Jonathan Gleason
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 Sep 12 revised Derivation of the Polyakov Action deleted 56 characters in body Sep 12 comment Derivation of the Polyakov Action . . . Of course, I guess you might just say that the simplest one that works is the way to go (in this case, an action with just one extra field). Still, I have to admit, I'm not completely satisfied with this answer. Sep 12 comment Derivation of the Polyakov Action . . . And hell, if we're going to add in yet another field, we could probably find get another symmetry along with it. Sep 12 comment Derivation of the Polyakov Action I like this, but this method doesn't really convince me that putting in another auxiliary field is the way to go. Let's say in QFT we want a theory with a complex scalar field with a $U(1)$ symmetry. Of course, we could always introduce other fields into the theory, but that's not what one usually does unless you wanted the extra fields to begin with. And even if you decide that introducing a new field is the way to go, why stop at one? Surely we could add two new fields that respected all the symmetries we wanted . . . Sep 12 revised Derivation of the Polyakov Action added 78 characters in body Sep 12 asked Derivation of the Polyakov Action Sep 6 revised Fine-Tuning, the Hiearchy Problem, and Mass in the Standard Model edited body Sep 4 asked Symmetry Breaking and Vacuum Expectation Values Sep 3 revised A rock connected to one end of string in circular motion gets released.. and what happens? added 156 characters in body Sep 3 accepted Divergence of One and Two Graviton Exchanges Sep 3 comment Divergence of One and Two Graviton Exchanges One last question, and then I think I got it. How do we determine the kinetic term for $h_{\mu \nu}$? (I ask because this will allow me to determine the appropriate dimensions of a spin $2$ field, and hence the appropriate dimensions of the coupling constant.) Sep 3 answered A rock connected to one end of string in circular motion gets released.. and what happens? Sep 2 comment Divergence of One and Two Graviton Exchanges Are you assuming that the particle is a scalar, so that the interaction is of the form $\partial ^\mu \phi \partial ^\nu \phi g_{\mu \nu}$? This is the only way I could see you getting momenta coming into what appears to be your vertex factors. But if my dimensional analysis is correct, $[\partial ^\mu \phi \partial ^\nu \phi g_{\mu \nu}]=L^{-4}$, so that the corresponding coupling constant would be dimension-less, and in particular, could not be proportional to $1/M_P$. What's going on? Sep 2 revised Divergence of One and Two Graviton Exchanges deleted 1 characters in body Sep 1 comment Vacuum Expectation Value and the Minima of the Potential . . . In practice, you do this by writing the Lagrangian in terms of $\phi :=\phi _0-v$, where $v$ is some minimum of the potential and $\phi _0$ is the original field. My question could then be equivalently phrased as "Why does this guarantee that $\langle 0|\phi (x)|0\rangle =0$?". Sep 1 comment Vacuum Expectation Value and the Minima of the Potential @MichaelBrown Indeed, I was under the impression that you had to have this as a re-normalization condition in order to apply LSZ (that is, a hypothesis require for the LSZ Reduction Formula to hold was that $\langle 0|\phi (x)|0\rangle =0$). In fact, I thought this was the entire idea behind the symmetry breaking: you must re-write your Lagrangian in terms of the re-normalized field (with vanishing VEV), and if the bare field had a non-vanishing VEV, this will 'break' the symmetry . . . Sep 1 comment Vacuum Expectation Value and the Minima of the Potential So then what is the proof of this leading-order approximation? Sep 1 asked Divergence of One and Two Graviton Exchanges Aug 31 asked Vacuum Expectation Value and the Minima of the Potential Aug 31 revised The vacuum in quantum field theories: what is it? Expanded the question