Parcial 3 Mecnica Analtica II Gabriel Sandoval
[email protected] Junio 2018
1
Complex conjugate of the the Schr Schr¨ ¨ odin odinge ger r Equa Equa-tion L=
The derivatives with respect to x˙ µ and and x x µ are ∂ L q µ = m ˙ x + Aµ 0 ∂ ˙ x˙ µ c ∂ L q ∂ µ = (x˙ Aµ (xν )) µ µ ∂x c ∂x q µ µν = ˙x δ ∂ µ Aν c q ν = ˙x ∂ µ Aν c
i (ψ ( ψ ψ˙ − ψ ψ˙ ) 2 ∗
−
∗
2
∗
(∇ψ )(∇ψ ) − V ψψ
∗
2m The derivatives with respect to ∂ A ψ and ψ are ∂ L ∂ L i = = ψ ∂ (∂ t ψ ) 2 ∂ ψ˙ ∂ L ∂ L 2 = =− ∇ψ ∂ (∂ i ψ ) ∂ (∇ψ) 2m ∂ L i = − ψ˙ − V ψ ∂ψ 2 Then, by the Euler-Lagrange equations ∂ L ∂ L ∂ L + ∂ i = ∂ t ∂ (∂ i ψ ) ∂ψ ∂ ψ˙ i ˙ 2 i ψ − (1) ∇2 ψ = − ψ˙ − V ψ 2 2m 2 Simplifying the equation above, gives
Applying the Euler-Lagrange equations
d ∂ L = dτ ∂ ˙ x˙ µ q m0 x ¨µ + ∂ ν ν Aµ ˙xν = c ⇒ m0 x ¨µ =
∂ L ∂x µ q ν ˙x ∂ µ Aν c q ν q ˙x ∂ µ Aν − ∂ ν ν Aµ ˙xν c c q ν m0x¨µ = ˙x F µν µν c
∗
∗
∗
∗
Writing this explicitly, for µ = 1 q m0 x ¨µ = (x˙ 0F 10 ˙ 1 F 11 ˙ 2 F 12 ˙ 3 F 13 10 + x 11 + x 12 + x 13 ) c q = γ (cE x + x˙ y Bz − x˙ z By ) c = qγ [(E)x + ( V × B)x ] q γ [(
∗
∗
2
∗
˙ ∇2 ψ + V ψ = −i ψ ∗
∗
∗
∗
(2) 2m Whi Which is the the comp compllex conj conjug ugat atee of the the Schr¨ odinger odinger equation. equation. −
2
3
Lorentz Force from a relativistic Lagrangian
Damped tion
3.1 3.1
Wave
Equa-
Find Find a Lagr Lagran angi gian an
The damped wave equation is given by,
m0 q L= x˙ µ ˙xµ + ˙xµ Aµ 2 c 1
∂ 2 1 ∂ 2 − ∂x 2 v 2 ∂t 2
y(t, x) = b
∂y( ∂y (t, x) ∂t
3.2
Or in a more compact notation, 1 ¨ y = b y˙ v2 Where y = ∂y/∂x, y = ∂y/∂t. ˙ Supposing the Lagrangian
y −
(3)
From the equations (6) the generalized velocities, in terms of the momenta are as follows
L = e
γt
Hamiltonian and Hamilton’s Canonical Equations
1 2 1 y − 2 ˙y 2 2 2v
y = ˙ −v 2 pt e y = p x e btv
Where γ is a constant. The derivatives of the Lagrangian with respect to y , y˙ and y are
btv 2
−
−
(7)
2
The Hamiltonian es given by the Legendre’s transformations of the Lagrangian
∂ L = e γt y ∂y 1 ∂ L = −eγt 2 y˙ ∂ y˙ v ∂ L =0 ∂y
H = p x y + pt ˙y − L
btv 2
H = p 2x e
−
−ebtv
It follows that ∂ ∂ L = e γt y ∂x ∂y ∂ ∂ L 1 1 = − 2 γe γt y˙ − 2 eγt y¨ ∂t ∂ y˙ v v
2
1 2 p e 2 x
2btv
−
2
− v 2 p2t e
btv 2
−
1 − v 2 p2t e 2
2btv
−
2
H=
1 e 2
btv 2
−
p2x − v 2 p2t
(8)
The Hamilton’s canonical equations are given by
Then, by Euler-Lagrange equations ∂ ∂ L ∂ ∂ L ∂ L + = ∂t ∂ y˙ ∂x ∂y ∂y 1 ˙ y¨) + eγt y = 0 − 2 eγt (γ y + v Since e γt = 0 for any value of γ and t, the equation above can be written as follows.
