查看完整版本 : Advanced Derivation of E=mc^2

LT3648 2018-9-12 01:34 PM

[url=http://www.emc2-explained.info/Emc2/Deriving.htm#.W5iJafmuyUk]http://www.emc2-explained.info/Emc2/Deriving.htm#.W5iJafmuyUk[/url]

[img]http://www.emc2-explained.info/Emc2/Deriving_htm_files/2024.jpg[/img]

[size=5][color=#0000ff][b]更新：[/b][/color][/size]

K=m_0v^2(1-v^2/c^2)^-1/2 - m_0∫v(1-v^2/c^2)^-1/2 dv
let u = 1-v^2/c^2
du = -2v/c^2 dv
dv = c^2/-2v du
K=m_0v^2(1-v^2/c^2)^-1/2 - m_0∫v(u)^-1/2 dv
K=m_0v^2(1-v^2/c^2)^-1/2 - m_0∫v(u)^-1/2 c^2/-2v du
K=m_0v^2(1-v^2/c^2)^-1/2 - m_0 [2(u)^1/2 c^2/-2]
K=m_0v^2(1-v^2/c^2)^-1/2 + m_0c^2 [(u)^1/2]
K=m_0v^2(1-v^2/c^2)^-1/2 + m_0c^2 [(1-v^2/c^2)^1/2]
K=m_0v^2(1-v^2/c^2)^-1/2 + m_0c^2 [(1-v^2/c^2)(1-v^2/c^2)^-1/2]
K=m_0v^2(1-v^2/c^2)^-1/2 +[size=4][color=#0000ff][b] [(m_0c^2-m_0v^2)(1-v^2/c^2)^-1/2] [/b][/color][/size]v上下限=v,0
K=m_0v^2(1-v^2/c^2)^-1/2 + (m_0c^2-m_02v^2)(1-v^2/c^2)^-1/2 - m_0c^2
K=m_0c^2(1-v^2/c^2)^-1/2 - m_0c^2
K=m_c^2 - m_0c^2

[[i] 本帖最後由 LT3648 於 2019-2-13 11:28 AM 編輯 [/i]]

Zzlaz 2018-9-12 05:05 PM

use substitution ja wor

LT3648 2018-9-12 05:13 PM

[quote]原帖由 [i]Zzlaz[/i] 於 2018-9-12 05:05 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487209948&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]
use substitution ja wor [/quote]

topochu 2018-9-12 08:48 PM

[quote]原帖由 [i]LT3648[/i] 於 2018-9-12 01:34 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487198360&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]
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[url=http://www.emc2-explained.inf][img]http://www.emc2-explained.info/Emc2/Deriving_htm_files/2024.jpg[/img][/url] ... [/quote]

提示 1: 先做後面嘅積分，設 u = 1 - v^2/c^2 ，搵 du 。

Zzlaz 2018-9-12 09:22 PM

[quote]原帖由 [i]LT3648[/i] 於 2018-9-12 05:13 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487210414&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]

[attach]8763381[/attach]

LT3648 2018-9-12 09:22 PM

[quote]原帖由 [i]topochu[/i] 於 2018-9-12 08:48 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487222560&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]

提示 1: 先做後面嘅積分，設 u = 1 - v^2/c^2 ，搵 du 。

du = -2v/c^2 dv

LT3648 2018-9-12 09:27 PM

[quote]原帖由 [i]Zzlaz[/i] 於 2018-9-12 09:22 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487224454&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]

8763381

Zzlaz 2018-9-12 09:30 PM

[quote]原帖由 [i]LT3648[/i] 於 2018-9-12 09:27 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487224762&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]

LT3648 2018-9-12 09:38 PM

[quote]原帖由 [i]Zzlaz[/i] 於 2018-9-12 09:30 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487224908&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]

[[i] 本帖最後由 LT3648 於 2018-9-12 09:41 PM 編輯 [/i]]

topochu 2018-9-12 09:47 PM

[quote]原帖由 [i]LT3648[/i] 於 2018-9-12 09:22 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487224499&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]

du = -2v/c^2 dv [/quote]

