quantum field theory - Hermitian adjoint of 4-gradient in Dirac equation

I'm having issues deriving the Dirac adjoint equation, $$\overline{\psi}(i\gamma^{\mu}\partial_{\mu}+m)=0.\tag{1}$$ I started by taking the Hermitian adjoint of all components of the original Dirac equation, giving me $$\psi^{\dagger}(-i\gamma^{\mu\dagger}\partial_{\mu}^{\dagger}-m)=0.\tag{2}$$ The adjoint of the gamma matrices is defined to be $\gamma^{\mu\dagger}=\gamma^0\gamma^{\mu}\gamma^0$, so no issues there. Now intuitively, I would think that the adjoint of the 4-gradient would be $\partial_{\mu}^{\dagger}=-\partial_{\mu}$. In non-relativistic quantum mechanics, it can be shown that first derivative operators are anti-Hermitian, so for example, $\frac{d}{dx}^{\dagger}=-\frac{d}{dx}$. So I would think this would the same case for the 4-gradient, but apparently it isn't. Among the many derivations I've gone over, for example, on page 77 here, it claims that the 4-gradient is self-adjoint. Could someone please explain why my intuition is incorrect?

Answer

Expanding on my comment.

The basic idea is that what you mean by adjoint depends on the vector space being considered. For example, we might have $\mathbb{C}^n$ as our space, with the usual inner product; in that case, the adjoint of a vector or matrix is the transpose conjugate. Note that technically, taking adjoint of a vector doesn't return a vector, because row vectors and column vectors belong to different spaces.

We could also use $L^2(\mathbb{R}^n)$ as our vector space. Its elements are functions, and the inner product is defined by

$$(f,g) = \int d^n x\ f^* g$$

You can take adjoints here too, using the inner product defined above. It can be shown that the derivative operator, which is a linear transformation on $L^2$, is anti-Hermitian.

Now to the Dirac equation. The vector space (that is, spinor space) being considered here is $\mathbb{C}^4$, not $L^2$. That is, $\psi$ is a vector because it has four components, not because it's a function. The fact that its components are functions is irrelevant here. When we take adjoints, we transpose and conjugate vectors and matrices. The derivative is an operator if you think about what it does to functions, but it is not a $4\times4$ matrix; it does nothing to spinors. Therefore, the particular adjoint we're doing here doesn't affect it.

What physical principle determines microstates in quantum mechanics?

If we want to calculate mean magnetisation of an equilibrium two-level-system, we know that we can resolve the identity $ \mathbf{1} = \sum_i | E_i \rangle \langle E_i |$ and giving us a uniform measure over the states of the system.

I then remember my classical stat mech, shove in some Boltzmann factors and obtain the gibbs state $$\rho = Z^{-1}\sum_i \mathrm{e}^{- \beta E_i}| E_i \rangle \langle E_i | $$ and can calculate the mean magnetisation, for a two level system with $E_0 = -B/2$ and $E_1=B/2$ we then get $\langle m \rangle = \mathrm{Tr}[\sigma_z \rho]/2 = \tanh \left(\beta B/2 \right)/2$. Alls seems very good.

However if I was ignorant to this result, I might also say I can write down a state $$ | \psi \rangle = \cos \frac{\theta}{2} | 0 \rangle + \mathrm e^{\mathrm i \phi}\sin \frac{\theta}{2} | 1 \rangle$$ which hase energy $\langle \psi | H | \psi \rangle = - B \cos \theta/2$.

So since I can write down a measure over my states $$ \mathbf{1} = \frac12\iint \sin \theta \, \mathrm d \phi \, \mathrm d \theta \, | \psi \rangle \langle \psi |$$ I can also contruct some sort of Gibbs state by weighting them all by a Boltzmann factor, this gives $$\rho = \frac1{2Z}\iint \sin \theta \, \mathrm d \phi \, \mathrm d \theta \, \mathrm e^{\beta B \cos \theta/2}| \psi \rangle \langle \psi |$$ However this yields $\langle m \rangle = \mathrm{Tr}[\sigma_z \rho]/2 = \coth (\beta B/2)/2-1/(\beta B)$, the classical result.

My question is: Is there a simple physical principal to which I can point to determine the correct process (one which is more satisfactory than simply observing that one of these approaches works and the other obtains a different answer)?

