I have been trying to imagine ways to visualise the gravitational field of the earth... It's near zero at the outer edges of the atmosphere, strong at sea level, strongest at the quarter point, then drops off to zero again towards the middle as the masses cancel each other out... Something like a 3D donut?
This got me thinking about black holes. Would they also have an area of zero g right at the centre where all the surrounding mass cancels each other out?
Why is the following equation true?
$$\frac{\partial \mathbf{v}_i}{\partial \dot{q}_j} = \frac{\partial \mathbf{r}_i}{\partial q_j}$$
where $\mathbf{v}_i$ is velocity, $\mathbf{r}_i$ is the displacement, and $q_j$ is the generalized coordinate into which $\mathbf{r}_i$ is transformed.
In reading further, I find that it's related to
$$ \mathbf{v}_i \equiv \frac{\mathrm{d}\mathbf{r}_i}{\mathrm{d}t} = \sum_k \frac{\partial \mathbf{r}_i}{\partial q_k}\dot{q}_k + \frac{\partial \mathbf{r}_i}{\partial t} $$
I know that in transforming the virtual displacement $\delta\mathbf{r}_i$ into generalized coordinates, I can use
$$ \delta\mathbf{r}_i = \sum_k \frac{\partial\mathbf{r}_i}{q_k}\dot{q}_k $$
The first equation above is of course equivalent to
$$\frac{\partial \dot{\mathbf{r}}_i}{\partial \dot{q}_j} = \frac{\partial \mathbf{r}_i}{\partial q_j}$$
I'm not sure why the dots vanish just like that. How do these all connect?
In a non-mathematical explanation, I can understand that as $q_j$ changes, $\mathbf{r}_i$ changes as well. In the same way, as $\dot{q}_j$ changes, $\mathbf{v}_i$ changes, too. I'd like to know, mathematically, how these changes (for $\mathbf{r}_i$ and $\mathbf{v}_i$) turn out to be equal.
Answer
Your question is very similar to a question I had asked previously on Physics.SE. If you understand how
$$\mathbf{v}_i \equiv \frac{\mathrm{d}\mathbf{r}_i}{\mathrm{d}t} = \sum_k \frac{\partial \mathbf{r}_i}{\partial q_k}\dot{q}_k + \frac{\partial \mathbf{r}_i}{\partial t}$$
is obtained, its relatively easy from there on. Clearly $\mathbf{v}_i$ is a function of $q_k$, $\dot{q_k}$ and $t$. So I can write:
$$\mathbf{v}_i \equiv \mathbf{v}_i(q_k, \dot{q_k}, t)$$
If I were to treat $q_k$ and $\dot{q_k}$ as independent variables, it turns out that I get some very nice expressions. So proceeding with $q_k$ and $\dot{q_k}$ as independent variables, if I were to differentiate $\mathbf{v}_i$ w.r.t $\dot{q_j}$, I would obtain:
$$\frac{\partial\mathbf{v}_i}{\partial\dot{q_j}} = \sum_k \frac{\partial^2\mathbf{r}_i}{\partial\dot{q_j}\partial q_k}\dot{q_k}+\sum_k\frac{\partial \mathbf{r}_i}{\partial q_k}\frac{\partial\dot{q}_k}{\partial \dot{q_j}} + \frac{\partial^2\mathbf{r}_i}{\partial\dot{q_j}\partial t}$$
Since the order of the partial derivatives in the first and third terms can be changed, this becomes:
$$\frac{\partial\mathbf{v}_i}{\partial\dot{q_j}} = \sum_k \frac{\partial}{\partial q_k}\left(\frac{\partial\mathbf{r}_i}{\partial\dot{q_j}}\right)\dot{q_k}+\sum_k\frac{\partial \mathbf{r}_i}{\partial q_k}\frac{\partial\dot{q}_k}{\partial \dot{q_j}} + \frac{\partial}{\partial t}\left(\frac{\partial\mathbf{r}_i}{\partial\dot{q_j}}\right)$$
But $\mathbf{r}_i \equiv \mathbf{r}_i(q_k,t)$ does not depend explicitly on $\dot{q_k}$. Thus the first and last term reduces to zero. And the only non-zero term in the second sum would be when $k=j$. Thus,
$$\frac{\partial\mathbf{v}_i}{\partial\dot{q_j}} = \frac{\partial\mathbf{r}_i}{\partial q_j}$$
The crux of the problem lies in the variables which you choose as independent.
