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2

The special state of motion you're talking about is often called the Hubble flow. (edit: oops, Ben Crowell already mentioned this.) I think that in modern slow-roll inflation the source of this asymmetry is an asymmetry in the tiny (Planck-scale?) seed of inflation, whatever it was, inflated by a factor of $e^{60}$ or more. The inflaton potential has to be ...


1

The structure of general relativity does not allow the theory itself to be classified in terms of global, discrete symmetries such as time-reversal. The Einstein field equations don't refer to a time coordinate; they're expressed tensorially, which means that they are completely independent of what coordinates you choose. Since there is no guarantee that you ...


4

Nice question. First off, there's a definitional problem because we can't apply a Lorentz boost to the universe as a whole. Lorentz symmetry is a local thing. So when we talk about "Lorentz symmetry" for the universe as a whole, I think we have to keep in mind that we mean something a little different. Basically if the "Lorentz symmetry" has already been ...


1

A pretty exhaustive summary in the context of Standard Model already exists in the following source: ''Dynamics of the Standard Model'' - Donoghue, Golowich, Holstein, Chapter 3 - Symmetries and Anomalies A limited preview can be found here. (Embarrassingly though, the very first page of the chapter is excluded from Google's preview!) But here's the ...


0

Noether's theorem in its usual form assumes that the system (in this case a fluid) is governed by an action principle. We assume for simplicity that the fluid consists of just one type of fluid particles. I) In the Lagrangian fluid picture, the (local) conservation of fluid particles is manifest from the onset, since the dynamical variables are the labels ...


3

I'd say that there is not a systematic summary of the status of symmetries on particle physics, but if any, it should be spread all over the PDG review. However, I'd like to comment on a few points. So far Lorentz symmetry is exact on all sectors.${}^\dagger$ Scaling (part of the conformal transformations) is broken once an energy scale is introduced in ...


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The inertia ellipsoid is computed from an integral about an axis - in other words you rotate the object. This will "smooth out" any symmetries and typically increase the symmetry. Sorry this is a "early morning" intuitive explanation - maybe someone else will give you a more formal answer.


1

In the region between two spherical plates, we have, $$\begin{array}{l} \nabla \times E = 0\,\,\, (1)\\ \nabla .\left( {\varepsilon \left( \theta \right)E} \right) = 0\,\,\, (2) \end{array}$$ The first equation leads to the definition of scalar potential $\Phi$, i.e. $E = - \nabla \Phi $. Therefore, from Eq. (2) we have, $$\nabla .\left( {\varepsilon ...


4

Congratulations, you made me look into this for the last hour! And, unfortunately, I believe the answer is: Nope We are looking for a Ricci-flat Riemannian symmetric space, since your isometry group is a Lie group. I spent some time trying to construct the Ricci-flat manifold from the irreducible symmetric spaces given there, but couldn't figure out a good ...


2

In a comment you write space time symmetries don't fit into the framework of the action since the action is a functional on the fields only not also on space time (space time here appears merely as a dummy variable This isn't quite right. A given spacetime transformation often induces a transformation on fields themselves, and in this way, spacetime ...


-1

The first half of this theorem is: you can substitute the total mass in the centre to get the net gravity. It probably don't stays, although it needs a little bit of integration. The second (there is no net gravity inside the shell), I think it passes, if the shell is convex. The proof is exactly the same as you can see in the wiki page.


3

You can easily convince yourself that this is not true. Just imagine a test particle on one side of a spherical shell. If the density is uniform, the theorem will apply. Now, however, decrease the density of that half of the shell until it's 0. Obviously this other half would attract the test particle. Alternatively, if changing density is not an option and ...


1

There are 5 standard model (SM) multiplets per generation of fermions. The SM gauge group is $\mathcal{G}_\text{SM} = SU(3)_C \times SU(2)_L \times U(1)_Y$. Various multiplets can then be written as $\mathcal{G}_\text{SM} \ni x = (C,T)_{(Y)}$, where $C$ denotes colour multiplet, $T$ weak isospin multiplet and $Y$ hypercharge value. Multiplets (1st ...


1

Yes! In fact, one of my (few) own research experiences was closely connected to random matrix theory! I did a small project under supervision of Enrico Pajer from Princeton University last January. The work I did was based on earlier work of his, which used random matrix theory in the context of inflationary theory, using a model where many fields ...


1

About the supposed paradox: $u$ and $\bar d$ have the same isospin quantum numbers, but not all the other properties. If you restrict your problem to only study the isospin space, you will not see that they have different charge and other different quantum numbers. About the charge: I don't know where your equation comes from, but it seems close to the ...


3

Group actions in classical field theory. Let a classical theory of fields $\Phi:M\to V$ be given, where $M$ is a ``base" manifold and $V$ is some target space, which, for the sake of simplicity and pedagogical clarity, we take to be a vector space. Let $\mathscr F$ denote the set of admissible field configurations, namely the set of admissible functions ...


1

Emmy Noether proved both the theorem and its converse. Look for the book "The Noether Theorems" for a precise and discussed formulation of her statements, as well as a translation of the original paper. It seems there is a link to the pdf in the princeton math website (I don't know about copyright issues, however).


1

Pick a point above the plane. From a point in the plane directly under the point above, draw a circle of some radius. Consider the contribution of the charge elements along the circle to the electric field at the point above the plane. Since the charge density is uniform, the horizontal components of the electric field from charge elements on opposite ...


1

An answer connected to Gauss law (I hope everything is correct, since it's long ago for me ... so no warranty): An infinite plane of uniform charge for example in the z-plane has the charge distribution: $\rho=q\,\delta(z)$ Thus, the electrostatic potential should be $\Phi=\frac{q\,|z|}{2\pi}$. Hence, the electric vectorfield is: ...


8

The answer by @NowIGetToLearnWhatAHeadIs is correct. It's worth learning the language used therein to help with your future studies. But as a primer, here's a simplified explanation. Start with your charge distribution and a "guess" for the direction of the electric field. As you can see, I made the guess have a component upward. We'll see shortly why ...


0

With respect to your test charge, there will always be an equal number of charges on your plane in all directions because the plane is infinite. So for every charge "in front" of your test charge there will be a charge "behind" your test charge. And for every charge to the left of your test charge will be a charge to the right. What this means is that no ...


8

You have to realize that the system is invariant under rotations about the normal to the plane. Then then electric field must also be invariant under these rotations. An electric field component in the plane does change under such a rotation, so such a component must not exist if we have this invariance. Thus the electric field is purely along the normal to ...



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