If \(q_{1}, q_{2}, q_{3}\) are orthogonal curvilinear co-ordinates, and the element of length in the \(q_{i}\) direction is \(h_{i} \mathrm{~d} q_{i}\), write down (a) the kinetic energy \(T\) in terms of the generalized velocities \(\dot{q}_{i}\), (b) the generalized momentum \(p_{i}\) and (c) the component \(\boldsymbol{e}_{i} \cdot \boldsymbol{p}\) of the momentum vector \(\boldsymbol{p}\) in the \(q_{i}\) direction. (Here \(\boldsymbol{e}_{i}\) is a unit vector in the direction of increasing \(\left.q_{i}\right)\)

Short Answer

Expert verified
In summary, given orthogonal curvilinear coordinates \(q_1, q_2, q_3\) with scaling factors \(h_i\), the kinetic energy \(T\) is given by: $$ T = \frac{1}{2} m (\dot{q}_1^2 h_1^2 + \dot{q}_2^2 h_2^2 + \dot{q}_3^2 h_3^2). $$ The generalized momentum \(p_i\) is given by: $$ p_i = m h_i^2 \dot{q}_i, $$ and the component of the momentum vector \(\boldsymbol{p}\) in the \(q_i\) direction is equal to the generalized momentum \(p_i\).

Step by step solution

01

Write down the element of length in Cartesian coordinates

First, write the element of length in Cartesian coordinates (\(x, y, z\)) and convert it into the orthogonal curvilinear coordinates (\(q_1, q_2, q_3\)). In Cartesian coordinates, the element of length is given by: $$ ds^2 = dx^2 + dy^2 + dz^2 $$ Expressing each differential element \((dx, dy, dz)\) in terms of orthogonal curvilinear coordinates and using the given scale factors \(h_i\), we can write: $$ ds^2 = (h_1 dq_1)^2 + (h_2 dq_2)^2 + (h_3 dq_3)^2. $$
02

Find the kinetic energy in terms of generalized velocities

The kinetic energy T of a particle is given by: $$ T = \frac{1}{2} m (\dot{x}^2 + \dot{y}^2 + \dot{z}^2), $$ where \(\dot{x}, \dot{y}\), and \(\dot{z}\) are the components of velocity in the Cartesian coordinates. Taking the time derivative of the orthogonal curvilinear coordinates and substituting them into the expression for kinetic energy, we have: $$ T = \frac{1}{2} m (\dot{q}_1^2 h_1^2 + \dot{q}_2^2 h_2^2 + \dot{q}_3^2 h_3^2). $$
03

Determine the generalized momentum

The generalized momentum \(p_i\) is obtained by taking the partial derivative of the kinetic energy T with respect to the generalized coordinates \(\dot{q}_i\): $$ p_i = \frac{\partial T}{\partial \dot{q}_i} = m h_i^2 \dot{q}_i $$ This gives us the generalized momentum components \(p_1, p_2\) and \(p_3\).
04

Find the momentum vector component in the \(q_i\) direction

Now, we need to find the component \(\boldsymbol{e}_{i} \cdot \boldsymbol{p}\) of the momentum vector \(\boldsymbol{p}\) in the \(q_{i}\) direction. To do this, we express the momentum vector \(\boldsymbol{p}\) as the sum of its components in the orthogonal curvilinear basis \(\boldsymbol{e}_i\): $$ \boldsymbol{p} = p_1 \boldsymbol{e}_1 + p_2 \boldsymbol{e}_2 + p_3 \boldsymbol{e}_3 $$ Taking the dot product of both sides with the unit vector \(\boldsymbol{e}_i\) in the direction of increasing \(q_i\), we get: $$ \boldsymbol{e}_i \cdot \boldsymbol{p} = (\boldsymbol{e}_i \cdot \boldsymbol{e}_1) p_1 + (\boldsymbol{e}_i \cdot \boldsymbol{e}_2) p_2 + (\boldsymbol{e}_i \cdot \boldsymbol{e}_3) p_3. $$ Since the basis vectors \(\boldsymbol{e}_i\) are orthogonal, their dot products with other basis vectors are zero, and the dot product with themselves is one. Thus, we have: $$ \boldsymbol{e}_i \cdot \boldsymbol{p} = p_i . $$ In the \(q_i\) direction, the component of the momentum vector is equal to the generalized momentum \(p_i\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy in Generalized Coordinates
Kinetic energy represents the energy that an object possesses due to its motion. In classical mechanics, we're accustomed to expressing kinetic energy in terms of Cartesian coordinates. However, many physical problems lend themselves more naturally to descriptions in different coordinate systems, such as cylindrical, spherical, or any set of orthogonal curvilinear coordinates.

In these alternative coordinate systems, the expression for kinetic energy involves 'generalized velocities' which are the time derivatives of the generalized coordinates. Instead of using plain \(x, y, z\) coordinates, we use \(q_1, q_2, q_3\) and their respective velocities \(\dot{q}_1, \dot{q}_2, \dot{q}_3\). These generalized velocities describe how quickly the position of the object is changing within the curvilinear framework.

To find the kinetic energy in terms of these generalized velocities, we consider each component's contribution. The mass \(m\) of the particle and the scale factors \(h_i\) for each coordinate direction, which account for the non-uniformity and curvature of the coordinates, are critical. The total kinetic energy \(T\) is then given by the sum of the kinetic energies in each of the generalized coordinates:
\[T = \frac{1}{2} m (\dot{q}_1^2 h_1^2 + \dot{q}_2^2 h_2^2 + \dot{q}_3^2 h_3^2).\]
This formula allows us to calculate the kinetic energy of a particle moving in a space defined by any set of orthogonal curvilinear coordinates.
Generalized Momentum
Momentum, fundamentally, is a quantity of motion. In Cartesian coordinates, we think of it as mass times velocity. But when dealing with curvilinear coordinates, the concept of generalized momentum becomes essential, especially in the field of analytical mechanics.

