prove = E=mc²
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Answer:
E = mc2, equation in German-born physicist Albert Einstein’s theory of special relativity that expresses the fact that mass and energy are the same physical entity and can be changed into each other. In the equation, the increased relativistic mass (m) of a body times the speed of light squared (c2) is equal to the kinetic energy (E) of that body.
Step-by-step explanation:
Answer:
This is the proof
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Step-by-step explanation:
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 xcomponents 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' ( 5) 'a + '
b = γ =
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
â
â
c
u' '
â
â
u /c â
u /c v
â
â
c
u' '
b
bx b
b
ax
ax
a
ax a
a
We are now ready to evaluate the x-component of the photon momenta
( ) ( )
( ) ( ) ϕ + β γ = −
+ β ϕ
ϕ + β −
+ β ϕ γ =
ϕ β γ =
β ϕ
ϕ β
− β ϕ γ =
cos
1 cos
cos (10b) P' 1 cos
cos - 1 - cos
cos - (10a) P' 1 cos
2c
E
2c
E
2c
E
2c
E
bx
ax
( ) ( ) 2 cos
2
cos
2 (11) c
ãEv â
c
ãE -â
c
ãE -ÄÄm' P' P' = ax + bx = ϕ − ϕ + = −
Upon substituting Eq. (5) we get
( ) ( ) 2 ax bx
c
E'v
2c
E
2c
E m'v P P ϕ + β = − γ ϕ β − γ (12) - ∆ = ' + ' = cos - cos
We thus arrive at our final result