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Why do certain mesons not self-destruct? As I understand it, particles and antoparticles annihilate one another. But some mesons (and pions?) consist of a quark and its antiquark. How can they exist without the two particles annihilating each other?
| But they do decay into the channels available from conservation laws.
Annihilation happens when all the quantum numbers cancel. In the pio the charge and baryon number of the quark antiquark cancel each other, and the decay particles add up to zero quantum numbers.
The $\pi^0$ goes into two photons as soon (electromagnetic interaction rates) as it is produced. The $\pi^+$ and $\pi^-$ survive because of charge conservation, and they decay with the weak interaction.
This is the same as with proton antiproton annihilation. The baryon number and charge cancel, and the decay products have to add up to zero for these quantum numbers
This is the case with all mesons.
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When only part of the surface an object is in contact with has friction, what is the normal force I should use? I have the following exercise:
A uniform rod of mass $M$ is given a horizontal velocity $v$ on a rough track as shown in the figure. The surface is rough on the right side of the origin $O$ and the surface is smooth and frictionless on the left side of the origin as shown in the figure. Express the velocity of the rod as a function of distance from the origin. Also find the distance before it comes to instantaneous rest.
I am not able to deduce what the force of friction on a small length $\mathrm{d}x$ of the rod should be. To get the friction on that part should I consider the normal reaction of that part only or of the whole rod ?
| Imagine there are two objects instead of one. One object has a certain mass, and experiences friction. The other object has another mass, and no friction. If those two objects were joined together, you would have no difficulty figuring out the equation of motion.
But actually the problem is harder than it looks: the way you have drawn it, the object has finite height; this means that there will be a torque as the object decelerates, and this torque will increase the normal force on the leading edge of the object. But unless the height of the object is given, you cannot solve for that.
| {
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Electric field inside a material I was thinking about the polarisation, and how the electric field behaves inside the material of permittivity greater than one.
I think to have understood what happens to D and P, but is not clear what happens to E.
Is the electric field inside the material bigger, remains constant, or is weaker?
| (For completely filled capacitors)
Q = CV
So, C = Q/V
So, C is charge stored per unit Potential Difference applied.
Now,
V = Ed
,where d is distance between plates.
$E = \dfrac{V}{d}$
Case 1) When you apply a constant V of 1V to capacitor
E across capacitor is $ \dfrac{1V}{d}$
which is constant independent of capacitance of capacitor or dielectric b/w plates.
So, E in capacitor is constant.
Case 2) You disconnect battery after applying a PD of 1V.
And then insert a capacitor.
So, C becomes C'.
Clearly Q = C' V'
So, since Q is constant and C' > C ** , **V' < V.
Since, $E = \dfrac{V'}{d}$ ,
E decrease.
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Centripetal force: if radius decreases, does ANGULAR or TANGENTIAL velocity change? Having conceptual trouble with this aspect of centripetal force. Say we have a puck on a frictionless table attached to a string that I am holding through a small hole, so that the puck moves in a circular path. So $F=m\;\dfrac{v^{2}}{r}=m\;\omega^{2}\; r$. If I pull on the string to change the radius of the puck's path, does angular velocity change, or linear? The math makes it look like it could be either, or both, which seems like it can't be right.
| As the puck is moving in uniform circular motion the centripetal force is normal to the velocity vector, no work is produced and the total energy of the puck, which is kinetic $\mathcal K=(1/2)mv^{2}$, remains constant.
Pulling the string, even slowly, the string tension becomes oblique to the velocity vector during the transition, work is produced transfered to the puck as kinetic energy. So the speed $v$ is increased, the radius is decreased and from $\omega=v/r$ the angular velocity is increased also.
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How black body absorbs light? I learn that black body absorbs light, but couldn't get the mechanism behind it. I wish I could get help.
| In physics one should have clearly defined terms.
Black body radiation is a mathematical model where a function can describe the effects of radiation for an object that radiates off , after absorbing, any radiation that falls on it:
"Blackbody radiation" or "cavity radiation" refers to an object or system which absorbs all radiation incident upon it and re-radiates energy which is characteristic of this radiating system only, not dependent upon the type of radiation which is incident upon it. The radiated energy can be considered to be produced by standing wave or resonant modes of the cavity which is radiating.
This is one of the lynch pins of quantum mechanics, because classical electrodynamics, as can be seen in the link, diverges for high radiation frequencies, and only the quantization hypothesis allows for a fit with the data. This is where photons and the energy they carry, proportional to their frequency, first appear.
It so happens that all solid bodies ot a given temperature, even though composed of complex quantum mechanical entities, fit this spectrum approximately. See the radiation from the sun for an approximate fit , and the microwave background radiation which fits the formula perfectly.
Now, black colored items are seen by our eyes as black because they absorb all visible frequencies. This does not mean that they absorb the invisible to our eyes also, as the spectrum of radiation is enormous, and the material may be reflecting a lot of frequencies, but our eyes would not preceive it.
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Ideal gas law: Will the piston move at all? We have the following experimental setup:
Before the experiment starts: $$p_1=p_2; \space V_1=V_2; T_1=T_2+\Delta T$$
The experiment starts and both the containers are heated so that the temperature difference $\Delta T$ remains constant (Edit: I wanted this to mean that there is no physical deformation, expansion or contraction of the containers. The piston is still allowed to move). The Volume of the containers also stays constant. Does the piston move to the left or to the right?
We have been discussing this question for an hour in our study group and we haven't really come to a conclusion. There are basically two hypothesis:
*
*Since the piston didn't move before the containers were heated, the piston won't move after the containers are heated because the tempereature will be increased by the same amount in both containers.
*Applying the ideal gas law $$pV=Nk_bT$$
and taking into consideration the fact that before the experiment started, both the containers had the same volume ($V_1=V_2$)and pressure ($p_1=p_2$) but different temperatures, it follows that there must be more particles/molecules in container 2 to "compensate" the higher temperature in the first one. Heating both containers by the same amount (equal temperature increase) implies more energy supplied to container 2 which will cause the piston to move left.
Can you give us a hint if we are going in the right direction with any of these hypothesis?
| Let $T_{20}$ be the initial temperature of tank 2 and $T_{10}=T_{20}+\Delta T$ be the initial temperature in tank 1. Let $\delta T$ be the equal rise in the temperature of both thanks. Assuming that the piston does not move, we would have $$p_{2f}=p_2\frac{T_{20}+\delta T}{T_{20}}$$and$$p_{1f}=p_1\frac{T_{20}+\Delta T+\delta T}{T_{20}+\Delta T}$$
Since $p_1=p_2$, if we divide one equation by the other, we obtain:
$$\frac{p_{2f}}{p_{1f}}=\frac{(T_{20}+\delta T)(T_{20}+\Delta T)}{T_{20}(T_{20}+\Delta T+\delta T)}=1+\frac{(\Delta T)( \delta T)}{T_{20}^2+T_{20}(\Delta T+\delta T)}$$
So, if the piston doesn't move, the final pressure in chamber 2 will be higher than in chamber 1. The piston must move in the direction from chamber 2 to chamber 1.
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When does a liquid 'wet' a solid surface? What is exactly meant when it is said that a liquid wets a solid surface. Has it got only to do with the contact angle ?
| The process of wetting of a solid surface is better explained with cohesive and adhesive forces. Wetting of a solid surface by a liquid means the liquid molecules succeeded in maintaining a contact with the surface through the inter molecular attractive forces. You should see that every liquid may not stick on to a given surface and similarly every surface may not get by a given liquid.
A liquid has it's own inter molecular force of attraction that maintains it's continuity. Such forces existing between same type of molecules is called cohesive force. Now if there is some inter molecular attractive forces between different molecules then it is called adhesive force. When the cohesive force between the water molecules is greater than the adhesive force of water molecules with the solid molecules, then it cannot stick on the surface. You can see a drop of water on such a surface (hydrophobic) forms a round ball and go in crazy directions. So there will be no wetting. Now in some hydrophilic surfaces, the adhesive force between the liquid and solid molecules dominate and the liquid molecules stick on to the surface. This makes the surface wet.
So the degree of wetting is determined by a force balance between adhesive and cohesive forces.
The contact angle which you mentioned and as shown in the figure depends on the balance between the adhesive and cohesive forces. So one could say that the degree of wetting depends on the angle of contact.If the adhesive forces dominate, then the drop spreads out on the surface, thereby decreasing the angle of contact. In such case, the wetting will be higher. If the cohesive forces dominate, the angle increases as the drop forms a bubble or round up into a spherical body which means the degree of wetting will be low. Thus the contact angle provides an inverse measure of wettability.
Hence if the contact angle between the liquid surface interface is zero, i.e., $\theta=0^0$, then the degree of wetting is maximum and is said to be perfect wetting. In such case the solid-liquid interactions are much stronger than the liquid-liquid interactions. If $0^0<\theta<90^0$, then the degree of wettingis high. If $90^0<\theta<180^0$, then the degree of wetting is low. If $|theta=180^)$, then the degree of wetting becomes minimum and the surface is said to be perfectly non-wetting. In such a case, the liquid-liquid interactions will be much higher than solid-liquid interactions.
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Why an impact exerts so much force? If an object of velocity $v$ and mass $m$ moves towards a resting object of mass $M$, then if the object which is hit might break. Why?
What is the reason that a collision has more power than a statical force $F$ acting on this object?
I haven't found any literature where such things are modeled. Can someone give literature where such things are described mathematically?
| Although the average force applied during a collision might be small enough that an object can take it, the peak force applied can be much higher. In physics this is called impulse.
Calculating the impulse for real world collisions (like a car crash) is very complicated. This is because cars have many structural members and the materials are not uniform. Depending on exactly how two objects collide, the forces can be applied to different members in different proportions and distributed around the object in different ways. This can cause forces to become concentrated on certain members, making them fail.
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Energy conservation around a black hole In the Schwarzschild black hole, the Killing vector "time translation" $k^a$, so that the following quantity is conserved along a geodesic:
$$E = -g_{ab}k^au^b = (1 - \frac{2GM}{r})\frac{dt}{d\tau}.$$
Which is interpreted as the total energy per unit mass measured by a static observer. However, a body orbiting a black hole will radiate part of its energy in gravitational waves.
What is the phyisical interpretation? Would this mean that energy is not conserved?
| The conservation of $\vec{k}\cdot\vec{u}$ only holds in the test particle limit. That is, it considers the metric to be unaffected by the motion of the particle. In this limit, there are no gravitational waves, since the metric has no time-varying quadrupole.
If you want to see gravitational waves, you need to allow the metric to evolve dynamically, considering the motion of the particle (whose mass must then enter into the problem). In this case that same $\vec{k}$ is no longer a Killing vector.
Since the spacetime is still asymptotically flat you can make some sense of a global energy quantity that is conserved, but it will be a sum of the particle's kinetic energy, the gravitational potential energy of the system, and the energy in gravitational waves.
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How do I find the relation between the accelerations of the ring and the disc (see image)? What is the relation between the accelerations of the ring and the disc (see image)?
Both the ring and the disc have mass $M$. The ring has a radius $R$ and disc has radius $2R$. They are connected by a light inextensible string. A force $F$ acts on topmost point of the disc.
The question actually asks the minimum value of the coefficient of friction for rolling without slipping to be possible. I have formulated the other equations, using torque or forces.
But, I need the relation between the acclerations of the ring and disc to solve them (I have 7 variables in my current set of equations, but only 6 equations). I have no idea how to relate the acclerations.
Also, I looked at the hint in my book, and it says $a_{disc} = 2 a_{ring}$. There is no explanation about how they arrived at this result.
| At the instantaneous moment shown in the diagram, we can write:
$$2R\alpha_{ring}=a_{disc}$$
as both are in pure rolling. This also tells us that the point on the ring where the thread is attached has an acceleration $=2R\alpha_{ring}=2a_{ring}$ so we find that:
$$a_{disc}=2a_{ring}$$
Note that when the string moves to another position this will not be true, but it might be enough for you to solve the question for now!
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Rotation of a vector Is a vector necessarily changed when it is rotated through an angle?
I think a vector always gets changed because its projection will change, and also its inclination with axes will always change. However the direction may remain same. Kindly make things clear to me.
| In general it changes although the reason is not exactly because its projections changes.
For example. You start with a vector (let us say the electric field of a parallel plate capacitor) on the plane $xy$. Then you rotate the coordinate system by an angle. The components of the vector on the new coordinate system is changed. But the vector did not change at all (you did not move the capacitor). This is called a passive rotation.
On the other hand, if you keep the axis fixed and rotate the vector (rotate the actual capacitor), it changed (unless you rotate by $2\pi$). This is an active rotation.
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Two Robertson-Walker observers, at what time will a light signal be received? Here is a question I have that is inspired by this question here.
The spacetime metric of a radiation-filled, spatially flat ($k = 0$) Robertson-Walker universe is given by$$ds^2 = - dT^2 + T[dx^2 + dy^2 + dz^2].$$Consider two "Robertson-Walker observers" [i.e., observers with $4$-velocity $(\partial/\partial T)^a$], the first one of which has spatial coordinates $(0, 0, 0)$ and the second of which has spatial coordinates $(x, 0, 0)$. At time $T = T_1$, the first observer sends a light signal to the second. At what time, $T_2$, will this signal be received?
| In the geometrical optics approximation light ray is represented by a null geodesic. Therefore you only need to find a null geodesic connecting points $(t_0,0,0,0)$ and $(t_1,x,0,0)$ for some $t_1$ (and this condition will determine $t_1$ uniquely). This is probably quite easy to do directly in this case, but in general for investigation of null curves in FLRW spacetime conformal time defined by $\mathrm d \eta = \frac{\mathrm d T}{a(T)}$ (with $a(T)=T^{1/2}$ in your case) is especially convenient.
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What does it mean that a Cooper pair behaves as a boson but respects the obligations of fermions? I refer to the fact that it has integer spin, but antisymmetric wavefunction. How is this possible?
| Well, $1/2\otimes1/2=0\oplus1$, so a system with two fermions has integer spin. But it is still a two fermion system, and therefore its wavefunction must be antisymmetric, as usual. This is not specific to Cooper pairs, but is basic Quantum Mechanics... [what is specific to Cooper pairs is that their size is $\gg a_0$, which means they are highly de-localised; this in turns means that they have mixed statistics: they have integer spin and yet they are not bosons].
The wavefunction of any system of any number of fermions must be antisymmetric (and this is one of the postulates of QM). The antisymmetry of the wavefunction is about the total wavefunction, that is, space wavefunction and spin wavefunction; you may have symmetric $\psi(r_1,r_2)$ and antisymmetric $\psi(s_1,s_2)$ or vice-versa. If you write $q=(\boldsymbol r,s)$, then the wavefunction of the pair is
$$
\psi(q_1,q_2)=\color{red}-\psi(q_2,q_1)
$$
If $\psi(q_1,q_2)=\psi_\mathrm{space}(\boldsymbol r_1,\boldsymbol r_2)\psi_\mathrm{spin}(s_1,s_2)$, then there are two possibilities:
$$
\begin{cases}\psi_\mathrm{space}(\boldsymbol r_1,\boldsymbol r_2)=+\psi_\mathrm{space}(\boldsymbol r_2,\boldsymbol r_1)\\
\psi_\mathrm{spin}(\boldsymbol s_1,\boldsymbol s_2)=-\psi_\mathrm{spin}(\boldsymbol s_2,\boldsymbol s_1)
\end{cases}\qquad\text{singlet}
$$
or
$$
\begin{cases}\psi_\mathrm{space}(\boldsymbol r_1,\boldsymbol r_2)=-\psi_\mathrm{space}(\boldsymbol r_2,\boldsymbol r_1)\\
\psi_\mathrm{spin}(\boldsymbol s_1,\boldsymbol s_2)=+\psi_\mathrm{spin}(\boldsymbol s_2,\boldsymbol s_1)
\end{cases}\qquad\text{triplet}
$$
In some cases the ground state is a singlet and in some others it's a triplet. It turns out that the Cooper pairs is a singlet.
For more details, see this SE post or this article (DOI: 10.1002/pssb.200461754).
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Intensity fluctuations at the output of a single mode fiber coupled to a He Ne laser I have coupled a Thorlabs HNL050L-EC - HeNe, 632.8 nm, 5 mW, Polarized Laser to a 2 meter long single mode fiber patch chord using a Thorlabs F230-FC-B aspheric lens. While I am certainly able to obtain a pure single mode Gaussian at the output, the total output intensity seems to be fluctuating over time scales of about a second. In some sense, the mode appears to be "breathing". The aspheric lens has been mounted on a stable mount, and the fiber is at the correct wavelength. I have also verified that the fluctuations are over and above the intrinsic fluctuations from the laser itself. Has anyone had this issue before? If so, what is the cause and what could be the best way to work around it to get a stable single mode Gaussian output?
P.S Please drop a comment if you require any further details to diagnose this issue.
| Check the pointing stability of the laser, which, together with mechanical vibrations, would make the coupling efficiency fluctuate. After making the setup as mechanically stable as possible, try to put small diameter tubes everywhere around the beam before the fiber. And/or enclose everything in a box. Air movement has an effect, and it helps to block it.
