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Frames of reference: Inertial and accelerated - and jerked, snapped, crackled and popped? There are inertial frames of reference and the accelerated frames of reference, but are there any frames of references w.r.t. higher order derivatives of velocity? [1] [2]
For example, jerked frames of reference, snapped frames of reference, crackled frames of reference and popped frames of reference and so on?
| Yes. For simplicity, consider an observer $O'$ moving in one dimension. Suppose that as measured by some other inertial observer $O$, the obsever $O'$ has the following position as a function of time
\begin{align}
x(t) = kt^n
\end{align}
where $n$ is an integer. When $n=2$ or higher, the observer has nonzero acceleration, so the a frame of reference in which he is at rest is accelerating. For $n=3$ or higher, the corresponding frame of reference will have nonzero jerk. For $n=4$ or higher...well you get the picture.
The only problem is that for $n=4$ or greater, rice crispies start appearing all over the place, and it gets really hard to make measurements.
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when we rub objects together, what determines which material will pick up electrons? For example We know glass when rubbed by silk will become positively charged while the silk will be charged negative.
What exactly makes glass appropriate for losing electrons in that experiment? (
| The process of transferring charge between two objects by bringing them into contact and/or rubbing them together is known as the triboelectric effect, and is historically, the first recorded observation of an electric phenomenon.
Wiki Page on Triboelectric effect
In general, there are too many competing factors such that there is little predictive power.
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Is such an orbit possible that allows a satellite on Earth and another on Mars to always be in direct line-of-sight? If not always, how much "most of the time" could it get?
Or would a multi-satellites setup be more suited?
| Yes, but the satellite(s) would be orbiting the sun instead of the earth. You place a satellite at either L4 or L5 Langrangian points on the Earth-Sun two-body system--these are points 60 degrees to either side of earth on her orbital plane around the sun where you can place a satellite. Throughout earth's entire orbit the satellite will maintain that 60 degree separation, even under the influence of earth's gravity. It does this by moving slightly faster than earth (tangential velocity) at a slightly larger orbit than earth, allowing the angular velocity to be equal to earth.
With the satellite in L4 or L5, mars will always be in line of sight to either earth or that satellite. You can add more satellites such as geostationary ones on both planets to allow to specific points on those planets to have line of sight 100% of the time.
| {
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Is inflation deterministic? In some theories inflation is supposed to be able to turn quantum fluctuations into macroscopic inhomogeneities.
I don't understand how an isolated system such as the universe can undergo such a random transformation : if at the beginning the universe is in a state $A$, quantum mechanics says that $A$ will evolve to $B=UA$ with $U$ being a unitary operator, and general relativity is also a deterministic theory.
So does inflation suppose that the universe is not isolated or does it use some modified theories which include randomness?
| In quantum mechanics, the state indeed evolves. But when you do an observation, it needs to choose one of the observable states - that's why we talk about superposition of states as long as you don't observe a quantum variable.
For inflation, as long as we were at high energies and short distances, the states could evolves, but once the inflation started, stretch distances and modes crossed the horizon, they had to pick up a state and then were fixed. What happen next is given by deterministic theories.
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What is meant when saying "the partial derivative operator $\partial/\partial x^\mu$ is a covariant vector"? Reading Weinberg's Gravitation and Cosmology, I came across the sentence (p.115, above equation (4.11.8))
The partial derivative operator $\partial/\partial x^\mu$ is a covariant vector, or in other words a 1-form, [...]
Now, a section of $T^*M$ is called a covector field or a 1-form. Elements of $T^*_pM$ are called tangent covectors. In books about differential geometry I read that tangent COvectors are called covariant vectors. How do I have to understand the quote from Weinberg's book?
| The way I understand it is that he is not referring to the tangent vectors $\frac{\partial}{\partial x^\mu}$, but to a differential operator. Notice that he doesn't use plural, he is talking about a single object. Namely the 1-form valued operator, which maps each function to its gradient. Then you can form the wedge product with this operator and a $p$-form to obtain a $(p+1)$-form, which is the exterior derivative of the original $p$-form. This is the way he defines exterior derivatives. The wedge product is introduced in the previous paragraphs. This seems like a strange choice of notations and definitions, but it may have been fairly standard at the time.
About Poincare's Lemma (since Qmecanic mentioned it): I have seen a number of places where $d^2=0$, for the exterior derivative, is called the Poincare's lemma, for example the reference, which Weinberg gives, Flanders, H. "Differential forms with Applications to the Physical Sciences".
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How does humidity affect the path of a bullet? Background
Last night, I was reading the FM 23-10 (The U.S. army official field manual for sniper training), and I've noticed that they're potentially teaching snipers incorrect information.
Generally speaking, when we say "impact goes up" it means that the bullet was either somehow made faster or its path was easier, therefore the curve in its ballistic trajectory is smoother. Thus, it will hit higher. When we say impact goes down, we mean the opposite.
For example, atmospheric heat will, loosely speaking, make the air "thinner" and therefore the impact will be higher. Cold weather will do the opposite. This part is correct.
What about humidity?
The FM 23-10 says:
The sniper can encounter problems if drastic humidity changes occur in
his area of operation. Remember, if humidity goes up, impact goes
down; if humidity goes down, impact goes up.
They're basically saying that when humidity goes up, then the bullet's travel will be more difficult-> steeper trajectory curve -> lower point of impact.
However, as far as I know, dry air is denser than humid air because air has higher molecular mass than water vapour. In humid air water vapour replaces other gases, thus bringing the whole density down. So, the point of impact should be higher with higher humidity.
So my question is:
All other factors being equal, does humid air pose less resistance to the bullet making the point of impact higher than in dry air?
| According to several sources, such as http://longrangebpcr.com/accuracy.htm , you are right: higher humidity - higher point of impact. But those sources also point out that humidity effect is small.
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How does a pressure suit work? I recently learnt that a suit called pressure suit is worn by fighter plane pilots to prevent red-outs and black-outs. And it seems to be work by -
"..applying pressure to selective portions of the body."
How do these suits work; i.e. by what means, selective portions of the body are pressurised?
Do astronauts wear these while takeoffs, and also F1 drivers?
| they provide mechanical constriction to the limbs and torso to prevent swelling in the event of decompression. hands, feet and face are not assisted.
astronauts wear full pressure suits for ascent and re-entry.
the main concern for F1 drivers is fire resistance not low air pressure.
https://airandspace.si.edu/collections/artifact.cfm?id=A19730755000
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What are the implications of the Holographic principle? What are the implications for the Holographic principle? I understand the basics of the principle, the relationship with black holes and string theory but what this is going to tell us? Does it help to explain quantum gravity? Is the model compatible with any multiverse theory? The Holographic principle seems to be proposed as a new theory but where does it fit with the fine-tuning problem?
| It should be noticed the difference between aplying the Holographic Principle in to a 3D or a 4D world.
First it avoids the idea of infinite dimensions or "reality is unknow (infinite) until you look at it" by allowing fast comunications.
Then if you start with a some sort of +4D world it returns to a 3D world except "inside the particles themselfs" wich are inded a loophole.
To my undertanding is a compromise between new findings, old and trusted theories, String Theory (wich requieres a single string) and the common sense of a 3D world.
Multiverse theory or island universes has nothing to do with it, I supose you are refering to paralel universes. Since the theory doesn't explain why light is scattered it's fully compatible with it, but I supect that it's not +4D space but some kind of interlaced 3D space (or N-3D space) the one which follows the idea.
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Dehn twists and topological order I am trying to understand notion of a "Dehn twist" and how it relates to topological order.
In particular refering to http://arxiv.org/abs/1208.4834 it is stated that
Xiao Gang Wen's paper on "Topological Order in Rigid States" (http://dao.mit.edu/~wen/pub/topo.pdf) is supposed to provide an introduction to "non-abelian adiabatic Berry phases associated with Dehn twists for a U(1) Chern Simons Theory". Skimming through the respective paper however I could not find the notion of a "Dehn twist" appearing at all?
Maybe it appears under a different name or it is not given a name at all?
I would be very happy for any support.
Best.
| In my paper, the Dehn twist is referred as modular transformation. See section V of http://arxiv.org/abs/1212.5121 which is available in arXiv. The unitary transformation generated by the Dehn twist is called the non-Abelian geometric phase.
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Determing time to complete known distance with constant acceleration Answer: $a=v^2/d$ is the formula I needed.
This is a problem in a programming assignment, I haven't taken physics.
I have a starting speed (0 m/s), an final speed (208.33m/s) and the distance it took to reach that speed (200km). I need to get the time it will take to travel any arbitrary distance lower than 200km.
From what I remember in highschool, I used basic calculus to get $a/2(t^2)=distance$ or for what I have $t=\sqrt(400000(m/s)/a)$. what I don't know is the acceleration, usually to calculate acceleration I'd need time. I'm not sure where to go from here.
| Converting all units to SI:
$$u=0$$ $$v = 208.33 \text{ m/s}$$ $$d =200000\text{ m}$$So: $$a=\frac{v^2-u^2}{2d}=0.1085\text{ m/s}^2$$
Then, for any distance,$d$ in meters:$$t=\sqrt{\frac{2d}{a}}$$This vehicle maintains a feeble, but constant, $0.01\text{ g}$ for a $200\text{ km}$ trip lasting a little more than 30 minutes.
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How to establish relation between flow rate and height of the water column of the tank? Suppose a water tank has 1" diameter drain at the bottom and is filled with water up to one meter height above the drain. What time it will take the tank to drain out completely. Now say, the tank is filled up to two meter height above the drain then what time it will take the tank to drain out? Will it be double or less than it? Can we establish a relation between flow rate for the given height of water column?
| Use Bernoulli's equation to derive Torricelli's Law (check any website for this) for the velocity out of the hole; $ v = \sqrt{2 g h(t)} $, where g is gravity and $ h(t) $ is the height of the fluid in the tank at any time.
Write a balance on the mass of the fluid in the tank as:
$$ \text{in - out + gen = accumulation} $$
$$ \rho Q_{in} - \rho Q_{out} = \frac{d(\rho V)}{dt} $$
where the generation term is zero, $\rho$ is the fluid density (constant here) and $Q_{in}$ and $Q_{out}$ is the flow rate in and out of the tank, respectively. $Q_{in}$ is zero so we get:
$$ \frac{dV}{dt} = -Q_{out} $$
The flow out is $v A = \sqrt{2 g h(t)} A$, where $A$ is the area of the hole which is calculated by knowing the diameter of the circular hole; given in the problem statement as 1 inch.
The volume of the tank, $V = a_t h(t)$, is the height, $h(t)$ times the area, $a_{t}$.
Putting it all together, we get the separable first order differential equation for the height of the fluid in the tank versus time:
$$ \frac{dh}{dt} = -\frac{A}{a_t}\sqrt{2g}\sqrt{h}$$
Prepare it for Integration
$$ \frac{dh}{\sqrt{h}} = -\frac{A}{a_t}\sqrt{2g}{dt}$$
Integrate the equation. The upper bound for $dh$ is $h(t)$. The lower bound is $h(0)=H$ . For $dt$ we integrate from $t$ to $0$:
$$ 2 (\sqrt{h(t)} - \sqrt{H}) =-\frac{A}{ a_t}\sqrt{2g} t$$
Solve for $h(t)$
$$ h(t)=[\sqrt{H} -\frac{A}{2 a_t}\sqrt{2g} t]^2 $$
To find the time when the tank empties, set $h$ equal to zero and solve for $t$:
$$ T= \sqrt{\frac{H}{2g}} \frac{2 a_t}{A}$$
Plug that back in the previous equation to clarify time dependence on t:
$$ h(t)=H[1 - t/T]^2 $$
The times to empty the same tank for two different starting heights, $H_1$ and $H_2$ is:
$$ \frac{T_1}{T_2} = \sqrt{\frac{H_1}{H_2}} $$
So, finally, if $H_2 = 2 H_1$ as in the problem statement, then the time to empty the tank is not double, but $\sqrt{2}$ times longer.
Make sense?
Paul Safier
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Quantum harmonic Oscillator analytic method I'm using a book from Griffiths, I got really stuck about how he arrived at the approximate solution, is it just by trying( trial solution method?), I really appreciate any help on this.
$$-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2} + \frac{1}{2}m\omega^2 x^2\psi~=~E\psi.$$
Change variables for convenience:
$$\begin{align}
\xi &\equiv \sqrt{\frac{m\omega}{\hbar}}x \\
K &\equiv \frac{2E}{\hbar\omega} \\
\frac{\mathrm{d}^2\psi}{\mathrm{d}\xi^2} &= (\xi^2 - K)\psi.
\end{align}$$
For very large $\xi$:
$$\frac{\mathrm{d}^2\psi}{\mathrm{d}\xi^2} \approx \xi^2\psi.$$
Approximate solution:
$$\psi(\xi) \approx A e^{-\xi^2/2} + B e^{+\xi^2/2}.$$
| I assume you have no qualms with the "large $\xi$" approximation - it's fairly obvious that $\xi^2-k^2\approx \xi^2$ for large enough $\xi$. After that you're left with the differential equation
$$\frac{d^2\psi}{d\xi^2}\approx-\xi^2\psi.\tag1$$
One way to solve this equation is by the method of divine inspiration: you somehow come up with two linearly independent functions you can write down which solve the equation, after which you know what the general solution is.
However, the two functions that Griffiths poses are not exact solutions of that equation, as you would know if you had done your proper diligences. Indeed,
$$
\frac{d^2}{d\xi^2}\left[e^{\pm \xi^2/2}\right]
=\frac{d}{d\xi}\left[\pm\xi e^{\pm \xi^2/2}\right]
=(\xi^2\pm1)e^{\pm\xi^2/2}.
$$
This means that we're looking for a solution that's approximately valid for the (already approximate) equation (1). As for how one might get such solutions, there's obviously a myriad different possible paths.
One really nice way of deriving the solutions is to factor the offending second-order differential operator into two different first-order operators:
$$
\left(\frac{d}{d\xi}-\xi\right)\left(\frac{d}{d\xi}+\xi\right)
\approx \left(\frac{d}{d\xi}+\xi\right)\left(\frac{d}{d\xi}-\xi\right)
\approx \frac{d^2}{d\xi^2}-\xi^2.
$$
These are of course approximate inequalities, and of course you must work these out to see what terms got dropped and why.
After that, you can simply work out solutions to the two equations $\left(\frac{d}{d\xi}-\xi\right)\psi=0$ and $\left(\frac{d}{d\xi}+\xi\right)\psi=0$, which are in fact the solutions given by Griffiths. These are first-order equations and therefore simply solvable by integrating. The (approximate) equalities above guarantee that a function in the kernel of either factor will be (approximately) in the kernel of the operator you do care about.
Of course, these factors are far from trivial inventions, and they are at the heart of the operator approach to solving the simple harmonic oscillator.
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Do orbitals overlap? Yes, as the title states: Do orbitals overlap ?
I mean, if I take a look at this figure...
I see the distribution in different orbitals. So if for example I take the S orbitals, they are all just a sphere. So wont the 2S orbital overlap with the 1S overlap, making the electrons in each orbital "meet" at some point?
Or have I misunderstood something?
| If you mean to ask "do the orbital radial probability distributions overlap?", the answer is yes:
Image Credit
making the electrons in each orbital "meet" at some point
As you can see from the image, the electron orbitals are not position eigentstates. If you're imagining two point-like electrons in different orbitals colliding, you're not thinking "quantum mechanically".
