미분 지오미터’s 접근
Prerequisites:
Stokes에 대한 기본적인 이해’ 미분 외장 텐서 대수학의 정리 n n n M M M
기본 Riemannian 기하학에 노출, 아인슈타인 / PAIN 표기법을 포함한 로컬 좌표의 esp.
Brownian Motion 및 Martingale Theory를 포함한 Smooth 및 Stochastic Dynamical Systems에 대한 관심.
부드러운 Hamiltonian 정의 H : T q ∗ M ⊕ R → R \mathcal H:T_q^{*}M\oplus\Reals\rightarrow\Reals H : T q ∗ M ⊕ R → R H ( p , q , t ) \mathcal H(p,q,t) H ( p , q , t )
θ : = p d q − H ( p , q , t ) d t ∈ T ∗ ( T ∗ M ⊕ R ) \theta := p\ dq - \mathcal H(p,q,t)\ dt\in T^*(T^*M\oplus\Reals) θ := p d q − H ( p , q , t ) d t ∈ T ∗ ( T ∗ M ⊕ R ) 정의 S H ( γ ) : = ∫ γ θ \mathcal S_\mathcal H(\gamma) := \int_\gamma \theta S H ( γ ) := ∫ γ θ γ : [ 0 , t ] → T ∗ M ⊕ R \gamma:[0,t]\rightarrow T^{*}M\oplus\Reals γ : [ 0 , t ] → T ∗ M ⊕ R
두 개의 커브가 γ 1 , γ 2 \gamma_1, \gamma_2 γ 1 , γ 2 빼기 를 정의합니다. γ 1 − γ 2 \gamma_1 - \gamma_2 γ 1 − γ 2 γ 1 \gamma_1 γ 1 γ 2 \gamma_2 γ 2 S S S γ 1 − γ 2 = ∂ S \gamma_1 - \gamma_2 = \partial S γ 1 − γ 2 = ∂ S
S H ( γ 1 ) − S H ( γ 2 ) = ∫ γ 1 − γ 2 θ = ∫ ∂ S θ = ∫ S d θ \begin{aligned}
\mathcal S_\mathcal H(\gamma_1) - \mathcal S_\mathcal H(\gamma_2) &= \int_{\gamma_1 - \gamma_2}\theta \\
&= \int_{\partial S}\theta\\
&= \int_S d\theta
\end{aligned} S H ( γ 1 ) − S H ( γ 2 ) = ∫ γ 1 − γ 2 θ = ∫ ∂ S θ = ∫ S d θ
의해 stokes’ 정리
이러한 표면의 여부와 상관없이 S S S S H \mathcal S_\mathcal H S H γ \gamma γ d θ d\theta d θ γ \gamma γ
ω H : = d θ = d p ∧ d q − d H ∧ d t ∈ ⋀ 2 T ∗ ( T ∗ M ⊕ R ) \omega_\mathcal H := d\theta = dp\wedge dq - d\mathcal H \wedge dt\in\bigwedge^2T^*(T^*M\oplus\Reals) ω H := d θ = d p ∧ d q − d H ∧ d t ∈ ⋀ 2 T ∗ ( T ∗ M ⊕ R ) ω H ∣ γ = p ˙ d t ∧ d q + q ˙ d p ∧ d t − ∂ H ∂ p d p ∧ d t − ∂ H ∂ q d q ∧ d t = ( − p ˙ i − ∂ H ∂ q i ) d q i ∧ d t + ( q ˙ i − ∂ H ∂ p i ) d p i ∧ d t \begin{aligned}
\omega_\mathcal H|_\gamma &= \dot{p}\ dt\wedge dq + \dot{q}\ dp\wedge dt - \frac{\partial \mathcal H}{\partial p}dp\wedge dt - \frac{\partial \mathcal H}{\partial q}dq\wedge dt \\
&= (-\dot{p}_i - \frac{\partial \mathcal H}{\partial q^i}) dq^i \wedge dt+ (\dot{q}^i - \frac{\partial \mathcal H}{\partial p_i}) dp_i\wedge dt
\end{aligned} ω H ∣ γ = p ˙ d t ∧ d q + q ˙ d p ∧ d t − ∂ p ∂ H d p ∧ d t − ∂ q ∂ H d q ∧ d t = ( − p ˙ i − ∂ q i ∂ H ) d q i ∧ d t + ( q ˙ i − ∂ p i ∂ H ) d p i ∧ d t
∴ ω H ∣ γ = 0 ⟺ γ ( t ) \therefore \omega_\mathcal H|_\gamma = 0 \iff \gamma(t) ∴ ω H ∣ γ = 0 ⟺ γ ( t ) p ˙ = − ∂ H ∂ q q ˙ = ∂ H ∂ p \begin{aligned}
\dot p &= -\frac{\partial \mathcal H}{\partial q} \\
\dot q &= \ \ \ \frac{\partial \mathcal H}{\partial p}
\end{aligned} p ˙ q ˙ = − ∂ q ∂ H = ∂ p ∂ H
⟺ γ : [ 0 , t ] → T ∗ M ⊕ R \iff \gamma:[0,t]\rightarrow T^*M\oplus\R ⟺ γ : [ 0 , t ] → T ∗ M ⊕ R 정상 곡선 입니다. S H ( γ ) = ∫ γ θ \mathcal S_\mathcal H(\gamma)=\int_\gamma \theta S H ( γ ) = ∫ γ θ
시기 H \mathcal H H p p p ∀ q ˙ ∈ T q M ∃ ! p = p m a x ( q ˙ ) \forall \dot{q} \in T_q M\ \exists !\ p=p_{max}(\dot q) ∀ q ˙ ∈ T q M ∃ ! p = p ma x ( q ˙ ) q ˙ = ∂ H ∂ p ( p m a x , q , t ) \dot{q} = \frac{\partial \mathcal H}{\partial p}(p_{max},q,t) q ˙ = ∂ p ∂ H ( p ma x , q , t ) L \mathcal L L H \mathcal H H
L ( q ˙ , q , t ) : = max p p q ˙ − H ( p , q , t ) = p m a x ( q ˙ ) q ˙ − H ( p m a x ( q ˙ ) , q , t ) S L ( π ( γ ) ) = ∫ π ( γ ) L ( q ˙ , q , t ) d t \begin{aligned}
\mathcal{L}(\dot q,q,t) &:= \max_p p\dot{q} - \mathcal H(p,q,t) \\&= p_{max}(\dot q)\dot q - \mathcal H(p_{max}(\dot q),q,t) \\
\mathcal S_\mathcal{L}(\pi(\gamma)) &= \int_{\pi(\gamma)} \mathcal{L}(\dot q, q, t)\ dt
\end{aligned} L ( q ˙ , q , t ) S L ( π ( γ )) := p max p q ˙ − H ( p , q , t ) = p ma x ( q ˙ ) q ˙ − H ( p ma x ( q ˙ ) , q , t ) = ∫ π ( γ ) L ( q ˙ , q , t ) d t
Lagrangian representation of the Action 의 약어입니다. π : T ∗ M ⊕ R → M ⊕ R \pi: T^*M\oplus\Reals \rightarrow M\oplus\Reals π : T ∗ M ⊕ R → M ⊕ R ( p , q , t ) ↦ ( q , t ) (p,q,t)\mapsto (q,t) ( p , q , t ) ↦ ( q , t )
최소 행동의 원칙은 단순히 자연의 고전 역학 자체가 최소화하는 궤적을 선택하는 경향이 있다고 주장한다. S L \mathcal S_\mathcal{L} S L
일반적으로 이 청구는 거짓 입니다. 상기 고정 곡선은 The fixed curves of S H \mathcal S_\mathcal H S H S L \mathcal S_\mathcal{L} S L 정지 궤적에 대한 차등 방정식은 동일합니다 , 등 S H = S L \mathcal S_\mathcal H = \mathcal S_\mathcal{L} S H = S L ( d L ∧ d t ) ∣ π ( γ ) = 0 : (d\mathcal{L}\wedge dt)|_{\pi(\gamma)} = 0: ( d L ∧ d t ) ∣ π ( γ ) = 0 :
∂ L ∂ q = d d t ∂ L ∂ q ˙ \frac{\partial \mathcal L}{\partial q} = \frac{d}{dt}\frac{\partial \mathcal L}{\partial \dot q} ∂ q ∂ L = d t d ∂ q ˙ ∂ L
두 번째 순서의 ODE t ↦ q ( t ) t \mapsto q(t) t ↦ q ( t ) 2 dim M + 1 2\dim M+1 2 dim M + 1 ( q ˙ 0 , q 0 , t 0 ) (\dot q_0, q_0, t_0) ( q ˙ 0 , q 0 , t 0 )
하지만 흥미로운 점은 S L ( π ∘ γ ) \mathcal S_\mathcal L(\pi\circ\gamma) S L ( π ∘ γ ) π ∘ γ \pi\circ\gamma π ∘ γ ( q 0 , t 0 ) (q_0, t_0) ( q 0 , t 0 ) ( q f , t f ) (q_f, t_f) ( q f , t f ) q ˙ 0 \dot q_0 q ˙ 0 ( q f , t f ) (q_f, t_f) ( q f , t f ) π ∘ γ \pi\circ\gamma π ∘ γ S = S ( q 0 , t 0 , q f , t f ) \mathcal S = \mathcal S(q_0,t_0, q_f, t_f) S = S ( q 0 , t 0 , q f , t f ) 전환 함수 로, 고정 옵션에 종속되지 않는다고 가정합니다. π ∘ γ \pi\circ\gamma π ∘ γ γ \gamma γ γ \gamma γ
허가’한 걸음 물러서서 더 간단한 것을 정의하십시오. “가로” 리프트 A = q ˙ ⊕ π − 1 : T q M ⊕ R → T q ∗ M ⊕ R \mathcal A=\dot q\oplus \pi^{-1}:T_{q} M\oplus \Reals \rightarrow T_{q}^{*}M\oplus \Reals A = q ˙ ⊕ π − 1 : T q M ⊕ R → T q ∗ M ⊕ R
( q ˙ , q , t ) ↦ ( p m a x ( q ˙ ) , q , t ) . (\dot q, q,t)\mapsto (p_{max}(\dot q), q, t)\ . ( q ˙ , q , t ) ↦ ( p ma x ( q ˙ ) , q , t ) .