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Is it true that $\frac{d}{dt}\int_S \mathbf{B} \cdot d \mathbf{a}$ goes to zero if the amperian loop delimiting $S$ contracts indefinitely? I suppose to have an ordinary magnetic field: in the answer I'm not interested to involve Dirac delta: the integral goes to zero. I want to focus on another point: an infinitesimal physical quantity can have a finite time derivative? Of course derivative of zero is zero, but this flux is never strictly zero, and this trouble me because the step
$$
\frac{d}{dt}\int_S \mathbf{B} \cdot d \mathbf{a} \to 0
$$
(when the surface connected to the amperian loop can be taken indefinitely small) is used when we exploit Maxwell equations to fix boundary conditions on the discontinuity between two media. Maybe I'm getting flustered in the slightest thing, but this confuse me and I can't get to the bottom of this problem. How could I see clearly this passage?
| As long as $\mathbf{B}$ is a continuous (once-differentiable) function, when you look at small enough sizes, $\mathbf{B}$ has a Taylor series, the first term of which is a constant. As you let the loop size shrink, only the constant term matters. But then $\int_S \mathbf{B}\cdot d\mathbf{a}\rightarrow \mathbf{B}\int_S d\mathbf{a}\rightarrow 0$ since the area of the enclosed surface goes to 0.
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Phases and sinusoidal waves When we're talking about a wave, just a singular sinusoidal wave, what exactly is a 'phase'?
I came across a question that gave values of frequency ($550$Hz), and speed ($330$m/s). The question then asked to find how far apart two points are that differ in phase by $\frac{\pi}{3}$ rad. The answer came out to be $0.1$m. I was a little confused by the result, because if the two points differ by a phase of $\frac{\pi}{3}$ rad, the distance between them should also be $\frac{\pi}{3}$ rad, right?
| The clue here is " how far apart". The question is asking for distance which must be in terms of the wave's wavelength. Phase measures fractions of wavelength. And you are given information of the wave's speed and the periodic time in which it propagates (frequency).
The fundamental "distance = rate * time" applies in terms of the wave speed, wavelength and period, but period is the reciprocal of frequency.
$$\lambda=\frac{c}{f}$$
where $\lambda$ is the wavelength, $c$ is the wave's velocity , and $f$ the wave's frequency.
Plugging the numbers in you get one wavelength of the wave equal to 0.6 meters. But you don't want to know a full wavelength ($2\pi$ radians), but rather the fraction $\frac{pi}{3}$ radians. So just apply a simple ratio
$$\frac{0.6}{2\pi}=\frac{x}{\frac{pi}{3}}$$
x is equal to 0.1 m
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Conflicting answers gained from different equations First time posting I hope this is the place.
if someone could help me sort out formatting for the equations that would be nice
"As a result of being hit from behind by a toy truck, a 50g toy car rolls 3.0m across a floor that applies a constant retarding force of 1.2N to it. The car stops 2.0s after being hit. If the truck was in a contact with the ball for 0.15s calculate the impulse given to the car"
I started off this question by calculating the acceleration acting on the ball after the impact
\begin{eqnarray*}
F &=& ma\\
\therefore F/m &=& a\\
\therefore -1.2/0.05 &=& a = -24ms^{-2}\\
\end{eqnarray*}
Now from this I attempted to calculate the initial velocity strait after the impact. This is there the problems began. Using the equasion
\begin{eqnarray*}
v^2 &=& u^2 + 2as\\
0^2 &=& u^2 + 2 * -24 * 3\\
u^2 &=& 144\\
u &=& 12ms^{-1}\\
\end{eqnarray*}
But
\begin{eqnarray*}
(v-u)/t &=& a\\
(0-u)/2 &=& -24\\
-u &=& -48\\
u &=& 48ms^{-1}\\
\end{eqnarray*}
Why do these two equasions yield different results
| The question is inconsistent. At least one of the numbers (mass, force, stopping distance, or stopping time) is wrong. Your calculation of the acceleration from force and mass is correct, but an acceleration of $24$ m/s$^2$ for $2$ seconds means that the toy car was initially traveling at $48$ m/s. This is over $100$ mph ($160$ kph) and there is no way the car is stopping in only $3$ meters. The actual stopping distance in $2$ seconds is $48$ meters. This isn't necessarily what the question writer had in mind, so we don't know what the real answer should be.
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Why does Special Relativity apply to more than just light? It is my understanding that time dilation is derived from the constancy of the speed of light in vacuum. I would assume this implies that the quirky consequences would therefore apply only to light. But they don't. They apply to all material objects. Why?
| The basic idea is that physical laws are same in all inertial frames. Framing your question in a different way: Why do we generalize a formula(which gives time-dilation) whose derivation is based on a light clock to physical clocks and even the biological clock?
A very interesting argument was given by Feynman in his Lectures on Physics, Vol:1.
""To answer the above question, suppose we had two other clocks made exactly alike with wheels and gears, or perhaps based on radioactive decay, or something else. Then we adjust these clocks so they both run in precise synchronism with our first clocks (the light clock). When light goes up and back in the first clocks and announces its arrival with a click, the new models also complete some sort of cycle, which they simultaneously announce by some doubly coincident flash, or bong, or other signal. One of these clocks is taken into the space ship, along with the first kind. Perhaps this clock will not run slower, but will continue to keep the same time as its stationary counterpart, and thus disagree with the other moving clock.
Ah no, if that should happen, the man in the ship could use this mismatch between his two clocks to determine the speed of his ship, which we have been supposing is impossible. We need not know anything about the machinery of the new clock that might cause the effect—we simply know that whatever the reason, it will appear to run slow, just like the first one.""
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Wormhole Construction & "Jump Conditions" Throughout the literature Wormholes are typically constructed by "Minkowski" or "Schwarzschild Surgery" (see e.g. Visser, Lorentzian Wormholes...), i.e. under quite simple and/or highly symmetric circumstances.
In the former case, regions are excised from a single manifold and their boundaries identified, and in the latter, two manifolds are joined.
In the Minkowski case, the joining is trivial (modulo orientational considerations) since the metric is constant (-1,1,1,1), and in the Schwarzschild case shell Jump Conditions (ref, Israel, Darmois, Lichnerowicz, O'Brien & Synge - of which I have only been able to track down a copy of Darmois online) are applied.
Question:
If one were to attempt similar surgery on a more general spacetime foliated by Cauchy hypersurfaces, adopting the approach that two regions are excised and identified on each slice, what conditions (metric, derivatives, etc.) should be imposed on the spacelike identifications within surfaces and what conditions on the timelike identifications between surfaces?
| Well general relativity doesn't really have any standard condition on the smoothness of the metric tensor, but generally the metric is assumed to be at least $\mathcal C^2$, so that the Levi Civitta connection will be $\mathcal C^1$ and the Riemann tensor $\mathcal C^0$. Minkowski surgery usually involves a bit of a weaker condition in the thin shell formalism, with the metric being $\mathcal C^0$, the connexion with some step discontinuities and the Riemann tensor some Dirac distributions. Since the computations done on the Riemann tensor are generally linear, this works out alright.
So the surgery should have the metric at least have the same value along the stitching, and perhaps better if you could identify up to second derivatives as well.
| {
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Normal reaction Consider a plank on a frictionless surface and a ball from a height H is dropped on this plank. There is no friction between the plank and ball. Can the plank jump up in air for any value of H?
I don't want to know the value of H for which would happen. I just want to know that is this even possible that the plank jumps due to the normal impulse being greater than the collision impulse. This is no numerical, just a conceptual doubt.
| This is an interesting question.
An ideal plank (rigid, much heavier than the ball) will not rebound, but a real plank has some elasticity - either in the material itself or as a structure (eg if the plank is supported at the ends and the middle is raised off the ground). If a heavy solid object is dropped onto it, it will rebound to some extent and jump off its supports.
This is an example of a multiple collision : body A collides with body B which collides with body C. Here body C is the Earth. Except for ideal situations and materials, it is very difficult to predict whether B will lift off the ground and how far. But you would need to know much more than simply the height from which A is dropped.
Apart from the masses of the ball m and plank M and the height H from which the ball is dropped, you also need to know something about the elasticity of each collision (A-B and B-C). This is related to how much kinetic energy is lost during the collision and is conveniently expressed in terms of the Coefficient of Restitution which is defined as :
e = relative speed of separation / relative speed of approach.
A similar problem (Bouncing Superballs) is solved in the following links (p 15 in the first link) :
http://web.mit.edu/8.01t/www/materials/modules/chapter15.pdf
http://hep.physics.wayne.edu/~harr/courses/2130/f99/demonstration1.htm
UPDATE after you addition to the question :
Yes, this is definitely possible. I have seen it happen myself when a brick has been dropped onto a plank. Try it for yourself using eg a tennis ball and a ruler.
| {
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Speed of gravitational waves vs speed of light I own an educational YouTube channel on physics and astronomy. I am currently working on a gravitational waves video extension to my "How Fast Is It" video book on relativity theory. I have a question on the speed of gravitational waves. I understand that the field equations show that it is equal to the speed of light. My question goes one level deeper. My audience knows that the speed of light is fixed by two key characteristics of 'empty space' namely permittivity and permeability. The speed of a gravitational wave would be related to the elasticity of 'empty space'. Is it just a coincidence that these give the same result, or is there a deeper physics in play here?
| A better way to think of it is "speed of causality". That's the fastest any cause-and-effect will spread over space.
With nothing to cause it to go slower, changes to electric and magnetic fields will occur at that speed. No coincidence that changes to spacetime (causing gravity) propigate at the same speed.
You really need to show how Minkowski spacetime results in such a speed limit as a basic principle. It's not a speed limit in the usual sense; it's a deep principle of what speed is.
| {
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Why doesn't orbital body keep going faster and faster? If we consider the change in velocity during an infinitesimal interval of an orbit:
where body B is orbiting body A, we can see that the magnitude of the resultant vector (the green arrow) is greater than the magnitude of the original tangential velocity. Why doesn't this magnitude keep increasing indefinitely?
As I understand elliptical orbits, they speed up and slow down, but according to the diagram, I would expect them to keep speeding up monotonically.
(The answers to the duplicate question do not answer my question).
| You have to consider the limit of infinitesimally short time, in which the (vertical on the paper) component of velocity is infinitely short, and thus also the angle changes for an infinitesimal amount. In this limit, the correction to the length is quadratic in the time step and vanishes exactly in the physical limit of continuous time. Pythagoras:
$$v_2=\sqrt{v^2+(adt)^2}\approx v + \frac{a^2}{2v}dt^2 +\cdots$$
| {
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Radiation-Glossy Black vs Matte White Well, studying the chapter Heat I know that a matte black body radiates heat more than a glossy white one but can anyone give me an answer about what happens when its a glossy black body and a dull colored white body?
| The spectral radiation rate depends on the material in question and the temperature of the material. You can start to learn about this by Googling "Black Body Radiation" and "Planck" . Your terms "glossy black" and "dull white" are far too vague (in a scientific or engineering sense) to be able to answer. Further, the visual color is not necessarily related to the emissivity in the infrared or other portions of the electromagnetic spectrum. The spectral emissivity, especially for objects whose temperature is less than roughly 1500 K, is of most interest in the IR band.
| {
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Derivation of Lorentz Transformations How can I derive the Lorentz transformations? I don't want to use hyperbolic functions and the fact that the light waves travel by forming spherical wavefronts. Is there a way to derive the Lorentz transformations applying the conditions I have mentioned. I was unable to understand the method given in Landau and lifshitz deeply. That's why I want a method other than the one using hyperbolic functions
| Here's a derivation that uses very basic properties of space and time (isotropy, homogeneity, the fact that two Lorentz boosts should compose into another valid Lorentz boost, etc.). The constant maximum speed through space (i.e., the speed of light) is a derived property, not an assumption.
One more derivation of the Lorentz transformation - Jean-Marc Levy-Leblond
Here's a similar one that uses linear algebra after deriving the fact that the transform is linear, with similar results.
Nothing but relativity - Palash B. Pal
These kinds of group-theory-based derivations go back to Vladimir Ignatowski in 1910.
| {
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Does a box containing photons have more inertia than an empty box? A box containing photons gravitates more strongly than an empty box, and thus the equivalence principle dictates that a box containing photons has more inertia than an empty box. The inescapable conclusion seems to be that we can ascribe the property of inertia to light. Is this a correct deduction?
| Yes, both the internal potential energy and the internal kinetic energy of a bound system (in the rest frame of its center of mass) contribute to the bound system's inertial mass according to $E=mc^2$. For a paper discussing the evidence that this is true for internal kinetic energy in particular, see Kinetic Energy and the Equivalence Principle.
| {
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Eddy currents in a superconductor Just had a lesson we just had our teacher introduced the concept of eddy currents, and showed us how a magnet moves slowly through a metal tube due to the opposite generated magnetic field.
If you dropped a magnet through a superconductor then, would the magnet just float there? (Because the superconductor's eddy currents would be exactly the right amount and would not decrease due to internal resistance)?
Thanks!
| Yes, you are correct. The magnet would just float, perhaps even before entering the tube. You anticipated correctly that the strength and lack of dissipation of the eddy currents keep the magnet in place. This is well illustrated in a clip about levitating superconductors.
| {
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Plane wave shift in a differential operator Does anyone can help me to prove the following equation
\begin{equation}
e^{-i\vec{k}\cdot\vec{x}}f(\partial_{\mu})e^{i\vec{k}\cdot\vec{x}} = f(\partial_{\mu}+ik_{\mu})
\end{equation}
Where $\vec{k}\cdot\vec{x}=k^{\mu}x_{\mu}, \mu = 0,1,2,3$.
| PART 1 : Taylor Expansion ( see @Prahar comment ).
In the following :
$$
\mathbf{x}=x^{\mu}, \quad \mathbf{k}=k_{\mu}, \quad \partial_{\mu}=\dfrac{\partial}{\partial x^{\mu}} , \quad \mu=0,1,2,3
\tag{1-01}
$$
$$
\mathbf{k}\cdot\mathbf{x}=k_{\mu}x^{\mu}=k^{\mu}x_{\mu}, \quad \text{Einstein's convention on}\: \mu
\tag{1-02}
$$
Suppose now that the function $\:f\left(z\right)\:$ could be expanded in Taylor series, see https://en.wikipedia.org/wiki/Taylor_series.
$$
f\left(z\right)=\sum_{n=0}^{n=\infty}\dfrac{f^{\left(n\right)}\left(0\right)}{n!}\:z^{n}
\tag{1-03}
$$
so
$$
e^{-i\mathbf{k}\cdot\mathbf{x}} f\left(z\right)e^{i\mathbf{k}\cdot\mathbf{x}} =\sum_{n=0}^{n=\infty}\dfrac{f^{\left(n\right)}\left(0\right)}{n!}\:e^{-i\mathbf{k}\cdot\mathbf{x}}z^{n}e^{i\mathbf{k}\cdot\mathbf{x}}
\tag{1-04}
$$
For $z=\partial_{\mu}$ equation (1-04) gives
$$
e^{-i\mathbf{k}\cdot\mathbf{x}} f\left(\partial_{\mu}\right)e^{i\mathbf{k}\cdot\mathbf{x}} =\sum_{n=0}^{n=\infty}\dfrac{f^{\left(n\right)}\left(0\right)}{n!}\:e^{-i\mathbf{k}\cdot\mathbf{x}}\partial_{\mu}^{\left(n\right)}e^{i\mathbf{k}\cdot\mathbf{x}}
\tag{1-05}
$$
In PART 2 we prove by induction the following equation :
$$
\bbox[#FFFF88,12px]{e^{-i\mathbf{k}\cdot\mathbf{x}}\partial_{\mu}^{\left(n\right)}e^{i\mathbf{k}\cdot\mathbf{x}}=\left( \partial_{\mu} + ik_{\mu}\right)^{n}}
\tag{1-06}
$$
so from (1-05)
$$
e^{-i\mathbf{k}\cdot\mathbf{x}} f\left(\partial_{\mu}\right)e^{i\mathbf{k}\cdot\mathbf{x}} =\sum_{n=0}^{n=\infty}\dfrac{f^{\left(n\right)}\left(0\right)}{n!}\:\left( \partial_{\mu} + ik_{\mu}\right)^{n}=f\left( \partial_{\mu} + ik_{\mu}\right)
\tag{1-07}
$$
QED.
