molecules - How are bond angles determined?

Electron microscopes cannot clearly depict the exact shape and structure of atoms and molecules, even though it does show a vague, cloudy image. In my AP chemistry class, I learned that the bond angle of some molecules is 109.5 degrees. How is this bond angle determined so precisely, if the bonds cannot be accurately observed through a microscope?

Answer

The positions of atoms respective to each other in a crystal lattice (solid) can be determined by X-ray crystallography. From these positions bond lengths and bond angles can also be calculated accurately.

Probably the most memorable case of solving the geometrical structure of a molecule was Franklin and Gosling's X-ray crystallography of DNA, information later used by Watson and Crick to solve the mystery of DNA's structure.

For many simple (binary) compounds molecular shapes and bond angles can also be determined theoretically (see link).

thermodynamics - Work done by isothermal expansion from two different viewpoints

Consider an adiabatic system as follows. It consists of a gas in a container and a piston. Initially, the system is at equilibrium and the gas inside it occupies a volume $V_i$ at a pressure $p_i$ which is equal to the outside pressure. Suddenly, the outside pressure changes and reduces to $p_{atm}$. The piston moves to equalize the pressure and the gas expands isothermally to obtain equilibrium. The gas now occupies a volume $V_f$ at a pressure of $p_f$ which is equal to $p_{atm}$

Now, two textbooks I have define the work done from two different viewpoints.

1: From the viewpoints of the surroundings:

The work done on the system by the surroundings equals $-p_{atm}\Delta V$. Since $p_{atm}$ is pretty much constant for the whole of the whole of the process, we can say that the work done equals:

$$W_1 = -p_{atm}(V_f - V_i)\tag1$$

2: From the viewpoint of the system:

We can write the internal pressure of the system as a function of its volume: $p_{in}(V)$. As during the expansion, the internal pressure changes, the work done by the system equals

$$W_2 = \int_{V_i}^{V_f}p_{in}(V)dV$$

Now, I don't know which definition to use. The work is done by the system (from definition 2) is done on the surroundings. But what about the negative work that the surroundings did on the system? Where did that energy go? Maybe, the two definitions express the same thing: the work done by the system. The negative sign in definition 1 signifies that the work is done by the system. But that would mean that

$$W_1 = W_2$$

We can simplify $W_2$ as follows. Clearly,

$$p_{in}(V) = \frac{p_iV_i}{V}$$

$$W_2 = \int_{V_i}^{V_f}\frac{p_iV_i}{V}dV = p_iV_i \ln{\frac{V_f}{V_i}} \tag2$$

The internal pressure when the volume is equal to $V_f$ is $p_{atm}$

$$\implies p_{atm} = p_{in}(V_f) = \frac{p_iV_i}{V_f}$$

$$\implies V_f = \frac{p_iV_i}{p_{atm}}$$

Putting this into $(1), (2)$ gives us,

$$W_2 = p_iV_i\ln{\frac{p_{i}}{p_{atm}}}$$

$$W_1 = -p_{atm}\left(\frac{p_iV_i}{p_{atm}} - V_i\right) = -V_i(p_{atm} - p_i)$$

Consider some values. Let $p_i = 5 \ \rm{Pa}, V_i = 1 \ \rm{m^3}, p_{atm} = 1 \ \rm{Pa}$.

$$W_1 = -1\cdot(1-5) = 4$$

$$W_2 = 1\cdot1\cdot\ln{\frac{5}{1}} = 1.609$$

Where am I making a mistake?

electromagnetism - Electric and Magnetic field's phase difference shift in linearly polarized electromagnetic waves

I am a high school student and we currently studying the electromagnetic theory. In my textbook i read that the oscillating electric the magnetic fields have phase difference equal to π/2 rad near the source (for example an antenna) while away from it they agree in phase.

Is this true? And if so, why and how is this happening.

Answer

Possibly you are talking about the difference between the "far field" and "near field" solutions for the simple oscillating electric dipole.

Often when dealing with such a system, if we are looking at the field more than a few wavelengths away from the dipole (or more formally, $kr \gg 1$ or $r \gg \lambda/2\pi$) then the solution looks like a spherically expanding electromagnetic wave; the E-field and B-field are in phase, mutually perpendicular and at right angles to the outward propagation direction. For a dipole moment aligned with the z-axis, the E-field is polarised in the $\theta$ (poloidal) direction and B-field is in the $\phi$ (toroidal) direction.

