Warp wrote:
> andrel <a_l### [at] hotmailcom> wrote:
>> Also not sure what Mueen means, but the m in E=mc^2 is the m that was 
>> used by einstein. IIRC the current definition would require a division 
>> by sqrt(1-v^2/c^2). The old definition was certainly not quantified for 
>> arbitrary velocities.
> 
>   I really can't understand what you are talking about. The 'm' which
> Einstein used (and others before him) is what is currently used. There's
> no "current definition of m".
> 
>   Don't confuse the 'm' in the E=mc^2 with 'm0' (m subscript zero), which
> is the rest mass of an object. 
Confuse implies that I don't know what I am talking about, so be careful 
with that word.
> The 'm' in E=mc^2 is the relativistic mass,
Correct.
> and equal to m0/sqrt(1-v^2/c^2). That was the definition back then, and
> that's the definition today. It hasn't changed.

Not entirely true. If we talk about the mass of a particle we mean the 
mass at rest, not the relativistic mass. Also when I studied physics we 
did nearly always write m where you would claim that we should have used 
m0. It is one of those cases where physicists are too lazy to use 
subscripts (at the minor expense of having to write various powers of 
that square root thing ;) ). The only time we used m0 is in exactly the 
equation that you quoted. After that it was 'and from here on m means 
the rest mass'. So we did write Einsteins formula as 
E=mc^2/sqrt(1-v^2/c^2) and I still would do that, even if you think that 
is wrong.

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