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In this answer we formally show that a (quasi)symmetry of an action implies a corresponding symmetry of its EOM$^{\dagger}$. The answer does not discuss form covariance of EOM. For further relations between symmetries of action, EOM, and solutions of EOM, see e.g. this Phys.SE post. Let us first recall the definition of a quasi-symmetry of the action ...


5

Fluctuations of density in the universe naturally become greater with time because matter is attracted to regions that are denser than average, and as a result they get denser still and the other regions less dense. So if there were even tiny density fluctuations in the early universe they would have grown into the density variations we see today - ...


4

The two notions are indeed related. Take for example the Weyl anomaly of bosonic string theory: the classical (Polyakov) action $S$ is invariant under Weyl rescalings of the worldsheet metric $\gamma_{ab}$, i.e. $$S[\gamma_{ab}(\tau,\sigma)]=S[\exp(2\omega(\tau,\sigma))\gamma_{ab}(\tau,\sigma)]=S[\gamma'_{ab}(\tau,\sigma)].$$ Since this is a conformal ...


4

There's (almost) no difference between matter and antimatter, and it would be very difficult to know if a galaxy far away is made of matter or not. The reason we know that the visible is made of matter, and not half-matter and half-antimatter is that it this were the case, then we would see a ton of events where anti-matter galaxies would smash into a ...


4

Neither the space between galaxies nor the space within galaxies is completely empty, but rather is filled with a diffuse gas. If there was an isolated galaxy, or region of galaxies that was made up of antimatter, there would be an interface between the two regions, and this region would be rich in annhilation events, which would be visible in X-Rays. We ...


4

First off, spherical symmetry isn't really the best description. Cosmological models usually assume that the universe is (approximately) homogeneous and isotropic. That's a higher degree of symmetry than spherical symmetry. Spherical symmetry would normally be used to describe something that has a lower degree of symmetry, so that there is a center. The ...


3

There is no contradiction since you should not be doing complex linear combination of generators that are already hermitian (indeed, you want the group transformation to be unitary). Hence, your linear combination with a complex coefficient (that brings you away from the $SU(2)$ group) doesn't imply a massless excitation.


3

(This is a largely a response prompted by your comment.) You can get the answer by just remembering the commutation/anti-commutation properties of the $\gamma$ matrices, and the fact that ${\bar \psi} = \psi^{\dagger} \gamma^0$. To see the following, you would have to expand the exponential factor, up to linear order $e^{M} = I + M + \ldots$. (I'm not ...


3

Newton's third law states that if object A acts on object B with force $\mathbf{F}_{AB}$, then object B must act on object A with force: $$\mathbf{F}_{BA}=-\mathbf{F}_{AB}$$ When expressed in terms of A and B's momentum, the same equation can be written as: $$\frac{\mathrm{d} \mathbf{p}_A}{\mathrm{d}t} = -\frac{\mathrm{d} \mathbf{p}_B}{\mathrm{d} t}$$ ...


3

If there is no external force with explicit time dependence, then the harmonic oscillator contains no explicit time dependence. Then the system has time translation symmetry, i.e. the result can only depend on the difference $T =t_b-t_a$, not on $t_a$ and $t_b$ individually.


1

The volume is defined as $V=|\vec{x}\cdot (\vec{y}\times \vec{z})|$, to avoid negative volumes. So under space inversion, volume would remain invariant, which is physically desirable since it should still take up the same amount of space. New: I guess you could define V to be a pseudoscalar, but that might have consequences. For example, mass is something ...


1

It has always struck me as odd that Newton's 3rd law was never explained in any detail in my physics degree lectures. I have quoted it glibly for decades, but it was only when a 10 year old asked me WHY? that I thought about how to understand this law. In the macroscopic world, we apply a force and there is an "equal" reaction force. What's actually going ...


1

Say we have two different coordinate systems $F=(x,y,z)$ and $F'=(x',y',z')$. Consider the basis spanned by the eigenvectors of the $L_z$ operator for a given $l$, $\{|l \; m\rangle\; ; \;m=-l,\dots,l\}$. Now, one can find the eigenvectors of the $L_{z'}$ on this basis. In general, the eigenvectors of $L_{z'}$ will be linear combinations of the ...


1

I assume to deal with an autonomous, first order (at least $C^1$ or smooth) system of ordinary differential equations and that the hypotheses sufficient for existence and uniqueness of maximal solutions are satisfied. You may always reduce to the case of a first order system by adding auxiliary variables, $\dot{x}$, to the initial system of differential ...


1

D. Hilbert derived the (same as Einstein's) equations of general relativity by demanding the invariance (form of symmetry) of the Einstein-Hilbert action under general differentiable coordinate transformations, i.e diffeomorphisms So this is the symmetry associated with General Relativity, also refered to as general covariance. UPDATE: Note that all ...


1

It is unclear what you mean by transformation involving space-time, well, at least, I found two possibilities. The mentioned transformations are not gauge, since $\Lambda$ does not depend on space-time. So one way to make it involve space-time is to set $\Lambda \rightarrow \Lambda(x)$. The given Lagrangian is not invariant under these gauge ...



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