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Do particles always flow from high to low pressure? In a recent question, it was stated that
particles in high pressure air always flow to lower pressure.
In a pipe with a constriction, fluid flows from from low to high pressure after the constriction. (From here.)
How are these concepts related?
| The net movement of particles is from high to low pressure however individual particles may not move this way.
Let us assume that the particles have negligible inter-molecular forces - ideal gas assumption but probably appropriate in most fluids. This would imply that a particle wouldn't 'know' if it was in a high or a low pressure region, it would just whizz along as would either way.
However, in a high pressure region there are more, faster, particles. If you had 2 regions, one with high pressure and one with low pressure, connected by a pipe, particles would move both ways. However, just out of statistics, more particles would move from high pressure to low pressure.
There is no extra force on the high pressure area, just more particles flowing.
Granted, for highly viscous fluids the inter-molecular forces will play a larger role and cannot be neglected like this.
| {
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Was everything in the Universe "created" from light? Is the following true?
The only matter existing directly after the big bang was electromagnetic radiation.
| I could start by saying that this depends on what your definition of the big bang is, but that would be pointless because the better question is how do you define matter? Or "directly after"?
Firstly, let me say the technical answer is no because electromagnetic radiation is not matter (in a cosmological context). But I assume that the quote meant the only type of energy. If that's the case, the answer is still no because dark energy existed at that time too. It's also possible the quote means to just say that there was no baryonic matter, which is true. If the quote meant there was no particles of any kind save photons, then that is both silly and false. Directly after the big bang, there were also gravitons and dark matter and other force mediating particles and some fundamental particles. At the time of the big bang (again depending on your definition of it), the energy scale was too high to be described by modern physics accurately, so it's tough to say. Some models have a universe where photons didn't even exist until a fraction of a second into the hot big bang era.
But the answer to what this question is most likely asking is no.
| {
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Maxwell's Equations, cast in terms of magnetic vector potential Derive $$ \nabla \times \frac{1}{\mu_r} \nabla \times A + \mu_0 \sigma \frac{\partial A}{\partial t} + \mu_0 \frac{\partial}{\partial t} \left( \epsilon_0 \epsilon_r \frac{\partial A}{\partial t} - D_r \right) = 0$$ where $D_r$ is a "remnant displacement field" similar to the material polarization, $$ D = \epsilon_0 \epsilon_r E + D_r $$
A numerical package (COMSOL Multiphysics 4.3b) solves maxwell's equations in the time domain in its transient electromagnetic wave module using this equation. The "remnant displacement field" is used to model the nonlinear polarization of a material.
My attempt...
$$ \nabla \times H = J_f + \frac{\partial D}{\partial t} $$
$$ \nabla \times \frac{1}{\mu_0 \mu_r} \nabla \times A = J_f + \epsilon_0 \epsilon_r \frac{\partial E}{\partial t} + \frac{\partial D_r}{\partial t} $$
$$ \nabla \times \frac{1}{\mu_r} \nabla \times A = \mu_0 J_f + \mu_0 \frac{\partial D_r}{\partial t} + \mu_0 \epsilon_0 \epsilon_r \frac{\partial}{\partial t}\left( -\nabla V - \frac{\partial A}{\partial t} \right) $$
$$ \nabla \times \frac{1}{\mu_r} \nabla \times A + \mu_0 \frac{\partial}{\partial t} \left( \epsilon_0 \epsilon_r \frac{\partial A}{\partial t} - D_r \right) = \mu_0 \left( J_f - \mu_0 \epsilon_0 \epsilon_r \frac{\partial}{\partial t} \nabla V \right) $$
$$ J_f = \sigma E = \sigma \left(-\nabla V - \frac{\partial A}{\partial t} \right) $$
Substitute $J_f$...
$$ \nabla \times \frac{1}{\mu_r} \nabla \times A + \mu_0 \sigma \frac{\partial A}{\partial t} + \mu_0 \frac{\partial}{\partial t} \left( \epsilon_0 \epsilon_r \frac{\partial A}{\partial t} - D_r \right) = \mu_0 \left( -\sigma \nabla V - \mu_0 \epsilon_0 \epsilon_r \frac{\partial}{\partial t} \nabla V \right) $$
Any ideas on why the right side goes to zero?
| Gauge transformation.
Let $\lambda$ be defined such that,
$$ \frac{\partial}{\partial t} \nabla \lambda = \nabla V $$
then,
$$ V^\prime = V - \frac{\partial \lambda}{\partial t} $$
$$ \nabla V^\prime = \nabla V - \frac{\partial}{\partial t} \nabla \lambda = \nabla V - \nabla V = 0 $$
You must also transform the magnetic vector potential.
$$ A^\prime = A + \nabla \lambda $$
However, as can be read in this link (page 577), electrodynamic problems in this software implicitly use this gauge. The $A$ used in the question's equation is untransformed; however, it is transformed in the software when implementing the routine.
| {
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Why does the diffusion pole universally appear in the two-particle Greens function (diffuson) I've been thinking about the calculation of the diffuson in the context of impurity-averaged Greens functions.
If you calculate the two-particle Greens function in the ladder approximation (for example, in Rammers' "Quantum Transport Theory, p.337" or any similar textbook) you obtain the famous Diffusion propagator
$$
\frac{1}{1- \zeta(q, \omega)} = \frac{1}{i \omega - D q^2}
$$
Here the ladder insertion $\zeta(q, \omega)$ should be calculated using the integral of advanced and retarded Greens functions, and the answer should seem to be dependent on the dimensionality, the density of states and so on. Yet we universally have $\zeta(q\rightarrow 0, \omega\rightarrow 0) = 1$ exactly. This is clearly not accidental. Is there some simple way to prove that $\zeta\rightarrow 1$ exactly in the diagrammatic technique, independent of dimension, dispersion and so on?
| This can be derived using diagrammagic technique (see eg Altland and Simons' book). The more fundamental reason is that the diffuson is a Goldstone mode, which has to be massless (also see Altland and Simons).
But to my knowledge, the direct answer to your question is "no" -- there is no "simple" way because the math underlying the above is quite involved. This is unfortunate.
| {
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Is it true that a processor with a higher temperature uses inevitably more power in comparison with a processor with a lower temperature? If a processor has a higher temperature, can you be sure it it consumes more power than a processor with a lower temperature?
And is the reverse relationship true?
A processorchip with a higher mean power consumption will always work at a higher temperature than a processer with a lower mean power consumption?
| No. Power and temperature are related, but not directly. Thermal environment and surface area also matter a great deal.
A tiny chip has less area to dissipate heat than a larger one. So for the same power draw, it may easily produce a higher temperature than a larger chip.
Imagine a 40W incandescent light bulb. Both the filament and the bulb as a whole are dissipating 40W of power. But the tiny filament can do so only by climbing to a very high temperature of a few thousand $K$. The larger bulb surface dissipates the same amount at a few hundred $K$.
Design affect things as well. If you have a chip that happens to be quite robust at high temperature, you might allow it to reach high temperatures in your product. But a more powerful one might need aggressive cooling, which keeps it at a lower temperature.
| {
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Is differential geometry used in solid state? I'm an undergraduate in physics interested in a career in solid state. While I know that any additional math is helpful--I am on time constraints, and can only take a few supplemental classes.
That said, is differential geometry used much in solid state physics? I'm aware of things like Fermi Surfaces, but wonder if much diff. geo. techniques are actually used.
| Yes, especially in research-level topics. There are several research groups that work with finding ways to apply differential geometry concepts to solid state systems (although condensed matter seems to be the preferred term nowadays). See for example the book by Altland and Simons, Condensed Matter Field Theory, Chapter 9 "Topology". This book is suitable for a masters degree level class with a solid state class and a quantum field theory as prerequisites. The specified chapter has a (very!) brief primer on differential geometry, but I would probably have found it unintelligible if I had not had a proper differential geometry class first. The chapter will give you the very basics of what kind of research people are doing.
Some keywords to start with are topological (insulators|superconductors|order|field theories), quantum hall effect, composite fermions. Maybe these two talks (not by me) can be a good introduction: What is topological matter & why do we care? I, II.
| {
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Centripetal force in this example? I have a general question regarding the centripetal force. In the example of a ferris wheel where there is a normal force pushing up against the person and the gravitational force pulling the person down, which force is centripetal? I know that the centripetal force counters the linear velocity, tangent to the circle of motion, which allows the object or person to stay in circular motion but which force is actually pulling it towards the center, the gravitational force or the normal force? Also, would it be correct to say that the net force equals zero (since the person is neither moving towards or away from the center) in this example or does the net force equal the centripetal force (since the centripetal force has to counter the linear velocity --- if this is correct, how would I compare the two since linear velocity is not a force)?
I know that if a car is moving around a banked curve, the horizontal normal force will be centripetal but what about in other examples such as the ferris wheel? Also would the net force of a car moving around a bank curved be zero since it is neither moving towards or away the center?
tl;dr - is the net force in a centripetal force example zero or is the net force equal to the centripetal force? Also, how would I relate this to the linear velocity that cancels it out?
Thanks for the help!
| The centripetal force is not a physical force but rather the component of the force which points towards the center during circular motion. For the example of the Ferris wheel, the centripetal force depends on the position. For instance, if the the person is on the top of the ferris wheel, the gravitational and the normal force combined is the centripetal force, but if the person is in the bottom of the ferris wheel, the normal force minus the gravitational force is the centripetal force.
The net force of a car moving around a bank curved is not zero, rather, because the net force is always pointing perpendicular to the velocity, the motion is circular and hence the car never moves toward or away from the center.
| {
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If an object falls - regarding air resistance - does the Potential or Kinetic energy get converted into thermal energy as it is falling? I read a paragraph on the transfer of potential energy to kinetic energy and heat from this website:
Even if air resistance slows down the ball, the potential energy is
the same (Mb x g x H). But if air resistance is in the way, not all of
the potential energy can be converted to kinetic energy. Some of the
energy has to be used to push the air molecules out of the way. When
that happens, the energy of the air molecules is increased. The air is
actually "heated" up by the falling ball.
This text indirectly mentions that the Potential Energy gets converted into heat. However my common sense (for lack of a better term) makes me think that the Potential Energy gets converted into Kinetic Energy which in turn gets converted into Thermal energy. Would someone be able to enlighten me on this please?
Also, as a side question, it isn't called Heat Energy right? A lot of websites seem to be saying that, but heat is just the transfer of Thermal Energy...
| Heat energy and thermal energy are pretty much the same thing. My science teacher taught me, that because potential energy is related to height, it would be the kinetic energy that is converted into thermal energy.
| {
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What position of the center of gravity can make the front wheels of the car lift off the ground? I have a question regarding the position of center of gravity required to just lift off the front wheel of a vehicle
Consider a vehicle of mass $m$ having a center of gravity at height $h = 0.5m$ from the ground. The coefficient of friction between the tire and the ground is 1. Assume that the engine supplies just enough torque to utilize all the friction force without causing the wheels to spin.
My question is where should the center of gravity of the vehicle should be located in relation to the rear wheels to make the wheels just lift off the ground. I have solved the question as shown, and I get a 'NEGATIVE' value for L2 meaning that COG should be 0.5m behind the rear wheel, but the solution says that it should be 0.5 m infront of the rear wheels.
Can someone help me out on this!
Since, my handwriting is not clear, I am writing the equations here too
Equilibrium in vertical direction
(1) $N_1 + N_2 = mg$
Equation of motion in horizontal direction
(2) $F_{tr} = ma$
Here $F_{tr}$ is the traction force on the rear tire which propels the vehicle. Also,
(3) $F_{tr} =\mu N_2$
The balance of the torque on the rear tire $ \to $ the net moment on the rear tire about the contact point is zero
(4) $mgl_2 + 0.5 \ ma = 0$.
Now, since the vehicle must just lift off the ground
(5) $N_1 = 0; \ N_2 = mg$.
Using the equations (2), (3), and (5),
(6) $a = \mu g$.
Now using the equation (6) in the equation (4),
(7) $l_2 = -0.5 \ \mu = -0.5$
Now, the final value of $l_2$ is negative, which means that it is opposite to the assumed direction. So the center of mass should be 0.5 m behind the rear wheels.
The only difference between my method and the solution manual which I am referring to is that they have considered inertia force(pseudo force) on the vehicle, and thus, they get the answers $l_2 = 0.5$, which means 0.5 m in-front of the rear wheels
| I think I understood the reason. I am incorrect in taking the torque of ma directly. This is because ma is the result of the forces acting on the system. Instead I should have balanced the torque about the center of mass, with the equation [4] looking like this
$$mg∗l2−Ftr∗0.5=0$$ [Torque balance about center of mass, assuming that there is no pitching moment]
Now I can replace
$$Ftr=m∗a$$
which on further solving will give
$$l2=.5$$
| {
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How much additional light does Earth receive from the Sun due to Earth's gravitational field? I was reading about how gravity affects light, and that got me wondering how much additional light is collected by the Sun due to the Earth's gravitational field.
Is it a significant amount of light (>1% of total light)? Is it significant enough to be considered when estimating the surface temperature of a planet?
| Assuming you mean "collected by the Earth"...
These effects are barely measureable when observing the Sun bending light towards itself. Earth bends light even less. The value will be much less than 1%.
The light is red-shifted as it leaves the Sun (effectively becoming less energetic)
It is blue-shifted to a lesser degree as it falls to Earth (becoming a little more energetic)
Of much greater import for changing Earth's temperature are the following for solar radiation:
The total energy emitted by the Sun varies
Sunspots raise Earth's temperature, possibly via cosmic ray formation of high altitude clouds
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/159103",
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Resultant frequency if 3 harmonic notes (a chord) is played If I know the frequency of individual notes being played (let's assume D, F# and A), how do I determine the final frequency if they are played (nearly) simultaneously as a chord.
To put the problem in context, I am writing a program where user inputs are classified and chords are played as output. My programming knowledge is limited (my degrees are in biochemistry and neuroscience), and the help file gave me the following options:
*
*Play a frequency
*Use a recording for playback
As I do not wish to use the second option, I was wondering how to calculate the resulting frequency or to represent resultant sound mathematically. If someone could explain the theory behind this or post a link for the same, I'd be especially grateful.
| Here's a minimum working example of a python program which generates a .wav file with a major triad of 440:550:660 Hz using sine waves. Your user input could be used to generate any frequencies for the chord.
import math, wave, array
duration = 1 # seconds
freq1 = 440 # tonic (Hz) (frequency of the sine waves)
freq2 = 550 # 2nd note
freq3 = 660 # 3nd note
amp1 = 0.2 # amplitude of freq1; sum of amplitudes should be <= 1
amp2 = 0.2 # amplitude of freq2
amp3 = 0.2 # amplitude of freq 3
volume = 100 # percent
data = array.array('h') # signed short integer (-32768 to 32767) data
sampleRate = 44100 # of samples per second (standard)
numChan = 1 # of channels (1: mono, 2: stereo)
dataSize = 2 # 2 bytes because of using signed short integers => bit depth = 16
numSamplesPerCyc = int(sampleRate / freq1)
numSamples = sampleRate * duration
factor=2*math.pi/sampleRate
for i in range(numSamples):
sample = 32767 * float(volume) / 100
sample *= amp1*math.sin(factor*freq1* i)+ amp2*math.sin(factor*freq2*i)+amp3*math.sin(factor*freq3*i)
data.append(int(sample))
f = wave.open('SineWave_' + str(freq1) + 'Hz.wav', 'w')
f.setparams((numChan, dataSize, sampleRate, numSamples, "NONE", "Uncompressed"))
f.writeframes(data.tostring())
f.close()
print "Done"
| {
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How to visualize the gradient as a one-form? I am reading Sean Carrol's book on General Relativity, and I just finished reading the proof that the gradient is a covariant vector or a one-form, but I am having a difficult time visualizing this. I usually visualize gradients as vector fields while I visualize one-forms with level sets. How to visualize the gradient as a one-form?
| In the context of general relativity it is also notable that the manifold is equipped with a metric tensor. This tensor provides a unique way (an isomorphism, s.a. answer by Phoenix87) to map covectors to vectors. The components are computed by index raising/lowering, e.g. for a given covector $w$ with components $w_i$ the corresponding vector reads $v^i=g^{ij}w_j$, where $g^{ij}$ are the components of the metric tensor and I use the summation convention over repeated indices.
