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--- theorems:work-energy_theorem [2021/04/24 23:28]
cleonis [Concrete] Fixed some yptos, added a sentence on release of potential energy linear with distance
+++ theorems:work-energy_theorem [2021/07/18 10:57] (current)
cleonis [Concrete] Reverted to the derivation with an intermediary stage, it is transparent whereas the version-with-less-steps wasn't
@@ Line 1: / Line 1: @@
 ====== Work-Energy theorem ======
-<tabbox Intuitive>
-<note tip>
-Explanations in this section should contain no formulas, but instead colloquial things like you would hear them during a coffee break or at a cocktail party.
-</note>
 <tabbox Concrete>
@@ Line 81: / Line 76: @@
 The following derivation is more abstract. This derivation is from the outset designed to accomodate an arbitrary acceleration profile.
-In the course of the derivation the following two relations will be used:
+The two relations that constitute the mathematics of the Work-Energy theorem:
-$$ ds = v \ dt  \qquad (2.1)  $$
+$$ ds = v \ dt \qquad (2.1) $$
+$$ dv = a \ dt \qquad (2.2) $$
-$$ dv = \frac{dv}{dt}dt  \qquad (2.2)   $$
 The integral for acceleration from a starting point $s_0$ to a final point $s$.
@@ Line 91: / Line 87: @@
 $$ \int_{s_0}^s a \ ds \qquad (2.3)  $$
-Use (2.1) to change the differential from *ds* to *dt*. Since the differential is changed the limits change accordingly.
+Intermediary step: change of the differential according to (2.1), with corresponding change of limits.
 $$ \int_{t_0}^t a \ v \ dt \qquad (2.4) $$
-Rearrange the order, and write the acceleration $a$ as $\tfrac{dv}{dt}$
+Change the order:
-$$ \int_{t_0}^t v \ \frac{dv}{dt} \ dt  \qquad (2.5) $$
+$$ \int_{t_0}^t v \ a \ dt \qquad (2.5) $$
-Use (2.2) for a second change of differential, again the limits change accordingly.
+Change of differential according to (2.2), with corresponding change of limits.
-$$ \int_{v_0}^v v \ dv  \qquad (2.6)  $$
+$$ \int_{v_0}^v v \ dv \qquad (2.6) $$
 Putting everything together:
-$$ \int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2  \qquad (2.7) $$
+$$ \int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \qquad (2.7) $$
-Combining with $F=ma$ gives the work-energy theorem
+Multiply both sides with $m$ and apply $F=ma$ to arrive at the work-energy theorem:
-$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2  \qquad (2.8) $$
+$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \qquad (8) $$
 \\
@@ Line 129: / Line 125: @@
 </note>
 <tabbox Why is it interesting?>
+As stated in the lead up to the derivation: the Work-Energy theorem is both old and new.
+The Work-Energy theorem is the basis of the formulation of mechanics that was developed by Joseph Louis Lagrange, called 'Lagrangian mechanics'. But at the time the Work-Energy theorem did not yet exist in its current form. The modern formulation of the Work-Energy theorem is not present in Lagrange's work. The form that Lagrange used is called d'Alembert's virtual work.
+**Kinetic energy and Lagrangian mechanics**
+Motion of objects is represented with coordinates of a coordinate system. In a system of cartesian coordinates a velocity vector can be decomposed in components along the axes of the cartesian coordinate system. In order to represent momentum in a cartesian coordinate system with three dimensions the momentum is decomposed along the three axes of the coordinate system. Likewise the force is decomposed along the three axes. Newton's second law is valid for each component of motion.
+$ \vec{F_x} = m \vec{a_x} $ \\
+$ \vec{F_y} = m \vec{a_y} $ \\
+$ \vec{F_z} = m \vec{a_z} $ \\
+Kinetic energy is proportional to the square of velocity. This means that the sum of the component kinetic energies (along the coordinate axes) is equal to the kinetic energy of the undecomposed velocity (pythagoras' theorem). This enables the following: in Lagrangian mechanics the direction of the velocity vector is discarded. The kinetic energy is treated as a //scalar//.
+The directional information of the velocity vector can be discarded because that information is still available in the expression for the //potential energy//. The potential energe is the integral of the force over distance. When the calculation is for motion in three dimensions of space then the expression for the potential energy has three components, one for each spatial dimension.
+Kinetic energy can be treated as a scalar without loss of expressive power; that is what Joseph Louis Lagrange capitalized on when he developed Lagrangian mechanics.
+The second innovation from Lagrange was systematic use of generalized coordinates. The importance of being able to use generalized coordinates cannot be overstated.
 /*<tabbox FAQ>*/

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