tensor calculus - Differential Forms and Densities

I've heard that differential forms are related to densities, however I'm still a little confused about that. I thought on the case of charge density and I came to that: let $U\subset\mathbb{R}^3$ be a region of $3$-space, and let $\rho : U \to \mathbb{R}$ give the charge density at every point of $U$. I can then create the $3$-form $\omega = \rho \ dx \wedge dy \wedge dz$, which in my understanding gives me the approximate amount of charge enclosed by a volume determined by $3$ vectors when they're given.

So, if I give the vectors $v, u, w$, the value of $dx \wedge dy \wedge dz(v,u,w)$ should be the volume enclosed by those vectors, and hence $\omega(u, v, w)$ should be an approximation of the charge enclosed. Is this correct?

My only problem is: in this point of view, the form isn't giving me the density, the density itself is being given by a scalar field, while the form gives me the charge instead of the density.

Is this correct? The form is always meant to give the charge instead of the density? The density should always be regarded as a scalar field?

Answer

By 'density' in this case I think you just mean "something on a manifold that can be integrated to give you a scalar". By this definition, on an $n$-manifold, a density would be an $n$-form (since if you integrate over a form of lower dimension you get zero). So in your 3d case, take 3 smooth functions $f,g,h:M^3\to \mathbb{R}$, the form

$df\wedge dg\wedge dh$

is a density. Now, in your example you are integrating over a scalar field multiplied by a 3-form, which is again a 3-form, which can be integrated over a 3-manifold to get you the change in the region. But the scalar field $\rho:M^3 \to\mathbb{R}$ is NOT a density (not a 3-form), so it cannot be integrated over to find the total charge. The charge density is

$\omega=\rho dx\wedge dy\wedge dz,$

and $\rho$ just tells us how 'big' this should be.

In other words, the mathematical term 'density' can be stated as '$n$-form on $M^4$', whereas the colloquial 'density' for 'something per unit length/area/volume' is shorthand for what we really mean ($n$-form).

estimation - Maximum Possible Information in the universe?

I remember hearing about this in one of the programs in discovery science. The physicist claimed that the maximum possible information in the universe is $10^{10^{123}}$ whereas the maximum possible information that can be known by man is $10^{10^{90}}$. Can anyone explain to me how can we arrive at such a specific number, and also how can information be represented by only numbers?

Answer

The number $10^{123}$ emerges as (roughly) the number of Planck areas contained within the boundary of the observable universe. If each Planck area can be (roughly) in two states, a total of $10^{123}$ yes/no questions suffice to describe the boundary of the universe and - via the (still speculative) holographic principle - the whole universe. In other words, if the universe is a hologram, about $10^{123}$ bits of information are needed to describe it.

Unit vector in displacement

When we use vectors in physics why does the unit vector (for displacement) equals magnitude of 1 or magnitude of 1m?

fluid dynamics - Rayleigh-Taylor instability with negative Atwood number?

I was reading a paper entitled "The Rayleigh—Taylor instability in astrophysical fluids" by Allen & Hughes (1984) that indicates the instability can occur for $ \rho_{01} < \rho_{02} $ which would indicate a negative Atwood number. But how is this possible? Does not the density gradient have to be opposite the direction of the effective gravity? Must not the Atwood number be necessarily positive for a Rayleigh-Taylor instability?

Answer

Your intuition is correct; there's no such thing as a Rayleigh-Taylor instability with a negative Atwood number. That would imply that the density of the upper fluid, $\rho_{01}$, is less than the density of the lower fluid, $\rho_{02}$, which is clearly a stable situation with respect to the R-T instability.

So how did $\rho_{01} < \rho_{02}$ appear in the Allen and Hughes paper? I'm pretty sure it was just a typo. I read through the paper and the only place I saw anything that looked like a negative Atwood number was in section 4.2.2, where there's a sentence: In conclusion, it may be seen that the growth of R—T instabilities saturates for large accelerations, except in the limit $\rho_{01} \ll \rho_{02}$ where the growth remains of the usual form

$$ \omega = (gk)^{1/2} $$

But this sentence refers to an earlier paragraph in the same section that says, "Again, for $\eta$ ~ 1, compressibility has little effect and $\omega^2 \approx gk$." Since in the authors' notation, the Atwood number $\eta$ is defined as

$$ \eta = \frac{\rho_{01} - \rho_{02}}{\rho_{01} + \rho_{02}} $$

it is obvious that $\eta$ ~ 1 implies $\rho_{01} \gg \rho_{02}$ rather than $\rho_{01} \ll \rho_{02}$.

cosmology - Did really everything begin with a state with very low entropy?

As emphasized by Penrose many years ago, cosmology can only make sense if the world started in a state of exceptionally low entropy. The low entropy starting point is the ultimate reason that the universe has an arrow of time, without which the second law would not make sense. However, there is no universally accepted explanation of how the universe got into such a special state. Are there some observations that would really tell us that the early universe was with small entropy? Is this claim really consistent with our theories?

homework and exercises - How can I convert a capacitor circuit to a simpler one?

In the given question we have to find equivalent capacitance across A and C .

I do not understand how can we convert the given circuit into a simpler one or if that can not be done how do we find the equivalent capacitance of such a complex circuit .

Answer

Perhaps the simplest method is to eliminate one of the nodes using the $Y$-$\Delta$ Transform. For a capacitor network it applies as follows (taken from section 5.7 of this document) :

$\Delta \to Y$ use $c_1=C/C_1$ etc where $C=C_1C_2+C_2C_3+C_3C_1$.

$Y \to \Delta$ use $C_1=c_2c_3/c$ etc where $c=c_1+c_2+c_3$.

---

You can re-draw your network with ABC in an equilateral triangle with D at the centre (fig 1 below). D is connected to each of ABC, forming a $Y$ configuration with D at the centre. Using the transform this central $Y$ configuration can be converted into an equivalent $\Delta$ configuration, eliminating node D. Each of the edges AB, AC and BC then consists of 2 capacitors in parallel : the original capacitor, plus another from the $Y$-$\Delta$ transformation (fig 2). You should be able to simplify each edge AB, AC, BC into a single equivalent capacitance. Finally the network consists of only the triangle ABC (fig 3), ie edge AC in parallel with the series combination AB+BC.

---

A more elementary method is to place charge -Q at A and +Q at C. These charges are distributed between the connected capacitors in the network such that :
1. the total charge connected to nodes A and C is -Q and +Q respectively, while the total charge at other nodes is 0, and the total charge on each capacitor is 0;
2. the charge and voltage across each capacitor are related by Q=VC.

From the resulting equations you need to eliminate everything except the voltage $V_{AC}$ between A and C, expressing this in terms of charge $Q$ and the capacitance C. Then the equivalent capacitance is $C_{AC}=Q/V_{AC}$.

newtonian mechanics - Non-Constant Acceleration due to Gravity

Recently, I had the first physics lab for my university physics course. This lab was fairly simple, as we were merely using a computer and a distance sensor to graph the position, velocity, and acceleration of a cart as it moved along a linear track.

One of the situations we captured data for involved starting the cart at the bottom of an inclined ramp and giving it a push upwards. As expected, it rolled up, came to a stop, and then came back down the track to its starting position. The position-vs-time graph was essentially parabolic, the velocity-vs-time graph was essentially linear, and the acceleration-vs-time graph was essentially linear. So far, so good.

At this point in the lab, the instructor pointed out that, if the data was examined closely, the acceleration of the cart was greater while the cart was traveling upwards than when the cart was traveling downwards (approximately $0.546\frac{m}{s^2}$ and $0.486\frac{m}{s^2}$, respectively), and asked us to determine why in our lab report.

Now, gravity was the only force acting upon the cart, and thus it's acceleration should be a constant, at least at the scale our experiment was conducted at, so these results are completely baffling my lab group. So far, we have proposed the following ideas, but none of them seem very plausible.

• 
  

  Doppler effect on the ultrasonic distance sensor

  
  


• 
  
  

  Friction

  
  


• 
  

  Air resistance

  
  


• 
  

  Human error

  
  

The first seems highly improbable, and the last three are more obfuscation and hand waving than actual theories.

Why does our experimental data show the acceleration due to gravity to change based on the direction the object is moving?

Answer

That the numerical value of the resulting acceleration is greater upwards than downwards is likely due to friction. Uphill friction and gravity pull in the same direction, while downhill friction and gravity pull in opposite directions.

You can use the difference in acceleration to estimate the friction. The average is the (projected) gravitational acceleration, from which you can calculate the inclination angle (assuming a constant slope).

optics - Seeing something from only one angle means you have only seen (what?)% of its surface area at most?

Is there a logical/mathematical way to derive what the very maximum percentage of surface area you can see from one angle of any physical object?

For instance, if I look at the broad side of a piece of paper, I know I have only seen 50% of its surface area (minus the surface area of the very thin sides). Is 50% always the maximum amount of surface area you can see of any object from one angle?

Assumptions: This is assuming we aren't considering transparent/semi-transparent objects or the extra parts we can see with the help of mirrors. Just looking at the very surface of an object from one particular angle.

Answer

There is no such upper bound.

