Differences

This shows you the differences between two versions of the page.

--- book:chap3:3.4_constants_of_motion_cm [2021/11/23 12:50] – [3.4 Constants of motion (CM)] abril
+++ book:chap3:3.4_constants_of_motion_cm [2024/12/16 15:32] (current) – jv
@@ Line 68: / Line 68: @@
     = \partial_{\mathbf q} \mathcal C
 \end{align*}))
-\begin{align}\label{eq:defGradient}
+<wrap #eq_defGradient></wrap>
+\begin{align}
   \nabla\!_{\mathbf q}\, \mathcal C
   = \begin{pmatrix} \partial_{q_1} \mathcal C \\ \vdots \\ \partial_{q_D} \mathcal C \end{pmatrix}
@@ Line 148: / Line 149: @@
     W
     = \sum_i \mathbf F_i \cdot \mathbf s_i
-    =\lim_{\Delta t \to 0}  \mathbf F_i \cdot \dot{\mathbf q} \; \Delta t
+    =\lim_{\Delta t \to 0}  \sum_i \mathbf F_i \cdot \dot{\mathbf q} \; \Delta t
     = \int_{t_0}^{t_1} \mathbf F (t) \cdot \dot{\mathbf q} (t) \; \mathrm{d} t
     = \int_{\mathbf q (t)} \mathbf F \cdot \mathrm{d}\mathbf q
 \end{align*}
 The last equality should be understood here as a definition of the final expression that is interpreted here in the spirit of the substitution rule of integration.
+Moreover, $\mathbf F (t)$ denotes here the force acting on the particle at time $t$, where the particle is at the position $\mathbf q (t)$ and moving with velocity $\dot{\mathbf q} (t)$,
+i.e., one may also have $\mathbf F \bigl( \mathbf q(t) \bigr)$, or $\mathbf F \bigl( \mathbf q(t), \dot{\mathbf q} (t) \bigr)$, or an explicit time dependence on top of the dependence on the particle position.
 <WRAP box round>**Definition 3.5** <wrap em>Work and Line Integrals</wrap> \\
@@ Line 170: / Line 174: @@
 \]
 </WRAP>
 <wrap lo>** Remark 3.5. **
+Here, $\mathbf F (t)$ denotes the force that is acting on the particle at time $t$, irrespective of how it emerges. Specifically, the time dependence of $\mathbf F (t)$ accounts for changes of the position $\mathbf q(t)$, the velocity $\dot{\mathbf q}(t)$, and any additional time dependences due to external incluences (if applicable).
+</wrap>
+<wrap lo>** Remark 3.6. **
 The scalar product $\mathbf F \cdot \mathrm{d}\mathbf q$ or $P(t) = \mathbf F (t) \cdot \dot{\mathbf q} (t)$
 singles out only the action of the force parallel to the trajectory.
@@ Line 186: / Line 194: @@
 </wrap>
-<wrap lo #remark_line-integral_length-parameterization>** Remark 3.6. **
+<wrap lo #remark_line-integral_length-parameterization>** Remark 3.7. **
 The result of the integral does not rely on the parameterization of the path by time.
 For instance mathematicians prefer to use the length $\ell$ of the path.
@@ Line 192: / Line 200: @@
 and one finds
 \[
-W = \int \mathbf F (t) \cdot \mathbf s
+W = \int \mathbf F (t) \cdot \mathrm{d}\mathbf q
 = \int \mathbf F (t(\ell)) \cdot \dot{\mathbf q} (t(\ell)) \; \frac{\mathrm{d} \ell}{\dot\ell}
-= \int \mathbf F (\ell) \cdot \frac{\mathbf q}{\mathrm{d} \ell} (\ell) \; \mathrm{d} \ell
+= \int \mathbf F (\ell) \cdot \frac{\mathrm{d}\mathbf q(\ell)}{\mathrm{d} \ell} \; \mathrm{d} \ell
 \]
-where $\mathrm{d}\hat{\boldsymbol q} / \mathrm{d}\ell$ is a unit vector pointing in the direction of the trajectory.
