Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
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Can an atom be split and put back together? I had recent came across this question when unintentionally tearing a piece of my journal paper. These atoms' bonds are pulled apart when the paper is torn, but is there a way to put them back together?
Now i know that, depending on the object, the method can vary. According to an article i read Franken-Physics: Atoms split then put back together this can be done. However i wasnt given enough info on what atom they split, as they can vary per object.
Though i am not well educated in the field of quantum physics i understand in some instances this will work. Can anyone clear this up for me regarding what types of materials can undergo this.
Edit: Forgive my ignorance when i said they are obviously split. i do realize the bonds were pulled apart. i just didn't word it as so
| I just wanted to point out that an atom cannot be split (or cut loose in such a way that it remains like that freely), they would have to recombine in order to achieve their stable electron configuration.
| {
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Why don't the electric field vectors cancel each other out in a non-conducting infinite plane sheet? Why do these vectors not cancel each other out in spite of their being in the opposite directions?
| This is a inside look at the atomic scale:
The field effect is radial and the intersecting field effects sort of cancles out. The direction lines are not forces on the plate but the field effect direction from the plate. You only do that sort of vector cancellation if the force vectors are rather on the plate. If another lone floating positively charged particle falls anywhere in this region, those field lines are the direction of force on the charge. We aren't vectorially resolving the field lines, but rather the force effect on another charged body.
| {
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Why is air pressure higher in winter than in summer? At the top of a mountain, say Mt Everest, atmospheric pressure is low.
So shouldn't the same thing be true for winter season.
I.e air pressure in winters should be lesser than that in summers.
But it's the opposite.
Can someone please explain why ?
| Air pressure is lower at the top of Mt. Everest because there is less atmosphere above it to compress the air, making the pressure less.
I can't be sure what you mean by "air pressure in winters should be lesser than that in summers", I have never heard of this being a straightforward observation before. But (to my knowledge) it is probably because where you live, winters tend to be drier. Dry air is denser than moist air (water molecules are lighter than nitrogen and oxygen molecules. When it is humid water molecules displace nitrogen and oxygen molecules making the air lighter). Also, cold air contracts, and is therefore more dense than hot air.
| {
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Why does the location of the north magnetic pole vary faster than that of the south magnetic pole? Noticing what Wikipedia asserts about the variation of the locations of the magnetic poles over time, e.g., 1998 $\sim$ 2000 to 2015, one would notice that the location of the north magnetic pole varies much faster than that of the south magnetic pole.
What logic would justify this dissimilar variation?
| Have a look on how the earth's magnetic field is modeled presently:
he Earth's magnetic field is attributed to a dynamo effect of circulating electric current, but it is not constant in direction.
So the question becomes why the circulating currents have a higher effect on the northern hemisphere than the southern one. It evidently depends on how the magma in the center of the earth is oriented and the dynamo effect appears.
Convection drives the outer-core fluid and it circulates relative to the earth. This means the electrically conducting material moves relative to the earth's magnetic field. If it can obtain a charge by some interaction like friction between layers, an effective current loop could be produced. The magnetic field of a current loop could sustain the magnetic dipole type magnetic field of the earth. Large-scale computer models are approaching a realistic simulation of such a geodynamo.
So the details are not known, but the difference will be due to the geological topology of the earth.
For an intuitive understanding, take a bar of magnet. The north and south are joined rigidly. Turn the middle portion to a fluid, with a different density from north to south, you could then in a rotation have the north turning in much larger circles than the south.
In a sense , if the dynamo is fitted successfully we will learn something about the composition of the earth, solid-fluid.
| {
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Is every open circuit a capacitor? I think that even open-ended wires can let AC current flow through them, just with a low capacitance. I also think an antenna could be a capacitor and open ended. Am I thinking correctly?
| If one were somehow able to try to draw current from any point while feeding current to any other point without involving anything else in the universe, the points would behave as though there was some capacitance between them. A network of three points, however, may behave as though there is capacitance between the first and second, and between the second and third, without behaving as though there is any meaningful capacitance between the first and third. This subject came up in a recent youtube video https://youtu.be/uGXQc-zanWg on the Mr. Carlson's Lab channel.
As explained by that video, if a half-wave rectified signal is fed into a 50uF cap to ground, and from there passes through a 10K resistor and into another 50uF cap to ground, the second cap could easily have orders of magnitude less ripple than the first. Although it would appear that there is a 25uF connection (two 50uF caps in series) between the two sides of the 10K resistor, the ground connection between the two caps may prevent unwanted coupling of any ground-referenced signal through that capacitance. If the two caps were disconnected from ground but remained connected to each other, however, then that 25uF capacitance would come into play. If that happened but one tried to fix the circuit by adding 50uF caps in parallel with the locations where the originals should have been, that 25uF capacitance would cause the second cap to have about 1/3 as much ripple as the first.
| {
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Quantum expressions for the Virasoro constraints I am trying to derive the quantum form of the Virasoro constraints.
$$ L_{m} = \frac{1}{2} \sum_{n} :\alpha_{m-n}.\alpha_{n}: $$
:...: meaans normal ordering.
Using the common commutator between modes of string when $ m=0 $
$$ [\alpha^{\mu}_{n}, \alpha^{\nu}_{-n}]= n \eta^{\mu\nu} $$
we easily get:
$$ L_{0}= \frac{1}{2}\sum_{n=-\infty}^{\infty} \alpha_{-n}.\alpha_{n}= \frac{1}{2} \alpha_{0}^2 + \sum_{n=1}^{\infty} \alpha_{-n}.\alpha_{n}\\ +\frac{D}{2} \sum_{n=1}^{\infty} n.
$$
My problem is here when calculating the last term. We know that it is infinite, forgetting it for a moment because text says we can use regularization to calculate its finite value. The simple calculation comes first
$$ \frac{d}{da}\sum_n e^{-na} ~=~\sum_n ne^{-na}~=~ \frac{d}{da} \frac{1}{1-e^{-a}}
~=~ \frac{e^{-a}}{(1-e^{-a})^{2}} ~=~ \frac{1}{a^2}-\frac{1}{12}.$$
First, I could not understand $(\frac{1}{a^2}-\frac{1}{12})$ part.
Second, the text concludes that:
$\frac{D}{2} \sum_{n=1}^{\infty} n= -\frac{D}{24} $ where I could not follow.
So my clear question is "how the regularization works"?
It is said that $ a= 2πε /\ell $.
References:
*
*David MacMahon, String theory demystified, p. 79-80.
| There are several typos in MacMahon's formula on the bottom of page 79. It should read
$$ \color{red}{-}~\frac{d}{da}\sum_{n=0}^{\infty} e^{-na} ~=~\sum_{n=0}^{\infty}ne^{-na}~=~ \color{red}{-}~\frac{d}{da} \frac{1}{1-e^{-a}}
~=~ \frac{e^{-a}}{(1-e^{-a})^{2}} ~=~ \frac{1}{a^2}-\frac{1}{12}\color{red}{+{\cal O}(a^2)}.$$
The main point is that the regular terms ${\cal O}(a^2)$ vanish as we remove the regularization $a\to 0^+$, while the singular term $\frac{1}{a^2}$ is cancelled by local counterterms. See e.g. this Math.SE post, this & this Phys.SE posts, and links therein.
| {
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How to calculate air resistance? What is the best way to calculate or take into account air resistance?
I have an assignment which goes as:
Drop an object by itself and with a parachute, calculate air resistance for the parachute. It is assumed the object itself has no air resistance
Im kinda confused how to proceed with this since we have no course text book and there are no lessons just assignments
| Your teacher or professor that assigned the practical experiment to you wants to help you put air resistance into perspective, so you'll have to make a clear distinction between just the fall and the fall with parachute. If you can access a height of about 40 feet, this will be enough to provide the clearance to record the time efficiently and also provide a good difference between the two cases. Weigh the combined mass of mass+parachute+strings with a precise and sensitive weight meter. Drop the ordinary mass first (for reference) couple of times and average the time observed. Now do the same for the mass plus parachute and average also.
Air resistance is a negating force on a moving object, hence the difference in net force on the body without parachute and that with parachute is the air resistance force.
Using $F=\frac {md}{t^2}$ you can calculate the net force on the mass with parachute and $F=9.8m$ for just the mass.
| {
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Why does the flame of a gas stove not spread backwards? On lighting a gas stove the flame is active only above the burner. However, the gas input from beneath the burner and all the way to the pipeline contains gas fuel. Why doesn't it all ignite up ? I am guessing it may be due to the pressurised flow of the fuel in one direction but then I am unable to relate it to combustion, i.e., if a combustible substance is in contact of a source of ignition then it should start to burn.
| The methane that goes to the stove contains little or no oxygen, so it will not burn until oxygen gets mixed into it. This occurs at the burner to the extent that a blue flame is formed in order to obtain complete combustion. At the point where air is mixed into the methane, the velocity of the flowing methane and air is kept high enough to exceed the methane flame velocity, so the flame cannot go backwards through the burner. In addition, the metal parts of the burner are substantially colder than flame temperature, and if the gas flow is decreased to the point where the flame tries to go backwards through the burner parts, the close clearances and cold metal temperature of the metal burner parts quenches the flame.
A good example of the type of behavior described above can be seen when a bunsen burner is lit. The chimney of the bunsen burner is adjusted such that the air register at the bottom of the burner admits enough air to obtain a blue flame. If you continue opening that air register, the air and gas flowing up the bunsen burner tube increase in velocity until the flame has difficulty burning fast enough to keep up, and eventually, the high velocity up the burner tube blows the flame away from the burner, and the flame goes out. On the other hand, if the gas and air flow are decreased to a very minimal amount, the flame cannot propagate down the tube because the tube diameter is small, and the flame gets quenched by the "cold" walls of the bunsen burner tube.
| {
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Why does a car's steering wheel get lighter with increasing speed I've noticed it is difficult to turn the wheels of a car when the car is stationary, especially cars without power steering, which is why the power steering was invented. However, I've noticed it becomes feather light when traveling at speed (some models even stiffen the steering wheel electronically at speed). So, why does a car's steering wheel get lighter with increasing speed?
| Imagine the car stationary. The tire sits on the ground with the contact patch touching.
As you start turning the wheel, the linkage to the wheels starts to rotate the contact patch on the ground. (There are also more complex motions because of the non-zero caster angle of the front wheel).
This rotation is opposed by the static friction of the tire. As you continue turning, portions of the tread on the contact patch are pulled over and stressed.
Now imagine holding the steering wheel at that angle and allowing the car to roll forward a bit. The tread at the rear of the contact patch lifts away from the road and the stress in that portion of the tire is released. Meanwhile new tread rolls onto the contact patch in front, but at the correct angle. Once the contact patch is covered by new tread, the stress from the turn is gone and steering wheel is back to a near-neutral force (again, modulo several effects from the suspension angles).
The faster the car is moving forward, the faster it can put tread into the contact patch with no side stress. So the steering becomes easier.
| {
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Magnitude of the cross product of two bra-kets? From the mathematical perspective, one can take the magnitude of a cross product:
$$
|a\times b|=|a| |b| \sin{\theta}\cdot n,
$$
where $\theta$ is the angle between a and b in the plane containing them, and n is the unit vector perpendicular to them.
Does this apply to the cross product of two bra-kets? I became curious from equation 10 on Dr. Berry's paper, where the Berry phase acquired by a state is the double integral of the following term:
$$
\langle \nabla n | m\rangle \times \langle m | \nabla n\rangle.
$$
So, does the following make sense?
$$
| \langle \nabla n | m\rangle \times \langle m | \nabla n\rangle| =
| \langle \nabla n | m\rangle|
| \langle m | \nabla n\rangle|
\cdot\sin{\theta},
$$
where
$$
\cos{(\theta)}=\langle \nabla n | m\rangle \cdot\langle m | \nabla n\rangle.
$$
To me, it doesn't make sense intuitively because cross products are usually given in tensorial notation in physics (for example: Expressing the magnitude of a cross product in indicial notation). Why is the application of the magnitude above incorrect, or when is it applicable?
| Actually, the entire equation is
$$ \langle \nabla n |\times | \nabla n\rangle \rightarrow \sum_{m \neq n} \langle \nabla n | m\rangle \times \langle m | \nabla n\rangle$$
EDIT: It's Stoke's Theorem using a differential 2-form.
Stokes Theorem can be generalized to higher dimensions using differential forms.
The $ m\neq n$ produces the wedge product.
And since $\langle \nabla n | m\rangle$ is the complex conjugate of $\langle m | \nabla n\rangle$ there are only 3 independent quantities.
See Differential Form
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Work when there is more than 1 force I know that for an object with an applied force, the work done is
$$W = Fd \cos \theta.$$
I was wondering what would happen when there is another force (e.g. friction)? Is it better to say that the work done for a general case is
$$W = F_{net} d \cos\theta.$$
| $\let\th=\theta \def\vF{\vec F}$
Let's begin by specifying the exact meaning of your formula
$$W = F\,d\,\cos\th.\tag1$$
Here it's understood:
1) that a force $\vF$ constant in magnitude and direction is given
2) that the point of application of $\vF$ undergoes a displacement of length $d$.
Note that it's not required that the motion of that point is
rectilinear; what is requested (if force is constant as we said) is that
3) distance between starting point A and arrival point B is $d$
4) angle between $\overrightarrow{\rm AB}$ and $\vF$ is $\th$.
Now if there is another force or more (and if your "object" isn't a single mass point) then each force will have its own point of application. All points will have their own displacements, generally different in magnitude and direction. Then there is only one reply: you have to compute works for each force using (1) and sum works together.
Using net force is only justified for a mass point, or in the special case when displacements are equal (as vectors) even if they apply to different points. This certainly happens if your object is undergoing a translational motion (i.e. with no rotation).
Net force would give a wrong result in all other cases. (Not to mention that if there are several forces, applied to different points undergoing different displacements, you had to determine which one to choose.)
What I've been saying holds if you are interested in total work. This is not always the case, however. But if you want to know work done by one or a subset of forces, nothing changes in the general rule of computing each work one by one and summing.
There are other important points concerning work done on a mechanical system and its relation with variation of kinetic energy. But I guess this is a rather advanced topic for your level of study, so I refrain from entering it. In case you're interested, ask again.
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Why are degenerate states more likely to be filled at a given temperature? Consider if we have a simple two-level toy model, where the ground state has energy $E_0 = 0$ and the excited state has energy $E_1 = \epsilon$ and degeneracy $g$. The partition function for this system is
$$Z = 1 + g e^{-\beta \epsilon}$$
where $\beta = 1/k_BT$. The probability of a particle being in any of the excited states at a temperature $T$ is given by
$$P(\epsilon) = \frac{ge^{-\beta \epsilon}}{1 + ge^{-\beta \epsilon}}$$
which for various $g$ looks like
My question is, why does having a higher degeneracy of the excited state mean it is more likely to be populated at a given temperature? What physical principle tells us this should be so?
| $P(\epsilon) = \frac{ge^{-b \epsilon}}{z} $
where, $z = \sum_\epsilon ge^{-b \epsilon}$
So, for two states with some difference in energy, if $g_1e^{-b \epsilon_1}>g_2e^{-b \epsilon_2}$, then state with energy $\epsilon_1$ is more likely to be filled even though $\epsilon_1$ might be less than $\epsilon_2$
Now, if we take $K_bT>>1$, probability difference in the above two states is proportional to:(I have taken the first order aprox for exponential)
$g_1e^{-b \epsilon_1} - g_2e^{-b \epsilon_2} =g_1 (1 - b\epsilon_1) - g_2 (1 - b\epsilon_2) = (g_1-g_2)(\frac{\epsilon_2}{K_bT} - \frac{\epsilon_1}{K_bT})$
as T is very high, one can see that in this limitng condition, the effect of $(g_1 - g_2)$ is much higher than the energy difference and thus the higher degeneracy cases are filled preferably.
| {
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How to interpret explicit moments in beams?
Consider the beam in the picture above. There are several forces that cause moments around all points in the beams. But there is also one "explicit" moment: 20kNm.
How do I interpret these kinds of moments? Here the arrow goes around the point 2m from A. So does this mean there is an extra moment around this point? But how does this moment affect the moment seen from other points, such as A? Or B? More specifically, if I wanted to form a moment equilibrium equation from A, would I simply add/subtract (depending on my definition of positive direction) 20kNm from/to the sum of moments about A? And if I were to form moment equilibrium equations around another point like B, what would I do with 20kNm then? Also, how would this kind of moment occur in real world? It is a moment without a force or a distance, just a moment. How could I produce something like this? Thank you!