But if $r \leq \lambda$ then the solution is more complicated. The E-field has both a $\theta$ and a radial component. The B-field is just toroidal, but contains two terms with differing radial dependencies.

In these extra terms for the nearby fields, the E-field becomes much more dominant (in transverse electromagnetic waves it is normally $c$ times bigger). Furthermore it is out of phase with the B-field by $\pi/2$. You can see this from the equations below - when $r$ is small, the first term in $B_{\phi}$ dominates the B-field, whereas it is the first (even stronger) terms in the $E_r$ and $E_{\theta}$ components that dominate the E-field. These are different in magnitude from the B-field by a factor that includes $i$ and hence are out of phase by $\pi/2$.

Perhaps this is what you mean?

The Maths:

The solutions for the E- and B-field from a simple oscillating dipole are $$E_{r} = \frac{p_0 \cos\theta}{4\pi \epsilon_0} \frac{k^2 \exp(ikr)}{r}\left[ \frac{2}{k^2r^2} - \frac{2i}{kr} \right]$$ $$E_{\theta} = \frac{p_0 \sin\theta}{4\pi \epsilon_0} \frac{k^2 \exp(ikr)}{r}\left[ \frac{1}{k^2r^2} - \frac{i}{kr} -1 \right]$$ $$B_{\phi} = \frac{p_0 \sin\theta}{4\pi \epsilon_0} \frac{k^2 \exp(ikr)}{r}\left[ - \frac{i}{kr} -1 \right] \left(\frac{\epsilon_0}{\mu_0} \right)^{1/2}$$

homework and exercises - How to show using Lorentz transformation the $vec{v}$ of one frame to another in either frame is $vertvec{v}vert$

I am trying to show that in the case of one frame at rest, S, and another frame moving in the +x direction, S', that:

1. The S frame see's the S' frame moving with $+\vec{v}$


2. The S' frame see's the S frame moving with $-\vec{v}$

First, we start in the S frame which see's the S' frame moving in the +x direction. We record two spacetime events of the location of the S' frame's origin and find the velocity:

$$Event 1: (t_1,x_1,0,0) $$

$$Event 2: (t_2,x_2,0,0) $$

$$\frac{x_2 - x_1}{t_2 - t_1}=\frac{\Delta x}{\Delta t}=+\vec{v}$$

Now, for the S' frame, they see the S frame's origin moving away in the -x direction. We can transform the events using time dialation and length contraction arguments:

1. 
   

   The length $\Delta x$ reported by the S frame will be longer than the S' frames length (because S' is at rest now), so $\Delta x'=- \frac{\Delta x}{\gamma}$.

   
   


2. 
   

   The time interval $\Delta t$ reported by the S frame will be shorter than the S' frames time interval, so $\Delta t'= \Delta t\gamma$.

   
   

$$\frac{\Delta x'}{\Delta t'}=\frac{\frac{-\Delta x}{\gamma}}{\Delta t\gamma} =\frac{-v}{\gamma^2}$$

Which is obviously wrong, but I am really stuck on how to set up this problem/system.

Answer

To measure length contraction you need to look at two parallel world-lines; you are looking at one individual world-line and therefore you've not been clear about what you're studying.

The coordinate transformation that you want is (with $w=ct$ and $\beta=v/c$ to show a certain symmetry of the equations)$$\begin{align} w'&=\gamma~(w - \beta~x),\\ x'&=\gamma~(x - \beta~w),\\ y'&=y,\\ z'&=z.\end{align}$$ Ignoring the $y$ and $z$ directions as trivial, length contraction comes when we consider the world-lines $(w, x) = (s, 0)$ for all $s$ and $(w, x)=(s, L)$ for all $s$. These both transform to:$$w'=\gamma~(s - \beta~L),\\ x'=\gamma~(L-\beta~s).$$Substituting for $s=w'/\gamma + \beta~L$ we can find the equation $$x'=\gamma~(L-\beta~(w'/\gamma+\beta~L) = -\beta~w' + \gamma~L~(1 - \beta^2).$$

Note that $\gamma~(1-\beta^2) = \gamma / \gamma^2 = 1/\gamma,$ giving our familiar length contraction factor; note also that this line which was stationary is now moving backwards with speed $-\beta~c=-v$.