Now the question is how this really looks like in the given case. The natural derivative of a function $f$ on a manifold $\mathcal{M}$ is $df$, with components $df_i = \partial_i f$, since this is possible without metric. As already mentioned, $df$ is visualized by level sets of $f$. Now the corresponding contravariant vector is computed by $X^i := g^{ij}\partial_i f\equiv (\nabla f)^i$, which we identify with the gradient of $f$. This vector is the vector that is everywhere perpendicular to the level sets of $f$. Note that the notion "perpendicular" can only be defined with a metric.
Many textbooks in physics also visualize the basis vector $dx^i$ with its gradient vector, i.e. the vector that is everywhere perpendicular to the levels sets of $x^i$. This is strictly speaking only possible when there is a metric present, but it simplifies the comparsion to $\frac{\partial}{\partial x^i}$ which is in general not the same direction.
| {
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Mass in special relativity? Is the mass of a object at rest defined by $$E=mc^2$$
where $m$ is the rest mass. I.e. does the rest mass include every thing from thermal to gravitational potential energy and every other possible energy that it could have at rest. And thus if we write the following:
$$total\ energy=mc^2+potential\ energy+thermal\ energy $$
are we double counting the potential energy and the thermal energy?
| The mass term includes all internal "energies". Heating up a body increases the internal kinetic energy. Binding energies also contribute to the mass (when nuclear fission occur, energy is freed and the products of the reaction are lighter than the original element), and this include any bond due to the fundamental forces of nature (which include, e.g., gravitational interaction, but only for parts within the body). Any extra energy coming from the interaction of the body as a whole with an external field doesn't contribute to the rest mass, which is a relativistic invariant.
| {
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Modeling the motion of a bouncing ball I'm writing a program that displays a line of text, and animates a ball that bounces from syllable to syllable (like a sing-along). The program knows the location of each syllable, and it knows at what time the ball should be at each syllable.
I have a set of equations that work OK, but not great. I came up with them a few years ago after much googling and stumbling about. They take the location of the previous syllable ($x_0$, $y_0$), the location of the next syllable ($x_1$, $y_1$), the time t (from start 0 to finish 1), and compute where the ball should be ($x$, $y$):
$$d = x_1 - x_0$$
$$v = d/t$$
$$ h = 5 + 0.3 |d|$$
\begin{align}
x(t) &= x_0 + v t\\
y(t) &= y_0 - h + \left[4 \frac{h}{d^2} \left(\frac{|d|}{2}- |v| t \right)^2 \right]
\end{align}
What I would like is a better set of equations that more accurately model the motion of a bouncing ball. A ball with a mind of it's own, I suppose, as it does need to change speed and direction with each new syllable.
| You need the position of the ball $(x(t)$, $y(t))$ for $0<t<1$ if at $t=0$, the ball was thrown with initial velocity $(v_x,v_y)$ at the position $(x_0,y_0)$ in a gravitational field of acceleration $\overrightarrow{a}=(a_x,a_y)=(0,-g)$. The velocity has to be calibrated in order to make the ball arrive the point ($x_1,y_1$) at $t=1$.
The position of the ball is given by
$$x(t)=x_0 + v_x t$$
$$y(t)=y_0 + v_y t - g\frac{t^2}{2}$$
We want to obtain $(v_x,v_y)$ to get $x(t=1)=x_1$ and $y(t=1) = y_1$, so,
$$x(t=1)=x_0 + v_x = x_1$$
$$y(t=1)=y_0 + v_y - \frac{g}{2} = y_1$$
and therefore,
$$ v_x = x_1 - x_0$$
$$ v_y = y_1 - y_0 + \frac{g}{2}$$
and finally, your movement equations are:
$$x(t)=x_0 + (x_1 - x_0) t$$
$$y(t)=y_0 + (y_1 - y_0 + \frac{g}{2}) t - g\frac{t^2}{2}$$
| {
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What would be the most general effective Lagrangian involving one Higgs and two gluons? Two different possibilities come into my mind
$\mathcal{L}\sim{}HG_{\mu}G^{\mu}$
where $G^{\mu}$ is the gluon field and $H$ the Higgs, or either
$\mathcal{L}\sim{}HG_{\mu\nu}G^{\mu\nu}$
Where $G_{\mu\nu}=\partial_{\mu}G_{\nu}-\partial_{\nu}G_{\mu}$.
I can't think of any argument to decide, so, which of this is best and why? or is there a better choice?
| First of all, since we are talking about an effective theory, there are infinitely many terms. Since the theory does not have to be renormalizable, we can include all operators with $D>4$. So I think you want the easiest one, namely only one additional operator giving you the $Hgg$-vertex.
This would, as you say correctly, be something like
$\delta \mathcal{L}\sim \frac{g}{\Lambda}H G_{\mu\nu}^a G^{\mu\nu\ a} + h.c.$
BTW: $G_{\mu\nu}^a=\partial_\mu A_\nu^a - \partial_\nu A_\mu^a+ g_s f^{abc}A_\mu^b A_\nu^c$. Gluons interact with each other.
Within the Standard Model, such an interaction could be realized through a, say, top-quark loop. In an effective theory, you would then let $m_{top}\rightarrow \infty$ to get the effective vertex. This can be done, when $m_{top}\gg \mu$, where $\mu$ is the mass of some other particle. Now in our case $\frac{m_{Higgs}}{m_{top}}\approx 0.72$. Though, it is $<1$, I don't know if $m_{top}\gg m_{Higgs}$ is an assumtion you can make.
| {
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Can a conservative field produce a torque? I am asking whether the following Lagrangian for a point moving in a conservative field, can be correct :
$L(r, v, \omega) = \frac {mv^2}{2} + \frac {I \omega^2}{2} - U(r)$.
$r$ is the distance between the equipotential surface on which the movement begins and the equipotential surface on which the movement ends, $v = \text d r/ \text d t$, $\omega$ is angular velocity of rotation around some fix point in space (see example on the bottom of the text), $I = m\rho^2$, where $\vec \rho$ is the vector connecting the fix point in the space with the current position of the moving point (see the example).
What I am not sure on, is the presence of the term $I \omega^2/2$. I think that $\omega $ can vary only if the potential energy can produce a torque ($\vec F \ \text x \ \vec \rho$), and in that case $U$ should also depend on a variable $\theta$, indicating the angle between the vector $\vec \rho$ and a fix axis in the rotation plane.
But, if there is a torque, if $U$ depends not only on the distance between equipotential surfaces, but also on an angle $\theta$, is this anymore a conservative field?
I know that in a conservative field the mechanical work doesn't depend on the path followed by the point, but on the distance between equipotential surfaces, however that doesn't help me in my question.
(As a simple example, one can think that the field is produced by an electric charge uniformly distributed on an ellipsoid. Then $d$ is the distance to the surface of the ellipsoid measured perpendicularly on the equipotential surfaces, and given a point $P$ in the field, $\vec \rho$ is the vector from the center of the ellipsoid to the pint $P$.)
| I'm not sure about the notation, so there will be a bit of guessing here. I assume $v = \dot r$, so the kinetic term can be interpreted to be that of a point moving on a plane, described by polar coordinates. Now take any radial potential, which by the rotational symmetry generates a central field. Let us consider Kepler's problem to be definite here. It is well known that closed orbits are elliptic in general, hence $\omega$ is not constant throughout the motion, but $\mathbf F\times\mathbf r = 0$ at every time.
| {
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Advection operator How are exactly $u_j\partial_ju_i$ and $u_i\partial_j u_i$ related?
And what is their relation to ($\boldsymbol{u}\cdot\nabla)\boldsymbol{u}$ and $\boldsymbol{u}\cdot(\nabla\boldsymbol{u})$ ?
I ask this because:
$$[\mathbf{u}\cdot(\nabla\mathbf{u})]_{i}=u_{j}\partial_{i}u_{j}=u_{x}\partial_{i}u_{x}+u_{y}\partial_{i}u_{y}$$
$$[(\mathbf{u}\cdot\nabla)\mathbf{u}]_i=u_{j}\partial_{j}u_{i}=u_{x}\partial_{x}u_{i}+u_{y}\partial_{y}u_{i}$$
from this it would seem they are different, but:
$$[(\mathbf{u}\cdot\nabla)\mathbf{u}]=(u_{x}\partial_{x}+u_{y}\partial_{y})\left(\begin{array}{c}
u_{x}\\
u_{y}
\end{array}\right)$$
$$[\mathbf{u}\cdot(\nabla\mathbf{u})]=\left(\begin{array}{c}
u_{x}\\
u_{y}
\end{array}\right)\left(\begin{array}{cc}
\partial_{x}u_{x} & \partial_{x}u_{y}\\
\partial_{y}u_{x} & \partial_{y}u_{y}
\end{array}\right)=\left(\begin{array}{cc}
\partial_{x}u_{x} & \partial_{y}u_{x}\\
\partial_{x}u_{y} & \partial_{y}u_{y}
\end{array}\right)\left(\begin{array}{c}
u_{x}\\
u_{y}
\end{array}\right)=\left(\begin{array}{c}
u_{x}\partial_{x}u_{x}+u_{y}\partial_{y}u_{x}\\
u_{x}\partial_{x}u_{y}+u_{y}\partial_{y}u_{y}
\end{array}\right)
$$
from this it would seem that they are the same. I am quite suspicious about my definition of $\nabla\boldsymbol{u}$. Could someone clarify this?
| The problem is in the way you wrote your last equation as a matrix multiplication.
You have
$$ [ \textbf u \cdot ( \nabla \textbf u) ]_i = u_j (\partial_i u_j) = (\partial_i u_j) u_j, $$
so if you want to write this in matrix form you have to multiply the vector $\textbf u$ at the right, as a column vector, i.e.
$$[\mathbf{u}\cdot(\nabla\mathbf{u})]=\left(\begin{array}{cc}
\partial_{x}u_{x} & \partial_{x}u_{y}\\
\partial_{y}u_{x} & \partial_{y}u_{y}
\end{array}\right)
\left(\begin{array}{c}
u_{x}\\
u_{y}
\end{array}\right),
$$
where the derivatives are intended to act only on the adjacent $u_i$.
Anyway, I don't see where the covariant derivatives comes in here. The $\nabla$ you are using is simply a gradient, not a covariant derivative.
| {
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How does a supersonic flight speedometer work? I'm sure today they can use GPS and radar, but I was pondering the queation when I saw a film clip of a vintage analog dial labeled in mach number. I'm supposing that the usual way of measuring the pressure drop of the air flow would not work in this case. So what does?
| If we consider a pitot-static probe in supersonic flow we get something that looks like (source):
The probe measures the stagnation pressure in the part of the probe normal to the flow and the static pressure in the part of the probe perpendicular to the flow. There is a small bow-shock around the tip of the probe. But we can go ahead and assume that the shock is normal directly in front of the inlet to the probe. This allows us to use the normal shock relations after some manipulations to get:
$$\frac{P_{stag}}{P_{static}} = \frac{\gamma+1}{2}M^2\left(\frac{\left(\gamma+1\right)^2M^2}{4\gamma M^2-2\left(\gamma-1\right)}\right)^\left(1/(\gamma-1)\right) $$
where $\gamma = 1.4$ unless you are flying at hypersonic speeds. This equation is non-linear and requires a solver, but once the two measurements are known from the probe you can determine the Mach number.
| {
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Reversible and Quasi-static processes Do we have any proof that reversible processes are always quasi-static or is it just a fact that hasn't been violated till date? If there is a proof then please provide a link.
| This comes from just definations.
In thermodynamics, a quasi-static process is a thermodynamic process that happens infinitely slowly.
Reversible process:Any process which can be made to proceed in the reverse direction by variation in its conditions such that any change occurring in any part of direct process is exactly reversed in the corresponding part of reverse process is called reversible process.
Conditions for Reversibility
:
1.The substance undergoing a reversible change must at all instances be in thermodynamic equilibrium with its surroundings. It means the pressure and temperatures of the working substance must never differ appreciably from its surroundings at any stage of the cycle of operation.
2.All the processes taking place in the cycle of operation must be infinitely slow.
3.There should be complete absence of frictional forces.
4.There should not be any loss of energy due to conduction, convection or radiation during the cycle of operation.
Now why a reversible process need to be quasi-static?
Ans:The process must be carried out infintesimally slowly so that the system remains in the thermal and mechanical equilibrium with sorroundings throughout.
further information:
http://www.gitam.edu/eresource/Engg_Phys/semester_1/THERMODYNAMICS/rev_and_irrev.htm
| {
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Why is an airplane able to increase thrust without moving? I was just watching some documentaries and saw planes building up power in the turbines without moving. I thought about it and remembered, that this happens before every take off.
So, why is this possible? A planes thrust isn't related to the ground, but to the air, so brakes would just increase friction but won't reliably prevent the plane from starting.
So, how is the thrust compensated? I just don't get the clue.
| In fact, for a short field take off, the pilot's operating handbook instructs you to do exactly that. The brakes are designed for that.
| {
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How does "pushing-start" a dead-battery manual car work? A few days ago the battery of my car went (almost) dead. As it is a manual car, my father once told me that the way to get it going without jumper cables was to push it or let it roll down a hill, sink the clutch, shift to 2nd gear and then let go the clutch.
After the joy of being able to 'revive' the car, I got to wonder of the reason for why this works. All I could think of was Electromagnetic Induction, however I couldn't find anything on the web to support this. I'm sure that there might be all sorts of engineering details, so I'm only looking for the physical principles involved and the basic process that make this work.
| In the days before Electric Rear Window Defrosters, cars still had generators, you could push start the car with a completely dead battery. Generators make electricity, whereas today's cars are equiped with alternators.
Alternators take electricity to make more electricity. They have a much higher output for today's cars but are useless completely flat. Perhaps if the battery has voltage but not enough to turn over the starter you could push start the car...
My first car had a hand crank and a generator. I do not miss generators, carburators, or points.
| {
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The velocity of the flowing coming out of the balloons? Last day , when i was working on two interconnected balloons , a question was kicking my brains !!! This is the explanation of the question:
First , suppose a system that composed of two spherical membranes filled with air (two balloons have different initial volumes {means that the pressure inside which balloons are different} and the air pressure is 1 a.t.m) . We connect them with hollow tube and a valve.
When we open the valve , one of them shrinks and the other one expands (It depends on their pressure) .
So how can we get the velocity of the flowing air between two balloons? (Consider everything but if you have reasons for not considering one of them -for example the ratio of friction in the tube- Don't consider them and just tell me the reason)
Thanks
| The air flow rate through a tube is approximately given by the Hagen-Poiseuille equation. If the pressure difference between the two ballons is $\Delta P$ then the HP equation gives the volumetric flow rate $Q$ as:
$$ Q = \frac{\pi r^2}{8\mu \ell} \Delta P $$
where $r$ is the radius of the pipe, $\ell$ is the pipe length and $\mu$ is the viscosity of the air. The equation actually only applies to incompressible fluids, but gives a reasonable answer even for compressible fluids like gases a long as the pressure differences aren't too high. In the case of two balloons it should be fine.
| {
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Which modes are excited when a drum is struck? I've been searching quite extensively for an answer to this question but I cannot find anything definitive. The most I can see is that "one or several" modes become excited, but this is never parameterized by any relevant factors. I imagine it depends on the location of the strike and the force applied. I do not need an exact formula, but I would appreciate some intuition here.
I have seen several videos now of the individual modes of a circular membrane being excited by oscillators at targeted frequencies. But when the drum is actually struck, does the membrane vibrate in some superposition of these modes? Are all possible modes - all infinitely many - excited simultaneously, just with all higher modes having infinitesimal amplitude? Or is there some finite fixed limit of the number of excited modes depending on say the force of the impact? (Further, I imagine if you strike it hard enough, the material will break, so if this is the case then some higher modes would maybe never become excited, but perhaps we can gloss over such details.)
| The question
mixes membranes and drums.
Hitting a membrane is different from a drum, because drums
have (more or less, depending on type of drum) a strongly
coupled air volume controlling the vibration of the drum.
A turkish kettle drum will have almost no higher modes,
whereas those membranes as used by Inuit people will
show Bessel modes.
| {
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Does the escape velocity of a black hole exceed $c$ *before* a singularity is created? As an offshoot of the question Can we have a black hole without a singularity? I'm curious if the point of no return at which the massive object is condemned to become a singularity happens before its escape velocity is greater than $c$?
