Amazing how things are simplified with a little simple math.

So clear so neat, then so easy to understand! I wish I could find theses videos last semester, but perhaps it’s the right time to review now.

Amazing explanation sir🙏Thanks a lot

I really appreciate your work sir And I loved your way of teaching ❤❤❤ continue

Well exlplained professor. Diametional analysis is powerful mathematical tool.

Amazing you are in concepting skills n building it... 🙏🍺🍺🍺❤

Wait a minute P=mv , how it =h/λ or can I say mv=h/λ ?

Albert Einstein initially wrote down the formula as m=E/C^2, basically proving mass = energy/infinity

Sir, what is meant by powerful relationship?

But to derive the energy-momentum-mass equation you need to set p = gamma*m*v and so you can’t set m=0 for a massless particle without also setting p=0. This means E = 0. Massless particles really have there own equation from QM E=h*f=p*c.

Very superficial. Is the Kinetic photon massless? Not made clear. If so then P = momentum = mass x velocity. So how is P "massless"? IOW, how can a product (P) be nonzero if one of its factors (m) is zero? Might as well call P a spirit or a ghost since they are allegedly massless.

Photon is a slice of emr wave yes, but the emr wave is not a wave of photons lmao. Photon is man made construct. No Such thing as a photon particle.

Minecraft

I can produce E = mc² using any physics. Newtonian derivation of E = mc² *In Newton’s law:* Force = mass x acceleration (Areal velocity = constant is included in Newton’s law) Newton’s gravity law F = - (G m M /r2) **r1** Acceleration in mathematical form is: (rʹʹ - r θʹ2) **r1** + (2 rʹ θʹ + r θʹʹ) **θ1** Force = mass x acceleration = m x [(rʹʹ - r θʹ2) **r1** + (2 rʹ θʹ + r θʹʹ) **θ1];** Force = F = mass x acceleration = - (G m M /r2) **r1** And m (rʹʹ - r θʹ2) = -G m M /r2 ---Newton’s gravitational force law. And (2 rʹ θʹ + r θʹʹ) = 0 -------------- Kepler’s constant areal velocity law. Divide Kepler’s law by r θʹ Visual distance: r = (distance) r₀ x e± i ω t Visual angular speed θʹ = (angular speed) θʹ ₀ x e± 2 i ω t = r₀ e± i ω t And θʹ = θʹ₀ e± 2 i ω t And (2 rʹ / r + θʹʹ/θʹ) = 0; rearrange: (2 rʹ/r) = - (θʹʹ/θʹ) = ± 2 i ω *In Kepler law*: Areal velocity = (r2θʹ) = constant Or (r2θʹ) = k; r = distance; θʹ = angular speed Take the derivative: d (r2θʹ)/ d t = 0 And (2 r rʹ θʹ + r2θʹʹ) = 0; divide by r2θʹ And (2 rʹ / r + θʹʹ/θʹ) = 0; rearrange (2 rʹ/r) = - (θʹʹ/θʹ) = ± 2 i ω And (rʹ/r) = ± i ω And d r/r = ± i ω d t And ∫ d r/r = ∫± i ω d t L n (r/r₀) = ± i ω t L n e = L n e± i ω t And r = r₀ e± i ω t And - (θʹʹ/θʹ) = ± 2 i ω And θʹʹ/θʹ = ± 2 i ω The solution is: θʹ = θʹ ₀ e± 2 i ω t If r₀ e i ω t; θʹ = θʹ₀ e- 2 i ω t If r₀ e - i ω t; θʹ = θʹ₀ e 2 i ω t E = mc² S = r ℮ ỉ ω t P = [v + ỉ ω r] ℮ ỉ ω t (P. P) = [v² - ω² r² + 2 ỉ ω r v] ℮ 2 ỉ ω t E = m (P. P)/2 = (m/2) [v² - ω² r² + 2 ỉ ω r v] ℮ 2 ỉ ω t E = (m/2) [c² - c² + 2 ỉ c²] ℮ 2 ỉ ω t With ω r = v = c E = (m/2) [2 ỉ c² ℮ 2 ỉ ω t] E = (m/2) │2 ỉ c² ││℮ 2 ỉ ω t│ E = (m/2) (2 c²) E = mc²