This is not a homework question; I am currently taking a course on statics but this specific example is one I found somewhere that illustrates my problem.
| I am very much aware that this thread has been a while but I am sure people keep coming here. If we look into the Euler-Bernoulli equation, this explicit moment is actually an imposed boundary condition (BC): the third derivative of the deflection wrt its horizontal axis (the x-axis).
In an Euler-Bernoulli, the input is only force/pressure. We can't apply torque. Hence, this explicit torque is a given BC for us to use as we solve the Euler-Bernoulli equation. If we perform a finite difference discretization, it is indeed a force couple but these force couple are not equal.
| {
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Is there a contradiction between isotropy and the Big Bang? Disclaimer: I'm not asking whether the Big Bang happened at a point. I'm asking whether the fact that the universe is isotropic and that the Big Bang happened contradict each other.
To be honest I am just starting to learn General Relativity and cosmology but I feel like this question has a profound basis. If we assume that the universe "started" by the Big Bang (which I naively imagine as just the expansion of a sphere) then surely it must have started out from a certain point, which is still in our sphere universe. However we also know that the universe is isotropic, that there are no "special" points. How can these two facts coexist since if the Big Bang really did happen then there must be a characteristic point in our universe, a special point, which contradicts the principle of isotropy. So what am I missing, if I'm missing anything?
| Einstein assumed space was isotropic and homogeneous because he believed the universe was closed.
And it simplified the model.
Also, an isotropic and homogeneous space is needed for constant curvature.
Cosmology principals assume the universe is isotropic and homogeneous - basically has a "stream line flow".
In fact, the cosmological principle reduce a 10 parameter theory to a single function and discrete parameter which characterizes Robertson-Walker space time.
However to determine the structure of the universe, we need an accurate measure of distance - which currently doesn't exist.
For instance, Google for the "Fingers of God" - which is a plot of all the measured red shifts in a quadrant of the sky - for instance the Virgo Galaxy. It looks like there's a big finger pointing back to the Earth.
And another interesting plot is the "Pancakes of God" - which suggests the universe is just a web of Machian objects.
| {
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Can this be considered the Newton's rings?
It can be made by pointing a laser pointer directly to a mirror at low angle and observing the reflection behind the laser pointer. No convex or concave lens needed unlike the Newton's ring experiment we know.
So far, Newton's ring experiments and theories I read only involves a combination of concave and convex lenses, as Wikipedia says: Newton's rings is a phenomenon in which an interference pattern is created by the reflection of light between two surfaces—a spherical surface and an adjacent touching flat surface. The pattern is created by placing a very slightly convex curved glass on an optical flat glass. The two pieces of glass make contact only at the center, at other points there is a slight air gap between the two surfaces, increasing with radial distance from the center.
None of Newton's ring theories I read in Wikipedia or somewhere else involves a laser pointer and a mirror. Is there the same fundamental physics involved in both experiments? Or could it be a completely different thing?
| It seems similar with Newton's experiment described in Second Book of Opticks, Part IV. He used concave mirrors of 1/4 and 5/62 inch thicknesses and light beam through a hole of the board which corresponds to the hole at the center of the rings on the screen in your drawing. I suppose that there is no special name for these rings but they would be made by the same way of interference.
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Maxwell Tensor Identity In Schawrtz, Page 116, formula 8.23, he seems to suggest that the square of the Maxwell tensor can be expanded out as follows:
$$-\frac{1}{4}F_{\mu \nu}^{2}=\frac{1}{2}A_{\mu}\square A_{\mu}-\frac{1}{2}A_{\mu}\partial_{\mu}\partial_{\nu}A_{\nu}$$
where:
$$F_{\mu\nu}=\partial_{\mu} A_{\nu} - \partial_{\nu}A_{\mu}$$
For the life of me, I can't seem to derive this. I get close, but always with an extra unwanted term, or two.
Anyone have a hint on the best way to proceed?
| Your expression is part of a Lagrangian. As the physics remains the same as long as the action remains the same, one can always do partial integration in the action integral over the Lagrangian to derive alternative Lagrangians describing the same physics.
| {
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Applying the Euler-Lagrange equations to Maxwell's Theory In Prof. David Tong's notes, specifically on page 10, he gives the Lagrangian of Maxwell's theory to be
$$
\mathcal{L} = -\frac{1}{2}(\partial_\mu A_\nu)(\partial^\mu A^\nu) + \frac{1}{2}(\partial_\mu A^\mu)^2
$$
and then he computes the following
$$
\frac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu)} = -\partial_\mu A_\nu + (\partial_\rho A^\rho)\eta^{\mu\nu}.
$$
I can see how the first term in the derivative is computed but am having problems with the second term. Any help is appreciated!
| We have $\frac12 (\partial_{\mu}A^{\mu})^2 = \frac12 (\partial_{\alpha} A^{\alpha})(\partial_{\beta}A^{\beta})= \frac12 (\partial_{\alpha} A_{\sigma}) \eta^{\sigma\alpha}(\partial_{\beta}A_{\rho}) \eta^{\rho\beta}$ so the derivative w.r.t. $\partial_{\mu} A_{\nu}$ is
$$\frac12\delta_{\alpha}^{\mu} \delta_{\sigma}^{\nu} \eta^{\sigma\alpha}(\partial_{\beta}A_{\rho}) \eta^{\rho\beta}+\frac12(\partial_{\alpha} A_{\sigma}) \eta^{\sigma\alpha}\delta_{\beta}^{\mu} \delta_{\rho}^{\nu} \eta^{\rho\beta}= \frac12 \eta^{\mu\nu} (\partial_{\beta} A^{\beta})+\frac12 (\partial_{\alpha}A^{\alpha}) \eta^{\mu\nu} = (\partial_{\rho}A^{\rho})\eta^{\mu\nu} $$
where I've freely labeled and relabeled dummy indices.
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Poisson's ratio in analytical beam deflection equations After looking at some of the analytical expressions for analyzing beams, I noticed that none of the equations depend on the material's Poisson ratio. Some analytical expressions can be found in https://www.linsgroup.com/MECHANICAL_DESIGN/Beam/beam_formula.htm
I believe these equations are derived from the Euler Bernoulli equations and not from 3-D Linear elasticity, and Poisson's ratio isn't included in Euler Bernoulli. What underlying assumption in Euler Bernoulli allows for the exclusion of the Poisson's ratio?
If I want to model a beam under the Euler Bernoulli assumptions in a 3-D linear elasticity code, where the Poisson's ratio is required as an input, how can this be done? Essentially, I am trying to perform a one to one comparison of the linear elasticity code I am using with the analytical beam expressions. To do this, I need to figure out what value to input for the Poisson's ratio in the code. When I model the beam with the code and vary the Poisson's ratio, the deflection changes as a function of the poisson's ratio. However, the analytical beam expressions do not depend on the Poisson's ratio, so it is not clear to me how this comparison can be performed.
| Poisson’s ratio, ν, is defined as
ν = -lateral strain/longitudinal strain.
It is applied when analyzing the lateral strain on members subjected to axial forces (compression and tension). None of the beams in the diagrams of your link involve axial loading (loads applied to the ends of the beams along the axis of the beam). Nor do they deal with strain resulting from shear forces and bending moments. Therefore the Poisson ratio would not apply.
UPDATE:
I ‘m not sure if this will help you, but the modulus of elasticity, $E$, is related to Poisson’s ratio, ν, by the following equation where $G$ is the shear modulus
$$G=\frac{E}{2(1+ν)}$$
Hope this helps
| {
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Does the band gap increase or decrease in this case? Suppose I deposit a thin layer of a material A (with band gap Ea, say) on a material B (having band gap Eb) to form the film AB. What will the band gap of AB be? Will it be equal to Ea or Eb? Or will it be lesser or greater than them?
| If you have distinct layers of material, they will each maintain their respective band gaps. Then you have a heterostructure, not a single “effective” band gap. There are caveats to this, for instance if the thin layer is strained due to a lattice mismatch with the other material, then the band gap within the thin layer will be modified somewhat but still distinct.
The physics of such a heterostructure is potentially complex, depending on the relative alignment of the bands and the types of excitations which exist.
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Effect of gravity at center of Earth Imagine I was a hypothetical ant man the size of an atom, and I position myself at the exact, down to the atom, center of mass of the earth. A move in any direction will move me out of the center. Would I experience gravity pulling outward on my body in all directions?
| There's actually a very useful way to solve this using the Gauss law for gravity, which is given by:
$$\oint\vec{g}\cdot d\vec{A}=-4\pi GM_{enc}$$
where $\vec{g}$ is the gravitational field, $\vec{A}$ the area enclosed in the surface of interest, and $M_{enc}$ the enclose mass of the object by the gaussian surface.
Assuming the Earth has an uniform volumetric density $\rho$, let's consider the two situations proposed:
a) You're at the exact center of the planet: in this case, all forces will cancel each other due to the symmetry of the object, so you will experience zero gravity. From Gauss Law, this is equivalent to having no enclosed mass.
b) You move away a distance $r$ from the object: in this case, the enclosed surface will be $4\pi r$, and assuming a constant density
$$\rho=3M/4\pi R^3=3M_{enc}/4\pi r^3 \ \ \rightarrow \ \ M_{enc}=4\pi r^3 \rho/3$$
Substituting in the Gauss equation,
$$g(4\pi r^2)=-4\pi G(4\pi r^3 \rho/3)$$
Simplifying,
$$\vec{g}=-\frac{4\pi G\rho}{3}\vec{r}$$
Or in terms of the radius of the Earth,
$$\vec{g}=-GM\left (\frac{r}{R^3}\right )\hat{r}$$
So you can see that only the area enclosed by your Gaussian surface will contribute to the net acceleration you feel towards the center.
Obviously the density of the Earth isn't constant (it's more concentrated on the nucleus than on the surface), so you can have a better approximation using a more empirical model of such density.
| {
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The Cosmic Microwave Background Paradox I was reading an article on Olbers' Paradox (why the universe isn't as bright as the sun) and the more I read on it, the more the same question came to mind...
We know the observable universe is approximately 93G LY across and we know the age of the universe is 13.8 B years. We also know that there are galaxies moving away from us faster than the speed of light due to ever-increasing expansion.
If we cannot see these galaxies due to the redshift of their light... how is it that we can see the CMB which is behind them (to our perspective)? Shouldn't it be moving even faster away from us?
If we can see it, why can we not see the galaxies furthest from us?
| If we can see a particular object at some moment, then we will always be able to see that object in the future, no matter how quickly the universe expands. It's analogous to how we can never see anything completely fall into a black hole. In both cases, the image you get just gets more and more redshifted over time. This is the point made in a comment: we could certainly see the CMB at some point, since it occupies the entire universe, so we will always be able to see it. It just continues to get more and more redshifted; it's already been redshifted by a factor of $1000$.
Here's another more direct argument. There are galaxies we can never see, because they form and begin emitting light when already outside of our horizon. But the CMB photons we are currently seeing have an enormous head start because they have been traveling towards us at the speed of light for billions of years, and furthermore doing so in an era when the universe was expanding relatively slowly. So it is certainly possible to be able to see these photons, but not ever see the photons from a newer galaxy that forms closer to us than the relevant CMB photons did.
| {
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The law of radioactive decay: explanation of a formula The law of radioactive decay can be expressed in terms of $\,\tau=1/\lambda$ (average life) as:
$$
N(t)=N_0e^{-t/\tau}, \quad \tag{1}
$$
Why deriving the (1) I have:
\begin{equation}
N'(t)=N_0(1-e^{-\lambda t})\, ?
\end{equation}
| It comes from solving the differential equation
$$\frac{dN}{dt} = -\lambda N(t). $$
This equation comes from observations of the number of decay events $N(t)$. It's found through experiment that the rate of decay over a given time interval is proportional to the number of events recorded during that time. You can arrive at this conclusion by plotting the rate vs the number of events on a log log plot and finding that it is linear.
Formally, this is a differential equation. But solving it is really just a fact which you know already.
Which function $N(t)$ can you take the derivative of and get itself back times a constant?
The answer is exponentials, and so the solution to this equation is
$$
N(t) = N(0) e^{-\lambda t}.
$$
Edit: I should also note that you took the derivative incorrectly. The correct derivative is
$$
N'(t) = \frac{d}{dt} N_0 e^{-\lambda t} = - \lambda N_0 e^{-\lambda t}
$$
| {
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Principle of friction force When two equal and opposite forces act on a body at rest or in motion we say it is rest or in its state of motion. But when a body kept on a table starts moving uniformly due to a force applied, it overcomes force of friction.Does it really overcome the force of friction or is just equal to it. ie for force F = ū.N if it's so then why would it start to move,why not stationary as the two forces balanced?
| You have two types of friction, static and kinetic (a.k.a sliding) friction. When you begin to apply a force on an object of mass $m$ on a table it is the static friction force that resists your effort to move it. As you keep increasing your force, the static friction force opposing you continues to increase, but only up to a point where your force equals the maximum possible static friction which is $μ_{s}mg$ where $μ_s$ is the coefficient of static friction.
When the maximum static friction force is reached the object starts sliding. Now the friction force that opposes your force is kinetic friction and is $μ_{k}mg$ where $μ_k$ is the coefficient of kinetic friction.
Now here’s the important point. Usually the coefficient of kinetic friction is less than static friction. So the kinetic friction force will be less than the force you applied to overcome static friction. Therefore, if you maintain that same level of force the object will accelerate based on the difference between your force and the opposing kinetic friction force.
On the other hand, if you reduce your force so as to equal the kinetic friction force, the net force will be zero and now the object will continue at constant velocity, the velocity it had just before you reduced your force.
The diagram in @Cal answer below shows how the friction force drops from static to kinetic. It is based on a diagram on the Hyperphysics web site.
Hope this helps.
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Propagation of light Suppose we are able to make any place free from the magnetic and electric field; then we turn on a light source at any place in that region (where there is no electric and magnetic field). Does light propagate in that region?
| Currently the accepted theory is the standard model, all the data from experiments supports that.
According to that model (and QFT), light consists of a herd of photons, and the photons are excitation of the EM (photon) field.
According to the model, these fields exist throughout space everywhere. This is the only way light can propagate through everywhere in the observable universe.
These fields are all part of the fabric of the universe, and there is no way we could create a part of spacetime where there is no EM field present.
But let's disregard that, and say we could (we can't) create a part of spacetime where there is no EM field present. Since photons themselves are excitation of the photon field, there could be no photons present at all, and no electrons (and atoms) to emit them, so light could not even exist in that part of spacetime.
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If you were to push a block in space in an ideal vacuum scenario with zero friction, what would be the final velocity of that object Say you push an object with 10 N force(just once and let it go) of mass 10kg object in space with zero friction and nothing in the way of its path. What would be the velocity of that object?
I would assume that if you are putting a "constant" force of 10N, then yes theoretically the object should continue accelerating forever right?
But what if you just give an initial push of 10N to the object and let it go so that it floats in space how would you calculate its definite velocity? I am assuming the time in which the object is in the contact with a pushing force will matter is that correct?
| If you don't have your space watch, but did bring a ruler, you could measure how far you pushed it (you pushed it $x$). Then the work you did is
$$ W = F\cdot x $$
where $F=10\,$N. That is equal to the object's kinetic energy:
$$ T = \frac 1 2 mv^2 = Fx $$
Of course, the average speed is $v/2$, so the time it takes to go $x$ is:
$$ t = \frac{2x} v $$
which you can plug in:
$$ \frac 1 2 m v^2 = \frac 1 2 v tF $$
or
$$ v = t\frac F m $$
Since $F=ma$, that's:
$$ v = at $$
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Parabolic motion of a projectile During a conference I have listened that an employee of an armed force, he has said that when a projectile fired from a gun upwards, it is very dangerous because when the projectile descends downwards it acquires a higher speed than the initial instant that is when the projectile was fired upwards.
In my opinion all this is false if we do not assume in the meantime that there is air resistance (friction) and from what height it is fired. Obviously if the air resistance is zero the initial velocity of the projectile $v_0$ will be equal to the final velocity of the same when it reaches the ground.