I envision the black hole creation process as a positive feedback loop of collapse and the escape velocity quickly shoots past $c$ on the way to becoming a singularity, but not necessarily at the very instant it becomes a singularity. This would mean that yes, a black hole (or perhaps better termed a "dark star") can exist for a brief moment in time (on the order of plank time?) without a singularity but positive feedback ensures a singularity will be created. Any truth to this statement as we know it?
| To add to John's very good answer, I'll emphasise a point that makes identifying the time when a black hole forms slightly strange: it depends on knowing the entire future evolution.
The black hole interior consists of the points from which you can't escape to infinity. But to be sure that you can escape requires knowing that nothing dramatic in the future will stop you.
For example, we could be sitting in a black hole right now: some aliens may be conspiring to collapse a huge shell of matter onto us, so great that even light we are sending out now will be pulled back in when the matter arrives. But there is no singularity, or even strong enough curvature to make Newtonian gravity invalid!
This is one manifestation of the idea that (at least in classical gravity) there is nothing special about what goes on at the event horizon.
| {
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If the solar system is a non-inertial frame, why can Newton's Laws predict motion? Since there is no object in the universe that doesn't move, and the solar system likely accelerates through space, how did Newton's Laws work so well? Didn't he assume that the sun is the acceleration-less center of the universe? Shouldn't there be many psuedo-forces to account for planetary motion?
| Newton's laws work well but if one considers the relativity theory, one finds things not explained by Newton's laws. A well-known example is the "anomalous" precession of the perihelion of Mercury, explained by the general relativity.
| {
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Textbook recommendation for computational physics Can anyone recommend some good textbooks for undergraduate level computational physics course? I am using numerical recipe but find it not a very good textbook.
| I like Bill Gibbs' book Computation In Modern Physics for a couple of reasons (aside from having taken the course from the author):
*
*After introducing basic tools (difference approximations to differential equations, numeric quadratures (i.e. integrals), and eigenvalue problems in a matrix form) it moves right on to problems of interest to me. The examples are all taken from nuclear and particle physics, so there neutron transport models and toy tomography problems and multi-electron atoms to solve and so on. Good stuff.
*It is at once a small book, and provides deep coverage of the problems that it spends time on.
*The sections on Monte Carlo methods are very good.
That said, it suffers from two issues that prospective users might want to know about:
*
*Bill is a Fortran 77 guy.
All the code in the book is in that language, and he uses some deep magic array slicing tricks that (1) will take considerable study for students to comprehend and (2) may set a bad example in the modern era where programming clarity is to be preferred over "optimization" until you've proven that the compiler can't optimize it enough.
*The level might be a little steep.
The course was taught to seniors and grad-students, but I would guess that it was conceived of as a graduate course. So the move from basic principles to problems is pretty demanding and the problems assume that the students know a fair amount of modern physics.
| {
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$v^2 = 2ax$ or $v^2 = ax$? As far as I am aware, $v^2 = 2ax$ is the formula to find the velocity in various questions.
If kinetic energy = work,
$$\frac{1}{2}mv^2=Fx$$
$$mv^2=2max$$
$$v^2=2ax$$
We use this formula to solve some questions in school.
But when i just fiddle around with basic formulas i get this.
$$x/v = t$$
$$v/a = t$$
$$t = t$$
$$x/v = v/a$$
$$v^2 = ax$$
And this just confuses me. I assume that $x/v=t$ and $v/a=t$ is actually kind of simplified or else I cannot see why $v^2$ equals $ax$ on one and $2ax$ on the other.
Can someone explain to me what am I doing wrong?
| In your second derivation, the correct formulas are
$$\begin{align}
\frac{\Delta x}{v} &\approx \Delta t &
\frac{\Delta v}{a} &\approx \Delta t
\end{align}$$
I'm sure you can easily find some examples to show you why $x/v = t$ and $v/a = t$ don't make any sense. Anyway, when you put these together, you get $v\Delta v \approx a\Delta x$, with the approximation becoming more accurate the smaller the $\Delta$s are.
Note that if you take the limit as $\Delta v$ and $\Delta x$ go to zero, then integrate, you get
$$\begin{align}
\int_{v_i}^{v_f} v\,\mathrm{d}v &= \int_{x_i}^{x_f} a\,\mathrm{d}x \\
\frac{v_f^2 - v_i^2}{2} &= a(x_f - x_i) \\
v_f^2 &= v_i^2 + 2a\Delta x
\end{align}$$
which is exactly the correct formula.
| {
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Mullins effect in the elastic material I was thinking on a experimental question, but I couldn't get the answer:
How can we measure how the Mullins effect influences in the pressure inside a balloon? (I mean that, if there is a formula, tell me!)
| A good way to measure pressure inside a balloon is to create a setup with a balloon, a manometer (long thin liquid filled U shaped tube), and a three way valve. You increase the pressure in the balloon by pumping it up, then measure the pressure with the manometer. Keep increasing the pressure and measure the diameter of the balloon as you go. Let the pressure out, and repeat. You will find that the pressure-diameter curve keeps shifting, just as it does in the curves in the article you linked.
Key to doing this experiment properly is having a good technique for measuring the volume. I would recommend placing the balloon in a darkened room with a small bright source of light, casting a shadow onto a piece of graph paper. Or cast the shadow onto a piece of tracing paper and take a digital picture of the shadow. Or just take a picture of the balloon against a differently-colored background. The volume will scale approximately with the area to the power $\frac32$, but if you plot area itself as a measure of "strain", it would show the effect. In fact, since the area of the balloon surface scales with the area of the projected volume (for an object with reasonable symmetry), just measuring the projected area and not converting to volume will actually be better...
There is one other thing to keep in mind - and that is the analysis given in this earlier answer to this question which relates to the fact that a balloon becomes thinner during inflation, and that this will affect the pressure as a function of diameter. But by repeated cycling and seeing that the curves shift you should be able to differentiate between Mullins effect and this other phenomenon.
| {
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Deutsch's Algorithm. Unitary Transform $U_f$ I'm studying Deutsch's algorithm and I keep coming across the phrase along the lines of "There is a unitary transform (a sequence of quantum gates) $U_f$ that transforms the state $|x\rangle |y\rangle \rightarrow |x\rangle |y \oplus f(x)\rangle$".
I was trying to figure out how this $U_f$ would be implemented as a sequence of quantum gates.
I originally thought that there would be some sort of transform that takes $|x\rangle \rightarrow |f(x)\rangle$ and then apply the CNOT transformation to obtain the result. However, I believe this way of thinking is incorrect and I wouldn't obtain the state desired.
So how is the transform $U_f$ realised or does it depend upon the function $f$?
| Your $U_f$ must depend on $f$. Let's consider the two trivial examples:
*
*$f$ is the zero-function. In this case, $U_f$ is just the identity.
*$f$ is the one-function ($x\mapsto 1$), then $|x\rangle|y\rangle \mapsto |x\rangle|y\oplus 1\rangle$, then $U_f$ is a NOT-gate on the second qubit.
Just a note: The whole idea of the Deutsch-Josza algorithm is that you don't need to worry about how to implement $U_f$ - it is given.
| {
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Metric tensor in SRT I just read on this webpage that we have (click me) $g_{\alpha \beta} = g_{\alpha}^{\beta} = g^{\alpha \beta}.$
Now, although I understand that the first and the last one are equal, I don't think that the term in the middle is the same as the other two, cause we should have $(g_{\alpha} ^{\beta}) = (g_{\alpha \alpha'})(g^{\alpha' \beta})$. This should be equal to the identity matrix.
What am I doing wrong?
| First a word on notation. In special relativity, the Minkowski metric is $\eta_{\alpha\beta}$. The general relativity curved metric is $g_{\mu\nu}$. A lot of texts that only use the Minkowski metric don't make that distinction for some reason. However, when you get to string theory and there are four different metrics floating around, it is important to keep things straight by calling $\eta_{\alpha\beta}$ the flat spacetime metric. This is a huge pet peeve of mine.
Second pet peeve: The statement that the co- and contravariant tensors are the same is rubbish. The components are the same. It doesn't even make sense to say that two tensors belonging to different tensor algebrae are equal.
Now to your question. That is an error on the author's part. You are completely correct. As you have shown,
$$\eta^\alpha_\beta=\delta^\alpha_\beta$$
| {
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Is $∣1 \rangle$ an abuse of notation? In introductory quantum mechanics it is always said that $∣ \rangle$ is nothing but a notation. For example, we can denote the state $\vec \psi$ as $∣\psi \rangle$. In other words, the little arrow has transformed into a ket.
But when you look up material online, it seems that the usage of the bra-ket is much more free. Example of this usage: http://physics.gu.se/~klavs/FYP310/braket.pdf pg 17
A harmonic oscillator with precisely three quanta of vibrations is
described as $|3\rangle$., where it is understood that in this case we
are looking at a harmonic oscillator with some given frequency ω, say.
Because the state is specified with respect to the energy we can
easily find the energy by application of the Hamiltonian operator on
this state, H$|3\rangle$. = (3 + 1/2)$\omega h/2\pi |3 \rangle$.
What is the meaning of 3 in this case? Is 3 a vector? A scalar? If we treat the ket symbol as a vector, then $\vec 3$ is something that does not make sense.
Can someone clarify what it means for a scalar to be in a ket?
| It is just a label. More conventional notation uses indices for the same purpose, but the latter gets unwieldy if you need more elaborate qualifiers.
One particular application is labeling states by occupation number (cf second quantization).
| {
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Laser, wire, and a circle of light So, when I point a laser at a piece of wire (pointed at a specific angle), a circle of light appears on a wall behind it (see image below). I am trying to see why this happens and if there are any readings on this. Could anyone give some tips on what I should search for or any references?
(screenshot from this youtube video)
So this is a problem for a physics tournament, but I haven't tried it myself yet. However I have found a video depicting what I mean.
| So I did it myself and figured it out. It's just really simple reflection.
Basically the wire acts as a mirror, albeit a curved one. Because the light from a laser is straight, we won't have messy light. When a laser reflects off of a flat surface, it keeps going straight. However when it reflects off a small curved surface (i.e a wire), the laser rays all go different ways, and create a circle of light.
| {
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Proof that all primitive cells have the same size A primitive cell of a crystal lattice is a set $A$ such that two copies of $A$ which are translated by a lattice vector do not overlap and such that $A$ tiles the entire crystal.
I have read (for example in the german “Festkörperphysik” by Gross, Marx), that all primitive cells have the same size/volume.
Intuitively, this seems plausible, but is there a proof?
My precise, measure theoretic, interpretation of this statement is:
If $a_1, \ldots, a_n$ is a basis of $\mathbf{R}^n$ and $A, B \subset \mathbf{R}^n$ are sets such that $(\cup (A+\alpha_1 a_1+\ldots+\alpha_n a_n))C$ and $(\mathbf{R}^n \cup (B+\alpha_1 a_1+\ldots+\alpha_n a_n))^C$ (where the union is over all $\alpha_1, \ldots,\alpha_n ∈ \mathbf{Z}$) are Lebesgue null sets and such that for all $\alpha_1,\ldots,\alpha_n∈\mathbf{Z}$: $(A+\alpha_1 a_1 + \ldots \alpha_n a_n) \cap A$ and $(B+\alpha_1 a_1 + \ldots \alpha_n a_n) \cap B$ are Lebesgue null sets, then $A$ and $B$ have the same Lebesgue measure.
| Let $\mathcal{L}\subseteq\mathbb{R}^n$ be a lattice with a basis $B\in\mathcal{R}^{n\times n}$ and $F\subseteq\text{span}(\mathcal{L})$ be measurable. $F$ tiles $\mathcal{L}$ iff
*
*$(x+F)\cap(y+F)=\emptyset\,\forall x\neq y\in\mathcal{L}$, and
*$\mathcal{L}+F=\text{span}(\mathcal{L})$
It is trivial to show (I'll leave it as an exercise) that 1. implies:
$$ |(\mathcal{L}+x)\cap F|\leq 1 $$
while 2. implies
$$ |(\mathcal{L}+x)\cap F|\geq 1 $$
and therefore
$$ |(\mathcal{L}+x)\cap F|= 1 $$
for $x\in\text{span}(\mathcal{L})$. Then we have
\begin{align} \text{vol}(F) &= \int_{\mathbb{R}^n} 1_F(x)dx \\
&= \int_{B[0,1)^n}\sum_{y\in\mathcal{L}} 1_F(x+y)dx\\
&=\int_{B[0,1)^n}|(\mathcal{L}+x)\cap F|dx=\text{vol}(B[0,1)^n) \end{align}
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What is the speed of sound in space? Given that space is not a perfect vacuum, what is the speed of sound therein? Google was not very helpful in this regard, as the only answer I found was $300\,{\rm km}\,{\rm s}^{-1}$, from Astronomy Cafe, which is not a source I'd be willing to cite.
| Just want to bring up that most answers seem to be taking "space" to be a nice uniform medium. However, even within our own galaxy, conditions vary wildly. Here are the most common environments in the Milky Way:
*
*Molecular Clouds, $\rho\sim 10^4\,{\rm atom}/{\rm cm}^3$, $T\sim 10\,{\rm K}$
*Cold Neutral Medium, $\rho\sim 20\,{\rm atom}/{\rm cm}^3$, $T\sim 100\,{\rm K}$
*Warm Neutral Medium, $\rho\sim 0.5\,{\rm atom}/{\rm cm}^3$, $T\sim 10^4\,{\rm K}$
*Warm Ionized Medium, $\rho\sim 0.5\,{\rm atom}/{\rm cm}^3$, $T\sim 8000\,{\rm K}$
*HII Region, $\rho\sim 1000\,{\rm atom}/{\rm cm}^3$, $T\sim 8000\,{\rm K}$
*Hot Ionized Medium, $\rho\sim 10^{-3}\,{\rm atom}/{\rm cm}^3$, $T\sim \;{>}10^6\,{\rm K}$
The sound speed is proportional to $\sqrt{T}$. Given that the temperature varies over about 7 orders of magnitude (maximum at about $10^7\,{\rm K}$, minimum at about $3\,{\rm K}$), the sound speed varies by at least a factor of $1000$. The sound speed in a warm region is on the order of $10\,{\rm km}/{\rm s}$.
Trivia: the sound speed plays a crucial role in many astrophysical processes. This speed defines the time it takes for a pressure wave to propagate a given distance. One place this is a key time scale is in gravitational collapse. If the sound crossing time for a gas cloud exceeds the gravitational free fall time (time for a gravity-driven disturbance to propagate), pressure is unable to resist gravitational collapse and the cloud is headed toward the creation of a more compact object (denser cloud, or if conditions are right, a star).
More trivia: space is a very poor carrier (non carrier) of high frequency sounds because the highest frequency pressure wave that can be transmitted has a wavelength of about the mean free path (MFP) of gas particles. The MFP in space is large, so the frequency limit is low.
| {
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Why would two protons repel? I understand that two protons would repel due to them both being positively charged, however, wouldn't the strong force act on the two protons pulling them together? Would this mean that in this case the electromagnetic repulsive force is greater than the strong force? If so why? If not why would they repel?
| Yea, a proton and neutron stick together, but two of the sam kind don't. You don't get neutron balls even with only the interneucleon force and no electric repulsion.
I asked about it some years ago in a physics on-line forum, long before StackExchange. Ended up getting a textbook and eventually learning that "the force is largely insensitve to species (whether proton or neutron), but highly dependent on spin."
I got two different answers as to whether the diproton exists as a (fleeting) bound state.
The answer, relative to species and spin combinations of neucleons, has to do with "singlet state vs triplet state". That is a unique enough phrase that you ought to be able to search on that (e.g. class notes ...Hence the force is attractive for isospin singlets (T = 0) and repulsive for isospin triplets (T = 1).; more class notes in a nice PDF with illustrations.
Bottom line: the interneucleon force is not so simple as what you think of with gravity or electric or magnetic but with faster fall-off. It has complex behavior involving just how the particles are parked and what its neighbors are doing.
| {
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What does the statement "the laws of physics are invariant" mean? In the first paragraph of Wikipedia's article on special relativity, it states one of the assumptions of special relativity is
the laws of physics are invariant (i.e., identical) in all inertial systems (non-accelerating frames of reference)
What does this mean? I have seen this phrase several times, but it seems very vague. Unlike saying the speed of light is constant, this phrase doesn't specify what laws are invariant or even what it means to be invariant/identical.
