Listas de exercício - RG Prof Marco

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Teoria da Relatividade Geral 
Prof. Dr. Marco André Ferreira Dias 
 
Camila Cezar de Lima 
Washington de Araujo Bandeira 
 
LISTA I 
 
a) 
Mostre que ∆𝑠 = ∆𝜌 √(𝜆2 + 1) + (𝜌
𝑑𝜙
𝑑𝜌
)² 
 
𝑑𝑠2 = 𝑑𝜌2 + (𝜌𝑑𝜙)2 + (𝑑𝑧)2 
(Δ𝑠)2 = (Δ𝜌)2 + 𝜌(Δ𝜙)² + (Δ𝑧)2 
(Δ𝑠)2 = Δ𝜌2 + [𝜌 
𝑑𝜙
𝑑𝜌
Δ𝜌] ² + (λΔ𝜌)2 
(Δ𝑠)2 = Δ𝜌2(1 + 𝜆2) + 𝜌2 (
𝑑𝜙
𝑑𝜌
)
2
Δ𝜌² 
(Δ𝑠)2 = Δ𝜌2 [1 + 𝜆2 + 𝜌2 (
𝑑𝜙
𝑑𝜌
)
2
] 
Δs = Δρ √1 + 𝜆2 + 𝜌2 (
𝑑𝜙
𝑑𝜌
)
2
 
Δs = Δρ √(𝜆2 + 1) + (𝜌
𝑑𝜙
𝑑𝜌
)² 
 
b) 
 
Seja 𝑓 = √(𝜆2 + 1) + (𝜌
𝑑𝜙
𝑑𝜌
)² ou ainda 𝑓 = √(𝜆2 + 1) + (𝜌𝜙′)² 
Assim, 𝑓 = 𝑓 (ρ, 𝜙′) então 
𝜕𝑓
𝜕𝜙
= 0 
Da equação de Euler-Lagrange, temos 
𝜕𝑓
𝜕𝜙
= 
𝑑
𝑑𝜌
(
𝜕𝑓
𝜕𝜌′
) 
Portanto, 
𝑑
𝑑𝜌
(
𝜕𝑓
𝜕𝜙′
) = 0 logo, 
𝝏𝒇
𝝏𝝓′
= 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕𝒆 
 
 
 
 
c) 
Mostre que 𝜙(𝜌) = 𝜙0 + √𝜆2 + 1 𝑐𝑜𝑠
−1 (
𝑐
𝜌
) 
 
Usando a equação 𝑓 (𝜌, 𝜙′) = √(𝜆2 + 1) + (𝜌𝜙′)² obtemos, 
𝜕𝑓
𝜕𝜙′
= 
1
2
 [(𝜆2 + 1) + (𝜌𝜙′)²]−
1
2(2𝜌²𝜙′) = 𝑐 
[(𝜆2 + 1) + (𝜌𝜙′)²]−
1
2(𝜌²𝜙′) = 𝑐 
𝜙′ =
𝑐
𝜌2
√(𝜆2 + 1) + (𝜌𝜙′)² 
𝜙′2 =
𝑐²
𝜌4
[(𝜆2 + 1) + (𝜌𝜙′)²] 
𝜙′2 =
𝑐²
𝜌4
(𝜆2 + 1) + 
𝑐²
𝜌4
 𝜌²𝜙² 
𝜙′2 =
𝑐²
𝜌4
(𝜆2 + 1) + 
𝑐²
𝜌2
 𝜙² 
𝜙′2 −
𝑐²
𝜌2
 𝜙² = 
𝑐2
𝜌4
(𝜆2 + 1) 
𝑑𝜙
𝑑𝜌
= √𝜆2 + 1
𝑐
𝜌2
 
[
 
 
 
 
 
1
√1 −
𝑐²
𝜌²]
 
 
 
 
 
 
Seja 𝑢 =
𝑐
𝜌
 então, 𝑑𝑢 = −
𝑐
𝜌²
 𝑑𝜌 
𝑑𝜙
𝑑𝜌
= √𝜆2 + 1 [
−𝑑𝑢
√1 − 𝑢²
] 
 
Lembrando que 
𝑑
𝑑𝜌
(𝑎𝑟𝑐 cos 𝑢) = −
1
√1−𝑢²
 𝑑𝑢 
Então, 
𝑑𝜙
𝑑𝜌
= √𝜆2 + 1 
𝑑
𝑑𝜌
(𝑎𝑟𝑐 cos 
𝑐
𝜌
) 
𝑑𝜙 = √𝜆2 + 1 [
𝑑
𝑑𝜌
(𝑎𝑟𝑐 cos 
𝑐
𝜌
)] 𝑑𝜌 
 
