mirror of
https://gitlab.gwdg.de/j.hahn02/university.git
synced 2026-01-01 06:44:25 -05:00
auto up 11:42:06 up 1:56, 2 users, load average: 0.03, 0.13, 0.17
This commit is contained in:
175
S3/ExPhyIII/VL/ExIIIVL8.typ
Normal file
175
S3/ExPhyIII/VL/ExIIIVL8.typ
Normal file
@@ -0,0 +1,175 @@
|
||||
// Main VL template
|
||||
#import "../preamble.typ": *
|
||||
|
||||
// Fix theorems to be shown the right way in this document
|
||||
#import "@preview/ctheorems:1.1.3": *
|
||||
#show: thmrules
|
||||
|
||||
// Main settings call
|
||||
#show: conf.with(
|
||||
// May add more flags here in the future
|
||||
num: 5,
|
||||
type: 0, // 0 normal, 1 exercise
|
||||
date: datetime.today().display(),
|
||||
//date: datetime(
|
||||
// year: 2025,
|
||||
// month: 5,
|
||||
// day: 1,
|
||||
//).display(),
|
||||
)
|
||||
|
||||
= Uebersicht
|
||||
|
||||
== Reflexion und Brechung
|
||||
|
||||
Wechselstroeme und das Zeigerdiagramm
|
||||
$
|
||||
I (t) = I_0 cos [omega t + phi_(I) ] \
|
||||
U (t) = U_0 cos [ omega t + phi _(U) ] \
|
||||
Z = a + i b = abs(z) exp(i phi).
|
||||
$
|
||||
|
||||
Wiederholung von Kombinantion von Wechselstromwiederstaenden und deren Impedanz
|
||||
$
|
||||
abs(Z) = sqrt(R^2 + (omega L - 1/ (omega C))^2 ) \
|
||||
tan phi = (omega L - 1/(omega C)) / (R) = (Im Z) / (Re Z).
|
||||
$
|
||||
|
||||
Es gilt
|
||||
$
|
||||
(diff V) / (diff x) = - ((diff L) / (diff x) )(diff I) / (diff t) \, space (diff I) / (diff x) = - ((diff C) / (diff x) )(diff V) / (diff t) \
|
||||
(diff ^2 V) / (diff t^2 ) = - 1/C_0 diff / (diff x) (diff I) / (diff t) = - 1/C_0 diff / (diff x) (- 1/L_0 (diff V) / (diff x) )= 1 / (L_0 C_0 )(diff ^2 V) / (diff x^2 ) \, space c = 1/sqrt(L_0 C_0 ) \
|
||||
partial _(x) I = - C_0 partial _(t) V ==> I_(+) = I_(0 +) e^(i (k x - omega t)) \, space V_(+) = V_(0 +) e^(i (k x - omega t)) \
|
||||
I_(0 +) i k = - C_0 (i omega) V_(0 +) \
|
||||
==> (V_(0 +) ) / (I _(0 +) ) = k/omega 1/C_0 = 1/c 1/C_0 = sqrt(L_0 C_0 )1/C_0 = sqrt((L_0 ) / (C_0 ) ) = Z_0.
|
||||
$
|
||||
|
||||
= Koaxialkabel
|
||||
Es gilt
|
||||
$
|
||||
E = 1/(2 pi epsilon_0 ) (lambda) / (r) = 1/(2 pi epsilon_0 ) (Q) / (L r) \
|
||||
Delta V = V_(b) - V_(a) \
|
||||
dif C = (2 pi epsilon_0 ) / (ln (R/r)) dif l \
|
||||
dif L = (mu mu_0 ) / (2 pi) ln (R/r) dif l.
|
||||
$
|
||||
|
||||
Wir haben Wellen der Form
|
||||
$
|
||||
I, U, arrow(E), arrow(B).
|
||||
$
|
||||
Fuer die Impedanz gilt dann
|
||||
$
|
||||
Z := sqrt(L/C) \, space [Z]= Omega.
