experimental physics - Capacitance dependence on separation between plates

I know that the capacitance of a parallel plate capacitor is $C = \frac{\varepsilon S}{d}$. I am supposed to perform a linear regression to obtain $\varepsilon$, however it turns to be a quadratic relation, .

My professor told us to justify it, so it might not be a mistake. It cannot be edge effect as the point that behave badly if we consider it a linear relation are those with smaller separation between plates. It cannot be caused by the saturation of the dielectric we used, as the electric field wasn't that strong. The problem isn't the absence of dielectric in a part of the separation between plates either.

We are using a parallel circular plate condensator. The surface of the plates is $S = \pi \left(\frac{0.255}{2}\right)^{2}$ m$^{2}$.

electricity - If Resistivity = $frac{RA}{L}$, why does it not depend on dimension?

The formula for resistivity is: $$\rho = \frac{RA}{L}$$ where $\rho$ is resistivity, $R$ is resistance, $A$ is cross-sectional area, and $L$ is the length of the conductor.

We can see from the formula that $A$ and $L$ are involved, why then does resistivity not depend on dimensions?

mathematical physics - Is the step of analytic continuation unavoidable or can you model around it?

One sometimes considers the analytic continuation of certain quantities in physics and take them seriously. More so than the direct or actual values, actually.

For example if you use the procedure for regularization, it sometimes seems like an ad hoc step.

Question: In cases where analytic continuation is applicable, does this suggest there is another formulation of that theory, which leads to these result directly?

That would be a theory where the "modified" interpretation of the mathematical quantities might be taken to be a starting point.

For example if you define some fundamental quantity of your theory as an integral or sum, and it doesn't converge somewhere, and you make an analytic continuation to get some valuable results. Could this imply there is a formulation where that value comes naturally, i.e. a formulation where there never is this sum object which makes problems?

special relativity - Four-velocity vector of light

Please note that my question is not a duplicate, it is not about the speed of light, my question is only technical about the four velocity vector for light, its definedeness, value and constantness.

I have read these questions:

What is the time component of velocity of a light ray?

Where Izhov says:

Four-velocity actually isn't well-defined for light.

And where ClassicStyle says in a comment:

The four velocity of light is perfectly well defined. You just can't use proper time to parameterize the world line. Four velocity is just the tangent vector to a world line

Are components of the velocity of light equal to $c$?

https://en.wikipedia.org/wiki/Four-vector

The four-velocity defined here using the proper time of an object does not exist for world lines for objects such as photons travelling at the speed of light

Why is light affected by time dilations in space-time curvatures

Where Safesphere says in a comment:

The magnitude of the 4-velocity of light is always zero (see my comment above).

The (always) non-zero time component of the 4-velocity of light does NOT mean that light moves in time. To calculate the 4-velocity of light, we have to use a different affine parameter instead of proper time, because the proper time of light is always zero.

Now this is confusing. Light must have a four velocity vector, but it seems to be either well defined or not, and it seems to have a magnitude of 0 or c and it seems to be always constant or not.

Questions:

1. 
   

   Which one is right, is the four velocity of light well defined or not?

   
   


2. 
   

   Is the magnitude of the four velocity vector for light always constant?

   
   


3. 
   
   

   Is the magnitude 0 or c?

   
   

Answer

So in relativity we have these things called four-vectors. Let $a$ be a four-vector, then in any given coordinate systems it has four components: $a^w$ in the time-direction $w=ct$, $a^{x,y,z}$ in the spatial direction.

Its squared-magnitude is defined by:$$a_\mu a^\mu = (a^w)^2 - (a^x)^2 - (a^y)^2 - (a^z)^2.$$ When this is negative we say that the four-vector is “space-like” or when it is positive we say that the four-vector is “time-like,” and when it is zero we say that it is null or “light-like.” And then if it's space-like we can either take the normal square root and get an imaginary number, or take $\sqrt{-a_\mu a^\mu}$ to get a positive number which we can call the magnitude; if it is time-like then the normal square root $\sqrt{a_\mu a^\mu}$ suffices to get a magnitude from a squared-magnitude. Like how rotations preserve vector lengths, Lorentz transformations preserve squared-magnitudes.

Four-velocity is one of these four-vectors. If we choose the right coordinate system so that your four-velocity lies along, say, the $z$-axis, then for particles it is defined as the vector $$v^w = c \cosh \phi,\\ v^{x,y}=0,\\ v^z = c\sinh \phi,$$ for some number $\phi$, and it corresponds to something moving with speed $c\tanh\phi.$ These functions, if you have not encountered them yet, are the hyperbolic functions.

Plugging this into the above equation gives $$v_\mu v^\mu = c^2(\cosh^2\phi - \sinh^2 \phi) = c^2,$$so that the four-velocity is always normalized to a constant value, it is timelike with magnitude $c$. Indeed you could imagine that it is the non-four-vector tangent vector to the particle's motion in spacetime, $(c, 0, 0, c\tanh\phi)$, but that has been normalized to have a constant magnitude $c$, so that Lorentz transformations can preserve this length.

Now whether you take this normalized value as a part of the definition of four-velocity, that is an aesthetic opinion rather than something which the mathematics forces on you. But there is a reason to be concerned.

See, you cannot normalize a light-like tangent vector into a normal four-velocity, because its squared magnitude is zero. So if you have the point in spacetime $(w,x,y,z) = (ct_0, x_0, y_0, z_0)$ and you add a little time $dt$ to this, a light ray moving from that point in the $z$-direction is now at point $(ct_0 + c~dt, x_0, y_0, z_0 + c~dt)$ and the difference between those two points is a four-vector, $$T^w = c~dt, T^{x,y} = 0, T^z = c~dt.$$ However we would find that $T_\mu T^\mu = 0$ and there is nothing you can multiply it by in order to give it constant magnitude $c$.

You might respond to this fact by saying “this is disastrous! let us say that light has no four-velocity!”—or you might instead respond by saying “ok, but this is actually a blessing in disguise, it does not mean that no normalizations are possible so much as all normalizations are trivial, I am free to choose whichever I want!”. Both responses have some merit. I personally tend towards the first one, for the following reason: my gut choice for the second is to normalize a lightlike tangent vector as $(c, 0, 0, c)$, so that the time component is constant. But, there is a problem with this normalization: the Lorentz transform will not preserve it. It will properly transform the vector, but I will have to renormalize it in the new context, too. I do not like that aspect very much, personally.

With that said there are occasionally reasons to do it. The easiest would be if you were thinking about what the universe looks like when you move through it: you would draw null rays in from all of the stars you can see at this very second, towards your face: and then it might be nice to project this all as if it had come from a sphere of fixed radius $R$ that you happen to be at the center of: a “celestial sphere.” And that is essentially the same normalization that I described above when I fixed the time component to a fixed value $c$. You might then perform a Lorentz boost in some direction, so the sphere gets mapped to some different sphere, and you might project the new sphere back onto a new sphere of fixed radius $R$, finding that the stars have all appeared to shift in the sky as a result of my boost (more specifically: they all appear to have shifted towards the direction that I was accelerating in). A similar argument about the light that I emit would suggest that it has also all crowded in that direction, leading to a well-known phenomenon called relativistic beaming where something which emits light preferentially emits it in the direction that it is travelling as it travels faster and faster.

What is the difference between thermodynamic and empirical temperature?

When I've studied Thermodynamics I did so in Callen's book and there the author talks about temperature as a single thing, which mathematically is simply defined as:

$$T = \dfrac{\partial U}{\partial S}.$$

Now, currently I'm needing to learn a more conceptual approach to Thermodynamics and in some books I see the authors talk about "two kinds of temperature".

One is called the empirical temperature and the other is called the thermodynamic temperature. As far as I understood the thermodynamic temperature is the one I've always known, which can be defined by the equation I've mentioned.

Now there's this empirical temperature and I have no idea about what it is. The authors introduce it by talking about lots of experiments relating properties of systems to thermodynamic equilibrium. In one of the books it is said:

Let $X$ represent the value of any thermodynamic property such as the emf $\mathcal{E}$ of a thermocouple, the resistance $R$ of a resistance thermometer, or the pressure $P$ of a fixed mass of gas kept at constant volume, and $\theta$ the empirical temperature of the thermometer or of any system with which it is in thermal equilibrium. The ratio of two empirical temperatures $\theta_2$ and $\theta_1$, as determined by a particular thermometer, is defined as equal to the corresponding ratio of the values of $X$:

$$\dfrac{\theta_{2}}{\theta_{1}}=\dfrac{X_2}{X_1}.$$

I don't really get what is going on here. What is this empirical temperature? What is its relation to the usual temperature? And why would anyone define something like that anyway?

astronomy - How much sky do we see at any one moment?

When we look at any particular point the sky, what percentage of the celestial sphere do we see?

This question arises from the notion that on average there passes one meteor per hour overhead. So assuming that one were to stare in one direction for one hour and during that hour exactly one meteor appears, what are the changes of the viewer having seen it?

Answer

According to this article:

Different animals have different fields of view, depending on the placement of the eyes. Humans have an almost 180-degree forward-facing horizontal field of view, while some birds have a complete or nearly-complete 360-degree field of view. In addition, the vertical range of the field of view in humans is typically around 100 degrees.

The range of visual abilities is not uniform across a field of view, and varies from animal to animal. For example, binocular vision, which is important for depth perception, only covers 120 degrees (horizontally) of the field of vision in humans[citation needed]; the remaining peripheral 60 degrees have no binocular vision (because of the lack of overlap in the images from either eye for those parts of the field of view). Some birds have a scant 10 or 20 degrees of binocular vision.

