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What is the relationship between a metric tensor and a metric?
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I have seen both terms many times, but never understood how the two are related (if they are). Could you explain the relationship? Concrete examples, and not just general explanations, would be appreciated.
I have a basic understanding of metrics and tensors, but would enjoy a thorough explanation nonetheless.
Edit:
Wolfram alpha (http://mathworld.wolfram.com/MetricTensor.html) mentions that, for an orthogonal coordinate system with (I presume) coordinate variables
$$q_1, q_2, q_3$$
it is the case that the line element in three-space becomes
$$ds^2 = g_11dq_1^2 + g_22dq_2^2 + g_33dq_3^2 tag1$$
where the $g$s follow a relationship labeled (31) on the linked site.
What I want to know is:
A) Do (1) and (31) imply that a metric tensor is a tensor with scale factors as its components, and that the tensor contains data on how the metric differs from the ordinary Euclidean metric?
B) What would (1) look like in a coordinate system without orthogonality?
metric-spaces tensors
edited Jul 19 at 18:15
user7530
33.4k558109
asked Jul 19 at 15:43
Tensor McTensorstein
627
add a comment |Â
up vote
2
down vote
favorite
I have seen both terms many times, but never understood how the two are related (if they are). Could you explain the relationship? Concrete examples, and not just general explanations, would be appreciated.
I have a basic understanding of metrics and tensors, but would enjoy a thorough explanation nonetheless.
Edit:
Wolfram alpha (http://mathworld.wolfram.com/MetricTensor.html) mentions that, for an orthogonal coordinate system with (I presume) coordinate variables
$$q_1, q_2, q_3$$
it is the case that the line element in three-space becomes
$$ds^2 = g_11dq_1^2 + g_22dq_2^2 + g_33dq_3^2 tag1$$
where the $g$s follow a relationship labeled (31) on the linked site.
What I want to know is:
A) Do (1) and (31) imply that a metric tensor is a tensor with scale factors as its components, and that the tensor contains data on how the metric differs from the ordinary Euclidean metric?
B) What would (1) look like in a coordinate system without orthogonality?
metric-spaces tensors
edited Jul 19 at 18:15
user7530
33.4k558109
asked Jul 19 at 15:43
Tensor McTensorstein
627
1
You're asking for a lot, essentially a textbook chapter. You are much more likely to get useful information if you can focus your question more narrowly. For example, you could put some definitions into your question and then ask specific questions about points of those definitions.
– Lee Mosher
Jul 19 at 16:11
Okay. I will try to do that. Would you be able to recommend any textbook? Thanks for helping me with this question too!
– Tensor McTensorstein
Jul 19 at 16:16
@LeeMosher I have specified what I want to know.
– Tensor McTensorstein
Jul 19 at 16:49
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have seen both terms many times, but never understood how the two are related (if they are). Could you explain the relationship? Concrete examples, and not just general explanations, would be appreciated.
I have a basic understanding of metrics and tensors, but would enjoy a thorough explanation nonetheless.
Edit:
Wolfram alpha (http://mathworld.wolfram.com/MetricTensor.html) mentions that, for an orthogonal coordinate system with (I presume) coordinate variables
$$q_1, q_2, q_3$$
it is the case that the line element in three-space becomes
$$ds^2 = g_11dq_1^2 + g_22dq_2^2 + g_33dq_3^2 tag1$$
where the $g$s follow a relationship labeled (31) on the linked site.
What I want to know is:
A) Do (1) and (31) imply that a metric tensor is a tensor with scale factors as its components, and that the tensor contains data on how the metric differs from the ordinary Euclidean metric?
B) What would (1) look like in a coordinate system without orthogonality?
metric-spaces tensors
edited Jul 19 at 18:15
user7530
33.4k558109
asked Jul 19 at 15:43
Tensor McTensorstein
627
I have seen both terms many times, but never understood how the two are related (if they are). Could you explain the relationship? Concrete examples, and not just general explanations, would be appreciated.
I have a basic understanding of metrics and tensors, but would enjoy a thorough explanation nonetheless.
Edit:
Wolfram alpha (http://mathworld.wolfram.com/MetricTensor.html) mentions that, for an orthogonal coordinate system with (I presume) coordinate variables
$$q_1, q_2, q_3$$
it is the case that the line element in three-space becomes
$$ds^2 = g_11dq_1^2 + g_22dq_2^2 + g_33dq_3^2 tag1$$
where the $g$s follow a relationship labeled (31) on the linked site.
What I want to know is:
A) Do (1) and (31) imply that a metric tensor is a tensor with scale factors as its components, and that the tensor contains data on how the metric differs from the ordinary Euclidean metric?
B) What would (1) look like in a coordinate system without orthogonality?
metric-spaces tensors
edited Jul 19 at 18:15
user7530
33.4k558109
asked Jul 19 at 15:43
Tensor McTensorstein
627
edited Jul 19 at 18:15
user7530
33.4k558109
edited Jul 19 at 18:15
user7530
33.4k558109
edited Jul 19 at 18:15
user7530
33.4k558109
33.4k558109
asked Jul 19 at 15:43
Tensor McTensorstein
627
asked Jul 19 at 15:43
Tensor McTensorstein
627
asked Jul 19 at 15:43
Tensor McTensorstein
627
627
1
You're asking for a lot, essentially a textbook chapter. You are much more likely to get useful information if you can focus your question more narrowly. For example, you could put some definitions into your question and then ask specific questions about points of those definitions.
