Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
Fuel and heated iron? If I have a pile full of fuel (gasoline) - and I also have heated iron ( very very hot - color orange) and I put the Iron in the pile ,
Would the gasoline will burn ?
I mean - does gasoline need heat or actual fire in order to burn ?
nb - I can't test it since I don't have the means to do it.
| I don't know what a "pile" of fuel is. I assume you mean a container full of it.
Gasoline needs oxygen to burn, and it needs the correct mixture. Too little oxygen and burning is impossible. Too much oxygen causes the same problem. To achieve ignition with Gasoline, you need between 1.4 and 7.6% petrol vapour (by volume) in the air. Ouside this range burning will not start.
If these conditions are met, as the flash point of gasoline (the minimum temperature at which it will burn) is $-40C$, your red-hot poker will do the job. In fact, gasoline and air will self-ignite at around $250C$. For more details, start with the Wikipedia article on the Flammability limit.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/105741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
astronomy & light pollution If you were staring to the sky in a big city and electricity is turned off in a big area around you, would you be seeing the sky with little light pollution instantly?
| If you include car lights and similar in what is turned off, then in a big city such as London, which has a radius of about 10 miles, there would be very little light pollution, instantaneously. However, there would still be a lot of atmospheric pollution, and so it wouldn't be possible to see stars as faint as can be seen from, say, the top of a mountain.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/105794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Differential equation for evolution of probability density in Quantum Mechanics? I have come up with this differential equation for the evolution of $\vert \Psi \vert^2$, the probability density in quantum mechanics.
Is there a name for this equation? Is the logic sound?
So I start from the conservation of the total probability:
$$ \frac{d}{dt} \int _{all space} \vert \Psi \vert^2 d^3\textbf{r} = 0$$
so that $$\int _{all space} \vert \Psi \vert^2 d^3\textbf{r} = {\rm const}$$
This means that
$$\int _{all space} \vert \Psi (\textbf{r},t) \vert^2 d^3\textbf{r} = \int _{all space} \vert \Psi (\textbf{r}+d\textbf{r},t+dt) \vert^2 d^3\textbf{r}={\rm same\,const}$$
So since the integration is over the same volume I can equate the intergrands?
This gives us:
$$\vert \Psi (\textbf{r},t) \vert^2 = \vert \Psi (\textbf{r}+d\textbf{r},t+dt) \vert^2$$
or
$$\Phi (\textbf{r},t) = \Phi (\textbf{r}+d\textbf{r},t+dt) $$
if we call $\vert \Psi (\textbf{r},t) \vert^2 = \Phi (\textbf{r},t)$. A Taylor expansion to first order gives us:
$$ \Phi (\textbf{r}+d\textbf{r},t+dt) =\Phi (\textbf{r},t) + \nabla\Phi\cdot d\textbf{r}+\frac{\partial \Phi}{\partial t} dt$$
Plugging this into the previous equation gives us:
$$\boxed{ \nabla\vert \Psi \vert^2\cdot \frac{d\textbf{r}}{dt} = -\frac{\partial \vert \Psi \vert^2}{\partial t} }$$
What is $\frac{d\textbf{r}}{dt} $ in the context of quantum mechanics? Is this equation correct? What is the physical meaning of this equation?
| This equation is wrong.
As it has been pointed out in the comments, I can't equate the integrands of two integrals just because the integration limits are the same.
The equation which is of relevance in the context of probability density and quantum mechanics is perhaps the well known continuity equation,
$$ \frac{\partial \rho}{\partial t} + \nabla \cdot \textbf{J} = 0 $$, where $\rho = \vert \Psi \vert^2$ and $ \mathbf{J}=\frac{\hbar}{m}\text{Im}\left[\psi^*\nabla\psi\right]$ is the probability flux.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/105854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How can point-like particles in an ideal gas reach thermodynamical equilibrium? Having learned that the particles of an ideal gas must be point-like (for the gas to be ideal) I wonder how they can reach thermodynamical equilibrium (by "partially" exchanging momentum and energy). First the probability of two point-like particles to collide is literally zero, and second, they can only collide head-on which implies that they can only "swap" their momenta and energies.
How is this puzzle to be solved?
| Ideal gas does only mean that there are no forces between the particles. They do not have to be point-like. For example 2-atomic gasses could have 3 translatory and 2 rotational degrees of freedom in kinetic gas theory, while still no forces act between the molecules. So ideal but not point-like. For one-atomic gasses the atoms are often taken to be hard (frictionless) balls, having small diameters compared to their free path lengths in order to allow for collisions between the atoms. However for certain considerations (pressure on wall) it is sufficient to assume point-like particles, as it doesnt matter if they collide with each other or not due to consevation of momentum.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/105918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 2
} |
Is speed of light ( Sun-to-Earth ) related to the movement speed of solar system? The speed of light has been measured to be 299 792 458 m/s.
Now, the Solar System is traveling at an average speed of 828,000 km/h (230 000 m/s). Summing up the numbers we get close to 300 000 000 m/s
Does it mean, that the speed of light can be actually a relative number based on the movement of the source?
Does it mean that it IS actually 300 000 000 m/s but is calculated incorrectly due to sun moving round the galaxy?
| No. Speed of light in Vacuum isn't dependent on Solar System's motion. It's a constant. It'd be same even if the motion wasn't there.
Due to your question type problems, we've even calibrated our scales to create Relativistic Physics.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/105974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
State of constant motion Why does an object remains in its state of constant motion if there are no forces acting on that object?
My understanding is that all the energy of the motion will be kept inside and a change in the speed needs a change in the amount of the energy stored by that object. But how is that energy stored in a form of speed?
Does anything changes inside the object's structure if it is moving?
| What is force? It is the transfer of momentum with respect to time. Momentum is the product of mass of a body, and it relative velocity with respect to something. If we imagine, for once, our world made up of tiny uniform particles, then we can take the mass of one particle as one unit mass. This simplifies the situation, as the momentum, 'mass x velocity' becomes, velocity x 1 = velocity. So, when two such particles collide, they exert force on each other.
Or, literally, the faster particle loses a bit of its velocity and the slower particle gains the same with respect to time.
You are asking- " why does an object remains in a state of constant motion if no force acts on it".
This question is same as asking " Why does a unit mass particle has a constant velocity if it is not gaining some velocity."
For some reason, relative motion of a particle in space is a conserved constant. A particle does not gains velocity until some other particle loses the same amount of it. This is the same as the fact that a shopkeeper gains money if some customer loses the same amount of money in his shop.
(Yes, of course money gets lost sometime, or damaged, with a net lose to humanity, but in the universe at least, it is not easy to lose your momentum with nobody gaining the same amount, instantly!)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/106158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Does mass affect velocity when travelling through frictionless medium? I found the following question on an standardized test, and was debating with some friends what the answer would be:
A car of mass M is travelling with a constant velocity through a plane in which friction is non-existent. An object of mass m (m = M/3) that is falling perpendicularly to the car lands inside of it. How will the velocity of the car be affected?
This illustration can help explain the problem.
My initial thought was that the velocity would be the same, given that friction is non-existent and that the momentum of the falling object is perpendicular to that of the car. However, some friends suggested that, since the mass of the car increases, the velocity should decrease.
| Let us first simplify a bit, and get rid of the vertical movement of $m$. It is much simpler to assume that $m$ is standing with zero velocity in the path of $M$, which will collide it at $t=0$. This will be the same problem along the horizontal axis, and avoids to superimpose it with another problem in the vertical direction.
The conserved quantities are both the kinetic energy and the momentum. If we call $v$ and $V$ the velocities of $m$ and $M$ respectively, and $V_0$ the initial velocity of $M$, we have that
$$
mv + MV = MV_0
$$
$$
mv^2 + MV^2 = MV_0^2
$$
I'll leave it for you to solve yourself, but you'll find that the velocities $V$ and $v$ are different: the masses do not travel together.
If you want to model a case where they'll travel together, you'll need to provide with a mechanism dissipating energy in the collision. In that case, the kinetic energy will not be conserved, and in the absence of external forces, you'll have $V=v=\frac{M}{M+m} V_0$. The system will have a lower kinetic energy than previously.
Thanks to @NeuroFuzzy for pointing my previous wrong answer (deleted).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/106236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Is an "infinitely sharp blade" possible? A staple of science fiction and fantasy is a blade (knife, sword, ...) that cuts through literally any solid object (wood, steel, concrete, skulls, ...) without effort, often even without the need to apply any pressure.
What (in theory) would be required to construct such a blade? Would, say, a diamond wedge that is only a single atom thick on one side have that property? Or is it physically impossible to make such a blade? If yes, what is the limiting factor?
| Pressure is a measure of the force applied to a given area. Blades are sharp because they have a small cross sectional area, allowing you to create very high pressure whilst applying only a modest force. This force generates so much of shear stress on the object getting cut that it crushes through the molecular bonds in that object. Cutting through something can tear its molecules apart but it does nothing to the atoms of that object.
A blade which is a single atom thick would firstly not be sturdy. Even if it was, it would get dull in a matter of seconds and will also be highly unstable (if it's metal).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/106390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Can we think of gravity as space itself moving? So if you move through space with a constant acceleration you experience longer time dilation than when you're at rest, but you also experience the same time dilation when you're under the effect of gravity like on earth, so is it possible that by standing on earth, space itself is moving relative to you at a constant acceleration, which causes the same time dilation as when you move in space with same acceleration of earth's gravity $9.8\tfrac{m}{s^2}$?
Maybe that's why under free fall you don't sense the acceleration and feel weightless, because you are moving along with space itself that is attracted to earth, neither you are moving relative to space, nor space is moving relative to you. By moving I mean accelerating and - constant speed of course.
Sorry if this question is silly or something that is maybe already suggested before.. I'm not a specialist here.
| Yes, in fact that's quite a good way to think of black holes.
Imagine instead of space think about the surface of a lake with a drain hole in the middle. And think of the water flowing into the hole. The nearer to the hole the faster the water will flow.
Imagine a ship on the water which has a maximum speed of 5 knots. When it gets to a certain distance to the drain hole the water is flowing into the hole at more than 5 knots so the ship can't escape.
In terms of a black hole this is the "event horizon".
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/106447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
The other side of the lever If I have a lever, but I can see only up to the hinge and not the other half, can I know whether the other half is 1 m long with a weight of 3 kg on it, or 3 m long with a weight of 1 kg on it?
| Just for fun let me suggest another rather impractical way to tell the difference.
The diagram shows the far side of the lever. It has a length $L$ that you don't know and there is mass $m$ on the end that you don't know. The torque is equal to $Fd = FL\cos\theta$, and the force is the gravitational force $F = GMm/r^2$, where $M$ is the mass of the Earth and $r$ is the radius of the Earth. So:
$$ T = \frac{GM mL\cos\theta}{r^2} $$
Only this isn't quite right because the distance from the centre of the Earth isn't $r$, but rather $r + L\sin\theta$. So a more accurate calculation of the torque is:
$$ T' = \frac{GM mL\cos\theta}{(r + L\sin\theta)^2} $$
The equation for the torque ignoring the change of gravity, $T$, only contains the product $mL$, so you can't tell the difference between $m = 1, L = 3$ and $m = 3, L = 1$ as in both cases the product $mL = 3$. However the second equation that includes the change in gravity has a separate dependance on just $L$ in the denominator, so you can tell the difference. In fact if you graph the difference between the equations, $T - T'$, you get:
The difference is very small, only around $10^{-5}$ Nm, but it is there. So by very precisely measuring the torque as a function of angle you can tell the difference between the two cases.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/106566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Does the Sun produce audible sound? Theoretically if I were able to build some sort of device that let me sit 1 foot away from the surface of the Sun (or any star for that matter) without being vaporized, would a star produce any sort of audible sound at that distance? From my understanding, sound waves need a medium to propagate through (air, water, etc). Does this mean that all stars are completely silent (in the audible spectrum of frequencies at least) as they fuse hydrogen into helium? I know stars emit radio, xray, infrared, ultraviolet electromagnetic waves, but how about audible (20 Hz to 20Khz) frequencies? For some reason it seems like this process would produce some sort of sound.
| helioseismology is what you need to learn about. yes, there are sound waves in Sun
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/106652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Derivation of Lagrangian? I know that the Lagrangian $L$ is defined to be $T-V$, i.e. the difference between kinetic energy and potential energy. Also the Action $S$ is defined to be $\int Ldx$ and from this we can derive Newton's 2nd law of motion.
*
*If we get Newton's second law out, does it mean that the formulation is correct? Couldn't it be just a coincidence?
*Where do we derive these expressions for the Action and for the Lagrangian from?
| Newton's second law, $\mathbf{F}_{net}=\dot{\mathbf{p}}$, is the definition of force. Lagrangian and action are defined to be $ T-V $ and $\int L\: \mathrm {d} t$ (and not $\mathrm {d} x $) respectively. You don't derive anything from anything here (however we can talk about how $ T $ and $ V $ come about).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/106786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Normalization of Source Terms in Large-N Gauge Theory Typically when you do the counting for large N gauge theory, you rescale fields so that the Lagrangian takes the form
\begin{equation}
\mathcal{L}=N[-\frac{1}{2g^2}TrF^2+\bar{\psi}_i\gamma^\mu D_\mu \psi_i]
\end{equation}
where I have chosen the original coupling of the theory to be $\frac{g}{\sqrt{N}}$. From this it is easy to see which vacuum diagrams contribute in the Large-N limit.
However, when you go on to consider connected correlators, people always add a source term $N\sum J_iO^i $ to the Lagrangian. The factor of N out front then determines the N-dependence of the correlators
\begin{equation}
\langle O_1...O_r \rangle=\frac{1}{iN}\frac{\partial}{\partial J^1}...\frac{1}{iN}\frac{\partial}{\partial J^r}W[J]
\end{equation}
The N-counting would be different if my source terms were instead just $\sum J_iO^i $.
So my question is, why are we forced to include the factor of N in the source terms? Is it because the original action has been written in terms of rescaled fields and is also proportional to N? If I instead worked with the action in terms of un-rescaled fields, would I not include the factor of N in the source term? Thanks.
| The operator $O^i$ in the source term will in general also contain fields that are rescaled, and the scaling behaviour is supposed to match the rest of the Lagrangian.
If you did not have a factor of $N$ in the source term, you would not need to divide by $N$ when taking functional derivatives. What matters is the result: functional derivatives of the generating functional should produce correlation functions of the operators without any multiplications by $N$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/106867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Can magnetic flux be negative I am studying magnetic flux linkage in an ac generator and it appears to be that magnetic flux linkage is negative half the time, how can this be?? Also with lenz's law why is emf defined as negative when magnetic flux is increasing and how does this relate to the direction of the current?
| Yes, magnetic flux can be negative. It just depends on where the field is going. Say there is a sheet and magnetic field is going through it from front to the back, we can call the flux there as positive and negative when it's the other way round.
It is pretty clear from the statement of Lenz's Law why the emf defined is taken as negative: An induced electromotive force (emf) always gives rise to a current whose magnetic field opposes the original change in magnetic flux. (wikipedia)
Basically, the magnetic field produced due to the induced current opposes the magnetic flux producing the current itself.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/106949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Are there any QM effects where charged particles are not intimately involved? Are there any QM effects that have been/could be measured from interactions involving non-charged particles?
Elementary QM is all about the electron energy levels in the atom, photon - atom interactions, etc.
When one looks at the nucleus, its all about quark interactions - which are also charged particles.
I can think of some theoretical ones - like neutrinos orbiting a mass, but they would be hard to impossible to measure. Another possibility is the strong / weak nuclear force - but that always happens with particles that are also charged.
In the end we always need matter built instruments to see a result - that's fine. You could for instance observe photons coming from some distant (metres to Mpc) away region where some interaction occurred.
| Weak and Strong interactions do not involve charge. Strong interactions involve color charge, which is a different property than "normal" charge. The weak interaction mostly involves flavor change.
As to your non-charge dependent Quantum Mechanical effects, we have tunnelling. See for example this link.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/107031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Lepton number conservation in standard model
*
*Why is it said that in standard model lepton number is conserved?
*How do I know that Lepton number is an abelian charge?
*Why is this conservation not as sacred as electric charge conservation.
*How does one mathematically distinguish between lepton number and electric charge?
| Let me address your questions one by one.
Why is it said that lepton number is conserved in Standard Model (SM)? How do I know that lepton number is an Abelian charge?
The SM Lagrangian is invariant under the fermion transformations,
$$
\psi \to e^{iL\theta}\psi
$$
where $L$ is assigned such that $e^-$, $\mu^-$ and $\tau^-$ leptons and lepton-neutrinos have $L=1$, whilst their antiparticles have $L=-1$, and everything else has $L=0$. This global $U(1)$ symmetry corresponds to lepton number conservation - $L$ is what we call lepton number. Lepton number is by construction the Abelian charge corresponding to that $U(1)$ global symmetry.
Why is this conservation not as sacred as electric charge conservation?
The lepton $U(1)$ global symmetry is accidental. We simply cannot write a gauge invariant, renormalizable operator in our Lagrangian that breaks conservation of lepton number. There is no reason to expect that physics beyond the SM respects lepton number conservation.
The electric charge $U(1)_{em}$ local symmetry was a principle on which the SM was built. The SM would not be renormalizable if $U(1)_{em}$ was explicitly or anomalously broken. It would be catastrophic if $U(1)_{em}$ were broken.