In explosive safety and stability testing, a drop test is commonly used to determine the sensitivity to impact. In the test, an impactor of known mass is dropped. The initial height varies throughout the experiment. The detonation status is recorded in the material and various techniques are used to convert the heights into metrics used to characterize the stability; for instance, a commonly used metric is $h_{50}$ or the height of the impactor that generates a detonation 50% of the time.
The explosive material being tested is typically in a rod or puck shape whose radius is considerably smaller than the impactor and anvil. The impact generates a shock wave through the material sample.
Since I am using an Eulerian code to simulate this, I need to know the pressure of the resulting shock wave in the material. Is there a way to determine this pressure?
Let's assume that I know all of the material properties (density, speed of sound, Youngs Modulus, etc), that the impactor is perfectly rigid, and that the impact is within the elastic limits of the material sample.
The only approach I can think of would be to treat the problem as 1D rod and assign an initial velocity (say in the $-x$ direction) to the rod such that the momentum is the same as that of the impactor used in the experiment. Then impose that the velocity is zero on the left edge of the rod. This would require some numerical work and is less than ideal. Perhaps there is an analytical or empirical relation that exists between impactor momentum and the resulting pressure wave magnitude?
Answer
Background
There is a good reference1 on the physics of sound/shock waves in solids (look at Chapter XI). I found the following (on page 688) very interesting and relevant to your question:
In a solid or liquid, a shock wave with a strength of even a hundred thousand atmospheres is regarded as weak. Such a wave differs little from an acoustic wave: it travels with a speed close to the speed of sound, compresses the material by only a few percent or perhaps of the order of ten percent, and imparts a velocity to the material behind the front which is of the order of a tenth the velocity of the wave itself... then a strong shock wave for condensed media is one whose pressure is not less than tens or hundreds of millions of atmospheres.
Let us define $P$ as the pressure and $\varepsilon$ as the internal energy of a solid material. These can be divided into two parts: an elastic (subscript $c$) and thermal part. $P_{c}$ and $\varepsilon_{c}$ depend only upon the density of the material, $\rho$, or the specific volume, $V$ = $1/\rho$. These are equal to the total pressure and specific internal energy at absolute zero or $T = 0 \ K$. Let us assume that the specific volume at $T = 0$ and $P = 0$ is given by $V_{oc}$, which is only ~1-2% smaller than the specific volume at STP, $V_{o}$, for most metals.
The potential energy curve, or curve defining $\varepsilon_{c}$, is qualitatively similar to the potential energy curve describing the interaction between two atoms as a function of the intranuclear distance, $\Delta x_{n}$. When $V > V_{oc}$, the attractive forces dominate but fall off rapidly as the intranuclear distances increase (e.g., as $T$ increases). In other words, when the atoms move further apart $\varepsilon_{c}$ will asymptotically increase to some value $U$, which is roughly the binding energy of the atoms in the body. Thus, $U$ represents the energy required to remove all atoms from the object to infinity, which is roughly equal to the heat of vaporization for the material (I wrote some more details on the heat of vaporization and provided several useful links in this answer). For instance, the heat of atomization (similar to vaporization) for iron is roughly $415 kJ mol^{-1}$ or ~4.3 eV/atom. Thus, $\varepsilon_{c} (V) \rightarrow U$ as $\Delta x_{n} \rightarrow \sim 2$.
Conversely, the repulsive forces dominate if $V < V_{oc}$. We can define this quantitatively by considering that the work done by compressing the material will be equal to the increase in internal energy. In other words: $$ P_{c} = - \left( \frac{ d \varepsilon_{c} }{ d V } \right)_{T = 0} $$ which is equivalent to saying it is the isothermic/isentropic equation for cold compression. The negative sign shows that if a tensile force were applied to the body, the binding forces between atoms would act as a restoring force. The slope of the $P_{c}$ curve at $P = 0$ (or 1 atm) defines the compressibility of the material under normal conditions (i.e., $T = T_{o} \sim 300 \ K$). This is given by: $$ \kappa_{o} = - \frac{ 1 }{ V_{o} } \left( \frac{ \partial V }{ \partial P } \right)_{T_{o}} $$ Note that the slope of $\kappa_{o}$ defines the speed of elastic waves within the object. Thus, let us define the speed of sound in the solid as this speed, given by: $$ C_{o} = \sqrt{ - V^{2} \left( \frac{ \partial P }{ \partial V } \right)_{S} } $$ where the subscript $S$ indicates an isentropic derivative and the partial derivative will be negative to avoid imaginary speeds of sound.