PART 2 : Proof by induction of equation (1-06) : $\:e^{-i\mathbf{k}\cdot\mathbf{x}}\partial_{\mu}^{\left(n\right)}e^{i\mathbf{k}\cdot\mathbf{x}}=\left( \partial_{\mu} + ik_{\mu}\right)^{n}$
Let a function $\:g\left(\mathbf{x}\right)\:$ infinitely differentiable. Then
(1). For $\:n=1\:$
\begin{align}
\left[e^{-i\mathbf{k}\cdot\mathbf{x}}\:\partial_{\mu}\:e^{i\mathbf{k}\cdot\mathbf{x}}\right]\:g\left(\mathbf{x}\right) & =e^{-i\mathbf{k}\cdot\mathbf{x}}\:\partial_{\mu}\:\left[e^{i\mathbf{k}\cdot\mathbf{x}}\:g\left(\mathbf{x}\right)\right]\\
& =e^{-i\mathbf{k}\cdot\mathbf{x}}\:\left[e^{i\mathbf{k}\cdot\mathbf{x}}\:\partial_{\mu}\:g\left(\mathbf{x}\right)+ik_{\mu}e^{i\mathbf{k}\cdot\mathbf{x}}g\left(\mathbf{x}\right)\right]
\tag{2-01}\\
& = \left( \partial_{\mu}+ ik_{\mu} \right)g\left(\mathbf{x}\right)
\end{align}
so
$$
e^{-i\mathbf{k}\cdot\mathbf{x}}\:\partial_{\mu}\:e^{i\mathbf{k}\cdot\mathbf{x}}= \left( \partial_{\mu}+ ik_{\mu} \right)
\tag{2-02}
$$
(2) Suppose that for a positive integer $\:n\:$
$$
e^{-i\mathbf{k}\cdot\mathbf{x}}\partial_{\mu}^{\left(n\right)}e^{i\mathbf{k}\cdot\mathbf{x}}=\left( \partial_{\mu} + ik_{\mu}\right)^{n}
\tag{2-03}
$$
then
(3) for $\:n+1\:$ we have
\begin{align}
\left[e^{-i\mathbf{k}\cdot\mathbf{x}}\:\partial_{\mu}^{(n+1)}\:e^{i\mathbf{k}\cdot\mathbf{x}}\right]\:g\left(\mathbf{x}\right) & = e^{-i\mathbf{k}\cdot\mathbf{x}}\:\partial_{\mu}^{(n)}\left(\partial_{\mu}\left[ e^{i\mathbf{k}\cdot\mathbf{x}}\:g\left(\mathbf{x}\right)\right] \right) \\
& = e^{-i\mathbf{k}\cdot\mathbf{x}}\:\partial_{\mu}^{(n)}\left[ e^{i\mathbf{k}\cdot\mathbf{x}}\partial_{\mu}g\left(\mathbf{x}\right)+ik_{\mu}e^{i\mathbf{k}\cdot\mathbf{x}}g\left(\mathbf{x}\right) \right]\\
& =\underbrace{\left[e^{-i\mathbf{k}\cdot\mathbf{x}}\:\partial_{\mu}^{(n)}\:e^{i\mathbf{k}\cdot\mathbf{x}}\right]}_{\left( \partial_{\mu} + ik_{\mu}\right)^{n}}\left( \partial_{\mu}+ ik_{\mu} \right)g\left(\mathbf{x}\right)\\
&= \left( \partial_{\mu}+ ik_{\mu} \right)^{n+1} g\left(\mathbf{x}\right)
\end{align}
so
$$
e^{-i\mathbf{k}\cdot\mathbf{x}}\:\partial_{\mu}^{(n+1)}\:e^{i\mathbf{k}\cdot\mathbf{x}}=\left( \partial_{\mu}+ ik_{\mu} \right)^{n+1}
\tag{2-04}
$$
proving finally (1-06)
$$
\bbox[#FFFF88,12px]{e^{-i\mathbf{k}\cdot\mathbf{x}}\partial_{\mu}^{\left(n\right)}e^{i\mathbf{k}\cdot\mathbf{x}}=\left( \partial_{\mu} + ik_{\mu}\right)^{n}}
\tag{1-06}
$$
| {
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Is Bohmian Mechanics incompatible with loop corrections? For those who continue to be unsatisfied with Quantum Mechanics (QM), Bohmian Mechanics (BM) is an alternative worth considering. It is sometimes claimed that BM is equivalent to QM, but Lubos Motl recently argued on his blog that this is true only for a limited class of quantum phenomena, namely those that do not belong to Quantum Field Theory (QFT) including loop corrections. I'm curious if anybody can provide any insight into how Bohmian Mechanics could incorporate what we know from QFT.
| For some counterarguments against Lubos Motl's argumentation against de Broglie-Bohm theory see http://ilja-schmelzer.de/forum/forumdisplay.php?fid=6 and http://ilja-schmelzer.de/realism/Motl.php
The first proposal for a Bohmian variant of a relativistic quantum field theory has been made in Bohm's original 1952 paper, for the EM field. For a possibility to handle fermion fields, see http://ilja-schmelzer.de/forum/showthread.php?tid=36
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Standing wave on a rope fixed at both sides: minus sign in the reflected wave I'm studying stationary waves on a rope fixed at both sides.
In some books I find that the wave function studied is the sum of incident wave $\xi_1(x,t)$ and of the reflected wave $\xi_2(x,t)$.
$$\xi(x,t)=\xi_1(x,t)+\xi_2(x,t)=A \mathrm{sin} (k x-\omega t)+ A \mathrm{sin}(kx+\omega t)=2 A \mathrm{sin}(kx)\mathrm{cos}(\omega t)\tag{1}$$
So this is the sum of two waves which differ only for the fact that one is progressive and one is regressive.
My doubts are on the fact that the fixed end of the rope cannot move, so there is a total reflection of $\xi_1(x,t)$ but the reflected wave $\xi_2(x,t)$ is in opposition of phase (i.e. uspide down), with respect to $\xi_1(x,t)$. So shouldn't $\xi_2(x,t)$ be
$$\xi_2(x,t)=- A \mathrm{sin}(kx+\omega t)$$
? The situation is the one in the picture.
Then if this was correct, $(1)$ would change to
$$\xi(x,t)=\xi_1(x,t)+\xi_2(x,t)=A \mathrm{sin} (k x-\omega t)- A \mathrm{sin}(kx+\omega t)=2 A \mathrm{cos}(kx)\mathrm{sin}(\omega t)\tag{2}$$
Am I missing something or is the reasoning somehow correct? If so, are $(2)$ and $(1)$ equivalent?
| If the phase difference between the wave is zero i.e lies in the plane of wave motion the resultant displacement is equal to zero,
Thus, $A = 0$, $(L,t)=0$, due to that fact, you can use $$A\sin(kx−\omega t)\to-A\sin(-(kx+\omega t))=A\sin(kx+\omega t)$$ for progressive wave, but nothing can happen when you use cosine rule because give us answer equal to zero for the maximum nodes
but $\sin Q = 1$.
| {
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Why do liquids exert pressure on the sides of a container? What makes a liquid push against the walls of a container if the liquid is completely static?
I was thinking a comparable situation would be a bin full of baseballs. Unless the balls were perfectly stacked they would be rolling off one another and the walls of the bin would stop them. Is it correct to assume the same is happening in liquid on a larger scale, or is something else going on?
It seems like if that was the case the pressure on the walls would be much less than on the bottom.
|
Unless the balls were perfectly stacked they would be rolling off one another and the walls of the bin would stop them.
Even in zero gravity fluids will exert an equal pressure on all walls. In zero gravity the balls would continue to remain stationary without exerting any force on the surroundings. So just thinking in terms of basketballs isn't enough.
The best answer I can give is extends on what @knzhou said, liquids do not have a preferred direction. If you press down slowly on a piston containing a fluid*, the particles will continue to move in all directions at the same velocity. However, the distance the particles move between the top and bottom surfaces decreases, hence the frequency of collisions with the sidewalls will increase and the force exerted, $F$ will increase. At the same time the area of the side-walls, $A$, decreases. Pressure = $F/A$ so the pressure on the side walls increases.
*(so that the temperature does not change)
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Vertical circular motion/How can tension be negative?
This is a rock tied to a string spinning vertically.
Here,
$T+mgsin\theta = mv_1^2/r => T = mv_1^2/r-mgsin\theta$
Suppose I give it a velocity $v$ at the bottom.
1) At what angle $\theta$ will the tension become zero?
2) If the velocity ends up $=0$ at $\theta = 0$, then the tension $T = m0^2/2-mg$ which would end up giving tension a negative value. How is this possible?
3) If the velocity at any point ends up zero, does the tension necessarily have to end up equalling zero as well?
| Tension in the string will never become zero, as long as rock is moving along the circular path.
Also, speed of the rock also, will never become zero in this case.
| {
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Wave speed derivation
The wave speed derivation approximates the wave as a circle. It uses that to know that $$a=\frac{v^2}{R}$$However, numerous functions can approximate the wave. A straight line, $x^2$, $x^3$, etc. If I used those I would get a different equation for a. So why is a circle the correct approximation choice?
| A single line isn't very useful for approximating a curve. You could use small segments, but then you'll need several, and the calculation would be more complicated. As noted in the comments, nothing stops you from using any other second order curve, i. e., a plane curve whose rectangular Cartesian coordinates satisfy an algebraic equation of the second degree, (a line is first order, since its corresponding equation is first degree, $y=ax+b$), except that doing the derivation with a circle is easier. Also keep in mind that you're approximating a single symmetrical pulse, not the whole wave.
| {
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Why does light act differently in miniatures? When painting miniatures (like RPG fantasy miniature soldiers)... why is it necessary to paint lights and shadows?
Being a 3D object, shouldn't the natural light be enough to create lights and shadows if the figure is simply painted with plain colours?
| When objects are very small, every source of illumination will appear to be "extended" - which softens the shadows and makes it harder to see the contours of the surface.
By painting highlights and shadows, you reduce the impact of the extended source. See for example Why don't fluorescent lights produce shadows?
| {
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The thickness of a puddled sheet of water If I pour water of different amounts into a puddle on a level surface, it appears by eye that both puddles have the same height. The water seems to retain itself into a puddle by surface tension. I have two questions:
(1) Does the height of a puddle of water on an infinite level surface depend on the material out of which the surface is made?
(2) Does the height of the puddle vary depending on how much water is in the puddle?
| Based on the page linked in the comments, the answer seems to be the following:
(1) The height of the puddle does not depend on the material of the surface so long as the surface is nonwetting.
(2) If the amount of water is large then it makes no difference, but if the amount is small, such as a droplet, then the contact angle will become significant and the height will depend on the contact angle.
| {
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Can we "see" into a black hole using gravity? I believe the "no hair" theorem means all black holes settle down into a state only determined by a few parameters, typically listed as mass, charge and angular momentum. But I don't think they can settle down instantly, which means the interior of the black hole may temporarily have an asymmetric mass/energy structure inside that may be observed from the outside using gravity.
If we had a huge massive black hole, say Milky Way mass, it would have a big radius, around a light year. Let’s assume it has no hair — it’s totally symmetric. We place plumb bobs on strings (let's call them pendulums) all around the black hole, a comfortable distance away. The symmetry means each pendulum points in a line through the common center of the black hole.
Now let’s drop in a more regular-sized black hole, say 100 solar masses. As it’s penetrating the horizon the situation is asymmetric. The pendulums wouldn’t point to the center of the big black hole anymore.
Would “no hair” require it to instantly becomes symmetric again once it penetrates? Isn’t it more likely it takes on the order of a year to become symmetric just because of the distances involved?
In other words, we can observe the interior structure of a black hole using gravity detectors.
| Yes. You're exactly right, deviations from no-hair do occur for example after BH mergers --- and hints of the "quasi-normal" mode ("ringdown") were observed in the LIGO detection. The no-hair theorem is constructed for a static, stationary BH (i.e. fully settled). In general, deviations from no-hair (magnetic fields, asymmetry, etc) will be radiated away on the scale of the light-crossing (or 'dynamical') time --- just like you suggest.
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Rest mass of phonon: is this concept definable? Phonons are obtaied by non-relativistic quantization of the lattice vibration. The dispersion relation is given by $\omega=c_s k$ where $c_s$ is the velocity of sound. What can we say about the mass of the phonon? I think it is not possible to compare this relation with the relativistic dispersion relation $E^2=p^2c^2+m^2c^4$ and conclude $m=0$. By mass, I do not mean the effective mass but the rest mass. Certainly, if the rest mass of phonon were zero it would have travelled with the velocity of light in vacuum.
I think in the non-relativistic approximation of the Einstein's energy momentum relation, the same $m$ appears in the non-relativistic kinetic energy $\frac{p^2}{2m}$. Therefore, we can still talk about rest mass in non-relativistic physics.
Moreover, phonon being a goldstone boson should have zero rest mass.
Edit: How does one define the rest mass of the phonon?
| Phonons are indeed massless, as you can see from their dispersion relation or from the fact that they are Goldstone bosons. The phonon dispersion relation that you wrote down tells us that we can excite a phonon mode, with some finite momentum, using an arbitrarily small amount of energy, hence they have no rest mass (in condensed matter language, they are not "gapped"). This does not mean that they travel with the speed of light; I guess one way to see that is that the lattice breaks Lorentz symmetry by giving us a preferred inertial frame. When formulating the theory of phonons we usually take the nonrelativistic limit $c \rightarrow \infty$ right from the start, so the speed of light never enters into any of the equations.
Phonons instead travel at the speed of sound $c_s$, which is the characteristic speed set by the lattice (if you compare the phonon dispersion to the relativistic dispersion relation you wrote down you see that $c_s$ replaces $c$, the speed of light).
Said differently, phonons are quasiparticles (=not true, elementary particles) that emerge in a theory with a lattice that breaks Lorentz symmetry, so your statement "if the rest mass of phonon was zero it would have traveled with the velocity of light in vacuum" does not apply to them.
| {
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What happens when two wavefunctions meet? Apologies for the over-broad question(s), but I'm having a hard time finding out where to look to answer these myself:
If a particle is a wavefunction describing a probability amplitude distributed through space, what happens when two wavefunctions meet? I imagine that their amplitudes begin to sum, so that as a photon's wavefunction approaches a wall (composed itself of many wavefunctions), eventually these two wavefunctions begin to overlap and sum (or multiply?), but what happens after this I'm still not clear on. Really I'm trying to figure out what is meant by an 'interaction'.
Is the better picture that it's all one wavefunction per field? In other words, is a light bulb is spraying trillions of 'individual' photon wavefunctions, or just one that describes all the potential individual photons?
| Wavefunctions combine trough tensor products, which is not the addition that one would expect naively. The reason for this is that a wavefunction contains the description of all possible futures of the system at once, so if there are multiple subsystems, then the wavefuntion of the entire system has to describe all possible futures for each part independently of the other parts.
Let's say a system can have two possible futures |1> and |2>. Then two copies of the system can have the possible futures |11>, |12>, |21> and |22>. Three copies can have the eight possible outcomes |111>, |112>,...|222> etc.. This is greatly complicated by the fact that if the subsystems are indistinguishable, then the permutations of them only can occupy one (further indistinguishable) future state, so instead of |12> and |21>, we can only talk meaningfully about the symmetric linear combination (|12> + |21>)/$\sqrt 2$. This gets even more complicated when fermions are involved which require antisymmetric combinations of this kind.
| {
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Fermi energy of electron gas with electrostatic interaction I have been given the following exam question and am unsure how I would go about solving it:
Consider the case of a one-dimensional metal, consisting of a chain of $N$
positive charges $+q$ separated by a distance $2R$ and immersed in a neutralizing background of electrons with density per unit length $n_e$. The electrostatic energy due to the interaction of the electrons with the ion cores and among themselves is:
$$\mathcal{E}_{el}=-\frac{\alpha q^{2}}{4\pi \epsilon_{0}}\frac{N}{R}$$
Assuming that the electrons form a non-interacting Fermi gas, calculate the Fermi energy. Write down an expression for the total kinetic energy of the electrons.
Now, ordinarily I would compute the Fermi energy as follows: First write down the density of states for a gas of Fermions:
$$g(\epsilon)=\frac{V m^{3/2}}{\sqrt{2}\pi^{2}\hbar^3}\epsilon^{1/2}$$
Then we have:
$$N = \int_{0}^{\infty}g(\epsilon)n_{F}(\epsilon)\:\mathrm{d}\epsilon = \int_{0}^{\infty}g(\epsilon)\Theta(E_{F}-\epsilon)\:\mathrm{d}\epsilon$$
Where $\Theta(\epsilon)$ is the Heaviside step function. We can thus calculate:
$$N=\int_{0}^{E_{F}}g(\epsilon)\:\mathrm{d}\epsilon = \frac{Vm^{3/2}}{\sqrt{2}\pi^{2}\hbar^{3}}\frac{2}{3}\epsilon_{F}^{3/2}$$
So we have the Fermi energy:
$$\epsilon_{F} = \left(\frac{N}{V}\frac{2\sqrt{2}\pi^{2}\hbar^{3}}{3m^{3/2}}\right)^{2/3}=\frac{2\hbar^{2}}{m}\left(\frac{\pi^{2}}{3}\right)^{2/3}n_{e}^{2/3}$$
To calculate the total energy, we have:
$$E = \int_{0}^{E_{F}}\epsilon g(\epsilon)\:\mathrm{d}\epsilon$$
But none of this takes into account the electrostatic energy, so I fear that I have terribly misunderstood something.