How does one mathematically distinguish between lepton number and electric charge?
They are the conserved charges associated with different $U(1)$ symmetries. In general, when you have multiple $U(1)$ symmetries, charge assignment is somewhat arbitrary, since one can pick different linear combinations of the original $U(1)$ generators as the symmetries. In this case, however, the $U(1)_{em}$ is local, so there is no mixing with the global lepton number $U(1)$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/107128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
What happens with a tunneling particle when its momentum is imaginary in QM? In classical mechanics the motion of a particle is bounded if it is trapped in a potential well. In quantum mechanics this is no longer the case and there is a non zero probability of the particle to escape the potential through a process call quantum tunneling.
This seems extraordinary from the point of classical mechanics because it implies the particle must cross a zone where it has imaginary momentum. I understand that from the point of view of quantum mechanics there is a non zero probability for the particle to be in such zones.
What is it know about the behaviour of the particle in this zone?
Links to research experiments or papers would be appreciated.
| There is a detailed description of the behaviour of quantum systems during tunnelling. They do not involve the particle travelling faster than light. The "group velocity" does not correspond to the velocity of anything in that situation. What happens instead is that the wave function undergoes interference inside the barrier in such a way that it decays exponentially and it does this slower than light. Physicists who have claimed otherwise have misinterpreted experimental results, see Herbert Winful's papers on this topic
http://sitemaker.umich.edu/herbert.winful/modest_contributions
and in particular
http://sitemaker.umich.edu/herbert.winful/files/physics_reports_review_article__2006_.pdf.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/107261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Does electric field cause changing magnetic field also? According to Faraday's law, changing magnetic field causes eletric field (imprecise wording, but generally accurate). My question is, can Faraday's law be interpreted in opposite way - that is, electric field causing changing magnetic field? Equation itself doesn't speak for the direction, so I ask.
| Yes. You can see this more clearly when working with the differential form of Maxwell's equations. You can consider the simple case of a wire of some radius with current flowing through it. If there's a small gap in the wire the current will accumulate some surface charge density at the ends of the wire at the gap. This creates a time varying electric field, similar to a parallel plate capacitor within the gap and it turns out this will create a magnetic field circulating around the wire axis, even at the gap, as if the wire was gapless with some current flowing through it everywhere.
So time varying electric fields act like currents and can create magnetic fields, i.e. act as sources for magnetic fields just as current densities do. This particular type of current is called a displacement current and was discovered by Maxwell to be a necessary modification to Ampere's law.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/107314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
How can I see this equation describes advection? My notes say that this is the equation for the "advection of a quantity A at speed v":
$$ \frac{\partial \mathbf{A}}{\partial t} = \nabla\times(\mathbf{v}\times\mathbf{A}) .$$
Is this true? How can I see (mathematically) that this equation corresponds to advection?
| This term looks like Faraday's law that is used in Ideal magnetohydrodynamics (MHD). So yes, it is true.
In order to see the mathematical advection, you'll need to apply some vector calculus:
$$
\nabla\times\mathbf a\times\mathbf b = \mathbf a\left(\nabla\cdot\mathbf b\right) - \mathbf b\left(\nabla\cdot\mathbf a\right) + \left(\mathbf b\cdot\nabla\right)\mathbf a - \left(\mathbf a\cdot\nabla\right)\mathbf b
$$
Then grouping the $\mathbf a$ and $\mathbf b$ terms,
$$
\nabla\times\mathbf a\times\mathbf b = \left(\nabla\cdot\mathbf b+ \mathbf b\cdot\nabla\right)\mathbf a - \left(\nabla\cdot\mathbf a+\mathbf a\cdot\nabla\right)\mathbf b
$$
which then reduces to give the advection equation in the form
$$
\frac{\partial\mathbf A}{\partial t}=\nabla\times\mathbf v\times\mathbf A=\nabla\cdot\left(\mathbf A\mathbf v^T - \mathbf v\mathbf A^T\right)\tag{1}
$$
This now gives us a time-derivative of vector $\mathbf A$ equal to the gradient of $\mathbf A$ times the vector velocity, $\mathbf v$, which is pretty much what your advection equation is supposed to be.
It's probably a little bit more easily seen in component form:
$$
\frac{\partial}{\partial t}\left(\begin{array}{c}A_x \\ A_y \\ A_z\end{array}\right) = \left(\begin{array}{c}\frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z}\end{array}\right)\left(\begin{array}{ccc}A_xv_x -v_xA_x& A_xv_y-v_xA_y & A_xv_z-v_xA_z \\ A_yv_x-v_yA_x & A_yv_y-v_yA_y & A_yv_z-v_yA_z \\ A_zv_x-v_zA_x & A_zv_y-v_zA_y & A_zv_z-v_zA_z\end{array}\right)
$$
which reduces to
$$
\frac{\partial}{\partial t}\left(\begin{array}{c}A_x \\ A_y \\ A_z\end{array}\right) = \left(\begin{array}{c}\frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z}\end{array}\right)\left(\begin{array}{ccc}0 & A_xv_y-v_xA_y & A_xv_z-v_xA_z \\ A_yv_x-v_yA_x & 0 & A_yv_z-v_yA_z \\ A_zv_x-v_zA_x & A_zv_y-v_zA_y & 0\end{array}\right)\tag{2}
$$
And, very neatly, you have yourself some divergence free terms in the zero's along the diagonal:
$$
\frac{\partial A_x}{\partial t}=\frac{\partial}{\partial x}0+\frac{\partial}{\partial y}\left(A_xv_y-A_yv_x\right)+\frac{\partial}{\partial z}\left(A_xv_z-v_xA_z\right)
$$
Now with regards to the actual question, how can we show that this is an advection equation. Advection simply is the transport of some material along a fluid flow. Usually this means that
$$D_t\psi=\frac{\partial}{\partial t} \psi+\mathbf u\cdot\nabla\psi=0,$$
where $\mathbf u$ is the velocity of the fluid flow. In the particular case that $\nabla\cdot\mathbf u=0$, then we can use
$$\frac{\partial}{\partial t}\psi+\nabla\cdot(\mathbf u\psi)=0\tag{3}$$
which is the more-familiar continuity equation.
But that is for scalar advection. In the case considered here, we have a vector to advect. This is easily done by replacing the scalar $\psi$ with the vector $\mathbf a$ in Equation (3)
$$\frac{\partial}{\partial t}\mathbf a+\nabla\cdot(\mathbf u\mathbf a)=0$$
But we have this tricky $\mathbf{ua}$ term here. It can't be the dot product because divergence of a scalar is meaningless, so it must mean some other geometric product. In this case it makes a dyad:
$$
\mathbf{ua}=\left(\begin{array}{ccc}u_xa_x & u_xa_y & u_xa_z \\ u_ya_x & u_ya_y & u_ya_z \\ u_za_x & u_za_y & u_za_z\end{array}\right)
$$
Taking the derivative of this matrix (a poor way to think of a tensor) gives you a vector. Thus, we should write this as
$$\frac{\partial}{\partial t}\mathbf a+\nabla\cdot(\mathbf u\mathbf a^T)=0$$
to specify that we mean a tensor product there.
This now gives the form of Equation (1) that I claimed was an advection equation. The extra minus term in Equations (1) & (2) comes from the fact that your equation from Faraday's law contains a vector product of two terms which leads to two tensor products, rather than just one.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/107579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
A zero gravitational potential and non zero gravitational field Give an example of a situation in which there is a non-zero gravitational field and a zero gravitational potential at the same point?
$$dV=-\vec E\cdot d\vec r.$$
The above equation implies that such a situation is possible.
| The gravitational field $\mathbf g$ equals, by definition, negative of the gradient of a correspondonding potential $\Phi$;
\begin{align}
\mathbf g = -\nabla\Phi.
\end{align}
Therefore, it suffices to produce a gravitational potential $\Phi$ whose value is zero at a point but whose gradient is non-zero at that point. This is straightforward to do.
Let a vectors $\mathbf x_0$ and $\mathbf c\neq \mathbf 0$ be given, and define
\begin{align}
\Phi(\mathbf x) = \mathbf c\cdot (\mathbf x - \mathbf x_0)
\end{align}
Then,
\begin{align}
\Phi(\mathbf x_0) = 0, \qquad \nabla\Phi(\mathbf x_0) = \mathbf c\neq 0
\end{align}
and this is a valid gravitational field for vanishing charge density since it satisfies Laplace's equation $\nabla^2\Phi = 0$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/107691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Lagrangian for relativistic massless point particle For relativistic massive particle, the action is
$$\begin{align}S ~=~& -m_0 \int ds \cr
~=~& -m_0 \int d\lambda ~\sqrt{ g_{\mu\nu} \dot{x}^{\mu}\dot{x}^{\nu}} \cr
~=~& \int d\lambda \ L,\end{align}$$ where $ds$ is the proper time of the particle; $\lambda$ is the parameter of the trajectory; and we used Minkowski signature $(+,-,-,-)$. So what is the action for a massless particle?
| It is conceptually possible to have a massless charged particle, although there are none that we know of. It is not true that the Lorentz force has to equal mass times acceleration. The momentum of a massless particle is a quantity indepenedent of its speed as all massless particles travel at the speed of light. The momentum $p$ is instead equal to $E/c$, the energy divided by the speed of light.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/107921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 1
} |
If the MH370 black box did sink to 15000 ft, how long would it have taken? I have been following MH370 on the news just as everyone and latest reports seem to indicate that the black-box could be found. A recent info-graphic http://t.co/lyBBE9C2hF shows the insurmountable depth of the oceans and how the black-box could have sunk 15,000 ft! I wonder how long it would have taken for it to sunk to the bottom of the sea-bed? What is the equation of motion for a sinking object at sea, ignoring under-water currents?
| You say, ignore currents and I assume, other extraneous factors.
If that is the case, then, considering a hydrostatic balance in the water column where z is the vertical coordinate. Then, the motion of a water parcel with density, $\rho$, displaced upwards by a distance, $\Delta z$, in a fluid with a reference density, $\rho_0$, is governed by
$\rho_0 \frac{d^2\Delta z}{dt^2} = g \frac{d \bar{\rho}}{dt^2} \Delta z$
I am confident you will be able to deduce the rest.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/107997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Why is the divergence of a magnetic field equal to zero? We know due to Maxwell's equations that:
$$\vec{\nabla} \cdot \vec{B}=0$$
But if we get far from the magnetic field, shouldn't it be weaker?
Shouldn't the divergence of the field be positive?
If we define the vector field as a function of distance, then if the distance increases then the magnitude of the vector applied to a distant point of the "source" should be weaker.
Is my reasoning correct?
| Your intuition about the meaning of the divergence operator is wrong.
In physics it's easiest to think intuitively about divergence by using the divergence theorem which states
$$\int_V dV \ \nabla \cdot \mathbf{B} = \int_{\partial V} \mathbf{B} \cdot d\mathbf{S}$$
where $\partial V$ is the surface area surrounding the volume $V$. The magnetic field has zero divergence, which means that
$$\int_{\partial V} \mathbf{B} \cdot d\mathbf{S}= 0$$
We can interpret this by saying there's no net flow of magnetic field across any closed surface. This makes sense because magnetic field lines always come in complete loops, rather than starting or ending at a point.
Put another way, the divergence-free condition is just saying that we don't have magnetic monopoles in Maxwell electromagnetism.
Let me know if you need any more help!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/108224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
Loop-the-loop question
Problem: The marble rolls down the track and around a loop-the-loop of radius R. The marble has mass $m$ and radius $r$. What minimum height $h$ must the track have for the marble to make it around the loop-the-loop without falling off?
Express your answer in terms of the variables $R$ and $r$.
I found this solution to be very reasonable:
$$mg = m a_c = m \frac{V^2}R $$
which leads to
$$V = \sqrt{g R} $$
The energy at the top of the loop KE = Delta PE
$$\frac12 m V^2 + m g (2R) = m g h \\
\frac12 (g R) + g (2R) = g h \\
\left(\frac12+2\right) R = h $$
so $h = 5/2 R $
However the correct answer is actually $\frac52(R-r)$, I think it's because the radius of the loop is measured from the center of the ball rolling on it, so the they subtracted $R$ from $r$, but how would you derive that? I tried the same steps just using $R-r$ instead of $R$ but I got a different answer.
| Your expression for the velocity looks right; but we have to get a few other things taken care of.
First - the center of the marble doesn't move from 0 to 2R, it moves from r to 2R-r - so the potential energy due to this is smaller than mg(2R) which is what you had in your expression.
On the other hand, you need to take account of the energy of the sphere rolling (which is stated explicitly). The moment of inertia for a solid sphere (the usual case for a "marble") is
$$I=\frac25 m r^2 \\$$
This leads to rolling energy
$$E=\frac12I\omega^2=\frac12\frac25 m r^2 (\frac v r)^2 = \frac15mv^2\\$$
Thus your energy equation has to be corrected to
$$\frac12 m v^2 + m g (2R-2r) + \frac15mv^2 = m g h \\$$
Also - note that the marble is moving in a path with radius R-r not radius R; you need to take that into account when you compute the limiting velocity ("fast enough to stick to the track") - you have to put R-r where you have R in your velocity equation:
$$\frac{mv^2}{R-r}=mg\\
v=\sqrt{g(R-r)}$$
Combining these:
$$\frac{7}{10}mg(R-r) +2mg(R-r)=mgh\\
h = \frac{27}{10}(R-r)$$
Note that it is sometimes said that "you can ignore the rotational energy of the marble if it is very small", but that is emphatically not true - the rotational energy for a solid sphere is always 2/5 of the (linear) kinetic energy, regardless of the size of the marble. It can therefore only be ignored for the case of a (frictionless) sliding object.
Finally - since no physics problem is complete without a diagram:
This shows clearly where the R-r term is coming from.
Thus if you ignore the rolling, the diagram explains the "correct" answer (which in a way was your question). If you include rolling, then you need to modify the solution as shown.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/108296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Battery Capacity vs. Battery draw Pardon my ignorance but I just can seem to find an Answer to my question and I hope that you can help me.
What is the equation for calculating how long a Lithium Battery can supply a device?
For instance a Lithium battery has a capacity of 5600mAh, and an Output of 5V 2A. The drawing device requires 350mA and 5Vdc.
Is it as simple as dividing 350 into 5600 which would give me 16 hours of charge with optimal conditions? (This seems too easy.) or am I missing something?
| Yes , that's easy because mAh is presented for a specific voltage that battery can output. Because the battery must output at a specified voltage. Your 5600mAh is for 5 volts. If it's not 5 volts. You must recalculate it. But the voltage shouldn't be change if the battery is not adjustable.
Hope you understand
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/108363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is an intuitive explanation using forces for the equatorial bulge? The earth is not a sphere, because it bulges at the equator.
I tried fiddling with centripetal force equations and gravity, but I couldn't derive why this bulge occurs.
Is there
(a) a mathematical explanation using forces (not energies) and
(b) a simple intuitive explanation to explain to others why the bulge occurs?
| If you have ever seen a pizza being made by hand, you will know that when the baker throws the disk of dough in the air, he makes it spin. As he does so, the pizza "disk" gets bigger because the dough on the outside experiences a larger centrifugal force (in the rotating frame of reference of the pizza. Don't start on "there is no such thing", you asked for an intuitive answer).
Now think of the earth as that pizza. The bits of the earth near the equator (a larger distance from the axis of rotation) feel a greater force, and are therefore trying to move outward. The force of gravity tries to pull it back. The balance is a slightly distorted sphere: the "bulge".
Simple enough, I hope.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/108640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
In what sense do Goldstone bosons live in the coset? Goldstone's theorem says that if a group, $G$, is broken into its subgroup, $H$, then massless particles will appear. The number of massless particles are given by the dimension of the coset, $G/H$. It is then often said that the Goldstone boson's live in the coset. In what sense is this statement true? The Lagrangian is not invariant under transformations of the coset so what does this "living" explicitly mean?
To be explicit we can consider the linear sigma model:
\begin{equation}
{\cal L} = \frac{1}{2} \partial _\mu \phi ^i \partial^\mu \phi ^i - \frac{m ^2 }{2} \phi ^i \phi ^i - \frac{ \lambda }{ 4} ( \phi ^i \phi ^i ) ^2
\end{equation}
We define,
\begin{align}
& \phi _i \equiv \pi _i \quad \forall i \neq N\\
& \phi _N \equiv \sigma
\end{align}
and give $\sigma$ a VEV.
