# Two metric tensors describing same geometry

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• arpon
In summary, the process for checking if two coordinate systems describe the same geometry (in this case spherical geometry) involves constructing the curvature tensor from the metric tensors of the two coordinate systems, finding the transformation from the first to the second metric, and showing that this transformation changes the first curvature tensor into the second. This can be done by making sure that the transformation equations satisfy the tensor transformation rules, using different symbols for the two coordinate systems to avoid confusion. The number of equations and unknowns can be determined based on the number of dimensions, and there may be an infinite number of solutions depending on the specific equations involved. Complex numbers are not allowed as the metric tensors are real numbers.
arpon
Consider two coordinate systems on a sphere. The metric tensors of the two coordinate systems are given. Now how can I check that both coordinate systems describe the same geometry (in this case spherical geometry)?
(I used spherical geometry as an example. I would like to know the process in general.)

arpon said:
The metric tensors of the two coordinate systems are given.
You mean that the components of the metric tensors are given? The metric tensor itself is not coordinate system dependent.

In either case, the way of checking whether the metric tensors are the same is to make sure that they act in the same way on all combinations of vectors. If you want to stay with components, you have to make sure that the components satisfy the transformation rules for coordinate transformations on the sphere.

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arpon said:
Consider two coordinate systems on a sphere. The metric tensors of the two coordinate systems are given. Now how can I check that both coordinate systems describe the same geometry (in this case spherical geometry)?
(I used spherical geometry as an example. I would like to know the process in general.)

Construct the curvature tensor from the first and second metric. Find the transformation from the first to second metric. Show that this transformation changes the first curvature tensor into the second.

Is this method correct?

No. You have multiplied with ##g^{lm}## but you already have ##l## as a summation variable in the right-hand side of your expression. Then you are changing which is the ##l## that actually belongs to the sum and which is a free index.

Sorry for the mistake.
Orodruin said:
If you want to stay with components, you have to make sure that the components satisfy the transformation rules for coordinate transformations on the sphere.
Is it necessary to know the transformation matrix? The metric tensor defines the geometry. Isn't it sufficient just to know the components of the metric tensor?
Moreover, if I know the components of metric tensor in the two different coordinate systems, I can actually find the transformation matrix by solving this equation:
$$g_{ij}=\frac{\partial x'^k}{\partial x^i}\frac{\partial x'^l}{\partial x^j}g'_{kl}$$
##n^2## equations, ##n^2## unknowns, thus solvable.

arpon said:
Consider two coordinate systems on a sphere. The metric tensors of the two coordinate systems are given. Now how can I check that both coordinate systems describe the same geometry (in this case spherical geometry)?
(I used spherical geometry as an example. I would like to know the process in general.)

Any method I can think of involves finding the transformation equations between the two coordinate systems. Though it seems like there should be a better way. In special cases there certainly may be, but I'm not sure if there is in general a better way or not.

Once you have those, there are several ways to proceed, but your idea of showing that said transformation equations do in fact transform the first metric into the second via the tensor transformation rules seems to be the most direct, if carried out correctly.

I'd start out by writing out the transformation equations. Primed/unprimed notation gets a bit confusing, I think, it would probably be better to use different symbols for the two coordinate systems.

So you might use ##x^1, x^2, x^3## and ##\chi^1, \chi^2, \chi^3## for the two sets of coordinates, then in the 3 dimensional case you'll have three functions for the transformation, for instance you might write ##\chi^1## as a function of x^1, x^2, x^3. Or you could write x^1 as a function of ##\chi^1, \chi^2, \chi^3##. Similarly for the other components.

We will get an overdetermined set of equations. Making use of the fact that the metric tensor is symmetric, in 2 dimensions we'll have 3 constraint equations in two unknowns, in 3 dimensions 6 constraints in 3 unknowns, in 4 dimensions 10 constraints in 4 unknowns.

pervect said:
We will get an overdetermined set of equations. Making use of the fact that the metric tensor is symmetric, in 2 dimensions we'll have 3 constraint equations in two unknowns, in 3 dimensions 6 constraints in 3 unknowns, in 4 dimensions 10 constraints in 4 unknowns.
Aren't there ##N^2## unknowns as the transformation matrix ##\frac{\partial \chi^i}{\partial x^j}## has ##N^2## components in ##N##-dimension?

arpon said:
Aren't there ##N^2## unknowns as the transformation matrix ##\frac{\partial \chi^i}{\partial x^j}## has ##N^2## components in ##N##-dimension?