[[i] 本帖最後由 topochu 於 2018-9-12 09:57 PM 編輯 [/i]]

Zzlaz 2018-9-12 09:56 PM

[quote]原帖由 [i]LT3648[/i] 於 2018-9-12 09:38 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487225403&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]

LT3648 2018-9-12 10:16 PM

u = 1 - v^2/c^2
du = -2v/c^2 dv
m_0 ∫ v dv/sqrt(1 - v^2/c^2) 寫做 du 同 u
m_0 ∫ [v/(u^2)(-2v/c^2)] du
m_0c^2 ∫ -2u^-2 du
m_0c^2 [2u^-1]

topochu 2018-9-12 10:18 PM

[quote]原帖由 [i]LT3648[/i] 於 2018-9-12 10:16 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487227661&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]

u = 1 - v^2/c^2
du = -2v/c^2 dv
m_0 ∫ v dv/sqrt(1 - v^2/c^2) 寫做 du 同 u
[color=#ff0000]m_0 ∫ [v/(u^2)(-2v/c^2)] du[/color]
m_0c^2 ∫ -2u^-2 du
m_0c^2 [2u^-1] [/quote]

LT3648 2018-9-12 10:19 PM

[quote]原帖由 [i]Zzlaz[/i] 於 2018-9-12 09:56 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487226399&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]

LT3648 2018-9-12 10:25 PM

[quote]原帖由 [i]topochu[/i] 於 2018-9-12 10:18 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487227810&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]

m_0 c^2 / 2 ∫ u^-1/2  du
m_0 c^2 / 2 [ 2 u^1/2]
m_0 c^2 [ u^1/2]

topochu 2018-9-12 10:31 PM

[quote]原帖由 [i]LT3648[/i] 於 2018-9-12 10:25 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487228231&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]

m_0 c^2 / 2 ∫ u^-1/2  du
m_0 c^2 / 2 [ 2 u^1/2]
m_0 c^2 [ u^1/2] [/quote]

topochu 2018-9-12 10:39 PM

[quote]原帖由 [i]topochu[/i] 於 2018-9-12 10:31 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487228625&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]

LT3648 2018-9-12 10:39 PM

[quote]原帖由 [i]topochu[/i] 於 2018-9-12 10:31 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487228625&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]

LT3648 2018-9-13 01:22 PM

[quote]原帖由 [i]topochu[/i] 於 2018-9-12 10:31 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487228625&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]

m_0 c^2 / 2 ∫ u^-1/2  du
m_0 c^2 / 2 [ 2 u^1/2]
m_0 c^2 [ u^1/2]

m_0 c^2 {[(1 - v'^2/c^2)^1/2] (上、下限係 v' = v ，同埋 v' = 0)
m_0 c^2 {[(1 - v^2/c^2)^1/2]  -  [ (1 - 0^2/c^2)^1/2]}
m_0 c^2 [(1 - v^2/c^2)^1/2  - (1)]
m_0 c^2 (1-v^2/c^2)^1/2  - m_0 c^2
m c^2 - m_0 c^2

[[i] 本帖最後由 LT3648 於 2018-9-13 04:21 PM 編輯 [/i]]

topochu 2018-9-13 04:26 PM

[quote]原帖由 [i]LT3648[/i] 於 2018-9-13 01:22 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487256928&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]
m_0 c^2 / 2 ∫ u^-1/2  du
m_0 c^2 / 2 [ 2 u^1/2]
m_0 c^2 [ u^1/2]

m_0 c^2 {[(1 - v'^2/c^2)^1/2] (上、下限係 v' = v ，同埋 v' = 0)
m_0 c^2 {[(1 - v^2/c^2)^1/2]  -  [ (1 - 0^2/c^2)^1/2]}
m_0 c^2 [(1 - v^2/c^2)^1/2  - (1)]
m_0 c^2 (1-v^2/c^2)^1/2  - m_0 c^2
[color=#ff0000]m c^2 - m_0 c^2[/color]