The spontaneously broken Lagrangian is,
\begin{equation}
{\cal L} = \frac{1}{2} \partial _\mu \pi _i \partial ^\mu \pi _i + \frac{1}{2} ( \partial _\mu \sigma ) ^2 - \frac{1}{2} ( 2 \mu ^2 ) \sigma ^2 - \lambda v \sigma ^3 - \frac{ \lambda }{ 4} \sigma ^4 - \frac{ \lambda }{ 2} \pi _i \pi _i \sigma ^2 - \lambda v \pi _i \pi _i \sigma - \frac{ \lambda }{ 4} ( \pi _i \pi _i ) ^2
\end{equation}
The Goldstone bosons, $\pi_i$, exibit a $O(N-1)$ symmetry, but this is not the coset group symmetry. So where in the Lagrangian do we see this symmetry?
| It is in fact a very simple matter if you use a different parametrization of the fields. Since we care about the Goldstone bosons only, just send $\lambda\rightarrow \infty$ so that the Higgslike state decouples. Moving to the following parametrization
$$
\phi_i(x)=U(x)\langle \phi_i\rangle \,,\qquad U(x)=e^{i \hat{T}^a \pi^a(x)}\,,\qquad \langle\phi_i\rangle=\left(0,0,0\ldots,v\right)^T
$$
(where $\hat{T}^a$ are the broken generators) you immediately see that there is a gauge redundancy in the definitions of the pion fields $\pi^a(x)$ since we are allowed rotate them with an $x-$dependent transformation $h(x)$ of the unbroken group $H$, namely
$$
\phi_i(x)=U(x)\langle \phi_i\rangle=U(x)h(x)\langle \phi_i\rangle\,.
$$
In other words, the pion field is defined only up to this equivalence $U(x)\sim U(x)h(x)$, which is the statement that they live on the coset space $G/H$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/108722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 2,
"answer_id": 1
} |
Does it make sense to open one window all the way when the other window is much smaller? I can't wrap my head around this idea because I don't know much about air flows.
Say we have this imaginary apartment with two windows, one of which is two times smaller than the other:
Will the first one let more air through than the second? Or is it the same because of the smallest window?
How does this work?
| If you are familiar with electric circuits in a quite loose analogy you can look at your windows as impedances for the air current. Since current is turbulent (take a look at the Reynolds number if you do not know it) the air does not directly go all the way down through all the windows, but each of them will create some impedance to the flow. The bigger the window the less the impedance so yes, fully open also the second one! You will also get more diffusion which is not included in the previous discussion.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/108802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Finding the vacuum which breaks a symmetry I will start with an example. Consider a symmetry breaking pattern like $SU(4)\rightarrow Sp(4)$. We know that in $SU(4)$ there is the Standard Model (SM) symmetry $SU(2)_L\times U(1)_Y$ but depending on which vacuum we use to break this symmetry, in a case you can totally break the SM symmetry, with the vacuum :
$$\Sigma_1 = \begin{pmatrix}
0& I_2 \\
-I_2 & 0
\end{pmatrix}$$
and in another case, these symmetry is preserved, with the vacuum
$$\Sigma_2 = \begin{pmatrix}
i\sigma_2& 0 \\
0 & i\sigma_2
\end{pmatrix}$$
In the first case (with $\Sigma_1$), the generators corresponding to the SM symmetry are part of the broken generators so the SM symmetry is totally broken.
In the second case ($\Sigma_2$), the SM generators are part of the unbroken generators then the SM symmetry is preserved. As you can read, I know the answers but not how to find them !
So, my questions are :
*
*How is it possible in general (not only for the $SU(4)\rightarrow Sp(4)$ breaking pattern) to construct the vacuum that breaks the symmetry ?
*Is it possible, when constructing the vacuum, to ensure that the vacuum will (or not) break a sub-symmetry like the SM symmetry in the previous example ?
| As I understand the question, it is: what are the possible unbroken subgroups when a symmetry group G is spontaneously broken? If we assume that Lorentz invariance is unbroken, then we can look at the possible vacuum expectation values of a scalar field that transforms under some representation R of the symmetry group G. This can be calculated for specific G and R, as in the examples already listed, but general results are few.
A scalar field transforming like a vector under SU(n) [or SO(n)] can break these symmetries
down to SU(n-1) $\otimes$ U(1) [or SO(n-1)], since the group transformations in the subspace orthogonal to the direction of the VEV leave the vacuum invariant. There are many other examples in the literature.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/108930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 2,
"answer_id": 1
} |
Converting between (abstract) linear operators and their position representations Just as we have an abstract state vector $|\psi\rangle$ and its position representation $\psi(\vec{x}) = \langle \vec{x} | \psi \rangle$, how do we transform between a linear operator, say $H$, that acts on state vectors thus: $H |\psi\rangle$, and the differential operator form of the Hamiltonian that can act on the position representations of these states?
I have read that we have something like this
$$\langle \vec{x} | H | \vec{x}' \rangle = -\frac{h^2}{2m} \nabla^2 \delta(\vec{x} - \vec{x}') \,.$$
If this is so, could somebody explain how the following leap is made --- it concerns the propagator, which I'm defining by (with $\theta$ the Heaviside function):
$$K(\vec{x},t;\vec{x}',t') = \theta(t - t') \langle \vec{x} | U(t-t') |\vec{x}'\rangle $$
So I can't see why this line is true:
$$ \langle \vec{x} | H U(0) | \vec{x}' \rangle = H K(\vec{x},t';\vec{x}',t') \,,$$
where I assume that on the left hand side, the Hamiltonian is acting as an abstract linear operator, whilst on the right it takes its position representation, differential operator form. $U$ is the time evolution operator.
Thanks.
| This simply follows from a resolution of the identity, written (formally) as $\int dy |y\rangle\langle y| = 1$:
$\langle x|HU(t)|x\rangle = \int dy\langle x|H|y\rangle\langle y|U(t)|x\rangle$, which after integrating out the delta function arising from the matrix element of $H$ gives you what you want.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/109082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can molten metal be suspended in air? I wondered if magnets could be used to hold a drop of molten liquid metal in air (not for any particular reason just because it could be done), but was disappointed when a quick Google search showed the metal would lose its magnetic traits before it melted.
Are there any other forces that could be used to suspend a drop of molten liquid metal in air such as sound waves, high pressure air, electric currents, or anything else?
| Molten glass is suspended (and spun) on air jets to form "pre-forms" for molding glass optics. I suspect that the same must be possible for metals.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/109163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Why does the Fermi Surface cross the Brillouin zone boundary at right angles? I'm not sure why the fermi surface crosses the Brillouin zone boundary at right angles. I understand that this is normally the case, but not necessarily always.
I'm aware that the fermi surface is a constant energy surface up to the filling point. The Brillouin zone is in reciprocal space.
| I believe it has to do with translation, inversion, and mirror symmetry?
In real space, the potential energy has to share the same periodicity as the lattice. Similarly in k-space, the Fermi energy surface has to share the same periodicity as the reciprocal lattice. Brillouin zone boundaries are like the markers in k-space that specify when one period ends and the next begins. Periodicity alone doesn't guarantee that the Fermi surface cuts the zone boundaries at 90 degrees. But with inversion and mirror symmetries it does! Not all crystal structures have inversion and mirror symmetries. But the common ones like BCC and FCC do.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/109416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Index Notation with Del Operators I'm having trouble with some concepts of Index Notation. (Einstein notation)
If I take the divergence of curl of a vector, $\nabla \cdot (\nabla \times \vec V)$ first I do the parenthesis:
$\nabla_iV_j\epsilon_{ijk}\hat e_k$ and then I apply the outer $\nabla$...
and get:
$\nabla_l(\nabla_iV_j\epsilon_{ijk}\hat e_k)\delta_{lk}$
I am not sure if I applied the outer $\nabla$ correctly. If I did do it correctly, however, what is my next step? I guess I just don't know the rules of index notation well enough. Can I apply the index of $\delta$ to the $\hat e$ inside the parenthesis? Or is that illegal?
| First some notation
$$\nabla \times \vec B \rightarrow \epsilon_{ijk}\nabla_j B_k$$
$$\nabla \cdot \vec B \rightarrow \nabla_i B_i$$
$$\nabla B \rightarrow \nabla_i B$$
Now, to your problem,
$$\nabla \cdot(\nabla \times \vec V)$$
writing it in index notation
$$\nabla_i (\epsilon_{ijk}\nabla_j V_k)$$
Now, simply compute it, (remember the Levi-Civita is a constant)
$$\epsilon_{ijk} \nabla_i \nabla_j V_k$$
Here we have an interesting thing, the Levi-Civita is completely anti-symmetric on i and j and have another term $\nabla_i \nabla_j$ which is completely symmetric: it turns out to be zero.
$$\epsilon_{ijk} \nabla_i \nabla_j V_k = 0$$
Lets make the last step more clear. We can always say that $a = \frac{a+a}{2}$, so we have
$$\epsilon_{ijk} \nabla_i \nabla_j V_k = \frac{1}{2} \left[ \epsilon_{ijk} \nabla_i \nabla_j V_k + \epsilon_{ijk} \nabla_i \nabla_j V_k \right]$$
Now lets interchange in the second Levi-Civita the index $\epsilon_{ijk} = - \epsilon_{jik}$, so that
$$\epsilon_{ijk} \nabla_i \nabla_j V_k = \frac{1}{2} \left[ \epsilon_{ijk} \nabla_i \nabla_j V_k - \epsilon_{jik} \nabla_i \nabla_j V_k \right]$$
Now we can just rename the index $\epsilon_{jik} \nabla_i \nabla_j V_k = \epsilon_{ijk} \nabla_j \nabla_i V_k$ (no interchange was done here, just renamed).
$$\epsilon_{ijk} \nabla_i \nabla_j V_k = \frac{1}{2} \left[ \epsilon_{ijk} \nabla_i \nabla_j V_k - \epsilon_{ijk} \nabla_j \nabla_i V_k \right]$$
We can than put the Levi-Civita at evidency,
$$\epsilon_{ijk} \nabla_i \nabla_j V_k = \frac{\epsilon_{ijk}}{2} \left[ \nabla_i \nabla_j V_k - \nabla_j \nabla_i V_k \right]$$
And, because V_k is a good field, there must be no problem to interchange the derivatives $\nabla_j \nabla_i V_k = \nabla_i \nabla_j V_k$
$$\epsilon_{ijk} \nabla_i \nabla_j V_k = \frac{\epsilon_{ijk}}{2} \left[ \nabla_i \nabla_j V_k - \nabla_i \nabla_j V_k \right]$$
And, as you can see, what is between the parentheses is simply zero.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/109628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Interpretation of Conjugate Momentum in Field Theory The conjugate momentum density, following as a conserved quantity with Noethers Theorem, from invariance under displacement of the field itself, i.e. $\Phi \rightarrow \Phi'=\Phi + \epsilon$, is given by $\pi=\frac{\partial L}{\partial ( \dot{\Phi})}$.
On the other hand the physical momentum density, following as a conserved quantity with Noethers Theorem, from invariance under translations, i.e. $\Phi(x) \rightarrow \Phi(x')=\Phi(x+\epsilon)$, is given by $\Pi = \frac{\partial L}{\partial ( \dot{\Phi})} \frac{\partial \Phi}{\partial x}$.
Does anyone know a enlightening interpretation of the conjugate momentum? Furthermore why do we, in quantum field theory, impose commutation relations with the conjugate momentum instead of the physical momentum?
(For brevity all possible indices are supressed)
| Conjugate momentum in field theory is really just an infinite particle generalization of conjugate momentum in classical mechanics. The reason we impose commutation relations using the conjugate momentum is due to Dirac's canonical quantization prescription. This is well explained by Qmechanic in this answer.
I'll briefly summarize the argument. To canonically quantize a classical theory you must first express it in Hamiltonian form. This involves defining the conjugate momentum, and imposing Hamilton's equations with Poisson brackets. You then promote Poisson brackets to commutators. It turns out that conjugate momentum is exactly the right variable that makes this approach work mathematically.
I'm not aware of a physical interpretation of conjugate momentum in field theory, but I'd be interested to hear other people's thoughts on this!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/109707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
What's the difference between average absolute error and relative error? I am quite confused by both these terms. I would like to know what's the exact difference between both these terms and which one is more accurate.
| The absolute error can be measured using this formula:
$$\varepsilon_a=\frac{x_{max}-x_{min}}{2}$$
That is the difference between the highest value and the lowest value that you get after some measurements. The Relative error is:
$$\varepsilon_r=\frac{\varepsilon_a}{\bar{x}}$$
where $\bar{x}$ is the average of all your measurements.
There is also there is the percent error (relative) that equals to:
$$\varepsilon_r\cdot100$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/109775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Does electricity flow on the surface of a wire or in the interior? I was having a conversation with my father and father-in-law, both of whom are in electric related work, and we came to a point where none of us knew how to proceed. I was under the impression that electricity travels on the surface while they thought it traveled through the interior. I said that traveling over the surface would make the fact that they regularly use stranded wire instead of a single large wire to transport electricity make sense.
If anyone could please explain this for some non-physics but electricly incline people, I would be very appreciated.
| I'll try to keep it short and sweet; Stranded wire is capable of delivering high amperage without overheating because the strands devide the load..I.E. battery cables on your car. stranded wire is superior to solid but to expensive for long runs, so solid wire is used for long runs like for your house (easy to snake or bend) solid but flexible electric company supply line. Yes it mabe true that on a solid conductor there will be less resistance in the center, it would be nomimnal. Take your home appliances for instance, 120v is supplied to your home as a wave length (keeps voltage constant & helps keep line from overheating) Now examine everything you plug into the wall, if it has an elecrtic motor it usually runs A/C ah! but everything else runs on DC. most devices transform A/C to DC because DC can handle short runs with high (Ampherage, Current, Resistance, or Load) To be a little technical solid wire carrying A/C as a wave means there is space between the waves where electricity isn't flowing which help in delivery and cooling however you would need a scope to observe it......GOOD LUCK RAD3
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/109897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "64",
"answer_count": 9,
"answer_id": 8
} |
Work done or not in this case? I have a very simple question. A motorboat directed upstream is seen to be at rest from the bank of a river. Is the engine doing any work? Is it right to say that since it is not causing any displacement, so it is not doing any work.
| It does do work: it's causing the water in the wake of the boat to move downstream faster than the rest of the current. The engine is doing work on the water, rather than doing work on the boat.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/109981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why is uncertainty divided by $\sqrt{3}$? Why do we sometimes have to divide uncertainty of the measurement by $\sqrt{3}$?
For example we have the uncertainty of the measurement with a ruler with the smallest scale of 0.01 cm. This 0.01 cm is not our uncertainty of measurement, but we first divide it by $\sqrt{3}$ and then the result is our final uncertainty value.
Why is that this way?
| I have found out the answer.
When you do a measurement (one measurement) you have many uncertainty sources. But if you want to have the combined uncertainty, you don't add like $1 + 1$, because that would give you uncertainty of $2$, but that doesn't have to be the case. The true value of the measurement may lie in between the uncertainty regions but it may lie on the left or on the right of the range.
As we don't know where exactly is the measurement we have to treat that situation like a rectangular distribution. We have $a=x-\Delta x$ and $b=x+\Delta x$, where $\Delta x$ is the uncertainty.
$$E(X)
= \int_a^b xp(x)\mathrm{d}x
= \int_a^b \frac{x}{b - a}\mathrm{d}x
= \left.\frac{1}{2}\frac{x^2}{b - a}\right|_a^b = \frac{1}{2}\frac{b^2 - a^2}{b - a} = \frac{b + a}{2}$$
$$\begin{align}
\sigma^2
&= \int_a^b \bigl(x - E(x)\bigr)^2 p(x)\mathrm{d}x \\
&= \int_a^b \biggl(x - \frac{b + a}{2}\biggr)^2\frac{1}{b - a}\mathrm{d}x \\
&= \left.\frac{1}{3(b - a)}\biggl(x - \frac{b + a}{2}\biggr)^3\right|_a^b \\
&= \frac{1}{3(b - a)}\Biggl[\biggl(\frac{b - a}{2}\biggr)^3 - \biggl(\frac{a - b}{2}\biggr)^3\Biggr] \\
&= \frac{1}{3}\biggl(\frac{b - a}{2}\biggr)^2
\end{align}$$
So we see that the standard deviation of this distribution is exactly our uncertainty divided by $\sqrt3$. Without that our uncertainty would be bigger than it needs to be. And so the final uncertainty of the single measurement is then: $u(x)=\Delta x/\sqrt3$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/110242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Number of planks required to stop the bullet A bullet looses (1/n)th of its velocity passing through one plank. The number of such planks that are required to stop the bullet can be?
Logically, to me the answer seems to be infinity, as always a fraction of velocity will get reduced. But in my book the answer is n^2/(2n-1) (that comes from energy balance). What is correct?
|
Ayush: Isn't the question telling that the bullet always loses 1/n th of its velocity no matter which plank?
Based on the answer provided, it seems the writer wanted you to assume that the energy loss per plank is constant. This is not the same as the bullet losing $1/n^\text{th}$ of its velocity per plank (however, the fact that the question does not mention this assumption arguably makes the question ambiguous).
With this assumption, the energy loss becomes
$$\Delta E=\frac{1}{2}mv^2-\frac{1}{2}m\left(v-\frac{v}{n}\right)^2$$
and the number of planks $N$ becomes
$$N=\frac{\frac{1}{2}mv^2}{\Delta E}=\frac{n^2}{2n-1}.$$
Otherwise, if you assume that the bullet loses $1/n^\text{th}$ of its velocity per plank, then the answer is $N=\infty$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/110401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Is frequency quantized in the black body spectrum? I'm aware that there're some questions posted here with respect to this subject on this site, but I still want to make sure, is frequency quantized? Do very fine discontinuities exist in a continuous spectrum like the black body spectrum?
