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14 Feb 2008 14:12:44 IST

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can you give derivation to e=mc.c

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can you give derivation to e=mc.c

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anchit saini

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14 Feb 2008 14:30:45 IST

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it is out of scope but still-
K.E.=integralF.dx
=integral F.dx/dt *dt
=integral Fvdt
=integral vdP
=integral v*d(mv)
=integral v*d(m0*v/(1-v square/c square)) where m0=restmass
relation -- m=m0/(1-vsquare/c square)
integrating it from 0 to v,
we get-
K.E.=mcsquare-m0csquare
or
energy in motion=mcsquare- energy at rest
energy in motion + energy at rest= total energy
=E=mcsquare

puneet

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14 Feb 2008 14:33:28 IST

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hii

well here is a derivation tat i read in my class 11 .. though i donot find this one very convincing myself .. but i was never able to find anything better ...

so we knw few things .. first is that lamda = h/p

now evergy = hc/lamda ... so energy = c.p ..

put p = mc u get answer .

cheers

The_Emperor

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15 Feb 2008 04:36:29 IST

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Sir this relation is used to derive lamda=h/p..........itself it was used by debroglie himself corelating einstien and planks equation........atleast thats wat the book says...........

Abhijith

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15 Feb 2008 16:11:35 IST

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Well this is the original proof....but this isnt included in JEE syllabi, so you dont really have to worry about it too much...Einstein considered a body at rest with mass M. If the body is examined in a frame moving with nonrelativistic velocity v, it is no longer at rest and in the moving frame it has momentum Mv...

Suppose now that the body emits two pulses of light to the left and to the right, each carrying an equal amount of energy E/2. Since the two pulses are equal, the object remains at rest after the emission since the two beams are equal in strength and carry opposite momentum.

But if we consider the same process in a frame moving with velocity v to the left, the pulse moving to the left will be redshifted while the pulse moving to the right will be blueshifted. The blue light carries more momntum than the red light, so that the momentum of the light in the moving frame is not balanced. The light is carrying some net momentum to the right.

But the object hasn't changed its velocity before or after the emission. Yet in this frame it lost some right-momentum to the light. The only wayit could have lost momentum is by losing mass.

The velocity is small, so the right moving light is blueshifted by an amount equal to the nonrelativistic Doppler shift factor (1-v/c). The momentum of the light is its energy divided by c, and it is increased by a factor of v/c. So the right moving light is carrying an extra momentum delta

P

given by:

$Delta P = {v over c}{E over 2c}. ,$

The left moving light carries a little less momentum, by the same amount

delta P

. So the total right-momentum in the light is twice

delta P

. This is the right-momentum that the object lost.

$2Delta P = v {Eover c^2} ,$

The momentum of the object in the moving frame after the emission is reduced by this amount:

$P' = Mv - 2Delta P = (M - {Eover c^2})v ,$

So the change in the object's mass is equal to the total energy lost divided by

c 2

. Since any emission of energy can be carried out by a two step process, where first the energy is emitted as light and then the light is converted to some other form of energy, any emission of energy is accompanied by a loss of mass. Similarly, by considering absorption, a gain in energy is accompanied by a gain in mass. Einstein concludes that all the mass of a body is a measure of its energy content.

Sunny Shah

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18 Mar 2008 20:41:27 IST

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Here you go:

dE=Fdx

dE=d(mv)/dt.dx = d(mv)/dt.dx/dt.dt

= v d(mv)

= v²dm + mvdv

m=m₀/(

1-v²/c²) [m₀=rest mass]

=>m.(

1-v²/c²) = m₀

On differentiating the above equation, we get

c²dm=v²dm + mvdv

Thus dE = v²dm + mvdv = c²dm

Thus dE= c²dm

On integrating,

E=mc²

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