 
∫ 𝑑𝜙
𝜙
𝜙0
= √𝜆2 + 1 [(𝑎𝑟𝑐 cos 
𝑐
𝜌
)] 
𝜙 − 𝜙0 = √𝜆2 + 1𝑎𝑟𝑐 cos (
𝑐
𝜌
) 
𝜙 = 𝜙0 + √𝜆2 + 1 𝑎𝑟𝑐 cos (
𝑐
𝜌
) 
 
d) 
𝜌 cos (𝜙(𝜌) − 𝜙0) = 𝑐 
Partindo de 𝜙 − 𝜙0 = √𝜆2 + 1 𝑎𝑟𝑐 cos (
𝑐
𝜌
) para 𝜆 = 0 temos, 
𝜙 − 𝜙0 = 𝑎𝑟𝑐 cos (
𝑐
𝜌
) 
cos(𝜙 − 𝜙0) = 
𝑐
𝜌
 
 
cos(𝜙 − 𝜙0) = 
𝑐
𝜌
 
ρ cos(𝜙 − 𝜙0) = 𝑐 
 
e) 
 
Para 𝜙0 = 0 temos: 
ρ cos(𝜙) = 𝑐 
Sendo 
ρ cos(ϕ) = x 
Logo, x = c 
 
 
 
 
f) 
 y 
 
 
 
 
  x 
 𝜌 ⋅ 𝑐𝑜𝑠𝜙 = 𝑐 = 𝑥 
 𝜌 ⋅ cos(𝜙 − 𝜙0) = 𝑐, 𝜙0 = 0 
 𝜌 ⋅ cos 𝜙 = 𝑐 ⇒ 𝜙(𝜌) = 𝑎𝑟𝑐 𝑐𝑜𝑠 (
𝑐
𝜌
) 
 
 reta perpendicular a linha com inclinação  –  
 a uma distância “c” da origem 
 
 . 𝜌 ⋅ cos𝜙 = 𝑐 
  
 
 
 
g) 
 
𝑧 = 𝜆𝜌 
𝑑𝑧 = 𝜆𝑑𝜌 
𝑑𝑙2 = 𝜆𝑑𝜌 + 𝑑𝜌2 reajustando, temos: 𝑑𝑙2 = (𝜆2 + 1)𝑑𝜌 
𝜂 = 𝜌𝜙 
𝑑𝜂
𝑑𝑙
= 
𝜕𝜂
𝜕𝜌
𝑑𝜌
𝑑𝑙
+
𝜕𝜂
𝜕𝜙
𝑑𝜙
𝑑𝑙
 
(
𝑑𝜌
𝑑𝑙
)
2
= 
1
𝜆2 + 1
 
𝑑𝜌
𝑑𝑙
= 
1
√𝜆2 + 1
 
 
 
𝑑𝜙
𝑑𝑙
= 0 
𝜕𝜂
𝜕𝜌
= 𝜙 e 
𝜕𝜂
𝜕𝜙
= 𝜌 
𝑑𝜂
𝑑𝑙
= 
𝜙
𝜆2 + 1
 
𝑑²𝜂
𝑑𝑙²
= 
1
𝜆2 + 1
𝑑𝜙
𝑑𝑙
 
𝑑²𝜂
𝑑𝑙²
= 0 
Se, 
𝑑²𝜂
𝑑𝑙²
= −𝑘𝜂 Logo, −𝑘𝜂 = 0 e 𝑘 = 0 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
LISTA II 
 
1) 
 
𝜕�̂�
𝜕𝑟
= 
𝜕
𝜕𝑟
(𝑥 ̂𝑠𝑒𝑛𝜃𝑐𝑜𝑠𝜙 + �̂�𝑠𝑒𝑛𝜃𝑠𝑒𝑛𝜙 + �̂�𝑐𝑜𝑠𝜃) = 0 (1) 
 
𝜕�̂�
𝜕𝜃
= 
𝜕
𝜕𝜃
(𝑥 ̂𝑠𝑒𝑛𝜃𝑐𝑜𝑠𝜙 + �̂�𝑠𝑒𝑛𝜃𝑠𝑒𝑛𝜙 + �̂�𝑐𝑜𝑠𝜃) 
𝜕�̂�
𝜕𝜃
= 𝑥 ̂𝑐𝑜𝑠𝜃𝑐𝑜𝑠𝜙 + �̂�𝑐𝑜𝑠𝜃𝑠𝑒𝑛𝜙 − �̂�𝑠𝑒𝑛𝜃 
𝜕�̂�
𝜕𝜃
= 𝑐𝑜𝑠𝜃(𝑥 ̂𝑐𝑜𝑠𝜙 + �̂�𝑠𝑒𝑛𝜙) − �̂�𝑠𝑒𝑛𝜃 (2) 
 