My Question
Can someone clarify the meaning of this statement?
(I obviously know what an inertial frame is)
| When constructing equations of motion which are the reflection of laws of nature so to speak, we must make them Lorentz invariant and invariant to spacial rotations. This means that they must have the same form under these transformations. One example is construction of a field theory, in which you begin by forming an action which is Lorentz invariant making sure from the very start that you will get it right. Action is a physical quantity with a dimension of Js (joule-second). This quantity is very important for the thing called Hamilton principal of stationary action...So laws of nature same in all inertial reff frames = equations that describe them invariant with form to Lorentz transformations.
| {
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Is the Liénard-Wiechert electric field conservative? I know that an accelerated charge should emit an e.m. field and loose energy. Therefore, the Liénard-Wiechert (L.W.) electric field of an accelerated charge should be non-conservative.
But I checked first what happens when the charge is not accelerated, i.e. moves with a constant velocity. I expected to find a conservative field as in the case when the charge is at rest. A charge moving with constant velocity doesn't radiate. But it seems that this is not what happened.
Given the scalar potential $\phi$ and vector potential $\vec A$, the electric field is
$$ \ (1) \ \vec E = - \nabla \phi - \frac {∂ \vec A}{∂t},$$
where
$$ (2) \ \phi (r, t) = \frac {1}{4 \pi \epsilon _0} \left( \frac {q}{(1 - \vec n \vec \beta _s)|\vec r - \vec r_s|} \right)_{t_r},$$
$$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3) \ \vec A = \frac {\mu _0 c}{4 \pi} \left( \frac {q \ \vec {\beta} _s}{(1 - \vec n \vec \beta _s)|\vec r - \vec r_s|} \right)_{t_r} = \frac {\vec \beta _s (t_r)}{c} \phi (r, t).$$
see the article.
I assume that for constant velocity of the charge, $t_r = t$.
A field that obeys
$$ \ (4) \ \vec F(\vec r) = \nabla V(r)$$
is conservative, i.e.
$$ \ (5) \ \int_{\vec {r_1}}^{\vec {r_2}} \vec F \ d \vec {\ell} = V(\vec {r_2}) - V(\vec {r_1}).$$
So, I expected that for the constant velocity the formula (1) will turn into (4), i.e. that I would get that $\vec A$ does not depend on time. But this doesn't happen. Why? A charge in movement with constant velocity shouldn't radiate, its electric field should be conservative.
Do I make a confusion, do I make a mistake?
| An Electric Field is only conservative if it is static. The propagation of E with a L-W field contradicts this, so it is not conservative.
| {
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Is $ds^2$ just a number or is it actually a quantity squared? I originally thought $ds^2$ was the square of some number we call the spacetime interval. I thought this because Taylor and Wheeler treat it like the square of a quantity in their book Spacetime Physics. But I have also heard $ds^2$ its just a notational device of some sort and doesn't actually represent the square of anything. It is just a number and that the square sign is simply conventional.
Which is true?
| It is a notational device. Note that in $(-+++\cdots)$ the proper length
$$ds^2=g_{\mu
\nu}dx^\mu dx^\nu$$
is negative for timelike $dx$. Thus $ds\equiv \sqrt{ds^2}\in\mathbb{C}$. It (the square root) thus has no physical meaning.
| {
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selection of p substrate as wafer in typical cmos process flow why is p-substrate typically used as wafer in the typical cmos process flow? why not n substrate?with respect to memories, Has it got anything to do with the aplha-paritcle radiation induced errors (soft errors) ? please explain.
| The mobility for electrons is generally higher than for holes, considering typical doping profiles, etc. Therefore, it is easier to achieve higher signal speeds when the conduction flow travels through an n-type silicon. Hence, you would want the substrate to be p-type since you will be doping in the conduction channel.
For the math which allows you to compute electron and hole mobilities in Si, and plots, take a look here.
| {
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What physical evidence is there that subatomic particles pop in and out of existence? What physical evidence shows that subatomic particles pop in and out of existence?
| This phenomenon is called Quantum Fluctuations or vacuum energy and it could be described theoretically by Heisenberg uncertainty relation with the energy term.
One of the physical evidences of such phenomenon is ''Casimir effect'' .
when two uncharged plates are put close to each other they exhibit a repulsive force, this force is explained by quantum fluctuations (subatomic particles popping in an out of existence).
| {
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Probability wave intuition Basically I am really new with this term. It all came before me when I was reading the Standard Double-Slit experiment.
An excerpt from Principles of Physics (by Resnick,Halliday,Walker) :
. . .by probability wave, to every point in the light wave, we can attach a numerical probability per unit time interval that a photon can be detected in any small volume centered on a point.
Now, this is not a theoretical definition; just an operational one and is applicable when the wave is about to interact with a matter.
By knowing the intensity, we can find the number of photons associated with the wave. Then is the probability wave saying that we cannot specify the specific location of photons in the wave? But photons do appear only at the time of interaction,right? What does it , then want to convey actually?? What is actually the meaning of probability wave intuitively?
| I think the answer you are looking for is from "Probability Amplitude" on Wikipedia:
Probability amplitudes have special significance because they act in quantum mechanics as the equivalent of conventional probabilities, with many analogous laws, as described above. For example, in the classic double-slit experiment, electrons are fired randomly at two slits, and the probability distribution of detecting electrons at all parts on a large screen placed behind the slits, is questioned. An intuitive answer is that P(through either slit) = P(through first slit) + P(through second slit), where P(event) is the probability of that event.
There is also a video that shows a probability wave forming as a mapping of a particle's motions over time, from a theory known as "Pilot-wave". For reference, this theory is still being studied, and is not accepted as the final answer to the wave-particle duality. However, it does an impressive job of visualizing the how a particle, or set of particles, move in patterns that appear to mimic a probability wave.
| {
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Ball's opposite rotation caused by friction Assume that you push a ball like in the picture (along the red line) with your hand with a some force.
The ball will move forward while its rotating in this way:
And after some movement, the ball will stop translating(still rotating) and it again moves back towards you. How does friction work here? How is this motion possible?
| According to the picture, the ball has acquired a spin which can be described by an angular velocity $\boldsymbol\omega$ emerging from the screen. The contact point is then moving with velocity $\mathbf u=\boldsymbol\omega\times\mathbf r + \mathbf v_{cm}$. As long as this is not zero (i.e. the ball is not rolling), you can model friction as that force that acts on the centre of mass of the ball while it is not rolling, opposite in direction to $\mathbf u$ and modulus that depends on the mass (and probably the geometry of the ball), so something of the form
$$\mathbf F_f = -\mu mg\frac{\mathbf u}{\Vert\mathbf u\Vert},$$
where $\mu$ is encoding the geometry and the mechanical materials involved. Hence the essence of this force is due to electromagnetic interactions (as usual for the most common types of frictions). Observe also that from Newton's third law the plane is exerting a torque on the ball, so the rotation is slowing down in the process.
For the case in the picture above, both $\boldsymbol\omega\times\mathbf r$ and $\mathbf v_{cm}$ are initially directed towards the right. Then the rotation slows down the ball, as well as its spin, so three things can happen in principle, depending on how strong the friction is (that is, the value of $\mu$):
*
*the ball slows down but starts rolling towards the right;
*the ball halts completely with no residue spin;
*the ball halts, reverts its motion and starts rolling towards the left.
| {
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Could dark energy just be particles with negative mass? The title speaks for itself.
Dark matter: We see extra attractive force, and we posit that there are particles which create such a force, and use the measure of that force to guess their locations.
Dark Energy: We see extra repulsive force.
Only thing is, dark energy is uniform. So I suppose the stuff would have to be (at least somewhat) uniformly distributed throughout the universe. How uniform do we know it to be? Could the "stuff" be somehow a part of empty space itself?
| “If dark energy would consist of particles, it would dilute with the growing radius of the universe to the third power, since the total number of particles would stay the same while the volume increases. What observations found was that dark energy rather behaves like a constant which does not thin out, that's why it is also known as the cosmological constant. That means even if the universe expands, the amount of dark energy per cubic meter stays (at least approximately) the same.”
The answer above was conventionally correct. Modern science has now developed further since then. There was a result last year by an Oxford professor Jamie Farnes, widely discussed in international media, about possible creation of negative mass particles.
https://arxiv.org/abs/1712.07962
When negative mass particles are created by a creation field, they do not thin out and can behave as a cosmological constant. Apparently the idea is also testable.
| {
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What information do $|\psi(0)\rangle$ and $|\psi(t)\rangle$ represent? I am starting to feel comfortable with the role of the unitary operator in quantum mechanics. For instance, one of the equations I have seen is
\begin{equation}
|\psi(t)\rangle = U(t) |\psi(0)\rangle
\end{equation}
I understand what a unitary operator is in that
\begin{equation}
U^*U=UU^*=I
\end{equation}
I also understand that if we have a vector space containing our wave function $|\psi(0)\rangle$, then the operator maps to $|\psi(t)\rangle$.
My question:
What information do $|\psi(0)\rangle$ and $|\psi(t)\rangle$ represent?
If the answer is probability amplitude, then perhaps someone can clarify what exactly that is. As far as I understand, the $\|\psi\|^2$ represents the probability density. But that the amplitude is more fundamental for some reason.
I ultimately hope to understand why we are using unitary operators in the first equation, but I think I need to figure out this question first before tackling that.
| The state vector ket describes all that can be known of the system at future time. Unlike classical mechanics where two quantities velocity and position are required to describe the future here a single quantity does it, done at the price of requiring the wave function to be complex. The representation in which you describe the state vector can range from many. If done using the x representation then you get the usual wave function but momentum representation can also be used. However wave function obtained from momentum representation can be used just like the position representation in obtaining the values of various dynamical variables.
| {
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Is rate of change of velocity wrt distance and rate of change of velocity wrt time the same thing? Is rate of change of velocity wrt distance and rate of change of velocity wrt time the same thing?If both are same can we define acceleration in the former way?
Please explain using calculus.
| No they aren't. Suppose we have some velocity $v(t)$. The differential with respect to time is just the acceleration:
$$ \frac{d}{dt}v(t) = a(t) $$
Now differentiate it with respect to distance $s$:
$$ \frac{d}{ds} v(t) = \frac{dt}{ds}\frac{d}{dt} v(t) = \frac{dt}{ds} a(t) = \frac{a(t)}{v(t)} $$
| {
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Why is the bottom part of a candle flame blue? What’s the explanation behind the bottom part of a candle flame being blue? I googled hard in vain. I read this. I don’t understand how it’s explained by the emission of excited molecular radicals in the flame. I read that a radical is a molecule or atom which has one unpaired electron. That made me more confused. I want a more detailed, clearer explanation.
| In a lit candle, when gaseous candle wax reacts with the oxygen in the air, the atoms will be unstably excited. To be stable, the excited electrons will relax to the ground state by emitting photons with energy equal to the energy difference between the 2 states. The photons’ energy doesn’t change much, so the wavelength doesn’t change much. The chemical reactions yield light with this spectrum. That light is blue to humans.
source – @gigacyan
| {
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Spacelike to timelike four vectors First at all, let me just say that I'm not a Physicist, I study mathematics. So, I have this question. If you have a spacelike four vector, is there any transformation that could change it to be a timelike four vector? I mean, I know that every Lorentz Transformation (LT) preserves this properties (timelike $\rightarrow$ timelike, spacelike $\rightarrow$ spacelike, etc.), but I was thinking in another frame $S'$, different from the former $S$, where a spacelike four-vector (in $S$) will be timelike (in $S'$). If it is possible to have this other frame then, the way to relate events between frames is not a LT? or I'm missing something?
| Let's think about this in terms of light cones. At any given point in space-time, we can consider the set of all possible light rays which pass through that particular point. If one now considers the set of all possible tangent vectors to those light rays (these are of course, null) then they form the light cone at that point.
If any one of those vectors is time-like then they will lie inside the cone, and space-like vectors will lie outside the cone. Null vectors lie on the cone.
I have not come across a transformation that would send one to the other, there may be some scope for considering a non-null foliation, where an integrable distribution of hypersurfaces can be either time-like or space-like but I concede that this isn't really answering your question.
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What can take kinetic energy, transform it into potential energy when pressed on, and put back out as kinetic energy when released (besides a spring)? A spring can only hold so much of the kinetic energy. For example, a 1 cm spring can hold less than 5 J. Is there anything that can hold a large amount of energy but be fairly small?
| A 1 kg object on the surface of the Earth can store up to $G\frac{M_e}{r_e}\cdot 1$ $kg = 6.3 \times 10^7$ $J$ in gravitational potential energy as long as your pressing direction is up and you press it up far enough.
A compressible thermodynamic system can theoretically hold an infinite amount of energy. See for example the apparati in this image, which are boxes with pistons: http://chemwiki.ucdavis.edu/@api/deki/files/10077/STEP_4-1.jpg
Assume for example that our box is initially 1 cm^3 (0.000001 m^3), and is initially filled with air (which we'll assume to be an ideal gas) at room temperature (~300 K), and atmospheric pressure (~100,000 Pa). Then this box will initially contain ~0.15 J of thermal energy. Then by the adiabatic relation for an ideal gas, which states that $PV^{5/3}$ is constant under any adiabatic process, and the (energy form of the) ideal gas law $\tfrac{3}{2}PV = U$ where $U$ is total internal energy, we can do some algebra and derive this formula for the change in energy as a function of added pressure (in SI units):
$\Delta U =\frac{3}{2}\cdot 0.001 \cdot (100000+\Delta P)^{2/5} - 0.15$ $J$.
Unfortunately, realistically this means you'd need to apply more than a million atmospheres of pressure in order to store more than a couple dozen Joules of energy, and there do not currently exist materials that can take those kind of pressures, as far as I know.
Finally, you could store the energy by compressing a charged parallel plate capacitor, but if the capacitor is fairly small then electrons are liable to jump from the negative end to the positive end when the plates get close enough, which will limit the amount of energy you can store.
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Wheel slowing down with constant acceleration A wheel of radius $R$ spins about its center with a centripetal acceleration of $v^2/R$.
I get that the acceleration at all points on the rim of the wheel point towards the center of the wheel. But, what happens when the acceleration of the wheel is slowing down?
Let's say the wheel is spinning clockwise.
So, at the top, in the case where the wheel is not spinning, there is only one acceleration vector (pointing down towards the center of the wheel). When the wheel starts to slow down, I would guess that the centripetal acceleration remains at $v^2/R$ but now there's another component - the acceleration vector that points to the left, tangent to the circle, and perpendicular to the centripetal acceleration. So... ←↓
Adding these up gives me the new acceleration, right? How does this relate to the new period of the wheel though? Since the wheel is slowing down, I cannot just use $T = 2 \times \pi \times R / v$... Or can I?
| If the wheel is undergoing some sort of angular acceleration, then the magnitude of its velocity - its speed - would be expressed as a function of time, $v\equiv v(t)$.
So what does this mean for the period of rotation? That means the period also becomes a function of time. However, since the period still represents the time required for the wheel to rotate through $2\pi$ radians, the equation won't change.
$$T(t)=\frac{2\pi R}{v(t)}$$
An alternate way of thinking about this is that the period is defined as the inverse of the frequency, $$T=\frac{1}{f}$$ And we can also define the angular frequency as $\omega=2\pi f$ to find that $$T=\frac{2\pi}{\omega}$$
However, it's clear that the angular frequency must be the same as the angular velocity, which is $\omega=\frac{v}{R}$, thus we get the original equation for period as a function of speed and when $v$ changes with time, the equation for period must remain the same but become a function of time as well.
If you want to know what the new period is at any time, you must figure out the function $v(t)$ and substitute that into the equation for $T(t)$.
For example, if the speed is changing such that $v(t)=v_0-at$, then the equation for the period would become
$$T(t)=\frac{2\pi R}{v_0-at}$$
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"Find the Lagrangian of the theory" I've heard a few of my professors throw around the term "finding the Lagrangian of a theory". What exactly is this referring to. From what I understand it seems that you determine invariances (symmetries) and they give you a hint for what your Lagrangian is. Furthermore there is more to the story because I know:
$L=T-U$ is only one of the forms the Lagrangian can take in classical mechanics.
