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Theory of Relativity
https://www.youtube.com/watch?v=i0CYIMoH_Hk
𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆
𝒕𝒓𝒂𝒗𝒆𝒍𝒍𝒆𝒅 𝒃𝒚 𝒍𝒊𝒈𝒉𝒕 𝒂𝒔 𝒔𝒆𝒆𝒏
𝒇𝒓𝒐𝒎 𝒆𝒂𝒓𝒕𝒉 𝒊𝒔 , 𝒙 = 𝒄𝒕
𝒄𝒕
𝒕𝒉𝒆 𝒑𝒍𝒂𝒏𝒆 𝒊𝒔 𝒎𝒐𝒗𝒊𝒏𝒈 𝒘𝒊𝒕𝒉 𝒂 𝒔𝒑𝒆𝒆𝒅 𝒐𝒇 𝒔𝒊𝒈𝒏𝒊𝒇𝒊𝒄𝒂𝒏𝒕
𝑭𝑹𝑨𝑪𝑻𝑰𝑶𝑵 𝒐𝒇 𝒕𝒉𝒆 𝒔𝒑𝒆𝒆𝒅 𝒐𝒇 𝒍𝒊𝒈𝒉𝒕
𝒊𝒏 𝒉𝒊𝒔 𝒑𝒐𝒊𝒏𝒕 𝒐𝒇 𝒗𝒊𝒆𝒘 𝒕𝒉𝒆 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒕𝒓𝒂𝒗𝒆𝒍𝒍𝒆𝒅 𝒃𝒚 𝒍𝒊𝒈𝒉𝒕 𝒊𝒔,
𝒙′ = 𝒄𝒕 − 𝒗𝒕
𝒎𝒆𝒂𝒔𝒖𝒓𝒆𝒎𝒆𝒏𝒕𝒔 𝒅𝒐𝒏𝒆 𝒐𝒏 𝒔′
𝒘𝒊𝒍𝒍 𝒃𝒆 𝒔𝒉𝒐𝒓𝒕𝒆𝒓 𝒃𝒚 𝒂 𝒇𝒂𝒄𝒕𝒐𝒓 𝒗𝒕
𝒕𝒉𝒊𝒔 𝒊𝒔 𝒕𝒉𝒆 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒕𝒓𝒂𝒗𝒆𝒍𝒍𝒆𝒅 𝒃𝒚 𝑺′
𝒄𝒐𝒎𝒑𝒂𝒓𝒆𝒅 𝒕𝒐 𝒍𝒊𝒈𝒉𝒕
𝒄𝒐𝒏𝒔𝒊𝒅𝒆𝒓 𝒕𝒉𝒂𝒕 𝒕𝒉𝒆 𝒑𝒍𝒂𝒏𝒆 𝒊𝒔 𝒓𝒂𝒄𝒊𝒏𝒈 𝒂𝒈𝒂𝒊𝒏𝒕𝒔 𝒍𝒊𝒈𝒉𝒕
𝒙 = 𝒄𝒕
𝒙′ = 𝒄𝒕 − 𝒗𝒕
𝒙′ = 𝒙 − 𝒗𝒕
𝒗𝒕
𝒄𝒕
𝒙 = 𝒄𝒕
𝒙′ = 𝒙 − 𝒗𝒕
𝒙′
= 𝒙 − 𝒗𝒕
𝒗𝒕
𝒙 = 𝒄𝒕
𝒔𝒐 𝒕𝒐 𝒆𝒙𝒑𝒍𝒂𝒊𝒏 𝒐𝒏 𝒐𝒃𝒔𝒆𝒓𝒗𝒆𝒓 𝒐𝒏 𝒕𝒉𝒆 𝒑𝒍𝒂𝒏𝒆 𝒕𝒉𝒂𝒕 𝒍𝒊𝒈𝒉𝒕 𝒉𝒂𝒔 𝒓𝒆𝒂𝒍𝒚
𝒕𝒓𝒂𝒗𝒆𝒍𝒍𝒆𝒅 𝒄𝒕 𝒂𝒏𝒅 𝒉𝒆′𝒔 𝒋𝒖𝒔𝒕 𝒈𝒆𝒕𝒕𝒊𝒏𝒈 𝒇𝒂𝒔𝒕𝒆𝒓 𝒂𝒏𝒅 𝒂𝒍𝒔𝒐 𝒊𝒏 𝒉𝒊𝒔
𝒑𝒆𝒓𝒑𝒆𝒄𝒕𝒊𝒗𝒆 𝒕𝒉𝒆 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒕𝒓𝒂𝒗𝒆𝒍𝒍𝒆𝒅 𝒄𝒐𝒏𝒕𝒓𝒂𝒄𝒕𝒆𝒅 𝒂𝒏𝒅 𝒕𝒊𝒎𝒆 𝒊𝒔
𝒂𝒍𝒔𝒐 𝒅𝒊𝒍𝒂𝒕𝒆𝒅, 𝒘𝒆 𝒊𝒏𝒕𝒓𝒐𝒅𝒖𝒄𝒆
𝒕𝒉𝒆 𝒇𝒂𝒄𝒕𝒐𝒓 𝓴
𝒙 = 𝒄𝒕 → 𝒆𝒒. 𝟏
𝒙′ = 𝒙 − 𝒗𝒕 → 𝒆𝒒. 𝟐
𝒙′
= 𝒙 − 𝒗𝒕
𝒗𝒕
𝒙 = 𝒄𝒕
𝒘𝒆 𝒓𝒆𝒘𝒓𝒊𝒕𝒆 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟐 𝒂𝒔 𝒙′ = 𝓴(𝒙 − 𝒗𝒕) → 𝒆𝒒. 𝟑
𝒐𝒓 𝓴𝒙′ = (𝒙 − 𝒗𝒕) → 𝒆𝒒. 𝟒
𝒕𝒐 𝒂𝒄𝒄𝒐𝒖𝒏𝒕 𝒇𝒐𝒓 𝒕𝒉𝒆 𝒇𝒍𝒐𝒘 𝒐𝒇 𝒕𝒊𝒎𝒆
𝒊𝒏 𝒕𝒉𝒆 𝒑𝒆𝒓𝒑𝒆𝒄𝒕𝒊𝒗𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒐𝒃𝒔𝒆𝒓𝒗𝒆𝒓
𝒊𝒏 𝒕𝒉𝒆 𝒑𝒍𝒂𝒏𝒆 𝒘𝒆 𝒘𝒓𝒊𝒕𝒆,
𝓴𝒙′
= 𝒙 − 𝒗𝒕
𝓴𝒙′ − 𝒗𝒕′ = 𝒙
𝒙 = 𝓴𝒙′ − 𝒗𝒕′ ⟶ 𝒆𝒒. 𝟓
𝒙 = 𝒄𝒕 → 𝒆𝒒. 𝟏
𝒙′ = 𝒙 − 𝒗𝒕 → 𝒆𝒒. 𝟐
𝒙′
= 𝒙 − 𝒗𝒕
𝒗𝒕
𝒙 = 𝒄𝒕
𝒙′ = 𝓴(𝒙 − 𝒗𝒕) → 𝒆𝒒. 𝟑
𝓴𝒙′ = (𝒙 − 𝒗𝒕) → 𝒆𝒒. 𝟒
𝒙 = 𝓴𝒙′
− 𝒗𝒕′
⟶ 𝒆𝒒. 𝟓
Laws of physics are the same for any
reference frame, hence those on the
opposite does not need a new
equation but Only a changed in sign.
