particle physics - Are there massless bosons at scales above electroweak scale?

Spontaneous electroweak symmetry breaking (i.e. $SU(2)\times U(1)\to U(1)_{em}$ ) is at scale about 100 Gev. So, for Higgs mechanism, gauge bosons $Z$ & $W$ have masses about 100 GeV. But before this spontaneous symmetry breaking ( i.e. Energy > 100 GeV) the symmetry $SU(2)\times U(1)$ is not broken, and therefore gauge bosons are massless.

The same thing happens when we go around energy about $10^{16}$ GeV, where we have the Grand Unification between electroweak and strong interactions, in some bigger group ($SU(5)$, $SO(10)$ or others). So theoretically we should find gauge bosons $X$ and $Y$ with masses about $10^{16}$ GeV after GUT symmetry breaks into the Standard Model gauge group $SU(3)\times SU(2)\times U(1)$, and we should find massless X and Y bosons at bigger energies (where GUT isn't broken).

So this is what happened in the early universe: when temperature decreased, spontaneous symmetry breaking happened and firstly $X$ & $Y$ gauge bosons obtained mass and finally $Z$ & $W$ bosons obtained mass.

Now, I ask: have I understood this correctly? In other words, if we make experiments at energy above the electroweak scale (100 GeV) we are where $SU(2)\times U(1)$ isn't broken and then we should (experimentally) find $SU(2)$ and $U(1)$ massless gauge bosons, i.e. $W^1$, $W^2$, $W^3$ and $B$ with zero mass? But this is strange, because if I remember well in LHC we have just make experiments at energy about 1 TeV, but we haven't discovered any massless gauge bosons.

Answer

I think you have understood it almost well.

The masses do not change, they are what they are; at least at colliders. At high energy, it is true that the impact of masses and, more generally, of any soft term, becomes negligible. The theory for $E\gg v$ becomes very well described by a theory that respects the whole symmetry group.

Notice that to do so consistently in a theory of massive spin $-1$, you have to introduce the Higgs field as well at energies above the symmetry breaking scale. For the early universe, the story is slightly different because you are not in the Fock-like vacuum, and there are actual phase transitions (controlled by temperature and pressure) back to the symmetric phase where in fact the gauge bosons are massless (except perhaps for a thermal mass, not sure about it).

EDIT

I'd like to edit a little further about the common misconception that above the symmetry breaking scale gauge bosons become massless. I am going to give you an explicit calculation for a simple toy mode: a $U(1)$ broken spontaneously by a charged Higgs field $\phi$ that picks vev $\langle\phi\rangle=v$. In this theory we also add two dirac fields $\psi$ and $\Psi$ with $m_\psi\ll m_\Psi$. In fact, I will take the limit $m_\psi\rightarrow 0$ in the following just for simplicity of the formulae. Let's imagine now to have a $\psi^{+}$ $\psi^-$ machine and increase the energy in the center of mass so that we can produce on-shell $\Psi^{+}$ $\Psi^{-}$ pairs via the exchange in s-channel of the massive gauge boson $A_\mu$. In the limit of $m_\psi\rightarrow 0$ the total cross-section for $\psi^-\psi^+\rightarrow \Psi^-\Psi^+$ is given (at tree-level) by $$ \sigma_{tot}(E)=\frac{16\alpha^2 \pi}{3(4E^2-M^2)^2}\sqrt{1-\frac{m_\Psi^2}{E^2}}\left(E^2+\frac{1}{2}m_\Psi^2\right) $$ where $M=gv$, the $A_\mu$-mass, is given in terms of the $U(1)$ charge $g$ of the Higgs field. In this formula $\alpha=q^2/(4\pi)$ where $\pm q$ are the charges of $\psi$ and $\Psi$. Let's increase the energy of the scattering $E$, well passed all mass scales in the problem, including $M$ $$ \sigma_{tot}(E\gg m_{i})=\frac{\pi\alpha^2}{3E^2}\left(1+\frac{M^2}{2E^2}+O(m_i^2/E^4)\right) $$ Now, the leading term in this formula is what you would get for a massless gauge boson, and as you can see it gets correction from the masses which are more irrelevant as $m_i/E$ is taken smaller and smaller by incrising the energy of the scattering. Now, this is a toy model but it shows the point: even for a realistic situation, say with a GUT group like $SU(5)$, if you scatter multiplets of $SU(5)$ at energy well above the unification scale, the masses of the gauge bosons will correct the result obtained by scattering massless gauge bosons only by $M/E$ to some power.

fluid dynamics - Why is exhaling more forceful than inhaling?

