A Why is this matrix symmetric here?

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The discussion centers on the symmetry of the matrix ##\mathbf{T}## in the context of Lagrangian mechanics, specifically in relation to kinetic energy expressed as a quadratic form. It is established that the kinetic energy can be represented in a form that leads to a symmetric matrix ##\mathbf{A}##, which is derived from the particle velocities and their dependencies on generalized coordinates and time. The symmetry arises from the properties of the quadratic form ##\dot{\mathbf{q}}^T \mathbf{A} \dot{\mathbf{q}}##, which is homogeneous and reflects the nature of kinetic energy in the system. The discussion also references the implications of moving constraints on the kinetic energy formulation. Understanding this symmetry is crucial for analyzing the dynamics of systems described by such Lagrangians.
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Goldstein 3rd Ed, pg 339

"In large classes of problems, it happens that ##L_{2}## is a quadratic function of the generalized velocities and ##L_{1}## is a linear function of the same variables with the following specific functional dependencies:
##L\left(q_{i}, \dot{q}_{i}, t\right)=L_{0}(q, t)+\dot{q}_{i} a_{i}(q, t)+\dot{q}_{i}^{2} T_{i}(q, t)##

where the ##a_{i}^{\prime} s##and the ##T_{i}##'s are functions of the ##q## 's and ##t##.

Under the given assumptions the Lagrangian can be written as##L(q, \dot{q}, t)=L_{0}(q, t)+\tilde{\dot{q}} \mathbf{a}+\frac{1}{2} \tilde{\mathbf{q}} \mathbf{T} \dot{\mathbf{q}}##"

It's then said that ##\mathbf{T}## is symmetric.
Why is it symmetric? Is it because it's always diagonal? The author didn't say that,instead says it's just a n by n square matrix.

Please give me a hint or an answer. Thank you
 
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With moving (i.e. time-dependent) constraints ##\mathbf{r}_a = \mathbf{r}_a(\mathbf{q},t)##, the kinetic energy is indeed generally of the form ##T = \dot{\mathbf{q}}^T \mathbf{A} \dot{\mathbf{q}} + \mathbf{B}^T \dot{\mathbf{q}} + C##, where ##\mathbf{A}##, ##\mathbf{B}## and C are all functions of ##\mathbf{q}## and ##t##.

This follows from the expression ##\dot{\mathbf{r}}_a = \sum_i \dfrac{\partial \mathbf{r}_a}{\partial q^i} \dot{q}^i + \dfrac{\partial \mathbf{r}_a}{\partial t}## for the particle velocities. Using this to write out the kinetic energy ##T = \frac{1}{2} \sum_a m_a \dot{\mathbf{r}}_a \cdot \dot{\mathbf{r}}_a## of the system, do you see why ##A_{ij} = A_{ji}## (i.e. ##\mathbf{A}## is a symmetric matrix and the quadratic form ##\dot{\mathbf{q}}^T \mathbf{A} \dot{\mathbf{q}}## is homogeneous?).
 
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ergospherical said:
With moving (i.e. time-dependent) constraints ##\mathbf{r}_a = \mathbf{r}_a(\mathbf{q},t)##, the kinetic energy is indeed generally of the form ##T = \dot{\mathbf{q}}^T \mathbf{A} \dot{\mathbf{q}} + \mathbf{B}^T \dot{\mathbf{q}} + C##, where ##\mathbf{A}##, ##\mathbf{B}## and C are all functions of ##\mathbf{q}## and ##t##.

This follows from the expression ##\dot{\mathbf{r}}_a = \sum_i \dfrac{\partial \mathbf{r}_a}{\partial q^i} \dot{q}^i + \dfrac{\partial \mathbf{r}_a}{\partial t}## for the particle velocities. Using this to write out the kinetic energy ##T = \frac{1}{2} \sum_a m_a \dot{\mathbf{r}}_a \cdot \dot{\mathbf{r}}_a## of the system, do you see why ##A_{ij} = A_{ji}## (i.e. ##\mathbf{A}## is a symmetric matrix and the quadratic form ##\dot{\mathbf{q}}^T \mathbf{A} \dot{\mathbf{q}}## is homogeneous?).
Thankyou sir.
 
This has been discussed many times on PF, and will likely come up again, so the video might come handy. Previous threads: https://www.physicsforums.com/threads/is-a-treadmill-incline-just-a-marketing-gimmick.937725/ https://www.physicsforums.com/threads/work-done-running-on-an-inclined-treadmill.927825/ https://www.physicsforums.com/threads/how-do-we-calculate-the-energy-we-used-to-do-something.1052162/
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