이제 우리는 모든 “예상” 부드러운 곡선(정지 곡선뿐만 아니라) γ ~ : [ 0 , t ] → M ⊕ R \tilde\gamma:[0,t]\rightarrow M\oplus\R γ ~ : [ 0 , t ] → M ⊕ R
S L ( γ ~ ) = S H ( A ∘ γ ~ ) . \begin{aligned}
\mathcal S_\mathcal{L}(\tilde\gamma) &= \mathcal S_\mathcal H(\mathcal A\circ \tilde\gamma) \ .
\end{aligned} S L ( γ ~ ) = S H ( A ∘ γ ~ ) .
Note: 볼록성 제약 조건 H \mathcal H H p m a x ( 0 ) p_{max}(0) p ma x ( 0 ) q ˙ = 0 \dot q = 0 q ˙ = 0 γ \gamma γ π − 1 \pi^{-1} π − 1 γ ~ \tilde \gamma γ ~ A \mathcal A A
시기 H ( p , q , t ) \mathcal H(p,q,t) H ( p , q , t ) p p p [ g i j ] : M ⊕ R → T M ⊙ T M [g^{ij}]: M\oplus\Reals\rightarrow TM\odot TM [ g ij ] : M ⊕ R → TM ⊙ TM [ g i j ] : M ⊕ R → T ∗ M ⊙ T ∗ M [g_{ij}]: M\oplus\Reals\rightarrow T^{*}M\odot T^{*}M [ g ij ] : M ⊕ R → T ∗ M ⊙ T ∗ M
H V ( p , q , t ) = 1 2 g i j ( q , t ) p i p j + V ( q , t ) ⟹ L V ( q , q ˙ , t ) = 1 2 g i j ( q , t ) q ˙ i q ˙ j − V ( q , t ) . \begin{aligned}
\mathcal H^\mathcal V(p,q,t) &= \frac{1}{2}\ g^{ij}(q,t)\ p_ip_j + \mathcal V(q,t) \implies\\
\mathcal{L}^\mathcal V(q,\dot q, t) &= \frac{1}{2}\ g_{ij}(q,t)\dot{q}^i\dot{q}^j - \mathcal V(q,t)\ .
\end{aligned} H V ( p , q , t ) L V ( q , q ˙ , t ) = 2 1 g ij ( q , t ) p i p j + V ( q , t ) ⟹ = 2 1 g ij ( q , t ) q ˙ i q ˙ j − V ( q , t ) .
Levi-Civita 연결’s 크리스토펠 기호 g g g
Γ i j k = 1 2 g k a ( ∂ i g j a + ∂ j g i a − ∂ a g i j ) Γ k i j = 1 2 g k a ( ∂ i g j a + ∂ j g i a − ∂ a g i j ) . \begin{aligned}
\Gamma^k_{ij} &= \frac{1}{2} g^{ka}(\partial_i g_{ja} + \partial_j g_{ia} - \partial_a g_{ij})\\
\Gamma_k^{ij} &= \frac{1}{2}g_{ka}(\partial^ig^{ja} + \partial^j g^{ia} - \partial^ag^{ij}).
\end{aligned} Γ ij k Γ k ij = 2 1 g ka ( ∂ i g ja + ∂ j g ia − ∂ a g ij ) = 2 1 g ka ( ∂ i g ja + ∂ j g ia − ∂ a g ij ) .
포함 ∂ i : = ∂ ∂ q i \partial_i := \frac{\partial}{\partial q^i} ∂ i := ∂ q i ∂ ∂ i : = g i j ∂ j \partial^i := g^{ij}\partial_j ∂ i := g ij ∂ j ∇ \nabla ∇
∇ a i ∂ i b j ∂ j = d b j ( a i ∂ i ) ∂ j + Γ i j k a i b j ∂ k , or ∇ ∂ i ∂ j = Γ i j k ∂ k , and contravariantly ∇ ∂ i ∂ j = Γ k i j ∂ k , so ∇ = d + Γ \begin{aligned}
\nabla_{a^i\partial_i} b^j\partial_j &=d b^j(a^i\partial_i)\partial_j + \Gamma_{ij}^k a^ib ^j\partial_k\ ,\text{ or}\\
\nabla_{\partial_i}\partial_j &= \Gamma_{ij}^k\partial_k \text{ , and contravariantly}\\
\nabla_{\partial^i}\partial^j &= \Gamma^{ij}_k\partial^k \text{, so} \\
\nabla &= d + \Gamma
\end{aligned} ∇ a i ∂ i b j ∂ j ∇ ∂ i ∂ j ∇ ∂ i ∂ j ∇ = d b j ( a i ∂ i ) ∂ j + Γ ij k a i b j ∂ k , or = Γ ij k ∂ k , and contravariantly = Γ k ij ∂ k , so = d + Γ
모든 텐서 필드. 특히 Γ \Gamma Γ ( i , j ) (i, j) ( i , j ) ∇ [ g i j ] = ∇ [ g i j ] = 0 \nabla [g_{ij}] = \nabla [g^{ij}] = 0 ∇ [ g ij ] = ∇ [ g ij ] = 0
일화적으로, Riemann-Christoffel 곡률 텐서는
R ρ σ μ ν = ∂ μ Γ ρ ν σ − ∂ ν Γ ρ μ σ + Γ ρ μ λ Γ λ ν σ − Γ ρ ν λ Γ λ μ σ \mathcal R^{\rho }{}_{\sigma \mu \nu }=\partial _{\mu }\Gamma ^{\rho }{}_{\nu \sigma }-\partial _{\nu }\Gamma ^{\rho }{}_{\mu \sigma }+\Gamma ^{\rho }{}_{\mu \lambda }\Gamma ^{\lambda }{}_{\nu \sigma }-\Gamma ^{\rho }{}_{\nu \lambda }\Gamma ^{\lambda }{}_{\mu \sigma } R ρ σ μν = ∂ μ Γ ρ ν σ − ∂ ν Γ ρ μ σ + Γ ρ μ λ Γ λ ν σ − Γ ρ ν λ Γ λ μ σ
H \mathcal H H L \mathcal L L 또한 H = H B \mathcal H = \mathcal H_B H = H B B ( q , t ) ∈ T q M \mathcal B(q,t)\in T_qM B ( q , t ) ∈ T q M p ∈ T q ∗ M p\in T^{*}_qM p ∈ T q ∗ M 광장을 완료 할 수 있습니다 다시 계산 L B \mathcal{L}_B L B L \mathcal L L
H B ( p , q , t ) = H + B p ⟹ L B ( q ˙ , q , t ) = max p p ( q ˙ − B ) − H = L ( q , q ˙ − B , t ) H B = 1 2 g i j p i p j + p B + V ⟹ L B = L − g i j B i q ˙ j + 1 2 g i j B i B j \begin{aligned}
\mathcal H_\mathcal B(p,q,t) &= \mathcal H + \mathcal Bp \implies\\
\mathcal L_\mathcal B(\dot q,q,t) &=\max_p p(\dot q - \mathcal B) - \mathcal H\\
&\ \begin{equation}
\tag{A}= \mathcal{L}(q,\dot{q}-\mathcal B, t)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{equation}\\
\mathcal H_\mathcal B &= \frac{1}{2}\ g^{ij}p_ip_j + p\mathcal B + \mathcal V \implies\\
\mathcal L_\mathcal B &\ \begin{equation}\tag{B}= \mathcal L - g_{ij}\mathcal B^i\dot q^j + \frac{1}{2}\ g_{ij}\mathcal B^i\mathcal B^j\ \ \ \ \ \ \end{equation}
\end{aligned} H B ( p , q , t ) L B ( q ˙ , q , t ) H B L B = H + B p ⟹ = p max p ( q ˙ − B ) − H = L ( q , q ˙ − B , t ) ( A ) = 2 1 g ij p i p j + p B + V ⟹ = L − g ij B i q ˙ j + 2 1 g ij B i B j ( B )
여기서 Lagrange Multiplier 간의 연결을 확인합니다. B \mathcal B B H \mathcal H H L \mathcal L L B \mathcal B B 두 표현식 모두 ( A ) (A) ( A ) ( B ) (B) ( B ) L B \mathcal{L}_\mathcal B L B
A \mathcal A A 시작일 ∂ H ∂ p i ( p , q , t ) = g i j p j ⟹ ∂ 2 H ∂ p i ∂ p j = g i j \frac{\partial \mathcal H}{\partial p_i}(p,q,t) = g^{ij}p_j \implies \frac{\partial^2\mathcal H}{\partial p_i \partial p_j} = g^{ij} ∂ p i ∂ H ( p , q , t ) = g ij p j ⟹ ∂ p i ∂ p j ∂ 2 H = g ij
p m a x i = g i j q ˙ j = ∂ L ∂ q ˙ i ⟹ A ( q ˙ , q , t ) = ( [ g ] q ˙ , q , t ) . \begin{aligned}
{p_{max}}_i &= g_{ij}\dot q^j = \frac{\partial \mathcal L}{\partial \dot q_i}\implies \\
\mathcal A(\dot q, q, t) &= ([g] \dot q, q, t)\ .
