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How experiments differentiate between valence quarks and sea quarks The picture that baryon consist of three quarks and mesons consist of a quark,anti-quark pair is inaccurate.However this simple picture is enough to explain many properties of the hadrons as mentioned in the answer here What is the experimental evidence that the nucleons are made up of three quarks?
A more accurate picture is that baryons are made up of three valence quarks plus a sea of quarks,anti-quarks and gluons.It is the sea of quark,anti quark pairs that determine their masses.What is the experimental evidence of the presence of sea quarks ?experimentally how can we prove the existence of sea quarks as opposed to valence quarks?
| Well in a baryon there are three quarks that are not virtual. Then there is the quark and the gluon field which constantly creates quarks and gluons and those are destroyed and created again. For a pion(neutral) there is not even these valence quarks. There is only the quark field. The charged pions are a little different and they can not have virtual quarks only. They have vitual quarks and real quarks(to make sure the charge is all perfect).
| {
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How many images are formed when an object is placed between two plane mirrors with angle $72^\circ$? I'm a little confused here since there are varying answers on the internet, and I cannot find any legitimate sources explaining this problem.
According to what I've seen, the formula is simply
$$
N = \frac{360^\circ}{A} - 1
$$
However, other sources say that $N$ needs to be an odd number (I do not know why), so when $N$ is even, the answer is actually $N+1$.
If I used the first method, then the answer would be $N=\dfrac{360^\circ}{72^\circ}-1=4$. If I followed the second method, then the answer would be $N+1=5$. (I have actually found some people saying it's 4 and others saying it's 5.)
Could anyone clarify this for me?
| The images formed and the object will always lie on a circle. This interesting property can be utilised here.
| {
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Does the wavelength of a particle depend on the relative motion of the particle and the observer? The de Broglie wave equation states:
$$\lambda = \frac{h}{p},$$
where $\lambda$ is the wavelength of the “particle”, $h$ is Plank's constant, and $p$ is the momentum of the particle.
Momentum is usually written $\,p=mv$, where $m$ is the mass and $v$ is the velocity of the particle. But presumably $v$ is the relative velocity between the observer and the particle.
So does this mean that the wavelength of a particle depends upon the relative motion between the particle and the observer?
Or, perhaps more accurately, when a particle is incoming to another particle, in as much as an interaction between the particles depends on their relative speed, or the energy of impact, it thus also has something to do with their relative wavelengths.
Is that a conclusion, or simply a restatement of the premise, using different words that mean the same thing?
| Yes, the de Broglie wavelength of a particle depends on the relative velocity between the particle and an observer. I find it easier to think about the de Broglie frequency instead of the wavelength. They are related by $v=f\lambda$ or $f=v/\lambda,$ where $v=p/m.$
If the particle is moving towards the observer, the frequency appears higher, and if the particle is moving away from the observer, the frequency appears lower. This is just the Doppler effect. Therefore, if the particle is moving towards the observer, the wavelength appears shorter, and if the particle is moving away from the observer, the wavelength appears to be longer. As you said, this will have implications for interactions because the energy is related to the wavelength.
| {
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Question about changing mass under conservation of momentum In class today, my professor was teaching conservation of momentum. One example she used was an open cart rolling on a frictionless track while in the rain. As the cart collects water, the mass increases; and due to conservation of momentum the velocity must decrease, in order to keep momentum constant. However, the way I understand it, since velocity is decreasing, that means it must have negative acceleration, implying net force is no longer 0, implying momentum is not constant. What am I misunderstanding here?
| This is really the same situation as an inelastic collision between two objects, the cart and the water. Before the "collision", the cart has mass $M$ and horizontal velocity $v$, and the falling rain has mass $m$ and horizontal velocity $0$. The horizontal momentum of the system is therefore $P = Mv$.
After the "collision", the cart and the rain have the same velocity $v'$. There were no external horizontal forces acting on the cart and the rain, so the momentum is still $P$, and the velocity $v'$ is given by $P = (M+m)v'$.
Momentum is taken away from the cart and added to the falling rain by the internal forces acting between the water and the cart.
Think about a different situation where the cart collides with a number of small rocks spaced along the track, and each rock sticks to the cart in a totally inelastic collision. In that case, its should be clear that momentum is transferred from the cart to each rock by the equal and opposite impact forces on the cart and the rocks.
In the original problem, it is hard to visualize exactly what is the force between the cart and the water, but the good thing about conservation of momentum is you don't need to work out all the local details of how the rain water sloshes around inside the cart to find the global behaviour of the system - i.e. the change in speed of the cart.
| {
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How does the tiling pattern of magnets influence the effective flux? Problem: A flat irregular shaped surface (A) need be affixed securely to a flat magnetic surface (B). The available means is small disc magnets of uniform dimensions and flux (the flat sides are N and S) to be glued onto (A). For a given thickness of the magnetic surface (B), what pattern of distributing the magnets will produce the strongest adhesive force? (Imagine you want to stick a clothes iron on the fridge door.)
Some example patterns:
1 NNNSSS
NNNSSS
NNNSSS
2 NSNSNS
SNSNSN
NSNSNS
3 NNNNNN
SSSSSS
NNNNNN
*
*The surface is covered in uniform halves.
*Every magnet is surrounded by opposite poles.
*Rows of alternating poles.
Keeping in mind that the shape my prohibit effectively accomplishing a particular pattern (especially ex. 2), will certain patterns be more effective in certain shapes?
For example:
1 N
SS
NNN
as opposed to:
2 N
SN
SNS
| Definitely number '2'.
The flux from multiple magnets is not a constant. For example, if I place two magnets against each other N to N, much of the flux is cancelled. But if I place them N to S, they both add to the external flux. You can then move one of those magnets along the flux lines of the other magnet, keeping the flux aligned, such that the two magnets end up side by side and in opposite directions (the magnet we moved also rotated 180 degrees). So when magnets are placed side by side, you get more external flux when they are pointing opposite directions than you do when they are pointing the same direction, and this is what we see empirically. In fact, the effect is pretty dramatic. Alternating magnets can produce upwards of ten times the near field force than same-direction magnets. Of course, the far field is weaker, but you don't care about that.
Another way to look at this is: When you bring magnets together, if you must add energy to do that then you are reducing the flux, but if bringing them together yields energy, then you are increasing the flux.
| {
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Why does a chain or rope move the way it does when suspended and rotated on a vertical axis? I have always been interested in why objects like chains, ropes, etc. move the way they do when "rotated" around a vertical axis while being held only where it is suspended. It forms a shape if you will, resemblant of a "C" or a wave depending on the length. I am curious to know what laws and effects of physics are at play.
| I think it is hard to study your problem, however, I try to justify why a chain/rope hanging from the ceiling makes a $C$ shape as it dangles freely. For a better perception, simulate each ring of the chain as a separate tiny ball which tends to move on a circular frictionless surface. Due to different radii of these surfaces, the period of oscillation would be different for each of the balls. $(T\approx 2\pi \sqrt{r/g})$ That is, the farther balls need more time to reach the lower part of their assigned circles because farther surfaces have greater radii. This difference in periods causes the distant balls (rings of the chain) from the center (pivot) to lag behind the nearer ones making a C shape.
| {
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In adiabatic expansion does the internal energy of an ideal gas decrease? By First Law of thermodynamic, for an ideal gas, if there isn’t heat transfer, work done by the gas is equal to decrease in internal energy of the gas.
Suppose that I have a perfectly-insulated syringe closed at one end and a frictionless piston on the other. The syringe initially contain ideal gas of volume $V$. If I pulled the piston outward, the volume of gas would increase. Since I am the one applying force, work is done by me instead of by the contained gas. So, in this case, does the internal energy of gas remain constant?
| The ideal gas inside the syringe is always applying a force on the piston, because of its pressure. Therefore, for the piston to remain motionless, there must be another force from the outside that contrasts the push of the gas inside. From the text of your question it seems that in the initial situation, e.g. when the piston is still, you are not applying any force on it (correct me if I'm wrong). This means that there must be something else doing it, otherwise the gas would expand without need of your intervention.
I suppose that your hypothetical experiment takes place on Earth in ordinary conditions, so that there is the air of the atmosphere outside of the syringe. If this is the case that you mean (again, correct me if I am wrong), then the piston is still at first because there's the atmospheric pressure outside that equilibrates it. In this case, your confusion may derive from forgetting the atmospheric pressure.
When you pull the piston you are obviously doing work, but not on the gas inside the syringe. Indeed, its pressure is pushing in the same direction in which you're pulling, so both you and the internal gas are doing work. But against what? Against the atmospheric pressure, which is opposing the expansion. Indeed, as the volume of the ideal gas increases, the volume of the atmosphere decreases. Therefore, both you and the gas are losing energy, in favor of the atmosphere.
| {
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What is the volume of spacetime we can survey? How large is the fraction of spacetime volume we can observe within a sphere with the radius of the present particle horizon distance?
| After studying the many answers (related or not - but many thanks anyway) to my question, I can now summarize my understanding: As the 3D (= 2D surface + time) past lightcone shell has zero spatial depth, its fractional extent within a 4D (= 3D space + time) spacetime object like the particle horizon "volume" must be also zero or infinitly small. -
Cosmic surveys can only map light-emitting events that lie on our past lightcone.
| {
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Why standing wave is needed in microwave oven? In microwave oven, standing wave is created in chamber by reflecting from metal surface. Do we create it intentionally? Is there any special advantages of it?
| The chamber is designed to be a microwave resonator . The first and foremost reason for this would be that it would not be desired to radiate these frequencies out into your living space, as they can heat biological material. For the same reasons, the door has a window with a mesh over it. The wavelengths of the radiation can not escape and enter our eyes, which is more sensitive to this radiation than our skin is. The mesh however allows optical wavelengths to pass through and we can observe how our cooking is going.
Secondly, it would be wasteful to radiate this energy out of the chamber. With a resonator, the energy is kept inside and thus the efficiency of the microwave oven will go up.
| {
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Confusion about expressing an inner product using the Einstein summation convention I think this likely comes down to the following expression,
$$g’^{ab}e’_a e’_b = \delta ^a_b $$
Is this in agreement with the Einstein summation convention? Because even though the two indices are summed over, they still appear on the right.
This leads to issues when trying to rearrange for $g’$
In this case how would you rearrange for $g’$?
| I assume that you're using Latin indices to mean abstract indices, whereas Greek would imply concrete indices. This would be the modern convention (as opposed to in older publications where Latin vs Greek would indicate timelike vs spacelike coordinates).
Why are the primes there?
This is a notational clash between the use of subscripts for two different purposes: (1) to identify what basis vector we're talking about, and (2) as an abstract index. If you just have a list of basis vectors, in no particular order and with no connotation of being unit vectors along a certain coordinate axis, then an index to state which vector you're using from the list is neither a concrete index nor an abstract index in the sense of Einstein summation notation. Let's use $i$ and $j$ for this type of index.
Then a way to notate this without the clash would be $g^{ab}e_{i,a}e_{j,b}=\delta_{ij}$. (Since the $i$ and $j$ aren't concrete or abstract indices, it doesn't make sense to write them as superscripts.)
| {
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Hermiticity of spin-orbit coupling in real space In the Kane-Mele model, the spin-orbit coupling is defined in real space as
$$\sum_{\langle \langle i j \rangle \rangle \alpha \beta} i t_2 \nu_{ij} s^z_{\alpha \beta} c_{i \alpha}^\dagger c_{j \beta}$$
where the sum is over next-nearest-neighbor sites on a honeycomb lattice, and $\nu_{ij} = - \nu_{ij} = \pm 1$ depends on the orientation of the next-nearest-neighbor bonds (I don't believe the details of how $\nu_{ij}$ is calculated is relevant for Hermiticity). I am having difficulty understanding how this term is Hermitian. Taking the Hermitian conjugate seemingly gives
$$\left(\sum_{\langle \langle i j \rangle \rangle \alpha \beta} i t_2 \nu_{ij} s^z_{\alpha \beta} c_{i \alpha}^\dagger c_{j \beta}\right)^\dagger = \sum_{\langle \langle i j \rangle \rangle \alpha \beta} (-i) t_2 \nu_{ij} s^z_{\alpha \beta} c_{i \alpha} c_{j \beta}^\dagger = \sum_{\langle \langle i j \rangle \rangle \alpha \beta} i t_2 \nu_{ij} s^z_{\alpha \beta} c_{j \beta}^\dagger c_{i \alpha}
\\
= -\sum_{\langle \langle i j \rangle \rangle \alpha \beta} i t_2 \nu_{ij} s^z_{\alpha \beta} c_{i \alpha}^\dagger c_{j \beta} $$
where in the final line we have relabeled indices and used that $\nu_{ij} = - \nu_{ji}$, $s^z_{\alpha \beta} = s^z_{\beta \alpha}$. I must be missing something obvious, but this seems to show that the term is anti-Hermitian, instead of Hermitian. What am I missing here?
| In the first expression after the equals sign on the first line you have $c_{i\alpha}c^\dagger_{j\beta}$. It should be $c^\dagger_{j\beta}c_{i\alpha}$ since
$(AB)^\dagger = B^\dagger A^\dagger$.
| {
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What is the $R$-symmetry group for ${\cal N}=6$ supergravity in $D=4$ dimensions? What is the $R$-symmetry group for ${\cal N}=6$ supergravity in $D=4$ dimensions?
| In $D=4$ the $R$-symmetry group is $U({\cal N})$, cf. Ref. 1, which also lists $R$-symmetry groups in other spacetime dimensions $D$.
References:
*
*D.Z. Freedman & A. Van Proeyen, SUGRA, 2012; Table 12.1 p. 240.
| {
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How do I experimentally measure the surface area of a rock? I hope this is the right place to ask this question.
Suppose I found a small irregular shaped rock, and I wish to find the surface area of the rock experimentally. Unlike for volume, where I can simply use Archimedes principle, I cannot think of a way to find the surface area. I would prefer an accuracy to at least one hundredth of the stone size.
How can I find the surface area experimentally?
| *
*Fully wrap stone really tight in aluminium foil. (Of course it will crinkle; press the crinkles tightly down.)
*Soot the whole thing with a candle, just enough so it's completely black.
*Carefully unwrap foil.
*Photograph flattened foil together with a reference scale square. Make sure there's a light background (e.g. white ceiling) opposite the foil, so it'll appear bright on the photo in the un-sooted area.
*Measure area of soot, using image processing software. This can be done by first using a perspective-correction tool, noting the size of the reference square, then cropping the relevant area and showing a histogram of the brightness values.
Instead of soot, you could also use spray paint, but it would likely sip into the crinkles more. Or you could wrap in paper instead of alu and use a pencil, but that would smear and be harder to see on the photo.
I don't think this method will get 1/100 accuracy, but it gives at least a decent estimate and doesn't require special equipment.
| {
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When the direction of a movement changes, is the object at rest at some time? The question I asked was disputed amongst XVIIe century physicists (at least before the invention of calculus).
