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Is the classification of particles into matter and anti-matter arbitrary? It is well known that every fundamental particle has a corresponding antiparticle, and that – except for particles which are their own antiparticle – for practically every pair of particles and antiparticles, one type of particle predominates. We call this particle matter, and its antiparticle is considered to be antimatter.
However, is there any intrinsic difference between matter and antimatter, or are these terms defined based on nothing more than the balance that we so happened to end up with? For example, is there any reason that we should place an up quark and a charm quark in the same category, and not an up quark and a charm antiquark? Is there any reason we should group together up quarks and electrons rather than up quarks and positrons?
| The definitions have a pure historic reasoning which has to do with the order of verification of the existence of particles. The names stuck in the scientific community because a radical redefinition of names to suit language would be impractical and confusing.
As for the grouping of particles, it has strong mathematical grounding. The particles are grouped according to the similarities they possess between each other, and the mathematics are described by Group Theory.
| {
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Why does mechanical equilibrium depend only on potential energy? As far as I understand, for a system to be considered in equilibrium, the sum of the forces that is applied to it must be $0$:
$\vec F = 0$
which is
$\partial \frac{E_p}{\partial x}\bigg\rvert _{x=x_0} = 0 $
But it only depends on the potential energy, what about kinetic energy?
| Equilibrium here just means that if a particle is placed at the location $x_0 = 0$ with $0$ kinetic energy, it will stay at that location. Imagine another location, say $x_1$, such that $\nabla V(x_1)\not = 0$, if a particle is placed there, it will start moving, if the kinetic energy is zero initially.
A simple example of this is a pendulum. Imagine a bar with ends points labeled $A$ and $B$ suspended from one of its end-points ($A$). If the bar is placed vertically with $B$ lower than $A$, the bar will stay at that configuration forever, unless you add kinetic energy. This is also the case if you place $B$ vertically above $A$. The difference between these two configurations is the type of equilibrium (stable vs unstable), but both of them are equilibrium points.
| {
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Schwarzschild geometry, what is physical meaning of coordinates? A past exam has a question:
For the Schwarzschild metric external to a non-spinning spherical mass, what is the physical significance to the coordinates $t,r,\theta,\phi$?
Not sure how to answer this question, I am thinking there is some obvious canonical answer, but it feels very non-specific.
Is the answer something like, an observer at $r\to \infty$ has that $t$ and $\tau$ are the same, and $r,\theta,\phi$ are all to some extent arbitary?
| Well, since the metric is assymptotically flat, in $\infty$, $t$ and $\tau$ do indeed coincide, so you can view the time coordinate $t$ as the time measured by an inertial observer at infinity.
The radial coordinate $r$ is actually more of an "areal" coordinate. Consider 2-surfaces of constanat $t=T_0$ and $r=R_0$. Then the induced metric on the 2-surfaces are just the spherical metrics $$ ds^2=R_0^2(d\vartheta^2+\sin^2\vartheta d\varphi^2), $$ which implies the area of the 2-surfaces are $$ \text{Area}(t=T_0,r=R_0)=\iint R_0^2\sin\vartheta\ d\vartheta d\varphi=2\pi R_0^2\cdot[-\cos\vartheta]_{0}^{\pi}=4\pi R_0^2, $$ which is of course, the surface area of 2-spheres in euclidean space.
Therefore, we can say that the $r$ coordinate denotes the points occupied by origin-centric spheres whose surface areas are $4\pi r$.
Because the part of the metric that contains the angular coordinates $\vartheta$ and $\varphi$ is the same as the spherical metric in euclidean space, the angular coordinates have the usual meaning.
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Where is a classical computer better than a quantum one? Where is a classical computer better than a quantum computer? Is there any known domain where classical algorithms always beat quantum ones, say, both in terms of time and space complexity?
If yes, could you please give me examples?
If no, could you please provide me with a link to the prove?
| As fs137 answered, a quantum computer can simulate a classical computer, and so from a purely complexity theory perspective, the classical computer is never superior to the quantum in an asymptotic sense (assuming $P\subset BQP$, currently an open question).
However, quantum computers currently operate with very low numbers of qubits (ruling out adiabatic QC like D-Wave) relative to classical computers with classical bits. Thus, we are not currently in a time where quantum computers can work at a scale where these asymptotics take over. Since quantum computers have a very large overhead to perform an operation that's comparatively simple on a classical computer, they have very large constant factors that dominate for small computations. Any single operation that a classical computer can perform, it will likely be much slower on a quantum computer.
With this in mind, classical computers dominate at small numbers of bits from a practical perspective. However, as we begin to scale the number of bits up, and we are solving a problem with a known quantum algorithm that improves upon the best known classical algorithm, we will see that a quantum computer can finish computations faster because it is executing an entirely different algorithm than the classical computer.
| {
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Probability depends on volume dependence of the multiplicity function? The formula for the multiplicity of a monoatomic ideal gas is
$$\qquad \Omega(U,V,N)=f(N)V^NU^{3N/2}$$
where $V$ is the volume of position space, $U$ is the molecule's kinetic energy, $N$ is the number of indistinguishable gas molecules and $f(N)$ is a complicated function of N
And from Schroeder's An Introduction to Thermal Physics:
Sometimes you can calculate probabilities of various arrangements of molecules just by looking at the volume dependence of the multiplicity function. For instance, suppose we want to know the probability of finding the configuration where all the molecules in a container of gas are somewhere in the left half. This arrangement is just a macrostate with the same energy and number of molecules, but half the original volume. Looking at equation we see that replacing $V$ by $V/2$ reduces the multiplicity by a factor of $2^N$. In other words, out of all the allowed microstates, only one in $2^N$ has all the molecules in the left half. Thus the probability of this arrangement is $2^{-N}$
Why do we just simply take $2^{-N}$ as our probability? Shouldn't we consider the macrostate for other volumes as well and make the probability be
$$Pr=\frac{\Omega(U,V/2,N)}{\sum_{V=0}^{100} \Omega(U,V,N)} \ ?$$
| Not really.
You want to calculate the probability to find all the molecules in the left half of the volume.
Since the probability density is
$$\rho(\{q,p\}) = c_N \frac{e^{-\beta H(\{q,p\})}}{\Omega(U,V,N)}\ ,$$
where, $c_N$ is a constant and, to ensure the normalization, $$\Omega(U,V,N ) = c_N\int e^{-\beta H(\{q,p\})} dq^{3N} dp^{3N}$$
what you need to consider is
$$\text{Pr} =\int_{\{q,p\}_{V/2}} \rho (\{q,p\}) dq^{3N} dp^{3N} = c_N \frac{ \int_{\{q,p\}_{V/2}} e^{-\beta H(\{q,p\})} dq^{3N} dp^{3N}}{\Omega(U,V,N)}$$
where $\{q,p\}_{V/2}$ is the set of coordinates such that all the molecules are in the left half of the volume.
You can immediately see that
$$c_N \int_{\{q,p\}_{V/2}} e^{-\beta H(\{q,p\})} dq^{3N} dp^{3N} = \Omega(N,V/2,U)$$
So that you finally obtain the desired result.
| {
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Why are we able to break a vector along it's components or in other words why is it that a vector exists along $x$, $y$ and $z$ axis? Does a 3 d vector exist in three dimensions at once? It seems to me that a vector always changes axis along which it is to fit into the scenario. For example: Electric field in $x-y$ plane when passes through $y-z$ plane only uses it's $x$ component.
|
Does a 3 d vector exist in three dimensions at once?
Absolutely, a general vector is $$\vec r=x \hat i+y\hat j+z\hat k$$ in Cartesian coordinates. This vector has a component in each of the three dimensions and all components are mutually orthogonal. Since the vector $\vec r$ is equal to the sum of these 3 components then it must exist in three dimensions at once.
| {
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A question related to time and motion I have a theory, I think that we cannot travel at speeds faster than light because, as we know,when you travel at speeds near light time passes slowly and that if we go further than light time may pause perhaps and that if time pauses its impossible to have motion because in 0 seconds ( I mean no time has passed) you cannot travel any distance.
I have also another theory, I think that if we go a little high than light speed except pausing time may reverse, and if time reverses an object will never exists in space but continuously go back in time and reach big bang. The object will never exist in space but actually in a TIME dimension ( I know dimension word is wrong but I couldn't think of a word except this).
Can anyone one of these 'theories' be true, even very tiny bit true? Please point out my mistakes. I am just a kid though of 9th grade. But, I really wonder could these be true or not.
| It is a common misconception to believe that if you move close to the speed of light you cannot move because time is slow. You do not experience anything different, other than how the rest of the universe looks to you. Everything will looks shorter and running very slow in time. But you will see the objects past you at high speed. Also, objects that move along you at the same speed will be normally experienced.
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Why can we tear a newspaper neatly one way but not the other way? When I try to tear the newspaper from top to bottom (or bottom to top), it's torn pretty neatly and like a line, but when I try to tear it from side to side, it goes all over the place and tries to lead the tear towards the top or bottom.
Why does this happen?
| Newspaper is made out of cellulose fibres (linear unbranched ones) bonded entangled together.
The fibre structure is anisotropic. The orientation of most of the fibres is along the direction of the movement of the machine.
In the direction of this orientation, it is relatively easier to tear a newspaper because it's just a matter of prying two fibres apart. (without significant tearing of fibres)
In the direction perpendicular to this, fibres have to be broken to tear the paper, and this requires a greater force.
The same asymmetry account for why tears are neat/messy depending on the direction you take.
If a piece of paper was isotropic, with random orientation of fibres, tearing would roughly take the same effort in any direction.
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It's easy to 3d model a room via binocular vision, can it be done with 2 microphones and a beeper? Many algorithms exist to construct a 3d model from two cameras, is it possible to 3d model of a room via a speaker that produces a beep, and two microphones with arbitrarily good hearing skill (ie can sample continuously not in discrete "snapshot"s per second)?
A) assuming all object's in the room are made of the exact same material
B) Is it doable to create the map even if there are multiple types of material in the room?
| http://www.pnas.org/content/110/30/12186.abstract
That's a link to a paper that claims to be able to do that with a "few" microphones. Of course, if the room is not convex, it has the potential to fall prey to the illumination problem.
With 2 microphones, you should be able to get an arbitrarily good map, but there will be no way of determining its orientation around the axis created by the line through both microphones other than through adding parameters accounting for gravity or differences in pickup sound due to the 3d orientation of the microphones.
| {
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What is Gray, from a physics POV? Quora explains how white and black colors fit into the spectrum of visible light. It explains that white is all colors together while black is the lack of color.
So, where is Gray? Gray is the mix of all-colors and no-color!! What does that mean? Can somebody explain Gray, the mysterious color?
| All colors are combinations of three wavelengths of visible light, red, green and blue.
*
*White = All Red + All Green + All Blue
*Black = No Red + No Green + No Blue
*Grey = unsurprisingly Half Red + Half Green + Half Blue!
Read this Wikipedia Page: https://en.wikipedia.org/wiki/RGB_color_model
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Tipping point on fence panels? I deal in temporary fence panels - and my concern is the tipping point of our product out in the field.
Panels are 6' tall x 12' long (63#)
Stands are 23" long x 6" wide
We use sand bags (30#-40#) on the bases, but for whatever reason, they don't always stay on (or people take them).
Is a 23" base an optimum base for this height - or rather the "minimum effective dose" to achieve a stable panel? I understand that increasing it to 30" (15" on each side) would make it stable, but is it necessary for that height and still remain stable?
We are reviewing this as we are looking at a taller product (8') and I don't believe the same stands can be used on both heights, as the added height will greatly increase the chances of a tip over.
So, maybe my actual question how do I determine my center of gravity on the upright fence - so that I can accurately define the necessary base width?
| What you are dealing with is countering the torque induced by a wind load on the fence section by the torque that your sand bags apply to your base when the section tips. It is hard to estimate the wind load, but it should be proportional to the area of the section. If your 6' fence is stable with a 23" base and 40# bags, then for a 8' fence you should incease your base by the same fraction as the height increase (33%) as well as increase your bag weight by the same fraction to maintain the same level of stability.
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Meanings of the word "phase" I have been confused at points due to multiple uses of the word "phase".
*
*Mainly, when I think of a phase diagram, I think of a graph relating temperature to pressure, and segments the possible combinations of these values into regions in which a particular substance is "solid", "liquid", etc.
*This is something completely different from phase space, in the dynamical systems sense, where each point in the space represents the state of a dynamical system.
*There are also notions of phase and phase velocity.
Am I correct in assuming that the words "phase" in these contexts have nothing to do with each other, and are there other meanings that I shouldn't conflate?
| The world phase comes from the greek phasys (ϕάσις), meaning "appearence", from the verb ϕαίνομαι, "to appear", "to show one self".
Therefore, you are correct when you say that the meaning of the word "phase" is different in the three cases you cited, but they all have in common the concept of appearence, state: the thermodynamic state in the phase diagram, the dynamical state un the phase space, and the "oscillation state" in a wave.
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What is the ratio of gluons to baryons? Gluons bond quarks into baryons (i.e., protons and neutrons). For example, two up quarks and one down quark form a proton while one up quark and two down quarks form a neutron. Is there one gluon per one baryon or two gluons per one baryon or what is the ratio of gluons to baryons?
|
Is there one gluon per one baryon or two gluons per one baryon or what is the ratio of gluons to baryons?
Gluons are elementary particles in the current standard model of particle physics.
The first three columns from the left are fermions. Fermions obey Fermi-Dirac statistics, and carry charge. the lower two rows of these columns are leptons, and in addition carry lepton number; the upper two rows carry baryon number.
Conservation of charge and lepton and baryon number conservation for each set make sure that the currently total baryon and lepton number under observation is conserved.
The last two columns are bosons, obey Bose-Enstein statistics and do not carry conservable quantum numbers as they are neutral and (, before electroweak symmetry breaking, zero mass.). The only conservation laws that have to be obeyed are energy and momentum conservation laws. Following the interaction possibilities allows them to multiply themselves with no constraints.
A single charge accelerating or decelerating can create any number of photons.
Gluons are in our everyday world bound within a baryon and because they are bosons, their numbers can only be limited by energy and momentum conservation in the overall modelling of a proton..
So one cannot have a ratio of quarks ( they carry either +/-1/3 or +/- 2/3 , charge, and1/3 baryon number)to gluons . Over all the baryon number of a proton sums up to 1, and there is no definitive sum of gluons, as they also depend on the energy of the observing interaction, as the other answer explains.
The question is analogous to asking "is the black body radiation leaving a body countable so that a ratio between photons and baryons in that body can be calculated?"
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In summer, should I close curtains during the day? In summer, that is when it's warmer outside the house than inside, one want to prevent the air in the house to warm up too much. Let's consider that all the windows are kept closed, that they are double panes and made of "Low-E" glass. During the day, does closing the curtains on the windows reduce the heating of the house? If so, does the color of the curtains have a significant impact?
Intuitively, I would think that once the sunlight goes through the window, it's too late to prevent the greenhouse effect. However, the curtains could create a layer of warm air that would improve the isolation, but maybe the convection would prevent this layer to form.
| The heat from direct sunlight is about 1000 watts per meter squared in hot countries. If sunlight passes the glass, the percentage not reflected will heat the room .
During the day, does closing the curtains on the windows reduce the heating of the house?
Yes , depending on the color of the curtain , its reflectance/albedo.
If so, does the color of the curtains have a significant impact?
Yes. Black curtains will have very low albedo, this means all the energy coming in will be absorbed by the curtain and transformed to infrared radiation, a heat source at the window.
Here is a link for reflectance/albedo of various materials.
White curtains would be close to the reflectance of" white paper sheet - 0.6-0.7".
