So in 3 dimensions, Riemann tensor has 6 independent terms. So we can fully describe it in terms of the Ricci tensor.
How do I show that $R_{abcd}=T_{ac}g_{bd}+T_{bd}g_{ac}-T_{ad}g_{bc}-T_{bc}g_{ad}$ for $$T_{ab}=R_{ab}-\frac{R}{4}g_{ab}?$$
What I thought:
I know that $R_{ab}=g^{mn}R_{namb}=g^{mn}(T_{nm}g_{ab}+T_{ab}g_{nm}-T_{nb}g_{am}-T_{am}g_{nb})$, but how do I continue from here?
Why do we have an elementary charge $e$ in physics but no elementary mass? Is an elementary mass ruled out by experiment or is an elementary mass forbidden by some theoretical reason?
Answer
Let me add two references to points already mentioned in this discussion:
Today, there is no reason known why the electric charge has to be quantized. It is true that the quantization follows from the existence of magnetic monopoles and the consistency of the quantized electromagnetic field, which was shown first by Dirac, you'll find a very nice exposition of this in
AFAIK there is no reason to believe that magnetic monopoles do exist, there is no experimental evidence and there is no compelling theoretical argument using a well established framework like QFT. There are of course more speculative ideas (Lubos mentioned those).
AFAIK there is no reason why mass should or should not be quantized (in QFT models this is an assumption/axiom that is put in by hand, even the positivity of the energy-momentum operator is an axiom in AQFT), but a mass gap is considered to be an essential feature of a full fledged rigorous theory of QCD, for reasons that are explained in the problem description of the Millenium Problem of the Clay Institute that you can find here:
Can one derive Newton's
second and third laws from the first law or
first and third laws from the second law or
first and second laws from the third law
I think Newton's laws of motions are independent to each other. They can not be derived from one another. Please share the idea.
Answer
The modern interpretation of Newton's First Law is about the existence of inertial reference frames, mainly to solidify the idea that such coordinate systems exist and are important.
However, I sincerely doubt this is what Newton himself had in mind when he postulated his laws. Historically, Newton probably introduced his first law to put emphasis on the fact that moving bodies do not slow down by their own accord (as was common wisdom at the time). So yes, the first law in this context can certainly be derived from the second law by setting F=0. In fact I doubt Newton even had an idea of what an inertial frame was (though it probably could have been explained to him with relative ease).
What is the relationship between the individual spin states of photons in a light beam and the polarization of the light beam, if any? What about the spin states determines the degree of polarization, if anything?
Hey guys, please someone help me with this exercise :)
For our exercise, assume that the bird’s body resistance (from leg to leg) is 50kΩ. Given is a high voltage power line with a current of 100A where the bird takes a rest. The high voltage power line is made out of aluminium, which has a conductivity of $3.79\times10^7 \,\rm S/m$ and a cross section of $100\rm\, mm^2$. Determine the theoretical distance between the two bird legs.
I know, the formula is V = R x I
I = 100 A, right?
R = ?
Please someone explain step by step?
Thanks.
I know that you can use induction to create a current in a superconducting loop, but this only works as long as the coil that induces the field has a current flowing through it. And obviously, this cannot work for the really big magnets that are used in MRI scanners or NMR machines.
How does it work?
Answer
Actually, induction works, although it is often used a bit differently than you described. You can place a warm superconductor loop into a normal coil. As you switch the coil on, there will be some current inside the superconductor, but since it is not cold yet, this current quickly dies down. Then you cool the superconductor below its critical temperature. When you now switch off the normal coil, the flux-change induces a current in the superconducting loop. This current is there to stay.
With the proper coil windings, you can get quite large currents, but you won't be able to get to field energies that are needed in NMR or MRI machines. These magnets are energized (sometimes called "charged") differently. The trick is to make part of the loop normal conducting during charging, and then "close the loop" when the coil is energized.
If you want to know details, this is the electrical schematic on how it is done:
I used values from a typical 5 Tesla NMR magnet that I energized a few years ago.
I will leave out all those pesky details that make charging a real superconducting magnet a hassle. The procedure boils down to this:
With the loop closed, the magnet is now in "persist" mode, so the current goes on forever. Well, not quite forever, there usually is a little bit of resistance that leads to "flux creep". For the magnet I am using this is an unusually low value of $10^{-11}$ Ohm, which makes the current decay with a time constant of a few 100 000 years. The designer of the magnet speculates that this resistance is caused by the weld that is needed to turn the wire into a loop (see the link above, near the end of the paper). Other reasons that are often given for this effect involve details that are specific to type-II superconductors.
Harmonic Oscillator
Particle in a box
Similarly, the cone potential $V(x)=|x|$ and the exponential potential $V(x)=\exp(|x|)$ have been shown to have $\Delta x\Delta p_x$ grow linearly with $n$.
We notice that for small n the product is of the same order of magnitude as $\hbar$ and for large n it grows linearly with it:
I've read and heard that quantum electrodynamics is more fundamental than maxwells equations. How do you go from quantum electrodynamics to Maxwell's equations?
Answer
Disclaimer: This is answer is given from a mathematical physics point of view, and it is a little bit technical. Any comment or additional answer from other points of view is welcome.
The classical limit of quantum theories and quantum field theories is not straightforward. It is now a very active research topic in mathematical physics and analysis.
The idea is simple: by its own construction, quantum mechanics should reduce to classical mechanics in the limit $\hslash\to 0$. I don't think it is necessary to go into details, however for QM this procedure is now well understood and rigorous form a mathematical standpoint.
For QFTs, such as QED, the situation is similar, although more complicated, and it can be mathematically handled only in few situations. Although it has not been proved yet, I think it is possible to prove convergence to classical dynamics for a (simple) model of QED, describing rigid charges interacting with the quantized EM field.
The Hilbert space is $\mathscr{H}=L^2(\mathbb{R}^{3})\otimes \Gamma_s(\mathbb{C}^2\otimes L^2(\mathbb{R}^3))$ ($\Gamma_s$ is the symmetric Fock space). The Hamiltonian describes an extended charge (with charge/mass ratio $1$) coupled with a quantized EM field in the Coulomb gauge: \begin{equation*} \hat{H}=(\hat{p} - c^{-1} \hat{A}(\hat{x}))^2+\sum_{\lambda=1,2}\hslash\int dk\;\omega(k)a^*(k,\lambda)a(k,\lambda)\; , \end{equation*} where $\hat{p}=-i\sqrt{\hslash}\nabla$ and $\hat{x}=i\sqrt{\hslash}x$ are the momentum and position operators of the particle, $a^{\#}(k, \lambda)$ are the annihilation/creation operators of the EM field (in the two polarizations) and $\hat{A}(x)$ is the quantized vector potential \begin{equation*} \hat{A}(x)=\sum_{\lambda=1,2}\int \frac{dk}{(2\pi)^{-3/2}}\;c\sqrt{\hslash/2\lvert k\rvert}\;e_\lambda(k)\chi(k)(a(k,\lambda)e^{ik\cdot x}+a^*(k,\lambda)e^{-ik\cdot x})\; ; \end{equation*} with $e_\lambda(k)$ orthonormal vectors such that $k\cdot e_\lambda(k)=0$ (they implement the Coulomb gauge) and $\chi$ is the Fourier transform of the charge distribution of the particle. The magnetic field operator is $\hat{B}(x)=\nabla\times \hat{A}(x)$ and the (perpendicular) electric field is $$\hat{E}(x)=\sum_{\lambda=1,2}\int \frac{dk}{(2\pi)^{-3/2}}\;\sqrt{\hslash\lvert k\rvert/2}\;e_\lambda(k)\chi(k)i(a(k,\lambda)e^{ik\cdot x}-a^*(k,\lambda)e^{-ik\cdot x})\; $$
$\hat{H}$ is a self adjoint operator on $\mathscr{H}$, if $\chi(k)/\sqrt{\lvert k\rvert}\in L^2(\mathbb{R}^3)$, so there is a well defined quantum dynamics $U(t)=e^{-it\hat{H}/\hslash}$. Consider now the $\hslash$-dependent coherent states \begin{equation*} \lvert C_\hslash(\xi,\pi,\alpha_1,\alpha_2)\rangle=\exp\Bigl(i\hslash^{-1/2}(\pi\cdot x+i\xi \cdot\nabla)\Bigr)\otimes\exp\Bigl(\hslash^{-1/2}\sum_{\lambda=1,2}(a^*_\lambda(\alpha_\lambda)-a_\lambda(\bar{\alpha}_\lambda))\Bigr)\Omega\; , \end{equation*} where $\Omega=\Omega_1\otimes\Omega_2$ with $\Omega_1\in C_0^\infty(\mathbb{R}^3)$ (or in general regular enough, and with norm one) and $\Omega_2$ the Fock space vacuum.
