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

[Keita]

TL;DR

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.

 

[Keita]

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.

Sorry for the delay of my response. I appreciate your explanation. Still, I would like to stand to my argument.

In the original x-y coordinate,
$$
z = \frac{1}{2}ax^2 + cxy + \frac{1}{2}by^2 = \frac{1}{2} \vec{x}^{T} M \vec{x}.
$$

$$
M = \begin{pmatrix}
a & c \\
c & b
\end{pmatrix}.
$$
(pages 83 and 84)

In the rotated u-v coordinate,
$$
z = \frac{1}{2}a'u^2 + c'uv + \frac{1}{2}b'v^2 = \frac{1}{2} \vec{x'}^{T} M' \vec{x'} = \frac{1}{2} (R\vec{x})^{T} M' (R\vec{x}) = \frac{1}{2} \vec{x}^{T}R^{T}M'R\vec{x} = \frac{1}{2} \vec{x}^{T}R^{-1}M'R\vec{x}.
$$

$$
\vec{x'} = \begin{pmatrix}
u \\
v
\end{pmatrix}.
$$

$$
M' = \begin{pmatrix}
a' & c' \\
c' & b'
\end{pmatrix}.
$$

$$
\vec{x'} = R \vec{x}.
$$
(page 84)

Since $z = z$, we have the followings.

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

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

I hope you would response to my argument.

 

[ergospherical]

@Keita what you have noticed is the distinction between what are sometimes called "active" vs "passive" transformations.

Passive: If you view the transformation as rotating the (x-y axes of) the coordinate system, but keeping the vectors themselves fixed, then the position vector and the matrix have new components in the new coordinate system and $z' = \tfrac{1}{2}(\mathbf{x}')^T M' \mathbf{x}' = \tfrac{1}{2} \mathbf{x}^T R^T M' R \mathbf{x}$, leading to $M' = R M R^T$ after setting $z' = z$.

Active: if you view the transformation as rotating the vector $\mathbf{x} \mapsto R\mathbf{x}$, but keeping the coordinate system fixed, then obviously $\mathbf{x}'$ has different components to before but the matrix $M$ is unchanged. Then $z = \tfrac{1}{2} (\mathbf{x}')^T M \mathbf{x}' = \tfrac{1}{2} \mathbf{x}^T R^T M R \mathbf{x}$. You could view this instead as a transformation of the operator itself as $M \mapsto M' = R^T M R$.

The two different versions of $M'$ are related by transpose (because as you should be able to see: rotating the coordinates $n$ degrees clockwise is the same as rotating the vectors $n$ degrees anti-clockwise. )

 

[TSny]

In the original x-y coordinate,
$$
z = \frac{1}{2}ax^2 + cxy + \frac{1}{2}by^2 = \frac{1}{2} \vec{x}^{T} M \vec{x}.
$$

$$
M = \begin{pmatrix}
a & c \\
c & b
\end{pmatrix}.
$$
(pages 83 and 84)

In the rotated u-v coordinate,
$$
z = \frac{1}{2}a'u^2 + c'uv + \frac{1}{2}b'v^2 = \frac{1}{2} \vec{x'}^{T} M' \vec{x'} = \frac{1}{2} (R\vec{x})^{T} M' (R\vec{x}) = \frac{1}{2} \vec{x}^{T}R^{T}M'R\vec{x} = \frac{1}{2} \vec{x}^{T}R^{-1}M'R\vec{x}.
$$

$$
\vec{x'} = \begin{pmatrix}
u \\
v
\end{pmatrix}.
$$

$$
M' = \begin{pmatrix}
a' & c' \\
c' & b'
\end{pmatrix}.
$$

$$
\vec{x'} = R \vec{x}.
$$
(page 84)

Since $z = z$, we have the followings.

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

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

I agree with your argument and your result $M' = RMR^{-1}$.

