax + b" . The solution of geodesic equation in coordinate system with component of metric as identity matrix gives y = ax + b. So I can take any metric in the coordinate system where its components are identity matrix and solving geodesic equation I will get y = ax + b. if I draw the equation I will see a straight line given any starting point and direction. Is this right?

Somebody may have already made this point, but in a curved space, you can only get the metric to be the identity at a single point. You can't get it to be the identity everywhere (in general).

One way to see this is by picking a coordinate system $x^\mu$ and considering the matrix in the neighborhood of the origin. Then we can expand the metric in a power series.

$g_{\mu \nu} = A_{\mu \nu} + B_{\mu \nu \lambda} x^\lambda + C_{\mu \nu \lambda \tau} x^\lambda x^\tau + ...$

By picking your coordinate system carefully, you can make it so that $A_{\mu \nu}$ is diagonal with entries $\pm 1$. You make it so that $B_{\mu \nu \lambda}$ is zero. But you can't pick a coordinate system (in general) where $A_{\mu \nu}$ is diagonal and both $B_{\mu \nu \lambda}$ and $C_{\mu \nu \lambda \tau}$ are zero.

 

[Orodruin]

Somebody may have already made this point, but in a curved space, you can only get the metric to be the identity at a single point. You can't get it to be the identity everywhere (in general).

Clarification: You can (at least locally), but not using a holonomic basis.

 

[VladZH]

Note that what you have written down is a basis transformation on the tangent space - not a coordinate transformation!

So how coordinate transforamtion is different from the example in my first post?
For coordinate transform of metric tensor we have
$$J^T*g'*J=g$$
where J is Jacobian matrix.

 

[Orodruin]

That is how the tensor transforms under a coordinate transformation. In itself, it is not a coordinate transformation. A coordinate transformation expresses the new coordinates as functions of the old coordinates. Your expression does not do that.

 

[VladZH]

That is how the tensor transforms under a coordinate transformation.

That is what I meant.
Given coordinates xⁱ and metric components g_ij I find such cordinates x^'i=f(x¹,...,xⁿ) so that having
Jacobian $$J_j^i=\frac{\partial x_i}{\partial x'_j}$$ I am getting new metric components
$$J^T*g*J=g'= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$
Than I solve geodesic equaion for the new coordinate system
$$\frac{\mathrm{d}^{2} x^{'i}}{\mathrm{d} \lambda^{2}} + \Gamma^{i}_{\mu\nu}\frac{\mathrm{d} x^{'\mu}}{\mathrm{d} \lambda}\frac{\mathrm{d} x^{'\nu}}{\mathrm{d} \lambda} = 0$$
$$x^{'1}=C_1\lambda+C_2$$
$$x^{'2}=C_3\lambda+C_4$$
that actually gives me a straight line $$x^{'2}=ax^{'1}+b$$
Sorry that I am repeating myself but those steps show that given any non-identity metric components in any coordinate system I can find such a coordinate system where metric components are identity matrix that would mean geodesics are straight lines and any metric has no curvature at all!

 

[Orodruin]

That is what I meant.
Given coordinates xⁱ and metric components g_ij I find such cordinates x^'i=f(x¹,...,xⁿ) so that having
Jacobian $$J_j^i=\frac{\partial x_i}{\partial x'_j}$$ I am getting new metric components
$$J^T*g*J=g'= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$
Than I solve geodesic equaion for the new coordinate system
$$\frac{\mathrm{d}^{2} x^{'i}}{\mathrm{d} \lambda^{2}} + \Gamma^{i}_{\mu\nu}\frac{\mathrm{d} x^{'\mu}}{\mathrm{d} \lambda}\frac{\mathrm{d} x^{'\nu}}{\mathrm{d} \lambda} = 0$$
$$x^{'1}=C_1\lambda+C_2$$
$$x^{'2}=C_3\lambda+C_4$$
that actually gives me a straight line $$x^{'2}=ax^{'1}+b$$
Sorry that I am repeating myself but those steps show that given any non-identity metric components in any coordinate system I can find such a coordinate system where metric components are identity matrix that would mean geodesics are straight lines and any metric has no curvature at all!

No. While you can always find a coordinate system such that the Christoffel symbols vanish at a particular given point, there is no guarantee that there will exist a coordinate system such that this will hold true in an extended neighbourhood of that point. In fact, such a coordinate system will not exist in a curved space.

 

[stevendaryl]

Sorry that I am repeating myself but those steps show that given any non-identity metric components in any coordinate system I can find such a coordinate system where metric components are identity matrix that would mean geodesics are straight lines and any metric has no curvature at all!

The distinction between coordinate transformation and basis transformation is important here. If you have a curved spacetime, you can pick a coordinate system, and in terms of that coordinate system, you can pick a basis at each point in spacetime. In terms of the basis at a point, you have a corresponding connection coefficient, $\Gamma^\mu_{\nu \lambda}$. But in general, $\Gamma^\mu_{\nu \lambda}$ will not be a constant. In general, the components will change from point to point.

For any particular point, you can find a coordinate transformation that makes $\Gamma^\mu_{\nu \lambda}$ equal to zero, but it will only be zero at that one point.

It would help, perhaps to look at a specific example: Let the metric be:

$g_{\theta \theta} = 1$
$g_{\phi \phi} = sin^2(\theta)$
(all other components are zero)

This is the metric for the surface of a sphere of radius 1.