– Lee Mosher
Jul 19 at 16:11
Okay. I will try to do that. Would you be able to recommend any textbook? Thanks for helping me with this question too!
– Tensor McTensorstein
Jul 19 at 16:16
@LeeMosher I have specified what I want to know.
– Tensor McTensorstein
Jul 19 at 16:49
add a comment |Â
1
You're asking for a lot, essentially a textbook chapter. You are much more likely to get useful information if you can focus your question more narrowly. For example, you could put some definitions into your question and then ask specific questions about points of those definitions.
– Lee Mosher
Jul 19 at 16:11
Okay. I will try to do that. Would you be able to recommend any textbook? Thanks for helping me with this question too!
– Tensor McTensorstein
Jul 19 at 16:16
@LeeMosher I have specified what I want to know.
– Tensor McTensorstein
Jul 19 at 16:49
1
1
You're asking for a lot, essentially a textbook chapter. You are much more likely to get useful information if you can focus your question more narrowly. For example, you could put some definitions into your question and then ask specific questions about points of those definitions.
– Lee Mosher
Jul 19 at 16:11
You're asking for a lot, essentially a textbook chapter. You are much more likely to get useful information if you can focus your question more narrowly. For example, you could put some definitions into your question and then ask specific questions about points of those definitions.
– Lee Mosher
Jul 19 at 16:11
Okay. I will try to do that. Would you be able to recommend any textbook? Thanks for helping me with this question too!
– Tensor McTensorstein
Jul 19 at 16:16
Okay. I will try to do that. Would you be able to recommend any textbook? Thanks for helping me with this question too!
– Tensor McTensorstein
Jul 19 at 16:16
@LeeMosher I have specified what I want to know.
– Tensor McTensorstein
Jul 19 at 16:49
@LeeMosher I have specified what I want to know.
– Tensor McTensorstein
Jul 19 at 16:49
add a comment |Â
2 Answers
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active
oldest
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up vote
2
down vote
accepted
Suppose I have a surface---the surface of the Earth, say. It's annoying to do calculations directly on the surface of the Earth, since it's a 2D object sitting in 3D space, so that not all points $(x,y,z)$ correspond to valid points on the surface. For calculations it's easier to work on a chart of part of the Earth: for example I can represent the northern hemisphere of the Earth by a unit disk in the plane, together with a map
$$phi_1: D_1 subset mathbbR^2 to mathbbR^3 quad (x,y) mapsto left(Rx, Ry, Rsqrt1-x^2-y^2right)$$
(where I've idealized the Earth as a sphere of radius $R$).
I could draw a second chart, of a region near the equator, using spherical coordinates:
$$phi_2: D_2 subset mathbbR^2 to mathbbR^3, (theta,phi) mapsto left(Rsintheta cos phi, Rsinthetasinphi, Rcosthetaright).$$
The following concept is absolutely critical: the object I actually care about is the Earth. It exists independent of any charts I draw of parts of the Earth. But because it's more convenient to compute on a 2D parameterization of the Earth rather than on the Earth itself, I can create these maps $phi_1$ and $phi_2$. But there is nothing natural or canonical about these charts: just as in real-world charts of the Earth, I can draw many different charts in addition to these two, and they could all have different kinds of area distortions, angle distortions, etc.
If I draw a curve on $D_1$, I can't really tell just by looking at the chart $D_1$ how long the curve is on the sphere. In fact it's not even true that a straight line on $D_1$ corresponds to a "straight" curve (geodesic) on the Earth! If I want to compute anything geometric, like angles, or lengths, I need to (1) copy what I'm trying to measure from $D_1$ onto the actual Earth, using the $phi_1$ map; (2) measure things on the Earth; (3) copy the result back to $D_1$.
For example, let's say I draw a vector at some point $(x,y)$ on $D_1$. I want to calculate its length: not the length of the arrow I've draw on the piece of paper with the chart of the Earth, but the length of the corresponding tangent vector on the actual surface of the Earth. There is some function $L(mathbfv)$ which takes in tangent vectors on the Earth and returns their lengths: and this is a function that's intrinsic to the Earth, independent of any chart.
The metric is a way of encoding lengths: the metric $g$ is a bilinear function $g(mathbfv,mathbfw)$ associated to the tangent plane at every point on the Earth, which when given two copies of the same vector, computes the squared length of the vector,
$$g(mathbfv,mathbfv) = L(mathbfv)^2.$$
It turns out this property uniquely determines $g(mathbfv,mathbfw)$ even when $mathbfvneq mathbfw$, by the parallelogram law.
