User:Akanksh

= Heisenberg uncertainty derivation explanation =

Subtracting eqn.(32) from (31), we get

$$ \left ( \frac{\delta k}{2} (x_2 - x_1) = \pi \right ) \Rightarrow \left ( x_2 - x_1 = \frac{2 \pi}{\delta k} \right ) $$

Therefore, Error in the measurement of the particle:

$$ \Delta x = \frac{2 \pi}{ \Delta k} = \frac{2 \pi}{\Delta \left ( \frac{2 \pi}{\lambda} \right )} $$

Now, based on deBroglie’s matter wave description, we can define a relatonship between the wavenumber [k] and the momentum [p] of the matter wave-particle:

$$ p = \hbar \Delta k = \frac{h}{2 \pi } \frac{2 \pi }{ \lambda } \Rightarrow \frac{2 \pi}{\lambda} = \frac{2 \pi p}{h} $$

Thus frome the above we have:

$$ \frac{2 \pi}{\Delta \left ( \frac{2 \pi}{\lambda} \right )} = \frac{2 \pi}{\Delta \left ( \frac{2 \pi p}{h} \right )} = \frac{1}{\Delta \left ( \frac{p}{h} \right )} $$

and because 'h' is a constant, we have:

$$ \frac{1}{\Delta \left ( \frac{p}{h} \right )} = \frac{\Delta h}{\Delta p} = \frac{h}{\Delta p} $$

So now we have:

$$ \Delta x. \Delta p = h $$

Note that this is true only if the probability distribution is normal. If it isn't $$\Delta x. \Delta p$$ will be greater, as the normal distribution turns out to have the minimum possible product. This means that we can write our more general physical law as:

$$ \Delta x. \Delta p \geq \hbar \left [ = \frac {h}{2 \pi} \right ] $$

References:
 * http://www.tjhsst.edu/~2011akessler/notes/hup.pdf [page 3]
 * http://www.mysearch.org.uk/website1/html/543.Uncertainty.html [eqn 16]

= Proof of Determinant to Dot and Cross Product =

$$ M = \begin{pmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{pmatrix} = \begin{pmatrix} \overrightarrow{A} \\ \overrightarrow{B} \\ \overrightarrow{C} \end{pmatrix} $$ $$ \parallel M\parallel = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} $$ $$ = a_1 \begin{vmatrix} b_2 & b_3 \\ c_2 & c_3 \end{vmatrix} - a_2 \begin{vmatrix} b_1 & b_3 \\ c_1 & c_3 \end{vmatrix} + a_3 \begin{vmatrix} b_1 & b_2 \\ c_1 & c_2 \end{vmatrix} $$ $$ = a_1 \times (b_2 c_3 - b_3 c_2) - a_2 \times ( b_1 c_3 - b_3 c_1 ) + a_3 \times ( b_1 c_2 - b_2 c_1 ) $$ $$ = a_1 \times (b_2 c_3 - b_3 c_2) + a_2 \times ( b_3 c_1 - b_1 c_3 ) + a_3 \times ( b_1 c_2 - b_2 c_1 ) $$ $$ = \overrightarrow{A} \cdot \langle b_2 c_3 - b_3 c_2, b_3 c_1 - b_1 c_3, b_1 c_2 - b_2 c_1 \rangle $$ $$ \langle b_2 c_3 - b_3 c_2, b_3 c_1 - b_1 c_3, b_1 c_2 - b_2 c_1 \rangle = \langle M_1 \rangle $$ $$ \Rightarrow \parallel M\parallel = \overrightarrow{A} \cdot \langle M_1 \rangle $$ $$ \overrightarrow{B} \times \overrightarrow{C} = \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = (b_2 c_3 - b_3 c_2) \times \widehat{i} + ( b_3 c_1 - b_1 c_3 ) \times \widehat{j} + ( b_1 c_2 - b_2 c_1 ) \times \widehat{k} = \langle b_2 c_3 - b_3 c_2, b_3 c_1 - b_1 c_3, b_1 c_2 - b_2 c_1 \rangle = \langle M_2 \rangle $$ $$ \langle M_1 \rangle = \langle M_2 \rangle \Rightarrow \parallel M\parallel = \overrightarrow{A} \cdot \langle M_2 \rangle $$ $$ \Rightarrow \parallel M \parallel = \overrightarrow{A} \cdot ( { \overrightarrow{B} \times \overrightarrow{C} } ) $$

= Special Functions =

Complete Fermi Dirac Integrals
$$F_j(x) := (1/\Gamma(j+1)) \int_0^\infty dt (t^j / (\exp(t-x) + 1))$$

