2022-06-17 12:40:07
قوانين نظرية المجال الكمي
مقدمة :- Overture
t'=γ(t - vx/c²) , x'=γ(x - vt) , y'=y , z'=z (1)
γ=(1- β²)⁻½ and β=v/c
∂ᵩ =∂/∂χᵠ=(∂/∂t , ∇)=(∂/∂t , ∂/∂χ , ∂/∂y , ∂/∂z)
a'ᵠ =Σᵦ₌₀³(∂x'ᵠ/∂xᵝ) aᵝ
a'ᵠ =(∂x'ᵠ/∂xᵝ) aᵝ (3)
∂Φ/∂x'ᵠ =(∂xᵝ/∂x'ᵠ) (∂Φ/∂xᵝ) (4)
It'I=Iγ -βγ 0 0I ItI
Ix'I I-βγ γ 0 0I IxI
Iy'I I0 0 1 0I IyI
Iz'I I0 0 0 1I IzI (5)
x'ᵠ=Λᵠᵦ xᵝ (6)
p'ᵠ=Λᵠᵦ pᵝ (7)
a'ᵩ=Λᵝᵩ aᵦ (8)
x|²=x . x=(x⁰) - (x¹)² - (x²)² - (x³)² (10)
a · b = a⁰b⁰ - a• · b• (11)
a · b = gᵩᵦ aᵠbᵝ (12)
gᵩᵦ =I1 0 0 0I
I0 -1 0 0I
I0 0 -1 0I
I0 0 0 -1I (13)
aᵩ= gᵩᵦ aᵝ (14)
a⁰= a₀ , aⁱ= -aᵢ (15)
a · b=gᵩᵦ aᵠbᵝ=aᵩbᵠ=aᵠbᵩ (16)
p . p= (E ، p• ) · (E ، p• ) = E² - p• ² = m² (17)
a . b=gᵠᵝ aᵩbᵦ
∂²=∂ᵠ∂ᵩ=∂²/∂t² - ∂²/∂x² - ∂²/∂y² - ∂²/∂z² (18)
∂²=∂ᵠ∂ᵩ=∂²/∂t² - ∇² (19)
Tᵠ'···ᵝ' ᵥ،...ₖ،=(∂xᵠ'/∂xᵠ)...(∂xᵝ'/∂xᵝ)(∂xᵛ/∂xᵛ')...(∂xᵏ/∂xᵏ')Tᵠ···ᵝ ᵥ...ₖ (20)
δ'ᵠᵦ=(∂x'ᵠ/∂xᵅ)(∂xᵞ/∂x'ᵝ)δᵅᵧ (21)
f'(k)dk =∫d⁴x eⁱᵏ · ˣ f(x) (22)
∫d⁴x = ∫dx⁰dx¹dx²dx³ (23)
f(x)dx =∫(d⁴k/(2π)⁴) e⁻ⁱᵏ · ˣ f'(k) (24)
f(w ، k• ) = ∫d³xdt exp[i(ωt - k• . x• )] f(t , x) (25)
∫ dᵈx δ⁽ᵈ⁾(χ)=1 (26)
∫ dᵈx f(x) δ⁽ᵈ⁾(χ)= f(0) (27)
δ'⁽ᵈ⁾(k)= ∫ dᵈx e⁻ⁱᵏ · ˣn δ⁽ᵈ⁾(χ)=1 (28)
δ⁽⁴⁾(x)= ∫ (d⁴k/(2π)⁴) e⁻ⁱᵏ · ˣ (29)
∇ . E=ρ/ε₀ , ∇ x E = -∂B/∂t
∇. B = 0 , ∇ x B = µ₀ J + (1/c²) ∂E/∂t (30)
V(x)=q/(4π|x|) (31)
∇ . E=ρ , ∇ x E = -(1/c)∂B/∂t
∇. B = 0 , ∇ x B = (1/c)[J+ ∂E/∂t] (32)
α = e²/(4π) (33)
α = e/(4πħc)=1/137
q=Q|e|
الفصل الاول : اللاغرانجيان :- Lagrangians .