We modify equation 5 .
𝒙 = 𝓴𝒙′ + 𝒗𝒕′ ⟶ 𝒆𝒒. 𝟔
𝒙′ = 𝒄𝒕′ → 𝒆𝒒. 𝟐. 𝟏
𝒃𝒚 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒕𝒊𝒂𝒍 𝒄𝒂𝒍𝒄𝒖𝒍𝒖𝒔
𝒙′ = 𝓴(𝒙 − 𝒗𝒕) → 𝒆𝒒. 𝟑
𝒙′
= 𝒙 − 𝒗𝒕
𝒗𝒕
𝒙 = 𝒄𝒕
𝒘𝒆 𝒄𝒂𝒏 𝒏𝒐𝒘 𝒔𝒐𝒍𝒗𝒆 𝒕𝒉𝒊𝒔 𝒕𝒘𝒐 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏𝒔 𝒔𝒊𝒎𝒖𝒍𝒕𝒂𝒏𝒆𝒐𝒖𝒔𝒍𝒚
𝒙 = 𝓴𝒙′
+ 𝒗𝒕′
⟶ 𝒆𝒒. 𝟔
𝒘𝒆 𝒔𝒖𝒃𝒔𝒕𝒊𝒕𝒕𝒖𝒕𝒆 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟑 𝒊𝒏𝒕𝒐
𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟔 𝒕𝒐 𝒓𝒆𝒅𝒖𝒄𝒆 𝒕𝒉𝒆 𝒏𝒖𝒎𝒃𝒆𝒓
𝒐𝒇 𝒗𝒂𝒓𝒊𝒂𝒃𝒍𝒆𝒔
𝒙 = 𝓴(𝓴(𝒙 − 𝒗𝒕) + 𝒗𝒕′)
𝒙 = 𝓴𝟐𝒙 − 𝓴𝟐𝒗𝒕 + 𝓴𝒗𝒕′
𝒘𝒆 𝒔𝒐𝒍𝒗𝒆 𝒇𝒐𝒓 𝒕′
𝒙 = 𝓴𝟐𝒙 − 𝓴𝟐𝒗𝒕 + 𝒗𝒕′
𝓴𝒗𝒕′ = 𝒙 − 𝓴𝟐𝒙 − 𝓴𝟐𝒗𝒕
𝓴𝒗𝒕′ = 𝒙(𝟏 − 𝓴𝟐) − 𝓴𝟐𝒗𝒕
𝓴𝒗𝒕′
𝓴𝒗
=
𝒙(𝟏 − 𝓴𝟐
)
𝓴𝒗
−
𝓴𝟐
𝒗𝒕
𝓴𝒗
𝓴𝒗𝒕′
𝓴𝒗
=
𝒙(𝟏 − 𝓴𝟐)
𝓴𝒗
−
𝓴𝟐𝒗𝒕
𝓴𝒗
𝒕′ =
𝒙(𝟏 − 𝓴𝟐)
𝓴𝒗
− 𝓴𝒕
𝒘𝒆 𝒔𝒐𝒍𝒗𝒆 𝒇𝒐𝒓 𝓴 𝒖𝒔𝒊𝒏𝒈 𝒆𝒒 𝟐. 𝟏 𝒂𝒏𝒅 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟑
𝒕′ =
𝒙(𝟏 − 𝓴𝟐)
𝓴𝒗
− 𝓴𝒕
𝒙′ = 𝒄𝒕′ → 𝒆𝒒. 𝟐. 𝟏
𝒙′ = 𝒄[
𝒙 𝟏 − 𝓴𝟐
𝓴𝒗
− 𝓴𝒕]
𝓴(𝒙 − 𝒗𝒕) = 𝒄[
𝒙 𝟏 − 𝓴𝟐
𝓴𝒗
− 𝓴𝒕]
𝓴𝒙 − 𝓴𝒗𝒕 = 𝒄[
𝒙 𝟏 − 𝓴𝟐
𝓴𝒗
− 𝓴𝒕]
𝒓𝒆𝒂𝒓𝒓𝒂𝒏𝒈𝒊𝒏𝒈 𝒕𝒆𝒓𝒎𝒔
𝓴𝒙 − 𝓴𝒗𝒕 = 𝒄[
𝒙 𝟏 − 𝓴𝟐
𝓴𝒗
− 𝓴𝒕]
𝓴𝒙 − 𝓴𝒗𝒕 = 𝒄𝒙(
𝟏 − 𝓴𝟐
𝓴𝒗
) − 𝒄𝓴𝒕
𝓴𝒙 − 𝓴𝒗𝒕 − 𝒄𝒙
𝟏 − 𝓴𝟐
𝓴𝒗
= 𝒄𝓴𝒕
𝓴𝒙 − 𝒄𝒙
𝟏 − 𝓴𝟐
𝓴𝒗
− 𝓴𝒗𝒕 = 𝒄𝓴𝒕
𝒘𝒆 𝒓𝒆𝒂𝒓𝒓𝒂𝒏𝒈𝒆 𝒕𝒆𝒓𝒎𝒔 𝒂𝒏𝒅 𝒔𝒐𝒍𝒗𝒆 𝒇𝒐𝒓 𝓴
𝓴𝒙 − 𝒄𝒙
𝟏 − 𝓴𝟐
𝓴𝒗
− 𝓴𝒗𝒕 = 𝒄𝓴𝒕
𝒙(𝓴 − 𝒄
𝟏 − 𝓴𝟐
𝓴𝒗
) = 𝒄𝓴𝒕 − 𝓴𝒗𝒕 𝑤𝑒 𝑟𝑒𝑤𝑟𝑖𝑡𝑒 𝑡ℎ𝑒 𝑅𝐻𝑆
𝒙(𝓴 − 𝒄
𝟏 − 𝓴𝟐
𝓴𝒗
) =
𝒄
𝒗
(𝓴𝒕 − 𝓴𝒕)
𝒘𝒆 𝒓𝒆𝒂𝒓𝒓𝒂𝒏𝒈𝒆 𝒕𝒆𝒓𝒎𝒔 𝒂𝒏𝒅 𝒔𝒐𝒍𝒗𝒆 𝒇𝒐𝒓 𝒙
𝒙(𝓴 − 𝒄
𝟏 − 𝓴𝟐
𝓴𝒗
) =
𝒄
𝒗
(𝓴𝒕 − 𝓴𝒕)
𝒙 =
𝒄
𝒗
(𝓴𝒕 − 𝓴𝒕)
(𝓴 − 𝒄
𝟏 − 𝓴𝟐
𝓴𝒗
)
𝒘𝒆 𝒎𝒂𝒏𝒊𝒑𝒖𝒍𝒂𝒕𝒆 𝒕𝒉𝒆 𝒏𝒖𝒎𝒆𝒓𝒂𝒕𝒐𝒓 𝒐𝒇 𝒕𝒉𝒆 𝑹𝑯𝑺
𝒙 =
𝒄
𝒗
(𝓴𝒕 − 𝓴𝒕)
(𝓴 − 𝒄
