field theory - Is it enough to assume $F_{munu}to 0$ at infinity but not $A_mu$ to derive the equation of motion?

Suppose the the Lagrangian $\mathscr{L}$ of the free electromagnetic field is augmented with the term $$F_{\mu\nu}\tilde{F}^{\mu\nu}=\partial_{\mu}(\epsilon^{\nu\nu\lambda\rho}A_\nu F_{\lambda\rho}).$$ Since this term is a total divergence it can be converted to a surface integral over $A_\nu F_{\lambda\rho}$.

Now, if $F_{\mu\nu}$ is assumed to vanish at the boundary (at infinity) it is sufficient to argue that such a term drops out in the equation of motion (EOM) this abelian theory even if $A_\mu$ is nonzero.

Questions

1. 
   

   Is $F_{\mu\nu}\to 0$ at infinity sufficient to derive the EOM in absence of the $F\tilde{F}$ term? Did anywhere in the derivation of the EOM we need to use that $A_\mu$ itself have to vanish at infinity/boundary?

   
   


2. 
   

   As a related question, physically, why do we need $F_{\mu\nu}$ to vanish at infinity?

   
   

Next, if we augment the Lagrangian of a nonabelian Yang-Mills (YM) theory with a $F\tilde{F}$ term, unlike the previous case, it does affect the EOM unless $A^a_\mu$ itself are assumed to vanish at infinity.

3. Do we assume $A_\mu^a=0$ at infinity only for a nonabelian YM theory or do we always have to assume $A_\mu=0$ irrespective of whether the theory is abelian or nonabelian?

group theory - Lie algebra in simple terms

My question is regarding a vector space and Lie algebra. Why is it that whenever I read advanced physics texts I always hear about Lie algebra? What does it mean to "endow a vector space with a lie algebra"?

I'm assuming it is the same Lie from the Lie groups? My current knowledge is that the Lie groups are "to do with rotations of molecules".

I'm not after lots of detail but would like a basic understanding of what this means and why it is such a prevalent idea.

Answer

It's an enormous subject, but briefly:

Lie groups are smooth groups. Technically, Lie groups are sets that are both a smooth manifold, like a sphere for instance, and also have a group structure (multiplication operator, inverses, and an identity). The group multiplication and inverse must be smooth (differentiable) functions on the manifold.

As you mentioned, the group of rotations in 3-dimensional space, called SO(3), is an example of a Lie group. Rotations have a group structure because you can compose or invert rotations and get other rotations, and they are also a smooth manifold because you can smoothly vary the axis or angle and so move continuously from one rotation to another.

There are many other examples of Lie groups. Many types of geometric transformations on different spaces form Lie groups. They show up in physics in transformations of spacetime (the Poincaré group, most generally), and in so-called "internal symmetries" that transform different quantum fields into each other (often special unitary groups of various dimensions). Another example is diffeomorphism groups, which show up in general relativity and string theory and are also Lie groups.

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As for Lie algebras, they are closely related to Lie groups. A Lie algebra basically consists of the "infinitesimal elements" of a Lie group, i.e. the "elements infinitesimally close to the identity". (I put that in scare quotes because in standard analysis, infinitesimal elements don't really exist—technically, a Lie algebra is defined on the tangent space of the Lie group at the identity. Still, the picture of infinitesimal elements is a useful and intuitive way of thinking about this.)

For example, in the case of rotations, we would be talking about rotations about any axis by infinitesimal angles.

When you multiply two group elements that are very close to the identity, the group multiplication looks like a vector sum—basically the same way that $(1+\delta)(1+\epsilon) \approx 1 + (\delta + \epsilon)$ when $\delta$ and $\epsilon$ are small. Similarly, $(1+\epsilon)^{-1} \approx 1 - \epsilon$ and so group inversion looks like vector negation. So the Lie algebra inherits its operations from those of the underlying Lie group, but it doesn't itself look like a group—instead, it looks like a vector space.

The Lie algebra also has, in addition to the standard vector space operations, a bilinear operation called the Lie bracket, denoted $[x, y]$ (where $x, y$ are two vectors in the Lie algebra and $[x, y]$ generates another vector). This operation measures "how noncommutative" the Lie group is; roughly speaking, it corresponds to the commutator $[A, B] = AB - BA$ of the Lie group.

Now, the funny and interesting thing about a Lie algebra is that even though it's derived from just an infinitesimal slice of a Lie group, it contains, encoded within it, almost everything there is to know about the group that it came from! You can actually reconstruct the entire Lie group, starting from just the Lie algebra, by using the exponential map—a generalization of the ordinary exponential function.

(It's almost because there are some cases where different Lie groups have the same Lie algebra, but different global structures—for example, SO(3) and SU(2).)

And since Lie algebras are vector spaces, and the Lie bracket is a bilinear operation, all you really need to have is a set of basis vectors for the Lie algebra and know what the Lie bracket does to each pair of basis vectors.

