So this is one of the questions on my physics assessment and I would like to know how to calculate how much work is done when pushing against a brick wall.
My teacher has told me the calculation is: weight(kg)x10 (this gives me the force)x distance.
But what is the weight of and the distance of? As when you push against a brick wall the wall doesn't move and you can stop yourself from moving.
Please answer as soon as possible, also I'm kind of younger so could you explain clearly(sorry)
Emsee
In quantum entanglement when something acts on one particle the other one reacts also, just in reverse (more or less). From what I've read though, anything acting on either particle will collapse the entanglement, right? So how do we know they were ever linked? Or is it just measurements that collapse it?
The reason I ask is because given the public impression of the topic it would suggest on of two things. That the information causing the reaction is superluminal or that the particles are occupying the same space since they are in different states. If the second were true then distance must be a human construct and somehow they must still be in the same place regardless of the virtual distance between them.
To clarify, I don't mean information in regards to communication such as QC. Just particle information.
The retarded electric field is given by: $$\mathbf E= \frac{q}{4\pi\epsilon_0}\left[\frac{e_{r'}}{r'^2} + \frac{r'}{c}\frac{d}{dt} \left(\frac{\mathbf e_{r'}}{r'^2}\right) + \frac{1}{c^2}\frac{d^2}{dt^2} e_{r'}\right]$$
Regarding the second term, Feynman remarks:
[...] The other terms tell us that the laws of electricity do not say that all the fields are the same as the static ones, but just retarded (which is what people sometimes like to say). To the “retarded Coulomb field” we must add the other two terms. The second term says that there is a “correction” to the retarded Coulomb field which is the rate of change of the retarded Coulomb field multiplied by $r′/c$, the retardation delay. In a way of speaking, this term tends to compensate for the retardation in the first term. The first two terms correspond to computing the “retarded Coulomb field” and then extrapolating it toward the future by the amount $r′/c$, that is, right up to the time $t\;!$ The extrapolation is linear, as if we were to assume that the “retarded Coulomb field” would continue to change at the rate computed for the charge at the point (2′). If the field is changing slowly, the effect of the retardation is almost completely removed by the correction term, and the two terms together give us an electric field that is the “instantaneous Coulomb field”—that is, the Coulomb field of the charge at the point (2)—to a very good approximation.
I'm having problem in comprehending what Feynman wants to say. There is no doubt that the first term is retarded Coulomb field; but what about the second term?
Feynman tells, it is a correction, a compensation to the retardation of the first term.
My questions are:
$\bullet$ Why do we need a correction, a compensation to the retardation of the first term?
$\bullet$ How does the second term make the correction? How does it compensate the retardation?
Okay, I got the point that the second term, being a derivative, when multiplied by the retardation time, gives the net change in electric field at that point during the retardation time. so, when I add it to the first term, I get the field at that point at time $t\;_,$ isn't it?
Thus, the two term adds to give the electric field at $t\;.$
But, since cause precedes effect, wouldn't the electric field at time $t$ be caused by the charge at $t-r/c\;?$ But the two term seems to give the condition at $t$, not $t-r/c\;.$ Doesn't this violate causality?
I am really missing something and messing the retarded field; but wouldn't the electric field at time $t$ be caused by the charge configuration at time $t-r/c\;?$ But the second term in the definition of the electric field adds the change in electric field in $r/c$ to the first term to give the configuration of the charge at time $t$ and not $t-r/c$ - this violates causality as the field at $t$ would be caused by the charge at $t-r/c\;.$
So, can anyone please explain where I'm mistaking?
Answer
the second term, being a derivative, when multiplied by the retardation time, gives the net change in electric field at that point during the retardation time. so, when I add it to the first term, I get the field at that point at time $t\;_,$ isn't it?
No. If the retarded Coulomb changed linearly in time, then the first two terms added together would give you the Coulomb now. Instead they merely together give a linear time extrapolation. Like using $x(0)+tv(0),$ in general it isn't giving you the current value. Only when it actually changed linearly in time.
But, since cause precedes effect, wouldn't the electric field at time $t$ be caused by the charge at $t-r/c\;?$
If you did that to the potential then all would be fine.
But the two term seems to give the condition at $t$, not $t-r/c\;.$ Doesn't this violate causality?
I'm not sure anyone knows what you mean. You are evaluating the field at time $t$ and it is based on the prior times; isn't that exactly what the the expression is saying? Isn't everything in the expression evaluated at the retarded time and the result is the current field?
wouldn't the electric field at time $t$ be caused by the charge configuration at time $t-r/c\;?$
It is. And you either base in on the position, velocity and acceleration at the retarded time, or you base it on those three terms of Feynman which are based on other things, but things also evaluated at the retarded time.
this violates causality as the field at $t$ would be caused by the charge at $t-r/c\;.$
Firstly, the field is caused by the charge at the retarded time. That's what you want. And Again, you are just making a mathematical error on your part. The derivative is not a magic thing that can see what you average rate of change over some future tine interval is going to be. It's like an instantaneous velocity versus an average velocity. The instantaneous velocity is just about the now and doesn't care if tomorrow you start moving at half the speed of light. But if you talked about your average velocity over the next coming week, that would care if you start moving at half light-speed starting tomorrow. But derivatives are instantaneous changes, not average changes.
The derivatives in Feynman's equation are all instantaneous things. They don't know what happens later. In particular they don't know if someone is going to grab that charge tomorrow and stick it on a train going at half light speed. They are just about what the charge is doing at that one instant in retarded time. And that's where and when the derivative is taken.
Let's start by noting that Coulomb is wrong. When charges far away accelerate at time t-r'/c then you feel a totally different field than the Coulomb field. And sure, when something is moving at a constant velocity then in another frame they are at rest, and when at rest. Then you expect Coulomb. So since you expect Coulomb when at rest: go ahead and include a term for that. And you expect a different frame to get a Coulomb force when it moves at constant velocity, so include a term for that so that the first two terms give that force. And finally those terms still aren't enough when it accelerates so add another term to get it to actually match experiments.
Matching experiments is the key, and its what we want to do ultimately.
this is my mental picture for how they travel without a medium, how (like water waves) some can't stay still, why they have wave and particle properties, energy/mass equivalence, conservation, etc. it might capture uncertainty too -- i've heard that all waves have an uncertainty relation (say in their power spectrum), but i don't get why -- it seems like we can discuss waves with absolute precision.
Answer
Actually that's not too far off the mark, although I'm not sure "knot" or "kink" is the best word. Quantum field theory, the best theory we currently have to describe particles, says that particles correspond to excitations of a field, which are kind of like waves in water; you could consider the surface of a pond "excited" whenever it's not flat. Just as with water waves, there's an infinite variety of "shapes" you can have for these excitations. For example, you could have a repeating wave, in which the surface of the water cycles up and down over a large area, or you could have just one wave front that just propagates across the water without spreading out very much. The former case is pretty typical for things like light waves, and the latter case is pretty typical for particles of matter, although technically any kind of field (whether it's the electromagnetic field for light, or a quark field in matter, or whatever) can have any of the different types of excitations.
By the way, according to special (and general) relativity, even an object that is standing still is moving through time. So all of these excitations move through spacetime in one way or another. But only certain ones (the excitations in fields corresponding to massless particles) can move through space in such a way that they appear to us to be traveling at the speed of light.
I am trying to understand why normal shocks are caused in Converging Diverging Nozzles. My textbook explains that since the exit pressure must match the back pressure, the normal shock is the only mechanism that can make this happen, which I understand.
However, why does the exit pressure have to match the back pressure? What if it did not? Then simply the jet of fluid exiting the converging diverging nozzle would expand or contract under the pressure driving force induced with the back pressure.
Any clarifications welcome.
Answer
Why does the exit pressure have to match the back pressure?
Source: Converging/Diverging Nozzle Applet.
Gas flows from the chamber into the converging portion of the nozzle, past the throat, through the diverging portion and then exhausts into the ambient as a jet. The pressure of the ambient is referred to as the 'back pressure' and given the symbol p$_b$.
Oversimplified: The back pressure is set to match the exit pressure to avoid turbulence and inefficiency, much as an aircraft wing is more efficient when it is a particular shape; though a variety of less efficient shapes are also sufficient to allow, for example, toy planes to fly. In the case of a rocket one can control the burn to cause the flow to operate in the most efficient zone, much as a diesel engine operates better at a particular rpm. In the case of a rocket it is not gearing in the nozzle but turbulence, though gearing in the pumps becomes a consideration with pumped liquid fuels.
[Non-simplified] You can only lower the back pressure to a certain point, after which the nozzle becomes choked; no further lowering of the back pressure, even to a vacuum, will increase the flow.
As the pressure increases, either when burning solid fuel or liquid or simply pumping water, the shock waves act upon the insides of the nozzle creating a non-optimal flow and wear.