𝟏 − 𝓴𝟐
𝓴𝒗
)
𝒙 =
𝒄𝒕 (𝓴 −
𝒗
𝒄
𝓴)
(𝓴 − 𝒄
𝟏 − 𝓴𝟐
𝓴𝒗
)
𝒙 =
𝒄𝒕𝓴(𝟏 −
𝒗
𝒄
)
(𝓴 − 𝒄
𝟏 − 𝓴𝟐
𝓴𝒗
)
𝒘𝒆 𝒎𝒂𝒏𝒊𝒑𝒖𝒍𝒂𝒕𝒆 𝒕𝒉𝒆 𝒏𝒖𝒎𝒆𝒓𝒂𝒕𝒐𝒓 𝒐𝒇 𝒕𝒉𝒆 𝑹𝑯𝑺
𝒙 =
𝒄𝒕𝓴(𝟏 −
𝒗
𝒄
)
𝓴 − 𝒄
𝟏 − 𝓴𝟐
𝓴𝒗
𝒙 =
𝒄𝒕(𝟏 −
𝒗
𝒄
)
𝟏 − 𝒄
𝟏 − 𝓴𝟐
𝓴𝒗
𝒘𝒆 𝒎𝒂𝒏𝒊𝒑𝒖𝒍𝒂𝒕𝒆 𝒕𝒉𝒆 𝒅𝒆𝒏𝒐𝒎𝒊𝒏𝒂𝒕𝒐𝒓 𝒐𝒇 𝒕𝒉𝒆 𝑹𝑯𝑺
𝒙 =
𝒄𝒕(𝟏 −
𝒗
𝒄
)
𝟏 − 𝒄
𝟏 − 𝓴𝟐
𝓴𝒗
𝟏 − 𝒄
𝟏 − 𝓴𝟐
𝓴𝒗
𝓴(𝟏 − 𝒄
𝟏 − 𝓴𝟐
𝓴𝒗
)
𝓴 −
𝒄𝓴
𝒗
−
𝒄𝓴 𝟑
𝒗
𝓴
𝓴
−
𝒄𝓴
𝒗𝓴
−
𝒄𝓴 𝟑
𝒗𝓴
𝟏 −
𝒄
𝒗
−
𝒄𝓴 𝟐
𝒗
𝒘𝒆 𝒇𝒂𝒄𝒕𝒐𝒓 𝒐𝒖𝒕
𝒄
𝒗
𝒙 =
𝒄𝒕(𝟏 −
𝒗
𝒄
)
𝟏 − 𝒄
𝟏 − 𝓴𝟐
𝓴𝒗
𝟏 −
𝒄
𝒗
−
𝒄𝓴 𝟐
𝒗
𝟏 − (𝟏 − 𝓴 𝟐
)
𝒄
𝒗
𝟏 − (
𝟏
𝓴 𝟐
− 𝟏)
𝒄
𝒗
𝒏𝒐𝒘 𝒘𝒆 𝒄𝒂𝒏 𝒘𝒓𝒊𝒕𝒆 𝒆𝒒. 𝟕
𝒙 =
𝒄𝒕(𝟏 −
𝒗
𝒄
)
𝟏 − 𝒄
𝟏 − 𝓴𝟐
𝓴𝒗
→ 𝒆𝒒. 𝟕
𝟏 − (
𝟏
𝓴 𝟐
− 𝟏)
𝒄
𝒗
𝒙 =
𝒄𝒕(𝟏 −
𝒗
𝒄
)
𝟏 − (
𝟏
𝓴 𝟐 − 𝟏)
𝒄
𝒗
→ 𝒆𝒒. 𝟕
𝒙 = 𝒄𝒕[
𝟏 −
𝒗
𝒄
𝟏 −
𝟏
𝓴 𝟐 − 𝟏
𝒄
𝒗
] → 𝒆𝒒. 𝟕
𝒕𝒉𝒊𝒔 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒘𝒊𝒍𝒍 𝒏𝒐𝒕 𝒗𝒊𝒐𝒍𝒂𝒕𝒆 𝒕𝒉𝒆 𝒍𝒂𝒘 𝒐𝒇 𝒑𝒉𝒚𝒔𝒊𝒄𝒔 𝒊𝒇 𝒕𝒉𝒆 𝒕𝒆𝒓𝒎
𝒊𝒏 𝒑𝒂𝒓𝒆𝒏𝒕𝒉𝒆𝒔𝒊𝒔 𝒊𝒔 𝒆𝒒𝒖𝒂𝒍 𝒕𝒐 𝟏
𝒙 = 𝒄𝒕[
𝟏 −
𝒗
𝒄
𝟏 −
𝟏
𝓴 𝟐 − 𝟏
𝒄
𝒗
] → 𝒆𝒒. 𝟕
𝒘𝒆 𝒆𝒒𝒖𝒂𝒕𝒆 𝒊𝒕 𝒕𝒐 𝟏 𝒂𝒏𝒅 𝒔𝒐𝒍𝒗𝒆 𝒇𝒐𝒓𝓴
𝟏 −
𝒗
𝒄
𝟏 −
𝟏
𝓴 𝟐 − 𝟏
𝒄
𝒗
= 𝟏 (𝟏 −
𝟏
𝓴 𝟐
− 𝟏
𝒄
𝒗
)
𝟏 −
𝒗
𝒄
𝟏 −
𝟏
𝓴 𝟐 − 𝟏
𝒄
𝒗
= 𝟏(𝟏 −
𝟏
𝓴 𝟐
− 𝟏
𝒄
𝒗
)
𝟏 −
𝒗
𝒄
− 𝟏 = (𝟏 − 𝟏) −
𝟏
𝓴 𝟐
− 𝟏
𝒄
𝒗
)
𝟏 − 𝟏 −
𝒗
𝒄
= (𝟏 − 𝟏) −
𝟏
𝓴 𝟐
− 𝟏
𝒄
𝒗
)
𝒘𝒆 𝒆𝒒𝒖𝒂𝒕𝒆 𝒊𝒕 𝒕𝒐 𝟏 𝒂𝒏𝒅 𝒔𝒐𝒍𝒗𝒆 𝒇𝒐𝒓𝓴
𝟏 −
𝒗
𝒄
𝟏 −
𝟏
𝓴 𝟐 − 𝟏
𝒄
𝒗
= 𝟏
𝒗
𝒄
= −
𝟏
𝓴 𝟐
− 𝟏
𝒄
𝒗
)
𝒗
𝒄
[
𝒗
𝒄
= −
𝟏
𝓴 𝟐
− 𝟏
𝒄
𝒗
)]
𝒗
𝒄
𝒗𝟐
𝒄𝟐
= −
𝟏
𝓴 𝟐
− 𝟏
𝒗𝟐
𝒄𝟐
= −
𝟏
𝓴 𝟐
+ 𝟏
𝟏
𝓴 𝟐
= 𝟏 −
𝒗𝟐
𝒄𝟐
𝒘𝒆 𝒆𝒒𝒖𝒂𝒕𝒆 𝒊𝒕 𝒕𝒐 𝟏 𝒂𝒏𝒅 𝒔𝒐𝒍𝒗𝒆 𝒇𝒐𝒓𝓴
𝟏 −
𝒗
𝒄
𝟏 −
𝟏
𝓴 𝟐 − 𝟏
𝒄
𝒗
= 𝟏
𝟏
𝓴 𝟐
= 𝟏 −
𝒗𝟐
𝒄𝟐
𝟏 = (𝟏 −
𝒗𝟐
𝒄𝟐
)𝓴𝟐
𝟏
(𝟏 −
𝒗𝟐
𝒄𝟐)
= 𝓴𝟐
𝓴𝟐 = (
𝟏
(𝟏 −
𝒗𝟐
𝒄𝟐)
)