|
||||
$
|
||||
Die Grenzflaeche ist dann zwischen $Z_(1) $ und $Z_2 $. Es gilt fuer die drei Stroeme
|
||||
$
|
||||
I = I_(1 + ) + I _(1 - ) = I _(2 + ) \
|
||||
U = U_(1 + ) + U_(1 - ) = Z_(1) I _(1 + ) - Z_(1) I_(1 - ) = U _(2 + ) = Z_(2) I_(2 + ) \
|
||||
==> I_(2 + ) = I _(1 + ) + I _(1 - ) \, space Z_(2) I_(2 + ) = Z_(1) I _(1 + ) + Z_(1) I _(1 - ) \
|
||||
==> 1 + (I_(1 - ) ) / (I _(1 + ) ) = (I _(2 + ) ) / (I _(1 + ) ) := underbrace(t, "Transmission") = 1 + underbrace(r, "Reflexion") \
|
||||
Z_(1) - Z_(2) (I_(1 - ) ) / (I _(1 + ) ) = Z_(2) (I_(2 + ) ) / (I _(1 + ) ) \
|
||||
==> r = (Z_(1) - Z_(2) ) / (Z_(1) + Z_(2) ) \, space t = (2 Z_(1) ) / (Z_(1) + Z_(2) ).
|
||||
$
|
||||
|
||||
= Dopplereffekt
|
||||
|
||||
Fuer Schall
|
||||
$
|
||||
f' = (1 +- v_(e) /c) / (1 minus.plus v_(s) /c) f_0,
|
||||
$
|
||||
wobei $+ $ aufeinander zu und $- $ voneinander weg ist. Auch ist $v_(e) $ die Geschwindigkeit des Empfaengers und $v_(s) $ die des Senders.
|
||||
Lorentztransformation impliziert
|
||||
$
|
||||
f = sqrt((c +- v) / (c minus.plus v) ) f_0.
|
||||
$
|
||||
Fuer die Wellenlaenge $lambda$
|
||||
$
|
||||
lambda = c T \, space f_0 = c /T \
|
||||
lambda' = (c + v_(s) ) T \, space lambda'' = (c - v_(s) ) T \
|
||||
f' = c/lambda' = (c) / ((c + v_(s) )T) = c / (v + v_(s) )f_0 ==> f' = 1/(1 +- v_(s) /c) f_0.
|
||||
$
|
||||
|
||||
Fuer den bewegten Empfaenger
|
||||
$
|
||||
T' = (lambda) / (c + v_(e) ) = T_0 /(1 + v_(e) /c) \
|
||||
==> f'= (1 + v_(e) /c) f_0 \
|
||||
f' = (1 +- v_(e) /c) / (1 minus.plus v_(s) /c) f_0.
|
||||
$
|
||||
Fuer Relativitaet
|
||||
$
|
||||
x' = gamma (x - v t) \, space y' = y \, space z' = z \, space t' = gamma (t - (x v) / (c^2 ) ) \, space gamma = (1) / (sqrt(1 - v^2 /c^2 )) \
|
||||
( c Delta t - v Delta t)/N \, space f = c/lambda = (c N) / (c Delta t - v Delta t) \
|
||||
N = f_0 Delta t' \, space Delta t' = Delta t = (Delta t)/gamma \
|
||||
f = (c) / (c Delta t - v Delta t) f_0 (Delta t) / (gamma) = 1/ (1 - v/c) f_0 /gamma = sqrt((1 + v/c)/(1 - v/c)) f_0
|
||||
$
|
||||
|
||||
= Brechung und Reflexion
|
||||
Betrachte Wellen
|
||||
$
|
||||
psi_(1) = a_1 e ^(i (arrow(k)_(1) arrow(r) - omega t)) \
|
||||
psi_(r) = a_(r) e^(i (arrow(k)_(r) arrow(r) - omega _(r) t))
|
||||
psi_(2) = a_(2) e^(i (arrow(k)_(2) arrow(r) - omega _(2) t)).
|
||||
$
|
||||
Das Ziel ist die Bestimmung der Geometrie. Also $Theta_(2) , Theta_(1) "und" Theta_(r) $ und $R = abs((a_(r) ) / (a_(1) ) )^2 \, space R = R (n_1, n_2, Theta_(1) )$ sowie $T = abs(a_2 /a_1 )^2 \, space T = T (n_1, n_2, Theta_(1) ) $. Es gilt dabei $R = abs(r)^2 \, space T = abs(t)^2 $.
|
||||
Alles folgt aus den Randbedingungen. So ist $psi$ hier stetig und
|
||||
$
|
||||
arrow(nabla) psi "stetig an der Grenzflaeche" z = 0.
|
||||
$
|
||||
|
||||
Stetigkeit von $psi$ impliziert
|
||||
$
|
||||
a_1 e^(i (arrow(k)_(1) arrow(r) - omega t)) + a_(r) e^(i (arrow(k)_(r) arrow(r) - omega_(r) t)) = a_(2) e^(i (arrow(k)_(2) arrow(r) - omega_(2) t)) space forall arrow(r) = vec(x, y, 0) space forall t \
|
||||
==> omega := omega_(1) = omega_(r) = omega_(2).
