User:Manishearth/Incubator

To see my major contributions, go to User:Manishearth/Bigcontribs

Working on a script, list of script pages at User:Manishearth/GadgetUS  DO STUFF BELOW THIS LINE PLEASE

=(Just for the latex syntax)= $$ \vec{E}=\begin{cases} \frac{\mu_0Ni_0\omega\sin(\omega t)r}{2l}\hat{\tau}, & rR \end{cases} $$ where r is radial distance of pont in polar coordinates, $$\hat{\tau}$$ is unit tangential vector, R is radius of solenoid $$ \vec{E}=\plusmn\begin{cases} \frac{\mu_0Ni_0\omega\sin(\omega t)}{2l}(x\hat{j}-y\hat{i}), & \sqrt{x^2+y^2}R \end{cases} $$ $$ F_0\hat{k} $$

=Rewrite of Reduced Mass=

Reduced mass is the "effective" inertial mass appearing in the two-body problem of Newtonian mechanics. This is a quantity with the unit of mass, which allows the two-body problem to be solved as if it were a one-body problem. Note however that the mass determining the gravitational force is not reduced. In the computation one mass can be replaced by the reduced mass, if this is compensated by replacing the other mass by the sum of both mass.
 * $$m_\text{red} = \mu = \cfrac{1}{\cfrac{1}{m_1}+\cfrac{1}{m_2}} = \cfrac{m_1 m_2}{m_1 + m_2},\!\,$$

Uses
Given two bodies, one with mass $$m_{1}\!\,$$ and the other with mass $$m_{2}\!\,$$, they will orbit the barycenter of the two bodies. The equivalent one-body problem, with the position of one body with respect to the other as the unknown, is that of a single body of mass $$\mu$$where the force on this mass is given by the gravitational force between the two bodies. The reduced mass is always less than or equal to the mass of each body and is half of the harmonic mean of the two masses.


 * {| class="toccolours collapsible collapsed" width="80%" style="text-align:left"

!Derivation
 * This can be proven easily. Use Newton's second law, the force exerted by body 2 on body 1 is
 * $$F_{12} = m_1 a_1. \!\,$$
 * $$F_{12} = m_1 a_1. \!\,$$

The force exerted by body 1 on body 2 is
 * $$F_{21} = m_2 a_2. \!\,$$

According to Newton's third law, for every action there is an equal and opposite reaction:
 * $$F_{12} = - F_{21}.\!\,$$

Therefore,
 * $$m_1 a_1 = - m_2 a_2. \!\,$$

and


 * $$a_2=-{m_1 \over m_2} a_1. \!\,$$

The relative acceleration between the two bodies is given by
 * $$a= a_1-a_2 = \left({1+{m_1 \over m_2}}\right) a_1 = {{m_2+m_1}\over{m_1 m_2}} m_1 a_1 = {F_{12} \over m_\text{red}}. $$

So we conclude that body 1 moves with respect to the position of body 2 as a body of mass equal to the reduced mass.

Alternatively, a Lagrangian description of the two-body problem gives a Lagrangian of


 * $$L = {1 \over 2} m_1 \mathbf{\dot{r}}_1^2 + {1 \over 2} m_2 \mathbf{\dot{r}}_2^2 - V(\vert \mathbf{r}_1 - \mathbf{r}_2 \vert ) \!\,$$

where $$m_i, \mathbf{r}_i $$ are the mass and position vector of the $$i$$ th particle, respectively. The potential energy $$V$$ takes this functional dependence as it is only dependent on the absolute distance between the particles. If we define $$\mathbf{r} \equiv \mathbf{r}_1 - \mathbf{r}_2 $$ and let the centre of mass coincide with our origin in this reference frame, i.e. $$ m_1 \mathbf{r}_1 + m_2 \mathbf{r}_2 = 0 $$, then


 * $$ \mathbf{r}_1 = \frac{m_2 \mathbf{r}}{m_1 + m_2}, \mathbf{r}_2 = \frac{-m_1 \mathbf{r}}{m_1 + m_2}.$$

Then substituting above gives a new Lagrangian


 * $$ L = {1 \over 2}m_\text{red} \mathbf{\dot{r}}^2 - V(r), $$

where $$m_\text{red} = \frac{m_1 m_2}{m_1 + m_2} $$, the reduced mass. Thus we have reduced the two-body problem to that of one body.

The reduced mass is frequently denoted by the Greek letter $$\mu\!\,$$; note however that the standard gravitational parameter is also denoted by $$\mu\!\,$$.

In the case of the gravitational potential energy $$V(\vert \mathbf{r}_1 - \mathbf{r}_2 \vert ) = - G m_1 m_2 / \vert \mathbf{r}_1 - \mathbf{r}_2 \vert\!\,$$ we find that the position of the first body with respect to the second is governed by the same differential equation as the position of a body with the reduced mass orbiting a body with a mass equal to the sum of the two masses, because


 * $$m_1 m_2 = (m_1+m_2) m_\text{red}\!\,$$

"Reduced mass" may also refer more generally to an algebraic term of the form


 * $$x_\text{red} = {1 \over {1 \over x_1} + {1 \over x_2}} = {x_1 x_2 \over x_1 + x_2}\!\,$$

that simplifies an equation of the form


 * $$\ {1\over x_\text{eq}} = \sum_{i=1}^n {1\over x_i} = {1\over x_1} + {1\over x_2} + \cdots+ {1\over x_n}.\!\,$$


 * }

In such a two body problem, the moment of inertia of the system about the center of mass is given by: $$I = \mu r^2$$ Where r is the distance between the two bodies.
 * {| class="toccolours collapsible collapsed" width="80%" style="text-align:left"