For example if $v_0=2$ m/s is the velocity of the projectile when it is fired upwards, the speed of descent $v_d$, for the opinion of this officer, is such that:
$$v_0<v_d$$
I simply don't agree because if I was facing a balcony and someone shoots me from the ground at an altitude $h>h_0=0$ it hurts less if the same projectile falls on my head at the time of the descent?
| Can you please add some context on what are you trying to compare exactly? I might be oversimplifying the problem, but the way I see it, I would agree with the employee:
*
*If the maximum height of the projectile fired upwards is equal to the height the other projectile is dropped, then we would have around the double of time to run away (thus less dangerous in that sense);
*If the time to run away is equal in both cases, it means that the maximum height is smaller in the former case, thus the impact velocity if lower, being again the first case, less dangerous.
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Self-interaction in spin-orbit coupling? In spin orbit coupling, in an atom motion of electrons about nucleus generates magnetic field and we consider this field to interact with magnetic moment of electron. It sound strange as in electrostatics a field is generate by a charge particle but this field does not interact with its origin charge. Doesn't this contradict ?
| An electron has a magnetic moment. When the electron moves it is perceived as having an electric dipole moment as well. This interacts with the nuclear electrostatic field.
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Difference between voltage, electrical potential and potential difference I am having hard time to visualize these two concepts in my mind seriously.
First of this confusion came from two parallel plates that was connected to a power supply, charged then disconnected from power supply and then separated from each other, strangely potential difference increased but why? I have learn that the electrical potential is$$ V= k \frac{q}{d} $$and when the distance increases the potential of a point must drop but why when we are talking for potential difference it is increasing, doesn't every point between these plates feel less stress as the plates moves apart and doesn't this mean potential is dropping and so the potential difference as well?
| You are mixing the properties of a point charge and that of a capacitor.
The formula you've written works perfectly for point charges and intuitively for two entities holding the same nature of charge and to be very precise, you should use $r$ instead of $d$ when dealing with point charges as the electric field from a point charge has spherical symmetry. Note that this formula won't work for charged plates. It makes sense that when two positive point charges of same polarity (not necessarily of same magnitude) are moved farther from each other the potential energy of the system decreases. But in your question you are talking about what are known as capacitors in the business. You might be knowing that when two oppositely charged plates are moved away from each other the potential energy of the system increases as you are doing work against the electric field and a capacitor comprises two plates placed in close proximity which hold equal and opposite charges. So here, the potential difference between the plates is given by $$E=\frac{V}{d}$$
As you might be knowing that the electric field in the region between the two plates is always constant no matter how far you move the plates provided the plates are parallel to each other and one must perfectly block the view of the other. This way the ratio $$\frac{V}{d}$$ will always be constant. So you may conclude that whenever distance between the plates increases the potential energy of the system also increases.
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Does an electric field penetrate through glass? I was just wondering something ridiculously simple; does an electric field penetrate through glass? Say I have a proton or charged cotton pulp ball inside a Pyrex glass container and apply an electric field on the outside, through the whole thing, do the particles 'feel' the field and will they be drawn to either of the electrodes? I was afraid of asking this for fear of being ridiculed, but here it is!
| Assume that a charge is completely surrounded by glass of any thickness. Gauss's law, in integral form, still applies to a surface surrounding the glass. The presence of the glass makes no difference for the outcome of the integral.
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Finding the eigenstates of an operator I am currently taking a course in QM and can't see how the eigenstates have been found for examples like this one:
Question
Let $\phi _1$ and $\phi _2$ be two normalised wavefunctions orthogonal onto each other. Let the action of the operator $\hat{A}$ on these states be:
$$\hat{A} \phi _1 = \phi _2$$
$$\hat{A} \phi _2 = \phi _1$$
Then the states $\Psi _1 = \phi _1 + \phi _2$ and $\Psi _2 = \phi _1 - \phi _2$ are eigenstates of $\hat{A}$ corresponding to the eigenvalues $+1$ and $-1$ respectively, that is:
$$\hat{A} \Psi _1 = \Psi _1$$
$$\hat{A} \Psi _2 = -\Psi _2$$
So I can see the states $\Psi _1$ and $\Psi _2$ satisfy the eigenfunction equation ($\hat{A}\Psi = \lambda \Psi$) with the given eigenvalues but I can't see how the states were derived, they seem to have been plucked from thin air but just work! Lots of QM questions seem to utilise states in the form $(\alpha \phi _1 \pm \beta \phi _2)$ but I'm not sure why?!
Appreciate any help!
| gabe has already given an answer in terms of matrices and using idempotency. I shall exhibit a rather dry approach.
Assuming that the $\phi$’s form a basis, then any vector can be expressed as $\alpha\phi_1+\beta\phi_2$.
Now let $\psi=\alpha\phi_1+\beta\phi_2$ be an eigenstate of $A$. Then we have,
$A\psi=A(\alpha\phi_1+\beta\phi_2)$
$\implies \alpha A\phi_1+\beta A \phi_2$
$\implies \alpha\phi_2+\beta\phi_1=\lambda\alpha\phi_1+\lambda\beta\phi_2$
$\implies \lambda\alpha=\beta , \lambda\beta=\alpha$
$\implies \lambda^2\alpha=\alpha$
$\implies \lambda = \pm 1$
Plugging these back into $\alpha-\beta$ relation gives us $\alpha=\pm\beta$
Finally $\psi = \phi_1 \pm \phi_2$
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Perpetual Rotation of Rigid Body Assuming no external torque or forces acting can a rigid body be set in perpetual rotation motion about an axis which is not its principal axis?
If no then does the earth continuously change its axis of rotation? as the principal axes of the earth changes changes continuously because of tectonic plate movements, animals humans moving etc.
|
Assuming no external torque or forces acting can a rigid body be set
in perpetual rotation motion about an axis which is not its principal
axis?
In a nutshell? No.
If a body rotates about a principal axis (an axis through its centre of mass) and no external torques act on it, then it will forever keep rotating.
Now look at the case where the rotation in NOT about a principal axis:
In order to keep the rotation going, a force has to act on it, called the centripetal force, $F_c$. It's a vector pointing from the centre of mass of the object to the axis of rotation.
In vector notation we can write:
$$\mathrm{F_c}=-m\omega^2\mathbf{r}$$
where $m$ is the object's mass, $r$ the radius of the orbit and $\omega$ its angular velocity.
If this force ceases to act then the object will leave its orbital trajectory.
If no then does the earth continuously change its axis of rotation? as
the principal axes of the earth changes changes continuously because
of tectonic plate movements, animals humans moving etc.
It doesn't really. The effects you list are minimal, compared to the inertial moment/inertial momentum of the Earth.
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What are equations of state in thermodynamics? So I am having real trouble understanding what equations of state are and how we form them. My issue stems from reading multiple sources. So I understand that an equation of state is used to build a relationship between variables to describe a state of a system.
For example $P=P(V,T)$, where $P=NKT/V$, is a function of state, but then I started reading about Gibbs, Helmholtz, enthalpy, etc and suddenly am very confused.
So looking at Gibbs, for example, we have two equations
$$G=E+PV-TS$$
and
$$dG=-SdT+VdP$$
but both describe the state of the system. I did not think an equation of state could be a differential equation but according to one book I have read, $dE$ is related to equation of state and this has caused me great confusion.
| Equations of state are literally just equations relating state variables. State variables are values that just depend on the current state of the system and not on how the system got to that state. Contrast this with things like work done on the system or heat leaving the system, which depend on the process a system undergoes.
So, if values in your equation depend on the "path" or history of the process, then it isn't an equation of state.
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How can a day be exactly 24 hours long? The longest solar day of year is approximately 24 hours 0 min 30 seconds (occurs at mid to late December) while the shortest solar day of year is approximately 23 hour 59min 38 seconds. If I average out both of these I come up with average solar day of 24 hour +4seconds. Why then it is said it is 24 hours 0min 0 seconds exactly??
Wouldn't using a 24 hour solar day as a definition of day cause the offset of 4 seconds each solar year
| The average value of a distribution is not the average of its minimum and maximum. For example, the average value of (0,0,0,4) is 1, not 2. Earth's orbit eccentricity is not 0, nor the Moon's one, so your distribution of day duration is probably slightly asymetric, hence the 4 seconds discrepancy. Sum all day duration of a year, divide by the number of days, you will get a better value.
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How can I calculate the speed at which a gas flows from one volume to another? Suppose that there are two rigid Volumes A and B. Each of these contains a gas with know properties (pressure, temperature, number of moles, volume, composition). Suppose also that there is a pipe with a valve that connects the two Volumes.
How can I calculate the speed at which gas goes from one volume to the other. Using the ideal gas laws, I can calculate the final properties of the gases in each volume, but that isn't what I'm looking for.
I'd imagine that the rate of flow is dependent on the difference of pressure in the two volumes or the ratio of pressure; in real life, when a high pressure volume is punctured, gases flow out of it quickly, and when a volume at room temperature is punctured, the gases flow out of it more slowly.
| Ah, in my link provided from my blog, I do have an equation which describes the flow in and out of a system which may prove useful:
$\dot{S}_2 - \dot{S}_1 = \dot{Q}(\frac{1}{T_2} - \frac{1}{T_1}) = \frac{\dot{Q}(T_1 - T_2)}{T_1T_2}$
It only applies to classical systems though. We can state the mass remains constant (m=1) and so the variation can be found by differentiating the velocity, I get
$\dot{S}_2 - \dot{S}_1 = 2v\dot{v}(\frac{1}{T_2} - \frac{1}{T_1}) = 2v\frac{\dot{v}(T_1 - T_2)}{T_1T_2}$
And also take into account we can write
$\frac{\dot{S}}{2}(\frac{1}{v_2} - \frac{1}{v_1}) = \dot{v}(\frac{1}{T_2} - \frac{1}{T_1}) = \frac{\dot{v}(T_1 - T_2)}{T_1T_2}$
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Can someone provide to me an intuitive explanation of the second integral of position with respect to time? I am aware of what the first integral of position, absement means (at least to a very superficial level).
However, I can find nothing regarding the physical intuitive meaning of absity, the second integral of position.
If anyone is able to explain to me, that would be great.
| Consider the gas pedal in your car. When you push the pedal down and hold it there, the car accelerates. Push it a bit further and the acceleration is larger. In other words, one's acceleration depends on the other one's position - which means that one's position (double integral) depends on the other one's absity (double integral).
*
*Specifically, the car acceleration $a_{car}$ relates to the pedal position $s_{pedal}$:
$$a_{car}=k\,s_{pedal}$$
$k$ is the proportionality constant (we assume a simple, linear and thus proportional relationship for simplicity). For a stiffer pedal (where less displacement is needed for the same acceleration) $k$ is larger.
*Therefore, the car velocity $v_{car}$ (integral) relates to the pedal absement (or absition) $p_{pedal}$:
$$\int a_{car}\;\mathrm dt=k\int s_{pedal}\;\mathrm dt\quad\Leftrightarrow\quad v_{car}=k\,p_{pedal}$$
*and the car position $s_{car}$ (double integral) relates to the pedal absity $q_{pedal}$:
$$\int\int a_{car}\;\mathrm dt \;\mathrm dt=k\int\int s_{pedal}\;\mathrm dt\;\mathrm dt\quad\Leftrightarrow\quad s_{car}=k\,q_{pedal}$$
If someone asks you, where the car is after a certain time, then you can answer it if you know the absity function of the gas pedal.
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Momentum conservation in Quantum Mechanics Most quantum mechanical potentials are not translationally invariant and therefore the expectation value of momentum varies.
The question is then where has this momentum been transferred to? Because as a whole the system must conserve momentum.
In electrodynamics of course there is a similar question with for example a particle in a constant electric field. This is resolved by saying the momentum has been transferred to the field.
So in the quantum mechanical case does the momentum transfer to the source of the potential?
|
Most quantum mechanical potentials are not translationally invariant...
... because those potentials are externally imposed from some overarching system whose dynamics you're not explicitly considering. The momentum goes to that overarching system.
As an example, consider the hamiltonian of the electron in an $\rm H_2^+$ molecular ion, which reads
$$
\hat H_\mathrm{electronic} = \frac{\hat{\mathbf p}^2}{2m} - \frac{e^2}{4\pi\epsilon_0}\left[\frac{1}{|\hat{\mathbf r}-\mathbf R_1|}+\frac{1}{|\hat{\mathbf r}-\mathbf R_2|}\right],
$$
in the Born-Oppenheimer approximation, with the two protons clamped at positions $\mathbf R_1$ and $\mathbf R_2$. This is not translation-invariant in $\mathbf r$, so its conjugate $\mathbf p$ is not conserved. To see where that momentum goes, simply extend your description of the dynamics so that the overarching system that imposes the potential (in this case, the protons) is fully included within the dynamics, i.e.
$$
\hat H_\mathrm{full} = \frac{\hat{\mathbf p}_1^2}{2m} + \frac{\hat{\mathbf p}_2^2}{2m} + \frac{\hat{\mathbf p}^2}{2m} - \frac{e^2}{4\pi\epsilon_0}\left[\frac{1}{|\hat{\mathbf r}-\hat{\mathbf R}_1|}+\frac{1}{|\hat{\mathbf r}-\hat{\mathbf R}_2|}\right].
$$
Then the hamiltonian does become translationally invariant (where the translation must move all of $\mathbf R_1$, $\mathbf R_2$ and $\mathbf r$ by the same amount), and the conjugate variable (the center-of-mass momentum, $\mathbf P = \mathbf p_1 + \mathbf p_2 + \mathbf p$) is conserved.
Identical considerations hold in every other such situation.
| {
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Why the time period of pendulum with infinite length is $84.6$ minutes? In a book I was reading about SHM it stated:
If the length of a simple pendulum is increased to such an extent that $\ell\to\infty$, then its time period is given by,
$$T=2\pi\sqrt\frac{R}{g}\approx84.6\text{ min}$$
Now I have many confusions like:
*
*Doesn't the gravity change with such a large length and amplitude?
*How can an infinitely long pendulum have a confined time period?
*Some parts near the end may have a velocity greater than light.
| I have also studied this relation. May be things get clearer in my explanation:
For a general pendulum, it is obvious that if the l is substituted to a large value the time period corresponding also increases. For a pendulum of very large length(comparable to radius of earth), we have to make a change in derivation. An assumption which is valid only for small lengths. That is:
For small pendulums, on small angular displacement, the corresponding linear displacement is very small. The force of gravity (causing restoring torque) can be assumed to be vertically downwards. But for a hypothetical large pendulum, this is no more true. Even for a small angular displacement the corresponding linear distance on the ground becomes too large(if $\theta$ is the angular displacement the corresponding linear displacement is $L\theta$ where L is length of string). The force of gravity acts along line joining the Bob and centre of earth, and it is no more vertically downwards. (The angle moved by Bob with respect to centre of earth is no more negligible. It is given by $R_e \beta= L\theta. R_e$ is radius of earth ) So the problem is that the torque due to gravity is no more given by the same expression.
You can see the picture for clarification. The infinite is a hypothetical case by substituting $L=\infty$ .
Gravity as you said , is taken to be constant
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Can we know when rolling occurs without slipping? In school we solve a lot of physics problems involving rolling cylinders/hoops/wheels. Often it is specified in the question statement that "rolling can be assumed to occur without slipping". Then we know we can use the relationship between translational acceleration and rotational acceleration: $a = \alpha r$.
But can we know from a particular set up when the object can be assumed not to slip when rolling? Assume we know the usual macroscopic details of the set up such as the mass and radius of the rolling object, the angle of the plane, friction coefficients etc.. without deep insight into specifics of the materials of the wheel and the plane.
| This is how I have approached these problems
*
*Assume there is no slipping and calculate the required friction force to enforce this constraint.
*Check that the friction force is less than the available traction. This is usually done in scalar form $\| \vec{F} \| \leq \mu \| \vec{N} \|$ where $\mu$ is whatever static coefficient of friction is appropriate for the materials.
*If the friction exceeds traction, then the problem changes, and you de-couple the rotational and translational degrees of freedom, but friction force is no longer an unknown, but it is set to oppose the relative motion $ \vec{F} = - \mu \| \vec{N} \| \frac{\vec{v}}{\| \vec{v} \| }$. Re-solving is needed at this point.
| {
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Why doesn't Newton's third law mean a person bounces back to where they started when they hit the ground? When we drop a ball, it bounces back to the spot where we dropped it, due to the reaction forces exerted on it by the ground. However, if a person falls down (say, if we push them), why don't they come back to their initial position where they started their fall?
