Chapter 24, Problem 26Q

(a)

To determine

The speed at which the blazar seems to be drifting away from us when a blazar at $z=1.00$ faces variation in brightness that lasts 1 week (168 hour) as seen from Earth.

Answer to Problem 26Q

Solution:

$v=1.8\times {10}^8\text{m/s}$.

Explanation of Solution

Given data:

Red shift of the blazar is $z=1$.

Formula used:

The expression for Hubble’s Law red shift is written as,

$z=\sqrt{\frac{1+\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$c$}\right.}{1-\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$c$}\right.}}$

Here, $z$ is the red shift, $v$ is the speed of the blazar, and $c$ is speed of the light.

Explanation:

Write the expression for Hubble’s Law for red shift.

$z=\sqrt{\frac{1+\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$c$}\right.}{1-\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$c$}\right.}}$

Rearrange this for $v$.

$v=c\left[\frac{{\left(z+1\right)}^2-1}{{\left(z+1\right)}^2+1}\right]$

Substitute $1$ for $z$ and $3\times {10}^8\text{m/s}$ for $c$.

$\begin{array}{l}v=\left(3\times {10}^8\text{m/s}\right)\left[\frac{{\left(1+1\right)}^2-1}{{\left(1+1\right)}^2+1}\right]\\ v=1.8\times {10}^8\text{m/s}\end{array}$

Conclusion:

Therefore, the speed of the blazar is $v=1.8\times {10}^8\text{m/s}$.

(b)

To determine

The time period for which the variation in brightness lasted as measured by an astronomer within the blazar’s host galaxy by using the idea of time dilation if a blazar at $z=1$ goes through fluctuation in brightness that lasts 1 week (168 hour) as seen from Earth.

Answer to Problem 26Q

Solution:

$\Delta t=134\text{h}$.

Explanation of Solution

Given data:

Red shift of the blazar is $z=1$.

Speed of the blazar is $1.8\times {10}^8\text{m/s}$

Formula used:

The expression for time dilation in relativity is written as,

$\Delta t\text{'}=\frac{\Delta t}{\sqrt{1-\raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}$

Here, $\Delta t$ is the time interval for an observer, $\Delta t\text{'}$ is the time interval for the person moving with speed $v$, $c$ is speed of the light and $v$ is the relative velocity between the observer and moving object.

Explanation:

Write the expression time for dilation in relativity.

$\Delta t\text{'}=\frac{\Delta t}{\sqrt{1-\raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}$

Rearrange this for $\Delta t$.

$\Delta t=\Delta t\text{'}\left(\sqrt{1-\raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}\right)$

Substitute $1.8\times {10}^8\text{m/s}$ for $v$, $168\text{h}$ for $\Delta t\text{'}$ and $3\times {10}^8\text{m/s}$ for $c$.

$\begin{array}{c}\Delta t=\left(168\text{h}\times \frac{3600\text{s}}{\text{h}}\right)\left(\sqrt{1-\raisebox{1ex}{${\left(1.8\times {10}^8\text{m/s}\right)}^2$}\!\left/ \!\raisebox{-1ex}{${\left(3\times {10}^8\text{m/s}\right)}^2$}\right.}\right)\\ =483840\text{s}\times \frac{\text{1}\text{h}}{3600\text{s}}\\ =134\text{h}\end{array}$

Conclusion:

Therefore, the fluctuation lasted for $\Delta t=134\text{h}$.

(c)

To determine

The maximum size $(\text{in au})$ of the region of the blazar that emits energy.

Answer to Problem 26Q

Solution:

The maximum size of the region of the blazar that emits energy is $\text{2}\text{.06}\times {\text{10}}^2\text{au}$.

Explanation of Solution

Introduction:

Blazars are an example of AGN. All the AGNs are powered by the material which falls into the black hole at the center of a host galaxy. This black hole generates a large amount of energy from the gas and dust.

Explanation:

Blazars emits energy from the surface. This surface is small with respect to the quasar. This is, approximately, ${10}^{-3}\text{parsecs}$, and $1\text{parsec}=\text{2}\text{.06}\times {\text{10}}^5\text{au}$. So, the maximum size of the region of the blazar that emits energy is found out as follows:

$\begin{array}{l}{10}^{-3}\text{parsecs=}\left(\text{2}\text{.06}\times {\text{10}}^5\text{au}\right){10}^{-3}\\ =\text{2}\text{.06}\times {\text{10}}^2\text{au}\end{array}$

Conclusion:

Therefore, the maximum size of the region of the blazar that emits energy is $\text{2}\text{.06}\times {\text{10}}^2\text{au}$.