Again, this object exists on the Earth itself. It doesn't depend on which chart you use for navigating on the Earth. Of course, it would be nice to actually calculate lengths, given a vector $(v_x,v_y)$ at $(x,y)$ on the chart $D_1$. Fortunately, we can do this easily: the tangent vector on the surface of the Earth corresponding to the vector on the chart is given by
$$left[Jphi_1(x,y)right]beginbmatrixv_x\v_yendbmatrix = beginbmatrixR & 0 \ 0 & R\ -fracR xsqrt1-x^2-y^2 & -fracR ysqrt1-x^2-y^2endbmatrixbeginbmatrixv_x\v_yendbmatrix,$$
where $Jphi_1$ is the Jacobian of $phi_1$. It follows that the metric evaluated on two vector $mathbfv, mathbfw$ at a point $(x,y)$ on $D_1$ can be written in terms of matrix multiplication (as can all bilinear functions, of course):
$$g(mathbfv,mathbfw) = beginbmatrixv_x & v_yendbmatrix^Tbeginbmatrix fracR^2(y^2+1)1-x^2-y^2 & fracR^2xy1-x^2-y^2\ fracR^2xy1-x^2-y^2 & fracR^2(x^2-1)1-x^2-y^2endbmatrixbeginbmatrixw_x\w_yendbmatrix = beginbmatrixv_x & v_yendbmatrix^Tg_1 beginbmatrixw_x & w_yendbmatrix.$$
Observe that the left-hand side involves the intrinsic metric, and does not involve any charts. The right-hand side expresses the same thing in coordinates of the chart $D_1$, with the matrix $g_1$, which is a coordinate representation of $g$, the metric tensor.
Notice, incidentally, that $g_1$ depends on the position $(x,y)$. This makes perfect sense: the farther from the north pole, the more length distortion on your chart $D_1$.
Different charts will give you different expressions for the metric tensor. For $phi_2$ you will get
$$g(mathbfv,mathbfw) = beginbmatrixv_theta & v_phiendbmatrix^Tbeginbmatrix R^2 & 0\ 0 & R^2sin(theta)^2endbmatrixbeginbmatrixw_theta\w_phiendbmatrix = beginbmatrixv_theta & v_phiendbmatrix^Tg_2 beginbmatrixw_theta & w_phiendbmatrix.$$
As you note, some authors like to write
$$g_2 = R^2 dtheta^2 + 0 dtheta dphi + R^2sin(theta)^2 dphi^2$$
but don't be fooled: it's just different notation for the exact same concept: a representation of the metric in coordinates as a $2times 2$ bilinear operator. Some people just prefer quadratic form notation over matrix notation.
I hope that clears things up. In your question you also ask about orthogonal coordinate systems, but that's a red herring: all of the above applies regardless of whether the metric tensor turns out to be diagonal or not.
answered Jul 19 at 19:00
user7530
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In short:
A metric is "macroscopic" in that it gives a distance between points however far away they are, while a metric tensor is "microscopic" in that it only gives a distance between (infinitesimally) close points.
The metric tensor $g_ab$ defines a metric in a connected space,
$d(p_1, p_2) = inf_gamma int_gamma ds,$
where $ds = sqrtsum_a,b g_ab , dx^a , dx^b = sqrtsum_a,b g_ab dot x^a(t) , dot x^b(t) , dt$ for some parametrization $x(t) = (x^a(t))_a=1^n$ ($n$ being the dimension of the space), and the infimum is taken over all smooth enough paths between $p_1$ and $p_2$.
Like many tensors, the metric tensor is often given in some coordinate system. For example, $ds^2 = dx^2 + dy^2$ and $ds^2 = dr^2 + r^2 , dtheta^2$ are both representations of the same metric in different coordinate systems (Cartesian and polar, respectively).
In a non-orthogonal coordinate system there are non-diagonal terms, e.g.
$$ds^2 = underbracedr^2 + r^2 , dtheta^2_textdiagonal - underbrace2 r , dr , dtheta_textnon-diagonal$$
answered Jul 19 at 19:25
md2perpe
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Suppose I have a surface---the surface of the Earth, say. It's annoying to do calculations directly on the surface of the Earth, since it's a 2D object sitting in 3D space, so that not all points $(x,y,z)$ correspond to valid points on the surface. For calculations it's easier to work on a chart of part of the Earth: for example I can represent the northern hemisphere of the Earth by a unit disk in the plane, together with a map
$$phi_1: D_1 subset mathbbR^2 to mathbbR^3 quad (x,y) mapsto left(Rx, Ry, Rsqrt1-x^2-y^2right)$$
(where I've idealized the Earth as a sphere of radius $R$).
I could draw a second chart, of a region near the equator, using spherical coordinates:
$$phi_2: D_2 subset mathbbR^2 to mathbbR^3, (theta,phi) mapsto left(Rsintheta cos phi, Rsinthetasinphi, Rcosthetaright).$$
The following concept is absolutely critical: the object I actually care about is the Earth. It exists independent of any charts I draw of parts of the Earth. But because it's more convenient to compute on a 2D parameterization of the Earth rather than on the Earth itself, I can create these maps $phi_1$ and $phi_2$. But there is nothing natural or canonical about these charts: just as in real-world charts of the Earth, I can draw many different charts in addition to these two, and they could all have different kinds of area distortions, angle distortions, etc.
If I draw a curve on $D_1$, I can't really tell just by looking at the chart $D_1$ how long the curve is on the sphere. In fact it's not even true that a straight line on $D_1$ corresponds to a "straight" curve (geodesic) on the Earth! If I want to compute anything geometric, like angles, or lengths, I need to (1) copy what I'm trying to measure from $D_1$ onto the actual Earth, using the $phi_1$ map; (2) measure things on the Earth; (3) copy the result back to $D_1$.
