Thank you for your suggestion. Let me confirm I understood your explanation correctly.haushofer said: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 RT = R-1. As such the rotation on x can be identified as a transformation of the matrix M (like in quantum mechanics).
Yes.Keita said:##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 ##.
Thank you for your answer. I understood your suggestion correctly. Now, let me show you my argument.haushofer said:Yes.
You say that after the coordinate transformation, you getKeita said: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?
Sorry for the delay of my response. I appreciate your explanation. Still, I would like to stand to my argument.haushofer said: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.
I agree with your argument and your result ##M' = RMR^{-1}##.Keita said: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.
$$
M = R^{-1}M' R.
$$
$$
M' = R M R^{-1}.
$$
ergospherical said:@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.ergospherical said:@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 for your suggestion. I also appreciate your note on Zee’s equation(17) on page 72.TSny said: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##.