Or is it a matter of accepting $\rho = \exp (-\beta H)/Z$ as the definition of thermal equilibrium? (and hence the von Neumann entropy as the correct entropy?)

Is gravity a fundamental law that is taken for granted?

The question might seem silly (and, maybe it is ), but I would like to know if gravity is really something you can not explain in some other terms and have just to take for granted (i.e. it simply works as it works and we, currently, do not know why), kind of like magnetic force?

Or is there an explanation?
Here are some, exact, points that seem puzzling to me:

1. 
   

   If we would build a spaceship resembling Earth in shape and size, would it have gravity, just because of mentioned properties (although, as I understand it, the Earth is round, because of gravity)?

   
   


2. 
   

   If we would take the Earth and cut off (kind of like from an apple) 1/10 of it the greater part would still have gravity, while the lesser won't?

   
   
   
   
   
   • What about if we would split Earth in half?
   
   
   • Or 40% : 60%?
   
   
   
   


3. 
   

   Are we currently (or, theoretically, will ever be) capable of creating artificial objects, that have a gravity of their own?

   
   

Answer

Physics models exist in many frameworks, a number of them emerge from an underlying framework, which is considered fundamental.

What you are asking about comes from the classical gravitational model, which is based on a gravity law and the laws of Newtonian mechanics.

Laws in physics are like axioms in mathematics, necessary to pickup the correct mathematical model that will describe data, and predict new situations.

All of your questions are answerable within classical gravitational theory.

answers for

1. 
   

   All masses interact gravitationally according to the law, $(G.m_1M_2)/{r^2}$. Shapes and densities and in general mathematical analysis allows the calculation of the force.

   
   


2. 
   

   The various cuts will follow the law

   
   
   


3. 
   

   ALL masses have a gravitational field

   
   

These answers are within the classical framework

Classical gravitation emerges from an underlying General Relativity mathematical model, but that is another story.

electromagnetism - How are forces "mediated"?

I hope this is the right word to use.

To me, these forces seem kind of fanciful (except for General Relativity and Gravity, which have a geometric interpretation).

For example, how do two charged particles know that they are to move apart from each other?

Do they communicate with each other somehow through some means?

I've heard some people tell me that they bounce together messenger photons. So does one electron receive a messenger photon, go, "Oh hey, I should move in the direction opposite of where this came from, due to the data in it", and then move?

Aren't photons also associated with energy, as well? Does this type of mediation imply that electrons give off energy in order to exert force on other electrons?

Every electron is repelled by every other electron in the universe, right? How does it know where to send its force mediators? Does it just know what direction to point it in? Does it simply send it in all directions in a continuum? Does that mean it's always giving off photons/energy?

I'm just not sure how to view "how" it is that electrons know they are to move away from each other.

These questions have always bugged me when studying forces. I'm sure the Standard Model has something to shed some light on it.

Answer

Brief answer: Read only the bold part (and ignore grammar then).

The answer you already mentioned lies in Quantum Field Theory (QFT). But to fully understand it, you must give up a particle as a point-like thing that is well-localized. There is one Quantum Field per sort of particle, e.g. the electron field for all electrons, and the photon field for all photons. (The fact that there is a single field for all electrons also results in the Pauli exclusion principle.)

What you consider a particle is basically just a local peak in the respective particle field, but one cannot even say "This peak corresponds to electron A, this one to B". Now QFT, more specifically Quantum Electrodynamcis (QED), describes the local interaction between the electron field and the photon field. But since the fields have a dynamic, a local change induced in the photon field by the electron field will propagate with the speed of light (flat space assumed) and interact with the electron field in another place, thus creating the impression "Electron A emitted a photon that told electron B to interact electromagnetically".

It's similar for the other interactions, there's a gluon field for the strong interaction (Quantum Chromodynamics), and for the electroweak interaction there's kind of a combination of the photon field and the weak-interaction-bosons.

cosmology - As the universe expands, the wavelengths of photons are stretched, and energy is lost. What about electrons?

Will electrons, and other particles, also loose energy as they travel through the cosmos? They have wavelengths. Do they get "stretched"? My guess is that the EM force, somehow, counteracts this effect. What about neutrinos?

