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/(2.0*TMath::Pi()*hbarc),
3)/(exp(E/k*T) - 1.0);
}

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

\\

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);
c1.cd();
f1.Draw(); 