Superluminal jets. Figure 37-29a shows the path taken by a knot in a jet of ionized gas that has been expelled from a galaxy. The knot travels at constant velocity $\overrightarrow{\mathbf{v}}$ at angle$\mathbf{\theta }$ from the direction of Earth. The knot occasionally emits a burst of light, which is eventually detected on Earth. Two bursts are indicated in Fig. 37-29a, separated by time$\mathbf{t}$ as measured in a stationary frame near the bursts. The bursts are shown in Fig. 37-29b as if they were photographed on the same piece of film, first when light from burst 1 arrived on Earth and then later when light from burst 2 arrived. The apparent distance$\stackrel{}{{\overline{\mathbf{D}}}_{\mathbf{app}}}$ traveled by the knot between the two bursts is the distance across an Earth-observer’s view of the knot’s path. The apparent time${\overline{\mathbf{T}}}_{\mathbf{app}}$ between the bursts is the difference in the arrival times of the light from them. The apparent speed of the knot is then ${\overline{\mathbf{V}}}_{\mathbf{app}}\mathbf{=}{\overline{\mathbf{D}}}_{\mathbf{app}}\mathbf{/}{\overline{\mathbf{T}}}_{\mathbf{app}}$. In terms of $\mathbf{v}$, $\mathit{t}$, and$\mathbf{\theta }$ , what are (a)${\overline{\mathbf{D}}}_{\mathbf{app}}$ and (b)${\overline{\mathbf{T}}}_{\mathbf{app}}$ ? (c) Evaluate${\overline{\mathbf{V}}}_{\mathbf{app}}$ for$\mathit{v}\mathbf{=}\mathbf{0.980}\mathit{c}$ and $\mathbf{\theta }\mathbf{=}\mathbf{30}{\mathbf{.0}}^{\mathbf{\circ }}$. When superluminal (faster than light) jets were first observed, they seemed to defy special relativity—at least until the correct geometry (Fig. 37-29a) was understood.

The given data can be listed below as:

• The velocity of the knot is $\overline{\mathrm{v}}$.
• The angle of inclination of the knot is $\mathrm{\theta }$.
• The busts are separated by a time of $\mathrm{t}$.
• The knot travelled to a apparent distance of ${\overline{\mathrm{D}}}_{\mathrm{app}}$.
• The knot travelled to a apparent time of ${\overline{\mathrm{T}}}_{\mathrm{app}}$.
• The knot travelled to a apparent speed of${\overline{\mathrm{V}}}_{\mathrm{app}}={\overline{\mathrm{D}}}_{\mathrm{app}}/{\overline{\mathrm{T}}}_{\mathrm{app}}$

The apparent velocity is mainly measured from the time travel plot directly. It also equals true velocity when refractors are parallel to their surface.

(a)

The equation of the spatial separation amongst the busts is expressed as:

$\mathrm{s}=\mathrm{vt}$

Here,$\mathrm{s}$is the spatial separation amongst the busts,$\mathrm{v}$is the velocity of the knot and $\mathrm{t}$is the separation time of the busts.

The length has been projected into the direction that is perpendicular to the Earth’s light rays.

The equation of the apparent distance travelled by the knot is expressed as:

${\overline{\mathrm{D}}}_{\mathrm{app}}=\mathrm{ssin\theta }$

Here,${\overline{\mathrm{D}}}_{\mathrm{app}}$ is the apparent distance travelled by the knot and $\mathrm{\theta }$is the angle of inclination of the knot.

Substitute$\mathrm{vt}$ for$\mathrm{s}$ in the above equation.

${\overline{\mathrm{D}}}_{\mathrm{app}}=\mathrm{vtsin\theta }$ …(i)

Thus, the value of ${\overline{\mathrm{D}}}_{\mathrm{app}}$ is $\mathrm{vt}\mathrm{sin\theta }$.

(b)

The bust 2 has been emitted after time $\mathrm{t}$of bust 1 that shows that the bust 1 is needed to travel an extra distance and also additional time is needed.

