spectroscopy - Temperature dependence of spectra

I have a question that is short and sweet:

Are spectra (both fluorescence and absorbance) of any molecule dependent on temperature?

In particular, is the spectral lineshape function of any molecule dependent on the temperature (so, kinetic energy) of the surroundings and possibly itself?

definition - What is the difference between the actual distance covered by a projected object and the displacement?

What is the difference between the actual distance covered by a projected object and the displacement?

A stone was projected with an angle of projection of 30 and it covered a horizontal distance of 6 meter find the velocity of the projection.

quantum chromodynamics - Mathematically, what is color charge?

A similar question was asked here, but the answer didn't address the following, at least not in a way that I could understand.

Electric charge is simple - it's just a real scalar quantity. Ignoring units and possible quantization, you could write $q \in \mathbb{R}$. Combination of electric charges is just arithmetic addition: $ q_{net} = q_1 + q_2 $.

Now to color charge. Because there are three "components", I am tempted to conclude that color charges are members of $\mathbb{R}^3$. However, I've read that "red plus green plus blue equals colorless", which seems to rule out this idea. I can only think that either:

• red, green and blue are not orthogonal, or


• "colorless" doesn't mean zero color charge (unlikely), or


• color charges don't combine in a simple way like vector addition

In formulating an answer, please consider that I know some mathematics (vectors, matrices, complex numbers, calculus) but almost nothing about symmetry groups or Lie algebras.

operators - The formal solution of the time-dependent Schrödinger equation

Consider the time-dependent Schrödinger equation (or some equation in Schrödinger form) written down as $$ \tag 1 i \partial_{0} \Psi ~=~ \hat{ H}~ \Psi . $$ Usually, one likes to write that it has a formal solution of the form $$ \tag 2 \Psi (t) ~=~ \exp\left[-i \int \limits_{0}^{t} \hat{ H}(t^{\prime}) ~\mathrm dt^{\prime}\right]\Psi (0). $$ However, this form for the solution of $(1)$ is actually built by the method of successive approximations which actually returns a solution of the form $$ \tag 3 \Psi (t) ~=~ \hat{\mathrm T} \exp\left[-i \int \limits_{0}^{t} \hat{H}(t^{\prime})~\mathrm dt^{\prime}\right]\Psi (0), \qquad t>0, $$ where $\hat{\mathrm T}$ is the time-ordering operator.

It seems that $(3)$ doesn't coincide with $(2)$, but formally $(2)$ seems to be perfectly fine: it satisfies $(1)$ and the initial conditions. So where is the mistake?

Answer

I) The solution to the time-dependent Schrödinger equation (TDSE) is

$$ \Psi(t_2) ~=~ U(t_2,t_1) \Psi(t_1),\tag{A}$$

where the (anti)time-ordered exponentiated Hamiltonian

$$\begin{align} U(t_2,t_1)~&=~\left\{\begin{array}{rcl} T\exp\left[-\frac{i}{\hbar}\int_{t_1}^{t_2}\! dt~H(t)\right] &\text{for}& t_1 ~<~t_2 \cr\cr AT\exp\left[-\frac{i}{\hbar}\int_{t_1}^{t_2}\! dt~H(t)\right] &\text{for}& t_2 ~<~t_1 \end{array}\right.\cr\cr ~&=~\left\{\begin{array}{rcl} \lim_{N\to\infty} \exp\left[-\frac{i}{\hbar}H(t_2)\frac{t_2-t_1}{N}\right] \cdots\exp\left[-\frac{i}{\hbar}H(t_1)\frac{t_2-t_1}{N}\right] &\text{for}& t_1 ~<~t_2 \cr\cr \lim_{N\to\infty} \exp\left[-\frac{i}{\hbar}H(t_1)\frac{t_2-t_1}{N}\right] \cdots\exp\left[-\frac{i}{\hbar}H(t_2)\frac{t_2-t_1}{N}\right] &\text{for}& t_2 ~<~t_1 \end{array}\right.\end{align}\tag{B} $$

is formally the unitary evolution operator, which satisfies its own two TDSEs

$$ i\hbar \frac{\partial }{\partial t_2}U(t_2,t_1) ~=~H(t_2)U(t_2,t_1),\tag{C} $$ $$i\hbar \frac{\partial }{\partial t_1}U(t_2,t_1) ~=~-U(t_2,t_1)H(t_1),\tag{D} $$

along with the boundary condition

$$ U(t,t)~=~{\bf 1}.\tag{E}$$

II) The evolution operator $U(t_2,t_1)$ has the group-property

$$ U(t_3,t_1)~=~U(t_3,t_2)U(t_2,t_1). \tag{F}$$

The (anti)time-ordering in formula (B) is instrumental for the (anti)time-ordered expontial (B) to factorize according to the group-property (F).

III) The group property (F) plays an important role in the proof that formula (B) is a solution to the TDSE (C):

$$\begin{array}{ccc} \frac{U(t_2+\delta t,t_1) - U(t_2,t_1)}{\delta t} &\stackrel{(F)}{=}& \frac{U(t_2+\delta t,t_2) - {\bf 1} }{\delta t}U(t_2,t_1)\cr\cr \downarrow & &\downarrow\cr\cr \frac{\partial }{\partial t_2}U(t_2,t_1) && -\frac{i}{\hbar}H(t_2)U(t_2,t_1).\end{array}\tag{G}$$

Remark: Often the (anti)time-ordered exponential formula (B) does not make mathematical sense directly. In such cases, the TDSEs (C) and (D) along with boundary condition (E) should be viewed as the indirect/descriptive defining properties of the (anti)time-ordered exponential (B).

IV) If we define the unitary operator without the (anti)time-ordering in formula (B) as

$$ V(t_2,t_1)~=~\exp\left[-\frac{i}{\hbar}\int_{t_1}^{t_2}\! dt~H(t)\right],\tag{H}$$

then the factorization (F) will in general not take place,

$$ V(t_3,t_1)~\neq~V(t_3,t_2)V(t_2,t_1). \tag{I}$$

There will in general appear extra contributions, cf. the BCH formula. Moreover, the unitary operator $V(t_2,t_1)$ will in general not satisfy the TDSEs (C) and (D). See also the example in section VII.

V) In the special (but common) case where the Hamiltonian $H$ does not depend explicitly on time, the time-ordering may be dropped. Then formulas (B) and (H) reduce to the same expression

$$ U(t_2,t_1)~=~\exp\left[-\frac{i}{\hbar}\Delta t~H\right]~=~V(t_2,t_1), \qquad \Delta t ~:=~t_2-t_1.\tag{J}$$

VI) Emilio Pisanty advocates in a comment that it is interesting to differentiate eq. (H) w.r.t. $t_2$ directly. If we Taylor expand the exponential (H) to second order, we get

$$ \frac{\partial V(t_2,t_1)}{\partial t_2} ~=~-\frac{i}{\hbar}H(t_2) -\frac{1}{2\hbar^2} \left\{ H(t_2), \int_{t_1}^{t_2}\! dt~H(t) \right\}_{+} +\ldots,\tag{K} $$

where $\{ \cdot, \cdot\}_{+}$ denotes the anti-commutator. The problem is that we would like to have the operator $H(t_2)$ ordered to the left [in order to compare with the TDSE (C)]. But resolving the anti-commutator may in general produce un-wanted terms. Intuitively without the (anti)time-ordering in the exponential (H), the $t_2$-dependence is scattered all over the place, so when we differentiate w.r.t. $t_2$, we need afterwards to rearrange all the various contributions to the left, and that process generate non-zero terms that spoil the possibility to satisfy the TDSE (C). See also the example in section VII.

VII) Example. Let the Hamiltonian be just an external time-dependent source term

$$ H(t) ~=~ \overline{f(t)}a+f(t)a^{\dagger}, \qquad [a,a^{\dagger}]~=~\hbar{\bf 1},\tag{L}$$

where $f:\mathbb{R}\to\mathbb{C}$ is a function. Then according to Wick's Theorem

$$ T[H(t)H(t^{\prime})] ~=~ : H(t) H(t^{\prime}): ~+ ~C(t,t^{\prime}), \tag{M}$$

where the so-called contraction

$$ C(t,t^{\prime})~=~ \hbar\left(\theta(t-t^{\prime})\overline{f(t)}f(t^{\prime}) +\theta(t^{\prime}-t)\overline{f(t^{\prime})}f(t)\right) ~{\bf 1}\tag{N}$$

is a central element proportional to the identity operator. For more on Wick-type theorems, see also e.g. this, this, and this Phys.SE posts. (Let us for notational convenience assume that $t_1

$$ A(t_2,t_1)~=~-\frac{i}{\hbar}\int_{t_1}^{t_2}\! dt~H(t) ~=~-\frac{i}{\hbar}\overline{F(t_2,t_1)} a -\frac{i}{\hbar}F(t_2,t_1) a^{\dagger} ,\tag{O}$$

where

$$ F(t_2,t_1)~=~\int_{t_1}^{t_2}\! dt ~f(t). \tag{P}$$

Note that

$$ \frac{\partial }{\partial t_2}A(t_2,t_1)~=~-\frac{i}{\hbar}H(t_2), \qquad \frac{\partial }{\partial t_1}A(t_2,t_1)~=~\frac{i}{\hbar}H(t_1).\tag{Q} $$

Then the unitary operator (H) without (anti)time-order reads

$$\begin{align} V(t_2,t_1)~&=~e^{A(t_2,t_1)} \\ ~&=~\exp\left[-\frac{i}{\hbar}F(t_2,t_1) a^{\dagger}\right]\exp\left[\frac{-1}{2\hbar}|F(t_2,t_1)|^2\right]\exp\left[-\frac{i}{\hbar}\overline{F(t_2,t_1)} a\right].\tag{R} \end{align}$$