So far I only learned about the Lagrangian in classical mechanics and might be building up to a limited knowledge of Feynman's path integral in my QM course.
What other theories have Lagrangians and how you can tell?
Are all Lagrangians of a given theory equivalent?
| Comments to the question (v2):
*
*"Find the Lagrangian of the theory" typically means that you are given the (classical) equations of motion (EOMs) of some physical system, and are supposed to find the action functional $S$ so that the EOMs are (parts of) the Euler-Lagrange (EL) equations for $S$.
*Note that an action principle/Lagrangian formulation does not always exists, cf. e.g. this Phys.SE post and links therein.
*Two different actions that yield the same EL eqs. are called classically equivalent action formulations.
*Adding total divergence terms in the action does not change the EL eqs., but are typically associated with other boundary conditions for the theory.
*Quantum mechanically, in the the corresponding path integral formulation, two classically equivalent actions need not lead to equivalent quantum theories.
*In fact, already the same classical action can give rise to non-equivalent quantum theories because of different quantization procedures, different operator ordering prescriptions, etc.
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$\mathrm{\rho^0}$ meson decay via the weak interaction? Of course, the $\mathrm{\rho^0}$ meson can decay in $\mathrm{\pi^{+}\ \pi^{-}}$ through the strong interaction. Using Feynman diagrams, I cannot understand why the same decay couldn't happen through the weak interaction. I attach the diagram I've drawn.
Strong decay:
Weak decay:
| As far as I understand, the decay of an anti-u quark into an anti-d quark will emmit a $\mathrm W^+$ boson. But the anti-u and d quarks into which the boson would decay, have charges $-2/3$, $-1/3$ respectively. So their total charge will be $-1$. A $\mathrm W^+$ boson cannot decay into two particles whose sum equals $-1$, the sum of the particles should be $+1$. So this decay cannot be a weak interaction because the conservation of charge would be violated.
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Why doesn't the speed of the wind have an effect on the apparent frequency? A boy is standing in front of stationary train. The train blows a horn of $400Hz$ frequency . If the wind is blowing from train to boy at speed at $30m/s$, the apparent frequency of sound heard by the boy will be?
The answer: The frequency remains the same at $400Hz$
MY QUESTION:
Why doesn't the speed of the wind have an effect on the apparent frequency?
| Because that is the result when you examine the process in detail.
For example:
The boy and the train are in a static relationship. The train could sound its whistle for as long as the power source held out. If the boy received more waves per second than the train produced, where would the extra waves come from?
Or:
The wind is snatching the waves from the train and speeding them up in the direction of the boy, while at the same time increasing the velocity of the wave relative to the ground, stretching out each wave by the same amount. Since:$$f\times \lambda=v$$increasing $\lambda$ and $v$ by the same factor must leave $f$ unchanged...
| {
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Conservation and forces/energy Are there really non-conservative forces in actuality ?
Feynman states in his book that in fact, all forces are conservative ( originating from conservative vector-fields ), provide we look close enough ( microscopic level ). The reasoning is that we can't allow non-conservative forces in order for Conservation of Energy to follow.
But at the same time, physicists who seem to really know the subject in an advanced-level, assert that most forces refuse to be conservative.
For instance, see acepted answer of Locally every force admits a potential?.
So, are all forces conservative forces and conservation of energy is not violated, or are there non-conservative forces and conservation of energ is violated , or finally, Law of Conservation of energy can cohexist with non-conservative forces ?
| There are macroscopic forces that admit no description in terms of a potential, for example, any friction force proportional to the velocity of a moving object as path-dependent integral, and is hence non-conservative.
But we know the macroscopic description is not the fundamental description. In terms of the interaction of the constituents of matter, all fundamental forces known - gravity, electromagnetism, the strong and the weak force - are conservative forces in the sense that they are descended from a (gauge) potential. It is highly non-trivial (and indeed, not done for the general case as far as I know) to derive the appearance of superficially non-conservative forces from this fundamental Lagrangian description.
Nevertheless, in the spirit of reductionism, Feynman and most other physicists believe the description in terms of the fundamental forces is more or less complete - all other forces emerge in some sense from them, and so, since the underlying microscopic description conserves energy, so must the emergent macroscopic description.
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Can one define wavefunction for Bogoliubov quasiparticle excitation in a superconductor? Wavefunction is essentially a single particle concept. It is easily extended to multiparticle system as follows- if one has say five electrons the wavefunction of this five electron state is any completely antisymmetric function of five coordinates which is square integrable in the five dimensional space. Given a five electron ket in fock space $|K\rangle$, its wavefunction is denoted as $\langle x_1 x_2 ...x_5|K\rangle$. But for a superconductor its effective Hamiltonian doesn't conserve the particle number. Then can one come up with any reasonable defination of a wavefunction for a single quasiparticle excitation of the superconductor over its ground state denoted by $\gamma_i^{\dagger} |G\rangle$ where $|G\rangle$ is superconducting groundstate composed of Cooper pairs and $\gamma_i^{\dagger}=\sum_k u_i^kc_k+v_i^kc_k^{\dagger}$ is Bogoliubov quasiparticle creation operator and $c's$ being electron operators and $u's$ and $v's$ being some complex numbers.
In Kitaev chain and it's solid state realisation one usually talks about Majorana fermion(Bogoliubov excitation) being localised at the two ends. How can one do that without a reasonable definition of wavefunction for superconducting states? The papers usually interpret eigenvectors of $H_{BdG}$ in coordinate space as representative wavefunctions. Is it justified?
| The localization of Majorana zero modes has a well-defined meaning: consider a Kitaev chain with two ends. Because of the zero modes, there are two nearly degenerate ground states, let us call them $|0\rangle$ and $|1\rangle$, which have opposite fermion parities. They are localized as "single-particle wavefunctions" in the following sense: if one computes the matrix element
$\langle 1|c^\dagger(x)|0\rangle$ where $c^\dagger(x)$ is the creation operator for fermions, the result is an exponentially decaying function of $x$ away from the edge. This definition works even when the system is interacting. Intuitively it means that the weight for creating a single fermion excitation is localized near the edge, and in the bulk there is a finite gap to the single particle excitations.
| {
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Eigenstates into which a system can be projected after a measurement I'm currently reading Dirac's Principles of Quantum Mechanics, on page 36, he says:
Another assumption we make connected to the physical interpretation of the theory is that, if a certain real dynamical variable $\xi$ is measured with the system in a particular state, the states into which the system may jump on account of the measurement are such that the original system is dependent on them.
On what physical basis can we make this assumption and why is it reasonable?
| When you measure an observable over a certain state, any possible outcome lies within the physical spectrum of the observable itself (which can be shown to coincide with the algebraic notion of spectrum for linear operators). So after the measurement has given you a value, say, $\lambda$, any other measurement on the system will give you $\lambda$ again with probability 100%. This is just the probabilistic interpretation of Quantum Mechanics, where the way the expectation value of an observable on a given state is defined as the average over a large enough ensemble of exact copies of the system in the given state. The outcomes must be in the spectrum of the observable, but they way the state is then defined gives you an average over all the possible outcomes. Clearly, once you have collected all the copies of the system in the ensemble that have given you the value $\lambda$, you now know that if you measure the same observable again on this subensemble, you will find $\lambda$ on all of them.
A typical example is that of polarisation. Once you have polarised light, the intensity through a polariser with the same axis will be 100%.
| {
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How could a cord withstand a force greater than its breaking strength? How could a 100 N object be lowered from a roof using a cord with a breaking strength of 80 N without breaking the cord?
My attempt to answer this question is that we could use a counter weight. But I don't really understand the concept behind counterweights so I hope someone can clear that up for me and if there is a better answer I'll love to know it.
| The simplest approach (and what the person asking this question probably was getting at) is to use a pulley like so:
The weight of the object is now shared between the two sides, with each carrying a 50 N load. You end up using twice as much cord. The other advantage is that you now have a "mechanical advantage" and you only need to use a force of 50 N to lower the object. You do need a point where you fix the other end, unless you hold both sides in your hands. In that case the load is evenly balanced between the two halves of the cord by the pulley.
Alternatively you could simply double up the cord - but the tricky thing there is to ensure that they share the load evenly. This is done in practice by twisting the ropes together (yes, that is one reason why ropes since time immemorial are twisted): if one strand carries a larger fraction of the load it tends to straighten out - which makes the other strand "take a longer path" (it becomes more twisty) until it starts carrying more of the load. In this way, twisting ensures sharing of the load. Twisting of the strands also makes a rope more flexible (since strands spend "equal amounts of time" on the inside of the bend and the outside - I put that expression in quotes since it is only approximately true but you get the idea).
| {
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What is the correct way to treat operators that has "time" in QM? I don't know if this question has already been resolved but considering that $i\hbar\partial_t$ is the energy operator, and $\partial^2_t$ is the waves operator (or helmholtz), I can't accept that $t$ itself isn't an operator
What is the argument here that says $t$ is not an operator?
| You are touching on the subject of relativistic quantum mechanics where time and space $(t,x)$ are handled on the same footing as operators. The accepted description is to not use quantum wavefunctions as describing one particle but rather the state of a quantum field. Doing this turns into the subject of quantum field theory and is the basis of modern quantum experimentation/theory.
| {
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What is basically the difference between static pressure and dynamic pressure? What is basically the difference between static pressure and dynamic pressure?
While studying Bernoulli's theorem, I came before these terms. The law says:
When the fluid flows through a small area, its pressure energy decreases & kinetic energy increases and vice versa.
Now that's wierd as I know due to having KE, ie. having momentum, one can impart pressure. Then why distinction ? What is then pressure energy??
In order to understand that I went to wikipedia & quora & others; there I found fluid exerts two pressure: Static & dynamic. But really nothing could be understood more than that. What are they actually?
| Under common assumptions and ignoring potential energy, static pressure is the expression of the fluid's temperature (internal energy) and dynamic pressure is the expression off the fluid's velocity, so if the fluid is brought to a rest adiabatically, their sum is equal to the stagnation pressure. The stagnation pressure represents the total energy of the fluid.
| {
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Calculate the approximate number of conduction electrons So i have the following problem: A cube of gold 0.1 meters on an edge, calculate the approximate number of conduction electrons whose energies lie in the range from 4.0 ev to 4.025 ev.
But I'm not clear on how to start. could someone offer any help?
| mmm. I don't think that's quite right. I'd rather use the equation for the number of electrons in any given energy state,
$N(E) = \int_0^{\infty} \frac{1}{e^{(E - E_F) / K_B T}} \frac{V (2 m) ^{3 / 2}}{2 \hbar ^3 \pi ^2}\sqrt{E} dE$ where $E_f$ is the fermi energy and $E$ is the energy of the electron. To do this, you'd need to know the temperature of the gold, but without that the number of electrons in any given energy isn't really hugely meaningful anyway.
Once you have a temperature, if it's sufficiently small (and if they're asking you this question, it probably will be) $(E - E_F) / K_b T$ will become either infinity or negative infinity (depending on whether E is greater or less than the fermi energy). Then you can just integrate.
| {
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When does Pauli's exclusion principle kick in? Imagine that I prepare a fermion in the $\left|\uparrow \right\rangle$ state and a second one far away in the $\left|\downarrow \right\rangle$ state and set them in a path for collision.
According to Pauli's exclusion principle, the composite wave function must be anti-symmetric. Does the wave function become anti-symmetric as they collide or was it like this from the start? Can one predict if the composite wave function will correspond to a singlet or a triplet state from the moment we prepare the separate fermions?
| Let's ignore the position-momentum observables. Your definition of far away has a precisely definition. We can treat one electron and ignore the another. So we don't have to worry about pauli principle. When we want to collide this two electrons, and no more work in far away paradigm, we need to define some length when pauli principle is applied. We can think in the length when one electron can reach the another in space by quantum fluctuations.If this electrons are confined by cells for example, is the tunneling length.
This can be done because don't make sense to have a far away systems in the same state! If the systems are far away one of another then exist some subspace in Hilbert space of the whole system that permits the definition of "far away". And is precisely the complementar of the subspace generated by $(\left|\uparrow \right\rangle\left|\downarrow \right\rangle,\left|\downarrow \right\rangle\left|\uparrow \right\rangle,\left|\downarrow \right\rangle\left|\downarrow \right\rangle,\left|\uparrow \right\rangle\left|\uparrow \right\rangle)$ that accounted the Pauli principle. Is the complementar space that has position-momentum observables and other degrees of freedom.
Here for more
For proceed with this issues we can substitute the pauli principle for some exchange energy-interaction.
$$
H=g(l_{tp})S_1.S_2
$$
where the $g(l)$ is the coupling in function of the typical lenght $l_{tp}$. Locality may tell us that if $l_{tp}\rightarrow \infty$ then $g(l_{tp})\rightarrow 0$ .
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Understanding magnetic force on charged particle if we put any charged motionless or static particle in the constant magnetic field, then why does it don't feel a magnetic force?
Mechanism by which electric and magnetic fields interrelate
I have read the above article which suggested that the magnetic field is the relativistic effect of the electric field then why does the static charge particle does not feel the magnetic force and the magnetic field is also an one type of electric field, and any static charged particle always feel the electric force in electric field.then what really happens there?
| Magnetic field is not a type of electric field, though the 2 are intimately related. Otherwise, every charged body kept in a magnetic field would experience a magnetic force and basically a lot of stuff would go out of control.
(For example, if you kept a magnet close to a circuit, then there would be a electric field created and hence a current flow, out of nowhere. In that petty example alone, you'd be breaking Maxwell's laws of electromagnetism, energy conservation and so on...)
As pointed out by Ocelo7, the Lorentz force includes a velocity dependence. However a particle with a velocity in one frame might be stationary in another and therefore there would be no magnetic force on that particle in that frame. It would be all electric force. The effect of the magnetic field would be taken care of by the electric field, as is shown in Griffith's Introduction to Electrodynamics.
However it is crucial to realise that magnetic force does not act on stationary charges. The field due to stationary charges is called Electrostatic field. So, if you have some arbitrary distribution of stationary charges in a particular frame of reference, the force on them is electrostatic of course, since that is how the term is defined and magnetic forces play no role in that frame.
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Are electrons miniature black holes? For something to be a blackhole, it must have gravity and the radius must be smaller than the schwarzschild radius for its mass.
-Electrons have gravity
-Electron are theoretically believed to be infinitely small points
Since it has gravity it is capable of being a black hole. Since its radius is infinitely small, it must have a schwarzschild radius and thus be a black hole.
| Well, according to the wild ER=EPR conjecture by Maldecena and Susskind, two entangled electrons are connected by a quantum wormhole. The mechanism and details of this quantum wormhole are left unspecified by these authors, though.
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Do transferring energy and applying force to a body imply same? Do transferring energy and applying force to a body imply same meaning? When we say, "I throw a ball using my pushing force so on the other hand, can I say that I transferred my kinetic energy to the ball therefore it became moving.
| No. In a uniform circular orbit the orbiting body maintains constant energy while a constant force, only changing in direction, operates on that body.
Kinetic energy changes when a net force is applied in the direction that an object is moving. It will reduce if the net force opposes velocity, or increase if net force supports velocity.
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Can mass be uncertain? If you can have uncertainty in momentum, then wouldn't you have uncertainty in mass and velocity?
Why can't mass be uncertain?
| Uncertainty is a property of observables. Mass is not normally taken to be an observable, so it does not obey uncertainty relations.
Why isn't mass an observable? There is a superselection rule that forbids it in the presence of reasonable symmetry assumptions. See the discussion here for more.
EDIT: In "true" relativistic QFT one wouldn't even talk about "mass" but "mass-energy", and Bargmann's superselection rule doesn't hold. In that context mass-energy is a well-defined observable, and it obeys an energy-time uncertainty relation.
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What observations would be needed to falsify the law of conservation of energy? I've been doing some thinking, and began to wonder: What observations have led us to the conclusion that ‘energy can neither be created nor destroyed’?
Essentially, this means that the big bang supplied our universe with all the matter that currently exist in it.
My question is, if this were untrue for some reason (despite what math tells us) - and there were more ‘atoms’, ‘quarks’ or ‘bits’ being added to the system, how would we tell? What would that look like?
What experiment would be used to detect ‘more matter’ existing now than during the big bang?
