Did I Get These Metric Tensors Right?

[The Floating Brain]

Homework Statement

For a function f( x, y ) = x^2 + y^2 find the metric tensor at
(0, 0), (0, 1), (1, 0), (1, 1)

Relevant Equations

( DirectionalDerivitive( v, f( x, y ) ) )^2

I have been teaching myself general relativity and wanted to see if I got these metric tensors right, I have a feeling I didn't.For the first one I get all my directional derivatives
(0, 0): (0)i + (0)j
(0, 1): (0)i + 2j
(1, 0): 2i + (0)j
(1, 1): 2i + 2j

Then I square them (FOIL):
(0, 0): (0)i + (0)j + (0)ij + (0)ij
(0, 1): (0)i² + (0)ij + (0)ij+ 4j² = (0)i + (0)ij + (0)ij + 4j
(1, 0): 4i² + (0)ij + (0)j² = 4i + (0)ij + (0)ij + (0)j
(1, 1): 4i² + 4j² + 16(i²)(j²) + 16(i²)(j²) = 4i + 4j + 16ij + 16ij

Then I put the products into a matrix

(0, 0):
g = [ [ 0, 0 ]
[ 0, 0 ] ]

(0, 1):
g = [ [ 0, 0 ]
[ 0, 4 ] ]

(1, 0):
g = [ [ 4, 0 ]
[ 0, 0 ] ]

(1, 1):
g = [ [ 4, 16 ]
[ 16, 4 ] ]I tried searching for "metric tensor calculator" but couldn't find anything to verify my results.

I have been following these tutorials they are really good and exactly the format I want, I have also done a little looking into non - euclidean geometry.

 

[Orodruin]

Where did you get that problem statement and is it reproduced exactly as stated? The metric is a property of the Riemannian manifold, not of a function on the manifold.

 

[The Floating Brain]

Where did you get that problem statement and is it reproduced exactly as stated? The metric is a property of the Riemannian manifold, not of a function on the manifold.

Thanks for the reply.
I made my own problem.

 

[Orodruin]

Thanks for the reply.
I made my own problem.

Then, unfortunately, the problem itself does not make much sense. What do you mean by ”finding the metric tensor for a function”?

 

[haushofer]

Then, ungortunately, the problem itself does not make much sense. What do you mean by ”finding the metric tensor for a function”?

I guess the "function" here is the (space)interval,

$$(ds)^2 = (dx)^2 + (dy)^2$$

and he is integrating it to get the Cartesian notion of length.

But then the metric tensor is just the identity matrix, which doesn't make this exercise really insightful. Maybe I'm not understanding it either.

 

[Ibix]

Do you mean, what is the metric of the 2d surface $f(x,y)$ embedded in the usual Euclidean 3-space?

 

[Orodruin]

Do you mean, what is the metric of the 2d surface $f(x,y)$ embedded in the usual Euclidean 3-space?

$z = f(x,y)$? That is a different question.

 

[The Floating Brain]

Thank you all for your replies.

As far as I know the metric tensor helps define intrinsic coordinates at any point on a surface by getting the identity matrix of the surface at a certain point. The intent was do create a surface with f( x, y ) = x^2 + y^2 then find the metric tensor at points (0, 0), (0, 1), (1, 0), (1, 1).

I understand the process of finding the metric tensor on a surface at a given point to be

Step 1: Get the directional derivative of the surface at that point.
Step 2: Square the directional derivative
Step 3: Put the products into a matrix (most convenient method is to FOIL)

Does this help?

 

[Orodruin]

The intent was do create a surface with f( x, y ) = x^2 + y^2

It is not clear what you mean by this. Your f is just a function of x and y. If you intend to construct an embedded surface in 3d Euclidean space you need to say so.

Step 1: Get the directional derivative of the surface at that point.
Step 2: Square the directional derivative
Step 3: Put the products into a matrix (most convenient method is to FOIL)

Does this help?

This does not make sense. It is unclear what you mean by ”directionalderivative of the surface”.

 

[The Floating Brain]

It is not clear what you mean by this. Your f is just a function of x and y. If you intend to construct an embedded surface in 3d Euclidean space you need to say so.This does not make sense. It is unclear what you mean by ”directionalderivative of the surface”.

As far as I could tell, that's how you define a surface, I thought the context made the meaning of f clear, sorry about that.

As for your second question, I mean the directional derivative of f, the gradient at a specific point multiplied by a vector in a particular direction as a unit vector. I am not sure what is not clear is it the fact that you said that f is not clearly defined as the surface in a 3d Euclidean space and therefore it is unclear what the surface is?

 

[Orodruin]

As far as I could tell, that's how you define a surface, I thought the context made the meaning of f clear, sorry about that.

It is just one possibility of defining an embedded surface in $\mathbb R^3$.

As for your second question, I mean the directional derivative of f, the gradient at a specific point multiplied by a vector in a particular direction as a unit vector. I am not sure what is not clear is it the fact that you said that f is not clearly defined as the surface in a 3d Euclidean space and therefore it is unclear what the surface is?

There is no such thing as the ”gradient of a surface”. You take gradients of functions in $\mathbb R^3$, but your function is defined on two dimensions. If you instead define a surface as the level surface of a 3D function, eg, $g(x,y,z)=z - f(x,y)=0$, then the gradient of $g$ is normal to the surface, not tangential to it.

The way of finding the metric on the surface using coordinates x and y is to find out how the position vector changes when the coordinates change. It seems to me you have only considered how the z-component changes with x and y but this ignores the fact that x and y also change.

 

[The Floating Brain]

I guess the "function" here is the (space)interval,

$$(ds)^2 = (dx)^2 + (dy)^2$$

and he is integrating it to get the Cartesian notion of length.

But then the metric tensor is just the identity matrix, which doesn't make this exercise really insightful. Maybe I'm not understanding it either.

It is just one possibility of defining an embedded surface in $\mathbb R^3$.
There is no such thing as the ”gradient of a surface”. You take gradients of functions in $\mathbb R^3$, but your function is defined on two dimensions. If you instead define a surface as the level surface of a 3D function, eg, $g(x,y,z)=z - f(x,y)=0$, then the gradient of $g$ is normal to the surface, not tangential to it.

The way of finding the metric on the surface using coordinates x and y is to find out how the position vector changes when the coordinates change. It seems to me you have only considered how the z-component changes with x and y but this ignores the fact that x and y also change.

Sorry I appreciate your patience.
So should I take a directional directional derivatives of √(x² + z) and √(y² + z) at the given points as well?EDIT:

One thing I probably should have done was write z = f(x, y) = x² + y², I thought it would be implied from the context. However another thing I should clarify is I am trying to define a 2d surface in a 3d euclidean space like in the video I posted where the gradient is tangent to the surface.

EDIT 2:

I should have also been clear that when I am "FOIL"ing I am taking a dot product.

Sorry for all this, thank you for your patience