Derivation of E=mC2

Iraides Belandria

¿How did Einstein derived that E=mC2?. ¿ Can I find an english translation of his original paper?.

Rebel

Quote by Iraides Belandria

¿How did Einstein derived that E=mC2?. ¿ Can I find an english translation of his original paper?.

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)

dextercioby

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.

Iraides Belandria

thanks Rebel and dextercioby for the required information

zeronem

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]

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Iraides Belandria	Derivation of E=mC2 Tweet ¿How did Einstein derived that E=mC2?. ¿ Can I find an english translation of his original paper?.

dextercioby Blog Entries: 9 Recognitions: Homework Help Science Advisor	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.

Iraides Belandria	Derivation of E=mC2 thanks Rebel and dextercioby for the required information

zeronem	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]