For example, let's say I draw a vector at some point $(x,y)$ on $D_1$. I want to calculate its length: not the length of the arrow I've draw on the piece of paper with the chart of the Earth, but the length of the corresponding tangent vector on the actual surface of the Earth. There is some function $L(mathbfv)$ which takes in tangent vectors on the Earth and returns their lengths: and this is a function that's intrinsic to the Earth, independent of any chart.
The metric is a way of encoding lengths: the metric $g$ is a bilinear function $g(mathbfv,mathbfw)$ associated to the tangent plane at every point on the Earth, which when given two copies of the same vector, computes the squared length of the vector,
$$g(mathbfv,mathbfv) = L(mathbfv)^2.$$
It turns out this property uniquely determines $g(mathbfv,mathbfw)$ even when $mathbfvneq mathbfw$, by the parallelogram law.
Again, this object exists on the Earth itself. It doesn't depend on which chart you use for navigating on the Earth. Of course, it would be nice to actually calculate lengths, given a vector $(v_x,v_y)$ at $(x,y)$ on the chart $D_1$. Fortunately, we can do this easily: the tangent vector on the surface of the Earth corresponding to the vector on the chart is given by
$$left[Jphi_1(x,y)right]beginbmatrixv_x\v_yendbmatrix = beginbmatrixR & 0 \ 0 & R\ -fracR xsqrt1-x^2-y^2 & -fracR ysqrt1-x^2-y^2endbmatrixbeginbmatrixv_x\v_yendbmatrix,$$
where $Jphi_1$ is the Jacobian of $phi_1$. It follows that the metric evaluated on two vector $mathbfv, mathbfw$ at a point $(x,y)$ on $D_1$ can be written in terms of matrix multiplication (as can all bilinear functions, of course):
$$g(mathbfv,mathbfw) = beginbmatrixv_x & v_yendbmatrix^Tbeginbmatrix fracR^2(y^2+1)1-x^2-y^2 & fracR^2xy1-x^2-y^2\ fracR^2xy1-x^2-y^2 & fracR^2(x^2-1)1-x^2-y^2endbmatrixbeginbmatrixw_x\w_yendbmatrix = beginbmatrixv_x & v_yendbmatrix^Tg_1 beginbmatrixw_x & w_yendbmatrix.$$
Observe that the left-hand side involves the intrinsic metric, and does not involve any charts. The right-hand side expresses the same thing in coordinates of the chart $D_1$, with the matrix $g_1$, which is a coordinate representation of $g$, the metric tensor.
Notice, incidentally, that $g_1$ depends on the position $(x,y)$. This makes perfect sense: the farther from the north pole, the more length distortion on your chart $D_1$.
Different charts will give you different expressions for the metric tensor. For $phi_2$ you will get
$$g(mathbfv,mathbfw) = beginbmatrixv_theta & v_phiendbmatrix^Tbeginbmatrix R^2 & 0\ 0 & R^2sin(theta)^2endbmatrixbeginbmatrixw_theta\w_phiendbmatrix = beginbmatrixv_theta & v_phiendbmatrix^Tg_2 beginbmatrixw_theta & w_phiendbmatrix.$$
As you note, some authors like to write
$$g_2 = R^2 dtheta^2 + 0 dtheta dphi + R^2sin(theta)^2 dphi^2$$
but don't be fooled: it's just different notation for the exact same concept: a representation of the metric in coordinates as a $2times 2$ bilinear operator. Some people just prefer quadratic form notation over matrix notation.
I hope that clears things up. In your question you also ask about orthogonal coordinate systems, but that's a red herring: all of the above applies regardless of whether the metric tensor turns out to be diagonal or not.
answered Jul 19 at 19:00
user7530
33.4k558109
add a comment |Â
up vote
2
down vote
accepted
Suppose I have a surface---the surface of the Earth, say. It's annoying to do calculations directly on the surface of the Earth, since it's a 2D object sitting in 3D space, so that not all points $(x,y,z)$ correspond to valid points on the surface. For calculations it's easier to work on a chart of part of the Earth: for example I can represent the northern hemisphere of the Earth by a unit disk in the plane, together with a map
$$phi_1: D_1 subset mathbbR^2 to mathbbR^3 quad (x,y) mapsto left(Rx, Ry, Rsqrt1-x^2-y^2right)$$
(where I've idealized the Earth as a sphere of radius $R$).
I could draw a second chart, of a region near the equator, using spherical coordinates:
$$phi_2: D_2 subset mathbbR^2 to mathbbR^3, (theta,phi) mapsto left(Rsintheta cos phi, Rsinthetasinphi, Rcosthetaright).$$
The following concept is absolutely critical: the object I actually care about is the Earth. It exists independent of any charts I draw of parts of the Earth. But because it's more convenient to compute on a 2D parameterization of the Earth rather than on the Earth itself, I can create these maps $phi_1$ and $phi_2$. But there is nothing natural or canonical about these charts: just as in real-world charts of the Earth, I can draw many different charts in addition to these two, and they could all have different kinds of area distortions, angle distortions, etc.