𝜕�̂�
𝜕𝜙
= 
𝜕
𝜕𝜙
(𝑥 ̂𝑠𝑒𝑛𝜃𝑐𝑜𝑠𝜙 + �̂�𝑠𝑒𝑛𝜃𝑠𝑒𝑛𝜙 + �̂�𝑐𝑜𝑠𝜃) 
𝜕�̂�
𝜕𝜙
= − 𝑥 ̂𝑠𝑒𝑛𝜃𝑠𝑒𝑛𝜙 + �̂�𝑠𝑒𝑛𝜃𝑐𝑜𝑠𝜙 
𝜕�̂�
𝜕𝜙
= 𝑠𝑒𝑛 𝜃 (− 𝑥 ̂𝑒𝑛𝜙 + �̂�𝑐𝑜𝑠𝜙) (3) 
 
�̂� = 
�̂� × �̂�
𝑠𝑒𝑛𝜃
 = 
1
𝑠𝑒𝑛𝜃
⌈
�̂� �̂� �̂�
0 0 1
𝑠𝑒𝑛𝜃𝑐𝑜𝑠𝜙 𝑠𝑒𝑛𝜃𝑐𝑜𝑠𝜙 𝑐𝑜𝑠𝜃
⌉ 
�̂� = �̂�𝑠𝑒𝑛𝜃𝑐𝑜𝑠𝜙 − 𝑥 ̂𝑠𝑒𝑛𝜃𝑠𝑒𝑛𝜙 = −𝑥 ̂𝑠𝑒𝑛𝜙 + �̂�𝑐𝑜𝑠𝜙 (4) 
 
𝜃 = �̂� × �̂� = ⌈
�̂� �̂� �̂�
−𝑠𝑒𝑛𝜃 𝑐𝑜𝑠𝜙 0
𝑠𝑒𝑛𝜃𝑐𝑜𝑠𝜙 𝑠𝑒𝑛𝜃𝑠𝑒𝑛𝜙 𝑐𝑜𝑠𝜃
⌉ 
𝜃 = 𝑥 ̂𝑐𝑜𝑠𝜃𝑐𝑜𝑠𝜙 − 𝑧 ̂𝑠𝑒𝑛𝜃𝑠𝑒𝑛2𝜙 − 𝑧 ̂𝑠𝑒𝑛𝜃𝑐𝑜𝑠2𝜙 + �̂�𝑠𝑒𝑛𝜙𝑐𝑜𝑠𝜃 
𝜃 = 𝑐𝑜𝑠𝜃 (𝑥 ̂𝑐𝑜𝑠𝜙 + �̂�𝑠𝑒𝑛𝜙) − 𝑠𝑒𝑛𝜃𝑧 ̂ (5) 
 
Portanto, 
De (1), 
𝜕�̂�
𝜕𝑟
= 0 
De (2) e (5), 
𝜕�̂�
𝜕𝜃
= 𝜃 
 
 
De (3) e (4), 
𝜕�̂�
𝜕𝜙
= 𝜃 𝑠𝑒𝑛𝜃 
 
E, se �⃗� = 𝑟�̂� e �̂� =
𝑟
𝑟
 Portanto, �̂� = 𝑟�̂� 
 
Logo, �⃗� = 𝑟 
Portanto, 𝑑�⃗� = 𝑑𝑟 
 
Dado que 𝑑𝑟 = 𝑑(𝑟𝑑�̂�) 
𝑑𝑟 = 𝑟𝑑�̂� + �̂�𝑑𝑟 
𝑑𝑟 = 𝑟 (
𝜕�̂�
𝜕𝑟
𝑑𝑟 +
𝜕�̂�
𝜕𝜃
𝑑𝜃 +
𝜕�̂�
𝜕𝜙
𝑑𝜙) + �̂�𝑑𝑟 
𝑑𝑟 = 𝑟(0𝑑𝑟 + 𝜃𝑑𝜃 + �̂�𝑠𝑒𝑛𝜃𝑑𝜙) + �̂�𝑑𝑟 
𝑑𝑟 = 𝑟(𝜃𝑑𝜃 + �̂�𝑠𝑒𝑛𝜃𝑑𝜙) + �̂�𝑑𝑟 
𝑑𝑟 = 𝑟(𝜃𝑑𝜃 + �̂�𝑠𝑒𝑛𝜃𝑑𝜙) + �̂�𝑑𝑟 
𝑑𝑟 = �̂�𝑑𝑟 + 𝑟𝜃𝑑𝜃 + �̂�𝑟𝑠𝑒𝑛𝜃𝑑𝜙 
 
Como 𝑑�⃗� = 𝑑𝑟 temos: 
𝑑�⃗� = �̂�𝑑𝑟 + 𝜃𝑟𝑑𝜃 + �̂�𝑟𝑠𝑒𝑛𝜃𝑑𝜙 
 
2) 
 
