Tag Archives: mechanics

What If Everyone JUMPED At Once?

Published on Aug 18, 2012

Follow Michael Stevens http://www.twitter.com/tweetsauce

Tell Geek and Sundry that Vsauce says “hi!” and SUB!:http://www.youtube.com/watch?v=6-Wef0…

http://www.Facebook.com/VsauceGaming
music courtesy of http://www.SoundCloud.com/JakeChudnow

Thanks to:
http://www.youtube.com/phoenixsaintly
and
http://www.youtube.com/theINENIvlogs
For helping me with this video at Summer in the City!!
theINENIvlogs behind the scenes Summer in the City vid with me:http://www.youtube.com/watch?v=F_Rd35…

Why videos views freeze in the 300s on YouTube (sub to this channel FTW, seriously): http://www.youtube.com/watch?v=oIkhga…

Japan Earthquake and Earth’s rotation:http://en.wikipedia.org/wiki/2011_T%C…

All people in one place LIVING:http://persquaremile.com/2011/01/18/i…

shoulder-to-shoulder: http://news.nationalgeographic.com/ne…

BBC Jump Video: http://www.bbc.co.uk/learningzone/cli…

SCALE OF UNIVERSE AWESOME: http://htwins.net/scale2/

STRAIGHTDOPE article on a jump:http://www.straightdope.com/columns/r…

Dot Physics on the jump: http://www.wired.com/wiredscience/201…

Interactive scale of the universe: http://scaleofuniverse.com/

Decimate: http://www.etymonline.com/index.php?a…

Dunbar’s Number: http://en.wikipedia.org/wiki/Dunbar&#…

NPR story on Dunbar’s number:http://www.npr.org/2011/06/04/1367233…

Life Expectency: http://en.wikipedia.org/wiki/Life_exp…

How many people you meet in your life:http://eclectic24.wordpress.com/2009/…

Newton’s Third Law: http://www.physicsclassroom.com/class…

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Stable Orbits

Published on May 3, 2012

If gravity is so attractive, why doesn’t the earth just crash into the sun? Or the moon into the earth?

The answer: Stable Orbits

hyperbolic funnel video: http://bit.ly/r5xhng

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Minute Physics provides an energetic and entertaining view of old and new problems in physics — all in a minute!

floating pyramids

Think floating pyramids are more metaphysics than physics? Think again. Results just in from an experiment that levitated open-bottomed paper pyramids on gusts of air reveal a curious phenomenon: When it comes to drifting through the air, top-heavy designs are more stable than bottom-heavy ones. The finding may lead to robots that fly not like insects or birds but like jellyfish.

Physics of Floating Pyramids

Animaciones

1: Motor radial de un avión

2: Distribución oval

3: Principio de la máquina de coser

4: Movimiento de Cruz de Malta – de la mano del segundero, que controla al reloj

5: Mecanismo de cambio de velocidades (automóvil)

6: Junta universal para velocidad constante automática

7: Sistema de carga de proyectiles

8: Motor giratorio – motor de combustión interna, el calor y no el movimiento del pistón, causa el movimiento giratorio

9: Motor en línea – cilindros alineados en forma paralela

Poisson bracket

mathematics and classical mechanics, the Poisson bracket is an important operator in Hamiltonian mechanics, playing a central role in the definition of the time-evolution of a dynamical system in the Hamiltonian formulation. It places mechanics and dynamics in the context of coordinate-transformations: specifically in coordinate planes such as canonical position/momentum, or canonical-position/canonical transformation. (A so-called “canonical transformation” is a function of the canonical position and momentum satisfying certain Poisson-bracket relations). Note that one example of a canonical transformation is the Hamiltonian itself:  H = H(q,p;t)\,. Namely: the Hamiltonian-canonical-transformation transforms canonical position/momenta into the conserved (constant-of-time-integration) quantity known as “energy”.