𝒘𝒆 𝒆𝒒𝒖𝒂𝒕𝒆 𝒊𝒕 𝒕𝒐 𝟏 𝒂𝒏𝒅 𝒔𝒐𝒍𝒗𝒆 𝒇𝒐𝒓𝓴
𝟏 −
𝒗
𝒄
𝟏 −
𝟏
𝓴 𝟐 − 𝟏
𝒄
𝒗
= 𝟏 𝓴𝟐 = (
𝟏
(𝟏 −
𝒗𝟐
𝒄𝟐)
)
𝓴 =
𝟏
(𝟏 −
𝒗𝟐
𝒄𝟐)
𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟑 𝒄𝒂𝒏 𝒏𝒐𝒘 𝒃𝒆 𝒘𝒓𝒊𝒕𝒕𝒆𝒏 𝒂𝒔
𝓴 =
𝟏
(𝟏 −
𝒗𝟐
𝒄𝟐)
𝒙′ =
𝟏
(𝟏 −
𝒗𝟐
𝒄𝟐)
(𝒙 − 𝒗𝒕)
𝒙′ = 𝓴(𝒙 − 𝒗𝒕) → 𝒆𝒒. 𝟑
𝒙′ =
(𝒙 − 𝒗𝒕)
(𝟏 −
𝒗𝟐
𝒄𝟐)
→ 𝒆𝒒. 𝟖
𝒘𝒆 𝒅𝒆𝒓𝒊𝒗𝒆 𝒇𝒐𝒓 𝒂𝒏 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒇𝒐𝒓 𝒕𝒊𝒎𝒆
𝒙′ =
(𝒙 − 𝒗𝒕)
(𝟏 −
𝒗𝟐
𝒄𝟐)
𝒄𝒕′ = 𝓴(𝒄𝒕 − 𝒗
𝒙
𝒄
)
𝒄𝒕′
𝒄
=
𝓴(𝒄𝒕 − 𝒗
𝒙
𝒄
)
𝒄
𝒕′ = 𝓴(𝒕 − 𝒗
𝒙
𝒄𝟐
)
𝒘𝒆 𝒅𝒆𝒓𝒊𝒗𝒆 𝒇𝒐𝒓 𝒂𝒏 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒇𝒐𝒓 𝒕𝒊𝒎𝒆
𝒕′ = 𝓴(𝒕 − 𝒗
𝒙
𝒄𝟐
)
𝒕′ =
𝒕 − 𝒗
𝒙
𝒄𝟐
(𝟏 −
𝒗𝟐
𝒄𝟐)
𝒕′ = 𝓴(𝒕 − 𝒗
𝒙
𝒄𝟐
)
𝒕𝒉𝒆 𝒇𝒍𝒐𝒘 𝒐𝒇 𝒕𝒊𝒎𝒆 𝒂𝒄𝒄𝒐𝒓𝒅𝒊𝒏𝒈 𝒕𝒐 𝑬𝒊𝒏𝒔𝒕𝒆𝒊𝒏
𝒕′ ≠ 𝒕
𝑬𝒊𝒏𝒔𝒕𝒆𝒊𝒏 𝒕𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎𝒆𝒅 𝒕𝒉𝒊𝒔 𝒊𝒏𝒕𝒐 𝒂 𝒈𝒆𝒐𝒎𝒆𝒕𝒓𝒚 𝒑𝒓𝒐𝒃𝒍𝒆𝒎
𝑳𝒐
𝒗(
𝒕
𝟐
)
(𝒄(
𝒕
𝟐
))
𝟐
= (𝑳𝒐)𝟐 + (𝒗(
𝒕
𝟐
) )𝟐
𝒄𝟐𝒕𝟐
𝟒
=
𝒄𝟐𝒕𝒐
𝟐
𝟒
+
𝒗𝟐𝒕𝟐
𝟒
𝒄𝒕𝒐
𝟐
= 𝑳𝒐 𝟏 𝒕𝒊𝒄𝒌 𝒖𝒑
(𝒄(
𝒕
𝟐
))
𝟐
= (
𝒄𝒕𝒐
𝟐
)𝟐+(𝒗(
𝒕
𝟐
) )𝟐
𝒄𝟐𝒕𝟐
𝟒
−
𝒗𝟐𝒕𝟐
𝟒
=
𝒄𝟐𝒕𝒐
𝟐
𝟒
𝑳𝒐
𝒗(
𝒕
𝟐
)
𝑵𝑶𝑾 𝑳𝑬𝑻𝑺 𝑺𝑶𝑳𝑽𝑬 𝑻𝑯𝑬 𝑷𝒀𝑻𝑯𝑨𝑮𝑶𝑹𝑬𝑨𝑵 𝑻𝑯𝑬𝑶𝑹𝑬𝑴
𝒄𝟐𝒕𝟐
𝟒
−
𝒗𝟐𝒕𝟐
𝟒
=
𝒄𝟐𝒕𝒐
𝟐
𝟒
𝒕𝟐
𝟒
𝒄𝟐 − 𝒗𝟐 =
𝒄𝟐
𝒕𝒐
𝟐
𝟒
𝟒
𝒕𝟐 𝒄𝟐 − 𝒗𝟐
𝟒𝒄𝟐
=
𝒄𝟐𝒕𝒐
𝟐
𝒄𝟐𝟒
𝟒
𝒕𝟐(𝟏 −
𝒗𝟐
𝒄𝟐
) = 𝒕𝒐
𝟐
𝑬𝒊𝒏𝒔𝒕𝒆𝒊𝒏 𝒕𝒊𝒎𝒆 𝑫𝒊𝒍𝒂𝒕𝒊𝒐𝒏 𝒇𝒐𝒓𝒎𝒖𝒍𝒂
𝑳𝒐
𝒗(
𝒕
𝟐
)
𝒕𝟐(𝟏 −
𝒗𝟐
𝒄𝟐
) = 𝒕𝒐
𝟐
𝒕𝟐
=
𝒕𝒐
𝟐
(𝟏 −
𝒗𝟐
𝒄𝟐)
𝒕 =
𝒕𝒐
(𝟏 −
𝒗𝟐
𝒄𝟐))

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Relativity