If I draw a curve on $D_1$, I can't really tell just by looking at the chart $D_1$ how long the curve is on the sphere. In fact it's not even true that a straight line on $D_1$ corresponds to a "straight" curve (geodesic) on the Earth! If I want to compute anything geometric, like angles, or lengths, I need to (1) copy what I'm trying to measure from $D_1$ onto the actual Earth, using the $phi_1$ map; (2) measure things on the Earth; (3) copy the result back to $D_1$.
For example, let's say I draw a vector at some point $(x,y)$ on $D_1$. I want to calculate its length: not the length of the arrow I've draw on the piece of paper with the chart of the Earth, but the length of the corresponding tangent vector on the actual surface of the Earth. There is some function $L(mathbfv)$ which takes in tangent vectors on the Earth and returns their lengths: and this is a function that's intrinsic to the Earth, independent of any chart.
The metric is a way of encoding lengths: the metric $g$ is a bilinear function $g(mathbfv,mathbfw)$ associated to the tangent plane at every point on the Earth, which when given two copies of the same vector, computes the squared length of the vector,
$$g(mathbfv,mathbfv) = L(mathbfv)^2.$$
It turns out this property uniquely determines $g(mathbfv,mathbfw)$ even when $mathbfvneq mathbfw$, by the parallelogram law.
Again, this object exists on the Earth itself. It doesn't depend on which chart you use for navigating on the Earth. Of course, it would be nice to actually calculate lengths, given a vector $(v_x,v_y)$ at $(x,y)$ on the chart $D_1$. Fortunately, we can do this easily: the tangent vector on the surface of the Earth corresponding to the vector on the chart is given by
$$left[Jphi_1(x,y)right]beginbmatrixv_x\v_yendbmatrix = beginbmatrixR & 0 \ 0 & R\ -fracR xsqrt1-x^2-y^2 & -fracR ysqrt1-x^2-y^2endbmatrixbeginbmatrixv_x\v_yendbmatrix,$$
where $Jphi_1$ is the Jacobian of $phi_1$. It follows that the metric evaluated on two vector $mathbfv, mathbfw$ at a point $(x,y)$ on $D_1$ can be written in terms of matrix multiplication (as can all bilinear functions, of course):
$$g(mathbfv,mathbfw) = beginbmatrixv_x & v_yendbmatrix^Tbeginbmatrix fracR^2(y^2+1)1-x^2-y^2 & fracR^2xy1-x^2-y^2\ fracR^2xy1-x^2-y^2 & fracR^2(x^2-1)1-x^2-y^2endbmatrixbeginbmatrixw_x\w_yendbmatrix = beginbmatrixv_x & v_yendbmatrix^Tg_1 beginbmatrixw_x & w_yendbmatrix.$$
Observe that the left-hand side involves the intrinsic metric, and does not involve any charts. The right-hand side expresses the same thing in coordinates of the chart $D_1$, with the matrix $g_1$, which is a coordinate representation of $g$, the metric tensor.
Notice, incidentally, that $g_1$ depends on the position $(x,y)$. This makes perfect sense: the farther from the north pole, the more length distortion on your chart $D_1$.
Different charts will give you different expressions for the metric tensor. For $phi_2$ you will get
$$g(mathbfv,mathbfw) = beginbmatrixv_theta & v_phiendbmatrix^Tbeginbmatrix R^2 & 0\ 0 & R^2sin(theta)^2endbmatrixbeginbmatrixw_theta\w_phiendbmatrix = beginbmatrixv_theta & v_phiendbmatrix^Tg_2 beginbmatrixw_theta & w_phiendbmatrix.$$
As you note, some authors like to write
$$g_2 = R^2 dtheta^2 + 0 dtheta dphi + R^2sin(theta)^2 dphi^2$$
but don't be fooled: it's just different notation for the exact same concept: a representation of the metric in coordinates as a $2times 2$ bilinear operator. Some people just prefer quadratic form notation over matrix notation.
I hope that clears things up. In your question you also ask about orthogonal coordinate systems, but that's a red herring: all of the above applies regardless of whether the metric tensor turns out to be diagonal or not.
answered Jul 19 at 19:00
user7530
33.4k558109
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Suppose I have a surface---the surface of the Earth, say. It's annoying to do calculations directly on the surface of the Earth, since it's a 2D object sitting in 3D space, so that not all points $(x,y,z)$ correspond to valid points on the surface. For calculations it's easier to work on a chart of part of the Earth: for example I can represent the northern hemisphere of the Earth by a unit disk in the plane, together with a map
$$phi_1: D_1 subset mathbbR^2 to mathbbR^3 quad (x,y) mapsto left(Rx, Ry, Rsqrt1-x^2-y^2right)$$
(where I've idealized the Earth as a sphere of radius $R$).
I could draw a second chart, of a region near the equator, using spherical coordinates:
$$phi_2: D_2 subset mathbbR^2 to mathbbR^3, (theta,phi) mapsto left(Rsintheta cos phi, Rsinthetasinphi, Rcosthetaright).$$
The following concept is absolutely critical: the object I actually care about is the Earth. It exists independent of any charts I draw of parts of the Earth. But because it's more convenient to compute on a 2D parameterization of the Earth rather than on the Earth itself, I can create these maps $phi_1$ and $phi_2$. But there is nothing natural or canonical about these charts: just as in real-world charts of the Earth, I can draw many different charts in addition to these two, and they could all have different kinds of area distortions, angle distortions, etc.