According to Newton's 3rd law of motion, to every action there is always an equal but opposite reaction. If we take the example of ball then it comes back with the same force as it falls down. But in the case of a human body, this law seems not to be applicable. Why?
| Newton's third law states that when a particle applies a force on another particle then the former experiences an equal but opposite force from the other.When a ball hits the ground it comes back due to the fact that it experiences an elastic collsion(that is does not undergo any deformation).Think about a fur ball,will it come back,obviously no,why is that so?This is because when a fur ball collides with the ground(assuming a concrete ground)it undergoes deformation (due to the reaction force from the ground) and the kinetic energy of the ball is used up in deforming the ball.Same is the case for a human being who undergoes deformation on hitting the ground and hence does not come back to it's initial position.
| {
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If atmospheric pressure is 76 cm of $\text{Hg}$ , why won't 76 cm of mercury stay in an open tube when suspended in air? If we keep an hold a tube in air with the closed end up and open end downwards, containing mercury upto a length of 76 cm, why does the mercury not stay in place? Shouldn't atmospheric pressure exert a force equal and opposite to its weight and balance it?
| In an open tube, air will press down on the top of the column of mercury, as well as up from the bottom via fluid pressure. The net force will be the weight of the mercury in the column.
In a closed evacuated tube, ther is no air pressure pushing down from the top; only the pressure from the bottom and the weight of the mercury. The mercury column needs to be about 76 cm high for those two forces ato be in equilibrium (the pressure due to the weight of the mercury equals the air pressure transmitted via the liquid at the bottom).
See the articles Understanding Vacuum Measurement Units, Pressure and the simple mercury barometer, and this SE post.
I misread the OP's question; it was not directed at why the mercury stays in a closed tube in a barometer with the open bottom in a mercury reservoir; it was directed at why mercury will stay in a closed tube that is open to the air at the bottom. The answer and comment by @MaxW are correct.
| {
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Minkowski space
*
*In Minkowski space, coordinates which satisfy
$x^2 = t^2 - X^2 > 0$
are in the region of spacetime that is time-like.
*If it's
$x^2 = t^2 - X^2 < 0$
the region is space-like.
*But if
$x^2 = t^2 - X^2 > 0$
then its "trajectory of light-like particles".
I have understood the first two points about time- and space-like regions but I could not get the third one about "light-like particles".
My confusion is - why just light-like particles? There are many other particles at quantum level.
|
My confusion is about why just light like particles? there are many other particles at quantum level.
You are correct. The terminology is historical in nature. Light was the first massless particle to be discovered. The terminology “lightlike” was established before any other massless particles were discovered. Once other massless particles were discovered it was shown that they also travel along lightlike geodesics, but by then the term “lightlike” was well established.
An alternative term with the same meaning as “lightlike” is “null”. If you prefer then you can always use “null” and just understand that people saying “lightlike” mean the same thing.
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Geometrical Optics: Infinite Rays Normally in ray optics, we draw a parallel line from the top of the image to the lens and stop when this line intersects an angled line (drawn from the height of the real object) and intersects. However, why do we stop? We can draw infinite rays from this object and they should be able to go out for infinitely far distances. Thus, we should be able to get a huge image (albeit a bit darker).
| An image is formed when there is a one to one correspondence between a point on the object and a point in space where all light rays emitted from that point on the object meet up. In other words, if we find that different light rays from the same point of the object end up meeting up at different points in space after passing through the lens, then we don't have an image there (or at least it will be pretty blurry).
This is why you typically see one arrow representing the object and then we look at light rays coming from the top. We pick rays that are easy to use and find where those rays intersect after going through the lens. This then means that the rest of the object must be in focus since it is represented by just a line that is perpendicular to the optical axis.
The problem with going out farther than where the image is formed is that all of the rays from that point on the object won't all go to that farther point. Hence no image will be out there.
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Rapidity in 4-vector Transformation In Lorentz transformation we have a concept of rapidity as related to boost. Rapidity is defined as a hyperbolic angle α such that $$\tanh(α)=v/c .$$ This further defines a matrix for Lorentz transformation in term of sinh(α) and cosh(α). This is also said to represent LT under a rotation of the frame of reference by an imaginary angle. I could not understand how this idea of using hyperbolic angle struck to Einstein or for that matter whosoever proposed it.
| I will show you the idea with a simple example: a 1+1 space-time.
In ordinary Euclidean 2-d space $\mathbb{R}^2$, the rotation is all operation on points $(x^1,x^2)$ which holds the quantity
$$(x^1)^2+(x^2)^2=C$$
invariant. So with the fact that $\cos^2 \theta+\sin^2 \theta=1$ one can simply write down the form:
$$x^1=\sqrt{C} \cos\theta \ ; \ x^2=\sqrt{C}\sin\theta$$
And the variation of $\theta$ satisfies our requirement automatically.
Then for a Minkowski 1+1 space, the invariance of light speed show that the "rotation", or the transform between frames should hold:
$$(x^1)^2-(x^2)^2=C$$
invariant. So with the fact that $\cosh^2\theta-\sinh^2\theta=1$ one can directly see how Lorentz boost connects to the hyperbolic functions.
| {
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Why is normal force at the bottom dependent on normal force on top?
Why does the normal force on bottom of the track have anything to do with the normal force on top of the track? Why isn't the normal force at the bottom simply $mg$?
| This really is a centripetal acceleration equation, compounded by gravity. You know when you get to a1g acceleration, the normal force at the bottom is double it's resting Force and would be a normal force of zero at the top. To get to point six at the top you would have to add 75% more centripetal acceleration, and I believe you would have a normal force of 2.2 kg at the bottom. If my math is correct your velocity would have to be 9.3 meters per second.
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Why does $\sqrt{\frac km}$ represent angular velocity and not frequency? When I break down $\omega = \sqrt{\frac km}$ (angular velocity for a simple harmonic oscillator) into its units, I get:
$$\omega = \sqrt{\frac{kg * \frac {m}{s^2}}{kg *m}}$$
which simplifies to:
$$\omega = \frac 1s$$
If I'm not mistaken, that is the unit for frequency (hz). I know that radians are considered "unitless", but I do not understand intuitively how it translates to angular velocity ($\frac{rad}{s}$) or how radians are involved.
| Angular frequency (rad/s) and frequency (1/s=Hz) have the same physical dimensionality, as OP correctly figured out, but are related by a factor of $2\pi$.
The inverse of this quantity is time, in one case the time it takes to go one radian in angle and in the other to go the full revolution. (There are $2\pi$ radian in a full circle.)
If you are given some physical quantity like $\sqrt{k/m}$ and you are not sure which of the two is it, there is only one way to find out, and it is to solve the equations exactly and find out the period of one oscillation and then compare. In other words, the quantity $\omega$ in the formula $\cos(\omega \, t)$ is an angular frequency, and $\nu$ in $\cos(2\pi\nu \, t)$ is the ordinary frequency, and obviously $\omega=2\pi\nu$ and the period $T=1/\nu$.
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Does the Central Limit Theorem hold for position measurements? A friend asked me recently if the Central Limit Theorem holds for quantum systems: i.e., if the distribution of measurements (e.g., of position) for any wavefunction would prove approximately normal, given enough samples.
My gut response was no, because I've encountered plenty of position wavefunctions in introductory QM that wouldn't lead to a normal distribution via Born's rule. But the CLT's requirements seem fairly permissive. What am I missing?
| In the measurement of position the probability of finding a given value is given by one probability distribution. The position is a - in the sense of one - random variable.
The Central Limit Theorem refers to the sum of random variables each of which follows a given probability distribution (in the classical fomrulation of the theorem, the probability distributions of different random variables are taken to be identical, there are formulations in which the probability distributions do not need to be identical).
For more information you can read the Wikipedia article.
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Can we replicate the sound of thunder with just the lightning? Scientists are able to extract sound of a two black holes merging from the data they collected without actually hearing it, but we don't know how it actually sounds, that's just a representation of the data in a sound format. That made me think, given that we can collect the data of a lightning (and we know what it sounds like), can we replicate or even predict with a high certainly the sound a thunder will make before we actually hear it?
|
That made me think, given that we can collect the data of a lightning
(and we know what it sounds like), can we replicate or even predict
with a high certainly the sound a thunder will make before we actually
hear it?
Yes, but not for that reason.
I want you to think about the sound of thunder. It starts high and then gets lower and lower in frequency. Ptchowwwwww.....
That's because the higher frequencies travel slightly faster. But only slightly, so to even hear this effect the strike has to be some distance off.
So you can very accurately predict the sound simply by measuring the distance to the lightning, how long it lasted, and how much total energy it had.
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Why the source function is equal to Planck function when we have a local thermodynamic equilibrium? I understand that the source function $ S_λ $ for the special case of blackbody radiation is equal to the Planck function $B_λ $.
However, in the broader case of a local thermodynamic equilibrium (and not the special case of a blackbody) I would expect that
$$ S_λ=εB_λ $$ where $ε$ the emissivity
and the equation of radiative transfer to be:
$$ \frac{dI_λ}{k_λρds}=-I_λ+εB_λ $$
and not
$$ \frac{dI_λ}{k_λρds}=-I_λ+B_λ $$
Where do I make a mistake?
| Remembering my lessons...
In LTE, the collisions dominate over the radiative transitions, then the probability of an emitted photon to be "destroyed" by a collision is much higher than to be scattered (abosrbed and re-emitted) by the atom. The $\epsilon$ parameter defines this concept ($\epsilon=\frac{C_{ul}}{C_{ul}+A_{ul}}$). For this reason, in the LTE regime $\epsilon=1$.
In the case of NLTE (non-LTE), the source function is defined as $S_{\lambda}=\epsilon B_{\lambda} + (1-\epsilon)J_{\lambda}$, where the $J_{\lambda}$ term considers the scattering role of the radiation field.
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Limitations of a vacuum pump Can a vacuum pump create a vacuum with a pressure lower than the product of it's pressure ratio and the ambient pressure? If you have a pump evacuating a tank to the ambient, it seems like we wouldn't be able to go below the pressure ratio times the ambient pressure.
Also, what influences the pressure ratio?
| If the pressure ratio is a truly independent of the inlet pressure, then you should be able to get an arbitrarily low pressure as follows:
*
*Pump from some volume A and exhaust to the atmosphere until the pressure in A is as low as the vacuum pump allows
*Pump from some volume B and exhaust to A until the pressure in B is as low as the vacuum pump allows
*Repeat with volume C and so on.
At each stage the pressure falls by less, but in the limit of repeating this infinitely you should theoretically be able to reach a true vacuum.
If your pump is a simple volume displacement pump (e.g. a piston), then (for ideal gases) the pressure ratio would depend on the volume ratio only, and would be independent of inlet pressure. Attempting the infinite process above would eventually reduce the inlet pressure to a point where the gas stopped behaving like an ideal continuum and started behaving like a collection of particles, at which point the pressure ratio would no longer be constant and the process could stall.
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Arrow of time and CPT symmetry It's been a few years since my physics degree. But I've been wondering: if you look at pictures of collisions at CERN, isn't it obvious which way time flows - simply on a probability basis? The likelihood of a huge cascade reversing into a proton and antiproton is vanishingly small. CPT seems violated to me. Am I missing something?
| This may not be the answer you’re looking for, but your sense of what is probable may have something to do with the way in which you categorize the states. Sure, it’s more likely to end up with a huge cascade, but there are many many different outcomes which you would call a “huge cascade”. The probability of one specific cascade, with a particular distribution of energy between all the particles, is also vanishingly small. The notion of many similar final states is encapsulated in the notion of “density of states”. CPT symmetry, on the other hand relates to the matrix element, or “amplitude” for a given process. Fermi’s golden rule tells us that the rate for a given process is the matrix element times the density of states.
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Why does quantum mechanics become unnecessary at sufficiently high temperatures? In my statistical mechanics intro class, we are taught that at sufficiently high temperatures, the quantum treatment of things becomes unnecessary. Why is this? Can this be shown using certain equations?
| Qualitatively, quantum effects can be ignored if the interchange properties of boson or fermions can be ignored, which is to say if the system is dilute. When will a system be dilute? A single particle occupies a volume of its thermal de Broglie wavelength cubed (in three dimensions). For massless particles, the thermal de Broglie wavelength must go as $\lambda_{th}\sim1/T$ by dimensional analysis. For massive particles, the thermal de Broglie wavelength goes as $\lambda_{th}\sim1/\sqrt{mT}.$ Thus for both massless and massive particles, as the temperature increases, the volume occupied by any one particle decreases and quantum effects become less and less important.
To flesh this out further, I'll essentially follow 8.4 and 8.5 of Baierlain's Thermal Physics book.
The number distribution of particles is either Fermi or Bose, $$n(\varepsilon)\propto \frac{1}{e^{(\varepsilon-\mu)/T}\pm1},$$ where, for simplicity, I'm using units in which Boltzmann's constant $k_B=1$.
The "quantumness" of the distributions is in the $\pm1$, which is to say that the quantumness can be ignored when $e^{(\varepsilon-\mu)/T}\gg1$. Then these two distributions can be well approximated by the classical number distribution of particles, the Maxwell distribution, $$n(\varepsilon)\propto e^{-(\varepsilon-\mu)/T}.$$
In the case that $e^{(\varepsilon-\mu)/T}\gg1$ we see that $n(\varepsilon)\ll1$, which is to say that the system has low occupation number.
When will a system have low occupation number? Qualitatively, when the system is dilute.
Baierlein derives the following expression for the average energy of a semi-classical system, i.e. one in which quantum effects are only a small correction. In three dimensions, $$\langle E \rangle = \frac{3}{2}NT\Big(1\pm\frac{1}{2^{5/2}}\frac{N \lambda_{th}^3}{(2s+1)V}\Big),$$ where $N$ is the number of particles, $s$ is their spin, $V$ is the volume of the system, and $\lambda_{th} = 1/\sqrt{2\pi mT}$.
One can see, then, that quantum effects can be ignored when $\frac{1}{2^{5/2}}\frac{N \lambda_{th}^3}{(2s+1)V}\ll1$, i.e. when $$T\gg\frac{1}{2\pi m}\Big( \frac{N}{2^{5/2}(2s+1)V} \Big)^{2/3}.$$ The denser the system---the large is $N/V$---the higher the temperature must be for one to ignore quantum effects.
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How to make an object stay under water? How to make an object submerged under water? Neutrally buoyant?
| Here is another way that I have used to ensure that the submarine dives- and also returns to the surface when the rubber power is spent.
As described by others here, you "ballast" the sub, but not for neutral buoyancy: you add slightly too much air so it barely floats. Then you add fins to the body of the sub, and bend them slightly down so that when it is being propelled, the sub experiences a downforce which makes it sink.
Down it goes- until it runs out of rubber, and then it rises back to the surface.
While we are on the subject, having a single propeller at the rear of the sub means that the rubber band applies a countertorque to the body of the sub which urges it to rotate in the opposite direction of the prop. This reduces the propulsive efficiency of the rubber motor. This torque reaction can be easily cancelled by installing a second prop in the nose of the sub, connected to the opposite end of the rubber band. You pitch the blades of this prop backwards relative to the rear prop so it produces forward thrust while rotating in the opposite sense of the rear prop. No torque reaction!
Finally, note that a small diameter prop will spin up to great speed in water and furnish a lot of thrust for a short period of time. the slower the prop turns, the less thrust it will develop but the longer the sub will run. But trimming the prop diameter, you can obtain tradeoffs between maximum speed and maximum range.
One last thing: the plastic propellers used in rubber-powered model airplanes work well for sub propulsion- as long as you can get a pair of them in left- and right-hand rotation. Hobby shops sell these. And be sure to have as much fun as possible in your experiments!
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Role of AdS/CFT correspondence in the context of integrability I was wondering how the AdS/CFT correspondence fits in the context of integrability. As I understand, the AdS/CFT correspondence postulates a duality between gravity theories and CFT's. If one theory has a strong coupling, the other has weak coupling. AdS/CFT would then be used to study the strongly coupled theory by analyzing the weakly coupled dual theory instead. However, with integrability, (I think) one can directly analyze the strongly coupled theory. Where would AdS/CFT correspondence fit in this situation?
| Having a weakly coupled gravity dual does not imply integrality. To the contrary black holes in classical gravity are the most chaotic objects there can be! See this article:
https://arxiv.org/abs/1503.01409
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Light in classical electrodynamics I am starting to learn elementary electrodynamics with Griffiths.