The 30% absorbed will add to the heat of the room. In hot countries it is wise to have shutters, shadowing the window, which also reduce heat entering the house even if there is no direct sunlight. Of course if the glass is double glaze or particularly reflective the improvement will be smaller, but there will always be improvement in using external shutters, particularly of white colors.
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Difference between de Broglie wavelength and electromagnetic wavelength What is the difference between de Broglie wavelength and wavelength of electromagnetic radiation? Is there any relation between $\lambda=\dfrac{hc}{E}$ and $\lambda= \dfrac{h}{mv}$? (E stands for energy of electromagnetic radiation.)
| Yes, there is a relationship. The base relationship is between an object's momentum and its de Broglie wavelength:
$$
\lambda = \frac{h}{p}.
$$
For a particle that is not moving at relativistic velocities ($v \ll c$), we have $p = mv$ and so the relationship becomes $\lambda = h/mv$. However, for a photon, its momentum is not equal to $mv$; instead $p = E/c$, where $E$ is the photon's energy. If you plug this in to the above relationship, you obtain $\lambda = hc/E$.
You may be wondering why a photon has a different relationship between its momentum and its velocity than a conventional particle does. That's probably a separate question, and one that I'm confident has been answered many times on this site; I would encourage you to search this site for "photon momentum" for answers to this.
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Why are the absolute magnitudes in M5 galaxy so puny? Wikipedia gives the following as the HR diagram for M5:
The stars at the base of the red giant branch have absolute visual magnitudes of 15? That seems way, way too dim. The sun's absolute magnitude according to wikipedia is 4.83, although it doesn't state in what filter that measurement was taken. What is going on here?
| It is a simple mistake. According to Layden et al. (2005), the distance to M5 is 7.76 kpc and has a V-band extinction of 0.11 mag. You need to subtract 14.56 mag from the y-axis to get the absolute magnitude.
As an aside, I did eventually find the incorrectly labelled diagram here. The "author", Lithopsian, gives no reference to where the data came from and claims it as their own work! Caveat emptor. I would stick to diagrams published in reputable journals.
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Why a propagating pulse has no DC spectral component? I found in a text that : "Because the laser pulse represents a propagating electromagnetic wave packet,
the dc component of its spectrum vanishes. Hence the time integral over the
electric field is zero."
Why is it so? I mostly don't understand why should a pulse not have any DC component.
| The pulse having a DC component would imply that there is a constant electric field forever after and before the pulse arrives. Because light obeys a dispersion relation $\omega = c k$, a field with $\omega=0$ does not propagate. Hence, the DC component is spatially and temporally uniform. In that case, it is meaningless to attribute the DC component to the laser pulse because it will be there far after the pulse leaves and before it even arrives.
Edit: OP sounds like he is still confused. The essential point here is that if the pulse satisfies the following:
$$\lim_{t \to \pm \infty}\mathbf{E}(\mathbf{x},t)=0$$
$$\lim_{\mathbf{x} \to \pm \infty}\mathbf{E}(\mathbf{x},t)=0$$
Then there is no DC component. The proof is that at $t\to\pm\infty$ and $\mathbf{x}\to\pm\infty$ any DC component will no go to zero strictly speaking.
Provided $\mathbf{E}(\mathbf{x},t)$ satisfies these conditions (as the usual definition of a pulse should) then if you integrate $\mathbf{E}$ over all time, you average to zero.
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Why is the Schwarzschild radius associated with the tiniest micro black hole formed by a Planck mass twice the Plank length? If one calculates the Schwarzschild radius, $r_s$, of a Plank mass $m_p=2,18*10^{-8} (m)$ one gets:
$$r_s=2{\frac{G{m_p}}{c^2}}=1,48*10^{-27}*2,18*10^{-8}=3,22*10^{-35}(m)$$
Now the Planck length $m_p$ is $1,61*10^{-35}(m)$, which is exactly half the Schwarzschild radius of the black hole associated with a Plank mass. Thus $l_p=\frac{G{m_p}}{c^2}$. Filling in the expression for $m_p=\sqrt{\frac{\hbar c} G}$ in the expression for $l_p$ gives the more familiair form for $l_p$, namely $\sqrt{\frac{\hbar G} {c^3}}$.
Why is it that the Schwarzschild radius of a Planck mass micro black hole (the tiniest that exists) is exactly twice the Planck length?
You can of course answer this question by saying that the formulae tell us that this is the case. But is the factor two by which the two lengths are connected just a coincidence, or has this connection some deeper significance?
| Planck mass black hole has to store 1 bit of black hole entropy. This requires 4 Planck areas. The 4Pi for the physical surface area of the sphere is "built-in" to the definition of BHE so you only need r^2.
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The definition of the Lorenz gauge condition The inner product of two vectors in space-time is:
$$(x_1, y_1, z_1, t_1) \cdot (x_2, y_2, z_2, t_2) = x_1 x_2 + y_1 y_2 + z_1 z_2 - t_1 t_2$$
So
$$(\frac{\partial }{\partial x}, \frac{\partial }{\partial y}, \frac{\partial }{\partial z}, \frac 1c \frac{\partial }{\partial t}) \cdot (A_1, A_2, A_3, \phi) = \text{div}(\vec A) - \frac 1c \frac{\partial \phi}{\partial t}$$
is Lorentz invariant, where $\vec A=(A_1, A_2, A_3)$. But the [Lorenz gauge condition] (https://en.wikipedia.org/wiki/Lorenz_gauge_condition) is defined by $\text{div}(\vec A) + 1/c\ \partial_t \phi=0$. Why has the minus changed into plus? So there is apparently no longer invariance.
| The Lorenz gauge condition is written as, $\partial_\mu A^\mu = 0$ which can be expanded as,
$$\partial_\mu A^\mu = \frac{\partial A^0}{\partial t} + \nabla \cdot \vec A = 0$$
in natural units, where we are simply doing what the Einstein summation convention instructed us to do, take a sum through the index, $\mu = 0, \dots, 3$. You can also write this as,
$$\partial_\mu A^\mu = \eta^{\mu\nu}\partial_\mu A_\nu$$
in which case you would get a minus sign, but notice the sum involves the co-vector $A_\nu$, a different quantity, related to $A^\mu$ by a change of sign, in flat spacetime since $\eta = \mathrm{diag}(\mp1,\pm1,\pm1,\pm1).$
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What is the purpose of baking a ultra high vacuum chamber? I don't understand the purpose for baking a UHV. I know it's to remove contaminants, but I don't know how baking would do that. Sure, heat may dislodge trapped dirt particles or may even break some down, but the particles would still be in a UHV. Therefore, I'm not sure how baking benefits a UHV.
| You don't just bake it, you bake it while pumping on it with a vacuum pump.
A solid contaminant such as a grain of sand may actually not be a problem at all. The reason a contaminant is a problem is if it contains substances that are at least somewhat volatile, and that will gradually evaporate when the system is in normal use. For example, suppose I leave a sweaty cotton sock in a beamline. The cotton fibers may have little or no effect on the vacuum, but my sweat will be gradually evaporating off -- this is why you can smell the sock normally. If I bake, the process of evaporation goes faster, and if I'm pumping while that evaporation occurs, the volatiles will gradually be removed from the system.
| {
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Frames of reference of Maxwell's Equations The Maxwell's Equations are one of the most famous sets of equations physics have ever known. But just as different sets of equations are applicable to different frames of reference, where are Maxwell's Equations applicable? To be more specific:
*
*Are Maxwell's Equations valid in only inertial frames of reference?
*If not, then how can we reformulate the equations so that it is valid in any accelerating frame with an arbitrary acceleration?
| As they are conventionally written Maxwell's equations are valid only in inertial frames of reference in flat spacetime. This is because the derivatives in the equation are not covariant derivatives and therefore don't apply when the coordinate system is curved.
It is possible to write Maxwell's equations in arbitrary coordinate systems though it gets somewhat complicated. The trick is to note that Einstein's equivalence principle tells us that acceleration is locally indistinguishable from gravity, and therefore the treatment of Maxwell's equations in accelerating frames is the same as formulating them in curved spacetime.
In principle all we need to do is replace all physical quantities by tensors, and replace normal derivatives by covariant derivatives. However the process of doing this makes the equations look very different. The details are described in the Wikipedia article Maxwell's equations in curved spacetime. Specifically note that the introduction to this article states:
The electromagnetic field also admits a coordinate-independent geometric description, and Maxwell's equations expressed in terms of these geometric objects are the same in any spacetime, curved or not. Also, the same modifications are made to the equations of flat Minkowski space when using local coordinates that are not Cartesian. For example, the equations in this article can be used to write Maxwell's equations in spherical coordinates.
So this approach is just as useful for curved (e.g. non-inertial) coordinates in flat spacetime as it is for curved spacetimes.
| {
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Differences between optical laser and amplifier I am preparing for an applied optoelectronics exam and I am having some trouble with telling the differences between optical fiber laser and optical amplifier. For now I only came up with following differences:
*
*No Bragg reflectors and optical resonators in amplifiers
*No signal conversion in optical amplifiers
I'm pretty sure there's more to it but I couldn't find anything that would answer this question fully.
| The main difference between a fiber laser and a fiber amplifier is the cavity.
A laser (fiber laser or solid state laser) is an gain media, a pump and a cavity whereas a amplifier has no cavity. The cavity allow selecting the oscillating mode so there is no need to seed with a signal to generate a single frequency. Because it's initiated by noise. Whereas an amplifier need a signal seed.
You can make an analogy with Larcen effect. An audio amplifier acts just as its optical counterpart : it needs an input signal to amplify it. Now, if you get a microphone closer of the speaker plugged to the amplifier you form a loop and if unlucky you get that really annoying noise at a selected frequency.
For fiber laser, the cavity can be a ring cavity, a bragg cavity, or mirrors.
| {
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Riemann tensor definition for non vanishing torsion From the definition of Riemann tensor we have:
$$
\mathbf{R}\left( \mathbf{z},\mathbf{v},\mathbf{w}\right)=\nabla_{\mathbf{[v}}\nabla_{\mathbf{w}]}\mathbf{z}-\nabla_{[\mathbf{v},\mathbf{w}]}\mathbf{z}
\label{riemannnew}
$$
and computing the coordinates of $\mathbf{R}$ in a coordinate basis we obtain:
$$
R^a_{\hphantom{a}bcd}=\partial_c\Gamma^a_{\hphantom{a}bd}-\partial_{d}\Gamma^a_{\hphantom{a}bc}+\Gamma^a_{\hphantom{a}\mu c}\Gamma^\mu_{\hphantom{a}bd}-\Gamma^a_{\hphantom{a}\mu d}\Gamma^\mu_{\hphantom{a}bc}
$$
I find another way to compute the coefficient fo Riemann tensor with not vanishing torsion:
$$
[\nabla_c,\nabla_d]V^a=2\nabla_{[c}\nabla_{d]}V^a = 2\partial_{[c}\nabla_{d]}V^a-2\Gamma^e_{\hphantom{e}[dc]}\nabla_eV^a+2\Gamma^a_{\hphantom{a}e[c}\nabla_{d]}V^e \nonumber \\
= 2\partial_{[c}(\partial_{d]}V^a+\Gamma^a_{\hphantom{a}|e|d]}V^e)+2S^e_{\hphantom{a}cd}\nabla_eV^a+2\Gamma^a_{\hphantom{a}e[c}(\partial_{d]}V^e+\Gamma^e_{\hphantom{a}|b|d]}V^b) \nonumber \\
= 2 \partial_{[c}\Gamma^a_{\hphantom{a}|b|d]}V^b-2 \Gamma^a_{\hphantom{a}e[c}\partial_{d]}V^e+2S^e_{\hphantom{a}cd}\nabla_eV^a+2\Gamma^a_{\hphantom{a}e[c}\partial_{d]}V^e+2\Gamma^a_{\hphantom{a}e[c}\Gamma^e_{\hphantom{a}|b|d]}V^b= \nonumber \\
=2(\partial_{[c}\Gamma^a_{\hphantom{a}|b|d]}+\Gamma^a_{\hphantom{a}e[c}\Gamma^e_{\hphantom{a}|b|d]})V^b + 2S^e_{\hphantom{a}cd}\nabla_eV^a
\tag{1}
$$
where the first bracket is the Riemann-Cartan tensor and second term is the part due to the non vanishing torsion tensor.
My question is:
The first term of the first definition $\nabla_{\mathbf{[v}}\nabla_{\mathbf{w}]}\mathbf{z}$ is the second equation (1) but only the first term of the second equation is the Riemann tensor. How can I solve this problem? Is the definition of the Riemann tensor incomplete?
| In the invariant notation $\nabla_X\nabla_Y$ corresponds to $X^a\nabla_a(Y^b\nabla_b)$, not $X^a Y^b\nabla_a\nabla_b$, eg. the vector field $Y$ also gets differentiated.
We can define $\nabla^2_{X,Y}Z=i_Xi_Y\nabla\nabla Z$, where here $i$ means "insert into the last empty argument", then we have $$ X^a\nabla_a(Y^b\nabla_b)Z^c=X^a\nabla_aY^b\nabla_bZ^c+X^aY^b\nabla_a\nabla_bZ^c, $$ so $$ \nabla^2_{X,Y}Z=\nabla_X\nabla_YZ-\nabla_{\nabla_XY}Z. $$
This gives then $$ R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z=\nabla^2_{X,Y}Z+\nabla_{\nabla_XY}Z-\nabla^2_{Y,X}Z-\nabla_{\nabla_YX}Z-\nabla_{[X,Y]}Z \\ =\nabla^2_{X,Y}Z-\nabla^2_{Y,X}Z+\nabla_{\nabla_XY-\nabla_YX-[X,Y]}Z=\nabla^2_{X,Y}Z-\nabla^2_{Y,X}Z+\nabla_{T(X,Y)}Z. $$
As you can see the $[\nabla_a,\nabla_b]X^c=R^c_{\ dab}X^d$ Ricci-identity corresponds to $R(X,Y)Z=\nabla^2_{X,Y}Z-\nabla^2_{Y,X}Z$, which is certainly true in absence of torsion.
In the presence of torsion, this gets modified to $$ R(X,Y)Z=\nabla^2_{X,Y}Z-\nabla^2_{Y,X}Z+\nabla_{T(X,Y)}Z, $$ but the definition of the curvature tensor, $$ R(X,Y)=\nabla_X\nabla_Y-\nabla_Y\nabla_X-\nabla_{[X,Y]} $$ doesn't depend on torsion at all.
| {
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Are time loops real? I'm new to physics and I was wondering if time loops (like those seen in the movies Groundhog Day or Edge of Tomorrow) possible?
| The closest thing in mainstream theoretical physics is "closed timelike curves", paths along which you can travel and thereby return to the same place and time as you started, provided your velocity varies as the path requires. (Shortcuts through spacetime called Einstein-Rosen bridges or "wormholes" can be a part of the setup.) Whether such paths exist without requiring faster-than-light travel depends on the topology of spacetime. It's easy enough to write down geometries containing CTCs, but the hard part is how such circumstances could arise, and proposals typically require conditions that aren't known to ever occur, such as matter with negative density.
| {
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Why does a surface always exert force normal to it? In whichever angle an object is thrown at a surface, the surface always exert force normal to it. But why? According to Newton's third law, if an object hits a surface at an angle, the reaction force provided by the surface must be equal and opposite to the applied force by the object. But why does the surface always exert force normal to it?
| A reaction force can surely be angled.
And it can then be split into a tangential part and a perpendicular part. The tangential part is called friction, and the perpendicular part normal force.