When you note that the metric tensor is symmetric, because ##g_{ij} = g_{ji}## there are fewer independent equations. For instance, ##g_{12}## is trivially equal to t ##g_{21}##. So there are in general N diagonal terms, which generate unique equations, and N^2 - N off-diagonal terms, which result in only half the number of unique equations, i.e. (N^2-N)/2 due to symmetry. So the total number of unique equations is ##N + (N^2-N)/2 = (N+N^2)/2##

pervect said:
When you note that the metric tensor is symmetric, because ##g_{ij} = g_{ji}## there are fewer independent equations. For instance, ##g_{12}## is trivially equal to t ##g_{21}##. So there are in general N diagonal terms, which generate unique equations, and N^2 - N off-diagonal terms, which result in only half the number of unique equations, i.e. (N^2-N)/2 due to symmetry. So the total number of unique equations is ##N + (N^2-N)/2 = (N+N^2)/2##
So, there are ##\frac{N+N^2}{2}## equations and ##N^2## unknowns (the ##N^2## components of ##\frac{\partial \chi^i}{\partial x^j}##). As ##N^2>\frac{N+N^2}{2}##, there will be infinite number of solutions.

arpon said:
So, there are ##\frac{N+N^2}{2}## equations and ##N^2## unknowns (the ##N^2## components of ##\frac{\partial \chi^i}{\partial x^j}##). As ##N^2>\frac{N+N^2}{2}##, there will be infinite number of solutions.

I don't think it as simple as that. Here is one equation ##x^2+y^2=0## in two unknowns, which doesn't have an infinite number of solutions even though ##2>1##.

martinbn said:
I don't think it as simple as that. Here is one equation ##x^2+y^2=0## in two unknowns, which doesn't have an infinite number of solutions even though ##2>1##.
Complex numbers are not allowed? Why?

arpon said:
Complex numbers are not allowed? Why?

Because ##g_{ij}## are real.

martinbn said:
Because ##g_{ij}## are real.
The components of transformation matrix can be complex number, can't they?
Here, I am looking for the transformation matrix between two coordinate systems while the components of metric tensor are given for the two coordinate systems.

There are a whole family of invariants (scalars) that can be constructed from the curvature tensor and the metric. Certainly a necessary (not sufficient) condition for same geometry is that all of these are identical. This test does not involve solving for any coordinate transforms. These can be computed directly in each case and must come out the same however different the coordinate components are.

Having passed this test, you might then solve for a coordinate transform. This is very hard, in general, and should not be possible if the metrics describe different geometry. In practice, it could be exceedingly difficult to distinguish between solution difficulty and non-existence of solution.

[ edit: oops, this only works in trivial cases, where the invariant is constant. Otherwise you need the transform to know which coordinate values correspond, to compare the invariants. ]

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If complex numbers are allowed, then there always exists a transformation matrix between two metric tensors. So I guess complex numbers are not allowed.
Could you explain me why?

arpon said:
If complex numbers are allowed, then there always exists a transformation matrix between two metric tensors. So I guess complex numbers are not allowed.
Could you explain me why?
A coordinate transform is supposed to be a set of 4 (for D=4) real (differentiable) functions of 4 real variables. The transform matrix is defined to be a matrix of partial derivatives of these. The latter must obviously be real under these definitions.

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arpon said:
So, there are ##\frac{N+N^2}{2}## equations and ##N^2## unknowns (the ##N^2## components of ##\frac{\partial \chi^i}{\partial x^j}##). As ##N^2>\frac{N+N^2}{2}##, there will be infinite number of solutions.
These are partial differential equations with 4 unknown functions. You have 10 equations for 4 unknown functions (in the 4 d case; in 2d, 3 equations for 2 unknown functions). For random metrics given as components, P of solvability is zero.

As I mentioned in an an earlier post, computing curvature invariants can help in simple cases. For example, in 2d, computing the Ricci scalar will distinguish between metrics in arbitrary coordinates on a sphere versus a saddle shape versus a plane or cylinder. It will even distinguish the case of a larger sphere from a smaller sphere.

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## 1. What are metric tensors?

Metric tensors are mathematical objects used to describe the geometry of a space. They are used in the field of differential geometry to measure distances and angles in curved spaces.

## 2. How do two metric tensors describe the same geometry?

Two metric tensors can describe the same geometry if they are related by a coordinate transformation. This means that the two tensors may have different components, but the underlying geometry they describe is the same.

## 3. Why would two metric tensors be used to describe the same geometry?

In some cases, it may be more convenient to use one metric tensor over another for certain calculations. Additionally, different coordinate systems may be used to describe the same geometry, resulting in different metric tensors.

## 4. Can two metric tensors describe different geometries?

Yes, two metric tensors can describe different geometries if they are not related by a coordinate transformation. This means that the underlying spaces they describe have different properties, such as curvature or dimensions.

## 5. Are there any real-world applications of two metric tensors describing the same geometry?

Yes, metric tensors are used in various fields such as physics and engineering to describe the curvature of spacetime and the geometry of objects. In these applications, different metric tensors may be used to describe the same physical system, depending on the chosen coordinates or reference frame.

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