\end{aligned} p ma x i A ( q ˙ , q , t ) = g ij q ˙ j = ∂ q ˙ i ∂ L ⟹ = ([ g ] q ˙ , q , t ) .
시기 g i j g^{ij} g ij S L ( γ ~ ) = S H ( A ∘ γ ~ ) \mathcal S_\mathcal L(\tilde\gamma) = \mathcal S_\mathcal H(\mathcal A\circ\tilde\gamma) S L ( γ ~ ) = S H ( A ∘ γ ~ ) 최소화된 로컬입니다.
방정식에 의하여 (12) ( A ) , ( B ) (A), (B) ( A ) , ( B ) L B V \mathcal L^\mathcal V_\mathcal B L B V
1 2 ∂ i g j k ( q ˙ j − B j ) ( q ˙ k − B k ) − ∂ i V = d d t g i j ( q , t ) ( q ˙ j − B j ) = p ˙ m a x B i , − ∂ i V = 1 2 ( ∂ i g j k ) ( q ˙ j − B j ) ( q ˙ k − B k ) − g j k ( ∂ i B j ) ( q ˙ k − B k ) + d d t ( q ˙ i − B i ) + g i j ∂ g j k ∂ t ( q ˙ k − B k ) − ∇ V ( q , t ) = ∇ q ˙ − B ( q ˙ − B ) − ∂ t B + ( ∂ t [ log g ] ) ( q ˙ − B ) . \begin{aligned}
\frac{1}{2}\partial_i g_{jk}(\dot q^j-\mathcal B^j)(\dot q^k -\mathcal B^k)-\partial_i \mathcal V &= \frac{d}{dt}g_{ij}(q,t)(\dot q^j - \mathcal B^j) = {\dot p^\mathcal B_{max}}_i\ ,\\
- \partial^i \mathcal V &= \frac{1}{2} (\partial^i g_{jk})(\dot q^j-\mathcal B^j) (\dot q^k-\mathcal B^k) - g_{jk}(\partial^i\mathcal B^j)(\dot q^k - \mathcal B^k) + \frac{d}{dt} (\dot q^i - B^i) + g^{ij}\frac{\partial g_{jk}}{\partial t}(\dot q^k -\mathcal B^k)\\
-\nabla\mathcal V(q,t)&\begin{equation}\tag{C}=\nabla_{\dot q-\mathcal B} (\dot q - \mathcal B) -\partial_t \mathcal B +(\partial_t [\log g])(\dot q-\mathcal B) \ .\ \ \ \ \ \ \ \ \ \end{equation}
\end{aligned} 2 1 ∂ i g jk ( q ˙ j − B j ) ( q ˙ k − B k ) − ∂ i V − ∂ i V − ∇ V ( q , t ) = d t d g ij ( q , t ) ( q ˙ j − B j ) = p ˙ ma x B i , = 2 1 ( ∂ i g jk ) ( q ˙ j − B j ) ( q ˙ k − B k ) − g jk ( ∂ i B j ) ( q ˙ k − B k ) + d t d ( q ˙ i − B i ) + g ij ∂ t ∂ g jk ( q ˙ k − B k ) = ∇ q ˙ − B ( q ˙ − B ) − ∂ t B + ( ∂ t [ log g ]) ( q ˙ − B ) . ( C )
이것들은 정확히 **뉴턴’모션의 법칙 (Law of Motion) F / m = a F/m = a F / m = a ∂ t : = ∂ ∂ t \partial_t := \frac{\partial}{\partial t} ∂ t := ∂ t ∂ V \mathcal V V B \mathcal B B
공칭 매니폴드 N N N T ∗ ( M ⊕ R ) T^*(M\oplus \Reals) T ∗ ( M ⊕ R ) ω ∈ ⋀ 2 T ∗ N \omega \in \bigwedge^2T^*N ω ∈ ⋀ 2 T ∗ N N N N ω \omega ω
필요 d ω = 0 d\omega = 0 d ω = 0 θ \theta θ d θ = ω d\theta = \omega d θ = ω ω \omega ω
우리가 역학에 관심이있는 것은 행동입니다. S ( γ ) = ∫ γ θ \mathcal S(\gamma) = \int_\gamma \theta S ( γ ) = ∫ γ θ θ \theta θ H \mathcal H H N N N θ \theta θ N N N phase 의 통합성 조건(즉, θ \theta θ N N N e S e^\mathcal S e S N N N
ω n / n ! \omega^n/n! ω n / n ! 고전 역학이 최소 동작의 원리를 충족하는 곡선을 찾는 경우 Quantum Dynamics는 전체 클래스의 (일반적으로) 고정되지 않은 곡선에 대한 값을 통합하면서 동작의 지수에 관한 것입니다. “무한 치수 Lebesgue 측정” D d t γ ~ \mathcal D_{dt}\tilde \gamma D d t γ ~
실제로는 오직 Gaussian “커플링”
∫ { γ ~ } e − S L B V ( γ ) D d t γ ~ \int_{\set{\tilde \gamma}} e^{-\mathcal S_{\mathcal L_\mathcal B^{\mathcal V}}(\gamma)}\mathcal D_{dt}\tilde \gamma ∫ { γ ~ } e − S L B V ( γ ) D d t γ ~
일부에서 a(복잡한 값) 측정으로 해석이 필요합니다. { γ ~ } \set{\tilde \gamma} { γ ~ } 그러나이 건설은 점점 더 정교한 사례의 시리즈로서 앞으로 우리의 초점이 될 것입니다. 그것이 무엇이 되었든 간에, 실제 가치는 S \mathcal S S ∞ \infty ∞ ∞ \infty ∞ d t dt d t D d t γ ~ \mathcal D_{dt}\tilde \gamma D d t γ ~
그 위에 너무 좋은 점을 두지 말고, 고전적인 역학은 액션을 끝의 수단으로 정의합니다. 그것은 자신의 실제 가치 의 의미 에 대한 어떤 이해에 오는 것에 결코 관심이 없습니다. 우리는 단지 우리가 생각할 수 있도록 필요한 미분 방정식을 구성하기 위해 그것을 사용합니다. S \mathcal S S stationary curves 를 통해 끝점 사이의 전환 기능입니다. 고정 요구 사항을 통해 해석 가능 S \mathcal S S θ ↦ θ + d f \theta \mapsto \theta + df θ ↦ θ + df f ∈ C ∞ ( T ∗ M ) f \in C^\infty(T^*M) f ∈ C ∞ ( T ∗ M ) f f f
우리는 Quantum Dynamics에 있습니다!
광장과 Lebesgue 측정 (섬유의 섬유)의 번역 불변을 완료함으로써 T ∗ M T^*M T ∗ M
ℏ n ∫ R n e p i q ˙ i Δ t − 1 2 ℏ 2 g i j p i p j Δ t d p 1 … d p n = ℏ n ∫ R n e − 1 2 ℏ 2 g i j ( p i − g i k q ˙ k / ℏ ) ( p j − g j k q ˙ k / ℏ ) Δ t d p 1 … d p n ⋅ e 1 2 ℏ 2 g i j q ˙ i q ˙ j Δ t = e 1 2 ℏ 2 g i j q ˙ i q ˙ j Δ t ( 2 π Δ t ) n det g i j \hbar^n\int_{\Reals^n} e^{ p_i\dot q^i\Delta t - \frac{1}{2}\hbar^2 g^{ij}p_ip_j\Delta t}dp_1\dots dp_n = \hbar^n\int_{\Reals^n} e^{-\frac{1}{2}\hbar^2 g^{ij}(p_i - g_{ik}\dot q^k/\hbar)(p_j - g_{jk}\dot q^k/\hbar)\Delta t} dp_1\dots dp_n \cdot e^{\frac{1}{2\hbar^2}g_{ij}\dot q^i\dot q^j\Delta t} =
\frac{e^{\frac{1}{2\hbar^2}g_{ij}\dot q^i\dot q^j\Delta t}}{\sqrt{(2\pi\Delta t)^n \det g^{ij}}} ℏ n ∫ R n e p i q ˙ i Δ t − 2 1 ℏ 2 g ij p i p j Δ t d p 1 … d p n = ℏ n ∫ R n e − 2 1 ℏ 2 g ij ( p i − g ik q ˙ k /ℏ ) ( p j − g jk q ˙ k /ℏ ) Δ t d p 1 … d p n ⋅ e 2 ℏ 2 1 g ij q ˙ i q ˙ j Δ t = ( 2 π Δ t ) n det g ij e 2 ℏ 2 1 g ij q ˙ i q ˙ j Δ t
Feynman Path Integral 표현식은 quadratic Kinetic Energy 사례에서 도덕적으로 동등합니다(아직 공식적으로 무한함).