So, the reaction force is not at all always perpendicular - but there always is a perpendicular component (the one called normal force). This arises because, since the wall doesn't break, it must be holding back. A normal force is a "holding back" force that objects create when being pushed upon to avoid breaking.
| {
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Why would an infinity corrected objective lens produce an image without a tube lens? I'm having a hard time wrapping my head around this situation I've come across. I'm essentially recreating a homemade microscope that my understanding would suggest shouldn't work. This setup uses a 10X Olympus PLAN N objective attached to a beam splitter to provide illumination. Then, 55 mm worth of extension tubes connect to a camera (Point Grey Grasshopper3 with a 1/1.2" sensor). There is no tube lens between the infinity corrected objective and the camera sensor (other than the beam splitter).
My understanding of how infinity corrected microscope systems works would seem to suggest that this shouldn't work, but a correctly oriented and seemingly undistorted image shows up on the camera.
Presumably, we aren't actually realizing the actual magnification this objective is designed to provide (in fact, its about half what it should be the sensor is 8 mm tall and a ruler placed in the view shows about 1.5 mm across the short dimension of image).
Can someone clarify what the light path looks like here and how we are actually able to see an image? Is it possible that the image quality is actually really poor, but we just seem to be getting results that are more than suitable for how we are using it?
| Are you sure it's an infinity objective and not an RMS standard 160mm one ;-)
It's also possible that you don't have it focussed at exactly it's working distance. Remember the objective is just a (complex) lens. It only produces an image at infinity when the object is at a particular distance. If you put the object further away from the lens the image will be formed nearer than infinity.
| {
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How electrostatic charge is distributed in a tube? If a tube has an electrostatic charge, is the charge uniformly distributed across its body or do the charge distributions vary from inside walls and the head edges? If so, how is the charge distributed?
I appreciate if you can name some references elaborating the charge distribution over objects with geometrical (or even physical) non-uniformity.
| As pointed out in sammy gerbil's answer, the charge density is quite large near the corners of a conducting surface. If we look near the edges of the cylinder, at length scales much less than the cylinder's radius, the corner will "look like" two planes meeting with an "interior angle" of $3 \pi/2$. It is a general result (see, e.g., Section 2.11 of Jackson) that the charge density at a location where two conducting planes meet with an interior angle of $\beta$ is
$$
\sigma(\rho) \propto \rho^{(\pi/\beta) - 1},
$$
where $\rho$ is the distance from the edge. In particular, this implies that the charge density near the edges of the cylinder will diverge:
$$
\sigma \propto \rho^{-1/3}.
$$
In reality, the charge density is only divergent to the extent that we have the faces meeting at an infinitely sharp edge. Realistic cylinders will have some slight rounding to their edges; and if nothing else, we cannot think of the conductor as a continuum when we're at scales smaller than the atomic spacing of the metal in question. Still, it can be deduced that the charge density will in fact be very large near the edges of a charged conducting cylinder.
(Aside #1: The case of a conical point is also treated in Jackson; see Section 3.4. However, there isn't a nice closed-form expression in this case; rather the solution is expressed in terms of the zeroes of Legendre functions $P_\nu(x)$ with non-integer $\nu$.)
(Aside #2: Although the charge density is infinite along an infinitely sharp edge, the actual amount of charge "on the edge" is negligible. Specifically, the amount of charge within a distance $\delta$ of such an edge will be
$$
Q_\delta = 2 \int_0^\delta \rho^{-1/3} d\rho \propto \delta^{2/3},
$$
which goes to 0 as $\delta \to 0$.)
| {
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Is there any qualitative difference between the WZW $SO(2)_1$ and the WZW $SU(2)_1$ CFT? Consider the anisotropic spin-$\frac{1}{2}$ Heisenberg chain $$H = \sum_{n=1}^N S^x_n S^x_{n+1}+S^y_n S^y_{n+1} + \Delta S^z_n S^z_{n+1}$$
which for $\Delta = 0$ realizes the Wess-Zumino-Witten (WZW) $SO(2)_1$ conformal field theory (CFT), whereas at the isotropic point $\Delta = 1$ we have the WZW $SU(2)_1$ CFT. Moreover, the line $\Delta \in [0,1]$ is said to be a fixed point line connecting the two critical models.
So on the level of the central charge and scaling dimensions, both models continuously connect to one another, and hence on that level there is no qualitative difference. However, I was just wondering: is there any qualitative difference between them? For example, even though the scaling dimensions are not qualitatively different, I could imagine that (for example) one model has log-like contributions whereas the other doesn't. After all: even though the above model has a line of fixed points connecting both extremes, $\Delta=1$ is the end point of such a line of fixed points, and hence one might expect something funny to happen there.
EDIT: rereading this question I posed two years ago, I regret its poor formulation. The question attempts to ask whether the $\Delta = 1$ point has any singular behavior on the compact boson line, e.g., are there loglike corrections at this point which are not present for $|\Delta|<1$ etc?
| Well, the names say it all... the symmetries are different. At the $SU(2)$ point, with the additional symmetry (for the Heisenberg chain it is an explicit symmetry, in the bosonized theory it is emergent) there are additional current operators. It is at a self-dual point of the T-duality.
| {
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Electric field at surface of a surface charge distribution I've heard this statement in class:
"Electric field is discontinuous across the surface of a surface charge distribution."
Could someone please explain why this is so? I understand why electric field is not defined at the location of a discrete charge, but I don't understand why the above statement is true.
I'd be very glad if someone could explain the matter to me clearly.
| Suppose there is an interface between two media, $A$ and $B$, and that there is a surface charge density $\sigma$ on this interface. Consider Gauss' law applied to a pill box across this interface:
$$\iint_S \vec E \cdot d\vec S = \frac{1}{\epsilon_0}\iiint_V \rho \, dV.$$
In the limit as the pill box's horizontal length shrinks, the only contribution is the flux out of the two ends, with cross-sectional area $A$, for which we have,
$$\iint_S \vec E \cdot d\vec S = (E^\perp_A - E^\perp_B) A.$$
Now by Gauss' law this flux is simply proportional to the total charge which is $\sigma A$, and so we have that the perpendicular components of the electric field are discontinuous,
$$E^\perp_A - E^\perp_B = \frac{\sigma}{\epsilon_0}$$
and the discontinuity is proportional to the surface charge density.
| {
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How is it that in a car crash, four 8mm bolts can anchor the seat to the car? In a car crash at for example twenty metres per second. I used suvat equations and newtons second law to work out the force as as body accelerates(negatively). I estimated that the distance travelled in the crash by the body would be roughly 0.4 metres.Even using average mass of a human and car seat the force calculated was way too large to be accurate as the tensile strength of steel would be easily exceeded. I concluded that a large portion of energy is transferred by the front of the car before it affects the body. My question is how could I find an accurate but rough figure for the force in newtons acting on each individual bolt and if anybody has any data or estimates.
Thank you
| You suggest $0.4\ \mathrm m$ stopping distance from $20\ \mathrm{m/s}$ velocity, which with $s=u^2/(2a)$ is a deceleration of $500\ \mathrm{m/s^2}$, or more than $50g$.
My guess is that is really the maximum deceleration your body might survive.
For $80\ \mathrm{kg}$ mass that is $40\ \mathrm{kN}$ of force.
The ultimate tensile load (from http://www.amesweb.info/Screws/Metric_Bolt_Grades_Strength.aspx) for an M8 bolt of the lowest strength class is $15.7\ \mathrm{kN}$, and can rise to $47.8\ \mathrm{kN}$ for suitably chosen bolts.
As already said by Sammy it is more the seatbelt that slows you down, so the fixing points of them to the chassis are more important, but a three-point seatbelt will have at least three of them (one per fixing point), so I would say three suitable chosen M8 bolts will be on the limit, but about right.
| {
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Estimate of the effect of quantum disturbances on a macroscopic object I am self-studying P. Davies, D. Betts, Quantum Mechanics. Exercise 4 of Chapter 1 says:
"A snooker ball of mass $0.1$kg rests on top of an identical ball and is stabilized by a dent $10^{-4}$m wide on the surface of the lower ball. Use the uncertainty principle to estimate how long the system will take to topple, neglecting all but quantum disturbances." (the answer is "About $10^{27}$s.")
I have a hard time thinking about how the uncertainty principle can be applied here. Is the ball to be considered as a "macroscopic particle", or should the principle be applied to a specific particle (or a group of particles) in the ball itself?
I would prefer to not receive the solution straightforwardly. I am looking for suggestions on the correct way to look at this problem.
| The arrangement of two hard smooth spheres balanced one on top of the other is doubly unstable. I agree with Bob Knighton : probably the lower ball should be assumed to be fixed in place while the upper ball is balanced on it.
The question asks for a time, so you should use the Uncertainty Relation between energy and time. The uncertainty in energy is the increase in potential energy required to topple out of the dent. The dent is presumably in the shape of a spherical cap, into which the upper ball fits snugly. So the CM of the upper ball has to rise by the depth of the dent in order for it to topple. The depth of the dent is much smaller than its diameter. This probably accounts for the extra 2 orders of magnitude.
| {
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How is the equation of Mach number derived? Wikipedia states that for a pitot-static tachometer, the mach number for subsonic flow equates to
$$M = \sqrt{5\left[\left(\frac{p_t}{p_s}\right)^\frac{2}{7}-1\right]}.$$
How did they get to that result? Is there a derivation, or is it just from a polynomial fit of a tabulated set of data?
Update
I accepted J.G's answer after glancing at the referenced flight test document (a treasure in itself) and realising that $\frac {7}{5}$ is the same as 1.4, but there remains an issue.
Sadly I don't have my uni books anymore with Bernoulli's equation for compressible flow. The issue is with dynamic pressure: for incompressible flows we can take $p_d = \frac {1}{2} \cdot \rho \cdot V^2$, for compressible flow this is $p_d = \frac {1}{2} \cdot \gamma \cdot p_{static} \cdot M^2$.
Right? If I substitute this I don't get to the equation above. So the answer is unfortunately not accepted anymore.
| $M$ is not the speed of sound. It is the Mach number -- the ratio of speed of the aircraft to the speed of sound. The equation is derived from Bernoulli's equation together with a suitable choice of $\gamma=C_P/C_V$ for air.
| {
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At what temperature do the laws of physics break down? I heard that as approaching the temperature of a kugelblitz the laws of physics break down, I saw this in the video The Kugelblitz: A Black Hole Made From Light, by SciSchow Space.
| Adding to @Emilio's answer, what happens during Planck's temperature is unknown. Our laws of physics does not seem to work at that temperature(@EvilSnack) for e.g. gravitational force. At that temperature, gravitational force seems to become as strong as other fundamental forces like electromagnetic forces, strong and weak nuclear forces leading to research in quantum nature of gravitational forces. On the Wikipedia article of Planck's temperature:
At temperatures greater than or equal to $\mathrm{T_P}$, current physical theory breaks down because we lack a theory of quantum gravity.
This statement is explained in details in this site:
The Planck temperature is the highest temperature in conventional physics because conventional physics breaks down at that
temperature. Above $\mathrm{10^{32}~K}$, Planck time-calculations
show that strange things, unknown things, begin to happen to space and
time. Theory predicts that particle energies become so large that
the gravitational forces between them become as strong as any other
forces. That is, gravity and the other three fundamental forces of the
universe—electromagnetism and the strong and weak nuclear
forces—become a single unified force. Knowing how that happens, the
so-called "theory of everything," is the holy grail of theoretical
physics today.
"We do not know enough about the quantum nature of gravitation even to
speculate intelligently about the history of the universe before this
time," writes Nobel laureate Steven Weinberg about this
up-against-a-brick-wall instant in his book The First Three Minutes.
"Thus, whatever other veils may have been lifted, there is one veil,
at a temperature of $\mathrm{10^{32}~K}$, that still obscures our view of the
earliest times." Until someone comes up with a widely accepted quantum
theory of gravity, the Planck temperature, for conventional physicists
like Steven Weinberg, will remain the highest temperature.
Basically, if $\mathrm{0~K}$ is absolute cold that Planck's temperature is absolute hot(i.e a body cannot get any hotter than Planck's temperature).
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Why is Higgs particle detected much later than top quark when it's lighter? The Higgs boson is lighter than the top quark. But the top quark was discovered in the mid-1990s where the Higgs boson escaped detection for two more decades. So if the energy has already been achieved to produce Higgs boson, why did it escape detection so far?
I understand that the couplings of Higgs boson to fermions is small and doesn't interact with the detector appreciably.
Does it mean that in LHC, with the increase in energy, the Higgs coupling increased and we finally detected Higgs?
| With both particles you cannot detect them within their own lifetimes, only look at what they decay into. The top decays to a b jet and W (which can then become fermion anti-fermion or leptons) and is fairly distinctive. The dominant Higgs decay, however, is to two b jets. B jets are very common within the LHC and we cannot infer from two b jets that a Higgs detection has been made.
It's all about the statistics. With the top, its decay mode was distinctive enough that fewer events were needed before a statistically significant signal was seen above the background, whereas the Higgs needed many more events.
| {
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Will overlapping two different beams of coherent light with different wavelength cause interference? If I use two different wavelength lasers to transmit light into a single mode optic fiber will they interfere with each other? If so, how much will be that interference.
| Yes, they will of-course overlap but it won't be the same pattern for when waves with same wavelength overlap. The resulting wave can be found from graphing simple addition of sine waves indicating the superposition at each point.
To explore this visually you can try the graphing calculator Desmos. Try changing the slider values of $a$ and $b$ in the resultant wave,$y=sin(ax)+sin(bx)$ to change the wavelength of each wave and see what happens to the wave after interference(producing resultant wave).
Note: They will only interfere if they meet/cross paths in the optical fiber. Also you may try change the phase difference by a third slider e.g. $k$ in $sin(ax-k)$.
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Is there a case when it is better to use the integral form of the Maxwell equations rather than the differential form? I was wondering if there is a case where the integral form of the Maxwell equations is preferred over the differential form?
If you could provide with an example for each one of the equations I would really appreciate that.
| The integral forms are useful in (typically static) situations where the charge/current distribution is symmetric enough that you can use a symmetry argument to replace the surface/line integrals with a simple product of a uniform field strength times an area/length of an imaginary closed "Gaussian surface" or "Amperian loop".
They are also useful for figuring out the far-field behavior of a localized charge/current source - e.g. we can use them to show that that, no matter how complicated a static localized charge distribution, very far away it produces an electric field of the form ${\bf E} = (Q/r^2)\, \hat{{\bf r}}$, where $Q$ is the distribution's total electric charge.
A friend of mine once attended a lecture titled "A Defense of the Integral Forms of Maxwell's Equations" in which a distinguished elderly physicist argued that the integral forms are underappreciated.
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"timestamp": "2023-03-29T00:00:00",
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Conjugate variables in thermodynamics vs. Hamiltonian mechanics According to Wikipedia, the canonical coordinates $p, q$ of analytical mechanics form a conjugate variables' pair - not just a canonically conjugate one.
However, the "conjugate variables" I directly think of are the quantities of thermodynamics - e.g. Temperature and Entropy, etc.
So, why both these classes of variables are called "conjugate"? What is the relation among them?
| In thermodynamics, conjugate pairs are related by the Legendre transform (like $T$ and $S$, or $P$ and $V$). In classical mechanics, you use the Hamiltonian to get the conjugate variable in a slightly different way, although the Lagrangian and Hamiltonian are related by the Legendre transform as well.
In general, conjugate variables are those which are related by some sort of transform, be it Legendre, Fourier, etc. That's why you will see the term used in a variety of contexts. I can't comment about the link you provided, and invite someone else to do so.
| {
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What is the difference between the valence shell and the valence band? What is the difference between the valence shell and the valence band?
