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Need proper interpretation of imploding/exploding self-gravitating sphere While searching for some nice and simple problems in Newton's theory of gravity, I've made this trivial calculation for which a proper physical interpretation is lacking. Some pieces are missing and I'm puzzled by the result, so here it is.
Consider a self-gravitating sphere of uniform density, of conserved mass $M$ and initial radius $R_0$. The sphere is initialy rotating with angular velocity $\omega_0$. The total mechanical energy is thus
\begin{equation}\tag{1}
E_0 = K_{\text{rot}} + U_{\text{sphere}} = \frac{1}{2} \, I_0 \, \omega_0^2 - \frac{3 G M^2}{5 R_0},
\end{equation}
where $I_0 = \frac{2}{5} \, M R_0^2$ is the initial moment of inertia of the uniform sphere. The spin angular momentum of the sphere is simply
\begin{equation}\tag{2}
S_0 = I_0 \, \omega_0.
\end{equation}
For some reason, the sphere change its radius and angular velocity (imploding or exploding). The radius is now $R \ne R_0$ and angular velocity $\omega \ne \omega_0$. The end mechanical energy and angular momentum are now these (the sphere is still of uniform density for simplicity) :
\begin{align}
E &= \frac{1}{2} \, I \, \omega^2 - \frac{3 G M^2}{5 R}, \tag{3} \\[12pt]
S &= I \, \omega. \tag{4}
\end{align}
If I impose strict conservation of BOTH mechanical energy and angular momentum : $E = E_0$ and $S = S_0$, I then get two constraints for two variables : $R$ and $\omega$. Solving these is trivial and give this :
\begin{align}
R &= \frac{R_0}{\frac{3 G M}{\omega_0^2 \, R_0^3} - 1}, \tag{5} \\[12pt]
\omega &= \frac{R_0^2}{R^2} \, \omega_0 = \Big( \frac{3 G M}{\omega_0^2 \, R_0^3} - 1 \Big)^2 \, \omega_0. \tag{6}
\end{align}
Since $R > 0$, this solution is possible only if
\begin{equation}\tag{7}
\omega_0 < \sqrt{\frac{3 G M}{R_0^3}},
\end{equation}
which is the same as saying $E < 0$. We can distinguish two cases for the final radius of the sphere, according to (5) :
\begin{align}
&R < R_0 \quad \text{if} \quad \omega_0 < \sqrt{\frac{3 G M}{2 R_0^3}}, \\[12pt]
&R > R_0 \quad \text{if} \quad \sqrt{\frac{3 G M}{2 R_0^3}} < \omega_0 < \sqrt{\frac{3 G M}{R_0^3}}.
\end{align}
This result is puzzling to me. It says that if (for some unspecified reason) a rotating uniform sphere change its size, conservation of energy and conservation of spin angular momentum gives a unique solution for its final state. If I put in the actual values for the Sun, for $M$, $R_0$ and $\omega_0$, I get a size $R$ and angular velocity $\omega$ of the same order of magnitude as a pulsar (very roughly).
The change must be brutal (no gradual or continuous change of $R$ and $\omega$), which is hard to accept !
In this case, my maths are okay but my physics is terrible (as Einstein once told Lemaître, but the other way around !)
So what is the physical interpretation of all this ? What am I
missing ? Where are the mistakes in this reasoning ? What the maths are actually saying ? I don't get it and I'm confused !
| My physics is less terrible this morning. I have a solution to my own question :
Yes, energy as defined by (3) cannot be conserved. If the sphere changed its radius, it's because something changed internally (obviously !). Some internal force is resisting gravity, until it changed for some reason. So there are at least 2 ways to show this.
First solution (mechanical) : Suppose there's a kind of repulsive internal force (pressure, springs, wathever) that counter-balance gravity. Then its energy potential should be added, if it can be described by a potential. I'm using a "spring" energy potential as an example :
\begin{equation}\tag{1}
E = \frac{1}{2} \, I \, \omega^2 - \frac{3 G M^2}{5 R} + \frac{1}{2} \, k R^2.
\end{equation}
If the spring constant $k$ continuousy change for some reason : $k_0 \Rightarrow k_1$, then the sphere need to collapse/expand to adapt to the new situation. Energy (1) is conserved. Susbsituting the conservation of angular momentum to eliminate $\omega$ (and put in $\omega_0$), then we can find the final radius $R_1$ as a complicated function of $R_0$, $\omega_0$, $k_1$ and $k_0$ (the algebraic equation is of the fourth order !). Then the radius can change continuously since there's a new variable that enters the game (the constant $k$), that can also change continuously.
Second solution (thermodynamical) :
In place of an internal potential energy, we can simply state that some "heat" is produced by the change of state of the sphere. Keeping mechanical energy (3) from my question, we simply have to write
\begin{equation}\tag{2}
E_1 = E_0 + Q,
\end{equation}
where $Q$ is a new continuous variable that enters the game. That heat may stay inside the sphere after the state change, or be released outside, it doesn't matter. We then can solve the energy conservation (which is now an algebraic equation of the second order, if $Q$ is independant of $R$) :
\begin{equation}\tag{3}
E_1 = \frac{1}{2} \, \Big( \frac{2}{5} \, M R_1^2 \Big) \Big( \frac{R_0^2}{R_1^2} \, \omega_0 \Big)^2 - \frac{3 G M^2}{5 R_1} = \frac{1}{2} \, \Big( \frac{2}{5} \, M R_0^2 \Big) \, \omega_0^2 - \frac{3 G M^2}{5 R_0} + Q.
\end{equation}
I'll define the following dimensionless parameters to simplify things :
\begin{align}
p &= \frac{3 G M}{\omega_0^2 \, R_0^3}, \tag{4} \\[12pt]
q &= \frac{5 Q}{M \omega_0^2 R_0^2}. \tag{5}
\end{align}
Parameter $q$ can be negative or positive, depending of the case. Then, the solution to (3) is this :
\begin{equation}\tag{6}
R_1 = \frac{p \pm \sqrt{p^2 - 4 (p - q - 1)}}{2 (p - q - 1)} \, R_0.
\end{equation}
Since $q$ can take any value, the final radius $R_1$ is a continuous function of $R_0$, $\omega_0$ and $Q$. If the dimensionless parameter $q$ is small, then to first order :
\begin{equation}\tag{7}
R_1 \approx \Big( \frac{1}{p - 1} - \frac{q}{(p - 2)(p - 1)^2} \Big) \, R_0.
\end{equation}
In the simpler case of a non-rotating sphere : $\omega_0 \rightarrow 0$, then the solution to (3) is simply this :
\begin{equation}\tag{8}
R_1 = \frac{R_0}{1 - \frac{q}{p}} = \frac{R_0}{1 - \frac{5Q R_0}{3 G M^2}},
\end{equation}
which can take any value, depending of $Q$ (positive or negative heat).
I think this solution solves all the issues. It even has great pedagogical values, in my opinion.
| {
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"timestamp": "2023-03-29T00:00:00",
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First law of thermodynamics, steady flow energy equation (SFEE) and $Vdp$ work Can first law of thermodynamics defined for a closed system be applied to the steady flow energy equation? Why?
I came across the derivation of $Vdp$ work and Every book applied the first law defined for closed system to steady flow energy equation.
Please clarify
| OH , i get it now. You are right.We can't apply the first law if you don't consider the entire system and if the energy can leak in or out by other means ,say a more energetic fluid comes in. But in steady state flow ,no particular trait of a fluid changes with time.This means that the property of a fluid at a particular spatial point is constant in time(to be more mathematical,it's partial derivative with respect to time is zero).This means that the only way the energy of fluid can change in a constraint volume is by change in kinetic energy,change in potential energy or by work done.
For detailed explanantion,go to https://wiki.ucl.ac.uk/display/MechEngThermodyn/First+law+applied+to+flow+processes
| {
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Deviation when light passed through optical centre I recently get to know that when light pass through optical centre then it shows a very very slight deviation but why? Why doesnt it pass through optical centre extremely straight?
And can i conclude this that when an light is travelling in direction of optical centre then it is actually travelling along the normal as in both cases light passes almost undeviated?...
I am really confused ...a help from you will be greatly appreciated..
| This is simply because real lenses have nonzero thickness, and we must one of several methods for dealing with them, such as those presented on the Hyperphysics Website. But this answer is really a slight generalization of Emilio Pisanty's comment:
Why do you find this that surprising? The same thing happens in a slab of glass with straight parallel surfaces.
For a nonzero thickness, rotationally symmetric imaging system, the system can be modelled by two principal planes.
Principal planes work as follows: you can calculate the paraxial behavior of any ray using them by the following recipe:
*
*Ray entering system propagates to the nearest principal plane. In the diagram below, for a ray propagating from left to right, it would meet plane $P_1$ first;
*The ray then "teleports" to the other principal plane $P_2$ and begins at the transverse same position relative to the optical axis as it met the plane $P_1$;
*The deviation of the ray is calculated as though a thin lens of the same focal length as the whole system's focal length were present at plane $P_2$.
For light propagating from right to left, we have the analogous process on the bottom diagram: propagate to $P_2$, teleport to $P_1$ preserving the transverse position, then the ray is acted on by a thin lens with focal length given by the system's focal length.
The two focal lengths, whether travelling from left to right or right to left, are equal if the refractive indices of the two mediums at either end of the lens are equal. Otherwise, the ratio between the focal lengths is the ratio of the corresponding refractive indices.
| {
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What's the conversion between apparent magnitude and lux? I know the apparent magnitude of the sun, but I'd like the units in lux. Please also direct me to a reliable source if possible.
| We can convert from magnitude to lux using the equation
$$
E_v=10^{(-14.18-M_v)/2.5}
$$
where $M_v$ is the apparent magnitude in the visible band, and $E_v$ is the illuminance in lux. You can read more about it on this page: http://stjarnhimlen.se/comp/radfaq.html#7.
| {
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Why are elements with even atomic number more abundant? In reading this article about the origins of elements, I found the following diagram:
What strikes me about this image is the very consistent zig-zagging of the line that appears to indicate that elements/isotopes with an even number are more abundant.
Am I correct? What's going on here?
| I will add to the answer from @BowlofRed https://physics.stackexchange.com/a/158270/36194 that the nuclear pairing interaction lowers the energy in nuclei where the number of like nucleons is even: thus for instance there are more isotopes with even rather than odd number of neutrons. This also favors the formation of even-proton-numbered nuclei over the neighbouring odd ones.
| {
"language": "en",
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Traveling close to the speed of light, would a person (or anything else) have a longer existence or would the existence be passing in slow motion? If something, let's say, an electron or a person, travels at some speed close to the speed of light, time would slow down, right?
But would it be passing in slow motion or would it have a bigger amount of time to 'live'?
| With respect to you, i.e, in your reference frame, you would be the same. You would not see length contraction or experience time dilation. However,say, from the frame of the earth, the people would see you at slow-mo.
| {
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Why is my solar panel more efficient as the temperature increases, rather than less efficient? I have recently performed an experiment in which I placed a solar panel inside a closed box, together with a heater and a lamp (60W). I increased the temperature using a heater from 20ºC to 50ºC. The lamp remained switched on the whole time.
I´ve read on the internet that the optimum temperature at which solar panels work is 25ºC. After that, the efficiency starts to decrease. Nonetheless, my data shows the opposite. The higher the temperature, the more efficiency it has. I don´t understand how this is possible. The current is supposed to increase, and the voltage to decrease, but in my data both of them increase linearly.
I don´t know if it´s because I have a small solar panel, which has different purposes, or if there´s a systematic error. I used a milliammeter and a millivoltmeter.
If anyone could help, or explain why this happens I would really appreciate it.
Thank you!
| Your voltages are increasing with temperature. This is in contraction to what one expects!
Was optical input power kept constant? Solar cells do get more efficient as sunlight is concentrate on to them. The voltage increase as you are seeing here. That’s one possible explanation. If the input power was kept constant, then this effect cannot be due to light concentration. So the next most likely explanation is you are improving the diode characteristic of the cell, by increasing the average carrier density and saturating trap states? Likely if you solar cell is a-Si (i.e. the reddish looking type often used in calculators).
| {
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The definition of the Lorenz gauge condition The inner product of two vectors in space-time is:
$$(x_1, y_1, z_1, t_1) \cdot (x_2, y_2, z_2, t_2) = x_1 x_2 + y_1 y_2 + z_1 z_2 - t_1 t_2$$
So
$$(\frac{\partial }{\partial x}, \frac{\partial }{\partial y}, \frac{\partial }{\partial z}, \frac 1c \frac{\partial }{\partial t}) \cdot (A_1, A_2, A_3, \phi) = \text{div}(\vec A) - \frac 1c \frac{\partial \phi}{\partial t}$$
is Lorentz invariant, where $\vec A=(A_1, A_2, A_3)$. But the [Lorenz gauge condition] (https://en.wikipedia.org/wiki/Lorenz_gauge_condition) is defined by $\text{div}(\vec A) + 1/c\ \partial_t \phi=0$. Why has the minus changed into plus? So there is apparently no longer invariance.
| The inner product of two spacetime vectors is given by
$$V^{\mu}W_{\mu}=V^{0}W_{0}+\textbf{V}\cdot\textbf{W}.$$
Note that there is no inherent minus sign in this definition. The minus signs only come in when your vector $W$ is naturally described with an upstairs (contravariant) index. In ths case, we would write
$$V^{\mu}W_{\mu}=-V^{0}W^{0}+\textbf{V}\cdot\textbf{W},$$
because our signature requires $W^{0}=-W_{0}$. Your example of $x\cdot y$ only has a minus sign because coordinate vectors are naturally contravariant.
Now on to your question: the Lorenz gauge is defined as
$$\partial_{\mu}A^{\mu}=\frac{\partial}{\partial t}A^{0}+\boldsymbol{\nabla}\cdot\textbf{A}=0.$$
Now, since the vector potential $A$ naturally has an upstairs index, we can write $A^0=\phi$. Thus, we have
$$\frac{\partial\phi}{\partial t}+\boldsymbol{\nabla}\cdot\textbf{A}=0.$$
No need for all of the finicky minus signs (they're annoying as hell)!
I hope this helped!
Note: I have used units where $c=1$ in this answer. Also note that I have a tendency to mix up the terms "covariant" and "contravariant." Please correct me if I've made a mistake.
| {
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What is a $D^{0 \ast}$ meson? What are the quark contents of $D^{0 \ast}$ meson?
How to distinguish between $D^{0 \ast}$ meson and $D^{0}$ meson as I guess both have same quark content.
| There are many excited states of flavoured mesons ($K$s, $D$s and $B$s), which are identified by their mass, spin and parity. The mass is usually written in parentheses in MeV (omitted for ground states), the subscript contains the spin (and the lighter quark if it isn't a $u$ or $d$), and the superscript contains the charge and sometimes a star.
The star tells you what the parity is in relation to the spin: positive for even spin, negative for odd. If there's no star then it's the other way around.
From the PDG naming scheme for hadrons in the section on flavoured mesons:
*If the spin-parity is in the "normal" series, $J^P = 0^+, 1^-, 2^+,...$, a superscript "$^*$" is added.
| {
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Differences between optical laser and amplifier I am preparing for an applied optoelectronics exam and I am having some trouble with telling the differences between optical fiber laser and optical amplifier. For now I only came up with following differences:
*
*No Bragg reflectors and optical resonators in amplifiers
*No signal conversion in optical amplifiers
I'm pretty sure there's more to it but I couldn't find anything that would answer this question fully.
| An optical laser without a feedback mechanism is essentially an optical amplifier. The feedback mechanism can be made using 2 reflectors, mirrors, or gratings that creates an optical resonator cavity.
Another difference between optical amplifier and laser is that amplifier integrates isolators to avoid feedback, backreflection or lasing effects. The tunability of laser comes from bandwidth of its amplifier (active gain medium)
| {
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Is there a more accurate form of the mirror equation $\frac{1}{f}=\frac{1}{u} + \frac{1}{v}$? In the mirror equation $$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$$
Q1: Are $u$ and $v$ the distances from the object to the mirror surface or the distance from the object along the principal axis to the pole?
These distances differ by a very small amount but they exist all the same. I am familiar with the usual derivation of the formula using a diagram, alternate angles, an isosceles triangle and then a small angle approximation.
Q2: Is there a formula that doesn't make any approximations?
| There is a more general (exact) formula for a spherical mirror. This formula was discovered by H. A. Elagha and was published in the journal of the optical society of america in 2012. The paper has the title: "Exact ray tracing formulas based on a nontrigonometric alternative to Snell's law" . This formula has the form:
$$\dfrac{1}{R-S_0}+\dfrac{1}{R-S_1} = \dfrac 2R\sqrt{1-\left(\dfrac hR\right)^2}$$
where $S_0$ and $S_1$ are the the distances of the object and the image from the mirror vertex respectively. h is the height of the point of incidence at the mirror and $R$ is the radius of curvature.
| {
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Fall of the number density $n$ with the scale factor $a(t)$ for a relativistic particle species in equilibrium? Consider the thermal dark matter (DM) scenario. Before the dark matter got frozen out, it was both in chemical and thermal equilibrium with the other particles in the early universe. At this point of time, when the dark matter was in equilibrium, the number densities of both the DM and the relativistic particles with which it were equilibrium depleted with the fall of temperature as $$n\propto T^3\tag{1}$$ where $T$ is the temperature of the universe.
Intuitively, the depletion in the number density with the expansion of the Universe should be like $$n\propto a^{-3}(t)\tag{2}$$ where $a(t)$ is the scale factor.
Now, my question is, whether the relations (1) and (2) are related. And intuitively, they should be because the fall in temperature is due to the expansion of the Universe. And if they are, how do we get relation (1) from (2)?
| The basic principle that is useful for addressing the issue you are talking about is conservation of entropy. In thermal equilibrium, the comoving entropy is conserved, and this can be used to find out how the temperature changes with the expansion of the universe.
