User talk:F=q(E+v^B)/Mathematical summary of physics

Examples?

I think the enumerated steps in the Classical mechanics that basically outline how problems are done are great. They appear to distill what happens over the course of a physics class. I think they describe the "big picture" that I never saw when learning physics for the first time. If you ever found the time, I would be really interested in seeing key examples worked out in according to the agenda. That goes for wherever you have a list like that. Rschwieb (talk) 17:46, 12 April 2012 (UTC)

Actually this is only supposed to be a summary, no examples were intended. I'll give you examples of how to do classical and quantum mechanics at some point using Lagrange's/Hamiltonian's/Schrodinger's equations, in a seperate place (here, or your talk page/physics subpage etc, just point it out). F = q(E+v×B) ⇄ ∑_ic_i 20:49, 12 April 2012 (UTC)

Interesting, looks like a formula-list article! Its a subpage - a nice one also, I'll let you off this one... Maschen (talk) 22:26, 12 April 2012 (UTC)

Sure, if you were willing to do the examples you're of course free to put them wherever . (No sense in breaking the flow of the summary.) It would just be nice to have two or three "canonical" examples that everyone should understand (because I could not for the life of me name what they should be). It would also be nice if the same examples could be used for each of the increasingly sophisticated levels, so that one can see how the more complex levels compare to the simpler ones. There's no rush and I'm not getting my hopes up that you'll have time. Rschwieb (talk) 00:13, 13 April 2012 (UTC)

Lets keep everything here in one place then - easier. F = q(E+v×B) ⇄ ∑_ic_i 16:22, 13 April 2012 (UTC)

Examples (LH Mech)

Let’s do classical Lagrangian and Hamiltonian mechanics first ("LH Mech"). Quantum can come later. It may appear contradictory that when the problems are set up, the standard Cartesian or spherical/cylindrical polar coords will be used even though not all sets of coordinates are needed.

Simple pendulum

Simple pendulum. To start, consider the simple pendulum. A mass m initially at (x, y) = (0, 0) is suspended by a rigid (can't bend) massless (negligible really..) rod of length l, placed at (x, y) = (0, l) is free to oscillate (no forced oscillations). Let θ be the angle between the pendulum rod and the vertical. Find the equation of motion. Well, in the Lagrangian formalism, you look for constraints (what limits the motion of the particle? which regions can the particle only move in?). Here, the particle is a pendulum "bob" constrained by the rod only to oscillate only in one vertical plane, back and forth along one arc. So the bob only has one degree of freedom (important tech term, abbreviate to "d.f", which is not a differential!). Just to be sure: d.f's are basically the number of ways the particle can move subject to the constraints. Each degree of freedom can be described by a number: a generalized coordinate, usually denoted by q. Instead of using all coordinates in the space (for this example just 2), we only need one coordinate to define and describe the position of the bob. More: its completely arbitrary: - we can use the arc length the bob traverses (relative to some initial position), the distance moved along some axis (x or y, no need for both), or the angle θ, which is the most convenient to use for the problem. So lets use "q = θ" as our generalized coordinate. To solve for the equation of motion we need the kinetic energy T and potential energy V of the particle is, then can calc. the Lagrangian L, then use Lagrange's equations and tidy up, solve etc. The problem was set up in Cartesian coords, so the x and y coords, and time derivatives, are: $x=\ell \sin \theta \quad y=\ell (1-\cos \theta )\Rightarrow {\dot {x}}=\ell \cos \theta {\dot {\theta }}\quad {\dot {y}}=\ell \sin \theta {\dot {\theta }}$ Notice how x and y are amalgamated into θ. The potential energy is easy (using the approximation mgh rather than GmM/r where M = mass of Earth and r = distance from Earth's c.o.m to pendulum bob c.o.m): $V=mgy=mg\ell (1-\cos \theta )$ the kinetic energy is: $T={\frac {1}{2}}mv^{2}$ where the speed (squared) of the bob is: (you may be able to tell already, just radius l times time derivative of angle θ): $v^{2}={\dot {x}}^{2}+{\dot {y}}^{2}=\ell ^{2}{\dot {\theta }}^{2}\Rightarrow T={\frac {1}{2}}m\ell ^{2}{\dot {\theta }}^{2}$ so the Lagrangian is: $L=T-V={\frac {1}{2}}m\ell ^{2}{\dot {\theta }}^{2}-mg\ell (1-\cos \theta )$ Lagrange's equations for this problem are (see the main page): ${\dfrac {d}{dt}}\left({\dfrac {\partial L}{\partial {\dot {q}}}}\right)={\dfrac {\partial L}{\partial q}}\quad \Rightarrow \quad {\dfrac {d}{dt}}\left({\dfrac {\partial L}{\partial {\dot {\theta }}}}\right)={\dfrac {\partial L}{\partial \theta }}$ This is important: The gen. coords and their time derivatives are necessarily treated as independent variables. Do NOT use the chain rule on the ${\dot {q}}$'s !!! Calculating the derivatives: ${\dfrac {d}{dt}}\left({\dfrac {\partial L}{\partial {\dot {\theta }}}}\right)=m\ell ^{2}{\ddot {\theta }},\quad {\dfrac {\partial L}{\partial \theta }}=-mg\ell \sin \theta$ so the equation of motion is: $\ell {\ddot {\theta }}=-g\sin \theta$ which is the familiar non-linear pendulum differential equation. You can take this from here yourself (apparently elliptic integrals are needed for exact solutions, I don't understand them in detail...). Of course for small amplitudes its easy to approximate, since $\sin \theta \approx \theta \quad \Rightarrow \quad {\ddot {\theta }}\approx -{\frac {g}{\ell }}\theta \quad \Rightarrow \quad \theta \approx A\sin \left({\sqrt {\frac {g}{\ell }}}t\right)+B\cos \left({\sqrt {\frac {g}{\ell }}}t\right)$ Since at t = 0, θ = 0, so B = 0, this reduces to $\theta \approx A\sin \left({\sqrt {\frac {g}{\ell }}}t\right)$