This effect that when you accelerate, clocks in front of you tick faster the further they are in front of you, and clocks behind you tick slower the further they are behind you, is called the "relativity of simultaneity." It is a very important effect: in fact I have argued before that it is the most important effect, in that for small $\beta\ll 1$ it is the only effect which survives: and we can reconstruct the general coordinate transformation above from this $\gamma\approx 1$ form.

quantum field theory - Does Dirac's idea of filled negative energy states make sense?

Please bear with me a bit on this. I know my title is controversial, but it's serious and detailed question about the explanation Dirac attached to his amazing equations, not the equations themselves.

Imagine for a moment that someone signed onto this group and proposed the following:

1. 
   

   Empty space is a lot like a metal, or maybe a semiconductor, because like those materials its chock full of negatively charged electrons occupying different velocity states, only...

   
   


2. 
   
   

   Unlike metals or semiconductors, the density of electrons in any one region of space is infinite, because there is no limit to how fast the electrons can move. That is because these are negative energy states in which an electron can always move faster simply by emitting a photon, so there's not "bottom" to how far they can drop and how dense they can become, and...

   
   


3. 
   

   Unlike metals or semiconductors, there is no exactly balancing sea of positive atomic charges, well, unless maybe there are infinite numbers positively charged atoms too, and...

   
   


4. 
   

   The resulting infinite negative charge density of real electrons not only doesn't matter but is in fact completely and totally invisible for some reason, and...

   
   


5. 
   

   The resulting infinite mass density of electrons (recall that these are quite real electrons, only in odd negative-energy kinetic states) also doesn't matter, and...

   
   
   


6. 
   

   Unlike the Fermi sea of a metal conduction band, removing an electron from this infinitely dense sea of electrons for some reason doesn't cause other electrons to collapse into it and fill it, even though the negative kinetic velocity electrons are pushed by exactly the same Pauli exclusion forces as the ones in a Fermi band; in short, for reasons not clear, semiconductor-style hole stabilization applies while metal-style hole filling does not (is there a band gap going on here?), and...

   
   


7. 
   

   Since the infinitely dense negative charges become invisible for no particular stated reason when the electrons fall into negative energy states, these unexpectedly stable open states in the negative energy sea have net positive charge, even though...

   
   


8. 
   

   ... such missing states categorically should have zero charge, since in sharp contrast to the positive ionic background of metallic and semiconductors, the vacuum has no background charge at all, which should leave holes in the mysteriously invisible negative kinetic energy just as uncharged and invisible as that sea for some reason is, and...

   
   



9. 
   

   Even if you do assume that the negative kinetic state electrons have visible charge, their infinite density would make the "comparatively" but infinitesimally smaller positive charge of such a hole invisible, and...

   
   


10. 
    

    Repeat this process for every other kind of particle in existence, and...

    
    


11. 
    

    If you have done all of this and done it correctly, congratulations: You now understand conceptually what anti-electrons (positrons) and other anti-particles are.

    
    

First question: Have I misrepresented any of the implications of Dirac's explanation of positrons as holes in an infinite sea of negative-kinetic-energy electron states? What I have tried very hard to do is nothing more than make a list of the implications of a physics idea, just as people do all the time on this group. Who said it should not really be the issue, not if we are talking about an unelaborated explanation rather than the math itself.

Second question: If someone had proposed a theory in this forum like the one I just described, and you had never heard of it before, what would you have thought of it? Please be honest.

My point in all of this obviously is this: While Paul Dirac's amazing equations (they really are) managed to predict antimatter, his explanation for why his equations require antimatter is... shall we say incompletely analyzed, to put a nice spin on it?

A final thought: Has anyone ever seriously tried to make Dirac's conceptual negative energy sea ideas, the ones that he espoused in his Nobel Lecture, into a real, working theory? And if so, how did they deal with the various issues I described above?

(Me, I just think antiparticles are regular particles moving backward in time. Yeah, that's a pretty weird idea too, I know...)

Answer

Dirac's explanation of the emergence of antiparticles such as positrons out of the Dirac sea, and the Dirac sea itself, is completely valid and legitimate, and you have described some non-quantitative aspects of it and differences between it and some condensed-matter situations.

Dirac just began with the assumption that the Dirac spinor field $\Psi$ is a pure combination of annihilation operators only while $\bar\Psi$ is only made of creation operators. However, it may be seen that the creation/annihilation operators create/annihilate electrons into states with both positive and negative energy.