𝑑𝑠2 = 𝐷𝑑𝑡2 − 𝐴 𝑑�⃗�𝑑�⃗� − 𝐵(𝑥.⃗⃗⃗ ⃗ 𝑑�⃗�)2 − 𝐶𝑑𝑡(�⃗�. 𝑑�⃗�) 
𝑑�⃗� = �̂�𝑑𝑟 + 𝜃𝑟𝑑𝜃 + �̂�𝑟𝑠𝑒𝑛𝜃𝑑𝜙 
�⃗� = 𝑟�̂� 
𝑑�⃗�𝑑�⃗� = 𝑑𝑟2 + 𝑟2𝑑𝜃2 + 𝑟2𝑠𝑒𝑛2𝜃𝑑𝜙² 
�⃗�. 𝑑�⃗� = 𝑟𝑑𝑟 
𝑑𝑠2 = 𝐷𝑑𝑡2 − 𝐴 ( 𝑑𝑟2 + 𝑟2𝑑𝜃2 + 𝑟2𝑠𝑒𝑛2𝜃𝑑𝜙2) − 𝐵(𝑟𝑑𝑟)2 − 𝐶𝑑𝑡(𝑟𝑑𝑟) 
𝑑𝑠2 = 𝐷𝑑𝑡2 − 𝐴[𝑟2(𝑑𝜃2 + 𝑠𝑒𝑛2𝜃𝑑𝜙2)] − 𝐴𝑑𝑟2 − 𝐵𝑟²𝑑𝑟² − 𝐶𝑟𝑑𝑡𝑑𝑟 
𝑑𝑠2 = 𝐷𝑑𝑡2 − 𝐴[𝑟2(𝑑𝜃2 + 𝑠𝑒𝑛2𝜃𝑑𝜙2)] − (𝐴 + 𝐵𝑟2)𝑑𝑟2 − 𝐶𝑟𝑑𝑡𝑑𝑟 
 
 
 
Considerando, 𝐴 + 𝐵𝑟2 = 𝐸 e 𝐶𝑟 = 𝐹 
 
𝑑𝑠2 = 𝐷𝑑𝑡2 − 𝐴[𝑟2(𝑑𝜃2 + 𝑠𝑒𝑛2𝜃𝑑𝜙2)] − 𝐸𝑑𝑟2 − 𝐹𝑑𝑡𝑑𝑟 
 
Renomeamos os termos E  B e F  C: 
 
𝑑𝑠2 = 𝐷𝑑𝑡2 − 𝐴[𝑟2(𝑑𝜃2 + 𝑠𝑒𝑛2𝜃𝑑𝜙2)] − 𝐵𝑑𝑟2 − 𝐶𝑑𝑟𝑑𝑡 
 
3) 
 
Seja t  t’ = t + f(r) ou ainda, t’ – t = f(r) 
Temos que; 
𝑑𝑡′ = 𝑑𝑡 +
𝜕𝑓(𝑟)
𝜕𝑟
𝑑𝑟 
 
𝑓(𝑟) =
1
2
 ∫
𝐶
𝐷
𝑑𝑟  𝑑𝑡′ = 𝑑𝑡 +
1
2
𝐶
𝐷
𝑑𝑟 
𝑑𝑓(𝑟)
𝑑𝑟
=
1
2
𝐶
𝐷
  
 
𝑑𝑠2 = 𝐷(𝑑𝑡′)2 − 𝐴[𝑟2(𝑑𝜃2 + 𝑠𝑒𝑛2𝜃𝑑𝜙2)] − 𝐵𝑑𝑟2 − 𝐶𝑑𝑟(𝑑𝑡′) 
𝑑𝑠2 = 𝐷 (𝑑𝑡 +
1
2
𝐶
𝐷
𝑑𝑟)
2
− 𝐴[𝑟2(𝑑𝜃2 + 𝑠𝑒𝑛2𝜃𝑑𝜙2)] − 𝐵𝑑𝑟2 − 𝐶𝑑𝑟 (𝑑𝑡 +
1
2
𝐶
𝐷
𝑑𝑟) 
𝑑𝑠2 = 𝐷 [𝑑𝑡2 +
𝐶
𝐷
𝑑𝑟𝑑𝑡 +
1
4
𝐶
𝐷
𝑑𝑟²] − 𝐴[𝑟2(𝑑𝜃2 + 𝑠𝑒𝑛2𝜃𝑑𝜙2)] − 𝐵𝑑𝑟2 − 𝐶𝑑𝑟𝑑𝑡 − 𝐶𝑑𝑟
1
2
𝐶
𝐷
𝑑𝑟 
 