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Why does voltage remains same over Parallel Circuit Why does voltage remains same over parallel circuit. If a resistor is connected in the circuit some of the charge should be transformed into heat and make a lack of charge after the resistor (in my sense). So, what's the reason in it?
| A single resistor wired in parallel with voltage source like a battery may give confusing reasons for a novice, so a few other things need to be taken into account.
Any battery also has an internal resistance (it is like a resistor in series with the voltage source), and this resistance also varies with the state of charge of the battery, which is to say, the internal resistance gets to be a quite large value when the battery is "low". This internal resistance of a battery can be complicated to figure out; don't try connecting an ammeter directly across the terminals of a battery!
So, when the battery is fresh and the value of the external (test) resistor in parallel with it is in a range that will not cause it to overheat (which would abruptly change its value to something much larger than marked), it is easy to understand that a voltmeter connected across the resistor will hopefully be the same as if the resistor were not there, and the voltmeter is connected directly across the battery.
By the same token, adding additional resistances in parallel to the first one should also not materially change the value of the voltage measured in the first case unless the voltage source has been overloaded.
Does this help? Some of the other examples seemed not to take into account the idea that the question being asked seemed to involve a (single) resistor and a (single) voltage source, which should be thoroughly understood before adding more stuff in parallel, to say nothing of the characteristics of real voltage sources, particularly a battery.
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Vanishing of the Ricci tensor in higher spacetime dimensions I understand how, if the Riemann tensor is 0 in all its components, since we construct the Ricci tensor by contracting the Riemann, Ricci tensor would be 0 in all components as well.
I've read that vanishing of the Ricci tensor in 3 spacetime dimensions implies the vanishing of the Riemann curvature tensor, but that in higher dimensions that does not hold.
Can somebody explain why is that so? Is it because we have more independent components of the Riemann tensor in 4 spacetime dimension, than in 3 (20 vs 6)?
Also if the number of independent components of Riemann tensor in $n$ spacetime dimensions is
$$N(n)=\frac{n^2(n^2-1)}{12}$$
and since we know that the Riemann tensor has 256 components, does that limit the spacetime dimensions of it's usage? Or does that mean, that for example in 10 spacetime dimensions, there won't be any independent components of Riemann tensor?
| The Riemann tensor can be decomposed into a trace part and a trace-free part:
$$R_{abcd}=C_{abcd}+\frac{2}{n-2}(g_{a[c}R_{d]b}-g_{b[c}R_{d]a})-\frac{2}{(n-1)(n-2)}Rg_{a[c}g_{d]b},$$
where the traceless object $C_{abcd}$ is the Weyl (conformal) tensor and the remainder is given by the Ricci tensor and its contraction, the Ricci scalar, where $n$ is the number of dimensions and $g_{ab}$ is the metric. In three dimensions, the Weyl tensor vanishes: this implies that when the Ricci tensor is zero in all components, the Riemann tensor is zero as well. However, for $n$ larger than 3, the Ricci tensor can be zero while the Weyl tensor is nonvanishing.
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Why is moment dependent on the distance from the point of rotation to the force? The formula for moment is:
$$M = Fd$$
Where F is the force applied on the object and d is the perpendicular distance from the point of rotation to the line of action of the force.
Why? Intuitively, it makes sense that moment is dependent on force since the force "increases the intensity". But why distance? Why does the distance from the line of action of the force to the point of intensity affect the moment?
I am NOT looking for a derivation of the above formula from the cross product formula, I am looking for intuition. I understand how when I am turning a wrench, if the wrench is shorter its harder to turn it but I don't understand WHY.
Thanks.
| The reason why torque (rotational force) depends on the distance $d$ from the pivot of rotation (i.e. why torque is a moment) is the following:
Torque is defined as change in angular momentum; if mass is constant that means change in angular velocity.
To achieve a change in angular velocity using a tangential force $F$, we need to travel a greater distance when we are farther away from the center. Or in other words, force $F$ only changes linear velocity; achieving a change in angular velocity requires more the farther out we are.
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Accelerated charge inside sphere (again!) Sorry to go on about this scenario again but I think something is going on here.
Imagine a stationary charge $q$, with mass $m$, at the center of a stationary hollow spherical dielectric shell with radius $R$, mass $M$ and total charge $-Q$.
I apply a force $\mathbf{F}$ to charge $q$ so that it accelerates:
$$\mathbf{F} = m \mathbf{a}$$
The accelerating charge $q$ produces a retarded (forwards in time) radiation electric field at the sphere. When integrated over the sphere this field leads to a total force $\mathbf{f}$ on the sphere given by:
$$\mathbf{f} = \frac{2}{3} \frac{qQ}{4\pi\epsilon_0c^2R}\mathbf{a}.$$
So I apply an external force $\mathbf{F}$ to the system (charge + sphere) but a total force $\mathbf{F}+\mathbf{f}$ operates on the system.
Isn't there an inconsistency here?
As the acceleration of charge $q$ is constant there is no radiation reaction force reacting back on it from its electromagnetic field - so that's not the answer.
Instead maybe there is a reaction force back from the charged shell, $-\mathbf{f}$, to the charge $q$ so that the equation of motion for the charge is given by:
$$\mathbf{F} - \mathbf{f} = m\mathbf{a}\ \ \ \ \ \ \ \ \ \ \ (1)$$
This reaction force might be mediated by an advanced electromagnetic interaction going backwards in time from the shell to the charge so that it acts at the moment the charge is accelerated.
Now the total force acting on the system is the same as the force supplied:
$$ \mathbf{F} - \mathbf{f} + \mathbf{f} = \mathbf{F}.$$
If one rearranges Equation (1) one gets:
$$\mathbf{F} = (m + \frac{2}{3} \frac{qQ}{4\pi\epsilon_0c^2R}) \mathbf{a}$$
Thus the effective mass of the charge $q$ has increased.
| There is no obvious inconsistency, whether we use retarded, advanced, or any other field.
If we use only retarded fields, things go as follows. At the time $t=0$, we begin to exert force $\mathbf{F}$ on the charge $q$. It will move with acceleration $\mathbf{F}/m$ for the time interval $R/c$, where $R$ is the radius of the sphere. At the time $t = R/c$, the sphere will be acted upon by the field emanated from the accelerated charge at the time $t=0$.
The fact that the system charge+sphere may experience force greater than the force exerted by us on the charge $q$ is alright. This can happen in EM theory. The total momentum is still conserved, because when sphere gets certain amount of momentum in one direction through its surface charges, the same but oppositely oriented momentum is acquired by the total field at the place where the interaction occurs (the vicinity of these sphere charges).
After the wavefront of the acceleration field of the center charge reaches and moves away from the sphere, the region where the negative momentum is contained will move away with it too.
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Energy source for phase transformation-latent heat of vapourisation As per my knowledge, phase transformation require energy gain or loss. If we are changing a liquid to gas we heat it and that is how molecules of liquid gains energy and lattice spacing increases and bonds becomes weaker and thus gets converted to gas.
But we know that, if we lower the pressure of water (a closed container and evacuate air and so lowering pressure) and it start boiling even at 20degree C. from where it is getting energy for phase transformation?please help me if I am completely wrong.
| In your example of lowering the pressure above water, the water will only boil if you reduce the external pressure to below the vapour pressure of the water i.e. the water is superheated. The energy required to boil off the steam comes from the internal energy of the water, and as a result the water cools down. As the water cools, the vapour pressure will reduce until it matches the external pressure and at that point the water will stop boiling. If you reduce the pressure again the water will resume boiling, and will cool further.
This is basically the same process as evaporative cooling as invented by evolution when it developed sweat glands some millions of years ago (how many million depends on exactly what you mean by sweat gland).
| {
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Why does Newton's third law exist even in non-inertial reference frames? While reviewing Newton's laws of motion I came across the statement which says Newton's laws exist only in inertial reference frames except the third one. Why is it like that?
| The cutest way to see this is to restate Newton's third law as "no interaction can change the total momentum of the universe." Then, note that since an accelerating reference frame is accelerating with respect to whatever "base" inertial reference frame you're using, everything else seems to be accelerating away. Therefore, the net momentum of the universe is changing. Therefore, Newton's Third Law does not hold in this reference frame.
| {
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do lenses with curved focal planes exist I know about spherical, aspherical, cylindrical lenses, but are there lenses that could have a curved focal plane? for example, the retina of the eye is curved, and still we see sharp images of distant objects.
In a recent problem I studied, I wanted to have a lens that would transform a flat object (a book cover, for example) into a curved image. That is, each point of my book cover would be imaged correctly (i.e. the image would be sharp enough) by my lens - only, on a curved screen, like a spherical screen for example.
Would there be a specific form of lens that could do exactly this, the same way that aspheric lenses do "perfect" images of distant objects on flat screens, better than any spherical lens?
I am asking this because, in my application, the screen will need to be strongly curved, and I still want the best possible stigmatism in these conditions, without reducing the aperture too much...
| Surely lenses with curved focal place exist - simplest spherical lens have curved focal plane. Human eye is a bad example here - field curvature of an eye does not match retina curvature. We just used to "see" using central part of retina.
You definitely can design your lens for more-or-less symmetrical field curvature you need, although I never did that. I've seen a design which was optimized for imaging on almost half-cylindrical focal plane. I have no links on how one should design such systems.
Designing such lens for such non-linear surface is probably quite complicated. This might require using exotics like freeform or computational holography elements i.e. quite expensive to design / manufacture.
| {
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Infinite Wells and Delta Functions In considering a delta potential barrier in an infinite well, I can just enforce continuity at the potential barrier-it doesn't have to go to zero. Why then does it need to go to zero at the walls of the infinite well? These two cases seem to be very similar to be, I even feel like the well wall is equivalent to a summation of delta functions... Where is my logic faulty?
| When you consider a potential barrier, you will see that the wave function of a particle with definite energy will decay exponentially at a rate that depends on the difference of the potential and the energy. Precisely, the exponent is proportional to $\sqrt{V - E}$.
In the limit, for $V\to\infty$, it will decay to 0 over a width 0.
A delta function doesn't exist as a function, but it is a limit of functions (in a suitable topology). For this problem it is easiest to consider block functions: functions of width w and height 1/w. The delta distribution is the limit as $w\to 0$. You can not first take the limit of the height going to infinity, followed by the width going to 0, something that you implicitly are doing.
Now for a fixed $w$ you get a value for the wavefunction on the other side of the barrier depending on the value on this side. When the height goes up, the decay will be faster, but it will also fall over a shorter interval, and in fact the decay will be less because the decay exponent contains a square root. In the limit there is no decay in height at all, in particular you do have continuity, but no reason to end at 0.
| {
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Expansion of the Universe: Conversion of gravitational potential energy to kinetic energy? Suppose there is an object floating in space which over time begins to fall toward the source of a gravitational field. As it falls, its motion happens to be such that it gets locked in orbit around the source with a greater velocity than it had before it 'began to fall'. So it's gravitational potential has been converted to kinetic energy. According to relativity, this increased speed should increase the gravitational potential of the object (and therefore the object + the original source of the field), correct?
Does this mean that as a result of this, the expansion of the Universe should slow down slightly (because there is now a slightly greater gravitational potential in the Universe)? And if so, would that imply that the conversion from gravitational potential to kinetic energy is in a sense a conversion between the bulk kinetic energy of the expanding Universe and the local kinetic energy of a test mass?
| There's a blog post that may be of some interest to you here: http://motls.blogspot.com/2013/11/the-expansion-is-accelerating-due-to.html
Basically, the universe is constantly expanding at an accelerated rate due to "negative pressure". This is better understood with the Second Friedmann Equation:
$\frac{\ddot a}{a} =-\frac{4\pi G}{3} (\rho + 3p)$
The same principles can be applied, I guess, to your problem.
Let me know if you need anything else!
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Determining if a semiconductor is n-type, p-type or intrinsic The probability that an energy state in the conduction band is occupied by an electron is 0.001. Would this semiconductor then be n-type, p-type, or intrinsic?
Notation that I use: $E_F$ represents the fermi energy level, $E_C$ represents the energy level of the conduction band, and $E_V$ represents the energy level of the valence band. $E_{Fi}$ represents the fermi level of an instrinsic semiconductor.
We know that the $E_F$ is where the majority of the electrons are. Because the probability that an electron is in the conduction band is 0.001, we know that there are very few electrons in the conduction band. In that case, that means that $E_F$ is very far from $E_C$ and is instead close to $E_V$. Because $E_F$ < $E_{Fi}$, the semiconductor is p-type. The semiconductor is actually n-type. What is the flaw in my understanding?
| Since the Fermi dirstribution is obviously either close to 1 or close to zero except at the fermi level, the answer to you question can be easily answered by looking at the figure below.
Would this semiconductor then be n-type, p-type, or intrinsic?
P-TYPE!
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Intuitive understanding of the entropy equation In thermodynamics, entropy is defined as $ d S = \dfrac{\delta q_{\rm }}{T}$. This definition guarantees that heat will transfer from hot to cold, which is the second law of thermodynamics. But, why do we denote entropy as$\dfrac{\delta q_{\rm }}{T}$ other than $\dfrac{\delta q_{\rm }}{T^2}$,$\dfrac{\delta q_{\rm }}{e^T}$,or something else?
Is there an intuitive explanation for this $\dfrac{\delta q_{\rm }}{T}$?
| First off, temperature is an intensive quantity, i.e., not additive. For example, two cups of coffee don't have twice the temperature of one cup. For an extensive (additive) quantity, such as mass, we can't just redefine $m\rightarrow m'=f(m)$, where $f$ is a nonlinear function, because then $m'$ wouldn't be additive. This constraint doesn't apply to temperature, because temperature is intensive.
The relation $dS=dq/T$ is really the definition of temperature, not entropy. (Entropy is really defined as the log of the number of accessible states.) But this still allows us to take $T\rightarrow T'=f(T)$, where $f$ is some nonlinear function, and then we would just define temperature as $T'=f(dq/dS)$. We would need $f$ to be a one-to-one function, because we don't want objects that aren't in equilibrium to have the same temperature.
There is really nothing wrong with this, and for example this question discusses the possibility $f(x)=1/x$. Some equations (e.g., the partition function) actually come out simpler if written using this definition, although others get more complicated (the heat capacity of an ideal gas is no longer constant). However, most possibilities for $f$ result in all equations looking more complicated.
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Charged spheres - help with method to work these out? Can anyone demonstrate how to get the answers to these questions? I'm just interested in the method I need to use in order to obtain the correct answer no matter what the values are.
Three small spheres are placed at fixed points along the x-axis, whose positive direction points towards the right.
Sphere A is at x = 36.0 cm, with a charge of –8.00 μC.
Sphere B is at x = 41.0 cm, with a charge of 9.00 μC.
Sphere C is at x = 46.0 cm, with a charge of –3.00 μC.
a) Calculate the magnitude of the electrostatic force on sphere B.
b) Sphere B is now removed:
What would be the magnitude of the electric field at the point where sphere B was located?
c) Sphere B is still missing. Give the x-coordinate of the point on the x-axis where the field due to spheres A and C is zero
| The spheres are effectively point charges, so the force between any pair of spheres is just given by Coulomb's law. The forces are additive so in (a) the net force on $B$ is just the vector sum of $F_{AB}$ and $F_{BC}$.
The field strength is the force (from Coulomb's law) on a point charge of 1 Coulomb, so in q (b) proceed as before replacing $B$ by a charge of 1 Coulomb.
(c) is just the point between $A$ and $C$ where the force on a test charge sums to zero.