The valence band is usually defined as the highest filled band whereas Wikipedia defines the valence shell as the outermost shell of an atom in its uncombined state, which contains the electrons most likely to account for the nature of any reactions involving the atom and of the bonding interactions it has with other atoms.
This seems contradictory.
| The valence shell is the outermost electron shell. An isolated atom's valence shell contain electrons with certain energy levels.
When atoms are brought into close proximity, repulsion of their electrons causes the energy levels corresponding to the shells to split into discrete energy bands. Thus the electrons will be separated by a short distance and have different energies within the band.
We consider the valence band in situations where the interatomic distance is short such as in crystalline structures.
The valence band is usually defined as the highest filled band
The valence band isn't necessarily filled.
| {
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Pumping up bicycle tires with helium instead of air If I pumped up my bicycle tires with helium instead of plain air, what would happen if the applied force on my pedals was constant?
Would I go faster because of the reduced ground friction? Would I go slower because I would have less contact with the surface I'm riding on? Or would it make no difference?
| Others mentioned that the weight difference will be small, on the order of 10g (one should take into account the pressure in the tires). You should have in mind though that lighter tires also have smaller moment of inertia (rotational inertia), which should facilitate acceleration. Some people also mentioned helium's higher thermal conductivity, which supposedly might help prevent heat buildup (http://www.bikeforums.net/road-cycling/331014-helium-your-tires-truth-myth.html)
| {
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About light in the universe As a light source in the universe (e.g. sun) emits light in different directions, some of the light emitted reaches places like Earth, and some doesn't. So does the light that reaches the Earth disappear or it is reflected in other directions? And for the light that doesn't reach any place, does it keep on going forever? If it does keep on going forever, will the universe become brighter and brighter? Thanks!
| As you probably know, the light that you mention is an electromagnetic radiation, so it is part of a large spectrum in which $\textit{visible light}$ makes only a small part. Even the sun emits tons of radiation, but not all of it is visible light. When you say about the brightness of the Universe (referring only to the visible light which is radiated by the cosmic objects), you need to be careful here: the brightness of the Universe can be calculated actually if you know how much light each cosmic object emits. Of course if you know the brightness of say Andromeda galaxy, then you probably know enough details about its components (i.e. nebulas, stars and so on).
Scientists want to measure the brightness of the Universe ( see here). Probably they will obtain a value which of course will be large, but of course, finite. It can also change in value due to normal cosmic events: supernovae, quasars, death of stars (so it can decrease or increase) but I guess the change in value is NOT noticeable in one day :D
| {
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Why surface normal is used while defining flux through an open surface? What is the significance behind defining normal to any surface? Why we do it?
| Geometrically, it's only the component normal to the surface that "pushes the stuff through" the surface.
As illustrated below, the "stuff" that "moves" parallel to the surface does not pass through this surface and so does not contribute to the flux through that surface. In the figure on the far right, the "stuff" "flows" with a velocity parallel to the surface so no "stuff" actually goes through the surface.
| {
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How can we show Power = $\mathbf{F}\cdot \mathbf{v}$? How can we say that $$\text{Power} = \mathbf{F}\cdot \mathbf{v}$$
We know that small work done by a force $\mathbf{F}$ to displace an object by '$\mathbf{x}$' is
$$W = \mathbf{F}\cdot \mathbf{x}$$
So derivating wrt time, we get
$$\begin{align}
P=\dfrac{dW}{dt}&=\frac{d\mathbf{F}}{dt}\cdot \mathbf{x}+\mathbf{F}\cdot\dfrac{d\mathbf{x}}{dt}\\
&=\frac{d\mathbf{F}}{dt}\cdot \mathbf{x}+\mathbf{F}\cdot\mathbf{v}
\end{align}$$
We get this wrong result. How actually can we show $P=\mathbf{F}\cdot\mathbf{v}$ ?
Edit
Actually I know that total work $W$ is $\int \mathbf{F}\cdot d\mathbf{x}$.
Infinitesimal work done by $\mathbf{F}$ to displace body by $d\mathbf{x}$ will be $dW = \mathbf{F}\cdot d\mathbf{x}$, so dividing by $dt$ on both sides gives $$P =\dfrac{dW}{dt} = \mathbf{F}\cdot \frac{d\mathbf{x}}{dt}$$
But I wanted a proper proof not involving differentials!
| See that work is done only when there is a displacement. So if there is no displacement then even having a variable force will not account to any Power. Hence the first term is excluded.
$$ W = \int \mathbf F \cdot {\mathbf v} dt $$
$$P=\dot{W} $$
When applied to the integral along with fundamental theorem of calculus it gives
$$P=\mathbf{F\cdot v}$$
| {
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Total and static pressure: which one is measured? Which pressure is measure using measuring device in a pipe flow. My first intuition that it is the static pressure. This is confirmed by this Wikipedia article (link)
The concepts of total pressure and dynamic pressure arise from Bernoulli's equation and are significant in the study of all fluid flows. (These two pressures are not pressures in the usual sense - they cannot be measured using an aneroid, Bourdon tube or mercury column.)
But I came across another document from MIT course (link) where the author says:
The dynamic pressure
is the difference between the total pressure—that is, the pressure you would
actually measure at the given point in the moving fluid, with some appropriate
instrument—and the static pressure.
To the best of my knowledge the total pressure can be only measure if we bring the fluid to rest, e.g. using a pitot tube. That's why it's sometimes called stagnation pressure.
So, the question know is which quote is true? or there is a specific device that can measure the total pressure while the fluid is flowing?
Another question: in the MIT document the author says:
the dynamic pressure is zero in a stationary
fluid, and also in a fluid that is in uniform motion, in the sense that there are no
accelerations anywhere in the fluid (Figure 1-3).
So, why the dynamic pressure is zero in uniform flow? What is the link between dynamic pressure an acceleration?
Note that in Figure 1-3 this isn't evident!!!!
| The dynamic pressure represents the volumic kinetic energy of the fluid, so $P_{dyn}=\frac{1}{2}\rho v^2$, and the total pressure is $P_{tot} = P_{stat} + P_{dyn}$. Thus, a device that take the speed of the fluid into account measures $P_{tot}$, and one which does not will only measure $P_{stat}$. For the first case, you can think about a Pitot tube, while for the second case you can simply use a tube which is perpendicular to the fluid flow, see this image for example.
However, one can only measure presure differences, or, in other words, when you measure a pression, you get it within a constant. Now, if the fluid is in uniform motion, $v=C^{te}$, so if you measure the total pressure, you will get $P_{mes} = P_{tot} + C^{te} = (P_{stat} + C^{te}) + C^{te}$. Finally, it is as if the dynamic pressure was zero in a stationnary fluid.
| {
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How much "earth" does it take to create an "earth ground" for measuring voltage? I was contemplating the idea of a multi-meter that had a "built in" Earth ground to it, and if it did have this ability how much earth/dirt would it take to register as a true Earth ground to register voltage of 120v when touching a standard household outlet. I imagine it'd be either a) a massive resistor or b) a ton of 'earth/dirt'. If it was a ton of dirt, how much exactly would create a ground such that you read 120v from a typical 120v hot wire and the earth. I've attached a picture to show that there is indeed a specific (unknown?) amount of earth to create an earth ground if you begin to consider it incrementally in terms of the amount of earth needed to be a ground.
| The ground/earth works because all the electricity suppliers agree to use the electrical potential of the earth as a reference point of zero voltage. Since the Earth is a conductor, if not a terribly good one, all electrical installations are in principle connected together to establish a uniform zero potential.
If you take your meter and connect one terminal to a large bucket of soil you aren't achieving anything useful because that bucket of soil isn't connected to the rest of the Earth. In effect you've just connected a rather small capacitance to one terminal of your meter.
Increasing the size of the bucket of soil would increase the capacitance, but it still wouldn't be very useful until you electrically connect it to the rest of the Earth.
| {
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Why does the Sun appear larger during the sunrise and sunset? Why does the Sun appear larger during sunrise and sunset compared to its size during midday even though its distance from the Earth remains fixed all the day? It is only during the Winter when the Sun is away from the Earth and that too is due to the motion of Earth around the Sun, not due to the diurnal motion of the Earth. So is it some optical effect? If yes, what is it really? I apologize if the question has already been asked.
| Yesterday we had a super moon full moon. At moonrise I held a transparent ruler at arm's length and measured the size of the moon against the ruler. It was 7.5 mm. At 11 pm I held the same ruler at arm's length against the moon which was now high in the sky. It was only 4 mm. This is not imaginary. It is real and observable. The larger size is caused by a refraction of light when the moon is at the horizon because the light has to pass through greater amounts of atmosphere to reach you compared to when it is overhead. The same is true of the setting sun. In fact, I recall reading years ago (I believe in my daughter's physics book) that we are actually seeing below the horizon at sunset because of light is being bent by the atmosphere. This phenomenon is worldwide. It is not imaginary. As far as I know it is sound science.
| {
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Why can a regular infrared camera not show temperature (thermography)? There are a lot of questions here dealing with infrared cameras and thermographic cameras. I think I understand the reason why a thermographic camera is able to retrieve the temperature values from any object and convert them to a falsecolor representation, but why is a "regular" infrared camera not able to retrieve this information? What are the differences between these cameras? Is it just the sensor within the camera?
| This is a common confusion, because both thermographic cameras and "normal" cameras with some IR capability are called IR cameras often.
The typical video camera with IR capability has a solid state semiconducting camera sensor normally used for capturing visible light, which relies on the photons interacting with electrons and electron-"holes" inside the semiconductor to convert the incoming light into electric charge which is subsequently measured. These photons are in the wavelength range of 300-800 nm or so, but the sensor technology is typically responsive up to 1000 nm or more. As the eye is not sensitive to the energy in the 800-1000 nm band, an IR cut filter is normally inserted in cameras to make the resulting photo seem similar to what the eye sees.
But if you remove the IR filter, you can get some "nightvision" capability by bathing the scene with light in the 850-950 nm range which is invisible to the eye.
On the other hand, thermal radiation is peaked at a much longer wavelength, typically at 8000 nm or longer, and is much more difficult to work with in a direct photon -> charge process, so the typical thermal camera uses a completely different and more mundane physical process - it actually uses an array of thermometers!
These are nothing else than a grid of small metal squares that are heated by the incoming thermal radiation, and their temperature can be read out because their resistance changes by their temperature (they are called micro-bolometers).
So, very different physical processes are used and the radiation is of an order of magnitude different wavelengths.
The thermal cameras need optics that can bend these longer wavelengths, they are often made of germanium for example and are opaque to visible light.
| {
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How can I use Newton's laws of motion to determine the force acting on the rope? Imagine there is a painter, weighing $180~\rm lb$, that is working from a bosun's chair hung down the side of a tall building.
Suppose that he pulls down on a fall rope with such a force that he presses against the chair with a force of $100~\rm lb$. You can assume that the chair's weight is $30~\rm lb$.
For finding the acceleration of the painter and the chair, I took into account that the weights of the painter and the chair are $180~\rm lb$ and $30~\rm lb$ respectively. I used this idea to perform the following step:
$$\text {Total mass of the painter and the chair} = \left(\frac{(180 + 30)~\rm lb}{g}\right) $$
He exerts a downward force of $100~\rm lb$ on the chair.
His net motion will be upwards.
I think the $100~\rm lb$ force the person exerts on the chair is transferred to the rope he is pulling on.
But that is just the string he is pulling on. The diagram shows that only one end of the rope is attached to the bosun chair. That end will have an upwards force of $(100 + 180 + 30)~\rm lb$ (as shown in the picture). This way, one end will have a $y~\rm lb$ force and other a force of $(100 + 180 + 30)~\rm lb$. I don't know if this is possible and I am not totally convinced that the rope is experiencing a force of $100~\rm lb$ due to the painter pulling on it.
How can I properly use Newton's third law to determine the impact of the $100~\rm lb$ downwards force on the overall system (the painter and bosun's chair)?
| There are a variety of ways to solve this problem, but I solved it pretty easily by analyzing the painter, and the whole system (painter + chair), giving 2 equations and 2 unknowns.
The painter's force on the chair (100 lbs) $= M_p(g+a)-T$, where $M_p$ is the painter's mass (100 lbs / g), $a$ is the upward acceleration, and $T$ is the rope's tension.
The force on the combined system is $2T = (M_p+M_c)(g+a)$, because twice the rope's tension is pulling up on the system. $M_c$ is the mass of the chair (30 lbs / g).
With a little bit of algebra, it's pretty easy to solve from there.
| {
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Can a cycloidal pendulum be extended to make a full swing? A cycloidal pendulum is isochrone, meaning it's period is independent of it's amplitude. But a cycloidal pendulum - as usually depicted - doesn't do a full 360 degrees swing. Why is that? Is there a limit on how high a pendulum can swing, so that it is still isochrone? Or is it possible to build a isochrone pendulum that does a full 360 degrees swing? And if so, what kind of curve would it need to follow? Is it just an extension of a cycloid as shown in the picture below.
Or would a 360 degree isochrone pendulum maybe follow a different curve, e.g. maybe something like a Cardioid?
| This sketch became too long to be a comment so I will post it as an answer anyway.
Let us say we have a wire forming a cycloid in the vertical plane as in the figure bellow
A friction-less bead into this wire will oscillate with known period $T$ independently of its releasing point. Assume we can symmetrically continue the wire above the dashed line. If we release the bead from $A'$, which is infinitesimally close to $A$ but above the dashed line, then it will take a time $\Delta t_{AB}<T$ to complete the segment ${AB}+BA$ since it travels faster. If we want the same period $T$ for the bead to oscillate between $A'$ and $B'$, then the time taken along $A'A$ shall be $d t_{A'A}=(T-\Delta t_{AB})/2$. The idea is to use energy conservation and the parametric equation of the cycloid to compute $\Delta t_{AB}$ and then obtain $d t_{A'A}$.The last step is to suitably chose the slope of the straight segment $A'A$ so that the interval $dt$ for the bead to slide from $A'$ to $A$ matches $dt_{A'A}$. I think you will find that $dt_{A'A}<dt$ even when $A'$ is vertically above $A$ (which gives the least $dt$). This means that the tautochrone curve cannot be continued above the dashed line. You can also recall that this construction can be actually used to obtain the tautochrone. The slope of the infinitesimal segments are always increasing, from zero at $C$ to $\pi/2$ at $B$. One increment further and we can no longer increase the slope.
| {
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Renormalization and canonical commutation relations My question is whether canonical commutation relations hold for renormalized quantum fields. Below I show reasoning which caused by doubts.
Consider a relativistic scalar QFT. We have spectral decomposition of two-point function
$$ \langle \Omega | \phi(x_1) \phi(x_2) | \Omega \rangle = \int \frac{\mathrm d m^2}{2 \pi} \rho(m^2) \Delta_+(x_1-x_2,m^2), $$
where $\rho \geq 0$ is called spectral density function and distribution $\Delta_+$ is defined as
$$ \Delta_+ (x,m^2) = \int \frac{\mathrm d ^3 p}{(2 \pi)^3 2p^0} e^{-ipx}, $$
with integral evaluated over the positive frequency ($p^0 \geq 0$) mass-shell $p^2=m^2$. I assumed above that field $\phi$ has no vacuum expectation value. If we take the difference of the first formula with itself with $x_2$ and $x_1$ interchanged, set $x_2 = 0$, take derivative with respect to $x_1^0$ and set $x_1^0=0$ we get canonical commutator on the left hand side. By comparing with the right hand side one obtains the Weinberg sum rule for the spectral density:
$$ \int \frac{\mathrm d m^2}{2 \pi} \rho(m^2) = 1. $$
What bothers me is that value of this integral depends on the values of finite parts of renormalization constants. Hence it is not renormalization scheme and scale independent. I checked some simple examples and it turned out to be possible to enforce this relation as renormalization condition and fix the value of wavefunction renormalization. However, I don't think this is what is usually done.
| The relevant axiom.