The equation of the extra distance travelled by bust 1 is expressed as:

$\mathrm{L}=\mathrm{vtcos\theta }$

Here,$\mathrm{L}$is the extra distance travelled by bust 1.

The equation of the additional time needed by bust 1 is expressed as:

${\mathrm{t}}^{\text{'}}=\frac{\mathrm{L}}{\mathrm{c}}$

Here, ${\mathbf{t}}^{\mathbf{\text{'}}}$is the additional time needed by bust 1.

The equation of the apparent distance travelled by the knot is expressed as:

${\overline{\mathrm{T}}}_{\mathrm{app}}=\mathrm{t}-{\mathrm{t}}^{\text{'}}$

Here,${\overline{\mathrm{T}}}_{\mathrm{app}}$is the apparent distance travelled by the knot and$\mathrm{t}\text{'}$is the additional time needed by bust 1.

Substitute$\frac{\mathrm{L}}{\mathrm{c}}$for$\mathrm{t}\text{'}$in the above equation.

${\overline{\mathrm{T}}}_{\mathrm{app}}=\mathrm{t}-\frac{\mathrm{L}}{\mathrm{c}}$

Substitute $\mathrm{vtcos\theta }$for $\mathrm{L}$in the above equation.

$\begin{array}{c}{\overline{\mathrm{T}}}_{\mathrm{app}}=\mathrm{t}-\frac{\mathrm{vtcos\theta }}{\mathrm{c}}\\ =\mathrm{t}\left(1-\frac{\mathrm{vcos\theta }}{\mathrm{c}}\right)\end{array}$ …(ii)

Thus, the value of ${\overline{\mathrm{T}}}_{\mathrm{app}}$is$\mathrm{t}\left(1-\frac{\mathrm{vcos\theta }}{\mathrm{c}}\right)$ .

(c)

The equation of the apparent speed travelled by the knot is expressed as:

${\overline{\mathrm{V}}}_{\mathrm{app}}={\overline{\mathrm{D}}}_{\mathrm{app}}/{\overline{\mathrm{T}}}_{\mathrm{app}}$

Here,${\overline{\mathrm{V}}}_{\mathrm{app}}$ is the apparent speed travelled by the knot.

Substitute the values of the equation (i) and (ii) in the above equation.

$\begin{array}{c}{\overline{\mathrm{V}}}_{\mathrm{app}}=\frac{\mathrm{vtsin\theta }}{\mathrm{t}\left(1-\frac{\mathrm{v}}{\mathrm{c}}\mathrm{cos\theta }\right)}\\ =\frac{\mathrm{vsin\theta }}{\left(1-\frac{\mathrm{v}}{\mathrm{c}}\mathrm{cos\theta }\right)}\end{array}$

Substitute $0.980\mathrm{c}$for $\mathrm{c}$and ${30.0}^{\circ }$for $\mathrm{\theta }$in the above equation.

${\overline{\mathrm{V}}}_{\mathrm{app}}=\frac{\left(0.980\mathrm{c}\right)\sin {30.0}^{\circ }}{(1-\left(0.980\right)\cos {30.0}^{\circ })}$

Substitute$3\times {10}^8\text{\hspace{0.17em}m/s}$ for $\mathrm{c}$in the above equation.

$\begin{array}{c}{\overline{\mathrm{V}}}_{\mathrm{app}}=\frac{(0.980\times 3\times {10}^8\text{\hspace{0.17em}m/s})\sin {30.0}^{\circ }}{(1-\left(0.980\right)\cos {30.0}^{\circ })}\\ =\frac{(2.9\times {10}^8\text{\hspace{0.17em}m/s})\left(0.5\right)}{(1-\left(0.980\right)\left(0.86\right))}\\ =\frac{(1.4\times {10}^8\text{\hspace{0.17em}m/s})}{(1-\left(0.8428\right))}\\ =9.3\times {10}^8\text{\hspace{0.17em}m/s}\end{array}$

Thus, the value of ${\overline{\mathrm{V}}}_{\mathrm{app}}$is$9.3\times {10}^8\text{\hspace{0.17em}m/s}$ .