Here the last expression in (R) displays the normal-ordered for of $V(t_2,t_1)$. It is a straightforward exercise to show that formula (R) does not satisfy TDSEs (C) and (D). Instead the correct unitary evolution operator is

$$\begin{align} U(t_2,t_1)~&\stackrel{(B)}{=}~T\exp\left[-\frac{i}{\hbar}\int_{t_1}^{t_2}\! dt~H(t)\right] \\~&\stackrel{(M)}{=}~:\exp\left[-\frac{i}{\hbar}\int_{t_1}^{t_2}\! dt~H(t)\right]:~ \exp\left[\frac{-1}{2\hbar^2}\iint_{[t_1,t_2]^2}\! dt~dt^{\prime}~C(t,t^{\prime})\right] \\ ~&=~ e^{A(t_2,t_1)+D(t_2,t_1)}~=~V(t_2,t_1)e^{D(t_2,t_1)}\tag{S}, \end{align}$$

where

$$ D(t_2,t_1)~=~\frac{{\bf 1}}{2\hbar}\iint_{[t_1,t_2]^2}\! dt~dt^{\prime}~{\rm sgn}(t^{\prime}-t)\overline{f(t)}f(t^{\prime})\tag{T}$$

is a central element proportional to the identity operator. Note that

$$\begin{align} \frac{\partial }{\partial t_2}D(t_2,t_1)~&=~\frac{{\bf 1}}{2\hbar}\left(\overline{F(t_2,t_1)}f(t_f)-\overline{f(t_2)}F(t_2,t_1)\right) \\ ~&=~\frac{1}{2}\left[ A(t_2,t_1), \frac{i}{\hbar}H(t_2)\right]~=~\frac{1}{2}\left[\frac{\partial }{\partial t_2}A(t_2,t_1), A(t_2,t_1)\right].\tag{U} \end{align}$$

One may use identity (U) to check directly that the operator (S) satisfy the TDSE (C).

References:

1. Sidney Coleman, QFT lecture notes, arXiv:1110.5013; p. 77.

newtonian mechanics - Why is the simple harmonic motion idealization inaccurate?

While in my physics classes, I've always heard that the simple harmonic motion formulas are inaccurate e.g. In a pendulum, we should use them only when the angles are small; in springs, only when the change of space is small. As far as I know, SHM came from the differential equations of Hooke's law - so, using calculus, it should be really accurate. But why it isn't?

Answer

The actual restoring force in a simple pendulum is not proportional to the angle, but to the sine of the angle (i.e. angular acceleration is equal to $-\frac{g\sin(\theta)}{l}$, not $-\frac{g~\theta}{l}$ ). The actual solution to the differential equation for the pendulum is

$$\theta (t)= 2\ \mathrm{am}\left(\frac{\sqrt{2 g+l c_1} \left(t+c_2\right)}{2 \sqrt{l}}\bigg|\frac{4g}{2 g+l c_1}\right)$$

Where $c_1$ is the initial angular velocity and $c_2$ is the initial angle. The term following the vertical line is the parameter of the Jacobi amplitude function $\mathrm{am}$, which is a kind of elliptic integral.

This is quite different from the customary simplified solution

$$\theta(t)=c_1\cos\left(\sqrt{\frac{g}{l}}t+\delta\right)$$

The small angle approximation is only valid to a first order approximation (by Taylor expansion $\sin(\theta)=\theta-\frac{\theta^3}{3!} + O(\theta^5)$).

And Hooke's Law itself is inaccurate for large displacements of a spring, which can cause the spring to break or bend.

operators - Quantum non-unitary transformation?

Let us say that I apply a non-unitary transformation $\def\ket#1{| #1\rangle} \def\braket#1#2{\langle #1|#2\rangle} \hat A$ to the ket's: $$\ket{\psi} \rightarrow \hat A \ket{\psi}$$ $$\ket{\phi}\rightarrow \hat A \ket{\phi}$$ Clearly in this case the probability: $$P=|\braket{\phi}{\psi}|^2$$ Will change. What physically is going on here? i.e. why for unitary operators we can perform such a transformation but for non-unitary operators we can't?

resource recommendations - Textbooks for Dyslexics: Atomic Physics

I am severely dyslexic (e.g. my reading ability is in the lowest 1st percentile) and as such I like textbooks with information clearly laid out and easily findable. An example of which is Nuclear and Particle Physics by W.S.C. Williams. To clarify I am looking for a book with (thanks to knzhou for helping me clarify this):

1. Boxes with important results (both mathematical and not).


2. Definitions clearly labeled (e.g. in a box or like Definition 1.2:).


3. Examples clearly labeled etc.

I am in general looking for books on a range of subjects, but to prevent the question been closed as too broad here I am looking for books on atomic physics.

---

_{This question stems from Books on Atomic Physics in the mathematical style? which was more ambiguous then I had intended it to be and was (rightly) interpreted differently from what I had intended. This is the question I had intended to ask. It is probably worth pointing out that $\text{dyslexic } \ne \text{ stupid}$ and book suggestions can still be very rigorous.}

cosmology - What if the size of the Universe doubled?

My question has a silly formulation, but I want to know if there is some sensible physical question buried in it:

1. Suppose an exact copy of our Universe is made, but where spatial distances and sizes are twice as large relative to ours. Would this universe evolve and function just as ours?

Since mass is proportional to volume which is distance cubed, but strength of a rope is only distance squared, I think not everything would scale proportionally.

2. 
   

   Does this mean that a universe with our physical laws that evolves like ours can only have one particular size?

   
   


3. 
   

   Or would the physical laws scale proportionally so that it evolves in the same way?

   
   


4. 
   
   

   If it has a particular size, what is it relative to?

   
   


5. 
   

   Another variation: suppose another exact copy of our Universe is made, but where everything happened twice as fast relative to ours. Would it evolve in the same way as ours?

   
   

Answer

If you think for a moment about how lengths and speeds in our universe are set (that is, independent of how we choose to measure them, by meters or seconds or whatever), you'll see that these must ultimately come from different ratios of fundamental constants.

I don't know and would be very surprised if there's a way to change these ratios so that all lengthscales or all timescales would change, since so much of what we observe actually comes from very complicated interacting systems acting at different timescales and lengthscales.

So the answer is probably no, there is no sensible physical question behind what you've asked, at least the way you've asked it.

You could ask of course, what would happen if some (dimensionless) physical constant was doubled or halved, and then we could chat some more.

Why do I emphasize dimensionless here? What if you asked "What happens if the speed of light was doubled?" Well, because the speed of light is what sets our time and length scales, we would observe no changes at all! In other words, think about what you can compare the speed of light to that doesn't depend on the speed of light itself. Just saying the numerical quantity changes is meaningless because the units we use to measure it rely (via a possibly long chain of dependencies) on the speed of light itself!

quantum mechanics - Where does de Broglie wavelength $lambda=h/p$ for massive particles come from?

I'm curious where the de Broglie relation $p=\frac{h}{\lambda}$ comes from?

I know that for light (which has no rest mass), the following is true:

$E=pc$ and $E=hf$

so,

$$pc=hf \Rightarrow p=\frac{hf}{c}=\frac{h}{\lambda}$$

But how is the expression $p=\frac{h}{\lambda}$ obtained for a massive particle where $E\neq pc$? I've read some people claim that the expression can be derived, and others saying it's an experimentally verified relationship.

Answer

This is covered nicely in the first (historically oriented) chapter of Weinberg's first book on quantum field theory. Here, he explains the motivation behind De Broglie's hypothesis as follows:

Of course, the main clue was provided by the analogy with radiation. However, there was more: If we want to try to implement wave-particle duality, it must be possible to describe particles my means of a wave with a phase which depends on position and time as:

$$\varphi(\vec x,t)=2\pi (\vec k \cdot \vec x -\nu t)$$

where $\vec k$ is the wave-vector and $\nu$ the frequency of the wave. Now, the key is Lorentz invariance. For this phase to stand a chance to be Lorentz invariant, we must have that $\vec k$ and $\nu$ transform under boosts like $\vec x$ and $t$, and hence like $\vec p$ and $E$. This forces them to be proportional, with an identical proportionality constant $\alpha$. From the Einstein relation $E=h\nu$ it is then natural to guess at $\alpha^{-1}=h$, and indeed this gives the correct De Broglie relations, including the ones you quote. Once we accept this, everything follows directly.

As we see, the De Broglie hypothesis can be made plausible, but it remains a hypothesis that needs to be verified experimentally (and cannot be proven).

lagrangian formalism - Caldeira-Leggett Dissipation: frequency shift due to bath coupling

I am trying to understand the Caldeira-Leggett model. It considers the Lagrangian

$$L = \frac{1}{2} \left(\dot{Q}^2 - \left(\Omega^2-\Delta \Omega^2\right)Q^2\right) - Q \sum_{i} f_iq_i + \sum_{i}\frac{1}{2} \left(\dot{q}^2 - \omega_i^2q^2\right)$$

where $Q$ is the generalised coordinate of the macro variable (an oscillator with natural frequency $\Omega$), $q_i$ are variables related to an array of harmonic oscillators each with natural frequency $\omega_i$.The first term describes the potentail and kinetic energies related to the macro degree of freedom, the second term describes the coupling using constants $f_i$, the third again describes potential and kinetic energies of the array of oscillators,

$$\Delta \Omega^2 = -\sum_i \left( \frac{f_i}{\omega_i} \right)^2$$

is the ad hoc term my first question relates to. The explanation I found goes

the quantity is inserted to cancel the frequency shift $$\Omega^2 \to \Omega^2 - \sum_{i} \left(\frac{f_i}{\omega_i}\right)^2$$ [...] the shift arises because a static Q displaces the bath oscillators so that $$f_i q_i = - \left(f_i^2 / \omega_i^2\right)Q$$ Substituing these values for the $f_i q_i$ into the potential terms shows that the effective potentail seen by $Q$ would have a "shifted" frequency.

I regretfully do not get it. I tried to get to the equation

$$f_i q_i = - \left(f_i^2 / \omega_i^2\right)Q$$

by considering the equation of motion in equilibrium, without success. Why would $\Omega$ be affected by the coupling? Any hint would be so appreciated.

general relativity - What is the relation between the metric tensor and the graviton?