At the design condition the back pressure should equal the pressure at the nozzle exit. In this case, the waves in the jet disappear altogether (figure 3f), and the jet will be uniformly supersonic.
What if it did not?
If you lower p$_b$ enough you can extend the supersonic region all the way down the nozzle until the shock is sitting at the nozzle exit (figure 3d). While that provides a good flow velocity within the nozzle the flow in the jet (outside the nozzle) will still be subsonic.
Lowering the back pressure enough so that it is now equal to the pressure at the nozzle exit causes the waves in the jet disappear altogether (figure 3f), and the jet will be uniformly supersonic. This situation is referred to as the 'design condition'.
Assuming that I learn Quantum Mechanics first, and then I approach Classical Mechanics as a special case of Quantum Mechanics, I will definitely find the relationship between Quantum Mechanics and Classical Mechanics very confusing. I don't know how to make sense of what happens when $\hbar \rightarrow 0$.
For one, you can't recover classical mechanics from quantum theory by setting $\hbar \rightarrow 0$. However, it is possible to recover classical mechanics from Schrodinger equations.
So, does limit $\hbar \rightarrow 0$ in Quantum Mechanics mean anything? How should we interpret it? Or does the above contradiction reveal yet another flaw in the fundamentals of Quantum Mechanics?
Answer
The question duplicates $\hbar \rightarrow 0$ in QM where it has been soundly answered. The most direct bridge is through deformation quantization, the phase-space formulation of QM, where operator observables are mapped injectively onto their Wigner transforms, c-number phase-space functions, just like their classical counterparts. It is then evident that QM laws are deformations of the classical laws, and include those in their $O(\hbar^0)$ pieces. So, for example, the Wigner transforms of the density matrix, the Wigner functions, go to Dirac deltas in phase space in that $\hbar \rightarrow 0$ limit. These δs, in turn, transcribe to the Liouville theorem for single particles, and hence Hamilton's dynamics, out of Heisenberg's equations of motion. The Ehrenfest theorem structures of course replicate and parameterize this injection. Note this is a mere shift in point of view: $\hbar \rightarrow 0$ means considering (macroscopic) phenomena at much larger scales, which thus dwarf the scale of $\hbar$ and make it insignificant. It is mere sloppy shorthand of the emergence of our classical world out of QM at large scales (the correspondence principle).
For the limit to make sense, however, you must always consider angular momenta and actions S much larger than $\hbar$: it makes no sense to consider $\hbar \rightarrow 0$ limits for a single particle of small actions and spins---$\hbar$ is a dimensionful quantity, and its scale matters. I gather there is no interest in microscopic physics without $\hbar$.
The least tortured limit then, is, as above, consideration of the $O(\hbar^0)$ part of macroscopic quantities. For details, see, e.g., Ref. 1. For instance, for a Freshman lab oscillator with maximum oscillation amp 10cm, m=10g, and ω=2Hz, the characteristic action is S=E/ω= $10^{-4}$ Js and so $\hbar / S= 10^{-30}$, suppressing any and all quantum terms and validating the limit. (One might be interested in thinking about the Wigner function for the nth excited state for n ~ $10^{30}$, a very spikey cookie-cutter function, indeed!)
References:
Some speculation here:
http://www.youtube.com/watch?v=_NMqPT6oKJ8
Is there a possibility it would pass 'undetected' through your hand, or is it certain death?
Can you conclude it to be vital, or only loose your hand?
Would it simply make a small cylindrical hole through your hand, or is there some sort of explosion-effect?
Assume your hand has a cross section of 50cm², and a thickness of 2cm, how much of the beam's energy would be transferred to your hand?
Answer
A mis-steered beam at CEBAF simply cut a hole thought the niobium wall of the cavity and flooded half the accelerator with helium (super-conducting cavities need a liquid helium jacket to work...).
We were down for more than a week.
That is an electron beam machine, and very high current (up to $400\,\mathrm{\mu A}$!), so the details would be rather different than the LHC beam.
Likewise, about $60\,\mathrm{\mu A}$ of $5.5 \,\mathrm{GeV}$ beams from that machine partially melted one of my iron targets (thick enough that about 6% of the beam interacted with the target) despite a raster spreading the beam over roughly $2\,\mathrm{mm}^2$, because we didn't have good enough thermal contact with the water-cooled frame of the target ladder.
High energy, high current beams can carry a lot of power.
Back to the question as asked:
Treat your hand as water. The particle data booklet puts the energy loss per proton at around $2.5\,\mathrm{MeV/g/cm}^2$ or something like $4$–$8\,\mathrm{MeV}$ though your hand, depending on how chubby you are. (We're only a few orders of magnitude above the minimum ionization energy, so this is not very sensitive to the actual beam energy.)
Phillip says $1.2 \times 10^{11}$ protons per beam in the ring, about 11,000 passes per second ($3.0 \times 10^8\,\mathrm{m/s} / 27\,\mathrm{km}$), so $1.3 \times 10^{15}$ protons per second is $8 \times 10^{15}\,\mathrm{MeV/s} = 1300\,\mathrm{J/s}$ is a fair bit of heat, and results in heating of about $300\,\text{K/s/(cubic cm exposed)}$. To finish up here we will have to know something about the beam diameter.
Your reflex time to move your hand is on order of $0.1$–$0.2$ seconds.
It's going to hurt: you will get badly burned, and the damage will extend though the whole depth of the exposed flesh, rather than being limited to the surface as with the contact burns we are all familiar with.
recently i was watching a video on quantum computing where the narrators describes that quantum entanglement information travels faster than light!
Is it really possible for anything to move faster than light? Or are the narrators just wrong?
Regards,
Answer
Collapsing an entangled pair occurs instantaneously but can never be used to transmit information faster than light. If you have an entangled pair of particles, A and B, making a measurement on some entangled property of A will give you a random result and B will have the complementary result. The key point is that you have no control over the state of A, and once you make a measurement you lose entanglement. You can infer the state of B anywhere in the universe by noting that it must be complementary to A.
The no-cloning theorem stops you from employing any sneaky tricks like making a bunch of copies of B and checking if they all have the same state or a mix of states, which would otherwise allow you to send information faster than light by choosing to collapse the entangled state or not.
On a personal note, it irks me when works of sci-fi invoke quantum entanglement for superluminal communication (incorrectly) and then ignore the potential consequences of implied causality violation...
I've read that you won't get electrocuted if you jump and touch an electric fence because you aren't closing the circuit with the ground. Which is also why birds don't get electrocuted when they're standing on power lines.
The way I understand voltage is that it's basically "a pressure of electrons" in the wire where electrons would like to escape from each other because they are the same charge, and they would love to reach the ground where there is plenty of space (i.e "low pressure"). When you touch an electric fence, the electrons see an opportunity to reach the ground (to "fill the space") through you and you get electrocuted because your body resists the current which makes heat.
Alright, but I don't understand why don't electrons want to fill up your body even you're not touching the ground? Especially at high voltages; I mean there is plenty of space in your body. So when you jump and touch a fence, even though you're not touching the ground, why don't electrons see an opportunity and rush to fill up your body and cause you a shock?
To put it another way, let's say a bird lands on the ground and discharges all the excess electrons (becomes neutral with the ground). Now let's say the bird flies off and lands on a power line. Why don't electrons in the wire rush to fill up the bird and electrocute it?
Please help me understand.
Answer
Electrons do "fill up your body" when you jump up and hit a high voltage wire - there is a property called the capacitance of the body that determines how much the voltage increases when you add a certain amount of charge - mathematically, $C = \frac{Q}{V}$.
But it's not charge that kills you, it is current: charge flowing per unit time. Since it takes relatively few electrons to bring the body up to 30,000 V or so, there is not much charge flowing and nobody gets killed. But you may have noticed a static "shock" when (especially in winter) you walked across a carpet, then touched a metal door and got a shock. As you walked across the carpet you built up static charge (with an associated potential that could reach several 10's of kV); and all that charge "leaks away" when you touch a grounded (conducting) surface. But while you can "feel" the current it's not enough to kill you.
So how much charge is there on your body when you are charged to 30,000 V? It's a bit hard to estimate the capacitance of a human body, so we'll use the physicist's trick of the "spherical cow": we approximate the human body as a sphere with 1 m diameter. The capacitance of a sphere is given by
$$C_{sphere} = 4\pi\epsilon_0 R = 0.11 nF$$
At 30 kV, that gives a charge of 3.3 µA; if that charge comes out of your body in 1 µs, it would result in a peak current of 3.3 A which is why it feels like quite a jolt; however, the total amount of energy is only $\frac12 C V^2 = 0.05 J$ - and that is not enough to kill you. It's enough to kill sensitive electronic circuits, which is why you have to be careful how you handle "bare" electronics, especially in winter (low humidity = build up of static electricity as conductivity of air is lower).