𝑑𝑠2 = 𝐷𝑑𝑡2 + 𝐶𝑑𝑟𝑑𝑡 +
1
4
𝐶𝑑𝑟² − 𝐴[𝑟2(𝑑𝜃2 + 𝑠𝑒𝑛2𝜃𝑑𝜙2)] − 𝐵𝑑𝑟2 − 𝐶𝑑𝑟𝑑𝑡 −
1
2
𝐶²
𝐷
𝑑𝑟² 
𝑑𝑠2 = 𝐷𝑑𝑡2 − 𝐴[𝑟2(𝑑𝜃2 + 𝑠𝑒𝑛2𝜃𝑑𝜙2)] − 𝐵𝑑𝑟2 −
1
2
𝐶²
𝐷
𝑑𝑟² +
1
4
𝐶𝑑𝑟² 
𝑑𝑠2 = 𝐷𝑑𝑡2 − 𝐴[𝑟2(𝑑𝜃2 + 𝑠𝑒𝑛2𝜃𝑑𝜙2)] − 𝐵𝑑𝑟2 −
1
4
𝐶²
𝐷
𝑑𝑟² 
 
 
 
 
4) 
 
Da questão 2: 
𝑑𝑠2 = 𝐷𝑑𝑡2 − 𝐴[𝑟2(𝑑𝜃2 + 𝑠𝑒𝑛2𝜃𝑑𝜙2)] − 𝐵𝑑𝑟2 − 𝐶𝑑𝑟𝑑𝑡 
Sendo d = d = 0 : 
𝑑𝑠2 = 𝐷𝑑𝑡2 − 𝐵𝑑𝑟2 − 𝐶𝑑𝑟𝑑𝑡 
Se a métrica é isotrópica, C = 0: 
𝑑𝑠2 = 𝐷𝑑𝑡2 − 𝐵𝑑𝑟2 
Sendo 
 𝑑𝑠2 = 𝑔𝜇𝜈 ⋅ 𝑑𝑥
𝜇 ⋅ 𝑑𝑥𝜈 , para 𝑔𝜇𝜈 = 𝑔𝜇𝜈(�⃗�, 𝑡) 
Então, 
𝐷 = 𝑐2 ⋅ 𝑔𝑡𝑡 e 𝐵 = 𝑐
2 ⋅ 𝑔𝑟𝑟 
Assim: 
𝑑𝑠2 = 𝑐2 ⋅ 𝑔𝑡𝑡 ⋅ 𝑑𝑡
2 − 𝑐2 ⋅ 𝑔𝑟𝑟 ⋅ 𝑑𝑟
2 
Mas, como a luz caminha por geodésicas nulas... 
𝑑𝑠2 = 0 ⟹ 𝑐2 ⋅ 𝑔𝑡𝑡 ⋅ 𝑑𝑡
2 = 𝑐2 ⋅ 𝑔𝑟𝑟 ⋅ 𝑑𝑟
2 
(
𝑑𝑦
𝑑𝑥
)
2
= 𝑐2 ⋅
𝑔𝑡𝑡
𝑔𝑟𝑟𝑑𝑟
𝑑𝑡
= 𝑐 ⋅ √
𝑔𝑡𝑡
𝑔𝑟𝑟
 
Renomeando gtt para g00 : 
𝑑𝑟
𝑑𝑡
= 𝑐 ⋅ √
𝑔00
𝑔𝑟𝑟
 
 
 
 
 
 
LISTA 3 
 
1) 
 
 𝐴𝑖;𝑗 − 𝐴𝑗;𝑖 = 𝐴𝑖,𝑗 − 𝐴𝑗,𝑖 
 
𝑇𝑖𝑗;𝑘 + 𝑇𝑗𝑘;𝑖 + 𝑇𝑘𝑖;𝑗 = 𝑇𝑖𝑗,𝑘 + 𝑇𝑗𝑘,𝑖 + 𝑇𝑘𝑖,𝑗 
 
Dado que: 
 𝐴𝑖;𝑗 = 𝐴𝑖,𝑗 − 𝛤𝑗𝑖
𝑎 ⋅ 𝐴𝑎 
e 
𝐴𝑗;𝑖 = 𝐴𝑗,𝑖 − 𝛤𝑖𝑗
𝑎 ⋅ 𝐴𝑎 
Temos: 
𝐴𝑖;𝑗 − 𝐴𝑗;𝑖 = (𝐴𝑖,𝑗 − 𝛤𝑗𝑖
𝑎 ⋅ 𝐴𝑎) − 𝐴𝑗,𝑖 − (𝛤𝑖𝑗
𝑎 ⋅ 𝐴𝑎) 
𝐴𝑖;𝑗 − 𝐴𝑗;𝑖 = 𝐴𝑖,𝑗 − 𝛤𝑗𝑖
𝑎 ⋅ 𝐴𝑎 − 𝐴𝑗,𝑖 + 𝛤𝑖𝑗
𝑎 ⋅ 𝐴𝑎 
 