In Zee's quantum theory in a nutshell, at the end of chapter I.10, he states that

the graviton is of course the particle associated with the field $g_{\mu\nu}$.

My understanding of quantum field theory is fuzzy, but I think I understood that each field assumes its values in an irreducible respresentation of the Lorentz group, the internal state space (or a representation of its universal covering, which is equivalent to a projective representation). Such representations are classified by their restriction to the $\text{SO}(3)$ or $\text{Spin}(3) = \text{SU}(2)$ subgroup, which is a spin representation and is classified by a single number called the spin. The graviton field has spin 2.

The metric tensor $g_{\mu\nu}$ is a section of the symmetric square of the (real) tangent bundle, which can be seen as a representation of the Lorentz group by the action $g\mapsto \Lambda g\Lambda^{T}$.

1. My first question would be: is this essentially the graviton field (as Zee seems to say, just to double check)?

The metric tensor is 6-dimensional, whereas a spin 2 field is 5-dimensional. Since for the Minkowski metric $\eta$ we have $\Lambda\eta\Lambda^T = \eta$, multiples of $\eta$ form a 1-dimensional subrepresentation, whose complement/quotient will be a 5-dimensional representation.

My other questions would be:

2. 
   

   Is this complement indeed the actual graviton field?

   
   


3. 
   

   What is its meaning or interpretation? Is it something like the non-flat part of the metric?

   
   



4. 
   

   If the preceding two were correct, does the fact that it is factored out of the graviton field mean that the graviton doesn't see unform dilations of spacetime? If so (or if not), what does that mean?

   
   

If my question is misguided, please point out why!

Answer

A spin-2 field in 4D is not five-dimensional. The standard spin-2 object transforms in the $(1,1)$-representations of the Lorentz group where the $(m,n)$-labels are half-integers labelling an equivalent representation of $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$, see the Wikipedia article on representations of the Lorentz group for more information. The spin of the $(m,n)$-representation is $m+n$ because the actual rotation algebra embeds diagonally into the two copies of $\mathfrak{su}(2)$ we are representing. A pure spin two field $h_{\mu\nu}$ is given by traceless and symmetric 2-tensor, which has 9 d.o.f. in 4D.

If we want to conceive of GR as a QFT (on ordinary flat space) in which we can speak of gravitons (an approach usually called Pauli-Fierz theory, we are essentially quantizing linearized gravity. I will not discuss the various inconsistencies and their fixes this approach needs to actually yield a somewhat consistent QFT), then you just write $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$ and treat $h$ as the dynamical field whose associated particles are gravitons. However, not all components of $h$ are physical - it enjoys a gauge symmetry $h_{\mu\nu}\mapsto h_{\mu\nu} + \partial_\mu X_\nu + \partial_\nu X_\mu$ for an arbitrary vector field $X$ - this is the infinitesimal version of the diffeomorphism invariance of GR.

At the end of the day, the gauge symmetry means that two d.o.f. of $h$ remain as physical - corresponding to the actual graviton and its possible polarizations.

---

JamalS's answer to a similar question explicitly counts the d.o.f. of $h$ and describes an alternative way to deduce the spin of the particle associated to $h$ in the quantum theory by looking at the Noether currents for the Lorentz symmetry applied to the part of the Lagrangian containing $h$.

electromagnetic radiation - Two possible directions of the B field in relation to the E field in radio waves. How induce the left-handed direction?

In vacuum the two field components of a radio wave are directed perpendicular on each over. $\mathbf{E}$, $\mathbf{B}$ and $\mathbf{k}$ form a right hand system ($\mathbf{k}$ is the direction of propagation).

Source

This matches the right hand rule for a current carrying wire and the induces magnetic field around the wire.

Now, for the reasons of symmetry, a radio wave also could be drawn with $\mathbf{E}$, $\mathbf{-B}$ and $\mathbf{k}$. Aplicated to the image above, the green fieldlines (which showing the direction of the magnetic field) simply would show in the opposite direction. How induce a left hand radio wave?

What is the definition of particle-hole symmetry in condensed matter physics?

People often talk about particle-hole symmetry in solid state physics. What are the exact definition and physics picture of particle-hole symmetry? How to define the density of particles and holes?

atoms - Is Palladium an exception?

I have been taught in school that atoms cannot have more than 8 electrons in the outer shell. Palladium atom's electron configuration is 2,8,18,18. Why isn't it 2,8,18,17,1 like the case of Platinum 2,8,18,32,17,1 or Nickel 2,8,17,1 and they are all in the same group?

Answer

As suggested by M.Sameer I convert my comment into an answer:

Dear M.Sameer: It seems that you are missing that the $n+\ell$ Madelung rule is not an exact result derived from first principles, but rather a rule of thumb, that holds for, say, approximately 95 percent of all the elements, with important exceptions, cf. this wikipedia page. Nevertheless, there are semi-rigorous theoretical arguments why such a rule of thumb should hold to a good approximation.

optics - The skin effect and the reflectivity of gold

I am simulating a waveguide in COMSOL, a FEM solver. My model looks like this (it is similar to a standard Quantum Cascade Laser geometry):

Therefore there is a very thin (30nm) layer of gold sandwiched by two regions consisting of Gallium Arsenide ($ \epsilon_r \simeq 13$). When I perform a mode analysis around 3THz (the frequency of my laser, $\simeq 100 \mu m$), the mode most likely to propagate (lowest losses, matching effective mode index) is the one below (E field norm is plotted here):

Now, I would expect this behaviour if the gold layer in between was thick enough, because the plasma frequency of the gold, according to the Drude model, is well above 3THz. The absorption coefficient is equal to

$$\alpha = \frac{2 \kappa \omega}{c}$$

and the skin depth, defined as $\delta = 2/ \alpha$, is around 40-50nm at this frequency. Therefore, due to the skin effect, the reflectivity of the thin gold layer would be very low and the wave would leak out into the bottom layer.

The E field does penetrate the gold, but decays very rapidly. This is the norm of E in the gold layer (0 to 0.03 on the y-axis), zoomed in a lot and the color range adjusted (notice $|E_{max}|>300$ on the original plot):

This is what I got from the support:

The reflection appears to be the result of the normal flux conservation boundary condition $Dy_1=Dy_2$. Since in the gold Er is much larger than in the material, $Ey_1<

But when I look at the displacement current (y component), I get a clearly discontinuous plot:

I got another reply, commenting on this issue:

This discontinuity is attributed to transition between a conductor and a dielectric. The D1-D2=pho_s condition is derived from the charge conservation law, J1-J2=dt phos_s, where J1 and J2 are the sum of displacement and conduction currents. In case that the conductivity in both domains is zero, you get the electric flux conservation. When passing through a conductor-dielectric boundary, the induction current becomes discontinuous and therefor, the displacement current will too be discontinuous. The displacement current is directly related to the flux density. Thus you see the jump in the normal flux component. If you plot emw.Ex or emw.Jy you will see that they are continuous. The different current components (emw.Jiy, emw.Jdy) and emw.Dy will be discountinuous.

If I understand correctly, this just means that there will be charge accumulation on the Au/GaAs interface due to the existence of a Schottky junction.

Am I wrong in my assumptions, am I misunderstanding the skin effect, am I plotting the wrong thing? If not, why am I getting the wrong result? From the technical point of view, the mesh in the software is small enough to resolve the thin layer, so it can't be the problem.

electromagnetism - Relationship between surface density and volume density

Often in an E&M problem, I'm having to "chop" an extended object into an infinite sum of smaller extended objects which I know more about to find a potential or electric field or whatever. The part I'm having trouble with is how to convert the surface density (line density) to volume density (surface density) and vice versa.

In one problem I did recently, I just thought maybe I should just collapse a dimension in volume density $\rho$ to get surface density $\sigma$ via: $\rho dr = \sigma$. It gave me the right answer, but I doubt that's the right way to think about it and that that formula will always work.

Right now I'm trying to "cut" a cylinder of uniform volume density $\rho$ into disks of uniform surface density $\sigma$. I thought maybe the right approach would be to relate the total charges. I've got $$Q_{\text{cylinder}}=\int \rho d\tau=\rho \pi r^2 h\\ \text{and}\\ Q_{\text{disk}}=\int \sigma dS=\sigma \pi r^2\; .$$ However then I'm at a loss as to how I should relate $Q_\text{cylinder}$ and $Q_\text{disk}$. What's the general process here?

Answer

I have personally also contemplated this issue, and have come up with a simple solution that is satisfactory, to me at least. I'm sure this can also be found in many textbooks. In general, we have

$$\tag{$\star$} Q=\int \rho\ d\tau$$

because we are considering a three dimensional space. Intuitively, we feel it should be possible to talk about a three dimensional charge distribution in every case. The question is how to conceptualize this when discussing surface, line or point charges. The solution comes in the form of the Dirac delta distribution (or function, depending who you ask).

Let's take a look at an example: Consider a 2-sphere of radius $R$, with some charge distribution $\sigma(\theta,\phi)$ on it. What is the three dimensional charge distribution $\rho(r,\theta,\phi)$ corresponding to this situation? Like I said, we have to use the Dirac delta:

$$\rho(r,\theta,\phi)=\delta(r-R)\sigma(\theta,\phi)$$

Now, $(\star)$ gives us: $$ Q=\int\rho\ d\tau=\int_{0}^{2\pi}\int_{0}^\pi\int_0^\infty \rho\ r^2\sin\theta\ dr\ d\theta\ d\phi=R^2\int_{0}^{2\pi}\int_{0}^\pi\sigma\ \sin\theta\ d\theta\ d\phi$$

Similarly, when considering a line or point charge, one uses two or three Dirac delta's to describe the distribution in 3-space.

newtonian mechanics - Slinky base does not immediately fall due to gravity

Why does the base of this slinky not fall immediately to gravity? My guess is tension in the springs is a force > mass*gravity but even then it is dumbfounding.