From all the texts I've read, it's always stated that the magnetic field would either curl or warp around the the current flowing within a conductive element, yet, I never was clarified as to why that is. Why would the field "curl" around it? I'm curious if there is an explanation for that nature.
Amazingly, it seems like a perfect circle too, similar to the diagrams that I've studied.
Answer
If you ignore all what you learned about the B-field being a "vector" and you treat it as what really is a 3D skew-symmetric tensor then the mystery goes away. In this view the B-field is a bi-vector, a surface-like quantity whose source is the current element from which it radiates outward, so to speak. Unlike the E-field that has lines of force coming out or ending in charges, the B-field has planes of force coming out or ending in currents and these "planes" form surfaces in which the magnetic action, i.e., attraction-repulsion and torque take place. This is analogous to the way the E-field acts along its lines of force. What conventionally is called the lines of force of the B-field are the orthogonal rays to these surfaces. A uniform current generates a uniform set of planes of action whose orthogonal rays are in fact circles, and they stay so approximately even when the source is curled up into a loop. You can see nice pictures of this in Roche: "Axial vectors, skew-symmetric tensors and the nature of the magnetic field", Eur. J. Phys. 22 (2001) 193–203.
I am going through Tong's lecture notes on String Theory and came across the following irrep decomposition (Chap 2, p.43) of the bosonic string first excited states:
$$\text{traceless symmetric} \oplus \text{anti-symmetric} \oplus \underbrace{\text{singlet}}_{=\text{trace}}$$
He then goes on and claims that the traceless symmetric tensor is the spin-2 graviton.
What is the reason behind that claim? Is there a relationship between degrees of freedom and the spin of a particle in any number of dimensions? I remember from the $SU(2)$ irrep decomposition that the $\ell=1$ irrep has 3 d.o.f. just like a massive spin-1 particle would have. But what about massless particles living in 26 dimensions?
According to Einstein it should be greater than $2 \pi R$ for a co-rotating observer, i.e. $L' = \gamma L$ where $L = 2 \pi R$ in a non-rotating frame and $\gamma$ is the usual Lorentz factor, which would make the spatial geometry hyperbolic - from wikipedia:
Imagine a circle drawn about the origin in the x'y' plane of K' and a diameter of this circle. Imagine, further, that we have given a large number of rigid rods, all equal to each other. We suppose these laid in series along the periphery and the diameter of the circle, at rest relatively to K'. If U is the number of these rods along the periphery, D the number along the diameter, then, if K' does not rotate relatively to K, we shall have $U/D=\pi$. But if K' rotates we get a different result. Suppose that at a definite time t, of K we determine the ends of all the rods. With respect to K all the rods upon the periphery experience the Lorentz contraction, but the rods upon the diameter do not experience this contraction (along their lengths!). It therefore follows that $U/D>\pi$.
On the other hand, there's a nice explanation here on SE that I find intuitively appealing but it concludes the opposite: $L' = L/\gamma$.
So which shall it be?
Due to obvious reasons I put more weight on the first possibility, although I don't actually understand the reasoning, so if someone could describe it in a more detail it'd be great.
There's also this paper I've been looking at, which arrives at Einstein's result in a general relativistic framework, but I'd rather just stay at measuring rods and the like if it's possible...
Answer
You can get the result you want by looking at the "circumference" in different ways. In the circle's rest system, there is no ambiguity of circumference. It is the sum of length's of the rods laid out along the circle. This is the total length of black rods in the following picture:
Now if you rotate very fast w.r.t. the circle, the rods will seem shorter. This is depicted by the green lines (they actually do not stay straight, but will bend a little, however for the sake of the argument, let's assume they stay nearly straight). The leftmost rod is placed where it was before. The next rod is directly adjacent to that, there no gap that suddenly appears, if you start to rotate. So the second rod is a little off w.r.t. its non-rotating counterpart (keep in mind, that this is just a "snapshot").
If you continue to lay down rods that way, you will run out of green rods before you complete the circle. The missing part is the dashed blue line.
Now you can say: "Well, completing a full circumference means to go around and counting rods. If you count eight rods, you completed the circle." By this argument, the circumference gets smaller, as in the linked SE post.
You could also say: "F**k the number of rods... I have plenty! I will lay down rods until I complete the circle." Then you would count how many rods are needed. Clearly more than before. So, you argue that the circumference has gotten longer. This is what happened in the wikipedia article.
You could say: "Well I want a very rigid definition of circumference! I will just do X and everyone will agree that it measures the circumference by any standard and unambiguously!" You would be wrong.
Rotating systems have a strange property: If you placed many clocks around the circumference of a rotating circle, you cannot synchronize all of them. You could start with one clock, synchronize it with the next and so on. Then, if you come back to your first clock you will see a time difference between the first and the last one, even though you set them all to the same time!
This is why I put the word "snapshot" in quotation marks. There is no "snapshot" of a rotating disk, because there is no way to define simultaneity along the whole circumference. And that is why you can define the circumference however the hell you like.
Consider the following (thought) experiment, where an electron is emitted, then deflected by a magnetic field, and then detected:
Because the momentum of the electron changes when it gets deflected, it seems intuitively clear that there should be a reaction force on the magnet, which could in principle be detected.
But now consider the following modified setup:
Here electrons are emitted, with their spin aligned in some arbitrary direction. The first Stern-Gerlach magnet, A, can deflect electrons either up or down. B and C deflect the paths of the two beams, and the final magnet, D, combines them back into a single beam, effectively reversing the action of A. This system of magnets is followed by a detector, which can measure the spin of the electron in the $z$ direction, perpendicular to the plane of the rest of the diagram.
According to my understanding of quantum mechanics, the measurements from this detector should be the same as if the magnets were not there. That is, if the electrons are emitted with their spins always pointing in the $z$ direction, the detector should also find that their spins are always 'up' in the $z$ direction.
The problem is that if we can measure the reaction force on any of the four deflecting magnets then we can work out which of the two paths the electron took, and thus we would have measured its spin in two directions simultaneously. Clearly this isn't possible, so where have I gone wrong? Is there a reaction force on the magnets? If there isn't, what happened to the conservation of momentum? Is it just that the reaction force does exist but we can't measure it? (And if so, what prevents us from doing so?) Or have I just made a mistake in reasoning about how this setup would behave?
Answer
Yes, you may arrange the magnets in such a way that the splitting of the original beam is "undone". The two parts of the wave functions "reinterfere".
The reason why you can't measure both $j_y$ and $j_z$ is pretty much the same reason as the reason why you can't see the interference pattern as well as the "which slit" information in a double slit experiment and your setup may actually be closer to the thought experiments that were intensely discussed during the Bohr-Einstein debates etc. Conceptually, however, the two situations are isomorphic.
The uncertainty principle and/or the destructive effect of your apparatus will always prevent you from learning $j_y,j_z$ at the same moment.
If you want to measure the impulse $\Delta p$ transferred from the electron to the magnets and see it's there, it implies that you must know the magnets' initial momentum with precision $\Delta p$ or better (or something of the same order). The uncertainty principle then guarantees that the vertical position of the magnets is undetermined, with $\Delta x\geq \hbar / \Delta p$.
The electrons also get a kick $\Delta p$ from the magnets – by the momentum conservation, it's the same $\Delta p$ as above, the kick to the magnets – and because you want to be able to distinguish the two beams, you must have $\Delta x\leq \hbar/\Delta p$ for the electrons. That directly contradicts the previous condition. So you either make the split beams focused enough to the two magnets in the middle i.e. small enough $\Delta x$ or you will have a small enough $\Delta p$ which is needed to measure the change of the momentum of the magnets but you can't have both, by the uncertainty principle.
You could also invent a more invasive way to measure which path the electron took. But to claim that your apparatus preserved the original state, especially the relative phase of "up" and "down" that is needed to predict the probabilities of various values of spin $j_z$ (which you untypically take orthogonal to the screen), you need the position of the magnets to be non-intrusive enough. Again, in analogy with the double slit experiment, you will find out that if you can find out about the path or momentum of the electrons, you will modify the electrons' wave functions so that the relative phase will be modified and the results for $j_z$ will be different than if the apparatus weren't there in the middle. You will change the rules of the game.
Fundamentally I want to know: How do holograms work?
The problem with that question is that normally you will end up with pages and pages talking about:
Even Wikipedia is heavy on how to make a hologram, rather than how does a hologram work.
Other people have mentioned stereoscopic vision; how having two eyes gives the illusion of a 3d object. That is also irrelavent, since someone with one eye (or, in my case, one eye closed) can still experience a hologram.