So, at the point $\phi = 0, \theta = \frac{\pi}{2}$, the metric is diagonal:

$g_{\theta \theta} = 1, g_{\phi \phi} = 1$

and the connection coefficients are all zero at that point. Does that mean that a geodesic will be a "straight line"? No. The nonzero connection coefficients are:

$\Gamma^\theta_{\phi \phi} = - sin(\theta) cos(\theta)$
$\Gamma^\phi_{\theta \phi} = \Gamma^\phi_{\phi \theta} = cot(\theta)$

Note that the $\Gamma$ are only zero at $\theta = \frac{\pi}{2}$, not everywhere.

 

[VladZH]

The distinction between coordinate transformation and basis transformation is important here. If you have a curved spacetime, you can pick a coordinate system, and in terms of that coordinate system, you can pick a basis at each point in spacetime. In terms of the basis at a point, you have a corresponding connection coefficient, $\Gamma^\mu_{\nu \lambda}$. But in general, $\Gamma^\mu_{\nu \lambda}$ will not be a constant. In general, the components will change from point to point.

For any particular point, you can find a coordinate transformation that makes $\Gamma^\mu_{\nu \lambda}$ equal to zero, but it will only be zero at that one point.

It would help, perhaps to look at a specific example: Let the metric be:

$g_{\theta \theta} = 1$
$g_{\phi \phi} = sin^2(\theta)$
(all other components are zero)

This is the metric for the surface of a sphere of radius 1.

So, at the point $\phi = 0, \theta = \frac{\pi}{2}$, the metric is diagonal:

$g_{\theta \theta} = 1, g_{\phi \phi} = 1$

and the connection coefficients are all zero at that point. Does that mean that a geodesic will be a "straight line"? No. The nonzero connection coefficients are:

$\Gamma^\theta_{\phi \phi} = - sin(\theta) cos(\theta)$
$\Gamma^\phi_{\theta \phi} = \Gamma^\phi_{\phi \theta} = cot(\theta)$

Note that the $\Gamma$ are only zero at $\theta = \frac{\pi}{2}$, not everywhere.

Thanks for reply
I understand the idea that connection can be zero at one point but not in others. But how I can derive it from the equation I wrote? Since that transformation will turn metric components to diagonal in any point.
I just want to find an error in my steps

 

[Orodruin]

Since that transformation will turn metric components to diagonal in any point.

This is wrong. Any given coordinate transformation will generally fail in doing so apart from in a particular set of points. It will not hold in the entire manifold. For any given point, you can find a coordinate transformation that puts the metric on the given form, but this will generally be different coordinate transformations for different points.

 

[stevendaryl]

Thanks for reply
I understand the idea that connection can be zero at one point but not in others. But how I can derive it from the equation I wrote? Since that transformation will turn metric components to diagonal in any point.
I just want to find an error in my steps

What coordinate transformation can take $\theta, \phi$ to a new coordinate system where the metric components are diagonal at every point? There is no such transformation.

If you've ever taken logic, you'll know that quantifier ordering is important. There is a distinction between:

1. There exists a transformation such that forall points P, the metric is diagonal at P.
2. Forall points P, there exists a transformation such that the metric is diagonal at P.

The first is false. The second is true.

 

[stevendaryl]

Since that transformation will turn metric components to diagonal in any point.
I just want to find an error in my steps

Go through the example I gave you: $\theta, \phi$ with the metric:

$\left( \begin{array} \\ 1 & 0 \\ 0 & sin^2(\theta) \end{array} \right)$

What coordinate transformation is going to turn that into
$\left( \begin{array} \\ 1 & 0 \\ 0 & 1\end{array} \right)$?

 

[stevendaryl]

Go through the example I gave you: $\theta, \phi$ with the metric:

$\left( \begin{array} \\ 1 & 0 \\ 0 & sin^2(\theta) \end{array} \right)$

What coordinate transformation is going to turn that into
$\left( \begin{array} \\ 1 & 0 \\ 0 & 1\end{array} \right)$?

You can diagonalize the above matrix using the transformation:

$A^T g A$

with $A = \left( \begin{array} \\ 1 & 0 \\ 0 & -\frac{1}{sin(\theta)} \end{array} \right)$

But that doesn't make any sense for a coordinate transformation. Going back to definition of the components of $A$, this implies that your new coordinates, call them $x$ and $y$, are related to $\theta$ and $\phi$ by:

$\frac{\partial \theta}{\partial x} = 1$
$\frac{\partial \theta}{\partial y} = 0$
$\frac{\partial \phi}{\partial x} = 0$
$\frac{\partial \phi}{\partial y} = -\frac{1}{sin(\theta)}$

These equations have no solution.

 

[VladZH]

You can diagonalize the above matrix using the transformation:

$A^T g A$

with $A = \left( \begin{array} \\ 1 & 0 \\ 0 & -\frac{1}{sin(\theta)} \end{array} \right)$

But that doesn't make any sense for a coordinate transformation. Going back to definition of the components of $A$, this implies that your new coordinates, call them $x$ and $y$, are related to $\theta$ and $\phi$ by:

$\frac{\partial \theta}{\partial x} = 1$
$\frac{\partial \theta}{\partial y} = 0$
$\frac{\partial \phi}{\partial x} = 0$
$\frac{\partial \phi}{\partial y} = -\frac{1}{sin(\theta)}$

These equations have no solution.

So basically if solution did exist that would mean the metric is flat?

 

[Orodruin]

So basically if solution did exist that would mean the metric is flat?

Since such a metric would lead to the curvature tensor being zero, yes.