Since the entropy density of relativistic species usually dominate the total entropy, it is useful to define the total entropy in terms of an effective number of relativistic degrees of freedom (for entropy), $g_{*\text{S}}$
\begin{equation}
s_\text{tot} = g_{*\text{S}} \frac{2 \pi^2}{45}T^3,
\end{equation}
where
\begin{equation}
g_{*\text{S}} = \sum_{\text{bosons}} g_i \left(\frac{T_i}{T}\right)^3 + \frac{7}{8} \sum_{\text{fermions}} g_i \left(\frac{T_i}{T}\right)^3,
\end{equation}
where $T$ is the temperature of the heat bath, and we have allowed for the possibility that some relativistic species have decoupled from the heat bath and have a different temperature, $T_i$. Note that we have also assumed here that the chemical potentials are negligible.
Assuming that the comoving entropy density is constant in time we get
\begin{equation}
\frac{d }{d t} \left(s a^3\right) = 0.\tag{1}
\end{equation}
This means that we can directly relate the temperature, $T$, and the scale factor, $a$
\begin{equation}
T \propto \frac{g_{*\text{S}}^{-1/3}(T)}{a}.\tag{2}
\end{equation}
So we see that as long as the number of relativistic degrees of freedom, $g_{*\text{S}}$, does not change then we have $T \sim 1/a$, and your (1) and (2) are compatible.
If the number of relativistic degrees of freedom changes, then your equation (2) is not valid any more. This is because heat is either added to or subtracted from the heat bath, increasing or decreasing the comoving number of particles of any given species.
(1) will always be valid, however, as long as a species is in thermodynamic equilibrium (with no chemical potential).
| {
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Why does a surface always exert force normal to it? In whichever angle an object is thrown at a surface, the surface always exert force normal to it. But why? According to Newton's third law, if an object hits a surface at an angle, the reaction force provided by the surface must be equal and opposite to the applied force by the object. But why does the surface always exert force normal to it?
| From a purely Newtonian-dynamics perspective the origin of the normal force (or any force) is not explained. Instead, we infer the existence and direction of the normal force by observations of acceleration: if you see two solid objects not sinking into one another despite the influence of gravity, they must be exerting a relative force upon one another that counteracts gravity - otherwise the net force would be nonzero, so there would be a relative acceleration.
What generates that force microscopically? Well, it's a bit complicated, depending on the details of the structure of the matter in question. At the end of the day this question requires a detailed explanation of the structure of matter, since for example liquids, solids, and gasses will exert different "normal forces". But roughly, it is a combination of electrostatic repulsion and Pauli exclusion between electrons in matter.
| {
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Why would an infinity corrected objective lens produce an image without a tube lens? I'm having a hard time wrapping my head around this situation I've come across. I'm essentially recreating a homemade microscope that my understanding would suggest shouldn't work. This setup uses a 10X Olympus PLAN N objective attached to a beam splitter to provide illumination. Then, 55 mm worth of extension tubes connect to a camera (Point Grey Grasshopper3 with a 1/1.2" sensor). There is no tube lens between the infinity corrected objective and the camera sensor (other than the beam splitter).
My understanding of how infinity corrected microscope systems works would seem to suggest that this shouldn't work, but a correctly oriented and seemingly undistorted image shows up on the camera.
Presumably, we aren't actually realizing the actual magnification this objective is designed to provide (in fact, its about half what it should be the sensor is 8 mm tall and a ruler placed in the view shows about 1.5 mm across the short dimension of image).
Can someone clarify what the light path looks like here and how we are actually able to see an image? Is it possible that the image quality is actually really poor, but we just seem to be getting results that are more than suitable for how we are using it?
| Just like any other lens, Objective lens is a lens, without the Tube lens also it will form an image by following the Lens equations, Lens Equations.
It is just that if working distance of the objective lens would be changed, the camera can image at different distances.
The above link gives an equation for single lens but this can be extrapolated to a multi-lens system.
| {
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what is difference between region around a charge and region out of it? Basically I want to know what makes a region around a charge particle so affective that another charge will have energy when brought in this region or it will experience a force when enters into this region
| A Charge experiences a force when brought close to another charge because the charges have a field of force which is the electric field around them. For why these fields exist, no one really knows exactly why as some explanations include quantum field theory as that's how our universe works. For point charges the force between them is given by Coulomb's law:
$$F=\frac{Q_1Q_2}{4\pi\epsilon_0r^2}$$
Where $Q_1$ and $Q_2$ are the charges and $r$ is the distance between charges, while the other terms are constants.
The charges get electric potential energy when one is moved closer to another because work/energy is required/found when a force( here the electric force) acts on a charge through a distance( work done = force x distance moved in direction of force) , hence these charges have an electric potential energy in an electric field formed from them.
| {
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How is it that in a car crash, four 8mm bolts can anchor the seat to the car? In a car crash at for example twenty metres per second. I used suvat equations and newtons second law to work out the force as as body accelerates(negatively). I estimated that the distance travelled in the crash by the body would be roughly 0.4 metres.Even using average mass of a human and car seat the force calculated was way too large to be accurate as the tensile strength of steel would be easily exceeded. I concluded that a large portion of energy is transferred by the front of the car before it affects the body. My question is how could I find an accurate but rough figure for the force in newtons acting on each individual bolt and if anybody has any data or estimates.
Thank you
| You could apply a pseudo force and treat this a static problem.
The pseudo force is the mass of the seat and passenger times the deceleration. It acts forward on the centre of mass, applying a clockwise torque (assuming the car moves to the right). The seat pivots about the forward 2 bolts, the rear 2 bolts provide a counter-clockwise torque. Balancing the torques will give you the force in each of the rear bolts.
However, this assumes that the passenger is attached rigidly to the seat. If the passenger is not wearing a seat belt, this assumption does not hold. And seat belts are usually attached to the chassis rather than the seat, providing an additional counter-clockwise torque which relieves the stress on the bolts.
| {
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Ising anyon topological order and its edge $c=1/2$ CFT We know that conformal field theories are closely related to two-dimensional topological orders via edge-boundary correspondence. An Ising topological order can be obtained by gauging the fermion parity from a $p+ip$ superconductor. The anyon fusion rule $\sigma\times \sigma = 1+ \psi$ (where $\sigma$ is the vortex excitation that binds a Majorana zero mode, from which the above fusion rule is easily identified) indicates its relation with the free (Majorana) fermion CFT with $c=1/2$.
Indeed, on its edge, there is a chiral mode with $c=1/2$. Besides the fermion mode, such a CFT has a twist operator $\sigma$ with $h=1/16$. My question is what is this operator in the context of $p+ip$ superconductor? What is its relation with the Majorana-binding vortex in the bulk? How do I understand its fusion rule $\sigma\times \sigma = 1+ \psi$ in the edge CFT sense?
| p+ip superconductor is an invertible topological order whose intrinsic bulk excitations are fermions. There is no non-abelian anyons. The vortex with Majorana zero mode is not an intrinsic bulk excitation.
The $SU(2)_2$ QH state $\chi_1(z_i)\chi_2^2(z_i)$ and the Paffian QH state have Ising topological order. They have $\sigma$ non-abelian particle as intrinsic bulk excitations.
(Here $\chi_n$ is the many-fermion wave function with $n$ filled Landau levels.)
| {
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How is the equation of Mach number derived? Wikipedia states that for a pitot-static tachometer, the mach number for subsonic flow equates to
$$M = \sqrt{5\left[\left(\frac{p_t}{p_s}\right)^\frac{2}{7}-1\right]}.$$
How did they get to that result? Is there a derivation, or is it just from a polynomial fit of a tabulated set of data?
Update
I accepted J.G's answer after glancing at the referenced flight test document (a treasure in itself) and realising that $\frac {7}{5}$ is the same as 1.4, but there remains an issue.
Sadly I don't have my uni books anymore with Bernoulli's equation for compressible flow. The issue is with dynamic pressure: for incompressible flows we can take $p_d = \frac {1}{2} \cdot \rho \cdot V^2$, for compressible flow this is $p_d = \frac {1}{2} \cdot \gamma \cdot p_{static} \cdot M^2$.
Right? If I substitute this I don't get to the equation above. So the answer is unfortunately not accepted anymore.
| The factors are not obvious, I agree.
For instance, for a polytrope index, $\gamma$, of 7/5 the exponent of 2/7 corresponds to a term of the form $\left( \tfrac{\gamma - 1}{\gamma} \right)$, which is our first hint. The second hint is that the pitot tube system can be applied to a Bernoulli system. The third thing to note is that for subsonic speeds, which is where a pitot tube actually functions, one can get away with assuming incompressible flow (I know it seems odd since things obviously do compress a little, but the effects can be considered secondary for most intents and purposes).
For a polytropic ideal gas, we know that $P \propto \rho^{\gamma}$. Thus, we can say that:
$$
P = \kappa \ P_{s} \ \rho^{\gamma} \tag{1}
$$
where $P_{s}$ is the static pressure (also can be considered the pressure at infinity). We can rewrite this equation in terms of density to find:
$$
\rho = \kappa^{-\frac{1}{\gamma}} \ \left( \frac{P}{P_{s}} \right)^{\frac{1}{\gamma}} \tag{2}
$$
The differential form of Bernoulli's equation can be given as:
$$
u \ du + \frac{ 1 }{ \rho } \ \frac{ d P }{ d \rho } \ d \rho = 0 \tag{3}
$$
and we know that the speed of sound is given by:
$$
\begin{align}
C_{s}^{2} & = \frac{ \partial P }{ \partial \rho } \tag{4a} \\
& = \gamma \ \kappa \ P_{s} \ \rho^{\gamma - 1} \tag{4b} \\
& = \frac{ \gamma \ P }{ \rho } \tag{4c}
\end{align}
$$
If we replace the $\rho$ in Equation 4b with the form shown in Equation 2, one can show that the 2nd term in Equation 3 can be rewritten as:
$$
\begin{align}
\frac{ 1 }{ \rho } \ \frac{ d P }{ d \rho } \ d \rho & = \frac{ \gamma \ \kappa \ P_{s} }{ \rho } \ \rho^{\gamma - 1} \ d \rho \tag{5a} \\
& = \frac{ \gamma \ \kappa \ P_{s} }{ \rho } \ \kappa^{-\frac{ \gamma - 1 }{ \gamma }} \ \left( \frac{ P }{ P_{s} } \right)^{\frac{ \gamma - 1 }{ \gamma }} \ d \rho \tag{5b} \\
& = \frac{ \gamma \ \ P_{s} }{ \rho } \ \kappa^{\frac{ 1 }{ \gamma }} \ \left( \frac{ P }{ P_{s} } \right)^{\frac{ \gamma - 1 }{ \gamma }} \ d \rho \tag{5c}
\end{align}
$$
If we differentiate Equation 2, we find:
$$
d \rho = \left( \frac{ \rho }{ \gamma \ P_{s} } \right) \ \left( \frac{ P }{ P_{s} } \right)^{-1} \ dP \tag{6}
$$
so that Equation 5c can be rewritten as:
$$
\frac{ \gamma \ \ P_{s} }{ \rho } \ \kappa^{\frac{ 1 }{ \gamma }} \ \left( \frac{ P }{ P_{s} } \right)^{\frac{ \gamma - 1 }{ \gamma }} \ d \rho = \kappa^{\frac{ 1 }{ \gamma }} \ \left( \frac{ P }{ P_{s} } \right)^{-\frac{ 1 }{ \gamma }} \ dP \tag{7}
$$
We define $u \ du \rightarrow C_{s}^{2} \ M \ dM$, thus we rewrite Equation 3 as:
$$
C_{s}^{2} \ M \ dM + \kappa^{\frac{ 1 }{ \gamma }} \ \left( \frac{ P }{ P_{s} } \right)^{-\frac{ 1 }{ \gamma }} \ dP = 0 \tag{8}
$$
We also define $\alpha = \tfrac{ P }{ P_{s} }$ so that $dP \rightarrow P_{s} \ d\alpha$. If we integrate Equation 8 with the limits ranging from $P_{s}$ to $P$, the change of variables makes the 2nd term go to:
$$
\kappa^{\frac{ 1 }{ \gamma }} \ P_{s} \int_{\alpha}^{1} \ d\alpha \ \alpha^{-\frac{ 1 }{ \gamma }} = \left[ \frac{ \gamma \ \kappa^{\frac{ 1 }{ \gamma }} \ P_{s} }{ \gamma - 1 } \ \alpha^{\frac{ \gamma - 1 }{ \gamma }} \right]_{\alpha}^{1} \tag{9}
$$
Thus, Equation 8 can be rewritten as:
$$
0 = \frac{ 1 }{ 2 } C_{s}^{2} \ M^{2} - \frac{ \gamma \ \kappa^{\frac{ 1 }{ \gamma }} \ P_{s} }{ \gamma - 1 } \left[ \alpha^{\frac{ \gamma - 1 }{ \gamma }} - 1\right] \tag{10}
$$
which after some algebra reduces to:
$$
M^{2} = \frac{ 2 }{ \gamma - 1 } \left[ \alpha^{\frac{ \gamma - 1 }{ \gamma }} - 1\right] \tag{11}
$$
As stated above, for $\gamma$ = 7/5, this results in the form about which you are concerned.
| {
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At what temperature do the laws of physics break down? I heard that as approaching the temperature of a kugelblitz the laws of physics break down, I saw this in the video The Kugelblitz: A Black Hole Made From Light, by SciSchow Space.
| Hank Green is describing the concept of the Planck temperature,
$$
T_\mathrm{P} = \sqrt{\frac{\hbar c^5}{Gk_B^2}}\approx1.4\times 10^{32}\:\mathrm K,
$$
which is defined as $\frac{1}{k_B}$ times the Planck energy $E_\mathrm{P}=\sqrt{\hbar c^5/G}\approx 1.9\times 10^{9}\:\mathrm J$.
As with all the Planck units, we don't really know what happens at those scales, but we're pretty sure that the laws of physics as we know them are likely to require modifications to continue describing nature at some point before you reach that regime.
What doesn't happen at the Planck scale is that "the laws of physics break down", which is a meaningless catchphrase that shouldn't be used. Unless, in fact, the world changes so much that there is no regularity to physical phenomena and no way to predict how an experiment will pan out, even in principle, then what you have is not a breakdown of the laws of physics, it's just that you've left the region of validity of the laws you know, and you need to figure out what the laws are on the broader regime.
| {
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Why is Higgs particle detected much later than top quark when it's lighter? The Higgs boson is lighter than the top quark. But the top quark was discovered in the mid-1990s where the Higgs boson escaped detection for two more decades. So if the energy has already been achieved to produce Higgs boson, why did it escape detection so far?
I understand that the couplings of Higgs boson to fermions is small and doesn't interact with the detector appreciably.
Does it mean that in LHC, with the increase in energy, the Higgs coupling increased and we finally detected Higgs?
| The Higgs being a scalar uncharged particle has the same quantum numbers as the vacuum, making it much harder to detect.
| {
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Will overlapping two different beams of coherent light with different wavelength cause interference? If I use two different wavelength lasers to transmit light into a single mode optic fiber will they interfere with each other? If so, how much will be that interference.
| Interference is a concept that only has true meaning when comparing two signals of the same wavelength/frequency.
For waves with different wavelength, it is true that the snapshots of the electric field (or magnetic field for that matter) will change because of the two signals, but there is nothing coherent about such addition.
By coherent, it means that the interference persists spatially and/or temporally.
Consider two plane waves with frequency $\omega_1,\omega_2$ and phases $\phi_1,\phi_2$. Examine their total intensity:
$$I_{\text{coherent}}=|e^{i(\omega_1t+\phi_1)}+e^{i(\omega_12+\phi_2)}|^2=2(1+\,\text{cos}(\Delta\omega t+\Delta\phi))$$
$$I_{\text{incoherent}} = |e^{i(\omega_1t+\phi_1)}|^2+|e^{i(\omega_12+\phi_2)}|^2=2$$
Notice that the intensity for an incoherent intensity, the sum is simply "$2$", but if interference is taken into account it deviates from that value. However, that deviation will rapidly vary in time if there is a frequency difference between the two beams $\Delta\omega\neq0$, and will be washed away if you average in time. This is why you don't get "interference" between two wavelengths of different values, the interference averages to zero.
However, if you were to look at a single moment in time, yes the intensity will fluctuate on the scale of $1/\Delta \omega$, so this does matter when you do time-resolved measurements.
| {
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Determining mass radius and charge radius of electrons First the mass radius problem:
Why can't the mass radius of electrons be determinded by shooting neutral particles on it. Similar to Rutherford's gold model only a bit more sophisticated.
Secondly the charge radius problem:
I often hear the term charge radius and the charge radius for protons has been calculated with accuracy for example described in this article:
https://phys.org/news/2016-08-deuterium-nucleus-proton-radius-puzzle.html
Why can't the charge radius of electrons be determinded in a similar way?
| One reason why a Rutherford-like experiment would't work in the case of the electron is because electron is an elementary particle i.e. it is not made of something else, so shooting particles at it won't have any effect in this sense (of course you can create another particle by colliding for example an electron and a positron, but this will not give you information about the structure of the electron). On the other hand, a proton is made of quarks, so you can do a Rutherford experiment in order to see it's inner structure (and it was done with a very high accuracy). However, in both cases, it is hard to define a precise radius, as both particles have a wave-like behavior so you can define a region in space where you can find the particle with 99% (let's say) certainty, but you can't define it as a solid sphere with a certain radius.
| {
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Conjugate variables in thermodynamics vs. Hamiltonian mechanics According to Wikipedia, the canonical coordinates $p, q$ of analytical mechanics form a conjugate variables' pair - not just a canonically conjugate one.
However, the "conjugate variables" I directly think of are the quantities of thermodynamics - e.g. Temperature and Entropy, etc.
So, why both these classes of variables are called "conjugate"? What is the relation among them?
| *
*Conjugate variables $(q^i, p_i)$ are given in thermodynamics via contact geometry as the first law of thermodynamics $$\mathrm{d}U~=~ \sum_{i=1}^np_i\mathrm{d}q^i,\tag{1}$$ where $U$ is internal energy. See also Ref. 1 and this & this Phys.SE posts.