Uma vez que 𝛤𝑗𝑖
𝑎 = 𝛤𝑖𝑗
𝑎 , temos: 
𝐴𝑖;𝑗 − 𝐴𝑗;𝑖 = 𝐴𝑖,𝑗 − 𝐴𝑗,𝑖 
 
Para o tensor antissimétrico, definimos: 
𝑇𝑖𝑗;𝑘 = 𝑇𝑖𝑗,𝑘 + 𝑇𝑎𝑗 ⋅ 𝛤𝑖𝑘
𝑎 − 𝑇𝑖𝑎 ⋅ 𝛤𝑗𝑘
𝑎 
𝑇𝑗𝑘;𝑖 = 𝑇𝑗𝑘,𝑖 + 𝑇𝑏𝑘 ⋅ 𝛤𝑗𝑖
𝑏 − 𝑇𝑗𝑏 ⋅ 𝛤𝑘𝑖
𝑏 
𝑇𝑘𝑖;𝑗 = 𝑇𝑘𝑖,𝑗 + 𝑇𝑐𝑖 ⋅ 𝛤𝑘𝑗
𝑐 − 𝑇𝑘𝑐 ⋅ 𝛤𝑖𝑗
𝑐 
Então: 
𝑇𝑖𝑗;𝑘 + 𝑇𝑗𝑘;𝑖 + 𝑇𝑘𝑖;𝑗
= (𝑇𝑖𝑗,𝑘 + 𝑇𝑎𝑗 ⋅ 𝛤𝑖𝑘
𝑎 − 𝑇𝑖𝑎 ⋅ 𝛤𝑗𝑘
𝑎) + (𝑇𝑗𝑘,𝑖 + 𝑇𝑏𝑘 ⋅ 𝛤𝑗𝑖
𝑏 − 𝑇𝑗𝑏 ⋅ 𝛤𝑘𝑖
𝑏)
+ (𝑇𝑘𝑖,𝑗 + 𝑇𝑐𝑖 ⋅ 𝛤𝑘𝑗
𝑐 − 𝑇𝑘𝑐 ⋅ 𝛤𝑖𝑗
𝑐) 
 
Devido à antissimetria do tensor Tij e à simetria dos símbolos de Christoffel: 
𝑇𝑖𝑗;𝑘 + 𝑇𝑗𝑘;𝑖 + 𝑇𝑘𝑖;𝑗 = 𝑇𝑖𝑗,𝑘 + 𝑇𝑎𝑗 ⋅ 𝛤𝑖𝑘
𝑎 − 𝑇𝑖𝑎 ⋅ 𝛤𝑗𝑘
𝑎 + 𝑇𝑗𝑘,𝑖 + 𝑇𝑏𝑘 ⋅ 𝛤𝑗𝑖
𝑏 − 𝑇𝑗𝑏 ⋅ 𝛤𝑘𝑖
𝑏 + 𝑇𝑘𝑖,𝑗 + 𝑇𝑐𝑖 ⋅ 𝛤𝑘𝑗
𝑐 −
𝑇𝑘𝑐 ⋅ 𝛤𝑖𝑗
𝑐 
Portanto: 𝑇𝑖𝑗;𝑘 + 𝑇𝑗𝑘;𝑖 + 𝑇𝑘𝑖;𝑗 = 𝑇𝑖𝑗,𝑘 + 𝑇𝑗𝑘,𝑖 + 𝑇𝑘𝑖,𝑗 
 
 
2) 
 
𝑑𝑠2 = 𝑑𝑠2 + 𝑟2 ⋅ 𝑑𝜙2 + 𝑑𝑧2 , 
são: 
𝛤𝜙𝜙
𝑟 = −𝑟, 𝛤𝑟𝜙
𝜙
=
1
𝑟
, 𝛤𝜙𝑟
𝜙
=
1
𝑟
 
 
 
Temos: 𝑔𝑎𝑏 = (
1 0 0
0 𝑟2 0
0 0 1
) e 𝑔𝑎𝑏 = (
1 0 0
0
1
𝑟2
0
0 0 1
) 
 
Tomando a definição dos símbolos de Christoffel de segunda espécie, para os não nulos temos: 
 
𝛤𝑏𝑐
𝑎 =
1
2
⋅ 𝑔𝑎𝑑 ⋅ (
𝜕𝑔𝑏𝑑
𝜕𝑥𝑐
+
𝜕𝑔𝑐𝑑
𝜕𝑥𝑐
−
𝜕𝑔𝑏𝑐
𝜕𝑥𝑐
) 
 