Any (canonical) field, renormalised or not, satisfies, by postulate,
$$
[\phi,\pi]=\delta
$$
where $\pi$ is the field conjugate to $\phi$. In Lagrangian field theory,
$$
\pi\overset{\mathrm{def}}=\frac{\partial \mathcal L}{\partial\dot\phi}
$$
Case 1.
If $\phi$ is an unrenormalised field,
$$
\mathcal L=\frac12\dot\phi^2_{\mathrm{un}}+\cdots
$$
then
$$
[\phi_{\mathrm{un}},\dot\phi_{\mathrm{un}}]=\delta
$$
Case 2.
On the other hand, if $\phi$ is a renormalised field,
$$
\mathcal L=\frac12Z\dot\phi^2_{\mathrm{re}}+\cdots
$$
then
$$
[\phi_{\mathrm{re}},Z\dot\phi_{\mathrm{re}}]=\delta
$$
The Upshot.
In conclusion, the canonical commutators, when expressed in terms of (canonical) phase-space variables, are independent of the normalisation of the fields. When expressed in terms of, say, configuration-space variables, they depend on the normalisation of the fields.
| {
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Why does Newton's Third Law actually work? My father explained to me how rockets work and he told me that Newton's Third Law of motion worked here. I asked him why it works and he didn't answer. I have wasted over a week thinking about this problem and now I am giving up.
Can anyone explain why Newton's Third Law works?
For reference, Newton's third law:
To every action there is always opposed an equal reaction: or the
mutual actions of two bodies upon each other are always equal, and
directed to contrary parts.
| I know i am too late for this answer but I couldn't stop myself from answering :). Also I am going to use the contradiction method which we generally use in mathematics.
Let us assume that it doesn't work .
So , nothing around you follows Newton's third law.
Now , you take a spring and try to compress it by applying a force on that spring. Since there is no Newton's third law, someone can easily argue that you can compress that spring to a very high extent or say to a point size . Since you are not feeling any opposing force, it should be a piece of cake for you to compress the spring . But from daily experience , can you really compress it to a point size ? No !!! You do feel an outward push on you which is opposing you from compressing that spring.
Also the moment you leave that spring, you no longer feel any push which means that what you were feeling was a result of your action.
This contradicts our assumption. So our assumption is wrong !!!
So it is truly said that when you apply a force on any object , the object also applies a force on you in the opposite direction and through thousands of experiments you can deduce that both the forces are equal in magnitude.
Hope it helps ☺️.
| {
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Do commuting Hermitian operators correspond to compatible observables? As far as I know, two compatible observables have a complete set of common eigenvectors, and using this fact, one can prove that their corresponding operators are commutative. Well now is the converse true? Do any two commutative hermitian operators correspond to compatible observables?
Another point I have in mind is that commutativity is not transitive. For example, $[x,y]=0$, $[y,p_x]=0$, but $[x,p_x]\neq0$. Is compatibilty transitive? It seems for me that it has to be so, since a single observable can not have two different complete sets of eigenvectors. Isn't that true?
| This question concerns more Mathematics than Physics, so it should be handled rigorously in order to avoid to generate even more confusion (I personally find quite confused this page Complete Set of Commuting Observables since it deals with the finite-dimensional case in the proofs and supposes that the statements are valid for the infinite-dimensional case, where instead things are much more subtle).
First of all compatibility of two observables represented by a pair of (generally unbounded) self-adjoint (not just Hermitian or symmetric) operators $A: D(A) \to H$ and $B:D(B) \to H$ in a (generally infinite-dimensional) Hilbert space $H$ means that their projection-valued measures commute.
In other words, if we focus on the spectral decompositions of the operators
$$A = \int_{\sigma(A)} \lambda dP(\lambda)$$ and $$B = \int_{\sigma(B)} \lambda dQ(\lambda)$$ it must be $$P_EQ_F=Q_FP_E\quad \mbox{for all Borel measurable sets $E,F \subset \mathbb{R}$.
}$$
Compatibility is the mathematical statement equivalent to the physical statement that the observables can be simultaneously measured.
If at least one, say $A$, of $A$ and $B$ is bounded (so that its domain coincide to the whole Hilbert space), compatibility is equivalent to commutativity
$$AB\psi = BA\psi \quad \mbox{for every $\psi \in D(B)$.}$$
When both $A$ and $B$ are unbounded (which physically means that the outcomes of their measurements can be arbitrarily large) commutativity on every common invariant domain is not equivalent to compatibility. There are famous counterexamples due to Nelson.
Finally, compatibility is by no means transitive, and this is one of the most interesting features of quantum theory. It gives rise to several no-go theorems regarding possible classical interpretations in terms of hidden variables (think of Kochen-Specker theorem for instance).
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Does gravity have anything to do with Van Der Waals forces? Does gravity have anything to do with Van Der Waals forces? Just throwing this out there, I was wondering if they do because gravity is such a weak force and the VdW forces at the molecular level could seem to be a good intermediary force between gravity and the forces acting within atoms. Given that there are so many atoms and molecules within objects like the earth doesn't it seem possible that an extrapolation of the VdW forces could make a good candidate for a theory of gravity?
| There is a connection. According to
R.L.Jaffe (2005). "The Casimir Effect and the Quantum Vacuum". Physical Review D. 72 (2): 021301. arXiv:hep-th/0503158.
the Casimir force is simply the (relativistic, retarded) van der Waals force between the metal plates.
Another interpretation of Casimir force is that it is due to quantum vacuum. That same vacuum is said to produce inertia, according to
Inertia as a zero-point-field Lorentz force,
Bernhard Haisch, Alfonso Rueda, and H. E. Puthoff,
Phys. Rev. A 49, 678 – Published 1 February 1994
and later
Inertial mass and the quantum vacuum fields,
Bernard Haisch, Alfonso Rueda, York Dobyns,
First published: 26 February 2001, Annalen der Physik, 10 (5), 393, 2001
With the principle of equivalence the connection to gravity appears. Some of the authors also wrote papers on that subject later.
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Lorentz transformation of the four velocity Can we derive the velocity addition rule by directly transforming the four velocity?
| Theory-wise, the easiest way to use this is to use $w=ct$ and then your Lorentz matrix in 1+1 dimensions can be written as $$\begin{bmatrix}w'\\x'\end{bmatrix} = \begin{bmatrix}\cosh\alpha & -\sinh\alpha\\-\sinh\alpha&\cosh\alpha\end{bmatrix}\begin{bmatrix}w\\x\end{bmatrix}.$$
This is because the hyperbolic cosine and sine obey the relation $\cosh^2\alpha - \sinh^2\alpha = 1,$ and therefore if you naturally just choose $\cosh\alpha = 1/\sqrt{1 - \beta^2}$ you naturally find $\sinh^2\alpha = \beta^2/(1 - \beta^2),$ so this is a valid way to write the usual Lorentz Matrix $[\gamma, -\gamma\beta; -\gamma\beta, \gamma].$
Since a 4-velocity in its rest frame is $c~[1; 0]$ it is not hard to see straight from this that a 4-velocity in any other reference frame is $c~[\cosh r; \sinh r]$and therefore that $r = \tanh^{-1}(v/c)$ where $v$ is the ordinary velocity and $r$ is this new quantity called the rapidity. Plugging that in, we can find that it's Lorentz transform is $$\begin{bmatrix}\cosh\alpha & -\sinh\alpha\\-\sinh\alpha&\cosh\alpha\end{bmatrix}\begin{bmatrix}\cosh r\\\sinh r\end{bmatrix} = \begin{bmatrix}\cosh r~\cosh\alpha - \sinh r~\cosh \alpha\\\sinh r~\cosh\alpha-\cosh r~\sinh\alpha\end{bmatrix} = \begin{bmatrix}\cosh (r -
\alpha)\\\sinh(r-\alpha)\end{bmatrix}.$$So if you transform to its rest frame for example you choose $\alpha=r$ and then the 4-velocity takes this $[c; 0]$ form directly, but rapidities add linearly in one dimension.
Of course for this particular question all of that theory-which-makes-things-simple can be a little bit overkill. If you do not want to use the hyperbolic sines and cosines, you can write this out in terms of the more conventional components as $$\gamma_2\begin{bmatrix}1\\\beta_2\end{bmatrix}= \gamma_0~\gamma_1~\begin{bmatrix}1&-\beta_0\\-\beta_0&1\end{bmatrix}\begin{bmatrix}1\\\beta_1\end{bmatrix}.$$
Notice that the noisy prefactors $\gamma_{0,1,2}$ do not need to enter our consciousness at all because the ratio of the second component to the first will be $\gamma_2 \beta_2 / \gamma_2,$ and they will just cancel out on either side So all you're left with when you take this ratio is,$$\beta_2 = \frac{\beta_1 - \beta_0}{1 - \beta_0\beta_1}.$$
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Prove there is an equipotential sphere between two point charges Given two point charges of opposite sign I need to prove that inside the electric field they create there is an equipotential sphere.
I'm very positive that this is more geometry than anything else and I really question why I've been given this exercise. Here's my thoughts anyways and everything I can remember from my background in math and geometry.
The potential for a point charge is given by $$V=\frac{q}{4\pi ε_0r}$$
Let's assume the charges are correlated by the following ratio $$\frac{q_1}{q_2}=-a$$ What we want is $$V_1+V_2=0=>\frac{a}{r_1}=\frac{1}{r_2}=>\\r_1=ar_2$$
The distances $r$ are the distance from the charge to any point in the equipotential surface.
The equation of a sphere is the following: $$(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=R^2$$
I can also write $$r_1=\sqrt{(x_1-x_0)^2+(y_1-y_0)^2+(z_1-z_0)^2}\\r_2=\sqrt{(x_2-x_0)^2+(y_2-y_0)^2+(z_2-z_0)^2}$$and then use r1=ar2. But then I really get lost.
Is there another way? And if not how do I show that this is a sphere?
| The electric field around the charges will have rotational symmetry about the line joining them, so the problem can be reduced to 2D and the task of proving that the locus is a circle.
The only physics here is getting $r_1=ar_2$. The rest is geometry. One proof relates to the Apollonian Circles Theorem.
Using co-ordinate geometry :
Suppose the charges are located at A(0,0) and B(d,0) where d=AB is the fixed distance between them. The distances AP, BP of some point P(x,y) from A,B are given by
$AP^2=x^2+y^2$
$AB^2=(d-x)^2+y^2$.
Set $a^2AP^2=BP^2$ to get an equation for the locus of points P. This has the form
$x^2+y^2-2fx-2gy+c=0$
which is the equation of a circle.
This problem is the inverse of finding the image charge A of a point charge B in a grounded conducting sphere. See Griffiths 2007, Problem 3.7.
I think Icchyamoy is wrong : the non-zero potential surfaces are not spherical. However, if a 3rd charge of suitable magnitude and position is introduced, the surface with any non-zero potential can be made spherical by a suitable choice of the magnitude and position of the 3rd charge. See A point charge near a conducting sphere.
| {
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Why does acceleration due resulting force depends on mass while acceleration due gravity doesn't? Objects intrinsically resist to be accelerated due to their masses. A clear example would be kicking a soccer ball vs kicking a bowling ball. The latter ball will resist much more to be accelerated than the first one due to its greater mass (intrinsic property).
What if we position them in a inertial frame of reference in space? If we push both previous balls with the same force we will obtain different accelerations due to the balls' different masses, isn't it?
| Although your question isn't clear, I'll try answering what I've understood from your question. In your question, you have assumed force is constant. In case of constant force, yes acceleration will vary inversely with mass, i.e., as you say, kicking a bowling ball will produce lesser acceleration than a soccer ball.
However, in the case of acceleration due to gravity, the force is not the same for both the balls. Instead, it's greater for the ball with greater mass and lesser for the ball with lesser mass. (The acceleration is given by GM$_e$/R$^2$ and as you can see it doesn't depend on the mass of the ball.) The net effect is that the acceleration is the same for both balls. Why this is so can easily be derived from Newton's laws of Gravitation - which I'll leave to you. Hope this answers your question.
| {
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Does increasing the resistance in a branch of a parallel circuit decrease the overall current?
In the above question, why does R3 increase? If R2 increases, wouldn't the parallel combination's resistance increase? If so, wouldn't the circuit have less current? Then why would the voltage across R3 increase?
| The voltage drop across the parallel combination is equal to-
$ V - current*(R_1)$
where $V$ is the voltage across the terminals of the battery.
This is so as the sum of the voltage drops across $R_1$ and the parallel combination is equal to $V$.
Hence if current decreases, voltage across the combination increases as $V$ and $R_1$ are constant.
| {
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Definition of symmetrically ordered operator for multi-mode case? As I know, Wigner function is useful for evaluating the expectation value of an operator. But first you have to write it in a symmetrically ordered form. For example:
$$a^\dagger a = \frac{a^\dagger a + a a^\dagger -1}{2}$$
For single mode case where there is only one pair of creation and destroy operator the symmetrically ordered operator is defined. But for multi-mode case,how is it defined? For example, how would we write
$$a_1^\dagger a_1 a_2^\dagger a_2$$
in a symmetrically ordered form (such that we could easily evaluate its expectation value using Wigner function)?
| Symmetrically order expansion of the ladder operator is written as follows;
$$a_1 b_1 = (a_1 b_1 + b_1 a_1)/2= a_1 b_1 + 1/2$$
where $a$ is the creation operator and $b$ is the annihilation operator, also
$$a_1 b_1 a_2 b_2 = \frac{1}{2} (a_1 b_1 + b_1 a_1) \frac{1}{2}(a_2 b_2+b_2 a_2) = (a_1 b_1+ \frac{1}{2})(a_2 b_2+ \frac{1}{2}) = a_1 b_1 a_2b_2+ \frac{1}{2}(a_1 b_1)+\frac{1}{2}(a_2 b_2) + \frac{1}{4}$$
| {
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Moment of a force about a given axis (Torque) - Scalar or vectorial? I am studying Statics and saw that:
The moment of a force about a given axis (or Torque) is defined by the equation:
$M_X = (\vec r \times \vec F) \cdot \vec x \ \ \ $ (or $\ \tau_x = (\vec r \times \vec F) \cdot \vec x \ $)
But in my Physics class I saw:
$\vec M = \vec r \times \vec F \ \ \ $ (or
$\ \vec \tau = \vec r \times \vec F \ $)
In the first formula, the torque is a triple product vector, that is, a scalar quantity. But in the second, it is a vector. So, torque (or moment of a force) is a scalar or a vector?
| It is obviously a vector, as you can see in the 2nd formula.
What you are doing in the first one is getting the $x$-component of that vector. Rememebr that the scalar product is the projection of one vector over the other one's direction. Actually you should write $\hat{x}$ or $\vec{i}$ or $\hat{i}$ to denote that it is a unit vector. That's because a unit vector satisfies
$\vec{v}\cdot\hat{u}=|v| \cdot |1|\cdot \cos(\alpha)=v \cos(\alpha)$
and so it is the projection of the vector itself.
In conclusion, the moment is a vector, and the first formula is only catching one of its components, as noted by the subindex.
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Special Relativity: Does non inertial frame of reference work in SR? I started on my own learning about GR and SR two months ago, and I still do not have clear if it is possible or not. The following example was explained to me by someone who affirmed: "SR applies only on inertial reference frames":
Let's imagine we have two different reference frames : A' and A. Reference frame (RF) A' is moving with constant velocity (v), meanwhile RF A has no velocity (A' moves relative to A with constant v).