Thanks!
| Kinetic energy in 1D, method 1. Free electrons. Assume no potential energy at the moment. Zero temperature.
\begin{equation}
n_e=\int_0^{E_F}g(\epsilon)d\epsilon=\int_0^{E_F}\frac{1}{\pi\hbar}\sqrt{\frac{m}{2\epsilon}}d\epsilon=\frac{\sqrt{2mE_F}}{\pi\hbar}
\end{equation}
\begin{equation}
E_F=\frac{\pi^2\hbar^2n_e^2}{2m}
\end{equation}
Kinetic energy, method 2. Number of filled states
\begin{equation}
N=\frac{kL}{\pi}
\end{equation}
where $L=\frac{N}{n_e}$ is the chain length. This gives the wave vector
\begin{equation}
k=\frac{\pi N}{L}=\pi n_e
\end{equation}
and thus the kinetic energy
\begin{equation}
E_{kin}=\frac{\hbar^2}{2m}k^2=\frac{\hbar^2\pi^2n_e^2}{2m}
\end{equation}
similarly to method 1. The total (Fermi) energy is the sum of kinetic and potential (electrostatic) energy:
\begin{equation}
E_F=E_{kin}+E_{el}=\frac{\hbar^2\pi^2n_e^2}{2m}-\frac{\alpha q^2}{4\pi\epsilon_0}\frac{N}{R}
\end{equation}
Hope this make sense. In your solution, DOS is taken 3D, but the problem says the system is 1D metal. This, at least, should be corrected.
| {
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Is there any atom which is dia-electric? Take an atom. Suppose we impose some magnetic field on it. For some atoms, the energy increases---this is a phenomenon of diamagnetism.
The question is, how about an electric field? Can the energy of the atom increase when the electric field is turned on?
Put in a different way, can the electric field induce a dipole anti-parallel to it? If not, why can a magnetic field?
| No, there can't be atoms like that, at least not in the real world.
In the magnetic case, diamagnetism means that the magnetic susceptibility may be negative (so the permeability may be lower or higher than in the vacuum, the magnetic susceptibility may have both signs).
But in the electric case, the electric susceptibility is always positive, and the permittivity of a material is always greater than the permittivity of the vacuum.
The asymmetry arises because atoms are full of charged particles that move in the expected direction. On the other hand, they don't contain any magnetic monopoles, just dipoles, and their behavior is harder to predict, if I omit the detailed explanations of paramagnetism, diamagnetism etc.
| {
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Transversal wave speed derivation for small amplitudes
The above is a derivation for the wave speed equation in my physics textbook. However, I've read online that this equation is only true for waves with small amplitudes. I do not see where this assumption is made in the derivation, so why is the equation only true for small amplitudes?
The above picture shows the vertical restoring force should be 2*T*sin(phi)
| The explanation is not a very full one. As you correctly note, you're taking a limit, so the assumption $\sin\theta \to\theta$ as $\delta z\to0$ becomes exact. So Eq 16-23 contains no approximation.
The assumption creeps in subtly when one assumes that the force calculated in Eq 16-23 is at right angles to the $z$ axis. That is, that $\mathrm{d}y/\mathrm{d} z$ is small, so that the normal to the tangent to the curve stays approximately vertical in the diagram. The best way to understand all this is to work out a more accurate equation; then the vertical component of the force restoring the small length $\mathrm{d}\,s$ of string is
$$T\,\partial\theta \,\cos\theta = \mathrm{d}s\, T\,\frac{\partial\theta}{\partial s}\,\cos\theta = \mathrm{d}s\, T\,\frac{\partial^2y}{\partial z^2}\,\frac{1}{\left(1+\left(\frac{\partial y}{\partial z}\right)^2\right)^2}$$
(recalling that $\frac{\partial\theta}{\partial s}$ is the curvature of the string and then using the formula for the curvature) and THEN you approximate that $\frac{\partial y}{\partial z}\ll 1$ and, equivalently, that $\mathrm{d}z = \mathrm{d}s$. The small amplitude approximation is then indirect: we're directly assuming small gradients, which imply and are implied by small amplitudes, given that we know the wavelength is limited.
| {
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Analysis of a system consisting of a leaking tank of water The departure point is this problem:
A water tank on wheels is moving over an horizontal trail with negligible friction. There is a small opening in one of the walls, at a depth of $h$ below the tank's water level. The cross-section area of the opening is $A$. The initial masses of the tank and the water are $M$ and $m_0$. What is the initial acceleration of the cart?
Can we consider the water at the top of the tank to be stationary? If so, then it is pretty straightforward to find the velocity at which the fluid exits the opening. Then I would guess the acceleration could be estimated by looking at momentum variations. However, this is a varying mass system, so the mass also varies. This ends up being similar to the rocket equation, which involves solving a system of differential equations.
Is there a simpler way to solve this kind of problem (in other words, can you obtain the value of the acceleration without having to solve differential equations)?
| For a particular setup, the equations may get very simple: the tank should be massless ($M=0$) and the hole is all the way at the bottom of the container. Then $h$ is proportional to the mass of the water in the container: $m=m_0 h/h_0$, with $h_0$ the initial height. It's straightforward to derive that the force generated by the water jet is $F=2\rho g h A$, where $g$ is the gravitational acceleration and $\rho$ the density of water. The acceleration is then
$$\frac{dv}{dt} = \frac Fm = \frac{2\rho g A h_0}{m_0},$$
which is a constant. Technically, this is still a differential equation, but a rather trivial one.
Update With nonzero initial mass $M$, it is
$$ \frac{dv}{dt} = \frac{2\rho g A h}{M + m_0 h/h_0}, $$
which is a bit more unpleasant to solve, since $h(t)$ will become an exponential function of time. It's still a trivial differential equation in the sense that the velocity $v$ does not appear in the right-hand side; you just need to take the primitive.
| {
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A Question about a $U(1)_{B-L}$ I know I can write the QCD lagrangian like this:
$$ \mathcal{L} = (i\bar{q}_{R} \gamma_{\mu}\partial_{\mu} {q}_{R} + i\bar{q}_{L}\gamma_{\mu}\partial_{\mu} {q}_{L}) + \text{other terms} $$
When written this way we say it is invariant under $SU(3)_{R}\times SU(3)_{L} \times U(1) \times U(1)$.
But in a book, "Stefan Scherer & Matthias Schindler - A Primer for Chiral Pertubation Theory", it says: $SU(3)_{R}\times SU(3)_{L} \times U(1)_{B-L}$.
Why that?
| I searched through the whole book and didn't find a single instance of "$U(1)_{B-L}$", so a page number reference would be helpful. But for the purposes of pure QCD, $U(1)_V$ is the same as $U(1)_B$, because all quarks carry the same baryon number, and that's the same as $U(1)_{B-L}$ since nothing carries lepton number.
One might prefer to write $U(1)_{B-L}$ rather than $U(1)_V$ if one has the rest of the Standard Model in mind. If you include the other Standard Model gauge fields, it turns out that $U(1)_V$ is anomalous, and $U(1)_L$ is anomalous as well; the unique non-anomalous combination is $U(1)_{B-L}$. This is nice because it gives us an absolutely conserved quantity, as well as a symmetry that may be gauged.
That still doesn't fully explain it, because if they were only going to write down symmetries that were non-anomalous in the SM, they shouldn't have written down $SU(3)_L \times SU(3)_R$ either, since this symmetry is anomalous as well. I suspect they were just writing out of habit without looking too closely at what they were doing.
| {
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Can air be considered incompressible as long flow velocities are less than 100 m/s (Ma = 0.3)? Multiple sources state that steady air flow (in open systems) can be considered incompressible at velocities less than 100 m/s (Ma = 0.3). Example:
Deborah A. Kaminski, Michael K. Jensen, "Introduction to Thermal and Fluids Engineering", ISBN: 0-471-26873-9, 2005
Do you have a source which gives more detailled information on this?
Specifically I am interested in the following:
*
*Any fluid is compressible. But what is the extent of compression under above stated conditions?
*How is the abovementioned statement justified? Is there a mathematical derivation? Is experimental data available?
| As far as I catch the question (without going digged in math), pressure doesn't go instantly from a point to a point, but has certain speed, that is sound speed. No matter if your system is opened or closed, the air is the same. If you compress air in cylinder slowly, then the air well compressible, but if you try to move the piston quickly, let's say with Ma just less than 1, then you face much resistance because pressure wave doesn't have enough time to go away. It means in that case you are making too high pressure in a thin layer close to the piston while rest of cylinder still has low pressure. So I think there is some confusion with the matter. You can consider the air as well compressible at low speed, but more hard compressible at higher speed, and incompressible at all (behave like water) at Ma > 1. As is already said, there is no clear criteria for transition from "compressible" to "incompressible" speed, apparently the sources suggest Ma=0.3 as rule of thumb for practical engineering problems. Below 0.3 you can simply consider air as well compressible, while starting from 0.3 you can't.
| {
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Parity conservation in second harmonic generation? The second harmonic arises from susceptibility of third rank tensor $X^{(2)}$ which have (-1) parity.
page 28
Let say two photons are absorbed and one is emitted, so the total change in parity is $(-1)^{(2+1)}$. The initial state equals the final state so $(-1)^0=1$.
Where is the mistake here and how to conserve parity?
| The requirement is that $\chi^{(2)}$ be non-centrosymmetric. That's a bit different than having a particular parity. The states involved must be neither odd nor even; the parity must be mixed. That way the dipole matrix element exists between all three intermediate states involved in calculation of the susceptibility.
| {
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Two contradictory groups of statements from two different books on quantum physics There are two contradictory groups of statements from two different famous books on quantum physics.
Which one is correct?
Group (1) : Following statements are from Berkeley Physics Course Vol. 3, "Quantum Physics" by Wichmann, 1967
Page 204:
"The de-Broglie wave and the particle are the same thing; there is
nothing else. The real particle found in nature, has wave properties
and that is a fact."
Group (2): Following statements are from "An Introduction to Quantum Physics" by French & Taylor, 1978.
Page 234:
"When we come to particles other than photons, the wavelength again is
a well-defined property, but only in terms of a large statistical
sample. And for these other particles, we do not even have a seemingly
concrete macroscopic property to associate with the wave, equivalent
to electric and magnetic field of a beam of light. We arrive at the
conclusion that the wave property is an expression of the
probabilistic or statistical behavior of large number of identically
prepared particles -- and nothing else!"
EDIT:
According to 1st group, there is wave-particle duality. According to 2nd group, there are only particles (there are no waves) but the distribution of these particles (when they are detected) is wavy.
So which one is correct?
| It's neither a classical wave nor a classical particle. I think any attempts to describe it as either of those need to be qualified like this. It might look like one or the other, but both are only approximations.
The best theories we have describe quantum fields, and a particle is a field quantum. I don't really know how to describe a field quantum in classical terms other than "sometimes it can look like a classical particle, sometimes like a classical wave, and sometimes it's not really like either of those".
| {
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How to calculate number of degenerate states? For example if we need to get number of degenerate states for a particle confined in a 3D box that have energy
$$E=41\frac{\pi ^2 \bar{h}^2}{2m_e L^2}$$
I know that
$$E=\frac{n^2 \pi ^2 \bar{h}^2}{2 m_e L^2} $$
and $$n^2=n_x ^2+ n_y ^2 +n_z ^2$$
$$\Rightarrow n_x ^2+ n_y ^2 +n_z ^2=41$$
But how do I get all the different values of $ (n_x, n_y, n_z)$?
| What you have here could be described as a subset sum problem. Given $n$ can take any integer value (not including zero), you have the set of squares up to $36$,
$S = \{1,4,9,16,25,36\}$
and you wish to find subsets of three which sum to $41$. Looking at the subset sum problem this can not be solved analytically but algorithms can be employed.
To do this vigorously you need to form a 3D matrix $6 \times 6 \times 6$ by summing together the relevant squares and then read off the indices where $41$ is achieved, a task made easier with a bit easier with a computer script.
You must also note that any summation where $n_x \neq n_y \neq n_z$ will form a 6-fold permutation, sets with 2 distinct values will form a 2-fold permutation, whereas for $n_x = n_y = n_z$ the solution is unique.
| {
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What type of instrument can I use to determine my current distance from the center point of Earth? Immediately, I think of a scale, but is there better way? I can only imagine weighing an object of known mass with an extremely precise scale.
I am asking because I would like to be able to address absolute elevations relative to the center of Earth as ternary component of the geographic coordinate system.
| Sounds like GPS is best here. The position of the antenna is determined first in coordinates relative to the center of the earth, and then translated onto the ellipsoid and geoid. So finding the position relative to the center of the earth should actually be more accurate than finding its altitude relative to sea level. (Survey-grade can be centimeter resolution).
However, a consumer receiver isn't going to give you any of the information from that intermediate step. I think you'll need to find some sort of survey-grade unit that can provide such data.
If there were a local measurement that would give distance to the center of the earth, it would make an easy way to determine the size of the planet. There is no simple local measurement that can do so.
| {
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Why does time-independent Hamiltonian not depend on angle variable? In Landau and Lifshitz Mechanics, $\S50$ Canonical variables a time-independent Hamiltonian is considered, and a canonical transformation is done such that adiabatic invariant $I$ becomes the new momentum. Then the angle variable is found as
$$w=\frac{\partial S_0(q,I;\lambda)}{\partial I},$$
where $S_0$ is abbreviated action (and generating function for the canonical transformation), $q$ is old position variable and $\lambda$ is a constant parameter.
Now L&L say:
Since the generating function $S_0(q,I;\lambda)$ does not depend explicitly on time, the new Hamiltonian $H'$ is just $H$ expressed in terms of the new variables. In other words, $H'$ is the energy $E(I)$, expressed as a function of the action variable. Accordingly, Hamilton's equations in canonical variables are
$$\dot I=0,\;\;\;\dot w=\frac{\mathrm dE(I)}{\mathrm dI}.\tag{50.4}$$
Now my question is: why does $H'=E$ not depend also on $w$? $H$ does in general depend on old position variable $q$ (even if it does not depend explicitly on time), so why shouldn't $H'$ depend on angle variable?
| This is more or less an exercise in chasing definitions. The adiabatic invariant $I$ is defined as
$$ I\equiv \oint p \frac{\mathrm{d}q}{2\pi}\tag{49.7}$$
where the integral is taken over the path for given $E$ and $\lambda$. The external parameter $\lambda(t)$ is a slowly varying function of time $t$ in $\S49$, but is assumed to be a constant in $\S50$. Let us suppress the role of $\lambda$ in what follows to keep formulas simple. Then $I=f(E)$ is only a function of the energy $E$. Combined with the fact that the Hamiltonian $H$ does not depend explicitly on time, it show that the Kamiltonian ($\equiv$ the new Hamiltonian) $$H^{\prime}~\equiv~ K~=~H~=~E~=~f^{-1}(I)$$ is only a function of $I$, which is also the new momentum.
| {
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How much noise is in the Cosmic Background Radiation, especially from Cosmic Rays Do we have an estimate of how much noise, if any, say caused by cosmic rays in particular, is present in the CMB datasets and the maps based upon them?
Can we extrapolate a figure from the cosmic ray flux estimated to enter our atmosphere?
My knowledge on this subject is limited so my apologies if I have made a wrong assumption regarding the interaction, if any, between these two sources of radiation.
| Cosmic Rays are most often high-energy particles, mostly protons and alpha particles accelerated to high velocities by cosmic magnetic fields.
They do not show up in the microwave wavelength range that comprise the CMB.
As @ACuriousMind says in the comment, there is contamination in the CMB, but this is mainly due to Galactic dust and Bremsstrahlung from electrons in the Galactic magnetic field. This is also the reason why a broad band around the galactic plane has been masked out of the data used to calculate the anisotropies in the CMB.
| {
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Why do quasi-free states satisfy the positivity condition? In LQFT, a state, $\omega$, is a linear map $\omega:A=:CCR({\cal{S}},\Omega)\rightarrow \mathbb{C}$ satisfying:
*
*$\omega(aa^{*})\geq 0$ for all $a\in A$.
*$\omega(I)=1$ where $I$ denotes the identity element of $A$.
where $CCR({\cal{S}},\Omega)$ is the Weyl-algebra that comes from the symplectic vector space $({\cal{S}},\Omega)$.
Now let $\mu:{\cal{S}}\times{\cal{S}}\rightarrow\mathbb{R}$ be an arbitrary (real) inner product on $\cal{S}$ that satisfies
\begin{equation}
\frac{1}{4}\Omega(u^{1},u^{2})\le\mu(u^{1},u^{1})\mu(u^{2},u^{2}) .
\end{equation}
where $\Omega$ is a symplectic structure.
Then $\omega_{\mu}: A\rightarrow\mathbb{C}$ defined by
\begin{equation}
\omega(W(u))=e^{-\frac{\mu(u,u)}{2}}
\end{equation}
for all $u\in {\cal{S}}$, is a called a quasifree state.