What it should be at least possible to prove is that ($\alpha_\lambda$ is the classical correspondent of $\sqrt{\hslash}a_\lambda$, and it appears inside $E(t,x)$ and $B(t,x)$ below): \begin{gather*} \lim_{\hslash\to 0}\langle C_\hslash(\xi,\pi,\alpha_1,\alpha_2),U^*(t)\hat{p}U(t)C_\hslash(\xi,\pi,\alpha_1,\alpha_2)\rangle=\pi(t)\\ \lim_{\hslash\to 0}\langle C_\hslash(\xi,\pi,\alpha_1,\alpha_2),U^*(t)\hat{x}U(t)C_\hslash(\xi,\pi,\alpha_1,\alpha_2)\rangle=\xi(t)\\ \lim_{\hslash\to 0}\langle C_\hslash(\xi,\pi,\alpha_1,\alpha_2),U^*(t)\hat{E}(x)U(t)C_\hslash(\xi,\pi,\alpha_1,\alpha_2)\rangle=E(t,x)\\ \lim_{\hslash\to 0}\langle C_\hslash(\xi,\pi,\alpha_1,\alpha_2),U^*(t)\hat{B}(x)U(t)C_\hslash(\xi,\pi,\alpha_1,\alpha_2)\rangle=B(t,x)\; ; \end{gather*} where $(\pi(t),\xi(t),E(t,x),B(t,x))$ is the solution of the classical equation of motion of a rigid charge coupled to the electromagnetic field: \begin{equation*} % \left\{ \begin{aligned} &\left\{\begin{aligned} \partial_t &B + \nabla\times E=0\\ \partial_t &E - \nabla\times B=-j \end{aligned}\right. \mspace{20mu} \left\{\begin{aligned} \nabla\cdot &E=\rho\\ \nabla\cdot &B=0 \end{aligned}\right.\\ &\left\{\begin{aligned} \dot{\xi}&= 2\pi\\ \dot{\pi}&= \frac{1}{2}[(\check{\chi}*E)(\xi)+2\pi\times(\check{\chi}*B)(\xi)] \end{aligned}\right. \end{aligned} % \right. \end{equation*} with $j=2\pi\check{\chi}(\xi-x)$, and $\rho=\check{\chi}(\xi-x)$ (charge density and current).
To sum up: the time evolved quantum observables averaged over $\hslash$-dependent coherent states converge in the limit $\hslash\to 0$ to the corresponding classical quantities, evolved by the classical dynamics.
Hoping this is not too technical, this picture gives a precise idea of the correspondence between the classical and quantum dynamics for an EM field coupled to a charge with extended distribution (point charges cannot be treated mathematically on a completely rigorous level both classically and quantum mechanically).
It is known that there is solution of Einstein's equations for charged black hole. Reissner–Nordström metric in case of non-rotating charged black hole and for rotating charged black hole is a Kerr–Newman metric.
In Reissner–Nordström metric I can calculate electric field, it has following form $$ F_{0r}=-F_{r0}=\frac{Q}{r^2} $$ But I can not understand the following thing. In my point of view when two charged particle are interact, they exchange photons between each other. If I apply this argumentation in case on interacting between charged black hole and probe charge, I will get zero force due to black hole can not emit photon and absorb all photon. Can our explain how it works in quantum point of view(I mean QED and not Quantum gravity!!).
When a top is spun, it will precess in some direction, either clockwise or counterclockwise. It's possible to find out which way using $\boldsymbol{\tau} = d\mathbf{L}/dt$ and $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$, where $\mathbf{F}$ is the force of gravity.
However, I was never totally satisfied with this because I couldn't "see" exactly why the conclusion was true. Intuitively, I don't get, in my gut, why a downward force $\mathbf{F}$ can push the rotation axis to the left or right.
In principle, one can explain this by just applying non-rotational mechanics to different pieces of the gyroscope. I think this would really help me visualize what's going on. Is there such an explanation?
Note: I already know about the math of rotational mechanics, so please don't rewrite it. I am not interested in any answer that contains a $\boldsymbol{\tau}$, $\mathbf{L}$, $\boldsymbol{\omega}$, or $\times$ anywhere in it.
Answer
The reason why a gyroscope does behave in this strange way is that if you try to rotate it's axis in some direction, the "endpoints" of this axis have to be pushed perpendicular to what our first intuition would say.
In order to verify why the axis starts rotating in this strange way, let's make some simplifications: the gyroscope consists of two identical rotating particles and the particles rotate much fasten than we rotate the axis.
To visualize how the particles move, I've made the following drawing: the circle depicts a sphere on which the particles revolve, the center of the particles is always the center of the sphere and the axis of rotation precesses counterclockwise. The rotation axis is initially in up-down direction. Blue and red correspond to different particles, continuous and non-continuous lines correspond to the particle being on our side or the back side of the sphere. Blue starts at B and red at I.
On the figure, you can notice that because of the curvatures of the trajectories, the particles on our side must be pushed up while those of the back side must be pushed down (when the axis hasn't yet rotated much). This force has to be compensated - in order to keep the axis rotating counterclockwise, the axis has to be pushed away from the downside and pulled from the upside.
Going back to the gyroscope, if the axis is tilted in the beginning, gravitation "tries" to make it fall down, but instead this axis will rotate in the direction perpendicular to it. Given that information, you can probably figure out yourself in which direction the gyroscope will preceed. I think it should do this in the same direction as it rotates, if I visualized this process correctly with a pencil.
The action for the free relativistic particle with worldline $\gamma : I\subset \mathbb{R}\to M$ is
$$S[\gamma]=-m\int d\lambda\sqrt{-\dot{\gamma}^a(\lambda)\dot{\gamma}_a(\lambda)}\tag{1} $$
Now, one may postulate a second action
$$S'[\gamma,\eta]=\frac{1}{2}\int d\lambda \bigg(\eta(\lambda)^{-1}\dot{\gamma}^a(\lambda)\dot{\gamma}_a(\lambda)-\eta(\lambda)m^2\bigg).\tag{2}$$
These are classically equivalent actions.
My question is: usually what we have is (1) and we have a problem both with the square root and with the massless limit. Given this, how could we think about postulating (2)? In other words, how can we reach (2)?
Usually some people answer this by saying: "it doesn't matter, actions are postulated, you postulate it, compute the equations, prove it works and its over".
Now I beg to differ. I want to know how could someone reason exactly in order to know what to postulate.
I am perfectly comfortable with computing the equations of motion. I want to know is how given (1) we would have the idea to postulate (2).
Is it some special case of some general procedure that deals with constraints?
Answer
It's a lore in physics that any Lagrangian system has an equivalent Hamiltonian formulation. If we start from OP's square root Lagrangian $$ L_0~:=~ -m\sqrt{-\dot{x}^2}, \qquad \dot{x}^2~:=~g_{\mu\nu}(x)~ \dot{x}^{\mu}\dot{x}^{\nu}~<~0,\tag{1}$$ it is natural to ponder what the Hamiltonian formulation is? The momentum reads $$ p_{\mu}~=~\frac{\partial L_0}{\partial\dot{x}^{\mu}}~\stackrel{(1)}{=}~\frac{m\dot{x}_{\mu}}{\sqrt{-\dot{x}^2}},\tag{2}$$ and the energy $$ H_0~=~p_{\mu}\dot{x}^{\mu}-L_0~\stackrel{(1)+(2)}{=}~0\tag{3}$$ vanishes, cf. e.g. this Phys.SE post. Now when we try to perform the Legendre transformation, we discover that the momenta $p_{\mu}$ are not all independent. They have to satisfy a mass-shell constraint $$ p^2+m^2~\stackrel{(2)}{\approx}~0, \qquad p^2~:=~g^{\mu\nu}(x)~ p_{\mu}p_{\nu}~<~0.\tag{4}$$ The Dirac-Bergmann prescription then tell us to impose this constraint in the Hamiltonian $$H~=~\frac{e}{2}(p^2+m^2)\tag{5}$$ via a Lagrange multiplier field $e$. It is easy to check that this is indeed the Hamiltonian formulation of a relativistic point particle. The corresponding Hamiltonian Lagrangian becomes $$ L_{H}~=~p_{\mu}\dot{x}^{\mu}-H~\stackrel{(5)}{=}~p_{\mu}\dot{x}^{\mu}-\frac{e}{2}(p^2+m^2) .\tag{6}$$ Ok, cool, but what does this have to do with OP's question?, the reader may ask. Wait for the punch-line: Now we can ask the opposite question: What happens if we perform the inverse Legendre transformation, namely eliminate/integrate out the momenta via their EL equations $$ p_{\mu}~\stackrel{(6)}{\approx}~ \frac{1}{e}~\dot{x}_{\mu}~?\tag{7}$$ Surprisingly, we don't get quite back to where we started. Instead we get OP's non-square root Lagrangian $$L~\stackrel{(6)+(7)}{=}~\frac{\dot{x}^2}{2e}-\frac{e m^2}{2}.\tag{8}$$ This is one possible answer to OP's title question. For more information about relativistic point particles, see this Phys.SE post. For the analogous question for strings, see this Phys.SE post.
In the following equation of a reaction
$$\mathrm{p}^0 + \mathrm{n} \to \mathrm{K}^+ + \Sigma^-$$
What is the quark composition of the $\mathrm{p}^0$ particle? Or is it supposed to be $\rho^0$?
For me it certainly looks more like $\mathrm p$, not $\mathrm \rho$.
(Source: K. A. Tsokos, Physics for the IB Diploma, Sixth Edition, Cambridge University Press)
Answer
It looks like a typo for $$ \rm \rho^0 + n \to K^+ + \Sigma^- $$ where $\rho^0$ is the uncharged member of the isospin triplet with mass 770 MeV. According to the particle data group, the quark content of the light, unflavored mesons with isospin $I=1$ is $u\bar d, (u\bar u - d\bar d)/\sqrt2, d\bar u$.