Similarly, color vision and the ability to perceive shape and motion vary across the field of view; in humans the former is concentrated in the center of the visual field, while the latter tends to be much stronger in the periphery. This is due to the much higher concentration of color-sensitive cone cells in the fovea, the central region of the retina, in comparison to the higher concentration of motion-sensitive rod cells in the periphery. Since cone cells require considerably brighter light sources to be activated, the result of this distribution is that peripheral vision is much stronger at night relative to binocular vision.

So a human observer should be able to see roughly half of the visible sky, or a quarter of a full sphere. If you look about 50° above the horizon, you should be able to see an area of the sky extending from directly to your left to directly to your right horizontally, and from the horizon to about 10° past the zenith.

You won't be able to see as well around the edges of your field of view. Depth perception isn't relevant when looking at the sky, but being able to see with both eyes is probably better than seeing with just one.

This can vary from person to person; some people have much better peripheral vision than others.

In non-metallic solids w/ just atoms or ions (no molecules), are bonds (vibrations) and electronic transitions the sole cause of blackbody radiation?

Since there wouldn't be a conduction band filled with any electrons in a non-metallic solid made of just atoms or ions (no molecules), it's hard to imagine any other type of movement and dipole moment than the stretching bonds between the atoms, or ions, as the cause of blackbody radiation (specifically infrared) in such solids. (And, yes, I'm aware that electronic transitions contribute, as well, in the range of visible light.) If there is another cause for such solids, please help clarify. I've heard of there being oscillations in electron density in the valence band of non metals, but not sure how that works if the valance band is full.

special relativity - What does it mean to say that mass "approaches infinity"?

What does it mean to say that mass "approaches infinity"?

I have read that mass of a body increases with the speed and when the body reaches the speed of light, the mass becomes infinity.

What exactly does it mean to say that the mass "approaches infinity" or "becomes infinity"? I am not able to get a picture of "infinite mass" in my mind.

Answer

The answers given so far are fine, but to my surprise nobody's mentioned the most important point: in modern terminology, we generally don't say that the mass of an object increases with speed. "Relativistic mass increase" is outdated terminology, not used by most physicists anymore. In general, nowadays, "mass" means "rest mass" and is independent of velocity. Igor Ivanov's answer to this question says it all.

I haven't read the article by Lev Okun that he refers to, but I like the term "pedagogical virus" for this notion.

vectors - Why do we represent the axis of rotation using vectorial notation

When a body (in pure rotation) rotates along an axis passing through it, why do we represent the axis of rotation in vectorial notation? Wouldn't it be sensible enough to represent the angular velocity (or similar quantities) in the vectorial notation? Is it because the direction changes continuously and thus representing the vectors themselves, would be anomalous because a different vector in a way would be called as a vector similar to another?

Answer

It is vector because it has a magnitude and a direction. Also typically the components of $\vec{\omega}$ are evaluated based on an inetial coordinate frame and thus only represent the motion of the body and not of the measuring frame.

• Rotational speed $\omega = \| \vec{\omega} \|$


• Direction of rotation $\hat{z} = \frac{\vec{\omega}}{\omega}$

Additionally, if the linear velocity of the rigid body $\vec{v}$ is given at some point, the location of the instant center of rotation is given by $$\vec{r}_{ICR} = \frac{\vec{\omega} \times \vec{v}}{\| \vec{\omega} \|^2} $$

So there is a lot of information given about the motion of the body from the 3 components of $\vec{\omega}$.

electron in magnetic and electric field

I have this problem where i should find the direction and magnitude of the electric and the magnetic force on the electron. And then I'm supposed to find the direction and magnitude of the acceleration.
E = 1000 N/m B = 2,5 T v = 500 m/s

X |X X| X
 ¤->
X |X X| X
X |X X| X
  V   V

I've found the electric force by Fe = E*q. With direction out of the screen by the right hand rule. Magnetic force = qvb in the y direction (meaning up) by the right hand rule. Now Im supposed to find the acceleration but I'm a bit stuck here. What i'm struggling with is to combine the two forces on the electron. Can someone get me in the right direction?

Answer

You just use vector addition and Newton's Second Law. For example, if you have $$\overrightarrow{F}_E=qE\hat{x},\space\space\space\overrightarrow{F}_B=qvB\hat{y}$$ then your total force $F_{tot}=F_E+F_B$ is just $$\overrightarrow{F}_{tot}=qE\hat{x}+qvB\hat{y}$$ Since $\hat{x}$ and $\hat{y}$ are totally linearly independent, these terms cannot be combined. Then using $F=ma$ you can deduce that $$\overrightarrow{a}=\frac{q}{m}(E\hat{x}+vB\hat{y})$$ but if you wanted the magnitude of the acceleration, you just take $$a=\sqrt{\overrightarrow{a}\cdot\overrightarrow{a}}=\frac{q}{m}\sqrt{E^2+v^2B^2}$$

Incompatibility Between Relativity and Quantum Mechanics

Why does Gravity distort space and time while the electromagnetic, strong, and weak forces do not?

Does this have to do with why Quantum Mechanics and Relativity are incompatible?

Answer

Although it would be more precise to say that gravity is the manifestation of the effect of curved space-time on moving bodies, and it is mass that curves the space-time, so prof. Rennie is correct about this, there are differences of opinion, at least, about the other aspects. It is not at all clear that mass is a kind of charge analogous to electric charge, although some researchers are trying to make this idea work and unify gravity with the Standard Model or QFT.

Be that as it may, what is clear is that gravity or curvature is different from electromagnetism or charm etc., for one thing, because gravity is not a force. Einstein, Schroedinger, other pioneers in GR were quite explicit about this. See gravity is not a force mantra, https://physics.stackexchange.com/a/18324/6432 for a discussion of this.

So there are major differences between gravity and the (other) fundamental forces, and this may well be the reason why gravity has not yet been successfully quantised.

But there are even more incompatibilities between the whole spirit of GR and the spirit of QM. J.S. Bell was quite concerned about the seemingly fundamental incompatibilities between relativity and quantum theory, too. For me, I would point out that in QM, the wave functions live on configuration space, which for, say, two particles, is six-dimensional, and also QM treats other dynamical variables such as spin as being equal in right, this makes the space even larger. Also QM treats momentum as just as valid a basis for coordinates as position, and this, too, is alien to the spirit of relativity, which treats the actual four-dimensional Riemannian manifold as basic.

For precisely the need to overcome this incompatibility, passing to Quantum Field Theory replaced these wave functions over configuration space with operator-valued functions on space-time. But although this kinda works to overcome the incompatibility of special relativity with QM, it makes the foundations of QFT much murkier (the role of probabilities, for instance the Born rule) and introduces infinities. Thus although it might be a way to reconcile QM and relativity theory, it is still more of an unfinished project and because of the unsatisfactory foundations of QFT (compared to the clear foundations of QM), one can still suspect there is a missing idea to really reconcile the two or even that somebody has to budge and concede something or there will be no treaty...

angular momentum - Spin coherent state for general spin $S$

In Wen's book on Many-body QFT, he claimed that the coherent state for a spin-$S$ particle can be written as a tensor product of $2S$ spin-1/2 coherent states:

$$|\hat{n}\rangle=|z\rangle\otimes|z\rangle...\otimes|z\rangle$$ where each $|z\rangle$ is a spin-1/2 spinor.

If we look at the degrees of freedom of both sides of the above equation, the spin-$S$ coherent state has $2S+1$ free parameters, whereas the $2S$ spin-1/2 have $4S$ free parameters in total. How do we understand this?

Answer

By definition, a spin coherent state $\vert \theta,\phi\rangle_S$ is just a rotation of the $\vert S,S\rangle$ state.

What makes the state "coherent" is that all the individual spin states point in the same $(\theta,\phi)$ direction (are "coherently aligned"), thus \begin{align} \vert \theta,\phi\rangle_S &:= \vert \theta,\phi\rangle_1\otimes\ldots\otimes \vert \theta,\phi\rangle_{2S} \tag{1} \\ &=R_z(\phi)R_y(\theta)\vert S,S\rangle \\ & \; = [R_z(\phi)R_y(\theta)\vert +\rangle_1]\otimes \ldots \otimes [R_z(\phi)R_y(\theta)\vert +\rangle _{2S}] \end{align} $\vert \theta,\phi\rangle_S$ clearly depends on only 2 parameters, as it should, and indeed there is no freedom in the orientation of any of the particle spins (except the first) are they must all be aligned in the same $(\theta,\phi)$ direction as the first. Thus, the right hand side of (1) depends on only two angles also.

The state in Eq.(1) is clearly symmetric under permutation of the particle indices, so taking $\theta=\phi=0$ gives
$$ \vert S,S\rangle = \vert +\rangle_1\otimes \vert +\rangle_2\otimes\ldots \otimes \vert +\rangle_{2S} \tag{2} $$ This state is an eigenstate of $L_z$ $$ L_z= L_z^{(1)}+L_z^{(2)}+\ldots +L_z^{(2S)} $$ with eigenvalue $M=S$, and is killed by the raising operator $$ L_+= L_+^{(1)}+L_+^{(2)}+\ldots +L_+^{(2S)} $$ where $L_+^{(k)}$ acts on the state of particle $k$ alone.

baryogenesis - Matter-antimatter asymmetry problem

As we know that matter-antimatter asymmetry is one of unsolved problems in physics. One possible solution to this problem is given as baryogenesis which produce asymmetry in rate of creation between matter and antimatter particles. But doesn't alternate solution like "different regions of space with different type of particles " holds more ground..