LT3648 2018-9-13 04:41 PM

[quote]原帖由 [i]topochu[/i] 於 2018-9-13 04:26 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487266228&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]

topochu 2018-9-13 04:51 PM

[quote]原帖由 [i]LT3648[/i] 於 2018-9-13 04:41 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487267021&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]

[color=#0000ff]m_0 v^2/(1-v^2/c^2)^1/2 + m_0 c^2 (1-v^2/c^2)^1/2[/color] - m_0 c^2
[color=#0000ff]m_0/(1-v^2/c^2)^1/2 [ v^2 + c^2 (1 - v^2/c^2) ][/color] - m_0 c^2
[color=#0000ff]m_0 c^2/(1-v^2/c^2)^1/2[/color] - m_0 c^2

LT3648 2018-9-13 05:33 PM

[quote]原帖由 [i]topochu[/i] 於 2018-9-13 04:51 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487267560&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]

m_0 v^2/(1-v^2/c^2)^1/2 + m_0 c^2 (1-v^2/c^2)^1/2 - m_0 c^2
m_0/(1-v^2/c^2)^1/2 [ v^2 + c^2 (1 - v^2/c^2) ] - m_0 c^2
m_0 c^2/(1-v^2 ... [/quote]

topochu 2018-9-13 05:49 PM

[quote]原帖由 [i]LT3648[/i] 於 2018-9-13 05:33 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487269653&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]

LT3648 2018-9-13 05:55 PM

[quote]原帖由 [i]topochu[/i] 於 2018-9-13 05:49 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487270481&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]

[[i] 本帖最後由 LT3648 於 2018-9-13 05:56 PM 編輯 [/i]]

topochu 2018-9-13 06:14 PM

[quote]原帖由 [i]LT3648[/i] 於 2018-9-13 05:55 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487270734&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]

[color=#006400][b]第一步[/b][/color]
[quote]原帖由 LT3648 於 2018-9-12 09:22 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487224499&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]

[color=#0000ff]設 u = 1 - v^2/c^2 ，搵 du
du = -2v/c^2 dv[/color] [/quote]

[color=#556b2f][b]第二步[/b][/color]
[quote]原帖由 LT3648 於 2018-9-12 10:16 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487227661&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]

u = 1 - v^2/c^2
du = -2v/c^2 dv
[color=#0000ff]m_0 ∫ v dv/sqrt(1 - v^2/c^2) 寫做 du 同 u[/color]
m_0 ∫ [v/(u^2)(-2v/c^2)] du
m_0c^2 ∫ -2u^-2 du
m_0c^2 [2u^-1] [/quote]

[quote]原帖由 LT3648 於 2018-9-12 10:25 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487228231&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]
[color=#0000ff]m_0 c^2 / 2 ∫ u^-1/2  du
m_0 c^2 / 2 [ 2 u^1/2]
m_0 c^2 [ u^1/2][/color] [/quote]

[color=#006400][b]第三步[/b][/color]
[quote]原帖由 LT3648 於 2018-9-13 01:22 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487256928&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]
m_0 c^2 / 2 ∫ u^-1/2  du
m_0 c^2 / 2 [ 2 u^1/2]
m_0 c^2 [ u^1/2]

[color=#0000ff]m_0 c^2 {[(1 - v'^2/c^2)^1/2] (上、下限係 v' = v ，同埋 v' = 0)
m_0 c^2 {[(1 - v^2/c^2)^1/2]  -  [ (1 - 0^2/c^2)^1/2]}
m_0 c^2 [(1 - v^2/c^2)^1/2  - (1)]
m_0 c^2 (1-v^2/c^2)^1/2  - m_0 c^2   [/color]
m c^2 - m_0 c^2

[color=#006400][b]第四步[/b][/color]

LT3648 2018-9-13 06:26 PM

[quote]原帖由 [i]topochu[/i] 於 2018-9-13 06:14 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487271563&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]

[[i] 本帖最後由 LT3648 於 2018-9-15 11:28 AM 編輯 [/i]]