In a more general sense: the Poisson bracket is used to define a Poisson algebra, of which the Poisson manifolds are a special case. These are all named in honour of Siméon-Denis Poisson.

http://en.wikipedia.org/wiki/Poisson_bracket

Quantum particle in a box

In quantum mechanics, the particle in a box model (also known as the infinite potential well or the infinite square well) describes a particle free to move in a small space surrounded by impenetrable barriers. The model is mainly used as a hypothetical example to illustrate the differences between classical and quantum systems. In classical systems, for example a ball trapped inside a heavy box, the particle can move at any speed within the box and it is no more likely to be found at one position than another. However, when the well becomes very narrow (on the scale of a few nanometers), quantum effects become important. The particle may only occupy certain positive energy levels. Likewise, it can never have zero energy, meaning that the particle can never “sit still”. Additionally, it is more likely to be found at certain positions than at others, depending on its energy level. The particle may never be detected at certain positions, known as spatial nodes.

The particle in a box model provides one of the very few problems in quantum mechanics which can be solved analytically, without approximations. This means that the observable properties of the particle (such as its energy and position) are related to the mass of the particle and the width of the well by simple mathematical expressions. Due to its simplicity, the model allows insight into quantum effects without the need for complicated mathematics. It is one of the first quantum mechanics problems taught in undergraduate physics courses, and it is commonly used as an approximation for more complicated quantum systems. See also: the history of quantum mechanics.

One-dimensional solution

In quantum mechanics, the wavefunction gives the most fundamental description of the behavior of a particle; the measurable properties of the particle (such as its position, momentum and energy) may all be derived from the wavefunction.[3] The wavefunction ψ(x,t) can be found by solving the Schrödinger equationfor the system

\mathrm{i}\hbar\frac{\partial}{\partial t}\psi(x,t) = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x,t) +V(x)\psi(x,t),

where \hbar is the reduced Planck constantm is the mass of the particle, i is the imaginary unit and t is time.

Inside the box, no forces act upon the particle, which means that the part of the wavefunction inside the box oscillates through space and time with the same form as a free particle:[1][4]

\psi(x,t) = [A \sin(kx) + B \cos(kx)]\mathrm{e}^{-\mathrm{i}\omega t},\;

where A and B are arbitrary complex numbers. The frequency of the oscillations through space and time are given by the wavenumber k and the angular frequency ω respectively. These are both related to the total energy of the particle by the expression

E = \hbar\omega = \frac{\hbar^2 k^2}{2m},

which is known as the dispersion relation for a free particle.[1]

Initial wavefunctions for the first four states in a one-dimensional particle in a box

The size (or amplitude) of the wavefunction at a given position is related to the probability of finding a particle there by P(x,t) = | ψ(x,t) | 2. The wavefunction must therefore vanish everywhere beyond the edges of the box.[1][4] Also, the amplitude of the wavefunction may not “jump” abruptly from one point to the next.[1] These two conditions are only satisfied by wavefunctions with the form

\psi_n(x,t) = \begin{cases} A \sin(k_n x)\mathrm{e}^{-\mathrm{i}\omega_n t}, & 0 < x < L,\\ 0, & \text{otherwise,} \end{cases}

where n is a positive, whole number. The wavenumber is restricted to certain, specific values given by[5]

k_n = \frac{n \pi}{L}, \quad \mathrm{where} \quad n = \{1,2,3,4,\ldots\},

where L is the size of the box.[7] Negative values of n are neglected, since they give wavefunctions identical to the positive n solutions except for a physically unimportant sign change.[6]

Finally, the unknown constant A may be found by normalizing the wavefunction so that the total probability density of finding the particle in the system is 1. It follows that

\left| A \right| = \sqrt{\frac{2 }{L}}.

Thus, A may be any complex number with absolute value √(2/L); these different values of A yield the same physical state, so A = √(2/L) can be selected to simplify.

Energy levels

The energy of a particle in a box (black circles) and a free particle (grey line) both depend upon wavenumber in the same way. However, the particle in a box may only have certain, discrete energy levels.

The energies which correspond with each of the permitted wavenumbers may be written as[5]

E_n = \frac{n^2\hbar^2 \pi ^2}{2mL^2} = \frac{n^2 h^2}{8mL^2}.

The energy levels increase with n2, meaning that high energy levels are separated from each other by a greater amount than low energy levels are. The lowest possible energy for the particle (its zero-point energy) is found in state 1, which is given by[8]

E_1 = \frac{\hbar^2\pi^2}{2mL^2}.

