A hard way to make things dark

In the laboratory frame, the time the ion takes to pass by two wavefronts is: $t=\dfrac{\lambda}{c+v}$ But $c=\lambda f$ , then $\lambda=\dfrac{c}{f}$ and: $t=\dfrac{c}{(c+v)f}=\dfrac{1}{(1+\beta)f}$

Where $\beta=\dfrac{v}{c}$ .

In the ion frame: $t'=\dfrac{t}{\gamma}=\dfrac{1}{(1+\beta)f\gamma}$

Where $\gamma=\dfrac{1}{\sqrt{1-\beta^2}}$ .

Is easy to see that the time $t'$ is the period of the wave in the ion frame. Then: $t'=\dfrac{1}{f'}=\dfrac{1}{(1+\beta)f\gamma}$ $\dfrac{f}{f'}=\dfrac{\sqrt{1+\beta^2}}{1+\beta }=\sqrt{\dfrac{1-\beta}{1+\beta}}$

As the speed of light is invariant, $c=\lambda' f'=\lambda f\implies\dfrac{f}{f'}=\dfrac{\lambda'}{\lambda}$ : $\dfrac{\lambda'}{\lambda}=\sqrt{\dfrac{1-\beta}{1+\beta}}$ $\dfrac{\lambda'^2}{\lambda^2}={\dfrac{1-\beta}{1+\beta}}$ $\beta=\dfrac{\lambda^2-\lambda'^2}{\lambda^2+\lambda'^2}$

In the laboratory frame, the ion is accelerating with an acceleration $a=\dfrac{q}{m}\cdot E$ .

As this acceleration is constant and the ion is initially at rest, the velocity of the ion when it absorbs the infrared radiation is $v=a\cdot t_{abs}$ .

So:

$\beta=\dfrac{v}{c}=\dfrac{a\cdot t_{abs}}{c}=\dfrac{\lambda^2-\lambda'^2}{\lambda^2+\lambda'^2}$ $t_{abs}=\dfrac{c\cdot m}{q\cdot E}\cdot \dfrac{\lambda^2-\lambda'^2}{\lambda^2+\lambda'^2}$

Where $\lambda'=\lambda_0$ (because the ion will absorb the radiation when the wavelenght of this radiation is, in the ion frame, equal to the wavelenght of the first absorption line of the helium) and $q$ is positive and with module equal to the elementary charge (because the ion is obtained by removing one electron of the helium atom).

Solving, we obtain:

$t_{abs}\cong\dfrac{3\cdot 10^8\cdot 6.65\cdot 10^{-27}}{1.6\cdot 10^{-19}\cdot 2}\cdot\dfrac{(2\cdot 10^{-6})^2-(1.083\cdot 10^{-6})^2}{(2\cdot 10^{-6})^2+(1.083\cdot 10^{-6})^2}$ $t_{abs}\cong3.407\ s$

In order for the ion to absorb the light, the ion must observe the wavelength of the light to be less than or equal to the wavelength of its first absorption line, i.e., $\lambda_{obs} \leq \lambda_0$ . In order for the ion to observe the $\lambda=2000 \ \mathrm{nm}$ light at a lower wavelength, the ion must be travel at a speed close to the speed of light. The observed wavelength is given by the relativistic Doppler shift equation

$$\frac{\lambda}{\lambda_{obs}}=\sqrt{\frac{1+\beta}{1-\beta}},$$

where $\beta=v/c$ . When the observed wavelength is just at the threshold of $\lambda_0$ , we can calculate what the ion's speed must be.

$$\frac{\lambda}{\lambda_0}=\sqrt{\frac{1+\beta}{1-\beta}}$$

Solving for $\beta$ yields

$$\beta=\frac{\lambda^2-\lambda 0^2}{\lambda^2+\lambda 0^2}.$$

So the ion's speed $v=\beta c$ is $\frac{\lambda^2-\lambda_0^2}{\lambda^2+\lambda_0^2} c$ .

