screen capture for fixed area in Mac

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The build-in screen capture shortcut shift-command-4 is cool. But human is just not reliable. For nice presentation, I searched a while, in Mac, there is a terminal command

screencapture -c -R x,y,w,h

 

where x, y, w ,h are x-position, y-position, width and heigh respectively. the option -c is for capture to clipboard.

 

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HELIOS – an inverse problem

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When give incident energy, the transfer reaction (or 2-body reaction in general) in a magnetic field is a transform:

\displaystyle \begin{pmatrix} E_x \\ \theta_{cm} \end{pmatrix} \rightarrow \begin{pmatrix} E \\  z \end{pmatrix}

We already give the constant E_x line and constant \theta_{cm} line is previous post. Now, we study an inverse problem that, given the energy and z-position, or a point on the EZ plot, what is the excitation energy and center of mass angle?

The solution is

\displaystyle \begin{pmatrix} E_x \\ \theta_{cm} \end{pmatrix} = \begin{pmatrix} - m_B + \sqrt{E_t^2 + m_b^2 - 2\gamma E_t (E - \alpha \beta z)} \\  \frac{\gamma \beta E - \alpha \gamma z}{ \sqrt{\gamma^2 (E-\alpha \beta z)^2 - m_b^2}} \end{pmatrix}


The above solution is for infinite detector, which is not really practical.

Assume the rotation radius is much larger than the detector size a. The coupled equations are:

\displaystyle E = \gamma E_{cm} - \gamma \beta p \cos(\theta_{cm})

\displaystyle \alpha \beta \gamma z = (\gamma E - E_{cm})\left( 1- \frac{1}{2\pi} \frac{a}{\rho} \right)

\displaystyle \rho = \frac{ p \sin(\theta_{cm}} {2\pi a}

eliminate \theta_{cm} ,

\displaystyle \alpha \beta \gamma z = (\gamma E - E_{cm})\left( 1- \frac{\alpha \gamma \beta a}{\sqrt{2\gamma E E_{cm} - E^2 - m^2 \gamma^2  - p^2}} \right)

Since E_{cm}^2 = m^2 + k^2

Change of variable

\displaystyle k \rightarrow m \tan(x) ,  0 < x < \pi/2

\displaystyle z = \frac{\gamma E - m \sec(x)}{\alpha \beta \gamma }\left( 1- \frac{\alpha \gamma \beta a}{\sqrt{2\gamma E m \sec(x) - E^2 - m^2 \gamma^2  - m^2 \tan^2(x)}} \right)

The square root can be rearranged as,

\displaystyle 2\gamma E m \sec(x) - E^2 - m^2 \gamma^2  - m^2 \tan^2(x) \\ = (E^2-m^2) \gamma^2 \beta^2 - (\gamma E - m \sec(x))^2  = H^2 - K^2

Use

\alpha \gamma \beta z \rightarrow Z \\ \beta \gamma \alpha a \rightarrow G

We have,

\displaystyle Z = K \left( 1 - \frac{G}{\sqrt{H^2 - K^2}} \right)

Next, change or replace,

\displaystyle K \rightarrow H \sin(\phi), -\frac{\pi}{2} < \phi < \frac{\pi}{2}

\displaystyle Z = H \sin(\phi)  - G \tan(\phi)

Since H , G > 0 , and G < H , as

\displaystyle \frac{G}{\sqrt{H^2 - K^2 }} = \frac{a}{2\pi\rho} < 1

The function

f(\phi) = H \sin(\phi) - G \tan(\phi)

looks like the orange line below

f_phi.PNG

The blue line is f(\phi) = Z . It seems that there are multi-solution for \phi. When \theta_{cm} >> 0 ,

f'(\phi) = H \cos(\phi) - G \sec^2(\phi) > 0

Thus, there is only one solution. However, when \theta_{cm} is small, there is multi-solution. That need to be careful.