The particle, therefore, always has a positive energy. This contrasts with classical systems, where the particle can have zero energy by resting motionless at the bottom of the box. This can be explained in terms of the uncertainty principle, which states that the product of the uncertainties in the position and momentum of a particle is limited by

\Delta x\Delta p \geq \frac{\hbar}{2}

It can be shown that the uncertainty in the position of the particle is proportional to the width of the box.[9] Thus, the uncertainty in momentum is roughly inversely proportional to the width of the box.[8] The kinetic energy of a particle is given by Ep2 / (2m), and hence the minimum kinetic energy of the particle in a box is inversely proportional to the mass and the square of the well width, in qualitative agreement with the calculation above.[8]

Spatial location

In classical physics, the particle can be detected anywhere in the box with equal probability. In quantum mechanics, however, the probability density for finding a particle at a given position is derived from the wavefunction as P(x) = | ψ(x) | 2. For the particle in a box, the probability density for finding the particle at a given position depends upon its state, and is given by

P_n(x) = \begin{cases}   \frac{2  }{L}\sin^2\left(\frac{n\pi x}{L}\right); & 0 < x < L \\   0; & \text{otherwise}. \end{cases}

Thus, for any value of n greater than one, there are regions within the box for which P(x) = 0, indicating that spatial nodes exist at which the particle cannot be found.

In quantum mechanics, the average, or expectation value of the position of a particle is given by

\langle x \rangle = \int_{-\infty}^{\infty} \psi^*(x) x \psi(x)\,\mathrm{d}x.

For the particle in a box, it can be shown that the average position is always \langle x \rangle = L/2, regardless of the state of the particle. In other words, the average position at which a particle in a box may be detected is exactly in the center of the quantum well; in agreement with a classical system.

Higher-dimensional boxes

If a particle is trapped in a two-dimensional box, it may freely move in the x and y-directions, between barriers separated by lengths Lx andLy respectively. Using a similar approach to that of the one-dimensional box, it can be shown that the wavefunctions and energies are given respectively by

\psi_{n_x,n_y} = \sqrt{\frac{4}{L_x L_y}} \sin \left( k_{n_x} x \right) \sin \left( k_{n_y} y\right),
E_{n_x,n_y} = \frac{\hbar^2 k_{n_x,n_y}^2}{2m},

where the two-dimensional wavevector is given by

\mathbf{k_{n_x,n_y}} = k_{n_x}\mathbf{\hat{x}} + k_{n_y}\mathbf{\hat{y}} = \frac{n_x \pi }{L_x} \mathbf{\hat{x}} + \frac{n_y \pi }{L_y} \mathbf{\hat{y}}.

For a three dimensional box, the solutions are

\psi_{n_x,n_y,n_z} = \sqrt{\frac{8}{L_x L_y L_z}} \sin \left( k_{n_x} x \right) \sin \left( k_{n_y} y \right) \sin \left( k_{n_z} z \right),
E_{n_x,n_y,n_z} = \frac{\hbar^2 k_{n_x,n_y,n_z}^2}{2m},

where the three-dimensional wavevector is given by

\mathbf{k_{n_x,n_y,n_z}} = k_{n_x}\mathbf{\hat{x}} + k_{n_y}\mathbf{\hat{y}} + k_{n_z}\mathbf{\hat{z}} = \frac{n_x \pi }{L_x} \mathbf{\hat{x}} + \frac{n_y \pi }{L_y} \mathbf{\hat{y}} + \frac{n_z \pi }{L_z} \mathbf{\hat{z}}.

An interesting feature of the above solutions is that when two or more of the lengths are the same (e.g. LxLy), there are multiple wavefunctions corresponding to the same total energy. For example the wavefunction with nx = 2,ny = 1 has the same energy as the wavefunction with nx = 1,ny = 2. This situation is called degeneracy and for the case where exactly two degenerate wavefunctions have the same energy that energy level is said to be doubly degenerate. Degeneracy results from symmetry in the system. For the above case two of the lengths are equal so the system is symmetric with respect to a 90° rotation.