You can tell that your "$p$" must be a meson, not a baryon, because both sides of the reaction must have the same baryon number, and the baryon number on the right side is $+1$.
Commenters on a duplicate question point out that it may also by a typo for $\pi^0$, if the printer's software produces the character "π" using the same code point as "p" but in some other typeface. Like the $\rho$, the $\pi$ also has zero baryon number and unit isospin; however the pion is spinless while the rho is a spin-one "vector" meson.
If I remember well, they said that it can't, but I do not know why.
Yes, I meant if gravity can be shielded using something like a Faraday cage (or something else?).
Thank you.
Lorentz force is in this form: $$\vec{F}=q[\vec{E}+\vec{u}\times\vec{B}]$$ As we know, it is Lorentz-invariant. Is there any way to justify or derive its form from relativity theory?
Answer
Although like any other physics, the Lorentz force law is experimentally measured, one could easily imagine an alternative universe where the discovery of relativity came before electromagnetic theory (suppose Michelson-Morely lived before Faraday, or look at the Ignatowskian approach to special relativity, where the form of the Lorentz transformation is argued without reference to light at all).
In this alternative history, two theoretical physicists might have been conversing:
Scene 1: A & B Talking About General Laws on "Electric Charge"
A: We know about this weird property recently discovered called 'electric charge'. What kind of laws would govern its motion?
B: Well, I've seen a charge sit pretty still in the laboratory or moving uniformly, so it's clearly possible to have an absence of electric effect. Let's postulate some electromagnetic field $F(\vec{X}, \vec{V},\,\cdots)$ ..."
A: Those of course all need to be four-vectors, or at least something to make the law Lorentz covariant ....
B: Of course, silly, I was just testing you. Anyhow, if $F$ to first order were linear, the law would have to be homogeneous ...
A: .... and you could get rid of $\vec{X}$ because, in the absence of other charges, the force doesn't depend on position ...
B: Of course: I was just testing you on that too ... anyhow, so we agree it's a linear, homogeneous function of velocity? ....
A: ... you mean four velocity ...
B: Of course: so a linear homogeneous function of four velocity is the simplest plausible thing to try. Hey quit interrupting my discovery of a new law will you ...
A: .. our law ...
B: OK. But you'll be second author, okay. Anyhow, here is what we have: our field has to be a rank two Lorentz covariant tensor to stand for a linear homogenous form of Newton's second law:
$$m\,\frac{\mathrm{d} v^\mu}{\mathrm{d}\tau} = q\, F^\mu{}_\nu\,v^\nu$$
where we'll write $q$ to measure the coupling strength between the 'charge' and the field.
A: Don't forget that a four velocity has a constant norm of unity .....
B:Oh DUUUH, I was just about to say that we can do better than this owing to obvious constraints like $\langle \vec{V},\,\vec{V}\rangle = 1$: so the acceleration has to be Minkowski-orthogonal to velocity, that gives us:
$$v^\mu\,\left(\eta_{\mu\,\sigma}\, F^\sigma{}_\nu + \eta_{\nu\,\sigma}\,F^\sigma{}_\mu\right)v^\nu = 0;\forall \vec{V} \Rightarrow \eta_{\mu\,\sigma}\, F^\sigma{}_\nu + \eta_{\nu\,\sigma}\,F^\sigma{}_\mu = 0$$
so the doubly covariant form of $F$ has to be skew symmetric to conserve the norm of the four velocity. Shall we publish now?
A: You do realize that that utter minging complete clot Heaviside is the editor of J. Modern Irreproducible Physics?
B: So ..?
A: We're never going to get that into print unless we write everything in his completely crap dotty crossy vectory notation! He's never going to go for that utterly indexxy $\eta_{\mu\,\sigma}\, F^\sigma{}_\nu + \eta_{\nu\,\sigma}\,F^\sigma{}_\mu = 0$, he'll upend his cup of tea into his own lap and choke on his sticky bun as soon as he reads it ...
B: Oh, so how do we do that then?
A: Only if I'm first authoooor .....!
B: Oh, alright ....
Exeunt A & B to go off and eat sticky buns whilst A derives the dotty crossy version
Scene Two: After Sticky Buns
A: Here's the dotty crossy version of $\eta_{\mu\,\sigma}\, F^\sigma{}_\nu + \eta_{\nu\,\sigma}\,F^\sigma{}_\mu = 0$. It can hold if and only if the force .....
B: the three-force .....
A: .... yes of course: we're talking dotty crossy here. Where was I? It can hold if and only if the three force acts according to:
$$\vec{F} = q\,\alpha\,(\vec{E} + \vec{V}\times \vec{B})$$
where $\alpha$ is an arbitrary scaling constant, which we can absorb into the definition of charge if we like, $\vec{E}$ is made of the off-diagonal zeroth row elements of $F$ and $\vec{B}$ is made of the three independent elements of the skew-symmetric $3\times 3$ lower right block of $F$.
(See my answer here for some more information).
during a lab experiment, i noticed that a metal ball has a much harder time rolling on metal rails when their is a current passing through it and the rails. I was wondering why and if there was a name for this force. Does the fact that a current passes "pull " the balls towards the rails and increases friction? Thanks
Consider two frames $S$ and $S'$, moving relative to each other. If I stand still in frame $S$ and watch frame $S'$, I can measure its speed to be $V$. However, why is this speed the same as the observer in $S'$ measured for $S$? (denote as $V'$?)
Can I have suggestions for some good book regarding tensors for physics.
In the derivation of Kelvin's circulation theorem, I take the material derivative of circulation, or
\begin{align} \dfrac{D\Gamma}{Dt} = \dfrac{D}{Dt}\oint_C \vec{u} \cdot d\vec{\ell}. \end{align}
Moving the material derivative inside the integral gives
\begin{align} \dfrac{D\Gamma}{Dt} = \oint_C \dfrac{D\vec{u}}{Dt} \cdot d\vec{\ell} + \oint_C \vec{u} \cdot \dfrac{D(d\vec{\ell})}{Dt}. \end{align}
Given that $d\vec{\ell} = ds \hspace{1mm} \vec{t}$, the second integral of the equation above becomes
\begin{align} \oint_C \vec{u} \cdot d\vec{u}. \end{align}
How does one arrive at this equation? My attempt would be to use the chain rule so that \begin{align} \dfrac{D(ds \vec{t})}{Dt} = ds\dfrac{D\vec{t}}{Dt} + \vec{t}\dfrac{D(ds)}{Dt}. \end{align}
From here, I expand each material derivative as
\begin{align} \dfrac{D\vec{t}}{Dt} = \dfrac{\partial \vec{t}}{\partial t} + u_t\dfrac{\partial \vec{t}}{\partial s}+ u_n\dfrac{\partial \vec{t}}{\partial n}\\ \dfrac{D(ds)}{Dt} = \dfrac{\partial ds}{\partial t} + u_t\dfrac{\partial ds}{\partial s}+ u_n\dfrac{\partial ds}{\partial n}. \end{align}
I am unsure how these two relations will simplify so that I get $$\dfrac{D(d\vec{\ell})}{Dt} = d\vec{u}.$$
I just started reading PCT, Spin and Statistics, and All That. Can someone explain why we use operator valued distributions to describe fields? I read somewhere that it would take infinite energy to measure an observable at a single point. Why don't we instead use functions from some collection of subsets (i.e. the open sets) of space to the operators to emulate the fact that measurements usually occur over some area? In other words, what is the physical meaning of the test functions used to define the operator valued distributions? Are some of these functions non-physical, possibly too narrow in width that they'd violate some uncertainty principle?
I know that moving charges produce both magnetic and electric fields. The image given below is of electromagnetic field. I was able to identify the magnetic field. But,I do not know where the electric field is? 
The planetary orbits have been studied as ellipses but the solar system is in motion in relation to the distant stars. Their path is along the tip of an helix and the ecliptic plane is a convenient plane of projection. I think that the studies were never conducted under this viewpoint.
The sunlight we see now was emitted more than 8 minutes ago when the Sun was ‘below’ the ecliptic but we see it centered in the plane. I wonder why we do not see any consequence of this.
I’m following a line of reasoning that the motion can have consequences and I revisited the anomalous precession of the perihelion of Mercury, settled long time ago by Einstein, and I found this perturbing equation:
$$\frac{43}{5557} = 2\pi\frac{369.2\ \text{km}\ \text{s}^{−1}}{299792.458\ \text{km}\ \text{s}^{−1}}$$
Where 5557 is the predicted theoretical value (in mathpages) for the advance and the 43 is the anomaly.
The simplicity of the formula $$\text{error}/\text{theoric}=2\pi\ V/c$$ and because $V$ is 0.054% off the central measured value of the speed of solar system - $369(\pm0.9$) - makes me wonder if this can be more than a coincidence.
Any kind of reasoning on the why's will be helpful.
First of all thank you to anyone who replies this question, I am not a physics major, but foolishly chose to take a physics for non-science majors at my local university I am going to.
I am trying to understand how cosmic inflation explains the uniformity of temperature (to around a thousandth of a degree) imprinted from the cosmic microwave background radiation: 2.725K, 2.724K, etc etc.