*Conjugate variables $(q^i, p_i)$ are given in Hamiltonian mechanics via symplectic geometry as Darboux coordinates, i.e. the symplectic 2-form takes the form $$\omega ~=~\sum_{i=1}^n\mathrm{d}p_i\wedge \mathrm{d}q^i.\tag{2}$$
Hamilton's principal function $S(q,t)$ satisfies
$$ \mathrm{d}S~=~ \sum_{i=1}^np_i\mathrm{d}q^i-H\mathrm{d}t,\tag{3}$$
cf. Ref. 2.
References:
*
*S. G. Rajeev, A Hamilton-Jacobi Formalism for Thermodynamics, Annals. Phys. 323 (2008) 2265, arXiv:0711.4319.
*J. C. Baez, Classical Mechanics versus Thermodynamics, part 1 & part 2, Azimuth blog posts, 2012.
| {
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About light in the universe As a light source in the universe (e.g. sun) emits light in different directions, some of the light emitted reaches places like Earth, and some doesn't. So does the light that reaches the Earth disappear or it is reflected in other directions? And for the light that doesn't reach any place, does it keep on going forever? If it does keep on going forever, will the universe become brighter and brighter? Thanks!
| When any source emits light it may hit any objects like Earth and it will be reflected back in universe and yes the light which doesn't hit any thing will travel forever but that doesn't mean that the universe is going to be brighter because as the universe expands the light with in it gets red shifted and eventually get out of visible spectrum
There is a good video which will help you to get it
https://m.youtube.com/watch?v=3tCMd1ytvWg
| {
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Why surface normal is used while defining flux through an open surface? What is the significance behind defining normal to any surface? Why we do it?
| Suppose that you want to find out the rate of a river using a circular hoop which has attached to it a device which will measure the mass of water flowing through the circular hoop per second - the flux of water.
With the plane of the circular hoop placed at right angles to the river flow you get a reading of $M\, \rm kg\,s^{-1}$ which represents the mass of water flowing through the circular hoop per second.
Now place the circular hoop so that its plane is parallel to the flow of water and no water will flow through the hoop.
If the angle between the plane of the hoop and the flow of water is $\theta$ then the rate of flow will be $M \sin \theta \, \rm kg\,s^{-1}$.
So the measured rate of flow depends on the orientation of the plane of the circular hoop and the direction of flow of water.
Instead of angle between the plane of the circular hoop and the direction of flow being used, the angle between the normal to the plane of the circular hoop and the direction of water flow can be used and this is what is usually done.
What you are really doing is using the projected area of the circular hoop $A \cos \theta$ which is at right angles to the flow of water where in this case $theta$ as the angle between the normal to the plane of the hoop and the direction of flow of water.
A good example of the use of such a normal is to estimate the rate at which energy from the Sun hits the Earth of radius $R$.
If the energy flux from the Sun is $E \,\rm W m^{-2}$ then the solar power arriving at the Earth is not
$W \times \text{area of a hemisphere} = W \,2 \pi R^2$
but is
$W \times \text{projected area of a hemisphere onto a plane at right angles to the incident sunlight } = W \, \pi R^2$
To find the projected area one needs to know not just the actual area but the angle between the normal to the area and the incoming radiation.
| {
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Maximum extension of a vertical spring when given a blow In the question mentioned above , i considered the length of the spring in the equilibrium position to be the natural length and the P.E. to be zero as we are free to consider any length as the natural length. After , giving a blow , we impart K.E. to the block equal to $ 0.5m v^2$. Lets say it moves down through a distance x , then , decrease in gravitational P.E. = $mgx$ and elastic P.E.= $0.5k x^2 $.
Hence , the equation becomes, . But it does not give the correct answer.
I think that we should neglect gravity but why ?
My textbook says:
| You should ignore the gravitational contribution since you are assuming that the initial position is the equilibrium position.
Adding the effect of gravity will shift the equilibrium position to a new height.
| {
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Why do hydrostatic fluid exert force on any wall normally with magnitude $ρgh$ always? I understand the derivation in middle of container, but for a container like an inverted cone the particle very near to the wall exerts pressure on wall of magnitude $\rho gh$ normally. I ask “why?” and tried to discover it, but I had some problems.
Just like in normal derivation I assumed that since very small volume of liquid is at rest, it experiences equal and opposite force from both the liquid side and wall side, but from the liquid side in the normal direction to the wall pressure is found to be $2\rho gh$ (integrating all the normal components of pressure), then the same amount of pressure (by normal force) is produced by wall, so that liquid is stationary and due to this a reaction force is developed in wall with $2\rho gh$ magnitude normally...
Where did I go wrong?
| Imagine a piston of small area, A inside an empty cylinder, close to its bottom end, which is open. The cylinder is immersed in liquid, at any angle, so that the piston is at depth h and orientated at any angle. When the piston is pushed down the cylinder by a small distance $\Delta x$, let's say a force $F_\bot$ normal to the piston is needed. Thus the work done by the piston on the liquid is $F_\bot \ \Delta x$. But the displaced liquid (volume $A \Delta x$) makes the liquid level rise; effectively the displaced liquid rises by height $h$ and gains gravitational potential energy $mgh = \rho A \ \Delta x \ gh.$ If the fluid is perfectly non-viscous, or the process is very slow (quasi-static), we can equate the work done to this gain in PE. Thus: $$F_\bot \ \Delta x=\rho A \ \Delta x \ gh.$$ Dividing both sides of the equation by $A \Delta x$ and writing $p=\frac{F_\bot}{A}$ we get $$p=h \rho g.$$
Once you've seen the power of this argument, you'll see that it works anywhere in the liquid, including at a wall of the container. Indeed you could make a small part of the wall the movable piston itself.
| {
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Work done by static friction on a car The tires of a car execute pure rolling. Therefore, the work done by friction on the tires (and hence the car) is zero. If no external work is done, how does a car's kinetic energy increase?
| The increase in the car's kinetic energy comes from the internal energy of the car, stored, for example, in its gasoline or batteries.
The engine exerts torque over the wheels, which are prevented by the friction from simply rotating in place. The reaction from the ground on the car (wheels) makes it move faster.
| {
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$2\to 3$ cross-section phase space simplification Suppose the $2\to 3$ cross-section:
$$
\sigma = (2\pi)^4\int \frac{d^{3}\mathbf p_{3}}{(2\pi)2E_{3}}\frac{d^{3}\mathbf p_{4}}{(2\pi)^{3}2E_{4}}\frac{d^{3}\mathbf p_{5}}{(2\pi)^{3}2E_{5}}|M(\mathbf p_{1}, \mathbf p_2, \mathbf p_3,\mathbf p_4,\mathbf p_5)|^2\frac{\delta^{4}(p_1+p_2-p_3-p_4-p_5)}{v_{\text{relative}}}
$$
Is there some way to simplify the phase space integration, i.e., whether some simplified form of the phase space integral exist, for some new variables with definite integration limits?
Edit. It turns out that relative simple phase space structure, without complicated angles, is given in terms of four kinematic invariants. It is not hard to evaluate the domain of definition for these invariants. Of course, the full cross-section not always can be expressed in terms of elementary functions, but typically the single cross-section, i.e., the differential cross-section with respect to one kinematic invariant, often exists in terms of elementary functions.
| I can recommend you to read the paper of Tord and Riemann Phase space integrals for 2,3 and 4 particle production. The Idea is to factor the phase space into 2 2-particle phase spaces times $ds'$ where $s'=(p_1+p_2)^2$.
https://www-zeuthen.desy.de/~riemann/Teaching/ws20132014/script-tord-riemann-sm-2013_12pt-appA.pdf
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How can I use Newton's laws of motion to determine the force acting on the rope? Imagine there is a painter, weighing $180~\rm lb$, that is working from a bosun's chair hung down the side of a tall building.
Suppose that he pulls down on a fall rope with such a force that he presses against the chair with a force of $100~\rm lb$. You can assume that the chair's weight is $30~\rm lb$.
For finding the acceleration of the painter and the chair, I took into account that the weights of the painter and the chair are $180~\rm lb$ and $30~\rm lb$ respectively. I used this idea to perform the following step:
$$\text {Total mass of the painter and the chair} = \left(\frac{(180 + 30)~\rm lb}{g}\right) $$
He exerts a downward force of $100~\rm lb$ on the chair.
His net motion will be upwards.
I think the $100~\rm lb$ force the person exerts on the chair is transferred to the rope he is pulling on.
But that is just the string he is pulling on. The diagram shows that only one end of the rope is attached to the bosun chair. That end will have an upwards force of $(100 + 180 + 30)~\rm lb$ (as shown in the picture). This way, one end will have a $y~\rm lb$ force and other a force of $(100 + 180 + 30)~\rm lb$. I don't know if this is possible and I am not totally convinced that the rope is experiencing a force of $100~\rm lb$ due to the painter pulling on it.
How can I properly use Newton's third law to determine the impact of the $100~\rm lb$ downwards force on the overall system (the painter and bosun's chair)?
| Due to Newton's 3rd law, when he exerts 100lbs downwards on the chair, the chair exerts the same upwards on him. That upwards force comes from the string.
So, considering the man+chair as the system, there is an upwards force (string) of 100lbs and a downwards force (gravity) of 210 lbs. It looks like he is accelerating downwards, and not upwards.
| {
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Photon interactions with photovoltaic cells I was wondering how different energy photons interact with the electrons in a semiconductor in a PV cell.
If the photon has less energy than the band gap, then the photon passes through and does not interact with the semiconductor, right? Does it just keep traveling until it hits a material beneath the PV cell that can absorb it?
If the photon has more energy than the band gap, then where does the excess energy go after the electron has been promoted to the conduction band? Is this heat energy?
| If the photon has less energy than the bandgap, it will not be absorbed and the material is theoretically transparent to this wavelength as you can see here:
http://www.pveducation.org/pvcdrom/materials/optical-properties-of-silicon
If the energy is higher than the bandgap energy (a little) it can still be absorbed as seen here
https://www2.pvlighthouse.com.au/resources/courses/altermatt/The%20PV%20Principle/Absorption%20of%20light.aspx
The absorption, although, is not fully efficient and the excess of energy will be lost, as you said, in heat, during the process in which it is transformed in electrical energy.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Why is there an upper energy limit on the photoelectric effect? I feel like I'm missing something really fundamental regarding the cross section of the photoelectric effect. I'm looking at this chart, and it seems that the lower the energy of the incident photon, the more likely a photoelectric interaction is to happen.
Higher energy interactions get less likely, until a cut-off point, at which they no longer occur.
If the energy of the ejected electron is equal to the incident photon energy minus the binding energy, I understand why a lower limit exists (since there must be enough energy to eject the e-), but why should there be an upper limit?
Why does the cross section reduce with energy at all?
| The cross section reduces for higher frequencies, because our material will get more and more transparent for the incoming light.
Think about it in this way: the electrons follow the oscillating E-field of the photons. Now imagine the field oscillates faster and faster. At some point the electrons will be unable to follow it hence they won't get any energy from it and the photon just passes by.
In this regime other processes set in and dominate for the higher frequencies (pair production).
| {
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"timestamp": "2023-03-29T00:00:00",
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Exerting Pressure I dip my finger into a container of water that is resting on a scale. Is it true that the force exerted on the scale by the container will increase because the finger creates a downward force?
| The surrounding water doesn't know that it is your finger that is filling the displaced space. It thinks that there is still water present there, or, more precisely, it develops a hydrostatic pressure distribution that is the same as if water was present in the submerged space occupied by your finger. This includes the pressure at the very base of the container, where the pressure is now higher (because your finger has raised the level in the container). Therefore, the reading on the scale will increase by the weight of a volume of water equal to the submerged volume of your finger.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why does Newton's Third Law actually work? My father explained to me how rockets work and he told me that Newton's Third Law of motion worked here. I asked him why it works and he didn't answer. I have wasted over a week thinking about this problem and now I am giving up.
Can anyone explain why Newton's Third Law works?
For reference, Newton's third law:
To every action there is always opposed an equal reaction: or the
mutual actions of two bodies upon each other are always equal, and
directed to contrary parts.
| If you imagine pushing a spring between two fingers, the spring presses in each finger the same amount. Similarly with your arm, if you push against something you are simply straightening your arm, your arm pushes you away the same as the object.
The same is true of a rocket, When the fuel explodes it pushes out in all directions. The explosion is pushing the rocket up as much as it pushes the exhaust gasses down.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "125",
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Do commuting Hermitian operators correspond to compatible observables? As far as I know, two compatible observables have a complete set of common eigenvectors, and using this fact, one can prove that their corresponding operators are commutative. Well now is the converse true? Do any two commutative hermitian operators correspond to compatible observables?
Another point I have in mind is that commutativity is not transitive. For example, $[x,y]=0$, $[y,p_x]=0$, but $[x,p_x]\neq0$. Is compatibilty transitive? It seems for me that it has to be so, since a single observable can not have two different complete sets of eigenvectors. Isn't that true?
| This is known as the compatibility theorem. The statement as well as a proof can be found on Wikipedia:
Complete Set of Commuting Observables
However, as Griffiths says in his book about Quantum Mechanics (3rd chapter, the one about formalism; subsection Eigenfunctions of a hermitian operator), the fact that eigenfunctions of an observable operator are complete (in the QM sense, i.e. they form a basis of the Hilbert space on which this operator is defined) is just an axiom. It is provable in some cases, but not in general.
Therefore I suppose that this is a hidden assumption in the theorem quoted earlier, namely that at least one of the commuting operators has a complete set of eigenfunctions.
If you'd like to know more about when one can be sure to find a basis of an operator, there is an open question about that:
https://math.stackexchange.com/questions/1074918/why-does-the-set-of-an-hermitian-operators-eigenfunctions-spans-the-functions
Short answer is Spectral Theorem.
| {
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What is the problem of having an inertia tensor not satisfying the triangle inequality? It is well known that rigid body inertia tensors are 3 by 3 positive semidefinite matrices, which is the same as saying that their eigenvalues are all non-negative.
A little less known is the fact that those eigenvalues also satisfy the triangle inequalities, which means that the sum of any two eigenvalues is always greater or equal than the remaining one. This arises from the definition of the tensor itself which involves integrals of an always non-negative distribution of mass. Per https://physics.stackexchange.com/a/48273/116038 :
In other words, if a semi-positive definite symmetric real $3×3$ matrix with non-negative eigenvalues [...] does not satisfy the triangle inequality (1), it doesn't represent a physically possible distribution of mass.
When estimating inertia tensors using, for example, regression techniques, one can obtain matrices that do not represent physically possible rigid bodies (if we do not constrain the regression solution to physically consistent values).
Using a non-positive semidefinite $3×3$ symmetric matrix in places where an inertia tensor is expected -- e.g., in some rigid-body formulations, in simulation, in control schemes, etc. -- would entail major problems. For example, a non-positive semidefinite matrix used as inertia tensor would give rise to negative kinetic energies,
$ E_k = \frac{1}{2} \omega^\top I \omega $.
Now, my question is:
What problems (in formulations, simulation, control, etc.), if any, can arise if we use as inertia tensor some matrix that, while satisfying the positive semidefinite condition, do not satisfy the triangle inequality conditions on the eigenvalues?
Update: I've checked that "inertia tensors" that are positive semidefinite but do not satisfy the triangle inequalities still verify the energy conservation law with respect to kinetic energy and work.
| A rigid body's principal moments of inertia are obtained from these equations :
$$I_1=\int_V\,(x_2^2+x_3^2)\,\rho\,dV$$
$$I_2=\int_V\,(x_3^2+x_1^2)\,\rho\,dV$$
$$I_3=\int_V\,(x_1^2+x_2^2)\,\rho\,dV$$
where $x=x_1~,y=x_2~,z=x_3$ and the inertia tensor is:
$$I= \left[ \begin {array}{ccc} I_{{1}}&0&0\\ 0&I_{{2}}&0
\\ 0&0&I_{{3}}\end {array} \right]
$$
with $$i_\alpha=\int_V\,x_\alpha^2\,\rho\,dV> 0~,\alpha=1,2,3$$
thus:
$$I_1=i_2+i_3$$
$$I_2=i_3+i_1$$
$$I_3=i_1+i_2$$
and
$$I_1+I_2=i_1+i_2+2\,i_3=I_3+2\,i_3 > I_3$$
$$I_2+I_3=2\,i_1+i_2+i_3=I_1+2\,i_1 > I_1$$
$$I_3+I_1=i_1+2\,i_2+i_3=I_2+2\,i_3 > I_2$$
thus the triangle inequality is a physical feature of a rigid body inertia tensor. If the rigid body is symmetric then the symmetry axes are principal axes and the principal moment of inertia must obey the triangle inequality, otherwise you don't describe the rigid body that you want to describe.
| {
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Lorentz transformation of the four velocity Can we derive the velocity addition rule by directly transforming the four velocity?
|
Instead of the well-known 1-space-dimensional Lorentz Transformation the more general 3-space-dimensional one for the configuration of above Figure (see its 3D version in the end) and with finite variables is(1)
\begin{align}
\mathbf{x}^{\boldsymbol{\prime}} & = \mathbf{x}+(\gamma_{v}-1)(\mathbf{n}\boldsymbol{\cdot} \mathbf{x})\mathbf{n}+\gamma_{v} \mathbf{v} t
\tag{01a}\\
t^{\boldsymbol{\prime}}& = \gamma_{v}\left( t+\dfrac{\mathbf{v}\boldsymbol{\cdot} \mathbf{x}}{c^{2}}\right)
\tag{01b}
\end{align}
where
\begin{equation}
\mathbf{n}\equiv \dfrac{\mathbf{v}}{\| \mathbf{v}\| }=\dfrac{\mathbf{v}}{v}\, , \qquad \gamma_{v} = \gamma\left(v\right)\equiv \left(1-\dfrac{v^{2}}{c^{2}}\right)^{-\frac12}
\tag{02}
\end{equation}
and because of linearity with differentials
\begin{align}
\mathrm{d} \mathbf{x}^{\boldsymbol{\prime}} & = \mathrm{d}\mathbf{x}+(\gamma_{v}-1)(\mathbf{n}\boldsymbol{\cdot} \mathrm{d} \mathbf{x})\mathbf{n}+\gamma_{v} \mathbf{v} \mathrm{d} t
\tag{03a}\\
\mathrm{d} t ^{\boldsymbol{\prime}}& = \gamma_{v}\left( \mathrm{d} t+\dfrac{\mathbf{v}\boldsymbol{\cdot} \mathrm{d} \mathbf{x}}{c^{2}}\right)
\tag{03b}
\end{align}
Dividing (03a) by (03b) we have
\begin{equation}
\mathbf{u}^{\boldsymbol{\prime}} = \dfrac{\mathbf{u}+(\gamma_{v}-1)(\mathbf{n}\boldsymbol{\cdot} \mathbf{u})\mathbf{n}+\gamma_{v} \mathbf{v}\vphantom{\dfrac12}}{\gamma_{v} \Biggl(1+\dfrac{\mathbf{v}\boldsymbol{\cdot} \mathbf{u}}{c^{2}}\Biggr)}
\tag{04}
\end{equation}
Equation (04) is a more general "rule" for the addition of velocities $\mathbf{v},\mathbf{u}$. The 1-dimensional version is the well-known equation
\begin{equation}
u^{\boldsymbol{\prime}} = \dfrac{u+v }{1+\dfrac{v\, u}{c^{2}}\vphantom{\dfrac12}}
\tag{05}
\end{equation}
Now, the 4-velocity
\begin{equation}
\mathbf{U} = \left(\gamma_{u}\mathbf{u} ,\gamma_{u} c\right)\, , \qquad \gamma_{u} = \gamma\left(u\right)\equiv \left(1-\dfrac{u^{2}}{c^{2}}\right)^{-\frac12}
\tag{06}
\end{equation}
as could be proved(2), is a 4-vector : it is Lorentz-transformed as the space-time position vector
\begin{equation}
\mathbf{X} = \left(\mathbf{x} ,c\,t\right)
\tag{07}
\end{equation}
I don't understand why do you want to reach (04) from the fact that (06) is a 4-vector when inversely this property is a consequence of (04).