∫ { γ } e S H B V ( γ ) D d t γ ≈ ∫ R 2 n e ( p q ˙ − H B V ( p , q , t ) ) Δ t ω 0 n / n ! = 1 ( 2 π Δ t ) n ∫ R n e L B V ( q ˙ , q , t ) Δ t det g i j d q 1 . . . d q n ≈ ∫ { γ ~ } e S L B V ( γ ~ ) D d t γ ~ . \begin{aligned}
\int_{\set{\gamma}} e^{\mathcal S_{\mathcal H^\mathcal V_\mathcal B} (\gamma)} \mathcal D_{dt}\gamma &\approx \int_{\Reals^{2n}} e^{(p\dot q - \mathcal H^\mathcal V_\mathcal B(p,q,t))\Delta t}\omega_0^n/n!\\
&= \frac{1}{\sqrt{(2\pi\Delta t)^n}}\int_{\Reals^n}e^{\mathcal L^{\mathcal V}_\mathcal B(\dot q, q, t)\Delta t}\sqrt{\det g_{ij}}\ dq^1...dq^n\\
&\approx \int_{\set{\tilde \gamma}} e^{\mathcal S_{\mathcal L^{\mathcal V}_\mathcal B}(\tilde \gamma)} \mathcal D_{dt}\tilde \gamma \ .
\end{aligned} ∫ { γ } e S H B V ( γ ) D d t γ ≈ ∫ R 2 n e ( p q ˙ − H B V ( p , q , t )) Δ t ω 0 n / n ! = ( 2 π Δ t ) n 1 ∫ R n e L B V ( q ˙ , q , t ) Δ t det g ij d q 1 ... d q n ≈ ∫ { γ ~ } e S L B V ( γ ~ ) D d t γ ~ .
따라서 윅 회전과 플랑크 상수에 의한 벡터 배율 조정 ℏ = h / 2 π \hbar = h/2\pi ℏ = h /2 π d t ↦ ℏ / i d t , p ↦ ℏ p , q ˙ ↦ q ˙ / ℏ , B ↦ B / ℏ dt\mapsto \hbar/i\ dt, \ p\mapsto \hbar p, \ \dot q\mapsto \dot q/\hbar,\ \mathcal B\mapsto \mathcal B/\hbar d t ↦ ℏ/ i d t , p ↦ ℏ p , q ˙ ↦ q ˙ /ℏ , B ↦ B /ℏ
∫ { γ } e ℏ / i S H B / ℏ V ( ℏ p , q , t ) ( γ ) D ℏ / i d t γ ≈ ∫ { γ ~ } e ℏ / i S L B / ℏ V ( q ˙ / ℏ , q , t ) ( γ ~ ) D ℏ / i d t γ ~ = ∫ { γ ~ } e − i ℏ S L B ℏ 2 V ( q ˙ , q , t ) ( γ ~ ) D ℏ / i d t γ ~ . \begin{aligned}
\int_{\set{\gamma}}e^{\hbar/i\ \mathcal S_{\mathcal H_{\mathcal B/\hbar}^{\mathcal V}(\hbar p, q, t)}(\gamma)}\mathcal D_{\hbar/i\ dt}\gamma &\approx \int_{\set{\tilde\gamma}} e^{\hbar/i \ \mathcal S_{\mathcal L_{\mathcal B/\hbar}^{\mathcal V}(\dot q/\hbar,q, t)}(\tilde \gamma)}\mathcal D_{\hbar/i\ dt}\tilde\gamma\\
&= \int_{\set{\tilde \gamma}}e^{-\frac{i}{\hbar}\mathcal S_{\mathcal L^{\hbar^2 \mathcal V}_{\mathcal B}(\dot q, q, t)}(\tilde \gamma)}\mathcal D_{\hbar/i\ dt}\tilde\gamma\ .
\end{aligned} ∫ { γ } e ℏ/ i S H B /ℏ V ( ℏ p , q , t ) ( γ ) D ℏ/ i d t γ ≈ ∫ { γ ~ } e ℏ/ i S L B /ℏ V ( q ˙ /ℏ , q , t ) ( γ ~ ) D ℏ/ i d t γ ~ = ∫ { γ ~ } e − ℏ i S L B ℏ 2 V ( q ˙ , q , t ) ( γ ~ ) D ℏ/ i d t γ ~ .
그래서 우리가 방정식의 오른쪽을 근사하고 싶을 때 ℏ ↓ 0 \hbar\downarrow 0 ℏ ↓ 0 ( C ) (C) ( C ) V ↦ ℏ 2 V ≈ 0 \mathcal V \mapsto \hbar^2 \mathcal V\approx 0 V ↦ ℏ 2 V ≈ 0
H ( p , q ) = T ( p , q ) + V ( q ) , where T = 1 2 g i j ( q ) p i p j ⟹ e − i t / ℏ H ^ ∣ ψ > : = e − i t / ℏ ( − ℏ 2 2 Δ M + V ) ∣ ψ > ⟹ i ℏ d d t ∣ ψ > = − ℏ 2 2 Δ M ∣ ψ > + V ∣ ψ > \begin{aligned}
\mathcal H(p,q) &\ = \mathcal T(p,q) + \mathcal V(q)\ \text {, where } \mathcal T = \frac{1}{2}g^{ij}(q)p_ip_j \implies \\
e^{-it/\hbar \hat{\mathcal H}}\ket{\psi} &:= e^{-it/\hbar(-\frac{\hbar^2}{2} \Delta_M + \mathcal V)} \ket{\psi} \implies \\
i\hbar \frac{d}{dt}\ket{\psi} &\ = -\frac{\hbar ^2}{2}\Delta_M \ket{\psi} + \mathcal V\ket{\psi}
\end{aligned} H ( p , q ) e − i t /ℏ H ^ ∣ ψ ⟩ i ℏ d t d ∣ ψ ⟩ = T ( p , q ) + V ( q ) , where T = 2 1 g ij ( q ) p i p j ⟹ := e − i t /ℏ ( − 2 ℏ 2 Δ M + V ) ∣ ψ ⟩ ⟹ = − 2 ℏ 2 Δ M ∣ ψ ⟩ + V ∣ ψ ⟩
(Δ M \Delta_M Δ M g g g 분석 이라는 것입니다. t t t d t ↦ i / ℏ d t , p ↦ p / ℏ dt\mapsto i/\hbar\ dt,\ p\mapsto p/\hbar d t ↦ i /ℏ d t , p ↦ p /ℏ
d d t e − t H ^ ∣ ψ > = ( 1 2 Δ M − V ) e − t H ^ ∣ ψ > . \frac{d}{dt}e^{-t\hat{\mathcal H}}\ket{\psi} = (\frac{1}{2}\Delta_M - \mathcal V) e^{-t\hat{\mathcal H}}\ket{\psi} \ .
d t d e − t H ^ ∣ ψ ⟩ = ( 2 1 Δ M − V ) e − t H ^ ∣ ψ ⟩ .
이것은 샘플 경로 기반 확률 분석에 amenable 형식이며, 타원 확산 방정식에 대한 솔루션의 분석 연속과 Feynman Path-Integrals를 전체 오른쪽 반평면에 정렬하는 의미있는 방법을 제공합니다. 본질적으로, 우리는 잘 정의되어있을 것입니다 “측정-이론” 오른쪽 절반 평면에서 경계 선형 연산자 세트로 분석 맵 H = L 2 ( M , g ) \mathscr H = L^2(M,g) H = L 2 ( M , g ) i t ℏ , t ∈ R it\hbar\ ,t\in\Reals i t ℏ , t ∈ R Δ M \Delta_M Δ M H \mathscr H H
즉, 방정식 (17)의 역학을 연구하기에 충분하며, 일단 제안 경로 적분 표현의 명시적 정의와 관련된 미묘함을 명확히 한 것입니다.
Itô/Stratonovich/Malliavin SDE semimartingale calculus를 전체 천에서 재설계하는 대신, 우리는 일반적인 이론으로 우리를 수행 할 일련의 간단한 (평면 미터법) 예제를 진행할 것입니다.