Para a = r, b = , c =  e d = r: 
 
𝛤𝜙𝜙
𝑟 =
1
2
⋅ 𝑔𝑟𝑟 ⋅ (
𝜕𝑔𝜙𝑟
𝜕𝑥𝜙
+
𝜕𝑔𝜙𝑟
𝜕𝑥𝜙
−
𝜕𝑔𝜙𝜙
𝜕𝑥𝑟
) =
1
2
⋅ (1) ⋅ (−2 ⋅ 𝑟) = −𝑟 
 
Para a = , b = r, c =  e d = : 
 
𝛤𝑟𝜙
𝜙
=
1
2
⋅ 𝑔𝜙𝜙 ⋅ (
𝜕𝑔𝑟𝜙
𝜕𝑥𝜙
+
𝜕𝑔𝜙𝜙
𝜕𝑥𝑟
−
𝜕𝑔𝑟𝜙
𝜕𝑥𝜙
) =
1
2
⋅ (
1
𝑟2
) ⋅ (2 ⋅ 𝑟) =
1
𝑟
 
 
Para a = , b = , c = r e d = : 
 
𝛤𝜙𝑟
𝜙
=
1
2
⋅ 𝑔𝜙𝜙 ⋅ (
𝜕𝑔𝜙𝜙
𝜕𝑥𝑟
+
𝜕𝑔𝑟𝜙
𝜕𝑥𝜙
−
𝜕𝑔𝜙𝑟
𝜕𝑥𝜙
) =
1
2
⋅ (
1
𝑟2
) ⋅ (2 ⋅ 𝑟) =
1
𝑟
 
 
Portanto: 
𝛤𝜙𝜙
𝑟 = −𝑟 
 
𝛤𝑟𝜙
𝜙
=
1
𝑟
 
 
𝛤𝜙𝑟
𝜙
=
1
𝑟
 
 
 
 
 
3) 
 
Seja a equação da geodésica: 
𝑑2𝑋𝑎
𝑑𝑠2
+ Γ𝑏𝑐
𝑎 ⋅
𝑑𝑋𝑏
𝑑𝑠
⋅
𝑑𝑋𝑐
𝑑𝑠
= 0 
 
Tomando os símbolos de Christofell não nulos e a métrica do exercício anterior: 
 
Para a = r: 
𝑑2𝑟
𝑑𝑠2
+ Γ𝜙𝜙
𝑟 ⋅
𝑑𝜙
𝑑𝑠
⋅
𝑑𝜙
𝑑𝑠
= 0 
 
Então: 
𝑑2𝑟
𝑑𝑠2
− 𝑟 ⋅ (
𝑑𝜙
𝑑𝑠
)2 = 0 
 
 
Para a = : 
𝑑2𝜙
𝑑𝑠2
+ Γ𝑟𝜙
𝜙
⋅
𝑑𝑟
𝑑𝑠
⋅
𝑑𝜙
𝑑𝑠
+ Γ𝜙𝑟
𝜙
⋅
𝑑𝜙
𝑑𝑠
⋅
𝑑𝑟
𝑑𝑠
= 0 
 
 
𝑑2𝜙
𝑑𝑠2
+
1
𝑟
⋅
𝑑𝑟
𝑑𝑠
⋅
𝑑𝜙
𝑑𝑠
+
1
𝑟
⋅
𝑑𝜙
𝑑𝑠
⋅
𝑑𝑟
𝑑𝑠
= 0 
 
Então: 
𝑑2𝜙
𝑑𝑠2
+
2
𝑟
⋅
𝑑𝑟
𝑑𝑠
⋅
𝑑𝜙
𝑑𝑠
= 0 
 
 
Para a = z: 
𝑑2𝑧
𝑑𝑠2
+ Γ𝑏𝑐
𝑧 ⋅
𝑑2𝑋𝑏
𝑑𝑠
⋅
𝑑2𝑋𝑐
𝑑𝑠
= 0 
 
 Termo nulo! 
 
Logo: 
𝑑2𝑧
𝑑𝑠2
= 0 
 
 
 
4) 
𝑉;𝛼𝛽
𝜇
= (𝑉,𝛽
𝜇
+ Γ𝛽𝜆
𝜇
⋅ 𝑉𝜆)
;𝛼
 
 
𝑉;𝛼𝛽
𝜇
= 𝑉,𝛽𝛼
𝜇
+ 𝑉,𝛽
𝜆 ⋅ Γ𝜇𝛼
𝜆 + 𝑉,𝜆
𝜇
⋅ Γ𝛽𝛼
𝜆 + 𝑉,𝛼
𝜆 ⋅ Γ𝜇𝛽
𝜆 + 𝑉𝜆 ⋅ Γ𝜇𝛽,𝛼
𝜆 − 𝑉𝜆 ⋅ Γ𝜈𝛽
𝜆 ⋅ Γ𝜇𝛼
𝜈 − 𝑉𝜆 ⋅ Γ𝜇𝜈
𝜆 ⋅ Γ𝛽𝛼
𝜈 
 