The ground state is defined as one in which Nature minimizes the energy. A good way to do so is to keep the positive-energy state empty but occupy all the negative-energy states (addition of a negative number is like a subtraction of a positive one). That's how we get the physical vacuum, one with the Dirac sea. On the contrary, we may create a hole, i.e. remove an electron from the Dirac sea of negative-electron states. A simple counting of signs shows that this is equivalent to adding a positive-charge, positive-energy particle, a physical positron.

So we may choose the convention in which the creation operators for the negative-energy states are relabeled as annihilation operators of positrons, and vice versa. The difference between the Dirac sea paradigm and the usual expansions taught in QFT courses is just a relabeling of $a$ as $b^\dagger$ and $a^\dagger$ as $b$. The physical convention with positrons aside from electrons is more physical because all annihilation operators actually annihilate the physical ground state (vacuum).

These days, we don't usually emphasize Dirac's construction, e.g. because it only applies to fermions (there is a symmetry between occupation numbers 0 and 1; no reflection of the spectrum is a symmetry for bosons whose occupation numbers are all non-negative integers) but the fact that the relativistic fields automatically predict antiparticles due to the negative-energy solutions is general and holds both for bosons and fermions. But that doesn't invalidate anything about Dirac's presentation.

If I didn't understand the actual maths and its connection with physics and someone gave me vague linguistic descriptions such as yours, I wouldn't know what to think and I would tend to think that the writer is confused. After all, I think you are confused even here, in the non-hypothetical world.

If I could check what's behind these statements, I would realize it's a valid theory whether it was written down by Bollinger or Dirac.

quantum mechanics - Wave/particle duality

Apologies if this has been asked before (I did check and I believe it wasn't). I have a question about the particle/wave duality of photons (or other particles). Depending on what and how we measure the photon turns out to be either a wave or a particle. Recently I saw some web page (and I can't remember where) that maybe neither is true. Both the wave perception and the particle perception are just that: perceptions, strengthened by our perhaps limited ability to observe. What if the reality of the photon is something else, something "on top" of our two perceptions?? Could someone direct me to a site that could tell me more about it (or maybe debunk the whole idea?)?

electromagnetism - Derive Lorentz Equation from Relativistic Hamilton-Jacobi Equation

Consider a ralativistic particle of rest mass $m$ and electric charge $e $ moving in electromagnetic field with four-potential ${\displaystyle A^{\mu}=(\phi ,\mathrm {A} )} $ in vacuum, then the Hamilton–Jacobi equation has the form

$$g^{\mu \nu}\left ( \frac{\partial S}{\partial x^{\mu}} + \frac {e}{c}A_{\mu} \right ) \left ( \frac{\partial S}{\partial x^{\nu}} + \frac {e}{c}A_{\nu} \right ) = m^2 c^2\tag{1}$$

or more compact expressed as Minkowski product

$$ \left( \frac{\partial S}{\partial x^{\mu}} + \frac {e}{c}A_{\mu} \right ) \left ( \frac{\partial S}{\partial x_{\mu}} + \frac {e}{c}A^{\mu} \right ) = m^2 c^2 \tag{2}$$

here we denote $g^{\mu \nu}$ the metric tensor with signature $(+ - - -)$ and $S$ is the action function from Hamilton-Jacobi-theory.

Especially $S$ satisfy the equation

$$p_{\mu}= \nabla_{\mu}S := \frac{\partial S}{\partial x^{\mu}}\tag{3}$$

where $p_{\mu}$ is the four momentum and $\nabla_{\mu}$ the four gradient.

Now I have following two questions:

1. 
   

   Does anybody have a reference for a rigorous derivation for $$ \left( \frac{\partial S}{\partial x^{\mu}} + \frac {e}{c}A_{\mu} \right ) \left ( \frac{\partial S}{\partial x_{\mu}} + \frac {e}{c}A^{\mu} \right ) = m^2 c^2 .\tag{4}$$

   
   


2. 
   