Reference: Spinoza, Principles of Descartes' philosophy ( Part II: Descartes' Physics, Proposition XIX). Here, Spinoza, following Descartes, denies that a body, the direction of which is changing, is at rest for some instant.
https://archive.org/details/principlesdescar00spin/page/86
How is it solved by modern physics?
If the object is at rest at some instant, one cannot understand how the movement starts again ( due to the inertia principle).
If the object is not at rest at some instant, it seems necessary that there is some instant at which it goes in both directions ( for example, some moment at which a ball bouncing on the ground is both falling and going back up).
In which false assumptions does this dilemma originate according to modern physics?
| After the invention of modern calculus and notions like continuity and differentiability, the answer is quite trivial in Newton's formulation of mechanics assuming the body is moving along a line. The second derivative of the position should be always defined as it equals the total force acting on the body. Therefore the first derivative must be continuous. This derivative is the velocity. You are assuming that it changes its sign passing from time $t$ to time $t'$. A continuous function defined on an interval which changes its sign at the endpoints of the interval must vanish somewhere in the interval. That is an elementary result of Calculus. In summary, the body must be at rest at some time, if accepting the modern version of Newton's formulation of classical mechanics.
The objection that if the body stops at a certain instant, then the direction of the motion immediately after that instant cannot be decided is untenable. The direction is actually decided by the acceleration, that is by the total force acting on the body at the said instant.
| {
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What determines whether we use a vector or scalar potential? I understand that electrostatic potential is scalar because the curl of the field is zero, and this implies the electrostatic field is the gradient of the scalar potential to satisfy this. Similarly the divergence of a magnetostatic field is zero so a magnetostatic field is the curl of the vector potential.
But what actually determines when you would use which potential? Is it purely to do with these definitions in electro and magneto statics, or is it something else?
| The good hint is in the scalar or vectorial properties of the sources.
For the magnetic field, the sources are currents, which are vectors since currents flows in specified direction: it’s no surprise the associated potential should be a vector.
For (time-independent) electric fields not induced by magnetic fields, the sources are charges, which are scalar (one only needs to specify the magnitude of the charge) so it’s no surprise the associated potential is a scalar.
Of course if you have induction, i.e. if the $\vec E$ is due to a changing $\vec B$, then $\vec E$ depends on the vector potential which is the source of $\vec B$.
| {
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Why do we feel weightlessness during free fall? An object in free fall accelerates towards the Earth at a acceleration equal to $g$ (the accleration due to gravity). Now if we ignore the air resistance, why do we feel weightlessness? I could not understand the reason that we could not feel the pulling force that the Earth exerts on the object.
| When in free fall, neglecting air drag, the only force acting on your body is gravity, which is a non contact force, and you feel weightless. In other words the feeling of “weightlessness” is that of not experiencing the sensation of any contact forces. Stand on the ground and you feel the upward contact force of the ground on your feet that opposes and equals the downward force of gravity, $mg$. Since our bodies are not rigid (they are deformable) they will undergo some degree of compression depending on its orientation and the location of the contact force.
The possibility of experiencing tidal forces in free fall has been mentioned. Tidal forces are due to changes in the force of gravity as a function of the distance from the center of mass of the gravitating body. In other words due to a gradient in the force of gravity.
For a person in vertical free fall near the earth, the force of gravity will be greater on the feet than the head, which could theoretically stretch the person. But for the earth the gradient in the gravitational force compared to the dimensions of the body is so small as to be negligible.
To put things into perspective. The acceleration due to gravity at the surface of the earth is about 9.81 m/$s^2$, depending on location. At 40 km above the equator it is about 9.67 m/$s^2$ or a 1.4% difference.
Of course the effect near a massive black hole is a different matter.
Hope this helps.
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Newton's 3rd law in lami's theorem I'm a high school physics student dealing with the following problem:
Despite solving the problem correctly, I don't understand the diagram (b) supplied with the problem:
These are my questions:
*
*Why are vectors ${ F }_{ RB }$ and ${ F }_{ RC }$ in opposite directions in diagram (b)? The textbook explains that it used Newton's 3rd law of motion, but I have no idea how.
*Newton's 3rd Law states that after the student applies a force on the rope, the rope would apply an equal but opposite force on the student. Where is that force in the diagram? Do we not consider it because it's acting on the student and not the rope?
| There's a tension in the rope that exerts a force in both directions.
Think about it this way: suppose you're participating in a tug-of-war with your friend. You pull him, and he pulls you. Although you're exerting a force on the rope, the rope is certainly also exerting a force on you (that's why you feel you are pulled forward). Similarly, your friend also feels a force pulling him forward. What's happening is that the tension in the rope is exerting a force in both directions.
You see something similar here. For the midpoint of the rope, it's being tugged 1) by the rock and 2) by the car, and it's being pushed by the student. Therefore there are three forces, all of them leading out of the midpoint. This explains the second diagram.
Again, the tension in the rope acts in both directions, which is why $F_{RB}$ and $F_{BR}$ can seem to act in opposite directions (but they are certainly of equal magnitude).
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Is there a way to construct a Hamiltonian from a set of DE? Let's say I have a set of first-order differential equations for set of position $x_i$ and conjugate momenta $p_i$, which might be complicated and time-dependent
$$ \dot{x}_i = f_i(x_j,p_j,t)$$
$$ \dot{p}_i = g_i(x_j,p_j,t)$$
and I know that these equations originate from some Hamiltonian (that is, they respect conservation of phase-space volume, for example). Is there a constructive way to find said Hamiltonian? If not, what are the minimal conditions that the set of equation have to fulfill for this to happen? I mean I know that if the equations are linear and time-independent I can in general invert them and find $H$ in a straight-forward manner but what is the general case?
| Well, given
$$
\dot{p} = - \frac{\partial H}{\partial x}
\quad
\text{and}
\quad
\dot{q} = \frac{\partial H}{\partial p}
$$
we have
$$
H = - \int dx_{i} \, g_{i}(x,p,t)
$$
and
$$
H = \int dp_i \, f_{i} (x, p, t).
$$
You deal with the constants of integration using the obvious constraint:
$$
- \int dx_{i} \, g_{i} (x, p, t)
=
\int d p_{i} \, f_{i} (x, p, t)
$$
I guess the condition to find $H$ is whether we are able to solve these integrals or not.
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Why does Heisenberg’s Uncertainty Principle not take wavelength into account? In class today we were taught about Heisenberg’s equation, $\Delta x\Delta p\ge\frac{h}{4\pi}$.
Why isn’t De Broglie wavelength a factor here - or is it, but it’s represented behind the deltas instead? After all, if we’re dealing with indeterminism, isn’t that where the wave part of particle-wave duality comes into play?
| It was precisely De Broglie who hypothesized that $p=h/\lambda$. So you can also write $\Delta x / \Delta \lambda = 1/4\pi$.
| {
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Do liquids continuously boil in a pressure cooker? First, I understand that the boiling point of water is increased as the pressure is increased. At 15 PSI, the boiling point is about 250F.
My question, which seems like nobody can answer with certainty, is this: Is there a continuous boil within the pressure cooker once it reaches temp/pressure?
I understand that there needs to be a boil at some point to create the steam. But once it reaches that pressure, then what? An electric PC will turn the heating element off until the pressure drops. A user will turn the heat down on an analog PC. Does the liquid just boil constantly while the heat is off? It probably maintains temp & pressure for a lot longer in a sealed environment. In my mind, if there were a continuous boil, that steam can only build so much until it exceeds the pressure that the PC is capable of maintaining and it begins to vent. So, sure you could have a constant boil, but that means it would also be constantly venting, right?
| It seems likely that the lid of a pressure cooker would be at a slightly lower temperature than the heated bottom. This could allow the steam from a very slow boil to condense without venting.
| {
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Is Newton's first law an "If and only if" On Wikipedia, Newton's first law is stated as:
In an inertial frame of reference, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force.
I read this as $$\mathbf{F}_{\textrm{net}}=0\Rightarrow \mathbf{a} = 0,$$ but does it also mean that $$\mathbf{a}=0\Rightarrow \mathbf{F}_{\textrm{net}}=0,$$ or do you need the second law for that? I would argue that you need the second law, as the first law doesn't say anything about what happens if there is a net force. A force could cause acceleration in some situations, but not in others. Wikipedia seems to disagree since the following can be read further down the page:
The first law can be stated mathematically when the mass is a non-zero constant, as, $$\sum \mathbf {F} =0\;\Leftrightarrow \;{\frac {\mathrm {d} \mathbf {v} }{\mathrm {d} t}}=0.$$
So, who is right?
| In my personal opinion, second law is more or less enough, as other two laws can be derived from it.
First law:
$$ \sum\vec{F} = m\frac{\textrm{d}\vec{v}}{\textrm{d}t} = 0 ,\\ \Rightarrow \vec{v}=\textrm{const}
$$
Third law
$$
\vec{F}_{A\to B} \neq 0, \vec{v_{_B}}=\textrm{const},\\
\Rightarrow \vec{F}_{A\to B} + \sum_i\vec{F_i} = 0 ,
\\ \Rightarrow \sum_i\vec{F_i} = - \vec{F}_{A\to B}
$$
In words: If body A exerts force on body B, but body B still keeps it's speed constant, then there must exist superposition of other forces which acts like original force, just in opposite direction. Otherwise, second Newton law would be crushed to peaces, because bodies must react to forces applied. Of course this formulation is not exactly the Third law, because superposition of other forces is involved, thus it's not clear from it if B exerts some force on A too, or just some external force(s) to B is applied or both. In any way, this "third law replacement" comes as a direct conclusion from second law.
| {
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Do the logarithmic corrections to black hole entropy imply corrections to their energy or temperature? Could somebody help guide the thinking in this situation?
Do the corrections to entropy $S$, like those of https://arxiv.org/abs/hep-th/0111001, affect the temperature of a black hole, or its mass, or both, or neither?
The Schwarzschild black hole entropy $S_{BH}$ with logarithmic corrections is given by
$$
S_{BH} = S_0 + (3/2) \ln S_0
$$
where $S_0= (c^3 A) / (4G\hbar )$ is the Bekenstein Hawking entropy, and $A$ the area of the black hole.
So the question is: is there also a logarithmic correction to mass (energy) or to temperature of the black hole? Why or why not?
The entropy of a black hole is thus not given exactly by "one quarter of the area". There are logarithmic corrections to that statement. The question is:
(1) Is the mass of a black hole exactly "proportional to its radius", or are there logarithmic corrections to this statement?
(2) Is the temperature of a black hole exactly "inversely proportional to its radius" or are there logarithmic corrections to this statement?
| It seems that the answer is that only entropy has logarithmic corrections. Neither energy nor temperature has such corrections. The logarithmic corrections of entropy are due to statistical fluctuations. These fluctuations have no effect on temperature or mass/energy.
| {
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Principle of stationary action vs Euler-Lagrange Equation I am a bit confused as to what I should use to derive the equations of motions from the lagrange equation.
Suppose I have a lagrange function:
$$L(x(t), \dot{x}(t)) = \frac{1}{2}m\dot{x}^2-\frac{1}{2}k(\sqrt{x^2+a^2}-a)$$
Method 1: Principle of least action
$$\delta L = \delta \dot{x}(m\dot{x})-\delta x \frac{kx(\sqrt{x^2+a^2}-a)}{\sqrt(x^2+a^2)}$$
$$\delta W = \int_{t_0}^{t_1} \delta L \ dt$$
After doing integration by parts, i obtain:
$$\delta W= -\int_{t_0}^{t_1} \delta x \biggl[m \ddot{x}+\frac{kx(\sqrt{x^2+a^2}-a)}{\sqrt(x^2+a^2)} \biggl] dt$$
for stationary points, $\delta W = 0$
Hence, inside the integral,
$$m\ddot{x}+\frac{kx(\sqrt{x^2+a^2}-a)}{\sqrt(x^2+a^2)} = 0$$
and this is the equation of motion.
Method 2: Euler-Lagrange Equation
Alternatively, we can consider the euler lagrange equation:
$$\frac{\partial L}{\partial x} - \frac{d}{dt}\biggl(\frac{\partial L}{\partial \dot{x}} \biggl) = 0$$
By substituting $L$ into the euler-lagrange equation, we get the same equation of motion:
$$m\ddot{x}+\frac{kx(\sqrt{x^2+a^2}-a)}{\sqrt(x^2+a^2)} = 0$$
So method 2 is a lot easier than method 1, but why do we arrive at the same answer? I have a hunch both methods are essentially calculating the same thing, but I am not sure if this hunch is right because the euler lagrange equations seems a bit too simple as compared to principle of least action. Is there something i'm missing here?
| First, I think there is something wrong with your partial derivative of the Lagrangian with respect to $x$.
Second, the Euler-Lagrange equations are nothing more than the process that you performed in Method 1, done without committing to a specific form for $L$ but leaving it generic. In your first step you took partial derivatives of $L$ with respect to its position and velocity terms, in your second step you took the velocity derivative and involved it in an integration by parts, where you took a total time derivative and then added a minus sign. If your Lagrangian also involved $\ddot x$ you would then have a $+\frac{\mathrm d^2~}{\mathrm dt^2}\frac{\partial L}{\partial\ddot x}$ term from two integrations by parts, for example.
| {
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Are the EM radiation categories objective or subjective? We were taught at school that EM radiation can be categorised as radio waves, microwaves, infrared waves, visible light, ultraviolet radiation, X-rays, and gamma rays, but how arbitrary is this categorisation?
I understand that for humans the 'visible light' category is an obvious, but how did scientists justify drawing a line in the sand between, say, the frequencies constituting radio waves and those constituting microwaves?
Would an advanced alien civilisation divide the EM spectrum in the same way that we do or could they have divided and categorised it in a whole different way than us?
| Any classification of electromagnetic radiation is probably going to have at least a few basic components, which can be derived purely from observing the way radiation interacts with atoms and molecules:
*
*Ionizing radiation vs. non-ionizing radiation: Radiation above a certain frequency has the ability to ionize atoms.
*Radiation emitted by nuclei vs. radiation not emitted by nuclei: Radiation above a certain frequency is generally seen only as gamma rays emitted during radioactive decay.
*Radiation corresponding to atomic transitions vs. radiation corresponding to molecular transitions: Higher-frequency radiation tends to excite electrons bound to individual atoms, whereas lower-frequency radiation tends to excite electrons in molecular bonds.
So, you can divide the electromagnetic spectrum pretty un-controversially into the following general bands, from lowest frequency to highest:
*
*Radiation not energetic enough to excite molecular transitions (radio to lower microwave).
*Radiation that can excite molecular transitions, but not atomic transitions (lower microwave to upper infrared).
*Radiation that can excite atomic transitions, but not ionizing radiation (upper infrared to visible/lower ultraviolet).
*Ionizing radiation not associated with nuclear transitions (ultraviolet to low-energy X-rays).
*Ionizing radiation associated with nuclear transitions and above (mid-energy X-rays and gamma rays).