What I do not understand is that I have been fed contradictory information, these are my following notes from lecture:
Here is where the contradiction comes in:
Then where did the radiation of the CMB come from?
One set of notes says, the uniformity of the temperature is from equilibrium before cosmic inflation. But the second says, that there was no matter/radiation until after the cosmic inflation. How can you have temperature equilibrium without matter/radiation?
Somehow during the process of the inflaton field going down its energy curve the Higgs field got turned on?
I feel like I would have learned more from watching tv documentaries. I've got little to no help from professor and from other students.
Edit: what I am asking is: if cosmic inflation causes space to expand rapidly via repulsive gravity (from negative pressure) when the inflaton field is at a high energy point. AND if the inflaton field falling down its energy curve forms matter/energy then How could the CMB have uniformity to the one thousandth degree? wouldnt the massive expansion of space make reaching equilibrium impossible ?
Answer
Lets start from the beginning :
Here is a comprehensive history of the universe:
Protons formed by 1microsend, matter as we know it now. At 380.000 years the density of matter is so low that there is no longer an equilibrium between photon absorption and emission with the rest of matter, and most photons escape without further interactions and, with continuous expansion give us the CMB and black body radiation measured in the experiments.
The uniformity of the temperature seen in the black body radiation curve it displays, when examined below the level of uniformity, agglomerations and depletions, which are correlated with the location of clusters of galaxies and galaxies.
You are correct to wonder:"wouldn't the massive expansion of space make reaching equilibrium impossible" .
The inflation period was proposed because the CMB uniformity could not be explained thermodynamically.
The microwave background radiation from opposite directions in the sky is characterized by the same temperature within 0.01%, but the regions of space from which they were emitted at 500,000 years were more than light transit time apart and could not have "communicated" with each other to establish the apparent thermal equilibrium - they were beyond each other's "horizon".
There could be no exchange of radiation thermodynamically through the whole universe, because of the light cone that interactions have to obey due to special relativity and general relativity. The introduction of the inflation period, solved this problem.
If the universe inflated by 20 to 30 orders of magnitude, then the properties of an extremely tiny volume which could have been considered to be intimately connected were spread over the whole of the known universe today, contributing both extreme flatness and the extremely isotropic nature of the cosmic background radiation.
The regions of the universe were "intimately connected" through the quantum mechanical solution to the same boundary conditions.
In addition, the quantum mechanical uncertainties can explain the "kernels" of anisotropy below the uniformity of the CMB as being the ones where matter coagulated into clusters of galaxies and galaxies.
The detailed, all-sky picture of the infant universe created from nine years of WMAP data. The image reveals 13.77 billion year old temperature fluctuations (shown as color differences) that correspond to the seeds that grew to become the galaxies. The signal from the our Galaxy was subtracted using the multi-frequency data. This image shows a temperature range of ± 200 microKelvin
With the above in mind,lets see your last question:
if cosmic inflation causes space to expand rapidly via repulsive gravity (from negative pressure) when the inflaton field is at a high energy point. AND if the inflaton field falling down its energy curve forms matter/energy then How could the CMB have uniformity to the one thousandth degree?
It is a fact that it does.
wouldnt the massive expansion of space make reaching equilibrium impossible
The uniformity is not due to an equilibrium, but to the initial conditions forming the inflation period.
An every day analogue: Shine a red laser on a piece of paper and take a picture. It is uniformly red not because the photons interacted between themselves, but because the were generated with the same boundary condition. If you enlarge the picture on the paper a lot, you will start seeing non uniformites due to the kind of paper you printed it on. or even quantum fluctuations in the lasing process. The low level non uniformities of the CMb are due to the quantum mechanical fluctuations proposed in the inflation model.
A is immersed in water, buoyancy must act on A; B is at the bottom of the container, there is no water between B and the container, that is to say, the bottom of B is not in contact with water, so will buoyancy act on B? Some people say buoyancy works on B, but I don't think so. B is like a suction hook. Water presses B on the bottom of the container. Buoyancy does not act on B. I'm not sure I'm absolutely right. 
I'm new to Quantum Field theory and that's something I've always seem people talk about but I've never understood.
On one hand we have the forces themselves. As I understood, the description within the framework of Quantum Field Theory is that each force is described by a Quantum Field. I've seem this done on Merzbacher's book on Quantum Mechanics, where in the last chapter he deals with Electromagnetic Fields and Photons.
The Quantum Field then is written in terms of creation and annihilation operators.
On the other hand, we have the theory of connections on principal bundles. The idea of Gauge Group fits into this mathematical framework. Indeed, if we have a principal bundle $\pi : P\to M$ with structure group $G$, we know that each fiber $\pi^{-1}(q)$ for $q\in M$ is isomorphic to $G$ and, furthermore, a choice of gauge is just a local trivialization $\sigma: M\to P$. The group $G$ is then the gauge group.
Now, I just can't understand how these two apparently totally different things relate.
So, just for the sake of an example, electromagnetic force is usually associated to the group $U(1)$.
We can indeed, pick any base space $M$ and consider a principal bundle $\pi : P\to M$ with structure group $U(1)$. But what this has to do with the Quantum Field Theory version of electromagnetism?
How to understand, in general, this connection established between the fundamental forces, described as Quantum Fields, and gauge groups?
Does the volume of a thermodynamic system always have to change for it to do work? If yes,could you explain why? And if no, could you provide the example of a system, where it is not neccesary.
Answer
Yes, because work is force times distance moved. It's not immediately obvious that this means work can only be done if the volume changes, but have a look at How much work is needed to compress a certain volume of gas? where I went into this in some detail. If any parts of this aren't clear comment here and I can go into them in more detail.
I have a paper which lies on a flat surface. The paper is fixed on one side and the opposite side can slide in the direction of the opposing side. As side end slides toward the other, a "bump" forms. I want to know what the solution is to the shape of this bump. There must be some standard solution for this case.
EDIT: Luboš Motl did provide a really nice answer for the case when the paper is assumed to lie flat against the surface at the ends (also with the assumption that the bump is small). I also found this interesting paper on a similar topic.
Answer
A very natural and interesting question.
If the paper stays nearly flat, we may use the linearized approximation. I will neglect the gravitational potential energy because it's probably much smaller than the bending energy (note that the shape of the bent paper is almost the same when the desk is vertical as when it is horizontal).
The paper doesn't want to bend much, so the energy contains a term that punishes the second derivatives of $y$ (the curvature) $$E = K \int_0^L {\rm d}x\,(y'')^2$$ where $y'\equiv dy/dx$, and so on. We want to keep the length of the paper fixed - it's prescribed by the way how you attach it at the end points. In the linear approximation, the length is a linear function of $$D = \int_0^L {\rm d} x\, (y')^2 $$ which may be seen, if you need it, by a Taylor expansion of the exact expression for the length of the graph, $\int (1+y^{\prime 2})^{1/2}{\rm d}x$. Add it with a Lagrange multiplier, requiring $$\delta (E+\lambda D) = 0.$$ I think that the resulting equations say $$ y'''' = \frac{\lambda}{K} y''.$$ Note that in the Euler-Lagrange equations, all the primes "clump". If you write $Y=y''$, it says that the second derivative of $Y$ is proportional to $Y$ itself. The physically relevant solution is $$ Y = Y_0 \cos (2\pi x n / L) $$ where only $n=1$ is possible if there is a table underneath the paper. Consequently, $$ y = y_0 [1-\cos (2\pi x / L)],$$ too. I added the term $1$ as an integration constant (while the other is zero) to guarantee that $y=0$ and $y'=0$ at the boundaries.
So what you get in practice is one wave of a cosine, from one minimum to the next one.
A more detailed discussion would be needed to eliminate the other cosine-like function with the same frequency, the sine, as well as linear functions, and to discuss whether the exponentially increasing/decreasing solutions may ever be relevant. Well, it wouldn't be too difficult. A more general solution for the same (cosine-like) sign of $\lambda$ would be $$ y = y_0 [1-\cos (\phi_0+ 2\pi x N / L)] + Ax^2+B,$$ and the four conditions $y=0$, $y'=0$ at $x=0,L$ as well as the fixed length of the paper would imply $N=1$, $\phi_0=0$, $A=0$, $B=0$, as well as the right normalization $y_0$. As suggested above, the conditions would have other solutions where $N=2,3,4\dots$ but these solutions wouldn't satisfy $y\geq 0$ for $0\leq x \leq L$, so the paper couldn't arrange itself above the table. If there were a hole in the table, these higher harmonics (several periods of the wave containing) solutions would be possible - but I guess that they would be unstable.
I am not sure whether I would be able to solve it analytically if the angle $y'$ were not infinitesimal. It seems clear that the exact solution for significant angles is not just a cosine: the bent paper tends to resemble a balloon - for which $y(x)$ is not even single-valued - when there's too much redundant paper in the middle.
On the other hand, if you sharply bend the paper at $x=0,L$ so that $y'$ may be arbitrary at these two points, the paper in between will be bent as an arc of a circle - see another answer below - and this result is accurate even if $y'$ is of order one. Note that in the linearized approximation, the arc gives $y$ being a quadratic function of $x$ so that $y''=0$ still solves the fourth-order equation above.