Means in one region matter particle dominates (the region we live ) and another region anti matter particle will dominate. When particle-antiparticle pair created from energy (at that energy is so much high that particle- antiparticle pair could be created from energy according to $E=mc^2$) before they meet and annihilate with each other they also has equal chance of meeting with same type of particle in neighborhood, because after some time of big bang the four force were united and gravity was as powerful as remaining three forces. So now in 50% of all pairs destroyed by annihilation and other 50% clumps together with same type of particle means matter with matter and antimatter with antimatter and then inflationary epoch throw these clumps from each other at very large distance so that they can not meet each other and have chance of annihilation. And we now can not see these regions because of accelerated expansion of universe because of dark energy.

Doesn't this hypothesis is more or at least equally valid with baryogeneis hypothesis.

electricity - Electric power and resistance dependance

According to the equations, $$P=VI =I^2R\,\text{ and voltage } V=IR$$ it seems clear that when the resistance is lower by fixing the voltage at constant, the current is therefore, higher, generating high power. But what confused me was when the resistance is higher by fixing the current at constant, the voltage is therefore, higher, which in turn lead to a higher power as well. Can anyone pull me out of this confusion?

Answer

The answer is "yes, that's what happens." There's no paradox. If you hold the current constant, and increase the resistance, the power increases. This is because of V=IR and P=VI (Ohms law and the power through a resistor). If you put these together, you can see that $P=I^2R$. If you hold the current constant, and increase the resistance, power goes up. That's just how the equations work.

What makes this confusing is that it's not intuitive how to hold current constant. We typically don't think that way. Usually we think in terms of voltages. So one way to think of this is our higher resistance forces the power supply to provide a higher voltage in order to push through the same current. Intuitively, it should make sense that a higher voltage supply can produce higher powers (though you would need the $P=I^2R$ equation to prove it).

You can use metaphors as well. Any metaphor where you can put a load on something works decently well. Take your own body. You can run at a fairly nice pace. The resistance on your body while running is quite low, so it doesn't take much effort. Now add resistance:

If you run at the same pace (the equivalent of keeping the current the same), you're going to have to push much harder (the equivalent of raising the voltage). And, if you notice, you get hot really fast (power is being dissipated). However, that's only because you kept the current the same. It would also be possible to slack off, not running as hard, in which case you could dissipate less power than before. But your original problem declares that you're keeping the current constant, so you're going to have to work harder and have more power!

quantum mechanics - How can one describe electron motion around hydrogen atom?

I remember from introductory Quantum Mechanics, that hydrogen atom is one of those systems that we can solve without too much ( embarrassing ) approximations.

After a number of postulates, QM succeeds at giving right numbers about energy levels, which is very good news.

We got rid of the orbit that electron was supposed to follow in a classical way ( Rutherford-Bohr ), and we got orbitals, that are the probability distribution of finding electron in space.

So this tiny charged particle doesn't emit radiation, notwithstanding its "accelerated motion" ( Larmor ), which is precisely what happens in real world.

I know that certain "classic questions" are pointless in the realm of QM but giving no answers it makes people asking the same questions over and over.

• If the electron doesn't follow a classic orbit, what kind of alternative "motion" we can imagine?


• Is it logical that while the electron is around nucleus it has to move in some way or another?


• Is it correct to describe electron motion as being in different places around nucleus at different instants, in a random way?

homework and exercises - Exponential decay of Feynman propagator outside the lightcone

In Chapter three (I.3) of A. Zee's Quantum Field Theory in a Nutshell, the author derives the Feynman propagator for a scalar field: $$ \begin{aligned} D(x)&=\int \frac{\operatorname{d}^4 \mathbf{k}}{(2\pi)^4} \frac{e^{ikx}}{k^2-m^2+i\epsilon} \\ &=-i\int \frac{\operatorname{d}^3 \mathbf{k}}{(2\pi)^3 2\omega_k} \left[e^{-i(\omega_kt-\mathbf{k}\cdot \mathbf{x})}\theta(t)+ e^{i(\omega_kt-\mathbf{k}\cdot \mathbf{x})}\theta(-t)\right] \end{aligned} $$ where $\omega_k=\sqrt{\mathbf{k}^2+m^2}$.

Without working through the $\mathbf{k}$ integral, the behavior of the propagator for events inside and outside the light-cone can be roughtly analyzed (or so the text states): for time-like events in the future cone, e.g., $x=(t,\mathbf{x}=0)$, with $t>0$, the propagator is a sum of plane waves $$D(t,0)=-i\int \frac{\operatorname{d}^3 \mathbf{k}}{(2\pi)^3 2\omega_k} e^{-i\omega_kt}$$ Likewise, for time-like events in the past cone ($t<0$) the propagator is a sum of plane waves with the opposite phase.

Now, for space-like events, e.g., $x=(0,\mathbf{x})$, after interpreting $\theta(0)=\frac{1}{2}$ and observing the propagator allows for the exchange $\mathbf{k}\rightarrow -\mathbf{k}$, we obtain $$D(0,\mathbf{x})=-i\int \frac{\operatorname{d}^3 \mathbf{k}}{(2\pi)^3 2\sqrt{\mathbf{k}^2+m^2}} e^{-i\mathbf{k}\cdot \mathbf{x}}$$

The author then states that "...the square root cut starting at $\pm im$ leads to an exponential decay $\sim e^{-m|\mathbf{x}|}$, as we would expect." It is left to the reader to verify this as a later problem.

The question is: how can I see that the above is true, without going through the $\mathbf{k}$ integral?

Secondarily, what does "the square root cut starting at $\pm im$" mean? I know that one must supply the complex square root with a branch cut, but said branch cut must be a whole ray of the plane, not just a segment.

I have tried going through the integral; by rotating the $\mathbf{k}$ so that $\mathbf{x}$ points along the $k^3$ direction and switching to spherical coordinates ($k=|\mathbf{k}|, x=|\mathbf{x}|$) the integral becomes:

$$ \begin{aligned} D(0,\mathbf{x})&=-i\int_0^\infty \operatorname{d}k \int_0^\pi \operatorname{d}\theta \int_0^{2\pi} \operatorname{d}\varphi \left( \frac{k^2 \sin{\theta}e^{-i kx\cos{\theta}}}{(2\pi)^3 2\sqrt{k^2+m^2}} \right)\\ &=-i\int_0^\infty \operatorname{d}k \int_0^\pi \operatorname{d}\theta \left( \frac{k^2 \sin{\theta}e^{-i kx\cos{\theta}}}{(2\pi)^2 2\sqrt{k^2+m^2}} \right)\\ &=\frac{-i}{(2\pi)^2}\int_0^\infty \operatorname{d}k \left( \frac{k}{2ix\sqrt{k^2+m^2}}\right)e^{ikx}-e^{-ikx}\\ &=\frac{-i}{(2\pi)^2}\int_0^\infty \operatorname{d}k \frac{k\sin{kx}}{x\sqrt{k^2+m^2}}\sim \frac{1}{|\mathbf{x}|} \end{aligned} $$ Which is not the desired result.

thermodynamics - Does placing a cold bottle next to an incubator cool the incubator through convection or radiation?

Context for question: My girlfriend who is studying to be a nurse asked me to explain the difference between radiation and convection in context of the above question.

My answer: Radiation is the emission of heat by all bodies as long as their temperature is above 0K. Which means that convection would be how the baby would be affected and not by radiation.

The problem: It seems there are several nursing books (such as this one ) and even a physics book that seems to agree that apparently there is "cold" radiation as well.

I don't get it, can you radiate "cold"? If someone can explain how a cold wall/cold bottle/ cold anything would affect the baby by radiation (and not by convection).

Answer

There is no such thing as cold radiation.

In the examples you link above, the issue is that the walls of the incubator are colder than the air or the baby. What actually happens in this situation is this: all objects radiate with a certain intensity which is based on their temperature (among other things). A baby in an incubator is constantly radiating energy away into the surrounding air and parts of the incubator; what keeps the baby warm is the fact that this same air and incubator radiate the same amount of heat back to the baby as the baby is losing. As long as this balance is maintained the baby can maintain its body temperature.

In the case of the cold incubator wall, the situation is not that the incubator wall radiates cold to the baby but that the baby continues to radiate energy away (which it would be doing no matter where in the incubator it was at) but because the wall of the incubator is cold (in some relative sense) the amount of energy being radiated by the incubator wall back on to the baby is lower than would be if the wall was warm. Because of this, the baby radiates away more energy than it is receiving from its surroundings (because the wall isn't radiating as much because of its lower temperature). Thus the baby cools down.

general relativity - If space warps distort moving objects' trajectories, does it mean that static objects are immune to gravity?

If gravity is just space distortion, which affects trajectories of moving objects, then a static object (not moving, thus no trajectory) will not suffer any type of accelerating force from gravity?

If so, this would also mean that we are always moving, even galaxies. The universe as a whole is moving and the definition of "static" has no physical manifestation.

Moreover if this holds true, it means that gravity just "bends" the kinetic energy (momentum?) that objects already contain for being in constant motion, which explains even further why gravity is not free energy. It just changes the direction of already existing kinetic energy.

So we are attracted to earth because the earth's mass bends space and we are moving in that space, so our trajectory gets bent down to earth..!

Why potential at infinity is 0? (sphere of constant electrostatic potential)

Suppose I have a sphere of radius $R$ with potential $V_o$. Since the volume inside the sphere is bounded, then the lack of curvature of the potential (i.e. $\nabla^2\phi = 0$) gives a potential $\phi(r\le R)=V_0$. Outside the sphere we have a choice of making $\phi(r)=V_0$ or $\phi(r)=V_0R/r$. Both are solutions of $\nabla^2\phi=0$. The constant potential does not produce an electric field, but the last one does.