Simple Zeroth Order Approximation
My knee-jerk assumption is that the simplest approach, given that you assume elastic collision relations, is to just approximate the $\Delta \varepsilon_{c}$ by the final kinetic energy of your impacting object, assuming that the impactee(?) does not move after impact.
First Order Approximation
[The following comes from Chapter XI, Sections 3.14-3.16 in Reference 1]
Below we will consider the effects on a cylindrical rod (used for symmetry and simplicity).
For small deformations, the relative change in volume, $\Delta V/V$, is given by: $$ \frac{ \Delta V }{ V } = - \kappa \ P = - \frac{ P }{ K } $$ where $K = 1/\kappa$ is the bulk modulus.
Let us define $C_{1}$ as the speed of a compression wave in the material due to the application of a constant pressure, $P$, applied to one end of the rod at some initial time. The material between the wave front and the end of the rod contracts at a constant speed, $u$. Under these conditions, we can use Hooke's law and show that for small loads and deformations we have: $$ \frac{ u }{ C_{1} } = \frac{ P }{ E } $$ where $E$ is Young's modulus. After some time, $t$, the mass of material encompassed by the wave will acquire a momentum $\rho \ C_{1} \ t \ u$, which must be equal to $P \ t$ from Newton's law, which gives us: $$ P = \rho \ u \ C_{1} $$
More Detailed Answer
Unfortunately, I do not have time to go through the full derivation but I suggest Chapter XI in Reference 1 and use Reference 2 for supporting information. Zel'dovich and Raizer basically spend all of Chapter XI discussing this topic and go into all the nuances that would apply to your problem (e.g., pressure wave induced by shock wave compression). I am guessing that much of it is going need a numerical analysis, but there are starting points and approximations that are analytical which would likely save you a lot of time.
References
15th group elements like As, Sb, P can be used for doping whereas N and Bi cannot be used for doping even though they too belong belong to 15th group. Why ?
Answer
For good doping you need two things: (1) get enough dopant in to be useful in changing carrier concentrations, and (2) having an energy level close to a band edge to generate electrons (holes) in the band, rather than making a mid-level recombination center. The below is assuming you are trying to dope Silicon. Data is generally from Sze's excellent 'Physics of Semiconductor Devices' text.
Bismuth certainly has a donor level not much lower than As, satisfying (2). However, the solid solubility of Bi is roughly 3 orders of magnitude less than As, peaking at below $10^{18}/cm^3$. This limits the utility of Bi in current device technology - you just can't get enough in.
Nitrogen has very low solubility in silicon. I cannot find quickly any info on energy levels in the gap, but oxygen has several levels, all pretty much near gap. I'd say nitrogen loses out on both (1) and (2).
I got a little puzzled when thinking about two entangled fermions.
Say that we have a Hilbert space in which we have two fermionic orbitals $a$ and $b$. Then the Hilbert space $H$'s dimension is just $4$, since it is spanned by \begin{align} \{ |0\rangle, c_a^\dagger |0\rangle, c_b^\dagger |0\rangle, c_a^\dagger c_b^\dagger |0\rangle\}, \end{align} where $c_i^\dagger$ are the fermionic operators that create a fermion in orbital $i$.
Say we have a state $c_a^\dagger c_b^\dagger |0\rangle$. Then if I partition my Hilbert space into two by looking at the tensor product of the Hilbert spaces of each orbital, i.e. $H = H_a \otimes H_b$, then my state can be written as $c_a^\dagger |0\rangle_a \otimes c_b^\dagger |0\rangle_b$, from which it is obvious that this state is unentangled ($|0\rangle = |0\rangle_a \otimes |0\rangle_b$).
Now I was thinking about writing the state in first quantized i.e. a wavefunction. Let $\phi_a(r), \phi_b(r)$ be the wavefunctions of the orbitals $a$ and $b$. Then \begin{align} \psi(x_1,x_2) = \langle x_1 x_2 | c_a^\dagger c_b^\dagger |0 \rangle = \phi_a(x_1) \phi_b(x_2) - \phi_a(x_1)\phi_b(x_2). \end{align} This is where I got confused. What object is $\psi(x_1,x_2)$, i.e. what Hilbert space does it belong to? What exactly are we doing when we do $\langle x_1 x_2 | c_a^\dagger c_b^\dagger |0 \rangle$? We seem to be changing/expanding our Hilbert space by taking the position representation?
