## Thursday, November 15, 2018

### Solutions To Special Relativity Problems (5)

1)   A student proposes to compute the kinetic energy of a particle relativistically by using the expression  1/2  m v with the 'relativistic mass of the particle.   Would this be correct? Explain why or why not.

Solution:    The student's proposal  must be incorrect given the relativistic formulation for kinetic energy is:

E K  =    mo  c2/ [(1 - u2/c2)½ ] -   mo c2

I.e. bearing a rest energy (m o c2 ) which is non-zero.  On the other hand, for the non-relativistic case:

E ' K  =    1/2  ( m v2 )   -   1/2  ( m v o  2 )

The rest energy (2nd) term is zero because the velocity is 0.  This also means the total energy is greater for the relativistic case, i.e.

E K   +   mo c2    >    E ' K       1/2  m v2

2) Determine the energy required to accelerate a proton from 0.25c to 0.50c.

Solution:  By the work -energy theorem:

W = K(f) - K(i)

K(i) = mo c2/ [(1 - v 2/c2)½ ]

v1 = 0.25 c

K(f) = mo c2/ [(1 - v 2/c2)½ ]

v 2 = 0.50c   (where: mo = 1.7  x 10 -27 kg )

Then:

K(f) - K(i) = mo c2/ [(1 - (0.50c)2/c2)½ ]    - mo c2/ [(1 - (0.25c)2/c2)½ ]

K(f) - K(i) = mo c2/[(1 - 0.25]½ - mo c2/[(1 - 0.0625]½

K(f) - K(i) =  1.15 mo c2 -    1. 03 mo c2     =  0.12 mo c2

But the proton rest mass energy in MeV is:   mo  c=  938 MeV

So that:  0.12 mo c2      =  (0.12) 938 MeV  =   112 MeV

3) Protons emerge from a particle accelerator with a kinetic energy equal to 0.49 mc2.   What is the speed of these particles? Compare the result to that obtained from the non-relativistic relation between mass and energy.

Solution:  Because the kinetic energy is relativistic, the velocity must be as well, so we use the relativistic form for KE:

E K  =    m  c2/ [(1 - u2/c2)½ ] -   m  c2

Whence:

0.49 m  c2   =   m  c2    /  [(1 - v 2/c2)½ ] -   m  c2

And:  1 .49 m  c2  =   m  c2/ [(1 - v 2/c2)½ ]

(1 - v 2/c2)½ =    m  c2/ 1 .49 m  c2         or:

(1  - v2/c2)   =     (1 / 1 .49) 2

And finally:  :   v     =  c Ö {1   -  (1 / 1 .49) 2 }

v = 2.2 x  10 8 m/s

For non-relativistic equation, we get an erroneous   v' =   Ö {2E K  /  m)  =   2.8 x  10 8 m/s

4)What is the speed of a particle whose kinetic energy is equal to its rest energy? What percentage error is made if the non-relativistic kinetic energy expression is used?

Solution:  Here we have in the KE equation:   E K  =    m  c2

So that:    m  c =    m  c2/ [(1 - u2/c2)½ ] -   m  c2

Hence:    2 m  c =    m  c2/ (1 - u2/c2)½

(1 - u2/c2)½    =    m  c2 /  2 m  c 2      =  1/2

u2/c2     =    ( 1   -  1/2 ) 2      =  (0.5) 2

u =  c Ö {1   -  (0.5)2 }=  c Ö (0.75) =  2.6 x  10 8 m/s

For non-relativistic case, we have, by the work-energy theorem:

K(f) - K(i)  = Wnet   =  1/2  m u' 2   -  0

Then:   u' =   Ö (2 Wnet  / m) =  Ö {2K  /   m)

u ' =   Ö {2 E K  /   m)  =  Ö {2  m  c2  /  m }     =  4.3 x  10 8 m/s

Percentage error =   [(u' -  u)/ u ]  x 100%  = 63%

5) Show that the relativistic  kinetic energy equation:

Ek   =   mo c2 / [(1 - v 2/c2)½]    -    mo c2

reduces approximately to 1/2  m v2   when  v << c.

Solution:

Apply binomial theorem to the relativistic factor, e.g.

(1 - v 2/c2-1/2    =    1  +    2/2 c2   +   (3 4/ 8c4  ) +    ....

Then, if we require v << c (non-relativistic case) all the terms containing v/c  or higher powers of this ratio can be neglected. Then we are left with (approximately) :

Ek   =  1/2  mo v 2