  • 2. 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒕𝒓𝒂𝒗𝒆𝒍𝒍𝒆𝒅 𝒃𝒚 𝒍𝒊𝒈𝒉𝒕 𝒂𝒔 𝒔𝒆𝒆𝒏 𝒇𝒓𝒐𝒎 𝒆𝒂𝒓𝒕𝒉 𝒊𝒔 , 𝒙 = 𝒄𝒕 𝒄𝒕 𝒕𝒉𝒆 𝒑𝒍𝒂𝒏𝒆 𝒊𝒔 𝒎𝒐𝒗𝒊𝒏𝒈 𝒘𝒊𝒕𝒉 𝒂 𝒔𝒑𝒆𝒆𝒅 𝒐𝒇 𝒔𝒊𝒈𝒏𝒊𝒇𝒊𝒄𝒂𝒏𝒕 𝑭𝑹𝑨𝑪𝑻𝑰𝑶𝑵 𝒐𝒇 𝒕𝒉𝒆 𝒔𝒑𝒆𝒆𝒅 𝒐𝒇 𝒍𝒊𝒈𝒉𝒕 𝒊𝒏 𝒉𝒊𝒔 𝒑𝒐𝒊𝒏𝒕 𝒐𝒇 𝒗𝒊𝒆𝒘 𝒕𝒉𝒆 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒕𝒓𝒂𝒗𝒆𝒍𝒍𝒆𝒅 𝒃𝒚 𝒍𝒊𝒈𝒉𝒕 𝒊𝒔, 𝒙′ = 𝒄𝒕 − 𝒗𝒕 𝒎𝒆𝒂𝒔𝒖𝒓𝒆𝒎𝒆𝒏𝒕𝒔 𝒅𝒐𝒏𝒆 𝒐𝒏 𝒔′ 𝒘𝒊𝒍𝒍 𝒃𝒆 𝒔𝒉𝒐𝒓𝒕𝒆𝒓 𝒃𝒚 𝒂 𝒇𝒂𝒄𝒕𝒐𝒓 𝒗𝒕 𝒕𝒉𝒊𝒔 𝒊𝒔 𝒕𝒉𝒆 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒕𝒓𝒂𝒗𝒆𝒍𝒍𝒆𝒅 𝒃𝒚 𝑺′ 𝒄𝒐𝒎𝒑𝒂𝒓𝒆𝒅 𝒕𝒐 𝒍𝒊𝒈𝒉𝒕 𝒄𝒐𝒏𝒔𝒊𝒅𝒆𝒓 𝒕𝒉𝒂𝒕 𝒕𝒉𝒆 𝒑𝒍𝒂𝒏𝒆 𝒊𝒔 𝒓𝒂𝒄𝒊𝒏𝒈 𝒂𝒈𝒂𝒊𝒏𝒕𝒔 𝒍𝒊𝒈𝒉𝒕
  • 3. 𝒙 = 𝒄𝒕 𝒙′ = 𝒄𝒕 − 𝒗𝒕 𝒙′ = 𝒙 − 𝒗𝒕 𝒗𝒕 𝒄𝒕
  • 4. 𝒙 = 𝒄𝒕 𝒙′ = 𝒙 − 𝒗𝒕 𝒙′ = 𝒙 − 𝒗𝒕 𝒗𝒕 𝒙 = 𝒄𝒕 𝒔𝒐 𝒕𝒐 𝒆𝒙𝒑𝒍𝒂𝒊𝒏 𝒐𝒏 𝒐𝒃𝒔𝒆𝒓𝒗𝒆𝒓 𝒐𝒏 𝒕𝒉𝒆 𝒑𝒍𝒂𝒏𝒆 𝒕𝒉𝒂𝒕 𝒍𝒊𝒈𝒉𝒕 𝒉𝒂𝒔 𝒓𝒆𝒂𝒍𝒚 𝒕𝒓𝒂𝒗𝒆𝒍𝒍𝒆𝒅 𝒄𝒕 𝒂𝒏𝒅 𝒉𝒆′𝒔 𝒋𝒖𝒔𝒕 𝒈𝒆𝒕𝒕𝒊𝒏𝒈 𝒇𝒂𝒔𝒕𝒆𝒓 𝒂𝒏𝒅 𝒂𝒍𝒔𝒐 𝒊𝒏 𝒉𝒊𝒔 𝒑𝒆𝒓𝒑𝒆𝒄𝒕𝒊𝒗𝒆 𝒕𝒉𝒆 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒕𝒓𝒂𝒗𝒆𝒍𝒍𝒆𝒅 𝒄𝒐𝒏𝒕𝒓𝒂𝒄𝒕𝒆𝒅 𝒂𝒏𝒅 𝒕𝒊𝒎𝒆 𝒊𝒔 𝒂𝒍𝒔𝒐 𝒅𝒊𝒍𝒂𝒕𝒆𝒅, 𝒘𝒆 𝒊𝒏𝒕𝒓𝒐𝒅𝒖𝒄𝒆 𝒕𝒉𝒆 𝒇𝒂𝒄𝒕𝒐𝒓 𝓴
  • 5. 𝒙 = 𝒄𝒕 → 𝒆𝒒. 𝟏 𝒙′ = 𝒙 − 𝒗𝒕 → 𝒆𝒒. 𝟐 𝒙′ = 𝒙 − 𝒗𝒕 𝒗𝒕 𝒙 = 𝒄𝒕 𝒘𝒆 𝒓𝒆𝒘𝒓𝒊𝒕𝒆 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟐 𝒂𝒔 𝒙′ = 𝓴(𝒙 − 𝒗𝒕) → 𝒆𝒒. 𝟑 𝒐𝒓 𝓴𝒙′ = (𝒙 − 𝒗𝒕) → 𝒆𝒒. 𝟒 𝒕𝒐 𝒂𝒄𝒄𝒐𝒖𝒏𝒕 𝒇𝒐𝒓 𝒕𝒉𝒆 𝒇𝒍𝒐𝒘 𝒐𝒇 𝒕𝒊𝒎𝒆 𝒊𝒏 𝒕𝒉𝒆 𝒑𝒆𝒓𝒑𝒆𝒄𝒕𝒊𝒗𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒐𝒃𝒔𝒆𝒓𝒗𝒆𝒓 𝒊𝒏 𝒕𝒉𝒆 𝒑𝒍𝒂𝒏𝒆 𝒘𝒆 𝒘𝒓𝒊𝒕𝒆, 𝓴𝒙′ = 𝒙 − 𝒗𝒕 𝓴𝒙′ − 𝒗𝒕′ = 𝒙 𝒙 = 𝓴𝒙′ − 𝒗𝒕′ ⟶ 𝒆𝒒. 𝟓
  • 6. 𝒙 = 𝒄𝒕 → 𝒆𝒒. 𝟏 𝒙′ = 𝒙 − 𝒗𝒕 → 𝒆𝒒. 𝟐 𝒙′ = 𝒙 − 𝒗𝒕 𝒗𝒕 𝒙 = 𝒄𝒕 𝒙′ = 𝓴(𝒙 − 𝒗𝒕) → 𝒆𝒒. 𝟑 𝓴𝒙′ = (𝒙 − 𝒗𝒕) → 𝒆𝒒. 𝟒 𝒙 = 𝓴𝒙′ − 𝒗𝒕′ ⟶ 𝒆𝒒. 𝟓 Laws of physics are the same for any reference frame, hence those on the opposite does not need a new equation but Only a changed in sign. We modify equation 5 . 𝒙 = 𝓴𝒙′ + 𝒗𝒕′ ⟶ 𝒆𝒒. 𝟔 𝒙′ = 𝒄𝒕′ → 𝒆𝒒. 𝟐. 𝟏 𝒃𝒚 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒕𝒊𝒂𝒍 𝒄𝒂𝒍𝒄𝒖𝒍𝒖𝒔