Why do we choose the last form of the potential? (I know that a sphere of potential $V_0$ represents a point charge in the middle of value $Q=4\pi\epsilon_0RV_0$, and that this charge should produce an electric field, but I never mentioned any charge at all...)

I appreciate if you can add something to this discussion.

Answer

Both solutions are valid, and the correspond to two very different spheres.

Choosing $\phi(r)=V_0$ corresponds to a sphere with no charge. Choosing $\phi(r)=V_0R/r$ corresponds to a sphere with total charge $4\pi\epsilon_0RV_0$.

Remember, when solving Laplace's equation, we don't just need $\nabla^2\phi=0$. We also need to match boundary conditions. On the surface of a conductor, $\phi$ must satisfy the boundary conditions $\frac{\partial}{\partial r}(\phi_{out}(r)-\phi_{in}(r))=-\frac{\sigma}{\epsilon_0}$. So you're not free to chose between the different solutions! You have to pick the solution that obeys your boundary conditions. $\phi(r)=V_0$ everywhere satisfies Laplace's equation, but not the boundary conditions.

supersymmetry - Which SUSY models are affected by the recent LHCb result?

The LHCb has recently published the observation of $B_s \rightarrow \mu^+ \mu^-$ with a branching ratio that agrees with the Standard Model (SM). There are many blog posts about it (See: Of Particular Significance). The result initiated a debate over its implication on Supersymmetry in general, which I don't want to discuss here. But it certainly affects some Supersymmetric models that may have a different prediction than that of the SM.

My questions are,

1. Which models in particular are affected (or variants of models)? What is the MSSM status? The NMSSM?


2. Does the result affect some of the parameter space of a particular model, or does it rule out an entire model/or variant of a model?

kinematics - Is the Coriolis acceleration formula just an approximation?

The Coriolis acceleration given in textbooks is:

$a_{cor} = -2\omega v$

When trying to understand the derivation, it seems as though this is just an approximation. And to illustrate this I give a simple example of a rotating platform with two people standing on it. Person A stands at the axis, so therefore will have no centripetal acceleration. Person B stands at the far edge of the platform, and they both face each other. A throws a ball to B at t=0. To keep the calculations simple, the angular velocity of the platform is 1 rad/s, the radius of the platform is 1m, and the velocity at which the ball is thrown is 1m/s.

After one second, the ball will travel radially outwards one meter, meaning that its position is now the same as B at t=0. B will now have moved an angular displacement of 1 rad to position B'.

According to the formula, $a_{cor} = -2ms^{-2}$. From this I integrate twice to find the displacement of the ball relative to B after this one second, so:

$s = -\frac 1 2 a_{cor}t^2 = $ -1m.

This doesn't seem accurate to me. B has travelled 1m from its original position, so I guess it's correct if you're measuring the distance along the circumference. But this is the arc length and not the actual displacement of the ball, which would be the chord length from B to B'. This is slightly less than 1 meter.

Am I right or am I not understanding this correctly?

Answer

The formula is exact. Your difference between the integral and the graph arises because the trajectory of the ball in the rotating reference frame is a curve, not a straight line.

That is, the coriolis force is always perpendicular to the radius, so you can see that the movevent of the ball is actually the composition of two movements: the linear one, along the radius, and the curved one, along the circumference. Since in 1 s the disc moved 1 radian, the curved trajectory is exactly 1 radian, and with length 1 meter, that is exactly 1 meter.

This picture shows the point of view of the rotating disk. Here B is static, the black line shows the real movement of the ball, the red line shows how that trajectory is the composition of the inertial movement and the coriolis accelerated one.

That you have to measure the movement along the circumference and not the chord should be obvious if you think: what happens if the ball goes a full turn around the disk? Surely you won't say that the coriolis force should be 0 because the ball ended up exactly at B...

biophysics - Why does light of high frequency appear violet?

When people are asked to match monchromatic violet light with an additive mix of basic colours, they (paradoxically) mix in red. In fact, the CIE 1931 color space chromaticity diagram shows this effect begins at about 510nm (greenish-cyan), where people mix in no red. From that point on, the higher the frequency of the light source, the more red they mix in.

This effect is reflected by the red curve of the CIE standard observer color matching functions, which has an additional bump in the area of blue light. However, that curve does not match the actual spectral sensitivity of red cones. So where does this additional perception of red at higher frequencies come from?

Answer

In the 19th century, the physicists Young and Helmholtz proposed a trichromatic theory of color, in which the eye was modeled as three filters with overlapping ranges. This is essentially a physical model of the pigments in the eye, and it predicts the response of the nerve cells at the retina. Helmholtz did related work on sound and timbre. Ca. 1950, Hering, Hurvich, and Jameson proposed significant modifications to the trichromatic theory, called opponent processing. This models a later stage in the processing of the signals, after the retinal response but before the more sophisticated stages of processing in the brain. Both the trichromatic model and opponent processing are needed in order to describe certain phenomena in human color perception.

The complete theory can be modeled by two functions depending on wavelength. I'll call these $RG(\lambda)$ and $BY(\lambda)$. These functions are drawn here. They both oscillate between positive and negative values. For any given pure wavelength $\lambda$, the net result of pigment-filtering plus the later neurological processing produces these two numbers, which can be thought of as the final signals that go on to later processing in the brain. I'm calling them $RG$ and $BY$ for the following reasons. Let's pretend, for the sake of simplicity, that these functions oscillated between -1 and +1. Then the pair $(RG,BY)=(1,0)$ produces the sensation of red, (-1,0) is green, (0,1) is blue, and (0,-1) is yellow. There is various psychological evidence for this model, e.g., no color is perceived as reddish-green or yellowish-blue. Roughly speaking, what seems to be happening is that the eye-brain system is taking differences between signal levels of different cone cells. This sort of makes sense because, for example, the red and green pigments have response curves that overlap a lot, so if you want to place a pure-wavelength color on the spectrum, the difference between them is more a more direct measure of what you want to know than the individual signals.

The $RG$ function actually has two different peaks, one at the red end of the spectrum and one, surprisingly, at the blue end. This implies that by mixing blue and red, you can produce an $(RG,BY)$ pair similar to what you would have gotten with monochromatic violet. If you look at other sources, e.g., this one (figure 3.3), they seem to agree on the secondary short-wavelength peak of the $RG$ function, but the details of how the two functions are drawn at the short wavelengths are different and seem to make for a less convincing explanation of the observed perceptual similarity between violet and a red-blue mixture.

I don't know if there is a valid reductionist explanation of the short-wavelength peak of the $RG$ function. Like a lot of things produced by evolution, it may basically be an accident that got frozen in. However, it's possible that it serves the evolutionary purpose of helping us to distinguish different shades of blue and violet. If the $RG$ function was simply zero over the whole short-wavelength end of the spectrum, then the $BY$ function would be the only information we'd get for those wavelengths. But the $BY$ function has a maximum, simply because the eye's sensitivity to light fades out as you get into the UV. Near this maximum, the ability of the $BY$ function to discriminate between colors becomes zero. In the York University graph, it appears that the short-wavelength extrema of the $RG$ and $BY$ functions are offset from one another, which would allow some color discrimination in this region. The physical information being preserved by the $BY$ function would then be the difference in response between the blue and green cones. But the Briggs graphs don't appear to show any such offset of the extrema, so it's possible that the explanation I'm giving is a bogus "just-so story."

There may be a good analogy here with sound. The sound spectrum is linear, but there is a psychological phenomenon of octave identification, which makes the spectrum "wrap around," so that frequencies $f$ and $2f$ are perceptually similar and can often be mistaken for one another even by trained musicians. Similarly, the predictive power of the "color wheel" model shows that to some approximation we can think of the trichromatic/opponent process model as resulting in a wrapping around of the visible segment of the EM spectrum into a circle. But in both cases, the wrap-around is only an approximation. In terms of pitch, $f$ and $2f$ are perceptually similar but not indistinguishable. For color, we have the 1976 CIELUV color color diagram, which is a modification of the 1931 diagram meant to represent at least somewhat accurately the degree of perceptual similarity between different points based on the distance between them. The monochromatic spectrum constitutes part of the outer boundary of this diagram, and is more of a "V" than a circle; there is quite a large gap between monochromatic violet and monochromatic red.

It is trivially true that any such diagram has a boundary that is a closed curve. If the diagram is not constrained to give any accurate depiction of the sizes of the perceptual differences between colors, then it can be distorted arbitrarily, and we can arbitrarily define it such that its boundary is a circle. In this sense, the success of the color wheel model is guaranteed, and it follows from nothing more than the fact that humans are trichromats, so that the color space is three-dimensional, and controlling for luminance produces a two-dimensional space. But this fails to explain why there is some degree of perceptual similarity between the red and violet ends of the monochromatic spectrum; for that you need the opponent processing model.

There is also a slight variation in the absorbance of the pigment in the red cones at the blue end of the spectrum. I don't think this is sufficient to explain the perceptual similarity between violet and red, or the even closer similarity between violet and a mixture of red and blue light, i.e., I don't think you can explain these facts using only the trichromatic theory without opponent processing. The classic direct measurements of the filter curves of cone-cell pigments were done with cone cells from carp by Tomita ca. 1965, but AFAIK the only direct measurement using human cone cells was Bowmaker 1981. Bowmaker's red-cell absorbance curve has a very slight rise at short wavelengths, but it's not very pronounced at all. You will see various other curves on the internet, often without any attribution or explanation of where they came from, and some of these show a much more pronounced bump rather than Bowmaker's slight rise. Possibly some of these are from people using the CIE 1931 curves, which were never intended to be physical models of the actual human cone-cell pigments. It should be clear, however, that the red and green pigments' curves must have some variation near the violet end of the spectrum. If they did not, then the dimensionality of the color space would be reduced there, and the human eye would be unable to distinguish different wavelengths in this region, which is contrary to fact.