하루가 끝나면 Feynman Path-Integral Quantization이 Schrödinger Quantization과 일치하거나 최소한 deviation 을 이해하기를 원할 것입니다. 특히, 우리는 Schrödinger PDE를 생성하기 위해 Semiclassical Approximation이 필요합니다 o ( t ) o(t) o ( t ) t ↓ 0 t\downarrow 0 t ↓ 0
그것이 밝혀지면, 여전히 논란이 있습니다. V \mathcal V V
대상 V ∈ C ∞ ( M ) V \in C^\infty(M) V ∈ C ∞ ( M )
e − i t / ℏ − Δ M ℏ / 2 e − i t / ℏ V = e − i t / ℏ ( − Δ M ℏ / 2 + V − i t / 4 ℏ [ Δ M ℏ , V ] + O ( t 2 ) ) e^{-it/\hbar -\Delta^\hbar_M/2} e^{-it/\hbar V} = e^{-it/\hbar(-\Delta^\hbar_M /2 + V - it/4\hbar [\Delta^\hbar_M,V] + O(t^2))} e − i t /ℏ − Δ M ℏ /2 e − i t /ℏ V = e − i t /ℏ ( − Δ M ℏ /2 + V − i t /4ℏ [ Δ M ℏ , V ] + O ( t 2 ))
Feynman-Kac Formula는 유클리드 공간에서 Brownian Motion의 경로 통합 공식에서 나옵니다. 이것의 핵심은 우리가 집중할 수 있다는 것입니다. V = 0 \mathcal V = 0 V = 0
Γ ^ \hat\Gamma Γ ^ 벡터 사용 위치 v ∈ T q M v \in T_qM v ∈ T q M Γ ^ t ( γ ) v ∈ T γ ( t ) M \hat\Gamma_t(\gamma)v \in T_{\gamma(t)}M Γ ^ t ( γ ) v ∈ T γ ( t ) M
v ( 0 ) = v ∇ γ ˙ ( t ) v ˙ = 0 \begin{aligned}
v(0) &= v \\
\nabla_{\dot \gamma(t)}\dot v &= 0
\end{aligned} v ( 0 ) ∇ γ ˙ ( t ) v ˙ = v = 0
특히 ∇ γ ˙ Γ ^ t ( γ ) = 0 \nabla_{\dot \gamma}\hat\Gamma_t(\gamma) = 0 ∇ γ ˙ Γ ^ t ( γ ) = 0 R ( X , Y ) = [ ∇ X , ∇ Y ] − ∇ [ X , Y ] \mathcal R(X,Y) = [\nabla_X,\nabla_Y] - \nabla_{[X,Y]} R ( X , Y ) = [ ∇ X , ∇ Y ] − ∇ [ X , Y ] Γ ^ \hat \Gamma Γ ^ γ \gamma γ R = 0 ⟺ Γ ^ t \mathcal R = 0 \iff \hat\Gamma_t R = 0 ⟺ Γ ^ t γ \gamma γ
다시 말해서, 우리가 병렬 전송을 무한한 움직임으로 분해하려고한다면 B ⊥ \mathcal B^\perp B ⊥ B \mathcal B B
Γ ^ ( γ ) = Γ ^ ( γ ∣ B ) Γ ^ ( γ ∣ B ⊥ ) − 1 2 R ( γ ˙ ∣ B , γ ˙ ∣ B ⊥ ) d t + O ( d t 2 ) ∇ γ ˙ Γ ^ ( γ ) = ∇ γ ˙ ∣ B Γ ^ ( γ ∣ B ⊥ ) + ∇ γ ˙ ∣ B ⊥ Γ ^ ( γ ∣ B ) − 1 2 R ( γ ˙ ∣ B , γ ˙ ∣ B ⊥ ) = 0 \begin{aligned}
\hat\Gamma(\gamma) &= \hat\Gamma(\gamma|_\mathcal B)\hat\Gamma(\gamma|_{\mathcal B^\perp}) - \frac{1}{2}\mathcal R(\dot{\gamma}|_\mathcal B, \dot{\gamma}|_{\mathcal B^\perp})dt + O(dt^2) \ \\
\nabla_{\dot \gamma}\hat\Gamma(\gamma) &= \nabla_{\dot \gamma|_\mathcal B}\hat\Gamma(\gamma|_{\mathcal B^\perp}) + \nabla_{\dot \gamma|{\mathcal B^\perp}}\hat\Gamma(\gamma|_{\mathcal B}) - \frac{1}{2}\mathcal R(\dot\gamma|_\mathcal B,\dot{\gamma}|_{\mathcal B^\perp}) = 0
\end{aligned} Γ ^ ( γ ) ∇ γ ˙ Γ ^ ( γ ) = Γ ^ ( γ ∣ B ) Γ ^ ( γ ∣ B ⊥ ) − 2 1 R ( γ ˙ ∣ B , γ ˙ ∣ B ⊥ ) d t + O ( d t 2 ) = ∇ γ ˙ ∣ B Γ ^ ( γ ∣ B ⊥ ) + ∇ γ ˙ ∣ B ⊥ Γ ^ ( γ ∣ B ) − 2 1 R ( γ ˙ ∣ B , γ ˙ ∣ B ⊥ ) = 0
방정식 (16)의 오른쪽은 일정한 계수를 위한 열 커널의 정확한 공식화입니다 (에서 q q q g i j g_{ij} g ij g i j = δ i − j g_{ij} = \delta_{i-j} g ij = δ i − j
이것은 **표준에 대한 히트 커널입니다. n n n
허가’이 경우 transition function 을 기억하십시오. S L ( q 0 , t 0 , q f , t f ) = ρ 2 ( q 0 , q f ) / 2 ( t f − t i ) \mathcal S_\mathcal L(q_0,t_0, q_f, t_f) = \rho^2(q_0, q_f)/2(t_f - t_i) S L ( q 0 , t 0 , q f , t f ) = ρ 2 ( q 0 , q f ) /2 ( t f − t i ) ρ \rho ρ q 0 q_0 q 0 q f q_f q f ∣ ∣ q ∣ ∣ 2 = q ⋅ q ||q||^2 = q\cdot q ∣∣ q ∣ ∣ 2 = q ⋅ q q q q
R H S t 16 ( q 0 , q f ) : = e − S L ( q 0 , 0 , q f , t ) ( 2 π t ) n g ( q f ) R = 0 ⟹ = e − ∣ ∣ q f − q i ∣ ∣ 2 2 t ( 2 π t ) n = ∫ R n R H S s 16 ( q i , q ) R H S t − s 16 ( q , q f ) d q 1 . . . d q n ∀ s ∈ ( 0 , t ) \begin{aligned}
RHS^{16}_t(q_0,q_f) &:= \frac{e^{-\mathcal S_\mathcal L(q_0, 0, q_f, t)}}{\sqrt{(2 \pi t)^n}} \sqrt{g(q_f)}\\
\mathcal R=0 \implies \\
&\ = \frac{e^{\frac{-||q_f - q_i||^2}{2t}}}{\sqrt{(2\pi t)^n}} \\
&\ = \int_{\Reals ^n}RHS^{16}_{s}(q_i, q)\ RHS^{16}_{t-s}(q, q_f)\ dq^1...dq^n\ \forall s\in (0, t)
\end{aligned} R H S t 16 ( q 0 , q f ) R = 0 ⟹ := ( 2 π t ) n e − S L ( q 0 , 0 , q f , t ) g ( q f ) = ( 2 π t ) n e 2 t − ∣∣ q f − q i ∣ ∣ 2 = ∫ R n R H S s 16 ( q i , q ) R H S t − s 16 ( q , q f ) d q 1 ... d q n ∀ s ∈ ( 0 , t )
왜 마지막 방정식은 사실입니까? 허가’s 경로 공간에서 사진을보고 : 우리는 연결하는 직선 지오데식이 있습니다 q 0 q_0 q 0 q f q_f q f t t t s s s g g g B \mathcal B B q ˙ − B \dot q-\mathcal B q ˙ − B R n \Reals^n R n
지정된 상수 벡터 필드 B t = ( q f − q 0 ) / t \mathcal B_t = (q_f - q_0) / t B t = ( q f − q 0 ) / t
q ( τ ) = B t τ + q 0 + q { τ / s 0 ≤ τ ≤ s ( t − τ ) / ( t − s ) s ≤ τ ≤ t q(\tau) = \mathcal B_t\tau + q_0 + q\begin{cases}
\tau/s & 0\leq\tau\leq s\\
(t - \tau)/(t-s)& s\leq\tau\leq t
\end{cases} q ( τ ) = B t τ + q 0 + q { τ / s ( t − τ ) / ( t − s ) 0 ≤ τ ≤ s s ≤ τ ≤ t
for fixed q ∈ R n q\in\Reals^n q ∈ R n s s s
By Equation (12) ( A ) (A) ( A ) ( B ) (B) ( B )
− L ( q ˙ , q , τ ) = − L ( q ˙ ( τ ) − B t , q ( τ ) , τ ) − B t ⋅ ( q ˙ ( τ ) − B t ) − 1 2 B t ⋅ B t ⟹ e − S L ( q 0 , q , τ ) = e − τ ∣ ∣ B t ∣ ∣ 2 / 2 e − ( B t − q 0 ) ⋅ ( q ( τ ) − B t τ − q 0 ) − S L B t ( q ˙ , q , τ ) = e − τ ∣ ∣ q f − q 0 ∣ ∣ 2 / 2 t 2 − S L ( q ˙ − B , q , τ ) e − ( q f − q 0 ) / t ⋅ q { τ / s 0 ≤ τ ≤ s ( t − τ ) / ( t − s ) s ≤ τ ≤ t ⟹ 1 ( ( 2 π ) 2 s ( t − s ) ) n ∫ R n e − S L ( q 0 , q , s ) e S L ( q , q f , t − s ) d q 1 . . . d q n = e − ( s + t − s ) ∣ ∣ q f − q 0 ∣ ∣ 2 / 2 t 2 ( ( 2 π ) 2 s ( t − s ) ) n ∫ R n e − t ∣ ∣ q ∣ ∣ 2 / 2 s ( t − s ) d q 1 . . . d q n = e − ρ 2 ( q f , q 0 ) / 2 t ( 2 π t ) n = R H S t 16 ( q 0 , q f ) . \begin{aligned}
-\mathcal L(\dot q, q, \tau) &= \begin{equation}\tag{D}-\mathcal L(\dot q(\tau) - \mathcal B_t, q(\tau), \tau) - \mathcal B_t\cdot (\dot q(\tau)-\mathcal B_t) - \frac{1}{2}\mathcal B_t \cdot B_t \ \ \ \ \ \ \ \ \ \end{equation}\\
\implies \\
e^{\mathcal -S_\mathcal L(q_0, q, \tau)} &= e^{-\tau||\mathcal B_t||^2/2}e^{-(\mathcal B_t -q_0)\cdot (q(\tau)-\mathcal B_t\tau - q_0) -\mathcal S_{\mathcal L_{\mathcal B_t}(\dot q,q,\tau)}} \\
&= e ^{-\tau||q_f-q_0||^2/2t^2 - \mathcal S_{\mathcal L(\dot q-\mathcal B, q, \tau)}}e^{-(q_f - q_0)/t\ \cdot q
\begin{cases}
\tau/s & 0\leq\tau\leq s\\
(t-\tau)/(t-s) & s\leq\tau\leq t
\end{cases}
}\\
\implies\\
\frac{1}{\sqrt{((2\pi)^2 s(t-s))^n}}\int_{\Reals^n} e^{-\mathcal S_{\mathcal L}(q_0,q,s)}e^{\mathcal S_{\mathcal L}(q,q_f,t-s)}dq^1...dq^n &= \frac{e^{-(s+t-s)||q_f - q_0||^2/2t^2}}{\sqrt{((2\pi)^2 s(t-s))^n}}
\int_{\Reals^n}e^{-t||q||^2/2s(t-s)} dq^1...dq^n \\
&= \frac{e^{-\rho^2(q_f, q_0)/2t}}{\sqrt{(2\pi t)^n}}\\
&= RHS^{16}_t(q_0,q_f) \ .