As a simple counter-example, consider a thin right-angled solid cone of base radius $r$ and height $h$, observed on-axis from some large(ish) distance $z$ away from the cone tip. You then observe the tilted sides, of area $\pi r\sqrt{r^2+h^2}$, and you don't observe the area of the base, $\pi r^2$, so you observe a fraction \begin{align} q &=\frac{\pi r\sqrt{r^2+h^2}}{\pi r^2+ \pi r\sqrt{r^2+h^2}} \\ &= \frac{\sqrt{1+r^2/h^2}}{r/h+\sqrt{1+r^2/h^2}} \\ &\approx 1- \frac rh \end{align} of the surface, in the limit where $r/h\ll 1$, and this can be arbitrarily close to $1$ so long as the cone is thin enough and long enough.

Landau's ambiguous statement about the existence of inertial frames

Landau writes "It is found, however, that a frame of reference can always be chosen in which space is homogeneous and isotropic and time is homogeneous."

Does he mean that we can prove the existence of an inertial frame or does he want to say that it is assumed by doing enough number of experiments?

Can we start with some axioms and definitions of properties of space and time and then deduce the existence of such a frame in which space is homogeneous and isotropic and time is homogeneous?

Answer

I believe this is just a restatement of the first Newton's law.

soft question - What is the difference between a philosopher of science and a theoretical physicist?

What is the characteristic objective difference in objectively observable attitudes and research styles between a philosopher of physics and a theoretical physicist?

Was Ernst Mach more of a philosopher of physics or a theoretical physicist? What about Niels Bohr? David Bohm? John Bell? Julian Barbour? John Wheeler? Max Tegmark? Albert Einstein?

The difference I am alluding to isn't about formal academic titles like which university department one belongs to, but the difference mentioned in the following quote by Albert Einstein:

This independence created by philosophical insight is—in my opinion—the mark of distinction between a mere artisan or specialist and a real seeker after truth.

i.e. between an "artisan" and a "seeker", between a "rote symbol-pushing calculator" and a "contemplator".

string theory - What is the difference between D branes, M branes, and NS branes?

I know that D$p$ branes are branes with $p$ spatial dimension with satisfying Dirichlet boundary condition. I want to confirm the following things.

First, I heard the D-branes are massive, and these massive D-branes is well described in $Ads$ geometry. Is it right? Why D-branes are massive?

In some context, D-branes are related with M branes. What is M-brane? Is it just some object describes in M-theory? I also heard M5 brane is related with D4 brane with some manipulation(dimensional reduction).

In the process of reading some papers, interestingly, M branes are also related with NS branes. What is NS-brane? Is it just some object describe in NS sector?

Is all branes are related with each other? What i mean is D brane and M branes seems have some similarity and M branes and NS brane have some relation. So there might be some relation about D-branes and NS branes.

Answer

When you try writing a quantu, theory of strings, you get supergravity in the classical limit. Branes are just solitonic solutions to those supergravity theories.

You can classify string/brane theories based on the type of SUGRA theory you get in the classical limit. That gives you four kinds in (9+1)dim and M-theory in (10+1)dim, all related by various kinds of dualities.

• NS5 brane and all the D branes are branes in Type2 theories.


• M-branes are just the solitonic solutions supported by SUGRA in (10+1)d. When you reduce that theory on a circle, you get Type2a (which is the same as Type2b, up to T-duality) so the M5 and M2 branes reduce to the various D-branes you know. Further, acting on the D5 by an S-duality will give you the NS5.

homework and exercises - How can static friction do work?

By definition, the work done by a force is $W = F\cdot d$, so how can static friction do work?

Can this force move the body a distance of $75~\text{m}$?

particle physics - Neutrino and electromagnetic forces

I learned from Wikipedia that neutrinos "are not affected by the electromagnetic forces". How was this identified experimentally?

quantum mechanics - What is a wave? What is a particle?

I am reading a David Bohm book on quantum theory. He says the idea that light is both a particle and a wave is incompatible:

(1) we know light has particle-like properties through the photoelectric effect

(2) we know light also has wave like properties because of slit experiments.

He then explains why they are incompatible.

But what is a "particle"? What is a "wave"? What do these terms mean precisely? I know what they mean loosely. A particle means something occupying a spatial position. A wave is like a density or something defined with peaks, troughs, and nodes over a spatial region. But I want something more rigorous and more accurate than these loose definitions so that I know what I mean when I use the term.

I'd like the definitions stated like we state axioms in math, clearly and specifically. In math, I say a vector space is closed under addition if $x,y\in V$ implies $x+y \in V$. Those are very specific claims.

Can someone do something similarly clear and specific with wave and particle?

Answer

The concepts "particle" and "wave" started from classical physics and from the everyday use of the terms, to begin with. A particle of dust got into one's eye, and the sea had huge waves.

Physics came into its reign when mathematics was seriously used to model observations.

For classical physics "particle" means an entity with small mass and a center of mass tracked at coordinates (x,y,z) at time t. Solutions of kinematic differential equations described the trajectory with accuracy determined by experimental errors.

For classical physics, waves are modeled by sinusoidal functions, i.e. functions that were the solution of "wave equations", could describe the behavior of sea waves, sound waves, and finally electromagnetic waves. Classically a wave is a variation of a measurable quantity like energy, or electric field, in space at a given time t, and the theoretical models were very successful in describing the observations of periodic energy distributions in bulk matter, and even in empty space ( electromagnetic waves).

Then quantum mechanics became necessary, from the discreteness of atoms, the black body radiation spectrum, the photoelectric effect it was finally understood that there were regions in the variables measured that displayed a quantization of energy.

It so happens that the equations that successfully describe the quantum mechanical state of matter were diferential equations with sinusoidal solutions, i.e. wave equations, like the Schrodinger equation. The solutions for the hydrogen atom were able to explain the spectral series adhoc assigned by the Bohr model, IF the postulate was assumed that the wavefunction squared did not represent the energy of the electron at (x,y,z) at time t, but a probability distribution. i.e. if one accumulated with the same boundary conditions a large number of measurements and plotted the (x,y,z) at time t distributions one would know how probable it would be to find the electron at that location.

As an example, this is similar to taking a census of the population of a city by age, and gauging how probable it would be that the first person you meet will be 8 years old. The wave function's function is just that, to give probabilities mathematically which are checked experimentally, and have been very accurate.

The "wave" part confused and continues to confuse people, because they think that the quantum mechanical entity, the electron for example, is spread out according to the solution of the Schrodinger equation. This is a misunderstanding, as the double slit interference experiments show with incoming single electrons:

Note the top photo, where the electron impinges on the screen, it is one whole electron . The probability pattern accumulated though shows clearly the interference effect that is expected by the sinusoidal form of the wave functions describing the electron when it hits the slits and goes through one or the other.

The "particle" facet of the electron is that it appears as a point at ( x,y,z_0) of the screen, and the "wave" facet is the probability distribution displayed in its trajectories.

If one is becoming a physicist it is simple to accept this fact, that the microcosm behaves differently than the macroscopic world we are used to. Bohm was stuck on classical frameworks and tried to derive the quantum mechanical probabilities from an underlying classical description. He succeeded in reproducing the same results as the usual quantum mechanical solutions, but afaik his model is complicated and limited, and cannot be extended into second quantization where the ball game has gone now.

differential geometry - Weitzenböck identity, and how to define the curved-space Laplacian of a spinor field?

I am following the notes by Freed about the Dirac operator in section 2. I am trying to understand the Weitzenbock equality that for a curved-space spinor's Dirac operator $D\!\!\!/\ $ and the associated 'Laplacian' $\Delta = \nabla^*\nabla $ satisfy

$$D\!\!\!/\ D\!\!\!/\ - \Delta = \frac{R}{4}$$

I'm trying to reproduce this formula using different notation, but I'm having difficulty defining the Laplacian $\Delta$. I know how to define the curved space Dirac operator in terms of an orthonormal frame $e^\mu_a$ and a basis of gamma matrices $\gamma^a$ as:

$$D\!\!\!/\ = \gamma^a e^\mu_a(\partial_\mu + \omega_\mu^I T^I) =: \gamma^a e^\mu_a \nabla_\mu$$

where I'm writing the $SO(n)$ spin connection as $\omega_\mu^I T^I$ for the Lie algebra generators $T^I$ and coefficents $\omega$. However, I'm confused on how the connection Laplacian should be defined. Using the naive definition of

$$\tilde{\Delta} = (\partial_\mu + \omega_\mu^I T^I)(\partial^\mu + \omega^{\mu I} T^I) = e^a_\mu(\partial_a + \omega_a^I T^I)e^{b\mu}(\partial_b + \omega_b^I T^I) $$

does not reproduce the right identity, and actually gives a differential operator rather than a scalar for $D\!\!\!/\ D\!\!\!/\ - \Delta$.

I have a feeling that the above expression for the Laplacian doesn't make sense for spinor fields, since implicitly I'm raising the spinor index using $g^{\mu\nu}$, whereas everything should in principle be doable via tetrads. What is the correct covariant way to write the operators $\nabla, \nabla^*$ and $\Delta$?

****EDIT****:

I figured it out, and typed it out below.