Double pendulum

Two simple penduli. Suppose we have a similar system, a simple pendulum with bob mass m1, rod length l1, angle θ1, with another pendulum suspended from the 1st bob (like a chain), of mass m2, rod length l2, angle θ2. The system is placed at (0, l1 + l2). There are now two generalized coordinates, the angle of each pendulum. So there will be two coupled (generally) 2nd order equations to solve. The kinetic energy of the system is (remember energy is additive): $T=T_{1}+T_{2}={\dfrac {1}{2}}(m_{1}\ell _{1}^{2}{\dot {\theta }}_{1}^{2}+m_{2}\ell _{2}^{2}{\dot {\theta }}_{2}^{2})$ the potential energy is: $V=V_{1}+V_{2}=m_{1}g[(\ell _{1}+\ell _{2})-\ell _{1}\cos \theta _{1}]+m_{2}g[(\ell _{1}+\ell _{2})-\ell _{1}\cos \theta _{1}-\ell _{2}\cos \theta _{2}]$ The Lagrangian is (notice how we can ignore the constant terms since their derivatives will be zero): $L=T-V={\dfrac {1}{2}}(m_{1}\ell _{1}^{2}{\dot {\theta }}_{1}^{2}+m_{2}\ell _{2}^{2}{\dot {\theta }}_{2}^{2})-m_{1}g[{\cancelto {}{(\ell _{1}+\ell _{2})}}-\ell _{1}\cos \theta _{1}]-m_{2}g[{\cancelto {}{(\ell _{1}+\ell _{2})}}-\ell _{1}\cos \theta _{1}-\ell _{2}\cos \theta _{2}]$ $L=T-V={\dfrac {1}{2}}(m_{1}\ell _{1}^{2}{\dot {\theta }}_{1}^{2}+m_{2}\ell _{2}^{2}{\dot {\theta }}_{2}^{2})-m_{1}g[-\ell _{1}\cos \theta _{1}]-m_{2}g[-\ell _{1}\cos \theta _{1}-\ell _{2}\cos \theta _{2}]$ $L=T-V={\dfrac {1}{2}}(m_{1}\ell _{1}^{2}{\dot {\theta }}_{1}^{2}+m_{2}\ell _{2}^{2}{\dot {\theta }}_{2}^{2})+m_{1}g\ell _{1}\cos \theta _{1}+m_{2}g[\ell _{1}\cos \theta _{1}+\ell _{2}\cos \theta _{2}]$ $L=T-V={\dfrac {1}{2}}(m_{1}\ell _{1}^{2}{\dot {\theta }}_{1}^{2}+m_{2}\ell _{2}^{2}{\dot {\theta }}_{2}^{2})+(m_{1}+m_{2})g\ell _{1}\cos \theta _{1}+m_{2}g\ell _{2}\cos \theta _{2}$ Lagrange's equations are: ${\dfrac {d}{dt}}\left({\dfrac {\partial L}{\partial {\dot {\theta }}_{1}}}\right)={\dfrac {\partial L}{\partial \theta _{1}}}\quad {\dfrac {d}{dt}}\left({\dfrac {\partial L}{\partial {\dot {\theta }}_{2}}}\right)={\dfrac {\partial L}{\partial \theta _{2}}}$ calculating the derivatives: ${\dfrac {d}{dt}}\left({\dfrac {\partial L}{\partial {\dot {\theta }}_{1}}}\right)=m_{1}\ell _{1}^{2}{\ddot {\theta }}_{1}\quad {\dfrac {d}{dt}}\left({\dfrac {\partial L}{\partial {\dot {\theta }}_{2}}}\right)=m_{2}\ell _{2}^{2}{\ddot {\theta }}_{2}$ ${\dfrac {\partial L}{\partial \theta _{1}}}=-(m_{1}+m_{2})g\ell _{1}\sin \theta _{1}\quad {\dfrac {\partial L}{\partial \theta _{2}}}=-m_{2}g\ell _{2}\sin \theta _{2}$ so the equations of motion happen to be decoupled: $m_{1}\ell _{1}^{2}{\ddot {\theta }}_{1}=-(m_{1}+m_{2})g\ell _{1}\sin \theta _{1},\quad m_{2}\ell _{2}^{2}{\ddot {\theta }}_{2}=-m_{2}g\ell _{2}\sin \theta _{2}$