If I draw a curve on $D_1$, I can't really tell just by looking at the chart $D_1$ how long the curve is on the sphere. In fact it's not even true that a straight line on $D_1$ corresponds to a "straight" curve (geodesic) on the Earth! If I want to compute anything geometric, like angles, or lengths, I need to (1) copy what I'm trying to measure from $D_1$ onto the actual Earth, using the $phi_1$ map; (2) measure things on the Earth; (3) copy the result back to $D_1$.
For example, let's say I draw a vector at some point $(x,y)$ on $D_1$. I want to calculate its length: not the length of the arrow I've draw on the piece of paper with the chart of the Earth, but the length of the corresponding tangent vector on the actual surface of the Earth. There is some function $L(mathbfv)$ which takes in tangent vectors on the Earth and returns their lengths: and this is a function that's intrinsic to the Earth, independent of any chart.
The metric is a way of encoding lengths: the metric $g$ is a bilinear function $g(mathbfv,mathbfw)$ associated to the tangent plane at every point on the Earth, which when given two copies of the same vector, computes the squared length of the vector,
$$g(mathbfv,mathbfv) = L(mathbfv)^2.$$
It turns out this property uniquely determines $g(mathbfv,mathbfw)$ even when $mathbfvneq mathbfw$, by the parallelogram law.
Again, this object exists on the Earth itself. It doesn't depend on which chart you use for navigating on the Earth. Of course, it would be nice to actually calculate lengths, given a vector $(v_x,v_y)$ at $(x,y)$ on the chart $D_1$. Fortunately, we can do this easily: the tangent vector on the surface of the Earth corresponding to the vector on the chart is given by
$$left[Jphi_1(x,y)right]beginbmatrixv_x\v_yendbmatrix = beginbmatrixR & 0 \ 0 & R\ -fracR xsqrt1-x^2-y^2 & -fracR ysqrt1-x^2-y^2endbmatrixbeginbmatrixv_x\v_yendbmatrix,$$
where $Jphi_1$ is the Jacobian of $phi_1$. It follows that the metric evaluated on two vector $mathbfv, mathbfw$ at a point $(x,y)$ on $D_1$ can be written in terms of matrix multiplication (as can all bilinear functions, of course):
$$g(mathbfv,mathbfw) = beginbmatrixv_x & v_yendbmatrix^Tbeginbmatrix fracR^2(y^2+1)1-x^2-y^2 & fracR^2xy1-x^2-y^2\ fracR^2xy1-x^2-y^2 & fracR^2(x^2-1)1-x^2-y^2endbmatrixbeginbmatrixw_x\w_yendbmatrix = beginbmatrixv_x & v_yendbmatrix^Tg_1 beginbmatrixw_x & w_yendbmatrix.$$
Observe that the left-hand side involves the intrinsic metric, and does not involve any charts. The right-hand side expresses the same thing in coordinates of the chart $D_1$, with the matrix $g_1$, which is a coordinate representation of $g$, the metric tensor.
Notice, incidentally, that $g_1$ depends on the position $(x,y)$. This makes perfect sense: the farther from the north pole, the more length distortion on your chart $D_1$.
Different charts will give you different expressions for the metric tensor. For $phi_2$ you will get
$$g(mathbfv,mathbfw) = beginbmatrixv_theta & v_phiendbmatrix^Tbeginbmatrix R^2 & 0\ 0 & R^2sin(theta)^2endbmatrixbeginbmatrixw_theta\w_phiendbmatrix = beginbmatrixv_theta & v_phiendbmatrix^Tg_2 beginbmatrixw_theta & w_phiendbmatrix.$$
As you note, some authors like to write
$$g_2 = R^2 dtheta^2 + 0 dtheta dphi + R^2sin(theta)^2 dphi^2$$
but don't be fooled: it's just different notation for the exact same concept: a representation of the metric in coordinates as a $2times 2$ bilinear operator. Some people just prefer quadratic form notation over matrix notation.
I hope that clears things up. In your question you also ask about orthogonal coordinate systems, but that's a red herring: all of the above applies regardless of whether the metric tensor turns out to be diagonal or not.
answered Jul 19 at 19:00
user7530
33.4k558109
Suppose I have a surface---the surface of the Earth, say. It's annoying to do calculations directly on the surface of the Earth, since it's a 2D object sitting in 3D space, so that not all points $(x,y,z)$ correspond to valid points on the surface. For calculations it's easier to work on a chart of part of the Earth: for example I can represent the northern hemisphere of the Earth by a unit disk in the plane, together with a map
$$phi_1: D_1 subset mathbbR^2 to mathbbR^3 quad (x,y) mapsto left(Rx, Ry, Rsqrt1-x^2-y^2right)$$
(where I've idealized the Earth as a sphere of radius $R$).