References

http://en.wikipedia.org/wiki/Particle_in_a_box

http://en.wikipedia.org/wiki/Quantum_harmonic_oscillator

Modern Quantum Mechanics (2nd Edition)

Quantum Mechanics Non-Relativistic Theory, Third Edition: Volume 3

Introduction to Quantum Mechanics (2nd Edition)

Introductory Quantum Mechanics (4th Edition)

Stark effect

The Stark effect is the shifting and splitting of spectral lines of atoms and molecules due to the presence of an external static electric field. The amount of splitting and or shifting is called the Stark splitting or Stark shift. In general one distinguishes first- and second-order Stark effects. The first-order effect is linear in the applied electric field, while the second-order effect is quadratic in the field.

The Stark effect is responsible for the pressure broadening (Stark broadening) of spectral lines by charged particles. When the split/shifted lines appear in absorption, the effect is called the inverse Stark effect.

The Stark effect is the electric analogue of the Zeeman effect where a spectral line is split into several components due to the presence of a magnetic field.

The Stark effect can be explained with fully quantum mechanical approaches, but it has also been a fertile testing ground for semiclassical methods.

Mechanism

Classical electrostatics

The Stark effect originates from the interaction between a charge distribution (atom or molecule) and an external electric field. Before turning to quantum mechanics we describe the interaction classically and consider a continuous charge distribution ρ(r). If this charge distribution is non-polarizable its interaction energy with an external electrostatic potential V(r) is

 E_{\mathrm{int}} = \int \rho(\mathbf{r}) V(\mathbf{r}) d\mathbf{r}.\,

If the electric field is of macroscopic origin and the charge distribution is microscopic, it is reasonable to assume that the electric field is uniform over the charge distribution. That is, V is given by a two-term Taylor expansion,

 V(\mathbf{r}) = V(\mathbf{0}) - \sum_{i=1}^3 r_i F_i \quad \hbox{with the electric field:}\quad F_i \equiv  -\left. \left(\frac{\partial V}{\partial r_i} \right)\right|_{\mathbf{0}},

where we took the origin 0 somewhere within ρ. Setting V(\mathbf{0}) as the zero energy, the interaction becomes

  E_{\mathrm{int}} = - \sum_{i=1}^3 F_i  \int \rho(\mathbf{r}) r_i d\mathbf{r} \equiv - \sum_{i=1}^3 F_i  \mu_i = - \mathbf{F}\cdot \boldsymbol{\mu}.

Here we have introduced the dipole moment μ of ρ as an integral over the charge distribution. In case ρ consists of N point charges qj this definition becomes a sum

 \boldsymbol{\mu} \equiv \sum_{j=1}^N  q_j \mathbf{r}_j.

Perturbation theory

Turning now to quantum mechanics we see an atom or a molecule as a collection of point charges (electrons and nuclei), so that the second definition of the dipole applies. The interaction of atom or molecule with a uniform external field is described by the operator

 V_{\mathrm{int}} = - \mathbf{F}\cdot \boldsymbol{\mu}.

This operator is used as a perturbation in first- and second-order perturbation theory to account for the first- and second-order Stark effect.

First order

Let the unperturbed atom or molecule be in a g-fold degenerate state with orthonormal zeroth-order state functions  \psi^0_1, \ldots, \psi^0_g . (Non-degeneracy is the special case g = 1). According to perturbation theory the first-order energies are the eigenvalues of the gg matrix with general element

 (\mathbf{V}_{\mathrm{int}})_{kl} = \langle \psi^0_k |  V_{\mathrm{int}} | \psi^0_l \rangle = -\mathbf{F}\cdot \langle \psi^0_k | \boldsymbol{\mu} | \psi^0_l \rangle, \qquad k,l=1,\ldots, g.

If g = 1 (as is often the case for electronic states of molecules) the first-order energy becomes proportional to the expectation (average) value of the dipole operator \boldsymbol{\mu},

 E^{(1)} = -\mathbf{F}\cdot \langle \psi^0_1 | \boldsymbol{\mu} | \psi^0_1 \rangle = -\mathbf{F}\cdot \langle  \boldsymbol{\mu} \rangle.

Because a dipole moment is a polar vector, the diagonal elements of the perturbation matrix Vint vanish for systems with an inversion center (such as atoms). Molecules with an inversion center in a non-degenerate electronic state do not have a (permanent) dipole and hence do not show a linear Stark effect.