The quantization of photon energies
| well, the frequency of a photon is not quantized, we observe all 'kinds of colors (frequencies)'...what is quantized is the quantity of photons of the exact same frequency we may produced, or otherwise exist...for example, lets say we have a photon of 633 nm, then the quantity of photons that may be 'produced' is 1, 2, 3, 4, 5,because the photon is quantized, or the energy of a photon of an exact frequency can be increased only by steps of hν, here ν is the not quantized frequency....quantum physics is pretty simple.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/110463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 2
} |
Why can't an excess of electrons or holes by themselves cause current flow? I am a beginner in electrical engineering. Often times (most cases actually), the underlying physics aren't really explained to us and we are just left to assume that it works "because it works." This is never enough for me in classes etc and I always end up doing a follow up of physics side of the spectrum.
My question is, you have a battery with excess electrons built up on the negative lead, and excess holes on the positive lead, why is it that in our universe then connecting a conductive compound (like.. copper) to just the negative lead does not produce current, or vice versa connecting the wire to just the positive lead.
After all, if we move a magnet passed a conductor, there is a tiny induced current. How can the magnet do this, but an excess of electrons or electron holes repelling each other cannot?
| An excess of electrons can cause a current flow.
Imagine two spheres of metal. One is neutrally charged, but the other is negatively charged with an excess of electrons. Connect a wire between them, and what happens? The charge re-distributes itself so that the charge on both spheres is equal. As this is happening, a brief current of electrons is flowing through the wire, but it stops as soon as the charge is equal.
My question is, you have a battery with excess electrons built up on the negative lead, and excess holes on the positive lead, why is it that in our universe then connecting a conductive compound (like.. copper) to just the negative lead does not produce current
It does produce a current. It's just very small, because the voltage of the battery and capacitance of the wire are very small, and it's very brief, because there's no path for the current to flow continuously, so after the charge has equalized there's no more current.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/110549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Exact diagonalization to resolve ground state degeneracies I am studying a perturbed Toric Code model that is not analytically solvable.
On a torus the ground state degeneracy of the unperturbed model is 4.
Once we turn on the perturbation there is a change in the ground state degeneracy.
I would like to detect this change in ground state degeneracy numerically using exact diagonalization techniques.
On my computer I have stored the action of the Hamiltonian on a set of basis states. So if you give me some state $\left|\psi\right\rangle$ I can give you
$\hat{H}\left|\psi\right\rangle$ in terms of the basis states. Now I used this information to compute the spectrum using the Lanczos algorithm and the Jacobi-Davidson algorithm. While the spectrum itself is correctly reproduced neither of these algorithms reproduces the degeneracy of the ground state correctly - not even in the unperturbed case.
Hence the following question: What are common exact diagonalization algorithms for this type of many-body system that correctly resolve the degeneracy of the ground state?
I am looking forward to your responses!
| You must use re-orthogonalization procedure (Gram-Schmidt) at every Lanczos Step in order to correctly capture this degeneracy. I have found that degeneracy will not be captured correctly if you do not orthogonalize your Lanczos vectors. I have used Lanczos on 2D antiferromagnetic Kitaev model that has 4-fold degenerate ground state, and Lanczos correctly capture this only after the implementation of Gram-Schmidt procedure.
While this will help to find the correct degeneracy in the eigen-values of the Hamiltonian, Lanczos still has difficulty with the actual state. Lanczos will end up returning the corresponding 4-degenerate states as some arbitrary linear combination. I do not know how to fix this issue.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/110626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
What is the relation between isothermal process and its surrounding?
In an isothermal process the temperature of a system remains constant but may or may not be similar to temperature of surroundings
How is this possible?
| Isothermal processes can happen in the real-world because of insulation, very little heat exchange, or very little temperature change as the process happens.
For instance, most substances undergo an isothermal process when changing phase. You could even force some processes to become practically isothermal by cooling it as it happens. If you, say, had a gas in a cylinder, and expanded the volume in the cylinder, the gas will have more room to move around in, but the average velocity (or temperature!) of the gas remains the same.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/110673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why does squeezing a water bottle make the water come out? This seems natural, but I can't wrap my head around it when I think about it.
When I squeeze an open bottle filled with water, the water will spill out. When I squeeze a bottle, the material collapses where I squeeze it, but expands in other areas, resulting in a constant volume. If the volume is constant, then I would think that the water shouldn't spill out.
If I were to guess, there is something related to the pressure my hand is creating inside the bottle, but I'm not entirely sure.
| Just another approach to visualize it. (two dimensional) If you look on a bottle from above, you can see the round shape. A circle is the optimal shape, where you can have the biggest amount of content per package. If you start to squeeze the circle (no matter how), the amount of stuff you can put into is decreasing. Cause the amount of package is not increasing, where should it come from. (If you have an inelastic package of course.)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/110747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Interference - the shortest way from the point of constructive one to the point of destructive one So this is a problem from Polish maturity exam.
The image shows 2 speakers (G1, G2) and point B. The wavelength of sound coming from both speakers is 0.155 m, and the wave coming from both speakers is in phase. So point B is where the constuctive interference appears (you can calculate this quite easily).
And the problem - "Draw an arrow in the direction of the shortest way possible from point B, where the sound intensity is high, to point A, where the sound intensity is low". Essentialy what they're asking is in which direction should we turn from point of constructive interference in order to, by going in a straight line, achieve point of destructive interference in the shortest way possible.
So any thoughts on how to calculate, solve this?
| Ignoring the $1/r$ decay, the point source fields can be modeled as
$$f_\mathbf{p}(\mathbf{r})=\exp\left(\frac{2\pi i|\mathbf{r}-\mathbf{p}|}{\lambda}\right)$$
where $\mathbf{p}$ is the speaker location.
The field intensity from the two speakers then becomes
$$I(\mathbf{r})=\left|f_{\mathbf{p}_1}(\mathbf{r})+f_{\mathbf{p}_2}(\mathbf{r})\right|^2$$
and the direction of increase becomes
$$\nabla I(\mathbf{B})=\pmatrix{-1.01788\\-0.18508}$$
by inserting $\mathbf{p}_1=(0,0),\mathbf{p}_2=(1.7,0),\mathbf{B}=(0,4.52),\lambda=0.155$.
Picture
The intensity pattern looks like this when plotted from $x\in[-4,4],y\in[0,6]$:
This gives visual confirmation that the previous value of $\nabla I(\mathbf{B})$ was correct.
Mathematica Code
f[x_] := Exp[2 \[Pi] I Sqrt[x.x]/0.155];
{X1, X2, B, X} = {{0, 0}, {1.7, 0}, {0, 4.52}, {x, y}};
conjugate[expr_] := expr /. Complex[x_, y_] -> x - I y;
Chop[D[(f[X - X1] + f[X - X2]) conjugate[f[X - X1] + f[X - X2]], {X,
1}] /. Thread[X -> B]]
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/110850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Static wave's equation
In my book equation for two advanced wave which created this static wave are
$$y_1=asin{\frac{2\pi}{\lambda}(vt+x)} $$
$$y_2=asin{\frac{2\pi}{\lambda}(vt-x)} $$
and equation for static wave is derived by adding this two equation: $\ Y=y_1+y_2$
Q1: Though this two wave comes from opposite direction why it is added rather then subtracted? Shouldn't it be $\ Y=y_1-y_2$
Q2: Why sign of $\ x$ is opposite for two equation though it is said that it's the distance of the particular point from the reference and whose displacement for the wave is $\ y$
| It doesn't matter whether you add or subtract $y_1$ and $y_2$, because all this does is introduce a phase shift in the resulting standing wave. To see this you need the identities:
$$ \sin(a + b) = \sin(a)\cos(b) + \cos(a)\sin(b) $$
$$ \sin(a - b) = \sin(a)\cos(b) - \cos(a)\sin(b) $$
Let's take your two equations and try adding them. Using the identity we rewrite $y_1$ and $y_2$ as:
$$ y_1 = a\sin\left(\frac{2\pi}{\lambda}vt\right) \cos\left(\frac{2\pi}{\lambda}x\right) + a\cos\left(\frac{2\pi}{\lambda}vt\right) \sin\left(\frac{2\pi}{\lambda}x\right) $$
$$ y_2 = a\sin\left(\frac{2\pi}{\lambda}vt\right) \cos\left(\frac{2\pi}{\lambda}x\right) - a\cos\left(\frac{2\pi}{\lambda}vt\right) \sin\left(\frac{2\pi}{\lambda}x\right) $$
So we get:
$$ y_1 + y_2 = 2a\sin\left(\frac{2\pi}{\lambda}vt\right) \cos\left(\frac{2\pi}{\lambda}x\right) $$
$$ y_1 - y_2 = 2a\cos\left(\frac{2\pi}{\lambda}vt\right) \sin\left(\frac{2\pi}{\lambda}x\right) $$
These don't immediately look the same, but note that:
$$\begin{align}
\sin(x) &= \cos(x - \tfrac{\pi}{2}) \\
\cos(x) &= \sin(x + \tfrac{\pi}{2})
\end{align}$$
and if we use these to rewrite $y_2$ we get:
$$ y_1 + y_2 = 2a\sin\left(\frac{2\pi}{\lambda}vt\right) \cos\left(\frac{2\pi}{\lambda}x\right) $$
$$ y_1 - y_2 = 2a\sin\left(\frac{2\pi}{\lambda}vt+ \frac{\pi}{2}\right) \cos\left(\frac{2\pi}{\lambda}x- \frac{\pi}{2}\right) $$
So the two functions are the same but just shifted a bit along the $t$ and $x$ axes.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/111143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Trajectory of a photon around a Schwarzschild black hole? Consider a photon coming from the infinity in a unbounded orbit to a Schwarzschild black hole (Schwarzschild radius $r_{s}$) (see this for illustration). Its impact parameter is $b$ and its distance of closest approach is $r_{0}$ with $$b^2=\frac{r_{0}^{3}}{r_{0}-r_{s}}$$.
Then its trajectory in polar coordinates is defined by :
$$\frac{d\varphi}{dr} = \frac{1}{r^2\sqrt{\frac{1}{b^2}-\left(1-\frac{r_s}{r}\right)\frac{1}{r^2}}}$$
Consequently :
$$\varphi\left(r\right) = \int_{r_{0}}^{r} \frac{dp}{p^2\sqrt{\frac{1}{b^2}-\left(1-\frac{r_s}{p}\right)\frac{1}{p^2}}}$$
and one can compute the total deviation using :
$\Delta\varphi = 2\times\left(\lim_{r\to+\infty}\varphi\left(r\right)-\frac{\pi}{2}\right)$
But my question is : how can I plot/draw the trajectory using the integral expression of $\varphi\left(r\right)$ ?
Because if I compute : $$f\left(r\right) = 2\times\left(\int_{r_{0}}^{r} \frac{dp}{p^2\sqrt{\frac{1}{b^2}-\left(1-\frac{r_s}{p}\right)\frac{1}{p^2}}}-\frac{\pi}{2}\right)$$
I obtain $f\left(r_{0}\right) = -\pi$, and then $f$ increases up to zero, crosses zero, and tends to its positive value at infinity $\Delta\varphi$. It does not make sense for me and I do not understand how to compute the trajectory from that.
If I compute : $$g\left(r\right) = \int_{r_{0}}^{r} \frac{dp}{p^2\sqrt{\frac{1}{b^2}-\left(1-\frac{r_s}{p}\right)\frac{1}{p^2}}}$$ it starts from $0$, and increase up to $\frac{\pi}{2}+\frac{\Delta\varphi}{2}$.
I would like to compute the trajectory in the $\left(x, y\right)$ plane, so how to use the values of $f\left(r\right)$ or $g\left(r\right)$ to compute the function $y\left(x\right)$ ?
| In fact, the problem was that computing the deflection is not very intuitive. So the trajectory in polar coordinates is :
$$\varphi\left(r\right) = \int_{r_{0}}^{r} \frac{dp}{p^2\sqrt{\frac{1}{b^2}-\left(1-\frac{r_s}{p}\right)\frac{1}{p^2}}}$$
and in cartesian coordinates it is nothing else than :
*
*$x=r\cos\left(\varphi\left(r\right)\right)$
*$y=r\sin\left(\varphi\left(r\right)\right)$
and it represents a photon starting from $\left(x_0, y_0\right) = \left(r_{0}, 0\right)$ and going up.
But what is non-intuitive (a drawing helps a lot), it that the half deflection $\alpha/2$ is in fact :
$$\frac{\alpha}{2} = \varphi\left(r\right)-\cos^{-1}\left(\frac{r_{0}}{r}\right)$$
Problem solved...
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/111272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Why fermions have a first order (Dirac) equation and bosons a second order one? Is there a deep reason for a fermion to have a first order equation in the derivative while the bosons have a second order one? Does this imply deep theoretical differences (like space phase dimesion etc)?
I understand that for a fermion, with half integer spin, you can form another Lorentz invariant using the gamma matrices $\gamma^\nu\partial_\nu $, which contracted with a partial derivative are kind of the square root of the D'Alembertian $\partial^\nu\partial_\nu$. Why can't we do the same for a boson?
Finally, how is this treated in a Supersymmetric theory? Do a particle and its superpartner share a same order equation or not?
| In fact, the Dirac equation can be expressed as a second-order differential equation of the form
\begin{equation}
\frac{1}{c^2}\frac{\partial^2\psi}{\partial t^2}-\nabla^2\psi\pm\frac{2mi}{\hbar}\frac{\partial\psi}{\partial t}-\left(\frac{mc}{\hbar}\right)^2\psi=0\,.
\end{equation}
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/111401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "46",
"answer_count": 7,
"answer_id": 6
} |
Earphones and electric guitars I remember being in my college dorm, watching a film with a friend. His computer wasn't loud enough and we couldn't hear properly. So he took his earphones and put them on a electric guitar coil. He then plugged his guitar to his amp and voilà, we could listen through the amp.
Why on earth did this work?
| Very basically, a speaker driver has a magnet and a coil which move relative to each other when electric current is applied. (The speaker cone is attached to the moving parts and this is what moves the air that causes the audible sound).
When placed close to a guitar pickup, the tiny coil movement itself (not the cone/diaphragm movement) and/or magnetic flux generated by the movement acts on the magnetic field of the pickup in a manner similar to the movement of the guitar strings themselves.
It is also possible, depending on the design of the headphones, to use them as microphones. Many years ago, I used a set of can-style headphones plugged into an reel-to-reel tape recorder as a sort of acoustic guitar pickup.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/111497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why doesn't this model plane fly? I have been designing a model plane for Design Technology for the past month or so, and today I laser cut my final design and assembled, it then tested it. Upon testing the plane does not get any lift, whereas the previous testing model which was virtually the same did.
The plane is built using Balsa Wood, and Assembled with hot glue (I used as little glue as possible to reduce weight :) )
Any Ideas?
Image:
| It's not easy to tell why a plane does not fly just looking at a couple of pictures of it!
My advice is to go to a soft ground (like >10cm tall grass) and try to launch it by hand like a paper plane and see how it behaves. If it does not get a decent glide there there are very few hopes for it to lift by itself.
In this way you will point out stability problems and you can get a starting point to correct the geometry, placement of the wings, tail elevators, COG position and so on.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/111746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Does electrical current have a measurable momentum? If I were to build a levitating super-cooled superconducting loop of wire with a electrically charged circuit that would discharge its stored energy to produce a large circulation of current around the superconducting loop levitating off the ground, would the loop of wire experience a counter torque, forcing the loop to rotate in the opposite direction that the electrical current rotates to have a total torque and angular momentum equal to zero? Is this experiment possible?
| Not only is the experiment possible, a version of the experiment at the atomic level was done by Einstein and de Haas[1]. Einstein and deHaas showed that the angular momentum inhering in the aligned electron spins in a ferromagnet can be exhibited on a macroscopic scale when the sample is demagnetize. Apparently, his is the only experiment Einstein ever performed himself[2].
References
1. Einstein and de Haas paper[pdf]
http://www.dwc.knaw.nl/DL/publications/PU00012546.pdf
*
*Wikipedia on the Einstein de Haas effect
https://en.wikipedia.org/wiki/Einstein%E2%80%93de_Haas_effect
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/111825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
The "replica trick" initial formula In Spin-glass theory for pedestrians by Castellani and Cavagna, the initial formula used to introduce the replica trick is written as:
$$\overline{\log Z}=\lim_{n\rightarrow0}\frac{1}{n}\log\overline{Z^{n}}\qquad(1)$$
where the overbar denotes average over quenched disorder. I don't know how to prove this formula.
