Q59PE - College Physics (Urone)

Question

(a) Show that \[{\left({pc}\right)^2}/{\left({m{c^2}{\rm{}}} \right)^2}{\rm{ }} = {\rm{ }}{\gamma ^2}{\rm{ }} - {\rm{ }}1\]. This means that at large velocities \[pc>>m{c^2}\]. (b) Is \[E{\rm{}}\approx{\rm{}}pc\] when \[\gamma{\rm{}}={\rm{}}30.0\], as for the astronaut discussed in the twin paradox?

Step-by-Step Solution

Answer

a. The value obtained is: \[{\gamma ^2}{\rm{ }}-{\rm{ }}1=\frac{{{{\left({pc} \right)}^2}}}{{{{\left( {m{c^2}} \right)}^2}}}\].

b. The value obtained is: \[E{\rm{ }}\approx{\rm{ }}pc{\rm{ }}={\rm{ }}30.0{\rm{ }}m{c^2}{\rm{ }}\].

Step 1: Define Special Relativity

The special theory of relativity, sometimes known as special relativity, is a physical theory that describes how space and time interact. Theoretically, this is known as STR theory.

Step 2: Evaluating the equation

With the help of the equation (\[{\rm{28}}{\rm{.43}}\]), the relativistic total energy is obtained by:

\[E=\gamma m{c^2}\]     …(i)                                                                                                        

We have the value of \[\gamma \] as the relativistic constant.

The value of \[m{c^2}\] is said to be the rest mass energy of the particle.

Then from the equation (\[{\rm{28}}{\rm{.63 }}\]), the relationship between energy and momentum is given by the equation:

\[{E^2}={\left({pc}\right)^2}+(m{c^2})\]                                                                                                  …(ii)

Here, the value of \[E\] is said to be the relativistic total energy.

The value of \[p\] is said to be the relativistic momentum.

As we know that the value of \[\gamma \] is the relativistic factor.

So, the equation obtained is:

\[\gamma=\frac{1}{{\sqrt{1-{{\left({\frac{u}{c}}\right)}^2}}}}\]                                                                                                          …(iii)

 

Step 3: Showing that \[{\left({pc} \right)^2}/{\left( {m{c^2}{\rm{ }}} \right)^2}{\rm{ }} = {\rm{ }}{\gamma ^2}{\rm{ }} - {\rm{ }}1\].

1. Squaring both the sides of the equation (i) ,

 we get

\[{E^2}{\rm{}}={\rm{}}{(\gamma{\rm{}}m{c^2})^2}\]                                                                                                      …(iv)

Now with the help of the equations (ii) and (iv), we get:

\[{\left( {\gamma m{c^2}} \right)^2} = {(pc)^2} + {\left( {m{c^2}} \right)^2}\]

                                                         OR, 

\[{\left( {\gamma m{c^2}} \right)^2} - {\left( {m{c^2}} \right)^2} = {(pc)^2}\]

                                                         OR, 

\[\left[ {{\gamma ^2} - 1} \right]{\left( {m{c^2}} \right)^2} = {(pc)^2}\]

                                                           OR, 

\[{\gamma^2}-1=\frac{{{{(pc)}^2}}}{{{{\left({m{c^2}}\right)}^2}}}\]                                                                                                 …(v) 

Therefore, the value is:

\[{\gamma^2}-1=\frac{{{{(pc)}^2}}}{{{{\left({m{c^2}} \right)}^2}}}\].

Step 4: Evaluating the solution for part b

(b) Here we have the relativistic factor as:

\[\gamma  = 30.0\]

With the help of the equation (i) , we have:

\begin{aligned}E = \gamma m{c^2}\\= 30.0m{c^2}\end{aligned}

…(vi)

Then, with the help of the equation (v), we have:

          

\begin{aligned} pc = \sqrt {{\gamma ^2}-1}m{c^2}\\=\sqrt{{{30.0}^2}-1}m{c^2}\\= \sqrt {899} m{c^2}\\= 29.983m{c^2}\\\approx 30.0m{c^2}\end{aligned}

…(vii)

At last, with the help of the equation (vi) and (vii), at the value \[\gamma  = 30.0\], we get:

\[E \approx pc = 30.0m{c^2}\]

Therefore, the value is: \[E \approx pc = 30.0m{c^2}\].