Small oscillations and spatial transformations | Part 1

[JD_PM]

TL;DR Summary

I am studying the theory of small oscillation (section 8 in the PDF I attached) and I do not fully understand it. That's why I want to discuss it in several parts.

Please note that the transformed quantities will be indicated by $'$.

Let me give some context first.

Let us assume here that the general approximate form of the potential energy $V$ and the kinetic energy $T$ are given to be

$$V^{app} = q^T V q \tag 1$$

$$T^{app} = \dot q^T V \dot q \tag 2$$

We assume that energies are unaffected by spatial transformations. Such a condition yields the following equations

$$q^T V q = V^{app} =V'^{app} = q'^T V' q'\tag{3a}$$

$$\dot q^T T \dot q = T^{app} =T'^{app}= \dot q'^T T' \dot q'\tag{3b}$$

Let us work with the following transformation:

$$q' = Sq \tag4$$

Where $S$ is a squared orthogonal (i.e. $S^T=S^{-1}$) matrix.

Let's plug $(4)$ into $(3b)$ to get that

$$T=S^T T' S \iff T'=(S^T)^{-1} T S^{-1} \tag5$$

Let us focus on $T'$; As it equals to the diagonalizable condition (i.e. $T'= STS^{-1}$), we can always find another transformation matrix that makes $T'$ a diagonal matrix (i.e. $q' = S_1q$ that makes $T'=D$, where $D$ is an unknown diagonal matrix).

I understand everything so far.

Then the notes I am studying state that we can apply another transformation

$$q'' = S_2q'=S_1S_2q \tag6$$

Where they use $q' = S_1q$ instead of $(4)$

But then they state that '$S_2=\sqrt{D}$ and then $T''= I$'

- How can I prove that the above sentence is indeed true?

- Why did they use $q' = S_1q$ instead of $(4)$?

Any help is appreciated.

EDIT: note I asked this question https://math.stackexchange.com/questions/3741710/transformation-of-spatial-coordinates but got little attention. If you'd like to have more information please let me know.

 

[patfinder]

Hello,
before answering your question I would like to draw your attention on a sloppy statement of the PDF file you put in attachement. Right before eq. (40) it says that all square matrices are diagonalisable, which is obviously not true. Take $\begin{pmatrix} 5 & 1 \\ 0 & 5 \end{pmatrix}$ for example, but in your case it doesn't matter since your are dealing with real symetric matrices which are diagonalisable by the spectral theorem.

For the question with $S_{1}q$, this is just a labelling issue. The text use $S$ to show you general properties. Then to go more precise in the problem it is usual to give special label to the matrix you use, especially when there will be more than one. So in short $S_{1}$ is a matrix verifying the same properties as $S$ and $P$ in your particular case, and with an arbitrary label.

For the proof with $S_{1}$ and $S_{2}$, you have $T = S_{1}^{T}T'S_{1} =S_{1}^{T}D S_{1}$, then if you define $S_{2} = \sqrt{D}$ you have $S_{2}^{T} S_{2} = S_{2}^{T} I S_{2} = D$ so by replacing $D$ in the previous expression by this one you have $T = S_{1}D S_{1}= S_{1}^{T} S_{2}^{T} I S_{2} S_{1} = \left( S_{2} S_{1} \right) ^{T} T'' S_{2} S_{1}$ , which corresponds to the transformation $q'' = S_{2} S_{1}$ and gives $T'' = I$.

 

[JD_PM]

Hi patfinder, thanks for your reply .

I've been working out this question so let's share and discuss.

real symetric matrices which are diagonalisable by the spectral theorem.

Here I found the proof and others related to real symmetric matrices.

For the proof with $S_{1}$ and $S_{2}$, you have $T = S_{1}^{T}T'S_{1} =S_{1}^{T}D S_{1}$, then if you define $S_{2} = \sqrt{D}$ you have $S_{2}^{T} S_{2} = S_{2}^{T} I S_{2} = D$ so by replacing $D$ in the previous expression by this one you have $T = S_{1}D S_{1}= S_{1}^{T} S_{2}^{T} I S_{2} S_{1} = \left( S_{2} S_{1} \right) ^{T} T'' S_{2} S_{1}$ , which corresponds to the transformation $q'' = S_{2} S_{1}$ and gives $T'' = I$.

I think we have done basically the same but let's see.