+where $\mathrm{d}{\boldsymbol q} / \mathrm{d}\ell$ is a unit vector pointing in the direction of the trajectory.
-</wrap>\\
+</wrap>
+<wrap lo #remark_length-of-lines>** Remark 3.8. **
+  Line integrals are also used to determine the length, $L$, of a path in space.
+  After all, the length amounts to the time integral of the speed, $\dot\ell(t)$,
+  and one has
+  \begin{align*} %  \label{eq:}
+    L = \int\mathrm{d}\ell
+    = \int \mathrm{d} t \: \dot\ell(t)
+    = \int \mathrm{d} t \: \frac{\dot{\mathbf q}^2(t)}{\dot\ell(t)}
+    = \int \mathrm{d} \vec q \cdot \frac{\dot{\mathbf q}(t)}{\dot\ell(t)}
+    = \int \mathrm{d} \vec q \cdot \frac{\mathrm{d} \vec q(\ell)}{\mathrm{d}\ell}
+  \end{align*}
+  This is further illustrated in [[3.6_problems#quest__lineIntegral_venyl-disc|Problem 3.22]].
+</wrap>
 The calculation of work simplifies dramatically
@@ Line 220: / Line 243: @@
 </WRAP>
-<wrap lo>** Remark 3.7. **
+<wrap lo>** Remark 3.9. **
 Conservative forces only depend on position, $\mathbf F = \mathbf F (\mathbf q)$.
 They neither explicitly depend on time nor on the velocity $\mathbf q$.
 </wrap>
-<wrap lo>** Remark 3.8. **
-Conservative forces only depend on position, $\mathbf F = \mathbf F (\mathbf q)$.
-They neither explicitly depend on time nor on the velocity $\mathbf q$.
-</wrap>
 <WRAP box round>**Example 3.8** <wrap em>Conservative forces: (counter-)examples</wrap> \\
@@ Line 272: / Line 291: @@
 **qed**
-<wrap lo>** Remark 3.9. **
+<wrap lo>** Remark 3.10. **
 The work performed along a closed path vanishes for conservative forces.
 After all, in that case $\mathbf q_1 = \mathbf q_0$ such that $W = \Phi (\mathbf q_0) - \Phi( \mathbf q_1) = 0$.
 </wrap>
-<wrap lo>** Remark 3.10. **
+<wrap lo>** Remark 3.11. **
 The potential in itself is not an observable. ((An //observable// is a quantity that can be measured by direct observation.))
 One can only observe the work, which is the potential difference between two positions,
@@ Line 328: / Line 347: @@
 </WRAP>
-<wrap lo>** Remark 3.11. **
+<wrap lo>** Remark 3.12. **
 According to [[#Thm_work |Theorem 3.3]] differences of the value of the potential between two positions
 amount to the work performed in the potential.
@@ Line 345: / Line 364: @@
 </wrap>
-<WRAP lo>** Remark 3.12. **
+<WRAP lo>** Remark 3.13. **
 One can make use of the properties of scalar products to reduce the computational work
 to determine the force for a given potential
@@ Line 460: / Line 479: @@
 ==== 3.4.3 Momentum ====
-<WRAP box round>**Theorem 3.5 <wrap em>Conservation of momentum</wrap>** \\
+<WRAP box round #thm_Newton-conserveP>**Theorem 3.5 <wrap em>Conservation of momentum</wrap>** \\
 The //momentum// $\mathbf P = \sum_{i=1}^N m_i \dot{\mathbf q}_i(t)$
@@ Line 556: / Line 575: @@
 In such settings it is often desirable to also consider the evolution of the angular momentum.
-<WRAP box round>**Theorem 3.6 <wrap em>Conservation of angular momentum</wrap>** \\
+<WRAP box round #Thm_Newton-conserveL>**Theorem 3.6 <wrap em>Conservation of angular momentum</wrap>** \\
 The //angular momentum// $\mathbf L = \sum_{i=1}^N m_i \mathbf q_i(t) \times \dot{\mathbf q}_i(t)$