RF A' has a wire underneath and RF A has an aerial above. When both interact, clocks start running in both RFs (clock A' and clock A) and a light ray emerges (from the wire-aerial interaction and with the same velocity vector direction RF A' has).
Then we agree distance can be determined from both RFs.
i.e. : x = x' + vt'
Then I asked myself: why would not be correct consider the case where A' is an accelerated RF and distance is determined from RF A (i.e.) as x = x' + at'?
My doubts about if "SR applies only on inertial reference frames" sentence was true increased when I checked out more sources and they affirmed accelerated reference frames were possible in SR.
| It is possible to use accelerated reference frames in special relativity. It is more advanced than many undergraduate texts cover. But, see for example chapter 7 of "Special Relativity", A.P. French, CRC Press, 1968. There it is shown that the direction of the acceleration is not necessarily equal to the direction of the force applied to a moving body. This is done using accelerated reference frames.
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Does this Nelson formula for Feynman integral have mistakes? In this paper (Maroun's PhD dissertation, 2013) at page 46 the following formula is given (apparently without a reference):
$$\int_0^{\infty } e^{i a x^s+i b x^p} \, dx=\sum _{n=0}^{\infty } \frac{\left(i b a^{\frac{1}{s}}\right)^n \exp \left(\frac{(i \pi ) (n p+1)}{2 s}\right) \Gamma \left(\frac{n p+1}{s}\right)}{n! a^{\frac{1}{s}} \left| s\right| }$$
Now I am trying to verify the formula. If I take $a=b=i/2$, $s=p=1$ the left hand side becomes $1$ while right hand side becomes $8/3$.
It had been suggested on MathOverflow, that the formula has a mistake, and the right-hand part should contain $ba^{-p/s}$ instead of $ba^{1/s}$. In this case with the above-mentioned data the equality holds, but still it does not work for $a=-b$; $p=s$.
Is there some error in the formula? Is this formula even well-known so to look somewhere for the correct form?
| It is correct that there is a typo in the right hand side series expression. The term in the numerator of the series inside of parenthesis raised to the n power should be
$$
\left(iba^{-\frac{p}{s}}\right)^n.
$$
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Why does my lampshade cast shadows right near the bulb? I've got an upside-down lampshade in my house on a post, and I noticed today that shadows were coming off the edge of it even at 6 inches from the bulb. Could someone explain the phenomena to me?
(I'm not at all Physics-oriented [lol I didn't do as I'd hoped on my high school physics final], so please try to give me some foundation before the explanation if possible)
Reference images:
| Looks like your lampshade is made of either glass or plastic, probably with frosted surface. If that is the case, just think about the geometry: light from the bulb reaches the edge at shallow angles, and more of its path would be in that frosted surface; it effectively becomes "thicker" and therefore allows less light to pass through, forming the shadow.
| {
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Perpetual Motion Machines If you go on YouTube, you will find a large number of machines that work for almost forever. But why do all of them stop working after some time?
Which Law (Other than Conservation of Energy) prevents a machine from running till eternity?
| Most don't work due to the laws of thermodynamics, usually the first law which states that energy can neither be created or destroyed as you mentioned in your question . This stops the opportunity for many perpetual motion machines as most lose heat (and therefore energy) in the process. Without putting more energy in the total energy of the system will run out.
Aside from conservation of energy there is the second law of thermodynamics, which states that the entropy of any isolated system always increases (entropy is defined as the measure of a system's thermal energy per unit of temperature that is unavailable for doing useful work). This means that over time less of the systems energy will be able to do the work, and the machine will eventually stop.
The ones that contain magnets and objects rolling up ramps, dropping down and rolling up again etc. don't work because eventually the magnets will stop being magnetised.
Other's I've heard of are things such as capillary action by water going up a tube and dropping down. The problem with ones such as that is that if the water's gone up against gravity, it's not going to drop out without external force.
Hope this answers your question - commennt other supposedly perpetual motion machines if you're not sure of their problems and I'll see if i can find one :)
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Why the excited states of an atom have an energy width? All my experience with textbook problems of quantum mechanics shows that the energy levels associated with the bound states of a confined quantum system are discrete and sharp. For example, the energy levels of the hydrogen atom. Why is it then said that excited states of an atom have an energy width? Where does the width come from? This fact doesn't match with the examples I know in quantum mechanics. If this question is asked before can some one give the links?
| Leaving aside Doppler broadening and the other main practical reasons for line broadening, the fundamental reason is that the excited atom is coupled to all modes of the EM field equally, or at least there is an extremely wide frequency band of modes that are coupled.
So the atom "tries" to couple its excess energy into all of the modes. As it does so, destructive interference hinders the process for modes that have large frequency separation from the center frequency defined by the energy gap. If you model this broadband coupling mathematically and assume truly equal coupling to all modes at once, you get a Lorentzian lineshape whose breadth is proportional to the coupling strength. This Lorentzian linewidth tends to be narrow compared to the other, more "practical" reasons I mentioned above. I show how to do this calculation in my answer here. This is essential Wigner-Weisskopf Theory. The transition rate, which is proportional to the frequency linewidth, when reckoned by a Fermi Golden Rule calculation, is given by:
$$\Gamma_{rad}(\omega) = \frac{4\, \alpha\, \omega^3\,| \langle 1|\mathbf{r}|2\rangle |^2}{3 \,c^2}$$
with $\alpha$ being the fine structure constant, $\omega$ the center frequency and $\langle 1|\mathbf{r}|2\rangle$ being the overlap between the two electronic states on either side of the transition.
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If centripetal acceleration varies with time then which velocity do I get on integrating the acceleration As the particle is in circular motion there is no radial component of velocity but which velocity is being given by integrating the time varying acceleration then?
| From the definition of acceleration, we know that
$$\vec a = \frac{d\vec v}{dt}$$
So, $$ \int_{t_1}^{t_2} \vec a\ dt = \int_{t_1}^{t_2} d\vec v$$
$$\Rightarrow \int_{t_1}^{t_2} \vec a\ dt = \vec v_{t_2} - \vec v_{t_1}$$
$$\Rightarrow \int_{t_1}^{t_2} \vec a\ dt = \Delta \vec v$$
which is the change in the velocity vector in the time $t_1$ to $t_2$. In other words, if the particle is in uniform circular motion, you don't get a velocity when you integrate centripetal acceleration (which is the only acceleration here), you rather get the change in the velocity vector.
Also, if the particle is not executing uniform circular motion, then integrating the centripetal acceleration (which is just a component of total acceleration in this case) will not give any meaningful result.
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What does the spikes and curves in the spectral graph for compact fluorescent lights represent?
I have read from sources that the curves are generated by the phosphors in the bulb, and the spikes are caused by the mercury vapor.
However, if the mercury vapor's release of uv particles combine with the phosphors to produce visible light, then how can they give out light in different wavelengths separately?
| Using google for help, the peak around 625 nm is from Europium which is added to the phosphor to produce red light. Several of the smaller peaks greater than 625nm are also attributable to Europium as well. The large peak at about 550 nm is due to mercury as are the smaller peaks at around 415 nm and 440 nm. the phosphor also contains terbium which accounts for several of the smaller peaks around 488nm to 600nm. The collection of small peaks around 575-600nm are also from Europium. This link https://commons.wikimedia.org/wiki/File:Fluorescent_lighting_spectrum_peaks_labeled_with_colored_peaks_added.png will help explain other peaks associated with your spectrum.
By having red, green and blue light, if they are of comparable intensity, you have produced white light.
| {
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How fast can you spin a proton or electron without breaking it? If you spin a single nucleus containing multiple nucleons fast enough it will fly apart.
Is there a speed limit to a spinning proton or electron assuming it's held at a fixed location with a strong magnetic field?
What speed would either have to attain before breaking up, assuming they can be broken up? What would they break up into?
|
How fast can you spin a proton or electron without breaking it?
Even in the link you give, spin is a quantum number which can increase if the energy input to the nucleus is increased, the individual nuclei going to higher energy bound states ( or quantum mechanically defined bands) in higher angular momentum quantum numbers.
A proton is composed out of quarks. When enough energy is supplied to the proton the quarks settle at higher energy states with higher angular momentum. These are called baryonic resonances and are, as stated in the comments by Jonathan Gross, and can have spins higher than 1/2 of the proton neutron. The energies are of order of Mev, the delta resonance, spin 3/2, is about 200 MeV heavier than the proton and energy has to be supplied in the scattering process higher than that order of magnitude.
The electron is an elementary particle , a point particle , in the very well validated standard model of particle physics. It has an intrinsic spin but it is not a composite , but a point particle. It cannot be "broken up".
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Does the earth has any electric field as the earth's magnetic field is changing over time?
Earth's magnetic field changes over time because it is generated by a geodynamo
-Wikipedia
A time-varying magnetic field can produce electric field. So does the earth has electric field due to changing magnetic field?
| There are various ways to relate the electric and magnetic fields. Semoi gives one example, but in this context perhaps it is simpler to use an alternative relation:
$$ \mathbf E = -\nabla\phi - \frac{\partial\mathbf A}{\partial t} $$
This expresses the electric field $\mathbf E$ as the sum of two contributions. The first term:
$$ \mathbf E = -\nabla\phi $$
is the contribution to the field from any charges present. The quantity $\phi$ is simply the electrical potential. As it happens the Earth has a net negative charge due to charge separation between the ground and the atmosphere, so it has an electric field regardless of what its magnetic field does.
The second term:
$$ \mathbf E = - \frac{\partial\mathbf A}{\partial t} $$
gives the contribution to the electric field from the changing magnetic field. The quantity $\mathbf A$ is a type of magnetic potential energy called the magnetic vector potential.
And you are quite correct that any changes in the magnetic field of the Earth will indeed result in a contribution to the total electric field. However the sorts of changes that Wikipedia is describing happen on very long timescales and the rate of change $\partial\mathbf A/\partial t$ is so small that any contribution it makes to the electric field is negligable.
On the other hand the Earth's magnetic field changes on much shorter timescales due to changes in the solar wind. Indeed a very large solar flare can cause large and rapid changes in the magnetic field that produce large electric fields. These fields are large enough to cause large currents to flow in electrical transmission lines and can cause power outages as a result.
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How many dimensions are there in the electric field? I am not a physicist. I am buying some polariser for my camera. Circular polariser intrigues me. Basically you pass light through a linear polariser, then through a waveplate, you get circular polarisation.
Wikipedia says the following:
By adjusting the thickness of the wave plate one can control how much the horizontal component is delayed relative to vertical component before the light leaves the wave plate and they begin again to travel at the same speed.
Does that mean the electric field in an electromagnetic wave is a 2D vector field? I am a bit confused. I thought in 3D space, the electric field should be 3D vector field.
| In an electromagnetic wave (light), the electric and magnetic fields are perpendicular to the direction of travel. If the wave is traveling in the $z$-direction, then $E_z = B_z = 0$. So, yes, the fields are two-dimensional.
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Quantum Mechanic Interpretations Context I was watching this video "Do we have to accept Quantum weirdness?" and it said how currently there are several interpretations of quantum mechanics that are consistent with the facts.
Question For quantum mechanics, how does one decide on which is the "correct" interpretation? What experiments could be done that could show one is correct and the other fails to predict such results?
Thanks
| That is precisely the problem. David Mermin said somewhere that every year a new interpretation of quantum mechanics is introduce and none are ever ruled out. This clearly points to the fact that interpretations of quantum mechanics are not Popper falsifiable. Thus, according to strict definition, interpretations of quantum mechanics do not qualify as a science. It lies in the domain of philosophy.
Now, this does not mean that somebody would not some day come up with a way to test (some of) these interpretations experimentally. However, I wouldn't hold my breath.
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What does the refractive index for e.g. alpha mean? When I look for some materials like https://en.wikipedia.org/wiki/Barium_sulfate and want to extract the refractive index then there is written: (nD)=1.636 (alpha). And sometimes also for beta and gamma.
What does this mean? The refractive index is mostly dependent on the wavelength so why is alpha, beta and gamma the only value given?
| Take RI in two of the three optic axis directions in Biaxial gemstones. ie. Alexandrites and tanzanite positive signs.
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Heat produced by a conductor I recently got to know about how the heat produced in a conductor. It's due to collision of electrons when drifting due to electric voltage But my question is Why doesn't a non current carrying conductor doesn't heat up due to collision of electrons in it as they are in random motion?
| In the end the energy that is converted to heat has to come from somewhere. In the sketched situation the electron has to be in an excited electron state so that it can relax into an energetically lower lying state due to the scattering event. If there is no net current in the conductor the electronic system is in thermal equilibrium with its environment. The electrons fill up the available states in the conductor according to the Fermi distribution. This means that the scattering event you sketch is just as probable as the opposite, i.e., an electron in a lower lying state gets excited due to its surroundings. If there is a net current in the system we are in a non-equilibrium situation. Electrons have to be in an excited state to contribute to this net current. Now the scattering process in which the electron motion is turned into heat is more probable than the opposite.
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If the bicep connects closely to the elbow joint, would the bicep need to exert more or less force to lift an object? I would like to know whether a less or large amount of force is exerted by the bicep to lift and hold an object if the bicep is connected close to the elbow joint (fulcrum) than if the bicep is connected more closely to the wrist. Why? And does torque play a role in this?
| To figure this out you can use the pivot formula. If the applied force $F_A$ is distance $r_A$ from the fulcrum and the exerted force, $F_E$, is distance $r_E$ from the fulcrum, then
$$
F_E=F_A\times r_A/r_E.
$$
If the bicep is connected close too the elbow joint it will have a smaller $r_E$ than if it is connected close to the wrist. Therefore, if the bicep is connected close too the elbow joint it will need a larger exerted force than if it is connected close to the wrist.
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Does work done depend on the frame of reference? Suppose I am sitting on a bench and looking at a moving car. Force is applied on the car by its engine, and it makes it displace, hence some work is done on the car. But what if I am sitting in the car and looking at the bench? The bench covers some displacement, but who has applied force to it? Is any work done on it?
| If your velocity is zero then due to work energy theorem the work done would be zero. But if you are accelerating then pseudo force would do the work
| {
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Can we derive Einstein-Hilbert action through action principle and Levi-Civita connection? Suppose that we take principle of least action as given. Also assume that any manifold allowed by the action would carry Levi-Civita connection (torsion-free characteristic). Also assume that the local symmetry imposed on the tangent space of each manifold point is that of Poincare group, via general covariance principle.
Would these be sufficient to derive Einstein-Hilbert action, and by corollary Einstein field equations? Or do we need extra conditions to derive the Einstein-Hilbert action?
Edit: If not, then what would be other extra conditions?
| No, any action that is a scalar would satisfy your requirements. For example, you could have various terms that are functions of the various scalar curvatures $R$, $R_{ab}R^{ab}$, or $R_{abcd}R^{abcd}$, to give a few examples; you could probably invent more. You need some further requirement to fix the Lagrangian to be just $R$.
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Photoelectric effect, low frequency light Let's say we have a emitter, emitting light that has frequency f, less than the threshold frequency of a metal.