Why does this state satisfy $\omega(aa^{*})\geq 0$?
| The condition on the modulus of $\Omega$ ensures that the complex bilinear form $\mu_c$ defined as $\mu_c(\cdot,\cdot)=\mu(\cdot,\cdot)+i\Omega(\cdot,\cdot)$ is a scalar product. Therefore $(\mathscr{S},\mu_c)$ is a complex pre-Hilbert space. Denoting by $\mathscr{H}_{\mu_c}$ its completion, it is then possible to define the symmetric Fock space $\Gamma_s(\mathscr{H}_{\mu_c})$ in the usual fashion:
$$\Gamma_s(\mathscr{H}_{\mu_c})=\bigoplus_{n=0}^\infty \mathscr{H}_{\mu_c}^{\otimes_s n}\; .$$
In addition, $$\omega(W(u))=e^{-\frac{1}{2}\langle u,u\rangle_{\mathscr{H}_{\mu_c}}}\; .$$
This is the generating functional of the vector state $\lvert \Omega_{\mu_c}\rangle\langle\Omega_{\mu_c}\rvert$, where $\Omega_{\mu_c}\in \Gamma_s(\mathscr{H}_{\mu_c})$ is the vacuum vector. This can be easily seen since $(\Gamma_s(\mathscr{H}_{\mu_c}),\pi_{\mu_c})$ with
$$\pi_{\mu_c}(W(u))=e^{\frac{i}{\sqrt{2}}\bigl(a^*_{\mu_c}(u)+a_{\mu_c}(u)\bigr)}$$
is a representation of the CCR on the Fock space ($a^{\#}_{\mu_c}$ are the usual creation/annihilation operators).
Now since $\omega$ is a vector state, then it is automatically positive, in fact:
$$\omega(A^*A)=\langle A\Omega_{\mu_c},A\Omega_{\mu_c}\rangle_{\Gamma_s(\mathscr{H}_{\mu_c})}\; .$$
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Tractrix - velocity pointing to pulling point It is said the tractrix is the curve described by a mass being pulled by a string, where the end of the string being pulled moves with constant speed, and the mass suffers a friction force. What is the physics explanation for why in the tractrix the velocity is always aligned with the string pulling the mass. Why if $h(t)=(h_x(t),h_y(t),h_z(t))$ is the position of the mass, and $j(t)$ is the position of the start of the string pulling the mass, then the velocity is always aligned with the string, that is $h'(t) = k (j(t) - h(t) )$ holds, where $k>0$ is some constant. Can this be derived by for instance stating the forces applied on the mass and then using $F = m a$ or some other physical argument ?
| UPDATE:
Many thanks for the update, Miguel. I am sorry, I misunderstood the description of the curve, which is misleading. The "mass" has no inertia (now I understand what philip_0008 meant in his comment) and is pulled infinitessimally slowly. So the "pulling" here is a "quasi-static" process, not a dynamic one. Although there is a force, it is at all times balanced by the drag, so there is no acceleration, and F=ma cannot be applied. There is also no velocity in the usual physical meaning of the term.
ORIGINAL ANSWER:
The velocity of the pulled object (or of the pulled end of the string) does not necessarily have to be in the same direction as the string. For example, the particle could be in circular motion. So there is no point trying to prove mathematically that velocity is in the direction of the string. I suggest that you provide a reference for this claim.
Perhaps what you have read is that if the only force on the particle is the tension in the string (eg the particle moves on a frictionless horizontal plane), then the vector acceleration is always in the direction of the string, because that is the direction of the net force. But if drag is included then this is not generally true either.
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How much of water's surface tension is entropic? Water molecules on the surface of an air-water interface have preferred orientations that lower their energy. This implies that these molecules are not uniformly distributed in orientation space, implying that the entropy is somewhat reduced when molecules are on the surface.
How much of water's surface tension is entropic? IE, the surface tension is the free energy change per unit area of surface created. How important is the $TS$ term compared to the $U$ term in this free energy? Can this be extracted from the surface tension as a function of temperature?
| If the water-air system is a closed system, when surface tension reshapes the interface shape between water and air, entropy of air increases due to increasing in volume and entropy of water decreases due to decreasing in volume. The heat to make this happen will reduce the air temperature. The water pressure will increases and air pressure will decreases. Because temperature change is very small, internal energy change can be neglected. TS is more important. So at high temperature, available free Gibbs energy is less than that at lower temperature.
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Calculating eigenvalues for operator Given relation $[a,a^\dagger]=I$. Operator $K$ is defined as $K=a^\dagger a+\lambda a^\dagger+\lambda^* a$. I need to find the eignevalues of operator $K$. How realtion that involves commutator could help me? Thanks for any suggestions.
| You may just complete the square:
$$ K = (a^\dagger+\lambda^*)(a+\lambda) - \lambda\lambda^* $$
Expand the product and subtract the last term to see that you get the same three terms. One may define $b=a+\lambda$. Then
$$ K = b^\dagger b - \lambda \lambda^* $$
and $[b,b^\dagger]={\bf 1}$, so these $b$ operators are isomorphic to $a$ and the spectrum of $K$ is the same as the spectrum of $a^\dagger a - \lambda\lambda^*$.
The spectrum of $a^\dagger a$ is $0,1,2,3,\dots$, by the usual raising-and-lowering operator solution of the harmonic oscillator (there is a state with the eigenvalue $0$ satisfying $a|0\rangle = 0$ and one may raise the eigenvalue by $1$ or an integer by acting with $a^\dagger$ on this ground state) so the spectrum of $K$ is $-\lambda \lambda^*, 1-\lambda \lambda^*, $ and so on.
In the language where $a\sim x+ip$, this subtraction of the linear terms from the quadratic $a^\dagger a$ is equivalent to simply moving the harmonic oscillator to a different mean value of $x$ and $p$ (the real and imaginary part of $\lambda$, up to some simple normalization factors).
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Relativity paradox with mirrors and light pulses
Consider two very short light pulses emitted from the centre (C) of two mirrors A and B (as shown in the diagram).
From the point of view of the lab frame, the apparatus is all moving to the left at velocity v.
Imagine there is also an electron near the centre of the apparatus, which is stationary in the apparatus frame and therefore also moving with velocity v to the left according to the lab frame.
The short light pulses (much shorter than the apparatus length) bounce off mirrors A and B and return and strike the electron.
This situation has similarities with the Michelson-Morley experiment.
According to the frame moving with the apparatus, the pulses take an equal time to bounce off the mirrors and arrive back at C. Therefore the EM waves cancel and there is no net radiation pressure exerted on the electron.
According to the lab frame, the light pulse emitted to the left has less distance to travel overall and so arrives at C before the pulse that was emitted to the right. Therefore the first pulse accelerates the electron by exerting a radiation pressure on it.
Does the electron accelerate or not? :)
(I'm looking for derivations/proofs showing both frames' interpretations)
| If $t_{CA}$ refers to the time it takes in the lab frame for the light to reach C from A, and the same with $t_{AC}$, $t_{CB}$ and $t_{BC}$ then we have:
$t_{CA}=\frac{L/2+v t_{CA}}{c}$
$t_{AC}=\frac{L/2-v t_{AC}}{c}$
$t_{CB}=\frac{L/2-v t_{CB}}{c}$
$t_{BC}=\frac{L/2+v t_{BC}}{c}$
Thus $t_{CA}=t_{BC}$ and $t_{AC}=t_{CB}$
Finally: $t_{CA}+t_{AC}=t_{CB}+t_{BC}$
| {
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How can I measure the amplitude of a light wave? Suppose I have a light wave and I want to measure its amplitude, or check to see if it has an amplitude of a certain value: how would one go about doing this?
| Direct measurement of the amplitude of the optical field requires interferometric techniques. One that works is the FROG - Frequency Resolved Optical Grating. There are many variations on it today, including from the original developers: Trebino Research Group.
These devices were designed for ultrafast pulses. For CW one usually just measures intensity and polarization, followed by a computation.
| {
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Can we write the wave function of the living things? If yes then how? In quantum mechanics we studied that everything has a wave function associated with it.My question is can we write down the wave functions of things. Then how we can write down the wave functions of the things like animals, human eye, motion of snake etc.
| My answer will be very non-technical, but hopefully will convey some basic ideas about what the quantum state (or wavefunction) is about.
One intuitive way to picture the nature of the quantum state of a system is to see it as the interference (hence the "wave" idea) of every different changes it could possibly undergo while it is not being messed with.
In other words, when a system is not observed, its dynamics takes into account absolutely all the behaviors it could have and somehow mixes them together in constructive and destructive ways (this is known as the "path integral").
There are two aspects here that make it impossible in practice to formulate the wavefunction of a living being:
First, if the system is not isolated one must take into account all interactions it has with its environment (which is made of other systems themselves subject to the same description). Living beings are obviously not isolated, they are very open systems continuously exchanging matter and energy with their environment.
Second, for anything else than simple systems there is no way to build the interference of all possible behaviors, because they can be very complex and there is an infinity of them.
| {
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Why does $\prod^n_{j=1}\sigma^{(j)}_x$ commute with this adiabatic Hamiltonian? In the section 4.1 of Quantum Computation by Adiabatic Evolution, Farhi et al proposes a quantum adiabatic algorithm to solve the $2$-SAT problem on a ring.
The adiabatic Hamiltonian is defined as
$$
\tilde{H} (s) = (1-s) \sum^n_{j=1}(1-\sigma^{(j)}_x) + s \sum^n_{j=1}\frac{1}{2} (1-\sigma^{(j)}_z \sigma^{(j+1)}_z )
$$
To prove the correctness of the algorithm, the authors consider an operator which negates the value of the bits.
$$
G = \prod^n_{j=1}\sigma^{(j)}_x
$$
Then on page 13, it is mentioned that $[G, \tilde{H}(s)] = 0$.
My question:
How do I prove that $\left[\prod^n_{j=1}\sigma^{(j)}_x, \left((1-s) \sum^n_{j=1}(1-\sigma^{(j)}_x) + s \sum^n_{j=1}\frac{1}{2} (1-\sigma^{(j)}_z \sigma^{(j+1)}_z )\right)\right] = 0$?
| The first term (sum) in $\bar H$ obviously commutes with all $\sigma_x$ variables because it's a function of $\sigma_x$ only and they commute with each other.
The second term (sum) in $\bar H$ also commutes with the product of all $\sigma_x$ because the first term in the summand is a $c$-number and the second term $\sigma_z^j \sigma_z^{j+1}$ anticommutes both with $\sigma_x^j$ and $\sigma_x^{j+1}$ (because $\sigma_x,\sigma_z$ anticommute), and it therefore commutes with the product of two $\sigma_x$ (two minus signs give a plus).
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Merging black holes makes them less dense, so According to What is exactly the density of a black hole and how can it be calculated? (more specifically, John's answer here made me think: if you merge a whole load of chunks of an element heavier than iron (to prevent them from fusing), the resulting object would either be more dense than a black hole of the same mass, or would become less dense by becoming a black hole.
So which one of these would happen, in this hypothetical situation? Or would neither happen, but some completely different situation? Both seem impossible to me, since such heavy objects would have no way to prevent gravity from crushing them down (which implies it must become a black hole), but if a black hole would form, it would require gravitational energy input in order to become less dense. So that would exclude both possibilities, right?
Of course this situation would never occur in real life, but this hypothetical situation would have no angular momentum in the system, so no mass would be ejected.
| There is nothing wrong in having something more dense than a black hole, large black holes can have densities less than water. If you put a lot of iron together it might or not become a black hole. An object of any density can be large enough to fall within its own Schwarzschild radius. The larger the black hole the lower the density, so you iron ball will become a black hole when it reaches the Schwarzschild radius: $r_s=\frac{2MG}{c^2}$, or $M=\sqrt{\frac{3c^6}{32 \pi \rho_{iron}G^3}}$
At that specific mass, the black hole has the same density as the iron, for larger masses the black hole will be less dense, and for less masses there will be no black hole. Thus, there is no way to create a black hole more dense than the mass used to make it. The created black hole, of any mass, will be less dense (or equal, if the mass satisfy the above equation) than the original material.
| {
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Check dimensions of the integral of a function I and a colleague are arguing about the dimensions of:
$$\int_0^x f(x) dx $$
in this particular case $[f(x)]=m^2/s^3$ and $[x]=m$.
Does it follow that $[\int_0^x f(x) dx]=m^2/s^3$ or $[\int_0^x f(x) dx]=m^2/s^3m$?
| It will be the latter case, $m^2/s^3m$ which is just $m^3/s^3$.
Remember that the integral is the sum of all the products $f(x)\;\text{times}\; dx$. $dx$ is a tiny piece of the path from $0$ to $x$, so it is in units of $m$ as well. Each of the products $f(x)dx$ have units $m^3/s^3$, and the sum of all these products keeps those units.
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How to pour water from a bottle as fast as possible? When one pours water out of a bottle, it first flows smoothly but then a pressure 'blockage' develops and the pouring becomes interrupted and turbulent, so that the water comes out in splashes. This seems to slow down the flow of water from the bottle.
What is the optimal way to pour the water so that it completely empties fastest?
Possible strategies:
*
*Holding the bottle at a certain angle
*Wildly shaking the bottle
*Squeezing the bottle
*Other...
It probably depends on the shape of the opening and/or the bottle itself, but we shall assume this beautiful example of a standard water bottle:
CLARIFICATION
The question is asking how to pour the water the fastest, so no straws, hole insertion and evaporating lasers allowed...
| A few years after asking this question I stumbled upon this super interesting video tweet, which is relevant for this question: link to tweet video.
The video shows two ways of emptying a certain water bottle. In one case, the bottle is simply flipped upside down. In the other case, the bottle is also flipped upside down and quickly moved in a circular pattern. The latter causes a kind of tornado current in the liquid, speeding up the emptying from 24s to 17s.
The video backs up the accepted answer. It is also rather interesting that the rotation does not have to be continued until the bottle is empty, but it is enough to start off the tornado current by a few rotations.
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Friction-free rolling/sliding on an inclined plane Suppose a sphere is rolling down an inclined plane. There is no friction. The body will not roll and undergo just a translation. But why is this so?
If we consider the axis to be along the point of contact, then there would be a torque which will cause it to rotate but in reality the body won't rotate. Why is this so?
| The question answered by @SatwikPasani is related but not quite a duplicate.
The apparent paradox is resolved by realising that using a frame of reference relative to the sphere which is accelerating down the slope is a non-inertial frame of reference.
If there is friction and the no slipping condition is satisfied then the frame of reference attached to the point of contact $P$ is an inertial frame because momentarily the sphere at that point of contact is not moving relative to the Earth which is assumed in such examples to be a good approximation to an inertial frame.
When there is no friction the point of contact on the sphere $P$ is accelerating relative to the Earth and so a pseudo force $mg \sin \theta$ up the slope (green) and passing through the centre of mass of the sphere $C$ has to be introduced if Newton's laws are to be used in that non-inertial frame.
Remember that when you sit in that frame of reference the sphere is stationary relative to you and so the net force and torque on the sphere as you see it must be zero.
So with the introduction of pseudo force, the net force and the net torque acting on the sphere are zero.
If you take a stationary point $P$ on the slope as defining your frame of reference then there is a torque about $P$ on the sphere due to the component of the weight down the slope $mg \sin \theta$ and that torque causes the centre of mass of the sphere $C$ to have an angular acceleration about point $P$.
| {
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Uniformly charged rod electric field A uniformly charged rod of length $L$ and total charge $Q$ lies along the $x$ axis as shown in in the figure. (Use the following as necessary: $Q$, $L$, $d$, and $k_e$.)
(a) Find the components of the electric field at the point $P$ on the $y$ axis a distance $d$ from the origin. $E_x=\,?$, $E_y=\,?$
I think I am getting close to a solution (maybe) but I keep getting turned around and confused. This is what I have so far. So, I know that I need $\lambda=Q/L$ and $r=\sqrt{d^2+L^2}$. My first thought was that the formula I need is $$E=\int_0^L \dfrac{\lambda \,k_e}{x^2}dx$$ but then it becomes a problem when I integrate and have to evaluate $\dfrac{-\lambda k_e}{x}$ at $x=0$. So then I tried to integrate the $x$ and $y$ components separately with respect to $\theta$. I found the following formulas at this site on the second slide.
$$E_x=\frac{-k_e\lambda}{L}(\sin\theta_2-\sin\theta_1)$$
$$E_y=\frac{-k_e\lambda}{L}(\cos\theta_2-\cos\theta_1)$$
Where, for this problem, $\theta_1=0$ and $\theta_2$ is the angle when x=L, so that $\tan^{-1}(\frac{d}{L})$, and this is where I really got lost. Are those formulas right for this problem? I had a thought to use the inverse sin and cos instead of the tan for $\theta_2$ but that made the formulas really messy. This is an online submission, so I'm guessing it should reduce to something nice, but I can't get to it.
| First I have to ask: does the question mention anything about the distance d? The reason why I am asking is because if d is large enough, we can say that it is in the far-field and we can easily approximate the field values using electrostatic theory treating the rod as point charge Q. I will provide an edit later if you want to use a far-field approximation.