  • 7. 𝒙′ = 𝓴(𝒙 − 𝒗𝒕) → 𝒆𝒒. 𝟑 𝒙′ = 𝒙 − 𝒗𝒕 𝒗𝒕 𝒙 = 𝒄𝒕 𝒘𝒆 𝒄𝒂𝒏 𝒏𝒐𝒘 𝒔𝒐𝒍𝒗𝒆 𝒕𝒉𝒊𝒔 𝒕𝒘𝒐 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏𝒔 𝒔𝒊𝒎𝒖𝒍𝒕𝒂𝒏𝒆𝒐𝒖𝒔𝒍𝒚 𝒙 = 𝓴𝒙′ + 𝒗𝒕′ ⟶ 𝒆𝒒. 𝟔 𝒘𝒆 𝒔𝒖𝒃𝒔𝒕𝒊𝒕𝒕𝒖𝒕𝒆 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟑 𝒊𝒏𝒕𝒐 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟔 𝒕𝒐 𝒓𝒆𝒅𝒖𝒄𝒆 𝒕𝒉𝒆 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒗𝒂𝒓𝒊𝒂𝒃𝒍𝒆𝒔 𝒙 = 𝓴(𝓴(𝒙 − 𝒗𝒕) + 𝒗𝒕′) 𝒙 = 𝓴𝟐𝒙 − 𝓴𝟐𝒗𝒕 + 𝓴𝒗𝒕′
  • 8. 𝒘𝒆 𝒔𝒐𝒍𝒗𝒆 𝒇𝒐𝒓 𝒕′ 𝒙 = 𝓴𝟐𝒙 − 𝓴𝟐𝒗𝒕 + 𝒗𝒕′ 𝓴𝒗𝒕′ = 𝒙 − 𝓴𝟐𝒙 − 𝓴𝟐𝒗𝒕 𝓴𝒗𝒕′ = 𝒙(𝟏 − 𝓴𝟐) − 𝓴𝟐𝒗𝒕 𝓴𝒗𝒕′ 𝓴𝒗 = 𝒙(𝟏 − 𝓴𝟐 ) 𝓴𝒗 − 𝓴𝟐 𝒗𝒕 𝓴𝒗 𝓴𝒗𝒕′ 𝓴𝒗 = 𝒙(𝟏 − 𝓴𝟐) 𝓴𝒗 − 𝓴𝟐𝒗𝒕 𝓴𝒗 𝒕′ = 𝒙(𝟏 − 𝓴𝟐) 𝓴𝒗 − 𝓴𝒕
  • 9. 𝒘𝒆 𝒔𝒐𝒍𝒗𝒆 𝒇𝒐𝒓 𝓴 𝒖𝒔𝒊𝒏𝒈 𝒆𝒒 𝟐. 𝟏 𝒂𝒏𝒅 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟑 𝒕′ = 𝒙(𝟏 − 𝓴𝟐) 𝓴𝒗 − 𝓴𝒕 𝒙′ = 𝒄𝒕′ → 𝒆𝒒. 𝟐. 𝟏 𝒙′ = 𝒄[ 𝒙 𝟏 − 𝓴𝟐 𝓴𝒗 − 𝓴𝒕] 𝓴(𝒙 − 𝒗𝒕) = 𝒄[ 𝒙 𝟏 − 𝓴𝟐 𝓴𝒗 − 𝓴𝒕] 𝓴𝒙 − 𝓴𝒗𝒕 = 𝒄[ 𝒙 𝟏 − 𝓴𝟐 𝓴𝒗 − 𝓴𝒕]
  • 10. 𝒓𝒆𝒂𝒓𝒓𝒂𝒏𝒈𝒊𝒏𝒈 𝒕𝒆𝒓𝒎𝒔 𝓴𝒙 − 𝓴𝒗𝒕 = 𝒄[ 𝒙 𝟏 − 𝓴𝟐 𝓴𝒗 − 𝓴𝒕] 𝓴𝒙 − 𝓴𝒗𝒕 = 𝒄𝒙( 𝟏 − 𝓴𝟐 𝓴𝒗 ) − 𝒄𝓴𝒕 𝓴𝒙 − 𝓴𝒗𝒕 − 𝒄𝒙 𝟏 − 𝓴𝟐 𝓴𝒗 = 𝒄𝓴𝒕 𝓴𝒙 − 𝒄𝒙 𝟏 − 𝓴𝟐 𝓴𝒗 − 𝓴𝒗𝒕 = 𝒄𝓴𝒕
  • 11. 𝒘𝒆 𝒓𝒆𝒂𝒓𝒓𝒂𝒏𝒈𝒆 𝒕𝒆𝒓𝒎𝒔 𝒂𝒏𝒅 𝒔𝒐𝒍𝒗𝒆 𝒇𝒐𝒓 𝓴 𝓴𝒙 − 𝒄𝒙 𝟏 − 𝓴𝟐 𝓴𝒗 − 𝓴𝒗𝒕 = 𝒄𝓴𝒕 𝒙(𝓴 − 𝒄 𝟏 − 𝓴𝟐 𝓴𝒗 ) = 𝒄𝓴𝒕 − 𝓴𝒗𝒕 𝑤𝑒 𝑟𝑒𝑤𝑟𝑖𝑡𝑒 𝑡ℎ𝑒 𝑅𝐻𝑆 𝒙(𝓴 − 𝒄 𝟏 − 𝓴𝟐 𝓴𝒗 ) = 𝒄 𝒗 (𝓴𝒕 − 𝓴𝒕)
  • 12. 𝒘𝒆 𝒓𝒆𝒂𝒓𝒓𝒂𝒏𝒈𝒆 𝒕𝒆𝒓𝒎𝒔 𝒂𝒏𝒅 𝒔𝒐𝒍𝒗𝒆 𝒇𝒐𝒓 𝒙 𝒙(𝓴 − 𝒄 𝟏 − 𝓴𝟐 𝓴𝒗 ) = 𝒄 𝒗 (𝓴𝒕 − 𝓴𝒕) 𝒙 = 𝒄 𝒗 (𝓴𝒕 − 𝓴𝒕) (𝓴 − 𝒄 𝟏 − 𝓴𝟐 𝓴𝒗 )