\end{aligned} − L ( q ˙ , q , τ ) ⟹ e − S L ( q 0 , q , τ ) ⟹ (( 2 π ) 2 s ( t − s ) ) n 1 ∫ R n e − S L ( q 0 , q , s ) e S L ( q , q f , t − s ) d q 1 ... d q n = − L ( q ˙ ( τ ) − B t , q ( τ ) , τ ) − B t ⋅ ( q ˙ ( τ ) − B t ) − 2 1 B t ⋅ B t ( D ) = e − τ ∣∣ B t ∣ ∣ 2 /2 e − ( B t − q 0 ) ⋅ ( q ( τ ) − B t τ − q 0 ) − S L B t ( q ˙ , q , τ ) = e − τ ∣∣ q f − q 0 ∣ ∣ 2 /2 t 2 − S L ( q ˙ − B , q , τ ) e − ( q f − q 0 ) / t ⋅ q { τ / s ( t − τ ) / ( t − s ) 0 ≤ τ ≤ s s ≤ τ ≤ t = (( 2 π ) 2 s ( t − s ) ) n e − ( s + t − s ) ∣∣ q f − q 0 ∣ ∣ 2 /2 t 2 ∫ R n e − t ∣∣ q ∣ ∣ 2 /2 s ( t − s ) d q 1 ... d q n = ( 2 π t ) n e − ρ 2 ( q f , q 0 ) /2 t = R H S t 16 ( q 0 , q f ) .
특히, 우리는 건설했습니다. B t \mathcal B_t B t q ˙ − B t \dot q - \mathcal B_t q ˙ − B t s s s q 0 q_0 q 0 N ( 0 , s ∧ t − s ) \mathcal N(0,s\wedge t-s) N ( 0 , s ∧ t − s ) R n = < B t > ⊕ B t ⊥ \Reals^n=<\mathcal B_t>\oplus \mathcal B_t^\perp R n =< B t > ⊕ B t ⊥ < B t > <\mathcal B_t> < B t >
사용하려고 하면 어떻게 됩니까? “연속적인 신념” on the semi-classical expression in Equation (16) to construct Brownian Motion on a negative curved 매니폴드 M M M
우리’d는 무언가를 얻지만,’d be 거의 곡선 공간에서 Brownian Motion — 우리는 Feynman-Kac의 무한 발전기의 결함을 찾아야합니다. 그것은 효과적인 잠재적 인 기능 오류가있을 것입니다 − 1 6 R ˉ -\frac{1}{6}\bar{\mathcal R} − 6 1 R ˉ R ˉ \bar{\mathcal R} R ˉ g i j g_{ij} g ij 일반 좌표 . 하지만 그때 dim = 2 \dim = 2 dim = 2 V = − 1 6 ( R ˉ − 1 4 R ˉ ) = 1 16 R ˉ \mathcal V = -\frac{1}{6}(\bar{\mathcal R} - \frac{1}{4}\bar{\mathcal R}) = \frac{1}{16}\bar{\mathcal R} V = − 6 1 ( R ˉ − 4 1 R ˉ ) = 16 1 R ˉ 1 6 R ˉ \frac{1}{6}\bar{\mathcal R} 6 1 R ˉ B \mathcal B B 1 24 R i c ( B / ∣ ∣ B ∣ ∣ , B / ∣ ∣ B ∣ ∣ ) \frac{1}{24}\mathcal{Ric}(\mathcal B/||\mathcal B||, \mathcal B/||\mathcal B||) 24 1 R i c ( B /∣∣ B ∣∣ , B /∣∣ B ∣∣ )
더 정확하게는, 대략적으로 1 / 2 ∇ ∣ 0 g = − 1 / 6 R i c i j q i ∂ j + o ( ∣ ∣ q ∣ ∣ ) ⟹ 1 / 2 ∇ ⋅ ∇ ∣ 0 g = − 1 / 6 R ˉ ( 0 ) 1/2\ \nabla\vert_0\sqrt{g} = - 1/6\ \mathcal{Ric}_{ij}q^i\partial^j + o(||q||) \implies 1/2\ \nabla\cdot\nabla\vert_0 \sqrt{g} = -1/6\ \bar{\mathcal R}(0) 1/2 ∇ ∣ 0 g = − 1/6 R i c ij q i ∂ j + o ( ∣∣ q ∣∣ ) ⟹ 1/2 ∇ ⋅ ∇ ∣ 0 g = − 1/6 R ˉ ( 0 )
( 2 π t ) n ( 1 2 Δ M − V ) ∣ 0 e − L ( q ˙ , q , t ) g ( q ) = ( 1 2 Δ R n ∣ 0 e − 1 / 2 t g i j q i q j ) − V ( 0 ) − 1 6 R ˉ ( 0 ) \sqrt{(2\pi t)^n}(\frac{1}{2}\Delta_M-\mathcal V)\vert_0 e^{-\mathcal L(\dot q, q, t)}\sqrt{g(q)} = (\frac{1}{2}\Delta_{\Reals^n}\vert_0 e^{-1/2t\ g_{ij}q^iq^j}) - \mathcal V(0) - \frac{1}{6}\ \bar{\mathcal R}(0) ( 2 π t ) n ( 2 1 Δ M − V ) ∣ 0 e − L ( q ˙ , q , t ) g ( q ) = ( 2 1 Δ R n ∣ 0 e − 1/2 t g ij q i q j ) − V ( 0 ) − 6 1 R ˉ ( 0 )
원래 Dewitt Term은 파생 된 그대로입니다. 우리가 킬링 필드 (Killing Field)의 존재에서 양자화에 편향 할 때 B = ∂ ∂ x 1 \mathcal B = \frac{\partial}{\partial x^1} B = ∂ x 1 ∂
( 2 π t ) n ( 1 2 Δ M − V ) ∣ x 1 , 0 ⃗ e − L B ( q ˙ , q , t ) 1 d e t ∣ I − J B ∣ = ( 1 2 Δ R n ∣ x 1 , 0 ⃗ e − x 1 x 1 / 2 t ) − V ( x 1 , 0 ⃗ ) + 1 8 R i c 11 ( x 1 , 0 ⃗ ) . \sqrt{(2\pi t)^{n}}(\frac{1}{2}\Delta_M-\mathcal V)\vert_{x^1,\vec 0} e^{-\mathcal L_\mathcal B(\dot q, q, t)}\frac{1}{det |I-\mathcal J_\mathcal B|} = (\frac{1}{2}\Delta_{\Reals^n}\vert_{x^1,\vec 0}e^{-x^1x^1/2t}) - \mathcal V(x^1, \vec 0) + \frac{1}{8}\mathcal{Ric}_{11}(x^1, \vec 0). ( 2 π t ) n ( 2 1 Δ M − V ) ∣ x 1 , 0 e − L B ( q ˙ , q , t ) d e t ∣ I − J B ∣ 1 = ( 2 1 Δ R n ∣ x 1 , 0 e − x 1 x 1 /2 t ) − V ( x 1 , 0 ) + 8 1 R i c 11 ( x 1 , 0 ) .