Today I learnt about Ohm's Law(I had some basic knowledge earlier).When I came home I searched in the internet and found that at Ohm's Law, resistance should not change because then the circuit will not obey Ohm's Law.I googled about this "problem" but I found nothing. My physic professor says that resistance can change at Ohm's Law. I don't know which one to trust. Can anyone help me ?
Answer
Well, the thing with Ohm's law is that the ordinary expression $I=\frac{V}{R}$ (sorry, I'm on mobile and writing "nice math" here is a pain), works only for what is usually referred to as "static resistance", that is $R=\frac{V}{I}$. However, this implies a material whose resistance does not vary. There are some materials whose resistance varies inversely to its cross-sectional area and proportionally its length ($R=k\frac{l}{A}$, where $k$ is the coefficient of resistivity of a material, a constant really). For this materials you need a "dynamic resistance" which is given by $\frac{dV}{dI}$, simply the rate of change of $V$ with respect to $I$ (or the slope of said line).
I am currently studying differential cross sections for my Nuclear Physics module. I'm looking an experiment where muon-neutrinos are interacting with nucleons in a scintillator producing muons (which then cause the scintillation to occur).
I went ahead and drew a Feynman diagram for the interaction:
I've read that the direction of the boson (and whether it is $ W^+ $or $ W^- $) depends on the 4-momentum transfer. In this case high energy muon-neutrinos are interacting with low energy (static) neutrons.
So my question is: Should the boson be a $ W^+ $going from the neutrino to the neutron or a $ W^- $going doing the opposite? I was originally going to just write $ W^±$without a directionality arrow, is this too an accepted convention?
Any advice would be much appreciated! Thanks, Sean.
Edit: Here's a model example taken from http://danielscully.co.uk/thesis/interactions.html 
Answer
Let's just focus on the top vertex.
First, draw an arrow pointing down. Then, you can determine from charge conservation (or looking for the relevant term in the lagrangian) that the particle must be $W^+$. Because $(W^+)^\ast=W^-$, this vertex has precisely the same value if you reverse that arrow and replace the $W^+$ with a $W^-$. Here I am just using the fact that arrow reversal corresponds to replacing the particle by its antiparticle.
Thus, not only do the different diagrams have the same value, in this sense, they are the same diagram. In particular, you don't 'double count' by adding the value of a $W^+$ diagram to the value of a $W^-$ diagram.
For what it's worth, I think it's worthwhile to verify this at the level of the Lagrangian. There is only a single term in the lagrangian that corresponds to the upper vertex interaction, but that term has two names, one with a $W^+$ and one with $(W^-)^\ast$.
So yes, you can safely leave the direction off and simply label it with $W$.
If flow of electrons analogy can be imagined as flow of water, how to imagine electricity, that comes in this whole picture?
When plug from bottom of sink is taken out, gravity pulls water molecules, creating pressure difference, in sink there is larger pressure than in open hole, making water to move to place, where is less pressure, until pressure is equal in all places, water can flow.
As I have understood, electricity is some field around each electron. Power of field is called electric charge, mesured in columbs, every electron has some charge, eaven those, who are sitting tight, that is called static electricity? That charge is generated when electrons rub/hit against each other, when they are "made to move", when electrical potencial, or pressure difference is made, by connecting to medium, another medium, that has... What that connected medium has, that makes electrons flow? Less electrons? Less electrons with electric field around them? If that is field, that can have some maximum ammount of charge, can charge vary, how that affect flow? I don't think, it can be explained with water, as water isn't carrying... anything...
What is mesured in amperes? What is rate of flow, is it a speed of water/electrons flow?
Are electrons same as water, in manner of, they are just some sort of matter, that is flowing in another medium, but water has much larger particle size, so its hard to make is analogy in my mind - there is no medium, in which water can flow, water is medium by itself.
If I forget, that water is medium, but pretend, that water can only flow in... air, places where air can go in... Then how does wire connections work? If There is one wire, that divides in two, as some sort of connection, does in each wire, goes same ammount of electrons (if we pretend ideal/same conditions in both wires)? Does that mean, there will be half of electrons in one wire, and other half in another wire? The more wires I connect, the less electrons will be in each wire? If one of wires will have damaged lattice or resistor sitting middle of it, will electrons "run way back" to connection place and go into other wires, where is less resistance or they will stay at same place, just building up pressure, constantly making denser electron cloud in resistance area... Ehh...
I know, I'm asking too many questions, that touches too many sides of electricity, but please, I want some analogy explanation, that would allow me better understand this whole "stuff". :) Maybe, someone can suggest some visual materials (video, pictures), for this?
Answer
To begin with, a good physical analogy matches physical quantities of one system to physical quantities of another system that act mathematically equivalent in some (limited) context.
The example of electric current (Amperes C/s) and electric potential (Volts or J/C) makes a good comparison to water flow (kg/s) and gravitational potential (J/kg). However, in order to use this comparison, you need to back off from the concept of pressure, as well as a few others. Consider a lazy river type situation where a pump imparts the gravitational potential to the water and allows it to flow through the resistance of the rest of the closed loop, and maybe a water wheel.
Alternatively, you could consider a loop of water employing the analogy with quantities of flow (kg/s) and hydrostatic pressure (J/kg or psi or many other units). An example of this could be something like a flow loop in a nuclear plant, for instance. There exist components that impart pressure to the flow, then there are components that have friction (again, like a resistor) that balance against the work of the pump. The reason this might not be as good of an example is because it requires a pressurized loop, which is further from our common everyday experience.
It is true that the water does not exist in a medium in the same way the electrons in an electric circuit do. This is just one of the many many reasons why the analogs are not perfect. The thing to focus on is that the equations can be the same with different quantities in place. Right now, I'm focused on the most simple laws possible. For instance:
$$V = I R$$ $$P = I V$$
The examples I discussed before can certainly be put in terms of these laws. All that these do is illustrate a case where there is a flow with a unit attached to it, and that flow has a per-unit energy value which dictates how much flow will occur over static components (the resistor case) and then concepts of power (among others) come naturally.
Consider the basic relation $$E=\sqrt{(pc)^2+(mc^2)^2}.$$
Every particle possesses a wave nature and it depends on the situation in which one among the two is perceptible...
Consider a particle with rest mass $m$. If we consider the speed of De-Broglie Waves, as usual for a wave $$v_{wave}=\nu \lambda.$$ And since we are taking relativistic effects into account, let's write $$\lambda =\frac{h}{\gamma mv}$$ where $\gamma$ denotes the Lorentz factor $\gamma =1/\sqrt{1-(v/c)^2}$, and $v$ the speed of the particle. Now clearly the energy of the wave could be written as $E=h \nu$. And for the particle, Energy is equal to $\gamma mc^2$. So clearly $$h \nu =\gamma mc^2.$$ Now plugging into $v_{wave}=\nu \lambda$, we get $$v_{wave}=\frac{\gamma mc^2}{h}\frac{h}{\gamma mv},$$ or $$v_{wave}=\frac{c^2}{v}.$$ Doesn't this seem to go against what we know, that the velocity of the wave is less than or equal to $c$?
So can anyone point out what's the mistake here? Does this have anything to do with phase or group velocity?
Answer
What you have calculated is the phase velocity, $v_p$, of the de Broglie wave associated with the particle. The phase velocity can be greater than $c$, and indeed it is always greater than $c$.
The velocity of the particle is the group velocity, $v_g$, and as you have demonstrated the two are linked by:
$$ v_p v_g = c^2 $$
The group velocity must always be less than $c$ and that implies the phase veocity must always be greater than $c$.
In some cases, according to Wikipedia, the envelope of a gaussian beam can go faster than speed of light hence leading to superluminal group velocity. However, the signal/energy still propagates at subluminal speed which is seen from the speed of the rising front of the pulse.
Do you know a practical example for which this situation arises? Is it possible to have a interactive picture of the corresponding wave?
I presume the pulse should distort quite significantly.
Answer
Superluminal group velocity can occur in near absorption peak, known as regions of anomalous dispersion. So-called "superluminal tunneling" experiments have bee conducted in thes regions, but when carefully analyzed there is no information transferred faster than light.
Some references are given here: https://www.rp-photonics.com/superluminal_transmission.html
I'm not familiar with any applications, but everything is good for something, certainly strong absorption lines are useful.
I have a doubt on the formulation of Newton third law.
When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.
On some textbooks I find that it follows from the law that the two forces have "the same line of action". What does this mean exactly?
In particular does it imply the collinearity of the two forces (i.e. the fact that the two forces acts on the very same line) (2)? Or two forces acting on two different but parallel lines still satisfy the law (1)?
Answer
Newton's third law comes in two versions:
The weak Newton's third law says that mutual forces of action and reaction are equal and opposite between two particles $i$ and $j$ at position $\vec{r}_i$ and $\vec{r}_j$, $$ \vec{F}_{ij}+\vec{F}_{ji}~=~\vec{0}.\tag{1}$$
The strong Newton's third law says besides eq. $(1)$ that the forces are also collinear, $$\vec{F}_{ij} ~\parallel ~\vec{r}_{ij},\tag{2}$$ i.e. parallel to the difference in positions $$\vec{r}_{ij}~:=~\vec{r}_j-\vec{r}_i,\tag{3}$$ or equivalently, act on the same line, as OP writes.