Is this simply a matter of creating instruments that measure more accurately, or would some clever experiment need to be devised?
I've read several answers on this and other sites that explain the law of conservation of energy, but none that nail the hypothetical question above.
| Conservation of Energy can be derived if one accepts that $F = ma$. I won't include the derivation here unless you ask. This means that to prove Conservation of Energy wrong, one must prove $F = ma$ wrong. This could be attempted in a variety of ways. One such way would be applying a force to an object and noticing the $F = ma$ doesn't give the correct acceleration.
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Flux linkage of 2 coils in series I have a question about magnetic circuits. I am an engineering student, therefore I will neglect some minor errors.
There is a simple magnetic circuit with 2 coils ($C_1$ & $C_2$), with $N_1$ & $N_2$ turns and $i_1$ & $i_2$ currents. Coils are in series and connected with 1 magnetic core of "O" shape (we may assume some reluctance in form of air gap = $R_c$). The flux linkage of $C_1$ is as follows:
$$
\lambda_1 = N_1 \Phi= N_1 \left(\frac{N_1 i_1}{R_c} + \frac{N_2 i_2}{R_c}\right)
$$
so in the end I got something like
$$
\lambda_1=N_1^2i_1\left(\cdots\right)+N_1N_2i_i\left(\cdots\right)
$$
The $(\cdots)$ are some constants regarding air gap and core geometry.
I know that magnetic circuit resembles electric circuits. If there would be similar circuit with 2 batteries in series and 1 resistor, the overall voltage would be $V = V_1 + V_2$
In above mentioned magnetic circuit the overall flux linkage would be
\begin{align}
\lambda_{tot}&=\lambda_1+\lambda_2\\
&=N_1^2i_1\left(\cdots\right)+N_2^2i_2\left(\cdots\right)+2N_1N_2i_2\left(\cdots\right)
\end{align}
My questions are then
*
*Why is there the term $2N_1N_2 i_2(...)$ in the last part?
*Why is the electric circuit $V = V_1 + V_2$ and no $V_{12}$?
|
I know that magnetic circuit resembles electric circuits.
Yes, but the analogy works between certain specific quantities, and you are considering the wrong ones: the electromotive force does not correspond to the flux linkage, but to the magnetomotive force.
For lumped magnetic and electric circuits, the main correspondences are as follows:
Magnetomotive force $\mathcal{M}$ $\qquad\longleftrightarrow\qquad$ Electromotive force $E$
Magnetic flux $\varPhi$ $\qquad\longleftrightarrow\qquad$ Electric current $I$
Magnetic reluctance $\mathcal{R}$ $\qquad\longleftrightarrow\qquad$ Electric resistance $R$
Only with these correspondences the magnetic analogues of the Kirchhoff's voltage and current laws hold (see also Magnetic circuit on Wikipedia).
In the case of your circuit, the total magnetomotive force is $\mathcal{M} = N_1i_1+N_2i_2 = \mathcal{M}_1+\mathcal{M}_2$, and there isn't any cross-term.
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Can an object appropriately isolated from its surroundings become colder than its surroundings? Consider a sealed box, well-insulated on all sides, except for the lid which is transparent to infrared. An object is placed inside the box and the box is evacuated (purpose being to thermally isolate the contents of the box from its surroundings). The box is placed outdoors (in an everyday atmosphere) on a clear night. Let's assume that at the start of this experiment, the box and its contents are in thermal equilibrium with its surroundings. The object inside the box will radiate infrared according to its temperature, which should escape through the lid of the box. With nothing but clear dark sky above, I assume there is nothing to radiate appreciable heat back into the box and maintain the object's temperature.
Question: will the object cool below the ambient temperature outside the box?
|
With nothing but clear dark sky above, I assume there is nothing to radiate appreciable heat back into the box and maintain the object's temperature.
In this case there wouldn't really be an "ambient temperature" though.
To elaborate: The inside of the box would cool down until thermal equilibrium is reached between the inside and the outside. At this point both systems will be at the same temperature and until one of the systems change they will remain there. Your assumption that there is no radiation to heat the inside of the box effectively means you assume the exterior to be at absolute 0.
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Predicting Polarity of capacitor in the given diagram I have to determine the polarity of the Capacitor in the given diagram.
I Approached the problem as follows:
When Magnet 1 moves with its North pole towards the coil, emf is induced in the coil as the magnetic flux through the coil changes. So, when seeing from the left hand side (i.e. from magnet 1) the direction of induced current appears to be Anticlockwise. Though, on seeing from the left hand side, the South pole of magnet is coming towards, according to Lenz's Law the coil will behave like a South pole, thus the direction of current is Clockwise. I am stuck at this point. How shall I proceed?
My textbook explained it this way which I did not understand:
The direction of induced current when seen from the left hand side is Anticlockwise, and its direction is Clockwise when seen from right hand side.Thus, direction of induced current is in Clockwise sense (why?) . This implies Plate A is positive plate and point B is negative one.
Please Help.
| The best way to think about this is imagine the capacitor plates A and B to be behind the paper plane and then think about the direction of current induced. Observe it from the right magnet's side whose South Pole is approaching so the face of coil facing that South Pole will itself produce clockwise current in the coil. So the current flows from high potential A to low potential B.
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What is the dyad corresponding to a stress tensor? (As I understand it ... qualifies every sentence in what follows).. a stress tensor is a rank 2 tensor that maps a unit vector normal to a surface to the stress (or traction) vector corresponding to that surface. A rank 2 tensor can be represented by a 3x3 matrix, and that matrix maps the components of the unit vector to the components of a stress (or traction) vector.
A rank 2 tensor can be written as a dyad, that is, the vector dyadic product of two vectors. Is there a geometric interpretation of the two vectors making up the dyad corresponding to the stress tensor?
Related question - the product of a dyad $UV$ and a vector $D$, say $UV$ dot $D$, corresponds to the matrix product of the matrix representing the dyad and the vector $D$, and is always a vector that equals $sU$ where $s$ is a scalar, so the stress (or traction) for any surface at a point always points in the same direction. T or F?
| A rank-2 tensor is a linear combination of dyadic products, simply because the space of all such tensors is spanned by the dyadic products of the basis vectors of the underlying vector space. Each dyadic product is also known as a rank-1 operator, where rank here refers to the matrix rank rather than the order of the tensor. On inner product vector spaces they are usually denoted as
$$\theta_{x,y}(z):=(y,z)x$$
but when the product is between a vector and a covector one can replace the inner product with the natural pairing between the vector space and its dual.
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If a bullet is fired vertically upwards, when it comes back does it fall to the same spot? What I'm basically asking is that if a body is projected with sufficiently high velocity so that it doesn't escape from the earth's gravitational field but reaches an appreciable height with respect to the radius of the earth, then when it comes back will it land on the same spot from which it was fired? You can neglect drag force and winds but do consider the rotation of the earth.
Basically what has to be considered is that the net force acts towards the centre of the earth and so I tried conserving angular momentum. That shows that the angular velocity of the object will decrease with increasing height above the earth. So basically the object moves with smaller angular velocity for some time in it's path.
That led me to believe that when the object finally lands back on earth it wouldn't do so at the place from which it was projected. Am I wrong?
| Even if I ignore wind and the drag forces and only consider the rotation of the earth the bullet will not hit the ground at the same place from where it was projected.
There will be Coriolis effect.
Coriolis effect: The Coriolis effect is a deflection of moving objects when the motion is described relative to a rotating reference frame.
I suggest this website,here is a derivation of the deflection due to Coriolis effect of a freely falling body.
http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node58.html
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Why is most probable speed not equal to rms speed for an ideal gas? The rms speed of ideal gas is $$\mathit{v_{rms}} = \sqrt{\dfrac{3RT}{M}}.$$
The most probable speed is the speed where $\dfrac{dP(\mathit
{v})}{dv} =0$ where $P(\mathit{v})$ is the probability distibution. Solving for $\mathit{v}$, we get $$ \mathit{v_p} = \sqrt{\dfrac{2RT}{M}}.$$
Now, $$\mathit{v_p} \neq \mathit{v_{rms}}.$$ Why? Why is it so?
| We're used to thinking of "most probable" and "mean value" as the same thing, but it need not be so. It's worth remembering that the "expectation value" of a six sided die is 3.5, but this is not a very probable result. You might object that this is due to discrete effects, but consider this example: you have two identical Gaussians, with width $\sigma$, but they are separated. One has mean value $m_1$ and the other has mean value $m_2 = m_1 + \delta$. If they're identical and we average between them, we get an expectation value of $(m_1 + m_2) /2 = m_1 + \delta/2$. But $\delta$ could be quite large, in particular perhaps the Gaussians are very separated $\delta >> \sigma$. Then the mean value could occur in a point with arbitrarily small probability of actually being selected!
So as a general principle, the most probable value of a distribution and the average value need not be together. Does that help, or would you rather talk more directly about Maxwell-Boltzmann distributions (of atomic velocity)?
| {
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Why are solar panels blue, rather than black, when black absorbs more light? This is an image of a solar panel array, courtesy of Wikipedia.
Some of these look rather black, but most of them are blue.
As far as I know, solar panels work by absorbing "light energy", and then converting this to "electrical energy". Some of the energy is also converted into "heat energy", as is natural; things put into sunlight will warm up. Lastly, some of the "light energy" will get reflected as "light energy". But only on specific wavelengths. That's how we can see colors...
Now, black objects reflect less light than blue objects. So, given a certain amount of light denoted by $x$, it should hold true that blue.reflected(x) > black.reflected(x). Inversed, it should hold that (black.heat(x) + black.electrical(x)) > (blue.heat(x) + blue.electrical(x)). Basically, because there's less light reflected, more energy is absorbed.
So if a black object (say, a black solar panel) absorbs more energy than a blue object (like a blue solar panel), why are blue solar panels still in use? Why aren't solar panels black, as to absorb the maximum amount of energy from the light?
| The colour you're seeing is from the very small fraction of light that the panels are reflecting. The vast majority of light is being absorbed to generate electricity.
Why some of the panels appear slightly blue while others don't I don't know. Presumably there must be small differences in the manufacturing process. The absorptance of solar panels does fall off at the extreme blue end of the spectrum, so you would expect the reflected light to have a blue tinge. A quick Google found this article that includes a typical absorption spectrum:
| {
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Distribution of dark matter in galactic halos Often dark matter around galaxies is referred to as a 'halo'. I've seen the galactic rotation curves, but I'm having trouble visualizing how the dark matter is distributed for a typical rotating galaxy.
I'm familiar with the expected relation $v \approx \frac{1}{\sqrt{r}}$ for the orbital speeds at distance $r$ from the center.
I simply cannot imagine what the distribution of additional gravitational sources should look like to make $v$ a constant (even ignoring the center).
Is there a function with respect to $r$ that can describe the distribution of dark matter in galactic halos?
Intuitively it seems at odds that adding more gravitational sources (anywhere) would increase the speeds of outer objects more so than inner objects; the reasoning for this is not clear to me.
Why / how does adding in additional sources of gravity allow for faster orbital velocities farther from the center of a galaxy?
|
Is there a function with respect to r that can describe the distribution of dark matter in galactic halos?
Yes, it is called the NFW-profile and it looks like this:
$$\rho_{(r)}=\frac{\rho_0}{\frac{r}{r_s}\left(1+\frac{r}{r_s}\right)^2}$$
where $\rho_{(r)}$ is the dark matter density inside the radius $r$, and $\rho_0$ and the scale-radius, $r_s$ are different for different galaxy types and sizes.
To integrate the mass inside the radius, $M_{(r)}$, you get
$$M_{(r)}=\int_0^{r} \{4\cdot \pi\cdot R^2\cdot \rho_{(R)}\} \, \text{d}R$$
The function for whole clusters is approximated by the function
$$M_{(r)}=4 \cdot \pi \cdot \delta \cdot \rho_c \cdot r_s^3 \int_0^{\frac{r}{r_s}} \frac{u^{2-\mu }}{\left(1+x^v\right)^{\lambda }} \, \text{d}x$$
where $\delta$ is the concentration parameter, $\mu$, $v$ and $\lambda$ some numerical values which may vary from cluster to cluster (for examples see this link) and $\rho_c$ is the critical density of the universe given by the equation
$$\rho_c = \frac{3 \cdot H_0^2}{8 \cdot \pi \cdot G} = 8.47\cdot 10^{-27} \, \text{kg}/\text{m}^3$$
with $H_0$ beeing the Hubble-constant and $G$ Newton's constant.
| {
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Electricity is the movement of electrons or ions? Electricity is the movement of electrons from one atom to another or the movement of charged negative particles (ions)?
| An electric current is the flow of electric charge. But electric charge is not an entity, it is a property that must be 'carried' by a charge carrier.
An electron current, the flow of electrons, contributes to an electric current since the electron 'carries' negative electric charge. However, an electric current is not necessarily an electron current.
The flow of ions (either positively or negatively charged) also contributes to an electric current in, for example, the electrolyte of an electrochemical cell.
As an aside, electricity is not identical to electric current. From the Wikipedia article "Electricity":
Electricity is the set of physical phenomena associated with the
presence and flow of electric charge.
| {
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Why can't we define a unique wavelength for a short wave train?
Here we encounter a strange thing about waves; a very simple thing . . .namely, we cannot define a unique wavelength for a short wave train. Such a wave train does not have a definite wavelength; there is an indefiniteness in the wave number that is related to the finite length of the train. . .$^\text{1}$
Now, why can't we define a unique wavelength for a wave packet?
$^\text{1}$ Lectures on Physics by Feynman, Leighton , Sands.
| I would say that an answer is that length of wave packet and width of spectrum are related by:
$\Delta\omega \Delta t\approx 1$
"Width of spectrum" here is characteristic range of frequencies that signal contains, that is width of Fourier transform of the signal. Infinite sine wave contains only 1 frequency, that is its spectrum/Fourier transform is infinitesimal thin delta function. Any other function will be constructed from many frequencies.
So that wave packet (wave train) in frequency space is defined by a bandwidth, rather than by delta function (like infinite sin wave).
See "wave packet" for more.
Simple example might demonstrate it. Delta function $\delta(t)$ in Fourier domain is constant, that is, contains all wavelengths/frequencies. Rectangular window turns into sinc function that occupies all spectrum as well.
Multiply any sine wave by rectangle (that is, create wave packet limited in time) and in Fourier space you will get convolution of delta function (defined by sine frequency) with sinc, which will give you a shifted sinc (with peak around sine function frequency).
| {
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Finding the appropriate coordinate transformation given two metrics Given the two-dimensional metric
$$ds^2=-r^2dt^2+dr^2$$
How can I find a coordinate transformation such that this metric reduces to the two-dimensional Minkowski metric?
I know that $g_{\mu\nu}=\begin{pmatrix}-r^2&0\\0&1\end{pmatrix}$ (this metric) and $\eta_{\mu\nu}=\begin{pmatrix}-1&0\\0&1\end{pmatrix}$ (Minkowski). Obviously, the matrix transformation is $\begin{pmatrix}1/r^2&0\\0&1\end{pmatrix}g_{\mu\nu}=\eta_{\mu\nu}$, but how is that related to the coordinate transformation itself?
EDIT: would the following transformation be acceptable?
$$r'=r\cosh t$$
$$t'=r\sinh t$$
Such that: $dr'=\cosh t\ dr+r\sinh t\ dt,\quad dt'=\sinh t\ dr+r\cosh t\ dt$
And: $ds'^2=-dt'^2+dr'^2=-r^2dt^2+dr^2=ds^2$
Where we have: $ds'^2=\eta_{\mu\nu}dx^{\mu}dx^{\nu}$ as requested.
Is that correct? Also, is there a formal way of "deriving" the proper change of coordinates (since mine is more of an educated guess)?
| You can also do the following, which may not be as general as you want it, but the idea might be usefull for other problems. You already know that the given metric is Minkowski metric in different coordinates, so look at the null geodesics. In the usual coordinates $(t',x')$ they are given by $x'\pm t'=const$. Then find the null geodesics in the given coordinates by setting the line element to zero i.e. $-t^2dr^2+dt^2=0$. Integrating gives you $re^{\pm t}=const$. You don't even have to prove that these null curves are geodesics, all you need is the transformations that will give the standard Minkowski metric. From here you take the transformations to be
$$x'+t'=re^t$$
$$x'-t'=re^{-t}$$
Solve for $x'$ and $t'$ and check that it works, that $dt'^2-dx'^2=-t^2dr^2+dt^2$.
| {
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White, is it a colour or absence of colours? Our chemistry sir and we had an argument today at the lab, he says that white actually is not a colour, it is the abscence of colour, but we say that it is a colour and we gave the following point to substanciate our point that white is a colour:
When we see an object in red colour, it actually reflects red colour and absorbs all the other colours, in this point of view, a white object reflects all colours which fall on it, so it is a colour.