Answer

First, my expression for the Laplacian above was off, and the definition of $\nabla^*$ and $\nabla^*\nabla$ is a bit more subtle than I thought. $\nabla^*$ is the adjoint of $\nabla$. In particular, some vector $X$, $\nabla^*_X$ is the adjoint of $\nabla_X = X^\mu \nabla_\mu$. This can be computed from trying to integrate $\langle w, \nabla_X v \rangle = \int_M d(vol) w_\nu X^\mu \nabla_\mu v^\nu$ by parts to get

$$\langle w, \nabla_X v \rangle = \int_M d(vol) w_\nu X^\mu \nabla_\mu v^\nu = -\int_M d(vol) v_\nu \nabla_\mu (X^\mu w^\nu) =: \langle \nabla^*_X w, v \rangle$$

so that $\nabla_X^* w^\nu = -\nabla_\mu(X^\mu w^\nu)$ in local coordinates.

Note that all of these definitions were covariant derivatives with respect to a fixed orthonormal frame field. The operator $\nabla^*\nabla$ is actually referring to $tr(\nabla^*\nabla) =: \Delta$, where the trace means given some local orthonormal basis $e^\mu_a$, we define the laplacian as

$$(\Delta w)^\mu = \sum_a(\nabla^*_{e_a}\nabla_{e_a}) = \sum_a -\nabla_\nu(e_a^\nu e_a^\rho \nabla_\rho w^\mu)$$

which we can rewrite as (using Freed's convention that $\{\gamma^a,\gamma^b\}= -2 \delta^{ab}$)

$$(\Delta w)^\mu = \frac{1}{2}\nabla_\nu(e_a^\nu \{\gamma^a,\gamma^b\} e_b^\rho \nabla_\rho w^\mu)$$

Now, we want to compare $D\!\!\!/\ D\!\!\!/\ $ with $\Delta$. If we work in Riemann normal coordinates centered at $x=0$, we note that it's possible to choose $e^\mu_a = \delta^\mu_a + O(x^2)$, so that $\omega_\mu = 0 + O(x)$. This means we can freely rewrite the above expression as

$$(\Delta w)^\mu = \frac{1}{2} e_a^\nu \{\gamma^a,\gamma^b\} \nabla_\nu(e_b^\rho \nabla_\rho w^\mu) = \frac{1}{2} \{\gamma^\nu,\gamma^\rho\} \nabla_\nu \nabla_\rho w^\mu $$

where we abuse notation and write $\mu = 0,...,d$ to represent the flat space coordinates locally. And we also rewrite $D\!\!\!/\ D\!\!\!/\ $ as

$$(D\!\!\!/\ D\!\!\!/\ w)^\mu = \gamma^\nu \gamma^\rho \nabla_\nu \nabla_\rho w^\mu$$

so that

$$D\!\!\!/\ D\!\!\!/\ - \Delta = -\frac{1}{2}[\gamma^\nu,\gamma^\rho] \nabla_\nu\nabla_\rho = -\frac{1}{2}\gamma^\nu \gamma^\rho [\nabla_{\nu},\nabla_{\rho}]$$

From here, it becomes straightforward, since $[\nabla_{\nu},\nabla_{\rho}]$ gives the curvature tensor. One tricky point is the eventual need for the identity $\gamma^i\gamma^j\gamma^k\gamma^l R_{ijkl} = 2R$ (which is explained in Freed's notes clearly). This can be shown by noting that the Riemann tensors symmetries require $i \neq j, k \neq l$. The case where $j \neq k \neq l$ are all distinct vanishes by the Bianchi identity $R_{ijkl} + R_{iklj} + R_{iljk} = 0$. The case where $j=k$ gives $\sum_j \gamma^i \gamma^j \gamma^j \gamma^l R_{ijjl} = - \gamma^i \gamma^l R_{il} = R$. the case $j=l$ similarly gives $R$, so the total sum is $2R$. Putting this all together will give the desired result after dealing appropriately with the factors of 2 from the spin connection.

particle physics - Mixing of quarks, neutrinos... and leptons?

This is a quite simple question: quarks do mix (through the CKM matrix), neutrinos do mix (through the PMNS matrix).

Then why do charged leptons not mix?

fluid dynamics - I disagree with the definition of Reynolds number. Why am I wrong?

My question is about the intuition behind the Reynolds number. For example they say that if a fluid is moving

1 10 mph past a sphere of radius 10m with viscosity v

2 100 mph past a sphere of radius 1 m also with viscosity v

then they will have the same reynolds number.

The formula is Re = LU/v where L is characteristic length and U is characteristic velocity. They say that 2 flows are "similar" if they have the same reynolds number.

I would consider the following flows

1 100 mph past a sphere of radius 10 m

2 10 mph past a sphere of radius 1 m

with the same viscosity similar. So my equation would be Re = L/Uv

What am I missing here? Why should the first 2 flows above be similar but the bottom 2 are not?

Answer

Maybe a theoretical treatment of the Reynolds number would be fruitful here. For some introduction see my answer to another question.

In a fluid flow there are typically two relevant characteristic forces present, i.e. inertial $\left(\rho UU\right)$ and viscous $\left(\mu U/L\right)$ forces. The ratio of these quantities $\left(\rho UU / \mu U/L= U L / \nu\right)$ is what is known as the Reynolds number and signifies the relative importance of inertial to viscous forces. When viscous forces are more important then $\mathrm{Re}\ll1$, known as the laminar regime. Vice versa, when inertial forces are more important the $\mathrm{Re}\gg1$, known as the turbulent regime.

A mathematical treatment of this is shown by transforming the Navier-Stokes equations which describe fluid flow:

$$\mathbf{u}\cdot\mathbf{\nabla}\mathbf{u}=-\frac{1}{\rho}\mathbf{\nabla}p + \nu\mathbf{\nabla}^2\mathbf{u}$$

to a non-dimensional form:

$$\bar{\mathbf{u}}\cdot\bar{\mathbf{\nabla}}\bar{\mathbf{u}}=-\bar{\mathbf{\nabla}}\bar{p} + \frac{1}{\text{Re}}\bar{\mathbf{\nabla}}^2\bar{\mathbf{u}}$$

Here we see that the Reynolds number occurs as a coefficient in front of the viscous term. When $\mathrm{Re}\gg1$, the viscous term becomes negligible compared to the pressure term and inertial term, which means inertial forces are more important than viscous forces. The same treatment can be done for $\mathrm{Re}\ll1$, but requires rescaling of the pressure term which is not really the point now.

The Reynolds number naturally occured as a result of non-dimensionalizing the Navier-Stokes equations. The equation went from being described by four parameters $\rho$, $\nu$, $U$ and $L$ to one degree of freedom $\mathrm{Re}$. Now two flows are called 'similar' when their Reynolds numbers are the same because it means that their flow is described by the exact same dimensionless differential equation which yields a solution only dependent on one degree of freedom, $\mathrm{Re}$.

The first two flows you indicate have the same Reynolds number (i.e $vL=10\cdot10=100\cdot1$ for the same viscosity) and are therefore described by exactly the same dimensionless equation and thus 'similar'. The bottom flows are not 'similar' as they have different Reynolds numbers (i.e. $vL=100\cdot10\neq10\cdot1$ for the same viscosity). Simply redefining Reynolds number is not physical as it then no longer is dimensionless.

Note that, in general, flows are 'similar' when all their dimensionless numbers are the same. If for example there is a body force included, this would introduce an additional dimensionless number known as the Froude number. Flows described by such an equation are only 'similar' when both the Reynolds and Froude numbers are the same.

vision - Why does the moon appear bigger close to the horizon, rising or setting?

Possible Duplicate:
Why does the moon sometimes appear giant and a orange red color near the horizon?

I made a little research about this and found this article http://en.wikipedia.org/wiki/Moon_illusion that states the explanation of the moon being bigger at the horizon is still debated.

I found this which looks pretty 'big' http://www.psychohistorian.org/img/astronomy/deep-sky/photos/ayiomamitis/20090804-moonrise.jpg but however this kind of repetitive shots http://apod.nasa.gov/apod/image/0706/UludagMoonrise_tezel.jpg show no decrease in the aparent moon radius.

What I am trying to understand is if it is a real illusion, if the atmosphere makes the image bigger or what other explanation could be possible and of course plausible.

Answer

As it turns out, this is not so much a physics question as it is a psychological one. If you use a ruler or some such held at a fixed distance from your eyes, you'll find that, as demonstrated in the repetitive shot image you linked, the moon has approximately the same apparent size across its entire path across the sky.

The optical illusion arises from the lack of reference points in the sky. Basically, when near the horizon, there are terrestrial objects (trees, hills/mountains, houses, etc.) whose size we recognize as large. The moon, looking larger than these objects, appears even bigger by comparison. When high in the sky, however, the only thing to compare the moon's apparent size to is the sky itself, which is much larger than the moon. The result is that the moon appears smaller.

cosmology - What was the major discovery on gravitational waves made March 17th, 2014, in the BICEP2 experiment?

The Harvard-Smithsonian Centre for Astrophysics held a press conference today to announce a major discovery relating to gravitational waves. What was their announcement, and what are the implications?

Would this discovery be confirmation of gravitational waves as predicted by general relativity (even though Sean Carroll links to the Nobel website implying that G-waves was detected decades ago, while my book on GR (B. Schutz) says they're still looking...I'm confused)

Also, regarding inflation theory, would this discovery confirm inflation or refute it? Or something else, a la string theory?