 

HELIOS constant thetaCM line

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somehow I confused the term “line”, because sometimes it means a “curve”, anyway~


There is a plot of the energy – z-pos plot (EZ-plot) in the last post. In the EZ plot, the constant excitation energy line is already given in the equation (3) in this post. And the finite-detector correction is presented in the last post. The constant \theta_{CM} line is

\displaystyle E = \frac{-\sin^2(\theta_{CM}) \alpha \beta \gamma^2 z + \cos(\theta_{CM})\sqrt{\alpha^2 z^2 + m_b^2 (1-\sin^2(\theta_{CM})\gamma^2)}}{1-\sin^2(\theta_{CM})\gamma^2}

where \alpha = \frac{cZB}{2\pi}

you can guess/see that, this is a solution of a quadratic equation. In fact, from the equation (1) and (2) in this post, by eliminating the \cos(\theta_{CM}) in \vec{\beta} \cdot \vec{p} , we got the constant E_x line. By eliminating the m_B (the mass of the heavier particle, or the excited energy) in E_{cm} (see E_{tot} in this post ), we got the above equation.

When \theta_{CM} = 0 , the equation reduces to

\displaystyle E = \sqrt{\alpha^2 z^2 + m_b^2}


OK, I know I skipped a lot of step, which is not my style. From the equation (1) and (2), eliminate E_{cm} gives you,

\displaystyle E = \frac{\alpha}{\beta}z - \frac{1}{\gamma}(\vec{\beta}\cdot\vec{p})

it is a linear! But if you look closely, the center of mass momentum has explicit E_x dependence, it is a constant of motion only when E_x is constant. The higher the excitation energy, the less energy of the recoil energy E of the proton! Thus, it is like p = p(E_x) = p(E_x(E)) . A more easy way to eliminate m_B or E_x, is expand E_{cm} into

\displaystyle E_{cm} = \frac{1}{E_{tot}}\left( E_{tot}^2 + m_b^2 - m_B^2\right)

Thus, eliminating m_B = (m_B)_{g.s.} + E_x by isolate m_B.

Finite detector effect in HELIOS

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I was very busy last month that I went to CERN to join two experiments. The experiments were using ISOLDE Solenoid Spectrometer, which is a HELIOS-type equipment. In fact, I planned to post a series about HELIOS spectrometer, mainly focus on the kinematics, to address different detail, like offset-beam effect, tilted-beam effect, the constant-thetaCM line, etc., but I was very busy, probably until Dec. Anyway~


The pervious post lay the foundation for HELIOS spectrometer. The energy and z-position is linear, that simplified a lot of things. And the Q-value or excitation energy is in the intercept and in most cases, can be linearly approximated, i.e. E_x \propto E_{cm}. We can derive the exact relation later.

Screen Shot 2018-10-26 at 23.41.44.png

In the above plot, 28Mg(d,p) at 9.43 MeV/u. The dashed line is for infinite detector,  the red line is for finite detector, and the blue line is 0 deg center of mass angle. We can see, for smaller energy, the change of the hit-position is larger and non-linear. It is because at smaller energy, the scattered angle is small, and the particle hits the detector at very small angle ( with respect to the detector plane, not detector normal), so that the particle will be stopped much earlier.

Today, I will show the algebraic solution for finite detector effect.

Finite detector effect, from it name, because the detector has size, so the hit position is difference from the on-axis position. Below is the x-y plane view, the detector plane is the orange line, with distance from the z-axis of a and make an angle \phi_p to the x-z plane. The blue line is the helix curve of the charged particle in the magnetic field. The radius of the helix curve is \rho, and the emittance angle of the particle is \phi.

Screen Shot 2018-10-26 at 23.07.48.png

If the particle hit back the z-axis, the \Delta \phi = 2\pi, but due to finite detector size, it will not complete a cycle.

The equation of the detector plane is

\displaystyle x\cos(\phi_p)+ y\sin(\phi_p) = a

The equation of the locus of the particle is

\displaystyle \begin{pmatrix} x \\ y \end{pmatrix} = \rho \begin{pmatrix} \sin(\tan(\theta) \frac{z}{\rho} - \sin(\phi)) \\ \cos(\phi) - \cos(\tan(\theta) \frac{z}{\rho} \end{pmatrix}

where, \theta is the particle emittance with respect to the z-axis. Solving these two equation for z, we have

\displaystyle z = \frac{\rho}{\tan(\theta)}\left( \phi_p - \phi + n\pi + (-1)^{n} \sin^{-1}\left( \frac{a}{\rho} + \sin(\phi-\phi_p) \right) \right), n=0,1,2,...

where n is the number of time that the particle crossed the detector plan. Note that, the length of a cycle is