(1)
The Lorentz Transformation (01) could be expressed in block matrix form as
\begin{equation}
\mathbf{X'}
=
\begin{bmatrix}
\mathbf{x'}\vphantom{\dfrac{\gamma \upsilon}{c}\mathbf{n}} \\
\\
c\,t'\vphantom{\dfrac{\gamma \upsilon}{c}\mathbf{n}}
\end{bmatrix}
=
\begin{bmatrix}
\mathrm{I}+(\gamma_{v}-1)\mathbf{n}\mathbf{n}^{\top} & \dfrac{\gamma_{v} \upsilon}{c}\mathbf{n}\\
&\\
\dfrac{\gamma_{v} \upsilon}{c}\mathbf{n}^{\top} &\hspace{5mm}\gamma\\
\end{bmatrix}
\begin{bmatrix}
\mathbf{x}\vphantom{\dfrac{\gamma \upsilon}{c}\mathbf{n}} \\
\\
c\,t\vphantom{\dfrac{\gamma \upsilon}{c}\mathbf{n}}
\end{bmatrix}
=\Lambda \mathbf{X}
\tag{fn-01}
\end{equation}
where $\:\mathbf{n}\:$ a $\:3\times 1\:$ unit column vector and $\: \mathbf{n}^{\top} \:$ its transposed $\:1\times 3\:$ unit row vector
\begin{equation}
\mathbf{n}=
\begin{bmatrix}
n_1\\
n_2 \\
n_3
\end{bmatrix}
\quad , \quad
\mathbf{n}^{\top}=
\begin{bmatrix}
n_1&n_2&n_3
\end{bmatrix}
\tag{fn-02}
\end{equation}
and $\:\mathbf{n}\mathbf{n}^{\top}\:$ a linear transformation, the vectorial projection on the direction $\:\mathbf{n}\:$
\begin{equation}
\mathbf{n}\mathbf{n}^{\top}=
\begin{bmatrix}
n_1\\
n_2 \\
n_3
\end{bmatrix}
\begin{bmatrix}
n_1&n_2&n_3
\end{bmatrix}
=
\begin{bmatrix}
n_1^{2} & n_1 n_2 & n_1 n_3\\
n_2 n_1 & n_2^{2} & n_2 n_3\\
n_3 n_1 & n_3 n_2 & n_3^{2}
\end{bmatrix}
\tag{fn-03}
\end{equation}
(2)
If a particle is moving with velocity $\mathbf{u}$ with respect to system $\mathrm{S}$ then between its proper time $\tau$ and times $t,t'$ we have
\begin{equation}
\dfrac{\mathrm{d}t}{\mathrm{d}\tau}=\gamma_{u}\, , \quad \dfrac{\mathrm{d}t'}{\mathrm{d}\tau}=\gamma_{u'}
\tag{fn-04}
\end{equation}
but from (03b)
\begin{equation}
\dfrac{\mathrm{d}t'}{\mathrm{d}t}=\gamma_{v} \Biggl(1+\dfrac{\mathbf{v}\boldsymbol{\cdot} \mathbf{u}}{c^{2}}\Biggr)
\tag{fn-05}
\end{equation}
so
\begin{equation}
\gamma_{v} \Biggl(1+\dfrac{\mathbf{v}\boldsymbol{\cdot} \mathbf{u}}{c^{2}}\Biggr)=\dfrac{\gamma_{u'}}{\gamma_{u}}
\tag{fn-06}
\end{equation}
Replacing this in the dominator of the rhs of (04) we note that the quantities $\gamma_{u}\mathbf{u}$ and $\gamma_{u}$ are transformed as $\mathbf{x}$ and $t$ in equations (01), that is finally
\begin{equation}
\mathbf{U'}
=
\begin{bmatrix}
\gamma_{u'}\mathbf{u'}\vphantom{\dfrac{\gamma \upsilon}{c}\mathbf{n}} \\
\\
\gamma_{u'}c\vphantom{\dfrac{\gamma \upsilon}{c}\mathbf{n}}
\end{bmatrix}
=
\begin{bmatrix}
\mathrm{I}+(\gamma_{v}-1)\mathbf{n}\mathbf{n}^{\top} & \dfrac{\gamma_{v} \upsilon}{c}\mathbf{n}\\
&\\
\dfrac{\gamma_{v} \upsilon}{c}\mathbf{n}^{\top} &\hspace{5mm}\gamma\\
\end{bmatrix}
\begin{bmatrix}
\gamma_{u}\mathbf{u}\vphantom{\dfrac{\gamma \upsilon}{c}\mathbf{n}} \\
\\
\gamma_{u}c\vphantom{\dfrac{\gamma \upsilon}{c}\mathbf{n}}
\end{bmatrix}
=\Lambda \mathbf{U}
\tag{fn-07}
\end{equation}
| {
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Prove there is an equipotential sphere between two point charges Given two point charges of opposite sign I need to prove that inside the electric field they create there is an equipotential sphere.
I'm very positive that this is more geometry than anything else and I really question why I've been given this exercise. Here's my thoughts anyways and everything I can remember from my background in math and geometry.
The potential for a point charge is given by $$V=\frac{q}{4\pi ε_0r}$$
Let's assume the charges are correlated by the following ratio $$\frac{q_1}{q_2}=-a$$ What we want is $$V_1+V_2=0=>\frac{a}{r_1}=\frac{1}{r_2}=>\\r_1=ar_2$$
The distances $r$ are the distance from the charge to any point in the equipotential surface.
The equation of a sphere is the following: $$(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=R^2$$
I can also write $$r_1=\sqrt{(x_1-x_0)^2+(y_1-y_0)^2+(z_1-z_0)^2}\\r_2=\sqrt{(x_2-x_0)^2+(y_2-y_0)^2+(z_2-z_0)^2}$$and then use r1=ar2. But then I really get lost.
Is there another way? And if not how do I show that this is a sphere?
| I guess this can be done if we use polar coordinates. In polar coordinates, the equation of a sphere looks like $R=C_0,$ where $C_0$ is a constant. Then the thing can be dealt in this way:
EDIT: $\mathbf {Z= r_0-r_1}$
| {
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Why does acceleration due resulting force depends on mass while acceleration due gravity doesn't? Objects intrinsically resist to be accelerated due to their masses. A clear example would be kicking a soccer ball vs kicking a bowling ball. The latter ball will resist much more to be accelerated than the first one due to its greater mass (intrinsic property).
What if we position them in a inertial frame of reference in space? If we push both previous balls with the same force we will obtain different accelerations due to the balls' different masses, isn't it?
| Well, the thing is that the gravitational force the earth or any other body exerts on another is given by gM,
where g is the acceleration due to gravity for that body, a constant for any given body and M is the mass of the other body.
In case of the earth g* equals g and hence the Force of gravity= Mg and hence, the acceleration of any body of mass M is Mg/M= g. You see, this does obey Newton's second law. Hope this helps you.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Mistake in eq. (10.7.19) Weinberg Vol I? In Weinberg Vol I, he writes in equation 10.7.19,
$$
\left\langle 0 | \phi (0) | {\mathbf{k}} \right\rangle = ( 2\pi ) ^{ - 3/2} \left( 2 \sqrt{ {\mathbf{k}} ^2 + m ^2 } \right) ^{ - 1/2} N,
\tag{10.7.19}$$
where $\phi $ is a unrenormalized scalar field, $ \left| {\mathbf{k}} \right\rangle $ is a one-particle state, and $ N $ is some constant. The $ {\mathbf{k}} $ dependence seems wrong. For one thing, this is often taken as the renormalization condition and set equal to $ 1 $. Furthermore, its easy to show the braket should be $ {\mathbf{k}} $-independent by performing a Lorentz transform:
\begin{align*}
\left\langle 0 | \phi (0) | {\mathbf{k}} \right\rangle & = \left\langle 0 | U ( \Lambda ) ^\dagger \phi (0) U ( \Lambda ) | {\mathbf{k}} \right\rangle \\
& = \left\langle 0 | \phi (0) | \Lambda {\mathbf{k}} \right\rangle
\end{align*}
Since I am free to choose $ \Lambda $ as I please, the result can't be dependent on $ {\mathbf{k}} $. Is this a mistake or am I overlooking something?
| I think the problem is you are taking the inner product to be Lorentz Invariant. In Weinberg's convention the inner products are not Lorentz Invariante, see Eqn. 2.5.19. For a covariant normalization the $\sqrt{2E}$ factor would be absent and your argument would go through.
| {
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Does increasing the resistance in a branch of a parallel circuit decrease the overall current?
In the above question, why does R3 increase? If R2 increases, wouldn't the parallel combination's resistance increase? If so, wouldn't the circuit have less current? Then why would the voltage across R3 increase?
| The voltage on all resistors in a parallel circuit is the same, so that means that the voltages on resistors $R_2,R_3,R_4$ in this example will always be the same. Since voltage is: $$U=I*R$$
this means that if we increase resistance value of element $R_2$, the voltage $U_2$ on that element will also increase, and since the voltages on all resistors in a parallel circuit are the same : $$U_2=U_3=U_4$$
that means that the voltage on $R_3$ will also increase.
| {
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How to approach estimating correction size in the BK equation? I am starting to do work in theoretical physics, and as a test, the professor I am working with asked me to estimate the size of a correction to an approximate solution to the BK equation. I am no really sure how to approach this, as I first tried to plug in the approximation, but found that the integral was still unsolvable. I was thinking using something along the lined of the Euler approximation, but I am still a little bit lost. I would appreciate any tips.
| The answer is by dimensional analysis
| {
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What is the name of the principle that replaces a history functional by internal/state variables? I just need to know the correct expression:
When narrowing down constitutive equations for the mechanics solids in continuum mechanics, one has in the very general case a Cauchy stress $\mathbf{T}$ as a result of the deformation history $\chi(\mathbf{x},t)$, written as a functional involving integrals over time and the body $\mathcal{B}$:
$$
\mathbf{T}(\mathbf{x}_0,t)=\int_\mathcal{B} \int_0^t f(\chi(\mathbf{x},\tau)-\chi(\mathbf{x}_0,\tau)) \ \mathrm d\tau \ \mathrm d \mathbf{x}
$$
where $f$ is the constitutive function (Peridynamics is an example).
Then, principles of material modelling are invoked to reduce the functional freedom of the material model. One generally replaces the time integral by an internal variable $\mathbf{v}$,
$$
\mathbf{T}(\mathbf{x}_0,\mathbf{v})=\int_\mathcal{B} g(\mathbf{\chi(x)}-\chi(\mathbf{x}_0),\mathbf{v}) \mathrm d \mathbf{x}\\
\dot{\mathbf{v}}(t,\mathbf{v})=h(...)
$$
the evolution of which is prescribed by another constitutive function $h$.
I would like to know if this reduction has a name. It is probably some kind of "principle of", like the "principles of determinism" or "principle of local action".
| The replacement of the whole history by some (finite dimensional) internal variables is called state space formulation.
| {
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Moment of a force about a given axis (Torque) - Scalar or vectorial? I am studying Statics and saw that:
The moment of a force about a given axis (or Torque) is defined by the equation:
$M_X = (\vec r \times \vec F) \cdot \vec x \ \ \ $ (or $\ \tau_x = (\vec r \times \vec F) \cdot \vec x \ $)
But in my Physics class I saw:
$\vec M = \vec r \times \vec F \ \ \ $ (or
$\ \vec \tau = \vec r \times \vec F \ $)
In the first formula, the torque is a triple product vector, that is, a scalar quantity. But in the second, it is a vector. So, torque (or moment of a force) is a scalar or a vector?
| There are some application, where we might want to quantify both the torque, which is a vector, and the component of the torque about a particular axis, which is a scalar.
I illustrate an example of this in the figure below, which is from 1 and provided here under fair use for the purpose of scholarship. The door is hinged so that it turns only around the $\mathbf{\widehat{k}}$ axis. Meanwhile, the door knob is located at a position $\mathbf{r}$ relative to the origin. A force $\mathbf{F}$ is applied to the door knob.
By $\boldsymbol{\tau}$, I denote the torque on the door knob, which is
$$\boldsymbol{\tau} = \mathbf{r}\times \mathbf{F}.$$
By $\tau_z$, I denote the scalar component of the torque vector about the axis of rotation. So,
$$\tau_z = \mathbf{\widehat{k}} \cdot \boldsymbol{\tau} = \left(\mathbf{r}\times \mathbf{F}\right)\cdot \mathbf{\widehat{k}}.$$
Bibliography
1 Mathematical Methods in the Physical Sciences, 3rd Edition, Mary L. Boas, ISBN: 978-0-471-19826-0 July 2005.
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Special Relativity: Does non inertial frame of reference work in SR? I started on my own learning about GR and SR two months ago, and I still do not have clear if it is possible or not. The following example was explained to me by someone who affirmed: "SR applies only on inertial reference frames":
Let's imagine we have two different reference frames : A' and A. Reference frame (RF) A' is moving with constant velocity (v), meanwhile RF A has no velocity (A' moves relative to A with constant v).
RF A' has a wire underneath and RF A has an aerial above. When both interact, clocks start running in both RFs (clock A' and clock A) and a light ray emerges (from the wire-aerial interaction and with the same velocity vector direction RF A' has).
Then we agree distance can be determined from both RFs.
i.e. : x = x' + vt'
Then I asked myself: why would not be correct consider the case where A' is an accelerated RF and distance is determined from RF A (i.e.) as x = x' + at'?
My doubts about if "SR applies only on inertial reference frames" sentence was true increased when I checked out more sources and they affirmed accelerated reference frames were possible in SR.
| There is something called The Clock Hypothesis that lets you deal with accelerated frames by assuming that you can break an accelerated frame into infinitesimal inertial frames, each of them measuring the proper time. Some authors like Goldstein in his Classical Mechanics book refer to it like an "stratagem" to deal with non inertial frames so it's not one of the postulates, but that's the usual way to deal with them on SR.
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Does this Nelson formula for Feynman integral have mistakes? In this paper (Maroun's PhD dissertation, 2013) at page 46 the following formula is given (apparently without a reference):
$$\int_0^{\infty } e^{i a x^s+i b x^p} \, dx=\sum _{n=0}^{\infty } \frac{\left(i b a^{\frac{1}{s}}\right)^n \exp \left(\frac{(i \pi ) (n p+1)}{2 s}\right) \Gamma \left(\frac{n p+1}{s}\right)}{n! a^{\frac{1}{s}} \left| s\right| }$$
Now I am trying to verify the formula. If I take $a=b=i/2$, $s=p=1$ the left hand side becomes $1$ while right hand side becomes $8/3$.
It had been suggested on MathOverflow, that the formula has a mistake, and the right-hand part should contain $ba^{-p/s}$ instead of $ba^{1/s}$. In this case with the above-mentioned data the equality holds, but still it does not work for $a=-b$; $p=s$.
Is there some error in the formula? Is this formula even well-known so to look somewhere for the correct form?
| Taylor series Expansion of $e^{ibx^p}$ yields
$\sum_{n=0}^\infty \frac{(ib)^n}{n!}x^{np}$.
Then you have to evaluate every term of the series by a Substitution. You can use the Substitution $u = -iax^s$ which yields $x = (i\frac{u}{a})^{1/s}$ and
$dx = e^{\frac{i \pi}{2s}}a^{-1/s}u^{1/s-1}/s$.
The integral $I_n = \int_0^\infty dx x^{np}e^{iax^s}$ turns to $I_n = \int_0^\infty \frac{du}{su} (i\frac{u}{a})^{np/s+1/s} e^{-u} = \int_0^\infty \frac{du}{su}(ia^{-1})^{np/s+1/s}u^{np/s+1/s} e^{-u} = e^{i \pi \frac{np+1}{2s}}a^{-np/s}/(a^{1/s}s) \int_0^\infty du u^{np/s+1/s-1} e^{-u} = \frac{e^{i \pi \frac{np+1}{2s}}{ \Gamma(\frac{np+1}{s})}a^{-np/s}}{a^{1/s}s}$.
