Question

For the passing reference frames in Fig. 37-25, events $\mathbf{A}$ and$\mathbf{B}$ occur at the following spacetime coordinates: according to the unprimed frame,$\mathbf{(}{\mathbf{x}}_{\mathbf{A}}\mathbf{,}{\mathbf{t}}_{\mathbf{A}}\mathbf{)}$ and$\mathbf{(}{\mathbf{x}}_{\mathbf{B}}\mathbf{,}{\mathbf{t}}_{\mathbf{B}}\mathbf{)}$ ; according to the primed frame,$\mathbf{(}{{\mathbf{x}}^{\mathbf{\text{'}}}}_{\mathbf{A}}\mathbf{,}{{\mathbf{t}}^{\mathbf{\text{'}}}}_{\mathbf{A}}\mathbf{)}$ and $\mathbf{(}{{\mathbf{x}}^{\mathbf{\text{'}}}}_{\mathbf{B}}\mathbf{,}{{\mathbf{t}}^{\mathbf{\text{'}}}}_{\mathbf{B}}\mathbf{)}$. In the unprimed frame, $\mathbf{\Delta t}\mathbf{=}{\mathbf{t}}_{\mathbf{B}}\mathbf{-}{\mathbf{t}}_{\mathbf{A}}\mathbf{=}\mathbf{1}\mathbf{.00}\mathbf{\mu }\mathbf{\text{s}}$and $\mathbf{\Delta x}\mathbf{=}{\mathbf{x}}_{\mathbf{B}}\mathbf{-}{\mathbf{x}}_{\mathbf{A}}\mathbf{=}\mathbf{400}\mathbf{\text{\hspace{0.17em}m}}$.

(a) Find an expression for$\mathbf{\Delta x}\mathbf{\text{'}}$ in terms of the speed parameter $\mathbf{\text{β}}$and the given data. Graph$\mathbf{\Delta x}\mathbf{\text{'}}$ versus$\mathbf{\text{β}}$ for two ranges of b: (b)$\mathbf{\text{0}}$ to$\mathbf{\text{0.01}}$ and (c) 0.1 to 1. (d) At what value of$\mathbf{\text{β}}$ is $\mathbf{\text{∆x'}}$minimum, and (e) what is that minimum?

Short answer

a. The expression for$∆x\text{'}$ is${\mathrm{\Delta x}}^{\text{'}}=\frac{400-300\mathrm{\beta }}{\sqrt{1-{\mathrm{\beta }}^2}}$ .

b. The plot of $∆\mathrm{x}\text{'}$for $0<\mathrm{\beta }<0.01$is

c. The plot of $∆\mathrm{x}\text{'}$for $0.1<\mathrm{\beta }<1$is

d. The minimum value of $∆\mathrm{x}\text{'}$lies at $\mathrm{\beta }=0.750$.

e. The minimum value of $∆\mathrm{x}\text{'}$is$265\text{\hspace{0.17em}m}$ .

Step-by-step solution

The length contraction equation

The length contraction equation states that ${\mathbf{\Delta x}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{\gamma }\mathbf{(}\mathbf{\Delta x}\mathbf{-}\mathbf{v\Delta t}\mathbf{)}$, where the expression$\mathbf{\gamma }$ is the Lorentz factor.

The expression of  ∆x'.

From the above formula, we can say that

$\begin{array}{c}{\mathrm{\Delta x}}^{\text{'}}=\mathrm{\gamma }(\mathrm{\Delta x}-\mathrm{v\Delta t})\\ =\mathrm{\gamma }(\mathrm{\Delta x}-\mathrm{\beta c\Delta t})\\ =\frac{400-300\mathrm{\beta }}{\sqrt{1-{\mathrm{\beta }}^2}}\end{array}$

Thus, the expression for${\mathrm{\Delta x}}^{\text{'}}$ is ${\mathrm{\Delta x}}^{\text{'}}=\frac{400-300\mathrm{\beta }}{\sqrt{1-{\mathrm{\beta }}^2}}$.

The plot of Δx' versus β for  0<β<0.01

(b)

The plot of ${\mathrm{\Delta x}}^{\text{'}}$as a function of $\mathrm{\beta }$for $0<\mathrm{\beta }<0.01$is shown below:

The plot of ∆x'versus β for 0.1<β<1

(c)

The plot of${\mathrm{\Delta x}}^{\text{'}}$ as a function of$\mathrm{\beta }$ for$0.1<\mathrm{\beta }<1$ is shown below:

Minimum value of ∆x'

To calculate the value of$\mathrm{\beta }$where the value of$∆\mathrm{x}\text{'}$is minimum, we have to find the derivative of$∆\mathrm{x}\text{'}$and then put it equal to zero to solve for$\mathrm{\beta }$.

$\begin{array}{c}\frac{\mathrm{d}\left({\mathrm{\Delta x}}^{\text{'}}\right)}{\mathrm{d\beta }}=\frac{\mathrm{d}}{\mathrm{d\beta }}\left(\frac{\mathrm{\Delta x}-\mathrm{\beta c\Delta t}}{\sqrt{1-{\mathrm{\beta }}^2}}\right)\\ =\frac{\mathrm{\beta \Delta x}-\mathrm{c\Delta t}}{{(1-{\mathrm{\beta }}^2)}^{\frac{3}{2}}}\end{array}$

Now, put it equal to zero:

$\begin{array}{c}\frac{\mathrm{\beta \Delta x}-\mathrm{c\Delta t}}{{(1-{\mathrm{\beta }}^2)}^{\frac{3}{2}}}=0\\ \mathrm{\beta \Delta x}-\mathrm{c\Delta t}=0\\ \mathrm{\beta }=\frac{\mathrm{c\Delta t}}{\mathrm{\Delta cx}}\\ \mathrm{\beta }=\frac{(3.00\times {10}^8)(1.0\times {10}^{-6})}{400}\\ \mathrm{\beta }=0.750\end{array}$

Thus, the minimum value of $∆\mathrm{x}\text{'}$lies at $\mathrm{\beta }=0.750$.

(e)

At $\mathrm{\beta }=0.750$, the value of $∆\mathrm{x}\text{'}$is:

$\begin{array}{c}{\mathrm{\Delta x}}^{\text{'}}=\frac{400-\left(300\right)\mathrm{\beta }}{\sqrt{1-{\mathrm{\beta }}^2}}\\ =\frac{400-300\left(0.75\right)}{\sqrt{1-{\left(0.75\right)}^2}}\\ =264.8\text{\hspace{0.17em}m}\end{array}$

Thus, the minimum value of$∆\mathrm{x}\text{'}$ is approximately$265\text{\hspace{0.17em}m}$ .