   It is known that applying method of characteristics to the PDE

   
   
   

$$F(S,\frac{\partial S}{\partial x^{\mu}} ,x^{\mu}):= \left( \frac{\partial S}{\partial x^{\mu}} + \frac {e}{c}A_{\mu} \right ) \left ( \frac{\partial S}{\partial x_{\mu}} + \frac {e}{c}A^{\mu} \right ) - m^2 c^2 =0\tag{5}$$

one can derive the relative Lorentz equation

$${\displaystyle {\frac {\mathrm {d} p^{\mu }}{\mathrm {d} \tau }}=eF^{\mu \nu }p_{\nu }}\tag{6}$$

with electromagnetic tensor $$F^{\mu \nu }:= \frac{\partial A_{\mu}}{\partial x^{\nu}}- \frac{\partial A_{\nu}}{\partial x^{\mu}}\tag{7}$$ and four momentum $p_{\mu}$.

Here I'm also looking for an explicit derivation of LE from the HJE using characteristics.

Indeed, the method of characteristics transform a PDE into a system of ODE with respect parametrizing variable $\tau$:

$$\frac{\partial p_{\mu}}{\partial \tau}= -\frac{\partial F}{\partial x^{\mu}} -\frac{\partial F}{\partial S} p_{\mu}\tag{8}$$

$$\frac{\partial x_{\mu}}{\partial \tau}= \frac{\partial F}{\partial p^{\mu}}. \tag{9}$$

Remark: HJ theory says $$p_{\mu}= \frac{\partial S}{\partial x^{\mu}}.\tag{10}$$

The problem is to derive from here the equation for Lorentz force

Answer

This answer does not address OP's specific question about the method of characteristics, but sketches a systematic derivation (of the various equations involved) starting from a Lagrangian formulation.

1. 
   

   A Lagrangian for a relativistic point particle of mass $m$ and charge $q$ in a EM background $A_{\mu}$ and gravitational background $g_{\mu\nu}$ is$^1$ $$ L~:=~L_0 - U,\qquad L_0~:=~\pm \frac{\dot{x}^2}{2e}-\frac{e m^2}{2}, \qquad \dot{x}^2~:=~g_{\mu\nu}(x)~ \dot{x}^{\mu}\dot{x}^{\nu}, \qquad \dot {x}^{\mu} ~:=~\frac{dx^{\mu}}{d\tau},\tag{A}$$ with Minkowski sign convention $(\mp,\pm,\pm,\pm)$ and speed-of-light $c=1$. Here $\tau$ is the world-line (WL) parameter (which is not necessarily proper time) and $e>0$ is an einbein field. The velocity-dependent Lorentz potential is $$ U~:=~ \mp q{\dot x}^{\mu} A_{\mu}, \tag{B} $$ with corresponding generalized Lorentz 4-force$^2$ $$ F_{\mu}~:=~\frac{d}{d\tau} \frac{\partial U}{\partial \dot{x}^{\mu}} - \frac{\partial U}{\partial x^{\mu}}~=~\pm qF_{\mu\nu}\dot {x}^{\nu}, \qquad F_{\mu\nu}~:=~\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}. \tag{C}$$

   
   


2. 
   

   The canonical/conjugate 4-momentum$^2$ becomes $$ p_{\mu}~:=~\frac{\partial L}{\partial \dot{x}^{\mu}}~=~\pm\frac{g_{\mu\nu}\dot{x}^{\nu}}{e}\pm q A_{\mu}.\tag{D}$$

   
   
   


3. 
   

   After a Legendre transformation the Hamiltonian reads $$ H~=~\frac{e}{2}\left(m^2\pm ( p\mp qA)^2 \right). \tag{E}$$ Note that the $e$ field is a Lagrange multiplier for the mass-shell constraint.

   
   


4. 
   

   The Hamilton-Jacobi (HJ) equation becomes essentially the mass-shell constraint $$ 0~=~E~=~\frac{e}{2}\left(m^2\pm ( \frac{\partial W}{\partial x}-qA)^2 \right), \tag{F}$$ where $W$ is Hamilton's characteristic function, and$^3$ $$ p_{\mu}~=~\pm \frac{\partial W}{\partial x^{\mu}}. \tag{G}$$ The fact that the energy $E$ is zero can be viewed as a consequence of WL reparametrization invariance $\tau\to \tau^{\prime}=f(\tau)$.

   
   

--

$^1$ To achieve the standard square root Lagrangian, simply integrate out the $e$ field, cf. e.g. this Phys.SE post.

$^2$ The usual notions of 4-momentum & 4-force correspond to the gauge where the world-line (WL) parameter $\tau$ is proper time.

$^3$ For sign conventions, see also this Phys.SE post.