We do not know who is correct, I am posting this question in hope that I will get the correct answer.
| we define color based on which range of wavelength it reflect. as u said a red object is red because it can only reflect red , so it is not any other color. maybe we could call black colorless, and i think our teacher just used a conventional form of speech (dont u think?)
ask our teacher what he would call the colour of water?
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Frequency dependence of the speed of light in air According to this link, the speed of light of different colors in a medium should be different. But if the refractive index of light in air is 1 then this means that the speed of light in air and vacuum should be the same. Could anyone help me out here?
Thanks
| You have two different concepts intertwined in your question. You begin by asking about the speed of light in a medium varying with color (i.e. wavelength). This phenomenon is called dispersion and it is present in all materials including air. Dispersion shows up in many places in the field of optics, but the case you are probably most familiar with is the separation of white light into its individual colors by a prism.
The other, somewhat unrelated concept that you ask about is the fact that the index of refraction of air is different from that of vacuum. In reality, the index of refraction of air is slightly different than vacuum, but in practice this small difference can often be neglected. Even air has some dispersion (see below), but not enough to separate the light from the sun into distinct colors like a prism.
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Time dependence of the Lagrangian of a free particle? I am working through Landau's book on Classical Mechanics. I understand the logic and physics of isotropy and homogeneity of space-time behind the derivation of the Lagrangian for a free particle, but I am confused regarding its time dependence. When we calculate the action as the integral of the Lagrangian for a wiggly trajectory, the velocity is obviously dependent on time and so should be the Lagrangian. Of course, we extremize the action to find the true trajectory of motion. But this time dependence of Lagrangian for wiggly trajectories is very confusing to me because this indicates to me that Lagrangian is dependent on time for a free particle. Where am I making the mistake here?
Also, when we have a particle in a position dependent potential, the velocity (and kinetic energy) is again dependent on time for any trajectory we choose and even for the true trajectory. But then again we write the velocity as independent of time in the Lagrangian. Why is that so?
| You say " When we calculate the action as the integral of the Lagrangian for a wiggly trajectory, the velocity is obviously dependent on time and so is the Lagrangian".
How exactly is the velocity dependent on time? Before applying the least action principle and find the trajectory of the object, we have a Lagrangian dependent on velocity (through the kinetic energy term), and position and (eventually) time (through the potential energy). We have no idea how the velocity depends on time. There is a continuum of forms of dependences, because there is a continuum of forms of trajectories that the object may follow in principle. This is why, before minimizing the action, we take in the Lagrangian the velocity as a variable in itself.
We don't know the trajectory before minimizing the action, s.t. we have no relationship between velocity and time.
| {
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Mass, energy, and entropy. I have a seemingly simple question about the relation between these three that for some reason doesn't make sense to me. If entropy is the disorder of a system, then a low entropy state is one of higher energy. As we know, mass is energy. From here we must say that the more mass something has, the lower its entropy because the mass can be converted to energy. Why then are black holes, the most massive things known, considered to be of such high entropy?
| A quote from comments:
how is a higher temp and faster moving particles not an increase in available energy and therefore a decrease in entropy?
Maybe higher temperature means more available energy and more unavailable energy! It definitely sometimes means that.
We have devices that decrease entropy of heat energy, in other words they make heat energy more "available". Heat pumps are those devices. They pump heat to higher temperature, in other words they make heat less entropic.
We also have devices that produce heat energy and entropy, Stoves and radiators are those kind of devices.
So here's advice to OP: Do not make too general conclusions, as entropy can be removed from heat energy, which will cause a rise of temperature. But also more heat and more entropy can be added to heat energy, which will cause a rise of temperature.
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A Proposed Improvement to the Diet Coke and Mentos Experiment I am sure most of you are aware of the Diet Coke and Mentos craze - put a few mentos in a bottle of diet coke and whoosh!
Equally, I am sure most of you are aware that this occurs because the gas is able to form bubbles in rapid succession at 'nucleation sites' - microscopic pits on the surface of the mentos.
It has been subsequently shown that using rock salt can extend the range of such rockets because of the high porosity, hence surface area, compared to mentos.
So, this got me thinking... could I used activated carbon instead of the salt/mentos?
It has a surface area of 500m2 per gram as measured by gas adsorption (funny that?) so to me it seems like it could have some pretty dramatic effects!
I haven't seen anything on the internet which indicated anyone has tried this?! Is there a reason why? Has anyone here done it? If not, it would be great if people could give it a go and post some results! I'm going to give this one ago myself and see what happens.
| Yes you can :) How did it work out for you?
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Spectral lines and QM In the various presentations I've seen so far in atomic physics of series such as the Balmer series, the wavelength of each spectral line is definite - but in QM, free particles have no definite energy if I understand correctly: none of the photons in a beam of photons that would interact with the electron of an hydrogen atom would carry an energy corresponding exactly to a wavelength of 656.3 nm for example (I'm saying "none of the photons" because if we have a distribution of energies for the photons in the beam, in the limit, there is no particle at a single discrete value of energy). What happens exactly? Should Heisenberg uncertainty be taken into account? Is there a quantum mechanical account of photon absorption by atoms?
| Of course the photon will have an amount of energy and that is radiation energy which if defined by Planck's equation. Radiation energy is also related directly to wavelength,frequency and wavelength number.The higher the energy, the less is the wavelength ( all this from Planck's equation). What Heisenberg equation is about, does not include the energy.It includes only the position and the rate of electrons ( you can't know both of them at the same time ). The absorbtion of radiaton by atoms is very well defined by absorbance formula A=abc.I hope I understood the question correct.
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Does the Earth revolve around the Sun? I am aware of this Phys.SE question: Why do we say that the earth moves around the sun? but I don't think this is a duplicate.
In a binary star system, where the masses of the 2 stars are not so different from each other, can we say that each star revolves around the other?
If yes, then couldn't the Sun-Earth system be an extreme case of such a system?
Therefore, strictly speaking, can we argue that the Sun revolves a tiny bit around the Earth as well?
| Both the the Earth and the Sun orbit around the solar system barycentre. This is defined as the centre of mass of all the bodies in the solar system. Because the Sun contains the vast majority of the mass of the solar system then the barycentre is very close to the Sun. The picture below, from the wikipedia entry on the solar system barycentre, has the barycentre stationary in the middle of the picture, and illustrates schematically the situation when one body is a lot more massive than the other (though is not to scale for the Earth-Sun system!).
The Sun executes a complicated orbit around this point (also illustrated here), pulled by the motions of, primarily Jupiter, but all the other planets also make a smaller contribution.
It is this "reflex motion" of a star, caused by planets in its solar system, that allows the detection of exoplanets by the doppler method. The orbit of the Sun around the barycentre would cause it to appear to a distant observer to be periodically redshifted and blue shifted with an amplitude of about 13 m/s, with a period of around 12 years (the orbital period of Jupiter). In isolation, the Earth would only cause the Sun to orbit the barycentre of the Earth-Sun system with a speed of 7 cm/s (which is one reason that finding Earth-like exoplanets is very difficult).
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If the Earth's atmopsphere spins with the earth due to friction, why is there no horizontal spiralling drag? Imagine a bucket of paint with a spinning ball in it. The paint would form a spiral and would not all move in synchronous movement with the ball.
To clairfy - In order for the Earth's atmosphere to appear to us to be still (as on a windless day) the upper atmosphere must be moving faster than the lower atmosphere - as it the case with any rotating spehere - the outer layers are moving faster than the inner layers.
What force is causing the upper atmosphere to move faster than the lower atmosphere?
If the atmosphere is being rotated soley by friction, then at best the upper layers would move at the same speed as the lower layers, thereby causing a spiralling effect.
We do not see this effect. Why?
| The same happens with the air around the earth, this phenomenon is called the Coriolis effect, and it affects our atmosphere. You also have to keep in mind that there are a lot of other variables to take into account when talking about atmosphere, like the angle of our axis towards the sun, etc...
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Why does Li+ ion move to cathode in Li-ion battery? When Li-ion battery is discharged or being used, the positive lithium (Li+) ions move from anode to cathode through the electrolyte. Meanwhile the electrons move in the same direction through the external circuit. Why does this happen? I mean, why does a Li+ ion get attracted to the positive electrode (cathode)?
|
Why does Li+ ion attract to the positive electrode (cathode)?
Let's first see how we define the Cathode and Anode based on electron movement.
A cathode is the electrode from which a conventional current leaves a polarised electrical device.
Now, importantly, Cathode polarity with respect to the anode can be positive or negative; it depends on how the device operates. Although positively charged cations always move towards the cathode (hence their name) and negatively charged anions move away from it, cathode polarity depends on the device type, and can even vary according to the operating mode. In a device which provides power, the cathode is positive.
Does it happens because of Redox Reactions?
Yes, you're correct. Another way to understand, the positive nature of Cathode, or build up of positive ions is, by considering the redox reactions occurring.
The anode is the electrode where the oxidation reaction
$Red \rightarrow Ox + e^{-}$ takes place while the cathode is the electrode where the reduction reaction
$Ox+e^- \rightarrow Red$
takes place.
That's how cathode and anode are defined.
Now at the cathode you have the reduction reaction which consumes electrons (leaving behind positive (metal) ions at the electrode) and thus leads to a build-up of positive charge in the course of the reaction until electrochemical equilibrium is reached. Thus this explains why positive ions move toward the cathode in a Lithium ion battery or more generally a galvanic cell.
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Is there a curve for which a particle restricted to move within it under the gravitational force will always exhibit a pure harmonic motion? A simple pendulum, for example, is not isochronous for large amplitudes (that is, the frequency will depend on the amplitude). So a particle confined in a circumference will not always exhibit a pure harmonic motion.
Is there a curve for which the frequency will always be independent of the amplitude?
| If you built a surface such that its height was proportional to a horizontal coordinate $x^2$, $h = k x^2$, then the potential energy at point x would be $mgh = m g k x^2$, which is a harmonic potential. In other words: yes, the curve exists and it's a parabola. This assumes uniform $g$, though, I guess if you want to be a stickler you could note that if the parabola were really big then that assumption wouldn't hold.
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Approximations of the kind $x \ll y$ I have an expression for a force due to charged particle given as
$$F=\frac{kQq}{2L}\left(\frac{1}{\sqrt{R^2+(H+L)^2}}-\frac{1}{\sqrt{R^2+(H-L)^2}}\right) \tag{1}$$ where $R$, $L$ and $H$ are distance quantities.
Now I want to check what happens when:
*
*$H\gg R,L$
*$R,H\ll L$
How can I work out the approximation of this force?
Do I have to write it slightly different into form (2) to get it right?
$$ ~F=\frac{kqQ}{2LR}\left(\frac{-1}{\sqrt{1+\left(\dfrac{H+L}{R}\right)^2}}+\frac{1}{\sqrt{1+\left(\dfrac{H-L}{R}\right)^2}}\right) \tag{2}$$ (which is the same expression just written out differently). Any explain about this subject would be very helpful.
| Your final expression is off by a minus sign, and is not what you want. For #1, $H$ is large, so I'd factor $H$ out. For #2, $L$ is large, so I'd factor $L$ out.
When you factor out a large thing, the remaining things are either numbers (which are what they are) or small things. And when something is small you can approximate it by comparing it to other things.
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100°C = 100 K =? I'm in first year. Our class is in lesson " Heat and Thermodynamics". While solving a numerical problem of a reversible engine he told us that 100 degree Celsius is equal to 100 kelvin. I inquired but could not get satisfactory answer. Pleas help me understand it.
Here is the numerical, please consider it: A reversibe engine works between two temperatures whose difference is 100c. If it absorbs 746J of heat from the source and rejects 546J to the sink, calculate the temperature of the source and the sink. Ans (100°C, 0°C)
| What your instructor has said is right. While taking Temperature differences (delta T), etc. we take x Celcius = x Kelvin, so we can take any delta T value in Celcius in our numericals interchangeably.
| {
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Does the wave/particle duality exist across the entire electromagnetic spectrum? Does the wave/particle duality exist across the entire electromagnetic spectrum?
If theory says so, then to what extent have physicists confirmed by experimental means?
| Wave/particle duality is present across all particles, an equation to show this is:
$$
p=h/\lambda
$$
where p is the momentum of the "particle", lambda is the wavelength and h is Planck's constant. From this it can be seen that anything can be considered a wave, but they must have a very small mass to have a wavelength that isn't negligible.
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What do spacelike, timelike and lightlike spacetime interval really mean? Suppose we have two events $(x_1,y_1,z_1,t_1)$ and $(x_2,y_2,z_2,t_2)$. Then we can define
$$\Delta s^2 = -(c\Delta t)^2 + \Delta x^2 + \Delta y^2 + \Delta z^2,$$
which is called the spacetime interval. The first event occurs at the point with coordinates $(x_1,y_1,z_1)$ and the second at the point with coordinates $(x_2,y_2,z_2)$ which implies that the quantity
$$r^2 = \Delta x^2+\Delta y^2+\Delta z^2$$
is the square of the separation between the points where the events occur. In that case the spacetime interval becomes $\Delta s^2 = r^2 - c^2\Delta t^2$. The first event occurs at time $t_1$ and the second at time $t_2$ so that $c\Delta t$ is the distance light travels on that interval of time.
In that case, $\Delta s^2$ seems to be comparing the distance light travels between the occurrence of the events with their spatial separation. We now have the following definitions:
*
*If $\Delta s^2 <0$, then $r^2 < c^2\Delta t^2$ and the spatial separation is less than the distance light travels and the interval is called timelike.
*If $\Delta s^2 = 0$, then $r^2 = c^2\Delta t^2$ and the spatial separation is equal to the distance light travels and the interval is called lightlike.
*If $\Delta s^2 >0$, then $r^2 > c^2\Delta t^2$ and the spatial separation is greater than the distance light travels and the interval is called spacelike.
These are just mathematical definitions. What, however, is the physical intuition behind them? What does an interval being timelike, lightlike or spacelike mean?
| Timelike is when an event is inside the lightcone (as you have mentioned) and as a result, one event CAN affect the other event (there can exist a causality between the two events. E.g. lets say there are two events, where I shoot a laser and another event where someone gets hit by a laser. If they are timelike seperated then the laser that hit the dud could have been from me).
Spacelike is when the two events are outside lightcone (as you have also mentioned) and as a result, one event CANNOT affect the other event. (For the previous example, it is impossible that my laser hit the dude and killed him, so I can safely conclude that someone else shot the laser to kill him.)
Lightlike is a special case which is like inbetween the two. For me to kill him, all the interval inside must have been vaccuum.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Can net torque $\sum_i\mathbf r_i\times\mathbf F_i$ be expessed as $\mathbf r\times$ (net force) for some $\mathbf r$?
Let $\mathbf F_i$ be forces each of which is applied on $\mathbf r_i$ of a rigid body. Then is there a position vector $\mathbf r$ that satisfies
$$\displaystyle\sum_i\mathbf r_i\times\mathbf F_i =\mathbf r\times\displaystyle\sum_i\mathbf F_i~ ? \tag{1}$$
Well, what I get is by letting $\mathbf r_i=(r_{i,x},r_{i,y},r_{i,z})$, $\mathbf F_i=(F_{i,x},F_{i,y},F_{i,z})$, $\mathbf r=(r_x,r_y,r_z)$ is
$$\begin{bmatrix}
0 & \sum F_{i,z} & -\sum F_{i,y}\\ -\sum F_{i,z} & 0 & \sum F_{i,x}\\\sum F_{i,y} & -\sum F_{i,x} & 0 \end{bmatrix}
\begin{bmatrix} r_x\\ r_y\\r_z \end{bmatrix}=
\begin{bmatrix} \sum (r_{i,y}F_{i,z}-r_{i,z}F_{i,y})\\\sum (r_{i,z}F_{i,x}-r_{i,x}F_{i,z})\\\sum (r_{i,x}F_{i,y}-r_{i,y}F_{i,x}) \end{bmatrix}, \tag{2}$$
and $$\begin{vmatrix}
0 & \sum F_{i,z} & -\sum F_{i,y}\\ -\sum F_{i,z} & 0 & \sum F_{i,x}\\\sum F_{i,y} & -\sum F_{i,x} & 0 \end{vmatrix}=0. \tag{3}$$
Since the matrix is singular, the system might not have a unique solution.
