¿How did Einstein derived that E=mC2?. ¿ Can I find an english translation of his original paper?.
May 17, 2005 #1 Iraides Belandria 55 0 ¿How did Einstein derived that E=mC2?. ¿ Can I find an english translation of his original paper?.
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May 17, 2005 #2 Rebel 53 0 Iraides Belandria said: ¿How did Einstein derived that E=mC2?. ¿ Can I find an english translation of his original paper?. Click to expand... See http://www.fourmilab.ch/etexts/einstein/E_mc2/e_mc2.pdf (there is translated in english the second article Einstein published in Annalen der physik about special relativity -thanks dex, i missed last- in german) Last edited: May 17, 2005
Iraides Belandria said: ¿How did Einstein derived that E=mC2?. ¿ Can I find an english translation of his original paper?. Click to expand... See http://www.fourmilab.ch/etexts/einstein/E_mc2/e_mc2.pdf (there is translated in english the second article Einstein published in Annalen der physik about special relativity -thanks dex, i missed last- in german)
May 17, 2005 #3 dextercioby Science Advisor Homework Helper Insights Author 12,962 536 It was his second article on SR in 1905 and the IV-th overall in that year. http://www.aip.org/history/einstein/chron-1905.htm Daniel.
It was his second article on SR in 1905 and the IV-th overall in that year. http://www.aip.org/history/einstein/chron-1905.htm Daniel.
May 19, 2005 #5 zeronem 110 1 You can simply take this integral and you'll get [tex] E = mc^2 [/tex] [tex] \int^c_0{P dv} ; [/tex] where [tex] P = mv\gamma [/tex] and [tex] \gamma = \frac{1}{\sqrt{1 - (\frac{v}{c})^2}} [/tex] [tex] \int^c_0 {\frac{mv}{\sqrt{1 - (\frac{v}{c})^2}} dv = mc^2 [/tex]
You can simply take this integral and you'll get [tex] E = mc^2 [/tex] [tex] \int^c_0{P dv} ; [/tex] where [tex] P = mv\gamma [/tex] and [tex] \gamma = \frac{1}{\sqrt{1 - (\frac{v}{c})^2}} [/tex] [tex] \int^c_0 {\frac{mv}{\sqrt{1 - (\frac{v}{c})^2}} dv = mc^2 [/tex]