∂ H ∂p i ∂ H ∂ i pi = − ∂y ∂ i y =
(9)
For this case i = t, x. Applying (9) to (8), it follows that
1 γ ¨ y = ˙y (4) v2 v2 By setting γ = bv 2 the equation, (4) is exactly the same as (3). So, a possible Lagrangian that reproduces the damped wave equation is
y −
L= 3.1.1
1 bv e 2
2
t
y2−
1 2 y˙ v2
(5)
∂ t
bv 2 t
∂ L e pt = = − 2 y˙ ∂ y˙ v px =
2
2
(10)
∂ t pt + ∂ x px = 0
(11)
2
ebv t − 2 y˙ v
⇒ −by˙ −
(6) ∂ L = e bv t y ∂y
−
−
The equations (10) are the same as (6). Replacing (10) into (11) gives,
Conjugate momenta
2
= y = ˙ −v 2 e bv t pt = y = e bv t px
∂ t y ∂ x y
⇒y −
2
2
+ ∂ x ebv t y
1 ¨ y + y = 0 v2
1 ¨ y = b y˙ v2
=0
4
The Sine-Gordon Equation
4.1
4.3
Given the function
S-G Equation from a Lagrangian
φ(t, x) = 4 arctan(ekγ (x
1 L = ρ − η µν ∂ µ φ∂ ν φ + k 2 (cos φ − 1) 2
1 L = ρ − ∂ µ φ∂ µ φ + k 2 (cos φ − 1) 2
1 arctan z = ln 2i
(12)
4 φ(η) = ln 2i
i − ekγη i + ekγη
And in terms of the new variable η
∂ L ∂ L ∂ µ = ∂∂ µ φ ∂φ ⇒ −ρ∂ µ ∂ µ φ = −ρk2 sin φ
v2 = 1− c2 1 ∂ 2 = 2 2 γ ∂η
Defining φ = ∂ µ ∂ µ φ, the equation above becomes
4.2
1 ∂ 2 ∂ 2 + =− c2 ∂t 2 ∂x 2
Therefore,
sin φ
i − z i + z
Where η = x − vt. The D’Alambertian in two dimensions is given by
∂ L = −ρ∂ µ φ ∂ (∂ µ φ) ∂ L = −ρk 2 sin φ ∂φ
2
Then, φ(t, x) can be written as
The derivatives with respect to ∂ µ φ and φ of the Lagrangian are
φ = k
)
Show that φ(t, x) satisfies the Sine-Gordon equation. The function arctan x can be written as
Or equivalently
vt)
−
Given the Lagrangian
Test a Solution
∂ 2 ∂η 2
Therefore, the D’Alambertian of φ(η) is
(13)
Canonical Momenta
φ(η)
=−
4k2 eγη k e2γη k − 1
The canonical momenta is given by
(e2γη k + 1)2
The function sin(z ) can be written as a combination of exponential functions as follows
∂ L = −ρ∂ µ φ ∂ (∂ µ φ)
eiz − e sin(z ) = 2i
iz
−
4.2.1
Stress-Energy Tensor
The Stress-Energy tensor is defined as follows T µν =
Then, applying this function to φ(η) and multiplying by k2, gives
∂ L ν ∂ − η µν L ∂ (∂ µ φ)
k2 sin(φ(η)) = −
For this particular case the tensor es given by 1 T µν = − ∂ µ φ∂ ν φ − ρk 2 η µν (cos φ − 1) (14) 2
Hence 3
φ(η)
4k 2eγη k e2γη k − 1
(e2γη k + 1)2
− k 2 sin(φ(η)) = 0
4.3.1
Plotting (15) gives the following
Energy Density
The energy density is the component T 00 of the tensor (14). 0
1 T = − ∂ 0 φ∂ 0φ − ρk 2η 00 (cos φ − 1) 2 1 = − (−∂ t φ)2 + ρk 2 (cos φ − 1) 2
−2 −4 −6 −8 −10
00
3
Then, the energy density is given by the following expression 1 ρenergy = − φ˙ 2 + ρk 2 (cos φ − 1) 2
−3
φ
(15)
0
0 3
−3
1: Plot of the energy density
4
φ˙