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Is there a heat transfer equation that takes conduction, convection, radiation and dT/dt into account? I see equations that take 2 or 3 of the listed parameters into account but I haven't been able to find one that is that complete. I'm seeking to solve this equation using matlab for a simulation project to study how the heat dissipates on a microheater given an electrical power or current as input and see how quickly the temperature rises, and so forth. But first, I need the proper equation.
| The nature of 'complete' heat equation will depend on your domain. If I understand correctly, you want to model the temperature rise/heat dissipation in a microheater (a solid?) when it is electrically powered. The heat transfer inside the heater in this case will be solely due to diffusion. Convection/radiation will not be a part of heat transfer internal to the heater. However, both these mechanisms would appear at the heater boundary i.e the boundary conditions. So the governing equations would be:
1) \begin{equation} \frac{\partial T}{\partial t} =a \nabla^2T + S \end{equation} in the heater interior. The term S represents heat generated per unit volume.
2) \begin{equation} -k\nabla T =h (T - T_0) + \sigma\epsilon(T^4 - T_0^4) \end{equation}
at the boundary.
You will need to know the heat transfer coefficient 'h' and emissivity in the problem. As a simplification (since the heater is 'micro') you can assume the heater to be a 'lumped' system so that you can neglect heat diffusion inside the heater. The first equation would then simply become
3) \begin{equation} \frac{\partial T}{\partial t} = S \end{equation}
Equation 2 remains unchanged. Be careful about the units and dimensions on 'S', the heat source...
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Algorithm to simulate impulse response of a solid? There are well-documented methods of calculating the impulse response of a room (ex. image method, ray-tracing method); however, I have not been able to find anything similar for the impulse response of an arbitrary solid given its size, density, etc. Can anyone point me in the right direction?
| The exact answer can easily be computed by solving the elastic wave equation using the finite-difference time-domain method.
You mentioned the image and ray-tracing methods for room acoustics. Theses are geometric methods, which would not be useful for solids since there would need to be separate rays to describe the compressional and shear waves, and the transfer of energy between the two kinds of waves which is complicated.
| {
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Have we found a Higgsino? In supersymmetry, for each particle (boson/fermion), there is a symmetric particle which is a fermion/boson.
The MSSM predicts five Higgs bosons: two neutral scalar ones (H and h), a pseudo-scalar (A) and two charged scalars.
Does this mean that there are five higgsinos?
My question arises because I read:
What is the difference between the SM Higgs and the supersymmetric H and h?
Nothing.
So I'm confused, is it saying that H and h are the supersymmetrical parters of the Higss boson?
|
Although I agree with everything that Mitchell Porter wrote in his
nice answer, I would like to add a few remarks.
With supersymmetry, there are equal numbers of bosonic and fermionic degrees of freedom for each type of field.
Prior to electroweak symmetry breaking, in the simplest supersymmetric models (e.g., the so-called MSSM), the Higgs scalars have $8$ real degrees of freedom. There must be (at least) two Higgs-doublets in supersymmetric models.
After electroweak symmetry breaking, there are indeed $5$ Higgs bosons ($h$, $H$, $A$, $H^\pm$), as you correctly write. The missing $3$ degrees of freedom are eaten by the $W^\pm$ and the $Z$-bosons, when they acquire masses.
There are $4$ higgsinos, labeled by their charge and by whether their scalar superpartner helps give mass to up-type or down-type quarks: $\tilde h_u^0$, $\tilde h_d^0$, $\tilde h_u^+$, $\tilde h_d^-$. Each higgsino has $2$ degrees of freedom - so in total higgsinos have $4\times2=8$ degrees of freedom, matching those of the Higgs scalars.
Nobody has observed direct evidence for higgsinos, or any other supersymmetric particle. Though there are strong theoretical hints and indirect evidence for their existence.
That is not the end of the story. The higgsinos mix with fermions with identical quantum numbers. The neutral higgsinos mix with the photino and zino, resulting in four neutral particles called neutralinos, and labeled $\chi_{i=1,2,3,4}^0$, where $\chi_1^0$ is the lightest neutralino etc. Similarly, the charged higgsinos mix with the charged wino, forming two charginos, $\chi_{i=1,2}^\pm$.
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Determining the spring constant in an oscillation problem
A 130g air-track glider is attached to a spring. The glider is pushed
in 10.4cm and released. A student with a stopwatch finds that 14.0
oscillations take 19.0s
I would like to know why the answer I get is wrong. I set up an equation to represent position as a function of time.
$$x(t) = \frac{52}{5}*cos(\frac{19\pi*t}{7})$$
$$a(t) = \frac{d^2x(t)}{dt^2} = -\frac{18772\pi^2}{245}*cos(\frac{19\pi *t}{7})$$
Also the force of the spring $$F_s = -kx$$
$$\Sigma F_x = ma_x$$
So plug in $t = 0$ $s$ and $x = 10.4$ $cm$ for simplicity
Then the acceleration is the constant $-\frac{18772\pi^2}{245}$ and multiply by the mass to get the force. We also know the position is $10.4$ $cm$ when the force is such.
$$k = -\frac{m*a(t)}{x(t)} = - \frac{0.13 * (-\frac{18772\pi^2}{245})}{10.4} = 9.45_\frac{N}{cm}$$
| Remember that an oscillation can be written as $\cos(\omega t) = \cos(2\pi f t) = \cos(\frac{2 \pi}{T} t)$. You have the period on the top instead of the bottom.
| {
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How does a Tesla Coil exactly work? How does a Tesla Coil exactly work? I am currently making a Tesla Coil in school but before I start, I want to know exactly how does a Tesla Coil work? I understand the various electrical components used in it, but what I don't understand is how do those components together work to make something like Tesla Coil.
| A Tesla coil is in many ways the same as a transformer and hence if you know the working of one you can understand the working of the other.
Now in common transformers, the coils are couple tightly, so that a large amount of energy transfers from the primary to secondary. This works well at low voltages, but at high voltages the insulating air gap may suffer a dielectric breakdown, thereby becoming the reason for heavy losses. The Tesla coil avoids this and hence can be used at very high voltages.
A Tesla coil consists of two LC oscillators very loosely coupled to each other. When a charged capacitor is connected to an inductor an electric current will flow from the capacitor through the inductor creating a magnetic field. When the electric field in the capacitor is exhausted the current stops and the magnetic field collapses. As the magnetic field collapses, it induces a current to flow in the inductor in the opposite direction to the original current. This new current charges the capacitor, creating a new electric field, equal but opposite to the original field. As long as the inductor and capacitor are connected the energy in the system will oscillate between the magnetic field and the electric field as the current constantly reverses. The figure below shows a common schematic of the Tesla coil.
When the switch is open, the cap is charged. Now when the switch is closed, the action discussed earlier will cause a magnetic field to be built up in the primary inductor and this in turn will set up a field in the secondary. Since the secondary has a large amount of turns, this will cause an extremely high E field to be built up thereby resulting in large voltages.
The new coils have some additional components, but this is the very basic working of the Tesla coil with minimal components.
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Doppler effect problem with moving mirror This was the given question:
A light beam of intensity $I$ and frequency $f$, directed along the
positive $z$-axis, is reflected perpendicularly from a perfect mirror which
itself is moving along the positive $z$-axis with a constant velocity $v$.
Find the reflected intensity $I'$ and the frequency $f'$ of the reflected light.
If a light beam is directed towards a mirror which is moving towards us with a constant velocity $v$, the apparent frequency of the reflected beam would be higher. However, I would like to know if there is any change in the observed intensity of the light beam.
Also, just rechecking, does the intensity of a light beam depend only on the amplitude?
Thanks in advance
| A real light source moving towards you gets more intense as the inverse square. Why should a virtual light source behave any differently?
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Could velocity be taken as fundamental instead of time? In physics time and length are taken as fundamental in the SI system and, as it seems, in the thinking of physicists. Could one instead take velocity, with c as its unit, together with length as fundamental and then understand time by dimensional analysis in terms of l/v (length divided by velocity)? If not, why not?
| Personally, I don't see why not. And I too believe it has the potential to open up new lines of inquiry and understanding. I have long thought the concepts of velocity and change have a good chance of in fact being most fundamental. As an example, what if the reality of the fundamental nature of our universe is such that, at an event, information is emitted at an angle, with the angle 0 being the minimum possible angle, so that the speed of light (c) is actually the minimum velocital angle 0. Slower speeds in our current way of thinking about such things would simply be an angle greater than 0 so that it would take longer, hence more "time", for that information to arrive at and influence some distant point in space.
| {
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Elliptical orbits passing 2 points I would like to find set of possible elliptical orbits which pass 2 points in plane. I was searching for some solutions in orbital mechanics texts but I didn't found any.
There are several possible approaches but I'm not sure which is the best - both looks quite difficult to solve algebraically.
*
*using polar equation relative to focus with $(R_1,\phi_1),(R_2,\phi_2)$, being coordinates of points
$$ R_1 = \frac{a(1-e^2)}{1-ecos(\phi_1-\theta)} $$
$$ R_2 = \frac{a(1-e^2)}{1-ecos(\phi_2-\theta)} $$
then for given $\theta$ solve for semimajor axis $a$ and eccentricity $e$
*using deffinition of elipse as a set of points of the same distance from both foci. Given 2 points of cartesian coordinates $(x_1,y_1),(x_2,y_2)$ and one focus in origin $(0,0)$. For each given distance parameter $L$ solve for coordinates of second focus $(x_f,y_f)$,
$$ L = \sqrt{x_1^2 + y_2^2} + \sqrt{(x_1 - x_f)^2 + (y_1 - y_f)^2} $$
$$ L = \sqrt{x_2^2 + y_2^2} + \sqrt{(x_2 - x_f)^2 + (y_2 - y_f)^2} $$
*I can also first rotate the coordinate system (or my input points) by given angle (which is my arbitrary parameter) and then use some simplified equation of ellipse which has major axis parallel to x-axis which has just 2 degrees of freedom. But even after this rotation I don't see much simplification of algebraic solution.
Nevertheless, the resulting equations are difficult to solve.I solved it using sympy, but the solution is very long expressing hard to simplyfy. I would like some more elegant solution if there is any.
I would also like to implement this into computer as a part of orbital transfer optimization, so I would prefer some explicit expression which is fast to evaluate numerically ( for example goniometric functions are quite slow to evaluate )
| I am not entirely sure if there is an explicit equation, but I do remember that a by the community made mod for the game Kerbal Space Program, called MechJeb, this is calculated by solving a Lambert's problem. The code they used to calculate this can be found on github.
Someone else also made a web application which can calculate Porkchop plots and its code can also be found on github.
Both calculate the solution by constraining the time between the two given points rather than your $\theta$. So this might be the cause the need for a numerical solver, rather than a explicit equation.
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Does antimatter curve spacetime in the opposite direction as matter? According to the Dirac equation, antimatter is the negative energy solution to the following relation:
$$E^2 = p^2 c^2 + m^2 c^4.$$
And according to general relativity, the Einstein tensor (which roughly represents the curvature of spacetime) is linearly dependent on (and I assume would then have the same mathematical sign as) the stress-energy tensor:
$$G_{\mu \nu} = \frac{8 \pi G}{c^4}T_{\mu \nu}.$$
For antimatter, the sign of the stress-energy tensor would change, as the sign of the energy changes. Would this change the sign of the Einstein tensor, causing spacetime to be curved in the opposite direction as it would be curved if normal matter with positive energy were in its place? Or does adding in the cosmological constant change things here?
| Antimatter is not the negative energy solutions to the energy-momentum relation! Even in the Dirac sea model it isn't. In the Dirac sea model, the negative-energy modes are all filled with electrons, and the absence of one of those electrons is a positron. Just as the absence of a positive-energy particle has (relatively speaking) a negative energy, so the absence of a negative-energy particle has (relatively speaking) a positive energy. So the Dirac sea predicts that positrons have the same positive energy as electrons.
We also know experimentally that positrons have positive energy and positive inertial mass. If positrons and electrons repel each other gravitationally, then positrons have a gravitational mass that is negative, and therefore not equal to their inertial mass. That violates the equivalence principle. Without the equivalence principle, the spacetime model of gravity doesn't work. There's no question at that point of positrons curving spacetime in the opposite direction, because the whole notion that gravity is spacetime geometry is dead. Needless to say, I don't consider this to be very likely.
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Will the Hubble parameter reach zero asymptotically in the far future? In the current accelerated expansion universe model will the Hubble parameter $H$ reach zero asymptotically in the far future?
| No, but it will asymptotically approach a particular value. In the Standard Model where dark energy is a cosmological constant, the Hubble parameter is a function of the present-day Hubble constant and the radiation (R), matter (M), curvature (K), and dark energy ($\Lambda$) densities as follows
$$
H(a) = \frac{\dot{a}}{a} = H_0\sqrt{\Omega_{R,0}\,a^{-4} + \Omega_{M,0}\,a^{-3} + \Omega_{K,0}\,a^{-2} + \Omega_{\Lambda,0}},
$$
with
$$
\begin{gather}
H_0 = 67.3\;\text{km}\,\text{s}^{-1}\text{Mpc}^{-1},\quad \Omega_{R,0}\approx 0, \\
\Omega_{M,0} = 0.315,\quad \Omega_{\Lambda,0} = 0.685, \quad \Omega_{K,0} = 0,
\end{gather}
$$
according to the latest Planck results. So in the far future, the scale factor $a$ goes to infinity, and we get
$$
H(\infty) = H_0\sqrt{\Omega_{\Lambda,0}} \approx 55.7\;\text{km}\,\text{s}^{-1}\text{Mpc}^{-1}.
$$
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How does energy transfer between B and E in an EM standing wave? I'm trying to understand how an electric field induces a magnetic field and vice versa, its associated energy, as well as relating it to my understanding of waves on a string.
Using a standing wave as an example, I came up with the equations
$\vec{E}=A\sin(\omega t)\sin(kx)\hat{y}$
$\vec{B}=\frac{Ak}{w}\cos(\omega t)\cos(kx)\hat{z}$
I checked them against Maxwell's equations, and they're self-consistent. At time 0, this reduces to:
$\vec{B}=\frac{Ak}{w}\cos(kx)\hat{z}$
Since the electric field is 0, based on the Poynting vector, there's no energy transfer at this time. At this time, at a node where $\vec{B}=0$, there's neither electric field nor magnetic field. If there's no energy transfer, and no energy stored in either field, then how can an electric field exist at this point at some time later? How is the energy stored, or transferred from elsewhere?
| The energy conservation is written $\dfrac{\partial u}{\partial t} + div \vec S=0$, where $u$ is the energy density $\vec E^2+\vec B^2$, and $\vec S$ is the Poynting vector $\vec E \wedge \vec B$ (skipping irrelevant constant factors).
If you choose $x,t$ such as $\sin \omega t=0$ and $\cos k x=0$, both $E$, $B$, the energy density $u$ and the Poynting vector $S$ will be zero.
We have $\vec S \sim \sin 2k x \sin 2\omega t ~\hat{x}$. The divergence of the Poynting vector will be $div \vec S\sim \cos 2k x \sin 2\omega t $, so it it zero too (because of the $t$ dependence ), and so $\dfrac{\partial u}{\partial t} = 0$. However, the first time derivative of $div(\vec S)$is not zero : $\dfrac{\partial (div \vec S)}{\partial t}\sim \cos 2k x \cos 2\omega t $, so, from the energy conservation equation, the second derivative of the density energy $\dfrac{\partial^2 u}{\partial t^2} = - \dfrac{\partial (div \vec S)}{\partial t}$ is not zero.
So, you may write :
$u(x,t+\epsilon) = \frac {\epsilon^2}{2} \dfrac{\partial^2 u}{\partial t^2} + o(\epsilon^2)$
So, at infinitesimal times after $t$, the energy density is not zero.
| {
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Does the moon affect the Earth's climate? So, this morning I was talking to a friend about astronomical observations, and he told me that lately there has only been good weather when there was a full moon in the sky, which was a shame.