If you leave light shining onto that metal, for long enough, does the energy of the individual photons accumulate, on the electrons, so eventually they will ionize, or does this not happen? What am I missing?
| For simplicity let's consider the photoelectric effect in a thin metal foil:
The first step in the photoelectric effect is when a photon strikes an electron in the metal and transfers all its energy to it. The electron energy is now equal to the photon energy $h\nu$. If this energy is greater then the work function $\phi$ the electron can escape the metal and will emerge with a kinetic energy:
$$ \tfrac{1}{2}mv^2 = h\nu- \phi $$
However the $h\nu \lt \phi$ the electron will in effect bounce off the metal-air interface back into the metal:
and the electron will start rattling around inside the metal. The trouble is that the metal has some resistance to the motion of electrons and the electron will very quickly lose its energy and come to a halt. By very quickly I mean less than a nanosecond.
So if a second photon strikes the electron before the electron has slowed to a halt, and while the electron is travelling in the right direction then yes the second photon could add enough energy to eject the electron. So in that case we would have photoelectrons ejected by absorbing two photons.
However this process is very unlikely as the two photons would have to be absorbed within a very short time. In practice the rate at which photoelectrons are ejected by two (or more) photon absorption is very slow though it can be observed in special cases. For example the paper Double-Quantum Photoelectric Emission from Sodium Metal by M. C. Teich, J. M. Schroeer, and G. J. Wolga, Phys. Rev. Lett. 13, 611, 1964 reports observation of exactly this effect in sodium.
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Why is the direction of magnetic field from South to North Pole inside a magnet? Since magnetic field lines are the path taken by a hypothetical North Pole when it is in range of a magnetic field of a magnet, it is clear that the direction of hypothetical North Pole would be from North Pole to South Pole of a magnet not even outside the magnet but inside the magnet too, because North Pole of magnet will repel the hypothetical North Pole inside and outside the magnet and would be attracted by the South Pole of the magnet.
| In nature, there is no magnetic monopole discovered yet. All of the magnets we have are created by certain kind of current (like the spin of an electron). Thus, the prototype of a magnet is a solenoid.
Now, there are a bunch of ways to argue the direction of the magnetic field in the solenoid. If you took introductory physics before, please use Biot-Savart. If not, the most intuitive way is probably assuming that magnetic field should be smooth (this is based on the assumption of no magnetic monopole). The north pole is defined as where the magnetic field comes out. Due to the smoothness assumption, even when you go into the solenoid a little bit (from north pole), the direction of the magnetic field should be the same, which is now "pointing toward north pole". Thus, it should be pointing from south pole to north pole inside.
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How is pion decay compatible with conservation of angular momentum? A pion is a spin-zero composite particle, so $S = S^z = 0$. A $\pi^-$ pion can decay into an antineutrino $\bar{\nu}$ and a negatively charged lepton $l$, each with spin-$1/2$. Let the direction of antineutrino motion be the positive $z$-axis. Since all antineutrinos are right-handed (neglecting neutrino masses), the neutrino must have $S^z = +1/2$. Conservation of $S^z_\text{tot} = 0$ then requires that the lepton have $S^z = -1/2$. But then the spin degrees of freedom
$$| \uparrow \rangle_{\bar{\nu}} | \downarrow \rangle_l = \frac{1}{\sqrt{2}} \left[ \frac{1}{\sqrt{2}} \left( | \uparrow \rangle_{\bar{\nu}} | \downarrow \rangle_l - | \downarrow \rangle_{\bar{\nu}} | \uparrow \rangle_l \right) \right] + \frac{1}{\sqrt{2}} \left[ \frac{1}{\sqrt{2}} \left( | \uparrow \rangle_{\bar{\nu}} | \downarrow \rangle_l + | \downarrow \rangle_{\bar{\nu}} | \uparrow \rangle_l \right) \right]$$
seem to have components in both the $S_\text{tot} = 0$ (singlet) and $S_\text{tot} = 1$ (triplet) sectors. How is this compatible with the conservation of $S_\text{tot} = 0$?
| Only the component of the spin wavefunction that has a nonzero projection onto $S=0$ will be present in the decay. That factor of $\frac{1}{\sqrt 2}$ gets factored in and reduces the decay probability by half (which is already taken into account).
| {
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Can anyone explain this condensation pattern? What is the hair-like "polymer" that condenses on the caps of the plastic cell vial? Inside vial are cells in a mixture of FBS and DMSO. The vials were frozen slowly in a special isopropanol-filled container (isopropanol does not touch the vials) to -80 degrees celsius, then put on dry-ice to move to a liquid nitrogen cryogenic freezer. The hair-like crystals appear to be water, but bend toward my finger when nearby, like static attraction between a comb and faucet stream.
I suspect the has something to do with DMSO/isopropanol leaking out, and the very low temperatures that causes this strange condensation pattern, but I'm not sure.
| The polystyrene container and your comments about the hairs moving suggest that they might be caused by static electricity.
This video shows ice crystals which formed on a plastic tube rack in a bucket of dry ice. They are attracted to a finger by static electricity, just as you described.
Water is made of polar molecules which can be aligned by an electric field. Water molecules in the air are attracted to the charged ends of the filaments of ice. The ends remains charged as the filaments grow. Separate filaments are kept apart from each other by the static charge, just like the hair of someone touching a Van de Graaff generator.
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Varying magnetic field, and a larger area? Considering this classical example:
The magnetic flux($\phi$) is increasing over some time(increase in $B$), and it's simple to indicate the direction of the induced current using Lenz law.
However, the magnetic field is only in a specific region, what if the magnetic field would increase, covering a larger area?
I can't find a current flow(pattern) for the loop that would resist the change,nor explain why that is.
| A conducting metal coil in a varying magnetic field opposes the change in magnetic flux through it. This opposition induces a current in the loop, the direction of which can be found out by Lenz's Law.
Now we know that magnetic flux is given by $$\phi_B = \int B.dA$$ which is the summation of the scalar products of the magnetic field and the differential area elements enclosed. Now, as said earlier, the induced current is generated by the opposing the change in flux and as the flux is calculated keeping in mind the area enclosed by the coil only and not its surroundings so we will not consider the larger area but only the area enclosed by the loop.
So as the change in magnetic field outside the loop does not affect current induced due to change in flux of the loop so in this case too the current will flow in anti clockwise direction and will have the same magnitude as in the first case, as illustrated below.
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Continuum mechanical analogous of Maxwell stress tensor Maxwell stress tensor $\bar{\bar{\mathbf{T}}}$ in the static case can be used to determine the total force $\mathbf{f}$ acting on a system of charges contanined in the volume bounded by $S$
$$ \int_{S} \bar{\bar{\mathbf{T}}} \cdot \mathbf n \,\,d S=\mathbf{f}= \frac{d}{dt} \mathbf {{Q_{mech}}}\tag{1}$$
Where $ \mathbf {{Q_{mech}}}$ is the (mechanical) momentum of the system of charges.
What theorem/relation is formally analogous to $(1)$ in continuum mechanics? I've read that also in continuum mechanic one can introduce a tensor such that the value of its components on a surface $S$ enclosing a system of masses determines the forces acting on the masses completely.
I could not find this analogy on Jackson or Griffiths, so what is the tensor that is similar to Maxwell stress tensor in mechanics? Is it the stress tensor? By which theorem does it determine the forces on a system of masses?
| Let's consider a point $\renewcommand{\vec}[1]{\mathbf{#1}}\vec{x}$ of the undeformed material, which moves to $\vec{x}'$ after deformation. We define the displacement vector
$$\vec{u} = \vec{x}'-\vec{x}.$$
The variation of $dl^2=d\vec{x}^2$ is then given by (implicit summation on repeated indices everywhere)
$$dl'^2 = dl^2 + 2u_{ik}dx_idx_k,$$
where $u_{ik}$ is the strain tensor,
$$\newcommand{\partialder}[2]{\frac{\partial{#1}}{\partial{#2}}}u_{ik}=\frac{1}{2}\left(\partialder{u_i}{x_k}\partialder{u_k}{u_i}+\partialder{u_l}{x_i}\partialder{u_l}{x_k}\right).$$
Then there is a stress tensor $\sigma_{ik}$ such that the force on a volume $V$ is given by
$$F_i = \int_{\partial V} \sigma_{ik}ds_k,$$
where $ds_k$ are the component of the infinitesimal element of surface. The key postulate to derive this formula is that an element of material exert a force on its immediate neighbourhood only, which precludes the existence of macroscopic electric field, as generated in piezoelectric materials for example, when the material is deformed.
Finally, a widely used model is Hooke's law, which gives the stress tensor knowing the strain tensor,
$$\sigma_{ik} = Ku_{ll}\delta_{ik}+2\mu(u_{ik}-\frac{1}{3}u_{ll}\delta_{ik}),$$
where $K$ is called the bulk modulus and $\mu$ is called the shear modulus.
Reference (shamelessly plundered above!): L.D. Landau and E.M. Lifshitz. Theory of Elasticity, volume 7 of Course of Theoretical Physics. Pergamon Press, 1970.
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What do they mean when they say that it does not require any work to move a charge from one point to another in an equipotential surface? In the textbook it says that no work is required to move a charge from one point to another on an equipotential surface. Do they mean work by the electric field or work by anything? Because clearly the object cant just magically move sideways with nothing.
| If the particle is on an equipotential surface, then that means there is no force from the electrostatic field on your charge, while moving along that surface. If there are no external forces, that means that locally momentum in that direction is conserved, just consider Newton's laws:
$$ \frac{\text{d}\boldsymbol{p}}{\text{d}t} = \boldsymbol{F} = 0 $$
However in moving your particle, you would change the velocity, so we require
$$\Delta \boldsymbol{p} = m_q \Delta \boldsymbol{v} \neq 0$$
Hence, to move your particle along an equipotential surface, you don't need to do any work, but you do need supply a change to the particle's momentum. Hence the particle can't spontaneously change its trajectory while on an equipotential surface. The issue here is when changing the momentum, in general you will change the energy too, and end up doing some work with respect to whatever external field you're using to push or pull your charge around.
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Do gases mix faster when of unequal temperature? Or,
Is my room aired quicker during winter?
I always feel that in winter, after opening the window, my study room becomes breathable quicker than in summer. Now, this is of course highly subjective, but even though I see no rational explanation supporting such a phenomenon, I neither can intuitively outrule it.
Surprisingly, Google was no help here.
| Probably yes.
The warmer air inside your room is less dense than the colder air from outside, so the latter can easily flow over the windowsill into the room, dislodging the warmer air, which escapes like smoke through the window.
If there are two openings allowing a strong draft to form, there shouldn't be much difference between summer and winter, but you'll probably still find it "becoming breathable quicker" in winter, since the additional temperature difference makes the fresh air more noticeable.
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Does Snell's window (the optical man-hole) work both ways? My physics book states: Just because you can see a fish under water (you are above; the fish is below) doesn't mean the fish can see you due to Snell's window.
Is this true? I would have assumed that someone standing in a boat surrounded by water would also have a Snell's window. Surely if light can travel from the fish to your eye then it can also make the reverse journey.
| Yes, light travels the reverse way. You cannot realize an optical diode with linear optical elements in absence of fields other the light itself. (Faraday effect with static magnetic field is needed for an optical diode for instance).
Most likely the textbook refers to the case where one practically cannot see the fish due to strong light coming through the (other) reflection path. But this is due to limited dynamic range of our vision. The light paths are still perfectly reversible.
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Does entropy really not increase here?
Two vessels separated by a partition have equal volume $V_0$ and equal temperature $T_0$. They both contain the same ideal gas, and the particles are indistinguishable. The left vessel has pressure $P_0$ and the right vessel has pressure $2P_0$. After the partition is removed and the system equilibrates, what is the net change in entropy?
At first my intuition tells me that $\Delta S=0$ since, for the whole system, $dQ=0$ and $dW=0$ throughout the entire process. Moreover, if we look at only one vessel - say, the left one - the change in entropy is given by:
$$\begin{align}\Delta S_1&=\frac{\Delta E_1}{T_0}\\
&=\frac{\Delta (c_V NT)}{T_0}\\
&=\frac{c_V T_0\Delta (N)}{T_0}\\
&=c_V\frac{N_0}{2} \,\,\,\,\textrm{(from ideal gas formula)}
\end{align}$$
If we do it for the right vessel, we simply get $\Delta S_2 = -\Delta S_1$, so once again $\Delta S=0$. But the accepted answer to this Physics SE question says otherwise. Is there something I'm missing?
[EDIT] To show where I got $\Delta N$, notice that if the left vessel has $N_0$ moles of gas at first, the right vessel has $2N_0$ moles of gas.
$$N_2=\frac{P_2T_0}{V_0}=\frac{2P_0T_0}{V_0}=2N_0$$
So in the end, the total system will have $N_0+2N_0=3N_0$ moles of gas. Since the volumes are equal, in the end each will have $\frac{3}{2}N_0$ moles of gas.
| The change in entropy is certainly not zero. It is greater than zero for this spontaneous process. Just because the Q in an irreversible process is zero does not mean that the entropy change is zero. The entropy change is the integral of dQ/T only for a reversible path.
I get $\frac{3}{2}P_0$ for the final pressure. The initial number of moles in the left container is $\frac{P_0V_0}{RT_0}$ and the initial number of moles in the right container is $\frac{2P_0V_0}{RT_0}$. If the initial moles in the left container goes from $P_0$ to $1.5P_0$ (compression) at temperature $T_0$, what is its change in entropy? If the initial moles in the right container goes from $2P_0$ to $1.5P_0$ (expansion) at temperature $T_0$, what is its change in entropy? What is the total change in entropy for the process?
You can check your result against my final answer of $$\Delta S=\frac{P_0V_0}{T_0}\ln{(32/27)}$$
EDIT: If you increase the pressure on an ideal gas isothermally and reversibly, then dU=0. So, $$TdS=PdV=d(PV)-VdP=-VdP=-\frac{nRT}{P}dP$$Integrating, we get:$$\Delta S=-nR\ln(P_2/P_1)$$
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How much time will it take to move an object whose length is equal to one light year? Suppose there's a stick whose length is one light year and I push it from one side by one centimeter. How much time would it take for its other side to move by one centimeter and why?
| It depends on the material. When you push one end of the stick, you move the atoms at the very end of the stick. Those atoms push the atoms next to them, those atoms push the next atoms, and so on down the stick. This is a sound wave that travels down the stick, so the time you have to wait for the other end to move is the length of the stick divided by the speed of sound in the material of the stick. If the stick is wooden, the speed of sound is about 4000 m/s (compared with 330 m/s in air). It would take $\frac{9.5\cdot10^{15}\,m}{4000\,m/s} = 2.4\cdot10^{12}$ seconds (74 000 years) for the other end of the stick that is a light-year away to move.
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Has the curvature of spacetime been measured at the human scale? The curvature of spacetime has been observed many times from the deflection of light around massive astronomical objects. But has it been observed around small objects in a lab?
In the Cavendish experiment, the gravitational attraction between two masses does sufficient effect for it to be measured on Earth. Thus, it raises the question whether light deflection from curved spacetime is also measurable in the lab.
If it has not been achieved, how far are we from it? How much precision would be needed?
| If you are willing to make gravitational time dilation count and if about a mile of elevation counts as human scale, then this is the coolest of my bookmarks I can share: http://www.leapsecond.com/great2005/
The author is an expert on atomic clocks and he set up an experiment to show they ran differently 1340 meters higher, by driving up Mt Rainier, for the education of his kids. Scores high in the category "my dad is a geek!" But cool.
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Question when comparing two experimental results In my textbook it says that if two experimental results vary less than 3$\sigma$ then they can be considered to have arrived at the same result. My question is how do you determine this "x$\sigma$".
For example if i did an experiment to calculate $g$ and my result was $g$=9.79$\pm$0.07 and i want to compare it to $g$$^´$=9.80$\pm$0.22 , should i just use $$\frac{g^´-g}{error}\,\,\,\,?$$
But then what error do i use?If i use o.o7 i get a difference of 1.4$\sigma$ but if i use 0.22 i get 0,45$\sigma$. Or should i use a combination of both?
| Given that both measurement have a 1 std. dev. random error estimate (all bets are off when systemantics and model dependencies rear their ugly heads!), then you are effectively comparing the difference of the measurements with zero.