Bowmaker, "Visual pigments and colour vision in man and monkeys," J R Soc Med. 1981 May; 74(5): 348, freely accessible at http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1438839/

optics - Principle of Reflection on atomic level

This well-observed phenomenon has, besides several others, always been a fascination to me. We are well aware of several theories, experiments, and practical applications of this well-known phenomenon, but is it established what goes on at the grassroots of reflection?

My high school teacher once told me that whenever light is incident on any reflecting surface, its electrons absorb the energy of the photon and release back the same energy. If this is actually true, then I have plenty of questions to continue, but if it is not, then what is reflection, and how and why does it take place?

Answer

The question What IS reflection? is really a duplicate of yours, but I suspect the answer may be a bit brief for you.

A light wave, like any electromagnetic wave, is a combination of an oscillating electric and magnetic field. These fields exert an oscillating force on the electrons in any material the wave hits, and those electrons start oscillating in response. However an oscillating electron emits electromagnetic radiation, and this radiation interferes with the incident light.

To calculate what happens at the interface, you have to take into account the incident light, the light reradiated away from the solid back into the vacuum, (i.e., the reflected light) and the light reradiated into the body of the solid, (i.e., the transmitted light). When you do this, you find light is reflected at the angle of incidence, and the light is transmitted at an angle given by Snell's law. The calculation is a bit messy, but if you are interested, lots of examples are yours for the cost of a quick Google.

Response to comment:

The oscillations of the electrons are driven by the incoming EM field, so the oscillation frequency is the same as the frequency of the incoming light. The phase need not be (which is the origin of the refractive index change).

In the real world, the reflection is not independent of colour (i.e., the frequency of the light), and this phenomenon is known as dispersion. It is also not independent of light intensity, which is known as non-linearity, though non-linearity is usually an extremely small effect.

The electrons do not react instantly when the light strikes. The fastest frequency at which they can react is the plasma frequency.

Response to response to comment:

Energy can be lost due to interactions with the lattice, and indeed this is the norm, so the sum of the reflected and transmitted waves is generally less than the incident wave. The lost energy ends up as heat i.e., lattice vibrations.

However the fact energy is lost does not change the frequency of oscillation of the electrons, because that is determined by the incident light. That means the frequency of the reflected and transmitted light is the same as the incident light i.e., the same colour.

If the incident light contains different frequencies, like white light, then the result can be a shift in the perceived colour. For example, if you shine white light onto gold, the reflected light is yellowish. However the process of reflection is not changing the frequency of the light, but rather it is changing the relative intensities of frequencies in the reflected light by absorbing some frequencies more strongly than others.

classical mechanics - Is there an equivalent of a scalar potential for torques?

For a given scalar potential $V$, it is known that the corresponding force field $\mathbf{F}$ can be computed from

$$ \mathbf{F} = -\nabla V $$

Suppose a rigid body is placed inside this potential. The torque on the body $\mathbf{T}$ exerted by the scalar field will be

$$ \begin{align} \mathbf{T} &= \int_M \left( \mathbf{r} \times \mathbf{F}(\mathbf{r}) \right) dm \\ &= \int_M \left( \mathbf{r} \times -\nabla V(\mathbf{r}) \right) dm \\ &= \int_M \left( \nabla V(\mathbf{r}) \times \mathbf{r} \right) dm \end{align} $$

with $\mathbf{r}$ the position vector to the mass element $dm$, and the integration carried out over the entire body $M$.

So, being not too familiar with rigid body dynamics, I was wondering -- does something like a (vector/scalar) potential $P$ exist, such that the local torque induced by the potential $V$ can be expressed as

$$ \matrix{ \mathbf{T} = I \cdot \nabla P & & &\text{(or some similar form)} } $$

with $I$ the moment of inertia tensor of the rigid body?

If such a thing exists:

• what is its name?


• where should I start reading?


• What is the proper expression for the torque $\mathbf{T}$?


• how does $P$ relate to $V$?

If such a thing doesn't exist:

• why not? :)

Answer

Great question. A little background first.

Note that any force $\boldsymbol{F}$ moment $\boldsymbol{M}$ system on a point A can be equipollently translated into the screw axis S leaving only the components of $\boldsymbol{M}$ that are parallel to $\boldsymbol{F}$. The location is found by

$$ \boldsymbol{r} = \frac{\boldsymbol{F} \times \boldsymbol{M}}{\boldsymbol{F}\cdot\boldsymbol{F}} $$

Also the moment components parallel to $\boldsymbol{F}$ are described by a scalar pitch value $h$ found by

$$ h = \frac{ \boldsymbol{M} \cdot \boldsymbol{F}}{\boldsymbol{F} \cdot \boldsymbol{F}} $$

In reverse, a moment is defined by a force vector $\boldsymbol{F}$ passing through an axis located at $\boldsymbol{r}$ with pitch $h$

$$ \boldsymbol{M} = \boldsymbol{r} \times \boldsymbol{F} + h \boldsymbol{F} $$

Have you noticed how difficult it is to apply a pure moment on a rigid body, without applying a force? This is because you cannot have one without the other. A moment is really a result of the line of action of forces. So the scalar potential of a moment is really the same as the one for forces with

$$ \boldsymbol{M} = - \boldsymbol{r} \times \nabla V - h \nabla V = -\left( \left[1\right] h + \boldsymbol{r}\times \right) \nabla V $$

The problem is that in rigid body mechanics forces are not treated as scalar fields, but spatially constant, and temporally varying. Furthermore, I cannot think of a case where spatially varying moments arise that are NOT due to a force at a distance. I suppose you can come up with a tensor pitch $h$ instead of a scalar which is spatially varying for a definition like $\boldsymbol{M} = -\left( H + \boldsymbol{r}\times \right) \nabla V$, but then you will be making things up that do not have any physical meaning that I know of.

gravity - Escape velocity for an arbitrary metric in General Relativity

It's a known calculation how to derive the escape velocity of a system following the Schwarschild Metric. It comes out to be, if I am not wrong $$v = c\sqrt{2GM/Rc^2 - (2GM/Rc^2)^2}$$

So, is there a general method of finding out the escape velocity for a given general metric, without assuming any symmetry considerations?

A mathematical help would be highly appreciated

quantum mechanics - Which particle aspect is required to explain photoelectric or Compton effect?

What do we mean when we say that it requires the particle nature of radiation i.e., photons, to explain photoelectric or Compton effect?

I don't understand which particle nature is used to explain these experiments. We cannot say that the momentum and energy are distinctive properties of a particle. Waves, too, have these properties.

So the question is which particle aspect is relevant for explaining the above-mentioned experiments? How do I make it understandable to a High school student?

thermodynamics - Why is $PV^gamma$ constant in an adiabatic process?

In non isolated systems where there is no adiabatic process, $PV$ is constant. But the graph gets steeper in adiabatic process because of the $\gamma$ over the $V$. Why is it there in adiabatic processes and why only over the $V$?

electricity - Current against the inverse of resistance graph, $I = V/R +c$

If I have a plot of current ($y$ axis) against 1/Resistance ($x$ axis).

The circuit it is measured from is a simply 2 resistors connected in parallel to battery, where the potential across the resistors is $1V$. One resistor remains fixed, and the second varies, the resistance plotted on the graph is only the resistance of the second resistor.

Does this mean that the gradient will be voltage, but will it be $1V$ or less, as the resistance on the graph is only one from one of the resistors?

Also what will the $y$ intercept signify? $I = V/R +c$, what does the $c$ represent?

Answer

Hint: Use Ohm's law $I=\frac{V}{R_p}$ and the formula $\frac{1}{R_p}=\frac{1}{R}+\frac{1}{R_0}$ for parallel resistors to derive a straight line in a $I$-$\frac{1}{R}$ diagram

$$ I~=~V \left(\frac{1}{R}+\frac{1}{R_0}\right). $$

Here $R$ and $R_0$ are the variable and the fixed resistor, respectively. To answer OP's two questions:

1. 
   

   The slope is $V=1{\rm Volt}$.

   
   


2. 
   

   The intercept of the straight line on the vertical $I$-axis is $c=\frac{V}{R_0}$. The intercept of the straight line on the horizontal $\frac{1}{R}$-axis is $\frac{-1}{R_0}$.

   
   

electromagnetism - Shape of electromagnetic wave

Please help me to clarify situation with shape of electromagnetic waves.

EDIT: From comments I see root of problem. I afraid I'm no so smart as all of you so I simply don't get what you talking about all these waveguides, near / far fields and other similar things. Unfortunately I don't know any of calculus also. Beside that I feel some language barier, so I rephrase my questions to be more specific trying to express what I exactly can't understand. And I would prefer answers in simple non-math terms.

EDIT2: I would like to thank you all for the answers and comments, they were very useful. At least I feel I've got some basic understanding.

Ok, let start with simplest EM radiation case - oscillating charge.

Here is a 3D animation of single pulse.

Full cycle would be like this:

I can't understand why $\vec{E}$ and $\vec{B}$ fields are take such orthogonal form and how it can satisfy Maxwell's equations and in particular Faraday's law.

There is usually mathematical proof in textbooks why it so with explanation considering some region of space in which such wave are exists. And I'm lost in this point.

Because I thought induced electric fields in case of time-varying magnetic fields looks like this:

Ok, maybe they not always take form of circles but at least loops are always closed.