Answer

What an awesome question! By the way, as far as I know, the original video is here for those interested.

One key to understanding this is the following fact from classical mechanics that is a version of Newton's second law for systems of particles:

The net external force acting on a system of particles equals the total mass $M$ of the system times the acceleration of its center of mass $$ \mathbf F_{\mathrm{ext},\mathrm{net}} = M\mathbf a_\mathrm{cm} $$ In the case of the slinky, which we can model as a system of many particles, the net external force on the system is simply the weight of the slinky. This is just given by its mass multiplied by $\mathbf g$, the acceleration due to gravity, so from the statement above, we get $$ M\mathbf g = M\mathbf a_\mathrm{cm} $$ so it follows that $$ \mathbf a_\mathrm{cm} = \mathbf g $$ In other words we have shown that

The center of mass of the slinky must move as if it is a particle falling under the influence of gravity.

However, there is nothing requiring that the individual particles in the system must move as though they are each falling freely under influence of gravity. This is the case because there are interactions between the particles that affect their motion in addition to the force due to gravity. In particular, there is tension in the slinky, as you point out.

You are absolutely correct that the bottom of the slinky does not move because the tension of the rest of the slinky pulling up balances the force due to gravity pulling down until the moment that the slinky is fully compressed and the whole thing falls with the acceleration due to gravity. Regardless, the center of mass is moving as though it is freely falling the whole time.

By the way, there are some nice comments about this experiment from the angle of wave propagation on physics.SE user @Mark Eichenlaub's blog which can be found here.

Interaction between joint qubit quantum system

Consider the following interaction Hamiltonian $$H = \hbar \mu \sigma_{x} \otimes \sigma_x = \hbar \mu ( |01 \rangle \langle 1 0 | + |10\rangle\langle 01|)$$ acting on the joint states of qubits $\rho_{prim} \otimes \rho_{aux}$ for $t = \frac{\pi}{2 \mu}$. It is stated that if the primary and auxiliary systems (respectively $\rho_{prim}$ and $\rho_{aux}$) are in the state $|0\rangle$ then the interaction doesn't change the primary but if the primary is in state $|1\rangle$ and auxiliary in state $|0\rangle$ then the primary flips to $|0\rangle$.

For the first case my revised working is as follows: We have $$e^{-i\frac{\pi \sigma_x \otimes \sigma_x}{2}}[|0\rangle \langle0 |\otimes|0\rangle \langle0|]e^{i\frac{\pi \sigma_x \otimes \sigma_x}{2}}$$ where the state of the primary is $$e^{-\frac{\pi \sigma_x}{2}}|0\rangle = \begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix} \begin{pmatrix} 1 \\0\\ \end{pmatrix} = \begin{pmatrix} 0 \\ -i \end{pmatrix}$$

Answer

Evolving a state $\rho$ according to an Hamiltonian $H$ does not work that way: $H\rho$ is not the evolved state (nor, in general, even a state at all).

The evolution with the Hamiltonian $H$ for time $t$ is described by the unitary operator $e^{-itH}$. To evolve a density matrix you have to compute $e^{-itH}\rho e^{itH}$.

quantum field theory - How does Hawking radiation grow as a black hole evaporates?

The temperature of Hawking radiation is inversely proportional to the mass of a black hole, $T_{\rm H}\propto M_{\rm BH}^{-1}$, and so as the black hole shrinks the temperature of the radiation should grow. What happens as the mass shrinks to zero?

Naive application of the above formula leads to arbitrarily energetic Hawking radiation. Do assumptions in the Hawking Radiation derivation just break down in this limit? If so, is it known how to handle this limit?

particle physics - Solar Neutrino Problem

I am not sure if this has been asked recently, but has there been any headway into solving the solar neutrino problem? I recently completed the Particle Physics for non-Physicists by Professor Stephen Pollack. It is from 2006 and I was wondering if the issue had been solved yet.

In it he stated that 66% of the expected neutrinos were being detected. He postulated that the neutrinos could possibly oscillate between types. However wouldn't a tau neutrino and muon neutrino be detected as well?

Answer

However wouldn't a tau neutrino and muon neutrino be detected as well?

Yes, and an experiment that led to a Nobel prize was done

. All of the solar neutrino detectors prior to SNO had been sensitive primarily or exclusively to electron neutrinos and yielded little to no information on muon neutrinos and tau neutrinos.

In 1984, Herb Chen of the University of California at Irvine first pointed out the advantages of using heavy water as a detector for solar neutrinos.Unlike previous detectors, using heavy water would make the detector sensitive to two reactions, one reaction sensitive to all neutrino flavours, the other reaction sensitive to only electron neutrino. Thus, such detector can measure neutrino oscillations directly.

....

On 18 June 2001, the first scientific results of SNO were published, bringing the first clear evidence that neutrinos oscillate (i.e. that they can transmute into one another), as they travel in the sun. This oscillation in turn implies that neutrinos have non-zero masses. The total flux of all neutrino flavours measured by SNO agrees well with the theoretical prediction. Further measurements carried out by SNO have since confirmed and improved the precision of the original result.

Italics mine.

homework and exercises - Inserting metal into parallel plate capacitor

The plates of an isolated parallel plate capacitor with a capacitance C carry a charge Q. The plate separation is d. Initially, the space between the plates contains only air. Then, an isolated metal sheet of thickness 0.5d is inserted between, but not touching, the plates. How does the potential difference between the plates change as a result of inserting the metal sheet?

The answer is: The potential difference will decrease. How do I get this result, it seems right but how do I reason it out?

$$V = Ed$$

There was mention about "E is the same outside the conductor. But the field inside is 0.". But in the formula, E is the field outside conductor right? If so it remains the same? Then $d = 0.5d$?

classical mechanics - Stiffness tensor

Let's have a stiffness tensor:

$$ a^{ijkl}: a^{ijkl} = a^{jikl} = a^{klij} = a^{ijlk}. $$

It has a 21 independent components for an anisotropic body.

How does body symmetry (cubic, hexagonal etc.) change the number of independent components of the tensor? For example, for cubiс symmetry it has three components. How to explain it?

Update.

Is the explanation a simple realization of idea $$ a_{ijkl}' = \beta_{im}a^{m}\beta_{jt}a^{t}\beta_{k f}a^{f}\beta_{ld}a^{d} = a_{ijkl}, $$ where $\beta_{\alpha \beta}$ is a components of a matrix $\beta$ for rotation around z-, x-, y-axis at the same time?

electric circuits - Capacitor charging, work done by battery on changing polarity

A uncharged capacitor $C$ is connected to a battery with potential $V$. It becomes fully charged and has a charge $Q=CV$ stored on it.

Now the polarity of the battery is reversed. The capacitor will have the charge $Q$ still but with polarity reversed too.

My question is: What is the work done by the the battery?

According to me, during 1st charging it will do a work of $QV$. The energy of the capacitor is also not changing, then what is the work done to change its polarity?

thermodynamics - How does a canvas water bag cool water?

I was reading about this water bottle by Botl that behaves like a canvas water bags to keep water cool. I found out that this idea is an old idea and cars would drive with water bags in front as shown below.

Here is what I found out about them: it seeps water through the bag and evaporates, causing the water to cool inside. Here is where I need some help:

1. 
   

   Why does the water seep through the bag? Is it because the water molecules have a Maxwell speed distribution and only the fastest molecules seep out?

   
   


2. 
   

   What mechanism is occurring for heat being removed from the water to evaporate the water on the bag’s surface? Doesn't radiation also contribute here?

   
   



3. 
   

   Why does moving increases the cooling effect faster?

   
   

Answer

As far as I understand the idea is very simple:

as the bottle is covered in a film of water (due to a leakage of the container), this can evaporate. Due to the Boltzman distribution this will also happen well below $100°C$. The process of evaporating though takes energy from the system (called latent heat). This cools down the bottle.

If the surrounding air is "filled with water" it is harder for the liquid to evaporate and thus slowing the process, so always bringing fresh air to the system helps.

This is basically the concept of sweating as well, where you can immediately feel the effect of wind blowing on your moist skin.

Hope this helps and I didn't miss something.

electromagnetic radiation - How can I create hindrances to radio waves?

How can I create hindrances to radio waves?

electromagnetic radiation - How does a receiving antenna get an induced electric current?

From this question, I've noted that an electromagnetic field carries no electric charge but it has two components:

• Electric field


• Magnetic field

Now what I failed to understand is how does the receiving antenna get an induced small voltage. Is it due to the electric or the magnetic field component of EMF? and How?

Answer

If you put an electron in an electric field it moves and this is how a radio wave induces a voltage in the aerial. The oscillating electric field of the waves causes electrons to move in the metal of the aerial and this generates the voltage.

Aerials are normally designed so they have a natural resonance at the frequency of the radio wave, so the strength of the potential gets amplified by the resonance.

Atom Theory vs Quantum Physics

This never really occurred to me until now, so maybe it does not categorize as a really important question, but, according to Quantum Mechanics, anything that is not observed exists as a probability until observed. So with that said, I would like to know if that theory has any any contradictions with atom theory. Atom theory is pretty straight forward but because you cannot see atoms on a macro or even a microscopic scale (only nano or smaller), how does that work with quantum physics. You cannot observe atoms with your eyes even if you're looking at something with mass, so does that mean (supposedly) that atoms are only probability until you can observe them with the proper equipment, even though on a smaller scale thats what the things you are observing are made up of?

homework and exercises - How do I use the signs in the spring force equation?

The spring force equation is F = - (k)(x). Assuming right direction of axis to be +ve, and left to be -ve, should we use the values of force and displacement, accordingly? And also, in this question -

"When a block is pulled out to x = +4.0 cm from 0.0 cm, we must apply a force of magnitude of 360 N to hold it here. Then find the spring constant."

So using F = - (k) (x) , we get (k) = - (F) (x), substituting values, k = - (360N) (4.0cm) . I.e., the spring constant comes out to be negative, why is it so? Please help , I am really confused with the sign convention.

homework and exercises - Write Equation of Motion in Polar Coordinates

a point of mass $m$ moves on the $xy-$Plane and has the potential energy $U(x,y)=\frac{A}{2}(x^2+y^2) + B$ with $A,B > 0$ being two constants.

