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Jun
18
awarded  Popular Question
May
24
comment Does Conformal Invariance of the Polyakov Action in Conformal Gauge imply Conformal Invariance of the Pre-gauge-fixed Polyakov Action?
Ah okay I think I get your meaning. Sorry to extend this discussion but I'm not sure my definition of 'conformal gauge' matches yours. In my definition a coordinate transformation (reparametrisation) is used to obtain the isothermal coordinates you describe, however in addition, a Weyl rescaling of the metric is used to take that metric to the flat space metric $\eta_{\alpha \beta}$. I think the Weyl rescaling means that the resulting action isn't simply the original action with a different choice of coordinates.
May
24
comment Does Conformal Invariance of the Polyakov Action in Conformal Gauge imply Conformal Invariance of the Pre-gauge-fixed Polyakov Action?
But doesn't coordinate invariance mean the same thing as reparametrisation invariance (i.e: $\sigma \rightarrow \tilde{\sigma}(\sigma)$)? And then, as you said in your original answer, the conformal invariance follows from this invariance under coordinate choice/reparametrisation?
May
24
comment Does Conformal Invariance of the Polyakov Action in Conformal Gauge imply Conformal Invariance of the Pre-gauge-fixed Polyakov Action?
Thanks for the reply @bechira. I think you might be using the term 'Weyl invariance' where I would use 'conformal invariance'. Just to be clear I use 'conformal' to mean reparametrisations of the coordinates that also result in a scaling of the metric. I take Weyl transformations to just be direct scalings of the metric. So, with these definitions, I see that invariance under general reparametrisations must mean conformal invariance also (as you describe). Is this correct?
May
23
comment Does Conformal Invariance of the Polyakov Action in Conformal Gauge imply Conformal Invariance of the Pre-gauge-fixed Polyakov Action?
On second thoughts, I think the answer is that the conformal transformations are just a special case of the reparametrisations. Since the pre-gauge-fixed action is reparametrisation invariant, it's also conformally invariant.
May
22
comment Does Conformal Invariance of the Polyakov Action in Conformal Gauge imply Conformal Invariance of the Pre-gauge-fixed Polyakov Action?
I may have found the answer to my own question: It is possible to show that the pre-gauge-fixed Polyakov action is invariant under Weyl transformations and diffeomorphisms (reparametrisations). Since conformal transformations are equivalent to combined reparametrisations and Weyl transformations it must follow that the pre-gauge fixed Polyakov action is conformally invariant. Some sort of feedback would be greatly appreciated however.
May
22
asked Does Conformal Invariance of the Polyakov Action in Conformal Gauge imply Conformal Invariance of the Pre-gauge-fixed Polyakov Action?
May
9
awarded  Tumbleweed
Apr
7
accepted Would Special Relativity Predict Time Dilation of a Geostationary Satellite Compared to an Observer on Earth?
Apr
7
comment Would Special Relativity Predict Time Dilation of a Geostationary Satellite Compared to an Observer on Earth?
Amazing, thank you.
Apr
7
asked Would Special Relativity Predict Time Dilation of a Geostationary Satellite Compared to an Observer on Earth?
Mar
4
accepted Is my Summary of a Spinor Bundle Associated with a String Worldsheet Correct?
Mar
4
comment Is my Summary of a Spinor Bundle Associated with a String Worldsheet Correct?
Awesome, thanks again!
Mar
4
comment Is my Summary of a Spinor Bundle Associated with a String Worldsheet Correct?
Thanks for the reply Tobias. Wouldn't $F_{x}$ be the space of oriented and pseudo-orthonormal frames of $T_{x}M$, just to be a little more precise? (Since $T_{x}M$ has Lorentzian metric signature). And I understand the $Spin(1,1)$-bundle now, thank you.
Mar
3
revised Is my Summary of a Spinor Bundle Associated with a String Worldsheet Correct?
grammar
Mar
3
asked Is my Summary of a Spinor Bundle Associated with a String Worldsheet Correct?
Feb
19
revised The Covariant Spinor Derivative in the Locally Supersymmetric Generalisation of the Polyakov Action and Potential Mistakes in the Literature
Explained how the Thesis and Wikipedia covariant derivatives don't match!
Feb
19
asked The Covariant Spinor Derivative in the Locally Supersymmetric Generalisation of the Polyakov Action and Potential Mistakes in the Literature
Feb
18
comment How to understand worldsheet fermion as a section?
Sorry for resurrecting this discussion. Just to make sure I understand the upshot of all this: Do the spinors $\psi_{+}$ and $\psi_{-}$ correspond to the sections of the spinor bundles $S=K^{1/2}$ and $S=\bar{K}^{1/2}$ respectively? Whilst $\psi^{i}_{+} \dfrac{\partial}{\partial \phi^{i}}$ and $\psi^{i}_{-} \dfrac{\partial}{\partial \phi^{i}}$ are sections of $K^{1/2} \times \phi*(TX)$ and $\bar{K}^{1/2} \times \phi*(TX)$ respectiveley?
Jan
10
comment Why can you re-write the functional measure of a real-valued field $\phi(x)$ as $\mathcal{D}\phi=\prod_{k_n^0>0}dRe \phi(k_n) d Im \phi(k_n)$?
Just to add to this discussion: I think the importance of the transformation being unitary might be the fact that unitary transformations don't alter the integration measure. So before the Fourier transformation we had $\mathcal{D} \phi(x)$ and after the transformation we still have $\mathcal{D} \phi(x)$, it just so happens that this can be expressed in terms of real and imaginary $\phi(k)$. Non-unitary transformations may have altered the integration measure so that $\mathcal{D} \phi(x)$ can't be used. It would be helpful if somebody could either confirm or reject this reasoning however.