Written in this way, and assuming the same partition $H_a \otimes H_b$, the unentangled nature of the original state is no longer manifest. I'm not sure what the partition $H_a \otimes H_b$ even means in this context. Would that be saying $\psi(x_1, x_2) = \psi_a(x_1, x_2) \times \psi_b(x_1,x_2)$ where $\psi_i(x_1,x_2)$ is a linear combination of $\phi_i(x_1), \phi_i(x_2)$? This does not seem right to me.
Regardless, now I have a state written in two different but supposedly equivalent ways, with the same partition of the Hilbert space, yet it is unentangled in one way and entangled in the other.
Help?
When solving the Einstein field equations in Schwarzschild metric for an observer falling into a black hole the radial coordinate r of the black hole and time t switch roles in the equations when r<2M.
If we transform into the resting coordinate system of an observer inside a black hole, the timelike geodesics will be along the radial dimension of the black hole. Would an observer inside the event horizon of a spherically symmetric black hole observer the radial dimension of the black hole as time? If yes, is it safe to assume that the laws of thermodynamics would hold inside the black hole, in which case the singularity of the black hole would as a zero entropy state be in the past along the radial "time" axis and the high entropy event horizon would be in the future along the same?
What would the cosmology of a spherically symmetric black hole look like from the perspective of an observer inside the black hole. It seems to me that from the perspective of an observer within the event horizon of the black hole:
How does time behave inside a black hole from the perspective of an observer inside the black hole? Could such an observer see the interior of the black hole as a universe relatively similar to ours (assuming the arrow of time would be along the radial axis of the black hole).
A couple years ago I ran upon a YouTube video demonstrating how researchers used x-rays given off by tearing tape off its spindle in hopes to miniaturized and cheapen future x-ray devices.
As of late I have been wondering what interactions are causing this. I understood from the video that the electrons are being ejected due to the added energy to the system (peeling). This instinctively pointed me towards QED where interactions between electrons produce a photon for short periods.
1)Are those same photons from electron interaction the resulting x-rays?
2)Are the frequencies of the photons being limited by the $\Delta$energy$\Delta$time uncertainty, and/or $\Delta$position$\Delta$momentum uncertainty (at the vertex of the ripping tape, where high densities of electrons may accumulate) ?
Note: I've also considered that the van der Waals force may come into consideration, not certain though.
Looking for a QED based descriptions/answers. Any elementary references to this work is appreciated.
Answer
There are two ways to get X-rays, :
1) acceleration and deceleration of charges, electrons/ions giving synchrotron radiation in one case and brehmstralung radiation in the second.
2) de-excitation of electrons falling from a higher energy level/band to a lower energy level/band with an X-ray magnitude energy difference
In 1953, Soviet scientists showed that triboluminescence caused by peeling a roll of an unidentified Scotch brand tape in a vacuum can produce X-rays.6 In 2008, American scientists performed an experiment that showed the rays can be strong enough to leave an X-ray image of a finger on photographic paper
and a recent research ;
Putterman’s group recorded X-ray emission in the form of intense bursts some billionth of a second long (with the width of the X-ray pulses calibrated using the well characterized radiowaves), and found that these bursts are correlated with very slight slippages (i.e. reductions in force) in the otherwise smooth removal of the tape from its reel.
The wiki entry attributes the phenomenon of light to triboluminescence
riboluminescence is an optical phenomenon in which light is generated through the breaking of chemical bonds in a material when it is pulled apart, ripped, scratched, crushed, or rubbed (see tribology). The phenomenon is not fully understood, but appears to be caused by the separation and reunification of electrical charges.
The more recent article speculates:
The group believes that as the tape peels the acrylic adhesive on the exposed tape becomes positively charged and the outer surface of the remaining polyethylene roll acquires a negative charge. This causes electric fields to build up to values that trigger discharges.
Breaking a bond gives kinetic energy to the constituents, and electrons and ions will be separated and left in excited states. The reorganization of charges happens with the emission of photons. For X-rays it means those bonds are strong enough to be left in breaking at energy levels high enough to release X-rays in the de-excitations.
The newer link explains it as :
The researchers say that at the reduced pressure in the experiment — about one millionth of an atmosphere — the discharges accelerate the electrons to energies that generate X-rays when they suddenly decelerate in the positive side of the tape. However, the researchers remain stumped as to how the diffuse mechanical energy needed to peel the tape is focused to the extent that it can produce X-rays, and even more strangely how it can do so in the form of nanosecond pulses. This, says Putterman, gives us a “new inspiring mystery to dig into”.
Searching on triboluminescence one finds that the exact mechanism is a research subject, explanations centering on excitations and de-excitations of the atoms in the material. See a spectrum of triboluminescence here, though not of x-ray energies.