By blowing at pencil, a piece of paper, or another object up to fifty centimeters away, I can cause it to move away from me significantly. But I can't move an object toward myself by inhaling sharply from that distance, even if it is extremely light. Why is that?

Answer

I think the reason is that when you are blowing on an object, you are making lots of air particles collide with it perpendicularly in one direction thus transferring a lot of momentum to the object. When you are sucking air in, the only force that's acting on the object is by the air particles that rush in to fill up the gap that you just created. These particles come in from all kinds of different directions failing to transfer momentum in a consistent way.

electromagnetism - What is the difference between induced emf and motional emf?

Are they two different phenomena or same phenomena having different properties in different situations? In what scenario should we consider induced emf and in what scenario should we consider motional emf? And which of the following does faraday's law and lenz law mention about? Are both caused due to induced electric field?

black holes - How does the Pauli exclusion principle create a force in degenerate matter?

My understanding is that when it comes to forming a white dwarf, it is the electron degeneracy pressure, due to the Pauli Exclusion Principle, preventing collapse in of the white dwarf. If the gravitational force is sufficiently large, then the electrons in the white dwarf will be forced to fuse with the protons to form neutrons, and the neutron star resists collapse in by neutron degeneracy pressure. If the gravitational force is even greater, then black hole will form.

How does the Pauli Exclusion Principle actually create a force? It seems to me from various things I have read that the force due to the Pauli Exclusion Principle increases as the fermions are squeezed closer together, although I am not sure why there is an increasing force and it is not simply the case that the fermions cannot be pushed into exactly the same position. It is as if the fermions know when they are approaching each other?

Answer

How does the Pauli Exclusion Principle actually create a force?

The Pauli exclusion principle doesn't really say that two fermions can't be in the same place. It's both stronger and weaker than that. It says that they can't be in the same state, i.e., if they're standing waves, two of them can't have the same standing wave pattern. But for bulk matter, for our purposes, it becomes a decent approximation to treat the exclusion principle as saying that if $n$ particles are confined to a volume $V$, they must each be confined to a space of about $V/n$. Since volume goes like length cubed, this means that their wavelengths must be $\lesssim (V/n)^{1/3}$. As $V$ shrinks, this maximum wavelength shrinks as well, and the de Broglie relation then tells us that the momentum goes up. The increased momentum shows up as a pressure, just as it would if you increased the momenta of all the molecules in a sample of air. A degenerate body like a neutron star or white dwarf is in a state where this pressure is in equilibrium with gravity.

everyday life - What's Optimal About Six Legs According to Physical Laws?

In many respects the insects can be regarded as the most successful class of animals in evolutionary terms. And one of the most common features of insects is that they (mostly) all have six legs.

Not discounting other traits, is there something about six legs that has helped insects achieve this success?

Can we use physical laws to analyze and determine an optimality of having six legs - perhaps such as stability?

Answer

I can think of two possible reasons: first, you can have half your legs up in the air at one time (as in walking - two on one side and one on the other, then change) and still be perfectly stable (3 legs = most stable, like a tripod); and second, if a predator chews off a leg on either side, you still have two legs (so you can still walk). I think those arguments are borderline biomechanical, rather than physical...

The first argument has some solid scientific backing - see for example http://web.neurobio.arizona.edu/gronenberg/nrsc581/powerpoint%20pdfs/cpg.pdf . It doesn't take a lot of brains to walk with six legs... I fact it can be done almost entirely with "local" neurons. That's a good thing when you don't have a lot of brains.

Quoting from https://answers.yahoo.com/question/index?qid=20090418111020AA75mgR :

Generalizing, insects walk with a metachronal gait and, with speed, a tripod gait - which involves a tripod stance - 2 legs on one side of the body and one on the other remain stationary while the other legs move forward, then the stationary legs walk as the others take a stance. In this way, walking involves maximum stability with a minimum of neural coordination. In fact, ganglia and other nerves and sensors located on each leg may contribute as much to the actual walking movement as the brain does. It's a very easy, stable and adaptable locomotory system which evolved from the basic arthropod body plan with 2 pairs of limbs on each body segment.

water - Do all phases of ice look the same visually?

I recently read about different phases of ice on Wikipedia. But I can't find any pictures of the different ice phases. Do they all look alike visually? If you weren't able to measure the pressure and temperature, would you be able to tell one phase from another in any way?

Answer

I don't know of any experimental results on the optical properties of ice at high pressures. I'm sure they must exist, but I couldn't find any relevent publications.