Let $\vec{r}$ denote the position of the mass on the $xy-$Plane.

We knoe that $m\ddot{\vec{r}}=-\nabla U(x,y)$

We first write $U(r,\varphi)=\frac{A}{2}r^2 + B$

so we get $\nabla U(r,\varphi)=\begin{pmatrix} Ar \\ 0 \end{pmatrix}$

We now calculate $\ddot{\vec{r}}$ in polar coordinates.

We have

$\vec{r}=r\begin{pmatrix}\cos(\varphi) \\ \sin(\varphi)\end{pmatrix}$

$\dot{\vec{r}}=\dot{r}\begin{pmatrix}\cos(\varphi) \\ \sin(\varphi)\end{pmatrix} + r\begin{pmatrix}-\sin(\varphi) \\ \cos(\varphi)\end{pmatrix}\dot{\varphi}$

$\ddot{\vec{r}}=\ddot{r}\begin{pmatrix}\cos(\varphi) \\ \sin(\varphi)\end{pmatrix} + 2\dot{r} \begin{pmatrix}-\sin(\varphi) \\ \cos(\varphi)\end{pmatrix}\dot{\varphi} - r\begin{pmatrix}\cos(\varphi) \\ \sin(\varphi)\end{pmatrix}\dot{\varphi}^2 + r\begin{pmatrix}-\sin(\varphi) \\ \cos(\varphi)\end{pmatrix}\ddot{\varphi}$

we put everything together:

$\ddot{\vec{r}}=\ddot{r}\begin{pmatrix}\cos(\varphi) \\ \sin(\varphi)\end{pmatrix} + 2\dot{r} \begin{pmatrix}-\sin(\varphi) \\ \cos(\varphi)\end{pmatrix}\dot{\varphi} - r\begin{pmatrix}\cos(\varphi) \\ \sin(\varphi)\end{pmatrix}\dot{\varphi}^2 + r\begin{pmatrix}-\sin(\varphi) \\ \cos(\varphi)\end{pmatrix}\ddot{\varphi}=\frac{-1}{m}\begin{pmatrix} Ar \\ 0 \end{pmatrix}$

(*)Since $\begin{pmatrix}\cos(\varphi) \\ \sin(\varphi)\end{pmatrix}=\hat{e}_r$ and $\begin{pmatrix}-\sin(\varphi) \\ \cos(\varphi)\end{pmatrix}=\hat{e}_\varphi$ and also $\begin{pmatrix} Ar \\ 0 \end{pmatrix}=Ar\hat{e}_r + 0\cdot \hat{e}_\varphi$ we get:

(**)$\begin{pmatrix}\ddot{\vec{r}}-r\dot{\varphi}^2 \\ 2\dot{r}\dot{\varphi} + r\ddot{\varphi}\end{pmatrix}=\frac{-1}{m}\begin{pmatrix} Ar \\ 0 \end{pmatrix}$

My lecture notes now write it down like this:

(***)$\begin{pmatrix}\ddot{\vec{r}} \\ r\ddot{\varphi}\end{pmatrix}=\begin{pmatrix} \frac{-Ar}{m} + r\dot{\varphi}^2\\ -2\dot{r}\dot{\varphi} \end{pmatrix}$

Question 1: Is $\begin{pmatrix}-\sin(\varphi) \\ \cos(\varphi)\end{pmatrix}=\hat{e}_\varphi$ true or is it rather $\begin{pmatrix}-\sin(\varphi) \\ \cos(\varphi)\end{pmatrix}=-\hat{e}_\varphi$?

Question 2: Is there a reason why we wrote ( ** ) as ( * ** )? Does the left side of (***) tell us something?

Question 3: Is it possible that my potential depends on $r$ and $\varphi$? And if so, could I write down the general formula like that:

$\begin{pmatrix}m\ddot{\vec{r}} \\ mr\ddot{\varphi}\end{pmatrix}=\begin{pmatrix} K_r(r,\varphi) + r\dot{\varphi}^2\\ K_\varphi(r,\varphi)-2m\dot{r}\dot{\varphi} \end{pmatrix}$

With $K(r,\varphi)$ being a conservative force field (so $K=-\nabla U$ applies)

Question 4: Why don't I have to transform the constants?

Answer

Question 1:

It's up to you to chose a basis to your problem. They can differ only by a global fase (such a minus sign, since $e^{i\pi}=-1$).

Question 2:

As you might see, both expressions are the same. It's up to you chose the one you prefer. However, when you get something equals to a constant is always a good thing, since means that a quantity is conserved (in your case, $\frac{d}{dt}2\dot{r}\dot{φ}+r\ddot{φ}$)

Question 3:

you can get (at least, theoretically) a potential $V=V(r,\varphi)$ and there will be a general formula (take into account that $\nabla U=\frac{\partial U}{\partial r}+\frac{1}{r}\frac{\partial U}{\partial \varphi}$)

visible light - Why does sunlight cause colors to fade?

If you leave something outside, its colors seem to inevitably fade or bleach due to exposure. Is this due to UV absorption? What sort of mechanism causes this - is it that man-made dyes deform on a molecular level? Are there notable materials that are exceptions?

To turn this string of questions into a focused query, let me refer back to the title:

Why does sunlight cause colors to fade?

Answer

This is more chemistry than physics... The color of a material is due to an interaction of the light with chemical bonds (usually double bonds) in the dye. The UV component of sunlight tends to knock electrons out of double bonds and can in time cause changes to the chemical composition which we experience as bleaching.

For a more thorough and authoritative answer I would recommend asking this question on the chemistry stack exchange.

Book covering differential geometry and topology for physics

I'm interested in learning how to use geometry and topology in physics. Could anyone recommend a book that covers these topics, preferably with some proofs, physical applications, and emphasis on geometrical intuition? I've taken an introductory course in real analysis but no other higher math.

Answer

If you want to learn topology wholesale, I would recommend Munkres' book, "Topology", which goes quite far in terms of introductory material.

However, in terms of what might be useful for physics I would recommend either:

• Nakahara's "Geometry, Topology and Physics"


• Naber's "Topology, Geometry and Gauge Fields: Foundations"

Personally, I haven't read much of Nakahara, but I've heard good things about it, although it may presuppose too many concepts. I've read selections of Naber and it seems fairly well written and understandable and starts from first principles, but again, it may not focus as much on the fundamentals, if that's what you're looking for.

homework and exercises - Why can I assume the force to be constant in this particular interval?

If I have force, or any function $f(z)$, I was told that I can assume it to be constant only in the interval $dz$.

However, in this case, I had to calculate the work done by the spring force as a function of $y$

Over here, I assumed the spring force, which is a function of its elongation $x$ ($F = -kx$) to be constant in the interval $dy$ and integrated and this gave me the correct answer

I want to know why the error vanished over here. Shouldn't spring force only be constant in the interval $dx$ and not $dy$?

I also want to know, in general, if I have a function, how to decide whether it is constant in some particular interval/in which cases the error will vanish as I take the limit and integrate.

Note: I do know I can assume a function $f(x)$ to be constant in the interval $[x,x+dx)$ while integrating, but over here I've assumed it to be constant in the interval $dy$. I want to know why I can do that and also if I can assume a force/function to be constant in any infinitesimal interval such as $Rdθ$, $dy\over cosϕ$,$dz$ etc.

Gravitational Constant in Newtonian Gravity vs. General Relativity

From my understanding, the gravitational constant $G$ is a proportionality constant used by Newton in his law of universal gravitation (which was based around Kepler's Laws), namely in the equation $F = \frac{G\cdot M\cdot m}{r^2}$. Later, Einstein set forward a different theory for Gravity (based around the equivalence principle), namely General Relativity, which concluded that Newton's law was simply a (rather decent) approximation to a more complex reality. Mathematically speaking, Einstein's Theory was completely different from Newton's Theory and based around his Field equations, which also included $G$ in one of it's terms.

How come two different theories that stemmed from completely different postulates end up having this same constant $G$ with the same numerical value show up in their equations? What exactly does $G$ represent?

Answer

Since in the limit of weak gravitational fields, Newtonian gravitation should be recovered, it is not surprising that the constant $G$ appears also in Einstein's equations. Using only the tools of differential geometry we can only determine Einstein's field equations up to an unknown constant $\kappa$: $$G_{\mu\nu} = \kappa T_{\mu\nu}.$$ That this equation should reduce to the Newtonian equation for the potential $\phi$, $$\nabla^2 \phi = 4\pi G\rho \tag{1}$$ with $\rho$ the density fixes the constant $\kappa = \frac{8\pi G}{c^4}. \tag{2}$

In detail, one assumes an almost flat metric, $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$ where $\eta_{\mu\nu}$ is flat and $h_{\mu\nu}$ is small. Then from writing down the geodesic equation one finds that if $h_{00} = 2\phi/c^2$, one obtains Newton's second law, $$\ddot{x}^i = -\partial^i \phi. \tag{3}$$ Using (3) and taking $T_{\mu\nu} = \rho u_\mu u_\nu$ for a 4-velocity $u_\mu$ with small spatial components, the $00$ component of the field equations (2) is $$2\partial^i \partial_i \phi /c^2 = \kappa \rho c^2.$$ In order to match this with (1), we must have $\kappa = \frac{8\pi G}{c^4}$. (The detailed calculations here are, as is often the case in relativity, rather lengthy and boring, so they are omitted.)

homework and exercises - EM vector potential

We can write the electromagnetic field tensor as $$\begin{bmatrix} 0 & -E_x/c & -E_y/c & -E_z/c \\ E_x/c & 0 & -B_z & B_y \\ E_y/c & B_z & 0 & -B_x \\ E_z/c & -B_y & B_x & 0 \end{bmatrix} = F^{\mu\nu}.$$ Erick J. Weinberg, Classical Solutions in Quantum Field Theory: Solitons and Instantons in High Energy Physics (p. 43), states:

For a static solution with vanishing electric fields $F_{0j}$, and hence $A_0 = 0$, [...]

