Polazimo od Dalamberovog principa da je rad sila reakcija podloge pri mogućem ili virtuelnom pomeranju tela jednak 0, tj. ako je:
m
ν
a
→
ν
−
F
→
ν
=
R
→
ν
{\displaystyle m_{\nu }{\overrightarrow {a}}_{\nu }-{\overrightarrow {F}}_{\nu }={\overrightarrow {R}}_{\nu }}
, gde je
m
ν
{\displaystyle m_{\nu }}
— masa ν-tog tela, aν — njegovo ubrzanje,
F
→
ν
{\displaystyle {\overrightarrow {F}}_{\nu }}
— rezultanta dejstvujućih sila na ν-to telo i
R
→
ν
{\displaystyle {\overrightarrow {R}}_{\nu }}
— reakcija podloge na ν-to telo, tada je:
∑
ν
=
1
N
R
→
ν
⋅
δ
r
→
ν
=
0
{\displaystyle \sum _{\nu =1}^{N}{\overrightarrow {R}}_{\nu }\cdot \delta {\overrightarrow {r}}_{\nu }=0}
, odnosno
∑
ν
=
1
N
(
m
ν
a
→
ν
−
F
→
ν
)
⋅
δ
r
→
ν
=
0
{\displaystyle \sum _{\nu =1}^{N}{\Bigl (}m_{\nu }{\overrightarrow {a}}_{\nu }-{\overrightarrow {F}}_{\nu }{\Bigr )}\cdot \delta {\overrightarrow {r}}_{\nu }=0}
1.
δ
r
→
ν
=
∑
i
=
1
n
∂
r
→
ν
∂
q
i
δ
q
i
{\displaystyle \delta {\overrightarrow {r}}_{\nu }=\sum _{i=1}^{n}{\partial {\overrightarrow {r}}_{\nu } \over \partial q_{i}}\delta q_{i}}
;
v
→
ν
=
d
r
→
ν
d
t
=
∑
i
=
1
n
∂
r
→
ν
∂
q
i
q
˙
i
{\displaystyle {\overrightarrow {v}}_{\nu }={d{\overrightarrow {r}}_{\nu } \over dt}=\sum _{i=1}^{n}{\partial {\overrightarrow {r}}_{\nu } \over \partial q_{i}}{\dot {q}}_{i}}
=>
∂
r
→
ν
∂
q
i
=
∂
v
→
ν
∂
q
˙
i
{\displaystyle {\partial {\overrightarrow {r}}_{\nu } \over \partial q_{i}}={\partial {\overrightarrow {v}}_{\nu } \over \partial {\dot {q}}_{i}}}
, sada izraz 1. postaje
∑
ν
=
1
N
(
m
ν
d
v
→
ν
d
t
−
F
→
ν
)
⋅
δ
r
→
ν
=
0
{\displaystyle \sum _{\nu =1}^{N}{\Bigl (}m_{\nu }{d{\overrightarrow {v}}_{\nu } \over dt}-{\overrightarrow {F}}_{\nu }{\Bigr )}\cdot \delta {\overrightarrow {r}}_{\nu }=0}
uz
d
v
→
ν
d
t
=
∑
i
=
1
n
(
∂
v
→
ν
∂
q
i
q
˙
i
+
∂
r
→
ν
∂
q
i
q
¨
i
)
{\displaystyle {d{\overrightarrow {v}}_{\nu } \over dt}=\sum _{i=1}^{n}{\Bigl (}{\partial {\overrightarrow {v}}_{\nu } \over \partial q_{i}}{\dot {q}}_{i}+{\partial {\overrightarrow {r}}_{\nu } \over \partial q_{i}}{\ddot {q}}_{i}{\Bigr )}}
, a
∂
r
→
ν
∂
q
i
q
¨
i
=
d
(
∂
r
→
ν
∂
q
i
q
˙
i
)
d
t
−
∂
v
→
ν
∂
q
i
q
˙
i
{\displaystyle {\partial {\overrightarrow {r}}_{\nu } \over \partial q_{i}}{\ddot {q}}_{i}={d{\Bigl (}{\partial {\overrightarrow {r}}_{\nu } \over \partial q_{i}}{\dot {q}}_{i}{\Bigr )} \over dt}-{\partial {\overrightarrow {v}}_{\nu } \over \partial q_{i}}{\dot {q}}_{i}}
∑
ν
=
1
N
m
ν
∑
i
=
1
,
j
=
1
n
(
d
(
∂
r
→
ν
∂
q
i
q
˙
i
)
d
t
∂
r
→
ν
∂
q
j
)
δ
q
j
−
∑
ν
=
1
N
F
→
ν
∑
j
=
1
n
∂
r
→
ν
∂
q
j
δ
q
j
=
0
