Energia Cinética e Energia de Repouso · Como essa energia 𝑚 2passou despercebida ?...
Transcript of Energia Cinética e Energia de Repouso · Como essa energia 𝑚 2passou despercebida ?...
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Energia Cinética e Energia de Repouso
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Os 𝐯𝑖(𝑡) mudam, mas 𝐏 = 𝑖 𝛾𝑖(𝑡)𝑚𝑖𝐯𝑖(𝑡)permanece constante
Os 𝐯𝑖(𝑡) mudam, mas 𝐸 = 𝑖 𝛾𝑖(𝑡)𝑚𝑖𝑐2
permanece constante
Sistema ISOLADOde partículas interagentes
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O que exatamente é 𝛾𝑣𝑚𝑐2 ?
1. É uma energia que depende da velocidade da partícula (a única presente quando a partícula está livre)
𝐸 =𝑚𝑐2
1 − 𝑣2/𝑐2
2. Em repouso, a partícula tem 𝐸 = 𝑚𝑐2
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Lembrando a origem de 𝐾 =1
2𝑚𝐯2
𝑚𝐚 = 𝐅𝑡𝑜𝑡𝐯 ∙ ∙ 𝐯
𝑚𝐯 ∙𝑑𝐯
𝑑𝑡= 𝐅𝑡𝑜𝑡 ∙
𝑑𝐫
𝑑𝑡
𝑑
𝑑𝑡
1
2𝑚𝐯2 = 𝐅𝑡𝑜𝑡 ∙
𝑑𝐫
𝑑𝑡 𝐅𝑡𝑜𝑡
𝑑𝐫
𝑑1
2𝑚𝐯2 = 𝐅𝑡𝑜𝑡 ∙ 𝑑𝐫
Teorema da energia cinética e do trabalho
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Não há arbitrariedade na energia da partícula livre!
𝐸𝑁𝑒𝑤𝑡 = 𝐶 +1
2𝑚𝐯2
𝐸 = 𝑚𝑐2 + (𝛾𝑣 − 1)𝑚𝑐2
𝐾 = (𝛾𝑣 − 1)𝑚𝑐2
(1 + 𝜖)−1/2= 1 −𝜖
2+3𝜖2
8− 𝑜(𝜖3)
≅1
2𝑚𝐯2 +
3
8𝑚𝐯4
𝑐2+⋯ (𝑣 ≪ 𝑐)
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𝑚 é a MASSA (de repouso) do objeto
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Ordens de grandeza
1 kg
1 m/s
𝐸 ≅ 𝑚𝑐2 +1
2𝑚𝐯2 +
3
8𝑚𝐯4
𝑐2+⋯ (𝑣 ≪ 𝑐)
9 × 1016 J0,5 J 4 × 10−18 J
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Como essa energia 𝑚𝑐2 passou despercebida ?
'Castle/Bravo' was the largest nuclear weapon
ever detonated by the United States. Before it
was set off at Bikini on February 28, 1954, it was
expected to explode with an energy equivalent of
about 8 million tons of TNT. Actually, it produced
almost twice that explosive power -- equivalent to 15 million tons of TNT.
15 megatons = 6,3 × 1016 J
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Alteração de 𝑚 em uma colisão
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Uma colisão 100% elástica
𝐯 −𝐯
𝑚 𝑚
𝐯
𝑚
−𝐯
𝑚
𝐏 = 0
𝐸 = 2𝛾𝑣𝑚𝑐2
𝐏′ = 0
𝐸′ = 2𝛾𝑣𝑚𝑐2
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Uma colisão 100% inelástica
𝐯 −𝐯
𝑚 𝑚
𝑀
𝐏 = 0
𝐸 = 2𝛾𝑣𝑚𝑐2
𝐏′ = 0
𝐸′ = 𝑀𝑐2
𝑀 = 𝛾𝑣(2𝑚)
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Essa mudança de massa é detectável?
𝑀 − 2𝑚
2𝑚= 𝛾𝑣 − 1
Resolução necessária:
𝛾𝑣 − 1 > 10−6
𝛾𝑣 ~ 1 +𝛽2
2+ …
𝑣 > 1.500.000 km/h
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Energia interna da matéria faz parte da massa de repouso
(frio)
(quente)
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200 ml @ 80℃
200 ml @ 10℃𝐸 = 𝑚10𝑐
2
𝐸′ = 𝑚80𝑐2
∆𝑚 = 6,5 × 10−7 g
∆𝐸 = 200 g 4186 J/g℃ 70℃ = 58,6 MJ = ∆𝑚 𝑐2
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Partículas Elementares(não têm estrutura interna)
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(2𝑚𝑢𝑝 +𝑚𝑑𝑜𝑤𝑛 ~ 9,6 MeV/c2)
(2𝑚𝑑𝑜𝑤𝑛 +𝑚𝑢𝑝 ~ 12 MeV/c2)
𝑀𝑛𝑒𝑢𝑡𝑟𝑜𝑛 = 939,5654133 58 MeV/c2
𝑀𝑝𝑟𝑜𝑡𝑜𝑛 = 938,272046 21 MeV/c2
doi:10.1
088/1
674-1
137/3
8/9
/09
000
1
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𝑚 = 0 implica 𝐸 = 0 ?