How can this be proven?

Answer

EDIT: Answer was incorrect. So it was removed

wavefunction - The role of the "observer" in quantum theory?

I've been reading about different ideas of getting the observer out of quantum theory. One Sean Carroll seems to like is Everett's many worlds concept.

I had a thought that I need someone here to shoot down. It could be something as basic as the double slit experiment that proves me to be wrong. Here's my thought:

As I see it, we never really "measure" the position of something. We sense a photon that is emitted by what we are "looking" at. Why couldn't the wave function of a particle just break down at the moment it emits the photon at a specific point in space-time regardless of whether there is an observer or not. After the emission, the particle location would go back to being a probability function.

What am I missing?

Answer

Quantum mechanics exists as a theory because there have been experimental observations that could not be modeled mathematically using Newtonian physics and Maxwell's equations.

It has a number of postulates intrinsic in every quantum mechanical theory,(theory= a general mathematical model which is descriptive and predictive) and in particle physics the symmetries and elementary particles in the table are also a priori assumptions equivalent to axioms. They are necessary to fit existing data and very successful in predicting future measurements.

In classical physics a measurement is an observation. It does not need an observer, most often is an instrument and data registered, now digitally in computers. Then the physicist thinking about those data is a live observer, but he/she has nothing to do with the origins of the data, except in setting up the original experiment.

This disposes of the "observer" as a conscious entity.

Quantum mechanics is a probabilistic theory. It can only predict probabilities of interactions . One gets scattering events accumulated so as to check a theoretically predicted distribution so as to validate or falsify the theory:

Here is one pi mu e event in the bubble chamber.

The beam , probably a K- 10GeV/c beam , hits a proton in the hydrogen bubble chamber , and several particles come out. One of them is identified as a pi mu e decay (by the ionization track the mass of the backwards curling track can be identified as a pi , etc).

This is one event, and probably thousands of people like us have observed it. In the film tapes where it is recorded there are probably tens of thousand like this.

From this one instance we may start building up a probability distribution for the decays of charged pions and muons. One instance. We would have to accumulate versus energy and momentum a statistically significant number of these decays to be able to check against energy and decay lifetimes as calculated/predicted quantum mechanically.

One is forced to the conclusion that , quantum mechanically, observer= recorded-interaction. Once an interaction is recorded it has been observed. The mathematics leading to the theoretical fit should not be anthropomorphized. One can visualize with Feynman diagrams the interactions and how their use will build up the probability distribution, but in the end the Lydian stone are the measurements, one by one building the probability distributions which fit the quantum mechanical predictions.

Beyond Hamiltonian and Lagrangian mechanics

Lagrangian and Hamiltonian formulations are the bedrock of particle and field theories, produce the same equations of motion, and are related through a Legendre transform. Are there more such mathematical objects that are equivalent, or are these two in some way unique? If so, why are there two equivalent systems, rather than a single (or more)?

Answer

There is also the Routhian formalism of mechanics which is described as being a hybrid of Lagrangian and hamiltonian mechanics. The Routhian is defined as $$R = \sum_{i=1}^n p_i\dot{q}_i - L$$ You can learn more about it by clicking this link for Wikipedia's description of it.

Reading more in regards to the routhian because I was bored, I realized it is defined as the partial Legendre transform of the Lagrangian and also in the language of differential geometry it is defined similarly to the Lagrangian as $$R^\mu : TM \to \mathbb{R}$$ where $$R^\mu(q, \dot{q}) = L(q, \dot{q}) - \langle A(q, \dot{q}), \mu\rangle$$ where $A$ is the mechanical connection term. You can read more about it in this pdf.

thermodynamics - Can a liquid boil in a closed container?

The title says it all. When one fills a container (with water for example) and closes it so no air can escape from it, then heats the container continuously, can the liquid boil or completely evaporate?

What does the continuum hypothesis of fluid mechanics mean?

I'm a bit confused by the continuum hypothesis stating that fluid are continuous objects rather than made out of discrete objects.

Say for $\rho (x,t)$ (density) is there more than one fluid particle at $x$ or less than one.

forces - Is gecko-like friction Coulombic? What is the highest known Coulombic $mu_s$ for any combination of surfaces?

Materials with large coefficients of static friction would be cool and useful. Rubber on rough surfaces typically has $\mu_s\sim1-2$. When people talk about examples with very high friction, often they're actually talking about surfaces that are sticky (so that a force is needed in order to separate them) or wet (like glue or the tacky heated rubber used on dragster tires and tracks). In this type of example, the usual textbook model of Coulombic friction with $\mu_s$ and $\mu_k$ (named after Mr. Coulomb, the same guy the unit was named for) doesn't apply.

I was looking around to try to find the largest $\mu_s$ for any known combination of surfaces, limiting myself to Coulombic friction. This group at Berkeley has made some amazing high-friction surfaces inspired by the feet of geckos. The paper describes their surface as having $\mu\sim5$. What's confusing to me is to what extent these surfaces exhibit Coulombic friction. The WP Gecko article has pictures of Geckos walking on vertical glass aquarium walls, and it also appears to imply a force proportional to the macroscopic surface area. These two things are both incompatible with the Coulomb model. But the Berkeley group's web page shows a coin lying on a piece of glass that is is nearly vertical, but not quite, and they do quote a $\mu$ value. This paper says gecko-foot friction involves van der Waals adhesion, but I think that refers to microscopic adhesion, not macroscopic adhesion; macroscopic adhesion would rule out the Coulomb model. (The WP Gecko article has more references.)

So my question is: what is the highest coefficient of Coulombic static friction ever observed, and does the Berkeley group's substance qualify?

quantum electrodynamics - Deriving the Coulomb force equation from the idea of photon exchange?

Since Newton's law of gravitation can be gotten out of Einstein's field equatons as an approximation, I was wondering whether the same applies for the electromagnetic force being the exchange of photons. Is there an equation governing the force from the exchange of photons? Are there any links which would show how the Coulomb force comes out of the equations for photon exchange? I know that my question is somewhat similar to the one posted here The exchange of photons gives rise to the electromagnetic force, but it doesn't really have an answer to my question specifically.

Answer

The classical Coulomb potential can be recovered in the non-relativistic limit of the tree-level Feynman diagram between two charged particles.

Applying the Born approximation to QM scattering, we find that the scattering amplitude for a process with interaction potential $V(x)$ is

$$\mathcal{A}(\lvert p \rangle \to \lvert p'\rangle) - 1 = 2\pi \delta(E_p - E_{p'})(-\mathrm{i})\int V(\vec r)\mathrm{e}^{-\mathrm{i}(\vec p - \vec p')\vec r}\mathrm{d}^3r$$

This is to be compared to the amplitude obtained from the Feynman diagram:

$$ \int \mathrm{e}^{\mathrm{i}k r_0}\langle p',k \rvert S \lvert p,k \rangle \frac{\mathrm{d}^3k}{(2\pi)^3}$$

where we look at the (connected) S-matrix entry for two electrons scattering off each other, treating one with "fixed" momentum as the source of the potential, and the other scattering off that potential. Using the Feynman rules to compute the S-matrix element, we obtain in the non-relativistic limit with $m_0 \gg \lvert \vec p \rvert$

$$ \langle p',k \rvert S \lvert p,k \rangle \rvert_{conn} = -\mathrm{i}\frac{e^2}{\lvert \vec p -\vec p'\rvert^2 - \mathrm{i}\epsilon}(2m)^2\delta(E_{p,k} - E_{p',k})(2\pi)^4\delta(\vec p - \vec p')$$

Comparing with the QM scattering, we have to discard the $(2m)^2$ as they arise due to differing normalizations of momentum eigenstate in QFT compared to QM and obtain:

$$ \int V(\vec r)\mathrm{e}^{-\mathrm{i}(\vec p - \vec p')\vec r}\mathrm{d}^3r = \frac{e^2}{\lvert \vec p -\vec p'\rvert^2 - \mathrm{i}\epsilon}$$

where Fourier transforming both sides, solving the integral and taking $\epsilon \to 0$ at the end will yield

$$ V(r) = \frac{e^2}{4\pi r}$$

as the Coulomb potential.

classical mechanics - Are the Hamiltonian and Lagrangian always convex functions?

The Hamiltonian and Lagrangian are related by a Legendre transform: $$ H(\mathbf{q}, \mathbf{p}, t) = \sum_i \dot q_i p_i - \mathcal{L}(\mathbf{q}, \mathbf{\dot q}, t). $$ For this to be a Legendre transform, $H$ must be convex in each $p_i$ and $\mathcal{L}$ must be convex in each $\dot q_i$.

Of course this is the case for simple examples such as a particle in a potential well, or a relativistic particle moving inertially. However, it isn't obvious to me that it will always be the case for an arbitrary multi-component system using some complicated set of generalised coordinates.

Is this always the case? If so, is there some physical argument from which it can be shown? Or alternatively, are there cases where these convexity constraints don't hold, and if so what happens then?

newtonian gravity - Where does tidal energy come from?

Kind of an odd, random question that popped into my head. Tidal energy - earth's ocean movement, volcanism on some of Jupiter's moons, etc. - obviously comes from the gravitational interaction between large bodies. On earth the interactions with the moon are pulling water around the surface, creating some amount of heat due to friction, etc.

My question is, where does that energy come from exactly? More specifically, what potential energy source is getting depleted to do that work? Is the earth minutely slowing down in its spin - or are the orbits of earth and the moon subtly altered over time by the counteractive movement and friction of liquids and gasses?

Answer

My question is, where does that energy come from exactly? More specifically, what potential energy source is getting depleted to do that work? Is the earth minutely slowing down in its spin - or are the orbits of earth and the moon subtly altered over time by the counteractive movement and friction of liquids and gasses?