{\displaystyle \sum _{\nu =1}^{N}m_{\nu }\sum _{i=1,j=1}^{n}{\Bigl (}{d{\Bigl (}{\partial {\overrightarrow {r}}_{\nu } \over \partial q_{i}}{\dot {q}}_{i}{\Bigr )} \over dt}{\partial {\overrightarrow {r}}_{\nu } \over \partial q_{j}}{\Bigr )}\delta q_{j}-\sum _{\nu =1}^{N}{\overrightarrow {F}}_{\nu }\sum _{j=1}^{n}{\partial {\overrightarrow {r}}_{\nu } \over \partial q_{j}}\delta q_{j}=0}
Kako je kinetička energija,
T
=
∑
ν
=
1
N
1
2
m
ν
v
ν
2
=
∑
ν
=
1
N
m
ν
∑
i
=
1
,
j
=
1
n
∂
r
→
ν
∂
q
˙
i
q
˙
i
∂
r
→
ν
∂
q
˙
j
q
˙
j
{\displaystyle T=\sum _{\nu =1}^{N}{\frac {1}{2}}m_{\nu }v_{\nu }^{2}=\sum _{\nu =1}^{N}m_{\nu }\sum _{i=1,j=1}^{n}{\partial {\overrightarrow {r}}_{\nu } \over \partial {\dot {q}}_{i}}{\dot {q}}_{i}{\partial {\overrightarrow {r}}_{\nu } \over \partial {\dot {q}}_{j}}{\dot {q}}_{j}}
=>
m
ν
∑
j
=
1
n
∂
r
→
ν
∂
q
j
∂
r
→
ν
∂
q
i
q
˙
j
=
∂
T
∂
q
˙
i
{\displaystyle m_{\nu }\sum _{j=1}^{n}{\partial {\overrightarrow {r}}_{\nu } \over \partial q_{j}}{\partial {\overrightarrow {r}}_{\nu } \over \partial q_{i}}{\dot {q}}_{j}={\partial T \over \partial {\dot {q}}_{i}}}
=>
∑
ν
=
1
N
m
ν
∑
i
=
1
,
j
=
1
n
(
d
(
∂
r
→
ν
∂
q
i
q
˙
i
)
d
t
∂
r
→
ν
∂
q
j
)
δ
q
j
=
∑
j
=
1
n
(
d
(
∂
T
∂
q
˙
j
)
d
t
−
∂
T
∂
q
j
)
δ
q
j
{\displaystyle \sum _{\nu =1}^{N}m_{\nu }\sum _{i=1,j=1}^{n}{\Bigl (}{d{\Bigl (}{\partial {\overrightarrow {r}}_{\nu } \over \partial q_{i}}{\dot {q}}_{i}{\Bigr )} \over dt}{\partial {\overrightarrow {r}}_{\nu } \over \partial q_{j}}{\Bigr )}\delta q_{j}=\sum _{j=1}^{n}{\Bigl (}{d{\Bigl (}{\partial T \over \partial {\dot {q}}_{j}}{\Bigr )} \over dt}-{\partial T \over \partial q_{j}}{\Bigr )}\delta q_{j}}
Pa ako su sile potencijalne, tj. važi
F
→
ν
=
−
∂
U
∂
r
→
ν
{\displaystyle {\overrightarrow {F}}_{\nu }=-{\partial U \over \partial {\overrightarrow {r}}_{\nu }}}
, to izraz
F
→
ν
∑
i
=
1
n
∂
r
→
ν
∂
q
i
=
−
∂
U
∂
q
j
{\displaystyle {\overrightarrow {F}}_{\nu }\sum _{i=1}^{n}{\partial {\overrightarrow {r}}_{\nu } \over \partial q_{i}}=-{\partial U \over \partial q_{j}}}
i konačno jednačina 1 postaje:
∑
j
=
1
n
(
d
(
∂
T
∂
q
˙
j
)
d
t
−
∂
T
∂
q
j
+
∂
U
∂
q
j
)
δ
q
j
=
0
{\displaystyle \sum _{j=1}^{n}{\Bigl (}{d{\Bigl (}{\partial T \over \partial {\dot {q}}_{j}}{\Bigr )} \over dt}-{\partial T \over \partial q_{j}}+{\partial U \over \partial q_{j}}{\Bigr )}\delta q_{j}=0}
Kako su moguća pomeranja
δ
q
j
{\displaystyle \delta q_{j}}
proizvoljna, to sledi:
−
d
d
t
(
∂
L
∂
q
˙
i
)
+
∂
L
∂
q
i
=
0
{\displaystyle -{\frac {d}{dt}}{\Bigl (}{\partial L \over \partial {\dot {q}}_{i}}{\Bigr )}+{\partial L \over \partial q_{i}}=0}