Zee uses the rotation matrix $R$ to induce a coordinate transformation $\mathbf{x}' = R \mathbf{x}$. So, this is an example of a passive transformation as described by @ergospherical. His $M' = RMR^T = RMR^{-1}$ agrees with your result.

Also, look at Zee's equation (17) on page 72 which shows how the matrix $g$ for the metric transforms under a general coordinate transformation $S$ : $$g'(x') = (S^{-1})^T g(x) S^{-1}.$$ For the special case where $S$ is a rotation $R$, this becomes $$g'(x') = (R^{-1})^T g(x) R^{-1} = Rg(x)R^{-1}.$$ This has the same form as you derived for the matrix $M$.

 

[Keita]

@Keita what you have noticed is the distinction between what are sometimes called "active" vs "passive" transformations.

Passive: If you view the transformation as rotating the (x-y axes of) the coordinate system, but keeping the vectors themselves fixed, then the position vector and the matrix have new components in the new coordinate system and $z' = \tfrac{1}{2}(\mathbf{x}')^T M' \mathbf{x}' = \tfrac{1}{2} \mathbf{x}^T R^T M' R \mathbf{x}$, leading to $M' = R M R^T$ after setting $z' = z$.

Active: if you view the transformation as rotating the vector $\mathbf{x} \mapsto R\mathbf{x}$, but keeping the coordinate system fixed, then obviously $\mathbf{x}'$ has different components to before but the matrix $M$ is unchanged. Then $z = \tfrac{1}{2} (\mathbf{x}')^T M \mathbf{x}' = \tfrac{1}{2} \mathbf{x}^T R^T M R \mathbf{x}$. You could view this instead as a transformation of the operator itself as $M \mapsto M' = R^T M R$.

The two different versions of $M'$ are related by transpose (because as you should be able to see: rotating the coordinates $n$ degrees clockwise is the same as rotating the vectors $n$ degrees anti-clockwise. )

@Keita what you have noticed is the distinction between what are sometimes called "active" vs "passive" transformations.

Passive: If you view the transformation as rotating the (x-y axes of) the coordinate system, but keeping the vectors themselves fixed, then the position vector and the matrix have new components in the new coordinate system and $z' = \tfrac{1}{2}(\mathbf{x}')^T M' \mathbf{x}' = \tfrac{1}{2} \mathbf{x}^T R^T M' R \mathbf{x}$, leading to $M' = R M R^T$ after setting $z' = z$.

Active: if you view the transformation as rotating the vector $\mathbf{x} \mapsto R\mathbf{x}$, but keeping the coordinate system fixed, then obviously $\mathbf{x}'$ has different components to before but the matrix $M$ is unchanged. Then $z = \tfrac{1}{2} (\mathbf{x}')^T M \mathbf{x}' = \tfrac{1}{2} \mathbf{x}^T R^T M R \mathbf{x}$. You could view this instead as a transformation of the operator itself as $M \mapsto M' = R^T M R$.

The two different versions of $M'$ are related by transpose (because as you should be able to see: rotating the coordinates $n$ degrees clockwise is the same as rotating the vectors $n$ degrees anti-clockwise. )

Thank you. I appreciate your suggestion.

 

[Keita]

I agree with your argument and your result $M' = RMR^{-1}$.

Zee uses the rotation matrix $R$ to induce a coordinate transformation $\mathbf{x}' = R \mathbf{x}$. So, this is an example of a passive transformation as described by @ergospherical. His $M' = RMR^T = RMR^{-1}$ agrees with your result.

Also, look at Zee's equation (17) on page 72 which shows how the matrix $g$ for the metric transforms under a general coordinate transformation $S$ : $$g'(x') = (S^{-1})^T g(x) S^{-1}.$$ For the special case where $S$ is a rotation $R$, this becomes $$g'(x') = (R^{-1})^T g(x) R^{-1} = Rg(x)R^{-1}.$$ This has the same form as you derived for the matrix $M$.

Thank you for your suggestion. I also appreciate your note on Zee’s equation(17) on page 72.