In the case of the Earth's oceans and of the volcanoes of Io and Enceladus, the source of the energy is the planet's rotational kinetic energy rather than orbital energy. I wrote extensively about Io in this answer to the question When a planet is heated through gravitational pull, where is the energy taken from at this site. Unless there are objections, I'll let that answer stand with regard to explaining the source of the sulfur volcanoes on Io and the cryovolcanoes on Enceladus.

In the case of the Earth's oceans, the Earth's rotation rate (one revolution per day) is much faster than the Moon's orbital rate (once revolution per month). Friction, viscosity, the Coriolis effect, and the sizes and shapes of the ocean basins means that the tides raised by the Moon are simultaneously slowly slowing down the Earth's rotation rate and are slowly making the Moon recede from the Earth.

The slowing of the Earth's rotation rate and the Moon's orbital rate are written in rock (hardened clay, actually) in eclipses of the Moon and the Sun recorded by ancient Babylonian astronomers. For example, the path of totality of the total solar eclipse observed in Babylon on 15 April 136 BC would have passed over Algiers rather than Babylon if the Earth's rotation rate and the Moon's orbital rate were constant (stephenson).

The slowing of the Earth's rotation rate and of the Moon's orbital rate are even more clearly written in rock (quite literally) in the form of some fossils and tidal rythmites, sedimentary rock formations that have extremely ancient records of daily/monthly/yearly variations in the tides. The day was a couple of hours shorter than it is now 450 million years ago, shorter yet 900 million years ago (williams).

References:

R. Stephenson, "Historical eclipses and Earth's rotation", Astronomy & Geophysics 44:2 (2008): 22-27.

G. Williams, "Geological constraints on the Precambrian history of Earth's rotation and the Moon's orbit," Reviews of Geophysics 38.1 (2000): 37-59.

forces - The Time That 2 Masses Will Collide Due To Newtonian Gravity

My friend and I have been wracking our heads with this one for the past 3 hours... We have 2 point masses, $m$ and $M$ in a perfect world separated by radius r. Starting from rest, they both begin to accelerate towards each other. So we have the gravitational force between them as:

$$F_g ~=~ G\frac{Mm}{r^2}$$

How do we find out at what time they will collide? What we're having trouble with is this function being a function of $r$, but I have suspected it as actually a function of $t$ due to the units of $G$ being $N(m/kg)^2$. I've tried taking a number of integrals, which haven't really yielded anything useful. Any guidance? No, this is not an actual homework problem, we're just 2 math/physics/computer people who are very bored at work :)

Answer

You should be able to use energy conservation to write down the velocities of the bodies as a function of time.

$$ \textrm{Energy conservation (KE = PE): } \frac{p^2}{2}\left( \frac{1}{m} + \frac{1}{M} \right) = GMm\left(\frac{1}{r} - \frac{1}{r_0}\right) $$

And

$$ \frac{dr}{dt} = -(v + V) = -p\left( \frac{1}{m} + \frac{1}{M} \right) $$

Momentum conservation ensures that the magnitude of the momenta of both masses is the same. Does this help?

Substituting into the second equation from the first you should be able to solve for: $$ \int_0^T dt = -\int_{r_0}^0 dr \sqrt{\frac{rr_0}{2G(M+m)(r_0-r)}} = \frac{\pi}{2\sqrt{2}}\frac{r_0^{3/2}}{\sqrt{G(M+m)}} $$

electric circuits - What's the physical meaning of the imaginary component of impedance?

As you know, impedance is defined as a complex number.

Ideal capacitors: $$ \frac {1} {j \omega C} \hspace{0.5 pc} \mathrm{or} \hspace{0.5 pc} \frac {1} {sC} $$

Ideal inductors: $$ j \omega L \hspace{0.5 pc} \mathrm{or} \hspace{0.5 pc} sL $$

I know that the reason why they 'invented' the concept of impedance is because it makes it easy to work with circuits in the frequency domain (or complex frequency domain).

However, since in real-life circuits both voltages and currents are real numbers, I'm wondering if there is any actual physical meaning behind the imaginary component of impedance.

Answer

The physical 'meaning' of the imaginary part of the impedance is that it represents the energy storage part of the circuit element.

To see this, let the sinusoidal current $i = I\cos(\omega t)$ be the current through a series RL circuit.

The voltage across the combination is

$$v = Ri + L\frac{di}{dt} = RI\cos(\omega t) - \omega LI\sin(\omega t)$$

The instantaneous power is the product of the voltage and current

$$p(t) = v \cdot i = RI^2\cos^2(\omega t) - \omega LI^2\sin(\omega t)\cos(\omega t) $$

Using the well known trigonometric formulas, the power is

$$p(t) = \frac{RI^2}{2}[1 + \cos(2\omega t)] - \frac{\omega LI^2}{2}\sin(2\omega t) $$

Note that the first term on the RHS is never less than zero - power is always delivered to the resistor.

However, the power for the second term has zero average value and alternates symmetrically positive and negative - the inductor stores energy half the time and releases the energy the other half.

But note that $\omega L$ is the imaginary part of the impedance of the series RL circuit:

$$Z = R + j\omega L$$

Indeed, via the complex power S, we see that the imaginary part of the impedance is related the reactive power Q

$$S = P + jQ = \tilde I^2Z = \frac{I^2}{2}Z = \frac{RI^2}{2} + j\frac{\omega L I^2}{2} $$

Thus, as promised, the imaginary part of the impedance is the energy storage part while the real part of the impedance is the dissipative part.

everyday life - Why do the big nuts always remain at top? The Brazil-nut Effect

Most of the time when I mixed different nuts in a bowl, I observed that the big Brazil nuts were always on top while the small peanuts settled near the base. No matter how you take it, if the big nuts are put at the bottom and then the small nuts are poured, then also after some time the big ones will come to the top. Will not the gravity attract those big nuts more and eventually will they not remain at the base? But it doesn't happen!

I googled it and found the term Brazil-nut Effect but couldn't find any proper explanation. What is the physical explanation for this effect?

Answer

The process you describe is called granular convection$^1$. It happens because under random motion it's easier for a small particle to fall under a big one than vice versa.

Let's assume that all the particles are made of the same material so there are no density differences in play. If you agitate the particles then temporary voids will open between particles as they randomly move. These voids will have a size distribution with lots of small voids and few large ones because it takes much less energy to create a small void than a large void. Small particles can fall into small voids, but a large particle can move downwards only if a large void appears. That means it's more likely for a small particle to move downwards than a large one. Over time the result is that small particles move downwards more often than large particles and hence small particles end up at the bottom and large particles at the top.

This is not the lowest energy state, because the greatest packing fraction is achieved with a mixture of particle sizes. Separating the mixture into layers of roughly similar particle size will decrease its density and hence increase its gravitational potential energy. The sorting is a kinetic effect and the sorted system is in principle metastable.

^{$^1$ I've linked the Wikipedia article, but actually I don't think the article is particularly rigorous and you should Google for more substantive articles.}

thermodynamics - Why does the air pressure at the surface of the earth exactly equal the weight of the entire air column above it

Why does the air pressure at the surface of the earth (resulting from collisions of molecules on the surface of the earth which has to do with the velocity of the particles) exactly equal the weight of the entire air column above it (which just has to do with the number and mass of the molecules in the air column)?

Here are the beliefs that give rise to my perplexity:

1. 
   

   The weight of the atmosphere is the mass of the molecules (in a column with cross section 1 square inch, let's say) times the force of gravity.

   
   


2. 
   

   Thus if we cooled the air column to the point where it was a solid and set the solid on a scale, we would get the weight of that column of air which, I am told, should read 14.7 lbs.

   
   


3. 
   

   The pressure from the air on a patch of ground 1 square inch in area is due to the collisions of the air molecules with that patch of ground. This is a function of the number of particles hitting the ground per unit time (which is itself a function of the density of the air at the ground) and the average kinetic energy of those molecules (their temperature). The air pressure I am told is 14.7 lbs/square inch.

   
   



4. 
   

   If I enclose some the air at the ground in a rigid container (say I screw the cap on to a glass jar on the ground), thus isolating it from the effects of the column of air above, the pressure of the gas inside the jar is 14.7 psi and would remain that (assuming the container doesn't change shape/volume and I keep it at the same temperature) even if I took it too the top of a mountain or in to space.

   
   


5. 
   

   The molecules of air are sufficiently far apart so that inter-molecular forces are negligible.

   
   

So why does the air pressure at the surface of the earth (resulting from collisions of molecules on the surface of the earth which has to do with the density and velocity of the particles) exactly equal the weight of the entire air column above it (which just has to do with the number and mass of the molecules in the air column)?

Answer

Suppose the pressure at the Earth's surface is $P$. Consider an air column of cross-sectional area $A$. The upward force on the column is $F_{\text{up}}=PA$. Denote the weight of the column as $W$. By definition of "weight", the downward force on the column is $F_{\text{down}}=W$.

Suppose the pressure is too low, such that $F_{\text{up}}

In other words, the pressure is such as to balance the weight of the column because that's the only situation which won't immediately change.

homework and exercises - Steady-State Current

I am in the process of completing some homework and have come across a question that says: "What is the steady-state current?" It is asking in reference to a circuit that includes two capacitors in parrallel and a DC source of 24V and an AC source given by $\epsilon = 20 \cos(120 \pi t)\ \text{V}$, where $t$ is in seconds. I assume that it is asking when the current is constant. Because naturally the current is not constant due to the AC source, would I just consider the DC source for this?

Answer

Steady state means, in this context, ignoring transients due to initial conditions.

For a capacitor, the steady state current due to a DC voltage source is zero.

The steady state current due to an AC source is simply the (constant) amplitude of the sinusoidal current.

To be clear, the current is changing in time and so is not constant. But, the amplitude of the current is constant, i.e.,

$$i_C = A \cos(\omega t + \phi)$$

where $A$ is a constant. This is called AC or sinusoidal steady state.

Credit

quantum field theory - What goes wrong if we add a mass term for gauge bosons without the Higgs mechanism?