Such a set of basis vectors is called a set of generators of the group. If you apply the Lie bracket to each pair of generators and write down the resulting vectors as coordinates in the same basis, the set of numbers you obtain are called structure constants.

From the generators and the structure constants, you can generate the Lie algebra and thence the entire Lie group (except for ambiguities of global structure as mentioned above)! This makes Lie algebras a very powerful tool for understanding the Lie groups that show up in physics. For example, in particle physics, the gauge bosons (photon, W, Z, gluons) are closely related to the generators of internal symmetry groups; momentum and angular momentum are related to the generators of the Poincaré group, and so on.

There's much more that could be said about this—I haven't even mentioned representations!—but this is probably enough for one answer, so I'll stop here. :)

quantum mechanics - Do protons exchange photons with electrons?

I'm sorry for this question but, I just don't get it. According to the electromagnetic field theory, electrons repel each other by exchanging photons. How do protons attract electrons, by photon exchange?

electromagnetism - Are there Optical Magnetic Mirrors (OMMs) which reflect via interaction with the magnetic field?

The most familiar mirror relies on reflection from a metal, where the very strong interaction of the electric field component of the EM wave with the conduction electrons produces a reflected wave.

Dielectric mirrors rely on the weaker interaction of the electric field with bound electrons, and a periodic modulation of the dielectric constant (alternating quarter wave thicknesses) builds up a reflected wave with high efficiency.

Both of those reflect due to the interaction of the electric field with electrons - be they bound or free.

However, the paper Optical Magnetic Mirrors (Schwanecke et al. 2007 J. Opt. A: Pure Appl. Opt. 9 L1) begins with:

Abstract: We report the first demonstration of an optical magnetic mirror achieved by nanostructuring a metal surface. It reverses the magnetic field of an incident wave upon reflection, acting as an ‘optical frequency superconductor’.

The introduction goes on to say:

A magnetic mirror imposes extremely unusual electromagnetic boundary conditions: it does not reverse the electric field of a light wave upon reflection, but reverses the magnetic field. This property renders it distinctly different from a normal mirror, which instead reverses the electric field of a reflected wave...

And so the working definition of a magnetic mirror here seems to be a surface which produces a reflection without reversing the direction of the electric field, but instead reversing the direction of the magnetic field.

However it's not clear to me if this is due to the interaction with the magnetic field directly.

In the paper Optical Magnetic Mirrors without Metals (also here Liu et al. 2014, Vol. 1, No. 4, 250) a nano-patterned purely dielectric surface is said to be an Optical Magnetic Mirror (OMM). I understand that the effect is due resonant modes within the dielectric patterns, and some of these modes are called 'electric' and some 'magnetic' as a naming convention, somewhat like there are TE and TM modes in a dielectric fiber waveguide. However, aren't these still due to the interaction of the electric field?

Question: Are there surfaces, or relatively thin nano-structures which reflect light via interaction with the magnetic field of the wave? Is there a physics distinction between that magnetic interaction, and a patterned surface that simply passes the test of not inverting the electric field?

classical mechanics - Is it more work to put more (apparent) effort to get the same outcome?

I was taking my dogs for a walk yesterday evening when this question occurred to me.

The two dogs were pretty enthusiastic about the walk and wanted to run on ahead, so the leads were taut and they were sort of gently pulling me along. But they're quite small so I could easily control the speed at which we walked. At this stage they were presumably using more energy than that required to simply walk at that same speed given the resistance I was providing.

Suddenly, a cat spotted us and ran away. The dogs, upon seeing this, tried to give chase, running and pulling the leads with all their might. But I held onto the leads and continued walking at the same speed as before. So the result was two dogs frantically trying to pull and run but not going any faster.

So my question is: Are the dogs doing any more work (or using any more energy) when trying to chase the cat than when they were gently tugging along? Intuitively I think yes, (especially given the gasping and panting coming out of both of them) but I can't figure out why: they are moving at the same speed and pulling the same weight (i.e. me) as before....

thermodynamics - Where do atoms of H and He go when they rise in the atmosphere?

What happens to atoms/molecules of H and He when we let them freely fly in atmosphere do they get into space? Is there any particular way to enumerate the 'force' that drives them upwards? Does anybody know how this force changes with their altitude/distance from Earth's surface?

quantum field theory - Renormalization and Conway/Surreal Numbers

In the final chapter of his book "An Interpretive Introduction to Quantum Field Theory", Paul Teller writes about three interpretations of renormalization in quantum field theory. In particular, Teller denotes one of these approaches "Real-Infinities Renormalization," which eschews regularization and as such, he concludes, doesn't make much sense. However, Teller suggests the "Real-Infinities" approach could possibly be made rigorous using Conway/surreal numbers. The relevant section of Teller's book can be found on Google Books here: Teller, pg. 162.

My question is this: has anyone tried to formulate renormalization in quantum field theories without the use of regulators, in terms of Conway/surreal numbers as suggested by Teller?