The h Bar

9:08 AM

@Relativisticcucumber Well, you don't directly expand the ket but you can write $\lvert x+dx\rangle=e^{-\frac{i}{\hbar}\hat{p}dx}\lvert x\rangle$

Then expanding the exponential $\simeq\lvert x\rangle-\frac{i}{\hbar}dx\hat{p}\lvert x\rangle$

9:50 AM

In Coulomb scattering for when we are dealing with relativistic particles, the following equation was written: $\vec p^2=\beta^2 E^2$. It is said that $\beta$ is the velocity, but isn't beta= \frac v c$ and how do we come up with this eq: $\vec p^2=\beta^2 E^2$ ?

10:10 AM

@imbAF in natural units $c=1$ i.e. $\beta=v/c=v$

But how we reach the equalisation

What?

$\vec p^2=\beta^2 E^2$

Start from $E=\gamma mc^2$ and $\vec{p}=\gamma m\vec{v}$

Square the second one and divide side by side

I usually know

$E=mc^2$ and $E^2=m^2c^4+p^2c^2$

10:20 AM

You are missing a $\gamma$ factor in the first one, otherwise it's just the second equation for stationary objects

What? I don't understand that

That's the relativistic energy. The correct formula is $E=\gamma mc^2$

Which is equivalent to your second equation

The first one you wrote is the rest energy for a massive object $p=0$

10:37 AM

But gamma, is the 4 dimensional vector, with elements matrcies

right?

ACuriousMind

...no.

it's the Lorentz factor

No, the gamma matrices are a separate thing

Have you studied some relativity before delving into Dirac's equation? :P

Yes I ave

have

but that is the gamma, as the lorentz factor

It is

ACuriousMind

it is very rare to see $\gamma$ mean "the vector of $\gamma$-matrices"

$\gamma$ is the Lorentz factor, $\gamma^\mu$ are the matrices, there's usually no danger of confusion

in particular since most equations can only make sense with one of these meanings :P

10:51 AM

I am a bit confused

I know the following eq. about energy

$E=$\frac{p^2}{2m}$ classic expression

and $E=mc^2$ when the particle has no velocity and $E^2=m^2c^4 +p^2c^2$ which is the relativistic dispersion equation

I don't see at which point a gamma is involved here

@imbAF this is the same as $E=\gamma mc^2$

Just written in terms of different variables

which of the two

cuz there are two expressions and

$E^2=p^2c^2+m^2c^4$

$\beta=\frac{v}{c}$ and $\gamma=\frac{1}{\sqrt{1-\beta^2}}$

Ok

ACM won't like this sentence but it's the same expression written in terms of the lagrangian or the hamiltonian variables :P

10:57 AM

But I wasn't aware of this : $\vec{p}=\gamma m\vec{v}$

$L(x,v)=-mc^2/\gamma$ is a relativistic lagrangian for a free particle

$\partial_v L=p$

Doing the calculation you'll find that expression for the momentum

in here c=1 right?

If you will you can set it equal to one, nothing will change

$E(x,v)=v\partial_v L-L=\frac{mc^2}{\sqrt{1-\beta^2}}$ is the energy

And if you express it in terms of $p$ you will get the hamiltonian $H(p)=\sqrt{p^2c^2+m^2c^4}$

Ok I'll try to derive it with all these tips

@Mr.Feynman I did, thanks

One more thing, why do we consider rest mass once relativistic speed is relevant ?

Like $pc>>>mc^2$ so why even bother considering the rest mass energy, when for the classical case, where one can make the claim that the kinetic energy can be in the range of rest mass energy, here we do not consider it. It feels that we are doing the reverse

More Anonymous

1:25 PM

Oh wise souls of stack exchange what is he talking about in these pages?

Relativisticcucumber

2:16 PM

@Mr.Feynman ahhh yup this gives me what i was looking for. nifty - thanks !!

Obliv

3:27 PM

whats the point of the support rods in a telescope? if you had them removed you wouldn't get the diffraction spikes

bobble

6:24 PM

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8:51 PM

What motivates the position of the spinor indicies when we write a Dirac spinor as $$\begin{pmatrix}\psi_\alpha \\ \bar{\chi}^{\dot{\alpha}} \end{pmatrix}$$

i.e. why is one down and one up?