I could draw a second chart, of a region near the equator, using spherical coordinates:
$$phi_2: D_2 subset mathbbR^2 to mathbbR^3, (theta,phi) mapsto left(Rsintheta cos phi, Rsinthetasinphi, Rcosthetaright).$$
The following concept is absolutely critical: the object I actually care about is the Earth. It exists independent of any charts I draw of parts of the Earth. But because it's more convenient to compute on a 2D parameterization of the Earth rather than on the Earth itself, I can create these maps $phi_1$ and $phi_2$. But there is nothing natural or canonical about these charts: just as in real-world charts of the Earth, I can draw many different charts in addition to these two, and they could all have different kinds of area distortions, angle distortions, etc.
If I draw a curve on $D_1$, I can't really tell just by looking at the chart $D_1$ how long the curve is on the sphere. In fact it's not even true that a straight line on $D_1$ corresponds to a "straight" curve (geodesic) on the Earth! If I want to compute anything geometric, like angles, or lengths, I need to (1) copy what I'm trying to measure from $D_1$ onto the actual Earth, using the $phi_1$ map; (2) measure things on the Earth; (3) copy the result back to $D_1$.
For example, let's say I draw a vector at some point $(x,y)$ on $D_1$. I want to calculate its length: not the length of the arrow I've draw on the piece of paper with the chart of the Earth, but the length of the corresponding tangent vector on the actual surface of the Earth. There is some function $L(mathbfv)$ which takes in tangent vectors on the Earth and returns their lengths: and this is a function that's intrinsic to the Earth, independent of any chart.
The metric is a way of encoding lengths: the metric $g$ is a bilinear function $g(mathbfv,mathbfw)$ associated to the tangent plane at every point on the Earth, which when given two copies of the same vector, computes the squared length of the vector,
$$g(mathbfv,mathbfv) = L(mathbfv)^2.$$
It turns out this property uniquely determines $g(mathbfv,mathbfw)$ even when $mathbfvneq mathbfw$, by the parallelogram law.
Again, this object exists on the Earth itself. It doesn't depend on which chart you use for navigating on the Earth. Of course, it would be nice to actually calculate lengths, given a vector $(v_x,v_y)$ at $(x,y)$ on the chart $D_1$. Fortunately, we can do this easily: the tangent vector on the surface of the Earth corresponding to the vector on the chart is given by
$$left[Jphi_1(x,y)right]beginbmatrixv_x\v_yendbmatrix = beginbmatrixR & 0 \ 0 & R\ -fracR xsqrt1-x^2-y^2 & -fracR ysqrt1-x^2-y^2endbmatrixbeginbmatrixv_x\v_yendbmatrix,$$
where $Jphi_1$ is the Jacobian of $phi_1$. It follows that the metric evaluated on two vector $mathbfv, mathbfw$ at a point $(x,y)$ on $D_1$ can be written in terms of matrix multiplication (as can all bilinear functions, of course):
$$g(mathbfv,mathbfw) = beginbmatrixv_x & v_yendbmatrix^Tbeginbmatrix fracR^2(y^2+1)1-x^2-y^2 & fracR^2xy1-x^2-y^2\ fracR^2xy1-x^2-y^2 & fracR^2(x^2-1)1-x^2-y^2endbmatrixbeginbmatrixw_x\w_yendbmatrix = beginbmatrixv_x & v_yendbmatrix^Tg_1 beginbmatrixw_x & w_yendbmatrix.$$
Observe that the left-hand side involves the intrinsic metric, and does not involve any charts. The right-hand side expresses the same thing in coordinates of the chart $D_1$, with the matrix $g_1$, which is a coordinate representation of $g$, the metric tensor.
Notice, incidentally, that $g_1$ depends on the position $(x,y)$. This makes perfect sense: the farther from the north pole, the more length distortion on your chart $D_1$.
Different charts will give you different expressions for the metric tensor. For $phi_2$ you will get
$$g(mathbfv,mathbfw) = beginbmatrixv_theta & v_phiendbmatrix^Tbeginbmatrix R^2 & 0\ 0 & R^2sin(theta)^2endbmatrixbeginbmatrixw_theta\w_phiendbmatrix = beginbmatrixv_theta & v_phiendbmatrix^Tg_2 beginbmatrixw_theta & w_phiendbmatrix.$$
As you note, some authors like to write
$$g_2 = R^2 dtheta^2 + 0 dtheta dphi + R^2sin(theta)^2 dphi^2$$
but don't be fooled: it's just different notation for the exact same concept: a representation of the metric in coordinates as a $2times 2$ bilinear operator. Some people just prefer quadratic form notation over matrix notation.
I hope that clears things up. In your question you also ask about orthogonal coordinate systems, but that's a red herring: all of the above applies regardless of whether the metric tensor turns out to be diagonal or not.
answered Jul 19 at 19:00
user7530
33.4k558109
answered Jul 19 at 19:00
user7530
33.4k558109
answered Jul 19 at 19:00
user7530
33.4k558109
answered Jul 19 at 19:00
user7530
33.4k558109
33.4k558109
add a comment |Â
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up vote
1
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In short:
A metric is "macroscopic" in that it gives a distance between points however far away they are, while a metric tensor is "microscopic" in that it only gives a distance between (infinitesimally) close points.
The metric tensor $g_ab$ defines a metric in a connected space,
$d(p_1, p_2) = inf_gamma int_gamma ds,$
where $ds = sqrtsum_a,b g_ab , dx^a , dx^b = sqrtsum_a,b g_ab dot x^a(t) , dot x^b(t) , dt$ for some parametrization $x(t) = (x^a(t))_a=1^n$ ($n$ being the dimension of the space), and the infimum is taken over all smooth enough paths between $p_1$ and $p_2$.