|
||||
$
|
||||
Es muessen also auch die Tangentialkomponenten von $arrow(k)$ gleich sein, also gilt auch
|
||||
$
|
||||
k_(1) ^((x)) = k_(r) ^((x)) = k_(2) ^((x)).
|
||||
$
|
||||
Also
|
||||
$
|
||||
k_(1) ^((x)) = k n_(1) sin Theta_(1) = k n_(1) sin Theta_(r) \
|
||||
==> Theta_(1) = Theta_(r) "Reflexionsgesetz".
|
||||
$
|
||||
Aus $k_(1) ^((x)) = k_(2) ^((x)) $ folgt
|
||||
$
|
||||
k n_(1) sin Theta_(1) = k_(2) n_(2) sin Theta_(2) \
|
||||
==> n_1 / n_2 = (sin Theta_(2) ) / (sin Theta_(1) ).
|
||||
$
|
||||
Mit $k = (2 pi )/lambda$ im Vakuum. Auch
|
||||
$
|
||||
a_1 + a_(r) = a_(2).
|
||||
$
|
||||
Aus der Stetigkeit des Gradienten folgt
|
||||
$
|
||||
arrow(k)_(1) psi_(1) + arrow(k)_(r) psi_(r) = arrow(k)_(2) psi_(2) \
|
||||
==> k n_1 a_1 sin Theta_(1) + k n_1 a_1 sin Theta_(1) = k n_2 a_2 sin Theta_2 \
|
||||
==> (a_1 + a_(2) ) n_1 sin Theta_1 = a_2 n_2 sin Theta_2 "wieder der Snellius".
|
||||
$
|
||||
Die senkrechte Komponenente liefert
|
||||
$
|
||||
- k n_1 a_1 cos Theta_1 + k n_1 a_(r) cos Theta_1 = - k n_2 a_2 cos Theta_2 \
|
||||
==> n_1 cos Theta_1 (a_1 - a_(r) ) = n_2 (a_1 + a_(r) )cos Theta_2 \
|
||||
( 1 - a_(r) /a_(1) ) = n_2 /n_1 (cos Theta_2 )/(cos Theta_1 ) (1 + a_(r) /a_(1) ) "mit" r := a_(r) /a_(1) ==> (1 - sigma) = n_2 /n_1 (cos Theta_2 ) / (cos Theta_1 ) space "?"
|
||||
$
|
||||
|
||||
Also
|
||||
$
|
||||
1 - n_2 /n_1 (cos Theta_2 ) / (cos Theta_1) = r (n_2 /n_1 (cos Theta_(2) ) / (Theta_(1) ) + 1) \
|
||||
==> r = (n_2 cos Theta_1 - n_2 cos Theta_2 ) / (n_1 cos Theta_(1) + n_2 cos Theta_(2) ) \
|
||||
==> t = a_(2) /a_(1) = (a_1 + a_(r) ) / (a_1 ) = 1 + r \
|
||||
==> t = (2 n_1 cos Theta_1 ) / (n_1 cos Theta_1 + n_2 cos Theta_2 ).
|
||||
$
|
||||
Jetzt als Spezialfall den senkrechten Einfall
|
||||
$
|
||||
Theta_1 = 0 degree \, space r = (n_1 - n_2 ) / (n_1 + n_2 ) \
|
||||
"fuer" n_1 = 1 \, space n_2 = 1.5 space ("Glas") \
|
||||
==> r = (- 0.5) / (2.5) = - 0.2 \, space R := r^2 = 0.04.
|
||||
$
|
||||
|
||||
Eine Grenzflaeche mit Normalen in zwei Medien mit Brechungsindizes $n_1 "und" n_2 $
|
||||
$
|
||||
Theta_1 = Theta_(c) <==> Theta_(2) = 90 degree.
|
||||
$
|
||||
Ist der Snellius fuer $Theta > Theta_(c) $ erfuellbar? Gibt es Verhaeltnisse fuer die der Reflexionskoeffizient $R = abs(r)^2 = 1$ sein kann?