The weak Newton's third law is Newton's original formulation. The strong Newton's third law is a non-trivial additional assumption, cf. e.g. this Phys.SE post. If the strong Newton's third law holds, it often facilitates arguments. E.g. when applying D'Alembert's principle or Lagrangian mechanics to a rigid body, one can then ignore all the internal forces, which is an enormous simplification, cf. this Phys.SE post.
Basic property of magnetic lines of force is that, they can never intersect each other. Among the two points given below, which one is correct?
Suppose I have two bar magnets. If I tend to make the repelling poles join together by exerting force (i.e, if I tend to make north pole of two magnets, or south pole of two magnets join together). Would the magnetic lines of force intersect each other?
What is the meaning of the phase of a wave and phase difference? How do you visualize it?
Suppose a spring with spring constant 1 N m^-1 is held horizontally. If a pull of 1 N is applied to its left end and a pull of 2 N is applied to its right end, how much longer would the spring be as compared to its unstretched length?
I came up with this question because the spring related qurstions I've seen all have one end fixed some where. I am totally stuck on this one and I really have no idea where to start.
All I can figure out is that the spring will accelerate due to a net force. It seems inappropriate to be taking net forces, since pulling both ends with a larger force makes the spring more elongated.
P.S. this is not homework
Answer
Solution 1---Guess-work: If the forces applied on the two ends are equal, say both $1.5\ N$, the spring will get stretched $1.5\ m$. A natural guess is that the stretch is determined by the average of the two forces at the ends, which in this case are both equal to $1.5\ N$. Therefore, for the case you mentioned ($1\ N$ applied to one end, and $2\ N$ to the other end), the answer is again obtained from the average: the spring is stretched $1.5\ m$.
Solution 2---Precise analysis: If the spring is massless, the question is ill-posed; the nonzero net force yields infinite acceleration.
If the spring has mass $1\ kg$ (you can extend the following analysis to arbitrary mass), then since the net force on it is $2\ N-1\ N =1\ N$, the spring will have $1\ m/s^2$ acceleration to the right.
Now let us, as observers, accelerate along with the spring so at all times we are in rest with respect to each other. So we (the observers) have $1\ m/s^2$ acceleration to the right.
Since we are accelerating to the right, we are no longer inertial observers; so if we want to use Newton's laws we should think everything we see is subject to a "gravitational field" with strength $1\ m/s^2$ to the left (this is the push-back force you experience when a car you sit inside accelerates). Therefore we see a spring with $1\ N$ force applied to its left end, $2\ N$ applied to its right end, and a gravitational pull of $1\ N$ on it to the left; the total force is zero in our (non-inertial) accelerated frame, as it should be, since we see the spring at rest.
Now, applying Newton's second law (we are still in the non-inertial frame remember) to the tiniest bit of spring at the left end, we learn that the tension of the spring is $1\ N$ at the left end; similarly the tension of the spring is $2\ N$ at the other end. The tension in the middle of the spring linearly interpolates between these two values. This linear gradient in spring's tension is due to the left-wards gravitational field; a similar (gravitational) effect results in linear vertical pressure gradient in liquids at rest.
Now you see why this problem is much more difficult than those with equal forces applied to both ends of the spring. Here the tension varies along the spring, leading to further complications. However it's not too difficult to overcome these complications ...
The tiniest bit of the spring at the left end is subject to $1\ N$ at its both ends; if every bit of the spring was like that, the spring would stretch $1 m$. The tiniest bit of the spring at the right end is subject to $2\ N$ at its both ends; if every bit of the spring was like that, the spring would stretch $2 m$. But the parts of the spring in the middle are subject to a tension which linearly interpolates these two values, therefore on average the tiny pieces of the spring stretch in a way that results in the whole spring stretching $1.5\ m$.
This confirms the guess-work in Solutions 1.
You can convince yourself that although the ($1\ kg$) mass we introduced for the spring is necessary to make the problem well-defined, the final answer doesn't depend on the mass of the spring.
In John Preskill's review of monopoles he states on p. 471
Nowadays, we have another way of understanding why electric charge is quantized. Charge is quantized if the electromagnetic $U(l)_{\rm em}$ gauge group is compact. But $U(l)_{\rm em}$ is automatically compact in a unified gauge theory in which $U(l)_{\rm em}$ is embedded in a nonabelian semisimple group. [Note that the standard Weinberg-Salam-Glashow (35) model is not "unified" according to this criterion.]
The implication of the third sentence is that, in some circumstances, the $U(1)_{\rm em}$ gauge group may not be compact. How could this be? Since $U(1)$ as a differentiable manifold is diffeomorphic to $S^1$ isn't it automatically always compact?
The following paragraph:
In other words, in a unified gauge theory, the electric charge operator obeys nontrivial commutation relations with other operators in the theory. Just as the angular momentum algebra requires the eigenvalues of $J_z$ to be integer multiples of $\frac{\hbar}{2}$, the commutation relations satisfied by the electric charge operator require its eigenvalues to be integer multiples of a fundamental unit. This conclusion holds even if the symmetries generated by the charges that fail to commute with electric charge are spontaneously broken.
is OK, but I don't follow what that has to do with the compactness of $U(1)$.
Answer
By the "noncompact $U(1)$ group", we mean a group that is isomorphic to $({\mathbb R},+)$. In other words, the elements of $U(1)$ are formally $\exp(i\phi)$ but the identification $\phi\sim \phi+2\pi k$ isn't imposed. When it's not imposed, it also means that the dual variable ("momentum") to $\phi$, the charge, isn't quantized. One may allow fields with arbitrary continuous charges $Q$ that transform by the factor $\exp(iQ\phi)$.
It's still legitimate to call this a version of a $U(1)$ group because the Lie algebra of the group is still the same, ${\mathfrak u}(1)$.
In the second part of the question, where I am not 100% sure what you don't understand about the quote, you probably want to explain why compactness is related to quantization? It's because the charge $Q$ is what determines how the phase $\phi$ of a complex field is changing under gauge transformations. If we say that the gauge transformation multiplying fields by $\exp(iQ\phi)$ is equivalent for $\phi$ and $\phi+2\pi$, it's equivalent to saying that $Q$ is integer-valued because the identity $\exp(iQ\phi)=\exp(iQ(\phi+2\pi))$ holds iff $Q\in{\mathbb Z}$. It's the same logic as the quantization of momentum on compact spaces or angular momentum from wave functions that depend on the spherical coordinates.
He is explaining that the embedding of the $Q$ into a non-Abelian group pretty much implies that $Q$ is embedded into an $SU(2)$ group inside the non-Abelian group, and then the $Q$ is quantized for the same mathematical reason why $J_z$ is quantized. I would only repeat his explanation because it seems utterly complete and comprehensible to me.
Note that the quantization of $Q$ holds even if the $SU(2)$ is spontaneously broken to a $U(1)$. After all, we see such a thing in the electroweak theory. The group theory still works for the spontaneously broken $SU(2)$ group.
I'm trying to understand the definition given on my electromagnetism course for the current density. More specifically, I want to know why, as defined below, the current density is given the name "current density."
On my course, the current density is $\vec{j}(t,\vec{x}):=\rho (t,\vec{x}) \vec{v} (t,\vec{x})$ where $\vec{v} (t,\vec{x})$ is the velocity field governing how the charged particles move. I'm trying to get some intuition for what this quantity is.
To give an example of what I'm talking about, in classical mechanics where you have momentum equal to mass multiplied by velocity, the definition makes sense intuitively because momentum is the oomph you will feel if an object hits you, and you feel that oomph more if either the mass or velocity of the object increases. So I have a really tangible idea of what momentum is.
Wikipedia describing current density: "In electromagnetism, and related fields in solid state physics, condensed matter physics etc. current density is the electric current per unit area of cross section." This justifies calling it a density (as it's an area density by defn.). I'm trying to understand what a current density could be, and in my head I've got an idea of a cross-sectional area with some fluid flowing through it (it's the same scenario in which I picutre Gauss' Law). I'm not sure what current at a point is, so I don't really understand what a current density could be! Then I need to relate this to the definition I've been given on my course somehow. Thanks for any help!
Answer
firtree is correct - I will just try to flesh out his answer a bit.
(1) Your last question first - charge (or current) at a point is like mass at a point.
For finite masses, if you want to see how much is contained in an infinitely small volume (i.e., at a point), the answer is zero. So instead, people consider the mass density which can have non-zero values at a point. You probably understand the relationship between mass and (mass) density quite well.
Similarly for a finite current, the amount of current at a point (i.e., in an infinitely small volume) is zero. The current density is the limit of the amount of current in a small volume around a point as the volume goes to zero - just like mass density, but with current instead.
So, just as one speaks of mass density at a point and not mass at a point (for extended bodies), one speaks of charge density at a point and not charge at a point or current density at a point and not current at a point (we're ignoring point particles for now - they do fit into this formalism, but you need Dirac delta functions).
