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3D common way Definition of angular momentum vector ($m = 1$) \begin{equation} L_l = \epsilon_{lij}x^iv^j \end{equation} ($L_l$ dual to $L^{ij} = x^iv^j - x^jv^i$ tensor) Rortational velocity of … particle \begin{equation} v^j = \epsilon^{jrk}\omega_rx_k. \end{equation} ($\omega_r$ dual to $\omega^{jk}$ antisymmetric tensor of angular velocity $\epsilon^{jrk}\omega_r = \omega^{jk … answered May 15 by Sergio This is not true. The linear energy$K_L$is given by \begin{equation} K_L=\frac{1}{2}mv^2 \end{equation} while the rotational energy$K_R$is given by \begin{equation} K_R=\frac{1}{2}I\omega^2 \end … {equation} where$m$is the mass of the ball,$I$the moment of inertia,$v$the linear velocity and$\omega$the angular velocity; in particular,$v$and$\omega$are related by \begin{equation} v … answered Jul 1 '17 by Saramago 1answer maybe very silly question :( In Magnetism: \begin{equation} \ B=H+4\pi M (SI) \end{equation} \begin{equation} \ B=\mu(H+M) (CGS) \end{equation} in these formulas M is Volume Magnetisation \begin{equation} \ M=d \mu/dV \end{equation} or \begin{equation} \ M=d \mu/dm \end{equation} … asked Mar 20 '15 by pan91 The chain is initially at rest, so \begin{equation} KE_{o} = 0 \end{equation} The force of friction is given by \begin{equation} f(x) = \mu \rho (L-x) g \end{equation} The net force on the chain is … \begin{equation} \sum F = F - \mu \rho (L-x) g \end{equation} Work done on the chain is the integral of force over distance, so \begin{equation} W = \int_{0}^{L}F - \mu \rho (L-x) gdx \end{equation answered Sep 2 '12 by yzernik the correct answer is \begin{equation} [Ĥ ,P̂ ]Ψ(r,t)=iĤγ̂_{0}Ψ(−r,t)−P̂ Ĥ Ψ(r,t)= \end{equation} \begin{equation} i((α̂⋅p̂)+γ̂_{0}m)γ̂_{0}Ψ(−r,t)−iγ̂_{0}(−(α̂⋅p̂)+γ̂_{0}m)Ψ(−r,t)= \end{equation … } \begin{equation} i[(α̂⋅p̂)γ̂_{0}+γ̂_{0}(α̂⋅p̂)]Ψ(−r,t)+i[γ̂_{0}mγ̂_{0}-γ̂_{0}mγ̂_{0}]Ψ(−r,t)= \end{equation} \begin{equation} i[(α̂⋅p̂)γ̂_{0}+γ̂_{0}(α̂⋅p̂)]Ψ(−r,t)= \end{equation} \begin{equation} =0 … answered Jul 29 '16 by Pedro Diego Silva Elastic Collisions: Conservation of momentum of object 1 and object 2: \begin{equation} (m_1v_1)_{before} = (m_1v_1 + m_2v_2)_{after} \end{equation} Conservation of kinetic energy (elastic collision … assumption): \begin{equation} (\frac{1}{2}m_1v_1^2)_{before} = (\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2)_{after} \end{equation} Since equal mass, momentum equation becomes: \begin{equation} (v_1 … answered Sep 25 '16 by Ali 1answer The masses of electron, proton and neutron (in SI units) are (approx.): \begin{equation} m_e=\text{electron mass}=9.109\times 10^{-31}\ \text{kg} \end{equation} \begin{equation} m_p=\text{proton … mass}=1.673\times 10^{-27}\ \text{kg} \end{equation} \begin{equation} m_n=\text{neutron mass}=1.675\times 10^{-27}\ \text{kg} \end{equation} The unified atomic mass unit ($u$) is defined as one twelfth … asked May 20 '17 by adiselann The point is that the response function should read \begin{equation} E(x,t) = \frac{1}{2c} \theta(ct - |x|). \end{equation} You can check this by calculating the second derivatives \begin{equation … } E_{xx} = \frac{1}{2c} \delta'(ct - |x|), \end{equation} \begin{equation} E_{tt} = \frac{c}{2} \delta'(ct - |x|), \end{equation} and taking your wave equation, \begin{equation} E_{tt} = c^2 E_{xx}. \end{equation} … answered Jul 8 by Ciruzz Broncio 1answer We have the Langevin equation, that describes the motion of a particle in a viscous medium, given by \begin{equation}\label{Langevin} \frac{dv}{dt} = -\gamma v + \zeta(t) \end{equation} With the … conditions that \begin{equation} \langle \zeta(t) \rangle = 0 \end{equation} \begin{equation} \langle \zeta(t)\zeta(t') \rangle = \Gamma \delta(t-t') \end{equation} And, if we make the time discrete … asked Oct 31 '16 by user78217 \begin{equation} \mathcal{L}\left( q,\dot{q},t\right)= \dot{q}^{2} - q\dot{q} \tag{01} \end{equation} \begin{equation} \dfrac{d}{dt}\left(\dfrac{\partial \mathcal{L}}{\partial \dot{q}}\right)-\dfrac … {\partial \mathcal{L}}{\partial q}=0 \tag{02} \end{equation} \begin{equation} \dfrac{d}{dt}\left[\dfrac{\partial \left(\dot{q}^{2} - q\dot{q}\right)}{\partial \dot{q}}\right]-\dfrac{\partial \left … answered Jul 22 '15 by user82794 A little trick required here. Perform a substitution first: \begin{equation} mv \gamma(v) = u , \end{equation} hence your integral becomes \begin{equation} \int \frac{du}{dt}dx , \end{equation} but … notice that \begin{equation} \frac{du}{dt} = \frac{du}{dx}\frac{dx}{dt}dx , \end{equation} but$\frac{dx}{dt} = v$- the definition of velocity! So the integral becomes: \begin{equation} \int \frac{du … answered Apr 29 '13 by Eugene B 1answer I have a problem calculating the electrostatic potential energy. I rely on these equations coming from mechanics: \begin{equation} U_{B}-U_{A} = -W_{A \ \rightarrow \ B} (done\ by \ the \ field … \ force) \end{equation} \begin{equation} U_{B}-U_{A} = W_{A \ \rightarrow \ B} (done\ by \ the \ opposite \ force) \end{equation} According to the next picture Work done by the coulomb force (field … asked Jul 5 '16 by Omar Apply: \begin{equation} F = ma \end{equation} to your particle to get a pair of acceleration vectors. You can then use the acceleration to work out the particle position at a given time. \begin … {equation} x = x1 + at^2/2 \end{equation} If you meant impulse instead of force then the process is similar but you equate the impulse to the total change in momentum and then: \begin{equation} impulse = m\Delta v \end{equation} \begin{equation} x = x1 + vt \end{equation} … answered Jan 22 '16 by Duke of Sam 1answer Given a lagrangian of a form: \begin{equation}\mathcal{L}=f(\phi,\partial_{\mu}\phi\partial^{\mu}\phi)\end{equation} where$f$is a function, I need to derive pressure and density in a FLRW universe … with$g^{\mu}_{\nu}=\delta^{\mu}_{\nu}\$. My approach is using: \begin{equation}T_{\mu\nu}=-\frac{2}{\sqrt{-g}}\frac{\partial}{\partial g^{\mu\nu}}(\sqrt{-g}\mathcal{L})\end{equation} \begin{equation