The precise location of the divisions between these bands is entirely subjective, but any intelligence that has sufficient experimental data to notice that molecules, atoms, and nuclei all respond differently to different types of electromagnetic radiation will probably have something similar to this classification in place.
| {
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Schrieffer-Wolff Transformation for conventional superconductors I was trying to follow the discussion in Radi A. Jishi's book (Feynman Diagrams in Condensed Matter Physics), Chapter 12 on superconductors. They basically have a Hamiltonian that comprises of a quadratic electron part and a quadratic phonon part, which is coupled through the usual electron-phonon coupling. They then perform a Schrieffer-Wolff transformation to conclude that there is an effective interaction between electrons that can become attractive for appropriate momenta.
It seems that in this procedure one does a canonical transformation on the Hamiltonian to eliminate the term linear in the electron-phonon coupling strength (and replace it by more complicated higher order terms). However, the phonons are not completely integrated out. Moreover, this is not the original Hamiltonian - but a canonically transformed Hamiltonian. How does then one conclude that the original Hamiltonian also results in an effective interaction between electrons that can become attractive for appropriate momenta?
(Note: I do appreciate the effective e-e interaction once one completely integrates out the phonons, eg. as done in Bruus-Flensberg or Altland-Simons.)
| As you describe, the Schrieffer-Wolff transformation does not "integrate out" the phonons in the sense of a path integral or partition function. Instead, the transformation returns an effective Hamiltonian where electrons and phonons are decoupled, up to 2nd order in perturbation theory. The phonons are still there, but they interact with the electrons only at 3rd order in perturbation theory, so the coupling can be ignored. This is especially true if your focus is on the electron dynamics, and the electrons are within the energy range where the effective interaction is attractive. In this regime, their attraction gives rise to the Cooper instability, which dominates their low-energy behavior (i.e., induces a superconducting transition).
| {
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Ground → ship Wi-Fi bandwidth in my fast moving spaceship I have a hypothetical question that hopefully makes sense considering only the rudimentary amount of knowledge I have on relativity.
Assume I am in a spaceship orbiting around a spherical satellite, at a certain fixed radius. The satellite sends data to the ship via radio (something like satellite internet or other long-range data radio, using electromagnetic signals).
Would the download speed (received data rate seen on-board ship) be faster or slower, compared to when I am stationary with respect to the satellite?
Also consider that attenuation is negligible.
What I initially thought was that time dilation would make "my" clock in the spaceship tick slower, hence the signal from the "satellite" would be received faster. I thought I should add this to clarify my train of thought that led to the question. Please do note that it is a hypothetical question.
(A comment pointed out that a spherically symmetric EM wave is impossible, therefore I have taken down the diagram and the statement regarding so.)
| If you are in a perfectly circular orbit, then your received signals are affected by the relativistic transverse Doppler effect.
The signals you would receive would be blueshifted by the Lorentz factor of the orbital speed. That is the information transferal rate would (potentially) go up (if you can tune to the new frequency).
If you are in a Keplerian orbit, then (in a Newtonian approximation)
$$ \frac{mv^2}{r} = G\frac{Mm}{r^2}$$
$$\frac{v}{c} =\sqrt{\frac{GM}{rc^2}} = \sqrt{\frac{r_s}{2r}},$$
where $r_s = 2GM/c^2$.
The transverse Dopper blueshift is by a factor of
$$ f_r = f_t\gamma = f_t\left(1 - \frac{v^2}{c^2}\right)^{-1/2} = f_t\left(1 - \frac{r_s}{2r}\right)^{-1/2},$$
where $f_r$ and $f_t$ are the received and transmitted frequencies - assuming that the transmitter is at $r=0$ or at least stationary with respect to the orbital speed.
However, in General Relativity there is also a gravitational redshift given by
$$ f_r = f_t \left( \frac{1 - r_s/r}{1 - r_s/r_t}\right)^{1/2},$$
where $r_t$ is the position of the transmitter (which has to be at $r_t > r_s$) and this may modify the conclusion.
Which will win out depends on the radius of the orbit compared with the radius of any "surface" you put your transmitter on.
| {
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Showing that the $A$-$j$ coupling in classical electromagnetism is gauge invariant I am attempting an early exercise from Altland's Condensed Matter Field Theory. The electromagnetic field's action is given as:
$$S[A]=\int d^4x(c_1F_{\mu\nu}F^{\mu\nu}+c_2A_\mu j^\mu),$$
and I wish to show that the second term is invariant under a gauge transformation $A_\mu\to A_\mu + \partial_\mu \Gamma$. It is hinted that I should use integration by parts and the continuity equation $\partial_\mu j^\mu$. Doing so, I can show that:
$$c_2\int d^4x(A_\mu+\partial_\mu\Gamma)j^\mu=c_2\int d^4x A_\mu j^\mu+c_2\int d^4x\partial_\mu(\Gamma j^\mu),$$
where I now must show that the second term on the right-hand side is zero to prove gauge invariance. I am not sure how to show this - how should I proceed? Is there some boundary condition that I'm missing?
| *
*Assuming the continuity equation $d_{\mu}J^{\mu}=0$, the gauge symmetry is more precisely a gauge quasi-symmetry, meaning that the action is only invariant up to boundary terms.
*It seems relevant to stress that no boundary conditions are imposed in Noether's theorems. In contrast, boundary conditions are necessary for the principle of stationary action. This point is also made in my Phys.SE answer here.
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How can the mechanism of electrons in an atom be explained? I am a high school student who takes both Physics and Chemistry.
Recently I learnt about the quantum mechanical point of view of looking at electrons or nuclei. I also learnt that the wave functions can be obtained by solving the Schrodinger's equation with various conditions specific to the problem (such as the particle in a box).
My shallow understanding of quantum mechanics is that we can only know the probability of an electron existing at a certain position and time, and the actual position can be determined when the 'observation' takes place.
The chemical bondings and chemical reactions are the results of electric interactions between nuclei and electrons. The Coulomb force is a function of the distance between two charges, so it is important that the exact locations of electrons should be known. But taking into consideration quantum mechanics, we don't even know where the electrons are, and we built up a subject called Chemistry, and most importantly, CHEMISTRY STILL WORKS VERY WELL.
So, what is going on?
| In addition to the comments mentioned, when you solve for 2 electron problems in quantum mechanics, you do include a term of the form $\frac{kq_1q_2}{r_{12}}$ which represent the interaction between the two wavepackets. For more information on how the 2 electron system is solved, see https://en.wikipedia.org/wiki/Two-electron_atom#Schr%C3%B6dinger_equation. Moreover, without quantum mechanics, you can't explain phenomena like superconductivity and superfluidity. Even the transistors in your computer require quantum mechanics to work.
| {
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"Contradiction" about Hubble parameter $H_0$ in the future? The Hubble parameter $H_0$ is expected to level off asymptotically in the far future as dark energy becomes dominant over matter, radiation, and space curvature. I think it's predicted to go down to 55 km/s/Mpc.
Yet dark energy is also the cause of a galloping acceleration of distant galaxies in the visible universe. Doesn't that imply the straight-line giving H0 today of 67-70 on a Hubble expansion graph will bend upwards to make a greater slope. Doesn't that mean H in the far future will increase dramatically?
I teach this stuff to adult amateur astronomers and still can't make sense of this "contradiction".
| The Hubble parameter does not measure the rate of change of expansion. It is the ratio of the expansion rate to the size of the universe. $H(t) = v(t)/d(t)$.
If the expansion is accelerating, then the size of the universe gets bigger faster than the expansion rate gets bigger - hence the Hubble parameter gets smaller.
e.g. If you jump off a cliff, what happens to the ratio of your velocity, to the distance you have fallen?
$$ v = gt, \ \ \ \ \ \ d = gt^2/2$$
and hence $v/d$ gets smaller as $1/t$, and ignoring pesky things like the ground and terminal velocity, would head towards zero.
The accelerating universe behaves slightly differently in that $v$ tends towards an exponential with time as the vacuum energy becomes dominant:
$$ v \rightarrow \exp( \alpha t)$$
Hence
$$ d \rightarrow \left(\frac{1}{\alpha} \right) \exp(\alpha t)$$
and
$$ H(t) = \frac{v}{d} \rightarrow \alpha$$
The net behaviour is similar. $v/d$ starts out big, but trends asymptotically to a fixed value as the velocity of expansion accelerates exponentially. That fixed value turns out to be $\sqrt{\Lambda/3}$.
| {
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What is orbit that could place satellite to be always under the Sun? I just thinking that, if we could place satellite to orbit earth in opposite direction of earth rotation, inverse of geostationary orbit. If we carefully choose a speed to sync with earth rotation, it would have that satellite stay exactly the same position in the sky as the sun
What is that altitude and are there any name of that orbit? Or is it just the same as geostationary orbit?
| It wouldn't be geostationary because we would see it move across the sky. I think you mean geosynchronous, which is what the orbit would almost be. If you wanted it to always stay in front of the sun from some special vantage point, then you would need to somehow have the satellite correct its orbit based on the changing tilt of the Earth.
| {
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How do physicists compare the relative strengths of the four forces? Since the four forces are different, with different force carriers, how are they (seemingly) directly compared? I often read that the weak force, for example, is many orders of magnitude stronger than gravity, and that electromagnetism is several orders of magnitude stronger than the weak.....
I do notice that different sources give different ratios when comparing the four forces' relative strengths, though....
| It is the history of how the SU(3)xSU(2)xU(1) standard model of physics was built up making a consistent whole of the grand majority of data up to now, and giving good predictions.
A whole lot of scattering experiments showed that there were different strengths and different lifetimes in the interactions, and these are described with quantum field theory, where each interaction and decay is a sum of Feynman diagrams, starting from the first order,which gives the gross behavior, and going as far as calculations can be made or are needed of higher orders.
Feynman diagrams have vertices, and these vertices in the integrals represented by the Feynman diagrams have different strengths, coupling constants, which are shown in the table here.
The column "strength" is the value used in Feynman diagram vertices.
The first two, electromagnetic and weak, are the ones working well with the series expansion of the interaction into feynman diagrams. The lowest order are the exchanges of the gauge particles, that is why they are identified with the force (first column). QCD the strong force, can only be figuratively shown in Feynman diagrams. The actual calculations need different tools, because the strong coupling constant is 1. Lattice QCD is used for that. Gravity is not yet definitively quantized but there are effective quantizations.
In electromagnetic and gravitational interactions there exist mathematical tools that connect the classical constants with the field theoretical ones. Strong and weak interactions belong only to the study of particle interactions.
So , in a nutshell, it is the fitting of data with a specific standard model that organizes the particle interactions in line with four forces. For example see this summary..
| {
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Different apparent brightness of a distant star from a moving frame - an apparent paradox Consider a stationary star at a distance $L$ from the Earth.
A spaceship at Earth moving with velocity $v$ towards the star will find the star to be at a distance of $\dfrac{L}\gamma$ in its own frame.
But it won't actually see the star at that distance since the speed of light is finite.
The light that it will get at that instant will be from a star that was at distance $\dfrac{Lc}{\gamma(c-v)}$ which is $>L$.
At $v=0.6c$, $\dfrac{Lc}{\gamma(c-v)}=2L$.
Therefore, the spaceship will see the star $0.25$ times as bright as an observer on the Earth.
But this is unexpected since both spaceship and the Earth are bathed in the same starlight and if anything, the observer on the Earth sees the spaceship gather more photons due to its motion.
How to resolve this apparent paradox?
| You can't do special relativity just using gamma! If you really want to know what stars look like when you are traveling quickly relative to them, you need to use the correct method.
I would suggest this source - the maths is as simple as it can be, but I'm not saying it will be an easy ride! The meat of the answer begins at the paragraph: "Meanwhile, Einstein's 1905 paper on the electrodynamics of moving bodies . . .", but it is well worth reading the whole thing. To summarize: using the correct theory, there is no paradox.
| {
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Might a cast iron pan set on top of a microwave oven affect the operation? Our microwave seems to take longer to get the job done these days. I notice that someone is storing a large, heavy cast iron pan on the top of the microwave. Is there any way at all that it is possible that the iron pan is interfering with the microwave?
I could try with/without the pan but it isn't that dramatic.
| Assuming that you have not drilled holes into your microwave (that can be really dangerous and should never be done) the enclosure forms a Faraday cage from which the microwaves cannot escape, therefore they can also not interact with the iron pan.
However, with some imagination one could certainly come up with weird reasons why a heavy weight on the top might influence, let's say, a part of the control circuit. Therefore I suggest you replace the iron pan with a stone of similar mass to assess if it is the weight or the ferromagnetic and conductive iron that causes the effect.
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Energy cannot be created nor destroyed, but can it be lost in expanding space? It is said that there are extremely distant stars and galaxies that we will never see, because overall, the space between us is expanding faster than their light travels towards us. Does this mean some radiation can travel forever through expanding space, and never reach a mass?
| I do understand your question and your confusion.
Photons are massless particles, they travel along null geodesics, meaning that the spacetime distance between emission and absorption is 0.
You are thinking about information (photon) being emitted somewhere, and then its very purpose should be to receive (absorb) it.
Though, expanding space can somewhat change that, since some photons might never be absorbed, received because the space as you say along their travel direction expands faster then the speed of light. In this case, and I think this is what you mean, this information carried by these photons might never fulfill its purpose. It will never be absorbed, received.
But, the information these photons carry, will not be lost physically. They will just in a way be lost in space, so you are correct. They will never be able to fulfill their very purpose to be received, absorbed.
Yes, the expansion of space itself is allowed to exceed the speed-of-light limit because the speed-of-light limit only applies to regions where special relativity – a description of the spacetime as a flat geometry – applies. In the context of cosmology, especially a very fast expansion, special relativity doesn't apply because the curvature of the spacetime is large and essential.
Can space expand with unlimited speed?
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Spin of higher string excitations I am trying to better understand spin in string theory.
There is already a good post here:
Ssecond and third level excitations of open string
But I have these questions:
1) Why would the second excited stated be spin-2?
If I use the spin operator then I find states which have
spin eigenvalues of +/-2, of +/-1 and of 0.
Take the spin operator $E^{\mu\nu}= -i\sum_{n=1}^{\infty}\frac{1}{n}(\alpha^{\mu}_{-n}\alpha^{\nu}_n-\alpha^{\nu}_{-n}\alpha^{\mu}_n)$ in some specific direction.
Spin 1: $[E^{ij},\alpha_{-2}^i+i\alpha_{-2}^j]=\alpha^{i}_{-2}+i\alpha^{j}_{-2}$
so that $E^{ij}(\alpha_{-2}^i+i\alpha_{-2}^j)|0>=(\alpha_{-2}^i+i\alpha_{-2}^j)|0>$
Spin 0: $[E^{ij},(\alpha_{-1}^i)^2+(\alpha_{-1}^j)^2]=0$ so that $E^{ij}((\alpha_{-1}^i)^2+(\alpha_{-1}^j)^2)|0>=0$
2) When you compactify from $D=10$ to $D=4$, how does that affect spin?
