Re: Past, present and future of E=mc2: A CRITICAL DISCUSSION
- From: "You smiled, you spoke, and I believed" <nospam@xxxxxxxxxx>
- Date: Wed, 25 Oct 2006 03:56:20 GMT
hhc314@xxxxxxxxx wrote:
Actually, the derivation is rather trivial, when you start with good
old Newton. (Most physicists study this by 2nd year college physics.
The transition from F = ma to E = mc^2 takes up only half a page in
most textbooks, and involves only simple calculus.
See for example page 132 of "Physics of the Atom", Wehr and Richards,
Addison-Wesley, 1960.
It begins by secaming the result of kinetic energy change as the result
of the application of a force over a distance.
dE = Fds
Next, comes Newtons famious F = ma, expressed in its general terms.
F = d(mv)/dt
The change in kinetic energy is given:
dE = d(mv)/dt X ds
But since dv/dt = v, we can write this as:
dE = v X d(mv) = v^2 dn + m dv
Now add in the Lorentz transformation:
m = km(o) = m(0) /(sqrt)(1 - v^2/c^2)
Then rearrrangine after squaring becomes:
m^2 c^2 = m^2 v^2 + m(0)^2 c^2
We differentiate, noting that both m(0) and c are contants to obtain:
2mc^2 dm = 2mv^2 dm + 2 m^2v dv
When 2m is divided out of this equation, we obtain:
dE = c^2 dm
Which put into middle school terminology is:
E = mc^2
Isn't it amazing how far we can go using only Newton, the Lorenze
tranform, and a bit of elementary calculus!
Physicist learn this as basic in their field of study. Then too, one
may wonder about how valid is the Lorentz transform. All I can say is
that it has been tested many thousands of times in nuclear particle
accelerators and other applications, and it has been found to work on
an empirical basis.
Harry C.
AJAY SHARMA wrote:
Past, present and future of E=mc2: A CRITICAL DISCUSSION
The link with detailed discussion
(i) How and When E=mc2 was derived ?
(ii) What are conditions of derivation ?
(iii) What is justification of dE =Ac2dm ?
and FAQs on the topic kindly see link
http://www.sulekha.com/blogs/blogdisplay.aspx?contributor=physicsajay
excellent answer
j.
.
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