I jokingly said: 'maybe there's a correlation!', but then I started thinking: wait, if the moon can affect the oceans, why shouldn't it also make an impact on the atmosphere, which is just another fluid. So... are there atmospheric tides? Does the moon affect the weather or the climate in a significant way?
| It might affect climate, but not on the time scale of a month, and does not significantly affect the weather.
The fact that the moon exists may significantly stabilise the inclination of the Earth relative to the Sun. This, in turn, affects climate in the long run. The debate is ongoing. For example, see long term axial tilt (Wikipedia):
The Moon has a stabilizing effect on Earth's obliquity. Frequency map analysis suggests that, in the absence of the Moon, the obliquity can change rapidly due to orbital resonances and chaotic behavior of the Solar System, reaching as high as 90° in as little as a few million years. However, more recent numerical simulations suggest that even in the absence of the Moon, Earth's obliquity could be considerably more stable; varying only by about 20-25°. The Moon's stabilizing effect will continue for less than 2 billion years. If the Moon continues to recede from the Earth due to tidal acceleration, resonances may occur which will cause large oscillations of the obliquity.
There are also atmospheric tides, but lunar atmospheric tides are very weak. To detect a lunar signal in weather patterns can be difficult, because other signals are so much larger, and there is noise too. So if your friend thinks it's only been good weather during full moons, that's either a coincidence or confirmation bias.
| {
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Thin lens formula Can someone help me or guide me how the thin lens formula:
$$\frac{1}{s_1}+\frac{1}{s_0}=\frac{1}{f}$$
can be proven?
I was trying to prove it on my own using similar triangles, only to fail.
| Another high school proof of the general lens equation:
Let's suppose you have a lens of refractive index $n_2$ seperating two media of refractive index $n_1$ and $n_3$, then by the equation of curved surface refraction, we can write image as refracted by first surface of lens:
$$ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R_1} \tag{1}$$
And, by the second surface of lens as:
$$ \frac{n_3}{v'} - \frac{n_2}{u'} = \frac{n_3-n_2}{R_2} \tag{2}$$
Variables used:
$v$ image formed by first surface
$u$ object of the first surface
$v'$ image formed by second surface
$R_1$ radius of curvature of first surface
$R_2$ radius of curvature of second surface
Fact: $u'=v$ because the image of first surface is object of second and hence (2) becomes:
$$ \frac{n_3}{v'} - \frac{n_2}{v} = \frac{n_3 - n_2}{ R_2} \tag{3}$$
Adding (3) and (1):
$$\frac{n_3}{v'}- \frac{n_1}{u}= \frac{n_2- n_1}{R_1} + \frac{n_3- n_2}{R_2}$$
If, $n_3=n_1=1$ that is the ambient mediums on both side are air then the previous equation becomes:
$$ \frac{1}{v'} - \frac{1}{u} = (n_2-n_1) \left[ \frac{1}{R_1} + \frac{1}{R_2}\right]$$
To find focal length, just send $u\to \infty$ which makes:
$$ \frac{1}{f} = (n_2-n_1) \left[ \frac{1}{R_1} + \frac{1}{R_2}\right]$$
Hence,
$$ \frac{1}{v'} - \frac{1}{u} = \frac{1}{f}$$
Generalization, let's say we have $j$ lens 'pieces' in series to each other and each has refractive index $n_j$ and radius of curvature $R_1$, then:
$$ \frac{n_2}{v_1} - \frac{n_1}{u_1} = \frac{n_2 - n_1}{R_1}$$
$$ \frac{n_3}{v_2} - \frac{n_2}{u_2} = \frac{n_3 - n_2}{R_1}$$
$$ \vdots$$
$$\frac{n_j}{v_j} - \frac{n_{j-1} }{u_{j-1} } = \frac{n_j - n_{j-1}}{R_j}$$
Noticing that $v_k = u_{k+1}$ and adding all the equatoins:
$$ \frac{n_j}{v_j} - \frac{n_1}{u_1}= \sum_{k=0}^n \frac{n_k -n_{k-1}}{R_k} \tag{4}$$
| {
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Classical Wave Equation - Approximations I don't understand the derivation of the wave equation given below -
$$T \sin (\theta _1) - T \sin (\theta ) = T\tan (\theta _1 )-T\tan (\theta ) = T \left. \left(\frac{\partial f}{\partial z} \right|_{z + \Delta z} - \left. \frac{\partial f}{\partial z}\right| _z \right) = T \frac{\partial ^2 f}{\partial z^2} \Delta z$$
I understand that the small angle approximation was used, but I'm at a loss for figuring out we turned $\tan$ into a derivative, and then after made it become a second derivative.
The derivative of $\tan \theta$ is of course $\sec \theta$ which is equal to $\frac{1}{cos \theta}$, which was taken with respect to $\theta$, maybe there's a way to use the chain rule to find $\partial _z f$?
| I just realized that the reason we're using $\tan \theta = \partial _z f$ is because $\tan \theta = \frac{df}{dz}$, which makes sense if we consider this a right triangle.
| {
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Chinook Helicopter Torque The Chinook Helicopter has 2 rotors to counteract the torque generated by spinning the blade.
Theoretically, could you use a smaller "back" rotor that is farther away from the main rotor to achieve the same result, ie no twisting?
| Assuming a mass moment of inertia of $I_1$ for the main rotor and $I_2$ for the secondary rotor, and a coefficient of drag of $\beta_1$ and $\beta_2$ respectively then the torque on the rotor shafts are
$$ T_1 = I_1 \dot \Omega + \beta_1 \Omega^2
\\ T_2 = I_2 (\gamma \dot \Omega) + \beta_2 (\gamma \Omega)^2 $$
where $\Omega$ is the main rotor speed, and $\gamma$ the gearing ratio for the small rotor.
To make those equal you need gearing of $$\gamma = \frac{I_1}{I_2}$$ and coefficient of drag $$\beta_2 = \beta_1 \frac{I_2^2}{I_1^2}$$
Since $I_1 > I_2$ this requires the drag to be $ \beta_2 \gg \beta_1 $ which is a) hard to do with a small rotor, and b) very inefficient.
FYI - The total torque on the motor is going to be
$$ T_E = \left( I_1 + I_2 \gamma^2\right) \dot \Omega + \left(\beta_1 + \beta_2 \gamma^3\right) \Omega
\\ = \left(I_1 + \frac{I_1^2}{I_2} \right) \dot \Omega + \left( 1 + \frac{I_1}{I_2} \right) \beta_1 \Omega^2 $$
| {
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Proper name for a thermodynamic process with constant internal energy $U$ Back in the day I learned that a few special thermodynamical processes have special names.
For example, if one keeps $P$ constant, the process is called isobaric, if one keeps $T, V$ or $S$ constant, one gets, correspondingly, isothermic, isochoric or isentropic processes. Similarly, if one keeps $\dfrac{\mathrm{d} \ln P}{\mathrm{d} \ln \rho}$ constant during the process, it is called polytropic, and if $\delta Q = 0$ at any time, the process is called adiabatic.
Now, the question: what is the process called, if one keeps internal energy $U$ constant?
| A process where the energy is kept constant is called isoenergetic (or, if you prefer, iso-energetic).
It also seems from the literature that a flow where the energy is constant when following a fluid particle is usually called an isoenergetic flow.
Similarly, when the enthalpy is kept constant, the process (or the flow) is said to be isenthalpic (or isoenthalpic). And so on.
Notice that if there is some subtlety and you keep a constant internal energy $U=\text{cte}$ but not a constant energy $E=U+E_{\text{m}}$, by modifying the mechanical energy $E_{\text{m}}$, you should refrain from using standard names like isoenergetic and explain precisely what happens.
| {
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One way insulation? I know from basic physics lessons that a box painted black will absorb heat better than a box covered in tin foil. However a box covered in tin foil will lose heat slower than a black box.
So what is the best way to conserve the temperature of a box? (aiming for 0 degrees Celsius inside the box when it's -60 outside).
I mean would painting the outside of the box black, and having tin foil on the inside work? So the box can absorb heat better (black paint) and the tin foil making it harder for heat to escape?
| A perfectly one way insulator would violate the law of conservation of energy. You could place it in a fluid filled box and let a temperature gradient develop. You could then use it to drive machinery. Bam! Energy for nothing. Therefore by the conservation of energy (and second law of thermodynamics: the entropy would decrease) such a one way insulator is impossible (although exceptions may exist if the insulator gets used up or something)
With respect to your question, covering the box with tin foil will prevent the energy from escaping better then the black one (the box's content will stay warmer for longer), however it won't make the box any warmer.
| {
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What spacetimes satisfy this identity?
*
*What spacetimes satisfy $R^{\mu\nu} R_{\mu\nu} =\alpha R^2$, where $R = g^{\mu\nu}R_{\mu\nu}$ is the Ricci scalar, and $\alpha$ is some constant?
*A follow-up question: in what spacetimes does $R^{\mu\nu\sigma\rho} R_{\mu\nu\sigma\rho} =\beta R^2$, for some constant $\beta$?
| I can give you some particular solutions, but not the most general solution.
*
*Any metric in 2 dimensions satisfies
$$
R_{\mu\nu\alpha\beta} = \frac{R}{2} \left( g_{\mu\alpha} g_{\nu\beta} - g_{\mu\beta} g_{\nu\alpha} \right) \implies R_{\mu\nu} = \frac{1}{2} g_{\mu\nu} R
$$
This implies
$$
R_{\mu\nu\alpha\beta} R^{\mu\nu\alpha\beta} = R^2,~~ R_{\mu\nu} R^{\mu\nu} = \frac{R^2}{2}
$$
Thus, in two dimensions, $\alpha = 1/2$ and $\beta = 1$.
*Maximally symmetric space-times in $d$ dimensions that satisfy
$$
R_{\mu\nu\alpha\beta} = \frac{R}{d(d-1)} \left( g_{\mu\alpha} g_{\nu\beta} - g_{\mu\beta} g_{\nu\alpha} \right) \implies R_{\mu\nu} = \frac{1}{d} g_{\mu\nu} R
$$
From this, we get
$$
R_{\mu\nu\alpha\beta} R^{\mu\nu\alpha\beta} = \frac{2 R^2}{d(d-1)},~~ R_{\mu\nu} R^{\mu\nu} = \frac{R^2}{d}
$$
Thus, for maximally symmetric space-times in $d$-dimensions, $\alpha = \frac{1}{d}$ and $\beta = \frac{2}{d(d-1)}$.
| {
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What is fundamental difference between wave and its 180 flip phase? I'm studying property of sound wave and I was wondering what is difference between two waves (one is original and one is 180 flip phase of original) ?
Amplitude and frequency remains same and also wavelength is same, so are they same?? I could not detect any difference from hearing two sounds. If different what is different and can human/computer detect it? What is basically difference between two waves?
| The difference is that at each instant in time, the 180-degree delayed wave has exactly the opposite value of the original wave.
If $y_1(t) = \sin(2\pi{}ft)$ and $y_2(t) = \sin(2\pi{}ft + 180^\circ)$, then $y_1(t)=-y_2(t)$ for all t.
If you are talking about a sound wave, then whenever one wave has a minimum of pressure, the other has a maximum and vice versa.
As other answers said, this difference is not perceptible to human hearing because the ear is not sensitive to phase differences or delays on the scale of a single cycle of a sound wave.
| {
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Question about the quantization of lattice vibration (phonons) In my syllabus about solid state physics they state that lattice vibration is quantized, analogous to the harmonic oscillator:
$$E = (n+\frac{1}{2})\hbar\omega$$
So the lattice vibration has zero-point energy $\frac{1}{2}\hbar\omega$.
I wonder what this actually means: are all possible vibration modes of the lattice quantized in this way? So do all the vibration modes (acoustical/optical and transversal/longitudinal) have nonzero energy for all possible wavevectors $\vec{k}$ in the Brillouin zone?
So If I interpret this quantization in this way, it means that the lattice is at all time vibrating in all possible vibration modes.
In the syllabus they state though (somewhat earlier (and translated to English)) :
"The vibration modes are purely longitudinal or transversal only in the case of sufficient symmetry, e.g. in some directions of a cubic crystal structure. Otherwise, the waves consist of a mixture of the two." (I thinks that this refers to f.i. the [100] direction in a cubic crystal)
This seems to be contradicting the idea that longitudinal and transversal modes should have zero point energy. I hope that someone can clarify this.
| In classical mechanics, you can describe a crystal (in some approximation) by a Hamiltonian that is a quadratic form in coordinates and momenta of atoms. After you diagonalize this quadratic form, you obtain a Hamiltonian of a set of independent, rather than coupled, oscillators (modes). Then you can quantize this system and you do get zero-point energy for each independent mode (by the way, this energy is $\frac{1}{2}$ℏω, not ℏω). However, not all independent modes are purely longitudinal or transversal. In other words, longitudinal modes are often coupled with transversal modes and, therefore, they are not independent modes that you get as a result of diagonalization. In other words, longitudinal and transverse modes are some linear superpositions of independent modes (which are also called eigen modes, or characteristic modes, or normal modes:-) ) .
| {
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Slow light and gravitational lensing It has been proposed that the Sun's gravitational lens be used to observe distant objects, but according to relativity a spacecraft would need to be 550 AU away in order to take advantage of the effect.
Would it be possible to instead slow down the light while it is being deflected using, for example, a Bose-Einstein condensate such as a large quantity of very cold sodium placed into orbit around an asteroid?
An answer to this question mentioned that it is useless to do the same thing using an ordinary material with a high refractive index because of dispersion, but as I understand it the mechanism for producing slow light changes the group velocity, which is the speed at which energy is carried, instead of the phase velocity.
| The limiting factor in optical observations isn't usually the magnification but rather how much light can be collected. The advantage of using the Sun as a lens isn't that it's especially powerful, but that it's exceptionally big!
So you could sit on an asteroid to make your observations, but there would be no benefit as you'd be getting little more of the light from the distant object than if the Sun wasn't there. The point of going out to 550AU is that all the light bent by the Sun's gravitational field converges at this point so you get a huge increase in the brightness of the image.
| {
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Why are high voltage lines “high voltage?” If I have two spheres of the same size and one sphere has a small amount of charge compared to the other that has a lot more charge, then clearly the sphere with the larger charge has a larger voltage (relative to the ground). My question is do high voltage power lines have a lot more charge that is placed on them? Is that what gives them the high voltage? I think I have a grasp of the step up stations that use transformers to kick up the voltage from power plants.
This question seems almost silly to me but I have been struggling with this for a long time. I’ve done several searches online and I am not able to find answers. If there is a link that someone can provide, I much appreciate it.
| They do carry a bit of extra charge, but it's sort of a side effect. Every conductor (such as a high voltage wire) has some capacitance. The capacitance of an object can be defined as the amount of charge added, per unit change in voltage on that object (keeping all other voltages constant). When we energize the high voltage line (say, increasing its voltage from zero to 1 megavolt), therefore, we must add some charge.
In the case of the high voltage line the capacitance is probably something on the order of 10 picofarads per meter (this is an educated guess based on approximate dimensions). This means that, in order to charge up a 100 km length of power line to 1 megavolt, we need to add about 1 coulomb of charge.
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Why don't metals bond when touched together? It is my understanding that metals are a crystal lattice of ions, held together by delocalized electrons, which move freely through the lattice (and conduct electricity, heat, etc.).
If two pieces of the same metal are touched together, why don't they bond?