If we apply the coordinate transformation $\vec q' = S_1 \vec q \$ to $T= \ \dot{\vec q'}^T T' \dot q '$ we get

$$T=(S_1 \dot{\vec q})^T T'(S_1 \dot{\vec q}) = \dot{\vec q}^T S_1^T T'S_1 \dot{\vec q} \tag1$$

Thus we have

$$T = S_1^T T'S_1 \iff T'=S_1^{-T} T S_1^{-1} \tag2$$

As $T'$ is a real symmetric matrix, it is diagonalizable. Thus we can assert that

$$T' = D \tag3$$

Next if we apply the coordinate transformation $\vec q'' = S_2 \vec q' \$ to $T'= \ \dot{\vec q''}^T T'' \dot q ''$ we get

$$T'=(S_2 \dot{\vec q'})^T T''(S_2 \dot{\vec q'}) = \dot{\vec q'}^T S_2^T T''S_2 \dot{\vec q'} \tag4$$

Thus we have

$$T' = S_2^T T''S_2 \iff T''=S_2^{-T} T' S_1^{-1} \tag5$$

If we set $S_2 = \sqrt{D}$ then, out of $(5)$ and using $(3)$, we get

$$T'' = \Big( \sqrt{D} \Big)^{-T} T'\Big( \sqrt{D} \Big)^{-1}=\Big( \sqrt{D} \Big)^{-T} D\Big( \sqrt{D} \Big)^{-1}=\Big( \sqrt{D} \Big)^{-1} D\Big( \sqrt{D} \Big)^{-1}=1 \tag6$$

Where I have used the fact that the inverse matrix of a non-singular symmetric matrix (in this case $\Big( \sqrt{D} \Big)^{-1}$) is symmetric (i.e. $\Big( \sqrt{D} \Big)^{-T}=\Big( \sqrt{D} \Big)^{-1}$).

Please let me know if you do not agree.

 

[JD_PM]

Now I am stuck in two basic aspects:

1) To justify $(3)$ I said 'As $T'$ is a real symmetric matrix, it is diagonalizable'; OK but I'd like to work it out explicitly. My idea is this: as we know that $T$ is a real symmetric matrix, we know that $T=D$. Thus:

$$T'= S_1^{-T} T S_1^{-1} = (S_1^{T})^{-1} D S_1^{-1} = (S_1^{-1})^{-1} D S_1^{-1} = S_1 D S_1^{-1} \tag7$$

Where I assumed that $S_1$ is an orthogonal matrix.

If we set $S_1=D$ we indeed get $T'=D$. I am not sure if this method is OK.

2) Something interesting happens if I try to apply the trick I used at 1. to $T''$

$$T'' = \Big( \sqrt{D}^{-1} \Big)^{T} T'\Big( \sqrt{D} \Big)^{-1}=\Big( \sqrt{D}^{-1} \Big)^{-1} T'\Big( \sqrt{D} \Big)^{-1}=\sqrt{D} D\Big( \sqrt{D} \Big)^{-1}=D \tag8$$

I get $T'' = D$ instead of $T'' = 1$.

So making the assumption that $(\sqrt{D})^{-1}$ is orthogonal seems to be incorrect.

Thus there has to be something wrong in my reasoning... Thinking...

 

[patfinder]

Hi,
I think you are confused and mixing things up. First :

To justify (3) I said 'As $T'$ is a real symmetric matrix, it is diagonalizable'

There is no justification to be presented. $T$ is diagonalisable, saying $T' = D$ is just a matter of identification in my previous message.
Then I think you are trying to proove the spectral theorem with T. If you want to see how to do so there is plenty of proofs for the real symetric matrices case on the internet.

Then you are making a big mistake :

My idea is this: as we know that $T$ is a real symmetric matrix, we know that $T = D$ .

No ! $T$ is diagonalisable, simply gives you thaht there exist two matrices $S_{1}$ and $D$ with $D$ diagonal such that $T = S_{1}^{-1} D S_{1}$. Here you are writing that $T$ is diagonal straight away, which is wrong in general, and moreover you are mixing up the previous notations, as we simply identified $D$ and $T'$ before.

-Second:

If we apply the coordinate transformation to $T= \ \dot{\vec q'}^T T' \dot q '$ we get

Writing this should give you a shock ! You are wirting $T$ a matrix, $= \ \dot{\vec q'}^T T' \dot q '$ the kinetic energy, which is a scalar, a number ! So you are starting from a wrong relation.

The good transformation law is given in equation (39) in your pdf file.
To derive it, it is done for $V$ but it is the same reasoning for $T$, you write the potential, ( or kinetic energy in your case), which is a scalar, and it is the same in the two coordinates systems. It is done in eq. (36) of your file. In case you are worrying, you know that the energy is the same in the two systems of corrdinates, as you are limiting yourself to a specific kind of transofrmations ( it is stated in the text). Then you inject the transformation to express $q'$ in function of $q$, eq. (37) in text, and indentify to have the relation between $T'$ and $T$.

I think you are lacking a bit of practice in pure linear algebra, and it is usual to be confused at first when you try to use it in physics, it is not trivial when you are studying it.