!Derivation Let r1 and r2 be the distances of the two bodies from the center of mass.  Taking the center of mass as the origin, the equation for center of mass becomes:
 * $$r_{cm} = 0 = \frac{m_1r_1 - m_2r_2}{m_1 + r_1}$$ 

From this, we get:
 * $$m_1r_1=m_2r_2$$ 

Now, moment of inertia about center of mass is: 
 * $$I = m_1r_1^2 + m_2r_2^2$$

As $$m_1r_1=m_2r_2$$,
 * $$I = m_1r_1^2 + m_1r_1r_2 = m_1r_1(r_1 + r_2) = m_1r_1r$$

But, as
 * $$ r_2 = \frac{m_1r_1}{m_2}$$, and $$r=r_1+r_2$$, 
 * $$r=r_1\left(1 + \frac{m_1}{m_2}\right)= \frac{r_1 (m_1+m_2)}{m_2}$$ <BR/>

Or,
 * $$r_1 = \frac{r m_2}{m_1 + m_2}$$ <BR />

Substituting in the expression for I, we get:
 * $$I=m_1r_1r = \frac{m_1m_2r^2}{m_1+m_2}= \mu r^2$$

The reduced mass is typically used as a relationship between two system elements in parallel, such as resistors; whether these be in the electrical, thermal, hydraulic, or mechanical domains. This relationship is determined by the physical properties of the elements as well as the continuity equation linking them.
 * }

Other uses
Reduced mass crops up in a multitude of two-body problems. In a collision with a coefficient of restitution e, the change in kinetic energy can be written as $$\Delta K = \frac{1}{2}\mu v^2_{rel}(e^2-1)$$, where vrel is the relative velocity of the bnodies before colission.

Todo
=Maxwell (Not to be published, just using WP's LaTeX)= Starting conditions: $$\vec B = \beta t \hat k, \vec J=0, \rho=0$$<BR/> <BR/> $$\text{Maxwell's laws:} \begin{cases} \nabla \cdot \vec E = \rho/\epsilon_0 & \dots (M_1) \\ \nabla \cdot \vec B = 0 & \dots (M_2) \\ \nabla \times \vec E = -\partial_t \vec B & \dots (M_3) \\ \nabla \times \vec B = \mu_0 \vec J + \mu_0\epsilon_0\partial_t\vec E & \dots (M_4) \end{cases} $$<BR> Verifying our choice of B with $$M_2$$,<BR> $$LHS=\nabla \times \vec B=\nabla \times (\beta t \hat k) = 0 = RHS$$<BR/> Now, using $$M_4$$<BR /> $$LHS=0, \vec J=0,$$<BR/> $$\Rightarrow 0=\mu_00 +\mu_0\epsilon_0\partial_t\vec E$$<BR/> $$\Rightarrow \partial_tE=0$$<BR/> Therefore, E is independant of time.<BR/ ><BR /> Now, by $$M_3, \nabla \times \vec E = -\partial_t \vec B = -\partial_t (\beta t \hat k) = -\beta \hat k$$<BR> Let $$\vec E=E_x \hat i + E_y \hat j + E_z \hat k$$<BR/> $$\therefore\begin{pmatrix} \hat i (\partial_y E_z - \partial_z E_y) \\ +\hat j (\partial_x E_z - \partial_z E_x) \\ +\hat k (\partial_x E_y - \partial_y E_x) \end{pmatrix} = \begin{pmatrix} \hat i (0) \\ +\hat j (0) \\ +\hat k (-\beta) \end{pmatrix} \qquad \dots (1) $$<BR/><BR/> Now, by symmetry of the original conditions, we can assume that:<BR /> $$|E_x|=|E_y|$$ at a point (x,y,z)<BR/> And also, $$|\partial_x E_y|=|\partial_y E_x|$$<BR/> $$=\beta/2$$, by z-component of (1)<BR/> <BR /></BR> $$\therefore E_x= \frac{\beta y}{2} + f_1(x,z) + c_1,$$<BR/> $$E_x= \frac{\beta x}{2} + f_2(y,z) + c_2$$<BR/> $$\text{let } E_x= f_3(x,y,z) + c_3$$<BR/> <BR /> By x and y components of (1),<BR/> $$\partial_yf_3=\partial_zf_2, \partial_xf_3=\partial_zf_1$$<BR/> For simplicity, we can assume:<BR />
 * Restructure
 * Derive moment of inertia of two bodies = mu*r^2, where r is dist betwn two bodies
 * Add more "other uses"
 * It does not only pertain to two body problem, it crops up in all two-body interactions.
 * Images
 * Explain spring connection with halfmuvsquared formula.

$$f_1 = ix + az$$<BR/> $$f_2 = my + bz $$<BR/> $$\text{let } f_3 = ax + by + nz$$<BR/> (Where l,m,n,a,b are constants)<BR /><BR/> $$\therefore \vec E = \hat i(\frac{\beta y}{2} + lx + az + c_1)$$<BR/> $$+\hat j(\frac{-\beta x}{2} + my + bz + c_2$$<BR/> $$+ \hat k (ax + by + nz + c_3)$$<BR/> <BR /> Applying $$(M_1)$$ to $$\vec E$$,<BR /> $$l+m+n=0$$<BR/> <BR /> $$\therefore \text{when } \vec B = \beta t \hat k, \vec J=0, \rho=0,$$<BR> $$\vec E = \hat i(\frac{\beta y}{2} + lx + az + c_1)$$<BR/> $$+\hat j(\frac{-\beta x}{2} + my + bz + c_2$$<BR/> $$+ \hat k (ax + by + nz + c_3)$$<BR/> This answer can be verified with the four Maxwell's Laws.