It seems to me the delocalized electrons would move from one metal to the other, and extend the bond, holding the two pieces together. If the electrons don't move freely from one piece to the other, why would this not happen when a current is applied (through the two pieces)?
| I think that mere touching does not bring the surfaces close enough. The surface of a metal is not perfect usually. Maybe it has an oxide layer that resists any kind of reaction. If the metal is extremely pure and if you bring two pieces of it extremely close together, then they will join together. It's also called cold welding.
For more information:
*
*What prevents two pieces of metal from bonding?
*Cold Welding
| {
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Could there have been two "Big Bangs"? A couple of years ago, I remember seeing a documentary on the big bang theory. The theory presented was that to explain the cosmic microwave background radiation, there needed to have been two big bangs. Is this theory legitimate? I've tried searching for details without success.
My question is essentially, is it possible for there to have been two swift metric expansions of space that were the "Big Bang"?
| The term Big Bang does not have rigorous physical definition. And if you mean specifically inflation, then the modern inflationary models (chaotic inflation with various scalar field potentials) predict that inflation is always present at certain regions of space.
Inflation is driven by a scalar field, the value of which lowers with time. When it drops below the certain value inflation stops. But the quantum fluctuations of the field are stretched out by the inflation itself, and start to affect the mean value of the scalar field in that region. In the regions where the stretched out fluctuation raises the value of the scalar field, inflation continues, and the cycle repeats itself, eternally.
References: Multiverse, bubble universe, eternal inflation.
| {
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Infinitesimal Lorentz transformation is antisymmetric The Minkowski metric transforms under Lorentz transformations as
\begin{align*}\eta_{\rho\sigma} = \eta_{\mu\nu}\Lambda^\mu_{\ \ \ \rho} \Lambda^\nu_{\ \ \ \sigma} \end{align*}
I want to show that under a infinitesimal transformation $\Lambda^\mu_{\ \ \ \nu}=\delta^\mu_{\ \ \ \nu} + \omega^\mu_{{\ \ \ \nu}}$, that $\omega_{\mu\nu} = -\omega_{\nu\mu}$.
I tried expanding myself:
\begin{align*}
\eta_{\rho\sigma} &= \eta_{\mu\nu}\left(\delta^\mu_{\ \ \ \rho} + \omega^\mu_{{\ \ \ \rho}}\right)\left(\delta^\nu_{\ \ \ \sigma} + \omega^\nu_{{\ \ \ \sigma}}\right) \\
&= (\delta_{\nu\rho}+\omega_{\nu\rho})\left(\delta^\nu_{\ \ \ \sigma} + \omega^\nu_{{\ \ \ \sigma}}\right) \\
&= \delta_{\rho\sigma}+\omega^\rho_{\ \ \ \sigma}+\omega_{\sigma\rho}+\omega_{\nu\rho} \omega^\nu_{{\ \ \ \sigma}}
\end{align*}
Been a long time since I've dealt with tensors so I don't know how to proceed.
| Since the Lorentz transformation is valid for any $x\in M_{4}$, it can be rewritten as $\Lambda_{\rho}^{\mu}\eta_{\mu\nu}\Lambda_{\sigma}^{\nu}=\eta_{\rho\sigma}$. Substituting the infinitesimal form of the Lorentz transformation into the previous formula we get
$$(\delta_{\rho}^{\mu}+\omega_{\rho}^{\mu})\eta_{\mu\nu}(\delta_{\sigma}^{\nu}+\omega_{\sigma}^{\nu})+o(\omega^{2})=\eta_{\rho\sigma}$$
after expanding
$$\eta_{\rho\sigma}+\omega_{\rho}^{\mu}\eta_{\mu\nu}\delta_{\sigma}^{\nu}+\omega_{\sigma}^{\nu}\eta_{\mu\nu}\delta_{\rho}^{\mu}+o(\omega^2)=\eta_{\rho\sigma}$$
and from this we can see that
$$\omega_{\rho\sigma}+\omega_{\sigma\rho}=0\Rightarrow\omega_{\rho\sigma}=-\omega_{\sigma\rho}$$
| {
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What is negative Energy/Exotic Energy? So I have been researching around a little as I am highly interested in Astrophysics and I came across an energy I have never heard of before; negative energy also commonly known as exotic energy. Now I started to research this however I found the concept rather hard to grasp due to a simple lack on information around on the Internet.
Could somebody kindly explain (if possible using real life analogies) what exactly negative energy is or at least the whole concept/theory behind it.
| Negative energy is a quite different than Anti-matter. If you collide Anti-matter with regular matter you get a result with positive energy (Gamma rays). If you were to collide Negative energy and matter you would get nothing. It has negative mass (Anti-matter has a positive mass.) It is a hot topic in physics as it allows the creation of (Warp drives, Traversable Wormholes, Time machines, etc.)
A good analogy is money.
The amount of money in a bank account is the amount of energy. Negative energy would be like having a deficit of money in a bank account.
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Difference between primitive unit cell and the associated basis? As I understand it, the basis is the group of atoms whilst the primitive unit cell is the unit space that fits the total space without any gaps, and only containing one lattice point? How do the two relate to each other? Thanks.
| They don't relate.
The primitive unitcell is a property of the lattice. The lattice has nothing to do with the basis. You can have a single atomic basis or a thousand atomic basis but both have the same lattice, and therefore the same primitive unit cell.
One of the most important facts you have to wrap your head around is that a lattice point has nothing to do with an atom. A lattice point just defines the origin of a local coordinate system. The position of the basis atoms is defined in this local coordinate system.
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Importance of MHV amplitudes Why are MHV amplitudes so important? How/where are they used and why do people keep trying to rederive them in many different ways?
| MHV amplitudes are not really any more important than next-to maximal helicity violating amplitudes ($NMHV$) or $N^kMHV$ amplitudes. You need all of them to compute a general scattering amplitude.
Basically, scattering amplitudes for non-Abelian Yang Mills theories are very complicated to compute for more than 4 particles, so people work on formulating easier ways to do the computations. The first step is usually to strip off the color dependence of the amplitude, so that the object you really need to compute is the color ordered amplitude $\mathcal{A}(\pm,\pm,...\pm)$. These objects get multiplied by traces of color matrices at the end of the calculation.
At tree level, if all particles have the same helicity or only 1 helicity is different, the color ordered amplitude vanishes. So the MHV amplitude is the first non-vanishing color ordered amplitude. For $2 \to 2$ scattering you only need the MHV, since you can have at most 2 $+$ and 2 $-$ helicities. But if you were to compute say $3 \to 3$ scattering, you would also need something like $\mathcal{A}(+++---)$ which is not MHV.
MHV probably gets a lot of attention because the formulas are extremely simple (see: Parke-Taylor formula). Moreover, the discovery of the BCFW recursion relations allow you to generate higher order amplitudes from the lower order ones. In general, people are deriving new methods to calculate the amplitudes, and they can check their methods by matching onto the MHV amplitudes which have a simple form. But to compute a general amplitude you need more than just MHV.
Hope this helps.
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Is there some special cutoff density after which spacetime "collapses" and forms a black hole? With crude calculations following densities can be approximated:
Given that radius of proton is $1.75×10^{−15} m$ and it's mass is $1.67 × 10^{-27}kg$, this gives density of proton to be $\dfrac {1.67 × 10^{-27}kg} {\frac{4}{3}(1.75×10^{−15} m)^3}=\dfrac {3}{4} \dfrac{1.67 × 10^{-27}}{5.36×10^{−45}} \dfrac{kg}{m^3}=2.34 \times 10^{15} \dfrac{kg}{m^3}$
Density of the heaviest naturally occurring solid $\mathrm U^{238}$: 1 cubic meter of Uranium is about $2 \times 10^4 kg$, and since there are 238 protons+neutrons in Uranium, this gives the density $8 \times 10^{-3} \times 10^4 \dfrac{kg}{m^3}=8\times10 \dfrac{kg}{m^3}$
This means the protons are dispersed by a factor of $2.92\times10^{13}$, in other words there are $2.92\times10^{13}$ empty space units to each unit of space filled with proton.
Density of water $=\frac{1}{20}\times10 \dfrac{kg}{m^3}=5\times10^{-1}\dfrac{kg}{m^3}$
The density in the core of neutron star is $8×10^{17} kg/m^3$, and the density of a black hole is supposedly infinite according to Wikipedia, but then again before transitioning to a black hole there must have been some finite mass distributed over a non zero volume of space therefore having a finite density.
So far on logarithmic scale (base 10) followings look obvious for $\operatorname {log}_{10}(density \dfrac {kg}{m^3}) = \operatorname {log}_{10} \dfrac{kg}{m^3} + \operatorname {log}_{10}density$
Let $ d =\operatorname {log}_{10}density,\quad \operatorname {log}_{10} \dfrac{kg}{m^3}= \text{what is one to make of this}?$
Then following crude observations can be made:
$-1\leq\mathcal O(d)\leq1$: Order of density of life as we know it.
$\mathcal O(d)\approx 15$: Order of density of matter packed space
$\mathcal O(d)\approx 17$:Order of density at the core of a Neutron star
$\mathcal O(d)\geq x$:Order of density of black holes, (collapse of space to contain matter?) what is $x$?
My question is: does these calculations make sense and is there anything more to the order of densities than these naïve observations?
| I think you're asking if there is some special cutoff density after which spacetime "collapses" and forms a black hole. If this is your question then the answer is no, there is no specific cutoff.
Density unites are $\frac{\mathrm{mass}}{\mathrm{volume}}$ but the size of black holes is dependent on the mass and the size is not proportional to the volume but to the radius. That is, the Schwarzchild radius $r_s$ of a black hole of mass $m$ is $$r_s = \frac{Gm}{c^2}$$
What this means is that as a black hole becomes more massive, the critical density $d_c$ needed to form a black hole is $$d_c = \frac{3m}{4\pi r_s^3}$$
That is, as the black hole mass increases the density needed to make it form is reduced. In principle you can make the density arbitrary small which means there is no fundamentally critical density.
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Detecting a photon without changing it: Does it break conservation laws? This is about an article published on ScienceMag: Nondestructive Detection of an Optical Photon. I don't have access to full text, but you can see a brief transcription in this link.
Basically, it says that a photon causes a phase shift in another system. This phase shift can be detected, and it does not change photon properties, such as frequency (pulse shape) and polarization.
How can that be true? I thought that for a photon to cause any change on a system, it must lose some energy, which is transferred to the detector. What am I missing?
| the simple answer is that a quantum state has several variables (degrees of freedom), so if you measure only one of them and leave the others unchanged, then you detect the photon , change its state but do not destroy it completely. this is what they say in introduction
Second, nondestructive detection can serve as a herald that signals
the presence of a photon without aecting its other degrees of
freedom, like its temporal shape or its polarization.
look at the full article in arxiv
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Photons from stars--how do they fill in such large angular distances? It would seem that far-away stars are at such a distance that I should be able to take a step to the side and not have the star's photons hit my eye. How do stars release so many photons to fill in such great angular distances?
| Allow me to channel something akin to the anthropic principle here. You can only see the stars that have a lot of photons reaching your eye. If a star were so far away that photons were reaching your eyes only occasionally then the star would be too dim for you to see in in the first place. Even if you could see the photons, the star would appear to blink.
So because you can see the star and it's relatively bright, that means there is enough of a continuous stream of photons reaching the Earth that stepping side to side doesn't change anything. Also, angular resolution isn't quantized so there is never a situation where stepping side to side (while maintaining the same radius from the star) ever changes the probability of receiving a photon.
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Nokia PureView microscope: fluorescent Imaging I've come across this article: Fluorescent Imaging of Single Nanoparticles and Viruses on a Smart Phone.
And what is the theoretical limit for such smartphone extension? And how that limit can be computed (same as for microscope $\sim 1.22\,\lambda / D$)?
| The imaging is not being done by focusing transmitted light as would be done in an optical microscope. Instead it's detecting light emitted by the nanoparticles as they fluoresce. This means there is no lower limit to the size of the particle detected, except that when the particles get very small they emit too little light, i.e. they are too faint, for the phone camera to distinguish them from background noise.
There still remains a limit to detecting structure in the particles. For large particles you'd expect to resolve differences in the fluorescence across the particle, while for small particles they will just appear to be a featureless blob.
It's a bit like seeing stars with the naked eye. A star is far too small for your eye to resolve but you can still see the light coming from it. The only limitation is that the star appears as a fuzzy blob rather than a disk.
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How do they draw collision pictures in Atlas? Is the picture below a simulation? How to they draw these pictures? The picture is from here.
| This picture is not a simulation, it is an actual collision event. But it could have been either: both real and simulated events can be visualized in this way. Everything is done mainly by using the GEANT4 software package.
(As partially mentioned in the comments, the small boxes are deposits of energy in detectors; different colour for each detector. The lines, both straight and curved, are the reconstructed tracks of particles. And the blue cones are hadronic jets, which are actually spread out.)
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Archimedes' principle for two liquid layers
Problem: I have a cylindrical vessel of height $H$ and radius $R$. There are two liquid layers in the vessel. The first has density $D_1$ and height $h_1$, the second has density $D_2$ and height $h_2$. The second liquid is floating on the first liquid (thus $D_2 < D_1$) and they are both entirely within the vessel ($h_1+h_2 \le H$).
Now I have a cube of density $D$ and edge length $n$. The cube fits into the vessel ($n \le \sqrt2R$) but it might not be completely inside it ($n>H$ is possible). I place it into the vessel such that one side is horizontal (the cube is not lopsided). Some liquid might overflow as a result. How do I use Archimedes' principle to calculate the topmost liquid level after placing the cube? I can ignore damping and other physical effects.
EDIT: Note that the only values I know are $H, R, D_1, h_1, D_2, h_2, n$ and their abovementioned constraints.
My workings: I tried using the formula directly, where $\text{weight of cube}=\text{weight of displaced liquid}$. I first get $Dn^3=D_2n^2s_2$ where $s_2$ is the height submerged in the second liquid. If $s_2 \le h_2$ then the cube is only submerged in the second liquid. So the answer is $\min(H, h_1+h_2+{n^2s_2\over\pi R^2})$, accounting for overflows. This case is simple.
But if $s_2>h_2$ then the cube can be submerged in the first liquid too. I thought of subtracting the weight of the second liquid displaced to obtain the height submerged in first liquid, but realize that I don't know that. The weight displaced depends on the final liquid height, which can be affected by overflows, whether the cube fits totally within the two liquids, etc. In fact, it can be the case that after the liquid level rises, the cube now displaces more of the second liquid, causing it to not be submerged in the first (is that even possible?) It seems very messy and I have no idea how to start. :(
Does anyone have a nice solution to this problem?
| You do no say what information you know and do not know. For example if the cube sinks and $h_1$ is big enough, it is possible that $s_2=0$. But if you know $s_1$ and $s_2$ then it is easy.
The volume of liquid displaced is $(s_1+s_2)n^2$ so the extra height (ignoring overflows) is $\dfrac{(s_1+s_2)n^2}{\pi R^2 }.$ So the final overall height of the liquid is $$\min\left(H,h_1+h_2+\dfrac{(s_1+s_2)n^2}{\pi R^2 }\right)$$ and if you know $s_1$ and $s_2$ then you do not need to use the densities.
It is possible to calculate $s_1$ and $s_2$ using the information available including the densities, and that is where there are several cases to consider.