3) What are explicitly the few first levels of the massive spin states for the superstring?
| You need more careful extract irreducible representation of little group.
See for example 4.2.2. in Gleb Arutunov notes Lectures on String Theory
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Phase transition on magnetic materials Is ferromagnetic to paramagnetic phase transition a reversible process?
If I start with a ferromagnetic material with a spontaneous magnetization below the Curie temperature, and then I start to heat it, it will become paramagnetic above the critical temperature. If I then start to drop the temperature slowly to below the Curie temperature then will I achieve the ferromagnetic behaviour with same spontaneous magnetization as before?
| Yes, locally it will have the same magnetization as before due to spontaneous ferromagnetic ordering within magnetic domains. But without an applied magnetic field, the domains will arrange in such a way that the large-scale magnetization will be zero.
I once made this Curie pendulum: https://www.youtube.com/watch?v=CvIGr3wFVgo
This can be compared with melting a nice single crystal of for example silver. Cooling it down will give an ordinary polycrystalline lump.
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Roche lobes in binary star systems While studying mass transfer in binary star systems, I came across the concept of Roche lobes and the role of the inner Lagrangian point $L_1$, as shown in the adjoining figure.
However, I am having a doubt in understanding the reason of the fact that one lobe is larger and the other is smaller. I think that the reason is due to the different masses of the two stars. Since the contours are equipotentials, it seems that larger the mass of a star, the larger is the size of the corresponding lobe. But I am not sure whether this is the exact reason.
So, I am looking for a proper reason for the different sizes of the two lobes.
| You are right: the size of the Roche lobe depends on the mass of the components. It scales linearly with the separation of the components and in addition is a function of the mass ratio. See https://en.m.wikipedia.org/wiki/Roche_lobe
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What happens when a weapon breaks on impact? Here is the situation:
You are attacking someone with a wooden pole (such as a pole arm or tree branch). You either (1) hit as hard as you can and the pole breaks into two pieces on impact OR (2) hit quite hard but the pole remains in tact. Assume that you are hitting the same spot with the same angle and everything, so the only difference is how much force was applied.
Which would cause more damage?
My gut instinct is that a weapon that breaks would transfer less of the impact to the person it hits, thereby causing less damage. However my boyfriend pointed out that if you are hitting someone hard enough that the pole breaks, you have used the maximum amount of force that the pole can withstand, so the most force you can transfer with a single blow has been done.
[Edit: I don't have an education in physics beyond high school, but it looks like this question has never been answered on this site before. That could be because I didn't know the physics terms to search for though! Any pointers in the right direction are appreciated.]
| The force applied on an object is proportional to the change in momentum:
$F = m (v-u)/t = dp/dt$
Momentum is a vector, so direction matters. The largest change in momentum happens when the object you're swinging reverses direction. For example, if you're swinging a stick at a pillar, if it bounces backwards after contact, that exerts the largest force.
If the stick breaks, however, the final momentum is no longer negative, but zero (or close to it). Therefore you exert less force and inflict less damage.
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Is time slower on the far galaxies? Since far galaxies move away faster, what would be the speed of their time relative to us?
If there is a difference:
*
*What determines whose time would be faster?
*(If I haven't understood it wrong) To resolve the twin paradox, acceleration is required. Is the expansion of the universe creates acceleration? If not how can we explain this difference?
I think following question is the same:
Imagine we have a some kind of machine that can bend the space-time, and create a gravitational pull in front of us. So we gain speed without feeling any acceleration. If one of the twin in the twin paradox would use such a machine to gain speed; what would happen?
I don't have a physics or math background, and I hope my question makes sense.
| Time measured in the frame of a galaxy that is moving relative to us would appear to us to be running slowly. The effect is reciprocal, so our time will appear to be running slowly to observers on the galaxy moving with respect to us. For both sets of observers time ticks away at the same rate in their respective reference frames- there is no difference that needs to be explained. The effect of the acceleration is to increase the relative speed at which the two sets of observers are moving, which will have the effecting of increasing the apparent time dilation effect.
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Witten's description of WZW conformal blocks I am reading this paper by Witten - Geometric Langlands From Six Dimensions. In section 4.1, he gives a description of the vector space of conformal blocks of the current algebra associated to a simply-laced and simply-connected Lie Group $G$ at level 1. He claims that on a Riemann Surface $W$, this vector space is the Heisenberg representation of the Heisenberg group associated to $H^1 (W, \mathcal{Z})$ where $\mathcal{Z} = \pi_1 (G)$. Later he shows that the dimension of this vector space is $(\#\mathcal{Z})^g$, which I have checked indeed coincides with the dimension computed through e.g. the Verlinde formula.
My questions are as follows - (1) What is the reason behind the appearance of $H^1 (W, \mathcal{Z})$ in this context?
(2) Does the vector space of conformal blocks have an interpretation in terms of a Heisenberg group associated to a cohomology group when the level is not equal to 1?
| I am no expert on this, but it sounds as though the blocks are labeled by the Wilson lines around the 1-cycles. This is what happens on a torus.
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How come Goldstone boson, PQWQ axion, be able to have mass at all? Quote:
Goldstone's theorem: For every spontaneously broken continuous symmetry, there is a massless particle created by the symmetry current.
However, under $U(1)_{PQ}$ symmetry, I read that PQWW axion can obtain mass from $G\tilde{G}$.
Both these sentence make sense, but, being a Goldstone boson, there seemed to be a contraction around PQWW axion.
I read that this might be connect with the approximation from t' Hooft's determinental interaction. Could you explain to me what the determinental interaction was?
Does that mean Goldstone's theorem only works for the first order of t' Hooft determinental interaction?
Does PQWW boson has mass or not?
Could you explain to me why a Goldstone could every obtain a mass?
| Like the pions, the axion is a pseudo-Golstone boson, so it has a mass. The $U(1)_{PQ}$, as a chiral symmetry, is broken by quantum corrections by the axion's coupling to gluons, and so is no longer an exact symmetry. Goldstone's theorem them does not apply. This is completely analagous to the so-called '$U(1)$ problem', where the $\eta '$ was seen to be too heavy to be a Goldstone boson. The resolution of this puzzle by 't Hooft made physical the CP-violating $G\tilde{G}$ term in the Lagrangian, which is what suggests a QCD axions in the first place.
| {
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Observed speed of a receding light source Let’s say there’s a planet 4 light years away from Earth and we send a rocket ship towards that planet at 99.9% light speed. We stay behind on Earth and watch the rocket ship travel towards the other planet.
Eventually we should be able to see our rocket ship reach it’s destination. How much time will have elapsed for us until we see that occur?
My intuition would say about 4 years. But I also know that when we observe such a far-away planet, we are ”seeing it as it was 4 years ago”. Well 4 years ago the rocket was still on Earth, so how can I be seeing it landing on the planet now?
Something has to give, but what? Will it appear as if the trip took 8 years to complete?
|
Will it appear as if the trip took 8 years to complete?
Yes.
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What is a Past Event Horizon? I've recently been informed about a few things from a reliable source:
1) Wormholes can't form via gravitational collapse
2) The event horizons in an ER bridge can't collide
I understand what a future event horizon is by realizing that once reaching any point on its "surface" there is no longer any trajectory but inward and thus all futures point inward. I understand this via the gravitational collapse mechanism. I don't understand how a past event horizon could form (and therefore what it is). I would very much appreciate answers that take the perspective of the past event horizon (without referring to black holes at all - to make it very clear)
| A past event horizon would be a surface such that one could only pass it on past-directed curves. Inside it, anything within would be forced out, and then once past the horizon, could not enter again. This object is a so-called 'white hole', the opposite of a black hole. We know of no physical processes that could create such an object and none have ever been observed, save for one contended event.
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Why do we use cross products in physics? We can define cross products mathematically like if we take two vectors, we can find another vector with certain properties but why do we use it in physics, if we consider a hypothetical physical quantity like force which is equal to cross product of certain vectors?
For example, the force exerted on a charge in motion in an uniform magnetic field.
Why is it so? Why does that force have to be a cross product of two vectors?
Is it possible to come up with them when what we do is just observe the nature?
| This is a great question. The dot and cross products seem very mysterious when they are first introduced to a new student. For example, why does the scalar (dot) product have a cosine in it and the vector (cross) product have a sine, rather than vice versa? And why do these same two very non-obvious ways of "multiplying" vectors together arise in so many different contexts?
The fundamental answer (which unfortunately may not be very accessible if you're a new student) is that there are only two algebraically independent tensors that are invariant under arbitrary rotations in $n$ dimensions (we say that they are "$\mathrm{SO}(n)$ invariant"). These are the Kronecker delta $\delta_{ij}$ and the Levi-Civita symbol $\epsilon_{ijk \cdots}$. Contracting two vectors with these symbols yields the dot and cross products, respectively (the latter only works in three dimensions). Since the laws of physics appear to be isotropic (i.e. rotationally invariant), it makes sense that any physically useful method for combining physical quantities like vectors together should be isotropic as well. The dot and cross products turn out to be the only two possible multilinear options.
(Why multilinear maps are so useful in physics is an even deeper and more fundamental question, but which answers to that question are satisfying is probably inherently a matter of opinion.)
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Do colors differ in terms of speed? Here is a very simple question about light. As far as I remember from the school program, each color is merely one of the frequencies of light. I also remember that each color's wave length is different. On the other hand, when talking about the speed of light, I've always heard only one value. Why is it so? Shouldn't it be like the red color's speed must be way higher (or lower) than, say, the purple color's speed? I am quite confused here.
(Sorry if my question is too foolish, but it has bugged me for years and I was quite bad at physics at school and have never touched it since I finished school)
| The speed of a wave is given by:
$$c=f\lambda$$
Where $c$ is the speed of the wave, $f$ is the frequency and $\lambda$ is the wavelength. You're right that different colours have different frequencies, but light with a lower frequency has a longer wavelength, in other words if $\lambda$ goes up then $f$ has to come down so they cancel each other out and the speed stays constant.
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Light beam vs sound beam Why is it that it's very common to have beams of light but not beams of sound? Laser beams are widely available, and I am aware that it is also possible to direct sound, however, we rarely see examples of it.
Is it more difficult to direct due to longer wavelength or is it more dispersive in air or something?
| The beam width is proportional to the wavelength $\lambda$ divided by the aperture width $L$. Audible sound frequencies are are in the KHz range with wavelengths between approximately 17 m and 17 mm. Whereas visible light wave lengths are in the micrometer range. So sound apertures would have to be vastly larger than light apertures to achieve similar beam widths--usually not practical.
Medical ultrasound imaging uses sound beams less than 1mm in width but their frequencies are in the MHz range. These frequencies are very strongly attenuated in air and only travel a few centimeters.
Re.
Is it more difficult to direct due to longer wavelength or is it more
dispersive in air or something?
Both are true if you replace 'dispersive' with 'attenuated'.
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Where does the power delivered to car's wheels go? Okay, so power is work/time. Most cases, when power is provided to something, energy is gained as kinetic energy or lost to friction.
But in a car, the engine puts power ( torque x rpm/5252) to wheels, but very little ends up in the wheels, assuming friction keeps them from spinning.
So where does the power go? Do the wheels thru the friction forces cause the energy to go to the car? Essentially the wheels do work on the car which transforms the rotational energy to kinetic energy of the car right?
|
So where does the power go?
Power is just like force and varies from one instant to another.(You cannot say that an object losses $3N$ force! )
What the car/object gains or loose with time energy (whose gain/loss is represented by work done). Power is the rate of work being done(i.e., it represents at what rate the energy is transferred to an object).
Do the wheels through the friction cause the energy to go to the car?
No the car doesn't gain any energy. The frictional force just acts as the converter of the internal energy stored in the car and converts it into kinetic energy( a bit of thermal too!).
Essentially the wheels do work on the car which transforms the rotational energy to kinetic energy of the car right?
Note that a net force is acting on the car which causes the car to gain translational kinetic energy. Now since the force isn't acting on the COM therefore a torque is generated in due process causing the car to gain rotational kinetic energy.
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Average value using partition function Let's say I have 4 particles with energy levels $0\,\rm{eV}$ , $1\,\rm{eV}$,and two particles with $3\,\rm{eV}$ energy levels. If I want to find the average value of energy I can say that
$$\bar{E}=\dfrac{(0+1+2\cdot3)\,\rm{eV}}{4}=1.75\,\rm{eV}$$
If I want to use the partition function to find the same average, I would say: $Z=e^{0}+e^{-\beta}+2e^{-3\beta}$ and to find average I can simply do:
$$\bar{E}=\dfrac{1}{Z}\sum_iE(i)e^{- \beta E(i)}=\dfrac{0+e^{-\beta}+3 \cdot2 e^{-2\beta}}{e^{0}+e^{-\beta}+2e^{-3\beta}}$$
which is not equal to the average that I found above.
This question shows that I am missing a key point about partition functions.
| You are missing a key point about averages.
In your first case you have assumed all energies are equally probable.
In your second case you have assumed energies have a probability distribution $P(E)$ of $e^{-\beta E}/Z$.
In general the average energy is $\sum E\cdot P(E)$.
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Do expectation values of quantum fields behave like classical fields? A famous result in quantum mechanics is Ehrenfest's theorem which states that the expectation values of observables are governed by the classical equations of motion.
Does a similar statement hold for quantum fields?
In particular, in the context of the Higgs mechanism we usually investigate the potential and its minima by treating the field as classical. Let's say we find that the potential has two minima at $\pm v$. The connection to the quantum description is then made by claiming that the vacuum expecatation value $\langle 0|\phi|0\rangle$ is equal to the classical minimum:
$$\langle 0|\phi|0\rangle = \pm v \, .$$
This can be calculated by taking the classical limit $\hbar \to 0$ of the path integral (c.f. page 563 in Schwartz's QFT book)
$$\langle 0|\phi|0\rangle = \lim_{\hbar \to 0} \int D\phi e^{\frac{i}{\hbar}\int d^4x\mathcal L [\phi] } \phi = v\, . $$
Can this be understood analogous to the Ehrenfest theorem in quantum mechanics?
| The QFT version of Ehrenfest's theorem are the Schwinger-Dyson equations, stating that the classical equation of motion (in presence of a source $J$) $\frac{\delta S}{\delta \phi} + J = 0$ holds as an operator equation $\langle \frac{\delta S}{\delta \phi} + J \rangle_J = 0$ in the quantum theory. If you evaluate this at $J=0$, you get that the classical equations of motions hold "up to contact terms" inside correlation functions, i.e.
$$ \langle \frac{\delta S}{\delta \phi(x)} \pi_i\phi(x_i)\rangle = \mathrm{i}\sum_i \langle \phi(x_1)\dots\phi(x_{i_1})\delta(x - x_i)\phi(x_{i+1}\dots \phi(x_n)\rangle$$
where the r.h.s. consists only of correlators in one field less than the r.h.s.
