Energy Mass Relation E = mc^2 | Physics with Professor Matt Anderson | M29-07
How does Einstein's famous equation come into play here? And how does it relate to energy and momentum.
Physics with Professor Matt Anderson
How does Einstein's famous equation come into play here? And how does it relate to energy and momentum.
Physics with Professor Matt Anderson
Пікірлер: 21
Amazing how things are simplified with a little simple math.
@yoprofmatt
2 жыл бұрын
Agreed. Cheers, Dr. A
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.
@yoprofmatt
2 жыл бұрын
Agreed. Glad you're finding these useful. Cheers, Dr. A
Amazing you are in concepting skills n building it... 🙏🍺🍺🍺❤
@yoprofmatt
Жыл бұрын
Thank you!Love the beer emoji. Cheers! Dr. A
Albert Einstein initially wrote down the formula as m=E/C^2, basically proving mass = energy/infinity
@yoprofmatt
2 жыл бұрын
Excellent historical reference. Cheers, Dr. A
Wait a minute P=mv , how it =h/λ or can I say mv=h/λ ?
Sir, what is meant by powerful relationship?
@qualquan
Жыл бұрын
Making momentum = mass x velocity = massless ghost. (or spirit or soul)
But P = momentum = mass x velocity. So how is P "massless? Might as well call P a spirit or a ghost since they are allegedly massless.
Minecraft
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.
@kenlogsdon7095
2 жыл бұрын
Yup, "photon" is a transtemporal interaction; there is only ever a full spin "boson" exchanged between two half spin "fermions", with zero distance and zero time between the two in the "boson" lightlike frame.
@hightttech
2 жыл бұрын
This type of discussion is what i love about YT comments 👍.
@yoprofmatt
2 жыл бұрын
Me too. Keep it up! Cheers, Dr. A
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