Square of orthogonal matrix vanishes

AI Thread Summary
The discussion centers on the kinetic energy expression involving an orthogonal rotation matrix, R, and its implications. It is noted that while R is orthogonal and its square equals the identity matrix, this property does not appear directly in the kinetic energy formula. The invariance of the dot product under rotation is highlighted, demonstrating that the transformation maintains the same value. Participants clarify the differentiation of the transformed coordinates and confirm the equivalence of their equations. The conversation concludes with a successful demonstration of the relationship between the transformed velocity and the original variables.
PhysicsRock
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Homework Statement
We consider a coordinate transform where ##\vec{x}^\prime(t) = R(t) (\vec{a} + \vec{x}(t))## with a constant ##\vec{a}##.
Write the lagrangian in terms of ##\vec{x}## and ##\dot{\vec{x}}##.
Relevant Equations
Velocity in terms of ##\dot{\vec{x}}## and ##\vec{x}##: ##\dot{\vec{x}}^\prime = R \left[ \dot{\vec{x}} + \vec{\omega} \times (\vec{a} + \vec{x}) \right]##.
Lagrangian function ##L = T - V##.
I found a the answer in a script from a couple years ago. It says the kinetic energy is

$$
T = \frac{1}{2} m (\dot{\vec{x}}^\prime)^2 = \frac{1}{2} m \left[ \dot{\vec{x}} + \vec{\omega} \times (\vec{a} + \vec{x}) \right]^2
$$

However, it doesn't show the rotation matrix ##R##. This would imply that ##R^2 = R \cdot R = I##. ##R## is an orthogonal matrix, but I'm pretty sure that the square of such is not always equal to the identity.

So then how come the matrix doesn't show up in the expression for the kinetic energy?
 
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PhysicsRock said:
So then how come the matrix doesn't show up in the expression for the kinetic energy?
The dot product ##\vec{z}\cdot\vec{z}## is shorthand for the matrix multiplication ##z^{T}z##, where ##z## is a column vector. So under a rotation by ##R##, $$\vec{z\,}'\equiv\overleftrightarrow{R}\cdot\vec{z}=R\,z\Rightarrow\vec{z}\,'\cdot\vec{z}\,'=\left(z'\right)^{T}z'=z^{T}R^{T}R\,z=z^{T}z=\vec{z}\cdot\vec{z}$$since ##R## is an orthogonal matrix. (This expresses the invariance of the dot-product under rotations.)
 
PhysicsRock said:
Homework Statement: We consider a coordinate transform where ##\vec{x}^\prime(t) = R(t) (\vec{a} + \vec{x}(t))## with a constant ##\vec{a}##.
Write the lagrangian in terms of ##\vec{x}## and ##\dot{\vec{x}}##.
Relevant Equations: Velocity in terms of ##\dot{\vec{x}}## and ##\vec{x}##: ##\dot{\vec{x}}^\prime = R \left[ \dot{\vec{x}} + \vec{\omega} \times (\vec{a} + \vec{x}) \right]##.
Lagrangian function ##L = T - V##.
Diffenciating ##\vec{x}^\prime(t) = R(t) (\vec{a} + \vec{x}(t))## by time, I get
\dot{\vec{x}^\prime(t)} = \dot{R(t)} (\vec{a} + \vec{x}(t))+R(t) \dot{\vec{x}(t)}
. Is it same as your relevant equation ?
 
renormalize said:
The dot product ##\vec{z}\cdot\vec{z}## is shorthand for the matrix multiplication ##z^{T}z##, where ##z## is a column vector. So under a rotation by ##R##, $$\vec{z\,}'\equiv\overleftrightarrow{R}\cdot\vec{z}=R\,z\Rightarrow\vec{z}\,'\cdot\vec{z}\,'=\left(z'\right)^{T}z'=z^{T}R^{T}R\,z=z^{T}z=\vec{z}\cdot\vec{z}$$since ##R## is an orthogonal matrix. (This expresses the invariance of the dot-product under rotations.)
That makes sense. Thank you.
 
anuttarasammyak said:
Diffenciating ##\vec{x}^\prime(t) = R(t) (\vec{a} + \vec{x}(t))## by time, I get
\dot{\vec{x}^\prime(t)} = \dot{R(t)} (\vec{a} + \vec{x}(t))+R(t) \dot{\vec{x}(t)}
. Is it same as your relevant equation ?
Yes. Allow me to demonstrate. We start with your expression and factor out an ##R##. Since it is orthogonal that leads us to

$$
\dot{\vec{x}}^\prime = R \left[ \dot{\vec{x}} + R^T \dot{R} ( \vec{a} + \vec{x} ) \right]
$$

Last semester, we derived that ##(R^T \dot{R})_{ij} = - \epsilon_{ijk} \omega_k##, where ##\omega_k## are the components of angular velocity. Now we plug that in and get

$$
\dot{x}^\prime_{i} = R_{ij} ( \dot{x}_j + (R^T \dot{R})_{jk} (a_k + x_k) )
= R_{ij} ( \dot{x}_j + (-\epsilon_{jkl} \omega_l) (a_k + x_k) )
$$

Recall the definition of the vector product ##(\vec{a} \times \vec{b})_i = \epsilon_{ijk} a_j b_k##. With that we obtain

$$
\dot{\vec{x}}^\prime = R ( \dot{\vec{x}} - (\vec{a} + \vec{x}) \times \vec{\omega} )
$$

Since the vector product is antisymmetric, we can alternatively write

$$
\dot{\vec{x}}^\prime = R ( \dot{\vec{x}} + \vec{\omega} \times (\vec{a} + \vec{x}) )
$$

And we're done.
 
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