  • 13. 𝒘𝒆 𝒎𝒂𝒏𝒊𝒑𝒖𝒍𝒂𝒕𝒆 𝒕𝒉𝒆 𝒏𝒖𝒎𝒆𝒓𝒂𝒕𝒐𝒓 𝒐𝒇 𝒕𝒉𝒆 𝑹𝑯𝑺 𝒙 = 𝒄 𝒗 (𝓴𝒕 − 𝓴𝒕) (𝓴 − 𝒄 𝟏 − 𝓴𝟐 𝓴𝒗 ) 𝒙 = 𝒄𝒕 (𝓴 − 𝒗 𝒄 𝓴) (𝓴 − 𝒄 𝟏 − 𝓴𝟐 𝓴𝒗 ) 𝒙 = 𝒄𝒕𝓴(𝟏 − 𝒗 𝒄 ) (𝓴 − 𝒄 𝟏 − 𝓴𝟐 𝓴𝒗 )
  • 14. 𝒘𝒆 𝒎𝒂𝒏𝒊𝒑𝒖𝒍𝒂𝒕𝒆 𝒕𝒉𝒆 𝒏𝒖𝒎𝒆𝒓𝒂𝒕𝒐𝒓 𝒐𝒇 𝒕𝒉𝒆 𝑹𝑯𝑺 𝒙 = 𝒄𝒕𝓴(𝟏 − 𝒗 𝒄 ) 𝓴 − 𝒄 𝟏 − 𝓴𝟐 𝓴𝒗 𝒙 = 𝒄𝒕(𝟏 − 𝒗 𝒄 ) 𝟏 − 𝒄 𝟏 − 𝓴𝟐 𝓴𝒗
  • 15. 𝒘𝒆 𝒎𝒂𝒏𝒊𝒑𝒖𝒍𝒂𝒕𝒆 𝒕𝒉𝒆 𝒅𝒆𝒏𝒐𝒎𝒊𝒏𝒂𝒕𝒐𝒓 𝒐𝒇 𝒕𝒉𝒆 𝑹𝑯𝑺 𝒙 = 𝒄𝒕(𝟏 − 𝒗 𝒄 ) 𝟏 − 𝒄 𝟏 − 𝓴𝟐 𝓴𝒗 𝟏 − 𝒄 𝟏 − 𝓴𝟐 𝓴𝒗 𝓴(𝟏 − 𝒄 𝟏 − 𝓴𝟐 𝓴𝒗 ) 𝓴 − 𝒄𝓴 𝒗 − 𝒄𝓴 𝟑 𝒗 𝓴 𝓴 − 𝒄𝓴 𝒗𝓴 − 𝒄𝓴 𝟑 𝒗𝓴 𝟏 − 𝒄 𝒗 − 𝒄𝓴 𝟐 𝒗
  • 16. 𝒘𝒆 𝒇𝒂𝒄𝒕𝒐𝒓 𝒐𝒖𝒕 𝒄 𝒗 𝒙 = 𝒄𝒕(𝟏 − 𝒗 𝒄 ) 𝟏 − 𝒄 𝟏 − 𝓴𝟐 𝓴𝒗 𝟏 − 𝒄 𝒗 − 𝒄𝓴 𝟐 𝒗 𝟏 − (𝟏 − 𝓴 𝟐 ) 𝒄 𝒗 𝟏 − ( 𝟏 𝓴 𝟐 − 𝟏) 𝒄 𝒗
  • 17. 𝒏𝒐𝒘 𝒘𝒆 𝒄𝒂𝒏 𝒘𝒓𝒊𝒕𝒆 𝒆𝒒. 𝟕 𝒙 = 𝒄𝒕(𝟏 − 𝒗 𝒄 ) 𝟏 − 𝒄 𝟏 − 𝓴𝟐 𝓴𝒗 → 𝒆𝒒. 𝟕 𝟏 − ( 𝟏 𝓴 𝟐 − 𝟏) 𝒄 𝒗 𝒙 = 𝒄𝒕(𝟏 − 𝒗 𝒄 ) 𝟏 − ( 𝟏 𝓴 𝟐 − 𝟏) 𝒄 𝒗 → 𝒆𝒒. 𝟕 𝒙 = 𝒄𝒕[ 𝟏 − 𝒗 𝒄 𝟏 − 𝟏 𝓴 𝟐 − 𝟏 𝒄 𝒗 ] → 𝒆𝒒. 𝟕
  • 18. 𝒕𝒉𝒊𝒔 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒘𝒊𝒍𝒍 𝒏𝒐𝒕 𝒗𝒊𝒐𝒍𝒂𝒕𝒆 𝒕𝒉𝒆 𝒍𝒂𝒘 𝒐𝒇 𝒑𝒉𝒚𝒔𝒊𝒄𝒔 𝒊𝒇 𝒕𝒉𝒆 𝒕𝒆𝒓𝒎 𝒊𝒏 𝒑𝒂𝒓𝒆𝒏𝒕𝒉𝒆𝒔𝒊𝒔 𝒊𝒔 𝒆𝒒𝒖𝒂𝒍 𝒕𝒐 𝟏 𝒙 = 𝒄𝒕[ 𝟏 − 𝒗 𝒄 𝟏 − 𝟏 𝓴 𝟐 − 𝟏 𝒄 𝒗 ] → 𝒆𝒒. 𝟕
  • 19. 𝒘𝒆 𝒆𝒒𝒖𝒂𝒕𝒆 𝒊𝒕 𝒕𝒐 𝟏 𝒂𝒏𝒅 𝒔𝒐𝒍𝒗𝒆 𝒇𝒐𝒓𝓴 𝟏 − 𝒗 𝒄 𝟏 − 𝟏 𝓴 𝟐 − 𝟏 𝒄 𝒗 = 𝟏 (𝟏 − 𝟏 𝓴 𝟐 − 𝟏 𝒄 𝒗 ) 𝟏 − 𝒗 𝒄 𝟏 − 𝟏 𝓴 𝟐 − 𝟏 𝒄 𝒗 = 𝟏(𝟏 − 𝟏 𝓴 𝟐 − 𝟏 𝒄 𝒗 ) 𝟏 − 𝒗 𝒄 − 𝟏 = (𝟏 − 𝟏) − 𝟏 𝓴 𝟐 − 𝟏 𝒄 𝒗 ) 𝟏 − 𝟏 − 𝒗 𝒄 = (𝟏 − 𝟏) − 𝟏 𝓴 𝟐 − 𝟏 𝒄 𝒗 )
  • 20. 𝒘𝒆 𝒆𝒒𝒖𝒂𝒕𝒆 𝒊𝒕 𝒕𝒐 𝟏 𝒂𝒏𝒅 𝒔𝒐𝒍𝒗𝒆 𝒇𝒐𝒓𝓴 𝟏 − 𝒗 𝒄 𝟏 − 𝟏 𝓴 𝟐 − 𝟏 𝒄 𝒗 = 𝟏 𝒗 𝒄 = − 𝟏 𝓴 𝟐 − 𝟏 𝒄 𝒗 ) 𝒗 𝒄 [ 𝒗 𝒄 = − 𝟏 𝓴 𝟐 − 𝟏 𝒄 𝒗 )] 𝒗 𝒄 𝒗𝟐 𝒄𝟐 = − 𝟏 𝓴 𝟐 − 𝟏 𝒗𝟐 𝒄𝟐 = − 𝟏 𝓴 𝟐 + 𝟏 𝟏 𝓴 𝟐 = 𝟏 − 𝒗𝟐 𝒄𝟐