Conclusion: You see that this series Expansion will take constants $(iba^{-p/s})^n$ that belong to the term $I_n$; the integral in your question has a small mistake.
The correct formula should be:
$$\int_0^{\infty } e^{i a x^s+i b x^p} \, dx=\sum _{n=0}^{\infty } \frac{\left(i b a^{-\frac{p}{s}}\right)^n \exp \left(\frac{(i \pi ) (n p+1)}{2 s}\right) \Gamma \left(\frac{n p+1}{s}\right)}{n! a^{\frac{1}{s}} \left| s\right| }$$
| {
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Why are light bulbs black body radiation emitters? If the definition of a black body is:
"A blackbody is an object that absorbs all of the radiation that it receives (that is, it does not reflect any light, nor does it allow any light to pass through it and out the other side). The energy that the blackbody absorbs heats it up, and then it will emit its own radiation."
Then why are light bulbs black bodies? Don't they generate heat themselves instead of absorbing radiation, then giving radiation out?
| "The energy that the blackbody absorbs heats it up, and then it will emit its own radiation." --- This part may be a bit misleading. It should read as
The energy that the blackbody abosrbs heats it up, and will be emitted as part of its own radiation.
In fact as long as all radiation is absorbed, i.e. not reflected/passed through, it is a black body. A black body does not need to absorb heat first, then to begin radiating out. A light bulb can be roughly viewed as a black body; as a result if you subject it to radiations with a temperature far higher than its own typical temperature, it would heat up to (possibly more than) that one.
| {
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weight of a canister of air Consider four scenarios:
(on a typical weighing scale)
*
*Measure the weight of a canister of air filled at atm pressure
*Measure the weight of a compressed canister of air
*Measure the weight of a canister with no air inside.
*Measure the weight of an empty canister on earth with no atmosphere
I would guess that 1 is the same as 4 and 2 and 3 weigh the same as the mass 2 plus the added or subtracted mass of air. Any explanations? This is just because a canister's volume doesn't or shouldn't change the force of it's weight being applied to the weighing scale. A weighing scale accounts for air pressure as it's applied equally on all sides and the machine is tared. When the canister is compressed the air is now heavier and tries to sink. I want to explain this more mathematically in terms of pressure resulting in a greater force on the scale. As you can probably tell, my reasoning doesn't hold much weight at the moment. (not sure if pun intended)
| Remember that air pressure acts in basically all directions on an object. Therefore a typical weighing scale would have the same reading weighing the same object in an atmosphere and outside of an atmosphere. There is less air in scenarios 3 and 4 so those canisters should be lighter than that in scenario 1 which is in turn lighter than that in scenario 2.
| {
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Perpetual Motion Machines If you go on YouTube, you will find a large number of machines that work for almost forever. But why do all of them stop working after some time?
Which Law (Other than Conservation of Energy) prevents a machine from running till eternity?
|
Which Law (Other than Conservation of Energy) prevents a machine from running till eternity
Why is the law of energy conservation not fulfulling for you to answer that? There is always energy conservation, and that is the sole reason that you can't draw more energy out of a machine than is put in.
That the output is always even less than that is a result of efficiency considerations (the Carnot limit e.g.)
So, drawing energy from a machine will cause it to eventually stop. But if you don't draw energy from a motion, then theoretically it will continue running forever. A rolling wheel will in theory never stop rolling if nothing stops it. A space ship drifting in space will never stop drifting but forever have constant velocity, until something stops it. But here on Earth we never see anything lasting forever. There is always friction, deformations in materials, air resistances etc. even in the smoothest gears, and all that cause small energy losses. Nothing is ideal; everything will stop eventually.
| {
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What is the Fabric of Spacetime? I always get the analogy of the pretend sun on a piece of cloth pulling everything down such as the image attached below:
But What I never really understood is what IS that piece of cloth. Like is it some sort of invisible force field or something? Sorry if I sound a little crazy, I only recently started reading a little bit of popular physics (fabric of the cosmos).
| The "fabric" you are asking about is the speed of time. Everything moves in time, but if the speed of time is different, then the motion is no longer straight and trajectories become curved. Why do things fall down in Earth? Because time there moves slower. Tragectories are always curved to where time moves slower. So the depth of how much the "cloth" in your picture is pulled down is how much the speed of time is slowed down near the heavy objects.
| {
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Newtonian limit of a perfect fluid In special relativity, with metric tensor $\eta_{\mu\nu}=\text{diag}(-c^2,1,1,1)$, take a perfect fluid stress-energy tensor : $T^{\mu\nu} = \left( \rho + \frac{p}{c^2} \right) \, U^\mu\otimes U^\nu + p \, \eta^{\mu\nu}$, where $U^\mu$ is the 4-speed, $\rho$ the volumic mass and $p$ the pressure of the fluid.
In the newtonian limit where $U^\mu \simeq (1,0,0,0)$, we find the newtonian fluid at rest $T^{\mu\nu}\simeq \text{diag}(\rho,p,p,p)$. However, if we want the more precise approximation $(U^t)^2=\gamma^2\simeq 1+\frac{v^2}{c^2}$, we get
$$ T^{tt} \simeq \left( \rho + \frac{p}{c^2} \right) \left( 1 + \frac{v^2}{c^2} \right) - \frac{p}{c^2} $$
Two things surprise me in this energy formula,
*
*The pressure term $\frac{pv^2}{c^4}$ remains. It is small but not zero.
*It figures $\rho +\rho\frac{v^2}{c^2}$ instead of the newtonian kinetic energy $\rho +\frac{1}{2}\rho\frac{v^2}{c^2}$.
Did I make a mistake in the approximation ?
EDIT:
On second thought, the newtonian limit would rather be $U^\mu \simeq (1,\vec{v})$ with $v\ll c$. In that case, the first line of the perfect fluid is $T^t=(\rho, (\rho+\frac{p}{c^2})\vec{v})$ and its zero divergence yields
$$ \frac{\partial\rho}{\partial t} +\text{div}(\rho\vec{v}) = -\text{div}\,\frac{p\vec{v}}{c^2} $$
On the right-hand we recognize the power received from the pressure forces (summed on the 6 faces of a small cube of mass). So it is an energy conservation equation, with energy approximately being $\rho c^2$, as usual. Still, I didn't get the newtonian kinetic energy, and now I have this strange $\frac{p}{c^2}\vec{v}$ in the momentum.
| Your approximations need to be consistent. In the non-relativistic limit, it's true that we should have $v/c<<1$, but that's not all. In particular, we should require that the mass energy of the fluid be much larger than the kinetic energy of the fluid.
Alternatively, one might demand that the sound speed in the fluid be much smaller than $c$, which would imply that $p/\rho c^2 << 1$.
To be consistent, we will assume that $\frac{\sqrt{p/\rho}}{c} \sim\frac{v}{c} \sim \delta$. To order $\delta$, we have that
$$U^\mu = (\gamma ,\gamma \vec v/c) \approx (1,\vec v/c)$$
$$T^{00} = (\rho + \frac{p}{c^2})U^0\otimes U^0 + p \eta^{00} \approx \rho $$
This is insufficient, so let's take it a step further and keep terms of order $\delta^2$:
$$U^\mu = (\gamma,\gamma \vec v/c) \approx (1+\frac{v^2}{2c^2},\vec v/c)$$
$$T^{00} = (\rho + \frac{p}{c^2})U^0\otimes U^0 + p \eta^{00} \approx (\rho +\frac{p}{c^2})(1+\frac{v^2}{2c^2})^2 - \frac{p}{c^2} $$
$$ = \rho + \rho\frac{v^2}{c^2} + \mathcal{O}(\delta^4)$$
The term proportional to $\frac{pv^2}{c^4}$ is too small; to keep it, but not incorporate terms of order $v^4/c^4$, is inconsistent, so we throw it away.
As you correctly state above, $T^{00}$ is the energy density in the rest frame of the fluid - $E = T^{00}d^3 x$. When we move to a frame where the fluid is moving with speed $v<<c$,
$$ E' = T'^{00} d^3 x' = T'^{00} \gamma^{-1} d^3x \approx T'^{00}(1-\frac{v^2}{2c^2})d^3x$$
when we keep terms of order $v^2/c^2$. Plugging in our calculated value for $T'^{00}$, we find that
$$ \frac{E'}{d^3x} = (\rho + \rho \frac{v^2}{c^2})(1 - \frac{v^2}{2c^2}) = \rho + \rho \frac{v^2}{2c^2} + \mathcal{O}(\delta^4)$$
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Does the earth has any electric field as the earth's magnetic field is changing over time?
Earth's magnetic field changes over time because it is generated by a geodynamo
-Wikipedia
A time-varying magnetic field can produce electric field. So does the earth has electric field due to changing magnetic field?
|
A time-varying magnetic field can produce electric field.
I think this statement is unclear. Rather: A time-varying magnetic field does produce an electric field. This is one of Maxwell's equation (maybe with some prefactors)
$$\nabla \times E = - \frac{dB}{dt}$$
If the right-hand-side does not vanish, the left-hand side does not vanish.
Therefore, the change in the earth's B-field will produce an E-field.
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Oscillations of what cause quantum waves? Electromagnetic waves are produced by oscillating electric charges. In quantum mechanics this fact is represented by photons mediating electromagnetic interactions of particles with an electric charge.
However, in quantum mechanics, any particle can be viewed as a wave. Can this wave also be viewed as caused by oscillations of some type of charges or quantum numbers?
For example, the electron can be viewed as a wave, just like the photon. So, if the electromagnetic wave that the photon represents is caused by oscillating electric charges, then oscillating of what causes the wave that the electron represents?
| Photons don't mediate electromagnetic interactions, Virtual photons do. They don't really exist like physical photons, they're just in mathematical formalisms. The photons emitted by accelerating charges are described by this.
AFAIK The 'waves' of matter in QM aren't waves in the same sense, depending on your interpretation they may not be physical at all, rather just information about what can be known about the system since the last measurement. Wavefunctions in QM just describe the different observable values and the probabilities tagged to them.
Hope this helps
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Free particle propagation amplitude (Peskin & Schroeder 2.1) In section 2.1 on p. 14 of Peskin and Schroeder QFT, they calculate the amplitude of a free non-relativistic particle to propagate from $\mathbf{x_0}$ to $\mathbf{x}$ as
$$U(t) = \left<\mathbf{x}|e^{-i\left(\mathbf{p}^2/2m\right)t}|\mathbf{x_0}\right>$$
$$ = \int \frac{d^3p}{\left( 2\pi\right)^3} \left<\mathbf{x}|e^{-i\left(\mathbf{p}^2/2m\right)t}|\mathbf{p}\right>\left<\mathbf{p}|\mathbf{x_0}\right>$$
$$ = \frac{1}{\left( 2\pi\right)^3} \int d^3p \ e^{-i(\mathbf{p}^2/2m)t} e^{i\mathbf{p}\cdot (\mathbf{x}-\mathbf{x_0})}$$
$$ = \ ...$$
using the nonrelativistic Hamiltonian $\mathbf{H} = \mathbf{p}^2/2m$. I think in going from the first to second line, they just inserted the identify operator as a sum over all the momentum projection operators, since the momentum eigenstates form a complete set. Could someone please explain the step from the second line to the third line?
| The transition from the 2nd line to the 3rd simply uses the fact that
$$\langle \mathbf{p}|\mathbf{x} \rangle = e^{-i \mathbf{p}\cdot\mathbf{x}}.$$
The exponent of the Hamiltonian merely picks up the eigenvalue $\mathbf{p}$ from the state $|\mathbf{p}\rangle$, so you can take it out of the inner product, and what you have left is
$$\langle \mathbf{x}|\mathbf{p} \rangle \langle \mathbf{p}|\mathbf{x}_0 \rangle = e^{i \mathbf{p}\cdot(\mathbf{x}-\mathbf{x}_0)},$$
since switching the order of states in the inner product gives you the complex conjugate, that is,
$$\langle \mathbf{x}|\mathbf{p} \rangle = \langle \mathbf{p}|\mathbf{x} \rangle^*.$$
Let me know if this is clear.
| {
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Average value of $\cos^2(\theta)$ in Malus' law When the Malus' law is applied to a beam of unpolarized light, I can understand that the incident light has all the possible polarizations, so that I should apply the law for all the angles.
If I had three angles $\theta_1$, $\theta_2$ and $\theta_3$, I guess the equation should be
$I = I_0\left(\cos^2\theta_1+\cos^2\theta_2+\cos^2\theta_3\right)$.
If the light is unpolarized, I won't have just three angles, but all the angles between $0$ and $2\pi$, so the sum turns an integral:
$I = I_0\displaystyle\int_0^{2\pi} \cos^2\left(\theta\right) d\theta$.
However, it seems that, instead of that integral, I should use the average value, i.e. the integral divided by $2\pi$. Why?
| The formula which you gave in the example is wrong, and I'll shortly show you why.
Unpolarized light can be taking as a mixture of light polarized in different directions, as shown in the figure. Thus, we must find the intensity due to each individual polarized light using malus law and add them up
Assume that there are $n$ different polarized light. For instance, in your example, you have taken $n=3$. Thus, each polarized light will have an intensity $\frac{I_0}{n}$ (This is where you went wrong in your example, it should have been $\frac{I_0}{3}$, not $I_0$ in your example). Additionaly, the angle between each polarized light is $\theta=\frac{\pi}{n}$. Thus, the net intensity transmitted is given by,
$$I = \frac{I_0}{n}(\cos^2{0}+\cos^2{\theta} + \cos^2{2\theta}+....+\cos^2{(\pi-\theta)}+\cos^2{\pi})$$
$$= \frac{I_0}{\pi}\frac{\pi}{n}\sum_{k=0}^{n} \cos^2{k\frac{\pi}{n}}$$
But since it is unpolarized light we are talking about, we must take the limit of $n$ becoming $\infty$. In that case, the sum reduces to a integral and we get,
$$I=\frac{I_0}{\pi} \int_{0}^{\pi} \cos^2{\theta}d\theta = \frac{I_0}{2}$$
| {
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Effect of humidity on temperature When temperature changes, relative humidity changes. Imagine I have a closed room at a certain temperature; then, I blow water vapor at a similar temperature. Can a change in relative humidity induce a (small) change in temperature? If it does, would the temperature raise or decrease?
I have tried to find information but all I find is about the influence of temperature on relative humidity, not the other way round.
| The point is that air at different temperatures can hold different amounts of water (the hotter, the more water). Therefore, the relative humidity decreases when heating up air and increases when cooling it. The absolute humidity, however, stays constant in an isolated setting, because water molecules are not leaving or entering the system.
Now to your question. If the added water is already gaseous and at the same temperature, $T$ will stay constant, because the mean kinetic energy per particle is the same as before. If the water you add is in liquid form like droplets, the temperature would decrease, because some of the kinetic energy of the air is needed to evaporate the water, decreasing the mean kinetic energy.
All this is assuming the relative humidity never reaches 100%.
| {
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Why do heavier isotopes of the same element have smaller atomic radii than lighter isotopes of the same element? I have been trying to figure out why higher-mass isotopes have higher melting and boiling points than lower-mass isotopes of the same element.
A Quora answer on this topic explored the idea that electron orbits in atoms with smaller nuclei behave as if the electron is lighter, contributing to larger orbits, whereas atoms with larger nuclei behave as if the electron is heavier, contributing to smaller orbits.
Could someone explain why the size of the nucleus affects electron orbits?
| To make it clear: I don't know the answer. However, here is how I would try to answer the why question:
*
*The nuclear radius increases with the number of nuclei. Assuming that the nucleus is a sphere, we would expect that the radius of the nucleus scales like $R_n = R_0 \cdot A^{1/3}$, where $A$ is the mass of the nucleus. You could cross-check this by looking at the so called liquid droplet model, but I am pretty sure that this is correct.
*If the electrons would go around the nucleus in circles, the radius of the nucleus should not influence the energy levels of the electrons. However, electrons do not merely circle the nucleus, but they have certain probability distributions depending on their orbits (s, p, d, ...). Some even have a non-zero probability to be within the nucleus. Therefore, the radius of the nucleus will affect the binding energy of the electrons. I would expect that the electrons are bound tighter if the nucleus is enlarged. However, I don't have a proper argument and you should validate this. Side mark: As can be seen from the muonic hydrogen experiments, a larger binding energy is equivalent to a heavier "electron".
Good luck.
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What is a potential? I am self-studying electrodynamics and am wanting to know what is meant by a potential. I understand the concept of potential energy but what is meant by a potential? Is it the same thing as a field, like gravitation or electromagnetic?
| Electric potential and electric potential energy are two different concepts but they are closely related to each other. Consider an electric charge $q_1$ at some point $P$ near charge $q_2$ (assume that the charges have opposite signs).
Now, if we release charge $q_1$ at $P$, it begins to move toward charge $q_2$ and thus has kinetic energy. Energy cannot appear by magic (there is no free lunch), so from where does it come? It comes from the electric potential energy $U$ associated with the attractive 'conservative' electric force between the two chages. To account for the potential energy $U$, we define an electric potential $V_2$ that is set up at point $P$ by charge $q_2$.
The electric potential exists regardless of whether $q_1$ is at point $P$. If we choose to place charge $q_1$ there, the potential energy of the two charges is then due to charge $q_1$ and that pre-existing electric potential $V_2$ such that:
$$U=q_1V_2$$
P.S. You can use the same argument if you consider chage $q_2$, in that case the potential energy is the same and is given by:
$$U=q_2V_1$$
| {
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Why is the Michelson-Morley experiment considered a null-result? The Michelson-Morley experiment did not seem to be a null-result. Here is what Michelson and Morey say in their 1887 paper:
the displacement to be expected was 0.4 fringe. The actual displacement was certainly less than the twentieth part of this, and probably less than the fortieth part.
