slug: particle-physics-2 datepublished: 2019-04-14T05:51:40 dateupdated: 2019-04-14T05:54:19 tags: English Posts, Acedemic Notes excerpt: "This week we prepare ourselves for the upcoming relativistic quantum mechanics." –-

This week we've been mostly talking about natural units and relativity, with some mentions of basic ROOT usage during sections.

Natural Units

As we all know, physics rely on the concept of *dimensions *quite often. For example, numbers with different units cannot be added, it makes no sense to add 1 s with 1 m. But SI units are human constructs, they are useful and convenient for describing scales close to our daily life – 1 meter, 1 second etc. But in the world of high energy physics, they're not what the nature meant us to use. As far as we know, nature offers at least two *constants *that we should try to use:

  • speed of light c, with dimension Length / Time

  • (reduced) Plank constant \(\hbar\), with dimension Energy * Time

In fact, now that kilogram has been redefined by fixing \(\hbar\) as a constant, humanity has officially got rid of 'human artifact' for defining SI units :)

The trick is to set \(c=1, \hbar=1\) when conducting calculation, and put them back as needed in the end. For example, a 4-momentum has:

\( p = (\frac{E}{c}, p _ x, p _ y, p _ z) \)

But \(E = \gamma m c ^2\), so if we let \(c=1\), at a rest frame, we can use energy unit (usually GeV or so) to represent mass, with an implicit \(\frac{1}{c ^2}\) omitted. Which is much easier to use. To put back the correct units, just give each m a \({c ^2}\) and then use \([c \hbar] = \text{Length * Energy}\) to fix things.

Relativistic Kinematics

There are many ways to 'derive' special relativity, after that, we call all the physical vectors that transform according to Lorentz Transformation a 4-vector, and we have to re-formulate physical quantities (such as velocity, momentum) with 4-vectors – because now they will be used in relativistic conservation laws.

The canonical 4-vector is the space-time 4-vector, which is what we used to derive special relativity in the first place, and they mark the coordinates of events in space-time. We call 4-vectors transform according to the normal LT (get \(t^\prime = ...\)) a contra-variant vector: \(\{X ^\mu\}\). **We call 4-vectors transform according to the inverse LT (get non-prime from prime) a **covariant vector: \(\{X _\mu\}\)

A big realization comes when we find that velocity needs to be replaced by 4-velocity because frame-dependent time \(dt\) is not good (because (dt, dx, dy, dz) is already a 4-vector so we can only divide by invariant quantity), instead, we use proper time \(d\tau\) to construct 4-velocity:

\( \{v ^ \mu\} = \frac{dx^ \mu}{d\tau} = \frac{dx^ \mu}{dv} \frac{dv}{d\tau} = \frac{dx^ \mu}{dv} \gamma\)

Following through some algebra, we find that:

  • \(\frac{\partial}{\partial x^ \mu}\) means taking derivative respect to contra-variant component but **itself transforms **as a co-variant component

  • \(\frac{\partial}{\partial x _ \mu}\)**itself transforms **as a contra-variant component

We found this by taking derivative of a Lorentz invariant function \(\Psi{x}\):

\( \begin{split} \frac{\partial \Psi}{\partial X^\mu} &= \frac{\partial \Psi}{\partial X' ^\nu} \frac{\partial X' ^\nu}{\partial X ^\mu}\ \end{split} \)

Let's just do one case of \(\mu = 0\), assuming boost is in x-axis:

\( \begin{split} \frac{\partial \Psi}{\partial X^0} &= \frac{\partial \Psi}{\partial t'} \frac{\partial t'}{\partial t} + \frac{\partial \Psi}{\partial x'} \frac{\partial x'}{\partial t} + \frac{\partial \Psi}{\partial y'} \frac{\partial y'}{\partial t} + \frac{\partial \Psi}{\partial z'} \frac{\partial z'}{\partial t}\\ &= \frac{\partial \Psi}{\partial t'} \gamma + \frac{\partial \Psi}{\partial x'} (-\beta\gamma) + \frac{\partial \Psi}{\partial y'} 0 + \frac{\partial \Psi}{\partial z'} 0\\ &= \gamma \frac{\partial \Psi}{\partial t'} -\beta\gamma\frac{\partial \Psi}{\partial x'} \end{split} \)

We can see this is a close resemble of \(t' = \gamma( t - \beta x)\), but the prime is switched, so \(\frac{\partial }{\partial X^\mu}\) transforms like \(\{X _ \mu\}\)!

ROOT basics

This week we're just learning how to use TF1 and TH1D, I will just paste some snippet:

Double_t BlackBodyFcn(Double_t* E, Double_t* T)
// E is the photon energy in eV
// T is the equilibrium temperature in K
// The function returns the Black Body spectrum (1/V)dE_tot/dE_gam
    Double_t hbarc = 1973; // units: eV Angstroms
    Double_t k = 8.617e-5; // units: eV K^{-1}
    return 8.0*TMath::Pi()*pow( E[0]/(2.0*TMath::Pi()*hbarc),
    3)/(exp(E[0]/k*T[0]) - 1.0);

void myFunc(){
    TF1* myFunc = new TF1("myFunc", "pow(x, 3)/(exp(x/[0]) - 1.0)", 0, 200);
    myFunc->SetParameter(0, 20.0);
    TF1 *fbb = new TF1("fbb", BlackBodyFcn, 0, 0.0001, 1);
    fbb->SetParameter(0, 50);


TF1 f1("f1", "-10.0*pow(x, 2) + 0.5*pow(x,4)", -5.0, 5.0);
TCanvas c1("c1","My Canvas 1",400,800);;