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Electromagnetism: Conductors Even though the thermal velocity of electron in a conductor is comparatively high, the thermal velocity is not responsible for current flow? Why is this the case?
| that's because thermal motion is random in nature, you'll find almost same number of electrons moving in any direction at some specific time, so on average the net motion of charges(i.e., current) in any direction is zero, so no current due to thermal motion,
now, if you apply external field to the metal the overall random motion of charges starts drifting in the direction of field(for positive charges), now since this gives a net motion to the charges you get a current flowing in the metal,, this drifting of charges is the order of mm/s while thermal velocity is the order of km/s,, way higher than drift velocity !
in short,, current is zero due to thermal velocity because of the randomness.
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The velocity of a cloud? I noticed an unusually fast moving cloud this morning.
My questions:
*
*What is the average velocity of a cloud on Earth?
*What is the greatest ever recorded cloud velocity?
*What factors affect the velocity of a cloud? (e.g. do they experience inertia?)
| Clouds move with the wind, so the cloud velocity is just the wind velocity. The recent storm in the Philipines reached wind velocities of 200 mph, though the higest speed reported is apparently 253 mph.
The fastest moving clouds known are on Neptune, where the winds reach 1340 mph.
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Steps involved in photon emission What are the detailed steps involved in the emissions of a photon, for example, when an electron drops to a lower energy level? How well do we understand the production of the photon in this example? Any good reference material I can study?
| The best account for photon emission when an electron drops to a lower eigenstate is the Wigner-Weisskopf Model for spontaneous emission, see this paper from the Photonics group at ETH Zürich and the co-efficients for this model can be calculated by standard quantum electrodynamics. I explain this model further and give references in my answer here.
A summary of what is going on: when one solves the first quantized Dirac or Schrödinger equation to calculate orbitals in an atom or molecule, one is assuming that the atom / molecule is sundered from the rest of the World, i.e. it has no interaction with its surroundings. So the "eigenstates" thus calculated are only energy eigenstates of the Hamiltonian (and thus stationary) in this highly idealized, one atom / molecule universe.
However, the electrons in real atoms / molecules are always coupled to the electromagnetic field. So the "eigenstates" as calculated above are no longer true eigenstate of the whole, coupled system (atom/molecule + the second quantized electromagnetic field), which is why the transition happens. The eigenstates of this system are quantum superpositions of the atom / molecule in its excited state with free photons in the field modes, and the superposition weights for the excited atom state in these new eigenstates are very small compared to those for the free photon weights. Once we "switch on" the electromagnetic field, the system will thus smoothly but inexorably evolve to one where the photon has been radiated. This coupling also begets the Lamb Shift: the splitting of energies of so-called "eigenstates" which are degenerate (have the same energy) in the isolated atom Dirac model.
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Why is the decay of a neutral rho meson into two neutral pions forbidden? Why is the decay of a neutral rho meson into two neutral pions forbidden? (Other modes of decay are possible though.)
Is it something with conservation of isospin symmetry or something else? Please explain in a bit more detail.
| If we look at isospin, $\rho = |1,0\rangle$ and $\pi^0= |1,0\rangle$.
Since SU(2) isospin is a really good symmetry in strong interactions, it must be conserved. Looking at the isospin of the final state:
\begin{equation}
|1,0\rangle \otimes |1,0\rangle = \sqrt{\frac{2}{3}} |2,0\rangle + 0 |1,0\rangle - \sqrt{\frac{1}{3}} |0,0\rangle
\end{equation}
That is, there is no $|1,0\rangle$ component in the final state, and therefore the process is not allowed by SU(2) isospin symmetry.
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Physics of the inverted bottle dispenser
When you invert a water-bottle in a container, the water rises and then stops at a particular level --- as soon as it touches the hole of the inverted bottle. This will happen no matter how long your water-bottle is. I understand this happens, because once the water level touches the hole, air from outside cant go inside and therefore there is nothing to displace the water that falls out of the container.
Now according to the laws of pressure ---- the pressure at the water level must be same everywhere --- whether it's inside the water bottle or outside. And that must be equal to the atmospheric pressure. Therefore the pressure of the water column + air column inside the inverted bottle must be equal to the atmospheric pressure.
What I dont understand is, no matter how long a bottle you take, the water level will always stop at the hole. So that means that no matter how long a bottle you take, the pressure of the water column + air column inside the water bottle will be equal to the atmospheric pressure. How could this be possible?
Also I'd like to let you know that, if you pierce the upper part of the bottle with a small pin, then the water level rises and overflows out of the container. I'm assuming air from outside rushes in and pushes the water out.
| Water stops draining from the jar into the dispenser once it forms an interface as draining of more water would result into the formation of a vacuum in the jar because no air can rush into the jar to displace the water as it has an interfacial-lock.
Consider the water level above interface $= h$, water level below interface $= x$
now
$$P_{surface}= P_{atm} + d\cdot g \cdot h $$
$$P_{dispenser~bottom} = P_{atm} + d\cdot g\cdot (h+x)$$
Now since $P_{bottom} > P_{surface}$! No further water drains (flow from lower to higher potential/pressure is not possible).
Also note that the air rushes in through the tap when you operate the system to take out water and not from the interface. And the pressure at the downside of the tap when you open it is just $= P_{atm}$.
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How to exchange light and matter? A while ago an experiment demonstrated that it is possible to stop a light pulse in a supercooled sodium cloud, store the data contained within it, and totally extinguish it, only to reincarnate the pulse in another cloud two-tenths of a
millimeter away.
Most of the papers discussing the details of this experiment are behind paywalls. I understand the setup that the light pulse can be revived, and its information transferred between the two clouds of sodium atoms, by converting the original optical pulse into a traveling matter wave which is an exact matter copy of the original pulse, traveling at a leisurely 200 meters per hour. The matter pulse is readily converted back into light when it enters the second of the supercooled clouds -- known as Bose-Einstein condensates -- and is illuminated with a control laser.
The key as I understand it is the Bose-Einstein condensates (BEC) become phase-locked. The light drives a controllable number of the condensate's roughly 1.8 million sodium atoms to enter into quantum superposition states with a lower-energy component that stays put and a higher-energy component that travels between the two BECs. The amplitude and phase of the light pulse stopped and extinguished in the first cloud are imprinted in this traveling component and transferred to the second cloud, where the recaptured information can recreate the original light pulse.
How does matter become "phased locked" to light and how do all these atoms contain the "information" of the light to reproduce it? And what is this talk of separating lower/higher energy components?
| Never worked with BEC but for ordinary matter it works like this: The incoming field makes the electrons oscillate with the same phase and frequency as the driving field (superposition state). If this state emits radiation before any kind of dephasing (usually takes femto/picoseconds) the outgoing field will be a copy of the incoming field.
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If a spaceship was pulled toward a sun, would it spin? I was watching a movie. A spaceship was forced into "warp speed". The co-ordinates could not be set. The spaceships trajectory was that of a nearby sun. Forcing the spaceship to power down was the solution. Now out of "warp speed" and with no computer aid (steering etc) the spaceship was seen to be spinning toward the sun trapped in its gravitation field.
My question is, would the spaceship (typically aerodynamically shaped) spin toward the surface? My opinion is no. The spaceship would just fall flat due to the surface area provided at the bottom of the fuselage
| The angular momentum of a massive sun may cause the freely falling spaceship to start spinning in the direction of the sun's angular momentum for an effect of frame dragging. You can take a look at the Kerr metric which describes the behaviour of the spacetime near a massive spinning object. If you're not familiar with general relativity it could be difficult to understand this effect. Anyhow, consider the Frame dragging section in the wikipedia article I pointed to. You can compare this effect with the one of a sea vortex; I guess you've seen sea vortex in many other movies and you know the effect they have on drifting ships. You can compare these two effects to understand the spinning effect of the spaceship towards the (spinning) star.
EDIT: Of course the fictional effect of a spaceship falling on the sun has been made similar to that of an airplane falling down. Typically, say in WWII movies, airplanes fall down for damages in their aerodynamics (such as broken wings and so on). None in a movie would have talked about the Kerr metric!
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Physical intuition for independence of components of velocity in derivation of Maxwell–Boltzmann distribution Maxwell derived the shape of the probability distribution of velocity of gas particles by starting with just two assumptions.
These are:
*
*The probability distribution is rotation invariant.
*The components (of velocity of a gas particle) in the direction of the coordinate axes are statistically independent.
And the rest is lovely deduction, but I found that as a layman I don't have any physical intuition as to why the second assumption is plausible. Is there an intuitive explanation behind the second assumption? If not, is there a way to derive the second assumption from a set of more plausible-looking assumptions?
| If I have a velocity which has some component $v_x$ in the $x$-direction, then is there any reason for you to assume you know anything anything about the component of my velocity which might be in a perpendicular direction, $v_y$?
No. So you can see that it is reasonable to assume that, if you know my $v_x$, my $v_y$ is still unconstrained, i.e. you have no information about it. The same holds for $v_z$, and there you have your statistically independent velocities!
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Calculate average speed with unknown variable accelaration I am in the middle of a vehicle tracking project where I have to calculate the distance traveled by the vehicle in a given amount of time.
Data I am getting:
Speed : 30.2 km/hr 12.7 km/hr 15 km/hr 21.8 km/hr
Time : 11:00:00 11:00:22 11:00:45 11:01:10
That is I am getting the speed of the vehicle every 20-25 seconds. So what is the best way to calculate the distance traveled by the vehicle during this whole duration? Is taking the median of two speeds the best way to calculate the average speed here?
| From this data, you don't actually know how the speed varies between steps. If you have more information about the acceleration then you could change the model from this but I would propose the following...
Assume a linear change in the speed between steps, you could take a simple graphical approach. Plot the speed (in km/s) on a vertical axis, against the total time elapsed since the beginning (in seconds) on a horizontal axis. Join up the points in your plot with consecutive straight lines. The total distance covered is given by the area under the line.
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Are coherent states of light 'classical' or 'quantum'? Coherent states of light, defined as
$$|\alpha\rangle=e^{-\frac{|\alpha|^2}{2}}\sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}}|n\rangle
$$
for a given complex number $\alpha$ and where $|n\rangle$ is a Fock state with $n$ photons, are usually referred to as the most classical states of light. On the other hand, many quantum protocols with no classical analog such as quantum key distribution and quantum computing can be implemented with coherent states.
In what sense or in what regime should we think of coherent states as being 'classical' or 'quantum'?
| If coherent state are indeed the most classical states (which means that the mean value of the EM fields obeys the classical Maxwell equations), the state used in the paper you mentioned are not coherent state (at least in the arXiv paper), but cat states !
The state $|\alpha\rangle+|-\alpha\rangle$ is not a coherent state ! It is the superposition of two classical state, which is really what we mean by quantumness.
Stated otherwise, coherent states form a basis which with you can write any quantum state, but that does not mean that all these states are as classical than a coherent state.
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Is a particular force different in different frames Can a particular real force have different magnitude in different frames?
| Well, take the electromagnetic force...it has been shown that the induced magnetic field around moving charges is a relativistic reference frame effect. See this post How Special Relativity causes magnetism
Therefore a static electric force in one frame becomes a magnetic force in another.
However, from a Newtownian view, we either allow pseudoforces and say "yes" or only allow forces present in an intertial refernce frame and say NO, it only appears that way to a person (e.g., falling vs floating in space)
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Dimension of the space of solutions in an electric circuit Consider an electric circuit with dc sources ( voltage and current) and resistors. Write down the equations. In the most general case, the solution of the system is not unique. The set of solutions can be empty or positive dimensional (simple example: 2 points in the graph and two batteries in parallel joining the points).
The dimension of the space of solutions can be computed with two different methods:
1) Mathematically : compute the determinants and the compatibility conditions
2) Physically. Give conditions on the graph to give a meaning for the dimension. For instance a cycle of batteries give an empty set of solutions or make the dimension increase by one.
I am interested in second method and I am looking for references in the litterature.
What I found is the following. In the student books, the unicity is always assumed to be true, eg. in the standard book by nilsson and riedel.
In more advanced books, I have found discusions only for particular cases, and with very technical tools. For instance, in Frankel (The geometry of physics), only purely resistive circuits are considered with source of currents in the nodes. And the proof uses ( a simplified version of) Hodge theory.
Now my question:
- Is there a book or an article where the very general case ( any graph with dc sources and resistors ) is considered and the dimension of the system described in terms of the graph ? I am interested both in sophisticated answers as above and in answers with basic tools of linear algebra. All references welcome.
I am looking for references, not for the solution. I have already written a solution for my students (an elementary one, with basic linear algebra). I want to compare my solution with the existing litterature. If this is useful and not a waste of time, I will make public my personal notes.
| This may not be a direct answer, but it can be shown by an elementary method that the relation between the number of nodes, branches and loops in a well-posed problem corresponds to Euler's polyhedron formula.
| {
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Momentum conservation problem Lets a plastic ball of mass m which is collided with steel. After collision the ball is coming back with the half initial speed. If the steel doesn't move then how can I interpretate this ?
Let the initial speed of the ball is $u_1$ and mass $m_1$ and mass of steel $m_2$ speed of steel before and after collision $0$.
Therefore we can write according to the conservation of momentum,
$$m_1 u_1 +m_2 u_2 = m_1v_1 +m_2 v_2$$
$$m_1 u_1 = m_1v_1 $$
$$ u_1 = v_1 $$
I have surmised $u_2 = v_2 = 0$.
Therefore speed is same then how can the speed of the ball can be
halved after the collision. Is this because of inelastic collision?
| If "the steel" that you're referring to is a fixed object or has a large mass, then the energy lost in the collision goes into increasing the internal thermal energy of the plastic ball and "the steel". This is an inelastic collision and kinetic energy is never conserved in inelastic collisions. Momentum, on the other hand is always conserved in the collision, but as your condition states, "the steel does not move". It means some other particle in the universe is gaining the momentum lost by the particle, or "the steel" moves so slowly that you are unable to measure it, or something else...
| {
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Question on energy mass conversion I have a question regarding the energy-mass conversion. Well, when a particle starts moving with a speed comparable to that of light, its (relativistic) mass increases that means some matter is created and that too of the same particle...energy being converted to mass is ok but how does energy perceive what atoms it has to form? Say I take a stone to a high speed, then constituents of stone is formed. And if I perform same thing with another substance, its constituents are formed..How? Energy can be converted to mass but a mass of what? Does that mean we can create matter of any desirable substance?
|
"how does energy perceive what atoms it has to form"
May i correct you here energy will not create new matter(or new atoms as you mentioned) in the case you mentioned. It will just increase the mass of the existing matter. For if you accelerates an electron from rest to a speed comparable to speed of light what you will get is the same electron with an increased mass. Similarly if you accelerates a stone to a speed comparable to $c$ the number of atoms in the stone will remain the same but the mass of each atom will be increased.
If you are interested in creating matter you should consider something like pair production. It would be more precise to use the word relativistic mass if you consider mass as an amount of matter
| {
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Why are roofs blown away by wind? Whenever there are high winds, such as in storms, thin metal roofs on sheds as well as concave roofs on huts are sometimes blown away.
One explanation provided to me said that the higher velocity of the air outside causes the air pressure above the roof to decrease and when it has decreased to a certain extent such that the air pressure above the roof is lesser than the air pressure beneath the roof and due to some kind of osmosis, the air particles move from the area of higher pressure (beneath the roof) to the area of low pressure. In this process, the roof is blown away.
Another explanation, specifically about the thin metal roofs, said that it was blown away due to the lift caused by the air and this is the same kind of lift you get when you blow on paper.
Both these explanations puzzle me.
What really bothers me is the basis of the first one, how can an increase in velocity cause pressure to drop? I can't seem to correlate that with the Force per unit area definition of pressure.
Please, oh great physicists of the internet, help me and every other ordinary person to understand how and why roofs get blown away.
| Refer Bernoulli's Theorem. Watch this video for demonstration http://dornsife.usc.edu/labs/lecture-support-lab/wind-storm/.
