Did I Get These Metric Tensors Right?

• The Floating Brain
In summary, the conversation discusses the process of finding the metric tensor for a function on a Riemannian manifold. The individual has been teaching themselves general relativity and has been trying to verify their results on metric tensors. They have also been following tutorials on non-Euclidean geometry. The conversation then delves into the meaning and purpose of finding the metric tensor for a function, with some confusion and clarification regarding the concept. The individual ultimately explains their process for finding the metric tensor at a given point on a surface.
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.

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.

DEvens and The Floating Brain
Orodruin said:
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.

The Floating Brain said:
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”?

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Orodruin said:
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.

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

Ibix said:
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.

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?

The Floating Brain said:
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.

The Floating Brain said:
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”.

Orodruin said:
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?

The Floating Brain said:
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##.

The Floating Brain said:
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.

haushofer said:
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.
Orodruin said:
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

Last edited:

1. What are metric tensors?

Metric tensors are mathematical objects used in the study of geometry and physics. They are used to measure distances and angles between points in a given space.

2. Why are metric tensors important?

Metric tensors are important because they allow us to define and understand the geometry of a space. They help us measure distances and angles, and are fundamental in many areas of mathematics and physics.

3. How do you calculate metric tensors?

The calculation of metric tensors depends on the specific space being studied. In general, they are calculated using a set of basis vectors and their corresponding dual basis vectors. These basis vectors define the geometry of the space and can be used to determine the metric tensor.

4. What is the difference between covariant and contravariant metric tensors?

Covariant and contravariant metric tensors are two different representations of the same mathematical object. Covariant metric tensors are used to measure distances along a given direction, while contravariant metric tensors are used to measure distances perpendicular to a given direction.

5. How are metric tensors used in physics?

Metric tensors are used extensively in physics, particularly in the theory of general relativity. They are used to describe the curvature of spacetime and the behavior of matter and energy in this curved space. They are also used in other areas of physics, such as electromagnetism and quantum field theory.

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