Question

Question: The energy of an electron at speed vin special relativity theory is$m{c}^2{\left(1-v^2/c^2\right)}^{-1/2}$, where m is mass of electron, and c is the speed of light. The factor $m{c}^2$ is called the rest mas energy (energy when v=0). Find two terms of the series expansion of ${\left(1-v^2/c^2\right)}^{-1/2}$ and multiply by $m{c}^2$ to get the energy" at speed v.

What is the second term in the energy series?(If v/cis very small, the rest of theseries can be neglected; this is true for everyday speeds).

Short answer

The total energy of an electron at speed ,Rest mass energy.The second term of expansionis known as the classical energy of the electron

Step-by-step solution

Determine the two terms of the series expansion , and multiply byto get energy at speed v, and also calculate the second term in the series

The 'rest mass' of an object is the inertial mass that an object has when it is at rest.

Total energy Eis defined to beand .Wherem is mass,c is the speed of light, and v=0is the velocity of the mass relative to an observer. There are many aspects of the total energy E that we will discuss—among them are how kinetic and potential energies are included in E , and how E is related to relativistic momentum. But first, note that at rest, total energy is not zero. Rather, when V=0 , we have, and an object has rest energy.

As the given information:

The energy of an electron at speed v in special theory of relativity is

, where m is mass of electron and c is the speed of light.

The factoris called the rest mass energy (energy when v=0 ).

Ifis very small the terms beyond second term can be neglected; this is true for everyday speeds.

A binomial series expansion ofis given by

Since,, therefore the factor can be expanded using formula (1).

Then, the expansion of the factoris:

${\left(1-\begin{array}{c}{v}^2\\ c^2\end{array}\right)}^{-1÷2}=1+\left[\begin{array}{c}\left(-\begin{array}{c}1\\ 2\end{array}\right)\\ 2\end{array}\right]\left(\begin{array}{c}-v^2\\ c^2\end{array}\right)+\left[\begin{array}{c}-1/2\left(-\begin{array}{c}1\\ 2\end{array}\right)\\ 2\end{array}\right]{\left(-\begin{array}{c}{v}^2\\ c^2\end{array}\right)}^2+....$ ....(2)

Now, multiplying both the sides of equation (2) byand neglecting the terms beyond the second term, it has

Thus, from equations (2) and (3) and it has

$m{c^2}^{2{\left(1-\begin{array}{c}{v}^2\\ c^2\end{array}\right)}^{-1/2}=m{c}^2\left(1+\begin{array}{c}{v}^2\\ 2c^2\end{array}+\frac{3v^2}{8c^2}+......\right)\approx m{c}^2}+\frac{m{v}^2}{2}$ .....(4)

Thus, from equations (2) and (3) and it has

…… (4)

The first term of equation (4) is rest mass energy of electron and the  second term is the energy of electron at speed  v

So, the total energy of an electron at speed V,

Rest mass energy

The second term of equationis known as the classical energy of electron.

The total energy of an electron at speed Rest mass energy

The second term of expansionis known as the classical energy of the electron.