What I am trying to figure out is how does a hologram work?. More to the point, how is it that rotating a flat holographic sticker allows the virtual object to change orientation - allowing me to see content that was not there a moment ago?
Wikipedia has an image that mentions reconstructing a virtual 3d object:
Some problems that that image, though, is that my credit card:
Assuming I have a holographic image of a simple cube. If I am looking at the holographic plate straight on, I will only see a square (i.e. the face of the cube closest to me):
If i rotate the holographic plate, so the right side of the plate is further away, the virtual cube will rotate, and I will actually be able to see the left face of the cube:
What is happening in the flat, 2-dimensional, holographic sticker that it can display continuously different information as I rotate it?
What is the mechanics of this holographic "paper" that it can present my eye different images?
Nowadays it seems to be popular among physics educators to present Newton's first law as a definition of inertial frames and/or a statement that such frames exist. This is clearly a modern overlay. Here is Newton's original statement of the law (Motte's translation):
Law I. Every body perseveres in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by forces impressed thereon.
The text then continues:
Projectiles persevere in their motions, so far as they are not retarded by the resistance of the air, or impelled downwards by the force of gravity. A top, whose parts by their cohesion are perpetually drawn aside from rectilinear motion, does not cease its rotation, otherwise than as it is retarded by the air. The greater bodies of the planets and comets, meeting with less resistance in more free spaces, preserve their motions both progressive and circular for a much longer time.
And then the second law is stated.
There is clearly nothing about frames of reference here. In fact, the discussion is so qualitative and nonmathematical that many modern physics teachers would probably mark it wrong on an exam.
I have a small collection of old physics textbooks, and one of the more historically influential ones is Elements of Physics by Millikan and Gale, 1927. (Millikan wrote a long series of physics textbooks with various titles.) Millikan and Gale give a statement of the first law that reads like an extremely close paraphrase of the Mott translation. There is no mention of frames of reference, inertial or otherwise.
A respected and influential modern textbook, aimed at a much higher level than Millikan's book, is Kleppner and Kolenkow's 1973 Introduction to Mechanics. K&K has this:
...it is always possible to find a coordinate system with respect to which isolated bodies move uniformly. [...] Newton's first law of motion is the assertion that inertial systems exist. Newton's first law is part definition and part experimental fact. Isolated bodies move uniformly in inertial systems by virtue of the definition of an inertial system. In contrast, that inertial systems exist is a statement about the physical world. Newton's first law raises a number of questions, such as what we mean by an 'isolated body,' [...]
There is a paper on this historical/educational topic: Galili and Tseitlin, "Newton's First Law: Text, Translations, Interpretations and Physics Education," Science & Education Volume 12, Number 1, 45-73, DOI: 10.1023/A:1022632600805. I had access to it at one time, and it seemed very relevant. Unfortunately it's paywalled now. The abstract, which is not paywalled, says,
Normally, NFL is interpreted as a special case: a trivial deduction from Newton's Second Law. Some advanced textbooks replace NFL by a modernized claim, which abandons its original meaning.
Question 1: Does anyone know more about when textbooks begain to claim that the first law was a statement of the definition and/or existence of inertial frames?
There seem to be several possible interpretations of the first law:
A. Newton consciously wrote the laws of motion in the style of an axiomatic system, possibly emulating Euclid. However, this is only a matter of style. The first law is clearly a trivial deduction from the second law. Newton presented it as a separate law merely to emphasize that he was working in the framework of Galileo, not medieval scholasticism.
B. Newton's presentation of the first and second laws is logically defective, but Newton wasn't able to do any better because he lacked the notion of inertial and noninertial frames of reference. Modern textbook authors can tell Newton, "there, fixed that for you."
C. It is impossible to give a logically rigorous statement of the physics being described by the first and second laws, since gravity is a long-range force, and, as pointed out by K&K, this raises problems in defining the meaning of an isolated body. The best we can do is that in a given cosmological model, such as the Newtonian picture of an infinite and homogeneous universe full of stars, we can find some frame, such as the frame of the "fixed stars," that we want to call inertial. Other frames moving inertially relative to it are also inertial. But this is contingent on the cosmological model. That frame could later turn out to be noninertial, if, e.g., we learn that our galaxy is free-falling in an external gravitational field created by other masses.
Question 2: Is A supported by the best historical scholarship? For extra points, would anyone like to tell me that I'm an idiot for believing in A and C, or defend some other interpretation on logical or pedagogical grounds?
[EDIT] My current guess is this. I think Ernst Mach's 1919 The Science Of Mechanics ( http://archive.org/details/scienceofmechani005860mbp ) gradually began to influence presentations of the first law. Influential textbooks such as Millikan's only slightly postdated Mach's book, and were aimed at an audience that would have been unable to absorb Mach's arguments. Later, texts such as Kleppner, which were aimed at a more elite audience, began to incorporate Mach's criticism and reformulation of Newton. Over time, texts such as Halliday, which were aimed at less elite audiences, began to mimic treatments such as Kleppner's.
Answer
I did not do more than read Newton, and a few commentators, so my insight on this is probably meager. But I am sure that you are right that the inertial frame interpretation of the first law is only a modern ex-post-facto justification for making it separate from the second law. Newton certainly never used the first law to define an inertial frame, he just assumed you had one in mind, since inertial frames were not the focus of his investigation.
I think that the statements of the laws of motion are unfortunately following Aristotle more than Euclid. Since physics is no longer regarded as philosophy, we value independence of axioms over clarity of philosophical expounding, and this makes the first law redundant. But if you are stating a philosophical position--- that things maintain their state of motion unless acted upon--- Newton's first law is a neat summary of the foundation of the world-system.
Note that Newton does not state it as "a body in linear motion continues moving linearly". He includes rotational motion too, even though this is a different idea. I think he conflates the two to fix in mind the philosophical position that uniform motion is the natural state of all objects. In Aristotle, the natural state of massive stuff like "earth" is to be down at the center of universe, and of light stuff like "fire" to be up in the heavens, leading to gravity and levity. Newton is replacing this notion with a different notion of natural state. Then the second law talks about deviations from the natural state, and is a separate philosophical idea (although not a separate axiom in the mathematical sense).
The influence of Aristotle has (thankfully) declined through the centuries, making Newtons laws a little anachronistic. I think that we don't have to be so slavish to Newton nowadays.
Newton was aware of the importance of linear momentum and angular momentum conservation. One other way of understanding and his first law can be thought of as making the conservation laws primary. This point of view is both closer to Newton's thinking (it is what makes his "natural states" natural), and it is also a better fit with modern understanding. So it might be nice to restate the first law as "linear momentum and angular momentum are conserved".
All this is based on personal speculation, not on sound historical research, so take with a grain of salt.
The classical description of electro-optic modulators is an index of refraction that depends on the applied voltage. For example, for a sine modulation $\sin(\Omega t)$, a monochromatic laser of frequency $\omega$ would get an additionnal phase $\varphi\propto\sin(\Omega t)$. This results in sidebands in the spectrum at $\omega-\Omega$ and $\omega+\Omega$.
Now, what is the interpretation of this phenomenon in terms of photons? A photon with initial frequency $\omega$ will end up at $\omega-\Omega$ or $\omega+\Omega$. How can the time-variation of the refractive index create new photon frequencies? Is it a non-linear effect similar to second-harmonic generation? If yes, it could be explained by an interaction such as $\hbar\omega + \hbar\omega \rightarrow \hbar (\omega-\Omega ) + \hbar(\omega+\Omega)$?
EDIT: A corollary to the original question. I shake my hand very fast in front of a laser beam, what happens to the photons? Do they get chopped in shorter photons? Instead of my hands, I could use a super-fast chopper. I would see photons with new frequencies (the sidebands) because of this modulation. How come the incident photons get a different energy?
We know that conserved quantities are associated with certain symmetries. For example conservation of momentum is associated with translational invariance, and conservation of angular momentum is associated with rotational invariance.
Now, if the particle position does not change, then the position of the particle is a conserved quantity. What is the symmetry that corresponds to conservation of position in this case?
Answer
Nature doesn't have this symmetry because your conservation law doesn't hold, either. According to the law of inertia, object keeps on moving with a constant velocity – which is however generically nonzero. In its own rest frame, it's zero, but in other frames, the velocity is nonzero.
If one studies the motion of the center-of-mass, it is indeed moving with a constant velocity. So the conserved quantity that is closest to your "conserved position" is the conserved velocity of the center-of-mass. This conservation law is directly linked, via Noether's theorem, to the Lorentz symmetry of the laws of physics – or, in the non-relativistic limit, to the Galilean symmetry. In the non-relativistic case, the generator of the Galilean symmetry is $\vec x_{\rm cm}$, the center-of-mass position, indeed: the generator of the symmetry is the conserved quantity itself.