9:20 PM

I'm guessing, it might be related to covariant&contravariant spinors

Due to the fact that left and right Weyl spinors have different transformation laws

It's diffcult to tell what is a convention or not

In the Lagrangian, the other side could have the top one up and the bottom one down

Section 56 of L&L 3 discusses contravariant and covariant spinors if it helps

Though I'm not sure that is the problem

I think maybe it may be convention because dotted indices are contracted conventionally through $\psi^{\dot{\alpha}} \xi_{\dot{\alpha}}$ and undotted through $\psi_{{\alpha}} \xi^{{\alpha}}$

Or maybe the other way round

So it would make sense to have one up and one down

But it seems like it's impossible to deduce what is a convention or not

Thinking about the way Weyl spinors transform, either with a matrix or its complex conjugate, I'm getting convinced it's about covariant/contravariant

ACuriousMind

it's co-/contravariant relative to the fundamental rep of $\mathrm{SL}(2,\mathbb{C})$, cf. physics.stackexchange.com/a/743759/50583

9:32 PM

Here's one (after the fact) way to see it. First, take the convention that $\psi_a' = M_a \, ^b \psi_b$, so that $\psi'^a = (M^{-1})_b \, ^a \psi^b$, as the $(1/2,0)$ representation. Then $\psi_{\dot{a}} = (M^*)_{\dot{a}} \ ^{\dot{b}} \psi_{\dot{b}}$ and $\psi^{\dot{a}} = (M^{-1})^*_{\dot{b}} \, ^{\dot{a}} \psi^{\dot{a}}$.

From $P = p_{\mu} \sigma^{\mu}$, which is Hermitian, and $P' = M P M^{\dagger}$. We thus see $M \sigma^{\mu} M^{\dagger} = M \sigma^{\mu} (M^T)^*$ has matrix indices $M_a \, ^c \sigma^{\mu}_{c \dot{c}} M^*_b \, ^{\dot{c}}$, where you need to be careful writing $P M^{\dagger}$ in indices due to the transpose.

Thus one Weyl equation is $\psi^a i \sigma^{\mu}_{a \dot{a}} \partial_{\mu} \psi^{\dot{a}}$. To figure out the second one this way, you need to determine what $\overline{\sigma}^{\mu}$ is, the position of it's indices, and write the second Weyl equation, which I'll leave as a way to check all this

There are other arguments e.g. involving parity, and how spinors transform under $\mathrm{SO}(3)$, and I can probably think of something better with some effort

Oh, since $\Lambda_L=\exp(\frac{1}{2}\sigma\cdot(-i\theta-\eta))$ transforms left spinors and $\Lambda_R=\exp(\frac{1}{2}\sigma\cdot(-i\theta+\eta))$, we have that $\Lambda_R=(\Lambda_L^\ast)^{-1}$. It's that inverse that gives rise to the notation

Ah okay thanks

Mhh wait, something's off with my message

Yes, the complex conjugation is wrong

I am off by a literal factor of $1/2$ in deriving the general BRST generator $Q$ from something sensible...

10:46 PM

How can I just define what $(\bar{\sigma}^m)^{\dot{\alpha}\alpha}$ is?

Can just define a random tensor to be anything e.g. $M_{\alpha \dot{\alpha}}=\begin{pmatrix} 4 & 5 \\31 &11 \end{pmatrix}_{\alpha \dot{\alpha}}$?

11:20 PM

Is
$$\psi^a i \sigma^{\mu}_{a \dot{a}} \partial_{\mu} \psi^{\dot{a}} = - \partial_{\mu} \psi^a i \sigma^{\mu}_{a \dot{a}} \psi^{\dot{a}} = \psi^{\dot{a}} i \sigma^{\mu}_{a \dot{a}} \partial_{\mu} \psi^a = \epsilon^{\dot{a} \dot{b}} \psi_{\dot{b}} i \sigma^{\mu}_{a \dot{a}} \epsilon^{ab} \partial_{\mu} \psi_b = \psi_{\dot{b}} i \epsilon^{ \dot{b} \dot{a}} \epsilon^{b a} \sigma^{\mu}_{a \dot{a}} \partial_{\mu} \psi_b = \psi_{\dot{b}} i \overline{\sigma}^{\mu,\dot{b} b} \partial_{\mu} \psi_b$$

(Ignoring a total derivative)

So you define $\bar{\sigma}^\mu = (\sigma^\mu)^T$?

where all indicies in the expression above are in the same place

@bolbteppa Why can't I just complex conjugate this. The indicies will just stay in the same place then

Something weird is going on