What I would recommend you is to pratice pure linear algebra, only the math side of it, without Physics, especially to train you in changing of coordinates and basis, reduction, diagonalisation and euclidean spaces. You have to master those notions to do this kind of physics in peace.

I'm sorry I cannot cite you good textbooks on the subject, as I didn't learn math in English, but I'm sure you will find plenty of them !

 

[patfinder]

which corresponds to the transformation $q'' = S_{2} S_{1}$ and gives $T'' = I$.

I'm sorry I did a typing mistake, I was trying to write which corresponds to the transformation $q'' = S_{2} S_{1}q$ ' off course.

 

[JD_PM]

Writing this should give you a shock ! You are wirting $T$ a matrix, $= \ \dot{\vec q'}^T T' \dot q '$ the kinetic energy, which is a scalar, a number ! So you are starting from a wrong relation.

Ups I was sloppy, thanks for pointing it out. My answers do not change though.

There's a discrepancy between your work and mine which I'll highlight once I rewrite my #3 and #4.

Rewriting my #3 (slightly modified)

If we assume that energies do not change under coordinate transformations we can write

$$\dot{\vec q}^T T \dot{\vec q} = \dot{\vec q'}^T T' \dot{\vec q'}$$

If we apply the coordinate transformation $\vec q' = S_1 \vec q \$ to $\dot{\vec q'}^T T' \dot{\vec q'}$ we get

$$\dot{\vec q}^T T \dot{\vec q}=(S_1 \dot{\vec q})^T T'(S_1 \dot{\vec q}) = \dot{\vec q}^T S_1^T T'S_1 \dot{\vec q} \tag1$$

Thus we have

$$T = S_1^T T'S_1 \iff T'=S_1^{-T} T S_1^{-1} \tag2$$

As $T'$ is a real symmetric matrix, it is diagonalizable. Thus we can assert that

$$T' = D \tag3$$

Next if we apply the coordinate transformation $\vec q'' = S_2 \vec q' \$ to $\dot{\vec q''}^T T'' \dot{\vec q''}$ we get

$$\dot{\vec q'}^T T' \dot{\vec q'}=(S_2 \dot{\vec q'})^T T''(S_2 \dot{\vec q'}) = \dot{\vec q'}^T S_2^T T''S_2 \dot{\vec q'} \tag4$$

Thus we have

$$T' = S_2^T T''S_2 \iff T''=S_2^{-T} T' S_1^{-1} \tag5$$

If we set $S_2 = \sqrt{D}$ then, out of $(5)$ and using $(3)$, we get

$$T'' = \Big( \sqrt{D} \Big)^{-T} T'\Big( \sqrt{D} \Big)^{-1}=\Big( \sqrt{D} \Big)^{-T} D\Big( \sqrt{D} \Big)^{-1}=\Big( \sqrt{D} \Big)^{-1} D\Big( \sqrt{D} \Big)^{-1}=1 \tag6$$

Where I have used the fact that the inverse matrix of a non-singular symmetric matrix (in this case $\Big( \sqrt{D} \Big)^{-1}$) is symmetric (i.e. $\Big( \sqrt{D} \Big)^{-T}=\Big( \sqrt{D} \Big)^{-1}$). Note I do not assume that $S_2$ is an orthogonal matrix here.

Rewriting my #4

Now I am stuck in two basic aspects:

1) To justify $(3)$ I said 'As $T'$ is a real symmetric matrix, it is diagonalizable'; OK but I'd like to work it out explicitly. My idea is this: let's use $(2)$

$$T'= S_1^{-T} T S_1^{-1} = (S_1^{T})^{-1} D S_1^{-1} = (S_1^{-1})^{-1} D S_1^{-1} = S_1 D S_1^{-1} \tag7$$

Where I assumed that $S_1$ is an orthogonal matrix.

My issue here is that I do not see why $S_1 D S_1^{-1}=D$. If we set $S_1=D$ we indeed get $T'=D$ sure. However, I do not think that's the whole story.

2) Something interesting happens if we assume $S_2$ to be an orthogonal matrix.

$$T'' = \Big( \sqrt{D}^{-1} \Big)^{T} T'\Big( \sqrt{D} \Big)^{-1}=\Big( \sqrt{D}^{-1} \Big)^{-1} T'\Big( \sqrt{D} \Big)^{-1}=\sqrt{D} D\Big( \sqrt{D} \Big)^{-1}=D \tag8$$

We get $T'' = D$ instead of $T'' = 1$.

So making the assumption that $(\sqrt{D})^{-1}$ is orthogonal seems to be incorrect based on my analysis. However, in your proof at #2, you assumed $S_2$ to be so and you ended up getting the expected answer.

Mmm what am I missing?