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Quantum field theory, interpretation of commutation relation Let $\phi$ be the quantum field
$$
\phi(x) = \int \frac{d^3\mathbf{p}}{(2\pi)^3} \frac{1}{\sqrt{2E_\mathbf{p}}} \Big[ b_\mathbf{p}e^{-ip\cdot x} + c_\mathbf{p}^\dagger e^{ip\cdot x} \Big]
$$
with commutation relations
$$
[b_\mathbf{p}, b_\mathbf{q}^\dagger] = (2\pi)^3\delta^{(3)}(\mathbf{p}-\mathbf{q}),
$$
$$
[c_\mathbf{p}, c_\mathbf{q}^\dagger] = (2\pi)^3\delta^{(3)}(\mathbf{p}-\mathbf{q}),
$$
all other commutators zero. Let $Q$ be the charge operator
$$
Q = \int \frac{d^3\mathbf{p}}{(2\pi)^3} \Big[c_{\mathbf{p}}^\dagger c_{\mathbf{p}} - b_{\mathbf{p}}^\dagger b_{\mathbf{p}} \Big].
$$
We calculate the commutator $[Q,\phi] = \phi$. The question is what is an interpretation of this commutation relation? We know that $Q$ is the number of antiparticles minus the number of particles.
| The commutation relation $[Q,\phi] = n \phi$ tells you that the field $\phi$ has charge (or eigenvalue) $n$ under $Q$ in our case $n=1$. An easy way to see this is to note
$$\left| \phi \right> := \phi \left| 0 \right> \implies Q \left| \phi \right> = Q \phi \left| 0 \right> = [Q,\phi] \left| 0 \right> = n \phi \left| 0 \right> = n \left| \phi \right> $$
where going from $Q\phi$ to commutator we used the fact that $Q \left| 0 \right> =0$.
In general an equation of the form $[Q^a,\phi] = q^a \phi$ for charge operators $Q^a$, which commute amongst each other, defines the charge of $\phi$ to be $q^a$.
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Fermi energy definition Ok, so I'm having a hard time understanding the definition of Fermi Energy. Several sites basically repeat each other, saying that it is the energy difference between the highest and lowest occupied single-particle states in a quantum system of non-interacting fermions at absolute zero 1, and others say that it is the highest energy that the electrons assumes at 0K 2. Are these 2 concepts the same, and I'm just not getting it? Is the top level of an electron at 0K equal to the difference in energy between highest and lowest occupied states? Any clarification would be greatly appreciated.
| One can derive the fermi-energy using quntumstatistics and is a more complex derivation. But in condensed matter you find a more vivid explanation about the meaning.
Let us assume a free fermi gas or electron gas. We make the following assumptions
*
*$N\gg 1$ conducting electrons move on a homogeneous charge background
*No interaction between particles
*Pauli-Principle, meaning $2$ electrons per energy state
*Describing crystal as cube with edge length $L$
*Choose periodic boundary conditions
We now start with the one particle Schroedinger equation
$$-\dfrac{\hbar^2}{2m_e}\Delta \psi_{\vec{k}} = \epsilon_{\vec{k}} \psi_{\vec{k}}.$$
We find plane waves as solutions
$$\psi_{\vec{k}} = \dfrac{1}{\sqrt{L^3}} \text{exp}(i{\vec{k}}\cdot {\vec{r}}).$$
Where $\epsilon_{\vec{k}} = \frac{\hbar^2k^2}{2m}$. The periodic boundary conditions are
$$\psi(x,y,z) = \psi(x+L,y+L,z+L)$$
and it must follow that for $a \in \{x,y,z\}$
$$k_a \cdot (a+L) = k_aa+2\pi n_a \Leftrightarrow k_a = \dfrac{2\pi}{L}$$
with $n_a \in \mathbb{N} $. As we can see the values of the wavevector are discrete, which is due to the finite volume. The k-space is also made up of discrete points where we have one point per volume $(2\pi/L)^3$. We now want to fill all these electrons in these states. Each state can contain two electrons due to the Pauli-Principle (one with spin-up, one with spin-down).
We start by fillig the states with the lowest energy at ${\vec{k}=0}$. For many particles the electrons will fill a sphere in the k-space, the so called Fermi-Sphere. The radius of this sphere is called the Fermi-Wavevector. One can calculate the radius as follows
*
*We have one state per volume $V_1 = (2\pi/L)^3$, which is filled by $2$ electrons
*The sphere of volume $V_2 = \dfrac{4}{3}\pi k^3_F$ contains all $N$ electrons
*The number $N$ of electrons then is:$$N = 2\cdot \dfrac{\dfrac{4}{3}\pi k^3_F}{\bigg(\dfrac{2\pi}{L}\bigg)^3} \Leftrightarrow k_F = \bigg(\dfrac{3\pi^2N}{V}\bigg)^{1/3} = (3\pi^2n)^{1/3}$$ with $V = L^3$ the volume of the cube in which the electrons are located and $n = N/V$ the charge density.
Now to the interesting part. In case of $T = 0\,$K the maximum occupied energy is given as the Fermi-Energy
$$\epsilon_F = \dfrac{\hbar^2k^2_F}{2m}.$$
As you can see the surface of the Fermi-Sphere is a surface of constant energy.
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Why can the masses of the Higgs and top quark determine whether we live in a false vacuum? I read in the paper linked below that the ground state of the universe depends on the potential of the Higgs field, particularly the graph of the Higgs potential with its minima and maxima. I think I understand why the Higgs is important, because it gives fundamental particles their mass, but what about top quarks?
Paper I read
| The Higgs potential receives quantum corrections from the Yukawa couplings with fermion doublets. The magnitude of the correction is determined by the Yukawa coupling constant, which is itself proportional to the fermion mass. As a result, the heaviest fermions make the greatest contribution to the effective action. Because the top quark is by far the heaviest fundamental fermion, quantum corrections due to top quark loops have the greatest effect on the Higgs effective potential, so to a good approximation only corrections due to top quark loops need to be accounted for.
The details of the Yukawa corrections and their effects on vacuum stability are reviewed in section 2 of this article (for transparency, it's written by my PhD supervisor).
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What is the minimum mass for gravity to form objects in a protoplanetary disk? I understand that at smaller scales the strong, weak, and nuclear forces, and chemical bonds have more influence on the formation of objects than gravity. At what amount of mass does gravity become the dominant force that holds together the matter that will become planetesimals, asteroids, or comets?
| It depends on what the dust is made of, specifically how many coulombs it has (a surplus or deficit of electrons). As mass can be measured in energy, electronvolts/speed-of-light2, and eV is the energy of 1 electron subject to 1 joule of electric potential difference (or one volt [force exerted over distance behind a coulomb's worth of electrons] with the coulombs canceling out, I think), if you know the number of missing or extra electrons and the distance and masses of the dust particles, you can multiply the masses by the gravitational constant and divide by the distance to get the gravitational energy which can be compared to the product of the two electron surpluses/deficits and the electrostatic constant in a vacuum divided by distance for the electric field energy. It's no coincidence that these two look the same, both gravity and electricity are fields described in classical field theory.
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What would the water pressure be at this point in a static incompressible fluid? I have a bowl full of water. I invert a glass and place it upside down in the water, leaving a small pocket of air.
My question: what would the water pressure at positions P1 and P2 be? I know P1 = P2 because there's no hydrostatic effect as the water is level, but what is this value?
*This may seem rather trivial but I can't decide between two lines of reasoning. From where my confusion arises:
The atmospheric pressure acts to push the water level in the bowl down and up into the glass. The small pocket of air in the glass has the opposite effect, acting to push the water level in the glass down and increase the level in the bowl. Since the air pressures are equal, the water is at the same level in the bowl as in the glass.
1) Now, since the fluid is in equilibrium, P0 must equal the pressure at P1 for no flow to occur. Likewise, then P2 must equal P0 in the glass. Then P1 = P2 = P0
2) Or the water pressure at P1 and P2 completely zero? And this would be because the pressure from the air in the glass counters the pressure from the air outside acting on the bowl?
Thank you in advance, sorry for the length of this.
| The water pressure at both points $P_1 = P_2 = 1 \mathrm{atm}$.
If you placed the entire glass under the water surface, inverted it, and then raised it, there would be no air in the pocket (the one above $P_2$). The water level would rise to $1 \mathrm{atm}$, or about 10 meters.
Conversely, if you inverted the glass and then put it into the water, then there would be air in the pocket, but the pressure of that air + the water column would also be $1 \mathrm{atm}$. In this scenario, it's quite possible the water level in the glass and outside of it are the same.
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Does sound propagate in the same way in both senses through a medium with obstacles? For instance, let's suppose that we are in a room and we hear a person speaking in the hallway. If we start to speak at the same volume as that person do, would he hear us as loud as we hear him?
| Mostly usually yes, in the same way that optical systems show the same attenuation in both directions.
but human mouths and ears are slightly directional and are slightly displaced, so tricks could be played using focusing structures to boost efficiency in one direction, eg: a speaking tube that leads from the mouth of one player to the ear of the other.
Also environment has an effect if it's more noisy in the hallway than in my room other player may not hear me.
| {
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Will state of water change in certain condition? Imagine I have an iron tank with a $20~\mathrm{pm}$ hole on it. Then I completely fill it with water and use a pump to get the water out of that hole. What will come out, water or gas?
| Your hole is too small to let even a single molecule through, but let's increase the hole to a size where molecules can come through one at a time. I'm sure this is what you had in mind. What then would be the result?
And the answer is that would have created a molecular beam. That is you would have a beam of isolated water molecules travelling away from the hole. The kinetic energy of the molecules would be about the thermal energy of $\tfrac32kT$. I make this a velocity of about 450 m/s.
| {
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Why are topological insulators interesting? Why are topological insulators interesting? Meaning, why should an undergraduate or graduate student start working on this? What are the technological applications? I am not sure how to answer these questions and wikipedia does not help since it does not explain why so many people work on this. I am especially interested in applications through photonics but any answer would be appreciated. Thanks in advance.
| Almost as interesting as sharks with laser beams on their heads are topological lasers. Topological protection can make microreasonators work in unison [1].
[1] Bandres, Miguel A., et al. "Topological insulator laser: Experiments." Science 359.6381 (2018): eaar4005.
| {
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Does stress depend on the intrinsic properties of the body? I'm trying to understand the difference between stress and pressure.
Assume a body, a box with 4 sides.
We apply pressure to this body, and because the body's right side is 'weak', it deforms.
Does this mean that the stress increases in this example because of the weak side, and therefore, is stress dependent upon the body's intrinsic properties? Had the box been completely rigid, would the stress had been lower?
| I think the way you are thinking about the weak point, it wouldn't be the stress that is changing in the material.
The weakness is often a material property where the material has a greater strain for the same amount of stress, so the weak point deforms more when the equal stress is applied throughout the body.
The stress may also depend on the shape of the body. Weak points can be created through stress concenrations. In this case the object might be weak because it's geometry concentrates the stress at particular locations when the external pressure/load is uniform.
What's important to notice is that stress typically describes the internal forces acting on the object; while pressure describes the force at the surface. The shape can have a large effect on how the stress develops based on applied pressure; and material properties will affect how the strain compares to applied stress.
| {
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Dimensionless expression for differential equation I am working through Nonlinear Dynamics and Chaos by Steven H Strogatz. In chapter 3.5 (overdampened beads on a rotating hoop), a differential equation is converted into a dimensionless form. I am trying to work out which dimensions the initial equations had, and why the converted form is dimensionless.
Initial equation:
$mr \ddot{\phi} = -b \dot{\phi} -mg \sin\phi + mr \omega^2 \sin\phi \cos \phi $
$m$ is mass, $r$ is radius, $\phi$ is an angle, $b,g$ are arbitrary, positive constants, and $\omega$ is angular velocity.
Using a characteristic time $T$, a dimensionless time $\tau$, with $\tau = \frac{t}{T}$ is introduced.
$\dot{\phi}$ and $\ddot{\phi}$ then become $\frac{1}{T}\frac{d\phi}{d\tau}$ and $\frac{1}{T^2}\frac{d^2\phi}{d\tau^2}$, respectively.
Then the initial equation becomes
$\frac{mr}{T^2}\frac{d^2\phi}{d\tau^2} = -\frac{b}{T}\frac{d\phi}{d\tau} - m g \sin\phi + mr \omega^2 \sin\phi \cos \phi$
This is made dimensionless by dividing through a force $mg$:
$(\frac{r}{gT^2})\frac{d^2\phi}{d\tau^2} = (-\frac{b}{mgT})\frac{d\phi}{d\tau} - \sin\phi + (\frac{r \omega^2}{g}) \sin\phi \cos \phi$
And all the expressions in the brackets are dimensionless.
I understand why the expressions in the brackets are dimensionless, but what about the differentials?
$\phi$ is dimensionless.
but would
$\dot{\phi}$ not have dimension $\frac{1}{s}$, and $\ddot{\phi}$ $\frac{1}{s^2}$?
| Because you take the derivative with respect to $\tau$. Since $\tau$ is dimensionless, the derivative is too.
| {
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Driving a nail with a light object? I was wondering if it is possible to drive a nail through, for example, concrete by dropping a light object on the head of the nail over many iterations.
I.e. is there a certain threshold of force that must be reached for the nail to make even the slightest progress into the concrete? Or is it possible for a small force applied thousands of times to drive the nail?
| Hammer is almost always able to transfer phonons, (this threshold is significantly lower than chemical bond energy).
To drill a chunk of concrete, put it in a thermally sealed container and use a non-conductive nail with a higher melting temperature.
Hit a nail, until heat dissipated in concrete melts it.
Disadvantages:
*
*material requirements, concrete changes properties on solidification
*time, to heat up an object of any significant size with small impacts may take millions of years
| {
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Spacelike separation, special case This is merely a terminology question. Consider two events A and B. Now suppose A and B happen in two different black holes, i.e. there is no way from A to B (or B to A). Is this fundamentally different from a "normal" spacelike separation, where the finite lightspeed hinders to get from A to B? (It does also in the black hole case...)
(Context: Trying to abuse physical notions for describing problem chess; repetition of position would correspond to a closed timelike curve)
| In GR we don't have displacement vectors (except for infinitesimal displacements). So we can't say, as in SR, that the vector from A to B is spacelike. There can also be things in GR like events that can be connected either by a purely timelike, future-directed curve or by a purely spacelike curve. To describe your situation, what we would typically say is that A is not inside the light cone of B.
| {
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Is Young's modulus a measure of stiffness or elasticity? Young's modulus seems like a modulus of stiffness. It tells us how difficult is it (how much stress is required) to produce longitudinal strain in a solid. It does not tell anything about how an object will react when the deforming force is removed.
How can one on the basis of young's modulus decide that steel is more elastic than a rubber band?
| It is not true that steel is more elastic than rubber. Not in common language.