In other treatments I have seen of the replica method (wiki, for example), one starts from:
$$\log Z = \lim_{n\rightarrow 0} \frac{Z^n-1}{n}\qquad(2)$$
which I understand. How are (2) and (1) connected? What's the proof of (1)?
| Try to look at Introduction to the Replica Theory of Disordered Statistical Systems by V. Dotsenko. In the following, I've written a possible answer to your question:
\begin{equation}
f=-\lim_{N\rightarrow\infty}\frac{1}{\beta N}\mathbb{E}\left[\ln Z_{J}\right]
\end{equation}
where:
*
*$\mathbb{E}\left[\mathcal{O}\right]=\left(\prod_{\left\{ i,j\right\} }\int dJ_{ij}\right)P\left[J\right]\mathcal{O}
$
*$Z_{J}=\sum_{\sigma}e^{-\beta H\left[J,\sigma\right]}
$
Then labelling with $a$ the replicas:
\begin{equation}
Z_{J}^{n}=\left(\prod_{a=1}^{n}\sum_{\sigma^{a}}\right)e^{-\beta\sum_{a=1}^{n}H\left[J,\sigma_{a}\right]}
\end{equation}
Thus, remember that $\ln x=\lim_{n\rightarrow0}\frac{1}{n}\left(x^{n}-1\right)$:
\begin{equation}
f=-\lim_{N\rightarrow\infty}\frac{1}{\beta N}\mathbb{E}\left[\ln\left(Z_{J}\right)\right]=-\lim_{N\rightarrow\infty}\lim_{n\rightarrow0}\frac{1}{\beta N}\mathbb{E}\left[\frac{\left(Z_{J}^{n}-1\right)}{n}\right]=-\lim_{N\rightarrow\infty}\lim_{n\rightarrow0}\frac{1}{\beta nN}\mathbb{E}\left[Z_{J}^{n}\right]
\end{equation}
but in general there are many issues concerning the commutation of the two limits.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/111950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Why are popcorn seeds soft after popping? When a seed of popcorn is heated up in oil, it pops like this:
You can take one of these popped pieces and eat it with little to no problem. However, if you get an un-popped seed and sink your teeth in, it is noticeably harder. Why is this?
| Normally I feel uncomfortable giving Wikipedia-based answers since it seems lazy, but the Wikipedia article does a good job explaining it. It appears that the heating itself, rather than the popping, causes the popcorn starch to soften:
Under these conditions, the starch inside the kernel gelatinizes, softens, and becomes pliable. The pressure continues to increase until the breaking point of the hull is reached: a pressure of about 135 psi (930 kPa)[10] and a temperature of 180 °C (356 °F). The hull ruptures rapidly, causing a sudden drop in pressure inside the kernel and a corresponding rapid expansion of the steam, which expands the starch and proteins of the endosperm into airy foam.
Additional evidence that it is the heating, and not the popping, which causes this softening is given in the following citation:
If heated too quickly, the steam in the outer layers of the kernel can reach high pressures and rupture the hull before the starch in the center of the kernel can fully gelatinize, leading to partially popped kernels with hard centers.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/112015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Twin Paradox: Still a Paradox? Alright, so David Griffiths in his "Introduction to Electrodynamics" states that the Twin Paradox is not a paradox at all since the traveling twin returns to Earth. By returning to Earth, the twin had to reverse direction, thus undergoes acceleration, and therefore cannot claim to be a stationary observer.
However, what if the traveling twin simply Skypes the twin that is on Earth. The twin on earth will still appear older, which would make no sense since in that case the rocket can be seen as the stationary frame of reference while the Earth "travels" at a speed close to the speed of light. No acceleration is undergone, yet the paradox remains.
Is Griffiths just completely glossing over important nuance again?
| I believe the twin paradox is truly a paradox in that the same problem approached different ways gives different answers. So if the travelling twin accelerates from near earth at 1g, he will be close to the speed of light (relative to earth) in one year. If he then turns 180 degrees and continues acceleration at 1g, he will return to earth in one year. Both twins will have experienced the same 1g acceleration. They are symmetrical and will see time slowing down for the other twin as they recede from each other and time speeding up as they approach each other. They are both the same age when they reunite. So there is no way we can use time dilation to explore the galaxy or ever get to other than the nearest stars. The indirect proof? We have not been visited by aliens. There are hundreds of billions of stars and many more planets in our galaxy. There must be millions of earth like worlds that have existed for as much time or longer than our earth. Surely we are not the most advanced civilization among these. If time dilation allowed travel throughout our galaxy, then why haven't we been visited? We have been transmitting radio and TV signals for more than 80 years now, and should be detectable to alien civilizations within 80+ light years. Where are they????
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/112158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 10,
"answer_id": 7
} |
How to calculate the energy freed in the reaction: $^{10}_5Be +\space ^2_1H \rightarrow \space^{11}_5B + \space ^1_1H$? I have the following reaction: $^{10}_5\mathrm{Be} +\space ^2_1\mathrm{H} \rightarrow \space^{11}_5\mathrm{B} + \space ^1_1\mathrm{H}$
And I know that I have to use the formula: $E = \Delta m\cdot c^2 = \Delta m \cdot \frac{931,5MeV}{u}$.
So I just need $\Delta m$ which is equal to:
$\Delta m = m_b - m_a$ where $m_b$ represents the mass "before the reaction" and $m_a$ the mass "after the reaction" so we have:
$m_b = m(^{10}_5\mathrm{Be}) + m(^2_1\mathrm{H})$
$m_a = m(^{11}_5\mathrm{B}) + m(^1_1\mathrm{H})$
The book which contains this problem contains the following table:
http://i.imgur.com/esoGDVf.png but from this table, I only know $ m(^1_1\mathrm{H})$ and $m(^2_1\mathrm{H})$ i.e.
$m_b = m(^{10}_5\mathrm{Be}) + 2.01410u$
$m_a = m(^{11}_5\mathrm{B}) + 1.00783u$
How do I calculate $m(^{10}_5\mathrm{Be})$ and $m(^{11}_5\mathrm{B})$ ?
P.S. I don't know if the tag is correct. The chapter in the book where I found this exercise is called "Basics of nuclear physics".
|
How do I calculate $m\left(^{10}_4\mathrm{Be}\right)$ and $m\left(^{11}_5 \mathrm{B}\right)$?
The masses and various other properties of isotopes are available freely at Wolfram Alpha. They are,
*
*$m\left(^{10}_4\mathrm{Be}\right)=10.013533818u$
*$m\left(^{11}_5 \mathrm{B}\right)=11.009305406u$
where $u$ denotes unified atomic mass units. Notice you are already given the mass number in the superscript of the isotope. As John Rennie noted, the reaction should probably be with $^{9}_4\mathrm{Be}$,
$$^{9}_4\mathrm{Be} + ^{2}_1 \mathrm{H} \to ^{10}_{5}\mathrm{B} + n$$
in which case the mass is $9.012182201u$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/112239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Gauss’s Law inside the hollow of charged spherical shell Use Gauss’s Law to prove that the electric field anywhere inside the hollow of a charged
spherical shell must be zero.
My attempt:
$$\int \mathbf{E}\cdot \mathbf{dA} = \frac{q_{net}}{e}$$
$$\int E \ dAcos\theta = \frac{q_{net}}{e}$$
$$E \int dA = \frac{q_{net}}{e}$$
$E\ 4\pi r^2 = \frac{q_{net}}{e}$ and since it is a hollow of a charged spherical shell the $q_{net}$ or $q_{in}$ is $0$ so: $E = 0$.
Is my reasoning on this problem correct? Essentially $E$ is $0$ because there is no charge enclosed.
| It isn't enough to show that the net field over some randomly chosen surface within the shell is zero. After all, this is also true of a closed surface between the plates of a capacitor, but the field is not zero there.
You need to use the spherical symmetry and choose a sperical shell centred at the centre of your sphere. Then you can use the symmetry to argue that the field must be everywhere normal to your surface and everywhere have the same value.
Apologies if this is already implicit in your working - it isn't clear to me whether you've considered this or not.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/112315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Clebsch-Gordan Coefficients for two spin-1 particles - Why is there a ∣0⟩∣0⟩ ket? I have used the rules for addition of angular momenta to work out the Clebsch-Gordan coefficients, which all seem right except for state $\lvert0,0\rangle$:
For n = 1
\begin{align}
\lvert1,1\rangle & = \frac{1}{\sqrt 2} \left( \lvert0\rangle\lvert1\rangle - \lvert1\rangle\lvert0\rangle \right) \\
\lvert1,0\rangle & = \frac{1}{\sqrt 2} \left( \lvert-1\rangle\lvert1\rangle - \lvert1\rangle\lvert-1\rangle \right) \\
\lvert1,-1\rangle & = \frac{1}{\sqrt 2} \left( \lvert0\rangle\lvert-1\rangle - \lvert-1\rangle\lvert0\rangle\right)
\end{align}
Now the state $\lvert0,0\rangle$ must be perpendicular to $\lvert1,0\rangle$ and is a linear combination of the basis kets of $\lvert1,0\rangle$:
$$\lvert0,0\rangle = \frac{1}{\sqrt 2} \left(\lvert-1\rangle\lvert1\rangle + \lvert1\rangle\lvert-1\rangle\right)$$.
But in the table, there is an extra ket $\lvert0\rangle\lvert0\rangle$; Why is this so?
(From the table):
$$\lvert0,0\rangle = \frac{1}{\sqrt 3} \left(\lvert-1\rangle\lvert1\rangle + \lvert1\rangle\lvert-1\rangle - \lvert0\rangle\lvert0\rangle\right).$$
My intuition tells me that you need to include the $\lvert0\rangle\lvert0\rangle$ state in order for the entire set of basis to be complete. But how do I show this?
| If you have two spin-1 particles, there are $three$ states where the projection of the total angular momentum is zero:
$$
\begin{align}
\left|2,0\right> &= \frac1{\sqrt6} \left(
\big|1,1\big>\big|1,-1\big>
~~+~~
\big|1,-1\big>\big|1,1\big>
~~+~~
\sqrt4\cdot
\big|1,0\big>\big|1,0\big>
\right)
\\
\left|1,0\right> &= \frac1{\sqrt2} \left(
\big|1,1\big>\big|1,-1\big>
~~-~~
\big|1,-1\big>\big|1,1\big>
\right)
\\
\left|0,0\right> &= \frac1{\sqrt3} \left(
\big|1,1\big>\big|1,-1\big>
~~+~~
\big|1,-1\big>\big|1,1\big>
~~-~~
\big|1,0\big>\big|1,0\big>
\right)
\end{align}
$$
The state $\left|1,0\right>\left|1,0\right>$ enters $\left|2,0\right>$ state with positive sign from applying the lowering operator to $\left|2,1\right>$. (If you haven't done this algebra yourself, do so — it's quite edifying.) Therefore the symmetric zero-spin combination must contain some $\left|1,0\right>\left|1,0\right>$ with a negative sign, for orthogonality.
Alternatively, you can operate on your prospective total-spin-zero state with the angular momentum raising operator for your two spin-1 particles:
$$
\begin{array}{rclr}
L_+ & \left|1,1\right>\left|1,-1\right>
&= & \sqrt2 \left|1,1\right>\left|1,0\right>\\
L_+ & \left|1,0\right>\left|1,0\right>
&= \sqrt2 \left|1,1\right>\left|1,0\right> &{}+{} \sqrt2 \left|1,0\right>\left|1,1\right>\\
L_+ & \left|1,-1\right>\left|1,1\right>
&= \sqrt2\left|1,0\right>\left|1,1\right>\\
\end{array}
$$
This should make it clear that a negative contribution from $\left|0\right>\left|0\right>$, is required to make to construct an $m=0$ state that vanishes when it sees a raising or lowering operator.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/112446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Constant power in rotational dynamics I am having trouble understanding and applying the concept of constant power (e.g. a motor) in rotational dynamics. We have that:
$$P=\tau\omega$$
Therefore if we imagine a physical system with a motor, supplied with a constant voltage and current such that $P=P_{0}$ is a constant, driving a disk with moment of inertia $I$. We surely have a differential equation of the following form:
$$I\dot{\omega}=\frac{P}{\omega}-\gamma \omega$$
Where $\gamma$ is some frictional drag coefficient. This is a separable ODE which yields (according to Wolfram Alpha):
$$\omega(t)=\pm\sqrt{\frac{P-e^{\frac{2\gamma(cI-t)}{I}}}{\gamma}}$$
This doesn't seem realistic as using initial conditions $\omega(0)=0$ we find that:
$$c=\left\{\frac{\ln(\pm i\sqrt{P})}{\gamma}\right\}$$
What am I misunderstanding here?
| Wolfram Alpha wont give you the coefficients of integration properly.
The solution is
$$ t = \int \frac{1}{\dot \omega}\,{\rm \omega} = \int \limits_{\omega_0}^\omega \frac{I \omega}{P-\gamma \omega^2}\,{\rm d} \omega $$
$$ t = - \frac{I}{2 \gamma} \ln \left( \frac{P-\gamma \omega^2}{P-\gamma \omega_0^2} \right) $$
where $\omega_0$ is the initial speed. When $\omega > \omega_0$ then the argument in the logarithm is less than one, but positive. This yields a positive time. The top speed is $\omega_f = \sqrt{ \frac{P}{\gamma}}$ and when $\omega \rightarrow \omega_f$ then $t \rightarrow \infty$.
The speed as a function of time is
$$ \omega(t) = {\rm e}^{-\left(\frac{\gamma}{I} t\right)} \sqrt{\omega_0^2 + \frac{P}{\gamma} ({\rm e}^{\left(\frac{2 \gamma}{I} t\right)} -1)} $$
Remember if initially at rest, the power is infinite. In reality there is a part of linear power (constant torque) before the constant power part.
To solve for angle use $\theta = \int \frac{\omega}{\dot \omega}\,{\rm d} \omega$ with appropriate limits.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/112544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Which clock is the fastest inside an accelerating body? The picture shows an accelerating spaceship with two clocks inside it. It is so far away from all other bodys that gravity is of no importance.
Will the bottommost clock be slower than the topmost one? Or will both clocks have the same speed?
| The clock ahead is behind in time by an amount depending on the speed of the inertial frame of reference theyare in. If the acceleration is thought of as moving through faster and faster inertial frames of reference then the amount that the clock ahead in distance is behind in time increases as the speed of the inertial frames increases.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/112645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 2
} |
Snells Law: Does the $k$ vector change on the boundary between mediums? I was using Waves - Berkley Physics Volume III, and in explaining Snell's Law the author claims that as a wave is on the boundary between glass and air (going from glass to air) that the number of wave crests per unit length along the $y$-axis must be equal in both mediums.
I still don't understand this claim and was wondering if anyone could help explain it.
Also the boundary runs along the $y$ axis, ie the boundary is vertical and so is the $y$-axis.
| Asuming I've understood your setup correctly it looks like this:
The red line shows the spacing between crests, and the number of crests per unit length is just the reciprocal of this distance. The red line is obviously constant on both sides of the interface because the wavelength otherwise there would be mismatches in the wave amplitude at the boundary i.e. crests on one side wouldn't line up with crests on the other.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/112879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Which side of wave-particle duality to choose in a given situation How does one know whether, in treating a certain problem, one should consider particles as waves or as point-like objects? Are there certain guidelines regarding this?
| It will depend on the results of your experiment, the results will tell you whether you are seeing the wave nature or the particle nature.
Take the scattering of an electron on a proton producing an electron and a proton and a pi0 meson. Your experiment measures "particle" interactions, in the form of classical particles, you can see the trajectories of the individual particles with your instruments.
If you measure a lot of scatters and plot the crossection versus energy, then the interpretation uses the quantum mechanical wavefunctions which by construction carry the wave nature of the "particles".
This two slit experiment of electrons one at a time
build up in time
shows clearly both natures. Each individual electron is a dot, i.e. a particle interacting with the screen. The accumulation though shows the probability distribution due to the wave function of the electron with the boundary condition of two slits, the wave form of the duality.
The classical type particle nature appears at a specific (x,y,z). The probability of appearing at the specific (x,y,z) has a wave nature.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/113021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
$\Delta^+$ decay in GZK process The dominant channels in the GZK process are
$$p+\gamma_{\rm CMB}\to\Delta^+\to p+\pi^0,$$
$$p+\gamma_{\rm CMB}\to\Delta^+\to n+\pi^+.$$
According to the pdg, $\Delta\to N+\pi$ makes up essentially 100% of the branching ratio (BR). It doesn't, however, say which process is favored: the proton and neutral pion or neutron and charged pion. My instinct is that they should each contribute about 50%, but I am not sure. So my question is, what are the BRs for each of the processes described above?
|
[...] $\Delta^+ \rightarrow p + \pi^0$, [...] $\Delta^+ \rightarrow n + \pi^+$,
which process is favored: the proton and neutral pion or neutron and charged pion [?]
Since the kinematics (and corresponding "phase space" factors) for the two final states are presumably as good as equal, the evaluation of the branching ratio
$$\text{BR} := \frac{\Gamma[ \Delta^+\rightarrow p+\pi^0 ]}{\Gamma[ \Delta^+\rightarrow n+\pi^+ ]}$$
simplifies to determining the ratio of "state constituent" transition probabilities
$$\text{BR} := \frac{\Gamma[ \Delta^+\rightarrow p+\pi^0 ]}{\Gamma[ \Delta^+\rightarrow n+\pi^+ ]} \simeq \frac{\left\lvert \langle p; \pi^0 \mid \Delta^+ \rangle \right\rvert^2}{\left\lvert \langle n; \pi^+ \mid \Delta^+ \rangle \right\rvert^2}.$$
Analyzing (or defining) the initial state $\Delta^+$ and the two distinct final states in terms of isospin leads to the expressions
$$ \lvert \Delta^+ \rangle \equiv \big\lvert \left(3/2, 1/2\right)_i \big\rangle, $$
where the first value represents the magnitude of $\mathbf I$, and the second value represents the magnitude of $I_3$, along with
$$ \lvert p; \pi^0 \rangle \equiv \big\lvert (1/2, 1/2)_f; (1, 0)_f \big\rangle \equiv \sqrt{ \frac{2}{3} }~\big\lvert (3/2, 1/2)_t \big\rangle - \sqrt{ \frac{1}{3} }~\big\lvert (1/2, 1/2)_t \big\rangle, $$ and
$$ \lvert n; \pi^+ \rangle \equiv \big\lvert (1/2, -1/2)_f; (1, 1)_f \big\rangle \equiv \sqrt{ \frac{1}{3} }~\big\lvert (3/2, 1/2)_t \big\rangle + \sqrt{ \frac{2}{3} }~\big\lvert (1/2, 1/2)_t \big\rangle, $$
where
*
*the coefficients of the linear combinations on the right-hand sides are Clebsch-Gordan coefficients (specificly those values listed in table "$1/2 \otimes 1$"),
*all states are normalized, and
*the indices $f$ and $t$ are to distinguish final states and "state representations to evaluate transition probabilities"; such that in particular the states $(1/2, 1/2)_f$ and $(1/2, 1/2)_t$ are (meant to be) distinct; and both are distinct, and indeed disjoint, from the initial state $\lvert \Delta^+ \rangle \equiv \lvert (3/2, 1/2)_i \rangle$.