 
𝑉;𝛽𝛼
𝜇
= (𝑉,𝛼
𝜇
+ Γ𝛼𝜆
𝜇
⋅ 𝑉𝜆)
;𝛽
 
 
𝑉;𝛽𝛼
𝜇
= 𝑉,𝛼𝛽
𝜇
+ 𝑉,𝛼
𝜆 ⋅ Γ𝜇𝛽
𝜆 + 𝑉,𝜆
𝜇
⋅ Γ𝛼𝛽
𝜆 + 𝑉,𝛽
𝜆 ⋅ Γ𝜇𝛼
𝜆 + 𝑉𝜆 ⋅ Γ𝜇𝛼,𝛽
𝜆 − 𝑉𝜆 ⋅ Γ𝜈𝛼
𝜆 ⋅ Γ𝜇𝛽
𝜈 − 𝑉𝜆 ⋅ Γ𝜇𝜈
𝜆 ⋅ Γ𝛼𝛽
𝜈 
 
𝑉;𝛼𝛽
𝜇
− 𝑉;𝛽𝛼
𝜇
= 𝑉,𝛽𝛼
𝜇
+ 𝑉,𝛽
𝜆 ⋅ Γ𝜇𝛼
𝜆 + 𝑉,𝜆
𝜇
⋅ Γ𝛽𝛼
𝜆 + 𝑉,𝛼
𝜆 ⋅ Γ𝜇𝛽
𝜆 + 𝑉𝜆 ⋅ Γ𝜇𝛽,𝛼
𝜆 − 𝑉𝜆 ⋅ Γ𝜈𝛽
𝜆
⋅ Γ𝜇𝛼
𝜈 − 𝑉𝜆 ⋅ Γ𝜇𝜈
𝜆 ⋅ Γ𝛽𝛼
𝜈 − 𝑉,𝛼𝛽
𝜇
− 𝑉,𝛼
𝜆 ⋅ Γ𝜇𝛽
𝜆 − 𝑉,𝜆
𝜇
⋅ Γ𝛼𝛽
𝜆 − 𝑉,𝛽
𝜆 ⋅ Γ𝜇𝛼
𝜆
− 𝑉𝜆 ⋅ Γ𝜇𝛼,𝛽
𝜆 + 𝑉𝜆 ⋅ Γ𝜈𝛼
𝜆 ⋅ Γ𝜇𝛽
𝜈 + 𝑉𝜆 ⋅ Γ𝜇𝜈
𝜆 ⋅ Γ𝛼𝛽
𝜈 
 
𝑉;𝛼𝛽
𝜇
− 𝑉;𝛽𝛼
𝜇
= 𝑉𝜆 ⋅ Γ𝜇𝛽,𝛼
𝜆 − 𝑉𝜆 ⋅ Γ𝜈𝛽
𝜆 ⋅ Γ𝜇𝛼
𝜈 − 𝑉𝜆 ⋅ Γ𝜇𝛼,𝛽
𝜆 + 𝑉𝜆 ⋅ Γ𝜈𝛼
𝜆 ⋅ Γ𝜇𝛽
𝜈 
 
𝑉;𝛼𝛽
𝜇
− 𝑉;𝛽𝛼
𝜇
= (Γ𝜇𝛽,𝛼
𝜆 − Γ𝜈𝛽
𝜆 ⋅ Γ𝜇𝛼
𝜈 − Γ𝜇𝛼,𝛽
𝜆 + Γ𝜈𝛼
𝜆 ⋅ Γ𝜇𝛽
𝜈 ) ⋅ 𝑉𝜆 
 
𝑉;𝛼𝛽
𝜇
− 𝑉;𝛽𝛼
𝜇
= (Γ𝜇𝛽,𝛼
𝜆 − Γ𝜇𝛼,𝛽
𝜆 + Γ𝜈𝛼
𝜆 ⋅ Γ𝜇𝛽
𝜈 − Γ𝜈𝛽
𝜆 ⋅ Γ𝜇𝛼
𝜈 ) ⋅ 𝑉𝜆 
 
𝑉;𝛼𝛽
𝜇
− 𝑉;𝛽𝛼
𝜇
= −R𝜇𝛽𝛼
𝜆 ⋅ 𝑉𝜆 
 
𝑉;𝛼𝛽
𝜇
− 𝑉;𝛽𝛼
𝜇
= 𝑅𝜇𝛼𝛽
𝜆 ⋅ 𝑉𝜆