Yes, steel has a larger modulus of elasticity, Young's modulus, the ratio of stress to strain $Y=\varepsilon/\sigma$. This is in the region of elastic response as long as the deformation $\sigma=\Delta \ell/\ell$ increases linearly with stress $\varepsilon =F/A$. The response is assumed to be immediate (fast, but slower than the speed of sound). For metals, this is on the order of 100 GPa, similar values for iron and for steel - it takes a lot of force to make a small elastic compression or elongation.
Steel also has a larger elastic limit of stress, a steel rod can support very large tensile loads without permanent changes in its length.
This does not say anything about the range of deformations that are reversible. For steel etc the elastic strain limit is usually on the order of $10^{-3}$. For rubber the range of elastic stretching is much larger. That is why rubber is elastic in ordinary language.
Compressibility is a word that has similar meanings in physics and in the common language. In physics, it is the reciprocal of the bulk modulus which is closely related to Young's modulus. In common language, compressibility is similar to elasticity.
| {
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Internal force disintegrating a solid body? Let $M$ be a block on a frictionless surface. Now let us mentally divide (not physically) the block into 2:1 ration (i.e $1/3$ of the left be called $M_1$ and $2/3$ right be called $M_2$). So $M_1$ applies force $F_1$ on $M_2$ and $M_2$ applies force $F_2$ on $M_1$ and by 3rd law they are equal. Hence acceleration of $M_1$ would be $2a$ and that of $M_2$ would be $a$. Shouldn't this deform the block?
| Most of your intuition is correct. The piece that you're missing is the constitutive relationship that describes how the force $F$ between the pieces depends on their relative position and velocity.
A decent model for most materials is that the constitutive forces act like some combination of a "spring" holding the pieces together, and a "friction damper" that makes their relative motion slow down (you can think of this combination like the mechanism on an automatic door closer).
The spring-damper forces are at equilibrium (zero force) when the block is undeformed and not changing shape, so that the block is moving like the rigid object you expect ($F=0$ gives $a=0$ in your model).
If something did disturb the pieces of the block (i.e. deform the whole block), then the spring-damper constitutive forces would pull it back into shape. The strength of the friction in the damper would determine whether it settles quickly back into shape (for large friction) or "rings" by having the parts of the block oscillate outward and inward (for small friction).
| {
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What caused quadrupole anisotropy at the time of recombination? The polarization of CMB requires a quadrupole anisotropy of the incident radiation field acting on the plasma. How did such a special pattern (i.e. hot and cold in two orthogonal directions) generate in the first place? Where does it come from?
| From Quadrupole Types and Polarization Patterns
Quadrupole anisotropies are associated with density, vorticity and gravitational wave fluctuations
Their projection determines the polarization pattern and may be distinguished by symmetry properties
The polarization probe more than just the density or temperature fluctuations at recombination. Because the polarization pattern is a projection of the quadrupole anisotropy, any source of quadrupole anisotropy leaves its imprint in the polarization.
n general there are three sources to the quadrupole anisotropy at recombination:
For example :
A passing gravitational wave causes an anisotropic stretching of space and correspondingly the frequency of CMB photons. This also produces a quadrupolar variation in the temperature. Importantly, it is not symmetric like the density quadrupole. This asymmetry causes a "handedness" to the pattern of polarization.
| {
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How was the mass of the moon first calculated? How was the mass of the moon first calculated? How do we measure it now?
| In the history of astronomy there was no correct concept of masses before Newton.
The Greeks had found reasonably accurate volumes of the Earth and Moon, but masses were unkwnown.
The planetary mass determination relies on Kepler's third law. Newton indirectly measured the mass of the Moon, trying to estimate the ratio between the solar and lunar masses looking at sea tides.
I suggest this lecture to clarify any doubts (in particular the first two pages).
http://articles.adsabs.harvard.edu//full/2002Obs...122...61H/0000061.000.html
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Could speed of light be variable and time be absolute? I get my "demonstration" of time dilation from the textbook thought experiment.
A laser is mounted on a cart with a reflective ceiling. At $t=0$ the cart starts moving and the laser is fired. When the laser is reflected back at the starting point the (thought) experiment stops.
Now, two different observers, one sharing the frame of the cart and another standing on the ground perpendicular to the cart will observe two different things. For the first one, the laser bounces back and then down in a straight line. For the second one, the light travels in a triangular pattern which is longer than the path observed by the first guy.
Given that the speed of light is constant, the time has to dilate/contract.
Why is the speed of light held constant here? Could we work out a physics where time is absolute but the maximum speed of light variable?
|
Why is the speed of light held constant here?
You are missing one important point of that thought experiment: it is meaningful after considering the Michelson-Morley experiment.
If our interpretation of the Michelson-Morley experiment is that the speed of light is constant with respect to the observer, then we can perform the thought experiment you described, using the assumption that the speed of light is constant (we justify this assumption because of the results of the Michelson-Morley experiment).
That thought experiment leads then to the formulae of time dilation and space contraction (and to the special relativity theory).
| {
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Is Griffiths simply wrong here? (Electrostatic Boundary Conditions)
In the above illustration, shouldn't $E_{above}$ and $E_{below}$ be in opposite directions? If not, how did Griffiths end up the following equation? From the above directions, shouldn't the flux add up?
| Two ways of seeing that it’s right:
*
*Consider the case of no charge. Then nothing interesting is happening at the sheet, so the fields should be equal: both sides of the equation are zero.
*Gaus’s law: the sum of the fields going away, which is the outward flux, is given by the charge. Since $E_{below}$ is defined as towards the charge, it enters that calculation with a minus sign.
He’s picked a sign convention where the field upward is positive everywhere. That means that $E_{below}$ is defined such that a positive value means "the E vector points up" and a negative value means "the E vector points down". Which way the E vector points is given by the physics:
*
*if there's positive charge on the surface and no external field, the E field below it will point down, hence $E_{below}$ will have a negative value.
*If there's a large upward going external field, then in that case the E field points upward everywhere, and $E_{below}$ will be positive.
To put it another way, the length of the $E_{below}$ arrow in the picture isn't showing you the absolute magnitude; that has to come from somewhere else and is written as "$E_{below}$", a number. And it's not even showing you the direction, because if $E_{below}$ is less than zero, the actual E vector is pointing the other way. That arrow there is just defining a direction, like $\hat{x}$, $\hat{y}$ and $\hat{z}$.
| {
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What is baking and what are the effects? In their experiment, Davisson and Germer had to bake the nickel mass because it was oxidized.
What is baking and what does it do to the lattice of the metal?
| Usually "baking" a vacuum chamber means heating up the whole installation to over 120 C during a day or so, to drive off adsorbed water etc from surfaces in order to achieve a good vacuum.
Davisson and Germer did something else. From their article:
The investigation reported in this paper was begun as the result of an
accident which occurred in this laboratory in April 1925. At that time
we were continuing an investigation, first reported in 1921, of the
distribution-in-angle of electrons scattered by a target of ordinary
(poly-crystalline) nickel. During the course of this work a liquid-air
bottle exploded at a time when the target was at a high temperature;
the experimental tube was broken, and the target heavily oxidized by
the inrushing air. The oxide was eventually reduced and a layer of the
target removed by vaporization, but only after prolonged heating at
various high temperatures in hydrogen and in vacuum.
So that was reducing the heavily oxidized surface in a hydrogen atmosphere in an oven. The high temperature caused crystals to grow to sizes larger than their electron beam.
| {
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I dont understand the work equation I don't understand how work = force * displacement as if a force of say 1 Newton was to be applied to two objects of different mass until the object reached a displacement of say 1 meter, surely the object of less mass would displace 1 meter in less time (due to faster acceleration) meaning the force would be applied for less time resulting in less work. I know there is something fundamentally wrong with my understanding of this but I'm not sure exactly what. any help would be greatly appreciated.
| One of the things that we can predict while applying work energy theorem is that the change in kinetic energy is the same for both the given case (faster as well as slower) given the fact that other forms of energy of the system change negligibly. Let's assume that these works aren't the same:
$$W=\Delta K$$
and $$W' = \Delta K'$$
[Note that mass doesn't show up in the above equation.]
Showing that time doesn't play a role (the way you think)
Let the same force $F$ act on two stationary objects of mass $m$ and $m'$ and there velocity becomes $v$ and $v'$ after some time $t$ and $t'$ respectively covering the same distance $d$.
$$a= \frac {F}{m}$$
$$a' = \frac {F}{m'}$$
Now since
$$d = ut + \frac {1}{2}at^2$$
$$\Rightarrow d = \frac {1}{2}at^2$$
and
$$d = \frac {1}{2}a't'^2$$
$$t = \sqrt {m\frac {2d}{F}}$$
and
$$t' = \sqrt {m'\frac {2d}{F}}$$
$$\Rightarrow t'= t \sqrt {\frac {m'}{m}} \tag 1$$
Therefore
$$v = t \frac {F}{m}$$ and $$v' = t' \frac {F}{m'}$$
Using eq. $(1)$ we get,
$$\Rightarrow v' =\frac {F}{ \sqrt {m m'}} t \tag {2}$$
Now the change in kinetic energy ($\Delta K$) is (clearly):
$$\Delta K = \frac {1}{2} mv^2 = \frac {F}{2m}t^2$$
and
$$\Delta K' = \frac {1}{2} mv'^2 $$
Now substituting $v'$ from eq. $(2)$ we get,
$$\Rightarrow \Delta K' = \frac {F}{2m}t^2$$
i.e.,
$$\Delta K = \Delta K' $$
$$\Rightarrow W = W'$$
This means that for same force acting for same distance the work done is the same.
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How did early radar determine range/ distance precisely? Wikipedia talks about precise timing of the returned radar pulse, with an animation of a clock.
But they didn't have atomic clocks and such before or during WWII.
So how did they determine distances and (possibly) velocities back rhen?
| There were WWII-era electronic circuits that were designed to readily slice time into fractions of a microsecond, for the purposes of timing pulse returns in the first radars and thus deducing range.
The imprecision in the time slicing process could be readily dialed out of the system by aiming the radar beam at a target a known distance away for calibration purposes.
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Finding orthnormal wavefunctions to given wavefunctions Consider a particle in an infinite square well in one dimension. The potential inside the well is 0. The length of the potential well be L. Let the wavefunctions of the lowest three energy states be $\psi_1(x)$, $\psi_2(x)$, $\psi_3(x)$ which are also orthonormal to each other.
When measurement was done the particle was found in a state described by :
$\psi(x)$ = $\alpha(\psi_1 + \psi_2 + \psi_3)$
Now my question is how to find other two wavefunctions that are orthonormal to each other and also to $\psi$ and expressed in terms of $\psi_1,\psi_2, \psi_3$.
So, I have tried to solve the question by first finding the value of $\alpha$ by normalization and is equal to $1/\sqrt3$.
Let the other two wavefunctions be $\lambda(x), k(x)$.
Assuming $\lambda(x) = a\psi_1 + b\psi_2 + c\psi_3 $
and $k(x) = A\psi_1 + B\psi_2 + C\psi_3 $.
According to question, $<\lambda|k>=0$
expanding term wise gives $a^*A + b^*B + c^*C=0 $
Similarly, $\alpha^*a+\alpha^*b + \alpha^*c=0 \\\alpha^*A + \alpha^*B + \alpha^*C=0$.
But now what? I am unable to solve further. Please help.
| HINT: View the wavefunctions as vectors and use the Gram-Schmidt process.
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Do colliding black holes violate time reversal symmetry? Two black holes can collide and merge into one bigger black hole, but not split into two. Does this mean colliding black holes violate time reversal symmetry?
Related: Do black holes violate T-symmetry? Based on the answer to that question, time-reversing a black hole yields a white hole. However, that seems to imply that white holes are very unstable because they can spontaneously split into two, which would then split into four, ad infinitum, and the universe would be covered with tiny white holes all over.
| When two black holes collide and merge into one, a lot of energy is sent out as gravitational waves. This energy spreading out without bound represents an increase in entropy, along with any increase due to the final BH surface area being greater than the sum of the original two.
If we had an exact solution to Einstein's field equation describing this, we could replace t with -t and have another valid solution. What this means in real life is that a black hole just sitting somewhere, when bombarded with converging gravitational waves arranged exactly the right way, could absorb those waves and split into black holes.
I don't know about ultra-advanced alien technology, but anything I can imagine like that isn't going to happen. It's just like the fragments and droplets of a dropped egg coming back together just right to make an unblemished whole egg.
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Local density in SSH model Considering the usual SSH model defined on N sites and parametrized with $\delta $:
$$H = \sum\limits_{j=0}^{N-1} (1-\delta)\ c_{j,A}^{\dagger}\ c_{j,B} +(1+\delta)\ c_{j,B}^{\dagger}\ c_{j+1,A} + h.c $$
In the fully dimerized limit, i.e $\delta=1$, how would one compute the local density for each state ($<c_{j,\alpha}^{\dagger}c_{j,\alpha}> $)?
I know the hamiltonian matrix and can numerically compute its eigenvector but don't manage to find the link with the local density. I would like to obtain the same kind of graph as in Fig 1.b from the paper "Topological phases of a dimerized Fermi-Hubbard model for semiconductor nano-lattices" which proves that the zero energy states are mostly localized at the edges.
| If you have obtained the Hamiltonian, and diagonalized it to get the ground state vector, you have presumably chosen a basis, and I would guess that it is labelling sites in real-space, as this is the most "natural" basis to take for a hopping Hamiltonian of this type. Is that right? In that case, the particle density at each point is given by $\rho(j) = |v(j)|^2$, where $v(j)$ is the $j$-th element of the vector you have. So typically $j=1$ is the left-most site, $j=2$ is its neighbor, etc, until you reach $j=N$ which is the rightmost site.
On a side-note, taking $\delta=1$ may give odd results, as the dimers are then completely decoupled from each other. I would start by taking increasingly small values of $\delta$ to see how the system evolves.
Hope this helps.
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Why does an open ended wire cause the load to the oscillator to increase? Using an ammeter between the wall and signal generator, I noticed that when the lead from a single channel is attached to the signal generator, the power into the signal generator increases by about 100 milliwatts.
When the wire leads are removed from the signal generator, the input power into the signal generator decreases by 100 milliwatts.
Why would an open ended wire cause a load? Is there a way for that wire not to add to the input load?
The lead wire from the signal generator is not attached to anything on the open end.
| There are waves in the wire, and they can be reflected back from the closure.
As an example: when I was working with preamplifiers to increase the signal measured in the oscilloscope, I had to use a 50 ohm closure at the end of every wire, to get a good enough reflection otherwise the signal kind of "leaked" , got noisy.
I guess some electrodynamics calculation could show you what's going on! :D
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Why does a weather vane arrow point in the direction of the wind? It seems that a weather vane will rotate in order to minimize energy and thus orient itself parallel to the wind.
What I do not understand is why it is implied that the weather vane arrow should point in the direction of the wind.