In order to obtain a non-zero matrix Vint for systems with an inversion center it is necessary that some of the unperturbed functions  \psi^0_i have opposite parity (obtain plus and minus under inversion), because only functions of opposite parity give non-vanishing matrix elements. Degenerate zeroth-order states of opposite parity occur for excited hydrogen-like (one-electron) atoms. Such atoms have the principal quantum number n among their quantum numbers. The excited state of hydrogen-like atoms with principal quantum number n is n2-fold degenerate and

 n^2 = \sum_{\ell=0}^{n-1} (2 \ell + 1),

where \ell is the azimuthal (angular momentum) quantum number. For instance, the excited n = 4 state contains the following \ell states,

 16 = 1 + 3 + 5 +7 \;\; \Longrightarrow\;\;  n=4\;\hbox{contains}\; s\oplus p\oplus d\oplus f.

The one-electron states with even \ell are even under parity, while those with odd \ell are odd under parity. Hence hydrogen-like atoms with n>1 show first-order Stark effect.

The first-order Stark effect occurs in rotational transitions of symmetric top molecules (but not for linear and asymmetric molecules). In first approximation a molecule may be seen as a rigid rotor. A symmetric top rigid rotor has the unperturbed eigenstates

 |JKM \rangle = (D^J_{MK})^* \quad\mathrm{with}\quad M,K= -J,-J+1,\dots,J

with 2(2J+1)-fold degenerate energy for |K| > 0 and (2J+1)-fold degenerate energy for K=0. Here DJMK is an element of the Wigner D-matrix. The first-order perturbation matrix on basis of the unperturbed rigid rotor function is non-zero and can be diagonalized. This gives shifts and splittings in the rotational spectrum. Quantitative analysis of these Stark shift yields the permanent electric dipole moment of the symmetric top molecule.

Second order

As stated, the quadratic Stark effect is described by second-order perturbation theory. The zeroth-order problems

 H^{(0)} \psi^0_k = E^{(0)}_k \psi^0_k, \quad k=0,1, \ldots, \quad E^{(0)}_0 < E^{(0)}_1 \le E^{(0)}_2, \dots

are assumed to be solved. It is usual to assume that the zeroth-order state to be perturbed is non-degenerate. If we take the ground state as the non-degenerate state under consideration (for hydrogen-like atoms: n = 1), perturbation theory gives

 E^{(2)} = \sum_{k>0} \frac{\langle \psi^0_0 | V_\mathrm{int} | \psi^0_k \rangle \langle \psi^0_k | V_\mathrm{int} | \psi^0_0 \rangle}{E^{(0)}_0 - E^{(0)}_k} =- \frac{1}{2} \sum_{i,j=1}^3 F_i \alpha_{ij} F_j

with the components of the polarizability tensor α defined by

 \alpha_{ij}\equiv -2\sum_{k>0} \frac{\langle \psi^0_0 | \mu_i | \psi^0_k \rangle \langle \psi^0_k | \mu_j | \psi^0_0\rangle}{E^{(0)}_0 - E^{(0)}_k}.

The energy E(2) gives the quadratic Stark effect.

Because of their spherical symmetry the polarizability tensor of atoms is isotropic,

 \alpha_{ij} = \alpha_0 \delta_{ij} \Longrightarrow E^{(2)} = -\frac{1}{2} \alpha_0 F^2,

which is the quadratic Stark shift for atoms. For many molecules this expression is not too bad an approximation, because molecular tensors are often reasonably isotropic.

Problems

The perturbative treatment of the Stark effect has some problems. In the presence of an electric field, states of atoms and molecules that were previously bound (square-integrable), become formally (non-square-integrable) resonances of finite width. These resonances may decay in finite time via field ionization. For low lying states and not too strong fields the decay times are so long, however, that for all practical purposes the system can be regarded as bound. For highly excited states and/or very strong fields ionization may have to be accounted for. (See also the article on the Rydberg atom).

References

Modern Quantum Mechanics (2nd Edition)

http://en.wikipedia.org/wiki/Stark_effect

The Theory of Atomic Spectra