Question: Why can't we add a mass term for the gauge bosons of a non-abelian gauge theory?

In an abelian gauge theory one can freely add a mass and, while this breaks gauge invariance, as long as the coupling current is conserved everything works fine (i.e., the scalar modes decouple and the theory is renormalisable).

In non-abelian gauge theories, it is often stated that the only way to introduce a mass term is through the Higgs mechanism. If we added a mass term without introducing the Higgs field, but the coupling current is still conserved, at what point would the theory break down? It seems to me that the scalar modes decouple as well, at least to tree level. I failed to push the calculation to one loop order, so maybe the theory breaks down here. Is this the most immediate source of problems, or is there any simpler observable which fails to be gauge invariant?

One would often hear that if we break gauge invariance the theory is no longer renormalisable. I may be too naïve but it seems to me that a (gauge-fixed) massive gauge boson has a $\mathcal O(p^{-2})$ propagator and therefore (as long as the current in the vertices is conserved) the theory is (power counting) renormalisable. Or is it?

To keep things focused, let us imagine that we wanted to give gluons mass, while keeping self-interactions and the coupling to matter (and ghosts) unchanged. Could this work without a Higgs?

---

There are many posts about that are asking similar things. For example,

Answer

What a great question OP! I have good news and bad news. The good news is that this exact same question is asked and answered in Quantum Field Theory, by Itzykson & Zuber, section 12-5-2. The bad news is that the answer is

If you introduce mass terms in non-abelian gauge theories by hand, the theory is non-renormalisable.

This means that one is forced to introduce the Higgs mechanism (or variations thereof, such as the Stückelberg mechanism), which for some people is rather inelegant (and plagued by problems of naturalness, etc). Oh well, that's the way the cookie crumbles.

Let me quote the first paragraph of the aforementioned section, so as to summarise the main point of the problem:

Is a gauge theory where mass terms are introduced by hand renormalizable?

In electrodynamics, the situation is favorable. After separation of the gauge field into transverse and longitudinal components, the longitudinal part $k_\mu k_\nu/M^2$ which gives rise to the bad behavior in the propagator does not contribute to the $S$ matrix. This results from the noninteraction of longitudinal and transverse components and from the coupling of the field to a conserved current. In a nonabelian theory, none of these properties is satisfied. Longitudinal and transverse parts do interact, while the current to which the gauge field is coupled is not conserved. On the other hand, unexpected cancellations of divergences at the one-loop level make the theory look like renormalizable. This explains why it took some time to reach a consensus, namely, that the theory is not renormalizable. The way out of this unpleasant situation is to appeal to the mechanism of spontaneous symmetry breaking, to be explained in the next subsection.

visible light - How can absorption spectra form if atoms can't remain in an excited state?

I have been tasked to write a research paper on stars. However, I know very little about physics in general. I am learning about how we can glean information about stars by analyzing the light that they emit. So, first, I am learning about how light interacts with matter.

I just learned about atoms and the fact that they normally exist in a grounded state. Either a collision with another atom or the absorption of a photon with the right wavelength can force the electron(s) in the atom up to a higher energy level. The atom is now in an excited state.

However, atoms can not remain in an excited state, as this state is not stable. So, 10^-6 to 10^-9 seconds later, a photon is emitted due to a new found surplus of energy as the electron drops back down to its ground level.

• Subquestion: which is the cause and which is the effect here? Is the electron dropping down because the photon is released? Or is the release of the photon the result of the electron being sucked back down by some fundamental force? If the latter is the case(which I suspect) what is this force? Is this the electromagnetic force?

It is my understanding that (assuming the excitation was caused by the absorption of a photon) the photon being released would have a wavelength equal to the photon that was absorbed.

If the above is true, I am confused as to how we notice absorption lines in light that passes through a gas.

It is stated that the atoms in the gas absorb some of the light that is passing through them, but under my current understanding of the interaction, this light would soon be re-emitted. So, I would think that we should still see a continuous spectrum. what am I missing here?

Answer

Basically, absorption lines exist because absorbed photon are not re-emitted in the same direction, so dark lines can be observed. There are various reason causing this.

For example, the extra energy can be dissipated as phonon in solid or strongly interacting system. Excited states can also emit multiple low frequency photon if there are meta-stable states. Lastly, even the atoms re-emit photon with same frequency, the photon direction is completely random. Therefore, the all re-emitted light can be ignored if the detector is sufficiently far away, hence, dark lines.

pressure - Will helium in the tires of a bike make it lighter?

I know that helium balloons float because it is less dense than air. I'm not expecting my bike to float, although that would be pretty cool. I just wanna know if replacing normal air with helium in the tires will produce a noticeable effect on its weight. Will the helium 'lift'/reduce the weight force on the bike?

Answer

It will make it lighter, but the effect will be very small. The volume of the tube is probably less than a liter. One mol of an ideal gas is 23 liters at atmospheric pressure. So you have about 0.2 mol of gas in there at 4 bar pressure. Helium weighs 4 g/mol, nitrogen about 28 g/mol. So for 0.2 mol, the weights are 0.8 g and 5.6 g. Cleaning off the dirt from the frame will have a greater effect.

Helium atoms are smaller than nitrogen molecules. Therefore there is a greater rate of diffusion through the bike tires. Your tires will become flat quicker than normal. Therefore it is not really a good idea to use helium.

waves - Question on open organ pipe

Although open organ pipe is open on both ends, how standing waves are produced in a open organ pipe. Can someone explain with more clarity?

Answer

You may find by starting from first principles, or by consulting external resources that pressure waves in air (in one dimension) are governed by the wave equation

$$\frac{\partial^2 p}{\partial x^2} - \frac{1}{v^2} \frac{\partial^2 p}{\partial t^2} = 0$$

where $x$ is a position and $t$ is the time, and $p$ denotes the pressure difference away from equilibrium. To solve this, let us write the solution as a Fourier transform$^{[a]}$

$$p(x,t) = \int_k \int_\omega \tilde{p}(k,\omega) \cos(kx - \omega t) \frac{d\omega}{2\pi}\frac{dk}{2\pi} \, .$$ Plugging this into the wave equation, we find

$$(-k^2 + \omega^2 / v^2) \tilde{p}(k,\omega) = 0 \, .$$

For each value of $k$ and $\omega$ one of two things must be true: either $\tilde{p}(k,\omega)=0$ or $\omega^2/v^2 - k^2 = 0$. The case where $\tilde{p}(k,\omega)=0$ are trivial: there is no wave. In the other case, for a particular choice of $k$ and $\omega$ we have a wave with the specific form $$p(x, t) = M \cos(kx - \omega t ) = M\cos(k(x \pm vt)) = \cos(kx)\cos(\omega t) \pm \sin(kx)\sin(\omega t)\, .$$ Since the wave equation is linear, these solutions can be summed together to produce other solutions. Furthermore, any solution can be written as a sum of these types of solutions.

Now consider an organ pipe which we stimulate at frequency $\omega$. In this case, the linear combination can be made up of only the following two parts: $$ \begin{align} p(x,t) &= p_{\text{left}}(x,t) + p_{\text{right}}(x,t) \\ \text{where} \qquad p_{\text{left}}(x,t) &= M_{\text{left}} \cos(kx + \omega t) \\ \text{and} \qquad p_{\text{right}}(x,t) &= M_{\text{right}} \cos(kx - \omega t) \\ \text{so} \qquad p(x,t) &= (M_{\text{left}} + M_{\text{right}})\cos(kx)\cos(\omega t) \\ &+ (M_{\text{right}} - M_{\text{left}}) \sin(kx)\sin(\omega t)\, . \end{align} $$

Suppose the pipe extends from $x=-L/2$ to $x=L/2$. The two ends are open to the outside air, so at those points $p \approx 0$. These are our boundary conditions. From the form of $p(x,t)$ you can convince yourself that the only possible solution is when $M_{\text{left}} = M_{\text{right}}$ and $k=\pi/L$.$^{[b]}$ Putting that in gives

$$p(x,t) \propto \cos(\pi x / L) \cos(\omega t)\, .$$

Stop and consider the meaning of this equation for a minute. It is a vibration of air pressure inside the pipe with cosine spatial profile and also cosine oscillation in time. In other words, the pipe, even with its open ends, can made a vibration at the frequency $\omega$. Note that the pressure difference at the ends of the pipe is zero by construction (i.e. our choice of $k$), yet there are still vibration modes!

$[a]$ I dropped a phase here which, if you put it in and follow everything through, doesn't change the argument.

$[b]$ Actually there are more possible values of $k$. They correspond to higher modes of the vibrating air wave. See if you can figure out what I mean :)

classical mechanics - What's wrong with Arnold's scaling argument on jumping height?

The following question was put on hold: Is it possible to prove that an elephant and a human could jump to the same height?

It reminded me of an exercise (24a) on that exact topic in Arnold's "Mathematical Methods of Classical Mechanics". The solution he gives goes like that:

For a jump of height h one needs energy proportional to $L^3h$, and the work accomplished by muscular strength $F$ is proportional to $FL$. The force $F$ is proportional to $L^2$ (since the strength of the bones is proportional to their section). Therefore, $L^3h$~$L^2L$, i.e. the height of a jump does not depend on the size of the animal. In fact, a jerboa and a kangaroo can jump to approximately the same height.

The comments of the above question tended to dismiss that argument. What's wrong with it?

Addendum: It seems obvious that not all animals jump exactly to the same height, given their different physiologies/shapes. Some of them can't jump at all.

The question is to be understood in the following spirit: if we plotted jumping height vs animal size for a lot of different species, would there be a correlation? I don't mean there is no spread; I totally expect a big spread due to the other factors involved.

Second addendum: Some interesting points have been raised in comments and answers. I will accept an answer that incorporates: the domain of validity of Arnold's argument (or explain why it is never valid), the effect of air drag on very small jumpers and the impossibility of very large animals.

Bonus points for documenting the yet elusive jumping elephant and plotting jumping height vs size for different species ;)