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Nuclear bomb power - myth? I'm not experienced in physics yet (if it helps I've covered as much as acceleration, momentum and energy transfer/chemistry ionic and covalent bonding) but I've heard that the way people compare destructive force of nuclear weapons by megatonnes or kilotonnes is wrong. This does seem to make some sense because the energy will turn into a mix of gamma (?) radiation, light radiation, heat radiation and other things, but is there an accurate way to compare nuclear weapon destructive force? Say, I wanted to compare today's weapons to Little Boy.
| The so-called TNT equivalent of a nuclear weapon is an unambiguous way of quantifying how much energy is released by the nuclear weapon. There's nothing 'wrong' about it.
The only caveat is that the damage caused by, say, Little Boy versus 15 kilotons of TNT would not be identical despite having an equivalent yield (for various practical reasons).
Generally, 10-20% of nuclear yield is emitted in the form of ionizing or residual radiation, unlike conventional weapons. Related: effects of nuclear explosions.
| {
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Special relativity: is this a known paradox, or one at all? Two ships of the same proper length $L$ move towards each other, as in the diagram below (which shows it in the reference frame where the ship at the left is at rest). The fronts (noses) are pointing to each other.
Now, when both noses pass each other, they synchronize their clocks to zero.
Then, when the ships are at the end of passing each other their backs meet, and they stop the watches.
From the point of view of the ship at rest, its clock shows
$\Delta t=(L+L')/v$
From the point of view of the moving ship, its clock shows the same:
$\Delta t'=(L'+L)/v$
This is expected because of the symmetry of the problem.
On the other hand, because of time dilation, we expect that each ship sees the other clock running slower that its own, so each expects that the clock on the other ship will stop showing a smaller time interval than its own. Which I believe it does not happen. None of the ships accelerates or decelerates during the process, thus none experience a change in inertial frames.
Question: which of the two arguments is wrong (the "symmetry" or the "time dilation") and why?
| The argument by Symmetry is correct and the argument of time-dilation is flawed. Time-dilation suggests that if the two events will have the least temporal interval in the proper frame. But none of the frames in this scenario is a proper frame and not just that, they are equally far off from being the proper frame (i.e. symmetry). So they would measure the equal amount of extra time between the considered events (as compared to the proper frame). Otherwise, how the observations of one person's clock will be observed and perceived by the other is very well said by WillO so I will not delve into that matter.
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Particle sliding on a sphere I believe most of you probably solved the following problem using energy conservation as shown here. It states
A particle starts from rest at the top of a frictionless sphere of radius R and slides on the sphere under the force of gravity. How far below its starting point does it get before flying off the sphere?
I've be trying to solve this problem using only Newton's laws without energy conservation. I would like to know if it is possible and, if it is, if you could give me some ideas of how to solve it. The problem I am currently having is that I believe that the Normal force in this problem is not a constant, but a function of the angle.
I believe it is clear that the block's trajectory is a curve before it falls from the sphere. If it is a curve, we have a centripetal force given by
$$
m\frac{v^2}{R} = mg\cos\theta - N(\theta)
$$
Where I believe $N$ is a function of $\theta$.
When the blocks gets off the sphere, there is no normal force anymore, so
at this instant the centripetal resultant is just
$$
m\frac{v^2}{R} = mg\cos\theta
$$
One can also see that in the $y$ axis, the resultant force is given by
$$
ma_{y} = P - N(\theta)\cos\theta
$$
And the acceleration is
$$
a_{y} = g - \frac{N(\theta)}{m}\cos\theta
$$
Now I could try to solve
$$
\frac{dv_y}{dt} = g - \frac{N(\theta)}{m}\cos\theta
$$
to get the velocity in the $y$ axis and somehow figure it the height
where the normal force is zero... Anyway, this is what I know from the problem and I am lost. Any tips on how to solve it?
| @dvij gave the equation $$g\sin \theta =R\frac{d^2\theta}{dt^2}=R\frac{d\omega }{dt}$$
If we multiply this by omega, we obtain:
$$g\sin \theta \frac{d\theta}{dt}=R\omega\frac{d\omega }{dt}$$
If we integrate this equation between 0 and t, we obtain:
$$g(1-\cos \theta)=\frac{R}{2}\omega^2$$
So we have $$mg\cos\theta-2mg(1-\cos \theta)=N=mg(3cos\theta-2)$$
I don't know whether this counts as an energy method or not.
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Vlasov equation, Maxwell distribution I have the Maxwellian distribution:
$$f(v)=n\left(\frac{m}{2\pi kT}\right)^{\frac{3}{2}}\exp\left(-\frac{mv^2}{2kT}\right)$$
I have to show that it is a solution to the Vlasov equation:
$$\frac{\partial f}{\partial t}+\vec{v}
\cdot \text{grad}(f)+\frac{q\vec{E}}{m}\cdot \text{grad}_v(f)=0$$
Since $f(v)$ depends on the velocity $v$ only, I assume that the first two terms are $0$. However, when I differentiate over $v$, I get something which is not $0$. So, am I on the right path? If not, any idea what can be done?
| A property of the Vlasov equation is that any distribution that is only a function of constants of motion is its solution. So if the velocity of the case you present is not a function of time, the distribution would trivially be a solution of Vlasov.
| {
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Double-Slit Experiment: Effect of Intensity Reduction on Fringes
Monochromatic light passes through a double-slit arrangement. The intensity of the monochromatic light passing through one of the slits of the double-slit arrangement is reduced. State, and explain, the effect of this change on the appearance of the bright fringes and of the dark fringes.
It makes sense that the brighter fringes will reduce in brightness due to the fact that the sum of the amplitudes will not be as high a number as before.
However, I'm a little confused as for the effect on the appearance of darker fringes. I thought that because darker fringes means 0 light intensity, I thought that even after the reduction in intensity of the monochromatic light, destructive interference will still take place at some point, therefore giving rise to no change in appearance to the dark fringes.
However, the answer to the above question says "dark fringes will be brighter / less dark because summing amplitudes no longer gives zero"
Could someone please explain the flaw in my conceptual understanding?
| Suppose the amplitude of the wave from slit $1$ arriving at a point is $A_1$ and the amplitude of the wave from slit $2$ arriving at the same point is $A_2$ and let $A_2>A_1$.
The relative phase between the waves from the two slits arriving at that point depends on the path difference between the slits and the point and not on the amplitude of the waves.
So if the path difference is an integer number of wavelengths then the two waves will arrive exactly in phase with one another and the resultant amplitude will be $A_2+A_1$ which means that the intensity will be proportional to $(A_2+A_1)^2$ at a maximum.
If the path difference is such that the waves arriving at a point are exactly out of phase with one another the resulting amplitude is $A_2-A_!$ which means that the intensity is proportional to $(A_2-A_1)^2$ at a minimum.
The minimum intensity is only zero if the amplitudes of the two waves is the same $(A_2=A_1)$.
So starting from equal intensity (amplitude) from each slit by decreasing the intensity from one slit you decrease the maximum intensities of the interference pattern but increase the minimum intensities of the fringe pattern - the contrast between the bright and dark fringes decreases.
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Can I recirculate water from an open reservoir to the bottom of a bigger, closed one, without a pump? A fountain head pumps water out of the main tank into a 'pond' reservoir. Can the water recirculate back into the main tank without the help of another pump?
I'm sorry if this a dumb question. I'm guessing it would not function as the diagram shows, as the pressure of the water in the main tank would not let any water in at the bottom,right? Any solutions? (not requiring additional pumps)
| How about:
open valve 2 to transfer water from the reseroir to the tank. Close valve 2 to create the 'fountain' feature. Open valve 3 to transfer water from tank to reservoir.
| {
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Why must a physical theory be mathematically self-consistent? I always read in modern physics textbooks and articles about the need for physical theories to be mathematically self-consistent, which implies that the theories must not produce contradictions or anomalies. For example, string theorists are proud of the fact that string theory itself is self-consistent.
But what does this really mean? Physical theories are not a collection of mathematical axioms, they are attempts at describing Nature. I understand the need for rigorous foundations in mathematics, but in physics, we have experiments to decide what is true and what isn't.
It's also weird (for me) to say that a theory is mathematically self-consistent. For example, Newton's Laws of Dynamics encode empirically known facts in a mathematical form. What does it mean to say that Newton's Laws are mathematically self-consistent? The same can be said for the Laws of Thermodynamics. There is no logical need for Nature to abhor perpetual motion machines, but from experiments, we believe this is true. Does it make sense to talk about thermodynamics as being self-consistent?
| To put it in a short way: Self-consistency is required because we expect nature to stick to laws that can be described mathematically. Mathematical descriptions by definition have to be self-consistent.
| {
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Does momentum increase with out of phase photons? This paper speculates that the EM drive produces thrust with out of phase photons:
http://scitation.aip.org/content/aip/journal/adva/6/6/10.1063/1.4953807
My question is this, do out of phase photons have more momentum than the same set of photons traveling in phase?
Edit: More specifically, I'm interested in whether all the force of the beam is applied in one direction, rather than half on the receiving and half on the sending.
| The momentum of the photon is $p=h\nu/c$, so it only depends on its frequency, not its phase. At constant frequency, all photons will have the same $p$ regardless of phase.
| {
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What is exactly the "progenitor bias"? I am taking a course in astrophysics and my teacher mentioned different biases that are present when taking a sample of galaxies: the progenitor bias and the Malmquist bias. I understand very well the Malmquist bias but I think that I don't yet really understand the progenitor bias.
What I get is that properties of low redshift galaxies (such as morphological type, ...) are not necessarily conserved when considering high redshift galaxies. What I mean is that, for example, an elliptical galaxy now might have been an elliptical, spiral or irregular galaxy in the past, while if we look at a high redshift elliptical galaxy it should have formed like that.
Am I right? Can you explain what the progenitor bias is in case I am mistaken?
| The progenitor bias arises in attempts to study early-type (elliptical) galaxies at higher redshift. The desire is to choose a sample of galaxies at high $z$ that are the analogs of the galaxies that evolved to form the low $z$ sample. The bias arises if one chooses a sample of only early-types at high $z$. Because some late-types eventually evolve into early-types, by selecting only early types, one excludes the high $z$ late-types that will become late-types by low $z$. This introduces a bias in the sample, which tends to underestimate the evolution of the population.
So it sounds like your explanation is, more or less, correct. Or at least along the right lines.
| {
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How do we know that an electron "spins"? As far as I know, you can't necessarily isolate an electron to observe it, you can only observe its effects on other particles due to fields. Moreover, we can't know an electron's exact location or how much space it occupies, although it has finite mass. It seems that the general model for the electron is either a wave or a rotating spherical particle. Being that we can only observe its mass/energy, how do we know that it is spherical and that some energy comes from a spin?
Edit: We have extensive comments on how the electron may not actually be spherical, and those models are purely for visual purposes. However, I am still wondering how someone determined that an electron spins. The Stern-Gerlach experiment was mentioned, but when I read up on that it seems that it is not a good experiment for charged particles as they will interact with the field.
| Electrons And Spin From Scientific American
Unfortunately, the analogy breaks down, and we have come to realize that it is misleading to conjure up an image of the electron as a small spinning object. Instead we have learned simply to accept the observed fact that the electron is deflected by magnetic fields. If one insists on the image of a spinning object, then real paradoxes arise; unlike a tossed softball, for instance, the spin of an electron never changes, and it has only two possible orientations. In addition, the very notion that electrons and protons are solid 'objects' that can 'rotate' in space is itself difficult to sustain, given what we know about the rules of quantum mechanics. The term 'spin,' however, still remains."
| {
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On the Stability of Circular Orbits Bertrand's Theorem characterizes the force laws that govern stable circular orbits. It states that the only force laws permissible are the Hooke's Potential and Inverse Square Law. The proof of the theorem involves some perturbation techniques and series expansion.
The most natural things that comes to my mind when thinking about such a problem is that the effective force should be a restoring force for circular orbits to be stable.
$f_{\mathrm{eff}}(r) = \dfrac{l^2}{\mu r^3}-f(r) = 0$, for orbit to be circular.
$f'_{\mathrm{eff}}(r)<0$, for orbit to be stable. Assuming a power law, $f=Kr^n$, for the central force, solving it gives me the solution $n>-3$.
This is very weak compared to the statement of Bertrand's Theorem. Could someone explain to me the rationale behind perturbation technique used, and what is missing from my interpretation of 'stable' in my derivation?
| What you just did was to find a condition for attractive power-law forces to have stable orbits where stable means they remain bounded when perturbed around the circular orbit. You got the correct result.
The Bertrand's Theorem though says something different: the only forces whose bounded orbits imply closed orbits are the Hooke's law and the attractive inverse square force. A closed orbit is one which the particle repeats their momentum and position after some finite time - it closes in the phase space.
The idea behind the proof of Bertrand's Theorem is to consider a perturbed orbit and then calculate the periods of the angular revolution and the radial oscillations. If these periods are commensurable then the orbit is closed. You can find the proof here.
| {
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The analytical result for free massless fermion propagator For massless fermion, the free propagator in quantum field theory
is
\begin{eqnarray*}
& & \langle0|T\psi(x)\bar{\psi}(y)|0\rangle=\int\frac{d^{4}k}{(2\pi)^{4}}\frac{i\gamma\cdot k}{k^{2}+i\epsilon}e^{-ik\cdot(x-y)}.
\end{eqnarray*}
In Peskin & Schroeder's book, An introduction to quantum field theory
(edition 1995, page 660, formula 19.40), they obtained the analytical
result for this propagator,
\begin{eqnarray*}
& & \int\frac{d^{4}k}{(2\pi)^{4}}\frac{i\gamma\cdot k}{k^{2}+i\epsilon}e^{-ik\cdot(x-y)}=-\frac{i}{2\pi^{2}}\frac{\gamma\cdot(x-y)}{(x-y)^{4}} .\tag{19.40}
\end{eqnarray*}
Question: Is this analytical result right? Actually I don't know
how to obtain it.
| The calculation of the propagator in four dimensions is as follows.
\begin{eqnarray*}
\int\frac{d^4 k}{(2\pi)^4}e^{-ik\cdot (x-y)}\frac{1}{k^2}
&=& i\int \frac{d^4 k_E}{(2\pi)^4}e^{ik_E\cdot (x_E-y_E)}\frac{1}{-k_E^2} \\
&=& \frac{-i}{(2\pi)^4} \left( \int_0^{2\pi}d\theta_3 \int_0^{\pi}d\theta_2 \sin \theta_2 \right) \int_0^{\infty} dk_E k_E^3 \frac{1}{k_E^2} \int_0^{\pi}d\theta_1 \sin^2 \theta_1 e^{ik_E | x_E-y_E | \cos \theta_1} \\
&=& \frac{-i4\pi}{(2\pi)^4} \int_0^{\infty} dk_E k_E \int_0^{\pi}d\theta_1 \frac{1-\cos 2\theta_1}{2} e^{ik_E | x_E-y_E | \cos \theta_1} \\
&=& \frac{-i}{4\pi^3} \frac{1}{| x_E-y_E |^2} \int_0^{\infty} ds s (\frac{\pi}{2} J_0(s)- \frac{\pi i^2}{2} J_2(s))
\end{eqnarray*}
where $s\equiv k_E\| x_E-y_E \| $, and $J_n(s)$'s are bessel functions and I made use of Hansen-Bessel Formula.
\begin{eqnarray*}
&=& \frac{-i}{4\pi^3} \frac{1}{| x_E-y_E |^2} \int_0^{\infty} ds s \frac{\pi}{2} \frac{2}{s} J_1(s) \\
&=& -\frac{i}{4\pi^2} \frac{1}{| x_E-y_E|^2} \int_0^{\infty} ds \, J_1(s) \\
&=& -\frac{i}{4\pi^2} \frac{1}{| x_E-y_E |^2} \\
&=& \frac{i}{4\pi^2} \frac{1}{(x-y)^2}
\end{eqnarray*}
| {
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Accessibility by reversible processes and the Second Law of Thermodynamics One common way of motivating the existence of Entropy as a state function is the following. Let us take the Clausius/Kelvin-Planck statement of the second law, from which we can deduce Clausius' theorem
$$\oint \frac{\delta Q}{T} \le 0,$$
where equality holds if and only if the cyclic process is reversible.
This of course means that the quantity
$$\int_C \frac{\delta Q}{T} $$
is path independent for reversible paths $C$, and so it defines a function of state which we call Entropy.
But this only seems to hold on the presumption that all states within our state space is mutually accessible through reversible processes, i.e. given any two states $A$ and $B$ in our state space, there exists some reversible process $A\rightarrow B$ and some reversible process $B\rightarrow A$. I don't see why this is necessarily true. Is this taken to be an additional (and apparently implicit) assumption? Or is this assumption provable? Or is it not actually needed to define entropy this way?
| So, you want to prove that between any arbitrary two states of a system, it exists at least one reversible path. You can prove this if you accept continuity of properties of substances. I.e. for example, if we have an ideal gas in equilibrium at initial state $(P_i,T_i)$ and final state $(P_f,T_f)$; then certainly there are infinite equilibrium states between initial and final states.