I do not understand why the arrow pointing in the opposite direction of the wind is also not a minimum energy solution.
| Wind vane will always point to the direction from which the wind is blowing.
I.E., if wind is blowing from East to West,
then the arrow of the wind vane points to 'East'.
The pointed end of the arrow offers least resistance to wind.
Therefore, the arrow achieves the state of equilibrium by pointing itself against the wind (Direction from which the wind is blowing).
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Properly reporting instrument readings As a first approximation, the uncertainty ($\delta X$) associated to a mensurand can be expressed as $\delta X= \Delta X / 2$ with $\Delta x$ being the resolution of the instrument.
There is also a recommendation indicating that the numerical expression of the result and its uncertainty saying that "Results should be rounded to be consistent with the uncertainty given" (EURACHEM/CITAC Guide CG 4).
So, if we have an instrument like a ruler with a resolution of 1 arbitrary unit ($\Delta X =1$ ) and when measuring an object, the end of said object is very close to the 3 unit mark, why should the reported value be 3 $\pm$ 0.5
and not 3.0 $\pm$ 0.5 or even 3.1 $\pm$ 0.5?
| The problem with your question (and figure) is that the resolution is well beyond the gradation of the instrument, and that is misleading.
So let's suppose can't really distinguish 3.1 from 3.5, all we know is that 3.1 is closer to 3 than 5, and while 3.5 is in the middle, we'll round it to 4.
In this case the uncertainty is the standard deviation of the process, not the extrema (or half the extrema). The standard deviation is that of a uniformly distributed random variable on [0, 1], which is:
$$ \delta X = \frac 1 {\sqrt{12}} $$
| {
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Is Snell’s Law valid in this case? When light travels in a perpendicular path from one medium to another medium of different optical density, is Snell’s law valid?
$\sin i$ and $\sin r$ are both 0, right?
So it isn’t valid.
Is this correct?
| Let me first give an example from Newton's laws. If there is no net force acting on an object, then it will not be accelerating. Therefore, the equation of Newton's second law $F=ma$ is a valid equation, as we have $0=m\cdot 0=0$. But what if we were looking at this scenario with the equation $F/a=m$? This gives us $0/0$ on the left side of the equation, and it's no longer valid. What happened?
Mathematically, $0/0$ is undefined, so it was invalid to divide by $a$ in the first place. However, physically this means here our mass is undefined. But this makes sense. For any mass, $0$ net force means $0$ acceleration, so in this scenario with only this information we cannot determine the mass of the object. Therefore, it makes sense that we get an undefined mass.
Moving onto your example, mathematically it is invalid to do what you propose, as you are dividing by $0$. However physically, for any two adjacent media, a light ray incoming perpendicular to their interface will not refract. So just knowing that both angles are $0$ cannot give us any information about the indices of refraction of the two media. Therefore, it makes sense that we get an undefined value in this version of the equation.
In any case, just like in the Newton's law example, it is better to just keep Snell's law without the division in this case. Then you have a valid equation that reads $0=0$.
| {
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Book suggestion about Neutrino effect on Cosmic Structure I am trying to find some nice explanatory books about neutrino effects on the cosmic structure. I did not take GR so I prefer sources that contain not much GR.
I prefer lecture note series or books rather than articles. I mean I want to understand the idea behind it.
For instance, I looked at Neutrino Cosmology but it contains a lot of GR and it's kind of hard to read as an undergrad. I am looking for something simpler but also explanatory as well.
| My favorite book on this topic is Soler, Froggatt & Muheim's Neutrinos in Particle Physics, Astrophysics and Cosmology. They start with neutrinos in the standard model and go over neutrino oscillation, and then go onto look at neutrinos emitted from stellar objects. Experimental neutrino detection is emphasized, and then finally leptogenesis. If I remember correctly, GR discussion is minimal.
Another good resource is Zuber's Neutrino Physics. It has a nice introduction to neutrinos in the standard model, and then goes onto discuss possible experimental ways of detecting them. The connection between neutrinos and several phenomena (neutrino oscillations, atmospheric neutrinos, neutrinos from supernovae, and connections to dark matter) are discussed in an introductory manner.
Finally, I would recommend the excellent lecture notes of Dolgov, Neutrinos in cosmology. Basic cosmology is introduced before neutrinos are formally introduced. Primordial nucleosynthesis, cosmological lepton asymmetry, dark matter, and neutrino oscillations in the early universe are discussed. It also contains over 800 references to relevant articles and additional books.
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Pitch of sound in half fill glass I read in my physics book that when we use a spoon and vibrate a glass, the pitch should be according to the air column inside. So, less filled the glass, longer the air column and lesser the frequency according to frequency = 1/4*( v/l) formula. But when I tried to do it practically I found the glass which has the least amount of water sounds shriller meaning high frequency. Why so? Am I doing something wrong in my exp?
Here is a link where this person does the same experiment but she does not explain the physics
https://www.youtube.com/watch?v=iFwtybB3R6Q
|
I read in my physics book that when we use a spoon and vibrate a
glass, the pitch should be according to the air column inside.
The book is wrong. The pitch depends on the vibration of the glass, not the air inside.
The glass changes shape as it vibrates, and when you add water, the water also has to move to match the shape of the glass. The extra mass of the water lowers the vibration frequency.
If you want to make the air inside a container vibrate, get a bottle with a fairly narrow neck and blow across the neck of the bottle. With a bit of practice you will be able to produce a tone. If you then partly fill the bottle with water, the vibrating air column will be shorter and the pitch will be higher.
| {
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Null Geodesics in Anti-de Sitter space time
Would anyone be able to explain how the step was taken in getting the final equation with $R \tan(t/R)$
I understand the steps before where we are finding the null geodesic equation for the AdS space time but not sure how the final equation is produced.
Note: This question is using a coordinate transformed metric of AdS with transformation $r=R \sinh \rho$ as shown at the top.
| Separate the differential relation
$$(\cosh\rho)\,\dot t=R\dot\rho$$
between $t$ and $\rho$ to get
$$\frac{dt}{R}=\frac{d\rho}{\cosh\rho}.$$
Then integrate to get
$$\frac{t}{R}=\tan^{-1}{(\sinh\rho)}$$
or
$$\sinh\rho=\tan\frac{t}{R}.$$
| {
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How does water vapour replace air molecules? I know that density of moist air is less than density of dry air becuase water molecules replace air molecules, and hence as the average molecular mass of water is less than that of air, the density decreases.
Now my doubt is why do the water molecules replace air molecules, why don't they just get mixed up with air molecules without replacing the already existing ones?
I assume the answer might be due to atmospheric pressure( to maintain it almost constant). But I am not able to find a logic to it.
Please explain. Also suggest if I have to add any extra tags relating to the topic.
| I'm not a professional. It is only my thought may you find it useful
If water molecules can dissolve in air then, their would be no change in volume. But as we know, water dosen't dissolve instead it only makes a mixutre with air. So water vapors fight for their volume too therefore increasing volume and decreasing density.
Now think about water vapors given off by evaporation. They contain relatively less energy as compared to vapors of boiling water. Therefore the vapors of evaporation must take some energy from air to stay in atmosphere. That's why these vapors cause only a little increase in volume of air than vapors of boiling do.
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60 kg on earth is 60 kg on the moon I'm writing a trivia quiz and intend this question, which dates from a high school physics test I took in 1972.
An astronaut tips the scale at 60 kg while on earth, what will she be if she steps on the scale on the moon? Answer 60 kg.
Kg measures mass, which is constant.
The question is not in pounds, or ask about weight. Yes, it is a trick question, but I think it is an entertaining one. The point is, I've looked around the internet and the discussion all focus on in common usage, weight is mass and nobody knows what a Newton is.
I'm certain of my answer, but I'd like to be ready for some blowback.
| The answer depends on the type of instrument you are allowed to use for the measurement. The two types are:
*
*Weighing Machine type
*Beam Balance type
This instrument (weighing machine) measures the downward force applied by the object and then divides it by $g$. That is if your body applies a force of $W$ on the scale then the scale would show a value of $W/g$. This is because $mg=W$ on earth. But say you are on moon and the machine is celebrated for earth then the body applies applies a force say $W'$ then the reading shown by the scale would be $W'/g$. Now on moon $a=g/6$. Therefore the scale would show a reading of about $m/6$ i.e., this machine on moon measures $10kg$ for a body of $60kg$ on earth.
This instrument (beam balance) works on the principal of moments and therefore for equal arm length it measures the mass of the body. So if you are using this instrument then the value would be same wethere you are on moon or earth.
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Understanding lenses for virtual reality and viewing a screen very close to eyes I'm trying to understand the physics of the technology involved in making a simple virtual reality headset.
More specifically, I just care how does one make it so they can see and focus on a screen right in front of their face?
I ask my question in picture format below:
Thanks in advance for any help with my understanding here.
| Two things need to be done.
*
*a sharp image needs to be formed on the retina,
*as perceived by the eye/brain the image needs to be upright.
Because the smartphone is so close to the eye, the rays entering the eye are highly divergent and they need to be made less divergent after passing through the lens.
This is achieved by using a convex lens of short focal length with the smartphone placed at or within the focal plane of the lens.
The image produced by the lens is upright.
This is a magnifying glass arrangement.
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Confused about what a wave is When a wave of something, let's say light or some electromagnetic wave is given, I am confused because I do not understand if shape of a wave represents projectile of it or some value that possess at certain positions. I researched a lot but I have no idea what a wave really is.
My question is: what is a wave?; what defines a wave?
| The most basic definition of a wave might be $f(x-ct)$ which is the equation of a function that moves to the right (translates) with a speed $c$. $f(x-ct)+g(x+ct)$ is the equation of two waves--one moving to the right and one moving to the left--and this equation is the general solution to the 1D wave equation.
So the most basic elementary waves are disturbances that move in space at some constant speed without changing their basic shape. (In 3D the waves are attenuated by spherical spreading but still retain their shape)
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How one pair of Schwarzschild coordinates can specify 2 different points on a manifold? My question is regarding Schwarzschild solution. I always heard that coordinate chart is a one to one map from a manifold to real numbers. But when we look inside the black hole using Kruskal coordinates we see that each pair (r,t) in Schwarzschild coordinates appears in 2 distinct places on the diagram. Each point in the interior of black hole with a certain value of (r,t) has its counterpart inside the white hole. How can a ceratin point (r,t,$\phi$,$\theta$ ) specify 2 different places on a manifold? Is this a result of coordinate singularities?
| Coordinate charts are always one-to-one from the manifold to some open subset to $\mathbb{R}^n$. This is because we use charts as local homeomorphisms to define what a manifold is (it can be defined in other, more abstract ways, e.g. locally ringed spaces, but let's not get into that).
Let's remark another important point: in general relativity we only know charts, we never ever "touch" the global manifold, and this is because physics knows about local stuff. For example, we don't know about the global topology of the universe, although some people think that it's a Poincarè sphere.
Going back to the question, when we have a change of charts with singularities it can mean one of two things: the manifold has a true (curvature) singularity or the change of charts is extended beyond the range of applicability, i.e. a coordinate singularity. This can easily be seen when we consider the change of charts between cartesian coordinates and polar coordinates: the origin is not really valid point in those coordinates, since one can arbitrarily change the angle(s) but nothing changes.
If this is so, why do we do this? Well because it's useful, and typically it's not a problem, at least when the set of troublesome points is "small enough" (measure zero).
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Derivation of combination of lenses using different sign conventions In the following derivation of combination of thin lenses, why no sign convention is applied? For the first step, image distance, $u=+v_e$, but shouldn't it be $-v_e$ as the object is to the left of the lens? Also, in the second step, since the virtual object is to the right of the lens, shouldn't both the image distance, $v$ and object distance, $v1$ be $-v_e$?
Combination of Thin Lenses in contact
Let us consider two lenses $A$ and $B$ of focal length $f_1$ and $f_2$ placed in contact with each other. An object is placed at $O$ beyond the focus of the first lens $A$ on the common principal axis. The lens $A$ produces an image at $I_1$. This image $I_1$ acts as the object for the second lens $B$. The final image is produced at $I$ as shown in figure. Since the lenses are thin, a common optical centre $P$ is chosen.
Let $PO = u$, object distance for the first lens ($A$), $PI = v$, final image distance and $PI_1 = v_1$, image distance for the first lens ($A$) and also object distance for second lens ($B$).
Combination of Thin LensesFor the image $I_1$ produced by the first lens $A$,
$$\frac{1}{v_1} – \frac{1}{u} = \frac{1}{f_1} \tag{1}$$
For the final image $I$, produced by the second lens $B$,
$$\frac{1}{v} – \frac{1}{v_1} = \frac{1}{f_2} \tag{2}$$
Adding equations (1) and (2),
$$\frac{1}{v} – \frac{1}{u} = \frac{1}{f_1} + \frac{1}{f_2} \tag{3}$$
If the combination is replaced by a single lens of focal length $F$ such that it forms the image of $O$ at the same position $I$, then
$$\frac{1}{v} – \frac{1}{u} = \frac{1}{F} \tag{4}$$
From equations (3) and (4),
$$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \tag{5}$$
This F is the focal length of the equivalent lens for the combination.
| The author did use a convention, probably just not the one you're used to. From the form of the well-known thin lens equation used ($\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$), we can conclude that the author considers both object and image distances positive towards the right and negative towards the left. It's a simple convention that many people like.
With the derivation you posted, it's easy to forget that $v_1$ (the image of the first lens) and $u_2$ (the object of the second lens) are not the same. In this case, they both have the same value because the lenses are in contact (zero distance between them) and because both objects and images are positive in the same direction with the convention the author used. So for the second lens, we can use $u_2 = v_1$ and $v_2 = v$ to turn $\frac{1}{v_2}-\frac{1}{u_2}=\frac{1}{f_2}$ into $\frac{1}{v}-\frac{1}{v_1}=\frac{1}{f_2}$.
Personally, I prefer the convention where object distances are positive towards the left and image distances positive towards the right. To my engineering brain it just seems more natural, and the thin lens equation is then nicely positive all over: $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$. I admit it does complicate things sometimes, but at least it's more difficult to just blindly plug the same numbers somewhere they don't logically belong.
With the convention I prefer, $u_2 = -v_1$ (because the images and objects are positive in opposite directions) and $v_2 = v$, so the equation for the second lens $\frac{1}{v_2}+\frac{1}{u_2}=\frac{1}{f_2}$ becomes $\frac{1}{v}-\frac{1}{v_1}=\frac{1}{f_2}$. Here's the derivation repeated with this convention:
$$\frac{1}{v_1}+\frac{1}{u}=\frac{1}{f_1}$$
$$\frac{1}{v}-\frac{1}{v_1}=\frac{1}{f_2}$$
Adding them together:
$$\frac{1}{u}+\frac{1}{v}=\frac{1}{f_1} + \frac{1}{f_2}$$
And substituting this full compound lens equation:
$$\frac{1}{u}+\frac{1}{v}=\frac{1}{F}$$
Gets us the same result:
$$\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}$$
| {
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Photon vs electromagnetic waves Suppose an electron makes a single transition from higher to lower energy level releasing energy. Would that energy be released in exactly one photon equal to $h\nu$? Also, is saying "one photon is released" equivalent to saying "one electromagnetic wave of $\nu$ frequency is released"?
| More than one photon can be emitted but these processes have lower probability. (Rule of thumb: A factor of $\alpha\approx 1/137$ for each extra photon.)