(2) Now, in analogy with mass flow, your picture of flow of charge is correct. Mass density times velocity gives a mass current density. $\vec{j}_{m}(t,\vec{x}):=\rho_{m} (t,\vec{x}) * \vec{v} (t,\vec{x})$. If you have a mass current density $\vec{j}_{m}$ and want to know the mass flow $\dot{m}$ through some area A, then you take
\begin{equation} \dot{m} = \int \vec{j}_{m} \cdot d\vec{A} \end{equation}
Similarly, charge density times velocity gives a charge current density. $\vec{j}_{q}(t,\vec{x}):=\rho_{q} (t,\vec{x}) * \vec{v} (t,\vec{x})$. If you have a charge current density $\vec{j}_{q}$ and you want to know the flow of charge $\dot{q}$ through some area A, then you take
\begin{equation} \dot{q} = \int \vec{j}_{q} \cdot d\vec{A} \end{equation}
So, the picture in your head is quite close - just picture charge or a fluid of charged particles flowing.
Although this question is going to seem completely trivial to anyone with any exposure to path integrals, I'm looking to answer this precisely and haven't been able to find any materials after looking for about 40 minutes, which leads me to believe that it makes sense to ask the question here. In particular I'm looking for an answer wherin any mathematical claims are phrased as precisely as possible, with detailed proofs either provided or referenced. Also my search for a solution has led me to think I'm actually looking for a good explenation of Wick rotation, which I can't really claim to understand in detail. Any good references about this would be very welcome as well.
I'm looking to make sense of the following integral identity:
$$\int_{-\infty}^{\infty} dx \ \exp\left(i\frac{a}{2}x^2+iJx\right)=\left(\frac{2\pi i}{a}\right)^{1/2}\exp\left(\frac{-iJ^2}{2a}\right), \qquad a,J\in\mathbb{R}$$
Wikipedia (and various other sources) say that "This result is valid as an integration in the complex plane as long as a has a positive imaginary part." Clearly the left hand side does not exist in Lebesgue sense for real $a, J$. An answer to the question "Wick rotation in field theory - rigorous justification?" claims:
"it is convergent as a Riemann integral, thanks to some rather delicate cancellations. To make the integral well defined -- equivalently to see how these cancellations occur -- we need to supply some additional information. Wick rotation provides a way of doing this. You observe that the left hand side is analytic in t , and that the right hand side is well-defined if Im(t)<0. Then you can define the integral for real t by saying that it's analytic continued from complex t with negative imaginary part."
I want to see the gory details and all known motivation for the validity of this procedure for the kinds of applications where such integrals occur. Suggestions such as "include an $i\epsilon$ in order to make it finite" seem arbitrary. In that case one would have to motivate that prescription from the very start, that is within the modeling procedure that ends up giving that integral expression (which is likely the correct way to approach this). I'm also not sure how to interpret the right hand side, since it involves the square root of an imaginary number, which should involve some choice of branch cut, which I have never seen specified in connection to this formula.
Answer
Proposition. Let there be given two complex numbers $a,b\in \mathbb{C}$ such that ${\rm Re}(a)\geq 0$. In the case ${\rm Re}(a)=0$, we demand furthermore that ${\rm Im}(a)\neq 0$ and ${\rm Re}(b)=0$. The Gaussian integral is well-defined and is given by $$ \underbrace{\int_{\mathbb{R}}\!dx~ e^{-\frac{a}{2}x^2+bx}}_{=:~ I_{\mathbb{R}}(a,b)} ~=~\lim_{\begin{array}{c} x_i\to -\infty \cr x_f\to \infty \end{array} } \underbrace{\int_{[x_i,x_f]}\!dx~ e^{-\frac{a}{2}x^2+bx}}_{=:~ I_{[x_i,x_f]}(a,b)} ~=~\underbrace{\sqrt{\frac{2\pi}{a}}e^{\frac{b^2}{2a}}}_{=:~ F(a,b)}, \tag{A}$$ where it is implicitly understood that the square root has positive real part.
Remark: The Riemann/Darboux integral is not defined for non-bounded sets, so it can only be used for the middle expression of eq. (A).
I) Sketched proof in case of ${\rm Re}(a)> 0$: The function $g(x)=e^{-\frac{{\rm Re}(a)}{2}x^2+{\rm Re}(b)x}$ serves as a majorant function for Lebesgue's dominated convergence theorem, which establishes the first equality of eq. (A). For the second equality of eq. (A), we divide the proof into cases:
Case $a>0$ and $b\in \mathbb{R}$. Complete the square. $\Box$
Case $a>0$. Complete the square. Shift the integration contour appropriately to a horizontal line in the complex plane in order to reduce to case 1, cf. Cauchy's integral theorem. Argue that contributions at infinity vanish. $\Box$
Case ${\rm Re}(a)> 0$. Rotate the integration contour to a line of steepest descent in order to reduce to case 2, cf. Cauchy's integral theorem. Argue that contributions at infinity vanish. $\Box$
II) Sketched proof in the oscillatory case ${\rm Re}(a)=0, {\rm Im}(a)\neq 0, {\rm Re}(b)=0$: The lhs. of eq. (A) is not Lebesgue integrable. It is an improper integral defined via the middle expression of eq. (A). It remains to prove the second equality of eq. (A). It is possible to give a proof using Cauchy's integral theorem along the lines of Jack's answer. In this answer we will instead give a proof in the spirit of an infinitesimal deformation prescription.
Given $\varepsilon>0$. As $x_i\to \infty$ and $x_f\to \infty$ it is not hard to see that $I_{[x_i,x_f]}(a,b)$ oscillates with smaller and smaller amplitude that tends to zero, and it is hence convergent without any regularization. The convergence improves if we let $a$ have a positive real part. In other words, the convergence is uniform wrt. ${\rm Re}(a)\geq 0$, i.e.
$$ \exists X_i,X_f\in \mathbb{R} ~\forall x_i\leq X_i~\forall x_f\geq X_f ~\forall {\rm Re}(a)\geq 0:~~ \left| I_{[x_i,x_f]}(a,b)- I_{\mathbb{R}}(a,b)\right| ~\leq~\frac{\varepsilon}{4}.\tag{B}$$
Next use Lebesgue's dominated convergence theorem with majorant function of the form $g(x)=C~1_{[x_i,x_f]}(x)$ (where $C>0$ is an appropriate constant) to argue that
$$I_{[x_i,x_f]}( i{\rm Im}(a),b) ~=~\lim_{{\rm Re}(a)\to 0^+} I_{[x_i,x_f]}(a,b) , \tag{C}$$
i.e. $\exists {\rm Re}(a)>0$ such that
$$ \left| I_{[x_i,x_f]}( i{\rm Im}(a),b)-I_{[x_i,x_f]}( a ,b) \right| ~\leq~\frac{\varepsilon}{4},\tag{D}$$
and
$$\left| \underbrace{F( a ,b)}_{=~ I_{\mathbb{R}}(a,b)}- F( i{\rm Im}(a),b) \right| ~\leq~\frac{\varepsilon}{4}.\tag{E}$$
In eq. (E) we used that the function $F$ is continuous. All together, eqs. (B), (D) & (E) yield $$ \begin{align} \left| I_{\mathbb{R}}(i{\rm Im}(a),b) - F( i{\rm Im}(a),b)\right| ~\leq~&\left| I_{\mathbb{R}}(i{\rm Im}(a),b)- I_{[x_i,x_f]}( i{\rm Im}(a),b)\right|\cr &+\left| I_{[x_i,x_f]}( i{\rm Im}(a),b) - I_{[x_i,x_f]}( a ,b)\right| \cr &+\left| I_{[x_i,x_f]}(a,b)- I_{\mathbb{R}}(a,b)\right|\cr &+\left|F( a ,b) - F( i{\rm Im}(a),b)\right| \cr ~\leq~&\varepsilon.\end{align} \tag{F}$$
Eq. (F) shows that the second equality of eq. (A) holds. $\Box$
''electric field always undergoes a discontinuity when you cross a surface charge $\sigma$'' GRIFFITHS
In the derivation; Suppose we draw a wafer-thin Gaussian Pillbox, extended just barely over the edge in each direction. Gauss law states that:
$$\int_{S} E \cdot A = Q_\text{enc}/ \epsilon$$
and so $$E_{\perp\;\text{above}} - E_{\perp\;\text{below} } = \sigma/ \epsilon . $$
My question is why not $2A$? $$\int_{S} E \cdot A = 2EA$$ because the top area of pillbox and the bottom area of pillbox, just as because the 2 parts of the flux...
SO.. WHY NOT : $$E_{\perp\;\text{above}}- E_{\perp\;\text{below} } = \sigma/ 2\epsilon ~? $$
And why there is tangencial component of electric field; not just perpendicular to the surface, which can be seen as flat just looking very close to the surface.
In high school, it was taught that formula for describing circular orbital velocity around a central body is derived by equating Newton's law of gravity with the centripetal force formula (under the logic that the inwards centeipetal force required is provided by the gravitational "force").
It was only recently that I discovered that gravity isn't actually a force but is actually a distortion of space time. (I came across this while wondering why light bends around large masses). Does the fact that gravity is not a force make the above derivation of orbital velocity any less valid? Because the above derivation assumes that gravity is a force.
Answer
You cannot longer use Newton's formalism $F = m a = -GMm/r^2$ if you introduce the fact that the geometry of space time is changed by the presence of the central body $M$.