Making precise the statement "particles are excitations in a quantum field"

I've been trying to self teach QFT lately. I find that the basic physical idea makes sense, and I can keep up with the mathematical formalism without too much trouble, but I'm having trouble connecting the two.

It is usually said that QFT is necessary because particles can appear and disappear, and so rather than treat a particle as a basic entity as in non-relativistic QM, we instead have a field, and particles are excitations or quanta in that field. In the canonical formalism, the field is a field of operators which creates and destroys particles. For example, if we have a real scalar field $\phi$, then we know that its Fourier components $a_\mathbf{p}$ and $a_\mathbf{p}^\dagger$ act on the vacuum as $a_\mathbf{p}^\dagger |0\rangle = |\mathbf{p}\rangle$. Also, if in the Schrödinger picture $\phi$ is independent of time, that is, $\phi=\phi(\mathbf{x})$, then in the Heisenberg picture we can get time dependence as $\phi(x) = e^{-iHt}\phi(\mathbf{x})e^{iHt} = e^{-ipx}\phi(0)e^{ipx}$.

Looking at the last expression, I don't see how the particles come into play. It seems to me that the current state of the world is "encoded" in the kets, so for example if we have two particles with momentum $\mathbf{p_1}$ and $\mathbf{p_2}$, the state is $|\mathbf{p_1}\ \mathbf{p_2}\rangle$. But it seems that the time evolution of the field depends only on the Hamiltonian and not on (for example) the number of particles. Is this true? If it is, then how does that statement "particles are excitations in a quantum field" translate into the formalism?

Perhaps more simply: is the field any different if we have one particle or if we have two particles?

Answer

The talk about "particles" being "excitations" is handwavy, and comes about precisely because of the analogy of the usage of Fourier components that you talk about.¹

Perhaps more simply, is the field any different if we have one particle or if we have two particles?

"The field" is inaccessible. Some fields aren't even observables (because they may not be (gauge) invariant in and of themselves), and even of these that are, you can only measure the expectation values w.r.t. some states. The statement "particles are excitations" is perhaps most "precise" in the LSZ formalism, where you take some definite-momentum particle-in/out states and obtain that you can calculate

$$\langle p_1,\dots,p_m \vert q_1,\dots,q_n \rangle$$

by looking at some integral whose integrand has $\langle \Omega \rvert T \phi(x_1)\dots\phi(x_m)\phi(y_1)\dots\phi(y_n)\lvert \Omega \rangle$ as a factor, so you could say that this "looks like the field creates the particles".

In QFT, you have to abandon the notion that there is a real value for the field at every point - that would be the classical picture of the field just being the solution of the classical equation of motion. All you can have are expectation values, and the idea that there is some kind of field actually being "excited" is...ill-defined. It isn't clear at all what one would want to say with that. It's just a cute picture that many like, not a mathematically rigorous statement.

Of course, the time evolution will act differently on many-particle states than it will on one-particle states. But these states are states in the Hilbert space of the theory, and only by analogy (since they are created by Fourier components of free fields) identified with "excitations", and analogy that breaks down if we would not consider the asymptotic Fock space but instead the full (but sady, mostly unknown) Hilbert space of the interacting QFT.

---

^{1I wrote about this some time ago. The analogy is not very helpful in my view. The mathematics are of the same form as that for an oscillator, but to say that particles are excitations is endowing the concept of field with an ontological weight that is unnecessary, and just tends to confuse people about non-perturbative, interacting QFTs, where it is not helpful at all to think in such terms.}

newtonian mechanics - Why doesn't the speed of Earth (moving through space) kill us?

I have read different speeds of Earth in different sources. $382\;{\rm km}/{\rm s}$, $12\;{\rm m}/{\rm s}$ and even $108,000\;{\rm km}/{\rm h}$. Basically, it's moving too fast around the Sun. And the Solar System is moving too. So why don't we feel it and why doesn't it harm us in any way? Inertia can only be a part of it. But what's the whole reason?

Answer

Speed doesn't kill us, but acceleration does.

When astronauts go into space at launch and when fighter pilots turn very tight turns at high speed they experience 'high g forces' - their bodies are accelerated very fast as they accelerate and gain speed to go into space or as the direction of their speed changes. One of the problems with this is that for fighter pilots the blood can rush to the feet (black out) or to the head (red out). Too much acceleration makes people pass out and could at extremes be fatal I guess.

To go around the sun in (nearly) a circular path we are acclerated by the gravity from the sun. The acceleration can be calculated by $v^2/r$ where $v$ is our speed and $r$ is the distance to the centre of the sun. This acceleration turns out to be $\sim~0.006~m/s^2$. By contrast the acceleration that we feel here at the surface due to the gravitational pull of the earth on us is $\sim~10~m/s^2$. So the acceleration due to travelling around the sun is so small we don't notice it. We do notice the pull of gratvity from the earth on us, but our bodies are used to it and can cope with it.

To think about it another way we can go very fast in a car on a motorway/highway without noticing it, the big danger is having to stop very quickly or crashing when we change speeds very rapidly - acceleration is the rate of change of speed so changing speed very rapidly is equivalent to a very high acceleration - in a car we might call this deceleration.

[for calculation above $v=3 \times 10^4~m/s$ and $r=1.5 \times 10^{11}m$]

after good comment from hdhoundt - For astronauts in orbit (e.g. in the space station) they can cope with the acceleration they experience, which holds them in orbit around the earth. Indeed they feel weightless because they are not held by the gravity of earth on the surface. Instead they and their surroundings are in 'constant free fall'. The speed of the space station in orbit is $7.71 km/s$, which is $\sim~ 17,000 ~mph$.

Full discussion of this topic might venture into relativity, but I think that is beyond the scope of the question.

after good comment from Mooing Duck -

Perhaps even more dangerous than acceleration is jerk, which is the rate of change of acceleration and other higher order terms. Jerk would be very severe in the case of car collisions. - But also if the driver of a car or bus has to 'brake' and slow down very suddenly it can be very uncomfortable for the passengers.

After good comment from Jim (and Cory)-

Good point raised about acceleration and/or jerk on a human body. If every part (and every particle) of the body experience the same acceleration or jerk then the body will suffer significantly less (possibly no) damage compared to when one part of the body is accelerated of jerked and the acceleration or jerk is transmitted to other parts of the body by the structure of the body. The classic example here is 'whiplash' neck injury, where a jerk on the body is transmitted to the head through the neck. To reduce the damage this may cause seats in cars generally hare head rests that will support the back of the head and for people who are involved in motor sports (e.g. car racing) may wear a neck brace/support that prevents the head from swinging backwards and forwards on the neck in the event of a collision.

Another aspect of acceleration to all part of the body concerns rocket launch for astronauts. The rockets will be designed so that as much as possible all part of the body are equally supported and the body lies 'flat with respect to the acceleration' so that the blood in the astronaut's body does not rush to the feet or head. This is a serious consideration and Memory Foam came from research by NASA into safety for aircraft cushions and helped cushion astronauts in rockets.

general relativity - Accounting for metric tensor derivatives in Einstein-Hilbert action

I'm puzzling over the canonical derivation of GR from the Einstein-Hilbert action; getting the derivation to gel with an explicit treatment of the functional derivative isn't working out. So the derivation (drawn here from Wikipedia, though other literature is similar) begins,

$$ I = \int{\sqrt{-g}d^4x} \left[\frac{1}{2 \kappa}R + \mathcal{L}\right] $$

and immediately proceeds to

$$ \delta I = 0 =\int d^4x \delta g_{\mu \nu}\left[\frac{1}{2\kappa}\frac{\delta\left(\sqrt{-g}R\right)}{\delta g_{\mu \nu}}+\frac{\delta\left(\sqrt{-g}\mathcal{L}\right)}{\delta g_{\mu \nu}}\right]. $$

But the Ricci scalar depends on the first and second derivatives of the metric tensor, so why do we not have factors

$$ \delta g_{\mu \nu, \alpha}~, \qquad \delta g_{\mu \nu,\alpha \beta}~, $$

against which we vary as well? Maybe there is some identity that in this case causes these terms to vanish, but I don't see it.

Answer

You absolutely do have these terms. Most people just always implicitly integrate by parts, and realistically, they hide these terms in other terms, because the algebra blows up into tons of terms very quickly. A very laborious version of this is worked out in the Classical Field Theory book in the Landau and Lifschitz series.

Alternately, you can use the Palantini form of the variation, and vary the Christoffel symbols, rather than the metric.

optics - Why does change in speed of a wave make it refract?

When a light wave enters a medium with a higher refractive index (e.g. from air to standard glass) and its speed decreases, why does that make it refract/bend?

I understand that wavelength decreases and frequency stays the same and therefore its speed decreases, but I can't find anywhere whatsoever why the speed decrease cause the wave to refract. So could someone please explain this?

Answer

The wave only refracts if it enters the medium at an angle. Follow a single wavecrest; if the wave is entering the medium at an angle, then part of the wavecrest enters the medium first, and starts to slow down, while the other part of the wavecrest is still going fast, and therefore the wavecrest must bend. If the wave enters at a right angle, then the entire wavecrest is slowed down simultaneously and no refraction occurs.

special relativity - is SR simultaneity symmetrical?

