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Welcome To Simply Lecture

Simply lecture is a platform that assists students on difficult lectures and help them in any area. This blog serves as a personal tutor for students in any part of the world..
All students, have to do is to complete the contact form with their most difficult topic and subscribe to the blog, to get the most understandable form of the topic requested.
One good thing, about this innovation is that students are not allowed to pay any cash! Sounds good right?  Now all you need to do is:
1. Complete the contact form
2. Subscribe to get notified when your simplified lectures are posted.
3. Done
Also, Simply lecture is also ready to surprise you with great features soon, such like:
1. Online step by step calculators.
2. Free Video lectures.
3. Slideshare review of power point lectures.
4. Download simplified lecture notes.
5. Help you with assignments for a little pay.
6. And a whole lot more.
All you have to do is to stay tuned on this blog and enjoy a whole lot of lecture fun..  Schooling don't have to be boring, right??  Now get on the track, subscribe and also post this to friends and family and let them benefit the joy of e-education.

Warm Regards, 
Simply lecture.

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Maxwell fourth equation

The Maxwell fourth equation is called "the modified Ampere's circuital law". Statement: It states that the line integral of the magnetic field H around any closed part or circuit is equal to the current enclosed by the path. Differential form (without modification): That is, $$\int H.dL = I$$ Let the current be distributed through the current with current density J, then: $$I = \int J. ds$$ This implies that: $$\int H.dL = \int J.ds$$ .........(9) Applying Stokes theorem to the LHS of equ(9) to change line integral to surface integral we have: $$\int_{s} (\nabla X H).ds = \int_{s} J.ds$$ Since, two surface integrals are equal only if their integrands are equal. Thus, $$\nabla X H =J$$ .........(10) Equ(10) is the differential form of Maxwell fourth equation (without modification) Take divergence of equ(10) $$\nabla.(\nabla X H) = \nabla.J$$ Since, the divergence of the curl of a vector is zero.  Therefore, $$\nabla.(\nabla X H) = 0$$ It means that $...
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Time dependent Schrodinger equation

Recall from the time independent Schrodinger equation: $$-\frac{\hbar^{2}}{2m} \nabla^{2} \psi(r) + V(r) \psi(r) = E \psi(r)$$ ........(1) Multiply both sides of equ (1) by \(\phi(t)\), we have: $$-\frac{\hbar^{2}}{2m} \nabla^{2} \psi(r) \phi(t) + V(r) \psi(r) \phi(t) = E \psi(r) \phi(t)$$ ..........(2) Recall: $$\psi(r) \phi(t) = \Psi(r,t)$$ .............(3) Substitute equ (3) into equ (2): $$-\frac{\hbar^{2}}{2m} \nabla^{2} \Psi(r,t) + V(r) \Psi(r,t) = E \Psi(r,t)$$ .........(4) Recall that the well known Planck-Einstein relation is given by: $$E = hv$$ ...........(5) And also, the reduced Planck's constant is given by: $$\hbar = \frac{h}{2\pi}$$ ................(6) Also recall that time dependence is given by: $$\phi(t) = C_+ exp(-i2\pi vt) + C_- exp(i2\pi vt)$$ ..........(7) Substituting equ (6) into (5), we have: $$V = \frac{E}{\hbar 2 \pi}$$ ..........(8) Substitute equ (8) into equ (7) $$\phi(t) = C_+ exp(-i2 \pi \frac{E}{\hbar 2 \pi} t)  + C_- exp(i2 \pi \f...
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Schrodinger equation as a law in physics

The unified theory of wave-particle duality has been used to derive the Schrödinger equations. The Schrodinger equations are generally accepted, by postulate rather than derivation, to be laws of physics. The Schrodinger equations provide a basis for analyzing many kinds of systems (molecular, atomic, and nuclear) in a particular inertial reference frame. The success of the Schrödinger equations constitutes a basis for accepting them, their derivations, and the unified theory of wave-particle duality which makes such derivations possible. This acceptance is completely justified in the favored inertial reference frame. In accord with the principle of relativity, all physical laws must be the same in all inertial reference frames, i.e., all physical laws must be Lorentz invariant. Recall, the relationship: $$\nabla^{2} \psi = \frac{\partial^{2} \psi}{\partial t^{2}}$$ ...........(1) Equ (1) is Lorentz invariant and reduces, by means of the procedure presented in the previou...
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