Now identifying
$$\big\lvert (3/2, 1/2)_t \big\rangle \equiv \big\lvert (3/2, 1/2)_i \big\rangle $$
we can evaluate
\begin{align}
\langle p; \pi^0 \mid \Delta^+ \rangle & \equiv \bigg\langle \sqrt{ \frac{2}{3} }~ (3/2, 1/2)_t - \sqrt{ \frac{1}{3} }~ (1/2, 1/2)_t \bigg\vert (3/2, 1/2)_t \bigg\rangle \\
& = \bigg\langle \sqrt{ \frac{2}{3} }~ (3/2, 1/2)_t \bigg\vert (3/2, 1/2)_t \bigg\rangle \\
& = \sqrt{ \frac{2}{3} }
\end{align}
and
\begin{align}
\langle n; \pi^+ \mid \Delta^+ \rangle & \equiv \bigg\langle \sqrt{ \frac{1}{3} }~ (3/2, 1/2)_t + \sqrt{ \frac{2}{3} }~ (1/2, 1/2)_t \bigg\vert (3/2, 1/2)_t \bigg\rangle \\
& = \bigg\langle \sqrt{ \frac{1}{3} }~ (3/2, 1/2)_t \bigg\vert (3/2, 1/2)_t \bigg\rangle \\
& = \sqrt{ \frac{1}{3} }
\end{align}
obtaining the sought branching ratio value as
$$\text{BR} := \frac{\Gamma[ \Delta^+\rightarrow p+\pi^0 ]}{\Gamma[ \Delta^+\rightarrow n+\pi^+ ]} \simeq \frac{ (\sqrt{ 2/3 })^2 }{ (\sqrt{ 1/3 })^2} = 2.$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/113252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Specific Heat equations - paradox! We know that enthalpy is $h = u + pv$. Where $u, p, v, h$ are internal energy, pressure, volume, and enthalpy respectively. Now specific heat at constant volume is calculated as $c_v = \frac{\partial u}{\partial T}$, and specific heat at constant pressure as $c_p = \frac{\partial h}{\partial T}$.
I am pretty confused at this point, in volume case it is internal energy, and in pressure case it is enthalpy. Because if $pv$ is a form of energy independent of internal energy, then it should figure in $c_v$ as well, because at constant volume when I am putting energy in the system, the pressure can not stay constant.
What is wrong in the above argument? Appreciate your thoughts!
| $pv$ is not "a form of energy". It is just product of pressure and volume. Specific heat is the amount of heat needed to increase temperature by one unit. This heat can be expressed as $Tds$ for quasi-static processes. In constant volume case this can be expressed as $du$, because $du = Tds -pdv$ generally and $dv=0$. In constant pressure case the heat is $dh$ because $dh = Tds+Vdp$ generally and $dp=0$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/113297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Is it possible to produce gamma radiaton using radio emitter? As in the title, I'm wondering is it possible.
I think it is possible, because we have powerful enough radiotechniques and gamma radiation are just EM waves, not particles. However I think is useless, because it costs too much.
Can anyone say something more about this?
*
*How much power would consume such emitter?
*Would it be helpful anywhere?
*How huge frequencies of current would it use in the antenna?
*How big device would it be?
| The 0.51 MeV photon from positron-electron annihilation is considerd a gamma ray and industrial X-Ray tubes emit photons that are over 1 MeV. So there is some overlap in the terminology. If your radio produces electric fields of 500 KeV or more, there is SOME way to make gamma rays come from something loosely called an antenna.
For example, your antenna could be a cyclotron or a modified Klystron tube. You can transmit X-Rays with an old improperly used vacuum tube and an automobile spark coil.
In the typical analysis of RF antennas, the electrons in the metal shake back and forth and create EM waves in the far field. If you shake the electrons harder and they bump into things, you can get X-Ray/Gamma-Ray radiation. Is there an energy gap between radio and X-Ray? I have never seen an antenna that shoots yellow light. However, a number of novel techniques have emerged in the last 20 years and I wold doubt any proof. I recall a proof that the X-Ray LASER was impossible, then Dr. Teller showed an arrangement of tungsten rods pumped by a nuclear explosion. You never know.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/113434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Measurement using Vernier Callipers
The diameter of a cylinder is measured using a Vernier Callipers with no zero error. It is found that the zero of the Vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The Vernier scale has 50 divisions equivalent to 2.45 cm. The 24th division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is ...
My attempt
1 division of vernier scale is equivalent to 50/2.45 cm.
Least count of main scale = 0.05cm Thus Vernier constant = 0.5-(50/2.45) cm
Surprisingly this is negative.?? I don't really know how to solve these kinds of problems. Please help me.
| When the zero lies between 5.10 and 5.15, it means that the diameter is more than 5.10 and less than 5.15. The length of the vernier scale is not important. Since the 24th mark (out of 50) lines up with another mark, then the additional amount is 24 parts of 0.05, since each mark is 0.05 (5.15 - 5.10) cm. The reading of the vernier is 5.124 cm. Google "reading a vernier" if you need more assistance.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/113506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Examples of Weyl transforms of nontrivial operators I've been able to find examples of Weyl transforms of operators like $\hat{x}$,$\hat{p}$, and $\hat{1}$, but not anything more complicated. Are there derivations of the Weyl transforms of more complicated operators, like the Hamiltonians of the Hydrogen atom or harmonic oscillator?
| The Wigner-Weyl transform of a function $f(x,p)$ is given by,
$$\Phi[f]= \frac{1}{4\pi^2}\iiiint f(x,p) \exp \left[ i \left( a(X-x)+b(P-p)\right)\right] \, dx\, dp\, da \, db$$
As you suggested, let us take the Hamiltonian of the harmonic oscillator, i.e.
$$\Phi[H]=\frac{1}{8m\pi^2}\iiiint p^2\exp \left[ i \left( a(X-x)+b(P-p)\right)\right] \, dx\, dp\, da \, db \\ + \frac{m\omega^2}{8\pi^2}\iiiint x^2\exp \left[ i \left( a(X-x)+b(P-p)\right)\right] \, dx\, dp\, da \, db$$
We concern ourselves with the last integral, as they are more or less analogous. As the first integration is over $x$, we may applying integration by parts and ignore $p$:
$$\frac{m\omega^2}{8\pi^2}\iiint \frac{1}{a^3}e^{ia(X-x)+ib(P-p)}(ia^2 + 2ax-2i) \, dp \,da \, db$$
Integrating with respect to $p$ is trivial:
$$\frac{m\omega^2}{8\pi^2}\iint \frac{b}{a^3} e^{ia(X-x)+ib(P-p)}(ax^2 +2-i2ax) \, da \, db$$
With the help of Mathematica 9, we may express the subsequent integral over $a$ in terms of a polyanomial, and the exponential integral function:
$$\frac{m\omega^2}{8\pi^2}\int \, b e^{ib(P-p)} \, \left[ (X-x)((ix^2+4x)-2X)\mathrm{Ei}(ia(X-x)) \\ -\frac{1}{a}e^{ia(X-x)}(i(X-x)+(x^2-i2x)+1) \right] \, db$$
The integral over $b$ is also trivial, as the integrand only features $b$ in the form $be^{b\dots}$ Hence,
$$-\frac{m\omega^2}{8\pi^2(P-p)^2} \left[ (X-x)((ix^2+4x)-2X)\mathrm{Ei}(ia(X-x)) -\frac{1}{a}e^{ia(X-x)}(i(X-x)+(x^2-i2x)+1) \right]e^{ia(X-x)+ib(P-p)}\left( ib(P-p)-1\right)$$
Apply the same procedure to the original first integral, combine the two, etc.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/113604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Photobombing without app used by Harley Davidson Here we see a photobombing trick, used by Harley Davidson.
They achieve to incrust a logo on a photo you take. The logo is invisible with naked eyes and only revealed on the photo you take.
How does it work?
| The camera in a smartphone is sensitive to wavelengths of light that the human eye cannot see. Have a look at this picture from this article:
This is actually a Canon EOS 40D SLR, but the idea is the same. The human eye can't see any wavelengths longer than about 700nm, but the graph shows the camera can detect light out to 1000nm.
So if you project your message using light with a wavelength of, for example, 900nm your eye won't be able to see the message but the camera will, and it will appear on the photo.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/113716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 1,
"answer_id": 0
} |
What would cause a spinning fluid to stop spinning? I once saw a demonstration where an electric current caused a drop of mercury to spin. The drop contained bits of iron, which could be seen flowing around in a circular pattern. As soon as the current was turned off, the spinning slowed fairly quickly. What caused the circular motion within the drop to slow? It seems to me that there would be very little friction within the drop, and that the motion should be similar to that of a gyroscope. Why was this not the case?
| Friction forces seem much stronger on small objects than expected from experience. If the momentum of the moving object is not larger than the average (thermal) momentum of the surrounding material, it will get stopped quickly.
Some funghi can eject their spores at supersonic speed, but still these spores have a reach of only about 20 cm. That is because their momentum is small due to their small mass.
My guess is a similar thing is going on in your drop of mercury. The iron bits are very small and therefore do not carry a lot of momentum. What little momentum got them spinning got quickly lost in thermal noise, once the driving force was turned off.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/113979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
What technological advance would allow LEP3 to surpass LEP2? I learned that for electron accelerators synchrotron radiation and acceleration are the limiting factors.
This article, that I found in one answer to this question mentions that one would not use the superconducting acceleration elements that were developed in the meanwhile, but rely on regular normal-conduction acceleration.
Then I wonder, at the same radius, what technological advance would allow LEP3 to surpass LEP2 in a significant ammount?
| There is nothing really surprising here. The limit is basically how much energy you can add to the beam in the space left after you have installed all your bending and tuning magnets.
Both magnets and cavities are better now than they were then. More over, the size of both items has gotten slightly smaller allowing more of them to be packed into the same distance.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/114077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Road to String Theory I have a question for our theoretic SuperUsers. How much knowledge and which fields of physics you have to know to start studying string theory? I am now on QFT, after I think I'll start study Supersymmetry or maybe special cases of QED. How much more should I know to start Strings? :)
| You need to master algebraic geometry (in particular derived functor cohomology of sheaves, say EGA 1-3), algebraic topology (characteristic classes, say the entire book of Spanier), commutative algebra and algebra in general (say Lang's book and A/M), and also homological algebra (the entire book of Weibel).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/114166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Which Force would help find the mass? Two masses, A and B, are connected to a rope. A constant upward force 86.0N is applied to box A. Starting from rest, box B descends 12.1m in 4.70s . The tension in the rope connecting the two boxes is 32.0N.
What is the mass of B?
What is the mass of A?
My work and what I am struggling with:
I am trying to find the acceleration experienced by B, with that I will find the mass.
I find the acceleration using:
$$2\Delta_y/t^2=a$$
$$a=1.0955m/s^2$$
Then I have a problem for B do I use this equation:
$$\sum F_y =32-M_b*g=M_ba$$
$$Or$$
$$\sum F_y =86-M_b*g=M_ba$$
| Split the rope and see what force are acting on B
$$T-M_b g = M_b a$$
and on A
$$F-T-M_a g = M_a a$$
Since you know $T=20\,{\rm N}$, $F=86\,{\rm N}$, and you have calculated $a=-1.09955\,{\rm m/s^2}$ it is trivial to find $M_a=\frac{F-T}{a+g}$ and $M_b = \frac{T}{a+g}$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/114354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What's the efficiency of real heat engines? Real heat engines always have lower efficiency than the Carnot efficiency.
I wonder how efficient real engines can be? Can their efficiency get anywhere near the Carnot-limit?
| The efficiency of heat engine cycle greatly depend on the individual processes that make up the cycle are executed. Thus cycle efficiency can be maximized by using the processes that require the least amount of work and deliver the most that is by using reversible process.
In 1824 French engineer Sadi Carnot proposed best reversible cycle is Carnot cycle.In case Carnot engine the heat addition and rejection from the reservoir or a heat source to the engine and from engine to sink is at constant temperature and reversible manner i.e. Reversible Isotherm condition due to which the total heat supplied is utilized or converted to the work with the help of engine and entropy generation is zero. Remaining Expansion and Compression process are also by reversible manner.
While in normal heat engine the process are become irreversible due to phenomenon like friction, vibration,heat losses etc. so the heat losses are more when it is compared with the carnot heat engine. Due to which Real heat engines always have lower efficiency than the Carnot efficiency.
A real engine efficiency can be increased by
1) Temperature at which heat is added i.e. source temperature so that losses are less as entropy generation is less.
2) Lowering the temperature of sink i.e temperature at which heat engine is rejecting heat
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/114539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
What does the work on a current carrying wire in a Magnetic Field? We consider that the force acting on a current carrying wire placed in a uniform magnetic field perpendicular to the length of the wire is given by $IBl$. If the wire moves by a distance $x$ in a direction perpendicular to its length the raise in its kinetic energy is $IBlx$
Now basically the magnetic forces are acting on the electrons moving inside the wire and is always perpendicular to the instantaneous velocity so it cannot perform any work on it and normals are also unable to provide any net work done on the system, so what provides the change in kinetic energy of the system?
| Your wire must be part of an electrical circuit (no closed loop = no current). Let us assume that the "return wire" is far away, where there is no magnetic field. So we have a large loop.
Now the motion of the wire in the magnetic field will induce a current that will oppose the magnetic flux change. What happens is that the magnetic field "seen" by the wire becomes less - and since the energy stored in the magnetic field goes as the integral of $B^2$ over the volume, there will be less energy in the magnetic field.
So the "source" of the energy of the wire is the energy that used to be stored in the magnetic field.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/114627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Is Newton's third law always correct? Newton's third law states that every force has an equal and opposite reaction. But this doesn't seem like the case in the following scenario:
For example, a person punches a wall and the wall breaks. The wall wasn't able to withstand the force, nor provide equal force in opposite direction to stop the punch.
If the force was indeed equal, wouldn't the punch not break the wall? I.e., like punching concrete, you'll just hurt your hand. Doesn't this mean Newton's third law is wrong in these cases?
| (As suggested in Classical Mechanics by H. Goldstein, 3rd edition in chapter 1 : Survey of the Elementary Particles )
No, the third famous law is not always valid. As pointed out above, in the case of electromagnetism, take for an example, two charged particles A and B are in motion.
B is just travelling perpendicular to the path of A and is right on the axis of A's motion.
You can calculate Coulomb's force for one due to another. But try finding the magnetic force due to one on the other. You will find the Lorentz force (sum of electric and magnetic forces) on one is not equal to the other.
Voila! Newton's third law violated!!
Well, if the fields concept is taken into account, and it has to be, then the third law is improved and protected: no violation. But excluding that, we can say it is not a strong law.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/114669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 14,
"answer_id": 9
} |
Why is the ratio of velocity to the speed of light squared in the Lorentz factor? Why is the ratio of velocity to the speed of light squared in the Lorentz factor?
$${\left( {{v \over c}} \right)^2}$$
My only guess is the value must be positive.
| It's due to the Lorentz invariance of the spacetime interval:
$$\Delta s^2 = (c\Delta t)^2 - (\Delta x)^2 = (c\Delta t')^2 - (\Delta x')^2$$
Assume that, for example, $\Delta x = 0$ such that $\Delta t$ is the elapsed time according to a clock stationary in the unprimed frame of reference.
Thus
$$(c\Delta t)^2 = (c\Delta t')^2 - (\Delta x')^2$$
With a little algebra, we have
$$ \frac{(c\Delta t)^2}{(c\Delta t')^2} = 1 - \frac{(\Delta x')^2}{(c\Delta t')^2} = 1 - \frac{v^2}{c^2}$$
where
$$v = \frac{\Delta x'}{\Delta t'}$$
is the velocity of the clock in the primed frame of reference.
A little more algebra yields
$$\Delta t' = \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} = \gamma \Delta t$$
which is the familiar time dilation formula
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/114913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 4
} |
Photons and perfect mirror A perfect mirror means, that all the photons which collided with the mirror will be reflected in the same amount, with the same energy and with the same - except sign - angle. Will the mirror get an impulse from the photons?
| Yes it will.
Assuming the light is incedent normally the change in the photon momentum is $2h\nu/c$, and consequently the momentum of the mirror will change by the same amount.