Reference found: http://books.google.com/books?id=AmlEt6TJ6jAC&pg=PA96&lpg=PA96&dq=redshift+of+de+broglie+wavelength&source=bl&ots=oAW0q9bmaj&sig=nAwE-ohARZ3YTCZrHDBBjZ_IYfQ&hl=en&sa=X&ei=0lQDVJjtLMKEjAL80IHQCw&ved=0CB4Q6AEwAA#v=onepage&q=redshift%20of%20de%20broglie%20wavelength&f=false

Answer

The answer is yes. The de Broglie wavelengths of freely propagating particles (i.e. forget the influence of interactions and gravity perturbations, just consider the Universe as a whole) are redshifted by the expansion of the universe. Another way of saying this is that their peculiar momenta with respect to a co-moving local volume decrease as the inverse of the scale factor.

Neutrinos are an example of a particle with a non-zero mass (maybe of order 0.1 eV - see http://adsabs.harvard.edu/abs/2014PhRvL.112e1303B ). They decouple from the rest of the Universe at about 1 second and freely propagate. The expansion then reduces their momenta to the extent that they should have a temperatures $<2$K in the present-day Universe, typical kinetic energies of 0.2 meV (e.g. see http://adsabs.harvard.edu/abs/2010PhRvD..82f2001K ) and may have speeds of only (depending on their actual masses) perhaps $\sim 10^3-10^4$ km/s and so are non-relativistic.

Electrons would behave in the same way, if they could be considered not to be strongly (or rather electromagnetically!) interacting with other particles and photons. I don't think this can be satisfied except in the very early universe and a typical free, intergalactic electron in the present day universe has an energy of $\sim 0.1$ keV due to heating by radiation from stars and galaxies.

electromagnetism - Would two Neodymium magnets stick together with twice the power?

Assume I use a coil to create B-flux and put ONE Neodymium cylinder magnet (1" diameter & 0.25" thick) close to the flux, it would create 10 lbs of force. Does that mean that putting TWO cylinder magnets (1" diameter & 0.25" + 0.25" = 0.5" thick) would create 20lbs of force?

Answer

Probably not, but it depends on the geometry of your coil.

For a couple of dollars at the hardware store you can get a big stack of those coin magnets. If the answer to your question were yes in general, it'd be harder to break apart the big stack of magnets that to separate two of them. That's not consistent with my experience.

In general for a dipole $\vec m$ in a field $\vec B$, the force is $\vec F=-\nabla(\vec m \cdot \vec B)$. If the dipole moment is a constant, and the dipoles are free to rotate, they will orient themselves so that $\vec m$ is antiparallel to $\vec B$. In that special case, the force simplifies to $$\vec F \approx m \vec\nabla B. $$ In other words, the dipoles "want" to align antiparallel to the field, then to move up the gradient into the strongest part of the field. A dipole in a uniform field feels a torque, but not a net force.

If your coil is set up to generate a uniform field, your two stacked magnets will feel the same force — but that force will be zero. If your coil is generating a magnetic field with some dipole component, your stacked magnets will rotate to align with the field, then (since we've already assumed there is a field strength gradient) one of them will see a weaker gradient than the other. The gradient $\vec\nabla B$ typically vary like $1/r^4$, where $r$ is the distance from the center of the coil, so the force on the two coin magnets can vary considerably over a short distance. This is why strong magnets like to suddenly "leap" together and pinch your fingers when you are playing with them.

Now you could put your two coin magnets next to each other, symmetrically about the axis of your coil, and they would see the same $|\vec\nabla B|$ in slightly different directions. But with the two coin magnets next to each other, they would exert torques on each other, since they prefer to be stacked pole-to-pole; you could engineer a solution like this, but there would be extra parts involved.

I actually happened to have some magnets with this shape on my desk. In response to Carl's comment, I built a little string lifting harness to measure the force on a single magnetized paper clip:

Adding a second magnet increased, but didn't double, the weight that the harness could hold:

stacked number of clips magnets that stayed up 1 11 2 17 3 23 4 24 5 24

The dominant effect in calculating the force is the field is right at the surface of the stack of magnets; it looks like that gets saturated, an effect I didn't consider in the first part of my answer.

newtonian mechanics - Why does a ball rolling without slipping stop due to friction?

I'm confused on a scenario. If we roll a ball on a horizontal surface and it rolls without slipping it should eventually stop due to friction. However rolling without slipping means the velocity at the point of contact is always 0. Since kinetic friction always opposes movement and the point of contact has v=0 there's no kinetic friction and static friction only arises when there is some force. Thus there is no friction? Can anyone clarify what is happening, and point to where I'm confused.