Recent content by p-brane

My opinion of [insert name of theory you don't really understand here but nonetheless don't like] is that [insert name of theory you don't really understand here but nonetheless don't like] is very aesthetically pleasing to those who understand [insert name of theory you don't really understand...

Marcus died?

There's not a single bit of experimental evidence that supersymmetry is wrong and an enormous amount of theoretical evidence that it's right. All experiment indicates right now is that the lightest supersymmetric particle is too heavy to for the LHC to produce. Get your facts straight.

By “isolated” is meant that there’s a gap between the mass m of any one-particle state and the mass of the lightest multiparticle state, which is the two-particle state with mass 2m. In other words, the mass of anyone particle state is isolated from the continuum of masses of multiparticle states.

7.36 is a "sum over singularities" indexed by λ. If the masses are those of one-particle states, the singularities are poles. If the masses are those of multiparticle states, the singularities are branch cuts. That's why I displayed the energy as a function of the index.

$${p^0} = {E_{\bf{p}}}\left( \lambda \right) = \sqrt {{{\left| {\bf{p}} \right|}^2} + m_\lambda ^2} \,\, \Rightarrow \,\,\,{E_{\bf{p}}}\left( \lambda \right){p^0} - E_{\bf{p}}^2\left( \lambda \right) = {\left( {{p^0}} \right)^2} - {\left| {\bf{p}} \right|^2} - m_\lambda ^2 = {p^2} - m_\lambda...

For a quantum theory to be correct, any result obtained using classical theory must somehow be obtainable using the quantum description. In the case of scattering in QFT, classical results are produced at tree-level. However this doesn’t mean that all tree-level results can be obtained or...

Most definitely NOT a stupid question!

As stated in the text, all external momenta have been set to zero. Thus the only momentum is the loop momentum k. If we take the gluon momentum to be k, by conservation of momentum one of the quark momenta must be -k.

They mean spacetime fermions and bosons, not world-sheet fermions and bosons.

Saying it slightly differently, M-theory in the original sense of Witten is the as yet unknown quantum theory having the 11d sugra as it's classical limit. By contrast, we already have quantum theories for the five standard string theories. The paper qsa refers to is a review of the earliest...

Let me restate my question in slightly modified form. "M-theory" as christened by Witten is a quantum theory. How is it related to the aforementioned 11-dimensional supergravity theory?

Do you know what the relation between the 11-dimensional supergravity theory discovered by Witten and the theory he christened "M-theory" actually is?

Forget about other systems. I was referring to your remarks about black holes. Can black holes have zero entropy and finite temperature or finite temperature and zero entropy? I posted the question for your benefit, not mine. What is "not at all clear" is what your point is. Are you saying that...

A system with entropy must have a temperature. Is the reverse true? I`m sure you meant the area bounding the region. What`s really interesting is that the dual system describing gravity involves only nongravitational interactions.