Now the question becomes how long does it take for the ion to reach this speed? To answer this, we need to find the acceleration of the ion, which can be found by using Newton's second law $F=ma$ and the Lorentz force equation $F=qE$ .

$$ma=qE$$

$$a=qE/m$$

Because the ion is initially at rest, the relationship between its acceleration and velocity is $v=at$ . Substituting for $v$ and $a$ gives

$$\frac{\lambda^2-\lambda 0^2}{\lambda^2+\lambda 0^2} c=\frac{qEt}{m},$$

and solving for $t$ gives

$$t=\frac{\left(\lambda^2-\lambda 0^2\right)mc}{\left(\lambda^2+\lambda 0^2\right) qE}.$$

Substituting in for the values given, and noting that the charge on the ion is the fundamental charge because it lacks one electron,

$$t=\frac{\left((2000 \ \mathrm{nm})^2-(1083 \ \mathrm{nm})^2\right)(6.65 \times 10^{-27} \ \mathrm{kg})(3.00 \times 10^8 \ \mathrm{m/s})}{\left((2000 \ \mathrm{nm})^2+(1083 \ \mathrm{nm})^2\right) (1.60 \times 10^{-19} \ \mathrm{C})(2 \ \mathrm{N/C)}}$$

$$=3.41 \ \mathrm{s}.$$

Firstly calculate the acceleration of the Helium ion:

$a = \frac{F}{m}$

$a = \frac{qE}{m}$

$a = \frac{(1.6\times10^{-19})(2)}{6.65\times10^{-27}}$

$a = 4.812\times10^7 ms^{-2}$

Now find the velocity of the ion when the ion absorbs the radiation by considering the Relativistic Doppler Effect:

\frac{\lambda^'}{\lambda} = \frac{\sqrt{1 + \frac{v}{c}}}{\sqrt{1 - \frac{v}{c}}}

$\frac{1083}{2000} = \frac{\sqrt{1 + \frac{v}{c}}}{\sqrt{1 - \frac{v}{c}}}$

$(\frac{1083}{2000})^2 = \frac{1 + \frac{v}{c}}{1 - \frac{v}{c}}$

Let $(\frac{1083}{2000})^2 = n$

Through simple rearrangement, it is found that:

$v = \frac{(n-1)(c)}{n+1}$

Substituting in the values of $n$ and $c=3\times10^8$ we get:

$v = -1.639\times10^8 ms^{-1}$

The magnitude of $v$ can then be used to find the time at which the ion absorbs the radiation:

$v = u + at$

$v = at$

$t = \frac{v}{a}$

$t = \frac{1.639\times10^8}{4.812\times10^7}$

$t = 3.41 s$ to 3 significant figures.

For particles travelling at a significant fraction of the speed of light (c), the relativistic Doppler equation must be used to determine the observed wavelengths of electromagnetic waves in its rest frame:

$\lambda_{o} = \sqrt{\frac{1+\beta}{1-\beta}} \lambda_{s}$

Where $\beta$ represents the velocity as a fraction of c ( $\beta = \frac{v}{c}$ ). In this problem, we must first determine how fast our particle must be travelling (relative to the lab frame) so that, in the particles rest frame, our light source of 2000 nm ( $\lambda_{s}$ ), coincides with the first absorption line of helium at 1083 nm ( $\lambda_{o}$ ). By using the above equation, we find:

$\frac{\lambda_{o}}{\lambda_{s}} = \frac{1083}{2000} = .542 = \sqrt{\frac{1+\beta}{1-\beta}}$

Solving for beta,

$\beta = .547$

Or, in meters per second

$v = \beta * c = 1.64 \rm{x} 10^{8} \frac{m}{s}$

Now that we know velocity, we need to look for a way to find out how long it took the particle to achieve that speed. The particle is experiencing only one force, caused by the electric field, E = 2 N/C. Using the Lorentz force law for electric fields ( $F = E*q$ ) and Newton's second law ( $F = m*a$ ), we can solve for the particles acceleration:

$F = E*q = m*a$

thus,

$a = \frac{E*q}{m} = 4.82 \rm{x} 10^{7} \frac{m}{s^{2}}$

From here it is a simple task to calculate, with a known constant acceleration, how much time, in seconds, it takes to reach the speed we calculated earlier. The integral of the acceleration equation with respect to time gives us velocity as a function of time, which is one of the three well known kinematic equations:

$v(t) = \int a dt = a*t + v_{0} = 4.82 x 10^{7} * t$

Since the particle started at rest, $v_{0}$ is zero and drops out. Thus we just solve v(t) for t knowing the final velocity calculated above.

$1.64 \rm{x} 10^{8} = 4.82 \rm{x} 10^{7} *t$

$t = 3.40 s$

When the ion moves in the direction of the (stationary) light source, it perceives that the light has a different wavelength due to the Doppler effect, namely

$\bar{\lambda} = \sqrt{\frac{c-v}{c+v}} \lambda$ ,

where $v$ is the speed of the ion, $c$ is the speed of light, $\lambda$ is the wavelength in the laboratory frame and $\bar{\lambda}$ as observed by the helium ion. We are given what $\lambda$ is and what $\bar{\lambda}$ needs to be, so we solve for $v$ and get

$v = \frac{ \lambda^2 - \lambda_0^2 }{ \lambda^2 + \lambda_0^2 } c$ .

