A question about Curved Spaces: Gauss and Riemann (Einstein Gravity in a Nutshell by Zee)

[Keita]

TL;DR Summary

A question about 1. 6. Curved Spaces: Gauss and Riemann

In p. 84, Zee says “In the new coordinates, M is replaced by M’ = R[-1]MR.” However, I figure out M is replaced by M’ = RMR[-1]. Why is M replaced by M’ = R[-1]MR?

 

[robphy]

Can you show your work?

In addition, you shouldn’t assume that your reference is easily accessible. Show more context (preferrably using MathJax).

 

[haushofer]

Indeed, how did you figure? Look at the expression for z on top of page 84, apply the rotation R to the vector x and use R^T = R^-1. As such the rotation on x can be identified as a transformation of the matrix M (like in quantum mechanics).

 

[Keita]

$R^T$

Indeed, how did you figure? Look at the expression for z on top of page 84, apply the rotation R to the vector x and use R^T = R^-1. As such the rotation on x can be identified as a transformation of the matrix M (like in quantum mechanics).

Thank you for your suggestion. Let me confirm I understood your explanation correctly.
Does your suggestion mean the following?

$$ z \sim \frac{1}{2} \vec{x}^{T} M\vec{x} $$
(Top of page 84)

$$
z \sim \frac{1}{2} \left(R\vec{x}\right) ^{T} M\left(R\vec{x}\right)
=
\frac{1}{2}\vec{x}^{T}R^{T}MR\vec{x}
=
\frac{1}{2}\vec{x}^{T}R^{-1}MR\vec{x}
$$
(Applying the rotation R to the vector x and using $R^{T} = R^{-1}$)

Therefore, $M' = R^{-1} M R$.

 

[haushofer]

$R^T$

Thank you for your suggestion. Let me confirm I understood your explanation correctly.
Does your suggestion mean the following?

$$ z \sim \frac{1}{2} \vec{x}^{T} M\vec{x} $$
(Top of page 84)

$$
z \sim \frac{1}{2} \left(R\vec{x}\right) ^{T} M\left(R\vec{x}\right)
=
\frac{1}{2}\vec{x}^{T}R^{T}MR\vec{x}
=
\frac{1}{2}\vec{x}^{T}R^{-1}MR\vec{x}
$$
(Applying the rotation R to the vector x and using $R^{T} = R^{-1}$)

Therefore, $M' = R^{-1} M R$.

Yes.

 

[Keita]

Yes.

Thank you for your answer. I understood your suggestion correctly. Now, let me show you my argument.

In the original coordinate, we have the following.

$$ z \sim \frac{1}{2} \vec {x}^{T} M\vec {x} (1)$$

(Top of page 84)

In the rotated coordinate, we have the following.

$$
z \sim \frac{1}{2} \vec{x'}^{T}M' \vec{x'}
=
\frac{1}{2} \left(R\vec{x}\right) ^{T} M'\left(R\vec{x}\right)
=
\frac{1}{2}\vec{x}^{T}R^{T}M'R\vec{x}
=
\frac{1}{2}\vec{x}^{T}R^{-1}M'R\vec{x}(2)
$$

(Applying the rotation R to the vector x and using $R^{T} = R^{-1}$)

From (1) and (2), we have the following.

$$
R^{-1}M'R=M (3)
$$

Therefore,

$$
M'=RMR^{-1}(4)
$$

What do you make of my argument?

 

[haushofer]

Thank you for your answer. I understood your suggestion correctly. Now, let me show you my argument.

In the original coordinate, we have the following.

$$ z \sim \frac{1}{2} \vec {x}^{T} M\vec {x} (1)$$

(Top of page 84)

In the rotated coordinate, we have the following.

$$
z \sim \frac{1}{2} \vec{x'}^{T}M' \vec{x'}
=
\frac{1}{2} \left(R\vec{x}\right) ^{T} M'\left(R\vec{x}\right)
=
\frac{1}{2}\vec{x}^{T}R^{T}M'R\vec{x}
=
\frac{1}{2}\vec{x}^{T}R^{-1}M'R\vec{x}(2)
$$

(Applying the rotation R to the vector x and using $R^{T} = R^{-1}$)

From (1) and (2), we have the following.

$$
R^{-1}M'R=M (3)
$$

Therefore,

$$
M'=RMR^{-1}(4)
$$

What do you make of my argument?

You say that after the coordinate transformation, you get

$$
z \sim \frac{1}{2} \vec{x'}^{T}M' \vec{x'}
$$

But why the prime on M? You apply the rotation to the coordinates, not to the matrix elements of M, right? So I'd say that afther the coordinate transformation,

$$
z \sim \frac{1}{2} \vec{x'}^{T}M \vec{x'}
$$

In other words: you should carefully think about on what the transformation is applied to.