Incomplete Fermi Dirac Integrals
$$F_j(x,b):= (1/\Gamma(j+1))\int_b^\infty dt(t^j / (\exp(t-x) + 1))$$

Airy Functions and Derivatives
$$Ai(x) = (1/\pi) \int_0^\infty \cos((1/3) t^3 + xt) dt$$ $$Bi(x) = (1/\pi) \int_0^\infty (e^(-(1/3) t^3) + \sin((1/3) t^3 + xt)) dt$$

Clausen Functions
$$Cl_2(x) = - \int_0^x dt \log(2 \sin(t/2))$$

Normalized Hydrogenic Bound States
$$R_n := 2 (Z^{3/2}/n^2) \sqrt{(n-l-1)!/(n+l)!} \exp(-Z r/n) (2Zr/n)^l L^{2l+1}_{n-l-1}(2Zr/n)$$ Where:$$L^k_n(x) = (-1)^k (d^k/dx^k) L_(n+k)(x)$$

Legendre Forms
$$ F(\phi,k) = \int_0^\phi dt 1/\sqrt((1 - k^2 \sin^2(t)))$$ $$ E(\phi,k) = \int_0^\phi dt  \sqrt((1 - k^2 \sin^2(t)))$$ $$ \Pi(\phi,k,n) = \int_0^\phi dt 1/((1 + n \sin^2(t))\sqrt(1 - k^2 \sin^2(t))$$

Carlson Forms
$$RC(x,y) = 1/2 \int_0^\infty dt (t+x)^(-1/2) (t+y)^(-1)$$ $$ RD(x,y,z) = 3/2 \int_0^\infty dt (t+x)^(-1/2) (t+y)^(-1/2) (t+z)^(-3/2)$$ $$ RF(x,y,z) = 1/2 \int_0^\infty dt (t+x)^(-1/2) (t+y)^(-1/2) (t+z)^(-1/2)$$ $$ RJ(x,y,z,p) = 3/2 \int_0^\infty dt(t+x)^(-1/2) (t+y)^(-1/2) (t+z)^(-1/2) (t+p)^(-1)$$

Gamma Functions
$$ \Gamma(x) = \int_0^\infty dt t^{x-1} \exp(-t)$$

Psi (Digamma) Function
$$ \psi^{(n)}(x) = (d/dx)^n  \psi(x) = (d/dx)^{n+1} \log(\Gamma(x)) \cdot $$

Riemann Zeta Function
$$\zeta(s) = \sum_{k=1}^\infty k^{-s}$$

Hurwitz Zeta Function
$$\zeta(s,q) = \sum_0^\infty (k+q)^{-s}$$

Eta Function
$$\eta(s)=(1-2^{1-s})\zeta(s) \cdot$$

Transport Functions
$$J(n,x) := \int_0^x dt t^n e^t /(e^t - 1)^2$$

Schwarzschild Radius Mass-Density relation
$$ \rho = \frac{3 c^6}{32 \pi G^3 m^2} \cdot \,\!$$

= MISC =

Lagrangian
The equations of motion are obtained by means of an action principle, written as:


 * $$\frac{\delta \mathcal{S}}{\delta \varphi_i} = 0\,.$$

where the action, $$\mathcal{S}$$, is a functional of the dependent variables $$\varphi_i(s)$$ with their derivatives and s itself


 * $$\mathcal{S}\left[\varphi_i, \frac{\partial \varphi_i} {\partial s}\right] = \int{ \mathcal{L} \left[\varphi_i [s], \frac{\partial \varphi_i [s]}{\partial s^\alpha}, s^\alpha\right] \, \mathrm{d}^n s }$$

and where $$s = \{ s^\alpha \} \!$$ denotes the set of n independent variables of the system, indexed by $$\alpha = 1, 2, 3, \ldots, n .$$

The equations of motion obtained from this functional derivative are the Euler–Lagrange equations of this action. For example, in the classical mechanics of particles, the only independent variable is time, t. So the Euler-Lagrange equations are
 * $$\frac{d}{d t}\frac{\partial\mathcal L}{\partial\dot \varphi_i} = \frac{\partial\mathcal L}{\partial\varphi_i} .$$

Dynamical systems whose equations of motion are obtainable by means of an action principle on a suitably chosen Lagrangian are known as Lagrangian dynamical systems. Examples of Lagrangian dynamical systems range from the classical version of the Standard Model, to Newton's equations, to purely mathematical problems such as geodesic equations and Plateau's problem.