가정 B \mathcal B B g g g 킬링 ) 벡터 필드 M M M
γ ~ = D q [ c ~ ] \tilde \gamma = \mathscr D_q[\tilde c] γ ~ = D q [ c ~ ] c ~ ∈ C ∞ ( [ 0 , t ] , T q M ) \tilde c\in C^\infty([0,t],T_qM) c ~ ∈ C ∞ ([ 0 , t ] , T q M ) 해결 대상 γ ~ \tilde \gamma γ ~
γ ~ ( 0 ) = q γ ~ ˙ = Γ ^ ( γ ~ ) c ~ ˙ \begin{aligned}
\tilde \gamma(0) &= q \\
\dot {\tilde \gamma} &= \hat\Gamma(\tilde \gamma)\dot{\tilde c}\\
\end{aligned} γ ~ ( 0 ) γ ~ ˙ = q = Γ ^ ( γ ~ ) c ~ ˙
c ~ ( τ ) = ∫ 0 τ Γ ^ s − 1 ( γ ~ ) γ ~ ˙ d s \tilde c(\tau)=\int_0^\tau\hat\Gamma_s^{-1}(\tilde\gamma)\dot{\tilde\gamma} ds c ~ ( τ ) = ∫ 0 τ Γ ^ s − 1 ( γ ~ ) γ ~ ˙ d s 노에테르’정리가 보장하는 d ( g − 1 B ) = 0 d({g^{-1}\mathcal B}) = 0 d ( g − 1 B ) = 0 g − 1 B g^{-1}\mathcal B g − 1 B B ^ \hat{\mathcal B} B ^ B = ∇ B ^ \mathcal B = \nabla \hat{\mathcal B} B = ∇ B ^ Γ ^ \hat \Gamma Γ ^ B \mathcal B B B ⊥ \mathcal B^\perp B ⊥
c ˙ ⋅ B = 0 ⟹ γ ˙ ⋅ B = 0 ⟹ d B ^ d t = 0 , \begin{aligned}
\\
\dot{c}\cdot\mathcal B &= 0 \implies\\
\dot{\gamma}\cdot \mathcal B &= 0 \implies\\
\frac{d\hat{\mathcal B}}{dt} &= 0\ ,
\end{aligned} c ˙ ⋅ B γ ˙ ⋅ B d t d B ^ = 0 ⟹ = 0 ⟹ = 0 ,
그래서 γ \gamma γ B ^ \hat{\mathcal B} B ^ c c c B ⊥ ⊂ T q M \mathcal B^\perp \subset T_qM B ⊥ ⊂ T q M γ ( t ) ≠ q \gamma(t) \ne q γ ( t ) = q
∣ ∣ B t ∣ ∣ = ρ ( q 0 , q f ) t ⟹ c ~ ( τ ) − c ( τ ) = t 2 ρ 2 ( q 0 , q f ) B t ∫ 0 τ c ~ ˙ ⋅ B t d s = B ∫ o τ c ~ ˙ ⋅ B d s = B ⋅ c ~ ( τ ) B = d B ^ ( c ~ ) B . \begin{aligned}
||\mathcal B_t|| &= \frac{\rho(q_0,q_f)}{t} \implies \\
\tilde{c}(\tau) - c(\tau) &= \frac{t^2}{\rho^2(q_0,q_f)}\mathcal B_t\int_0^\tau \dot{\tilde c} \cdot \mathcal B_t\ ds\\
&= \mathcal B\int_o^\tau\dot{\tilde c}\cdot \mathcal B \ ds\\
&= \mathcal B\cdot \tilde c(\tau)\ \mathcal B\\
&= d\hat{\mathcal B}(\tilde c)\ \mathcal B \ .
\end{aligned} ∣∣ B t ∣∣ c ~ ( τ ) − c ( τ ) = t ρ ( q 0 , q f ) ⟹ = ρ 2 ( q 0 , q f ) t 2 B t ∫ 0 τ c ~ ˙ ⋅ B t d s = B ∫ o τ c ~ ˙ ⋅ B d s = B ⋅ c ~ ( τ ) B = d B ^ ( c ~ ) B .
M M M D − 1 \mathscr D^{-1} D − 1 허가 Ω t B ( q ) \Omega^\mathcal B_t(q) Ω t B ( q ) M M M q q q exp t B q \exp{t\mathcal B}\ q exp t B q μ t ( ω ) \mu_t(\omega) μ t ( ω ) ω ∈ Ω t B : = { Ω t B ( q ) : q ∈ M } \omega\in \Omega^\mathcal B_t := \set{\Omega^\mathcal B_t(q): q\in M} ω ∈ Ω t B := { Ω t B ( q ) : q ∈ M } E t B ( f ∣ A ) : = ∫ Ω t B f ( ω ) d ( μ t ∣ A ) ( ω ) E_t^\mathcal B (f|A):=\int_{\Omega^\mathcal B_t} f(\omega) d(\mu_t|A)(\omega) E t B ( f ∣ A ) := ∫ Ω t B f ( ω ) d ( μ t ∣ A ) ( ω ) P μ t B ( A ) : = μ t ( A ) / μ t ( Ω t B ) ∀ A ⊂ Ω t B P^\mathcal B_{\mu_t}(A) := \mu_t(A)/\mu_t(\Omega_t^{\mathcal B})\ \forall A\subset\Omega^\mathcal B_t P μ t B ( A ) := μ t ( A ) / μ t ( Ω t B ) ∀ A ⊂ Ω t B ( D ) ⟹ (D) \implies ( D ) ⟹
k t H ^ ( q 0 , exp t B t q 0 ) g ( q 0 ) d q = e − ρ 2 / 2 t 2 π t det ∣ I − J B ( q 0 ) ∣ E t B ( e − ∫ 0 t V ( ω ( s ) ) d s + ∫ 0 t R ˉ ( ω ( s ) ) d s / 12 − ∫ 0 t R i c ( B ∣ ∣ B ∣ ∣ , B ∣ ∣ B ∣ ∣ ) ( ω ( s ) ) d s / 24 χ Ω t B ( q 0 ) ( ω ) ∣ d B ⊥ ) d B ^ ( q 0 ) \begin{equation}
\tag{E}
k^{\hat{\mathcal H}}_t(q_0,\exp{t\mathcal B_t}\ q_0)\sqrt g(q_0)\ dq = \frac{e^{-\rho^2/2t}}{\sqrt{2 \pi t\det{|I-\mathcal J^{\mathcal B}(q_0)|}}} E^\mathcal B_t({e^{-\int_0^t V(\omega(s))ds\ +\ \int_0^t \bar {\mathcal R}(\omega(s))ds/12\ -\ \int_0^t \mathcal {Ric}(\frac{\mathcal B}{||\mathcal B||},\frac{\mathcal B}{||\mathcal B||})(\omega(s))ds/24}\chi_{\Omega^B_t(q_0)}(\omega)}|d\mathcal B^\perp)d\hat{\mathcal B}(q_0) \ \ \ \ \ \ \
\end{equation} k t H ^ ( q 0 , exp t B t q 0 ) g ( q 0 ) d q = 2 π t det ∣ I − J B ( q 0 ) ∣ e − ρ 2 /2 t E t B ( e − ∫ 0 t V ( ω ( s )) d s + ∫ 0 t R ˉ ( ω ( s )) d s /12 − ∫ 0 t R i c ( ∣∣ B ∣∣ B , ∣∣ B ∣∣ B ) ( ω ( s )) d s /24 χ Ω t B ( q 0 ) ( ω ) ∣ d B ⊥ ) d B ^ ( q 0 ) ( E )
여기서 ρ = ∣ ∣ t B t ∣ ∣ = d i s t ( q 0 , q f ) \rho=||t\mathcal B_t||=dist(q_0,q_f) ρ = ∣∣ t B t ∣∣ = d i s t ( q 0 , q f ) J B ( q 0 ) \mathcal J^{\mathcal B}(q_0) J B ( q 0 ) M M M B ^ \hat{\mathcal B} B ^ γ ~ ( λ ) = exp λ B t ( q 0 ) \tilde \gamma(\lambda) = \exp{\lambda\mathcal B_t}\ (q_0) γ ~ ( λ ) = exp λ B t ( q 0 ) q 0 q_0 q 0 q f q_f q f λ \lambda λ 0 0 0 t t t J B \mathcal J^{\mathcal B} J B t t t I − J B I-\mathcal J^\mathcal B I − J B q 0 ≠ q f q_0 \neq q_f q 0 = q f B \mathcal B B − B -\mathcal B − B q 0 q_0 q 0 q f q_f q f
시작일 R \mathcal R R γ ~ \tilde \gamma γ ~ γ \gamma γ is 지오데식(최대 길이 변형) J B ( q 0 ) \mathcal J^\mathcal B(q_0) J B ( q 0 ) Jacobi Fields 의 관점에서 사소하게 호환됩니다. J ( λ ) \mathcal J(\lambda) J ( λ ) γ ~ \tilde \gamma γ ~ λ = t \lambda=t λ = t
이 방정식의 증거 V = 1 12 ( R ˉ − 1 2 R i c ( B / ∣ ∣ B ∣ ∣ , B / ∣ ∣ B ∣ ∣ ) \mathcal V = \frac{1}{12}(\bar{\mathcal R} - \frac{1}{2} \mathcal {Ric}(\mathcal B / ||\mathcal B||, \mathcal B/||\mathcal B||) V = 12 1 ( R ˉ − 2 1 R i c ( B /∣∣ B ∣∣ , B /∣∣ B ∣∣ )
좋은 코롤러는 상수에서 발생합니다. R i c ( B ∣ ∣ B ∣ ∣ , B ∣ ∣ B ∣ ∣ ) = R ˉ / dim M \mathcal {Ric}(\frac{\mathcal B}{||\mathcal B||},\frac{\mathcal B}{||\mathcal B||}) = \bar{\mathcal R} / \dim M R i c ( ∣∣ B ∣∣ B , ∣∣ B ∣∣ B ) = R ˉ / dim M − κ -\kappa − κ B \mathcal B B S 1 S^1 S 1 M M M