LT3648 2018-9-14 11:23 AM

[quote]原帖由 [i]Zzlaz[/i] 於 2018-9-12 09:30 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487224908&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]

[[i] 本帖最後由 LT3648 於 2018-9-15 11:20 AM 編輯 [/i]]

LT3648 2018-9-25 01:13 PM

[quote]原帖由 [i]LT3648[/i] 於 2018-9-14 11:23 AM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487302138&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]

[img]http://www.emc2-explained.info/Emc2/Deriving_htm_files/2024.jpg[/img]

K=m_0v^2(1-v^2/c^2)^-1/2 - m_0∫v(1-v^2/c^2)^-1/2 dv
let u = 1-v^2/c^2
du = -2v/c^2 dv
dv = c^2/-2v du
K=m_0v^2(1-v^2/c^2)^-1/2 - m_0∫v(u)^-1/2 dv
K=m_0v^2(1-v^2/c^2)^-1/2 - m_0∫v(u)^-1/2 c^2/-2v du
K=m_0v^2(1-v^2/c^2)^-1/2 - m_0 [2(u)^1/2 c^2/-2]
K=m_0v^2(1-v^2/c^2)^-1/2 + m_0c^2 [(u)^1/2]
K=m_0v^2(1-v^2/c^2)^-1/2 + m_0c^2 [(1-v^2/c^2)^1/2]
K=m_0v^2(1-v^2/c^2)^-1/2 + m_0c^2 [(1-v^2/c^2)(1-v^2/c^2)^-1/2]
K=m_0v^2(1-v^2/c^2)^-1/2 + [color=Blue][(m_0c^2-m_0v^2)(1-v^2/c^2)^-1/2][/color] v上下限=v,0
K=m_0v^2(1-v^2/c^2)^-1/2 + (m_0c^2-m_02v^2)(1-v^2/c^2)^-1/2 - m_0c^2
K=m_0c^2(1-v^2/c^2)^-1/2 - m_0c^2
K=m_c^2 - m_0c^2

[color=Blue]結論是：題目的第二行第二項印錯囉！:smile_03: [/color]

topochu 2018-9-25 06:27 PM

[quote]原帖由 [i]LT3648[/i] 於 2018-9-25 01:13 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=487893476&ptid=27709567][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]

[attach]8812288[/attach]

K=m_0v^2(1-v^2/c^2)^-1/2 - m_0∫v(1-v^2/c^2)^-1/2 dv
let u = 1-v^2/c^2
du = -2v/c^2 dv
dv = c^2/-2v du
K=m_0v^2(1-v^2/c^2)^-1/2 - m_0∫v(u)^-1/2 dv
K=m_0v^2(1-v^2/c^2)^-1/2 - m_0∫v(u)^-1/2 c^2/-2v du
K=m_0v^2(1-v^2/c^2)^-1/2 - m_0 [2(u)^1/2 c^2/-2]
K=m_0v^2(1-v^2/c^2)^-1/2 + m_0c^2 [(u)^1/2]
K=m_0v^2(1-v^2/c^2)^-1/2 + m_0c^2 [(1-v^2/c^2)^1/2]
K=m_0v^2(1-v^2/c^2)^-1/2 + m_0c^2 [(1-v^2/c^2)(1-v^2/c^2)^-1/2]
K=m_0v^2(1-v^2/c^2)^-1/2 + [color=Blue][(m_0c^2-m_0v^2)(1-v^2/c^2)^-1/2][/color] v上下限=v,0
K=m_0v^2(1-v^2/c^2)^-1/2 + (m_0c^2-m_02v^2)(1-v^2/c^2)^-1/2 - m_0c^2
K=m_0c^2(1-v^2/c^2)^-1/2 - m_0c^2
K=m_c^2 - m_0c^2

[color=Blue]結論是：題目的第二行第二項印錯囉！:smile_03:[/color] ... [/quote]

[[i] 本帖最後由 topochu 於 2018-9-25 06:33 PM 編輯 [/i]]