If the mirror is free to move it will be accelerated by the light and as a result the light will be slightly red shifted. There is more discussion of this in Can relativistic momentum (photons) be used as propulsion for 'free' after the initial generation? though the question is not an exact duplicate.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/115031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Why does magnet attract iron but not other metals? While searching, i found this page: http://timesofindia.indiatimes.com/home/stoi/Why-does-a-magnet-attract-iron/articleshow/4298171.cms but it does not have full explanation. So Please tell full explanation of why this happens?
| Materials (including some non-metals) that are strongly attracted to magnets are known as ferromagnetic. If you Google for this, or just search this site, you'll find lots of articles on this subject, thoughly surprisingly I don't think the question how does ferromagnetism arise has been asked before.
Electrons have a magnetic moment so they interact with magnets. However in most solids the electrons tend to line up in pairs so their magnetic moments cancel out. This means the solid has no net magnetic moment and doesn't interact strongly with magnets. However, in a small number of solids the outermost electrons of the atoms line up parallel with each other and their magnetic moments reinforce each other to give the solid a large net magnetic moment. These solids interact strongly with magnets, and we call them ferromagnetic.
Only solids having unpaired electrons can be ferromagnetic, but only a small fraction of these solids are actually ferromagnetic. For example iron is ferromagnetic but manganese isn't, even though both metals contain unpaired electrons. Whether a solid will be ferromagnetic is exceedingly hard to predict because it is controlled by a fine balance between forces in the solid. The Wikipedia articles on ferromagnetism and the exchange interaction go into some details, but be warned that this is a complex area for the non-physicist.
There is a list of known ferromagnetic materials here.
As Hasan mentioned in a comment all solids interact with magnetic fields to some extent. Non-ferromagnets may be diamagnetic or paramagnetic. However these interactions are several orders of magnitude weaker than ferromagnetism and the interaction is too weak to be measured outside a laboratory.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/115181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
Is it more efficient to stack two Peltier modules or to set them side by side? Is it more efficient to stack two Peltier modules or to set them side by side?
And why?
I have a small box that I want to cool down about 20 K below ambient -- cold, but not below freezing.
(I want to keep my camera cool, so I'm putting in this cool box.
The camera looks through a flat glass window on one side of the box).
The heatsink I have on hand is about twice as wide as the widest Peltier module I originally planned on using.
So there's room to put 2 Peltier modules side-by-side under the heatsink.
Or I could center a stack of 2 Peltier modules under the heatsink.
Which arrangement is more efficient?
I have to cut a bigger hole in the insulation for the side-by-side arrangement, so the unwanted heat that "back-flows" through the side-by-side arrangement is worse.
On the other hand, other effects are worse for the stacked arrangement.
(Is https://electronics.stackexchange.com/ a better place to post questions about Peltier coolers?)
| For a 20 degree Celsius differential, parallel devices will be more efficient, if the devices are operating in their normal range of heat removal. If your camera box is extraordinarily well insulated and you can wait for hours and hours for it to cool down, then a series arrangement with both devices running at less than about 10% of maximum power is most efficient.
The reason has to do with the graphs supplied by the manufacturer of your module. Typically you can get efficiencies of better than 200% for delta T of 10 degrees at low heat flow, so that's 100% for 2 in series. Double the heat flow and the required power almost quadruples, so efficiency is 110% per module or 55% overall.
In parallel, 2 devices with a delta T of 20 degrees might be 75% efficient at the total heat flow of the first case in the previous paragraph. With the doubled heat flow of the second case, the efficiency might be 85%.
Running Peltier devices at near their rated power is very inefficient and requires a lot of care to prevent thermal runaway and device destruction. Avoid it if possible.
Heat sinks on both sides need to be very effective, and you should consider a fan for the hot side, vibration isolated from the camera.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/116271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 5,
"answer_id": 3
} |
Inertial navigation system: am I doing it wrong? I'm trying to develop an inertial navigation system.
I can access data from an accelerometer sensor (acceleration on three axes) and gyroscope sensor (angular velocity on three axes).
First of all, I integrate my angular velocity data with respect of time, and get angles on all three axes at every moment ($x \to \phi, y \to \theta, z \to \psi$).
Then I feed the angles to this rotation matrix
and I use it to rotate my acceleration vector at that time, thus taking all my acceleration values on the same reference frame.
Finally, I integrate acceleration to get space travelled, using the simple formula
$$s(t) = \frac12 a(t) t^2 + v(t - 1)t + s(t - 1)$$
My method seems to work fine with fake data, but performs really bad when I plug in the real data, the output is almost meaningless.
Am I doing something wrong with the math or I have to search the problem in my implementation?
| One possibility is that your approach is hugely sensitive to measurement uncertainty: integrating noisy signals can be a huge problem. You might think about ways to average the measurements over time, so that they're (hopefully!) more stable.
Another possibility is that your instrument doesn't output the data in quite the format that you think. Have you verified that your inputs are sensible?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/116350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding out the Potential difference
Find the potential inside and outside a uniformly charged solid sphere whose radius is $R$ and whose total charge is $q$. Use infinity as your reference point. Compute the gradient of $V$ in each region, and check that it yields the correct result.
The above one is a question cropped from Griffiths page no 82.
When I try to solve the problem they used a potential equation inside the sphere
My problem is in the right part. Why do we need to consider the next part? we have been asked to solve potential inside a charged sphere. so can't I use only the limit $\infty $ to $ R$?
I am saying this because, i have considered the infinity as the reference point and R as the point where I will find the potential.
Is it because we need to add the potential outside of the sphere and inside?
| Considering $R$ as the radius of the sphere, The work done in taking a point charge from $\infty$ to $r(0\le r\le R)$ will involve moving the charge through the field outside the sphere and also indside from $R$ to $r$ (Note that is a solid sphere and field will also be present inside it).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/116478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Mechanical waves edge between material and vacuum I have been thinking about the propagation of EM waves vs. mechanical waves and some of their odd cases. One such case that I haven't been able to puzzle out is what happens when a mechanical wave reaches the end of a medium (such as the outer layer of the atmosphere) and the beginning of a vacuum - outer space itself.
Edit: for clarity, I created a simple image demonstrating what I thought should happen:
Basically, my theory was that the "edge" particles (or particles with a large amount of space between them) would continue into space with perpetual motion as the force applied to them is not counteracted by particles in front of them.
My question is, after the wave has affected the last molecules on the edge of a vacuum, what happens to it, and, do the molecules on the very edge continue to move indefinitely (or do they return to equilibrium)?
| the materials are elastic, so when it reaches the end it would reflect, because the last layer won't have other layer to collide and the elastic properties of the material would make it bounce back, making the wave reflect back. Unless the wave energy is sooo large that would break the material and in that case the particles would indeed leave the material. if you have a bucket of water and drop a tiny piece of something, it would make a wave, but no water would separate, now if you drop something really big, some part of the water would separate from the other part.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/116581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Normal reaction - force without acceleration When a body lies on the surface of the Earth it is under the influence of gravity. The force on the body due to gravity causes it to exert a force on the ground and the normal reaction acts in the opposite direction causing the resultant force on the body to be zero.
However, how can the body exert a force on the ground when it does not have any acceleration? Since force equals mass times acceleration how does a body without acceleration experience a force?
| Newton's second law of motion states this (verify on Wikipedia):
$$\vec F_{net} = m\vec a$$
Here, $\vec F_{net}$ is zero, so $\vec a$ is zero too. Going back in reverse (what you did in question), $\vec a = \frac{\vec F_{net}}{m}$ can only deduce that the body is experiencing no net force. That's it.
Feel free to use Law of Gravitation, Coulumb's Law etc when the force in question isn't the sole cause of the effect. Newton's second law of motion can't help here.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/116632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Dielectric boundary I am trying to determine why electric field may be confined to a certain region if there is a large difference in the permitivity for example if electric field flows through water and then reaches a water air boundary.
I have also been reading about EM waves, is it possible to model electric field as a wave because if so then the transmission T and reflection R coefficients given below in terms of n which is $\propto \epsilon_{r}^{1/2} $ and so if $\epsilon_{1}>>\epsilon_{2}$ then $n_{1}>>n_{2}$ ;
$\displaystyle R$ $\textstyle =$ $\displaystyle \left(\frac{n_1-n_2}{n_1+n_2}\right)^2, \rightarrow 1$
$\displaystyle T$ $\textstyle =$ $\displaystyle \frac{n_2}{n_1}\left(\frac{2 n_1}{n_1+n_2}\right)^2 \rightarrow 0.$
and so it is clear that the wave is reflected at the boundary, is this approach valid ?
http://farside.ph.utexas.edu/teaching/em/lectures/node103.html
| The answer to the second part of your question is yes your approach is OK. I have gone through the link you have posted and want to say that the inferences drawn from formulae written above are for non collisional case i.e. dielectric constant $\epsilon$ is real. However if $\epsilon$ is complex you can not draw the same conclusion from above formulae.
For the first part of your question when you deal with static fields there is no reflection because the fields are not propagating.
High dielectric constant (para-electric medium) enhance the electric field inside the medium via alignment of molecular dipoles. If the dielectric constant of one medium is much higher than that of other most of the electric field strength will appear to be trapped inside that medium. In reality the field inside the medium is enhanced via its para-electric properties.
I hope it will help
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/116702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What will happen after escaping earth's gravitational field? Suppose that I escaped the gravitational field of earth.
Then: am I going to be pulled by Sun's gravity?
| When at distance $R$ from the center of earth, the minimum velocity needed to escape from earth's gravity is $v_{esc} = \sqrt{2GM/R}$. Here, $G$ denotes Newton's gravitational constant, and $M$ earth's mass.
As should be obvious from this equation, regardless the distance $R$, the escape velocity never vanishes. So no matter how far you travel, you never escape earth's gravity!
Some figures to put things in perspective:
Using $2GM = 7.9 *10^{5} m^3 s^{-2}$, it follows that at one earth radius distance ($6.37 *10^6 m$) from earth's center (i.e. at earth's surface), the escape velocity equals $11 km/s$.
You have to fourfold this distance (be four earth radii away from earth's center) to half this escape velocity. Even when being a moon's distance (60 earth radii) away from earth, the escape velocity still is more than $1/8$ of that at earth's surface.
Although regardless your distance your escape velocity never falls to zero, you might define the 'escape distance' as the distance beyond which the Hubble expansion $H_0 R$ exceeds the escape velocity $\sqrt{2GM/R}$. Physically, this corresponds to the expansion of space trumping the gravitational attraction. It follows that the escape distance thus defined is given by $R_{esc} = {(2GM/H_0^2)}^{1/3}$.
Using $H_0 = 1.62 * 10^{-18} s^{-1}$, we find for earth an escape distance equal to $6.7 * 10^{16} m$ or $7$ lightyears. A considerable distance given that no human has travelled farther from earth than 1.3 light seconds!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/116772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Are diffeomorphisms a proper subgroup of conformal transformations? The title sums it pretty much. Are all diffeomorphism transformations also conformal transformations?
If the answer is that they are not, what are called the set of diffeomorphisms that are not conformal?
General Relativity is invariant under diffeomorphisms, but it certainly is not invariant under conformal transformations, if conformal transformations where a subgroup of diff, you would have a contradiction. Or I am overlooking something important?
| A general diffeomorphism is not part of the conformal group. Rather, the conformal group is a subgroup of the diffeomorphism group. For a diffeomorphism to be conformal, the metric must change as,
$$g_{\mu\nu}\to \Omega^2(x)g_{\mu\nu}$$
and only then may it be deemed a conformal transformation. In addition, all conformal groups are Lie groups, i.e. with elements arbitrarily close to the identity, by applying infinitesimal transformations.
Example: Conformal Group of Riemann Sphere
The conformal group of the Riemann sphere, also known as the complex projective space, $\mathbb{C}P^1$, is called the Möbius group. A general transformation is written as,
$$f(z)= \frac{az+b}{cz+d}$$
for $a,b,c,d \in \mathbb{C}$ satisfying $ad-bc\neq 0$.
Example: Flat $\mathbb{R}^{p,q}$ Space
For flat Euclidean space, the metric is given by
$$ds^2 = dz d\bar{z}$$
where we treat $z,\bar{z}$ as independent variables, but the condition $\bar{z}=z^{\star}$ signifies we are really on the real slice of the complex plane. A conformal transformation takes the form,
$$z\to f(z)\quad \bar{z}\to\bar{f}(\bar{z})$$
which is simply a coordinate transformation, and the metric changes by,
$$dzd\bar{z}\to\left( \frac{df}{dz}\right)^{\star}\left( \frac{df}{dz}\right)dzd\bar{z}$$
as required to ensure it is conformal. We can specify an infinite number of $f(z)$, and hence an infinite number of conformal transformations. However, for general $\mathbb{R}^{p,q}$, this is not the case, and the conformal group is $SO(p+1,q+1)$, for $p+q > 2$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/116853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
} |
Does the general topology of Minkowski space-time change under a Lorentz transformation? Does the general topology of Minkowski space-time change under a Lorentz transformation? Open balls in $\mathbb{R}^{4}$ (with the standard topology) are not invariant under Lorentz transformations. Does this mean for example that observers in one reference frame would have different notions of convergence, continuity etc?
Note: I'm asking from the perspective of a curious undergraduate who just finished introductory analysis and modern physics. Please answer appropriately.
| This is simply a short addendum to Robin Ekman's answer and response to your comment
So I could phrase it this way: Lorentz transformations are homeomorphisms, so even though they open sets not invariant, all topological notions are still preserved?
Homeomorphism is indeed the key concept here, and I wish to add a very slight nitpick with Robin's answer so that there is no risk of your being confused in contexts other than SR/GR (for example, in quantum mechanics): you also need the information that Minkowski spacetime is finite dimensional to infer continuity from linearity and homeomorphism from linear and invertible (which of course is a given in Minkowski spacetime). In infinite dimensions, not all linear maps are continuous: witness the Dirac delta on $\mathcal{L}^2(\mathbb{R})$ for example. The difference between the concepts of general linear and the strictly more specialised "linear continuous" in, say the standard, countably infinite dimensional Hilbert space $\mathcal{L}^2(\mathbb{R})$ is actually the very reason for being of distributions and the framework of rigged Hilbert space for talking about them; see my answer here for more information.
As I said, none of this is any worry in classical relativity. Linear and linear continous are the same notions in this field.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/116910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
A strange audio phenomenon, could there be a physical interpretation to it? https://mathoverflow.net/q/165038/14414
Motivation : Here is a motivation as to why this problem is so important.
Let $f(t)$ be an audio signal. We can safely asume it to be bandlimited to 0-20kHz as we cannot hear anything above that. Capture this signal in digital computer with appropriate sampling frequency and denote it as $f[n]$.
Now take Discrete Hilbert transform of $f[n]$ to get $f_h[n]$, (using the code $f_h$ = imag(hilbert(f)); in Matlab).
Compute the signal $f_{\theta}[n] = f[n]\cos\theta + f_h[n]\sin\theta$ for any value of $\theta$, then listen to the signal with different values for $\theta$.
They all sound exactly identical.
Similarly our $MI_{\omega_0,\omega_1}(t)$ is same for all $f_{\theta} = f\cos\theta + f_h\sin\theta$, for any value of $\theta$.
Question :
just try it. $<f,f_h> = 0$, they why do they produce same effect in the listner? Is it some quantum mechanical effect gone wrong?
Added :
Also see this metric space : metric space
I've recently filed a patent using this metric with a slight change, instead of arccos i used sqrt(2(1-cos(theta))), which makes it a Hilbertian metric. I had then embedded this metric space into an Hilbert space isometrically, to model using vectors.
MATLAB code :
[f,fs] = wavread('audio_file.wav');
fh = imag(hilbert(f));
theta = pi/4;
f_tht = fcos(theta) + fhsin(theta);
wavplay(f,fs);
wavplay(f_tht,fs);
| You can express your signal as the series expansion:
$$f(t) = \sum_k a_k \cos(kt) + b_k \sin(kt)$$
The Hilbert transform is a linear operator, so:
$$f_h = \sum_k a_k H[\cos(kt)] + b_k + H[\sin(kt)] = \sum -a_k \sin(kt) + b_k \cos(kt)$$
So, $f_h$ has changed the phases f the different frequencies, but leaving the mangitudes unchanged. You hear the same thing because your ear is doing a Fourier Transform of the input (each frequency is detected by a different part of the cochlea), that remains invariant (modulo some phase) under Hilbert Transform.
What your transformation is doing is shifting each frequency by a a certain phase, but not affecting the magnitudes.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/116957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why doesn't a block rotate due to friction? In a horizontal surface, a block (cube) is sliding due to a sudden push. When the block slides, there is frictional force which is acting on the block.
Frictional force will have a torque around the center of mass, so why does the block not rotate/roll around (a horizontal axis through) the center of mass instead of sliding?
| The image below shows a rough diagram of the forces on a box just as it would begin to rotate.
In this case we can consider the normal force to the surface as focused on the leading corner about which the box would rotate.
This results in 2 torques on the box. The first from friction which is causing the box to rotate $$\frac{L}{2}Fr$$
and the second due to the normal force which is in the opposite direction. $$\frac{H}{2}mg$$.