Talking about “one electromagnetic wave” is meaningless. An electromagnetic wave pulse with a small range of frequencies centered on $\nu$ is emitted. (The spread in frequencies depends inversely on the transition time.) The pulse has many wave crests and troughs.
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Using the uncertainty principle to estimate energies in ground states Suppose, for example, that we want to find the minimum energy of a particle undergoing simple harmonic motion. In classical mechanics, the energy is:
$$E = \frac{p^2_x}{2m} + \frac{1}{2} m \omega_0^2x^2$$
where $m$ is the mass of the particle and $x$ the position. I was taught that one can use Heisenberg's uncertainty principle to find an estimate (which actually turns out to be an exact answer) for the ground state energy of this system. Since the potential energy well is symmetric, one can deduce that the expectation value of $x$ and $p_x$ will be zero. The uncertainty in the momentum will be given by:
$$\Delta p_x \ge \frac{\hbar}{2\Delta x}.$$
Now then I was told that the energy of the system must obey the following:
$$E \ge \frac{1}{2m} \left(\frac{\hbar}{2\Delta x}\right)^2 + \frac{1}{2}m\omega_0^2(\Delta x)^2.$$
I can see that this inequality is obtained plugging in the value $p_x = \frac{\hbar}{2 \Delta x}$ and $x=\Delta x$, but I really don't get how this is "mathematically" possible. The first equation contains on both sides values of energy, momentum and position but the new inequality contains on the LHS a value for the energy and on the RHS uncertainties, how is this possible? Surely the equation must be the following:
$$\Delta E \ge \frac{1}{2m} \left(\frac{\hbar}{2\Delta x}\right)^2 + \frac{1}{2}m\omega_0^2(\Delta x)^2$$
| Building on the last equation in Semoi's answer:
\begin{equation}
\left\langle E \right\rangle = \frac{(\Delta p)^2}{2m} + \frac{m\omega^2 (\Delta x)^2}{2},
\tag{1}
\end{equation}
if you want to arrive at the minimum energy, all you need to do is substitute the uncertainty principle, $\Delta p = \hbar / (2 \Delta x)$, into Eq. (1), and minimize the energy with respect to $\Delta x$, i.e., setting the derivative equal to zero:
\begin{equation}
\frac{d\left \langle E \right\rangle}{d(\Delta x)} = 0.
\tag{2}
\end{equation}
You will arrive at the result:
\begin{equation}
(\Delta x)^2 = \frac{\hbar}{2m\omega}.
\tag{3}
\end{equation}
Substituting Eq. (3) in the original expression you will arrive at $\left \langle E \right\rangle=\frac{1}{2}\hbar\omega$, which is precisely the ground-state energy for the harmonic oscillator.
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Is the equation $g=GM_Em/R_E^2$ in Tipler incorrect? I was reading my textbook (Tipler et al.), and I am unsure of one of the expressions they used. On page 374, it says (near Figure 11-10) that $g = GM_Em/{R_E}^2$. Is this even dimensionally correct? I got a units of $m/s^2$ on the left hand side and Newtons on the left hand side. I don't think that they are in agreement. Do they mean to say $g = GM_E/{R_E}^2$?
| Yes, this is certainly an error. The gravitational acceleration is $g = G M_E/R_E^2$, consistent with the preceding sentence of the text. It would be inconsistent (and a violation of the equivalence principle) for $g$ to depend on $m$.
| {
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Gauge ghosts & unphysical states in gauge theory I have a general question about a statement from Wikipedia about ghost states as occuring in gauge theory:
"In the terminology of quantum field theory, a ghost, ghost field, or gauge ghost is an unphysical state in a gauge theory."
I learned gauge theory up to now with mainly mathematical beckground. My main reference is this https://arxiv.org/abs/1607.03089 paper by A. Marsh.
Question: What is concretely an "unphysical" state or field from viewpoint of gauge theory?
| There are different ways of defining what "unphysical" state is. Often states with negative norm $\langle \Psi | \Psi \rangle \lt 0$ are considered to be unphysical.
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Entropy change in the free expansion of a gas Consider the adiabatic free expansion of a gas since there is no external Pressure hence Work done on the system is 0 and since the walls are insulated (hence adiabatic) the heat absorbed is 0. However since this is a irreversible process then entropy change > 0 hence dQ > 0 . However there is no heat absorption. What am I missing ?
|
What am I missing ?
Entropy can be generated without there being heat transfer, i.e., when $Q=0$. That's the case for a free expansion into a vacuum. The classic example given is an ideal gas located in one side of a rigid insulated vessel with a vacuum in the other side separated by a rigid partition. An opening is created in the partition allowing the gas to expand into the evacuated half of the vessel. $W=0$, $Q=0$, $\Delta T=0$ (for an ideal gas) and therefore $\Delta U=0$. Although no heat transfer has occurred, the process is obviously irreversible (you would not expect the gas to be able to spontaneously return to its original location) and entropy increases.
You can calculate the entropy increase by assuming any convenient reversible process that can bring the system back to its original state (original entropy). The obvious choice is to remove the insulation and insert a movable piston. Then conduct a reversible isothermal compression until the gas is returned to its original volume leaving a vacuum in the other half. All properties are then returned to their original state. The change in entropy for the isothermal compression is then, where $Q$ is the heat transferred to the surroundings by the isothermal compression,
$$\Delta S=-\frac{Q}{T}$$
Since the system is returned to its original state, the overall change in entropy is zero, meaning the original change in entropy due to the irreversible expansion has to be
$$\Delta S=+\frac{Q}{T}$$
Hope this helps.
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What were the experiments and works that suggested kinetic theory of gases and proved that the static theories were incorrect? What are the experiments which could not be explained by static theory?
How can one conclude that these theories are insufficient to explain the experimental results?
| I have searched , and it is a sobering search. All reviews start and stay on the way the theory developed. It does not seem that there was a decisive need to go further than the thermodynamic theory, but the history shows that they did, and it was all theoretical! An example is in this link, ( also of course in wikipedia).
As far as I can see, the thermodynamic properties of fluids and gases were known and measured at the time, and the kinetic theory developed in order to fit the already there observations on temperature and pressure etc, with a new model that opened new possibilities.
One can find now a number of experiments that demonstrate molecular behavior etc, but the theory developed without more experimental observations than the thermodynamic ones.
One could say that the experiments for Avogadro's law were the ones really depending on the kinetic theory and validated it.:
Experimental studies carried out by Charles Frédéric Gerhardt and Auguste Laurent on organic chemistry demonstrated that Avogadro's law explained why the same quantities of molecules in a gas have the same volume. Nevertheless, related experiments with some inorganic substances showed seeming exceptions to the law. This apparent contradiction was finally resolved by Stanislao Cannizzaro, as announced at Karlsruhe Congress in 1860, four years after Avogadro's death. He explained that these exceptions were due to molecular dissociations at certain temperatures, and that Avogadro's law determined not only molecular masses, but atomic masses as well.
These organic chemistry experiments may be the first validating the kinetic theory.
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Why COM and COG are different things? In the formula for Newton's law of gravitation i.e.
$$F = G(m_1)(m_2)/r^2 $$
here $r$ is the distance between COM of the two objects , so we consider that the object is a single point located at it's COM and at this point gravitational force of attraction acts . Then why is sometimes COM and COG different for an object ? Please give some practical example .
| While your question is essentially a duplicate of the one I linked to, none of the answers have used the equations, so I will supply them here in a comparison.
The COM is weighed based on the mass density $\rho(\mathbf r)$:
$$\mathbf r_\text{COM}=\frac{\int\text dV\,\rho(\mathbf r)\cdot \mathbf r}{\int\text dV\,\rho(\mathbf r)}$$
Whereas the COG is weighed based on, well, the weight
$$\mathbf r_\text{COG}=\frac{\int\text dV\,\rho(\mathbf r)\cdot g(\mathbf r)\cdot\mathbf r}{\int\text dV\,\rho(\mathbf r)\cdot g(\mathbf r)}$$
When $g(\mathbf r)=g$ is uniform over the object, then the two definitions become equal.
| {
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What makes a wheel spin? I don't fully grasp what makes a wheel much easier to move than to push a solid block.
The pressure at the point of contact between a wheel and the ground must be pretty enormous compared to the pressure created by a block of same material and mass as the wheel.
Friction is defined as the product of normal force exerted on the object and the coefficient of friction between the object and ground. So I assume that for two identical objects of infinite masses this parameter does not make any difference.
Given these circumstances, I don't understand the physics behind it. Am I missing some other attributes of a wheel that makes it easier to move?
| Pushing a heavy block across the ground is going to be hard because as you say the frictional force between the block and ground is $F = \mu F_N = \mu mg$ and for a heavy block this is going to be a large force:
But suppose you split your block into two parts, and you put a layer of oil in between the two parts like this:
At the contact with the ground the friction $\mu_1$ is large, while at the oil layer between the two blocks the friction $\mu_2$ is very small because oil is a good lubricant. So now when you push the top block it's going to slide forward with very little force needed, while the bottom block is going to stay stationary relative to the ground.
Although it may not be immediately obvious this is exactly what happens with a wheel. The bottom block corresponds to the rim of the wheel and the top block corresponds to the axle. And in between the axle and wheel we have a layer of oil:
The friction between the wheel and the ground $\mu_1$ is high so the point of contact of the wheel with the ground doesn't slide. However the friction between the axle and the wheel $\mu_2$ is very low so the wheel slides easily relative to the axle. That means when we push on the axle the wheel rotates easily around the axle and rolls forward with very little force needed.
| {
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If a ball is thrown to a person standing on a frictionless surface, is the impulse of the thrower equal to that of the catcher? If a person throws a ball, exerting a given impulse does the person that catches the ball receive the same impulse assuming that the catcher moves. Is the impulse that the catcher receives less than the impulse that the thrower receives because the ball continues to move with the catcher or does the catcher receive all of the impulse, to begin with and then return momentum to the ball as they pull it along in their hand?
| As we know from the definition of impulse that $$ J = \int \vec{F} dt$$ and thus as thrower exert a Force for short time while the catcher have longer time for taking a force , So we concluded that Thrower have greater impulse than catcher. But there is also $\vec{F}$ sitting there, To make it simple let us suppose the thrower throws a ball vertically upward with a force say $\vec{F_1}$ for a short interval $dt$. While force exert by ball $mg$ on catcher,that is constant for short interval say $\Delta t$. Now which one is greater? Let's take the limit that the thrower exert a really high force in that case force exert by ball will be same as $mg$.Thus we concluded that Impulse by thrower is greater that catcher in all the cases.
| {
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Given two states which are close in trace distance, are there purifications of these states which are equally close? Given a pair of state $\rho_A$ and $\sigma_A$ such that they are close in trace distance i.e.
$$|\rho_A - \sigma_A| \leq \epsilon,$$
can one find purifications of these states which are also $\epsilon$ close in trace distance i.e.
$$|\phi_{AA'} - \psi_{AA'}| \leq \epsilon$$
We are free here to take our pick over all unitaries on the $A'$ subsystem to help achieve the bound. And if not $\epsilon$, can we say anything non-trivial about the distance between the best possible purifications?
| You might want to look at Uhlmann's Theorem: It is not about the trace distance, but a different distance measure, the fidelity, and states that the fidelity of two mixed states is equal to the maximum fidelity of their purifications - that is, their "fidelity distance" is the smallest "fidelity distance" over all purifications. (In addition, it is possible to relate fidelity and trace distance, so that way you can also get a relation for the latter.)
| {
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Generalising Cutting lens changes intensity of light of image We were told that if I cut a (say biconvex lens) in half, my intensity of light which forms the image is now half.I was wondering if there was a general formula for if I cut a lens horizontally (a variable,say 2/3 of the aperture),what intensity of light would I receive for the image formed?
| In terms of reducing intensity, there is no difference between adding an aperture to the lens (in front of it or behind it) or cutting out pieces of the lens. In a sense a cut lens is just an ordinary lens with a different type of aperture. In any case, the intensity is proportional to the area of the aperture/remaining lens.
As an interesting side note, the focused parts of the image would stay the same shape as before cutting. Defocused parts would have blur spots with the shape of the remaining lens, like in these two images (taken from here: https://en.m.wikipedia.org/wiki/Bokeh), instead of circular as with a whole lens (assuming the lens is circular, of course):
Maybe this helps visualise what's happening here: you're cutting off pieces of the image each object point makes. When the image is sharp, you don't see the shape because its size reduces to a point. But the shape is still there and you've still cut off pieces of it, removing those pieces' contribution to intensity.
| {
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What is the fastest moving macroscopic object people have made? The Parker Solar Probe will move at 690,000 km/h (430,000 mph), or 0.064% the speed of light, at its closest approaches to the sun. Parker Solar Probe top speed
Will that be the highest speed yet achieved by a manufactured, visible-size object? If not, what is?
I have tagged this question assuming the highest speed will by something we send to space -- but I do not know that. It could as well be some kind of thing I have not thought of at all.m
| According to this Wikipedia page, the current fastest macroscopic man-made object was a shell (mass about 0.1 mg) in an inertial confinement fusion capsule driven by the National Ignition Facility. This particle was measured to be travelling at a speed of $445000\text{m/s}$ or $0.0015c$.
| {
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On the "spectrum" of an operator in quantum mechanics Very simple question, I'm new to this. I'm reading Griffiths book on QM and have a question about the "spectrum" of an observable operator. Does the spectrum of an operator require specification of a particular system? Or is the spectrum of an operator just every possible eigenvalue that can be obtained by every possible eigenfunction of an operator?
| The spectrum of a particular operator is the set of all possible eigen values.
In regards to the other answer: since the Hamiltonian of a system has a different form for different systems, also the spectrum of that Hamiltonian will vary with different systems. So talking about the spectrum of a Hamiltonian is equivalent to “the set of all possible energies a system can have”. However, once you specify that it’s the Hamiltonian of a harmonic oscillator, this fixed the shape of $H$ and therefore its spectrum.
| {
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$V$-$I$ characteristic of a solar cell please explain the VI characteristics of a solar cell. The characteristics is given in my book without any explanation. How can the Voltage decrease on increasing current shouldn't it be opposite.
Solar Cell I-V characteristics
(Image from Electrical 4 U - Characteristics of a Solar Cell and Parameters of a Solar Cell)
| The characteristic is measured under illumination, when the solar cell is a source of electrical power, connected to a load. When the resistance of the load varies, one can get different points on the curve.
Compare with a battery: voltage decreases when a load is connected.
| {
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