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The Kaluza-Klein metric, by reduction, can be written as a ##(4+m) \times (4+m)##symmetric matrix, where ##m## is the dimension of the additional spacetime (if we decompose ##M_D = M_4 \times M_m##). It was show by Bryce de. Witt that, if the...
OBS: Ignore factors of ## (2 \pi) ##, interpret any differential ##dx,dp## as ##d^4x,d^4p##, ##\int = \int \int = \int ... \int##. I am using ##x,y,z## instead of ##x_i##.
Honestly, i am a little confused how to show this "triangle-star duality". Look, the propagators in positions space gives...
I would appreciate if someone could help me to understand what is happening in section 12.3 from the Howard George's book.
First of all, the propose of the section is to show how $SU(3)$ decomposes into $SU(2) \times U(1)$. But i can't understand what is happening. First of all, i can't get the...
We have a Lagrangian of the form:
$$
\mathcal{L} = \overline{\psi} i \gamma_{\mu} \partial^{\mu} \psi - g \left( \overline{\psi}_L \psi_R \phi + \overline{\psi}_R \psi_L \phi^* \right) + \mathcal{L}_{\phi} - V(|\phi|^2)
$$
Essentially, what we are studying is spontaneous symmetry breaking...
I joined this forum years ago at the beginning of my undergraduate program. Now, I am almost graduated. Sometimes, I think about where the other students and users who joined with me are (mainly, one genius boy that i can't renember the user, but people will know who i am talking about).
Ups, just tiped it wrong.
So it should be
$$C_{rb} \psi^b + \psi^{a}C_{ar}$$
By the way
"Why do you think it's zero?"
Well,
$$C_{rb} \psi^b + \psi^a C_{ar} = C_{rb} \psi^b + \psi^b C_{br} = \psi^b ( C_{rb} - C_{rb} ) = 0$$
Where i have used that ##C## is anti-symmetric, and that since we...
Suppose i have a term like this one (repeated indices are being summed)
$$x = \psi^a C_{ab} \psi^b$$
Such that ##C_{ab} = - C_{ba}##, and ##\{\psi^a,\psi^b\}=0##. How do i evaluate the derivative of this term with respect to ##\psi_r##?
I mean, my attempt g oes to here
$$\frac{\partial...
$$i \gamma^{\mu} \partial_{\mu} \psi = m \psi_c \\
i \gamma^{\mu} \partial_{\mu} \psi_c = m \psi
$$
Where ##\psi_c = C \gamma^0 \psi^*##
Show that the above equations can be obtained from the followong lagrangian
$$
L = \overline{\psi} i \gamma^{\mu} \partial_{\mu} \psi - \frac{1}{2} m \left...
I think it is cool to mention that, while seen as a particle, the graviton indeed provides non-renormalizable results. But, if avaliated as a string excitation, and then the Feynman diagrams being smooth 2-dimensional surfaces, there are no ultraviolet divergences at all.
Where
##:## really means normal ordered, in the sense that ##:A(w)B(z): = \lim_{w \to z} \left ( A(w)B(z) - \langle A(w)B(z) \rangle \right )##
##\partial X(z) = \frac{\partial X(z)}{\partial z}##
How do we go form the first line to the second one?? I am not understanding it!
it seems to me...
I don't want to be rude, but you are missing trivial points about elementary calculus, as chain rule and product rule.
May i ask, have you take those class? Don't try to skip steps in your learning process.