However there has been lots of work on theoretical calculations of the optical properties. See for example Ab initio investigation of optical properties of high-pressure phases of ice and Blueshifting the Onset of Optical UV Absorption for Water under Pressure. The results of these calculations are that the optical properties do not change in the optical spectrum, though you get big changes in the UV spectra.

So assuming you trust the calculations the answer is that all the different phases of ice look the same to the eye. All the phases are clear, and the refractive index doesn't change (much) so you wouldn't see a difference in the sparkle.

electrons - Confusion about calculating electrostatic energy using the electric field

Recently I learned that electrostatic potential energy of a system of charges can be calculated like so:

$$E = \int \frac{1}{2} \epsilon_0 \mathbf E^2 dV.$$

However, for a system of two point charges Q1 and Q2, we have

$$E = \int \frac{1}{2} \epsilon_0 (\mathbf E_1 + \mathbf E_2)^2 dV = \int_{}^{}\frac{1}{2}\epsilon_0 \mathbf E_1^2 dV + \int_{}^{}\frac{1}{2}\epsilon_0 \mathbf E_2^2 dV + \int_{}^{}\frac{1}{2}\epsilon_0 * 2 \mathbf E_1 \cdot \mathbf E_2 dV$$

from which we subtract the first two terms because they are already properties of the charges themselves.

Question:

Why do we not need to do something like this for charge distributions? Aren't charge distributions made up of many many little charges of magnitude $e$? Even if we treat the charge distribution as perfect continuous, won't there be infinitely infinitesimal charges?

Answer

Even if we treat the charge distribution as perfect continuous, won't there be infinitely infinitesimal charges?

It is instructive to apply a few vector calculus identities to put the electrostatic energy in another form: $$ U = \frac{\epsilon_0}{2} \int E^2 \, d\tau = \frac{\epsilon_0}{2} \int (-\vec{\nabla}V) \cdot \vec{E} \, d \tau = -\frac{\epsilon_0}{2} \oint V \vec{E} \cdot d\vec{a} + \frac{1}{2} \int V (\epsilon_0 \vec{\nabla} \cdot \vec{E}) \, d\tau \\= -\frac{\epsilon_0}{2} \oint V \vec{E} \cdot d\vec{a} + \frac{1}{2} \int V \rho \, d\tau $$ If $\rho$ is non-zero only over a bounded region of space (or in technical terms, $\rho$ has compact support), then it can be shown that the boundary integral vanishes in the limit of an integral over all of space. The second integral, meanwhile, will be finite so long as $\rho$ and $V$ are finite everywhere (since it's effectively an integral of a finite function over a finite region of space.) It's also possible to show that if $\rho$ is finite and of compact support, and if we require $V \to 0$ at infinity, then $V$ will be finite everywhere in space. So if you write this in terms of continuous charge distributions, the electrostatic energy will never diverge.

So that answers your question about whether the integral will diverge. But your question bespeaks a mindset that I would encourage you to move away from as you learn more about E&M: It is better to think of point charges as limits of smooth charge distributions, not the other way around.

It is instructive, when students are first learning E&M, to talk about point charges and infinitesimally thin wires; it's a lot easier to calculate fields, forces, energies, etc. when you only have to do a sum rather than an integral. However, Maxwell's equations are what really govern the behavior of electric and magnetic fields, and these are written in terms of charge and current densities $\rho$ and $\vec{J}$. If you try to model a "point charge" or a "line current" in terms of these quantities, they'll have to be infinite; and whenever you plug in an infinite quantity into an equation, you can't expect it to behave nicely.

Some of the calculations that students do in intro E&M can be rigorously re-derived for point charges by introducing the language of distributions, and in particular the Dirac delta function. You can go a long way by allowing your charge densities, fields, and potentials to be distributions rather than honest-to-god functions. But distributions can't do everything that functions can; and in particular, you're courting danger every time you try to multiply two distributions together, as you have to do to calculate electrostatic energy. Sometimes you can fix this problem with an ad-hoc subtraction (as you found above); sometimes you can't (the Abraham-Lorentz force.)

All of this means that it's better to think of Maxwell's equations as being well-defined only for smooth charge and current densities. So instead of thinking of a smooth charge density as "really" being a bunch of point charges, one should think of a point charge as being a limiting case of a series of smooth charge densities with a fixed total amount of charge as the occupied volume goes to zero. If you have this mentality, then the pathologies that sometimes when solving problems involving infinite charge or current densities will be a lot less confusing.