Square of orthogonal matrix vanishes

In summary, we discussed the expression for kinetic energy in terms of a rotation matrix ##R## and its invariance under rotations, and derived the relevant equation for velocity in terms of ##R## and angular velocity. We also demonstrated the equivalence between your derived equation and the relevant equation.
  • #1
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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  • #2
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.)
 
  • #3
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
[tex]\dot{\vec{x}^\prime(t)} = \dot{R(t)} (\vec{a} + \vec{x}(t))+R(t) \dot{\vec{x}(t)}[/tex]
. Is it same as your relevant equation ?
 
  • #4
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.
 
  • #5
anuttarasammyak said:
Diffenciating ##\vec{x}^\prime(t) = R(t) (\vec{a} + \vec{x}(t))## by time, I get
[tex]\dot{\vec{x}^\prime(t)} = \dot{R(t)} (\vec{a} + \vec{x}(t))+R(t) \dot{\vec{x}(t)}[/tex]
. 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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