If you designed boring laws in which the position has to be conserved, the symmetry would be generated by the conserved quantity $\vec x$. This symmetry generator generates translations in the momentum space. So the laws of physics (the Hamiltonian) would have to be effectively independent of the momentum. That would be pretty bad: you couldn't include the kinetic energy term to the total energy, among other things. That's related to the fact that the particles would have "infinite inertial mass", which would force them to sit at a single point. The whole term "dynamics" would be a kind of oxymoron because things wouldn't be changing with time.
Appendix
Consider the generator equal to the center-of-mass position $$ \vec x_{\rm cm} = \frac{m_1 \vec x_1 + m_2 \vec x_2 +\dots +m_N \vec x_N}{m_1+m_2+\dots +m_N} $$ How do physical observables transform under the symmetry generated by it? Compute the commutators. The commutators of the position above with positions $x_i$ vanish, so positions (at $t=0$) don't transform. However, the commutator with $p_i$ is equal to $m_i \delta_{mn} / M_{\rm total}$, and if this is added to $p_i$ with an infinitesimal coefficient $\vec \epsilon \cdot M_{\rm total}$, you see that all velocities are changed by $$ \vec v_i \to \vec v_i + \vec \epsilon $$ But if all velocities are just shifted by a constant, that's the Galilean transformation. let me emphasize that this simple transformation rule only holds at $t=0$. For $t\neq 0$, one would have to add extra terms proportional to $t$ to the generator (they would be similar for a Lorentz symmetry, too), namely $t\cdot \vec P_{\rm total}$. At any rate, the center-of-mass position is the generator of the Galilean transformations, the transformations switching from one inertial system to a nearby inertial system (which moves by a speed differing by $\delta \vec v$).
Note that the commutator of $\vec x_{\rm cm}$ with the Hamiltonian isn't quite zero, so according to some definitions, it isn't a symmetry. Instead, the commutator is proportional to the total momentum $\vec p$ which is a symmetry itself. So the commutators of various generators yield other generators – the standard form of a Lie algebra (Galilean/Lorentz in this case) in which the Hamiltonian isn't necessarily commuting with everyone else but is one of the generators of a non-Abelian group.
What would happen if every person in the world gathered at a minimum possible area and jumped all at once? Would that have any effect on Earth?
Answer
First, obligatory note: this was already answered by Randall Munroe of XKCD fame in his "What if blog" (this was his 8th question & answer posted).
The answer, as given by Randall (emphasis mine)
it doesn’t really affect the planet. Earth outweighs us by a factor of over ten trillion. On average, we humans can vertically jump maybe half a meter on a good day. Even if the Earth were rigid and responded instantly, it would be pushed down by less than an atom’s width
The problem is the 7 billion-ish people leaving said area.
I'm having trouble understanding how some conclusions are made in my book. I'm studying from a coursebook based on Goldstein's "Classical Mechanics", here's what's written in my book, with my problems under each section:
Assume a system with conservative forces and holonomic, time independent constraints, described by $n$ generalized coordinates $q_k$.
The Lagrangian is $L=T-V$, with the kinetic energy:$T=\sum_{kl}\dot q_k\dot q_l (\sum_i \frac{1}{2}m_i\frac{\partial\vec r_i}{\partial q_k}.\frac{\partial\vec r_i}{\partial q_l}) = \sum_{k,l}\frac{1}{2}m_{kl}(q1,...,q_n)\dot q_k \dot q_l$
...
...
a polynomial of the second degree in the generalized velocities. The potential energy $V(q_1, ..., q_n)$ only depends on the generalized coordinates.The system has a point of equilibrium ($q_k^0=q_1^0, ... q_n^0$) when all generalized forces are $0$ in this point:
$Q_k=-(\frac{\partial V}{\partial q_k})_0=0$
If the system is in a point of equilibrium at a certain starting time, with the starting velocities $0$, then the system will remain in that point of equilibrium. In other words: $q_k(t)=q_k^0$ is the solution to the Lagrange equations
$\sum_l m_{kl}\ddot q_l+\sum_l\sum_m\frac{\partial m_{kl}}{\partial{q_m}}\dot q_l\dot q_m=\sum_{lm}\frac{1}{2}\frac{\partial m_{lm}}{\partial q_k}\dot q_l \dot q_m -\frac{\partial V}{\partial q_k}$
with these initial conditions.
Answer
The $m_{kl}$ is a "mass matrix". In Cartesian coordinates the kinetic energy is $$T = \sum_i \frac{m_i}{2} v_i^2$$ with $v_i^2 = v_{i,x}^2 + v_{i,y}^2 + v_{i,z}^2$. For simplicity let us consider the case of one particle. We can write this in matrix form as $$T = \frac{1}{2} (v_x, v_y, v_z) \begin{pmatrix} m_i & 0 & 0 \\ 0 & m_i & 0 \\ 0 & 0 & m_i \end{pmatrix} \begin{pmatrix} v_x \\ v_y \\ v_z \end{pmatrix} = \frac{1}{2} \sum_{kl} m_{kl} v_k v_l.$$ The latter expression generalizes to the case of several particles, just imagine a bigger matrix.
But typically when using the Lagrangian formalism Cartesian coordinates are not the most convenient. Now, since $v_k = \dot{x_k}$, if we change coordinates to $\mathbf{q} = \mathbf{q}(\mathbf x)$, we have $$\dot{q}_k = \sum_k \frac{\partial q_k}{\partial x_j} \dot{x_j}$$ and since the coordinate change is invertible, $$\dot{x_k} = \sum_j \frac{\partial x_k}{\partial q_k} \dot{q}_j. $$ Therefore in the new coordinates the expression for the kinetic energy is $$T = \frac{1}{2} \sum_{ijkl} m_{kl} \frac{\partial x_k}{\partial q_i} \dot{q}_i \frac{\partial x_l}{\partial q_j}\dot{q}_j = \frac{1}{2} \tilde{m}_{kl} \dot{q}_k\dot{q}_l \tag{1} $$ where $\tilde{m}_{kl} = \frac{1}{2} \sum_{ij} m_{ij} \frac{\partial x_i}{\partial q_k} \frac{\partial x_j}{\partial q_l}.$ (This, of course, is the transformation law for a rank 2 tensor, because that is what the mass matrix is.)
Since we are allowed to do any coordinate transformations, the entries in the matrix $\tilde{m}_{kl}$ may be non-constant functions of the coordinates. Then we see that (1) is the first equation in your question.
For a concrete example where the mass matrix has non-constant entries you can try to find the mass matrix in spherical coordinates, then compare with the expression for the kinetic energy in spherical coordinates. (The latter should be in Goldstein if you don't remember it.)
For the second part of your question, it is really just Newton's first law. If the system is at rest in a position where $\partial V /\partial q_j = 0$, it is at rest with no forces acting on it. The more rigorous way to to understand it is to write down the Euler-Lagrange equations and check that $q_k(t) = q_k^0$ is a solution, then appeal to a uniqueness theorem for solutions to differential equations.
$$ L = \sum _ { i = 1 } ^ { N } \frac { 1 } { 2 } m _ { i } \left| \dot { \vec { x } _ { i } } \right| ^ { 2 } - \sum _ { i < j } V \left( \vec { x } _ { i } - \vec { x } _ { j } \right) $$
This is just a typical classical Lagrangian for $N$ particles. Since the Lagrangian does not explicitly depend on time, the energy must be conserved. Also, the linear and angular momentum seem to be conserved too.
However, if there is a change in the coordinate by the Galilean transformation $\overrightarrow{x}_i(t) \to \overrightarrow{x}_i(t) +\overrightarrow{v}t$, then the aforementioned quantities seeem clearly "variant". So, my question is that whether there exists a quantity that is invariant under this Galilean transformation. Could anyone please present me one? Or if there is no such quantity, could anyone please explain why?
In Volume II Chapter $28$ of the Feymann Lectures on Physics, Feynman discusses the infamous $\frac43$ problem of classical electromagnetism. Suppose you have a charged particle of radius $a$ and charge $q$ (uniformly distributed on the surface). If you integrate the energy density of the electromagnetic field over all space outside the particle, you'll get the total electromagnetic energy, which is an expression proportional to $c^2$. The energy divided by $c^2$ is what we usually call the mass, so if we calculate the "electromagnetic mass" in this manner we'll get $m = \frac{1}{2}\frac{1}{4\pi\epsilon_0}\frac{q^2}{ac^2}$. If, on the other hand, you took the momentum density of the electromagnetic field and integrated it over all space outside the particle, you'd get the total electromagnetic momentum, which turns out (for $v<
Feynman claims that this fundamental issue remains when we move to quantum electrodynamics. Was he right, and if so has the situation changed since the $1960's$ when he was writing? I've seen claims on the Internet (I don't have the links) that the $\frac43$ problem is still there in QED, but instead of $\frac{4}{3}$ the coefficient is something closer to $1.$ Is that true, and if so what's the coefficient? All of this is of course related to issues of self-energy and renormalization.