In the book, he has shown the natural correspondence between light and electromagnetic plane waves.
The problem that has agitated me is that plane waves are "global solutions", i.e. they have non-trivial EM field almost everywhere; while light seems to be localized phenomenon, when considered as stuff generated in region A and traveled to region B.
How is this paradox resolved classically(without QM)?
From Fourier Theory, it seems viable to create localized solutions by adding plane waves of different frequencies. However, this does not solve the paradox when monochromatic light is considered.
| @ErickShock's answer is perfectly correct, but let's look at some numbers. A HeNe laser spectrum shown:
with a peak at
$$\nu = c/\lambda = 473755.4646016\,{\rm GHz}$$
If we send out a $\tau = 1$ millisecond pulse, we can call that monochromatic pulse that 182 miles long--physically it is a very long pulse. It is not infinite (per the OP's complaint), but it is much larger than the lab.
Mathematically, we model it with:
$$ \sin{(2\pi\nu t)} \times {\rm rect(0,\tau)} $$
The Fourier spectrum of that then has a width of:
$$\Delta\nu \approx 1/\tau = 1\, {\rm kHz}$$.
So the modeled pulse has a frequency range:
$$ \nu = 473755.4646016 \pm 0.000001\,{\rm GHz}$$
(and hence a coherence length of 182 miles).
Meanwhile, an actual HeNe laser has a bandwidth on the order of 1.5 GHz ( 1 foot coherence length).
So, the deviations from "truly monochromatic" (in space or time) caused by the finite size of the apparatus or duration of operation are tiny compared with the inherent deviations in the device itself.
| {
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"url": "https://physics.stackexchange.com/questions/473133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Does the center of a black hole have a physical body? From science class, I'm led to belive that all matter breaks down once it's sucked into a black hole. I get that part, but doesn't all that matter still exist in the center of the black hole? yes it's no longer what i used to be, but it's still there in a different form right? If so, what's it look like? does it look like a pea? or maybe it would be bigger in some cases and it would look like a basketball? a planet sized sphere even? If light were "allowed to escape" would it have color and look like some mushed up piece of multicolored clay-dough? How would it look like if we could somehow bypass gravity and observe it? does anyone know?
| We don't know for certain what lies inside the event horizon of a black hole, since we can never observe what happens inside the event horizon. General relativity predicts a gravitational singularity at the centre of a black hole, but this may not be physically meaningful - how can the curvature of spacetime actually become infinite in reality ? We need a theory of quantum gravity to make a better prediction. It is possible that some form of extremely dense degenerate matter may exist at near the centre of a black hole - but we do not know for sure as our current state of knowledge only allows us to speculate.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/473286",
"timestamp": "2023-03-29T00:00:00",
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Why is the magnitude of friction acting on the rear wheels greater than the magnitude of friction acting on the front wheels? When a car is accelerating on a horizontal road , friction acts in the forward direction on the rear wheel and in the backward direction on the front wheel .
But I am not able to understand why the friction acting on the rear wheel is greater in magnitude than the friction acting on the front wheel ? Is this because the car is accelerating forward and the net force should act in the forward direction ?
| I think you have misunderstood how friction slows cars down. The friction between the tires and road allow the car to move (think about it). The friction between the axles and wheel are what slow the car down.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If a satellite speeds up, does that make it move farther away or closer? If a satellite is in a stable circular orbit and goes about 41% faster (escape velocity) then it leaves its host forever. I get that. However, what if it speeds up by less than 41%?
Intuitively, it would seem to make the satellite move farther away from the host and thus enter a higher (more distant) orbit.
However, according to my understanding, a stable orbit requires the satellite to move more slowly the farther away it is from the host. For example, the earth moves more slowly around than the sun than Venus because it is farther away from the sun than Venus.
So, if a satellite speeds up then the stable orbit would be closer to the host, not farther away. What am I missing here?
| If the satellite will reach a velocity somewhat smaller than the escape velocity then the satellite (coming out of the circular motion), the satellite will get both closer and farther away.
It will follow an elliptic trajectory, the earth being one of the ellipse's two focus points. This means the satellite will pass both the aphelion (the point that closest to earth, smaller than the radius of the circle) as the perihelion (the point which is the furthest distance removed from the earth, bigger than the radius of the circle). See the picture below, where the $F$'s are the two foci (the earth stays virtually put on one of these).
If the satellite will get out of the circular orbit with a velocity higher than the escape velocity it will escape us forever. Depending on the speed, the corresponding trajectory will be a parabola or a hyperbola. See the second picture where the ellipse is a the intersection of a flat plane with a cone (as are the parabola and the hyperbola).
fig.1
fig. 2
For a certain range of angles of the increase in speed the satellite will crash on the host. The majority of angles though will not make this happen. The range of angles for which a collision does appear depends, or not, on the mass of the host. I'm not sure if this range of angles is scale-independent. You have to calculate it. It's Sunday and I'm not in the mood to calculate...
| {
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"timestamp": "2023-03-29T00:00:00",
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Was it ever experimentally verified whether antimatter can form chemical bonds the same way matter does? It's "common knowledge" that antimatter only differs from matter in its charge, and if left alone, it would behave exactly as matter does. This would infer that atoms of antimatter would form similar chemical bonds as matter does. Two anti-hydrogen and an anti-oxigen would form anti-water, which would behave just as regular water does.
However, was this ever experimentally verified? Were chemical bonds between atoms of antimatter ever observed? If so, were their properties the same as the chemical bonds between regular atoms?
I'm not talking about antiprotons and positrons forming atoms. I'm talkin about anti-atoms forming molecules.
| So far, although a lot of positrons and antiprotons have been produced, only a very small number of anti-hydrogen atoms have been created for study.
The positrons and antiprotons are produced with much, much more kinetic energy than the atomic binding energy. It requires some neat technology to slow and cool them, trap them and finally combine them into an atom.
The current state of the experimental art is to look at the anti-atom’s spectra and compare to regular atoms.
Doing chemistry is a long way down the road.
| {
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Location of free charge in insulators I'm going through the introductory section to Electrostatics in Materials in Griffiths, and I have a question that I can't seem to find a satisfactory answer to.
If I have an insulator with free charge, is it necessarily confined to the surface?
In the case of a conductor, Gauss' law immediately gives a "yes", because no electric field can exist in a conductor, leading us to conclude that there is no free charge inside the surface. But insulators can have electric fields inside them. Does this mean that free charge can exist inside the volume? Or does the free charge still move to the surface of the insultator?
| Dielectrics can be both conductors and insulators. For a capacitor plate, an insulator is the best thing to use. It is due to this insulation property that means no free charges, that creates an electric field to counter the one between the plates due to some surface charge density.
If you are talking about conductors used as dielectrics, then surely, the internal field is zero (from Gauss' law as you said). But then, this is the reason why conductors are not used as dielectrics, because of the charge leakage as current in capacitors.
| {
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EMF generated by a rotating rod can be zero? I have taken a look at this question: Emf generated by a rotating rod
But my question is different:
A metallic rod is being rotated in a uniform magnetic field with a constant angular velocity $\omega$. The axis of rotation of the rod is vertical, and passes through its midpoint. If the specific charge of free electrons in the rod is $\frac{e}{m}$, then for what value of the vector $\omega$ will the potential difference between the tip of the rod and the midpoint be zero, considering the magnetic field to be upwards?
I'm having trouble understanding the physics behind it, much like the other question I linked.
Here's my attempt at this question:
Considering that the rod is moving perpendicular to the direction of the magnetic field, there should be a force applied on the electrons present in the conducting rod. This force will eventually get cancelled out when the charges get accumulated at the end of the rod and at its midpoint, which will generate some electric field that cancels out the magnetic force.
Thus: $q((\omega$ x$)\times$B)+q(E$_{in}$)=0
Here, x is the distance from the midpoint.
When we solve this equation, using $\Delta$V=$-\int$E$\cdot$dx, we get $\Delta$V=$\frac{B\omega l^{2}}{2}$. That can't be zero for any value of $\omega$ except zero. But, the answer is not zero...
Please guide me...
| The pd set up by the redistribution of electrons would indeed cancel the magnetic force on the free electrons in the rod, preventing further motion (apart from random thermal). The same applies inside an open-circuit battery when charges have built up on its terminals.
Your question (and it's one I've not seen presented this way before) is exploring a different effect. For the magnetic force to urge the electrons along the rod, even before the opposing electric field has built up, the magnetic force first has to provide the centripetal force on the electrons. I'll now leave it up to you...
| {
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Resistors parallel to short circuit I have a trouble solving this problem.
(Actually, each resistor is not parallel with the bottom wire since the voltage across each of them is different with that of wire.)
In the manual, the bottom wire(which is red circled) is simply ignored and thus can be easily solved.
I don't understand why the bottom wire can be ignored. At first, I thought it's because the node between the wire has the same voltage, making voltage difference across the wire to be 0 resulting current flowing through the wire to be 0.
I used mesh current method by myself, not ignoring the bottom wire and i got different answer, which indicated that the wire has some non-zero current flowing through it.
| The following three circuits are equivalent:
Resistances given: $(R_1, R_2, R_3, R_4, R_5)=(5,4,7,1,20)\ \Omega$;
batteries $(E_1, E_2)=(480, 168) \ Volt$.
The resistance in the middle $R_{345}=R_5+\frac{R_3 R_4}{R_3+R_4}$.
I think you want the voltage drop across $R_5$ which is $U_5=I_5 R_5$. Use Kirchhoff's laws to obtain a system of three equations.
$ \left\{ \begin{array}{l}
upper\ junction\\
left\ loop \\
right\ loop
\end{array} \right\} $
$ \left\{ \begin{array}{l}
I_1=I_2+I_5\\
E_1-R_1 I_1 -R_{345}I_5=0 \\
E_2+R_{345} I_5 - I_2 R_2=0
\end{array} \right\} $
Solving for $I_5$ gives $$R_5 I_5=R_5\cdot\frac{R_2E_1-R_1E_2}{R_1R_{345}+R_1R_2+R_2R_{345}}=103.909\ V$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Origin of 2π normalization factor in chemical $J$-coupling compared to the Heisenberg model In chemistry, particularly the field of NMR spectroscopy, the interaction between two (nucleic) spins (or so I guess?) is governed by the Hamiltonian:
\begin{align}
\mathcal{H}=2\pi\cdot J_{ij}{\vec {S_{i}}}\cdot {\vec {S_{j}}}
\label{eq1}
\end{align}
https://en.wikipedia.org/wiki/J-coupling
This interaction is called $J$-coupling and it is usually heuristically explained by interaction between two electronic spins which is governed by the Pauli principle and the electron-nucleus interaction which in the end favors antiparallel alignment of the nucleic spins over parallel and thereby leads to a splitting of energy levels. Note that mainly chemists cover this topic and their approach is usually very hands-on in the sense that the values of $J_{ij}$ are usually measured on a spectrometer.
Now the above equation looks very similar to the Heisenberg model, governing the general relationship between two spins:
\begin{align}
H_{\text{Heis}}=-J\sum _{\langle i,j\rangle }{\vec {S_{i}}}\cdot {\vec {S_{j}}}\qquad {\text{with }}i,j\,{\text{next neighbors}}
\label{eq2}
\end{align}
https://en.wikipedia.org/wiki/Heisenberg_model_(quantum)
This surely looks similar, here $J$ is the value of the exchange integral, which I would assume is also what defines the scalar coupling tensor $J_{ij}$ in the first equation. However, I am by no means sure those two $J$s are the same and if they are, where does the normalization factor of $2\pi$ coming from? Is it just common among chemists to divide their exchange integrals by $2\pi$? I would be very grateful if someone could shed a little light on that.
| Yes, chemists like J couplings in Hz. The NMR J or scalar coupling and the Heisenberg model occur in very different contexts. In NMR, the scalar coupling is typically more than 1000 times smaller than the Zeeman term in the Hamiltonian. Moreover, the temperature is more than 1000 times larger than the Zeeman term. The NMR J coupling is a residual magnetic dipole interaction, which remains after the main magnetic dipole interaction has been averaged away by motion. It can be of either sign. All of this is very different from the usual situation and concerns in applications of the Heidelberg model.
| {
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Why do we observe the nuclear force only in scatterings and decays? Why, at first glance, are the only forces we perceive to be gravity without quantizing, electromagnetism and nuclear forces only in disintegrations?
|
Why, at first glance, are the only forces we perceive to be gravity without quantizing, electromagnetism and nuclear forces only in disintegrations?
If we accept as given the standard model of particle physics,SM with the addition of an effective quantization gravity, it can be shown that it mathematically describes/fits all of our observations and data.
Within this model your why is answered with: because we, as biological units, live in dimensions where h, the Planck constant is effectively zero, so quantization effects are not easily observable in our immediate surroundings. That is why it took so long to discover quantization.
In the SM, electromagnetism emerges in our dimensions from the underlying quantum state, where the coupling constant is of infinite range. Gravity has a very weak coupling constant, and again of infinite range, and emerges from the underlying effective quantized frame ( still to be proven). The infinite range in space is why we immediately perceive these two forces.
The weak and strong can be perceived in our dimensions with specific experiments exploring their short range.
This is the answer within the SM model for "why".
Why do we observe the nuclear force only in scatterings and decays?
The nuclear force is a spill over force from the strong interaction, and is limited by the limited range of the interaction, as seen in the link given above:
Special experiments allow to measure cleanly specific scatterings at the small ranges that the strong force acts. The decays are not only due to the nuclear force, alpha decays, but also the weak interaction, beta decays, which when quantum numbers and energy conservation allow can appear in decays of nuclei, and the same for the electromagnetic one, gamma decay.
The ultimate why is "why this standard model with this specific mathematics", and the answer is : because it fits data and observations.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is thermodynamics only applicable to systems in equilibrium? So I was going through callen's thermodynamics book and their he says that thermodynamics is only applicable to systems which are in equilibrium and that naturally raised a few questions in my mind
Is thermodynamics really never applicable to systems which are not in equilibrium, if so why should such a restriction exist?
And also it might sound silly but why is the theory called "thermodynamics"- specifically the "dynamics" part?
| A "non-equilibrium" system held steady was coined "thermostaedics" by Prof. Ralph J. Tykodi (now deceased)who was my advisor on my M.S. Thesis under the title of a book that he published called, "Thermodynamics of the Steady State" at Illinois Institute of Technology. Thermostaedics seems to me best because it covers any system as it approaches equilibrium slowly.
| {
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"timestamp": "2023-03-29T00:00:00",
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The Electromagetic Tensor and Minkowski Metric Sign Convention I am trying to figure out how to switch between Minkowski metric tensor sign conventions of (+, -, -, -) to (-, +, +, +) for the electromagnetic tensor $F^{\alpha \beta}$.
For the convention of (+, -, -, -) I know the contravariant and covarient forms of the electromagnetic tensor are:
$$ F^{\alpha \beta} =
\begin{bmatrix}
0 & -\frac{E_{x}}{c} & -\frac{E_{y}}{c} & -\frac{E_{z}}{c} \\
\frac{E_{x}}{c} & 0 & -B_{z} & B_{y} \\
\frac{E_{y}}{c} & B_{z} & 0 & -B_{x} \\
\frac{E_{z}}{c} & -B_{y} & B_{x} & 0 \\
\end{bmatrix}
$$
and
$$ F_{\alpha \beta} = \eta_{\alpha \mu} F^{\mu v} \eta_{v \beta} =
\begin{bmatrix}
0 & \frac{E_{x}}{c} & \frac{E_{y}}{c} & \frac{E_{z}}{c} \\
-\frac{E_{x}}{c} & 0 & -B_{z} & B_{y} \\
-\frac{E_{y}}{c} & B_{z} & 0 & -B_{x} \\
-\frac{E_{z}}{c} & -B_{y} & B_{x} & 0 \\
\end{bmatrix}.
$$
Now for the convention of (-, +, +, +) are the contravariant and covariant forms of the electromagnetic tensor just switched from above along with signs?:
$$ F^{\alpha \beta}=
\begin{bmatrix}
0 & \frac{E_{x}}{c} & \frac{E_{y}}{c} & \frac{E_{z}}{c} \\
-\frac{E_{x}}{c} & 0 & B_{z} & -B_{y} \\
-\frac{E_{y}}{c} & -B_{z} & 0 & B_{x} \\
-\frac{E_{z}}{c} & B_{y} & -B_{x} & 0 \\
\end{bmatrix}
$$
and
$$ F_{\alpha \beta} = \eta_{\alpha \mu} F^{\mu v} \eta_{v \beta} =
\begin{bmatrix}
0 & -\frac{E_{x}}{c} & -\frac{E_{y}}{c} & -\frac{E_{z}}{c} \\
\frac{E_{x}}{c} & 0 & B_{z} & -B_{y} \\
\frac{E_{y}}{c} & -B_{z} & 0 & B_{x} \\
\frac{E_{z}}{c} & B_{y} & -B_{x} & 0 \\
\end{bmatrix}~?
$$
Basically, I am trying to figure out how to switch between the two sign conventions.
| Misner, Thorne, and Wheeler have a nice two-page summary of sign conventions in general relativity, in the front endpapers of the book, so that would be the first place I would turn for this kind of thing.
