So there isn't a contradiction there? If something is rotating, say earth, the east hemisphere could be at point x in space and after a few hours it will be at point x'. What do you say?? @Dale
Can metrics be stationary and rotating at the same time? Doesn't stationary here means that the metric is time-independent. Thus, if a metric is time-indepedent how could it be rotating?
Why do, we, in susy deal with "classical" solutions and "classical" eom? So is susy a classical theory? Because in this https://www.physicsforums.com/threads/scalars-and-special-geometry.840811/ fzero mentioned something about "quantum" theory in his first comment on that post.
In special relativity, we can prove that the metric is -+++ for all observers and that is by making use out of lorentz invariance. Some on this forum say that it comes as a result of constancy of light and others say that Minkowski predated einstein in making that metric, which was confusing...
I am afraid I can't but use the terminology. Maybe my question is simply, is it okay to call the term involving ##\Gamma## an inertial acceleration and still use your previous answers to understand things? If not, what is the exact reason that drove the author to say that this term involving...
@PeterDonis I am confused. I have to say. Then what happens to your answer here:
How can this answer (that answered my question about why sum and OP's question what experience does this tell us) be translated after knowing that the book we're using means different things?
Also, notice there...
@PeterDonis , I ve read some place that when we are in Minkowski space we have##\frac{d^2}{d\tau^2}x(\tau)=0## and then we we do general coordinate transformation, we get ##\frac{d^2}{d\tau^2}x^{\mu}+\Gamma^{\mu}_{\nu\lambda}\frac{dx^{ni}}{d\tau}\frac{dx^{\lambda}}{d\tau}=0##, the author calls...
@PeterDonis thanks very much for your answer. I understand your argument very nicely if I considered e.g. when you said " you can make "gravitational acceleration" vanish everywhere, globally, by choosing appropriate coordinates." is correct. However, I can't tell you that I know what you mean...
Hello again, I have a question please @haushofer
So, when you say
, you mean the metric (boson) transforms into "0" which is the same thing as ##\psi_{\mu}## (fermion) because the solution above was ##\psi_{\mu}=0##.
BUT, when you say
, the fermion (##\psi_{\mu}##) transforms into what?Should...
Thanks a lot for your explanation @haushofer ! I have read it but still I have to reread it with more attention, and will then get back to you next year if I have any questions :p. However, I wanted to wish you a happy new year.:oldbiggrin:
@haushofer from what I read the answer is not quite precise and does not answer the questions samuelphysics is asking about. What he's asking is in short "What does it mean to have a preserved supersymmetry". He's practically asking what does it mean that local supersymmetry variations must...
@nrqed I want to ask you about this because @samuelphysics's last question got me a little confused. We say that a Hermitian manifold is one
in which unmixed components of metric tensor vanish (##g_{\alpha\beta}=g_{\bar{\alpha}\bar{\beta}}=0##). Is that correct? So every Hermitian metric there...