NÃO. As partículas elementares com 𝑚 = 0 têm 𝑣 = 𝑐
𝐸 = 𝛾𝑐𝑚𝑐2 |𝐩| = 𝛾𝑐𝑚𝑐
? ?
Maxwell + Schrödinger𝐸𝛾 = ℎ𝑓
𝐩𝛾 =ℎ
𝜆(ℎ = 6,62 × 10−34 J s)
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Relação Energia Momento Relativística
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A “norma” invariante do 4-vetor 𝑝 de umapartícula
𝑝 = (𝑚𝑐𝛾𝑣, 𝑚𝑣𝑥𝛾𝑣, 𝑚𝑣𝑦𝛾𝑣, 𝑚𝑣𝑧𝛾𝑣) = (𝐸/𝑐, 𝐩)
(𝐸/𝑐, 𝐩) ≠ (𝐸/𝑐, 𝐩)
(𝐸/𝑐)2−𝐩2 = 𝐸/𝑐 2 − 𝐩2
= 𝑚2𝑐2𝛾𝑣2 −𝑚2𝐯2𝛾𝑣
2
= 𝑚2𝑐2porém,
𝛾𝑣 =1
1 − 𝑣2/𝑐2
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𝐸 = |𝐩|2𝑐2 +𝑚2𝑐4
(válido também para 𝑚 = 0)
(todos os referenciais verificam isso)
𝐸𝛾 = ℎ𝑓
𝐩𝛾 =ℎ
𝜆𝐸𝛾 = 𝐩𝛾 𝑐+ 𝜆𝑓 = 𝑐 ⟹
onda eletromag.
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Colisões Relativísticas
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Colisão Inelástica (= “𝐾 não se conserva”)
Partículas não elementares: 𝑚’s mudam
Partículas elementares: partículas são criadas/destruídas
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O referencial 𝐶 (onde 𝐏∗ = 0)
𝐏 ≠ 𝟎
𝐸 = 𝑚 +𝑚 𝑐2 + 𝐾
𝐏∗ = 𝟎
𝐸∗ = 𝑚 +𝑚 𝑐2 + 𝐾∗ + 𝐾∗
Laboratório Referencial 𝐶
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Obs: 𝐶 não é o referencial do CM
𝐑 = 𝑖𝑚𝑖𝐫𝑖𝑀
𝐕 = 𝑖𝑚𝑖𝐯𝑖𝑀
𝐏𝑁𝑒𝑤𝑡
No referencial do CM 𝐕∗ = 𝟎, e portanto 𝐏𝑁𝑒𝑤𝑡∗ = 𝟎,
porém 𝐏 = 𝑖 𝛾𝑖𝑚𝑖𝐯𝑖 …
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𝐶 mede a menor energia total dentre todos os referenciais
𝐢
𝐣
𝐢
𝐣
𝐶𝑆
𝐕∗
𝐏
𝐸/𝑐𝑃𝑥0
=𝛾∗ 𝛾∗𝛽∗ 0𝛾∗𝛽∗ 𝛾∗ 00 0 1
𝐸∗/𝑐00
𝐸 = 𝛾∗𝐸∗
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Produção de anti-prótons (ref. Lab)
𝑝 𝑝
𝑝
𝑝
𝑝
𝑝
Qual o menor valor possível de 𝐾 para produzir pares?
𝐸 = 2𝑚𝑐2 + 𝐾
𝐏 ≠ 𝟎
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(ref. 𝐶)
𝑝 𝑝
𝐏∗ = 𝟎
𝑝
𝑝
𝑝
𝑝
𝐸∗ = 4𝑚𝑐2 + 𝐾′∗
O menor valor possível de 𝐸∗ é 4𝑚𝑐2
Qualquer valor de 𝐾′∗
respeita conservação de energia e momento
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Relação entre 𝐾 e 𝐸∗ (caso geral)
𝑚𝑀
𝐩
𝐸 = 𝑀𝑐2 +𝑚𝑐2 + 𝐾
𝐏 = 𝐩
𝐸∗
𝐏∗ = 𝟎
(𝐸/𝑐, 𝐩)Lorentz
(𝐸∗/𝑐, 𝟎)
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(𝐸∗/𝑐)2= (𝐸/𝑐)2−𝐩2
𝐸∗2 = (𝑀𝑐2 +𝑚𝑐2 + 𝐾)2 − 𝐩2𝑐2
𝐸∗2 = (𝑀𝑐2)2 +2(𝑀𝑐2)(𝑚𝑐2 + 𝐾) + (𝑚𝑐2 + 𝐾)2 −𝐩2𝑐2
𝑚2𝑐4
Relação energia momento da partícula incidente
𝐸∗2 = (𝑀𝑐2 +𝑚𝑐2)2 + 2𝑀𝑐2𝐾
alvo
partícula incidente
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no nosso caso...
𝐸∗2 = (𝑀𝑐2 +𝑚𝑐2)2 + 2𝑀𝑐2𝐾
(4𝑚𝑐2)2= (𝑚𝑐2 +𝑚𝑐2)2 + 2𝑚𝑐2𝐾𝑚𝑖𝑛
𝐾𝑚𝑖𝑛 = 6𝑚𝑐2 = 5,63 GeV