The electromagnetic tensor is defined by the Lorentz force equation, which gives the four-force acting on a charged particle as $f^a=qF^a{}_bv^b$. The definition of the upper-index four-force and four-velocity have nothing to do with the choice of signature, so the components of the mixed-index electromagnetic tensor $F^\mu{}_\nu$ do not depend on the choice of signature. The forms $F_{\mu\nu}$ and $F^{\mu\nu}$ do have components that depend on the signature, and they can be found, if required, from the components of $F^\mu{}_\nu$.
| {
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Direction of $d\mathbf{l}$ A solid sphere has charge $q$ and radius $R$. Find the potential at a point a distance $r$ from the center of the sphere where $r>R$, using infinity as the reference point.
My attempt:
From Gauss' theorem we may deduce that $\displaystyle\mathbf{E}=\frac{1}{4\pi\epsilon_0}\frac{q}{r'^2}\hat{\mathbf{r}}$ where $r'$ is the distance of an arbitrary point from the center of the sphere provided $r'>R$.
$V=-\int_\infty^\mathbf{r}\mathbf{E}\cdot d\mathbf{l}$
My question: What is $d\mathbf{l}$? Since we are traversing in the direction opposite to $\hat{\mathbf{r}}$, I think it should be $-dr'\hat{\mathbf{r}}$. But when I use it to find $V$, I get a sign error. Please help!
| Potential is defined as the negative of the work done in moving unit charge at zero acceleration from reference to that point in field where the potential is being calculated. So dl represents a differential movement from reference (here infinity) towards r' (assumed straight line path, else tangential to path towards r'). On the other hand, since r' is being measured from origin so its differential dr' is directed in the incresing direction of r'--from r' towards reference point(infinity). Hence, if the path connecting the two points between which the test charge is being moved is a straight line, the differentials only differ in sign so that dr'=-dl.
Hence
$$V
=-\int_\mathbf{reference}^\mathbf{target}\mathbf{E(r').}\,\mathbf{dl}
$$
At this point instead of proceeding as
$$
\begin{align}
V
&=-\int_\mathbf{\infty}^\mathbf{r}\mathbf{E(r').}\,\mathbf{dl}\\
&=-\int_\mathbf{-\infty}^\mathbf{-r}\mathbf{E(r').}\,(-\mathbf{dr'})\\
&=\int_{-\infty}^{-r}\frac{1}{4 \pi \epsilon_0} \frac{q}{r'^2} \,dr'\\
&=\frac{1}{4 \pi \epsilon_0} \frac{q}{r} \\
\end{align}
$$
most books follow (as @Bio suggests)
$$
\begin{align}
V
&=+\int_\mathbf{target}^\mathbf{reference}\mathbf{E(r').}\,\mathbf{dr'}\\
&=\int_\mathbf{r}^\mathbf{\infty}\mathbf{E(r').}\,\mathbf{dr'}\\
&=\int_r^\infty\frac{1}{4 \pi \epsilon_0} \frac{q}{r'^2} \,dr\\
&=\frac{1}{4 \pi \epsilon_0} \frac{q}{r} \\
\end{align}
$$
This is imho, probably because in the former way there is an implicit substitution changing l to r' but without the proper use of limits(as in $lim_{}$), the negation in limits(as in $\int_a^b\,$) cannot be explained.
$$
\\
\\
\\
$$
This becomes clearer when one considers doing the integral this way--
$$
V=
-\int_\mathbf{reference}^\mathbf{target}\mathbf{E(l).}\,\mathbf{dl}
$$
Since there exists dl so must l. Hence it should be possible to do the RHS without converting to r' coords. Doing this is a bit tricky as the limits would be
$$
\begin{align}
\mathbf{reference}&=\mathbf{0}\\
\mathbf{target}&=\lim_{h\to \infty}(h-r)\mathbf{\hat{l}}\\
\end{align}
$$
while
$$
\mathbf{E(l)}=\lim_{h\to \infty}\frac{-1}{4\pi\epsilon_{0}}\frac{q\mathbf{\hat{l}}}{(h-l)^2}
$$
Then
$$
\begin{align}
V&=-\int_\mathbf{reference}^\mathbf{target}\mathbf{E(l).}\,\mathbf{dl}\\
&=-
\lim_{h\to \infty}
\int_
0^{h-r}
\lim_{h'\to h}
\frac{-1}{4\pi\epsilon_{0}}\frac{q\mathbf{\hat{l}.dl}}{(h'-l)^2}
\,\\
&=\frac{q}{4\pi\epsilon_{0}}\lim_{h\to \infty}\lim_{h'\to h}(\frac{1}{0-h'}+\frac{1}{h'-(h-r)})\\
&=\frac{1}{4 \pi \epsilon_0} \frac{q}{r}
\end{align}
$$
The integration performed in line 3 above is obtained from Mathematica as
$$
\int_a^b \frac{1}{(A-x)^2} \, dx=\frac{1}{a-A}+\frac
{1}{A-b}, \quad\quad\quad(a\geq A\lor A\geq b)\land a<b
$$
| {
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Expectation Value of displacement to the $n$th power quantum harmonic oscillator? Is there a closed-form expression for $\langle x^n\rangle$ for the ground state quantum harmonic oscillator, where $n =$ even integer $>0$? I am attempting to pursue this with rising and lowering operators but the foiling is getting out of hand.
I would like $$\langle0 | x^n |0\rangle = \underline{\qquad\qquad}\,, n={2,4,6,8,\dots}$$
Example:
$$\langle0 | x^2 |0\rangle = \langle0|\left[\sqrt{\frac{\hbar}{2m\omega}}\left(a^{\dagger} + a\right)\right]^{2}| 0\rangle = \frac{1}{2\alpha},$$ where $\alpha = m\omega/\hbar$
| You can get the answer without raising and lowering operators, by directly calculating the the overlap integral using the identity
$$\int_{-\infty}^{\infty}dx \ x^{2n} \ e^{-\alpha x^2} = \frac{(2n-1)!!}{(2\alpha)^n} \sqrt{\frac{\pi}{\alpha}}\ ,$$
which you can prove by induction or just look up in a table of integrals such as Gradshteyn and Ryzhik.
| {
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Partition function in renormalization When studying statistical mechanics, renormalization is understood from attempts to calculate partition function by simplifying. (For example, David Tong's lecture note)
While I understand that partition function allows one to calculate important thermodynamic quantities, what I do not get is why renormalization is understood as keeping partition function intact. Renormalization changes other quantities, such as temperature, free energy, so why should partition function be alone in being intact?
Edit/second question: also, texts confuse me on how renormalization is supposed to preserve partition function. Does it like map it exactly as $Z(K) = Z'(K')$ where $K'$ is a set of renormalized variables and $Z$ is partition function? Or is it more like $Z(K) = f(K')Z'(K')$? Texts seem to suggest both, and I am confused about that.
| The whole point of renormalization is that you are changing all of these parameters in the precise way that keeps the partition function intact. In a sense, you're trying to find all the solutions to the equation $Z(K_0)=Z'(K')$, where $Z(K_0)$ is whatever bare couplings you start with. Sometimes, when doing this, authors will drop terms that correspond to multiplying the overall partition function by a constant, since these terms don't affect the physics, but this is purely cosmetic and you should not think of this as a fundamental part of renormalization.
| {
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Question on two-neutrino double electron capture There was a fascinating paper in Nature recently, on the observation of two-neutrino double electron capture in xenon, with a half-life time of $1.8\times 10^{22}$ years.
The process described in the article is
$$^{124}\mathrm{Xe} + 2e^- \to {}^{124}\mathrm{Te} + 2 \nu_e.$$
According to Wikipedia, double electron capture can occur only when competing modes are strongly suppressed.
My Question: Why is the single electron capture so strongly suppressed? Why can't we have
$$^{124}\mathrm{Xe} + e^- \to {}^{124}\mathrm{I} + \nu_e$$
while the decay mode
$$^{125}\mathrm{Xe} + e^- \to {}^{125}\mathrm{I} + \nu_e$$
exists?
| This is explained by Scott Manley in Why a Dark Matter Search Also Observed The Rarest Radioactive Decays at around the 7:20 mark.
The short answer is that the process is electronically forbidden, because the iodine-124 nucleus has a higher binding energy than the xenon-124 nucleus. Using the data from Wikipedia, the masses for the nuclides involved are
\begin{align}
m({}^{124}\mathrm{Xe}) & = 123.905\,893(2) \:\mathrm{u} \\
m({}^{124}\mathrm{I}) & = 123.906\,2099(25) \:\mathrm{u} \\
m({}^{124}\mathrm{Te}) & = 123.902\,8179(16) \:\mathrm{u}.
\end{align}
This means that the ${}^{124}\mathrm{Xe}\to {}^{124}\mathrm{Te}$ decay is allowed, and releases
$$(m({}^{124}\mathrm{Xe}) -m({}^{124}\mathrm{Te}))c^2 = 2.86\:\mathrm{MeV}$$
of energy, whereas that same difference for the decay to iodine yields a negative mass difference,
$$(m({}^{124}\mathrm{Xe}) -m({}^{124}\mathrm{I}))c^2 = -0.295\:\mathrm{MeV},$$
which means that the beta decay as it occurs one mass unit up is energetically forbidden. (If you do the same calculation there, you get $(m({}^{125}\mathrm{Xe}) -m({}^{125}\mathrm{I}))c^2 = 1.64\:\mathrm{MeV},$ which is plenty of energy to fuel a beta decay.)
That said, though, this is not enough to rule out an electron-capture mechanism on energetic grounds, since the energy hill from xenon-124 to iodine-124 can be climbed with the annihilation of the electron,
$$(m(e^-) + m({}^{124}\mathrm{Xe}) -m({}^{124}\mathrm{I}))c^2 = +0.21\:\mathrm{MeV},$$
so there are definitely substantial details left to explain there, which can hopefully be explained by a nuclear physicist. Still, the difference in the energetics is definitely large enough that the two processes cannot be considered a priori to be roughly equivalent.
| {
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What is an integrator lens? I was looking into how photolithographic mask aligners works and came across this diagram, I am curious as to what an 'integrator' is. It seems like these lenses are used to focus the light beam, am I correct? Is the diagram correct in showing these lenses as being flat or is their curvature simply subtle?
| To minimize the influence of the arc spread and its actual position there is a diffusing element placed in the secondary focus of the mask aligner (Integrator 1). The Integrator 1 is followed by the so-called Integrator 2. The Integrator 1 has the function to create a radiation field as uniform as possible at the plane of the Integrator 2. A glass plate with pyramids or lenses on one side is used as Integrator 1 within the standard illumination system.
Here is a close up look to the integrator
It segments the incident spatial distribution. The fly’s eye lens array can be, and often is, split into two separate arrays. In either configuration, the
“lenslets” of the first array or the first surface focuses onto the second. The second is the pupil of an imaging system that images the input faces of the lenslets to the uniform plane. The field lens is the back half of the imaging system which overlays all the images to form a uniform plane.
| {
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Examples of non-sine waves? What would be a non-sine wave? AFAIK, all sound is a sine wave, equally to waves on the sea. What would be a common example of something in nature that's a wave but not a sine wave? Or, would we have to look at man made regularities like bus timetables or stock prices to find non-sine waves?
| It is not true that all sound is sine waves. What is true is that one can find sinusoidal solutions to the wave equation. This allows an arbitrary wave to be expresses as a linear superposition of sines. It looks like the Soliton has been mentioned above. While this is one counter example it relies on the non-linear KdV equation. I want to point out that even the linear wave equation allows for strange non-sine looking solutions. The decomposition of an arbitrary wave or pulse in terms of sines is related to the Fourier transform (also mentioned).
| {
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Adiabatic Approximation, Solving the Schrödinger equation In the adiabatic approximation one looks at the Hamiltonian
$$ H_0 = \sum_{i = 1}^{N_e} \frac{\vec{p}_i^2}{2m_e} + \sum_{i < j} \frac{e^2}{|\vec{r}_i - \vec{r}_j|} + \sum_{k < l} \frac{Z_k Z_l e^2}{|\vec{R}_k - \vec{R}_l|} + \sum_{i, k} \frac{- Z_k e^2}{|\vec{r}_i - \vec{R}_k|}. $$
Now typical statements are
"For a fixed ion configuration $ \{\vec{R}_1,\ldots, \vec{R}_n\}$ let $\Psi_{\alpha} \equiv \Psi_{\alpha}(\vec{r}_1,\ldots,\vec{r}_n,\vec{R}_1,\ldots\vec{R}_n)$ be a solution of the eigenvalue problem
$$ H_0 \Psi_{\alpha} = \varepsilon_{\alpha}(\vec{R}_1, \ldots \vec{R}_n) \Psi_{\alpha},$$
where $\alpha$ denotes a complete set of quantum numbers."
What is baffling to me is: How does one know about the existence of such solutions? (Not only, that one finds one solution of that eigenvalue problem, but also a complete set of solutions?!)
| So, the Hamiltonian of the question describe a set of atoms and electrons where the atoms are supposed to be fixed in space.
There are two answer to this question.
The mathematical answer is:
The existence of a complete set of solutions follows from the properties of the Hermitian operators, as stated in the other answer. One can show (it is complicated) that the Hamiltonian can be written as a matrix in the full basis of the Fock space. Therefore the completeness of the set of solutions follows from linear algebra.
The physical answer is:
Look around you! Solutions of Hamiltonians like this are all around us. They are called solids. Physically, this guarantees me that solutions exists. Otherwise quantum mechanics would fail spectacularly and it would be not worth considering.
| {
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Why hydrogen lines are less visible in the Sun spectrum than in supernovae clouds? Supernovae clouds are very colorful, and if I trust documentaries I watched, the colors are due to excitation of elements, as in fireworks. Since the Sun is mostly made of hydrogen, I suppose those lines should be very apparent but they are not so much, its light looking like a blackbody radiation. What contributes to the rest of the spectrum up to the point it masks hydrogen lines?
| The sun's spectrum is a blackbody spectrum with absorption lines superimposed. The blackbody spectrum comes from the photosphere, which is the highest elevation at which the sun is opaque. It becomes opaque because it's ionized to form a plasma. A plasma is opaque because it has free charges that interact strongly with electromagnetic waves and absorb their energy. The spectrum is a continuous blackbody spectrum rather than an emission spectrum because it's a dense plasma. Discrete emission and absorption spectra are typical of thin gases composed mainly of neutral atoms or molecules.
The absorption lines in the sun are created by layers farther out in which the gas is not ionized and less dense.
| {
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Unpolarised light falls directly on a half-wave plate Will an unpolarised light get polarised if it falls directly on a half-wave plate? If so, then what will be the state of polarisation?
| If you shine unpolarized light onto a half-wave plate, then the result will be similarly unpolarized.
It seems you really don't want to believe this answer, so let's address your two objections thus far:
*
*
But.... By the definition of half wave plate.... It says that an unpolarised light will be divided into two polarised lights... E-ray and o-ray.... And a phase difference of (π) will be there
Yes. But in unpolarized light, those two components are incoherent with respect to each other, and they don't have a well-defined relative phase. Adding $\pi$ phase to that undefined relative phase doesn't change anything.
*
But I found half wave plate will rotate plane of polarisation by 2theta....so won't it be 90degree rotation?
