Physics, asked by tallurijyothsna, 11 months ago

What is the third derivative in E=MC^2

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Answered by hemanth101
0

Answer:

A simple derivation of E = mc2

Peter M. Brown

E-mail: [email protected]

Abstract – Einstein’s 1905 derivation of E = mc2 has been criticized for being

circular. Although such criticism have been challenged it is certainly true that the

reasoning in Einstein’s original derivation is not at all obvious. Einstein’s original

derivation could be been made clearer. This article describes a clear way of doing

so.

I - Introduction

In Einstein’s 1905 paper On the Electrodynamics of Moving Bodies 1 he derived an

expression for the relationship for the energy of light wave as observed from two

different inertial frames of reference. In another paper in the same issue the

journal Einstein’s paper Does the inertia of a body depend on its energy content?

appeared 2. In it he used the aforementioned energy relationship to derive what

has become written as E = mc2 which relating to the change in the mass of the

body due to a decrease in. His arguments were later criticized, notably by H.

Ives, for being circular 3. However even though such criticisms have been

challenged they do demonstrate the difficulty in following Einstein’s logic. In

this paper I provide a simple derivation in the spirit of that given by Einstein but

which is much easier to understand. Einstein’s derivation was based on the

change in kinetic energy which results from a decrease in the bodies mass where

he mass is the m in K /2 2 = mv . This relation provides a secondary meaning for

the body’s mass. By this I mean that the m, the inertial mass, is defined as the m

in p = mv. The relation K /2 2 = mv is a derived relation and has a meaning not

directly associated with the body’s inertial mass. Although this is not an

objectionable usage of the inertial mass, a more direct usage would be more

satisfactory. Especially if the proof becomes easier to follow. I present below

what I believe to be a truly intuitively obvious derivation.

II - Derivation of E = mc2 based on conservation of momentum

Consider a body at rest in an inertial frame, S, as shown above. The body emits

two photons, a and b, of equal energy in opposite directions. The total energy of

the two photons, as measured in S, is E. Due to the conservation of momentum,

the emitting body must remain at rest in S. It follows that the velocity must be

constant in S’. As viewed from S’ before the photons are emitted there is only a

single body, the total momentum being that of the body which is defined as

i x (1) P' m' v e i = −

where m’i

is the inertial mass of the body before the emission. After the body

emits the two photons the total momentum will be still be conserved and will

have the value

m'i v ex m'f v ex P a P b m'f v ex ( ) P ax P bx ex ( ) P ay P by ey (2) - = − + ' + ' = − + ' + ' + ' + '

where m’f is the inertial mass of the body after the emission, P’a and P’a are the

momenta of photon’s a and b, respectively, as measured in S’. Equating the x-

components of Eq. (2) we find

m'v x ( ) m'i m'f v x P ax P bx (3) - ∆ e = − − e = ' + '

Einstein showed that the energies, as measured in S’, of photons a and b are 1

( )

( ) + β ϕ γ =

− β ϕ γ =

(4b) ' 1 cos

(4a) ' 1 cos

2c

E E

2c

E E

b

a

E E E E'

The x-components of the momenta in Eq. (3) are related to the energies in Eq. (4)

according to the relation E = Pc which is deduced from Maxwell’s equations. The

x-components of the photon momenta, in S’, are

b

a

(6b) P' cos '

(6a) P' cos '

= ϕ

= ϕ

c

E'

c

E'

b

bx

a

ax

The cosines are found be transforming the components of the velocity of light

from S to S’ using the velocity transformation relations. Referring to the diagram

below

c

u

c

u

bx

b

ax

a

ϕ = −

ϕ =

(7a) cos

( 7a) cos

where cosϕb = −cosϕa = − cosϕ . The velocity transformation relation for the x

component of velocity is 1

2

x

x

u v/c

u v

− =

1 (8) u'x

Using Eq. (6) and (7) gives

( )

+ ϕ

ϕ + = − − ϕ

ϕ − ϕ = =

− ϕ

ϕ − = −

− = − ϕ

ϕ − ϕ = =

1 cos

cos

1 cos

cos (9b) cos

1 cos

cos

1 cos 1

cos (9a) cos

â

â

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