Alice and Bob's spaceships are converging inertially at 0.99c. Alice knows enough about Bob's clock that she can calculate for any time of her clock what his clock would show "at that time" in her frame. Bob is similarly knowledgeable about Alice's clock. When Alice's clock shows Ta, Alice calculates Tb = what Bob's clock shows at that time in Alice's frame. When Bob's clock shows Tb, Bob calculates Tc = what Alice's clock shows at that time in Bob's frame. Question: does Ta = Tc ?

Answer

Short answer: No. If they did, there would be no relativity of simultaneity.

The simplest way to track the conflict is to add a buoy, say in Alice's frame, at some distance from Alice's own location. Let the buoy carry its own clock, synchronized to that of Alice, such that when Bob passes the buoy location, the buoy clock shows time $T_A$, just like Alice's clock. This means Alice observes 2 simultaneous events at non-coincident locations: her clock beating time $T_A$ at her location, and the buoy clock beating time $T_A$ at its own location.

Now look at Bob as he passes the buoy at his time $T_B$ and buoy time (Alice's time) $T_A$. Relativity of simultaneity means that Bob cannot see both Alice and the buoy marking time $T_A$ at the same moment of his time: spatially separated events that look simultaneous to Alice never appear simultaneous to Bob. So although he does see the buoy clock showing $T_A$ at his time $T_B$, he finds that Alice's clock shows a different time ${\tilde T}_A$.

Fun exercise: Apply the same reasoning to a symmetrical setup using buoys in both Alice's and Bob's frame.

Note following comments: As @dmckee observed, the phenomenon is independent of a particular choice of time origin, in either frame. Alice and Bob may choose their individual $t=0$ marks arbitrarily, yet they always arrive at the same conclusion regarding the specific time intervals. In more formal if obvious terms, time intervals are invariant under time translations in any frame.

interference - Diffraction grating vs single slit

When I think about it, the single slit interference experiment should resemble the diffraction grating experiment as it is almost like single slit is the equivalent of a grating with many many many slits in a small length. So I would have though that the more slits you add in a diffraction grating, the closer its pattern of interference would resemble that of the single slit. However, it does not! The maximums become more and more distinct when adding more slits. Why do these two methods produce such different results? Thanks

How does the Higgs boson give mass to other elementary particles like electrons?

So, the nucleus of an atom can be broken to protons and neutrons, and those can be broken down to quarks.

Electrons however are a different story, they can't be broken down since they are elementary particles but they do have mass.

So the higgs boson itself does not give mass to particles but the interaction between the Higgs field and the Higgs boson does? How does the electron for example interact with the Higgs field?

angular momentum - Derivation of nuclear spin $I$ for $^{87}$Rb and $^{40}$K

So I know that $^{87}$Rb has $I=3/2$, and has 37 protons and 50 neutrons.

I try to make sense of it form the nuclear shell model:

taken from here.

• 50 neutrons: they are a closed shell, so no net angular momentum


• 37 neutrons: 28 get stacked in a closed shell, the remaining 4 fill the 2p$_{3/2}$ sub-shell and contribute no net angular momentum, and the remaining 5 go into the 1f$_{5/2}$ sub-shell.
  4 of them are paired, so zero net angular momentum, and the fifth should have net 5/2?

What is wrong in the above argument?

How would I go on proving that $^{40}$K has $I = 4$?

Answer

In 87Rb, the Fermi level for spherical shape is very close to both f5/2 and p3/2. We do expect 87Rb to be spherical, because the neutron number is magic. Therefore we expect a 5/2- state and a 3/2- state, both of them very low in energy. One of them should probably be the ground state, and the other the first excited state. This is what we see when we look up the nucleus in ENSDF. So there is really no huge surprise here. You just can't take these diagrams of single-particle energy levels as if they were guaranteed to give exact predictions for all nuclei. A paper discussing the nuclear structure is here. The relevant discussion is on p. 3.

Because 40K has only a single proton hole and a single neutron outside of the closed shell at particle number 20, we again expect it to be spherical. However, odd-odd nuclei are a mess because there are lots of ways of putting together states for the two odd particles. 40K has a 4- ground state, but it also has a known 3- state at only 30 keV excitation energy. There may be a dozen more states within 100 keV above the ground state which just haven't been detected. There is no easy way to sort out which should be the actual ground state theoretically, short of doing a full shell-model calculation ("shell model" meaning you actually diagonalize the Hamiltonian for some interaction among the valence particles, not just assign particles to states up to the Fermi level). Even a shell-model calculation is unlikely to get things right to a precision of 30 or 100 keV. ENSDF says for the ground state:

Dominant configuration=π1d3/2−1⊗ν(1f7/2 or 2p3/2) (2014Pa45) from comparison with shell-model calculations.

However, this doesn't quite make sense because you can't couple spins 3/2 and 3/2 to make the observed spin 4 of the ground state, so I think the ENSDF evaluator made a mistake interpreting the paper they refer to. On p. 8 of that paper, they seem to be saying that it's π1d3/2−1⊗ν1f7/2. The spins 3/2 and 7/2 can couple to make anything up to spin 5, so spin 4 is possible.

particle physics - Can two colliding photons create a Higgs Boson?

The LHC discovered the Higgs using the following decay mode: $$ H^0 \rightarrow \gamma \gamma $$

This got me thinking: if we reverse this mode, will it be valid? In some annihilation/decay pairs, these will stay valid:

$$e^+e^- \rightarrow \gamma\gamma$$

$$\gamma\gamma \rightarrow e^+e^-$$

So if the Higgs will decay into two high energy photons, will the annihilation of two high energy photons create a Higgs by this mode?:

$$ \gamma \gamma \rightarrow H^0 $$

Answer

In principle, yes. You can reverse any decay process and the corresponding synthesis will be valid - in this case, since $H_0\to\gamma\gamma$ happens, then $\gamma\gamma\to H_0$ will also happen, assuming the kinematics work out.

However, the corresponding probability is very small. Out of all the possible things that could happen when two photons cross paths, turning into a Higgs boson is a relatively unlikely one. In fact, by far the most common option is nothing, because photons don't directly interact with each other. In order to get an interaction, you need one of the photons to quantum-fluctuate into a pair of charged particles, so that the other photon can interact with one of them - this makes it a second-order interaction at best. And Higgs production is even more unlikely because photons don't interact with the Higgs either, so you need another fluctuation to go from the charged particles to the Higgs boson. All the Feynman diagrams are third-order or higher, which makes for a very unlikely process indeed.

quantum mechanics - Can nowadays spin be described using path integrals?

In Feynmans book, "Quantum mechanics and Path Integrals" he writes in the conclusions (chapter 12-10)

With regards to quantum mechanics, path integrals suffer most grievously from a serious defect. They do not permit a discussion of spin operators or other such operators in a simple and lucid way. ... It is a serious limitation that the half-integral spin of the electron does not find a simple and ready representation.

This has been written in 1965. Has there been any progress on this problem? For example, it is possible nowadays to not only derive the Schrödinger equation, but also the Pauli equation from the path integral formulation of QM?

Answer

Reposting comment as an answer and expanding.

The answer is yes. You can find an exposition in Condensed Matter Field Theory by Altland and Simons, starting on page 134 in the second edition. The troubles come from that spin can't be described with a Hamiltonian that is a function of $q$:s and their conjugate $p$:s. However the more general formulation of Hamiltonian mechanics in terms of symplectic manifolds permits a description of spin. Altland and Simons cite Arnold's Mathematical Methods of Classical Mechanics as a reference for this. It's an under-appreciated gem of a book.

So, when we construct the path integral we consider the paths to be paths in phase space, $p$:s and $q$:s. I think to understand this in geometrical terms we have to back to "basics". The Lagrange formulation we have coordinates $q$ and velocities $\dot q$. The coordinates can be coordinates on any manifold -- this is why the Lagrangian formalism is so neat for constrained systems -- so the velocities are really tangent vectors. Lagrangian mechanics is thus naturally formulated on tangent bundles. But we can take the Legendre transform and pass to the Hamiltonian formalism with $p$:s and $q$:s. This takes us to the cotangent bundle, for $$p_i = \frac{\partial L}{\partial \dot q^i}$$ is the 1-form that is $\partial L /\partial \dot q^i$ on the vector field $\dot q^i$ and 0 on the other coordinate vector fields.

However you can do all of Hamiltonian mechanics on any manifold that comes with a symplectic structure. A symplectic structure is 2-form $\omega$ such that $d\omega = 0$ and for every vector $v$, $\omega(v, \cdot)$ is not the zero 1-form. (You can think of $\omega$ as a sort of anti-symmetric metric.) This is what Arnold does in his book. The cotangent bundle naturally comes with such an $\omega$: $$\omega = dp_1 \wedge dq_1 + \ldots dp_n \wedge dq_n$$ (this 2-form is independent of your choice of coordinates $q$). The Hamiltonian is a function on $M$, so in the particular case of a cotangent bundle, you can take it to be a function of the $p$:s and $q$:s.

Now you can couple orbital angular momentum to the vector potential just fine with $p$:s and $q$:s. But how to do it with intrinsic angular momentum, that is spin? We want a Hamiltonian like $$H = \mathbf B \cdot \mathbf S.$$ Since the spin of a particle has a definite magnitude, the dynamical part of the spin is its direction. So this Hamiltonian is defined on a manifold like $T^* M \times S^2$ (I put the first factor there because $\mathbf B$ could of course vary in space). Of course this rests on $S^2$ having an admissible $\omega$, but you can take the volume form $$\omega = \sin\theta\; d\theta \wedge d \varphi$$ with $\theta, \varphi$ the usual coordinates.