Any help would be greatly appreciated.
In Wikipedia it was mentioned Luminous disturbance so I Did get confused that this principle only works for light waves and not for all of the Waves. Like some mechanical waves example wave on string.
Wiki text:
In 1678, Huygens [1] proposed that every point which a LUMINOUS DISTURBANCE reaches becomes a source of a spherical wave; the sum of these secondary waves determines the form of the wave at any subsequent time.
Answer
The principle that every point on a wavefront can be thought of as an emitter of spherical (or, in 2D, circular) waves is applicable to any waves - see any introductory high school course on waves, where the demonstrations are typically done with water surface waves. Note that, as @ignacio pointed out, the construction is only exact in odd dimensions (in practice that means 3D), but even in 2D it is quite convincing:
Nice demonstration with water waves in 2D
And the link given by Ignacio:
Mathematical proof that Huygens construction is only valid in odd dimensions
What is difference between partial and ordinary time derivative? for example: what is difference between $\frac {\partial v}{\partial t}$ and $\frac {dv}{dt}$?
where the $v$ is velocity.
Suppose that $U(x)$ is an element of the gauge group say $SU(2)$ and suppose $U(x)=1$ as $|\vec{x}|\to\infty$. Then, why does space have the topology of $S^3$?
This is done in Srednicki page 571. Note that I'm not asking how to prove that $SU(2)\cong S^3$. What I'm asking is how to prove that when $U(x)=1$ as $|\vec{x}|\to\infty$ the space $\mathbb{R}^3$ is compactified to $S^3$ space.
First of all, I am unfortunately not an expert in physics, so please be indulge with me. I am trying to model a $2$-dimensional mass-spring system with $1$ mass and $3$ springs to solve a dynamics problem in frequency domain. I've been looking for a solution for a similar problem but I couldn't find anything useful. Are these classical newton equation of motion mass-spring systems limited to $1D$?
The mass $m$ is connected to $3$ springs $k_1, k_2, k_3$, which are fixed at their endpoints, rotations are possible. The springs are assumed linear and can be simplified $k_1=k_2=k_3$. In the equilibrium state, the angle between the springs is $120^{\circ}$.
Answer
The simplest setup is for small displacements. Suppose the spring rest lengths are $L_1,L_2,L_3$, the mass has mass $m$, the springs have constants $k_1,k_2,k_3$, the angle is 120 degrees between attachments, and the attachment points are set up so that at rest, the springs are all unstretched.
The potential becomes $$U(\mathbf{r})=\sum_{j=1}^3\frac{k_j}{2}(|\mathbf{r}-\mathbf{u}_j|-L_j)^2$$ where $$\mathbf{u}_j=\{L_j\cos(2\pi j/3),L_j\sin(2\pi j/3)\}.$$ It's easy to verify that $$\nabla U(\mathbf{0})=\mathbf{0}$$ which means that the system is at equilibrium when the mass sits at the origin.
Defining $$H=\nabla\nabla U(\mathbf{0})=\left( \begin{array}{cc} \frac{1}{4} \left(k_1+k_2+4 k_3\right) & \frac{1}{4} \sqrt{3} \left(k_2-k_1\right) \\ \frac{1}{4} \sqrt{3} \left(k_2-k_1\right) & \frac{3}{4} \left(k_1+k_2\right) \\ \end{array} \right)$$ we obtain eigenvalues $$\lambda_\pm=\frac{1}{2} \left(k_1+k_2+k_3\pm\sqrt{k_1^2-k_2 k_1-k_3 k_1+k_2^2+k_3^2-k_2 k_3}\right)$$ and the ordinary vibrating frequencies become $$\omega_\pm=\frac{1}{2\pi}\sqrt{\frac{\lambda_\pm}{2m}}.$$ Notice that the lengths are irrelevant, and that in the case $k_1=k_2=k_3$ the frequencies become identical, thus becoming like a 2D spherical oscillator.
When you add more masses to the system, things get interesting.
What manner of driving force are you planning on applying?
Imagine a bar
spinning like a helicopter propeller,
At $\omega$ rad/s because the extremes of the bar goes at speed
$$V = \omega * r$$
then we can reach near $c$ (speed of light) applying some finite amount of energy just doing
$$\omega = V / r$$
The bar should be long, low density, strong to minimize the amount of energy needed
For example a $2000\,\mathrm{m}$ bar
$$\omega = 300 000 \frac{\mathrm{rad}}{\mathrm{s}} = 2864789\,\mathrm{rpm}$$
(a dental drill can commonly rotate at $400000\,\mathrm{rpm}$)
$V$ (with dental drill) = 14% of speed of light.
Then I say this experiment can be really made and bar extremes could approach $c$.
What do you say?
EDIT:
Our planet is orbiting at sun and it's orbiting milky way, and who knows what else, then any Earth point have a speed of 500 km/s or more agains CMB.
I wonder if we are orbiting something at that speed then there would be detectable relativist effect in different direction of measurements, simply extending a long bar or any directional mass in different galactic directions we should measure mass change due to relativity, simply because $V = \omega * r$
What do you think?
Answer
Imagine a rock on a rope. As you rotate the rope faster and faster, you need to pull stronger and stronger to provide centripetal force that keeps the stone on the orbit. The increasing tension in the rope would eventually break the it. The very same thing would happen with bar (just replace the rock with the bar's center of mass). And naturally, all of this would happen at speeds far below the speed of light.
Even if you imagined that there exists a material that could sustain the tension at relativistic speeds you'd need to take into account that signal can't travel faster than at the speed of light. This means that the bar can't be rigid. It would bend and the far end would trail around. So it's hard to even talk about rotation at these speeds. One thing that is certain is that strange things would happen. But to describe this fully you'd need a relativistic model of solid matter.
People often propose arguments similar to yours to show Special Relativity fails. In reality what fails is our intuition about materials, which is completely classical.
Today, I tried creating a very basic Faraday cage by surrounding a radio with two baking trays made out of iron. It didn't seem to affect the radio's signal (AM was being used, not FM).
In theory, should what I used block the radio's signal?
I can't remember how thick the baking trays were. There was some gaps totalling a few square centimetres, because the two baking trays were not identical. I didn't attempt to "Ground" the baking trays.
As a planet moves through the solar system, a bow shock is formed as the solar wind is decelerated by the magnetic field of the planet. Presumably the creation of this shock wave would cause drag on the planet, certainly in the direction of orbit but possibly rotation as well.
Is there an estimate for the amount of drag on the Earth as it orbits the Sun? Based on the drag, how long would it take before the orbital velocity slows to the point that we spiral slowly into the Sun? Would any planets fall into the Sun prior to the Sun expanding into a Red Giant, gobbling them up?
Answer
This is a really rough calculation that doesn't take into account the realistic direction of the bow shock, or calculation of the drag force. I just take the net momentum flow in the solar wind and direct it so as to produce the maximum decceleration and see what happens.
Apparently the solar wind pressure is of the order of a nanoPascal. As I write this it's about $0.5\ \mathrm{nPa}$. You can get real time data from NASA's ACE satellite or spaceweather.com (click through "More data" under "Solar wind"). During periods of intense solar activity it can get up to an order or magnitude or so more than this. Let's take this worst case and assume, unrealistically, that all of the pressure is directed retrograde along the Earth's orbit. This will give the maximum deccelerating effect. I get a net force of $\sim 10^6\ \mathrm{N}$. Dividing by the Earth's mass gives a net acceleration $2\times 10^{-19}\ \mathrm{m/s^2}$. Let's fudge up again and call it $10^{-18}\ \mathrm{m/s^2}$. The time it would take for this to make a significant dint the the Earth's orbital velocity ($30\ \mathrm{km/s}$) is of the order of $10^{15}\ \mathrm{yr}$. I think we're safe.
For the other planets there is a $1/r^2$ scaling of the solar wind with the distance from the sun (assuming the solar wind is uniformly distributed) and an $R^2$ scaling with the size of the planet. So for Mercury the former effect gives an order of magnitude increase in drag and the latter effect takes most of that increase away again. There is an additional $R^{-3}$ increase in effect due to the decreased mass of a smaller body (assuming density is similar to the Earth). Then there is the $r^{-1/2}$ increase in orbit velocity due to being closer to the sun. So the total scaling factor for the time is $ R r^{3/2} $, which for Mercury is about 0.1. So the end result is not much different for Mercury.
This site always causes me to learn new Mathematica features. It made really quick work of this since it has all sorts of astronomical data built in:
Note that the number of digits displayed in the final column is ludicrous. :)
Can someone please explain what the event horizon of a black hole is? I mean is it the actual surface of the black hole or is it the point of no return where light can no longer escape?