Lorentz force

Here is a nice and important example - the force on a charged particle of mass m and charge q in an electromagnetic field (A, ϕ). Given an A field, the particle in motion has potential momentum $q\mathbf {A} (\mathbf {r} ,t)$, so its potential energy is $q\mathbf {A} (\mathbf {r} ,t)\cdot \mathbf {\dot {r}}$. For a ϕ field, the potential energy is $q\phi (\mathbf {r} ,t)$. The potential energy is: $V=q\phi -q\mathbf {A} \cdot \mathbf {\dot {r}}$ The kinetic energy is: $T={\frac {m}{2}}\mathbf {\dot {r}} \cdot \mathbf {\dot {r}}$ hence the Lagrangian: $L=T-V={\frac {m}{2}}\mathbf {\dot {r}} \cdot \mathbf {\dot {r}} +q\mathbf {A} \cdot \mathbf {\dot {r}} -q\phi$ $L={\frac {m}{2}}({\dot {x}}^{2}+{\dot {y}}^{2}+{\dot {z}}^{2})+q({\dot {x}}A_{x}+{\dot {y}}A_{y}+{\dot {z}}A_{z})-q\phi$ Lagrange's equations are ${\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {x}}}}={\frac {\partial L}{\partial x}}$ (same for y and z). So partial derivatives are ${\begin{aligned}{\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {x}}}}&=m{\ddot {x}}+q{\frac {dA_{x}}{dt}}\\&=m{\ddot {x}}+{\frac {q}{dt}}\left({\frac {\partial A_{x}}{\partial t}}dt+{\frac {\partial A_{x}}{\partial x}}dx+{\frac {\partial A_{x}}{\partial y}}dy+{\frac {\partial A_{x}}{\partial z}}dz\right)\\&=m{\ddot {x}}+q\left({\frac {\partial A_{x}}{\partial t}}+{\frac {\partial A_{x}}{\partial x}}{\dot {x}}+{\frac {\partial A_{x}}{\partial y}}{\dot {y}}+{\frac {\partial A_{x}}{\partial z}}{\dot {z}}\right)\\\end{aligned}}$ ${\frac {\partial L}{\partial x}}={\frac {\partial \phi }{\partial x}}+q\left({\frac {\partial A_{x}}{\partial x}}{\dot {x}}+{\frac {\partial A_{y}}{\partial x}}{\dot {y}}+{\frac {\partial A_{z}}{\partial x}}{\dot {z}}\right)$ equating and solving: $m{\ddot {x}}+q\left({\frac {\partial A_{x}}{\partial t}}+{\frac {\partial A_{x}}{\partial x}}{\dot {x}}+{\frac {\partial A_{x}}{\partial y}}{\dot {y}}+{\frac {\partial A_{x}}{\partial z}}{\dot {z}}\right)=q{\frac {\partial \phi }{\partial x}}+q\left({\frac {\partial A_{x}}{\partial x}}{\dot {x}}+{\frac {\partial A_{y}}{\partial x}}{\dot {y}}+{\frac {\partial A_{z}}{\partial x}}{\dot {z}}\right)$ ${\begin{aligned}F_{x}&=q\left({\frac {\partial \phi }{\partial x}}+{\frac {\partial A_{x}}{\partial t}}\right)+q\left[{\dot {y}}\left({\frac {\partial A_{y}}{\partial x}}-{\frac {\partial A_{x}}{\partial y}}{\dot {y}}\right)+{\dot {z}}\left({\frac {\partial A_{z}}{\partial x}}-{\frac {\partial A_{x}}{\partial z}}\right)\right]\\&=qE_{x}+q[{\dot {y}}(\nabla \times \mathbf {A} )_{z}-{\dot {z}}(\nabla \times \mathbf {A} )_{y}]\\&=qE_{x}+q[\mathbf {\dot {r}} \times (\nabla \times \mathbf {A} )]_{x}\\&=qE_{x}+q(\mathbf {\dot {r}} \times \mathbf {B} )_{x}\end{aligned}}$ Simalarly for the y and z directions. Hence the force equation is: $\mathbf {F} =q(\mathbf {E} +\mathbf {\dot {r}} \times \mathbf {B} )$