In the absence of further constraints on the upper limit, how is a 0.01 or 0.02 displacement in the fringes a null-result?
Did the following later experiments constrain Michelson and Morley's upper bound on the displacement of the fringes?
*
*Sagnac 1913
*Miller 1921-26
*Tomaschek 1924
*Kennedy 1926
*Piccard 1926
*Illingworth 1927
*Zeeman 1927
*Michelson 1926-29
*Joos 1930
cf. "Why do the Michelson-Morley (1887) & Michelson-Gale (1925) experiments both measure non-zero fringe shifts?"
| You can never say that anything is zero, you can only put an upper bound on it, because the effect can always just be made to be smaller and smaller until it is indistinguishable from zero to your apparatus. Saying "The expected result was $x$, but it was no more than $x/20$" is basically saying "I couldn't measure the effect.", but in more precise language.
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How is the 4-momentum, especially the transverse momentum, in a detector calorimeter measured? assuming a jet is produced in the detector at a hadron collider. Some of the jet particles are uncharged and will only interact with the hadronic calorimeter and therefore there is no information about their momentum from their trajectory in a magnetic field. The calorimeter measures the jet's kinetic energy by absorbing it, and its direction of flight can be inferred from the azimuthal and polar angle/pseudorapidity. But how do I get the jet's transverse momentum from it? I understand that I can get the transverse energy by looking at the projection of the energy onto the transverse plane, but to get the transverse momentum I still need information about the particles mass, which I don't have, do I?
| Just found in Tao Han's "Collider Phenomenology", that many objects such as photons and light-quark and gluon jets are assumed to be massless, and therefore the transverse energy is equal to the transverse momentum.
| {
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Does work done depend on the frame of reference? Suppose I am sitting on a bench and looking at a moving car. Force is applied on the car by its engine, and it makes it displace, hence some work is done on the car. But what if I am sitting in the car and looking at the bench? The bench covers some displacement, but who has applied force to it? Is any work done on it?
| If the frame of reference is on the accelerating car, we have a non inertial frame of reference, so a pseudoforce (~fictious force) is born on the bench, which makes it accelerates as to the non inertial frame of reference.
| {
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Orbital period and nodal precession & apsidal precession In classical orbital mechanics as per Newton/Kepler, the time taken to complete one orbit around a body of mass $M_e$ is:
$$
T = 2 \pi \cdot \sqrt { \frac {a^3} {G \cdot M_e} }
$$
But also, for orbits around an oblate spheroid, we also have precession of the orbital plane itself, and the precession of the orbit's major axis which, for every turn of the orbit, are respectively:
$$
\Delta\Omega =
-3\pi\cdot \cfrac {J_2} {G\cdot M_e} \cdot \left[\frac {1} {a \cdot \left(1-e^2\right)}\right]^2 \cdot \cos i
$$
$$
\Delta\omega =
-6\pi\cdot \cfrac {J_2} {G\cdot M_e} \cdot \left[\frac {1} {a \cdot \left(1-e^2\right)}\right]^2 \cdot \left(\frac 5 4 \cdot \sin^2 i - 1 \right)
$$
What I'm unclear is, in presence of these factors, how is the orbital period $T$ defined?
Can it be defined as the time elapsed between two consecutive passes through the periapses (or apoapsis)? How does that work when the periapsis (or apoapsis) is itself shifting? Do I count the time to where the periapsis was when I started, or where the periapsis is now?
Specifically, what happens in an equatorial circular orbit? (Or, if we like, an orbit where $i$ and $e$ are just very, very small?) Substituting $i=0$ and $e=0$ gets us:
$$
\Delta\Omega =
-3\pi\cdot \cfrac {J_2} {G\cdot M_e \cdot a^2}\qquad
\Delta\omega =
+6\pi\cdot \cfrac {J_2} {G\cdot M_e \cdot a^2}
$$
So if we assume the oblate spheroid is also rotating with time period $T$, and the orbiting body passed directly over a point $P$ on the spheroid's equator at time $t_0$, where will it be at time $t_0 + T$?
*
*Will it be directly over $P$?
*Or will it be $3\pi\cdot \cfrac {J_2} {G\cdot M_e \cdot a^2}$ ahead (East) of $P$ in longitude (considering $\Delta\Omega + \Delta\omega$)?
*Or will it be $3\pi\cdot \cfrac {J_2} {G\cdot M_e \cdot a^2}$ behind (West) of $P$ in longitude (considering $\Delta\Omega$ only)?
*Anything else?
Cheers!
Note: $a$ = semi-major axis, $e$ = eccentricity of the orbit, $i$ = inclination of the orbit, $G$ = gravitational constant, $J_2$ = a coefficient in the spherical harmonic expansion of the spheroid's gravitational potential field
| Since the period $T$ doesn't appear in either of those formulas of precession angle, I don't understand why you need a definition of $T$ (the formulas are self-consistent without $T$). In fact, we can not define period very well in some cases, since the orbit may not close on itself due to precession, which makes the movement aperiodic.
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Can we derive Einstein-Hilbert action through action principle and Levi-Civita connection? Suppose that we take principle of least action as given. Also assume that any manifold allowed by the action would carry Levi-Civita connection (torsion-free characteristic). Also assume that the local symmetry imposed on the tangent space of each manifold point is that of Poincare group, via general covariance principle.
Would these be sufficient to derive Einstein-Hilbert action, and by corollary Einstein field equations? Or do we need extra conditions to derive the Einstein-Hilbert action?
Edit: If not, then what would be other extra conditions?
| The way to get the Einstein-Hilbert action in 4 dimensions is to take those requirements along with the following ones :
*
*The stress-energy tensor has a null divergence
*The equation of motion has at most second derivatives in the metric
With those two extra conditions, the Einstein-Hilbert action, plus a cosmological term, is the unique solution in 4 dimensions.
| {
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Photoelectric effect, low frequency light Let's say we have a emitter, emitting light that has frequency f, less than the threshold frequency of a metal.
If you leave light shining onto that metal, for long enough, does the energy of the individual photons accumulate, on the electrons, so eventually they will ionize, or does this not happen? What am I missing?
| Energy from photon comes in packets and these do not accumulate. The point to be noted that a single photon transfers energy to the electron.
| {
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Why is the direction of magnetic field from South to North Pole inside a magnet? Since magnetic field lines are the path taken by a hypothetical North Pole when it is in range of a magnetic field of a magnet, it is clear that the direction of hypothetical North Pole would be from North Pole to South Pole of a magnet not even outside the magnet but inside the magnet too, because North Pole of magnet will repel the hypothetical North Pole inside and outside the magnet and would be attracted by the South Pole of the magnet.
| Take a large number of small magnets, all pointing in the same direction, and glue them together. You get something like:
This is a pretty good model for a typical bar magnet. The individual small magnets are individual atoms. Now place your hypothetical north pole inside this larger magnet, and you can see which way it is being pushed.
So far I quoted a picture also mentioned in the answer of HolgerFiedler. But I would like to take this a little further.
The next question might be, 'but what about those little magnets, what is inside them?' The answer to this question is that the magnetism of individual atoms is caused partly by intrinsic magnetism which particles such as electrons have, and partly by currents in the atom, associated with the charge on the electron and its motion. Both of these require a quantum physical description, but both respect the equations of classical electromagnetism called Maxwell's equations, and ultimately this lets you know that the lines of magnetic field $\bf B$ will always form continuous loops, never come to stop or a start. In the case of a single electron, the loops run right down to the electron itself. You can imagine the electron as spread out a little owing to its wavefunction being not perfectly point-like, and then the magnetism of the electron is also like the picture given above; that is, it is like a continuously spread-out bunch of tiny magnets all side by side. When the electron also moves, such as in the current loops found inside atoms, this adds a further contribution which you can model as like a solenoid.
| {
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Gravitational lenses and two slit experiment When an image of a distant galaxy is split by a gravitational lens are the photons from the two images capable of displaying interference? This may not be possible in real conditions because of magnetic fields/polarization/intervening matter etc. but in an ideal case it is possible? Is this split image inherently different from a two slit interference experiment?
There is a new observation cited of using correlations in intensity fluctuations of a quasar image split by a gravitational lens to determine the Hubble constant.
https://www.sciencedaily.com/releases/2019/01/190122171325.htm
What are the primary limiting criteria for observing interference with this example of a quasar split by a gravitational lens? Is this a small enough source? Is the time interval too large to create a delay line in one of the images to have interference? Is the correlation length too small?
| There is no interference in this situation. You need coherent light for interference. The distant galaxy is not a source of coherent light. It is an extended source emitting incoherent light.
| {
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Does photon behave both wave like and particle like at the same time? As we all know, photons show dual character, but do they behave both ways at the same time? Then how do they propagate just like other particles do and combined with wave motion?
| Well, according to Heisenberg in his essay named "The physical principles of the quantum theory", the answer is more complicated that it looks. He wrote, and I agree, that our language is conditioned by our macroscopic experience. In this scale the concepts of wave and particle are perfectly defined. In the quantum world, that is, with energy on the order of the $\hbar$ (the Planck's constant) or the respective longitudinal dimensions, "waves" and "particles" start to loose their meanings, the same with "trajectory" (a concept, by the way, fundamental in the definition of particle). The answer, once we understand this, is: there exist phenomena that in certain conditions seem like waves or particles: We can't just ensure that, in fact, the physical quantum objects are waves or particles.
| {
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If electron carries negative charge, why bother looking for it's electric dipole moment? I got confused after reading about spherical electron, would there really be some type of unknown particles popping in and out of existence within the virtual particle cloud around the electron that give electron it's property? Why is it important for the electron to have electric dipole moment?
| If you have two classical static electric charges and wish to compute potentials and electric field, you can approximate with multipole expansion.
Because, you can have separately charged particles. Contrast to magnetism, there is no such thing as magnetic monopole.
| {
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How the current moves in the battery and outside the battery? Does it move again to the negative terminal or stays at the positive terminal until the electrons all come .so the voltage will.be zero and current flow .I want simple clear explanation because i am confused .when I see a circuit that has two batteries one charging the other so I don't know how to think about it .i guess that electrons come from the negative terminal of the battery of higher emf to the negative terminal of the battery being charged and stay there but if this happen how the current moves to the positive terminal of the battery of higher emf.
| Generally charge (negative) moves from anode to cathode and the chemical reaction in the battery is mediated by a salt bridge (may be different for various battery types). The salt bridge keeps the charge of the total system from separating completely and, in effect, slows down the electro-chemical reaction so that the battery isn't depleted almost immediately. This means that once charge moves through the loop there exists an unbalanced charge distribution in the entire system. The salt bridge supplies ions to the electrolytic solution to maintain neutrality but does not move individual electrons. Without the salt bridge there would be no way for charge to continue to flow and supply power to any connected load. This means that although some electrons may be transported through the salt bridge to move through the loop again the majority complete the loop once and either remain at the cathode, or bind to an atom to become a anion and are absorbed by the salt bridge.
| {
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Continuum mechanical analogous of Maxwell stress tensor Maxwell stress tensor $\bar{\bar{\mathbf{T}}}$ in the static case can be used to determine the total force $\mathbf{f}$ acting on a system of charges contanined in the volume bounded by $S$
$$ \int_{S} \bar{\bar{\mathbf{T}}} \cdot \mathbf n \,\,d S=\mathbf{f}= \frac{d}{dt} \mathbf {{Q_{mech}}}\tag{1}$$
Where $ \mathbf {{Q_{mech}}}$ is the (mechanical) momentum of the system of charges.
What theorem/relation is formally analogous to $(1)$ in continuum mechanics? I've read that also in continuum mechanic one can introduce a tensor such that the value of its components on a surface $S$ enclosing a system of masses determines the forces acting on the masses completely.
I could not find this analogy on Jackson or Griffiths, so what is the tensor that is similar to Maxwell stress tensor in mechanics? Is it the stress tensor? By which theorem does it determine the forces on a system of masses?
| The classical-physics analog of the Maxwell stress tensor is the Cauchy stress tensor. It is a linear operator that inputs a unit vector at a point and outputs the stress density vector across an infinitesimal area normal to the unit vector at that point.
| {
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Expectation value of position of eigenstate of position Consider a quantum system in an eigenstate $|x\rangle$ of the position operator $\hat{x}$, which means that $\hat{x}|x\rangle=x|x\rangle$. I hope that the expected value of $\hat{x}$ will be $x$, since the state $|x\rangle$ is the one in which the system is located in the position $x$. To prove it we do the following:
\begin{equation}
\langle \hat{x} \rangle = \langle x|\hat{x}|x\rangle = \langle x|x|x\rangle = x\langle x|x\rangle
\end{equation}
Now, $\langle x'|x\rangle=\delta(x-x')$, so I think that $\langle x|x\rangle$ would be $+\infty$?, and then:
\begin{equation}
\langle \hat{x} \rangle = \pm\infty?\quad (\text{depending on the sign of $x$})
\end{equation}
Which is clearly wrong. How do you calculate this mean value?
| The formula for the expectation value $\langle A\rangle=\langle\psi|\hat{A}|\psi\rangle$ is given for the normalized states $\langle\psi|\psi\rangle=1$. You can generalize it as
\begin{equation}
\langle A\rangle=\frac{\langle\psi|\hat{A}|\psi\rangle}{\langle\psi|\psi\rangle}
\end{equation}
Of course this expression would still be ill-defined for $|x\rangle$ as it's not a proper quantum state, just as $\delta(x-x')$ is actually not a function but rather a distribution.
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What's the difference between muons and electrons in experiment? In the lecture note Track Reconstruction and Pattern Recongnition in High-Energy Physics written by Prof. Ivan Kisel, there is a figure (in page 7), which describes the different pieces of modern detectors.
He explained, Electromagnetic Calorimeter measures the total energy of $e^{+}$, $e^{-}$ and photons, and only muons and neutrinos can get to Muon Chambers.
Theoretically, muons and electrons are very similar except for their mass, both of them can interact with electromagnetic field. So my question is, why only muons can get to Muon Chambers, or why muons can get to Muon Chambers but not be detected in EM Calorimeter.
| Muons can more easily penetrate more material. Typically in most detectors there is a distinguishable pattern between muons and electrons. A clear example would be data from the Super Kamiokande detector where one detects Cerenkov light coming from electrons/muons.
The "fuzzines" of the right circle means that the light came from an electron which got scattered and emitted a few Bremsstrahlung photons.
The point of all this is to show you that just the difference in mass is enough to make the Muon penetrate a lot more material than the electron. This is due to the fact that the (simplified) formulas for Bremsstrahlung are:
or
and both are proportional to the acceleration SQUARED! The force applied on both particles from the material in the calorimeter for example is the same (just an E-field), but due to the different masses, the electron has a lot bigger acceleration, thus loses more energy. This of course is simplified, since if you are calculating it precisely you need to take relativistic effects into account, but the intuition is the same.
| {
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Gravitational Deflection of light I'm looking for an expression for the deflection of light in a static gravitational field. Referring to 'deflection of star light past the sun' in Sean Carroll's "Spacetime and Geometry" - equation 7.80 for the "transverse gradient":
$$
\nabla\perp\Phi = \frac{GM}{(b^2 + x^2)^{3/2}}\vec b
$$
Deflection angle is
$$
\alpha = {2GMb} \int {\frac{dx}{(b^2 + x^2)^{3/2}}} = \frac{4GM}{b}
$$
As far as I understand it, this transverse gradient is only valid for weak fields/small deflection. And I'm not looking for a general integral solution - I'd like to plot photon paths in strong fields, so I'm looking for the instantaneous deflection, which I'll plot/integrate numerically, based on mass, radial distance from mass, and initial angle of photon trajectory. It should not use the Schwarzschild solution/metric, because I don't want the singularity at r=Rs and it only needs to be in 2 dimensions, because of spherical symmetry. So, is there an expression for the polar coordinates r2, θ2 and trajectory a2, for a photon travelling from p1 to p2, using M, r1, θ1, a1, L? Below is a diagram which I hope illustrates it.
| As far as I can understand if you want a "general solution" you have to integrate to obtain the solutions for a null geodesic. That cannot be done in general and each static gravitational field will produce a different answer.
| {
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Why does the spectrum of blue LEDs appear to have two peaks? Due to having poor distance vision and thus strong corrective lenses, I'm "blessed" with the ability to see the spectrum of a light source whenever I look at it from an angle. One thing I've particularly noticed is that blue LEDs in appliances and such seem to have two distinct blue peaks to their spectrum - that is, I see 2 offset copies of the light source's shape, both blue, but with one appearing considerably dimmer and probably close to the UV range.
All the information I can find online about blue LEDs seems to indicate I should just see a narrow spectrum around a single wavelength somewhere around 460-480 nm. Are these perhaps just white phosphor-based LEDs behind a blue filter that's transmitting both the original blue and part of the spectrum of the yellow phosphor? Or is something else going on?
| The second image is probably the same colour, but due to internal reflection of the lens (off the near face and back off the far face).
I'm sorry to say that your blessing (which I share, btw) is not limited to chromatic abberation.
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Applying force on centre of mass of a body which lies outside the body The center of mass of a uniform ring lies at its geometrical center i.e.
outside the body. But we generally define center of mass as a point in which if we apply a force, the whole body will move in the same way as if the whole of the mass is assumed to be concentrated there and how that point mass moves under that force. So my question is how can we apply a force to a point not on body (here the point is the center of mass) and still see the force's effect on that body?
For example consider this: suppose a uniform ring hanging in mid air in a gravity free environment in $y$-$z$ plane. Now you have to move the ring along x-axis so that whole ring remains in a plain. No torque is to be applied and you can not put any charge on the ring and distribute it uniformly and apply an electric field along $x$-axis. How can we do it?
| If you apply a force to a point not on the body, there will be no effect, obviously.