Therefore the paths you use in your construction of the path integral for spin should be paths on the sphere, and you can use $\theta, \varphi$ as coordinates. There is a formal complication in the quantum case since our states also have phases. This means that we should really use paths in $SU(2)$, since an arbitrary spin state can always be written like $$|g\rangle := g\vert\uparrow \rangle$$ where $\vert\uparrow\rangle$ is some reference state. So the resolution of identity inserted to construct the path integral should be $$\operatorname{id} \int_{SU(2)} \vert g \rangle \langle g\vert$$ (the integration is with respect to an $SU(2)$-invariant volume form.) This leads to an extra term, the usual $\partial_\tau$, in the path integral. However it turns out that the phase is irrelevant to this term, so the path integral is really over paths on the sphere $S^2$. You can find the detailed calculation in Altland and Simons.

newtonian mechanics - Calculating wind force and drag force on a falling object

I'm trying to numerically integrate the motion of an object (say, a falling vertical cylinder). Here, there's a drag force: the wind "acting" on the body (presumably adding horizontal velocity) and the air itself slowing down the vertical motion.

Is it correct to calculate both forces with the drag equation, $F_D = \frac{1}{2} \rho v^2 C_D A$? Here I suppose the velocity would be the relative velocity between the falling object (i.e. initially $(0, -15)\ m/s$) and the "air" (i.e. $(5, 0)\ m/s$). If that's the case, where would the drag force point towards to? $-\hat{v}_{relative}$?

Thanks a lot for your help in advance.

Answer

Yes, the force points along the vector of the relative velocity between the object and the air.

Quadratic drag is an interesting phenomenon. You have to calculate the net velocity vector (which includes a horizontal and vertical component) and compute the force along that axis; when you then decompose it into horizontal and vertical components you will find that the vertical drag is greater because of the cross wind. This is not an intuitive result!

special relativity - Impulse from absorbing a photon? Is there an increase in rest mass?

I'm going through A P French's special relativity. In one chapter (6) the following is set up:

Suppose that a stationary particle of mass $M_0$ is struck by a photon of energy $Q$, which is completely absorbed. The combined system will have mass $M'$ and will recoil with a velocity $v=\beta c$.

It then goes on to say that conservation of energy implies: $$E=M_0 c^2+Q=M' c^2 $$ (1)

and conservation of momentum implies: $$p=Q/c=M' \beta c$$ (2)

Using the first equation to solve for $M'$, and plugging $M'$ into the second equation to solve for $\beta$, we wind up with: $$\beta=\frac{Q}{M_0 c^2+Q}$$

However, if I use what was developed earlier in the chapter, $M'=\gamma M_0$, and assume equation (1) [ignoring (2)], I get a completely different answer. Likewise, if I assume equation (2), I get yet a different answer, for $\beta$.

(So, the first solution would be solving $M_0 c^2+Q=\gamma M_0 c^2$ for $\beta$, with $\gamma=(1-\beta^2)^{-1/2}$. The second solution would be solving $Q/c=\gamma M_0 \beta c$ for $\beta$. I didn't include the solutions I got because they're [from what I gather] incorrect, and because the equations for them aren't too interesting/enlightening/short to write)

I figure the only way to reconcile these problems is if $M'$ doesn't only depend on $\beta$, but if instead the rest mass actually increases from before the collision, so that I have $M_0 c^2+Q=\gamma M_0' c^2$ and $Q/c=\gamma M_0 ' \beta c$. Then, I'd have two equations and two unknowns, so the result in the book would hold, and I can calculate the new rest mass.

My question is: Is this a real effect? If so, is there a name for it? It just seems a bit remarkable, and it's something that if I had ignored momentum and only looked at energy (or vice versa), I would have missed completely.

Answer

Yes there must be an increase in rest mass. Here's why:

Let's start from a position of maximal ignorance. If we assume nothing about the rest mass before and after, then conservation of energy and momentum read \begin{align} M_0c^2 + pc &= \gamma M c^2 \\ pc &= \gamma Mv \end{align} with $p\neq 0$. With some manipulation, this reduces to $$ \frac{M_0}{M} = 1- \frac{v}{c} $$ Now if $M=M_0$, meaning rest mass is conserved, then we get $v=0$, but this contradicts the assumption $p\neq 0$. So we see that rest mass cannot be conserved in this process.

Addendum. I've been thinking about this, and I realized that the increase in rest mass upon absorbing a photon is clear from some basic examples. Here's one:

Suppose that the particle of mass $M_0$ in this problem is a hydrogen atom in its ground state, and imagine that the photon has enough precisely energy to kick the atom's electron into the first excited state while also satisfying energy and momentum conservation. In this case, it is clear that the rest mass of the hydrogen at will have increased since the binding energy of the electron with the nucleus will have increased.

The main idea, then, behind the original though experiment is that in order for the photon to be absorbed, there must be some mechanism for the absorption (like in the Hydrogen atom example I just gave), and whatever that mechanism is, relativistic energy-momentum conservation will guarantee that it must result in an increase of the rest mass of the original stationary particle.

Hope that helps!

Cheers!

electromagnetism - does matter radiate energy?

Since we know that all accelerated charges radiate energy and we also know that all matter is made up of protons and electrons which are all the while doing accelerated motion.So from this can i conclude that every piece of matter radiates energy because of jiggling motion of atoms?

electromagnetism - How wrong are the classical Maxwell's equations (as compared to QED)?

Now, I don't really mean to say that Maxwell's equations are wrong. I know Maxwell's equations are very accurate when it comes to predicting physical phenomena, but going through high school and now in college, Maxwell's equations are seen as the equations of electricity and magnetism. Now, it's common knowledge among students that, while Newton's laws are generally accurate when applied to everyday experiences, they are also replaced at high velocities by special relativity (and general relativity for very large gravitational fields).

But this is less so the case with Maxwell's equations. I have read that Maxwell's equations are replaced with quantum electrodynamics (which to me has all the effect of mere buzzwords, since I don't know what quantum electrodynamics is) as a more accurate way of describing electromagnetic waves, but what are the limitations of Maxwell's equations?

Or let me phrase this differently. I'm currently an electrical engineering major. I know NASA scientists and engineers can still get away with using Newtonian physics for their calculations, because it's that accurate. I also know, however, that relativity does have to come into play with GPS. So, in what situation as an electrical engineer would Maxwell's equations ever fail me? When (assuming I'm working on such a sufficiently advanced project) would I have to resort to more accurate ways to describe electromagnetic waves?

Answer

Maxwellian electrodynamics fails when quantum mechanical phenomena are involved, in the same way that Newtonian mechanics needs to be replaced in that regime by quantum mechanics. Maxwell's equations don't really "fail", as there is still an equivalent version in QM, it's just the mechanics itself that changes.

Let me elaborate on that one for a bit. In Newtonian mechanics, you had a time-dependent position and momentum, $x(t)$ and $p(t)$ for your particle. In quantum mechanics, the dynamical state is transferred to the quantum state $\psi$, whose closest classical analogue is a probability density in phase space in Liouvillian mechanics. There are two different "pictures" in quantum mechanics, which are exactly equivalent.

• 
  

  In the Schrödinger picture, the dynamical evolution is encoded in the quantum state $\psi$, which evolves in time according to the Schrödinger equation. The position and momentum are replaced by static operators $\hat x$ and $\hat p$ which act on $\psi$; this action can be used to find the expected value, and other statistics, of any measurement of position or momentum.

  
  


• 
  

  In the Heisenberg picture, the quantum state is fixed, and it is the operators of all the dynamical variables, including position and momentum, that evolve in time, via the Heisenberg equation.

  
  

In the simplest version of quantum electrodynamics, and in particular when no relativistic phenomena are involved, Maxwell's equations continue to hold: they are exactly the Heisenberg equations on the electric and magnetic fields, which are now operators on the system's state $\psi$. Thus, you're formally still "using" the Maxwell equations, but this is rather misleading as the mechanics around it is completely different. (Also, you tend to work on the Schrödinger picture, but that's beside the point.)

This regime is used to describe experiments that require the field itself to be quantized, such as Hong-Ou-Mandel interferometry or experiments where the field is measurably entangled with matter. There is also a large gray area of experiments which are usefully described with this formalism but do not actually require a quantized EM field, such as the examples mentioned by Anna. (Thus, for example, black-body radiation can be explained equally well with discrete energy levels on the emitters rather than the radiation.)

This regime was, until recently, pretty much confined to optical physics, so it wasn't really something an electrical engineer would need to worry about. That has begun to change with the introduction of circuit QED, which is the study of superconducting circuits which exhibit quantum behaviour. This is an exciting new research field and it's one of our best bets for building a quantum computer (or, depending on who you ask, the model used by the one quantum computer that's already built. ish.), so it's something to look at if you're looking at career options ;).

---

The really crazy stuff comes in when you push electrodynamics into regimes which are both quantum and relativistic (where "relativistic" means that the frequency $\nu$ of the EM radiation is bigger than $c^2/h$ times the mass of all relevant material particles). Here quantum mechanics also changes, and becomes what's known as quantum field theory, and this introduces a number of different phenomena. In particular, the number of particles may change over time, so you can put a photon in a box and come back to find an electron and a positron (which wouldn't happen in classical EM).