So is it the case that generally such $\mathbf r$ may not be unique (or even nonexistant)? If so what is the criterion for uniqueness of $\mathbf r$?
| Let me introduce the notation
$$\sum F_{i,x} = F_x, \ \ \ \sum F_{i,y} = F_y, \ \ \ \sum F_{i,z} = F_z, \tag{i}$$
Since the determinant is zero, there may be indeed, no solution of the system. But if the system of equations has a solution, recall that the body doesn't rotate around a point, but around an axis. So, your $\vec r$ is bound to be on an direction perpendicular to the resultant force $\vec F$ and on the rotation axis. A direction in the space may be described by a simple system of equations, e.g.
$$\begin{cases} {y = a x + b} \\ {z = a' x + b'} \end{cases}. \tag{ii}$$
Now, for simplicity let's introduce one more notation
$$\begin{cases} {\sum (r_{i,y}F_{i,z}-r_{i,z}F_{i,y}) = \tau_x} \\ {\sum (r_{i,z}F_{i,x}-r_{i,x}F_{i,z}) = \tau_y} \\ {\sum (r_{i,x}F_{i,y}-r_{i,y}F_{i,x}) = \tau_z} \end{cases}, \tag{iii}$$
From your system $(2)$ and using the notations $\text {(i)}$ and $\text {(iii)}$ one obtains three equations, two of which of the form $\text {(ii)}$.
Well, for this system of equations to have a solution, between the constants $\tau_x, \tau_y, \tau_z$ has to be a linear and homogeneous relation. From now on the job is yours. Find the relation.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What is the mass of a sphere? A solid sphere of mass M is rotating along an axis.
We can consider it as a collection of large number of point masses, every point mass is moving with respect to center of mass with velocity which depends on its radius from rotating axis.
Then, according to relativity, the mass of every point increases and consequently the mass of the sphere increases.
But if we consider the overall sphere, it is not moving at all and its mass remains the same. which produces a contradiction.
Please tell me where I am wrong.
| In relativity (but often not Newtonian physics) there is a huge difference between a static mass distribution and a stationary one. A static distribution is one in which there is no velocity, whereas a stationary one is defined by looking the same at any given time. All static distributions are stationary, but your rotating sphere is an example of a stationary distribution that is not static.
The fact is, the rotating sphere has additional angular momentum and energy. Thus it does have a greater "relativistic mass" (i.e. total energy minus rest mass energy). In fact, there's nothing terribly relativistic about the misconception here; a rotating Newtonian sphere has a kinetic energy, even though the mass isn't going anywhere.
For concreteness, your approach of looking at each point mass works (ignoring material stresses). If a uniform sphere has rest mass $M$ and radius $R$, its rest mass density will be $\rho = 3M/4\pi R^3$. If it is rotating with angular velocity $\omega < c/R$, then each point with colatitude $\theta$ has velocity $R\omega \sin\theta$, and so the total relativistic mass is
$$ M_\mathrm{rel} = \int\limits_\text{sphere} \frac{\rho}{\sqrt{1-(R\omega/c)^2\sin^2\!\theta}} \, \mathrm{d}V = \frac{Mc}{2R\omega} \log\left(\frac{1+R\omega/c}{1-R\omega/c}\right) = M \left(1 + \mathcal{O}\left((R\omega/c)^2\right)\right). $$
This diverges as $R\omega \to c$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Hula-hoop physics What are the important parameters to consider when trying to keep a hula-hoop in the air by spinning it around the waist?
For instance, when shopping for hula-hoops, one can commonly find its diameter and weight, but how do they affect the actual ease of keeping it in the air? What's the physics behind it?
| The ratio of the circumference of the body and the hulahoop (HH) is an important parameter. The greater the ratio, the more the HH will rotate - and the more it rotates, the greater its angular momentum. When you have an object spinning with its axis vertically, and apply a torque, it will precess - instead of falling in the direction of the torque, its angle of rotation changes. As it does so, the angle of the HH relative to the body changes - and in fact, it will "climb" a little bit on the body.
That is the key of HH physics, I think: the axis of rotation is precessing. As long as the ring is "thin", the angular momentum scales with $r^3$ - assuming that mass is linear with size. The torque, on the other hand, scales with $r^2$. That means that the rate of precession will be slower for larger HHs - and that a larger hoop is easier. But also, the ratio of diameters of body and hoop will change the rate of rotation. The faster you rotate, the less far the angle of the hoop needs to tip in order to maintain precession.
| {
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"url": "https://physics.stackexchange.com/questions/169975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Biot-Savart law from relativity Few days ago I came by a derivation of STR from Bio-Savart law. Since then I have been trying to derive Bio-Savart law from STR. The derivation mentioned previously used two parallel current-carrying wires of unit length and used the change of their current density to achieve Lorentz contraction (for the entire derivation, go to http://www.andrijar.com/sr/sr.htm). Using the same idea, I have been able to derive ampere force law(F=2kII'/r) from STR. So,
Is it possible to derive Biot-Savart law from ampere force law? If yes, how?
(N.B. Please don't use Maxwell's equations for I have not yet learnt them)
| Consider a positive point charge moving in positive $z$ direction at a constant velocity, at that instant when it's at the origin. The magnetic-field lines generated in plane $z=a$ are closed loops, further we observe circular symmetry in the problem about $z$-axis, henceforth the loops are circles.
A image to develop a idea of how we would be going to proceed:
Let's apply Ampere's displacement law to calculate magnetic field. Intuitively we can think, as the particle moves towards the plane $z=a$, the electric flux passing through a surface bounded by any circle centered on the z-asix in that plane will increase.
So, starting with Ampere's Law we have:
$$\oint \vec{B}.d\vec{I} = \mu_o \epsilon_o \frac{d\phi}{dt}$$
We integrate the left hand-side around a circle of radius $b$ on $z$-axis in plane $z=a$, and this is our Amperian Loop.
so,
$$ \oint \vec{B}.d\vec{I} = 2 \pi bB$$
To calculate electric flux $\phi$ enclosed by the circle of radius $b$, we select a spherical surface, of surface area $A$, which is bounded by the circle, is symmetric about z-axis, and has a radius $r$. So it follows from the understanding, we have:
$$ \phi = \int \vec{E}.d\vec{A} = EA$$, where $$E= \frac{q}{4 \pi \epsilon_o r^{2}}$$
The area $A$ can be obtained using spherical, polar coordinates:
$$ A= r^{2} \int_0^2\pi d\phi \int_0^\theta \sin x dx = 2 \pi r^2(1-\cos \theta)$$
$$\cos \theta= \frac{z}{\sqrt{z^2+x^2}}$$
Using/combining the above three equations we can have:
$$\phi = \frac{q}{2 \epsilon_o}(1-\cos \theta)$$
Differentiating with time gives,
$$\frac{d\phi}{dt} = -\frac{q}{2 \epsilon_o} \frac{d \cos \theta}{dt}$$ where, $$\frac{d \cos \theta}{dt} = \frac{d \cos\theta}{dz} \frac{dz}{dt}$$
In the above equation, $z$ is the distance between the particle, which is moving in the positive $z$ direction, and the Amperian Loop, which is fixed in $z=a$ plane. Since $z$ is decreasing with time,
$$\frac{dz}{dt} = -v$$ where v is the speed of the particle. Now,
$$\frac{d \cos \theta}{dz}=\frac{y^2}{r^3}$$, where $r=\sqrt{z^2+y^2}$
Combining the first two and last three equations, we arrive at our awaited result as,
$$B= \frac{\mu_o}{4\pi} \frac{qv \sin \theta}{r^2}$$, which can be vectorially rewritten as, $$\vec{B} = \frac{\mu_o}{4\pi} \frac{q \vec{v} \times \vec{r}}{r^3}$$
Hence, we are done! :)
| {
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"url": "https://physics.stackexchange.com/questions/170095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Parity transformation is proper orthochronous? In 3+1 dimensional spacetime the parity transformation is $$P^\mu_{\;\,\nu}=\begin{pmatrix}+1&&&\\&-1&&\\&&-1&\\&&&-1\end{pmatrix}.$$ This is orthochronous but not proper and thus is not the result of compounding infinitesimals.
However, in $(2n)+1$ dimensional spacetime, the parity transformation will have determinant one and thus be proper orthochronous.
My (naive) question: what physical consequences does this have in evenly spatial dimensional spacetimes? I could not find any references to this seemingly profound phenomenon after preliminary Googling.
| Parity and Time reversal are by definition elements of the full lorentz group with which you need to supplement the proper orthochronous subgroup in order to be able to span the entire group.
As noted, the proper way to define parity in any dimension is to flip one of the spatial axes. In even space-time dimensions it so happens that flipping all spatial coordinates and flipping one are related through some sequence of rotations, as should be the case for any two improper transformations, since they both have det$=-1$ and should be connected.
For odd space-time dimensions, flipping all spatial dimension is actually just a rotation (or a sequence thereof). For example take $SO(1,2)$, then flipping all spatial coordinates
$$ \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{array}\right)$$
is equivalent to a rotation about the only axis by and angle of $\phi =\pi$ radians, since a rotation is
$$ \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \text{cos }\phi & -\text{sin }\phi \\ 0 & \text{sin }\phi & \text{cos }\phi \end{array}\right)$$
However either
$$ \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \pm 1 & 0 \\ 0 & 0 & \mp 1\end{array}\right)$$
cannot be achieved by a rotation, and has the correct determinant.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/170180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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A cup of water in ZERO gravity What will happen if I try to pour a cup of water in zero gravity, into another empty cup? Will the water come out of the cup? The adhesive force between the water molecules and the interior of the cup should prevent the water from coming out. Is it correct? Or is there something more to to it?
|
The adhesive force between the water molecules and the interior of the
cup should...
Even in absence of adhesive force, the water will never move in 0-gravity, because there is no up nor down, no force is acting on it.
You can clearly see in this video at 1:15 that in order to get the water out of a plastic cup you have to tap it on the bottom
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/170338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Deriving the equations for a moving inertial reference frame I assume $c=1$ in the following derivation:
In order to derive the equations for a moving inertial reference frame, I immediately wrote down the following:
$$ x'=Ax+Bt, \tag{1}$$
$$t'= Dx+Et. \tag{2}$$
In order to solve it I would need 4 independent equations. Here they are:
*
*Since the speed of light is constant in all reference frames, it follows that if $x = t$, also $x' = t'$, therefore
$$At+Bt=(Dt+Et) \overset{(x=t)}{\implies} A+B=D+E. \tag{3}$$
*I can reverse the reference frame and the physics should be the same hence
$$x=0 \implies x'=-vt' \implies B=-Ev. \tag{4}$$
*Finding the $x'$ component of the point $A(\frac{1}{1-v},\frac{v}{1-v})$ gives:
$$Av+B= -Av. \tag{5}$$
*Finally finding the $t'$ component of $A$ gives (I'll do this one explicitly):
$$t'= \frac{D+Ev}{1-v}. \tag{6}$$
From the diagram one can read off using the Pythagorean theorem that:
$$t'= \sqrt{\left( \frac{1}{1-v} \right)^2 +\left( \frac{v}{1-v} \right)^2 } = \frac{\sqrt{1+v^2}}{1-v}$$
$$\implies D+Ev= \sqrt{1+v^2}. \tag{7}$$
From these equations one easily arrives at the desired result ie
$$x'=\frac{x-vt}{\sqrt{1-v^2}} \; \text{and} \; t'=\frac{t-vx}{\sqrt{1-v^2}} \tag{8}$$
All this seems to be correct. However considering the equation $(6)$ and putting back the $c$'s in it one arrives at the equation
$$ D+Ev= \sqrt{1+v^2} \quad (!) \tag{9}$$
First of all this dimensionally doesn't make sense. Secondly if you calculate and find the coefficients you don't get the correct answer. Intuitively I know that this equation has to be $D+Ev= \sqrt{1+v^2/c^2}$ so that everything works perfectly but I don't know why this has to be so and I cannot show it by reasoning physically. I fell in my guts that there is something fishy about using Pythagorean theorem but I don't know what went wrong exactly. If I just say that the use of Pythagorean is wrong, then I cannot explain why it gives the correct answer when using $c=1$. Such a coincidence seems to be highly unlikely.
Edit: I've made a major typo in the diagram you should swap $x=0$ with $t=0$ and $x'=0$ with $t'=0$!
| I am sorry, but I have no time to find your mistake for you. However, I can give you a tip that I am 100% sure would work.
Your mistake can be very easily found by dimensional analysis. Start from the beginning of your derivation and check all formulas. The first one which makes no sense due to dimensions is wrong.
You can just say that because $c$ is one in your calculations, you define $v$ do be what is actually $v/c$. That would give correct results by simple substitution $v \rightarrow v / c$, but I am under impression that is not good enough for you, though I can't understand why.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Heuristics for the Hawking mass I have the following definition of Hawking Mass. Given a spacelike 2-surface $S$ embedded in a 3+1-dimensional Lorentzian Manifold $L$,
$$
M(S) := \sqrt{ \frac{\text{Area}(S)}{16 \pi}} \left(1- \frac 1 {16 \pi}\int_S H^2 d_{\sigma_S}\right),
$$
where $\sigma_s$ is the induced volume form on the surface $S$, $H$ is the mean curvature of the immersion $S \to L$.
From a heuristic point of view, I see how the second term can arise, as it encodes for an average of the rate of change of the area of $S$ when 'moved' in null directions. I would like to understand a formal way of saying this, if any.
My ultimate aim is to understand what the Hawking mass measures. Also, I'd like to understand if there is a relation with the ADM mass.
| A small note on the definition. It should be
$$
M(S) := \sqrt{ \frac{\text{Area}(S)}{16 \pi}} \left(1- \frac 1 {16 \pi}\int_S \theta^-\theta^+ d_{\sigma_S}\right),
$$
where $\theta^\pm$ are the divergences along the two null directions. It is equal to what you have written if the 2-surface $S$ lies in a space-like 3D submanifold with vanishing extrinsic curvature.
The Hawking mass is one of the quasilocal energies and it measures the energy within $S$. But there are a number of qualification to be made and I am not sure how well it is understood. It is certainly above my head.
If the space-time is flat at spatial infinity and the 2-surface is a coordinate sphere, then as the radius goes to infinity the Hawking mass will approach the ADM mass.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/170549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Do photons with a frequency of less than 1 Hz exist? A photon with a frequency of less than 1 Hz would have an energy below
$$
E = h\nu < 6.626×10^{−34} \;\rm J
$$
which would be less than the value of Planck's constant. Do photons with such a low energy exist and how could they be detected? Or does Planck's constant give a limit on the amount of energy that is necessary to create a single photon?
| The shortest answer for why such photons "exist" is that whether a given photon qualifies depends on your rest frame. Take your favourite high-frequency photon in the universe, say of frequency $n$ Hertz in your rest frame. With a Lorentz boost $-\beta$, I multiply this by $\gamma (1-\beta)=\cosh\phi-\sinh\phi=e^(-\phi)$ with $\phi$ the rapidity $\phi=\mathrm{artanh}\beta$. Setting $\phi>\ln n$ (equivalently, $\beta>\frac{1-n^{-2}}{1+n^{-2}}$) does the job.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/170828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Average Energy of a coherent state The question is relating to a previous problem concerning the harmonic oscillator.
Determine the average energy < E > in a coherent state |alpha>.
From my understanding the expectation of the energy would simply mean calculating
< alpha| H | alpha>. Where < H > = < T > + < V > for a harmonic oscillator. Is my approach valid?
I am confused as to what role the coherent state plays in this and whether this really is the average energy.
|
From my understanding the expectation of the energy would simply mean
calculating < alpha| H | alpha>. Where < H > = < T > + < V > for a
harmonic oscillator. Is my approach valid?
Yes. This is a valid approach. This is what is meant by the expectation value of the energy.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/170920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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