First of all, half-wave plates don't rotate the plane of polarization - they reflect it. (To rotate the polarization, you need two subsequent half-wave plates, with their slow axes at an angle.) In any case, if you want to analyze it like that, then the unpolarized light is the incoherent mixture of two orthogonal polarizations, which will be taken by the HWP's reflection to two different orthogonal polarizations, still with no relative coherence. Or, in other words, unpolarized light.
| {
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How do car rear-view mirrors work? I wonder, how does a car rear-view mirror work?
When there is a car behind me with high-beam, all I do is flip a tong at the bottom of the mirror to relax the lights!
Are there two mirrors in it, one darker than the other?
| For manual anti-glare mirrors, the glass is actually a prism with the silvered rear surface not parallel to the front surface
In day-time position, drivers are seeing reflections from the rear surface with large amounts of reflected light reaching their eyes
In night-time anti-glare position, drivers are seeing reflections from the front surface of the glass, with much less light going into their eyes; the brighter rear reflection goes elsewhere. This is still enough to distinguish headlights behind, but not much else, and substantially less than if the day-time position was used at night, so reducing the contrast which could be blinding if the following vehicles were foolish enough to use full-beam headlights
See https://en.wikipedia.org/wiki/Rear-view_mirror#Anti-glare for more (and the automated alternative) which has these two diagrams
| {
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Momentum matrix elements in a crystal I am trying to follow along a derivation (E. I. Blount, Solid State Phys. 13, 305 (1962)) in which he derives the matrix elements of the true momentum $p_{n,n'}(k,k')$ (not the crystal momentum). He arrives at the following expression:
$p_{n,n'}(k,k') = \delta(k-k')(\hbar k\delta_{n,n'}-i\hbar\int u_n^*\frac{\partial u_{n'}}{\partial x}\,d\tau)$,
where $n$ labels the band index, $k$ is the Bloch vector, $u_n$ is the Bloch function with the periodicity of the lattice, and the integral is over a unit cell. Can anyone help me to derive this result?
Crucially, the problem boils down to evaluating
$\int dx \, e^{-ikx}u_{nk}^*(x)e^{ik'x}u_{n'k'}(x)$,
which, according to Blount ought to equal $\delta_{n,n'}\delta(k,k')$. This makes sense if our wave functions
$\psi_{n,k}(x) = e^{ikx}u_{nk}(x)$
(i.e. Bloch waves) are to be normalized, but I just can't seem to figure out how to 'pluck' those two delta functions out from the integration.
| I agree with aljg up to the third line from the bottom of his derivation. I would argue that from there, the derivation should go as follows.
\begin{align}
\langle \psi_{n \mathbf{k}} | \frac{\mathbf{p}}{\hbar} | \psi_{n' \mathbf{k}'} \rangle
&= \frac{1}{V} \int d\mathbf{r} e^{i(\mathbf{k}'-\mathbf{k})\cdot \mathbf{r}} u_{n\mathbf{k}}^*(\mathbf{r}) \left[ \mathbf{k}' u_{n'\mathbf{k}'}(\mathbf{r}) -i \nabla_\mathbf{r} u_{n'\mathbf{k}'}(\mathbf{r})\right] \\ & = \frac{1}{V}\sum_\mathbf{R}e^{i(\mathbf{k'-k)\cdot R}}\int_{V_{UC}}d\mathbf{r} e^{i(\mathbf{k}'-\mathbf{k})\cdot \mathbf{r}} u_{n\mathbf{k}}^*(\mathbf{r}) \left[ \mathbf{k}' u_{n'\mathbf{k}'}(\mathbf{r}) -i \nabla_\mathbf{r} u_{n'\mathbf{k}'}(\mathbf{r})\right] \\ & = \frac{N}{V}\delta_{\mathbf{kk'}}\int_{V_{UC}}d\mathbf{r} e^{i(\mathbf{k}'-\mathbf{k})\cdot \mathbf{r}} u_{n\mathbf{k}}^*(\mathbf{r}) \left[ \mathbf{k}' u_{n'\mathbf{k}'}(\mathbf{r}) -i \nabla_\mathbf{r} u_{n'\mathbf{k}'}(\mathbf{r})\right] \\ & = \frac{N}{V}\delta_{\mathbf{kk'}}\int_{V_{UC}}d\mathbf{r} u_{n\mathbf{k}}^*(\mathbf{r}) \left[ \mathbf{k} u_{n'\mathbf{k}}(\mathbf{r}) -i \nabla_\mathbf{r} u_{n'\mathbf{k}}(\mathbf{r})\right] \\ & = \frac{N}{V}\delta_{\mathbf{kk'}}\left[\mathbf{k}V_{UC}\delta_{nn'}-i\int_{V_{UC}}d\mathbf{r}u_{n\mathbf{k}}^*(\mathbf{r})\nabla_\mathbf{r}u_{n'\mathbf{k}}(\mathbf{r}) \right] \\ & =\delta_{\mathbf{kk'}}\left[\mathbf{k}\delta_{nn'}-\frac{i}{V_{UC}}\int_{V_{UC}}d\mathbf{r}u_{n\mathbf{k}}^*(\mathbf{r})\nabla_\mathbf{r}u_{n'\mathbf{k}}(\mathbf{r}) \right].
\end{align}
However, it seems I have picked up an unwanted factor of $\frac{1}{V_{UC}}$ in the second term, which messes up the dimensions.
| {
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Binding energy of a molecular ion?
The protons in the $\text{H}_2^{+}$ molecular ion are $0.106 \, \mathrm{nm}$ apart, and the binding energy of $\text{H}_2^{+}$ is $2.65\,\mathrm{eV} .$ What negative charge must be placed halfway between two protons this distance apart to give the same binding energy?
Question: In the above quote, does "the binding energy of $\mathrm{H}_2^{+}$" refer to the energy required to disassemble the whole system (two protons and one electron) or just the protons?
| Just the protons, basically - it's the energy required to take the system apart into a single proton and an isolated hydrogen atom.
Separating the proton and electron from that hydrogen atom would take a fair bit more energy - an additional 13.6 eV.
| {
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Why is it that magnetic fields(or any field)not move in space? When I imagine a magnetic field produced by a magnet, or the electric field produced by a charge, I've learned that the fields are stationary, however, their value(across space) changes.
If I placed the magnet at a point $P$($0,0,0$), and then moved the magnet to $P_2$$(1,1,1)$ Why wouldn't it's associated magnetic field move with it?
| If the magnet is moved, the magnetic field changes to match the new position of the magnet. As the magnetic field changes, it also induces an electric field.
The details of this interaction depend heavily on how exactly the magnet is being moved, and may or may not involve the production of propagating electromagnetic waves (for example, an accelerating magnet will generate electromagnetic waves, while a magnet moving at a constant velocity will not).
| {
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Momentum average on phase space for free particle I'm studying from Greiner statistical mechanics, and he uses an approximation which I don't really understand.
On averaging over many phase-space points we have
$$\langle\vec{p}^2\rangle=3\langle p_x^2\rangle=3\langle p_y^2\rangle=3\langle p_z^2\rangle$$
since no direction in space is preferred, i.e.,
$$\sqrt{\langle\vec{p}^2\rangle}=\frac{\sqrt{3}}{3}\left(\sqrt{\langle p_x^2\rangle}+\sqrt{\langle p_y^2\rangle}+\sqrt{\langle p_z^2\rangle}\right).$$
Therefore, we make the approximation
$$\epsilon=c\left(p_x^2+p_y^2+p_z^2\right)^{1/2}\approx\frac{c}{\sqrt{3}}\left(\vert p_x\vert+\vert p_y\vert+\vert p_z\vert\right)$$
Can someone please explain this approximation?
| The claim on Greiner is
$$ \langle\vec{p}^2\rangle = 3\langle p_x^2\rangle = 3\langle p_y^2\rangle = 3\langle p_z^2\rangle $$
This claim follows from the fact that
$$ \text{E}[p^2] = \text{E}[p_x^2] + \text{E}[p_y^2] + \text{E}[p_z^2] $$
and that $(x,y,z)$ are indistinguishable:
$$ \text{E}[p^2] = 3 \text{E}[p_x^2] = 3 \text{E}[p_y^2] = 3 \text{E}[p_z^2] $$
The rest follows if you substitute all of $(x,y,z)$ with either one.
| {
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Regularising the Green's function in 2D The Green's function for the 2D Helmholtz equation satisfies the following equation:
$$(\nabla^2+k_0^2+\mathrm{i}\eta)\,{\mathsf{G}}_{2\mathrm{D}}(\mathbf{r}-\mathbf{r}',k_o)=\delta^{(2)}(\mathbf{r}-\mathbf{r}').$$
By Fourier transforming the Green's function and using the plane wave representation for the Dirac-delta function, it is fairly easy to show (using basic contour integration) that the 2D Green's function is given by
$${\mathsf{G}}_{2\mathrm{D}}(\mathbf{r}-\mathbf{r}',k_0)=\displaystyle\lim_{\eta\to0}\int\frac{\mathrm{d}^2\mathbf{k}}{(2\pi)^2}\frac{\mathrm{e}^{\mathrm{i}\mathbf{k}\cdot(\mathbf{r}-\mathbf{r}')}}{k_0^2+\mathrm{i}\eta-k^2}=\frac{1}{4\mathrm{i}}\operatorname{H}_0^{(1)}\left(k_0|\mathbf{r}-\mathbf{r}'|\right)$$
where $\operatorname{H}_0^{(1)}$ is the Hankel function of zeroth order and first kind.
However, this 2D Green's function diverges (logarithmically) at $\mathbf{r}=\mathbf{r}'$. Therefore, if we want it to be well-defined for $\mathbf{r}=\mathbf{r}'$, one can introduce a Gaussian cut-off function like so
$$\tilde{{\mathsf{G}}}_{2\mathrm{D}}(\mathbf{r}-\mathbf{r}',k_0)=\displaystyle\lim_{\eta\to0}\int\frac{\mathrm{d}^2\mathbf{k}}{(2\pi)^2}\frac{\mathrm{e}^{\mathrm{i}\mathbf{k}\cdot(\mathbf{r}-\mathbf{r}')}}{k_0^2+\mathrm{i}\eta-k^2}\mathrm{e}^{-\frac{a^2k^2}{2}}$$
where $a$ is some cut-off parameter.
Question: How do you evaluate this integral?
| The poles are still in the same place so what is stopping you using residue theorem?
| {
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Grassmann-even action I am currently studying supersymmetric quantum mechanics with the help of the book Mirror Symmetry by Kentaro Hori (and others). On page 155 where they introduce Grassmann variables they say that the action is Grassmann-even without an explanation. But i do not quite understand why this is the case and a Grassmann-odd action is not allowed?
| The Grassmann-odd path integrals are special cases of Grassmann-even path integrals. This follows since
$$
\int [d\psi]e^{-S}=-\int [d\psi]S=-\int[d\psi]d\chi (\chi S) = \int [d\psi]d\chi e^{-\chi S}
$$
for $S$ being Grassmann-odd and $\chi$ a fermion.
Let us work with an instructive example. If you have a Grassmann-odd action of the type:
$$
S=A^{ijk}\psi_i\psi_j\psi_k
$$
The number of fermionic integrations mus be equal or lower than $3$ such that the path integral does not vanishes. Note that
$$
\exp\left(A^{ijk}\psi_i\psi_j\psi_k\right)=1+A^{ijk}\psi_i\psi_j\psi_k
$$
since
$$
(A_{ijk})(A_{lmn})=+(A_{lmn})(A_{ijk}), \qquad (\psi_i\psi_j\psi_k)(\psi_l\psi_m\psi_n) = -(\psi_l\psi_m\psi_n)(\psi_i\psi_j\psi_k)
$$
There will be no products of $A_{ijk}$ like in the general Grassmann-even case since the expansion stops here. This will always happen for the Grassman-odd case. The result of the path integral will be proportional to $A_{ijk}$ with some indices contracted. In particular, no Pfaffians or determinants.
This type of action will not survive in the continuum limit $N\rightarrow\infty$ since the number of fermionic integrals increase in with $N$.
This Grassmann-odd path integral can be easily promoted to be a Grassmann-even by introducing an extra variable $\chi$:
$$
S=\chi A^{ijk}\psi_i\psi_k\psi_k
$$
and integrate over this variable.
| {
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Why aren't satellites disintegrated even though they orbit earth within earth's Roche Limits? I was wondering about the Roche limit and its effects on satellites.
Why aren't artificial satellites ripped apart by gravitational tidal forces of the earth?
I think it's due to the satellites being stronger than rocks?
Is this true?
Also, is the Roche limit just a line (very narrow band) around the planet or is it a range (broad cross sectional area) of distance around the planet?
| To add to other answers, also consider that artificial satellites are much smaller than natural satellites. This means that the difference between the gravitational force at the point the closest to the planet and at the point the furthest from the planet is much smaller in artificial satellites.
| {
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Convergence Property of Path-Integral Let the action be $$S= \int \bigg\{ \frac{1}{2} \big(\frac{dX}{dt}\big)^2 - V(X) \bigg\} d\tau$$
and the corresponding Path-Integral
$$Z= \int DX(t) e^{iS}.$$
Since the convergence is not clear we Euclideanize the time coordinate $t$ by the Wick rotation
$$ t \rightarrow -i \tau$$
and get the Path-Integral
$$Z_E=\int DX(\tau) e^{-S_E},$$
with $$S_E= \int \bigg\{ \frac{1}{2} \big(\frac{dX}{d\tau}\big)^2 + V(X) \bigg\} d\tau.$$
And now my question - the Euclideanized path-Integral allegedly has a better convergence property, but i do not quite see why this is the case?
| You just have to look at the exponential. So assuming your Lagrangian is bounded from below when written in its euclidean version, you can see that the Wick rotation is taking an imaginary exponential which is oscillatory and turning it into a decaying exponential which can be approximated more easily, for example summing over its extremal points.
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Physical intuition behind torque converter A torque converter (also here) is a device used in some cars. It uses several "fans" coupled through a liquid (transmission fluid) in order to perform the function of a clutch, but more importantly it acts as a liquid gear in the sense that it multiplies the torque going from the engine to the wheels.
Is there an intuitive way to explain what is happening in the liquid? In particular, is it possible to explain the torque multiplication effect without resorting to numerical analysis?
| A torque converter contains a propeller that is spun by the engine and another close by which is connected to the rest of the transmission. these two propellers face each other in a chamber filled with oil, so that when the engine-driven prop spins, it pumps oil through the second prop and causes it to spin as well.
Between these two props is a disc with blades in it which catches the oil after it has passed through the second prop, and diverts its flow back around to the engine-driven prop which then picks up that return oil and pumps it into the second prop again.
This disc is cleverly designed so the return oil flow strikes the engine-driven prop blades at such an angle that the energy contained in the return flow adds to the energy being pumped into the flow by the engine-driven prop, in an amount proportional to the difference in rotating speed (called the slippage) between the engine-driven propeller and the second prop.
Recovering the energy in the return flow multiplies the torque force exerted on the second prop by as much as ~3X in the high-slip limit where the second propeller is standing still and the engine is running at speed (which is called the stall torque condition)- as they would be when you put your foot on the gas from a standstill.
In the limit where both props are running at roughly the same speed (the no-slip condition), the torque multiplication factor falls to ~1X.
The action of that extra disc of blades is the key element in getting torque multiplication during the slip condition and is the hardest thing to grasp in the operation of a torque converter!
| {
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Does the ideal gas law apply to a moving vehicle? If for example you have a car travelling along a road. There is obviously a high pressure region in front of the car, where the air is forced around the vehicle. Likewise, I understand that cars will leave a region of low pressure air directly behind them.
Air is described as being incompressible below mach 0.3. Therefore, this would suggest that the density of the air would be the same at the front of the car as it is behind it.
However, this would contradict the ideal gas law which suggests that for there to be a change in pressure in the air, there must also be a change in density.
| "Incompressible" means different things in different contexts.
*
*In thermodynamics, an "incompressible substance" is one whose properties are completely insensitive to pressure. This is a good model for solids and liquids but not for gases.
*In fluid dynamics, "incompressible flow" means that the velocities present are not high enough to cause significant density changes.
The statement "air is incompressible below Mach 0.3" refers to incompressibility in the fluid dynamical sense; it means that, in air, velocities below Mach 0.3 are not capable of causing significant density changes. The statement does not imply that air is an incompressible substance; unless the temperature is very very low or the pressure is very very high, air is best modelled as an ideal gas and not an incompressible substance.
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Why does the potential difference across a type of parallel circuit not act like a potential divider?