\displaystyle z_0 = 2\pi \frac{\rho}{\tan(\theta)}

In the case when \phi = 0 , \phi_p = \pi , and we want the particle hit from the outside, i.e. n = 2m+1, m = 0, 1,2,.. , the solution becomes

\displaystyle z = z_0 \left( 1 - \frac{1}{2\pi}\sin^{-1}\left( \frac{a}{\rho} \right) \right)

When \rho >> a ,

\displaystyle z \approx z_0 \left( 1 - \frac{1}{2\pi} \frac{a}{\rho}\right)


In transfer reaction, when the beam energy is fixed, the degree of freedom is 2: the excitation energy (E_x) and center of mass angle (\theta_{cm}. After the HELIOS, the observables are energy (e) and z-position (z). With the finite detector effect, the mapping look like this

Screen Shot 2018-10-26 at 23.52.27.png

We can see, there are some regions that the manifold of E_x - \theta_{cm} is folded. So, there is no way to have an inverse transform to get back the E_x and $latex_{cm}$ from the observables. Unless, we can, somehow, extrapolating the z-axis position.

Kinematics in Magnetic field

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Generally, the best starting point of  any kinematics calculation for any reaction is from the CM frame. In the exit channel, the four-momentum in the CM frame is

P_{cm} = \begin{pmatrix} \sqrt{m^2 + p^2} = E_{cm} \\  \vec{p} \end{pmatrix}

the boost from CM frame to Lab frame is \vec{\beta} . Lets define the perpendicular vector on the plane of \vec{p} be \hat{n} , thus, the four-momentum in the Lab frame is

P = \begin{pmatrix} E \\ \vec{k} \end{pmatrix} =\begin{pmatrix} \gamma E_{cm} + \gamma \beta \hat{\beta}\cdot \vec{p} \\ \left(  \gamma \beta E_{cm} + \gamma (\hat{\beta} \cdot \vec{p}) \right) \hat{\beta} + (\hat{n} \cdot \vec{p} ) \hat{n} \end{pmatrix}

where

\displaystyle E_{cm} = \frac{1}{2E_t}\left( E_t^2 + m_b^2 -m_B^2 \right)

\displaystyle p^2 = \frac{1}{4E_t^2} \left(E_t^2 - (m_b + m_B)^2\right) \left(E_t^2 - (m_b - m_B)^2\right)

\displaystyle E_t^2 = (m_a+m_A)^2 + 2m_aT

\displaystyle \beta = \frac{\sqrt{(m_A+T)^2 - m_A^2}}{m_a + m_A + T}, \gamma = \frac{1}{\sqrt{1-\beta^2}}

We can see, the particle-A is moving with kinematic energy of T, hit particle-a, result in particle-b and particle-B.


Suppose the magnetic field is parallel to z-axis

\vec{B} = B \hat{z}

The rotation radius is

\displaystyle \rho = \frac{\vec{k} \cdot \hat{xy} }{cZB}

where Z is the charge state of the particle, and \hat{xy} is the perpendicular unit vector on xy-plane.

kinB.PNG

The time for 1 cycle is

\displaystyle T_{cyc} = \frac{2\pi \rho}{v_{xy}} = \frac{2\pi}{cZB} \frac{\vec{k}\cdot\hat{xy}}{v_{xy}} = \frac{2\pi}{c^2 ZB} E

And the length for 1 cycle is

\displaystyle z_{cyc} = v_z T_{cyc} = \frac{2\pi}{cZB} \vec{k}\cdot \hat{xy} \frac{v_z}{v_{xy}} = \frac{2\pi}{cZB} \vec{k}\cdot \hat{z}


When the Lorentz boost is parallel to the B-field direction,

\vec{\beta} \cdot \hat{z} = 1, \hat{n} \cdot \hat{z} = 0

\displaystyle z_{cyc} = \frac{2\pi}{cZB} \left(\gamma \beta E_{cm} + \gamma (\hat{\beta} \cdot \vec{p})\right) ………………. (1)

and the total energy is

\displaystyle E = \gamma E_{cm} + \gamma \beta (\hat{\beta}\cdot \vec{p}) …………….. (2)

By eliminating the \hat{\beta} \cdot \vec{p} ,

\displaystyle E = \frac{1}{\gamma}E_{cm} + \frac{cZB}{2\pi} \beta z_{cyc} …………….. (3)

We can see that, the energy and z_{cyc} have a linear relation. That simplified a alot of thing. For instance, the intercept is related to E_{cm}, which is further related to the Q-value of the reaction. If we can plot the E-z plot and determine the intercept, we can extract the Q-value, which is related to the excitation energy or some sort.