The helium ion accelerates due to the electric field. It experiences a force $F = q E$ , where $q = e$ is the net electric charge of the ion. Since the problem statement specifies that we should use Newtonian mechanics(*), we have a constant acceleration $a = F/m$ (where $m$ is the mass of the ion) and hence

$v = at = \frac{eEt}{m}$ .

Solving for $t$ gives

$t = \frac{mv}{eE} = \frac{mc}{eE} \frac{ \lambda^2 - \lambda_0^2 }{ \lambda^2 + \lambda_0^2 } = 3.4\ \mathrm{s}$ .

(*) Since $v = 0.55 c$ , we can expect that ignoring relativistic effects may not be such a good approximation. Relativistically, we'd have

$Ft = p = \frac{mv}{\sqrt{1-v^2/c^2}}$ ,

which ultimately leads to $t = 4.1\ \mathrm{s}$ .

We will assume all computations besides the Doppler effect to be classical. In particular, the acceleration of the ion can be seen to be $a = F / m = eE / m$ where $e$ is the charge of the electron.

As the time passes, the velocity $v$ of the ion increases linearly as $v = at$ . Because the ion is moving towards the source of the radiation it will experience the relativistic blue shift. The dependence of the observed wavelength $\lambda_1$ on velocity is $\lambda_1 / \lambda_0 = \sqrt{c - v \over c + v }$ (see wikipedia for derivation which hangs on classical Doppler ideas and time dilation) which we can rewrite as $v = c { \lambda_0^2 - \lambda_1^2 \over \lambda_0^2 + \lambda_1^2 }.$

Putting it altogether and noting that the ion absorbs the radiation when $\lambda_1 = 1032\, \rm{nm}$ we get

$t = v /a = {mc \over eE } \cdot { \lambda_0^2 - \lambda_1^2 \over \lambda_0^2 + \lambda_1^2 }$ so that $t = 3.4 \,\rm{s}.$

Since the helium ion is put into an electric field, the electric force will accelerate it and change its velocity in time. That means that the ion will "see" the light towards it blue shifted (the frequency of light will increase). Once the light is blue-shifted enough to reach the threshold for absorption, the intensity of the light behind the ion will drop, since part of it will be absorbed.

Let us first calculate how does the ion's velocity change: $eE = m a$ , $v = a t = \frac{eE}{m} t$ . Therefore, the ratio of the velocity of the atom to the velocity of light changes like $\beta(t) = \frac{eE t}{mc}$ .

The light is blue shifted due to the Doppler effect: $\nu = \nu_0 \sqrt{\frac{1+\beta}{1-\beta}}$ , $\lambda = \lambda_0 \sqrt{\frac{1-\beta}{1+\beta}}$ . Since we want $\lambda = 1083~\mbox{nm}$ and $\lambda_0 = 2000~\mbox{nm}$ , $\beta = 0.5466$ .

Plugging in the value for $\beta$ into $\beta(t) = \frac{eE t}{mc}$ , we get $t = 3.416~\mbox{s}$ .

the wave period observed by the ion is $\tau =\frac{\tau '}{\gamma}$ where $\tau$ is the time period observed by the ion so $f'= \gamma f$ then we can use the classical Doppler law $F=f'(1+\frac{v}{c})$ then

$F=f\frac{1+\frac{v}{c}}{\sqrt{1-\frac{v^2}{c^2)}}}$

so

$F=f\frac{1+\frac{v}{c}}{\sqrt{(1+\frac{v}{c})(1-\frac{v}{c})}}$ then $F=f\sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}$ and because the frequency equals to $f=\frac{c}{\lambda}$ then $\Lambda=\lambda\sqrt{\frac{1-\frac{v}{c}}{1+\frac{v}{c}}}$ Big lambda refers to the doppler wave length let $\frac{\Lambda}{\lambda}=\beta$ then $\beta=\sqrt{\frac{1-\frac{v}{c}}{1+\frac{v}{c}}}$ and $\beta^2=\frac{1-\frac{v}{c}}{1+\frac{v}{c}}$ so $(1+\frac{v}{c})\beta^2=1-\frac{v}{c}$ then $\frac{v}{c}=\frac{1-\beta^2}{1+\beta^2}$ $\sum{F}=ma$ then $a=\frac{Ee}{m}$ $v=at$ that yields $t=\frac{v}{a}$ then $t=\frac{\frac{1-\beta^2}{1+\beta^2}}{\frac{Ee}{m}}$ putting $\Lambda=1083nm$ and $\lambda=2000nm$ $t=3.416 sec$