  • 21. 𝒘𝒆 𝒆𝒒𝒖𝒂𝒕𝒆 𝒊𝒕 𝒕𝒐 𝟏 𝒂𝒏𝒅 𝒔𝒐𝒍𝒗𝒆 𝒇𝒐𝒓𝓴 𝟏 − 𝒗 𝒄 𝟏 − 𝟏 𝓴 𝟐 − 𝟏 𝒄 𝒗 = 𝟏 𝟏 𝓴 𝟐 = 𝟏 − 𝒗𝟐 𝒄𝟐 𝟏 = (𝟏 − 𝒗𝟐 𝒄𝟐 )𝓴𝟐 𝟏 (𝟏 − 𝒗𝟐 𝒄𝟐) = 𝓴𝟐 𝓴𝟐 = ( 𝟏 (𝟏 − 𝒗𝟐 𝒄𝟐) )
  • 22. 𝒘𝒆 𝒆𝒒𝒖𝒂𝒕𝒆 𝒊𝒕 𝒕𝒐 𝟏 𝒂𝒏𝒅 𝒔𝒐𝒍𝒗𝒆 𝒇𝒐𝒓𝓴 𝟏 − 𝒗 𝒄 𝟏 − 𝟏 𝓴 𝟐 − 𝟏 𝒄 𝒗 = 𝟏 𝓴𝟐 = ( 𝟏 (𝟏 − 𝒗𝟐 𝒄𝟐) ) 𝓴 = 𝟏 (𝟏 − 𝒗𝟐 𝒄𝟐)
  • 23. 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟑 𝒄𝒂𝒏 𝒏𝒐𝒘 𝒃𝒆 𝒘𝒓𝒊𝒕𝒕𝒆𝒏 𝒂𝒔 𝓴 = 𝟏 (𝟏 − 𝒗𝟐 𝒄𝟐) 𝒙′ = 𝟏 (𝟏 − 𝒗𝟐 𝒄𝟐) (𝒙 − 𝒗𝒕) 𝒙′ = 𝓴(𝒙 − 𝒗𝒕) → 𝒆𝒒. 𝟑 𝒙′ = (𝒙 − 𝒗𝒕) (𝟏 − 𝒗𝟐 𝒄𝟐) → 𝒆𝒒. 𝟖
  • 24. 𝒘𝒆 𝒅𝒆𝒓𝒊𝒗𝒆 𝒇𝒐𝒓 𝒂𝒏 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒇𝒐𝒓 𝒕𝒊𝒎𝒆 𝒙′ = (𝒙 − 𝒗𝒕) (𝟏 − 𝒗𝟐 𝒄𝟐) 𝒄𝒕′ = 𝓴(𝒄𝒕 − 𝒗 𝒙 𝒄 ) 𝒄𝒕′ 𝒄 = 𝓴(𝒄𝒕 − 𝒗 𝒙 𝒄 ) 𝒄 𝒕′ = 𝓴(𝒕 − 𝒗 𝒙 𝒄𝟐 )
  • 25. 𝒘𝒆 𝒅𝒆𝒓𝒊𝒗𝒆 𝒇𝒐𝒓 𝒂𝒏 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒇𝒐𝒓 𝒕𝒊𝒎𝒆 𝒕′ = 𝓴(𝒕 − 𝒗 𝒙 𝒄𝟐 ) 𝒕′ = 𝒕 − 𝒗 𝒙 𝒄𝟐 (𝟏 − 𝒗𝟐 𝒄𝟐)
  • 26. 𝒕′ = 𝓴(𝒕 − 𝒗 𝒙 𝒄𝟐 ) 𝒕𝒉𝒆 𝒇𝒍𝒐𝒘 𝒐𝒇 𝒕𝒊𝒎𝒆 𝒂𝒄𝒄𝒐𝒓𝒅𝒊𝒏𝒈 𝒕𝒐 𝑬𝒊𝒏𝒔𝒕𝒆𝒊𝒏 𝒕′ ≠ 𝒕
  • 27. 𝑬𝒊𝒏𝒔𝒕𝒆𝒊𝒏 𝒕𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎𝒆𝒅 𝒕𝒉𝒊𝒔 𝒊𝒏𝒕𝒐 𝒂 𝒈𝒆𝒐𝒎𝒆𝒕𝒓𝒚 𝒑𝒓𝒐𝒃𝒍𝒆𝒎 𝑳𝒐 𝒗( 𝒕 𝟐 ) (𝒄( 𝒕 𝟐 )) 𝟐 = (𝑳𝒐)𝟐 + (𝒗( 𝒕 𝟐 ) )𝟐 𝒄𝟐𝒕𝟐 𝟒 = 𝒄𝟐𝒕𝒐 𝟐 𝟒 + 𝒗𝟐𝒕𝟐 𝟒 𝒄𝒕𝒐 𝟐 = 𝑳𝒐 𝟏 𝒕𝒊𝒄𝒌 𝒖𝒑 (𝒄( 𝒕 𝟐 )) 𝟐 = ( 𝒄𝒕𝒐 𝟐 )𝟐+(𝒗( 𝒕 𝟐 ) )𝟐 𝒄𝟐𝒕𝟐 𝟒 − 𝒗𝟐𝒕𝟐 𝟒 = 𝒄𝟐𝒕𝒐 𝟐 𝟒
  • 28. 𝑳𝒐 𝒗( 𝒕 𝟐 ) 𝑵𝑶𝑾 𝑳𝑬𝑻𝑺 𝑺𝑶𝑳𝑽𝑬 𝑻𝑯𝑬 𝑷𝒀𝑻𝑯𝑨𝑮𝑶𝑹𝑬𝑨𝑵 𝑻𝑯𝑬𝑶𝑹𝑬𝑴 𝒄𝟐𝒕𝟐 𝟒 − 𝒗𝟐𝒕𝟐 𝟒 = 𝒄𝟐𝒕𝒐 𝟐 𝟒 𝒕𝟐 𝟒 𝒄𝟐 − 𝒗𝟐 = 𝒄𝟐 𝒕𝒐 𝟐 𝟒 𝟒 𝒕𝟐 𝒄𝟐 − 𝒗𝟐 𝟒𝒄𝟐 = 𝒄𝟐𝒕𝒐 𝟐 𝒄𝟐𝟒 𝟒 𝒕𝟐(𝟏 − 𝒗𝟐 𝒄𝟐 ) = 𝒕𝒐 𝟐
  • 29. 𝑬𝒊𝒏𝒔𝒕𝒆𝒊𝒏 𝒕𝒊𝒎𝒆 𝑫𝒊𝒍𝒂𝒕𝒊𝒐𝒏 𝒇𝒐𝒓𝒎𝒖𝒍𝒂 𝑳𝒐 𝒗( 𝒕 𝟐 ) 𝒕𝟐(𝟏 − 𝒗𝟐 𝒄𝟐 ) = 𝒕𝒐 𝟐 𝒕𝟐 = 𝒕𝒐 𝟐 (𝟏 − 𝒗𝟐 𝒄𝟐) 𝒕 = 𝒕𝒐 (𝟏 − 𝒗𝟐 𝒄𝟐))