Perhaps an important clarification on the whole center of mass thing: if for example you're dealing with gravity, you aren't saying "well it's as if there's a force $mg$ applied at the center of mass"! What you're saying instead is "it's as if the whole thing is just a 0-D point mass at the center of mass".
| {
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How much time will it take to move an object whose length is equal to one light year? Suppose there's a stick whose length is one light year and I push it from one side by one centimeter. How much time would it take for its other side to move by one centimeter and why?
| The answer to this depends on the material properties of the stick. If it were a mythical "rigid body," the entire stick would move all at once, but that "rigid body" is an approximation that isn't valid in situations like this. The rigid body assumptions assume that you can transmit information about movement instantaneously. In many cases, that's close enough to correct, but in situations like this, the speed at which you can propagate the force through the stick is not infinite, and matters.
Instead, you would need to set up a compression wave along the stick which transmits the information about you pushing it down the length of the stick. The speed of that compression wave depends on the material you use. Thus, in the real world, the "best case scenario" is dependent on the speed of sound in the medium you build the stick out of.
| {
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Why are photons not involved in $\beta -$ decay? Theory of $\beta -$ emission exclusively considers participation of one $W -$ boson, one electron and one electron-antineutrino, but no photons. However Maxwell's electrodynamics require that during any change of an electromagnetic field, EM-radiation (photons) must be emitted. Spontaneous separation of two oppositely charged particles generates a variable electric dipole field which should emit EM-radiation. Moreover, as the emitted electron is strongly decelerated in the Coulomb field this also should generate EM-radiation. However, spin angular momentum conservation would require pairwise emission of photons with mutually opposed spin (sum $+1 - 1 = 0$). To my knowledge such EM-radiation has never been reported or predicted.
| It has been reported, predicted, and even used to place an upper bound on the neutrino mass. My background is condensed matter so I may be very wrong about these sorts of particle-physics topics: but I believe the presence of a large nucleus nearby probably breaks your expectation of pairwise emission. (Pairwise emission is however a well-known and common feature of positron annihilation, where the opposite-directionality is used by PET scanners to filter out high-energy photons coming from non-positron sources.)
The reason that a first romp through the literature might not have revealed this to you is most likely that you do not have a particle physics background; this sort of electron-braking radiation is known to particle physicists by its German name bremsstrahlung even in English publications. More specifically the bremsstrahlung from the inside of the nucleus is known as "inner bremsstrahlung"; this search phrase leads to many discussions of the phenomenon, and even a mention on Wikipedia.
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Does the density parameter alone determine the fate of the universe? According to this: http://astronomy.swin.edu.au/cosmos/D/Density+Parameter
The density parameter alone determines the fate of the universe (if its less than 1 expands forever, etc.). But according to wikipedia that's not exactly the case because of dark energy. For example a universe with a density parameter of more than 1 with enough dark energy can actually expand forever.
So, which is correct?
| The first of your two sources is incorrect. The fate of the universe depends on the total density parameter and the contribution of dark energy to that total density.
Universes can be open, closed or flat and these will occur for cases where the total density parameter is $<1$, $1$ or $>1$ respectively. However, this just refers to the geometry of the universe. Only in the case of a universe with no dark energy are these terms synonymous with universes that will expand forever, expand asymptotically to a standstill or eventually recollapse.
The presence of dark energy complicates matters as shown below. This plot, from the supernova cosmology project, shows constraints on the density due to matter and that due to dark energy. The diagonal line marks where the total of the two is unity and we have a flat universe (strongly favoured by cosmic microwave background results). A curved locus is also shown that divides those universes that will expand forever and those that will eventually recollapse.
The thing to notice is that there are areas of parameter space where closed universes can expand forever and also regions where open universes can recollapse (but only if $\Omega_{\Lambda}<0$).
At present, the best bet is that we live in an almost flat universe that is accelerating and will expand forever.
The reason for this behaviour is that in an expanding universe, the behaviour of the matter and dark energy densities are different. Whilst the matter density decreases as the cube of the scale factor, the dark energy density may remain constant. Thus a universe can only recollapse if it stops it's expansion before dark energy becomes too dominant.
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What does Enthalpy mean? What is meant by enthalpy? My professor tells me "heat content". That literally makes no sense. Heat content, to me, means internal energy. But clearly, that is not what enthalpy is, considering: $H=U+PV$ (and either way, they would not have had two words mean the same thing). Then, I understand that $ΔH=Q_{p}$. This statement is a mathematical formulation of the statement: "At constant pressure, enthalpy change may be interpreted as heat." Other than this, I have no idea, what $H$ or $ΔH$ means.
So what does $H$ mean?
| A brilliant analogy by Daniel Schroeder:
*
*To summon a rabbit the magician must "build" it with all the energy it consists of. He must provide its internal energy $U$.
*But first he must push away all the air that is in the way. This requires some work, $W=pV$. In total, the energy he must spend is $U+pV$. Let's call that enthalpy $H$:
$$H=U+pV.$$
*
*But the surroundings might help him out a bit. The warm air might provide some energy, while he is working on the summoning, by adding heat $Q=TS$. The only energy he actually has to spend himself is therefore $U+pV-TS$. Let's call this the free energy needed, or Gibbs free energy $G$:
$$G=H-TS.$$
| {
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Does entropy decrease if time is reversed? Entropy increases if we let newton's equation work its magic.
Since newton's equation is time reversible, I would assume that in a closed isolated system, solving the differential equation and running time backwards would increase (and NOT decrease) the entropy of the system.
Is that true?
| Simple answer: in our universe, definitely no.
You're hitting here on an idea known as Loschmidt's Paradox[1]: given that microscopic laws are time reversible, entropy should have the same tendency to increase whether we run a system forwards or backwards in time, exactly as you understand.
The fact that this understanding is manifestly against experimental observation can be explained if we observe that the universe began (i.e. found itself at the time of the big bang) in an exquisitely low entropy state, so that almost any random walk in the universe's state space tends to increase entropy. Likewise, in the everyday world, things "happen" when a systems are not in its maximum entropy state: they spontaneously wander towards these maximum entropy states, thus changing their states and undergoing observable changes. Sir Roger Penrose calls this notion the "Thermodynamic Legacy" of the big bang and you could read the chapter entitled "The Big Bang and its Thermodynamic Legacy" in his "Road to Reality". In summary, we have a second law of thermodynamics simply by dint of the exquisitely low entropy state of the early universe.
[1] Loschmidt's own name for it is "reversal objection" (umkehreinwand), not "paradox". Paradoxes, i.e. genuine logical contradictions cannot arise in physics, otherwise they could not be experimentally observed.
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How can energy of oscillators be quantised but they can still vibrate at all frequencies? In Black Body radiation, Planck's law has a postulate saying
The wall of black body contains oscillators of all possible frequencies,ν.
There is one more postulate which says
The energy of these oscillators is not continuous but discrete valued.
Of course the second one is very well known but doesn't it contradict the first one? Shouldn't it mean that frequencies of vibration are quantised like maybe in case of standing waves?
Please correct me if I am wrong in stating the postulates itself.
| I think the source of confusion is relative sizes of the cavity and typical wavelengths of standing waves in the cavity. In the Plank's law derivation the size of the cavity is considered to be much bigger than the wavelengths, that is why the oscillator frequency spectrum is considered to be continuous.
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Do frame fields (tetrads) satisfy orthonormality vector field condition if orthogonal? Referencing https://en.wikipedia.org/wiki/Frame_fields_in_general_relativity#Relationship_with_metric_tensor.2C_in_a_coordinate_basis :
Suppose we start directly from $g^{\mu \nu}= e^{\mu}_{\ a} e^{\nu}_{\ b} \eta^{ab} \,$ (eq 1), where $g$ of course refers to metric tensor, and $e$ refers to tetrad, with $\nu$ representing Lorentz metric. Assuming some coordinate system, if tetrads are set to satisfy eq 1 and tetrads satisfy orthogonality condition, then wouldn't tetrads also satisfy orthonormality conditions? Or is there any other condition needed to make tetrads to be orthonormal vector fields instead of just orthogonal vector fields?
| Let $g,e,\eta$ denote the corresponding matrices. Equation is $$ g^{-1}=e\eta^{-1}e^T. $$
Invert everything here: $$ g=(e^T)^{-1}\eta e^{-1}. $$ Now express $\eta$: $$ e^Tge=\eta\Leftrightarrow e^\mu_ag_{\mu\nu}e^\nu_b=\eta_{ab}. $$ There's your ortonormality condition.
| {
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Confusion regarding gravity not being a force In high school, it was taught that formula for describing circular orbital velocity around a central body is derived by equating Newton's law of gravity with the centripetal force formula (under the logic that the inwards centeipetal force required is provided by the gravitational "force").
It was only recently that I discovered that gravity isn't actually a force but is actually a distortion of space time. (I came across this while wondering why light bends around large masses).
Does the fact that gravity is not a force make the above derivation of orbital velocity any less valid? Because the above derivation assumes that gravity is a force.
| You cannot longer use Newton's formalism $F = m a = -GMm/r^2$ if you introduce the fact that the geometry of space time is changed by the presence of the central body $M$.
It is true that the test mass $m$ still moves around $M$ because of gravity, but you should think of gravity not as a force any more, but as an emergent property of the curvature of space-time. Fortunately there's a whole body of mathematical tools that allow you solve this problem in particular.
Actually, it is one the most well known problems you can analytically solved using general relativity: the two body problem
| {
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Aliens from antimatter region If we established a radio contact with an extraterrestrial civilization, assuming the signal latency permits, what question would we ask them to know if they are made of matter or anti-matter? I realize that there may not be anti-matter regions in the universe. However, my question is not about what the universe consists of. My question is about conceptual differences between matter and anti-matter that could be tested and communicated. My thinking is that we would need to ask two questions. One would establish the chirality of weekly interacting leptons. However, this alone would not be enough, because we would not have a common reference to the concepts of left and right. Then we would ask for the results of an experiment with a CP symmetry violation. This result in relation to the chirality data should create a distinction between left and right and also between matter and antimatter. Am I on the right track or completely derailed?
| The question would be fairly simple. Anti-matter aliens would have nuclear physics with anti-protons and anti-neutrons. Weak interactions would produce then produce what would appear to be right handed CP violations. The question to ask is what is the parity violation in the beta decay of cobalt-60 nuclei found by by Chien-Shiung Wu. If these aliens given an answer to the parity violation that is the mirror image of what we know then they are anti-matter.
| {
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Quantum Mechanics and Schur's lemma Today i was studying on a textbook and i crossed a paragraph that confused me a little.
Suppose you have an algebra generated by $\hat{X}$ and $\hat{P}$ and a function $f(\hat{X},\hat{P})$ that commutes with $\hat{X}$ and $\hat{P}$. Then you can prove that this function is proportional to the identity.
Here comes the issue, i don't understand the proof of this.
Reasoning in Position representation (meaning that $\hat{X}=x$ and $\hat{P}=-i\hbar\frac{d}{dx}$), i don't get why $[x,f]=0$ implies that $f=f(x)$ and why $[-i\hbar\frac{d}{dx},f]=0$ implies that $i\hbar\frac{df(x)}{dx}=0$.
| Let $\psi(x)$ be any wave function, and assume $f$ can be formally expanded:$f=\sum_{km} a_{km} x^k p^m$. Then
\begin{align}
[x,f]\psi(x)&=x\left(\sum_{km} a_{km} x^k p^m\right)\psi(x)- \left(\sum_{km} a_{km} x^k p^m x\right)\psi(x)\, ,\\
&=\left(\left(\sum_{km} a_{km} x^{k} xp^m\right)- \left(\sum_k a_{km} x^k p^m x\right)\right)\psi(x)\, ,\\
&=\sum_{km}a_{km}x^k [x,p^m]\psi(x)\, .
\end{align}
If $[x,f]=0$, and since $[x,p^m]\ne 0$ unless $m=0$, it must hold that $a_{km}=0$ for $m\ne 0$, or
$$
f=f(x)=\sum_{k}\tilde a_k x^k\, .
$$
Now consider $g(x,p)$
\begin{align}
[p,g(x,p)]\psi(x)&= p\left(\sum_{km} b_{km} x^k p^m\right) \psi(x) -\left(\sum_{km} b_{km} x^k p^m\right) p\psi(x)\, ,\\
&=\sum_{km} b_{km} [p,x^k] p^m\psi(x)\, .
\end{align}
Again, if you assume $[p,g]=0$ then the rhs is never $0$ unless $k=0$, implying this time that $g=g(p)$ only.
| {
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Tensor in different coordinate system I have the tensors $F_{\mu\nu}$, $F^{\mu\nu}$ in coordinate system $(t,x,y,z)$ and want to transform these to coordinate system $(t',x',y',z')$ just by multiplicating matrices.
My idea was to calculate the Jacobians $J=(\frac{\partial x^i}{\partial x'^j})_{ij}$ and $J'=(\frac{\partial x'^i}{\partial x^j})_{ij}$.
Then I would find $$F'_{\mu\nu}=J^\top F_{\mu\nu}J$$ and $$F'^{\mu\nu}=J' F^{\mu\nu}J'^\top,$$ in matrix notation.
Is this correct?
My ultimate goal is to prove that $F_{\mu\nu}F^{\mu\nu}$ is the same in both systems, however calculating this explicitly does not give me this result.
| By definition, under a change in coordinate system a tensor's components transform(s)
$$F_{\mu'\nu'}=\frac{\partial x^{\mu'}}{\partial x^{\mu}}\frac{\partial x^{\nu'}}{\partial x^{\nu}}F_{\mu'\nu'}$$
So you could prove your statement by showing
$$F_{\mu'\nu'}F^{\mu'\nu'}=F_{\mu\nu}F^{\mu\nu}$$
by using the metric $\eta_{\mu\nu}$ to raise and lower the indices on the definition of the transformation.
| {
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QFT: Range of 'collision' If two particles approach each other, they can [provided that their properties add to those of other particle(s)] interact and go from, say, $$e + \bar e \to \gamma + \gamma$$
My question is how would one estimate the range of this? What distance is needed between $e$ and $\bar e$, is it less than or equal to some expectation value of an operator? Or do they particles just have to be delocalised enough to overlap, and then there's a finite probability of them interacting as a function of that overlap? If two particles like these interact, is there any restriction on where the two photons propagate from [i.e. the same point, or just anywhere in the overlap etc]?
| My understanding is as follows: "in" and "out" states are fully delocalized states with exact momentum (plane waves) whose history starts at time $-\infty$ and ends in $+\infty$. Everything in between is not a state but so-called "S-matrix evolution". So: how is you question truly answered in nature? My answer: I do not know. How is it answered in our perturbation-based quantum field description: there is no "typical range". Particle do not feel each other only in times $\pm \infty$ when they become states from the free theory (on which perturbations are applied). In times in between they feel each other always. Actually the mere existence of "asymptotic states" for long-range interactions (such as electrodynamics) is questioned.
| {
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Can I achieve BEC with electrons only? Suppose I traps some free electrons using magnetic field and cool them to subzero temperature, can I get a super-electron this way? and does it violate the exclusion principle?
| No because electrons are fermions not bosons.
It is possible for fermions to form a BEC if they can pair up with spins opposed to form a bosonic particle. For example this is what happens in superconductivity and superfluid helium-3. However this requires some attractive force to bind the particles together. In superconductivity deformation of the crystal lattice provides the attractive force while in helium-3 it's the London force.
However for a gas of free electrons there would be no such attractive force to pair up the electrons, so they could not form a BEC.
| {
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What is the scientific term for the photon emission caused by electrons jumping orbitals? Suppose an electron in an atom at the K-orbital is sufficiently energized (excited) that it jumps to the L-orbital. When this electron returns to its ground electronic state (it jumps back to the K-orbital), in order to respect the law of conservation of energy, the excess energy it has garnered from the excitation is shed through the emission of a photon, the frequency of which varies with the atom. This is why the burning of sodium chloride atop a near-colourless hydrogen flame produces an intense yellow light.
What is the scientific term for this photon emission? I thought it was Bremsstrahlung, but turns out that refers to a completely unrelated type of photon emission.
| Excitation of electrons to higher orbitals give spectral absorption lines. The consequent deexcitations give spectral emission lines.
The term is spectral lines, to contrast with the continuum produced by other emissions.
| {
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What is the hydrodynamical explanation for the 'footprint' of a diving whale? When a whale dives it leaves behind a so called 'footprint'. The water seems to be calmer or the surface is at least more smooth an shows less wrinkles.
Image source
I read some text which were talking about a 'wake of the diving whale' but I wanted to understand it from a fluid dynamic (two-phase) point of view, so I thought asking here might be a good idea.
| The whale creates a vortex ring with its tail, which moves upwards and creates an oval patch of outward current on the surface: short surface waves can't propagate (well) against the current, and that's why the patch is smoother than the ocean around it.
Source: The flow induced by a mock whale : origin of flukeprint formation.
| {
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Torque and its origin What's the origin of the definition of torque, that is, the moment of force about the point in consideration? - (1)
Is there a proof?
Why should torque be r x F, and not r² or r³ x F for that matter? I have tried searching many books but none gave the proof as to why torque should be r x F only, and not any other combination of the two.
I was hoping if it'd be possible to design a thought experiment, that clearly validates equation (1)
Thanks!
| By definition a moment in physics is the product of a physical quantity and a position. So the quantity $\vec{r}\times\vec{F}$ could by definition be considered a moment of force. The utility of this moment is found when we consider the moment of momentum, $\vec{L}=\vec{r}\times \vec{p}$ ($\vec{p}$ is momentum), also called the angular momentum. It's called this because moments are generally related to rotational or turning or angular behaviors.