Without the magnetic field, to extract the Q-value, we really need to know the scattering angle (the azimuthal angle can be skipped due to symmetry in some cases). And usually, the relation is not linear.

Magnetic Rigidity

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In ion optics, there is a concept called Magnetic Rigidity. When an charged ion passing through a magnetic field, it will bend and rotate. The magnetic rigidity is the B\rho, where B is the magnetic field strength in Tesla, and \rho is the rotation radius.

using basics physics, we have the centripetal force equals to the Lorentz force,

\displaystyle \frac{mv^2}{\rho} = Q \vec{v}\times \vec{B}

Simplify,

B\rho = mv/Q


The above is non-relativistic, and I always assume the mv can be replaced by relativistic momentum p = m \gamma \beta . Now I give a prove.

The relativistic centripetal force is

\displaystyle \vec {F} = m \gamma \vec{a_\perp} = - m \gamma\frac{ v^2}{\rho} \hat{\rho}

Thus,

\displaystyle m \gamma \frac{v^2}{\rho} = QvB

\displaystyle m \gamma \beta = p = cQB \rho

where c is speed of light. In the last equation, the unit of p is MeV/c, c = 299.792458 mm/ns, and \rho in meter.

Tunneling through the earth core

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I do something difference today, ha.


The problem is simple, How long does it take to tunneling through the center of the earth to the other side?

Assuming uniform density of the earth (\rho ), the acceleration inside the earth at radius r is

\displaystyle  - G \frac{4\pi}{3} \rho r = -k r

k = 1.5413\times 10^{-6} in 1/sec^2.

because only the mass within the radius matter. Thus, the equation of motion is

\displaystyle \frac{d^2 r}{dt^2} = - k r 

the solution is

\displaystyle r(t) = R \cos( \sqrt{k} t )

Thus, the time for a trip is T = \pi / \sqrt{k}   = 2530.5 sec or 42 min and 10.5 sec.

The maximum speed when passing the core. The speed is R \sqrt{k} = 7910 m/s


How about we use a realistic earth density?

The density, and acceleration can be found in the web, for example, here.

The travel time is 2291 sec, or 38 min 11 sec.

tunnelingEarth.PNG


It is interesting that, in the uniform density calculation, the travel time is independent of the radius, but density. The peak velocity is

\displaystyle R \sqrt{k} < c \rightarrow \rho < \frac{c^2}{R^2}\frac{3}{4\pi} \frac{1}{G}

Let compare with Schwarzschild Radius:

\displaystyle R_S = \frac{2GM}{c^2} \rightarrow \rho_S = \frac{1}{2}\frac{c^2}{R^2}\frac{3}{4\pi} \frac{1}{G}

The Schwarzchild density is half of the maximum density by classical argument.


In case of point mass.

The equation of motion is

\displaystyle \frac{d^2r}{dt^2} = -\frac{GM}{r^2}

change of variable v(r) = \frac{dr}{dt}

\displaystyle \frac{d(v^2/2)}{dr} = -\frac{GM}{r^2}

\displaystyle v^2(r) = 2 \frac{GM}{r} - 2 \frac{GM}{R}

The above solution assume v(R) = 0 . We can see that, at r = 0 , the speed go to infinity. Something wrong…..

Update: that is not wrong at all. imagine two neutrino with head-on collision, the released gravitational energy will be infinity! But, because of uncertainly principle, their separation distance can never to be zero.


Taking account of the earth rotation with the centripetal force, the equation of motion becomes,

\displaystyle \frac{d^2 r}{dt^2} = - k r + \omega^2\cos(\theta) r

where \omega is the angular velocity of earth, and \theta is the polar angle from the north pole. The travel time would be

\displaystyle T = \frac{\pi} {\sqrt{k - \omega^2 \cos(\theta)}}

\displaystyle \omega = \frac{2\pi}{24 \times 60\times 60} = 7.3 \times 10^{-5} rad/s

Thus,

\omega^2 = 5.3 \time 10^{-9} 1/sec^2.

which is a very small correction.

 

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