Deriving Hamilton's equations
We can derive Hamilton's equations by looking at how the Lagrangian changes as you change the time and the positions and velocities of particles

\mathrm{d} \mathcal{L} = \sum_i \left ( \frac{\partial \mathcal{L}}{\partial q_i} \mathrm{d} q_i + \frac{\partial \mathcal{L}}{\partial {\dot q_i}} \mathrm{d} {\dot q_i} \right ) + \frac{\partial \mathcal{L}}{\partial t} \mathrm{d}t \,.$$

Now the generalized momenta were defined as $$p_i = \frac{\partial \mathcal{L}}{\partial {\dot q_i}}$$ and Lagrange's equations tell us that

\frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial \mathcal{L}}{\partial {\dot q_i}} - \frac{\partial \mathcal{L}}{\partial q_i} =0 \,$$

We can rearrange this to get

\frac{\partial \mathcal{L}}{\partial q_i} = {\dot p}_i \,$$

and substitute the result into the variation of the Lagrangian
 * $$ \mathrm{d} \mathcal{L} = \sum_i \left [ {\dot p}_i \mathrm{d} q_i + p_i \mathrm{d} {\dot q_i} \right] + \frac{\partial \mathcal{L}}{\partial t}\mathrm{d}t

\,.$$

We can rewrite this as

\mathrm{d} \mathcal{L} = \sum_i \left [ {\dot p}_i    \mathrm{d}q_i + \mathrm{d}\left ( p_i {\dot q_i} \right ) - {\dot q_i} \mathrm{d} p_i  \right ] + \frac{\partial \mathcal{L}}{\partial t}\mathrm{d}t \,$$

and rearrange again to get

\mathrm{d} \left ( \sum_i p_i {\dot q_i} - \mathcal{L} \right ) = \sum_i \left [ -{\dot p}_i  \mathrm{d} q_i + {\dot q_i} \mathrm{d}p_i  \right] - \frac{\partial \mathcal{L}}{\partial t}\mathrm{d}t \,.$$

The term on the left-hand side is just the Hamiltonian that we have defined before, so we find that

\mathrm{d} \mathcal{H} = \sum_i \left [ -{\dot p}_i \mathrm{d} q_i + {\dot q_i} \mathrm{d} p_i  \right] - \frac{\partial \mathcal{L}}{\partial t}\mathrm{d}t = \sum_i \left [ \frac{\partial \mathcal{H}}{\partial q_i} \mathrm{d} q_i + \frac{\partial \mathcal{H}}{\partial p_i} \mathrm{d} p_i \right ] + \frac{\partial \mathcal{H}}{\partial t}\mathrm{d}t \,$$

where the second equality holds because of the definition of the partial derivatives. Associating terms from both sides of the equation above yields Hamilton's equations
 * $$\frac{\partial \mathcal{H}}{\partial q_j} =- \dot{p}_j \,, \quad \frac{\partial \mathcal{H}}{\partial p_j} = \dot{q}_j \,, \quad \frac{\partial \mathcal{H}}{\partial t } = - {\partial \mathcal{L} \over \partial t} \,.$$

Covariant Derivative
A covariant derivative is a (Koszul) connection on the tangent bundle and other tensor bundles. Thus it has a certain behavior on functions, on vector fields, on the duals of vector fields (i.e., covector fields), and most generally of all, on arbitrary tensor fields.

Functions
Given a function $$f\,$$, the covariant derivative $$\nabla_{\mathbf v}f$$ coincides with the normal differentiation of a real function in the direction of the vector v, usually denoted by $${\mathbf v}f$$ and by $$df({\mathbf v})$$.

Vector fields
A covariant derivative $$\nabla$$ of a vector field $${\mathbf u}$$ in the direction of the vector $${\mathbf v} $$ denoted $$\nabla_{\mathbf v} {\mathbf u}$$ is defined by the following properties for any vector v, vector fields u, w and scalar functions f and g:
 * 1) $$\nabla_{\mathbf v} {\mathbf u}$$ is algebraically linear in $${\mathbf v}$$ so $$\nabla_{f{\mathbf v}+g{\mathbf w}} {\mathbf u}=f\nabla_{\mathbf v} {\mathbf u}+g\nabla_{\mathbf w} {\mathbf u}$$
 * 2) $$\nabla_{\mathbf v} {\mathbf u}$$  is additive in $${\mathbf u}$$ so $$\nabla_{\mathbf v}({\mathbf u}+{\mathbf w})=\nabla_{\mathbf v} {\mathbf u}+\nabla_{\mathbf v} {\mathbf w}$$
 * 3) $$\nabla_{\mathbf v} {\mathbf u}$$ obeys the product rule, i.e. $$\nabla_{\mathbf v} f{\mathbf u}=f\nabla_{\mathbf v} {\mathbf u}+{\mathbf u}\nabla_{\mathbf v}f$$ where $$\nabla_{\mathbf v}f$$ is defined above.

