Yesterday we had the 4th weekly session of watching the string theory video lecture by Prof Shiraz Minwalla. Unfortunately the seminar room was occupied by some people and we had to move to an another room where I could not not configure my laptop with the projector and after trying for more than half an hour we decided to watch the video in the laptop only. In this part of his lecture he talked about tachyonic states in bosonic string theory and showed how 26 space-time dimensions appear naturally for the consistency of the theory. But the ground state of the theory still remains tachyonic. The first excited becomes massless for D=26. Since the 1st excited state behaves like a vector which has D-2 components and hence transform under SO(D-2) it can not be massive which transform under SO(D-1). I was wondering if we fix N=1 (1st excited state) in the the relation $$ m^2 = \frac{2}{\alpha}[ 2N-(D-2)/12]
$$ and requiring this state to be massless, arrive at D=26 how would the same value of D would make the second excited state massless. Then Prof Ramadevi cleared my doubt saying that the higher excited states are Higher dimensional representation of SO(D-2) and hence need not be massless. Then he talked about $$ N= \tilde{N} = 1 $$ where the tilde means the right moving ( in case of closed string we have both left and right moving fourier modes). This corresponds to $ 24 \times 24 $ representation of SO(24) which will give rise to a massless symmetric tensor field $G_{\mu \nu}$, a massless antisymmetric tensor field $B_{\mu \nu}$ and a scalar field $ \Phi$ called the dilaton. Thus gravity comes out automatically in closed string theory. After that he gave a brief introduction to path integrals which he will probably use in his next lecture.
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