Metric Tensor in Spherical Coordinates

In summary, the speaker derived a metric tensor in spherical polar coordinates in 3-D, with components g11 = sin2(ø) + cos2(θ), g12 = -rsin(θ)cos(θ), g13 = rsin(ø)cos(ø), g21 = -rsin(θ)cos(θ), g22 = r2sin2(ø) + r2sin2(θ), g23 = 0, g31 = rsin(ø)cos(ø), g32 = 0, and g33 = r2cos2(ø). They used a vector in their derivation and transformed it to spherical polar coordinates. They also differentiated the vector with respect to the
  • #1
space-time
218
4
I recently derived a matrix which I believe to be the metric tensor in spherical polar coordinates in 3-D. Here were the components of the tensor that I derived. I will show my work afterwards:

g11 = sin2(ø) + cos2(θ)

g12 = -rsin(θ)cos(θ)

g13 = rsin(ø)cos(ø)

g21 = -rsin(θ)cos(θ)

g22 = r2sin2(ø) + r2sin2(θ)

g23 = 0

g31 = rsin(ø)cos(ø)

g32 = 0

g33 = r2cos2(ø)

The above is what I derived, but when I tried to verify to see if my answer was correct by checking various websites, I did not see any site have what I derived.

Here is my work:

The axes were:

x1 = r

x2 = θ

x3 = ø

The vector that I differentiated was:

<rcos(θ)sin(ø) , rsin(θ)sin(ø) , rcos(θ)>

I then differentiated the vector with respect to the various axes in order to derive my tangential basis vectors.

Here were my basis vectors:

er = <cos(θ)sin(ø) , sin(θ)sin(ø), cos(θ)>

eθ = <-rsin(θ)sin(ø), rcos(θ)sin(ø) , -rsin(θ)>

eø = <rcos(θ)cos(ø), rsin(θ)cos(ø) , 0>

Finally, I did the dot product with these basis vectors to derive the components of my metric tensor.

That is how I got what I derived, but I don't see any confirmation of this online.

Can anyone please either verify if I am right with this metric tensor or tell me where I went wrong?
 
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  • #2
space-time said:
The vector that I differentiated was:

<rcos(θ)sin(ø) , rsin(θ)sin(ø) , rcos(θ)>

The transformation to spherical polar coordinates is

x = r sin(θ) cos(ø)
y = r sin(θ) sin(ø)
z = r cos(θ)

r = <r sin(θ) cos(ø), r sin(θ) sin(ø), r cos(θ)>
 
  • #3
Bill_K said:
The transformation to spherical polar coordinates is

x = r sin(θ) cos(ø)
y = r sin(θ) sin(ø)
z = r cos(θ)

r = <r sin(θ) cos(ø), r sin(θ) sin(ø), r cos(θ)>

So I made a careless mistake it seems. Thank you Bill K.
 

1. What is the metric tensor in spherical coordinates?

The metric tensor in spherical coordinates is a mathematical object that describes the relationship between distance and angles in three-dimensional space. It is used in the study of general relativity and other branches of physics to describe the curvature of spacetime.

2. How is the metric tensor used in spherical coordinates?

In spherical coordinates, the metric tensor is used to calculate the distance between two points in space, taking into account the curvature of spacetime. It is also used to calculate the components of vectors and tensors in spherical coordinates.

3. What are the components of the metric tensor in spherical coordinates?

The components of the metric tensor in spherical coordinates are given by gij = diag(1, r2, r2sin2θ), where r is the radial distance and θ is the polar angle.

4. How does the metric tensor change in different coordinate systems?

The metric tensor changes in different coordinate systems because the relationship between distance and angles may be different in each coordinate system. In spherical coordinates, the metric tensor takes into account the curvature of spacetime, while in Cartesian coordinates it does not.

5. What is the significance of the metric tensor in physics?

The metric tensor is of great significance in physics because it is used to describe the curvature of spacetime, which is a fundamental concept in general relativity and other theories of gravity. It is also used in other branches of physics, such as electromagnetism and quantum field theory, to describe the relationship between distance and angles in different coordinate systems.

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