In diagram above $(P_1,T_1)$ is an equilibrium state so that $P_1=P_i+\Delta P$ and $T_1=T_i+\Delta T$.
If, $(\Delta P,\Delta T)\to (0,0)$ then $(P_1,T_1)\to(P_i,T_i)$ So, we can reach from $(P_1,T_1)$ to $(P_2,T_2)$ through infinite number of equilibrium states so that for each two adjacent states $(\Delta P,\Delta T)\to(0,0)$ and if we do this quasi-static process by insulating the system; then that process will be an isentropic process and we know that an isentropic process is a reversible process.
| {
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Resistor in series We know that when a resistor is connected in series , the current flowing through that resistor will be constant.But how do we use a resistor to limit the current flowing through a circuit ,when resistor is connected in series with the circuit ?
| Resistance can be interpreted in various ways depending on the circuit. It can be used to cause a potential drop, or it can be used as a heating device, etc.
You are asking how resistance can change the current flowing through the circuit when connected in series. In that context, the resistance can be used to alter the total resistance of the circuit which affects the current flowing through the circuit.
Ohm's law states that for small voltages (up to 1000V - depends on the device actually), ohmic devices obey the following relation
$$I = \frac{V}{R}$$
For a simple circuit such as the one given below, the total resistance decides the amount of current flowing through the circuit.
Here the E.M.F (voltage of the source in an ideal battery) is fixed and what the turns up in the denominator decides how much current will flow through the circuit.
Real-life Application (Plugging an LED to your 12V battery)
Let's talk about a more practical application, say you have an LED bulb of which has a resistance of $2 \Omega$. If you plug it to your 12V battery, the LED is going to blow up because it would draw nearly 6A which is WAY too much.
So does this mean you can't use LED with your battery? No. You can use by connecting a resistor in series with the LED.
The LED needs just a little bit of current, let's say it is rated to operate at 1A of current.
You want the total current flowing through your circuit to be 1A. The total resistance you'll need in the circuit will be given by,
$$R = \frac{V}{I} = \frac{12V}{1A} = 12\Omega$$
$$R = 2\Omega(LED) + R_{resistor}$$
$$R_{resistor} = 10\Omega$$
So you will have to connect a resistor with the resistance of $10\Omega$ in series with the LED to ensure that nearly $1A$ of current passes through the LED.
| {
"language": "en",
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How much realistically usable energy can be obtained from matter/antimatter interaction? The high energy density of a matter/antimatter system is well known. However, depending on the nature of the material, most of the energy from the interaction is released in the form of photons (gamma rays), which are difficult to extract work from.
Is there a theory that describes the maximum amount of usable work extractable from a matter/antimatter system — something akin to the equations for calculating the maximum efficiency of a Carnot engine?
| There was a lot of hype $10$ to $15$ years ago over the hydrogen economy. It was of course rather odd that anyone could take this seriously. How much free hydrogen gas is available? Answer: virtually none. The problem is that you have to either put electrical energy into water to split it into $H_2$ and $O_2$, or if you chemically condition methane $CH_4$ you may strip the hydrogen atoms free of the carbon, but you still have that carbon. As a result hydrogen is either an energy storage system, or a way of banking another form of energy (fossil fuels etc) in a way that lacks the carbon.
Antimatter is a similar issue, though at much higher energy. There is no freely available antimatter in the universe, at least as far as we know. With accelerators the amount of energy required to generate antimatter is far larger than the mass-energy contained in it. The energy utility in generating antimatter is highly negative at this point.
| {
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Two qubits system in polar co-ordinates I know that I can write a single qubit state in terms of polar co-ordinates $(r,\theta,\phi)$ on a Bloch sphere.
\begin{equation}
\rho =
\begin{pmatrix}
\frac{1+r \cos\theta}{2} &\frac{r \exp(-i\phi)\sin\theta}{2} \\
\frac{r \exp(i\phi)\sin\theta}{2} &\frac{1-r \cos\theta}{2} \nonumber
\end{pmatrix}
\end{equation}
Are this kind of polar decomposition exist for two qubits system?
| A valid density operator is any Hermitian, trace 1, matrix (with complex entries) and all eigenvalues between 0 and 1. Any two qubit system may be represented therefore by a Hermitian, trace 1 4x4 matrix.
Your qubit representation could be rewritten, more suggestively as:
\begin{align}
\rho &= \frac{1}{2}\left(\operatorname{I} + a_1 \sigma_x + a_2 \sigma_y + a_3\sigma_z \right)\\
&= \frac{1}{2}\left(\operatorname{I} + \vec{a} \cdot\vec{\sigma}\right)
\end{align}
Where $\operatorname{I}$ is the 2x2 identity matrix and the $\sigma_i$ are the Pauli matrices which are a basis for the space of Hermitian, trace 0 2x2 complex matrices.
Similarly to qubits we can decompose any n-dimensional quantum system using the identity and any basis for the space of Hermitian, traceless nxn complex matrices, a good choice of basis is the Gell-Mann matrices, and a construction in dimension n is given here.
Since they are Hilbert-Schmidt orthogonal you can find the coefficients $a_i$ for your state $\rho$ by taking the inner product:
$$a_i = \operatorname{tr}\sigma_i \rho$$
If you like you can then rewrite the vector $\vec{a}$ in any coordinate system you like (including polars).
There is a problem however because although every valid density operator may be rewritten in this way (with $\vec{a}$ being a n-dimensional sub-normalised vector) it is not the case that every operator of that form will be a valid density operator, the problem is that some of your eigenvalues will become negative and others will become greater than 1 to balance this out. This doesn't happen for qubits (any operator of the form $\rho$ I wrote above will be a valid density operator) and it makes working in higher dimensions fairly annoying sometimes.
| {
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Projection of a tensor Consider the following tensor (abstract index notation, e.g. Wald's) $B_{ab}$ and timelike vector field $X^{a}$ such that $X^aX_a=-1$ and
\begin{equation}
B_{ab}=\nabla_bX_a
\end{equation}
Then one claims that $B_{ab}$ is purely spatial, i.e. $B_{ab}X^a=B_{ab}X^b=0$. I do not quite understand how this is interpreted as "spatial", though I presume it borrows the intuition that such operation is like dot product or projection (e.g. a vector is purely spatial with respect to timelike vector if it is orthogonal to the said timelike vector). [This example is from Wald chapter 9 section 9.2]
I have seen similar reasoning used in general relativity, e.g. when computing "transverse-traceless" component of a tensor. I can see "transverse" notion for waves, but to say a general tensor (even second rank) is transverse is still a bit confusing. Could anyone explain the concept physically? I would expect that the reason the terminology was borrowed from simpler physics is because they indeed have analogous interpretation.
| Take the covariant derivative of the equation $X^aX_a=-1$. The RHS becomes zero so we have $$2X_a\nabla_b X^a=0\implies X^aB_{ab}=0.$$
The other equation, $X^bB_{ab}=0$, is the geodesic equation, so it doesn't hold for just any $X^a$.
Let's consider the situation at some point $p\in M$. Then $X^a$ is a prime candidate for the timelike basis vector of $T_pM$. Let $E_1^a,E_2^a,E_3^a$ be a triple of orthonormal spacelike vectors in the subspace orthogonal to $T_pM$. Then $X^a,E_1^a,E_2^a,E_3^a$ forms a basis of $T_pM$. The two conditions means that the component expansion of $B^{ab}$ can only contain factors of $E_i^aE_j^b$, i.e. something like $X^aE_1^b$ is forbidden. It is in this sense that it is "purely spatial".
| {
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Prove that an electron in a hydrogen atom doesn't emit radiation According to electrodynamics, accelerating charged particles emit electromagnetic radiation.
I'm asking myself if the electron in an hydrogen atom emits such radiation. In How can one describe electron motion around hydrogen atom?, Murod Abdukhakimov says the total momentum of the electron is zero, hence it does not emit radiation. Could someone prove this statement ?
It may be an obvious question, but I can't figure out why the total momentum of the electron should always be zero, in any energy state.
| The existence of hydrogen atoms is enough to demonstrate that the electrons don't emit radiation.
If they did, that energy would have to come from somewhere. The only place it could come from would be a reduction of orbital radius until the electron finally reaches the nucleus.
If you accept that electrodynamics applies, then you have to accept that atoms cannot exist - since they do, electrodynamics must not be the whole story.
| {
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Does curved spacetime change the volume of the space? Mass (which can here be considered equivalent to energy) curves spacetime, so a body with mass makes the spacetime around it curved. But we live in 3 spatial dimensions, so this curving could only be visualized in a chart with 4 dimensions, and the living being will notice only the effect that this bend causes on matter and light, the gravity, or the curve will expand-contract also the space itself? and can be measured by the change in volume?
Let's take an analog on 2 dimensions:
The distortion on space is visible only by a 3d being (such as a human.) The 2d creatures can't see the distortion, as they are 2d creatures, but they can see that space is 'bigger' where the distortion occurred, because the total area of the squares (as viewed from the top--i.e., from a different coordinate in a third, orthogonal dimension--increases.
In the real world, the axis of distortion is trickier to define. The space needs to be expressed as cubes, not squares, and what changes with distortion is the volume of those cubes.
So, if you have two identical boxes, one with a very massive object, and another empty, the one with the massive object fits more things inside it?
| Within the Schwarzschild metric, the volume does change.
It is the rectangle formed by the radial dimension and time which is invariant:
The dilating effect of the Schwarzschild metric
$$ \mathrm ds^2 = -\left(1 - \frac{2GM}{c^2 r}\right) c^2 ~\mathrm dt^2 + \frac{1}{1 - \frac{2GM}{c^2 r} }~\mathrm dr^2 + r^2 (\mathrm d\theta^2 + \sin^2 \theta~\mathrm d\varphi^2)$$
compared to the corresponding flat metric
$$ \mathrm ds^2 = -c^2 ~\mathrm dt^2 + \mathrm dr^2 + r^2 (\mathrm d\theta^2 + \sin^2 \theta ~\mathrm d\varphi^2)$$
does impact only the radial direction $\mathrm dr$ and time $ct,$ the rest of the geometry ($\theta $ and $\varphi $) remains unchanged.
You can easily see that the factor by which $\mathrm dr$ is multiplied is the same as the factor by which $c~\mathrm dt$ is divided. The factor
$$ \sqrt{1 - \frac{2GM}{c^2 r}}$$
is equal to the gravitational time dilation.
By consequence, the rectangle formed by the radial dimension and time conserves its surface in Schwarzschild geometry.
If you now consider the additional dimensions you see that the spacetime volume must have changed. If you add e.g. one dimension, obtaining a cylinder, the 3D-volume is $\pi r^2 t$, that means that a reduced radial dimension r is appearing squared, while the time dimension remains the same.
| {
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Physical meaning of enthalpy I've been reading about thermodynamics and reached the topic about enthalpy . I've understood its derivation but I don't understand its physical meaning ... Also I don't understand why they have divided by the mass of gas to get to the specific enthalpy equation . what's the use of it? I know the meaning of all state variables the enthalpy contains but I can't see the benefit of combining them together to have the enthalpy ..
| Enthalpy is heat at constant pressure:
$ dH = dU + pdV + Vdp $
$ dU = \delta Q - pdV $
$ \delta Q = dU + pdV $
$ dH = \delta Q + Vdp $
$ dp = 0 $ for constant pressure
So on a psychrometric chart of air, enthalpy is heat (energy) content per unit weight.
| {
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Why don't the magnetic dipole moments in a neutron cancel out? This may be a silly question. I thought magnetic dipoles were dependent on electric charge, so why in a neutron do the dipole moments not just cancel each other out?
| I am presenting this classical description for simplistic understanding of the process.
In general (not for elementary particles) the Magnetic dipole moment is generated by a current loop. If you consider a current loop with current $I$ then the charge exiting from loop is same as charge entering into loop and hence there is no net charge on the loop.
If the area of the loop is $A$ then magnetic dipole moment
$\mu=IA$
Hence you can understand that requirement of excess charge is not necessary for the magnetic dipole moment (even classically).
The relative strength of the magnetic force to the electric force is very small
$F_{mag} = \frac{1}{c}F_{elect}$
If excess charge would be the necessary requirement then magnetic force would not be detected on the first place. This is just my humble opinion.
Regards,
| {
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How does the uncertainty product $\Delta x \Delta p$ behave for the bound states of the triangular potential? As has been remarked earlier, if you take an unbounded potential $V(x)$ (so that all the eigenstates are bound) and you look at the uncertainty product $\Delta x\Delta p$ as a function of the index $n$ of the eigenstate, then for the two usual unbounded potentials (the simple harmonic oscillator and the infinite square well) $\Delta x\Delta p$ grows linearly with $n$.
How does this product behave for other solvable potentials, such as the linear cone potential $V(x)=|x|$?
| The cone potential $V(x)$ is exactly solvable, with eigenstates of the form
$$
\psi(x)\propto\mathrm{Ai}(|x|-b)
$$
in terms of the Airy $\mathrm{Ai}$ function, so this is rather easy to test. It is probably possible to produce explicit analytic expressions for the uncertainty product, but simple numerical evidence is plenty to see the behaviour here. Mathematica has most of it already implemented, and the rest is details.
Thus, if one sets
ψ[n_?EvenQ, x_] := AiryAi[Sqrt[x^2] + b[n]]/(Sqrt[-2 b[n]] AiryAi[b[n]])
b[n_?EvenQ] := x /. FindRoot[AiryAiPrime[x], {x,
-(3 Pi/8 (Max[0, 2 n - 1]))^(2/3), -(3 Pi/8 (2 n + 3))^(2/3)}]
ψ[n_?OddQ, x_] := Sign[x] AiryAi[Sqrt[x^2] + AiryAiZero[(n + 1)/2]]/(
Sqrt[2] AiryAiPrime[AiryAiZero[(n + 1)/2]])
b[n_?OddQ] := AiryAiZero[(n + 1)/2]
with $\psi_n(x)$ set to a properly normalized wavefunction of the form $\mathrm{Ai}(|x|-b)$ with $b$ set to a zero of the Airy function of its derivative, then the eigenfunctions are obviously right,
Plot[
Evaluate[Join[{Abs[x]},
Table[-b[n] + 0.7 ψ[n, x], {n, 0, 16}]
]]
, {x, -12, 12}
, Frame -> True
, Axes -> False
, ImageSize -> 700
, PlotRangePadding -> None
]
and can be checked to be normalized using
Table[NIntegrate[ψ[n, x]^2, {x, -∞, ∞}], {n, 0, 25}]
The uncertainty product can be integrated numerically, via
Table[{n, Sqrt[
NIntegrate[x^2 ψ[n, x]^2, {x, -∞, ∞}]
NIntegrate[Evaluate[
-D[ψ[n, x], {x, 2}] /. {Sign'[_] -> 0, Sign''[_] -> 0}]
ψ[n, x], {x, -∞, ∞}]]}
, {n, 0, 25}];
and it looks like this,
i.e. obviously very, very linear. (As an interesting aside, the ground state obviously satisfies the uncertainty principle, at $\Delta x\Delta p = 0.50463\cdots$, but as a plus it is also rather close to the minimal value.)
| {
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Electric field dependence on distance How can it be proved that for a point charge, $E$ is proportional to $$1/r^2$$ using the concept of Electric field lines (or lines of force)? I tried to show that if field lines are close, then magnitude of Electric field is higher. But, I couldn't show the given dependence.
| As Anthony B said,the number of field lines cutting any sphere surrounding a point charge is the same(because any field line which passes through a sphere of radius 1 also Passes through a sphere of radius 200) given that, the flux = E 4pir^2 should be constant. That explains the 1/r^2 dependance theoretically
| {
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Question in Lagrangian formalism In lagrangian mechanics,
where $L=T-U$
and the lagrangian formulation is $ \frac{d}{dt}\big( \frac{\partial L}{\partial \dot{q_i}}\big)-\frac{\partial L}{\partial {q_i}}=F_i$,
where $F$ is the non-conservative force.
My question is if I want to find out the above equation for a given problem then the $q_i$ should be written for every term in which the system is expressed.
Like if I want to write the equation for a pendulum then the $q_i$ will be the angle displacement.
So for example in a double pendulum there will be two angles $\phi ,\theta $ for the respective rods than the equation in lagrangian formalism will be
$$\frac{dL}{dt}(\frac{\partial L}{\partial \dot{\phi}}+\frac{\partial L}{ \partial \dot{\theta}})+\frac{\partial L}{\partial \theta}+\frac{\partial L}{\partial \phi}=0$$
is this correct?
| No, you get a separate Euler-Lagrange equation for each individual degree of freedom, i.e. a system of simultaneous equations. So in your example,
\begin{align}
\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}}\right)-\frac{\partial L}{\partial \theta} &= 0, \,\mathrm{and} \\
\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\phi}}\right)-\frac{\partial L}{\partial \phi} &= 0
\end{align}
| {
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How can the energy released when bonds formed be harnessed to do work? I'm trying to understand how living organisms get energy from food. I've heard that "energy is released when bonds form", but what exactly does "released" mean?
