Einstein's Derivation of e=mc^2: Urgent Help Needed

[uraknai]

hi,

This has probably been asked and answered a million times before so sorry but here goes. I urgently need help with a Maths University project about the A-Bomb which will include a chapter on Special Relativity and e=mc^2. I have read loads of books but they treat special relativity from a modern view point whereas I need to know how Einstein "figured out" e=mc^2 as it's a projcent on the impact and development of maths. Can anyone explaine how Einstein derived e=mc^2 or recommend and books/websites. Also, does anyone have any useful mathematical info on the development of the A-Bomb that might help?

Thanks

 

[Doc Al]

Here are two sites to get you started. The first is an entry on mass-energy equivalence in the Stanford Encyclopedia of Philosophy (a good resource):
http://setis.library.usyd.edu.au/stanford/archives/fall2001/entries/equivME/

The second is a translation of Einstein's orginal (1905) derivation of E = mc^2: http://www.fourmilab.ch/etexts/einstein/E_mc2/www/

Have fun!

 

[Mk]

Where did E^2=m^2c^4+p^2^2 come from?

And why can't users view their own "warnings?"

 

[Tide]

Actually, it's

$$E^2 = m_0^2 c^4 + p^2 c^2$$

and it is the same as $$E = mc^2$$ which uses the "relativisitc mass" whereas the previous expression uses the "proper mass."

 

[Mk]

Ummm, which one is rest mass and which one is not? Rest mass is proper mass? Right?

 

[jtbell]

Actually, it's

$$E^2 = m_0^2 c^4 + p^2 c^2$$

...and you can derive it by starting with the relativistic equations for energy and momentum:

$$E=\frac {m_0 c^2} { \sqrt {1 - \frac {v^2} {c^2}}}$$

$$p=\frac {m_0 v} { \sqrt {1 - \frac {v^2} {c^2}}}$$

and combining them so as to eliminate v.

Rest mass is proper mass? Right?

Right. It's also known as "invariant mass".