|
||||
|
||||
#highlight[TODO: Totalreflexion herleiten]
|
||||
84
S3/MaPhyIII/VL/MaPhIIIVL7.typ
Normal file
84
S3/MaPhyIII/VL/MaPhIIIVL7.typ
Normal file
@@ -0,0 +1,84 @@
|
||||
// Main VL template
|
||||
#import "../preamble.typ": *
|
||||
|
||||
// Fix theorems to be shown the right way in this document
|
||||
#import "@preview/ctheorems:1.1.3": *
|
||||
#show: thmrules
|
||||
|
||||
// Main settings call
|
||||
#show: conf.with(
|
||||
// May add more flags here in the future
|
||||
num: 5,
|
||||
type: 0, // 0 normal, 1 exercise
|
||||
date: datetime.today().display(),
|
||||
//date: datetime(
|
||||
// year: 2025,
|
||||
// month: 5,
|
||||
// day: 1,
|
||||
//).display(),
|
||||
)
|
||||
|
||||
= Uebersicht
|
||||
|
||||
Hilbertraeume und ONS fuer PDEs. Fouriertrafo fuer die Loesung von PDEs.
|
||||
|
||||
Harmonischer Oszillator
|
||||
$
|
||||
(dif ^2 y) / (dif x ^2 ) + omega ^2 y = 0 \, space y = A e ^(i k x) \
|
||||
A e ^(i k x) (- k^2 + omega^2 ) = 0 ==> k = +- omega ==> y = A e ^(i omega x) + beta e^(- i omega x).
|
||||
$
|
||||
|
||||
Motivation der Fourertransformation. Sei $u$ eine Funktion
|
||||
$
|
||||
P: u |-> - i u^(1) \
|
||||
Q: u |-> x * u.
|
||||
$
|
||||
Wir suchen einen Funktionenraum und eine lineare Transformation $u |-> hat(u)$, sodass $hat(P u) = Q hat(u)$.
|
||||
|
||||
Aus einer DGL soll ein algebraischer Ausdruck werden. Mit dem Ansatz
|
||||
$
|
||||
hat(u) = integral K (x, xi) u (x) dif x.
|
||||
$
|
||||
Welche Anforderungen brauchen wir an $u$, sodass das Integral definiert ist. Es soll also partielle Integration durchfuehrbar sein
|
||||
$
|
||||
i xi hat(u) (xi) = integral i xi e ^(i x xi) u (x) dif x = integral e ^(i x xi) (+ u') (x) dif x = - underbrace([e ^(- i xi x) u (x)]_(- oo) ^(oo), =^(!) 0) + integral e ^(- i x xi) partial _(x) u dif x = hat(u)' (xi).
|
||||
$
|
||||
|
||||
Um den bei der partiellen Integration auftretenden Randterm zu eliminieren, wird die Fourertransformation auf $L^(1) $ definiert.
|
||||
|
||||
Q: Warum faellt dieser Term bei $L^(1) $ Funktionen weg?
|
||||
|
||||
#theorem[
|
||||
Riemann-Lebesgue-Lemma.
|
||||
|
||||
Es gilt
|
||||
$
|
||||
lim_(abs(xi) -> oo) hat(u) (xi) = 0.
|
||||
$
|
||||
]
|
||||
|
||||
Die Fourertransformation ist immer stetig. Im allgemeinen ist die Fourertransformation wie hier noch nicht integrierbar.
|
||||
|
||||
Einfuehrung der Multi-Index Notation mit einem Produkt aber keine Summe. Es gilt dann
|
||||
$
|
||||
P^(alpha) u (x) = (- i)^(abs(alpha)) partial^(alpha) u (x).
|
||||
$
|
||||
|
||||
#lemma[
|
||||
Falls $u in C^(k) (RR^(n) )$ und $P^(alpha) in L^(1) (RR^(n) ) space forall alpha : abs(alpha) <= k$, so ist
|
||||
$
|
||||
hat(p ^(alpha) u ) = Q^(alpha) u \, space abs(alpha) <= k \
|
||||
abs( hat(u) (xi)) <= C (1) / (1 + norm(xi) k).
|
||||
$
|
||||
Falls $u$ kompakt getragen ist, so kann $hat(u)$ in eine ueberall konvergente Potenzreihe entwickelt werden
|
||||
$
|
||||
hat(u) (xi) = sum c_(k) (xi - xi_0 )^(k).
|
||||
$
|
||||
]
|
||||
|
||||
#proof[
|
||||
Betrachte
|
||||
$
|
||||
hat(P^(alpha) u) = Q^(alpha) hat(u) = x ^(alpha) hat(u)
|
||||
$
|
||||
]
|
||||
Reference in New Issue
Block a user