It is true that the test mass $m$ still moves around $M$ because of gravity, but you should think of gravity not as a force any more, but as an emergent property of the curvature of space-time. Fortunately there's a whole body of mathematical tools that allow you solve this problem in particular.
Actually, it is one the most well known problems you can analytically solved using general relativity: the two body problem
Newton's Law of Universal Gravitation tells us that the potential energy of object in a gravitational field is $$U ~=~ -\frac{GMm}{r}.\tag{1}$$
The experimentally verified near-Earth gravitational potential is $$U ~=~ mgh.\tag{2}$$
The near-Earth potential should be an approximation for the general potential energy when $r\approx r_{\text{Earth}}$, but the problem I'm having is that they scale differently with distance. $(1)$ scales as $\frac 1r$. So the greater the distance from the Earth, the less potential energy an object should have. But $(2)$ scales proportionally to distance. So the greater the distance from the Earth, the more potential energy an object should have.
How is this reconcilable?
Answer
Your equation (2) is the change in potential energy when the object moves vertically by a distance $h$ i.e. when the object moves from $r$ to $r+h$. Let's use equation (1) to calculate this:
$$ \Delta U = GMm\left(\frac{1}{r}-\frac{1}{r+h}\right) $$
Subtracting the two fractions inside the bracket gives:
$$\begin{align} \Delta U &= GMm\left(\frac{r+h}{r(r+h)}\frac{r}{r(r+h)}\right) \\ &= GMm\frac{h}{r(r+h)} \end{align}$$
Since $h \ll r$ that means $r+h\approx r$ and our equation becomes:
$$\begin{align} \Delta U &\approx GMm\frac{h}{r^2} \\ &\approx \frac{GM}{r^2}mh \\ &\approx gmh \end{align}$$
Footnote:
I've just noticed that in your comment to Itachí's answer you ask if you can use a Taylor series. You can use a binomial expansion to make the aproximation more obvious. You rewrite:
$$ \Delta U = GMm\frac{h}{r(r+h)} $$
as:
$$ \Delta U = \frac{GM}{r^2}mh\left(1+\frac{h}{r}\right)^{-1} $$
then a binomial expansion gives:
$$ \Delta U = \frac{GM}{r^2}mh\left(1-\frac{h}{r} + O\left(\frac{h}{r}\right)^2 \right) $$
And as before since $h \ll r$ the term in the brackets is approximately one and we once again get:
$$ \Delta U = \frac{GM}{r^2}mh $$
I was listening to Quantum Mechanics lecture and there were wave explanation; to be exact it is periodic wave ..
its formula is v (speed) = lambda (wavelength) X f (frequency cycles/second)
the professor said that if for example wavelength is shorter wave, that means the frequency is becoming higher .. and if wavelength is longer wave, that means the frequency is becoming lower ...
I don't get it, I thought that if we have an object which has many cycles per second such as 50 c/s, then it has much more opportunity than the 30 c/s to reach as far place as I know...
please someone could explain what does professor means?
Answer
Imagine a wave that advances 10 meters per second.
if every second a wave crashes on the shore, the frequence is one per second (1 herz) the wavelength is then 10 meters since each second the wave advanced 10 meters and therefore the waves are 10 meters apart.
If we halve the wavelength (the wave still advances at 10 meters per second) then the waves are 5 meters apart. if waves 5 meters apart advance at 10 meters per second then 2 waves will crash on the shore per second. This means that the frequency of waves has doubled.
Conclusion : if the speed of a wave in an environment is constant, then doubling the frequency is equivalent to dividing wavelength by two.
Constant = frequency * wavelength
for electromagnetic waves : speed of light = frequency * wavelength
I'm wondering why we only feel the centrifugal force in a circular motion.
I did look at Why do we only feel the centrifugal force? question which is exactly that, but I'm not really satisfied with the answer. When you are rotating in a non-inertial reference frame the centripetal force pulls you towards the center, but this acceleration is balanced by the centrifugal force. But if that's the case, why do we feel a centrifugal force?
The other question says that it's a reaction force, but since the acceleration is balanced, why do we still feel this force. And in a related example, if a ball starts to spin in a circle, why doesn't it keep the same radius, but instead is pushed out because of centrifugal force? Aren't the centrifugal force and centripetal force balanced?
Answer
The thing is that net force of $0$ does not mean "no forces act on you". Indeed, you feel pushed into your chair as you sit on it, and if you were sitting in your chair on an upward accelerating elevator you would feel pushed into your chair even more. In either case the net force acting on you in your non-inertial frame is $0$ (once you take pseudo-forces into account), but you would "feel" more in the case of the upward accelerating elevator compared to just sitting in a chair on the ground.
This shows us the key to your question. We "feel in the opposite direction of acceleration (as observed from an inertial frame)." In the case of the elevator, we more we accelerate upwards, the more we feel pushed downwards. In circular motion, the acceleration is inwards, so we feel "pulled" outwards.
You can essentially link this to Einstein's equivalence principle: experiencing an acceleration is just like being in a gravitational field pointing in the opposite direction as the acceleration. Indeed, this is even why you feel pulled to the ground right now, you could argue that you are actually being accelerated upwards (approximately).
Another way to think about it. In non-inertial frames Newton's laws no longer hold. But which law/laws does/do not hold? Well, technically speaking Newton's second law no longer holds. In other words, we can feel forces without experiencing accelerations in our non-inertial frame. So it should be no surprise that we still feel an outwards force when we don't observe acceleration in our frame, if we felt nothing this law would hold!
However, this idea doesn't sit well with us. What if we still want Newton's second law to hold? Well then we introduce the idea of fictitious forces. However, by doing this we have now thrown Newton's third law out the window. Forces now longer have to have an equal and opposite partner. Forces do not have to arise from interactions anymore. So your claim that the centrifugal force is a reaction to the centripetal force is actually false! Remember, action-reaction pairs do not act on the same object. The centrifugal force arises from the non-inertial reference frame.
So I would argue that while we have introduced fictitious forces to help us keep N2L valid, I think it is better to say that we are "feeling" the effects of being in a non-inertial reference frame.
I have had this question in my mind for a long time, I thought you guys might enlighten me easily.
I am confused about some space photographs and claims like "this galaxy is 13 billions light years away from us.": how we can take the photograph of something that far, if it is 13 billion light years away wouldn't it take 26 billion light years to take those pictures?
today this post led me to ask this question, at last: a space picture
There certainly is something I don't know about photography or light years; if you could tell me the logic behind this, I would appreciate it.
I am not a physicist or any science guy, so please tolerate my ignorance.
Answer
The error is probably in this statement
if it is 13 billion light years away wouldn't it take 26 billion light years to take those pictures?
I think you are imagining that cameras send out light to the objects, and when this light comes back records the light as an image. Not really. Cameras merely record the light they see from that area. So if that area is 13 billion light years away (not sure how credible source is) then all that means is that the light you are capturing today is the light that galaxy emitted 13 billion years ago.
Imagine for instance Anna and Bob are playing catch with a ball. Anna throws the ball to Bob. Bob receives the ball, and says the ball came at 3:00pm sharp. But the ball was in the air for 1 minute (anna is a slow thrower). That means Anna threw the ball at 2:59, even if Bob recorded it at 3:00. In this scenario, Bob is acting much like a camera acts, by receiving information (in this case a ball, in a camera's case it would be light from galaxies) The reason that Hubble took photos for 4 months (this might be wrong, I'm no good with photography) is that the longer it receives the information, the more 'background' light that we don't want to capture can be removed.
Hoped this makes sense.
P.S. may have misunderstood the question. You say
if it is 13 billion light years away wouldn't it take 26 billion light years to take those pictures?
as if light years are a measure of time. A light year is a measure of distance, the distance light travels in a year in a vacuum.
When I am on earth, the weight of my body is countered by the reaction of the ground. So, there is no net force acting on me.
But I am spinning with earth. But if there is no centripetal force then why am I spinning? And the equal air pressure on both side of my body won't be enough for me to stay in the same angular velocity as the earth.
Is it just conservation of angular momentum?
Answer
It is easier to consider you standing on the Equator.
Assume that the gravitational field strength at the Equator is $g$. This would be the acceleration of free fall at the Equator with no air resistance if the Earth was not spinning.
If the reaction of the Earth is $N$ then assuming down is positive and using N2L, $mg-N=0$ if your mass is $m$.
If the Earth of radius $R$ is spinning with angular speed $\omega$ then using N2L one gets $mg-N'=mR\omega^2$.
So the reaction force due to the Earth $N'$ has decreased.
The acceleration of free fall would also decrease to $g-R\omega^2\;(\approx 0.03 \rm ms^{-2})$ as would your apparent weight $m(g-R\omega^2)$.
So measuring your "weight" at the Equator using a spring balance would yield a smaller value than that at the geographic poles where you your centripetal acceleration would be zero.
If it so happened that the period of rotation of the Earth was 84.5 minutes you would find that there was no reaction force due to the Earth and the acceleration of free fall would be zero.
Objects which you let go of would not fall closer to the Earth.
This would be a state of weightlessness.
It so happens that 84.5 minutes is the theoretical speed of a satellite of the Earth whose circular orbit had a radius equal to that of the Earth.
All this has ignored the effect of air resistance and the fact that if the Earth was made to spin that fast it would disintegrate due to the brittle crust not being very good at sustaining tensile stresses.
