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

[@fizixx]

Amazing how things are simplified with a little simple math.

[@yoprofmatt]

Agreed.
Cheers,
Dr. A

[@juniorcyans2988]

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.

[@itssherry7551]

Amazing explanation sir🙏Thanks a lot

[@anasbenhmida8350]

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

[@garffieldiscool1163]

Well exlplained professor. Diametional analysis is powerful mathematical tool.

[@yoprofmatt]

Agreed. Glad you're finding these useful.
Cheers,
Dr. A

[@sonukaushik]

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

[@yoprofmatt]

Thank you!Love the beer emoji.
Cheers!
Dr. A

[@3bdul-3ziz38]

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

[@SenneVorsselmans]

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

[@yoprofmatt]

Excellent historical reference.
Cheers,
Dr. A

[@BasitKhan-zk3jv]

Sir, what is meant by powerful relationship?

[@qualquan]

Making momentum = mass x velocity = massless ghost. (or spirit or soul)

[@qualquan]

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.

[@eclipse369.]

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]

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]

This type of discussion is what i love about YT comments 👍.

[@yoprofmatt]

Me too. Keep it up!
Cheers,
Dr. A

[@islamisverygood]

Minecraft

[@joenahhas4377]

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