Note that $$\nabla_{\mathbf v} {\mathbf u}$$ at point p depends on the value of v at p and on values of u in a neighbourhood of p because of the last property, the product rule.

Covector fields
Given a field of covectors (or one-form) $$\alpha$$, its covariant derivative $$\nabla_{\mathbf v}\alpha$$ can be defined using the following identity which is satisfied for all vector fields u
 * $$\nabla_{\mathbf v}(\alpha({\mathbf u}))=(\nabla_{\mathbf v}\alpha)({\mathbf u})+\alpha(\nabla_{\mathbf v}{\mathbf u}).$$

The covariant derivative of a covector field along a vector field v is again a covector field.

Tensor fields
Once the covariant derivative is defined for fields of vectors and covectors it can be defined for arbitrary tensor fields using the following identities where $$\varphi$$ and $$\psi\,$$ are any two tensors:
 * $$\nabla_{\mathbf v}(\varphi\otimes\psi)=(\nabla_{\mathbf v}\varphi)\otimes\psi+\varphi\otimes(\nabla_{\mathbf v}\psi),$$

and if $$\varphi$$ and $$\psi$$ are tensor fields of the same tensor bundle then
 * $$\nabla_{\mathbf v}(\varphi+\psi)=\nabla_{\mathbf v}\varphi+\nabla_{\mathbf v}\psi.$$

The covariant derivative of a tensor field along a vector field v is again a tensor field of the same type.

Christoffel symbols
The Christoffel symbols can be derived from the vanishing of the covariant derivative of the metric tensor $$g_{ik}\ $$:


 * $$0 = \nabla_\ell g_{ik}=\frac{\partial g_{ik}}{\partial x^\ell}- g_{mk}\Gamma^m {}_{i\ell} - g_{im}\Gamma^m {}_{k\ell}.\ $$

As a shorthand notation, the nabla symbol and the partial derivative symbols are frequently dropped, and instead a semi-colon and a comma are used to set off the index that is being used for the derivative. Thus, the above is sometimes written as
 * $$0 = \,g_{ik;\ell} = g_{ik,\ell} - g_{mk} \Gamma^m {}_{i\ell} - g_{im} \Gamma^m {}_{k\ell}. \ $$

By permuting the indices, and resumming, one can solve explicitly for the Christoffel symbols as a function of the metric tensor:
 * $$\Gamma^i {}_{k\ell}=\frac{1}{2}g^{im} \left(\frac{\partial g_{mk}}{\partial x^\ell} + \frac{\partial g_{m\ell}}{\partial x^k} - \frac{\partial g_{k\ell}}{\partial x^m} \right) = {1 \over 2} g^{im} (g_{mk,\ell} + g_{m\ell,k} - g_{k\ell,m}), \ $$

where the matrix $$(g^{jk}\ )$$ is an inverse of the matrix $$(g_{jk}\ )$$, defined as (using the Kronecker delta, and Einstein notation for summation) $$g^{j i} g_{i k}= \delta^j {}_k\ $$. Although the Christoffel symbols are written in the same notation as tensors with index notation, they are not tensors. Indeed, they do not transform like tensors under a change of coordinates; see below.

NB. Note that most authors choose to define the Christoffel symbols in a holonomic basis, which is the convention followed here. In an anholonomic basis, the Christoffel symbols take the more complex form
 * $$\Gamma^i {}_{k\ell}=\frac{1}{2}g^{im} \left(

g_{mk,\ell} + g_{m\ell,k} - g_{k\ell,m} + c_{mk\ell}+c_{m\ell k} + c_{k\ell m} \right) \ $$

where $$c_{k\ell m}=g_{mp} {c_{k\ell}}^p\ $$ are the commutation coefficients of the basis; that is,
 * $$[e_k,e_\ell] = c_{k\ell}{}^m e_m\,\ $$

where ek are the basis vectors and $$[,]\ $$ is the Lie bracket. An example of an anholonomic basis with non-vanishing commutation coefficients are the unit vectors in spherical and cylindrical coordinates.

The expressions below are valid only in a holonomic basis, unless otherwise noted.

-- $$ \alpha \kappa \alpha \eta \kappa \varsigma h \ \nu \alpha \varsigma h \iota \varsigma \tau h \,\!$$

-- &alpha;&kappa;&alpha;&eta;&kappa;&sigmaf;h &nu;&alpha;&sigmaf;h&iota;&sigmaf;&tau;h

=Akanksh Vashisth=