This question is about why we have a universal speed limit (the speed of light in vacuum). Is there a more fundamental law that tells us why this is?
I'm not asking why the speed limit is equal to $c$ and not something else, but why there is a limit at all.
EDIT: Answers like "if it was not.." and answers explaining the consequences of having or not having a speed limit are not--in my opinion-- giving an answer of whether there is a more fundamental way to derive it from law to explain this limit.
I'm pretty astounded that I did not hear about this sooner, but in my course on QFT our professor told us that the concept of spin can be used to mean three things:
Mechanical spin (apparently a relativistic effect giving rise to classical spin-orbit coupling)
Magnetic spin (purely quantum mechanical)
Classification of representations of the Lorentz group (the manner in which the particles transforms under Lorentz-transformations)
I take it that these meanings don't generally coincide since they don't seem to do so in the case of the photon: we describe this as a spin-1 particle (3rd meaning), though it has no intrinsic magnetic moment (2nd meaning).
However, despite being swiftly explained in a few words in class, I cannot remember what exactly the first meaning is about. Furthermore I'd like to find out the exact relations between these three meanings. As the case of the photon showed the last two at least don't generally seem to coincide. Can anyone clarify?
Related questions are:
What is spin as it relates to subatomic particles?
What are some useful ways to imagine the concept of spin as it relates to subatomic particles?
Is Angular Momentum truly fundamental?
What does spin 0 mean exactly?
Spin of a particle and spin quantum number
Why do many people say vector fields describe spin-1 particle but omit the spin-0 part?
but the answers there don't make the light go on in my head.
Answer
The three meanings of spin are in the Quantum world equivalent. What the professor means (I guess) is the following:
These three interpretations may seem different but are all three equivalent I believe. The mechanical spin is a way of giving a classical interpretation. While the magnetic and representational spin are both the same (so it seems from the quantum-field theories). Beware, altough the three representations are equivalent the three shouldn't necessarily coexist! This is shown by the spin of the photon.
For the case of the photon for example:
The photon has two polarisations, if we take the two linear polarisations and combine them we can get two different circular polarisations. This circular polarisation gives the mechanical spin.
The magnetic polarisation for a photon doesn't exist. This is because of the fact that photons don't interact and hence don't couple to the electromagnetic field.
Upon interactions the photon is able to interchange its momentum, upon interacting with other particles this way it has a spin-1 interpretation.
I do understand the intuitive logic behind the Higgs mechanism for parallel D-branes that are separated by a distance $x$, but I would like to see a more quantitative argument. Is it possible to explicitly find the potential of the scalar field and an exact expression of the resulting worldvolume theory?
What I do understand, is that if I separate $M$ D-branes from an initial stack of $N$ D-branes the gauge group will split in $U(N-M)\times U(M)$ as the gauge fields, that correspond to strings stretching from the remaining stack of $N-M$ branes to the stack of $M$ branes, become massive. In addition, the transverse scalars, corresponding to the position of the $M$ separated D-branes along the separated direction, will develop a vev (as is partially addressed in this thread).
I do understand the above logic sound like some 'Higs-like' mechanism breaking the original $U(N)$ gauge symmetry and giving mass to some of the gauge fields, but could someone show more quantitatively how it follows from the string description that this is indeed a Higgs mechanism? I cannot find such a derivation in any of the introductory books on String and Branes like Johnson, Kutasov and Giveon, Zwiebach or Becker and Schwarz.
Is it guaranteed that there exists a universal time on which all comoving observers will agree?
Special relativity teaches us that inertial observers in relative motion do not agree on the time they observe. Now in standard cosmology, the physical distance between the comoving observers changes with time because of the expansion of the Universe even though they sit at fixed coordinates in the comoving coordinate system.
Does this mean they do not agree on the cosmic time $t$? I'm not talking about the conformal time.
I'm a complete newbie to Quantum Theory, but I want to know more, so I've been watching few YouTube videos (an example below).
https://www.youtube.com/watch?v=ZuvK-od647c
All videos I've watched explain, that when an entangled particle has its spin measured, it will instantaneously communicate its measurement with its entangled partner, so that when it too is measured in the same direction, it will have the opposite spin.
Often these videos explain why these particles must be communicating with each other, rather than containing "hidden information" (Bell's theorem), but do not delve into why it must be instant.
Answer
There are three kinds of possibilities that can explain the COMBINATION of two types of below observations relating to entanglement.
Correlation of two particles of any single entangled pair. Perfect anti correlation (opposite spins when measured in same direction) is a specific case of this, which is most commonly sighted
Statistical correlation between measurement outcomes of numerous pairs when measured at any angles. 1.) can also be considered a special case of 2.)
In order to explain above combination of observations, three kinds of possibilities are there. These are just possibilities, no one knows how actually the correlations form.
1) There is an active link between the two particles of entangled pair that is able to signal at FTL, thus measurement of one particle influences the state of other and then it is easy to expect the observed statistical correlation. This possibility is the one you are talking about and it is also most commonly assumed by most public when they try to explain/understand the correlations.
By signal, here I mean the means by which the entanglement is supposedly collapsed on measurement of first particle. This signal can not be used/detected by any observer for any/information/communication purpose. It just a speculation on how the entanglement may be collapsing. see no-communication theorem on wiki.
2) Reality/Locality is not as we understand it. Meaning, two particles, being spatially far off can still be considered at same location in quantum sense. This allows measurement of one particle to influence that of the other without violating light speed and then it is easy to expect the observed statistical correlation. This appears to be the line taken by most main stream scientists.
3) There is a yet to be discovered classical mechanism that forms the statistical correlations over duration of the experiment. Local Hidden Variables alone is not sufficient to support such mechanism and that has been proven by Bell's inequality and supported by experimental outcomes.
In order for such a mechanism to be at work, it has to involve "Local Hidden Variables" PLUS some kind of global memory/accumulation/balancing/synchronization. Global means that the the natural environment in the vicinity of the experiment accumulates information about creation and measurement outcomes of previous pairs and steers the creation and measurement outcomes of subsequent pairs in such a way that it balances out per quantum mechanics, over large number of pairs.
I have not seen enough literature which would convince that this possibility (3) has completely, critically and honestly been ruled out. Most literature mentions Bell's inequality that only disproves local hidden variables.
Some people also speculate many worlds, which in my opinion is even more weird.
Some people also try to explain it in terms of random probability just like toss of coin eventually turns out ~50/50 percent heads and tails. This explanation becomes weak when you try to explain both kinds of observations listed in the beginning of this answer. You can not explain perfect anticorrelation as random probability. For some states, perfect anticorrelation is guaranteed. There are no guarantees in randomness.
I have a background in calculus but don't really know anything about physics. Forgive me if this is a really basic question.
The equation for distance of an accelerating object with constant acceleration is:
$$d=ut +\frac{1}{2}at^2$$
which can also be expressed
$$d=\frac{\mathrm{d}x}{\mathrm{d}t}t+\frac{\mathrm{d^2}x}{\mathrm{d}t^2}\frac{t^2}{2}$$
(where x(t) is the position of the object at time t)
That's fine for a canonball or something like that, but what about a car accelerating from 0 to cruising speed? The acceleration is obviously not constant, but what about the change in acceleration? Is it constant? I suspect not. And then what about the change in the change of acceleration, etc. etc.? In other words, how does one know how many additional terms to add in the series?
$$d=\frac{\mathrm{d}x}{\mathrm{d}t}t+\frac{\mathrm{d^2}x}{\mathrm{d}t^2}\frac{t^2}{2}+\frac{\mathrm{d^3}x}{\mathrm{d}t^3}\frac{t^3}{3}+\frac{\mathrm{d^4}x}{\mathrm{d}t^4}\frac{t^4}{4}\cdot etc. \cdot ?$$
Answer
Technically, the equation
$$d = \frac{\mathrm{d}x}{\mathrm{d}t}t + \frac{\mathrm{d}^2x}{\mathrm{d}t^2}\frac{t^2}{2}$$
is not right. Instead, for constant acceleration, you need
$$d = \left(\left.\frac{\mathrm{d}x}{\mathrm{d}t}\right|_0\right) t + \left(\left.\frac{\mathrm{d}^2x}{\mathrm{d}t^2}\right|_0\right) \frac{t^2}{2}$$
In other words, a quantity like $\mathrm{d}x/\mathrm{d}t$ changes in time, but you want to use the initial velocity only. I think this is what you probably intended to begin with, though.
If you wanted to solve the problem purely kinematically, then you could try to expand the position in a Taylor series as you wrote in your answer. However, this only works if the function is equal to its Taylor series. For simple functions like exponentials and trig functions this is true, but for a person driving a car it is not. If a function equals its Taylor series everywhere, then if you observe its position over any finite interval of time, no matter how short, you can completely determine what the car will do in the future. This is not realistic.