If the first is bigger the box will at least begin to rotate. Note it may not actually flip as the ratio can change as the box decelerates and begin to rotate. This will be true if $$LFr > Hmg$$ using the standard friction law we get $$L\mu mg > Hmg $$ $$H > L\mu$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/117010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Moment of inertia of a cylinder When I tried to calculate the moment of inertia ($I_C$) of a cylinder (mass M, height H, radius R) around the rotating axis going symmetrically through its middle, I came up with a different result than expected ($\frac{1}{2}MR^2$), but I do not spot my mistake, since my calculation makes perfect sense to me:
$$
I_C := \int_V{ρr^2dV} = \int_0^H{\int_0^{2π}{\int_0^R{ρr^2 dr dφ dh}}} = ρ \int_0^H{\int_0^{2π}{\frac{R^3}{3} dφ dh}} = ρ \cdot 2πH \frac{R^3}{3} = Vρ\frac{R^2}{3} = \frac{1}{3}MR^2
$$
Can anyone spot what's wrong?
| You are missing an r in the dV. So it's $dV=r dr d\phi dh$ and in the integration you get $I_C=\int_{0}^{H}dh\int_{0}^{2\pi}d\phi \int_{0}^{R}\rho r^3 dr =2\pi H \rho R^4/4=\frac{2}{4} \rho (\pi R^2 H) R^2=\frac{1}{2}M R^2$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/117151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Tritium decay is spontaneous even if the binding energy of tritium is higher than the binding energy of 3He. Why? Given this nuclear reaction:
$^3_1\mathrm H\to {}^3_2\mathrm{He}+e^-+\bar{\nu}$
and knowing the binding energies:
$BE(^3_1\mathrm H)=8.48 \,\mathrm{MeV}$
$BE(^3_2\mathrm{He})=7.72 \,\mathrm{MeV}$
If I calculate the mass defect (obviously considering the binding energies in the mass calculation) I obtain a positive value:
$M(^3_1\mathrm H)c^2=2809.08 \,\mathrm{MeV} > M( ^3_2\mathrm{He})c^2+M(e^-)c^2=2808.991 \,\mathrm{MeV}$
as expected for a spontaneous decay.
Considering the binding energies I have written above I expect the $^3\mathrm{H}$ to be more stable than $^3_2\mathrm{He}$.
My question is: why does this decay occur?
| Binding energy simply isn't the right metric (because it is calculated from different starting points on account of the differing masses of the constituent nucleons).
Proper energy (AKA mass) of the states is the right metric.
Wolfram Alpha gives the masses as
$$M_{\mathrm{T}} = 2809.432 \,\mathrm{MeV}$$
$$M_{^3\mathrm{He}} = 2809.413 \,\mathrm{MeV}$$
In other words, there is about 19 keV to be had in this decay.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/117245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
What is is a molecular/microscopic explanation for why a balloon rises in water? If we consider a balloon full of air submerged in water then we all know that it will rise rapidly. I am having trouble understanding this at the level of individual molecules of air and water. What is a molecular/microscopic explanation for this phenomenon?
| I will give it a try but remember I am new here.
Pressure in the end is brownian motion, therefore more pressure more motion, more total force. Therefore left and right forces are compensated, left and right recoils are compensated.
What is not compensated is on the vertical axis therefore the water above the balloon will push it less downwards, than the water below the balloon.
The total difference is gravity or better the weight of the water above the balloon minus the weight of the water below the balloon.
Note that even if we consider an infinite quantity of water above and below, the total difference is substantially given by the size of the balloon, but in water.
Therefore to recap the total difference is the weight in water of the same size of the baloon. Therefore there is less weight above and less molecules above that push down the baloon with their recoils.
Something similar is explained here http://en.wikipedia.org/wiki/Archimedes%27_principle
The subtraction of infinities may be an interesting curiosity for some readers.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/117349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is tea weaker if you make it in a half full cup of water? Imagine you put a tea bag in a cup and half fill it with boiling water. Then after one minute you take the tea bag out and then fill the cup up to fill with hot water. Will the tea be weaker than if you had filled the water up to the normal level with boiling water in the first place and then left it for one minute before taking the tea bag out?
Intuitively I can make an argument that it should be but is there a mathematical description for how much weaker it would be?
| In answer to your question - the math you seek is somewhere in the Noyes-Whitney equation as pointed out in the following link analysis of teabag dynamics unfortunately the author doesn't expressly calculate the rate of dissolution but the ammount of detail will either A/ give you a good start on figuring it out - or B/ help you decide you don't need to know the exact answer that bad.
My take on it is that the rate of dissolution starts the same in both cases and the value for the bulk mass concentration [Cb] will increase faster for the half cup slowing its dissolution rate faster.
To determin a definative answer you would need to integrate the rate of dissolution for both cups to find out the total ammount of tea compounds delivered into the water on your calculator.
My gut says that brewing in the full cup will be stronger.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/117414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Moon's pull causes tides on far side of Earth: why? I have always wondered and once I even got it, but then completely forgot. I understand that gravity causes high and low tides in oceans, but why does it occur on the other side of Earth?
| Let us try to find the acceleration at points A and B with respect to the centre of the earth O due to the influence of moon and earth, as shown in the figure.
O and X are the center of the earth and moon respectively. Let the radius of earth be $R_E$, distance between earth and moon be $d$, mass of earth and moon be $m_E$ and $m_M$ respectively. $O$ to $X$ is taken as the positive direction. I have assumed $R_E << d$.
Acceleration of point B $a_B$ is:
$$a_B = -\frac{Gm_E}{R_E^2} + \frac{Gm_M}{(d-R_E)^2}$$
$$= -\frac{Gm_E}{R_E^2} + \frac{Gm_M}{d^2}(1+ \frac{2R_E}{d})$$
Similarly, $a_A$ is:
$$a_A=\frac{Gm_E}{R_E^2} + \frac{Gm_M}{(d+R_E)^2}$$
$$=\frac{Gm_E}{R_E^2} + \frac{Gm_M}{d^2}(1-\frac{2R_E}{d})$$
And $a_O$ is:
$$a_O = \frac{Gm_M}{d^2}$$
Thus, the accelerations of point A and B with respect to O are:
$$a_{AO} = a_A-a_O = \frac{Gm_E}{R_E^2} - \frac{2Gm_MR_E}{d^3}$$
$$a_{BO} = a_B-a_O = -\frac{Gm_E}{R_E^2} +\frac{2Gm_MR_E}{d^3}$$
But now, we get $a_{BO}=-a_{AO}$, which means that on both the sides, water will be trying to move away from the centre of the earth, thus causing tides on both the sides of the earth.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/118460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "80",
"answer_count": 8,
"answer_id": 3
} |
What does an applied force on an already accelerating object do? I'm a big beginner having only taken Grade 11 high-school physics.
Imagine this situation: an object is accelerating north, and while it's accelerating, it's 'hit' by a northward force.
To me it would make sense that the acceleration of the object would increase - something like this:
aNew = aOld + (f/m)
I got the (f/m) from f=ma.
But then think of gravity - while an object is accelerating downward at 9.8m/s2, it has the force of gravity pulling it too. So then, based on my previous logic, it's acceleration would constantly be increasing!
aNew = aOld + (fGravity/mass)
And I know that you don't increase your acceleration while you fall, so now I'm confused.
In a nutshell: What exactly are the rules for how an already accelerating object reacts to forces? Am I wrong that your acceleration would increase?
EDIT: I realized I've made a big mistake; I never knew that the acceleration drops to 0 the second the net force is 0. This makes sense now.
| The acceleration of an object is a result of the sum of all the forces. If it was accelerating before you hit it, there must have been a force. If an object is falling in air, there are forces of gravity and air friction on it. When all forces even out the object continues with the same velocity.
Simply take the vector sum of all forces - that will tell you what is going on.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/118597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Entropy was created after inflation? I'm puzzeled by a statement in Big Bang Cosmology-review about the reheating phase subsequent to the exponential expansion during inflation:
In this reheating process, entropy has been created and the final value of $RT$ is greater than the initial value of $RT$.
(Taken from section 21.3.5. on page 17.)
How can I reconcile this with the first law of thermodynamics...
| A form of the first law of thermodynamics:
In a thermodynamic process involving a closed system, the increment in the internal energy is equal to the difference between the heat accumulated by the system and the work done by it.
It is a form of conservation of energy.
Thermodynamics developed and has been validated as a theory in the framework of Newtonian mechanics. The Big Bang model, particularly at the inflationary period is dominated by the General Relativity framework. In General Relativity energy conservation is moot,
In general relativity conservation of energy-momentum is expressed with the aid of a stress-energy-momentum pseudotensor. The theory of general relativity leaves open the question of whether there is a conservation of energy for the entire universe.
One can find a number of discussions on this, and the limits of how energy conservation can be defined.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/118674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Gravitational Constant in Newtonian Gravity vs. General Relativity From my understanding, the gravitational constant $G$ is a proportionality constant used by Newton in his law of universal gravitation (which was based around Kepler's Laws), namely in the equation $F = \frac{G\cdot M\cdot m}{r^2}$. Later, Einstein set forward a different theory for Gravity (based around the equivalence principle), namely General Relativity, which concluded that Newton's law was simply a (rather decent) approximation to a more complex reality. Mathematically speaking, Einstein's Theory was completely different from Newton's Theory and based around his Field equations, which also included $G$ in one of it's terms.
How come two different theories that stemmed from completely different postulates end up having this same constant $G$ with the same numerical value show up in their equations? What exactly does $G$ represent?
| Since in the limit of weak gravitational fields, Newtonian gravitation should be recovered, it is not surprising that the constant $G$ appears also in Einstein's equations. Using only the tools of differential geometry we can only determine Einstein's field equations up to an unknown constant $\kappa$: $$G_{\mu\nu} = \kappa T_{\mu\nu}.$$ That this equation should reduce to the Newtonian equation for the potential $\phi$, $$\nabla^2 \phi = 4\pi G\rho \tag{1}$$
with $\rho$ the density fixes the constant $\kappa = \frac{8\pi G}{c^4}. \tag{2}$
In detail, one assumes an almost flat metric, $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$ where $\eta_{\mu\nu}$ is flat and $h_{\mu\nu}$ is small. Then from writing down the geodesic equation one finds that if $h_{00} = 2\phi/c^2$, one obtains Newton's second law, $$\ddot{x}^i = -\partial^i \phi. \tag{3}$$
Using (3) and taking $T_{\mu\nu} = \rho u_\mu u_\nu$ for a 4-velocity $u_\mu$ with small spatial components, the $00$ component of the field equations (2) is $$2\partial^i \partial_i \phi /c^2 = \kappa \rho c^2.$$
In order to match this with (1), we must have $\kappa = \frac{8\pi G}{c^4}$. (The detailed calculations here are, as is often the case in relativity, rather lengthy and boring, so they are omitted.)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/118750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
"answer_id": 0
} |
Can we say there are 2 EM radiation types? EM radiation seems to come from two different sources:
*
*According to the Maxwell, by acceleration of electrons
*According to the Bohr, by jumping of electrons between energy levels?
Are these two, completely different things?
| These are different causes of radiation, but they both produce the same kind of radiation: electromagnetic radiation.
When an electron is accelerated, it loses energy and emits one or more photons. This is a purely classical view of what happens.
When an electron falls from a higher energy state in an atom to a lower energy state, it emits a photon. This isn't really viewed as an acceleration in quantum mechanics; the "jump" is instantaneous. The speed of the electron before and after the jump are determined entirely by the states the electron is in, rather than by any details of the jump. In addition, there really isn't any force present that would cause the jump; it just happens.
In addition to those two mechanisms for creating EM radiation, there is also matter-antimatter annihilation. If an electron and a positron collided, they would annihilate each other, producing gamma rays. I can't think of any other mechanisms for causing radiation, but there might be others.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/118840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
} |
Inverting the equation for $T_{\mu\nu}$ in terms of $F_{\mu\nu}$ The Stress-Energy Tensor for electromagnetism is given by:
$$ T_{\mu \nu} = F_{\mu}\,^{\alpha}F_{\nu\alpha}-\frac{1}{4}g_{\mu\nu}F_{\alpha\beta}F^{\alpha\beta} $$
How can I find $F_{\mu\nu}$ in terms of $T_{\mu\nu}$?
Rewriting the above equation using:
$$ T_{\mu\nu}=- F_{\mu \alpha} g^{\alpha\beta} F_{\beta\nu} + \frac{1}{4} g_{\mu \nu}g^{\alpha\beta}F_{\beta\delta}g^{\delta\gamma} F_{\gamma\alpha}$$
from which we can write the following $4\times4$ matrix equation for the three matrices $T,\,F,\,g$, where $T$ is symmetric, $F$ is anti-symmetric and $g$ is symmetric and invertible:
$$ T = -F g^{-1} F+\frac{1}{4}\left(\mathrm{Tr}\, \left[g^{-1}Fg^{-1}F\right]\right)\,g$$
The only way I can think of is writing down 10 equations (as there are free components in $T^{\mu\nu}$) and then trying to find the 6 unknowns (as there are free components of $F^{\mu\nu}$).
Is there a better way to do this?
| The easiest way I can think of in Minkowski space, short of doing the algebra in terms of matrices, is to use $$\begin{split} f^a &= \rho E^a + \epsilon^{abc} J_b B_c = \partial_b T^{ab} - \epsilon_0 \mu_0 \partial_t S^a\\ \frac{\partial T^{00}}{\partial t} &= - \vec{J}\cdot \vec{E}- \vec{\nabla} \cdot \vec{S} \end{split},$$ with $S^a \equiv \frac{1}{\mu_0} \epsilon^{abc} E_b B_c = T^{0a},$ $a,b \in \{1,2,3\}$, and hope that the fields will be easy to discern.
Perhaps the more symmetric form $$\frac{1}{2} (F_\mu{}^\rho F_{\nu\rho} + \star F_\mu{}^\rho \star\! F_{\nu\rho}),$$ with the star denoting the Hodge dual, will prove easier to handle in curved spacetime, if you manage to break $T_{\mu\nu}$ into a sum of matrices of similar structure.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/118975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Particle in a box: simultaneously bounded momentum and position While writing an answer to this question, I started doubting about the interpretation of the uncertainty principle for the particle in a box.
In the 1-dimensional particle in a box problem, explicit solutions for the energy eigenstates exist, and are essentially of the form $\sin nx$ outside the support of the potential, and 0 where the potential is infinite. These somehow feel as if they should have definite momentum (up to direction), since they are free where the potential is 0, so that their energy is kinetic energy, and they cannot be in the region where they are not free.
On the other hand, position is strictly restricted to some interval. Consequently, any state must have bounded support in positional representation, and one would say there can be no states with bounded support in momentum space (by Paley-Wiener if you want). Indeed, the energy eigenstates involve an infinite number of momentum eigenstates (as they are zero outside an interval).
What is the way out? I think it must be that part of the energy in the energy eigenstates is potential energy. To see that, we can think of the idealized infinite potential as a limit of some sequence of finite potentials, which also is necessary in order to be able to give a meaning to the second derivative appearing in the Hamiltonian. This second derivative in the boundaries of the well tends to a multiple of a Dirac mass which has to be compensated in a very essential way by the term $V(x)\psi(x)$ in the Schrӧdinger equation $\psi''(x) + V(x)\psi(x) = \psi(x)$ (up to constants), so that in any approximation to the infinite potential will have a non-negligible contribution from the potential energy.
Is this a correct interpretation?
| But the momentum is not definite.
The standing wave state is a mixture of waves with wave number $k$ and $-k$ (here I am assuming a square-well for simplicity of notation). In more than one dimension we have to write those as vectors, but the same principle applies: there is a mixture of multiple states with different momenta.
And the smaller the box the larger the ground state energy, which also implies a larger magnitude for $k$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/119035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Difference between matrix representations of tensors and $\delta^{i}_{j}$ and $\delta_{ij}$? My question basically is, is Kronecker delta $\delta_{ij}$ or $\delta^{i}_{j}$. Many tensor calculus books (including the one which I use) state it to be the latter, whereas I have also read many instances where they use the former. They cannot be the same as the don't have the same transformation laws. What I think is that since
$\delta_{j}^{i}$ = ($\delta_{j}^{i}$)$'$, but $\delta_{ij}$ doesn't, so the latter cannot be a tensor. But the problem is that both have the same value :- ($1,0$) depending upon the indices. So it makes me think that $\delta_{ij}$ is just the identity matrix $I$ and not a tensor, and $\delta^{i}_{j}$ is a function. But since $\delta_{j}^{i}$ also has the same output as $\delta_{ij}$, WHAT IS THE DIFFERENCE?
I think it could be matrix representations. IN GENERAL, is there a difference between matrix representations of $\delta_{ij}$, $\delta^{ij}$ and $\delta_{j}^{i}$ (or any other tensor for that matter). Please answer these (the difference between mixed indices AND matrix representations).
| These Kronecker symbols have the same matrix representations, as you said, just the unit matrix. The indices are placed at upper or lower positions to suit Einstein summation convention (http://en.wikipedia.org/wiki/Einstein_notation). They are always used together with covariant and contravariant vectors (http://en.wikipedia.org/wiki/Covariance_and_contravariance_of_vectors) in curvilinear coordinate systems. I learned these things in the context of electromagnetic theory. If you are familiar with electromagnetic theory too, I would recommend the sections 1.14 - 1.17 in the book Electromagnetic Theory by Stratton. There you can find a quite clear explanation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/119126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 4,
"answer_id": 1
} |
Subsets and Splits
No saved queries yet
Save your SQL queries to embed, download, and access them later. Queries will appear here once saved.