Again, here the problem is not EM itself, but rather the mechanics around it. QFT is built around a concept called the action, which completely determines the dynamics. You can also build classical mechanics around the action, and the action for quantum electrodynamics is formally identical to that of classical electrodynamics.

This regime includes pair creation and annihilation phenomena, and also things like photon-photon scattering, which do seem at odds with classical EM. For example, you can produce two gamma-ray beams and make them intersect, and they will scatter off each other slightly. This is inconsistent with the superposition principle of classical EM, as it breaks linearity, so you could say that the Maxwell equations have failed - but, as I pointed out, it's a bit more subtle than that.

thermodynamics - Joule Thomson Coefficient derivation

In deriving Joule Thomson effect coefficient, why do we take $dH=0$ and also why do we take $H$ as a function of $T$ and $P$ only and not $V$? i.e. why do we take $$(\partial H/ \partial T)dT+(\partial H/\partial P)dP=0$$ and not

$$(\partial H/\partial T)dT+(\partial H/\partial P)dP+(\partial H/\partial V)dV=0$$

Any links would also be appreciated.

Thanks.

Trying to understand what a gravitational wave is

It seems to me that there is a great deal of interest in the possibility of gravitational waves. Several gravitational-wave detectors have been built, and there is even a branch of science with that specific goal in mind, which is gravitational-wave astronomy.

What I don’t understand is the difference between gravitational wave as produced by, say an exploding supernova, versus the change in gravitation at a fixed point in space by some other effect such as a planet (or some other dense body) passing by. Wouldn’t a passing dense body produce a single wave pulse of stronger, and then weaker gravitation, which would travel in a wave to infinity?

For example isn't the effect of Jupiter moving along its orbit producing more or less a gravitational wave which could be detected by sufficiently sensitive instruments millions of miles away?

Answer

Accelerating masses generate gravitational waves, much like accelerating charges generate electromagnetic waves.

Masses in orbit are continually accelerating, so you're right, Jupiter will generate gravitational waves by orbiting around the Jupiter-Sun centre of mass. In this case there's also gravitational radiation from the sun orbiting around the centre of mass of the Jupiter-Sun system (which is at about the radius of the sun). However the power of the wave emitted is tiny.

Wikipedia gives the following formula for the gravitational radiation of two masses in orbit around each other:

$ P = -\dfrac{32}{5} \dfrac{G^4}{c^5}\dfrac{(m_1m_2)^2(m_1+m_2)}{r^5} $

For the Jupiter-Sun system this gives about 200 Watts - less power than my fridge uses!

For two stars orbiting very closely this power can be much higher - if the two objects are both neutron stars of one solar mass and are separated by 1.9$\times$10$^8$ m, then the power radiated is around 1$\times$10$^{28}$ W, which is quite a bit higher.

Using the inverse square law, the two gravitational wave sources will have the same intensity if their distances are in the ratio 1:10$^{14}$. Since we're about 1.5$\times$10$^{11}$ metres away from the sun, a binary neutron pair's gravitational waves will overwhelm that of the Sun if they're closer than around 10$^{25}$ m away, or about 1 billion light years!

How is quantum teleportation any better than regular transfer of information?

After the article that China "teleported" a particle into orbit, I'm trying to understand the applications of quantum teleportation. It sounds like a nice oddity of the universe, but has very little to no practical value. I'd love to be corrected, but from what I understand, the basic procedure is this

1. Alice and Bob split up a pool of entangled widgets



2. Alice and Bob move far away from each other


3. Alice does black magic to her pile of widgets, through blacker magic this affects Bob's pile of widgets instantly

This sounds all well and good, until I hit this part:

4. Alice sends the new configuration of her pile of widgets to Bob through traditional channels


5. After Bob gets the information from Alice he knows what state his pile of widgets should be in

This is great and all, but since Alice sends a the new configuration of her pile of widgets to Bob through traditional channels, what is the point of even using the widgets? It seems like we've added a bunch more complexity for no real reason other than "Hey, this is cool".

Answer

There are several reasons quantum teleportation is a useful way to move qubits, even if you already have a normal quantum channel.

1. 
   

   Latency reduction. Suppose the quantum channel has decent bandwidth but high latency, e.g. boxes of qubits put on trucks being driven across the country. If Alice uses the truck to send a qubit to Bob, it'll take days for the message to arrive. But if the truck is loaded with EPR halves, then the truck can be dispatched early (or in a steady stream of trucks) before Alice even knows what she'll be sending with those EPR halves. Then, by using the classical internet to send the measurement results needed to complete the teleportation, qubits can be sent in tens of milliseconds instead of over days.

   
   


2. 
   

   Error recovery. If Alice sends a qubit to Bob and some noise on the quantum channel (e.g. the truck full of qubits falling off a cliff) scrambles the message, that qubit is gone. Too bad so sad. But, when performing teleportation, the qubits you send over the channel are just EPR halves. If one gets lost, you can make more. Just keep trying until one of the EPR halves makes it through the channel, then use it to teleport the actually-precious qubit across.

   
   


3. 
   

   Backwards transmission. Suppose the quantum channel is one-way-only, so Alice can send qubits to Bob but Bob can't send qubits to Alice. No problem, just use the channel to build up entanglement (by sending EPR halves) then use those to power teleportation from Bob to Alice. Teleportation turns any mechanism for creating entanglement (plus a classical channel, which we have thanks to the internet) into a mechanism for sending qubits.

   
   
   


4. 
   

   Quantum Computation. A bunch of gate constructions and fault tolerant quantum computation schemes use slightly-tweaked teleportation to apply operations (e.g. read https://arxiv.org/abs/1310.7290). Some designs of ion trap quantum computers use teleportation to combine small modules into a single larger computer. Etc.

   
   

wavefunction - What does superposition mean in quantum mechanics?

What does superposition mean in quantum mechanics?

When I say $A+B=C$ (forces). I can mean push something with force $A$ + force $B$ together, and that is same as I push it with force $C$.

But when I say wavefunction $A$ + $B$ is also a solution of Schrodinger equation, what do I mean? The physics between them obviously is not same. Is it just something pure mathematical?

general relativity - What would happen if somebody dropped a small black hole into the Earth?

Suppose we drop a small black hole, but large enough that it accumulates mass more quickly than it radiates it, into the Earth. What happens? This question has already been asked here and here, but the answers seem to be wrong or non-authoritative.

Many of the answers assume that the small black hole will eventually consume the Earth, resulting in a black hole with the mass of the Earth. A calculation shows that this is impossible due to conservation of angular momentum. The maximum angular momentum an Earth-mass black hole can have is $7.9 \times 10^{30}$ kg m$^2$/s. This is much less than the angular momentum of the Earth: $7.1×10^{33}$ kg m$^2$/s.

So what really would happen? The conservation of angular momentum shows it can't consume the entire Earth, but it will consume at least a small portion of it. How much would it consume? And what happens to the rest?

Answer

What exactly would happen depends on a lot of factors. For instance, you cannot quite assume perfect spherical symmetry of the Earth, and how exactly this symmetry is broken changes the outcome. Also, if the black hole is as massive as the Earth (and thus 9 mm in size), half of the planet's material is suddenly attracted more to the black hole than to Earth, so setting the mass precisely is also important. There are, however, a few general statements that we can give (I am discussing small, but macroscopic black holes):

1) There is nothing that can "hold" the black hole, the only force that can push and pull the black hole around is gravity and gaining momentum by absorbing matter (I assume the black hole is uncharged). If you approach close enough, any Earth material will shred to pieces smaller than the black hole due to tidal forces. The first direction of the black hole will thus be to start drilling downwards unless convinced otherwise by gravitational interaction.

2) The black hole will accrete, but not as much or as fast as you might think. Black holes are not universal vacuum machines, and from a few tens of Schwarzschild radii away (which is the case we are considering here) it can really be treated as a Newtonian point mass. Then the accretion rate will be roughly given by the Bondi-Hoyle-Lyttleton accretion formula $$\dot{M}_{\rm BHL} = \frac{\rho G^2 M^2}{c_{\rm s}^3 + v^3}$$ where $c_{\rm s}, \rho$ is the sound speed and density of the surrounding material respectively, $M$ the black hole mass, and $v$ the speed of the bulk of the material with respect to the black hole. The BHL formula takes into account both the fact that moving matter tends to "miss" the black hole (the $v$ part), but also that there will be a certain "choked up" region in the accretion flow known as the Bondi radius where too much pressure builds up and only a little bit ends up flowing through.

3) The accretion and the gravitational forces end up exerting an effective frictional force that tries to force the black hole to synchronize the motion with the surrounding material as $\dot{p} = \dot{M}_{\rm BHL}v$. Notice that the "frictional acceleration" $\dot{p}/M$ still depends on $M$, so the size/mass of the black hole determines its "frictional coefficient".

Now for the influence of the black hole mass as compared to the mass of the Earth.

If the black hole is sufficiently light (much lighter than the Earth), it will tear a decent hole into the Earth, sink into its core (slowed down by the "accretion friction"), and may live there for a prolonged time accreting at the BHL rate before triggering some sort of instability in Earth's structure.