det ∣ I − J B ∣ = ( 2 κ sinh ( κ ρ / 2 ) ) 2 ⟹ μ t ( Ω t B ) = ∫ M / S 1 ⊕ S 1 k t − Δ / 2 ( q , exp t B q ) g ( q ) d q = e − ρ 2 / 2 t 2 π t ∫ M / S 1 ⊕ S 1 1 2 κ ( q ) sinh κ ( q ) ρ / 2 E t B ( e ∫ 0 t R ˉ ( ω ( s ) ) d s / 16 χ Ω t B ( q ) ( ω ) ∣ d B ⊥ ) d B ^ ( q ) = e − ρ 2 / 2 t − t κ / 8 2 π t 2 κ sinh κ ρ / 2 ∫ B ⊥ ⊕ [ 0 , ρ 0 ] P μ t B ( Ω t B ( d B ^ ) ∣ d B ⊥ ) , Bayes ⟹ = e − ρ 2 / 2 t − t κ / 8 2 π t 2 κ sinh κ ρ / 2 ∫ 0 ρ 0 P μ t B ( B ⊥ d B ^ ∣ Ω t B ( d B ^ ) ) P μ t B ( Ω t B ( d B ^ ) ) P μ t B ( B ⊥ d B ^ ) d B ^ = e − ρ 2 / 2 t − t κ / 8 2 π t 2 κ sinh κ ρ / 2 ρ 0 \begin{aligned}
\det |I - \mathcal J^\mathcal B| = (2 \kappa\sinh(\sqrt{\kappa}\rho/2))^2 \implies \\
\mu_t(\Omega_t^\mathcal B) = \int_{M/S^1\oplus S^1} k^{-\Delta/2}_t(q,\exp{t\mathcal B}\ q) \sqrt g(q)\ dq &= \frac{e^{-\rho^2/2t}}{\sqrt{2\pi t}} \int_{M/S^1\oplus S^1} \frac{1}{2\kappa(q) \sinh \sqrt{\kappa(q)}\rho/2}E^\mathcal B_t(e^{\int_0^t \bar{\mathcal R}(\omega(s))\ ds/16}{\chi_{\Omega^B_t(q)}(\omega)}|d\mathcal B^\perp)d\hat{\mathcal B}(q)\\
&=\frac{e^{-\rho^2/2t\ -\ t\kappa/8}}{\sqrt{2\pi t}\ 2\kappa\sinh \sqrt {\kappa}\rho/2}\ \int_{\mathcal B^\perp\oplus[0,\rho_0]}P^\mathcal B_{\mu_t}(\Omega_t^\mathcal B(d\hat{\mathcal B})|d\mathcal B^\perp), \ \text{ Bayes}\implies\\
&=\frac{e^{-\rho^2/2t\ -\ t\kappa/8}}{\sqrt{2\pi t}\ 2\kappa\sinh \sqrt {\kappa}\rho/2}\ \int_0^{\rho_0} P_{\mu_t}^\mathcal B(\mathcal B^\perp d\hat{\mathcal B}|\Omega_t^\mathcal B(d\hat{\mathcal B}))\frac{P^\mathcal B_{\mu_t}(\Omega_t^\mathcal B(d\hat{\mathcal B}))}{P^\mathcal B_{\mu_t}(\mathcal B^\perp d\hat{\mathcal B})}d\hat{\mathcal B}\\
&=\frac{e^{-\rho^2/2t\ -t\kappa/8}}{\sqrt{2\pi t}\ 2\kappa\sinh \sqrt {\kappa}\rho/2}\ \rho_0\\
\end{aligned} det ∣ I − J B ∣ = ( 2 κ sinh ( κ ρ /2 ) ) 2 ⟹ μ t ( Ω t B ) = ∫ M / S 1 ⊕ S 1 k t − Δ/2 ( q , exp t B q ) g ( q ) d q = 2 π t e − ρ 2 /2 t ∫ M / S 1 ⊕ S 1 2 κ ( q ) sinh κ ( q ) ρ /2 1 E t B ( e ∫ 0 t R ˉ ( ω ( s )) d s /16 χ Ω t B ( q ) ( ω ) ∣ d B ⊥ ) d B ^ ( q ) = 2 π t 2 κ sinh κ ρ /2 e − ρ 2 /2 t − t κ /8 ∫ B ⊥ ⊕ [ 0 , ρ 0 ] P μ t B ( Ω t B ( d B ^ ) ∣ d B ⊥ ) , Bayes ⟹ = 2 π t 2 κ sinh κ ρ /2 e − ρ 2 /2 t − t κ /8 ∫ 0 ρ 0 P μ t B ( B ⊥ d B ^ ∣ Ω t B ( d B ^ )) P μ t B ( B ⊥ d B ^ ) P μ t B ( Ω t B ( d B ^ )) d B ^ = 2 π t 2 κ sinh κ ρ /2 e − ρ 2 /2 t − t κ /8 ρ 0
여기서 ρ = min q ∈ M d i s t ( q , exp t B q ) \rho = \min_{q\in M}dist(q, \exp t\mathcal B \ q) ρ = min q ∈ M d i s t ( q , exp t B q ) S 1 S^1 S 1 ρ 0 \rho_0 ρ 0 ρ \rho ρ B ⊥ \mathcal B^\perp B ⊥ Horocycles 및 g g g S 1 S^1 S 1 B \mathcal B B
방정식을 고려하십시오 (32) ( E ) (E) ( E ) B B B g g g q 0 q_0 q 0 Ω t 0 \Omega^0_t Ω t 0
μ t ( Ω t 0 ) = v o l ( M ) 2 π t ∫ 0 ∞ e − ρ 2 / 2 t − t κ / 8 2 π t κ sinh κ ρ / 2 ρ d ρ \mu_t(\Omega_t^0) = \frac{vol(M)}{2\pi t}\int_0^\infty \frac{e^{-\rho^2/2t\ -t\kappa/8}}{\sqrt{2\pi t}\ \kappa\sinh \sqrt {\kappa}\rho/2}\ \rho d\rho μ t ( Ω t 0 ) = 2 π t v o l ( M ) ∫ 0 ∞ 2 π t κ sinh κ ρ /2 e − ρ 2 /2 t − t κ /8 ρ d ρ
이 방정식이 분석적이기 때문에 κ \kappa κ κ → − κ \kappa \rightarrow -\kappa κ → − κ sinh \sinh sinh sin \sin sin ∣ κ ∣ |\kappa| ∣ κ ∣
다시 말해서, 우리는 확률과 기하학을 통해 2 차원 Selberg 추적 공식을 다시 파생, 대칭 공간에 대한 일반적인 조화 분석 대신 .
매끄러운 진짜 가치 diffeomorphism를 위해 h : R → R h:\Reals\rightarrow\Reals h : R → R h ( 0 ) = 0 h(0)=0 h ( 0 ) = 0 d s 2 = ( 1 + h 2 ( y ) ) d x 2 + 2 h ( y ) d x ⊙ d y + d y 2 ds^2 = (1+h^2(y)) dx^2 + 2h(y) dx\odot dy + dy^2 d s 2 = ( 1 + h 2 ( y )) d x 2 + 2 h ( y ) d x ⊙ d y + d y 2 − ( κ ( y ) = d 2 d y 2 h 2 ( y ) / 2 ) -(\kappa(y)=\frac{d^2}{dy^2}h^2(y)/2) − ( κ ( y ) = d y 2 d 2 h 2 ( y ) /2 ) h ( y ) h(y) h ( y ) det ( d s 2 ) = 1 \det(ds^2)= 1 det ( d s 2 ) = 1 B ^ ( x , y ) = x \hat{\mathcal B}(x,y) = x B ^ ( x , y ) = x
μ t ( Ω t B ) = e − ρ 2 / 2 t 2 π t ∫ − ∞ ∞ ρ 0 2 sinh κ ( y ) ρ / 2 ∫ Ω t B ( 0 , y ) e − ∫ 0 t κ ( y ( s ) ) d s / 8 d ( μ t ∣ B ⊥ ) / d y d y
\mu_t(\Omega_t^\mathcal B) = \frac{e^{-\rho^2/2t}}{2\pi t}\int_{-\infty}^\infty\frac{\rho_0}{2 \sinh \sqrt{\kappa(y)}\rho/2} \int_{\Omega_t^\mathcal B(0,y)}e^{-\int_0^t\kappa(y(s))ds/8}d(\mu_t|\mathcal B^\perp)/dy \ dy
μ t ( Ω t B ) = 2 π t e − ρ 2 /2 t ∫ − ∞ ∞ 2 sinh κ ( y ) ρ /2 ρ 0 ∫ Ω t B ( 0 , y ) e − ∫ 0 t κ ( y ( s )) d s /8 d ( μ t ∣ B ⊥ ) / d y d y
만약 우리가 h ( y ) : = y 2 y 2 / 3 + κ ( 0 ) ⟹ κ ( y ) = 4 y 2 + κ ( 0 ) > 0 h(y) := y\sqrt{2y^2/3 + \kappa(0)} \implies \kappa(y) = 4 y^2 + \kappa(0) \gt 0 h ( y ) := y 2 y 2 /3 + κ ( 0 ) ⟹ κ ( y ) = 4 y 2 + κ ( 0 ) > 0 ( E ) (E) ( E )
μ t ( Ω t B ) = e − ρ 2 / 2 t − t κ ( 0 ) / 8 2 π t cosh t ∫ 0 ∞ ρ 0 e − y 2 / 2 t 2 π t sinh 4 y 2 + κ ( 0 ) ρ / 2 d y
\mu_t(\Omega_t^\mathcal B) = \frac{e^{-\rho^2/2t -t\kappa(0)/8}}{\sqrt{2\pi t \cosh t}}\int_0^\infty\frac{\rho_0e^{-y^2/2t}}{\sqrt{2\pi t}\sinh \sqrt{4y^2+\kappa(0)}\rho/2 }\ dy
μ t ( Ω t B ) = 2 π t cosh t e − ρ 2 /2 t − t κ ( 0 ) /8 ∫ 0 ∞ 2 π t sinh 4 y 2 + κ ( 0 ) ρ /2 ρ 0 e − y 2 /2 t d y
그대여, 그대여 0 0 0 κ ( 0 ) = 0 \kappa(0) = 0 κ ( 0 ) = 0 t → 0 t\rightarrow 0 t → 0 − κ ( 0 ) -\kappa(0) − κ ( 0 ) t → 0 t\rightarrow 0 t → 0
외부 시간 축은 인공, 때문에 시간 는 4차원 매니폴드 자체의 형상에 포함됩니다. 경로 적분 배합이 불어오기 때문에 -1 본질에 있는 Lorenzian 미터법의 서명 시간 방향. det g \det{g} det g 시간).