Brief explanation: When the velocity of the wind is great enough, the air pressure above the surface is lower compared to that underneath. This cause the roof to blow off. The aeroplane work in the same principle (lower pressure on top surface).
| {
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Question about heat engine efficiency I would like to know how much thermal energy is converted to kinetic energy in a steam engine, or a more efficient means if available.
I have done some research and found out that Carnot made some equations to determine efficiency. It is also mentioned that as part of the Carnot cycle it is important that the cold temperature reservoir be colder, but I don't understand how that would help increase efficiency of the conversion of thermal energy to kinetic energy. Here is a paragraph taken from Wikipedia at http://en.wikipedia.org/wiki/Carnot_cycle
"Lowering the temperature of the cold reservoir will have more effect on the ceiling efficiency of a heat engine than raising the temperature of the hot reservoir by the same amount."
| Look up something called the Carnot efficiency. That is the theoretical limit of how effecient any heat engine can be at converting heat power to some other form. This maximum possible efficiency is
Carnot efficiency = Tdiff / Thot = (Thot - Tcold) / Thot
By simple 8th grade algebra, you can see that you get a higher value by decreasing Tcold (the cold side temperature) than by increasing Thot (the hot side temperature) by the same amount.
For example, the Carnot efficiency of 100°C to 0°C is 100°K / 373°K = 26.8%. Adding 10 degrees to the hot side you get 110°K / 383°K = 28.7%, but decreasing the cold side by the same 10 degrees yields 110°K / 373°K = 29.5%.
| {
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How would I explain Ohm's Law in terms of Electrical Fields and Force? In terms of current, resistance, and voltage, it's easy: Ohm's Law is the relationship between current, voltage, and resistance of a circuit. Boom, simple as that. How could I put this in terms of $E$ and $F$? I can sort of see a way to do it by relating the formulas $E=F/q$ and $I=q/t$ to Ohm's Law, $V=IR$, but I'm not entirely sure how I could explain this in words.
| There are a number of ways you can examine the law in a microscopic view. One of them is this:
An applied voltage creates an electric field, which superimposes a small drift velocity on the free electrons in a metal conductor. This drift velocity is way smaller than the speed of transmission in a conductor.
Now, the basic relations are:
$$
I=\frac VR\\
J=I/A//
R=\frac{\rho l}A
$$
From the above, we can get:
$$
J=\frac V{RA}=\frac V{\rho l}=\frac {El}{\rho l}=\frac E{\rho}=E\sigma
$$
These relations can help you put the equation in terms of E, F or whatever else it is you want.
| {
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Why does $\ell=0$ correspond to spherically symmetric solutions for the spherical harmonics? In quantum mechanics why do states with $\ell=0$ in the Hydrogen atom correspond to spherically symmetric spherical harmonics?
| Suppose that there existed a spherically symmetrical wavefunction $\psi({\bf r})=f(r)$ for which $l\neq0$. This cannot be, for if we calculate $\langle \psi | L^2 | \psi \rangle$ we will always get zero, as each term in $L^2$ has derivatives with respect to $\theta$ and $\phi$.
Conceptually speaking, a spherically symmetric state gives the electron the option to be in orbit around any axis. In other words, it orbits around no axis.
| {
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Braiding statistics of anyons from a Non-Abelian Chern-Simon theory Given a 2+1D Abelian K matrix Chern-Simon theory (with multiplet of internal gauge field $a_I$) partition function:
$$
Z=\exp\left[i\int\big( \frac{1}{4\pi} K_{IJ} a_I \wedge d a_J + a \wedge * j(\ell_m)+ a \wedge * j(\ell_n)\big)\right]
$$
with anyons (Wilson lines) of $j(\ell_m)$ and $j(\ell_n)$.
One can integrate out internal gauge field $a$ to get a Hopf term, which we interpret as the braiding statistics angle, i.e. the phase gained of the full wave function of the system when we do the full braiding between two anyons:
$$
\exp\left[i\theta_{ab}\right]\equiv\exp\left[i 2 \pi\ell_{a,I}^{} K^{-1}_{IJ} \ell_{b,J}^{}\right]
$$
see also this paper and this paper.
I would like to know the way(s) to obtain braiding statistics of anyons from a Non-Abelian Chern-Simon theory? (generically, it should be a matrix.) How to obtain this braiding matrix from Non-Abelian Chern-Simon theory?
| The (unitary) "phase" factor for non-Abelian anyons satisfies the
(non-Abelian) Knizhnik-Zamolodchikov equation:
$$\big (\frac{\partial}{\partial z_{\alpha}} + \frac{1}{2\pi k} \sum_{\beta \neq \alpha} \frac{Q^a_{\alpha}Q^a_{\beta}}{z_{\alpha} - z_{\beta}}\big )U(z_1, ....,z_N) = 0 $$
Where $z_{\alpha}$ is the complex plane coordinate of the particle
$\alpha$ , and $Q^a_{\alpha}$ is the matrix representative of the $a-$th
gauge group generator of the particle $\alpha$ and $k$ is the level .
Please, see the following two articles by Lee and Oh (article-1,
article-2).
In the first article they explicitly write the solution in the case of
the two-body problem:
$$U(z_1, z_2) = exp( i\frac{Q^a_1Q^a_2}{2\pi k} ln(z_1-z_2))$$
The articles describe the method of solution:
The non-Abelian phase factor can be obtained from a quantum mechanical
model of $N$ particles on the plane each belonging possibly to a
different representation of the gauge group minimally coupled to a gauge field with a Chern-Simons term in the Lagrangian.
The classical field equations of the gauge potential can be exactly solved and
substituted in the Hamiltonian. The reduced Hamiltonian can also be exactly solved. Its solution is given by the action of a unitary phase factor on a symmetric wave function. This factor satisfies the Knizhnik-Zamolodchikov equation.
The unitary phase factor lives in the tensor product Hilbert space of
the individual particle representations. The wave function is a vector in this Hilbert space valued holomorphic function depending on the $N$ points in the plane.
| {
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Experimental evidence for non-abelian anyons? Since non-abelian anyons have become quite fashionable from the point of view of theory. I would like to know, whether there has actually been experimental confirmation of such objects.
If you could cite original literature, that would be great!
| Try http://arxiv.org/abs/1301.2639 (Phys. Rev. Lett. 111, 186401 (2013)): "Magnetic field-tuned Aharonov-Bohm oscillations and evidence for non-Abelian anyons at v=5/2" , although I am not sure this is a definite evidence.
"We show that the resistance of the v=5/2 quantum Hall state, confined to an interferometer, oscillates with magnetic field consistent with an Ising-type non-Abelian state."
| {
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When is quasiparticle same as elementary excitation, and when is not? Can anyone shed light on the comparison between these two concepts?
| In the context of condensed matter physics:
To make it short, and with the caveat that it is not a universally accepted definition, an elementary excitation may be called a quasiparticle if it is fermionic (e.g. dressed electron), and collective excitation if it is bosonic in nature (e.g. phonon, magnon).
But there is not clear cut and absolute divide between such terms, and you will not get into trouble for using them in a lax manner.
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Which surface to use in Ampere's law? In calculating the current enclosed by an Amperian loop, one must, in general evaluate an integral of the form
$$I_\text{encl} = \int \mathbf{J}\cdot\mathrm{d}\mathbf{a}$$
The trouble is, there are infinitely many surfaces that share the same boundary line. Which one are we supposed to use?
| The identity is correct for all of the infinitely many surfaces (isn't math amazing?).
In general you either have a surface you care about in the first place or are in a position to choose the easiest surface.
| {
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Why is a vacuum cleaner not as good heater as an electric radiator? I've read this question and answer: How efficient is an electric heater?
, but still don't understand.
If I have an electric radiator it heats the room with 1000 Watts of power. And I feel the room's getting warmer.
In contrast, if I turn on a vacuum cleaner which consumes 1000 Watts as well as the radiator, it doesn't seem to heat the room as well.
Why? Won't all kind of energy transform into heat ultimately?
| It is because you wouldn't hide in the corners like your kitty does!
A electric radiator is designed to be directional and therefore it doesn't heat the unnecessary part of your room. It makes you feel warming in front of it, but some part of the room don't get heated like those corner and the ceiling. In comparison, a vacuum cleaner heating the gas instead, so it is much more uniform, but less efficient from your point of view because you don't feel it (Your kitty might be happy about it though).
As what @JohnRennie, they dissipate the same amount of heat (and some become noise).
| {
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What gives an object its colour? My understanding of colour is that atoms in a particular object will absorb certain wavelengths of electromagnetic radiation, and the scattered wavelengths give the object its colour. The absorbed wavelengths contribute to lattice vibrations, increasing the kinetic energy and raising the objects temperature. Is this correct?
What is the sequence of events when a solution of particular atoms is sprayed through a Bunsen burner? I don't understand this. In this case is the colour seen a result of electrons moving up and then back down energy levels?
| Your coloured object is absorbing light, i.e. light is changing into mechanical energy, while the atoms in the Bunsen burner are emitting light, i.e. mechanical energy is changing into light.
If you have, for example, sodium atoms in a flame those atoms are continuously colliding with air molecules. The velocities of the air molecules are a function of temperature and are described by the Maxwell-Boltzmann distribution. At typical flame temperatures the average energies of gas molecules are well below the energies of electronic transitions in sodium, however the Maxwell-Boltzann distribution has a tail of very high velocities. The small percentage of gas molecules with high energies can collide with the sodium atoms hard enough to excite their electrons into a higher energy state. The atoms then relax and emit photons by spontaneous emission.
So in the flame the emission of the light is cooling the flame by turning the mechnical energy of the gas molecules into light. As you correctly say in your first paragraph, when light is absorbed it causes electronic transitions in the solid and these then transfer their energy into mechanical energy of the solid thereby heating it. The two processes are opposites of each other.
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Observations in the cathode ray tube experiement
1.One of the observations I learned was that the glass tube begins to glow with a brilliant green light. Many websites I read through refer to a fluorescent material. However, as shown in the above diagram there was no fluorescent material in the experiment carried out first on the cathode ray tube. So where does the green glow come from. Is this the color of the radiation itself?
2."Cathode rays travel in straight lines. That is why, cathode rays cast shadow of any solid object placed in their path. The path cathode rays travel is not affected by the position of the anode." I just can't seem to understand this explanation of the one of the observations.Also, different websites analyses this observation differently. For example, " The cathode rays consist of material particles because they produced shadow of objects placed in the way"
3.Two of the conditions of the experiment were air at very low pressure and secondly a very high potential difference. Could someone please tell me why these conditions were necessary?
I know the questions are very silly but because different websites refer to different things, I am becoming confused with something that should be simple to understand.
| Notice the "Air at very low pressure"? That thin air is what glows. And it has to be very thin or it disrupts the "ray" nature of the phenomena. I'm not sure what exactly was glowing in the earliest experiments, but it might well be the $\mathrm{N}_2$.
Later it was normal to put some mercury in the tubes because it works really well. In fact, that is what a florescent light is: a mercury vapor lamp (generally with some phosphors on the glass to get a nicer spectrum).
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Derivation of Lagrangian density for an infinite classical dielectric in interaction with the EM field I am tasked with reading and reproducing all the steps in J.J. Hopfield's 1958 paper "Theory of the Contribution of Excitons to the Complex Dielectric Constant of Crystals". Embarrassingly I am stuck on equation (3). Hopfield takes the Lagrangian density for an infinite classical dielectric in interaction with the electromagnetic field to be
Instead of taking this equation at face value, I would like to fully derive this Lagrangian density, but am unsure how to approach the problem. I have found many articles and sites tackling a similar problem, but never ending up with an expression as here. Any pointers on how to approach the problem or links to material that could be useful would be very much appreciated.
| I will set constants like $c$ equal to one.
Then he starts with the normal relativisitc lagrangian, $\mathcal{L} = -\frac{1}{4}F^{\alpha \beta} F_{\alpha \beta} - A^\alpha J_\alpha$. Translating this into non-relativistic language, we get $\mathcal{L} = \frac{1}{2}(E^2 - B^2) - \phi \rho + \mathbf{A} \cdot \mathbf{J} $.
Now at this point he seems to assume no free charges or currents, and no magnetization. Therefore the microscopic charge density is the bound charge density $\rho_b = -\mathbf{\nabla} \cdot \mathbf{P}$, and the microscopic current density is related to changes in the polarization: $\mathbf{J} = \mathbf{J}_b = \partial_t \mathbf{P}$ Then $\mathcal{L} = \frac{1}{2}(E^2 - B^2) + \phi \mathbf{\nabla} \cdot \mathbf{P} + \mathbf{A} \cdot \partial_t \mathbf{P}$.
Recognizing $\mathbf{E} = -\partial_t \mathbf{A} - \mathbf{\nabla} \phi$ (this is off by a minus sign from what he says, I don't know why) and $\mathbf{B} = \mathbf{\nabla} \times \mathbf{A}$ we now have $\mathcal{L} = \frac{1}{2}((\partial_t \mathbf{A} + \mathbf{\nabla} \phi)^2 - (\mathbf{\nabla} \times \mathbf{A})^2) + \phi \mathbf{\nabla} \cdot \mathbf{P} + \mathbf{A} \cdot \partial_t \mathbf{P}$
This gives us the first two and last two terms of his expression up to factors of $4 \pi$ coming from I think gauss's law, which I have been ignoring.
Now I do not get his middle two terms. I have not read the paper. What are $\beta$ and $\omega_0$?
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A question about Fermi-Dirac Distribution function It seems more like a mathematical question, about the property of Fermi-Dirac Distribution function $$f=\frac{1}{e^{(E-\mu)/k_BT}+1}$$
where $\mu$ is the chemical potential and $k_B$ is the Boltzmann constant.
I find that $\frac{\partial^nf}{\partial T^n}|_{T\to0}=0$, for any positive integer $n$. That is true for either $T\to0^+$ or $T\to0^-$.
This seems that we are unable to taylor expand $f$ near $T=0$. Or, say, we are unable to use any function of $T$ to approximate the Fermi-Dirac Function according to the order of T near $T=0$ point.
Are there any physical meaning or application of this property? Why nature gives this property to the widely used Fermi-Dirac Function?
|
Are there any physical meaning or application of this property? Why nature gives this property to the widely used Fermi-Dirac Function?
The property you've found (Taylor expansion $\neq$ original function on any interval around the point) has nothing to do with physics or nature, and is not particularly connected only to the Fermi-Dirac distribution. The failure may be limited only to some special points; if some expansion is really needed regardless of this, often you can expand the function around some other point, in the present case, say $T= 1~$K.
One often encounters functions that are equal to their Taylor expansion only on a small interval (disc if the function is complex), or only at one point (the point around which the expansion was made). Such functions have their place in physics. For example, to express that property of a signal that its intensity at time B is not determined by its properties at a different time A, on the interval A-B the signal has to be described by function whose Taylor expansion around A cannot reproduce its value at B.
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How to calculate velocities after collision? I'm currently writing a program for a particle simulator. One of the requirements is that the particles collide in a realistic way. However, I don't know how to calculate the final velocities.
For each collision, I have the $x$-component and $y$-component of each velocity, as well as the displacement and mass of each particle.
Is it possible to calculate the direction and magnitude of their velocities after the collision? If so, how?
| Note: The simplest method of calculating collisions use some method of applying impulses to objects that collide at some point in time. The simplest algorithm is to test if two objects are intersecting at some point in time, and if they are you fix it so that they stop intersecting each other, and then apply the calculated impulses on each object. However, this isn't completely correct. You really need to find out which collisions happened first, step the whole simulation back to the moment of the first collision, and calculate the new velocities right then. This works correctly if you have many momentary and interdependent collisions.