Now, as for me, in case of EM wave we got same situation and here is part of explanation confirming this:

Faraday's law relates a changing magnetic flux to a non-Coulomb electric field. Because the region of space in which E and B are nonzero is “moving,” the magnetic field at a particular location in space changes with time, and this time-varying magnetic field is associated with a curly electric field...

...With the right thumb pointing in the direction of -dB/dt, the fingers of the right hand curl clockwise, which is the sense of the electric field around the path.

Similar time-varying magnetic field but it looks completely different. I don't see any circular fields around changing $\vec{B}$, there just vertical perpendicular $\vec{E}$ field which "moves" to the right, no horizontal or any curly components (roughly speaking it looks like electrostatic but with no any charges). At same time there also is mention about right hand "rule" in text. I must be blind or something.

And I just think that same problem with magnetic fields - they not in forms circles as well. Maybe these orthogonal propagating fields just small part of main global induced fields which forms closed loops like it should be. Don't know. Feels like I can't see the whole picture.

Hence questions are:

Question 1:

Is this vertical $\vec{E}$ field in wave are actually induced ?

Question 2:

Why there is no any components of $\vec{E}$ field except for vertical if it induced?

Question 3:

In other words how such orthogonal constitution of fields can satisfy Faraday's Law which implies closed loops in induced $\vec{E}$ fields ?

Question 4:

Looks like these EM waves can take pretty different and complex forms. Does it means that in this case of oscillating charge induced electric and magnetic fields just take this simple orthogonal shape and this does not contradict Faraday's / Ampere-Maxwell's laws despite of this constitution ?

Thanks for help.

Here is a similar question, unfortunately without satisfying answer.

Answer

You've explicitly asked for answers that are simpler than the existing ones, so let me start off with a warning: Maxwell's equations are intrinsically mathematical statements. If you want any sort of detailed understanding of how they work, or of how some specific situation relates to the Maxwell equations, then there is a point at which there is simply no other choice than sitting down, buckling up, and doing the maths. If you're unwilling or unable to do the maths, then you just need to accept that there are situations in which you will not be able to understand things to the level of detail that you might wish for. If that sounds harsh, then sorry, but that's just the way things are.

---

Next up, there's an important point that's often not made strongly enough: plane waves are a model and nothing more. Plane waves are a possible solution of the Maxwell equations, and nothing more: they are useful in constructing some forms of general solution to the equation, but the fact that a given set of fields (say, a plane wave) is a solution to the equations does not imply that all solutions have that form.

The reason this is important is that several of your questions have the form "why are plane waves like this?", and they are essentially, fundamentally unanswerable. The fact that plane-wave solutions have a characteristic X is an existence statement ─ "there exist plane-wave solutions of the Maxwell equations with characteristic X" ─ and not a universal statement, i.e. it does not imply that all solutions of the Maxwell equations must behave like that.

Thus, for example, the answer to

Why there is no any components of $\vec{E}$ field except for vertical if it induced?

is that the general plane-wave solutions you've been exposed to are linearly polarized; those are fine because they're solutions, but they're not the only solutions; there are also e.g. circularly polarized plane waves that don't behave like that. The fact that you've only seen examples that behave one way does not mean that all examples behave like that.

---

Now, as I understand your long-and-rambling post, the core of your question is basically how do plane waves satisfy the Faraday induction law?, and the answer to that can fortunately be seen graphically. The diagram you presented is nice, but the statement of the Faraday law of induction is a bit more precise: it states that for any given loop inside of which there is a changing magnetic flux, the circulation of the electric field along the loop, i.e. the integral of its components along the loop, must be nonzero.

That's quite patently the case for the diagram you presented, but it's also obviously the case for a plane wave:

^{Image adapted from Wikipedia}

Note, in particular, that for the curve marked in red, the electric field is always along the direction of the curve, or orthogonal to it; that implies that the circulation $\oint_C \mathbf E\cdot \mathrm d\mathbf l$ is nonzero.

Now, that said, if you want any more details, then you just need to buckle up and learn the maths, or accept it as a given that plane-waves satisfy the Maxwell equations because they have a mathematical form that satisfies the mathematics of the equations.

---

And, finally:

Looks like these EM waves can take pretty different and complex forms. Does it means that in this case of oscillating charge induced electric and magnetic fields just take this simple orthogonal shape and this does not contradict Faraday's / Ampere-Maxwell's laws despite of this constitution ?

No. Electromagnetic fields can take a multitude of different forms, and no matter how many examples you see, that still doesn't imply anything about their general behaviour, i.e. it doesn't imply for any circumstance that the EM fields "just take" this or that form.

And, as I emphasized above, the $E$-orthogonal-to-$B$ configuration does not contradict either the Ampère-Maxwell or the Faraday induction laws. Electromagnetic fields always satisfy all four of the Maxwell equations; if they don't then they're not EM fields, period.

particle physics - If matter and antimatter were produced equally during the big bang, where is the rest of the antimatter?

As far as my understanding goes, during the 'Big Bang' equal amounts of matter and antimatter (matter's oppositely charged twin) were produced, and the physical matter that remains within this universe is what's left over from the equilibrium of matter and antimatter canceling one another out following the Big Bang. If matter and antimatter truly do cancel one another out (are transformed into pure energy) upon contact, than:

a) why is there still matter in existence?

b) if matter is still in existence then doesn't that mean that somewhere within the universe there is an equal amount of antimatter?

cosmology - Can Hubble red shift be interpreted as time dilation?

Can we interpret the de Sitter universe as a spherical cosmic horizon null surface of finite radius, centered at Earth, and containing the Hubble volume of space where time is dilated and radial dimensions contract closer to the edge in such a way that objects closer to the edge do not recognize that they are radially contracted?

Everything is attracted to the edge, but the total radius remains more or less constant and emits de Sitter radiation at finite temperature.

special relativity - Does the discreteness of spacetime in canonical approaches imply good bye to STR?

In all the canonical approaches to the problem of quantum gravity, (eg. loop variable) spacetime is thought to have a discrete structure. One question immediately comes naively to an outsider of this approach is whether it picks a privileged frame of reference and thereby violating the key principle of the special relativity in ultra small scale. But if this violation is tolerated doesn't that imply some amount of viscosity within spacetime itself? Or am I writing complete nonsense here? Can anybody with some background in these approaches clear these issues?

Answer

Here is as I see the problem. The best way to understand it is to think what happens with rotations and quantum theory. Suppose that a certain vector quantity $V=(V_1,V_2,V_3) $ is such that its components are always multiples of a fixed quantity $d $. Then one is tempted to say that obviously rotational invariance is broken because if I take the vector $V=(d,0,0) $ and rotate it a bit, I get $V=(\cos(\phi) d, \sin(\phi) d,0) $, and $\cos(\phi) d $ is smaller than $d $. Therefore, either rotational invariance is broken, or the vector components can be smaller. Right? No, wrong. Why? Because of quantum theory. Suppose now that the quantity $V$ is the angular momentum of an atom. Then, since the atom is quantum mechanical, you cannot measure all the 3 components together. If you measure one, you can get say either $0 $, or $\hbar$, or $2\hbar$ ..., that it, precisely multiples of a fixed quantity.

Now supose you have measured that the $x$ component of the angular momentum was hbar. Rotate slowly the atom. Do you measure them something a bit smaller that $\hbar$? No! You measure again either zero, or $\hbar$ ... what changes continuously is not the eigenvalue, namely the quantity that you measure, but rather the probabilities of measuring one or the other of those eigenvalues. Same with the Planck area in LQG. If you measure an area, (and if LQG is correct, which all to be seen, of course!) you get a certain discrete number. If you boost the system, you do not meqsure the Lorentz contracted eigenvalues of the area: you measure one or the other of the same eigenvalues, with continuously changing probabilities. And, by the way, of course areas are observables. For instance any CERN experiment that measures a total scattering amplitude amplitude is measuring an area. Scattering amplitudes are in $\mathrm{cm}^2$, that is are areas.

energy - What causes an electron to return from a higher orbit to a lower orbit

Ultraviolet light can cause electrons from Hydrogen gas to jump from a lower energy orbit to a higher energy orbit. What causes the electrons to return from the higher energy orbit to the lower energy orbit by releasing photons? Would it be possible to maintain the electrons in the higher energy orbit? How? Would placing the hydrogen gas in an electric field generated between the parallel plates of a capacitor have this effect?

torque - Parallel axis theorem non-uniform density

The parallel axis theorem says that if the moment of inertia of a body rotating about the body's centre of mass is $I_{cm}$, then the moment of inertia of the body rotating about an axis parallel to the original axis and displaced from it a distance $d$ is $I_{S}=I_{cm}+Md^2$, where $M$ is the body's mass.

I know for a fact that this applies in cases of uniform density. Does this same theorem apply if the body's density is not uniform?

kinematics - Are velocity and acceleration smooth quantities?

My thinking:

1. 
   
   

   acceleration corresponds to a force which is instantaneous, so the acceleration of a rigid body can be rather spiky (non-smooth)

   
   


2. 
   

   velocity (angular velocity) describes the ratio of change of the distance(angle), so it is smooth in the real world.

   
   

Conclusion, it makes sense to smooth (e.g., simple averaging) a velocity signal (temporal velocity), but it does not make so much sense to do smoothing on acceleration signal. Am I right?

