Question

Prove that if v and u' are less than c, it is impossible for a speed u greater than c to result from equation (2-l9b). [Hint: The product (c-u')(c-v) is positive.]

Short answer

It is showed that speed u can never be greater than c.

Step-by-step solution

Write the given data from the question.

The velocity v and u' are less than c.

The equation 2.19b is $u=\frac{u\text{'}+v}{1+{\displaystyle \frac{u\text{'}v}{c^2}}}$.

Determine the formulas to show that, it is impossible for a speed u   greater than c to result from equation (2-l9b).

The expression for the Lorentz velocity transformation is given as follows.

$\mathbf{u}\mathbf{=}\frac{\mathbf{u}\mathbf{\text{'}}\mathbf{-}\mathbf{v}}{\mathbf{1}\mathbf{-}{\displaystyle \frac{\mathbf{u}\mathbf{\text{'}}\mathbf{v}}{{\mathbf{c}}^{\mathbf{2}}}}}$

Here, u' is the velocity of the object in frame s' , u is the velocity of object in frame s , v is the velocity of frame s' relative to s and c is the speed of the light.

Show that, it is impossible for a speed u greater than c to result from equation (2-l9b).

Determine the fraction of u and c,

$$\begin{gathered} u=\frac{u\text{'}+v}{1+{\displaystyle \frac{u\text{'}v}{c^2}}} \\ u=\frac{u\text{'}+v}{{\displaystyle \frac{c^2+u\text{'}v}{c^2}}} \\ u=\frac{c^2\left(u\text{'}+v\right)}{c^2+u\text{'}v} \\ \frac{u}{c}=\frac{c\left(u\text{'}+v\right)}{c^2+u\text{'}v} \end{gathered}$$

Solve further as,

$\frac{u}{c}=\frac{cu\text{'}+cv}{c^2+u\text{'}v}$ ……. (i)

The denominator of the equation (1) can be expressed as,

$$\begin{gathered} \left(c-u\text{'}\right)\left(c-v\right)=c^2-vc-u\text{'}c+u\text{'}v \\ {c}^2+u\text{'}v=\left(c-u\text{'}\right)\left(c-v\right)+u\text{'}c+vc \end{gathered}$$

Substitute $\left(c-u\text{'}\right)\left(c-v\right)+u\text{'}c+vcfor{c}^2+u\text{'}v$into equation (1).

$\frac{u}{c}=\frac{cu\text{'}+cv}{\left(c-u\text{'}\right)\left(c-v\right)+u\text{'}c+vc}$

Divide the denominator and numerator of the above equation by u'c+vc .

$$\begin{gathered} \frac{u}{c}=\frac{{\displaystyle \frac{cu\text{'}+cv}{cu\text{'}+cv}}}{{\displaystyle \frac{\left(c-u\text{'}\right)\left(c-v\right)+u\text{'}c+vc}{cu\text{'}+cv}}} \\ \frac{u}{c}=\frac{1}{{\displaystyle \frac{\left(c-u\text{'}\right)\left(c-v\right)}{cu\text{'}+cv}}+1} \end{gathered}$$ ……. (ii)

Since the term (c-u') (c-v) is always be a positive number. Therefore, denominator of the equation (2)$\frac{\left(c-u\text{'}\right)\left(c-v\right)}{cu\text{'}+cv}+1$ always be grater that .

From the above discussion, it can be concluded that, the ratio of the right side of equation (ii) is always be less than 1. Therefore u can never be greater than c.

Hence it is proved that speed u can never be greater than c .