In the worst case, if the particles are dense and come into contact with each other, the easiest method I've found (which is used in this simulation and which I used to handle collision in this simulation) is to use forces instead of impulses. Again you reduce it to a 1-dimensional problem, and have the force of interaction between two particles be:
$$f=-k d-\mu v$$
where $\mu$ and $k$ are constants, $-k d$ is a Hookean spring repulsive force, $v$ is the relative velocity of the two particles along their axis of collision, and $-\mu v$ is a viscous friction force.
I believe that if you want more accurate contact physics (with accurate sliding point contacts instead of a spring force) you'll have to use a much more sophisticated method.
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Increase in velocity by loss of mass?
A trolley of mass 300kg carrying a sand bag of 25kg is moving uniformly with speed of 27km/h on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kg/s.
What is the speed of the trolley after the entire sand bag is empty?
I was so surprised when I read this question. It doesn't make sense to me. I can't comprehend how the loss of sand creates an external unbalanced force on the trolley such that it affects its velocity.
Maybe I haven't analyzed the question enough but I find this a bit conceptually challenging for me. Maybe I have to consider how the sand particles affect the back wheels of the trolley or maybe consider the sand to be a propellant?
| The answer is... 27 km/h.
It is a trick question, the net force on the trolley is always zero.
People might be tricked into blindly applying momentum conservation to find an increase in velocity but this would be incorrect. As the sandbag decreases in weight the momentum carried by the trolley-sandbag system decreases.
| {
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Quantum eraser double slit experiment In the quantum eraser double slit experiment, does the photon (or wavefunction) pass through one slit or both slits when different polarizers are placed over the slits?
| According to quantum physics, when certain different polarizers are placed over the slits in the double-slit experiment (for instance, one vertical and one horizontal polarizer, or one circular clockwise and one circular counter-clockwise), thus "marking" each photon with which-way information, the photon indeed passes through only one slit, resulting in no interference. If you cover up one of the slits, you'll observe the very same absence of interference.
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Stacking Shelves with Overhang I want to stack some boxes which are 14" x 10" with some 12" records inside. This means that there will be a 2" overhang outside of each box.
I know that to have an item to balance on a edge 50% of an item needs to be supported. So I think I'm safe to load the boxes in this way.
If I were to stack boxes loaded in this way on top of each other is there a limit to how high they can go?
| While the standard answer to this problem is usually given in terms of harmonic series,
(see for instance this page at MathWorld) which results in stack looking like this:
I would like to share the nonstandard answer I found following one of the MathWorld links.
A whole new class of solutions was found in the paper by Paterson and Zwick:
Paterson, Mike, and Uri Zwick. Overhang. Proceedings of the seventeenth annual ACM-SIAM symposium on Discrete algorithm. ACM, 2006. arXiv:0710.2357.
The paper is quite accessible, so I suggest to have a look, but the main idea
could be easily understood by looking at some of the solutions for various number of shelves:
So, by discarding assumption that only one shelf could be placed per level we
are able to produce much more solutions: overhang now scales like $c\,n^{1/3}$ with number of shelves, instead of $\frac 12 \ln n$ for harmonic series solution.
| {
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"url": "https://physics.stackexchange.com/questions/91472",
"timestamp": "2023-03-29T00:00:00",
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Transforming a lagrangian to hamiltonian and vice versa I am not refering to Legendre transform, but to something more simple.
In analytical mechanics, the Lagrangian can be described as $L=T-V$, and the Hamiltonian is if the Lagrangian doesn't explicitly depend on time, then $H=T+V$.
There a simple change of functions which I am contemplating here, basically if I write:
$U=i \sqrt{V}$, the the Lagrangian becomes: $L=T+U^2$, and the Hamiltonian becomes $H=T-U^2$.
I know it looks like meaningless, but also going from Minkowskian metric from Euclidean metric and vice versa doesn't seem like such a big deal to me, but physicist use it.
So is this change of variables between Lagrangian and Hamiltonian being used in theoretical physics?
Does it have any meaningful applications?
|
$U=i \sqrt{V}$, the the lagrangian becomes: $L=T+U^2$, and the hamiltonian becomes $H=T-U^2$.
Many Lagrangians/Hamiltonians are not of the $T\pm V$ form. If there are velocity dependent potential terms or similar, this breaks down and you have to use a Legendre transform to switch between the two.
For more complicated systems (such as the ones considered these days), it may not even be immediately evident (without carrying out a Legendre transform) if the Lagrangian/Hamiltonian can be written as $T\pm V$. Since this method involves first verifying that the system is of the $T\pm V$ form, which requires a Legendre transform, I don't really see this method being of any use without the Legendre transform.
In the end, the Langrangian and Hamiltonian are different quantities with different origins, behaviors, and differential equations. It is unreasonable to expect all but the simplest to follow from addition.
| {
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Is it possible to "cook" pasta at room temperature with low enough pressure? It is known fact, that boiling point of water decreases by decreasing of pressure. So there is a pressure at which water boils at room temperature.
Would it be possible to cook e.g. pasta at room temperature in vacuum chamber with low enough pressure?
Or "magic" of cooking pasta is not in boiling and we would be able to cook pasta at 100°C without boiling water (at high pressure)?
| If you hydrate pasta by keeping it in cold water, it won't be 'cooked' from a technical standpoint. Cooking requires heat (or the chemical equivalent, as in ceviche). It will taste like wet, raw flour.
| {
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Realistic calculation of heat loss for pipe Good day everyone,
I am new on this site and I hope to find here help, since I am not going anywhere with the literature I have found.
I try to calculate realistically the heat loss of a hot, uninsulated pipe. Let's say,
it is $170\,\,C$, 1 meter long, $8" (=0,203\,\,m)$ of diameter. I want to calculate the whole losses for the situation of no wind (only normal convection) and $20C$ of outside temperature.
The $170C$ is the temperature at the outside surface of the pipe, the surface is known, I also know the Stefan–Boltzmann constant ($\sigma$) and I take $\alpha=5$ for the convective coefficient of air at the surface of the pipe as well as $\epsilon=0.85$ for the steel pipe emissivity.
My problem is, that in literature, some parameter is always being rejected as negligible due to the fact, that most of the times the example is some kind of exercise of a heat transfer class. I want to calculate the real thing and then decide what is negligible and what is not.
So my main question is: do I just add the radiation losses to the convective losses of the pipe ?
$$Q_{loss} = Q_{conv}+ Q_{rad}$$
$$= (\alpha*Α*\Delta T) + \epsilon*Α*\sigma*((T_{pipe})^4-(T_{air})^4) \tag{with T in K}$$
According the above, I get $Q_{conv} = 479\,\,W$ and $Q_{rad} = 958\,\,W$.
Is there any mistake in my way of thinking or is it truly that simple ?
Thanks in advance.
Marcus
| According to http://www.engineeringtoolbox.com/steel-pipes-heat-loss-d_53.html , the heat losses are close to what you get. I'd note that the convective coefficient of air is rather tricky: it depends even on the orientation of the pipe (horisontal/vertical).
| {
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General relativity in terms of differential forms Is there a formulation of general relativity in terms of differential forms instead of tensors with indices and sub-indices? If yes, where can I find it and what are the advantages of each method? If not, why is it not possible?
| differential forms. It's not 100% clear how to read the question so I'll fire in all directions.
In general relativity, the field of interest is the symmetric metric tensor $g$ which you see written $g_{\mu\nu}$ or e.g. $g_{ab}$. The latter often implies abstract index notation.
The base vectors on the cotangential space are $\mathrm dx^i$ so there you have forms, but it's not like you want to write the symmetric tensors like the energy-momentum tensor $T$ as image of $\mathrm d$. You can drop the indices from objects like $g_{\mu\nu}$ any time you want. E.g. You can write down Einsteins field equations as $G=\kappa\ T$ and then it's in your face covariant per definition.
In this answer someone rants about why you wouldn't want to dismiss index notation anyway. (But the person has governed zero reputation on this site yet, so I would take his opinions with a grain of salt.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Is it possible to estimate the speed of wind by the sound emitted by a cable of an overhead power line? I was near ($\approx40m$) an overhead power line and I heard a sound coming from the cables of the power line; I think the sound was made by the vibrations of the power cables due to the wind but I am not sure. The wind was very light.
The sound was not the "buzz" asked about here.
My question is: assuming the sound was generated by the wind, is it possible to estimate the speed of wind from the sound properties (i.e. its spectrogram) and the mechanical properties of the cable?
If yes, how accurate will be the estimate?
If yes, can you provide some back-of-the-envelope calculation?
| The sound by the cable is produced because of the Kármán vortex shedding.
This empirical formula from the wikipedia page relates the frequency of the vortex shedding with the Reynolds number:
$$
\frac{fd}{V}=0.198\left (1-\frac{19.7}{\mathrm{Re}}\right),
$$
where $f$ is the frequency, $d$ cable diameter and $V$ is the flow velocity.
The Reynolds number $\mathrm{Re}$ in turn is defined for this system as $$\mathrm{Re}=\frac{Vd}{\nu},$$
where $\nu$ is the kinematic viscosity of the medium. For the air at $15\,{}^\circ \text{C}$ it is $1.48\times 10^{−5}\,\text{m}^2/\text{s}$.
So solving the equations for the velocity $V$ we obtain
$$
V = 5.05 \left(f d + 3.90 \frac{\nu}{d}\right).
$$
Of course, this formula implies idealized conditions, so in a more realistic situations (including for instance turbulence in the wind flow) extracting the vortex shedding frequency from the sound spectrum could be tricky.
| {
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"timestamp": "2023-03-29T00:00:00",
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How can space and time arise from nothing? Lawrence Krauss said this on an Australian Q&A programme.
"...when you apply quantum mechanics to gravity, space itself can arise from nothing as can time..."
Can you elaborate on this please?
It's hard to search for!
| This is not so much an answer as a comment and caution: ex nihilo nihil fit (from nothing nothing is produced).
From Wikipedia:
It is important, however, to recognize what a physicist may mean by
the word nothing. Some physicists, such as Lawrence Krauss, define
nothing as an unstable quantum vacuum that contains no particles.
This is incompatible with the philosophical definition of nothing,
since it can be defined by certain properties such as space, and is
governed by physical laws. Indeed, many philosophers criticize these
physical explanations of how the universe arose from nothing, claiming
that they merely beg the question.
| {
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What exactly is a bound state and why does it have negative energy? Could you give me an idea of what bound states mean and what is their importance in quantum-mechanics problems with a potential (e.g. a potential described by a delta function)?
Why, when a stable bound state exists, the energies of the related stationary wavefunctions are negative?
I figured it out, mathematically (for instance in the case of a potential described by a Delta function), but what is the physical meaning?
| If you have a copy of Griffiths, he has a nice discussion of this in the delta function potential section. In summary, if the energy is less than the potential at $-\infty$ and $+\infty$, then it is a bound state, and the spectrum will be discrete:
$$
\Psi\left(x,t\right) = \sum_n c_n \Psi_n\left(x,t\right).
$$
Otherwise (if the energy is greater than the potential at $-\infty$ or $+\infty$), it is a scattering state, and the spectrum will be continuous:
$$
\Psi\left(x,t\right) = \int dk \ c\left(k\right) \Psi_k\left(x,t\right).
$$
For a potential like the infinite square well or harmonic oscillator, the potential goes to $+\infty$ at $\pm \infty$, so there are only bound states.
For a free particle ($V=0$), the energy can never be less than the potential anywhere***, so there are only scattering states.
For the hydrogen atom, $V\left(r\right) = - a / r$ with $a > 0$, so there are bound states for $E < 0$ and scattering states for $E>0$.
Update
*** @Alex asked a couple questions in the comments about why $E>0$ for a free particle, so I thought I'd expand on this point.
If you rearrange the time independent Schrödinger equation as
$$
\psi''= \frac{2m}{\hbar^2} \left(V-E\right) \psi
$$
you see that $\psi''$ and $\psi$ would have the same sign for all $x$ if $E < V_{min}$, and $\psi$ would not be normalizable (can't go to $0$ at $\pm\infty$).
But why do we discount the $E<V_{min}=0$ solutions for this reason, yet keep the $E>0$ solutions, $\psi = e^{ikx}$, when they too aren't normalizable?
The answer is to consider the normalization of the total wave function at $t=0$, using the fact that if a wave function is normalized at $t=0$, it will stay normalized for all time (see argument starting at equation 147 here):
$$
\left<\Psi | \Psi\right> = \int dx \ \Psi^*\left(x,0\right) \Psi\left(x,0\right) = \int dk' \int dk \ c^*\left(k'\right) c\left(k\right) \left[\int dx \ \psi^*_{k'}\left(x\right) \psi_k\left(x\right)\right]
$$
For $E>0$, $\psi_k\left(x\right) = e^{ikx}$ where $k^2 = 2 m E / \hbar^2$, and the $x$ integral in square brackets is $2\pi\delta\left(k-k'\right)$, so
$$
\left<\Psi | \Psi\right> = 2\pi \int dk \ \left|c\left(k\right)\right|^2
$$
which can equal $1$ for a suitable choice of $c\left(k\right)$.
For $E<0$, $\psi_k\left(x\right) = e^{kx}$ where $k^2 = - 2 m E / \hbar^2$, and the $x$ integral in square brackets diverges, so $\left<\Psi | \Psi\right>$ cannot equal $1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Meaning of the chemical potential for a boson gas My lecturer told me that $\mu$, the Chemical potential, is zero or negative, and in the following example, mathematically it acts as a normalization constant. But is there any physical insight about why boson gas can be zero or negative?
I think it is due to the fact the photon gas can pop up from nowhere (i.e. vacuum fluctuation).
$$
f_{BE}(\varepsilon)=\dfrac{1}{e^{(\varepsilon-\mu)/(k_B T)}-1}$$
| To be a little more precise: the chemical potential of a non-interacting Bose gas must be exceed the energy of the ground state single particle energy of that gas. If there are (as say, in $^4\rm He$) repulsive interactions between the particles, the chemical potential can be anything.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/92314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 3,
"answer_id": 2
} |
At what point is the spin determined in a Stern-Gerlach Apparatus Consider a particle with spin that travels through a Stern Gerlach box (SGB), which projects the particle’s spin onto one of the eigenstates in the $z$-direction. The SGB defines separate trajectories for the particles that travel through it depending on their spin.
My Question: At what point is the spin determined when it is in superposition? When the particle starts to feel the magnetic field? Or only when the trajectory is measured in the detector?
This is a similar question, however it does not answer my question.
| The spin is determined when observed or measured. At this point the particle must take only one position. A big issue in experimental physics is that when you observe or measure something, you are actually changing it yourself for example when observing a stream of electrons, you are exerting photons onto it, thus changing the electrons' velocity, position etc. The spin is dependant on an environmental factor i.e measurement.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/92389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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If a Goldstone boson is an excitation moving between degenerate vacua, how do symmetries remain broken? In spontaneous symmetry breaking, moving around the circular valley of the Mexican hat potential doesn’t change the potential energy. These angular excitations are called Goldstone bosons. But doesn't the change of angle implies that the system moves from one vacuum to another because different points on the circular valley represent degenerate vacua?
If Goldstone excitations are like this, how does the symmetry remain broken? Goldstone excitation, by definition (because they represent variations in the coordinate on the circular valley), will then take the system from one vacuum to another. However, this doesn't happen.
| Not sure I understand the question. The symmetry is broken because you're in the valley. As you say correctly, moving around in the valley doesn't change anything about that.
| {
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