Generalized momentum is derived from the Lagrangian formulation of mechanics and is represented by the symbol \(p_i\). It is defined as the derivative of the kinetic energy \(T\) with respect to the corresponding generalized velocity \(\dot{q}_i\), for each generalized coordinate \(q_i\).
\[p_i = \frac{\partial T}{\partial \dot{q}_i} = m h_i^2 \dot{q}_i\]
Here, \(h_i\) are the scale factors that adjust for the dimensions and distortions introduced by different coordinate systems. Generalized momentum is a powerful concept as it blends the geometric properties of motion (mediated through the scale factors) with the physical properties, encapsulating the motion of a particle in a comprehensive mathematical term.
Momentum Vector in Curvilinear Coordinates
In an orthogonal curvilinear coordinate system, the momentum vector of a particle is intrinsically linked with the nature of the coordinate system itself. This vector can be decomposed into components that align with the coordinate axes represented by the unit vectors \(\boldsymbol{e}_i\).

The component of the momentum vector \(\boldsymbol{p}\) in the direction of a specific coordinate \(q_i\) is expressed as \(\boldsymbol{e}_{i} \cdot \boldsymbol{p}\), which is the dot product of the unit vector in that direction with the total momentum vector.
\[\boldsymbol{p} = p_1 \boldsymbol{e}_1 + p_2 \boldsymbol{e}_2 + p_3 \boldsymbol{e}_3\]
However, due to the orthogonality of the coordinate system, the dot product of different unit vectors is zero, leaving us with:
\[\boldsymbol{e}_i \cdot \boldsymbol{p} = p_i \]
Thus, in orthogonal curvilinear coordinates, the component of the momentum vector along a coordinate direction is exactly the generalized momentum \(p_i\) for that coordinate. Physically, this means that for determining the motion or predicting the future state of the particle along a specific curvilinear axis, we only need to consider the component of momentum parallel to that axis. This simplifies calculations in complex systems and is a cornerstone in the study of dynamics in non-Cartesian geometries.

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Most popular questions from this chapter

\({ }^{*}\) By expanding the logarithm in (3.17), find the approximate equation for the trajectory of a projectile subject to small atmospheric drag to first order in \(\gamma\). (Note that this requires terms up to order \(\gamma^{3}\) in the logarithm.) Show that to this order the range (on level ground) is $$ x=\frac{2 u w}{g}-\frac{8 \gamma u w^{2}}{3 g^{2}} $$ and hence that to maximize the range for given launch speed \(v\) the angle of launch should be chosen to satisfy \(\cos 2 \alpha=\sqrt{2} \gamma v / 3 g .\) (Hint: In the term containing \(\gamma\), you may use the zeroth-order approximation for the angle.) For a projectile whose terminal speed if dropped from rest (see Chapter 2, Problem 13) would be \(500 \mathrm{~m} \mathrm{~s}^{-1}\), estimate the optimal angle cand the range if the launch speed is \(100 \mathrm{~m} \mathrm{~s}^{-1}\).

Compute the work done in taking a particle around the circle \(x^{2}+y^{2}=\) \(a^{2}, z=0\) if the force is (a) \(\boldsymbol{F}=y \boldsymbol{i}\), and (b) \(\boldsymbol{F}=x \boldsymbol{i}\). What do you conclude about these forces? (Use the parametrization \(x=a \cos \varphi, y=\) \(a \sin \varphi, z=0 .)\)

Evaluate the force corresponding to the potential energy function \(V(\boldsymbol{r})=c z / r^{3}\), where \(c\) is a constant. Write your answer in vector notation, and also in spherical polars, and verify that it satisfies \(\nabla \wedge \boldsymbol{F}=\mathbf{0}\).

For the case of plane polar co-ordinates \(r, \theta\), write the unit vectors \(e_{r}\) \((=\hat{\boldsymbol{r}})\) and \(\boldsymbol{e}_{\theta}\) in terms of \(\boldsymbol{i}\) and \(\boldsymbol{j}\). Hence show that \(\partial \boldsymbol{e}_{r} / \partial \theta=\boldsymbol{e}_{\theta}\) and \(\partial \boldsymbol{e}_{\theta} / \partial \theta=-\boldsymbol{e}_{r} .\) By starting with \(\boldsymbol{r}=r e_{r}\) and differentiating, rederive the expressions for the components of the velocity and acceleration vectors.

Find which of the following forces are conservative, and for those that are find the corresponding potential energy function ( \(a\) and \(b\) are constants, and \(\boldsymbol{a}\) is a constant vector): (a) \(F_{x}=a x+b y^{2}, \quad F_{y}=a z+2 b x y, \quad F_{z}=a y+b z^{2}\); (b) \(F_{x}=a y, \quad F_{y}=a z, \quad F_{z}=a x\); (c) \(F_{r}=2 a r \sin \theta \sin \varphi, \quad F_{\theta}=\operatorname{arcos} \theta \sin \varphi, \quad F_{\varphi}=\operatorname{arcos} \varphi\); (d) \(\boldsymbol{F}=\boldsymbol{a} \wedge \boldsymbol{r}\); (e) \(\boldsymbol{F}=r \boldsymbol{a} ;\) (f) \(\boldsymbol{F}=\boldsymbol{a}(\boldsymbol{a} \cdot \boldsymbol{r}) .\)

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