Prikaz klatna sa dejstvujućim silama, vezama i generalisanom koordinatom
Kruto telo zanemarljive mase ograničava kretanje tela mase ṁ zanemarljivih dimenzija, tako da se kretanje prati zamo uglom θ — otklona štapa od vertikale, pa se dobije:
d
r
→
=
d
r
e
→
r
+
r
d
θ
e
→
θ
{\displaystyle d{\overrightarrow {r}}=dr{\overrightarrow {e}}_{r}+rd\theta {\overrightarrow {e}}_{\theta }}
=>
d
r
→
=
r
d
θ
e
→
θ
;
d
r
=
0
{\displaystyle d{\overrightarrow {r}}=rd\theta {\overrightarrow {e}}_{\theta };dr=0}
v
=
r
ϕ
˙
{\displaystyle v=r{\dot {\phi }}}
;
T
=
1
2
m
v
2
=
1
2
m
l
2
θ
˙
2
{\displaystyle T={\frac {1}{2}}mv^{2}={\frac {1}{2}}ml^{2}{\dot {\theta }}^{2}}
;
U
=
m
g
l
(
1
−
cos
θ
)
{\displaystyle U=mgl(1-\cos \theta )}
L
=
T
−
U
=
1
2
m
r
2
ϕ
˙
2
−
m
g
l
(
1
−
cos
θ
)
{\displaystyle L=T-U={\frac {1}{2}}mr^{2}{\dot {\phi }}^{2}-mgl(1-\cos \theta )}
, pa iz
−
d
d
t
(
∂
L
∂
θ
˙
)
+
∂
L
∂
θ
=
0
{\displaystyle -{\frac {d}{dt}}{\Bigl (}{\partial L \over \partial {\dot {\theta }}}{\Bigr )}+{\partial L \over \partial \theta }=0}
=>
l
2
θ
¨
=
−
l
g
sin
θ
{\displaystyle l^{2}{\ddot {\theta }}=-lg\sin \theta }
, pa za
θ
→
0
{\displaystyle \theta \rightarrow 0}
proizilazi rešenje
θ
=
A
cos
(
ω
t
+
ϕ
0
)
{\displaystyle \theta =A\cos {(\omega t+\phi _{0})}}
ω
=
g
l
{\displaystyle \omega ={\sqrt {\frac {g}{l}}}}
Kulonovo polje sila pripada tipu centralnih sila, kod kojih je moment impulsa jednak 0.
r
→
×
m
v
˙
→
=
r
→
×
F
(
r
)
r
→
r
{\displaystyle {\overrightarrow {r}}\times m{\overrightarrow {\dot {v}}}={\overrightarrow {r}}\times F(r){\frac {\overrightarrow {r}}{r}}}
, a po svojstvu vektorskog proizvoda
r
→
×
r
→
=
0
{\displaystyle {\overrightarrow {r}}\times {\overrightarrow {r}}=0}
, pa je
r
→
×
m
v
→
=
m
r
2
ϕ
˙
k
→
{\displaystyle {\overrightarrow {r}}\times m{\overrightarrow {v}}=mr^{2}{\dot {\phi }}{\overrightarrow {k}}}
konstanta kretanja.
Isti rezultat lako dobijamo iz Langraževog formalizma:
−
d
d
t
∂
L
∂
ϕ
˙
−
∂
L
∂
ϕ
=
0
{\displaystyle -{d \over dt}{\partial L \over \partial {\dot {\phi }}}-{\partial L \over \partial \phi }=0}
L
=
T
−
U
=
1
2
m
r
2
ϕ
˙
2
−
Z
e
2
4
π
ϵ
0
r
{\displaystyle L=T-U={\frac {1}{2}}mr^{2}{\dot {\phi }}^{2}-{\frac {Ze^{2}}{4\pi \epsilon _{0}r}}}
=>
∂
L
∂
ϕ
˙
=
m
r
2
ϕ
˙
{\displaystyle {\partial L \over \partial {\dot {\phi }}}=mr^{2}{\dot {\phi }}}
,jer je
∂
L
∂
ϕ
=
0
{\displaystyle {\partial L \over \partial \phi }=0}
Ẕ — broj protona u jezgri atoma ili redni broj atoma, ṁ — masa elektrona , e — naelektrisanje elektrona, ε0 — dielektrička propustljivost vakuuma.