Image credit (Q3)
In this attached circuit, when $R_1=0\Omega$, I am failing to understand how the two cells affect the potential difference across the central resistor R3. I understand that potential difference is constant across different strands in parallel, and that so 12 volts should be distributed between R2 and R3 as a potential divider, and so 9V should be across R3 from V2. Likewise, from V1, 10V will cross R3 from V1, and so the total potential difference across R3 should be 19V.
However, according to the answer sheet, the potential difference across R3 is 10V. Is my misunderstanding here conceptual, or something more basic, and why does potential difference act in this way?
| If you are trying to view the system as a linear superposition of the two voltage sources, you must "kill" each source separately. To kill a defined voltage source (like the batteries), you should replace them with a short circuit.
Killing the 12 V source leaves $R_3$ and $R_2$ in parallel with the $10~V$ battery: $10~V$ contribution to each.
Killing the $10~V$ source leaves $R_3$ in parallel with a short circuit, attached to the $R_2$ and $12~V$ source, so there is a $0~V$ contribution to $R_3$.
$10+0=10~V$
You can't throw away the short circuit on the left.
If you are trying to view $R_3$ as part of a voltage divider, you can't ignore that it's in parallel with something else. You must evaluate that effective resistance.
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Why is there no current between two capacitors connected in series? I want to ask a question about the figure below.
Why the current does not flow from $-Q$ term to $+Q$ term or from a to b since there is a voltage?
| I assume that the question is about the DC case. There is no current between a and b, as there is no voltage difference. The voltage difference between the plates opposite to a and b, say a' and b', is canceled out by the charges on a and b. The plates a and b are connected so form, a single conductor. The charges on a and b are an example of how a conductor cancels out any voltage across it by building up surface charge.
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Is velocity of a fluid the gradient of something physically significant? For incompressible flow,
$$\nabla\cdot \mathbf v=0.$$
That means $\mathbf v$ got to be the gradient of some scalar field.
How can I find the scalar field? Is it physically important?
| Suppose that the flow is potential, then we have $\vec {u}=\nabla \phi$ and $\nabla .\vec {u}=\nabla ^2\phi=0$. Then the equations of motion are reduced to the Bernoulli integral.The Laplace equation $\nabla ^2\phi=0$ with boundary conditions is widely used in aerodynamics to calculate the distribution of velocity and pressure in flows with a small Mach number. Figure 1 shows the velocity distribution in a potential incompressible flow (left) and in a viscous compressible flow with Mach number $M= 0.25$ (right) for the NACA 2415 airfoil with the angle of attack $\pi /16$.
| {
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What affects the propagation of secondary cosmic rays? Primary cosmic rays produce, upon entering the Earth's atmosphere, a whole load of secondary particles. These primary particles are necessarily stable particles such as protons, electrons, and neutrinos.
Wikipedia says: "When cosmic rays enter the Earth's atmosphere they collide with atoms and molecules, mainly oxygen and nitrogen. The interaction produces a cascade of lighter particles, a so-called air shower secondary radiation that rains down, including x-rays, muons, protons, alpha particles, pions, electrons, and neutrons.[64] All of the produced particles stay within about one degree of the primary particle's path."
Famously, muons produced this way can reach the Earth's surface and be detected.
Highly energetic photons above the threshold for pair production would spontaneously produce electron-positron pairs, I would guess.
My question, however is, what happens with neutrons in particular? They have a much longer half-life than muons, so they would have no problem with reaching the Earth's surface before decaying, but on the other hand they are much more massive. They are not charged, so would not be deflected by electromagnetic fields. Do they "bounce" multiple times between air molecules or other particles and never reach a target close to the primary's path?
|
My question, however is, what happens with neutrons in particular?
Found this paper which estimates the effects of cosmic air showers on electronics.
Cosmic rays at sea level consist mostly of neutrons, protons, pions, muons, electrons, and photons. The particles which cause significant soft fails in electronics are those particles with the strong interaction: neutrons, protons, and pions. At sea level, about 95% of these particles are neutrons.
So, due to the small density of the atmosphere and the fact that neutrons have to hit the nuclei in order to interact with the strong interaction, the neutrons produced in the cosmic showers still end up at sea level of the earth .
BTW photons have to interact with the electric fields of the molecules in the atmosphere and the interactions will generate pairs of particle once of high enough energy, and also scatterings and degrading of energy.
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Why do my books introduce the equation $\nabla \cdot \mathbf{E}=\frac{\rho}{\epsilon_0}$ without showing partial derivatives of $\mathbf{E}$ exist? In electromagnetism (electrostatics), we often come across the equation $\nabla \cdot \mathbf{E}=\frac{\rho}{\epsilon_0}$.
In order for this equation to be meaningful, $\mathbf{E}$ must be a differentiable vector field. Or at least we need to show that partial derivatives of $\mathbf{E}$ exist if we rotate the Cartesian coordinate system.
In my textbooks none of them has been shown. Yet they introduce the equation $\nabla \cdot \mathbf{E}=\frac{\rho}{\epsilon_0}$. How is it justifiable?
| The partial derivatives of the electric field do not always exist. They won't exist, for example, at a point charge, line charge, or charged plane -- situations where there is a singularity in the charge density. In these situations, Gauss's law still holds in integral form, if the Gaussian surface doesn't intersect any singularities.
None of this causes any foundational issues in physics, because we can only verify our physical theories down to some finite scale.
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Apart from the sine wave, are there any other waveshapes that could be thought of as commonly appearing "in nature"? I'm familiar with the sine wave being something that can be used to model many types of oscillation in nature (and the way that multiple sine waves can be seen as sum to produce complex repeating waveshapes, a la Fourier's theorem).
However, I'm struggling to think of any other waveshapes that can be associated with phenomena in nature. Are there any, or does the sinusoid stand alone as the basic 'shape' of most naturally-occurring cyclic phenomena?
(To give another perspective on my question - when it comes to static values, there are various well-known mathematical constants such as π, e, The imaginary unit i, the golden ratio φ - but are there any well-known mathematical or physical cycle shapes, apart from the sinusoid?)
|
However, I'm struggling to think of any other waveforms that can be associated with phenomena in nature.
The motion of an ordinary pendulum of length $L$ is cyclic, but non-sinusoidal (it is only approximately sinusoidal for small angles). The exact non-sinusoidal motion is governed by the non-linear equation:
$$
\frac{d^2\theta}{dt^2}=-\frac{g}{L}\sin(\theta)
$$
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Why haven't we yet tried accelerating a space station with people inside to a near light speed? Is that something we could do if we use ion or nuclear thrusters?
Wouldn't people in the station reach 0.99993 speed of light in just 5 years accelerating at 1g and effectively travel into the future by 83.7 years?
That would be a great experiment and a very effective way to show relativity theory in action. I mean, the people inside the station would have effectively traveled into the future, how cool is that? Why haven't it been done yet?
| It is not feasible because it would cost an enormous amount of energy
to accelerate the spacecraft.
To prove this let's calculate with some concrete numbers.
Very optimistically estimated, your spacecraft may have a mass of $m=1000\text{ kg}$ (enough for a few people and a small space capsule around them, but neglecting the mass of the fuel needed).
And you said you want a speed of $v=0.99993\cdot c$.
Now you can calculate the relativistic kinetic energy of it:
$$\begin{align}
E_{\text k} &= \frac{mc^2}{\sqrt{1-v^2/c^2}} - mc^2 \\
&= \left(\frac{1}{\sqrt{1-v^2/c^2}}-1\right) mc^2 \\
&= \left(\frac{1}{\sqrt{1-0.99993^2}}-1\right)\cdot 1000 \text{ kg}\cdot (3\cdot 10^8\text{ m/s})^2 \\
&= (84.5-1)\cdot 1000 \text{ kg}\cdot (3\cdot 10^8\text{ m/s})^2 \\
&= 7.5 \cdot 10^{21}\text{ J}
\end{align}$$
Now this is an enormous amount of energy.
It is comparable to the yearly total world energy supply.
(According to Wikipedia:World energy consumption
the total primary energy supply for the year 2013 was $5.67 \cdot 10^{20}\text{ J}$.)
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2 windows - will I see the reflections?
I have a question regarding photons nature. Let's say I have a single source of light - regular bulb and the observer - in the same room.
The observer looks through a glass window (normal glass window-nothing special about it) and sees his reflection, but some of the light is passing through the window.
Now I put a second window - as presented on the attached photo. Will the second window reflect some of the light back to the observer, or it will pass 100% of the light forward?
This may be pretty basic but this question sparked a discussion and there where no definitive answer.
| Classical electromagnetism predicts that the observer will see four first order reflections, one from each glass-air interface. The reflectivity at each interface is 4%. Multiple reflections will also occur, for which light is reflected back and forth. They are at least third order in intensity, that is, are 4% of 4% of 4% or 64 per million. There are 14 of these, so the total intensity is 0.1 per thousand or 160 times less than the first order total. The list goes on with fifth order etc.
Quantum mechanically these numbers mean the probability that an incoming photon is reflected. The same images that are predicted classically will be built up from single photons much in the way the paintings of French painter Seurat are built up from dots. The more photons that are observed the less noisy the images will be u til on the classical limit all shot noise is averaged out.
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Does pair production happen even when the photon is around a neutron? In order for a photon to decay into a pair of $e^+ e^-$, it must have at least $E_{\gamma}=1.022$ MeV and must be near a nucleus in order to satisfy the conservation of energy-momentum.
But would this happen even if the photon is near a neutron and not necesarily a nucleus? Does the fact that the nucleus is charged have anything to do with this decay? Who acts upon the photon to induce the interaction?
| Yes, pair production can occur even near a lone neutron. The presence of a mass for the photon to interact with is required for conservation of momentum (further explanation can be found here). To my knowledge, the charge of the nucleus is not significant to the process of pair production, though the probability of pair production increases approximately with atomic number squared. Experimentally, this would be hard to demonstrate as neutrons are hard to control and have a relatively short half life (~10.3 minutes).
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How strong are Wi-Fi signals? My family members dislike the idea of having many devices communicating wirelessly in our house, arguing that the signals have negative effects on our physical health.
I would like to tell them the EM signals are in fact weaker than the light from our lights but I could not really confirm this. Could someone tell me how strong the signals from the wireless devices are compared to those from lights, and perhaps those from the Sun as well? How about the signals for radio devices and handphones?
What is the scientific basis for claiming the radiation has or doesn't have effects on the human body?
| You only get out what you put in
This aims to be a layperson's answer, rather than specific complex formulae which are mind boggling to comprehend to a layperson, but the easiest way to prove the lightbulbs are more harmful than the wifi, is to consider the input power (as output can never exceed input; due to inefficiencies, it will always be less).
Consider how, after several hours, you can burn your hand on a lightbulb due to the heat it produces, but you cannot really burn your hand on a router.
Most routers only take 5v-12v at perhaps 5 amps; lightbulbs can take, anywhere between 12v to 240v at varying amperages. If you consider how a microwave requires 600 watts of power (110-240 volts depending on system) to take several minutes to boil water, a 5v to 12v device with maybe 5 amps tops (usually 0.8 amps) isn't going to nearly have as much power output as a lightbulb.
You could argue it's the continuous duration of exposure over a long period of time that potentially causes the harm with wifi, but it's worth noting wifi signals will already be in abundance (neighbour's wifi, public wifi, cell phone towers, other people's phones etc). Just short of insulating the entire house, refusing to use microwaves, cellphones, routers etc, they're going to be exposed anyway.
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Quantum field theory with only 3-point vertexes Given an arbitrary quantum field theory, can I always write it in terms of another (different) quantum field theory containing only operators with 3 fields? (i.e. vertexes with 3 legs)
I guess that should be possible introducing arbitrary additional fields that once integrated away bring to the original operators. Is that so?
Can it be done generally without having additional degrees of freedom in the new (3 vertexes) theory? I am thinking of something with maybe ghost fields.
Or maybe there is some other way?
I heard this has been done with General Relativity (in Hidden Simplicity of the Gravity Action), but I really did not check in detail the content.
Maybe the simplest case to think of is that of a theory containing only one interaction vertex (let's say 4 legs, $\phi^4$). We replace the theory with a theory with only the three leg vertex ($\phi^3$) and if we quotient the space of asymptotic states removing those with any odd number of external legs (you cannot have those in $\phi^4$), then we should obtain the starting theory. This said I don't see how to upgrade this trick in cases where the starting lagrangian contains more interactions, hence the question.
| Yes you can! I'm sure you can do this all sorts of ways, but here is a nice systematic one. Suppose you have a term in the lagrangian $\prod_{i=1}^{k} \phi_i\subseteq L$. We can start by introducing two new fields, $\lambda_1$ and $\sigma_1$. Add the term $i\lambda_1(\sigma_1-\phi_1*\phi_2)$. If we first integrate out the $\lambda$ variable, we get a delta function. Integrating out the $\sigma$ gives us unity. Notice that the terms added to the lagrangian are at most cubic in the vertices. We can then introduce $lambda_2, \sigma_2$, and add in the term $i\lambda_2(\sigma_2-\sigma_1*\phi_3)$. If we keep going, we get that the path integral contains a bunch of delta functions forcing $\sigma_{k-1}$ to be our our term in the lagrangian. So we can just remove the offending term by replacing it with a $\sigma_{k-1}$, and our higher-order term was replaced with a family of cubics.
You can show that this all works out for vector fields, and even fermionic fields. If you have multiple terms, just repeat the procedure multiple times. In this way, we can force every lagrangian to be cubic.
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Why doesn't the Earth's acceleration towards the Moon accumulate to create noticeable motion of the earth, towards the moon I get that Earth's mass is very large, so its acceleration is very tiny. But wouldn't the acceleration accumulate over a period of time and become noticeable?
| The moon is actually moving away from the Earth; 4 billion years ago it was much closer.
The moon raises tides which have the effect of slowing down the rotation of the Earth, so that the daylength is now much longer than it used to be, but some of the energy which has been lost by the Earth was captured by the moon, and it has boosted it into a higher orbit. It is still receding from us at the rate of a few centimetres per year.
| {
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Can a wire having a $610$-$670$ THz (frequency of blue light) AC frequency supply, generate blue light? We know that when we give alternating current across a wire then it will generate an electromagnetic wave which propagates outward.
But if we have a supply which can generate 610 to 670 terahertz of alternating current supply then does the wire generate blue light?
| An alternating voltage at that frequency is light. There's no 'generate' about it - the power supply is just a light source.
And if you have a wire, that is, a conductor made of metal, then the light won't propagate inside it at depths longer than the skin depth for that material at that particular frequency, which is generally tiny.
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Does rotation happen throughout the whole x axis of an object simultaneously? For example, if I draw a line on the side of a pencil top to bottom, then snap one end of it as in launching it due to the pressure of my fingers. Anyways, if I record the pencil launch in slow motion (perhaps it’s my phone that has to do with it) but it will focus on where the line was, and it appears that only some of the actual line is there, or out of focus. So that leads to the question, does rotation happen simultaneously down the pencil axis? Maybe I’m completely clueless and I’m missing something but figured I’d ask.
| No, the rotation would not occur instantaneously. When you flick one end of the pencil, the molecules (at the end) you imparted an impulse to will have gained momentum and would pull the molecules just beside them due to internal electrostatic forces holding the molecules together, (much like the propogation of a mechanical wave) and bring them to motion and this cascading effect will keep travelling down the pencil.
TL;DR, it will not be instantaneous and will take a small finite time that is usually negligible.
| {
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How much energy is needed to make fire? I'm so curious about fire. So I searched a lot in the internet. And now, I knew that fire is some kind of chain reaction and combustion energy make the other molecule hot and the other molecule makes other chemical reaction and so on...
Then does the first given energy make this reaction start and maintain chemical reaction(fire)? if so, if a long time passes, fire goes out? and how much energy is needed to make this chain reaction start?
to sum it up,
*
*when cigarette lighter make candlelight start how much energy is needed?
*if a long time passes, this candlelight goes out?
(Assuming that oxygen and material(will be burned) are supplied)
*how much energy go out with light or heat?
(I'm not good at english, so please understand me)
| The first given energy is called the activation energy. The chemical reactions of burning liberate energy in the form of heat and light (and probably some sound), and a small part of that energy activates further combustion. The amount of activation energy required to start the fire depends on what you're burning.
In your example, once the candle is alight, the bit of the candle that is currently burning provides the activation energy for the next part of the candle.
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