If the energy dissipates in the form of heat, then how can the cell use it for things?
If the energy is not in the form of heat, what form does it take?
| In most reactions the energy is released as kinetic energy of the reaction products. If you consider some reaction:
$$ A + B \rightarrow C + D $$
then if you add up the kinetic energies of $A$ and $B$ before the reaction and add the kinetic energies of $C$ and $D$ after the reaction you'll find that the kinetic energy after the reaction is higher than before it.
Kinetic energy is basically heat, and while this is great if you're trying to run a steam turbine it's pretty useless if you're a cell. Fortunately there is a type of reaction called a redox reaction where the energy goes into exciting electrons. Redox reactions all involve pumping electrons around, and it's the energy of the electrons that cells can use to drive other reactions.
My knowledge of biochemistry is too sketchy to say much more, but one type of redox reaction you'll certainly be familiar with is the battery. Batteries use redox reactions to raise the energy of electrons, and that's how they power electrical circuits. Cells aren't like electrical circuits, but they use the energy of redox reactions in the same way to power the reactions they need to live.
| {
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Measuring different components of spin simultaneously I'm reading Griffiths Introduction to QM and I'm having trouble understanding why you can't simultaneously measure the x,y and z components of spin. I know that the uncertainty principle prevents this but I still don't see why.
Griffiths' example is that if we have a particle in its up state, $\chi_+$ then we know the z-component of its spin is $\frac{\hbar}{2}$. If we then measure the x-component, then we're suddenly left with a 50-50 probability of the x-component being $\frac{\hbar}{2}$ or $-\frac{\hbar}{2}$. First off, why is it a 50-50 chance? If the state of the z-component is $\chi^z$ then $$ \chi^z=a\chi_+ ^z + b\chi_- ^z$$ and the x-component is $$ \chi^x =\dfrac{a+b}{\sqrt{2}}\chi_+ ^x +\dfrac{a-b}{\sqrt{2}}\chi_- ^x$$
If the z-component is in its upstate, does $\chi^z$ collapse to $$\chi^z = \chi_+ ^z$$
and so $a=1$ and then $b=1$. Therefore, there is a 50-50 chance of the x-component to be in its up or down state. Is this why it is 50-50 or am I understanding it wrong?
Next, if the particle is in its up state, then shouldn't the x-component also be in the x-component up state i.e $\frac{\hbar}{2}$ or does its up and down state 'reset' every time we measure? If it does reset, does it mean that once I measure for the x-component I lose knowledge about the z-component? So I'm left with a definite x-component but now I only have a 50-50 probability of knowing if it's spin up or down in the z-component? If it even does reset, what causes it? Is it just because of the uncertainty principle?
| In general you are getting it right: Non-commuting operators do not share eigenstates, thus measuring $S_x$ on an eigenstate of $S_z$ will result in a state that is not an eigenstate of $S_z$ anymore. The spin operators do not commute because they are defined via the Lie-algebra relation $[S_i, S_j] = i \hbar \varepsilon_{ijk} S_k$.
Next, if the particle is in its up state, then shouldn't the x-component also be in the x-component up state i.e $\hbar/2$ or does its up and down state 'reset' every time we measure?
The 'up' state is defined with respect to a direction, i.e. the $z$ 'up' state is not equivalent to the $x$ 'up' state. Think of the spin as a vector in a 3D cartesian space. Clearly, a vector pointing along the positive $z$ axis is not the same as the vector pointing along the positive $x$ axis.
| {
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What means that a pendulum system having saddle points? What means that a pendulum system having saddle points? I know when it haves drain, source but I can't find any information of when it has saddle points
| Very simply put: a saddle point is an unstable equilibrium. Without solving the equations, you can probably guess where there is an unstable equilibrium for a simple pendulum system: if the pendulum is above right the pivot, it will stay there. But give it even a the slightest notch, and it will not go back to this position.
In contrast, when the pendulum hangs straight down, and you give it a small push, it will restabilise. This is a stable equilibrium.
| {
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Two Rolling logs Suppose we have two logs rolling down a hill, one of gold and the other of wood; the acceleration for both will be equal, something which is unclear to me; I get that this may be due to their form, which is the same, but how come the mass of the objects doesn't matter?
| $$F = M A$$
Simply, more mass doesn't mean more acceleration, but more force. The logs will accelerate at equal speeds, but the heavier one will carry more force with it.
So, the gold one would have much more force behind it, and would take more force to stop it. If you put up bowling pins down the hill to attempt to stop the logs (Not the best method of halting I know), the gold one would plow through them more than the wooden one.
| {
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Why do we not use the SI system for distance in space? One of the closest stars to Sol is Alpha Centauri at 4.367 Ly according to wikipedia. Why do we not say that it is 41.343 Peta-meters rather? (4.367 Ly = 41.343 Pm)
Why does Light-years or Parsecs seem to be the standard rather than SI?
|
Why does Light-years or Parsecs seem to be the standard rather than SI?
In the solar system astronomy, the astronomical unit is much more widely used rather than meters for distance, days (86400 seconds), Julian years (365.25 days), or Julian centuries (36525 days) are used rather than seconds for time, and the solar mass is used rather than kilograms for mass. The latter is particularly problematic because of the uncertainty in the Newtonian gravitational constant G. Using the solar mass as the standard for mass rather than the kilogram eliminates this problem. One solar mass is the standard for mass throughout astronomy and cosmology.
The astronomical unit is only useful for solar system objects. The parsec (not the light year) is the scientific standard for nearby stars and galaxies. The reason the parsec is the scientific standard is because parallax is directly observable for nearby stars. A series of standard candles enable the use of parsecs as the unit of distance for stars and galaxies that are somewhat more remote. The distance in parsecs can easily be translated to light years for lay readers who may not have an intuitive grasp of what a parsec is. Expressing distance in some ridiculously large number of meters doesn't make sense, period.
For extremely remote objects, the concept of distance doesn't quite make sense. Redshift z is what is used in the scientific literature for extremely remote objects. This can be translated into the amount of time it took the light from those extremely remote objects to reach us, but there are certain cosmological assumptions that need to be made to perform that translation. The constants used in that translation are not perfectly known; they change as scientists improve their understanding of the universe. On the other hand, the observed redshift value is what it is.
Multiplying that time needed for light emitted by a remote object to reach us by the speed of light yields a distance value, but that value is highly suspect.
| {
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What is the difference between a physical constant, a scalar, an invariant, and a conserved quantity? I don't really know how to properly articulate this question. This question popped into my mind when pondering why the fact that a physical constant like the speed of light doesn't have an associated symmetry even though it's "conserved" in every frame of reference.
| *
*A conserved quantity is a quantity whose value remains the same over time.
*An invariant, or scalar quantity is a quantity whose value is the same in all reference frames.
These two properties are completely independent. Energy is conserved but not invariant. Mass (i.e. $E^2 - c^2 \mathbf{p}^2$) is invariant but not conserved. Charge is both, and lots of things are neither. All physical constants are invariant, otherwise they wouldn't be worth calling constants.
You are wondering if the invariance of the speed of light leads to a conserved quantity. The answer is no, because Noether's theorem deals with symmetries, not invariant quantities.
However, there is a deeper sense in which you are right. The invariance of the speed of light is one feature of Lorentz symmetry, and this symmetry does produce conserved quantities! Since Lorentz transformations include rotations, three of them are just angular momentum. The three boosts produce conserved quantities that are, heuristically, the "speed of the center of mass". A more detailed derivation can be found here.
| {
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Scattering from a step potential barrier Suppose a potential barrier of the form
$$
V(x) = \begin{cases} V_0 & x>0 \\ 0 & x<0 \end{cases}
$$
Then, for energy $E$ such that $E < V_0$, we have that the transmission and reflection coefficients for the probabilities are $R = 1, T = 0$. In case where $V_0$ is not enormously large with respect to $E$, wave function will decay in $x>0$, but in a somewhat not rapid fashion. This means there is some reasonable not negligible probability for the particle to tunnel through. How does this agree with the fact that $R=1$, which means all of the probability has been reflected?
| $T$ and $R$ are transmission and reflection coefficients for waves. They refer to the probability that an incident wave will penetrate the barrier and continue propagating infinitely far. Physically, you should think of it as sending a constant sine wave in from the far left and looking to see what amplitude of constant sine wave you get at the far right.
In this case, since $E < V_0$ no matter how far to the right you go, the amplitude keeps decaying exponentially for all positive $x$. It never recovers its sinusoidal form, as it would with a rectangular barrier of finite extent, so there is no transmission.
Now, if you think of this waves as probability waves, then yes, it seems like there is a chance that the particle will materialize inside the barrier. And there is. That doesn't count as transmission, though. The process of a wave "turning into a particle" (i.e. wavefunction collapse) is not taken into account by the calculation of the transmission and reflection coefficients.
| {
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Why Do Glueballs Have Mass, When Individual Gluons Are Massless? From Wikipedia Glueballs
Glueballs are predicted by quantum chromodynamics to be massive, notwithstanding the fact that gluons themselves have zero rest mass in the Standard Model. Glueballs with all four possible combinations of quantum numbers P (parity) and C (c-parity) for every possible total angular momentum have been considered, producing at least fifteen possible glueball states including excited glueball states that share the same quantum numbers but have differing masses with the lightest states having masses as low as 1.4 GeV/c2 (for a glueball with quantum numbers J=0, P=+, C=+), and the heaviest states having masses as great as almost 5 GeV/c2 (for a glueball with quantum numbers J=0, P=+, C=-).
Rather than going through a list of possible mechanisms that unfortunately I know next to nothing about, such as can the mass be attributed to virtual quarks, or binding energy between the gluons, I would rather leave the question as in the title to find out as much as I can.
Also, although the SM is firmly established, would the discovery of Glueballs buttress it further?
My apologies for not knowing more about the interior of hadron like particles or if the answer is readily available (or worse, blindingly obvious).
| Because glueballs have energy, and $E = m c^2$ says that energy is equivalent to mass. (Or another way to say it is that if you "zoom out" far enough that you can't see the constituent gluons that form the glueball, than you just lump all their energy into an effective glueball mass.) The energy can be thought of as just being the kinetic energy of the individual gluons, which are zooming around each other at highly relativistic speeds. (Strictly speaking, it's actually less than the sum of the individual gluon kinetic energies, because you have to subtract off the strong-force binding energy that holds the glueball together.)
| {
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Principle of Maximum Work for Different Paths The principle of maximum work states that for any process between two states, the work done by the system is maximised for a reversible process (and heat transfer is minimised), and that the work done by any reversible process between these two states is equal.
I don't see how that can be compatible with going between 2 states with equal entropy by different paths. For example:
-An adiabatic expansion between 2 states with the same entropy.
-An isovolumetric pressurisation followed by an isobaric expansion with the same initial and final states as the adiabatic expansion.
The area below of the second path in a P-V diagram is clearly higher, so we have a reversible process between two states doing less work than another process between those states. How can this apparent contradiction be resolved?
| It doesn't seem possible for the original premise to be correct. If you start at a certain state, and carry out a reversible Carnot cycle ending up at the same original state, you can design a great big Carnot cycle and you can design a tiny little Carnot cycle. Certainly the reversible work for these two cycles will not be the same.
| {
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Sliding sphere wear shape Please refer to the figure attached. Consider a normal force is acting on the top of sphere. A constant coefficient of friction causes frictional force throughout the sliding. I want to know after this sphere slides (pure sliding no rolling) for sometime and assuming that it wears as it slides, what should be the shape of sphere after sliding? The one shown in (a) or (b)? In short, I want to know whether the worn side of sphere will be a straight line or a curved one? What will happen if a sphere slides against a sphere? Also, I will highly appreciate if someone can refer to some good papers / books about this.
Edit: Additional assumptions
Lets assume that hardness of both surfaces is the same. Also assume that material is removed but is not attached to any of the surfaces (no adhesive wear). The phenomenon under consideration is abrasive wear (but no accumulation of wear debris). If a lubricant is in circulation, it is easy to realize this kind of wear
| It depends. Assuming the lower surface is much harder than the sphere then :
If the sphere does not rotate at all as it slides then the answer is (a).
If the sphere rotates, ie rocks to and fro, as well as sliding, then it is (b).
On the other hand, if the sphere is harder than the lower surface, or the two surfaces have comparable hardness, then I think it will be something in between, even if the sphere does not rotate.
As tfb says, the real situation is complicated. For example, abraded material will accumulate at the edges of the depression hollowed out of the lower surface. What effect will that have? Will it provide lubrication? Or enhance the amount of abrasion at the edges? Also, if abraded material from the harder surface gets embedded in the softer surface, this suggests to me that wear will gradually tend towards being equal on each surface.
This is a situation in which doing an experiment is essential if the outcome is crucial - by which I mean, if you will be making a decision which it could be costly to get wrong, either in time delay or money or even reputation. Don't rely on theory or speculation. Test it.
| {
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Does the speed of light in vacuum define the universal speed limit?
*
*Is light the thing causing the universal speed limit to be $299\,792\,458\,\mathrm{m/s}$? So the universal speed limit would be different if light travelled faster or slower?
*Or, is $299\,792\,458\,\mathrm{m/s}$ the universal speed limit anyway and light just goes that fast? Light is just something we commonly associate with it because it goes super fast.
| There is quite a bit of ambiguity in the question(s), so let me start by substituting electro-magnetic (EM) wave for "light." Then, the "universal speed limit," is the speed at which EM waves propagate in "space." The reason I use space (not vacuum), is because it is the characteristics of space ($u_o, \epsilon_o$) that determine the speed of propagation of the EM waves. If these characteristics were different, the value of EM wave propagation would be different (larger, smaller) but it would still be the universal speed limit.
As you can see, the correct option is #2, and since light happens to be an EM wave, it propagates at the universal speed limit.
| {
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Conceptual doubt in Tension force I recently studied that Tension in a string is a kind of force originated from electrostatic attraction between the atoms of the string in which the force is originating. My doubt was that:
Assume that I am pulling a rope with a force $F$, and the rope will develop a tension $T$ in itself and will pull me with $T$ but what about the force $F$ with which I started pulling the rope? Where would the reaction pair of this force would be felt? I know that this force is not included in the free body diagram of me (the one who is pulling the rope) as FBD only incorporates the forces acting on a body and not the ones exerted by it and hence I am kind of confused where I can find the reaction pair of $F$ being felt.
| If the rope is in static equilibrium then $T=F$. If $T\neq F$ then that section of the rope (where tension is $T$) must be accelerating, which may happen if the rope is slack or if it is extensible.
| {
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Wouldn't a photon disappear because of length contraction? I was experimenting with the formula for length contraction, when I realized that anything traveling at the speed of light shrinks out of existence. This is the formula for length contraction:
$$T=T'\sqrt{1-\frac{v^2}{c^2}}$$
Where $T$ is the observed length and $T'$ is the proper length. When the speed of light, $c$, is plugged in for $v$, then the formula simplifies to this:
$$T=T'\sqrt{1-1}$$
Which further simplifies to this:
$$T=T'(0)$$
Therefore, anything traveling at the speed of light will not be seen, regardless of actual length. So, wouldn't we not be able to see any particle traveling at the speed of light?
| I am only a layman, so don't take this answer seriously.
This length contraction formula, and the whole Special Relativity in its original form, is for macro-sized, non-quantummechanical objects. Thus, the formulas work if you want to calculate the size of a spaceship nearing the speed of the light. And not if you want to calculate the size of photons with exactly the speed of the light.
Photons are elementary particles. Or waves. Or both of them. They are quantummechanical objects, described by QED (quantumelectrodynamics), which is a field theory.
So, the theories seem so:
1. Classical (newtonian, non-QM) mechanics
/ \
2. Quantummechanics 3. Special relativity
\ /
4. Relativistic Quantummechanics (QED, QFT)
You are now thinking in (3), but your problem is answered by (4).
Essentially, photons are the waves of the electromagnetic field, and these waves propagate with c.
Unfortunately, QFT hasn't so simple and beautiful formulas as the SR, but it is tremendous fun even trying to understand them.
Extension: There is also a possibility, that you are calculating only with classical electrodynamics, without any QM. But there are no photons in classical electrodynamics, there is only the electromagnetic field described by the Maxwell Equations. Here it is possible to calculate light as a wave packet. In this case, you don't have any particles, there is only a propagating wave of the EM field.
| {
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Can we know the tension of the string that is connected to a pully?
If the pulley have mass, I learned that T1 and T2 isn't same. then what about the circled part of the string? can we calculate the tension with given T1 and T2?
| Somewhere in between $T_1$ and $T_2$ -- but exactly what depends on details about how the friction between the string and the pulley varies, how the string stretches under tension and the pulley deforms while being accelerated by the string ... All of these are things that cannot be deduced from an idealized picture such as this.
In one extreme, if the pulley is a gear and the string is a chain whose links are naturally slightly longer than the tooth spacing of the gear, the string tension at your yellow circle may approach zero, even if both $T_1$ and $T_2$ are higher. You probably won't call that a "string", but it's not easy to define a crisp distinction between the two cases.
| {
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