Like many tensors, the metric tensor is often given in some coordinate system. For example, $ds^2 = dx^2 + dy^2$ and $ds^2 = dr^2 + r^2 , dtheta^2$ are both representations of the same metric in different coordinate systems (Cartesian and polar, respectively).
In a non-orthogonal coordinate system there are non-diagonal terms, e.g.
$$ds^2 = underbracedr^2 + r^2 , dtheta^2_textdiagonal - underbrace2 r , dr , dtheta_textnon-diagonal$$
answered Jul 19 at 19:25
md2perpe
5,93011022
add a comment |Â
up vote
1
down vote
In short:
A metric is "macroscopic" in that it gives a distance between points however far away they are, while a metric tensor is "microscopic" in that it only gives a distance between (infinitesimally) close points.
The metric tensor $g_ab$ defines a metric in a connected space,
$d(p_1, p_2) = inf_gamma int_gamma ds,$
where $ds = sqrtsum_a,b g_ab , dx^a , dx^b = sqrtsum_a,b g_ab dot x^a(t) , dot x^b(t) , dt$ for some parametrization $x(t) = (x^a(t))_a=1^n$ ($n$ being the dimension of the space), and the infimum is taken over all smooth enough paths between $p_1$ and $p_2$.
Like many tensors, the metric tensor is often given in some coordinate system. For example, $ds^2 = dx^2 + dy^2$ and $ds^2 = dr^2 + r^2 , dtheta^2$ are both representations of the same metric in different coordinate systems (Cartesian and polar, respectively).
In a non-orthogonal coordinate system there are non-diagonal terms, e.g.
$$ds^2 = underbracedr^2 + r^2 , dtheta^2_textdiagonal - underbrace2 r , dr , dtheta_textnon-diagonal$$
answered Jul 19 at 19:25
md2perpe
5,93011022
add a comment |Â
up vote
1
down vote
up vote
1
down vote
In short:
A metric is "macroscopic" in that it gives a distance between points however far away they are, while a metric tensor is "microscopic" in that it only gives a distance between (infinitesimally) close points.
The metric tensor $g_ab$ defines a metric in a connected space,
$d(p_1, p_2) = inf_gamma int_gamma ds,$
where $ds = sqrtsum_a,b g_ab , dx^a , dx^b = sqrtsum_a,b g_ab dot x^a(t) , dot x^b(t) , dt$ for some parametrization $x(t) = (x^a(t))_a=1^n$ ($n$ being the dimension of the space), and the infimum is taken over all smooth enough paths between $p_1$ and $p_2$.
Like many tensors, the metric tensor is often given in some coordinate system. For example, $ds^2 = dx^2 + dy^2$ and $ds^2 = dr^2 + r^2 , dtheta^2$ are both representations of the same metric in different coordinate systems (Cartesian and polar, respectively).
In a non-orthogonal coordinate system there are non-diagonal terms, e.g.
$$ds^2 = underbracedr^2 + r^2 , dtheta^2_textdiagonal - underbrace2 r , dr , dtheta_textnon-diagonal$$
answered Jul 19 at 19:25
md2perpe
5,93011022
In short:
A metric is "macroscopic" in that it gives a distance between points however far away they are, while a metric tensor is "microscopic" in that it only gives a distance between (infinitesimally) close points.
The metric tensor $g_ab$ defines a metric in a connected space,
$d(p_1, p_2) = inf_gamma int_gamma ds,$
where $ds = sqrtsum_a,b g_ab , dx^a , dx^b = sqrtsum_a,b g_ab dot x^a(t) , dot x^b(t) , dt$ for some parametrization $x(t) = (x^a(t))_a=1^n$ ($n$ being the dimension of the space), and the infimum is taken over all smooth enough paths between $p_1$ and $p_2$.
Like many tensors, the metric tensor is often given in some coordinate system. For example, $ds^2 = dx^2 + dy^2$ and $ds^2 = dr^2 + r^2 , dtheta^2$ are both representations of the same metric in different coordinate systems (Cartesian and polar, respectively).
In a non-orthogonal coordinate system there are non-diagonal terms, e.g.
$$ds^2 = underbracedr^2 + r^2 , dtheta^2_textdiagonal - underbrace2 r , dr , dtheta_textnon-diagonal$$
answered Jul 19 at 19:25
md2perpe
5,93011022
answered Jul 19 at 19:25
md2perpe
5,93011022
answered Jul 19 at 19:25
md2perpe
5,93011022
answered Jul 19 at 19:25
md2perpe
5,93011022
5,93011022
add a comment |Â
add a comment |Â
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1
You're asking for a lot, essentially a textbook chapter. You are much more likely to get useful information if you can focus your question more narrowly. For example, you could put some definitions into your question and then ask specific questions about points of those definitions.
– Lee Mosher
Jul 19 at 16:11
Okay. I will try to do that. Would you be able to recommend any textbook? Thanks for helping me with this question too!
– Tensor McTensorstein
Jul 19 at 16:16
@LeeMosher I have specified what I want to know.
– Tensor McTensorstein
Jul 19 at 16:49