Answer

As far as we know and can test, space is continuous, not discrete. Since space is continuous, then the labels we associate with it (i.e., positions) are also continuous. Calculus requires continuous functions to do the derivative and integral, so this implies that velocities and accelerations are also continuous because they are derivatives of positions: $$ v(t)=\frac{dx(t)}{dt}\qquad a(t)=\frac{dv(t)}{dt}=\frac{d^2x(t)}{dx^2} $$

If you have a discrete spectrum (e.g., measurements at different times/positions), then interpolation (whether linear or some higher-order method) is a necessary and useful tool to reconstruct the smooth distribution that we expect.

kinematics - Problems in the relation $a=vfrac{mathrm{d}v}{mathrm{d}x}$

We all know $a = \frac{\mathrm{d}v}{\mathrm{d}t}$. A little application of the chain rule leads to the relation $$a=v\frac{\mathrm{d}v}{\mathrm{d}x}$$ But the above equation shows that $a=0$ whenever $v=0$. And this must be wrong, as when we throw something vertically from the earth surface it stops and then returns due to gravity. There is always an acceleration in the downward direction.

I have come up with two possible ways to solve this problem:

1. 
   

   We know that $v = \frac{\mathrm{d}x}{\mathrm{d}t}$, so whenever $v=0$ then $\mathrm{d}x=0$ (i.e. no displacement in that infinitely small time). Also, $$a = \frac{\mathrm{d}v}{\mathrm{d}t} = \frac{\mathrm{d}v}{\mathrm{d}x}\frac{\mathrm{d}x}{\mathrm{d}t}$$ only if $\mathrm{d}x$ is not equal to $0$, as multiplying both numerator and denominator with $0$ will make it $0/0$ (undefined). Or multiplying by $0/0$ is not equivalent to multiplying by $1$.

   
   
   

   Now since $\mathrm{d}x=0$ whenever $v=0$, therefore we can not write $a = v\frac{\mathrm{d}v}{\mathrm{d}x}$ when $v=0$.

   
   


2. 
   

   If we plot a $v$-$x$ curve for the motion, whenever $v=0$, as explained, $\mathrm{d}x=0$. Therefore, at that instant $\frac{\mathrm{d}v}{\mathrm{d}x}$ (the slope of the $v$-$x$ curve) will be not defined (as $\mathrm{d}x=0$), therefore the equation $a = v\frac{\mathrm{d}v}{\mathrm{d}x}$ will be undefined and we cannot determine acceleration at that instant with the $v$-$x$ curve.

   
   

   Now that actually means that by plotting $v$-$x$ curve we lose information about acceleration of the particle when $v=0$.

   
   

I want a method to find the acceleration at that point using $v$-$x$ curve. Also a explanation for this defect of the $v$-$x$ curve.

Also I would like to add that so far no physics book (that I had read) has explained this before writing this relation. Also they do not mention that this will not work for $v=0$.

---

I would like to add the situation where I observed this: when you throw something vertically upward with velocity such that it reaches to a height 10 then the v-x curve will be- the slope of above curve is clearly not defined at x=10(or the instant where velocity is 0) But we know that there was a constant acceleration throughout the flight. Then how can we find acceleration at that instant??? Also assume that we only have v-x curve. Or if we cannot then why??

Answer

Your error is simply that you are assuming that $v(x)$ is differentiable with respect to $x$ at $v=0$. The chain rule needs that all derivatives involved exist before you can apply it. In the case of just letting go of something, the function $v(x) = \sqrt{2gx}$ is not differentiable at $x=0$, which is where $v=0$, so you are not allowed to apply the chain rule there.

On the other hand, if $v(x)$ is differentiable where $v=0$, then applying the chain rule is valid. There is nothing fundamental about $v=0$ that prohibits applying the chain rule here, but there is something about your example of letting something fall down that prohibits it.

Lesson: Do not apply rules without checking whether their prerequisites are fulfilled.

orbital motion - Where does energy for high and low tides come from?

High and low tides are caused by Moon gravity attracting water. Now there's friction, waves cause erosion, their energy is used in power plants yet the tides work for millions of years and are perceived as a free source of energy.

Now that's impossible - energy can't just appear out of nowhere.

What is the energy source for the high and low tides?

Is entropy of all systems zero at absolute zero?

Is the entropy of every system zero at the absolute zero?

Or is it taken to be zero at the absolute zero?

Are there systems that doesn't reach zero entropy even till absolute zero?

cosmology - What is the universe expanding into?

Due to curiosity, it made me wonder what is outside of the universe, is it a new chemical or just empty space?

But empty space have protons appearing and disappearing in it, it basically still has something in it.

(It might have something to do with string theory.)

Answer

The universe isn't expanding "into" anything, it's the space itself that is expanding, or "stretching" if you will! Besides, it's the space between the galaxies that's expanding mostly. Inside them, gravity is overcoming the expansion, so it's really only on the larger scales.

electromagnetism - What are the Electric and Magnetic Properties of Aluminum Oxynitride?

Is Transparent Aluminum (Aluminum Oxynitride) a good conductor of electricity? Also, is it attracted by a magnet?

Answer

No and No.

There are very few transparent electricity conductors (like ITO), and this is not one of them.

Also, I don't see it in a list of common ferromagnetic materials (if that's what you was asking by 'attracted').

rotational dynamics - What gives energy for precessional motion of gyroscope?

Let us suppose we have an ideal friction-less gyroscope and it is suddenly given an angular momentum, and left as in the figure. It will start precessional motion.

Since there is no friction, the rotational speed of the gyroscope about its own axis should not decrease. Then what is responsible for the (extra) energy of the precessional motion of the gyroscope?

Answer

The centre of mass of the rod and the spinning disc drops a little.
Thus there is a loss of gravitational potential energy and a gain in the precessional kinetic energy.

You can think of it as the torque which is causing the precession rotating through an angle and so work is done by the torque which increases the rate of procession.

The change of angle of the rod is small and so the effect is difficult to see.
If you watch this Walter Lewin video at 2:30 he releases the gyro.
Does the inclination of the rod change?

At 2:40 a mass is added to the rod which increases the precessional kinetic energy and hence the rate of precession and there is a definite change in the angle of the rod. The inclination of the rod seems to decrease when the mass is removed at 2:47 with the rate of precession decreasing.

Update

The movement of the centre of mass down to ensure energy conservation also ensures angular momentum conservation.
The system starts with no precessional angular momentum in the vertical direction and then when released starts to precession with an angular momentum due to the precession in the vertical direction even though there are no external torques in that direction.
The precession angular momentum comes from the fact that when the centre of mass drops the component of spin angular momentum in the vertical direction changes (e.g. from having a horizontal spin angular momentum to the spin angular momentum pointing slightly down introduces a vertical component of spin angular momentum) and that change in spin angular momentum is exactly balanced by the increase in the precessional angular momentum so the net vertical component of the total angular momentum of the system stays the same.

quantum field theory - Dimension analysis in Derrick theorem

The following image is taken from p. 85 in the textbook Topological Solitons by N. Manton and P.M. Sutcliffe:

What I don't understand from the above statement:

• why $e(\mu)$ has minimum for $d=2,3$, whereas when $d=4$, $e(\mu)$ is scale independent and stationary points and vacuum solutions are possible?


• How $e(\mu)$ is a continuous function bounded by zero?

Answer

Generalized versions of Derrick's No-Go Theorem compare spatially scaled, non-trivial, time-independent, finite-energy, classical, field-configurations to exclude the existence of static solitons. (Since we are only considering time-independent field-configurations, there is no kinetic energy $T=0$. Therefore the stationary action principle $\delta S=0$ amounts to minimize the (potential) energy $T+V$. Here $S:=\int \! dt ~L$, and $L:=T-V$. See also this Phys.SE post.)

Let $d$ be the number of spatial dimensions. The (potential) energy of the $\mu$-scaled configuration is

$$ \tag{4.22} e(\mu)~=~\sum_{n\in\{0,2,4\}} \mu^{n-d} E_n\geq 0, $$

where we assume that the scale parameter

$$ \tag{A}\mu~\in~]0,\infty[ $$

is strictly positive, and that the energies

$$\tag{B} E_n~\geq ~0, \qquad n\in\{0,2,4\}$$

are non-negative. From this it already follows that the ($\mu$-scaled potential) energy $e:]0,\infty[\to [0,\infty[$ is a non-negative and continuous function, and in particular that it is bounded from below, cf. some of OP's subquestions (v1).

1. 
   

   Case $E_0,E_4>0$ and $d\leq 3$: Then $$\tag{C} e(0)~:=~\lim_{\mu\to 0^{+}}e(\mu)~=~\infty~=~\lim_{\mu\to \infty}e(\mu)~=:~e(\infty),$$ so that there must exist an interior$^1$ (relative) minimum, and therefore Derrick's No-Go conclusion does not apply.

   
   


2. 
   

   Case $d\geq 4$: The function $e$ is monotonically weakly decreasing. [The word weakly means here that it could be (locally) constant.] The only possibility to have an interior minimum (and therefore evade Derrick's No-Go conclusion) is if $E_0=0=E_2$ and if moreover either (i) $d=4$ or (ii) $E_4=0$. The former case corresponds to pure 4+1 gauge theory, which indeed has non-trivial static soliton solutions with $E_4>0$. The latter case corresponds to vacuum solutions $e\equiv 0$. The function $e$ is in both cases a constant function, i.e. independent of the scale $\mu$.

   
   


3. 
   

   Case $E_0=0=E_2$: The function $e$ is monotonic. The only possibility to have an interior minimum (and therefore evade Derrick's No-Go conclusion) is, if (i) $d=4$, or if (ii) $E_4=0$, i.e. we are back in the previous case (2).

   
   
   

References:

1. 
   

   N. Manton and P.M. Sutcliffe, Topological Solitons, 2004, Section 4.2.

   
   


2. 
   