قوانين الفصل الاول:
F = mx·· (1-1)
Tₐᵥ=(1/τ) ∫₀ᵞ (1/2)m[x·(t)]² dt (1-2)
Vₐᵥ= (1/τ) ∫₀ᵞ V[x(t)] dt (1-3)
F[f]= ∫₀¹ f(x) dx (1-4)
F[f]= ∫₀¹ x² dx=(1/3)|x³|₀¹=1/3 (1-5)
G[f]= ∫₋ₐᵃ 5[f(x)]² dx (1-6)
G[f]= ∫₋ₐᵃ 5[x²]² dx = ∫₋ₐᵃ 5 x⁴ dx =5(1/5)|x⁵|₋ₐᵃ=2a⁵ (1-7)
Fₓ[f]=∫₋ₐᵃ f(y) δ(y - χ) dy=f(x) (1-8)
df/dx= limₑ→₀ [f(x + e) - f(x)]/e (1-9)
δF/δf(χ)= limₑ→₀ [F{f(x') + eδ(χ - χ')} - F{f(x')}]/e (1-10)
δJ[f]/δf(x)=limₑ→₀ (1/e) ∫ [f(y) + eδ(y - x)]ᵖ Φ(y)dy + ∫ [f(y)]ᵖ Φ(y)dy]
δI[f]/δf(x₀)=limₑ→₀ ∫₋₁¹ δ(χ - x₀) dx
δI[f]/δf(x₀)=1 if -1≤ x₀ ≤ 1
or=0 otherwise (1-11)
δJ[f]/δf(x)= p[f(x)]ᵖ⁻¹ Φ(x) (1-12)
δH[f]/δf(x)= g'[f(x₀)] (1-13)
Vₐᵥ(x)=(1/T)∫₀ᵀ V[x(t)] dt
F[Φ]=∫(∂Φ/∂y)² dy
δVₐᵥ(x)/δχ(t)=(1/T)V'[x(t)] dt (1-14)
δg[f(χ)]/δf(x)=-(d/dx)(dg(f')/df') (1-18)
δF[Φ(χ)]/δΦ(x)= -2(∂²Φ/∂y²) (1-19)
Τₐᵥ(x)=(1/T)∫₀ᵀ(1/2) m[x·(t)]² dt
δTₐᵥ(x)/δχ(t)= (1/T)(1/2)(δt/δχ(t))(δ/δt)[m x·] , δVₐᵥ(x)/δχ(t)= - (1/T)(δ/δχ(t))V(x)
δTₐᵥ(x)/δχ(t)= (1/T)(1/2)(1/x·)[2m x·x··] , δVₐᵥ(x)/δχ(t)= - (1/T)(δ/δχ(t))V(x)
δTₐᵥ(x)/δχ(t)= (1/τ)(δ/δχ(t))[(1/2)m x·²] , δVₐᵥ(x)/δχ(t)= - (1/τ)(δ/δχ(t))V(x)
δTₐᵥ(x)/δχ(t)=m x··/T , δVₐᵥ(x)/δχ(t)=- V'(x)/T (1-21)
δTₐᵥ(x)/δχ(t)= - m x··/T
[δTₐᵥ(x)/δχ(t)]T= - [- δVₐᵥ(x)/δχ(t)]T
δTₐᵥ(x)/δχ(t)= δVₐᵥ(x)/δχ(t) (1-22)
δ/δχ(t))[Tₐᵥ(x) - Vₐᵥ(x)]=0 (1-23)
δTₐᵥ(x)/δχ(t) - δVₐᵥ(x)/δχ(t)=0 (1-24)
L=T - V (1-25)
S = ∫ L dt (1-25)
S = ∫ (T - V) dt =τ(Tₐᵥ[a] - Vₐᵥ[x])
δS/δχ(t)= 0 (1-26)
δS/δx(u)=δL/δχ(t) - (d/dt)(δL/δχ·(t)) (1-27)
δL/δχ(t) - (d/dt)(δL/δχ·(t)) =0 (1-28)
L = ∫ dx L° (1-29)
S= ∫ dt L= ∫dt dx L° (1-30)
391 viewsأَبُو إِلِيكْسِي (𝔓𝔯𝔬𝔣 𝔪𝔬𝔥𝔞𝔫𝔫𝔢𝔡), 09:40