There should, however, be an intermediate-range of masses such that the tidal forces are large enough to make the Earth break off a significant piece and the orbit of the black hole and the two or more pieces of the Earth correspond to a gravitational three(or more)-body problem. In such a problem, one of the masses can receive sufficient angular momentum to be ejected, the rest of the bodies to be left in a tighter orbit. This is especially true for lighter bodies in the system. So it is in principle possible that the black hole would be pushed away while leaving behind a spun-down shattered Earth. On the other hand, the same type of gravitational interaction can lead to a piece of the Earth to be ejected while making the black hole and the leftover piece in a tighter grip (which almost certainly involves shattering it to even smaller pieces).

At some point, only pieces of comparable or smaller mass are left in the vicinity of the black hole. This may have also been the case from the very beginning since the black hole might be about as massive as the Earth from the very start. The leftover pieces will then gradually be shattered and some of them may be ejected with higher angular momentum by collisions and many-body gravitational interaction. Eventually, the collisions cause the debris to flatten into a disk orbiting the central black hole. Occasional collisions between the pieces of debris will then cause further exchange of angular momentum and a fraction of the matter is continuously travelling outwards, away from the black hole, and other inwards, towards the black hole. Macroscopically, this is seen as spreading of the disk at its outside edge to higher radii, and further accretion onto the black hole at its inner edge, which is located roughly at the innermost stable circular orbit of the BH. In certain scenarios, the collisions within the disk may drop to such low rates that the disk "freezes out" into Saturn-type rings.

I have not talked about the radiative processes a lot. It is quite likely that at many points mentioned above, a considerable amount of heating may occur. This may come about A) purely from collisions, or B) from a triggered nuclear fission/fusion. I believe that scenario B) is unlikely, since we are starting from a black hole above the surface of the Earth, so the situation will be very asymmetric from the very start and there will thus be plenty of channels for the matter to find a way to escape such high compressions.

Either way, if a lot of radiation is released early on in the processes mentioned above, it can either heat up and evaporate many of the mentioned players, or completely blow them away by radiation pressure. Any gas component that is still bound to the system after that will very quickly circle towards the black hole and join an accretion disk around it. If an accretion disk has a large fraction of gas, it is unlikely it will ever "freeze out". Instead, it will keep the accretion process running while transforming the binding energy into heat and radiating it out as photons as we know it from active galactic nuclei or X-ray binaries.

---

In none of these scnearios, gravitational waves will be of any significance - unlike what has been stated in other answers. There are really quite general arguments that take you to the conclusion that if we are looking at systems say for as much as thousands of their orbital periods, gravitational waves can matter only for interacting systems of compact objects, that is, objects with radii not laughably far from their Schwarzschild radii at given mass. If all of the actors are not compact, they will simply smash into each other or get tidally disrupted before they can conspire to radiate enough gravitational waves to influence their dynamics in any appreciable manner.

What's the interpretation of Feynman's picture proof of Noether's Theorem?

On pp 103 - 105 of The Character of Physical Law, Feynman draws this diagram to demonstrate that invariance under spatial translation leads to conservation of momentum:

To paraphrase Feynman's argument (if I understand it correctly), a particle's trajectory is the path AB. Space is horizontal; time vertical.

Because of spatial translation symmetry, the path CD has the same action as AB. Because AB has stationary action, ACDB has the same action as well.

That means the action of AC and BD are the same. (Note that they are traversed in opposite directions on the path ACDB.) This is a conserved quantity, and it turns out to be the momentum.

My question is about the meaning of the action of AC and BD. These paths aren't physical trajectories; they represent infinite velocity. I tried thinking of the trajectory as if the velocity, as a function of time, has two delta functions in it. However, because the Lagrangian depends on $v^2$, I think this leads to infinite action for the horizontal segments.

Mathematically, I see that the symmetry here implies $\frac{\partial L}{\partial x} = 0$. Least action implies $\frac{\partial L}{\partial x} = \frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot{x}}$. Combining these shows that the momentum $\frac{\partial L}{\partial \dot{x}}$ is constant, but I don't quite understand the connection to the picture.

Also, Feynman doesn't describe how he knows the conserved quantity is the momentum. Is there a way to get this from the picture? Finally, if momentum is being conserved, why isn't the trajectory a straight line?

homework and exercises - Why can't transverse waves travel through a liquid?

Can someone explain why a longitude wave can pass through the liquid, but a transverse wave can't. And can someone recommend some good animation of these processes.

photons - $U(1)$ local gauge invariance in QED

While constructing Lagrangian of QED, we don't add the mass term for photon $\dfrac{1}{2} m^{2}A_{\mu}A^{\mu}$ because gauge invariance does not allow. I want to ask, whether "$\bf{Theoretically}$", is this the only reason we don't have mass term. I know why we need this term to vanish by using the experimental facts, but I feel like I am missing something while constructing the lagrangian for QED. Please stay within the domain of Quantum field theory and Lagrangian formulation while answering (if possible).

electromagnetism - Do magnets lose their magnetism?

I recently bought some buckyballs, considered to be the world's best selling desk toy. Essentially, they are little, spherical magnets that can form interesting shapes when a bunch of them are used together.

After playing around with these buckyballs for a while, I wondered: "Can these guys ever lose their magnetism?" Then I went a step further and thought, "How are magnets affected by the 2nd law of thermodynamics?"

So, how are magnets affected by the 2nd law of thermodynamics? Do they break down and lose their magnetism over time (like iron rusts over time)?

Answer

The second law of thermodynamics - about the increasing entropy - which is apparently what you're talking about - holds for any system. Permanent magnets are no exception.

A ferromagnet may look "more ordered" than a non-magnetic material because the spins are oriented in the same direction, rather than random directions. But physical systems may only try to maximize their entropy among configurations that conserve energy (much like the momentum, charge, and other conserved quantities). For ferromagnets, the configuration with spins oriented in random directions would have a much higher energy - because one reduces the energy by orienting the spins, elementary magnets, in the same direction.

So the spontaneous disappearance of the uniform electrons' spin would violate the energy conservation.

Among the configurations with the same energy, the magnet still tries to maximize its entropy. In particular, the heat is flowing from warmer pieces of the material to colder ones, and so on. More generally, the entropy never goes down, and that's the only general statement that follows from the second law of thermodynamics.

Ferromagnets are not special among physical objects that could have a higher entropy if you allowed the energy to increase. For example, any object would raise its entropy - the amount of disorder - if its temperature increased. But a higher temperature requires a higher energy, too. One can't violate the first law of thermodynamics (energy conservation) just because it would make it more straightforward to satisfy the second law. Both of them hold in Nature.

newtonian mechanics - Will momentum be conserved in case of electrostatic force?

If a charged particle approached another particle of same charge held by some external force, then by coulomb's law the particle will start electrostatic repulsion. Thus coming to a closest distance and stopping and thereafter, tracing its path back (the free particle). I know that energy is conserved here. The kinetic energy turning to Electrostatic Potential energy and thus back to its former state.

But my question is whether the momentum be conserved or not?

Edit:- Ignore Gravity, Radiation, Electromagnetic Field etc.

Answer

Momentum is definitely conserved, but “the system” you must consider includes:

1. the incident particle


2. the target particle


3. the object providing the “external force”


4. the electromagnetic field

I trust that you can understand why (3) is there, and (4) is there because accelerating charged bodies radiate and radiation carries momentum.

The need to deal with radiation makes the problem rather more difficult than the corresponding pure mechanics problem.

quantum mechanics - Don't electrons have discrete energy levels in semiconductors?

I am having a hard time understanding the relation between the fermi distribution of electrons in a semiconductor, and the fact the electron energy states are discrete.

The fermi distribution is supposed to give us the probability of an energy level being occupied by electrons. It looks like this:

yet electrons are supposed to have discrete energy states, and no electron is supposed to be in the band gap:

I must be missing something here, how is it that electrons have a fairly high probability of being in the band gap ?

Answer

There are two important contributions: First the density of states, which tells you the states, which can potentially be occupied. Then there is the Fermi-Dirac distribution, which tells you, which energies are occupied. The Fermi-Dirac distribution does not include allowed and forbidden states. You must fold it with the density of states, which is then called combined density of states. This function then tells you the distribution of occupied or unoccupied states.

newtonian mechanics - Where does the angular momentum of the solar system come from?

We inhabit a system with significant angular momentum:

http://www.zipcon.net/~swhite/docs/astronomy/Angular_Momentum.html

If our solar system formed by gravity gathering its material together to form the sun, proto-planetary disc, and eventually the planets, which all orbit in the same direction...

Where did this angular momentum come from in the first place, since angular momentum is conserved?

It does not seem possible to me that the formation of the solar system under gravity could impart this angular momentum on it, if it is a closed system. If it formed from a 'cloud of space dust', then it must have been present in that dust cloud, but where did the dust cloud get it from?

Answer

A collapsing gas cloud is an open system. It loses mass, energy and angular momentum as it collapses. Even if the net angular momentum of the cloud is zero, after the collapse the final planetary disk can have a significant net angular momentum, and the ejected material will have the opposite angular momentum. What can not happen, and that's where your intuition is correct, is that all the material in the original cloud collapses into the disk while rotating in the same direction.