Instead, you will want some way of determining either the velocity or the acceleration as a function of time or position. In physics, it is common to be able to determine the acceleration as a function of position. The reason is that acceleration comes from the equation $$F=ma$$ so that if you can determine the forces present, you know the acceleration, and higher-order derivatives are not necessary.
If you know the velocity as a function of time, you can simply integrate it to find the displacement. $$d(t) = \int_{t_0}^t v(t') \mathrm{d}t'$$
If you know the acceleration as a function of time, you can integrate that too, although this situation is less common.
$$d(t) = v_0(t - t_0) + t\int_{t_0}^t a(t')\mathrm{d}t' - \int_{t_0}^t t'a(t')\mathrm{d}t'$$
I found this expression by looking for something whose derivative with respect to time was the velocity
$$v(t) = v_0 + \int_{t_0}^t a(t')\mathrm{d}t'$$
If you know the velocity as a function of position, you have the differential equation
$$\frac{\mathrm{d}x}{\mathrm{d}t} = v(x)$$
which you can solve by separation of variables.
If you know the acceleration as a function of position, you have the differential equation
$$\frac{\mathrm{d}^2x}{\mathrm{d}t^2} = a(x)$$
which is not always easy to solve. In more realistic scenarios, the acceleration will depend not only on the object's own position, but also on the positions of the things it's interacting with. This gives coupled differential equations, which can be simplified in a special cases, but frequently can only be solved numerically.
The following was originally given to me as a homework question at my physics 2 course:
Consider the following circuit
The difference of potentials between the point $V_{1}$and the point $V_{2}$ is $4.4$ volts, the resistance of all the resistors is the same $R=1\Omega$.
Find the current between point $A$ and point $B$.
The answer given is simply $0$ and the argument was just the pair of words ``using symmetry''.
I don't really understand the answer:
First, it is not completely symmetric: There is a difference of potentials so the potential at the point $V_{1}$ is not the same as the potential at the point $V_{2}$.
Secondly: How can I see that the symmetric structure will give me that the current between $A$ and $B$ is $0$ ?
Also, I would appreciate to see a calculation of this current to get a better feel for whats going on, I know the rule $V=IR$ (which seems the most useful here, but I also know other rules that can be used), but I don't understand how to use this rule to find the current.
Answer
This is your circuit:
The current that comes from the source, when reaches the point that must choose it's way, sees no difference between the two paths (symmetry) , so half of it flows through one way and the other part flows in the second way. It means that, $I_1=I_2$ , So the potential difference across yellow resistors is the same. It means that the potential of point $\mathbf{A}$ is equal to potential of point $\mathbf{B}$ :
$$I_1=I_2\to V_A=4.4-I_1R \text{ , }V_B=4.4-I_2R\to V_A=V_B$$
So there isn't any potential difference across the blue resistor, and the current through it is 0, and it can be omitted from the circuit without any change in the behavior of the circuit.
I wish to understand the statement in this paper more precisely:
(1). Any 3d Topological quantum field theories(TQFT) associates an inner-product vector space $H_{\Sigma}$ to a Riemann surface $\Sigma$.
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(2) In the case of abelian Chern-Simons theory $H_{\Sigma}$ is obtained by geometric quantization of the moduli space of flat $T_{\Lambda}$-connections on ${\Sigma}$. The latter space is a torus with a symplectic form
$$ ω =\frac{1}{4π} \int_{\Sigma} K_{IJ} \delta A_I \wedge d \delta A_J.$$
(3) Its quantization is the space of holomorphic sections of a line bundle $L$ whose curvature is $\omega$. For a genus g Riemann surface $\Sigma_g$, it has dimension $|\det(K)|^g$.
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(4) The mapping class group of $\Sigma$ (i.e. the quotient of the group of diffeomorphisms of $\Sigma$ by its identity component) acts projectively on $H_{\Sigma}$. The action of the mapping class group of $\Sigma_g$ on $H_\Sigma$ factors through the group $Sp(2g, \mathbb{Z})$.
We are talking about this abelian Chern-Simons theory: $$S_{CS}=\frac{1}{4π} \int_{\Sigma} K_{IJ} A_I \wedge d A_J.$$
Can some experts walk through this (1) (2) (3) (4) step-by-step for focusing on this abelian Chern-Simons theory?
partial answer of (1)~(4) is fine.
I can understand the statements, but I cannot feel comfortable to derive them myself.
In books, it is usually said that this is a consequence of the fact that parallel transport preserves dot product. How ?
Answer
This is an axiom, not a result. When defining the covariant derivative, we choose for it to obey a number of properties. Carroll's Spacetime and Geometry summarizes these well -- look at the preprint for Chapter 3 here.
We of course want $\nabla$ to act linearly on its argument and to obey the product rule. We also demand that it commute with contractions and that it reduce to the well-known partial derivative on scalars. No one really debates these properties, but as it turns out they don't uniquely define a derivative. For that we add on the properties of being torsion-free and metric compatible. Summarizing Carroll, we have \begin{align} \nabla(T + S) & = \nabla T + \nabla S && \text{(linearity)} \\ \nabla(T \otimes S) & = (\nabla T) \otimes S + T \otimes \nabla S && \text{(product rule)} \\ \nabla_\mu ({T^\lambda}_{\lambda\rho}) & = {{(\nabla T)_\mu}^\lambda}_{\lambda\rho} && \text{(example of commuting with contractions)} \\ \nabla_\mu \phi & = \partial_\mu \phi && \text{(reduction to partial derivative)} \\ \Gamma^\lambda_{\mu\nu} & = \Gamma^\lambda_{\nu\mu} && \text{(torsion-free connection)} \\ \nabla_\rho g_{\mu\nu} & = 0 && \text{(metric compatibility).} \end{align}
You can have covariant derivatives that are not metric compatible. This just means the differential structure of spacetime provided by the derivative doesn't play nicely with the structure induced by the metric, and such constructions turn out to be not very useful for general relativity.
When people "derive" metric compatibility from parallel transport, they are taking as an axiom "parallel transport [which depends only on the covariant derivative] is compatible with inner products [which depend only on the metric]." It is the same as just saying $\nabla_\rho g_{\mu\nu} = 0$, but more complicated. Better would be to assume $\nabla_\rho g_{\mu\nu} = 0$ from the start and prove that this means parallel transport preserves inner products.
If all the lights in a city, and the area surrounding it were shut off. How long would it take for the light pollution to clear, so the light from stars could be seen?
I wanted to go to the depth of the discovery of classical mechanics, instead of just reading, accepting and learning things. Now my question is not a single question, but it can answer many of my doubts. I completely understand motion, but I got stuck with forces. First, I want to tell all the physicists out there that I have read many books in which while describing momentum, they simply write that it is $m*v$, or they give this proof that I am not able to understand:
$p \varpropto m$
$p \varpropto v$
So $p = Kmv$, where K is constant of proportionality.
By experiments, it was found that value of k is 1.
So $p = mv$;
After long searching, I found a book which talked about a very unique experiment and drew a table of values for it, and before telling Newton's second law, they told that a pattern exists within the table, and that pattern was that $m*v$ in every trial was the same, with which I completely agree, and they told that this $m*v$, which remains conserved, is defined to be momentum, for easier equations. Then through momentum conservation, they came out to the fact that $F = ma$, which was completely intuitive and I moved on.
But, (now moving to my original question) , the equipment's used in that experiment were very new, they didn't existed in Newton's period(example light gates), or did they?
So my second question: How did Newton came to the fact that $m*v$ remains conserved, or simply this pattern exists?
Did he conducted any experiments?
If yes, then what were they? If no, then how he led to momentum conservation?
Now my final doubt. There are many concepts in physics which include a constant of proportionality, and books just throw their values, instead of telling how these values were found.
Can someone please tell me just the method which is used to find the value of $k$ in my first question? It will be enough to satisfy me. I have asked many questions about this, but never got the answer which was intuitive.
Note: I request the readers to not give answers involving calculus, because I don't understand it right now, I will study it in the future.
Answer
The value of K was arbitrarily defined. Actually as momentum was to be defined here, Newton had a freedom of choosing any constant except 0. He could have chosen it to be 2 but that would make the calculations difficult as compared to K = 1. This is what gives you the definition of momentum as mass times velocity. In case Newton would have chosen it to be 2, your definition would have been twice of mass times velocity.
Actually Newton's second law is (F)(∆t) ∝ (m)(v). This was what was experimentally found by Newton by changing one of the parameters while the others remained fixed.
He used
This quantity of (m)(v) was defined as a new thing called Momentum. Now if F remains zero then (m)(v) doesn't change and hence momentum remains conserved.