   S. Coleman, Aspects of symmetry, 1985. Note that Sidney calls solitons for lumps.

   
   



3. 
   

   R. Rajaraman, Solitons and Instantons: An Introduction to Solitons and Instantons in Quantum Field Theory, 1987.

   
   

--

$^1$ An interior minimum point $\mu$ means that $\mu$ is different from the boundary $0$ and $\infty$.

electromagnetism - How can there be a voltage across an inductor if the voltage through any conductor is zero?

It's commonly taught that the electric field in a conductive material is zero. Hence the voltage through a perfectly conductive material is zero. I however when learning about inductors in physics assumed the presence of changing magnetic fields to be an exception to this rule. An ideal inductor is assumed to have a resistance of zero but has a voltage of $-L \frac{\mathrm{dI} }{\mathrm{d} t}$. This means the magnetic fields must produce electric fields inside the coil. We get this from: $\nabla \times E = -\frac{\mathrm{dB} }{\mathrm{d} t}$. I, however, read the Feynman Lectures on the topic and he says:

As we have seen before, there can be no electric fields inside a perfect conductor. (The smallest fields would produce infinite currents.) Therefore the integral from a to b via the coil is zero. The whole contribution to the line integral of E comes from the path outside the inductance from terminal b to terminal a. Since we have assumed that there are no magnetic fields in the space outside of the “box,” this part of the integral is independent of the path chosen and we can define the potentials of the two terminals.

My question is how does zero voltage through the inductor coil amount to a voltage between the ends? At what point in the wire is this voltage introduced?

Answer

This means the magnetic fields must produce electric fields inside the coil.

Yes, but this electric field, called induced electric field $\mathbf E_i$ , is only one component of total field in the conductor of the coil. It is not the total field.

Total electric field in an ideal wire of the coil is zero, but this does not imply that voltage on the terminals has to be zero. Voltage in AC circuits is not defined by total electric field, but only by its Coulomb ("electrostatic") component. This is because voltage comes from electrostatics and only makes sense for electrostatic field.

In an ideal conductor, total electric field has to be zero, so any induced electric field has to be counteracted and cancelled by some other contribution to the field, of same magnitude but opposite direction in space. Electric field that is always present in electric circuits is the Coulomb field $\mathbf E_{Coulomb}$ due to charges on conductors' surfaces. This field is zero inside conductors in the special case where everything is in static equilibrium, but as soon as charges accelerate (when electric current changes), the Coulomb field will be non-zero inside conductors, so that it can cancel the induced field. So, in the conductor of coil with changing current, the induced electric field due to accelerated charges in the coil is cancelled by the Coulomb field of all the charges in the circuit (mostly on the surfaces of the conductors in the circuit).

Voltage is integral of this Coulomb, or "electrostatic" component of the field. It is electrostatic in the sense it is the potential Coulomb field. Its integral is therefore independent of the path, it only depends on the endpoints. To calculate voltage between coil terminals A and B, one can use the path that the charge carriers actually follow, but if one chose a path that went out of that path and then back, as long as the endpoints are the same, the result would be the same.

Induced electromotive force (emf), on the other hand, is integral of the induced electric field, and does depend not only on the endpoints, but also on the path. However, usually we are interested only in the value of emf for the path the charge carriers in the coil go through.

Let motion "from A to B" be the positive sense of circulation in the circuit. Then the value of the emf for this oriented path is, in the common convention,

$$ emf(A~to~ B) = \int_A^B \mathbf E_i\cdot d\mathbf s = -L\frac{dI}{dt}.~~~(1) $$

Since induced electric field is everywhere in the ideal coil cancelled by the Coulomb electric field, the integral of this Coulomb field has to have the opposite value: $$ \int_A^B \mathbf E_{Coulomb}\cdot d\mathbf s = + L\frac{dI}{dt}.~~~(2) $$ This integral is also the difference of potentials $\varphi_B - \varphi_A$, i.e. voltage of B with respect to A.

In real coils, this cancellation of electric field components is not complete and total electric field is not necessarily zero. Thus voltage does not cancel emf exactly. Emf is still given by the general formula (1), but real case voltage has no such general formula as (2), that formula is valid only in the ideal case.

However, if the conductor is Ohmic, we can find a different relation between the emf and voltage. We can write the generalized Ohm law for the coil:

$$ \int_{A,~through~the~path~of~the~current}^B (\mathbf E_i + \mathbf E_{Coulomb})\cdot d\mathbf s = RI $$ where $R$ is Ohmic resistance of the coil conductor from $A$ to $B$. Using emf and voltage, this is

$$ emf(A~to~B) + (\varphi_B-\varphi_A) = RI $$

So emf and voltage no longer cancel each other completely, but their sum is "net active force" that pushes the current against the resistance; the greater the resistance, the greater the difference of emf and voltage magnitudes has to be to maintain the same current.

superfluidity - Are all bose-einstein condensates superfluid?

I feel like the answer should be "no" since all superfluids are not strictly BEC since they can undergo a Kosterlitz–Thouless transition in 2D, for example. I believe the ideal gas isn't superfluid, but is there any experimental evidence of a BEC without superfluid properties? I've been searching with no luck.

Answer

You can have superfluids that are not BECs and BECs that are not superfluid. Let me quote a text, "Bose-Einstein Condensation in Dilute Gases", Pethick & Smith, 2nd edition (2008), chapter 10:

Historically, the connection between superfluidity and the existence of a condensate, a macroscopically occupied quantum state, dates back to Fritz London's suggestion in 1938, as we have described in Chapter 1. However, the connection between Bose-Einstein condensation and superfluidity is a subtle one. A Bose-Einstein condensed system does not necessarily exhibit superfluidity, an example being the ideal Bose gas for which the critical velocity vanishes, as demonstrated in Sec. 10.1 below. Also lower-dimensional systems may exhibit superfluid behavior in the absence of a true condensate, as we shall see in Chapter 15.

classical mechanics - Lagrangian of an effective potential

If there is a system, described by an Lagrangian $\mathcal{L}$ of the form

$$\mathcal{L} = T-V = \frac{m}{2}\left(\dot{r}^2+r^2\dot{\phi}^2\right) + \frac{k}{r},\tag{1}$$

where $T$ is the kinetic energy and $V$ the potential energy, it is also possible to define the total energy $E$ of the system

$$ E = T + V =\frac{m}{2}\left(\dot{r}^2+r^2\dot{\phi}^2\right) - \frac{k}{r}.\tag{2}$$

If the angular momentum $M$ is defined by $$ M = m r^{2} \dot{\phi},\tag{3}$$ then $E$ can be written as

$$ E = \frac{m}{2}\dot{r}^2 + \underbrace{\frac{M^2}{2mr^2}-\frac{k}{r}}_{V_\textrm{eff}\left(r\right)},\tag{4}$$

where the last two terms are written as new, "effective" potential $V_\textrm{eff}\left(r\right)$.

In addition, using the definition of $M$, the Lagrangian $\mathcal{L}$ can be written as

$$ \mathcal{L} = \frac{m}{2}\dot{r}^2 + \underbrace{\frac{M^2}{2mr^2}+\frac{k}{r}}_{-V_\textrm{eff}\left(r\right)},\tag{5}$$

where the sign of $V_\textrm{eff}\left(r\right)$ has been changed, since $\mathcal{L} = T-V$. But from this argument, it appears that there are two different possible ways to construct the same effective potential. This seems to me contradictory. Where is my mistake?

Answer

1. 
   

   The underlying reason for OP's flawed argument is, that a premature use of EOMs in the stationary action principle $$ S~=~\int\!dt ~L(r,\dot{r};\theta,\dot{\theta}), \qquad L(r,\dot{r};\theta,\dot{\theta})~=~\frac{1}{2}m(\dot{r}^2 +r^2\dot{\theta}^2) -V(r),\tag{A}$$ invalidates the variational principle. Concretely OP is achieving the incorrect Lagrangian (5) by eliminating the angular coordinate from the original Lagrangian (1) via the fact that the angular momentum (3) is a constant of motion (which is the EOM for the angular coordinate).

   
   


2. 
   

   Now for OP's example, it turns out that a correct reduction can be perform via the Hamiltonian formulation $$ H(r,p_r;\theta,p_{\theta}) ~=~\frac{p_{r}^2}{2m}+ \frac{p_{\theta}^2}{2mr^2} + V(r). \tag{B}$$ Note that the canonical momentum $p_{\theta}$ (which is the angular momentum) is a constant of motion because $\theta$ is a cyclic variable.

   
   


3. 
   
   

   We next re-interpret the system (B) in a rotating frame following the particle with fictitious forces and only 1D radial kinematics. The Hamiltonian (B) becomes $$ H(r,p_r)~=~\frac{p_r^2}{2m}+ V_{\rm cf}(r)+ V(r), \tag{C}$$ where $$ V_{\rm cf}(r)~:=~\frac{p_{\theta}^2}{2mr^2} \tag{D}$$ is a centrifugal potential in a 1D radial world, cf. my Phys.SE answer here. Hence the Hamiltonian system (C) has effectively only one radial degree of freedom.

   
   


4. 
   

   Finally, perform a Legendre transformation on the Hamiltonian (C) to obtain the corresponding 1D Lagrangian system: $$ L(r,\dot{r})~=~\frac{m}{2 }\dot{r}^2\color{Red}{-}V_{\rm cf}(r)-V(r).\tag{E} $$ Note the crucial minus sign marked in red. One may check that the corresponding EL eq. for (E) is the correct EOM.

   
   


5. 
   

   The Lagrangian (E) is minus Routhian, cf. this Phys.SE post.