So the error you use is the error of the difference.
Which means propagating the error in the usual way (i.e. adding the errors in quadrature)
$$ \left[\frac{g' - g}{\sqrt{(\Delta g')^2 + (\Delta g)^2}}\right] \stackrel{?}{\le} 1 \;.$$
If the computed value is
*
*less than one the measurements are in agreement.
*more than a few the measurements clearly disagree.
*bewtween 1 and—say—3 is more ambiguous.
But your instructor may want you to treat them as disagreeing so that you can have a yes/no answer.
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How can energy of oscillators be quantised but they can still vibrate at all frequencies? In Black Body radiation, Planck's law has a postulate saying
The wall of black body contains oscillators of all possible frequencies,ν.
There is one more postulate which says
The energy of these oscillators is not continuous but discrete valued.
Of course the second one is very well known but doesn't it contradict the first one? Shouldn't it mean that frequencies of vibration are quantised like maybe in case of standing waves?
Please correct me if I am wrong in stating the postulates itself.
| Your first statement can be understood if you take into account that a blackbody, is -by definition- an object or system which absorbs all radiation incident upon it and emits energy which is characteristic of this radiating system only, not dependent upon the frequencies which are incident upon it. So, it has to contain osscillators of all possible frequencies, since it absorbs (and re-emits) all frequencies.
The meaning of the second statement, is that since the re-emited energy is produced by standing waves or resonant modes of the cavity which is radiating, it has to be discrete.
In other words, although we have oscillators at all possible frequencies, each one of them has a discrete spectrum.
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Does quantum mechanics allows absolute energies? When we solve Schrödinger equation:
$$H\Psi=E\Psi$$
for a system, is $E$ an absolute energy? I mean, an energy that is not a difference between states.
Does this equation implies something about absolute or non-absolute entropy?
| The problem is that the Hamiltonian operator contains a potential energy $V$:
$$ \hat{H} = \frac{-\hbar^2}{2m}\nabla^2 + V $$
and the potential energy isn't an absolute energy because we are free to set the zero of the potential energy wherever we want. So the energy in your equation is not an absolute energy but depends on where we take the zero for the potential energy.
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Confusion regarding gravity not being a force In high school, it was taught that formula for describing circular orbital velocity around a central body is derived by equating Newton's law of gravity with the centripetal force formula (under the logic that the inwards centeipetal force required is provided by the gravitational "force").
It was only recently that I discovered that gravity isn't actually a force but is actually a distortion of space time. (I came across this while wondering why light bends around large masses).
Does the fact that gravity is not a force make the above derivation of orbital velocity any less valid? Because the above derivation assumes that gravity is a force.
| TL:DR
Newton's law of gravity (used to describe velocity, and not force) is not wrong, it is just imprecise and has a limited scope for which it is accurate (I would not equate less widely applicable with less valid).
No, the fact that gravity is not a force, as Newton described it, does not mean that his calculations for describing circular orbital velocity are incorrect. Given that his formulae were derived based upon measurements of the same world as Einstein's formulae were, they are both designed to (and do a good job) describe the motion of large masses under the influence of gravity.
Newton's laws are, as you would expect given the time difference between Newton and Einstein, less precise than Einstein's, and they fail under particular circumstances, but overall, they do a pretty good job at describing a large chunk of gravitational effects at the precision that is necessary for a vast majority of application.
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What is the $\{q_i\}$-$t$ space called? In classical mechanics, the configuration space of a system of $f$ degrees of freedom is the $f$-dimensional space of the set of generalized coordinates $\{q_i\}=(q_1,q_2...q_f)$ of a system. While talking about the principle of extremum action, one draws a path in the $\{q_i\}$-$t$ "plane" where $t$ represents time. Does this space which contains both the set of generalized coordinate and time has any name?
| The name could vary from author to author. In a non-technical setting I think you would find it under the name "extended configuration space". In more specialized contexts (e.g. Hamiltonian mechanics on symplectic spaces, field theory on fiber bundles, etc.) they are called contact manifolds or even just configuration spaces.
| {
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"url": "https://physics.stackexchange.com/questions/357454",
"timestamp": "2023-03-29T00:00:00",
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Quantum Mechanics and Schur's lemma Today i was studying on a textbook and i crossed a paragraph that confused me a little.
Suppose you have an algebra generated by $\hat{X}$ and $\hat{P}$ and a function $f(\hat{X},\hat{P})$ that commutes with $\hat{X}$ and $\hat{P}$. Then you can prove that this function is proportional to the identity.
Here comes the issue, i don't understand the proof of this.
Reasoning in Position representation (meaning that $\hat{X}=x$ and $\hat{P}=-i\hbar\frac{d}{dx}$), i don't get why $[x,f]=0$ implies that $f=f(x)$ and why $[-i\hbar\frac{d}{dx},f]=0$ implies that $i\hbar\frac{df(x)}{dx}=0$.
| If $[x,f]=0$ then $f$ belongs to the commutant of $x$. Of course $x$ is in such a commutant, but $p$ isn't, therefore $f$ is a function of $x$ alone. Now, for any vector $\psi$ in the Hilbert space of the Schroedinger representation,
$$(f\psi)'(x) - (f\psi') = 0,\qquad\forall x$$
which implies that $(f'\psi)(x)=0$ for any $x$, hence $f'$ is the zero operator. Therefore $[p,f]=0$ implies $f'=0$, i.e. $f$ is a constant multiple of the identity operator.
| {
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"timestamp": "2023-03-29T00:00:00",
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Would mechanically moving electrons create a super-strong magnetic field? According to this Veritasium video, the magnetic field in a wire with a non-zero current is an artifact of special relativity. A moving charge sees a speed difference between the wire and the electrons in it (since the electrons are moving). Due to Lorentz contraction there appears to be a net positive charge and a repelling force arises. The physical speed of electrons supporting a current is very slow, a few micrometers per seconds, but due to the number of electrons and the strength of the electric field the effect is macroscopic.
What if you could mechanically increase this speed? Say you had a charged capacitor, and one of the plates was given a parallel velocity. E.g. two concentric cylinders with a charge imbalance, with one of them attached to a motor. Wouldn't that create a very strong apparent magnetic field?
|
What if you could mechanically increase this speed? Say you had a charged capacitor, and one of the plates was given a parallel velocity. E.g. two concentric cylinders with a charge imbalance, with one of them attached to a motor. Wouldn't that create a very strong apparent magnetic field?
Well how large is relativistic change if a capacitor plate changes its speed from 0 to 100 m/s? Think about it.
Here's my answer:
At non-relativistic speeds the magnetic force is a very tiny fraction of the electrostatic force. (It's equally small as the length contraction of objects at that speed)
At relativistic speed 0.87 c the magnetic force is 50% of the electrostatic force. If there is an electrostatic force of 1 N between capacitor plates, then when the plates move at speed 0.87 c, there is a 0.5 N magnetic force between those plates. The plates move together.
If only one plate moves, then there is no magnetic force between the plates, because there is only one electric current. (The Veritasium video may disagree with that, which means that the video is not perfectly correct)
| {
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"timestamp": "2023-03-29T00:00:00",
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Tensor in different coordinate system I have the tensors $F_{\mu\nu}$, $F^{\mu\nu}$ in coordinate system $(t,x,y,z)$ and want to transform these to coordinate system $(t',x',y',z')$ just by multiplicating matrices.
My idea was to calculate the Jacobians $J=(\frac{\partial x^i}{\partial x'^j})_{ij}$ and $J'=(\frac{\partial x'^i}{\partial x^j})_{ij}$.
Then I would find $$F'_{\mu\nu}=J^\top F_{\mu\nu}J$$ and $$F'^{\mu\nu}=J' F^{\mu\nu}J'^\top,$$ in matrix notation.
Is this correct?
My ultimate goal is to prove that $F_{\mu\nu}F^{\mu\nu}$ is the same in both systems, however calculating this explicitly does not give me this result.
| The idea is more or less correct but you need to be careful with the mixing of matrix and tensor notation. Let's say that $F_{\mu\nu}$ are the components of a matrix $F$, and $F^{\mu\nu}$ those of $F_U$ (for "upper"). Then $F_{\mu\nu}F^{\mu\nu}$ becomes $\operatorname{tr}(F F_U^T)$. Also you need to notice that $J'$ is the inverse of $J$.
With this notation, the transformations are $F' = J^T F J$ and $F_U' = J' F_U J'^T$, and the trace is transformed to $\operatorname{tr}(J^T F J J' F_U^T J'^T) = \operatorname{tr}(J^T F F_U^T J'^T) = \operatorname{tr}(F F_U^T J'^T J^T) = \operatorname{tr}(F F_U^T)$, using the cyclic property of the trace in the middle.
| {
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"timestamp": "2023-03-29T00:00:00",
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When is the heat transfer between a solid and fluid conduction or convection? I know that the heat transfer between solids and liquids occur via both conduction and convection.
However, I am not sure about the fine line that separates them. For example, what is the mode of heat transfer when a hot piece of steel is put out in the air. Does the wind change the situation?
Or is, for instance, the heat transfer between a hot solid and a cold static liquid, which are in the same container conduction or convection and why?
| If a hot metal object placed openly then heat can be transferred
*
*by means of conducting (the molecules in contact to the metal block)
*by convection (the flow of hot air in contact to the block)
*and by radiation.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Physical difference of waves produced by a deep and a treble I tried to find what the difference is on the pressure wave caused by a bass speaker and a wave produced by a treble speaker.
Off course the frequency is lower on the bass wave, but is that all? Why does the waves from the bass speaker appear to have a much stronger and bigger sound intensity than the treble speaker?
Thank you very much!
| Because the bass frequency is lower, its wavelength is longer, because the speed of sound is roughly constant, so the sound wave travels further in one full cycle.
But the efficiency of the speaker depends on the ratio of its diameter to the wavelength. So a bass speaker must be bigger than a treble speaker.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why are there two definitions for the functional derivative? I have seen two definitions for the functional derivative. Why are there two definitions?
*
*In Goldstein's Classical mechanics 3rd edition page 574 eq. (13.63), and also in a Student's Guide to Lagrangians and Hamiltonans by Patrick Hamill on page 55 eq. (2.10), the functional derivative of a function $\Phi(y,y',x) $, where $y = y(x)$, is given by
$$ \frac{\delta \Phi}{\delta y} = \frac{\partial \Phi}{\partial y}-\frac{d}{dx} \frac{\partial \Phi}{\partial y'} .$$
*The second definition of a functional derivative is given by $$\frac{\delta F[y(x)]}{\delta y(x')} = \lim_{\varepsilon \rightarrow 0}\frac{1}{\varepsilon} (F[y(x) + \varepsilon \delta(x-x')]-F[y(x)]).$$ This definition is found on wikipedia and is used in QFT. This definition tells us that for the functional
$$ S[y(x)] = \int \Phi(y,y',x)dx$$ where $y =y(x)$, the functional derivative is given by $$ \frac{\delta S}{\delta y} = \frac{\partial \Phi}{\partial y}-\frac{d}{dx} \frac{\partial \Phi}{\partial y'} $$
One definition is in terms of a function and the other in terms of a functional, but both give the same output?
| For a functional
$$S~=~\int d^nx ~{\cal L}(x) , \qquad {\cal L}(x)~\equiv~ {\cal L}(\phi(x), \partial \phi(x), \ldots, x),\tag{0}$$
the second definition with notation
$$\frac{\delta S}{\delta\phi^{\alpha} (x)}\tag{2}$$
is the traditional definition of functional/variational derivative (FD), while the first definition with notation
$$\frac{\delta {\cal L}(x)}{\delta\phi^{\alpha} (x)}\tag{1}$$
is the so-called 'same-spacetime' FD, which obscures/betrays its variational origin, but is often used for notational convenience. For more details, see e.g. my Phys.SE answers here and here.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/358417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is there a "square root" version of the Einstein field equation? It is well known that the Klein-Gordon equation have a kind of "square root" version : the Dirac equation.
The Maxwell equations can also be formulated in a Dirac way.
It is also well known that the metric of general relativity have a kind of "square root" version : the tetrad field (or vierbein) of components $e_{\mu}^a(x)$ :
\begin{equation}\tag{1}
g_{\mu \nu}(x) = \eta_{ab} \, e_{\mu}^a(x) \, e_{\nu}^b(x).
\end{equation}
Now, a natural question to ask is if the full Einstein equations :
\begin{equation}\tag{2}
G_{\mu \nu} + \Lambda \, g_{\mu \nu} = -\, \kappa \, T_{\mu \nu},
\end{equation}
could be reformulated for the tetrad field only (or other variables ?), as a kind of a "Dirac version" of it ? In other words : is there a "square root" version of equation (2) ?
| By taking the "Dirac square root" of the Hamiltonian constraint for GR, you naturally end up with Supergravity...so in some appropriate sense, SUGRA "is" a "square root" of GR. For more on this, see:
*
*Romualdo Tabensky, Claudio Teitelboim, "The square root of general relativity". Physics Letters B 69 no.4 (1977) pp 453-456. Eprint
*Claudio Teitelboim, "Supergravity and Square Roots of Constraints". Phys. Rev. Lett. 38 (1977) 1106. Eprint
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/358501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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"answer_id": 1
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Construct operator such that division of expectation values is equal to expectation value of the operator Is is possible to construct an operator $\hat{C}$ out of $\hat{A}$ and $\hat{B}$ such that:
$$\frac{\langle \psi|\hat{A}|\psi\rangle}{\langle\psi|\hat{B}|\psi\rangle} = \langle \psi|\hat{C}|\psi\rangle,$$
for any state $|\psi\rangle$?
| No. The r.h.s. is a quadratic form, while the l.h.s. is not. For example, under $\psi\to\lambda\psi$ the r.h.s. scales as $\lambda^2$ while the l.h.s. stays the same.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/359046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
The boundary conditions in a waveguide Suppose a cubic waveguide, made of perfect conductor, has only two open parallel sides. And the boundary conditions in this case are that the electric field at the surface must satisfy:
$$\vec{B} \cdot \vec{n}=0,$$
and magnetic field:
$$\vec{E} \times \vec{n}=0,$$
where the $\vec{n}$ is the normal vector pointing outwards from the conductor. These two relations come from the equations:
$$\nabla \cdot \vec{B}=0,$$
$$\nabla \times \vec{E}=0.$$
The question is how to derive the other boundary condition that at the surface the electric field must satisfy:
$$\frac{\partial{E_n}}{\partial n}=0.$$
$E_n$ means the electric field along normal direction.
| This simply follows from the Gauss' law:
$$\nabla\cdot\vec E=\frac\rho{\varepsilon_0}.\tag1$$
Since we know that $\vec E\parallel\vec n$ at the boundary inside the waveguide, the divergence of $\vec E$ reduces$^\dagger$ to $$\nabla\cdot\vec E=\frac{\partial E_n}{\partial n}.\tag2$$
Since the waveguide doesn't have any charges inside, $\rho=0$ in the whole internal region including the vicinity of the conductor. Inserting this into $(1)$ simplified by $(2)$, we get our boundary condition.
$^\dagger$ This is true because $\vec E$ is analytic at the interface (with $n=0^+$), in which case $\lim\limits_{n\to0}E_m\to0$, where $m$ is the coordinate along any tangent to the interface and $n$ – along the normal, implies $\frac{\partial E_m}{\partial m}\to0$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/359176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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