If we accept that angular momentum is a conserved quantity, then an important idea to consider is the time rate of change of momentum, which we call torque, often symbolized by $\vec{\Gamma}$ or $\vec{\tau}$. Conservation of angular momentum says
$$\vec{L}_{\mathrm{new}} = \vec{L}_{\mathrm{old}}+\int\Gamma\ \mathrm{d}t.$$
Calculating the time rate of change of $\vec{L}$ we get
$$\frac{\mathrm{d}\vec{L}}{\mathrm{d}t}= \frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\times \vec{p}+\vec{r}\times \frac{\mathrm{d}\vec{p}}{\mathrm{d}t}$$
The first cross product vanishes (I'll let you figure that out), and $\frac{\mathrm{d}\vec{p}}{\mathrm{d}t}=\vec{F}.$
So the time rate of change of momentum, which we name torque, is equal to the moment of a force. [Edit: added "equal to"]
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Calculations in the Relativity of Simultaneity Train Thought Experiment I have some confusions about proper time and the calculations involved in Einstein's famous thought experiment where two lightning strikes occur simultaneously at the ends of a train for an observer observing the train to be moving at velocity v (Observer 1) while an observer stationary with respect to the train (Observer 2) observes these strikes to non-simultaneous.
In a lecture, my teacher presented the two "events" as being (for each lightning strike):
*
*The lightning strikes the end of the train
*The light travels to the observer's eyes
To show that there was a discrepancy between Observer 1 and 2 wrt the train and the observer watching the train move at velocity v, the time dilation formula was used. My understanding is that the time dilation formula relates proper time and coordinate time. Aren't neither observers measuring the proper time between these two events? Proper time is the time measured by an observer stationary in a frame in which the events occur at the same spatial coordinates, and this does not happen for either observer here.
I will go through the derivation presented:
Let L be the length of the train from Observer 1's frame (it is dilated from the rest length observed in Observer 2's frame). Let t be the time for for Observer 1 to see light from the front of the train hit Observer 2, and t be that time for Observer 2.
In Observer 1's frame, Observer 2 is moving towards the incoming light from the front of the train, so we have $$vt = L/2 - ct \rightarrow t = \frac{L}{2(c+v)}.$$
In Observer 2's frame, the time they observe $\tau$ is given by $$t = \gamma \tau.$$
The last step is the source of my confusion as I mentioned above.
Another question I have concerns the Lorentz transformations, as I see in this thread. From this, it looks to be that doing the Lorentz transformations is equivalent to applying time dilation and length contraction together. Is this understanding correct in some sense?
| We have two events here :
\begin{align}
&\text{event f : A light pulse from the train front}
\tag{01}\\
&\text{event b: A light pulse from the train back}
\tag{02}
\end{align}
Let these two events have the following space-time coordinates
For Observer 1:
\begin{align}
\text{event f : } &\left(x^{f}_{1},t^{f}_{1}\right)
\tag{03}\\
\text{event b : } &\left(x^{b}_{1},t^{b}_{1}\right)
\tag{04}\\
x_{1} \text{ difference :} \quad & \Delta x_{1}=x^{f}_{1}-x^{b}_{1}=L_{1}
\tag{05}\\
t_{1} \text{ difference :} \quad & \Delta t_{1}=t^{f}_{1}-t^{b}_{1}=0
\tag{06}
\end{align}
For Observer 2:
\begin{align}
\text{event f : } &\left(x^{f}_{2},t^{f}_{2}\right)
\tag{07}\\
\text{event b : } &\left(x^{b}_{2},t^{b}_{2}\right)
\tag{08}\\
x_{2} \text{ difference :} \quad & \Delta x_{2}=x^{f}_{2}-x^{b}_{2}=L_{2}
\tag{09}\\
t_{2} \text{ difference :} \quad & \Delta t_{2}=t^{f}_{2}-t^{b}_{2}
\tag{10}
\end{align}
Lorentz Transformation :
\begin{align}
\Delta x_{2}&=\gamma\left(\Delta x_{1}-v\Delta t_{1}\right)=\gamma\left(L_{1}-v\cdot 0\right)=\dfrac{L_{1}}{\sqrt{1-\dfrac{v^{2}}{c^{2}}}}=L_{2}
\tag{11}\\
\Delta t_{2}&=\gamma\left(\Delta t_{1}-\dfrac{v\Delta x_{1}}{c^{2}}\right)=\gamma\left(0-\dfrac{v L_{1}}{c^{2}}\right)=-\dfrac{vL_{1}}{c^{2}\sqrt{1-\dfrac{v^{2}}{c^{2}}}}=-\dfrac{v}{c^{2}}L_{2}
\tag{12}
\end{align}
Nothing more, nothing less.
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Thoriated Tungsten filaments I have a question about doping a tungsten wire with thorium to improve the thermionic emission. I have found that the doping lowers the work function of the system, resulting in a thermionic current greater than the one produced from a pure W filament. The point is, WHY this happens? What are the mechanisms that lower the work function?
An idea that I have is that thorium (and most of its decay chain products) is an $\alpha$-emitter. The $^{4}$He nuclei then produce scattering sources for the electrons, enhancing the emission probability.
References are higly appreciated :)
| This seems to be well described here, which, in turn contains a reference to L.W. Turner,(ed), Electronics Engineer's Reference Book, 4th ed. Newnes-Butterworth, London 1976. I just might have a copy of that somewhere, but not to hand.
The mechanism agrees with what I remember being told it was: Thorium is added to the Tungsten filament, and then the whole thing is heated significantly hotter than its normal working temperature. This causes (why?) the Thorium to migrate to the surface of the filament, resulting in essentially a Thorium-plated Tungsten filament. Thorium has a lower work-function than Tungsten so the effective work-function of the filament is that of Thorium.
It should be fairly easy to convince yourself, based on the activity of Thorium compared with the number of electrons a hot cathode needs to emit, that its radioactivity can't be a factor.
I am not sure why Thorium is not just plated onto the filament, although I expect this is to make production simpler: if you can effectively plate the filament by heating it after production (which you can do electrically of course, since that's what you do anyway) then this makes it all cheaper and easier. Probably the initial heating of the thing also helps weed out infant mortality: filaments which are going to fail early probably fail during the initial heating.
| {
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Charged particles emiting radiation I tried to find the reason by couldn't get the right answer. I wished to know that if both proton and electron have same acceleration, do they radiate same amount/amplitude of EM radiation. Also, since quarks have electric charge, do they radiate photons along with gluons (which I have not heard much about)?
| Proton and electron have exactly the same charge so they would radiate the same amount of EM radiation as per Larmor Formula. Relativistic generalization of it could be calculated through Liénard–Wiechert potential, but it is still going to be independent of the mass.
Quarks are charged particles with a charges equal to the fraction of the electron charge ($\frac{1}{3}$, $\frac{2}{3}$, $\cdots$), and thus are obliged to interact with Electromagnetic field too.
This answer might be useful too.
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What feature of the Beryllium nucleus makes it such a good source of neutrons when hit with alpha radiation Be is often used as the neutron gun to supply neutrons to initiate nuclear reactions, particularly nuclear chain reactions. I always presumed the nucleus of lighter elements to be more stable than the heavier ones. Be with an atomic number of 4, meaning 4 protons, it is twice that of a helium nucleus. The helium nucleus is the same as an alpha particle, and is regarded as the most stable light nucleus.
Being that we can think of Be as two helium nuclei fused together, one would presume that the Be nucleus would be very hesitant to change its structure and give up neutrons. But this is what happens when it is struck with an alpha particle. Why is this so? I read that it has a neutron spin of 3/2. I don't yet understand the neutron spin property, but maybe this could explain it.
| As noted, the (sole) stable isotope of Be is $^{9}$Be, with 4 protons and 5 neutrons. $^{8}$Be spontaneously dissociates into two $\alpha$ particles since it is more favorable energetically (this was the basis for the first human-controlled nuclear reaction, $^{7}$Li(p,$\alpha$)$\alpha$ by Cockroft and Walton in 1932 where they varied the proton energy).
The $^{9}$Be($\alpha$,n)$^{12}$C reaction, with Q=5.7MeV, is detailed in Richard G. Miller and R.W. Kavanagh, Nuclear Physics 88 492-500 (1966). The fundamental take away is that there are a number of resonances between the excited $^{13}$C nucleus and available $^{12}$C+n states, for various neutron energies. Furthermore there is the possibility of a 3$\alpha$+n output state mentioned in the paper, although that does not appear on nuclear energy level diagrams available at, e.g., tunl.duke.edu.
I wouldn't say there is anything particularly special about the overall reaction, just lots of opportunities to end up with a neutron out.
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Inducing emf in a circular coil Magnetic field lines from a moving magnet induces emf in a circular coil.What does the velocity of the magnet have to do with the emf induced?Increasing the velocity of the magnet doesn't increase the amount of field lines entering the coil right?
| According to Faraday's Law (ignoring Lenz's law, which isn't relevant here):
$$\epsilon = \frac{\Delta N\phi}{\Delta t}$$
Therefore, the EMF ($\epsilon$) is proportional to the rate of change of magnetic flux linkage ($N\phi)$. Magnetic flux ($\phi$) is just the number of field lines (imaginary lines denoting the strength and direction of a magnetic field) through the coil and $N$ is the number of turns there are in your coil.
Anyway, this is what the flux around a bar magnet looks like:
Source (https://s3.amazonaws.com/classconnection/812/flashcards/1488812/gif/magnetic_fields-15319FD4B8B56187995-thumb400.gif)
If you move this bar magnet towards the coil of wire, the number of field lines passing through the coil will change. In other words, the total magnetic flux (and therefore magnetic flux linkage) through the coil is changing. This change is what induces an EMF.
Note: the field lines are stronger closer to the magnet and extend away from it. So when the magnet is far away only a few outer field lines are passing through the coil, but as the magnet comes closer more of the closer field lines pass through the coil as well, resulting in a change in flux linkage through the coil.
A change in flux may induce an EMF across each turn in the coil, but what determines the size of this induced EMF is the rate of change of flux linkage.
So now think logically: If the magnet moves towards the coil with a fast velocity, the rate at which the flux through the coil is changing will be very fast, right? That's why the velocity of the magnet is important. An EMF will be produced whether the magnet is moving quickly or slowly, but it's the magnitude of the EMF that depends on the rate of change of flux, which is dependant on many things, including the speed of the magnet (or the speed of coil moving towards the magnet - same difference).
If you want to know more about whether the EMF induced is positive or negative (remember, electromagnetic induction produces an AC current), then look up Lenz's law.
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Notion of anisotropic Fermi surface I would like to check something.
I know that the Fermi energy is the maximum energy occupied by a Fermion at $T=0$ (if I have $N$ fermion it will be the energy of the Fermion that has the highest single particle energy).
I would like to check something about the anisotropy of the Fermi surface.
Does the anisotropy of the fermi surface occurs only because of the geometry of the material?
Imagine I have a free electron gas, I know that the wavevectors allowed are of the form:
$$ k=2 \pi (\frac{n_x}{L_x},\frac{n_y}{L_y},\frac{n_z}{L_z}).$$
Thus, if $L_x \neq L_y \neq L_z$, the value of $k_F$ can be reached for vectors that do not lie on a sphere.
Thus, in a general case I would have $$k_F^2=4\pi^2(\frac{n_x^2}{L_x^2}+\frac{n_y^2}{L_y^2}+\frac{n_z^2}{L_z^2})$$
Which is the equation of an ellipsoïd.
In conclusion : the fermi surface is anisotropic only because of the geometrical structure of the crystal and it is always an ellipsoïd.
Am I right?
Also, can we define a fermi surface for an interacting system? Indeed, to define it we need to talk about particle wavevectors. And it is a good quantum number for free electrons. So how is it defined for an interacting system?
| Like you said, Fermi energy corresponds to the highest energy occupied by an electron at $ T = 0 $. For the free electron gas, $E = 1/(2m)\times\left(k_x^2+k_y^2+k_z^2\right)$. If you plot the momenta on the Cartesian axes, the constant energy will be a spherical shell.
Here's a more intuitive way of seeing this. Take a 3D box with $L_x\neq L_y\neq L_z$. The Schrodinger equation that you need to solve is
$$
-\frac{\hbar^2\nabla^2}{2m}\Psi\left(x,y,z\right) = E\Psi\left(x,y,z\right)
$$
for $x\in \left[0,L_x\right]$, $y\in \left[0,L_y\right]$, $z\in \left[0,L_z\right]$. The differential equation is separable so we get
$$
-\frac{\hbar^2}{2m}\Psi''_x\left(x\right) = E\Psi_x\left(x\right)\,,
\\
-\Psi''_x\left(x\right) = \frac{2mE}{\hbar^2}\Psi_x\left(x\right)\,,
\\
-\Psi''_x\left(x\right) = k^2\Psi_x\left(x\right)\,.
$$
Note that $k$ does not depend on the direction, only on energy and the mass. In $x$-direction, we have the unnormalized solution
$$
\Psi_x\left(x\right) \propto \sin kx\,.
$$
Because the wave vanishes at the boundaries, $k = n_x \pi / L_x$. Similarly, for $y$ and $z$ directions, we have $k = n_y \pi / L_y$ and $k = n_z \pi / L_z$.
If we set $E \rightarrow E_F$, $k\rightarrow k_F$ and
$$
n_x = \frac{k_F L_x}{\pi}\,,
$$
and similarly for $y$ and $z$. You see that the $n$'s are not the same in the three directions. $k_F$, on the other hand is the same so that your Fermi surface is a sphere.
Now, imagine if the mass were anisotropic. Then, the energy becomes
$$ E = \frac{k_x^2}{2m_x} + \frac{k_y^2}{2m_y} + \frac{k_z^2}{2m_z}\,.$$
In this case, the Fermi energy surface is no longer a sphere, but an ellipsoid.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/360204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How may sound waves behave inside the human body? in Vibroacoustic Therapy (VAT) sound is transferred to skin surface via transducers that are in direct contact with the skin. This means no energy loss to surrounding air. We are mostly using sinusoidal frequences between 30 Hz and 120 Hz. The different organs inside the body are of different density and the sound waves will therefore bounce in different directions and be differently absorbed by the internal organs. We know, empirically, that the effect of the internal sound massage usually is positive , relaxing and stimulating. as the density of different organs are different, sounds will be differently absorbed or penetrating different tissues. Have anyone done research how single cells and their internal structure are reacting to sounds? As acoustics describe auditively perceived sounds, the "endoacoustic" effects must be described by a totally different vocabulary, because we are looking for the medical effects of surface-to-surface transfer of phonons to the human body.
Do anyone have ideas of how we can describe the effect of VAT.
| sound waves of that low frequency will not be affected by the position or density or size of the internal organs. for organs to be targeted, the wavelength of the sound must be smaller than the dimensions of the organ. Note also that individual cells in the body that make up those organs are extremely small compared to your wavelengths and so they individually experience only extremely tiny effects when low frequency sound passes through them. those effects would be nearly impossible to measure.
Regarding phonons: the phonon concept only makes sense when describing the response of a crystalline solid to wavelengths of sound that are of similar order to the spacing between individual crystallites, or smaller. It's important on scale lengths where quantum effects become important and not when describing sound wave propagation through macroscopic bodies of varying composition.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/360351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Gravity and centrifugal force Assuming the radius of Earth as $6.36\cdot10^6\:\mathrm m$. Then i get for the centrifugal force of an object with the mass of $m=10,000\:\mathrm{kg}$:
$$r\cdot\omega^2 \cdot m = 6.36\:\mathrm m \cdot \left(\frac{1}{24\cdot3600\:\mathrm s} \right)^2\cdot 10,000\:\mathrm{kg} = 8.52\:\mathrm{N}$$
So when I want to know how big the gravitational force is on a mass $M$, which is the correct attempt?
*
*$F=m \cdot a = M \cdot g$ but here the centrifugal force is ignored. So the actual resulting force should be $M \cdot g - F_\mathrm{centrifugal}$.
or
*Is the value of $g$ actually given with the centrifugal force in mind? Would that mean that the gravitational force would actually be higher on a mass $M$ when the earth stops rotating?
| The centrifugal force does have an effect on the weight of objects on the Earth.
It is about half the reason things are lighter at the equator than the poles, the other being that they are further from the center of the Earth, owing to its oblate shape (itself a result of the centrifugal force). The reason you got such a low estimate of the influence of the centrifugal force is a miscalculation of $\omega$. It is not 1/day but 2$\pi$/day. So the force, using your numbers, is 336 N rather than 8. If the value of g without the centrifugal force were 9.81 $m/s^2$, this would reduce it to about 9.78.
In practice, yes, the value of g used always includes not just a purely gravitational force, but also the centrifugal contribution. This is more convenient than considering the forces separately, and is possible because the centrifugal contribution is always the same at any given place (in contrast, for example, to the Coriolis force, which depends on the motion of the the object experiencing it). So, for example, at locations away from the equator and the poles, the direction of "gravity" (i.e. the direction a plumb bob points) is not toward the center of the Earth, but rather a bit toward the opposite pole from there, always perpendicular to the local surface of the geoid. The true gravitational force on a mass m points toward the center of the earth (or very nearly), the centrifugal force points outward from the Earth's axis, and mg is the sum of the two.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/360525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why do radio waves spread out while higher frequency waves travel in beams? Why is it that radio waves spread out in proportion to the square of the distance, while higher frequency electromagnetic waves, like microwaves, infrared waves, light, etc are able to propagate as beams? What fundamental property allows higher energy waves to travel differently than lower energy?
| Due to diffraction, wave effects become more important as the size of the wave source becomes comparable to the length of the wave. Visible light has micrometer-scale wavelengths, so a millimeter-sized light source is thousands of wavelengths across and diffraction isn't a very big deal. But radio wavelengths can be many meters, producing similar collimation for a radio "beam" would require an emitting antenna hundreds or thousands of kilometers across.
You can use the same logic to think about shadows. A hair that's less than a millimeter across can cast a well-defined shadow, while radio waves diffract around buildings. However larger objects can cast well-defined radio shadows: for instance astrophysical radio sources disappear when they are covered by the Moon or the Sun, which are both very many wavelengths across.
Note that even "collimated" light undergoes dispersion.
Any sort of focusing optical system will produce a beam waist at some finite distance from the final focusing element (mirror or lens or whatever); beyond that beam waist the intensity of the light falls off like $r^2$ just as if a light source were at that location. A perfectly collimated beam of light is prohibited by the uncertainty principle, unless the beam is infinitely wide.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/361055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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} |
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