Metric Tensor in Spherical Coordinates

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space-time
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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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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)>
 
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.