Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
Are there other properties besides lower boiling point that make isobutane a better refrigerant than butane? Asked differently, if -1C is low enough for the application is there any reason not to use butane rather than isobutane as the working fluid in a refrigeration system?
| They have different pressure curves:
But to answer the question: yes, there are other properties that make isobutane preferred over n-butane. The most widely
used refrigerants in household appliances are isobutane and R134a according to "Evaluation of N-Butane as a Potential Refrigerant
for Household Compressors" by Preben Bjerre & Per Larsen (2006). See the following quote:
The theoretical evaluation shows at LBP CECOMAF that N-butane (R600) has:
• 2.8% higher theoretical COP compared to isobutane (R600a)
• About 70% of the capacity of isobutane
• 10% higher theoretical COP compared to R134a
The measurements show that the higher theoretical COP of N-butane compared to isobutane can not be found in the
measurements. This is caused by the higher sensitivity to clearance volume, suction gas pressure drop and heating,
compared to isobutane. However the COP level of N-butane in the present compressor design is at the same level as
isobutane, which today is the refrigerant giving the highest COP on the market.
Lifetime test show acceptable results. The wear tendency is comparable with isobutane.
N-butane is an option for reaching high COP levels in household appliances, but it does not offer significant
advantages to isobutane on existing isobutane stroke volumes. However it opens the possibility to extend the range
to lower capacity by using the existing compressor designs. The disadvantages are that the cost and size are
unchanged.
$\huge \text{Glossary:} $
LBP: Low Back Pressure. "The minimum evaporating temperature and the condensing temperature allows for the
identification of the compressor application (LBP, MBP, or HBP). Low Back Pressure
systems such as freezers have evaporator temperatures below -20ºC (-4ºF). "
CECOMAF: European Committee of Refrigeration Equipment Manufacturers. Or, in French Comité Européen des Constructeurs de Matériel Frigorifique.
COP: Coefficient of Performance. "For a refrigeration system a COP of 4 indicates that 1 kW of electricity is needed for a evaporator to extract 4 kW of heat."
R134a: refrigerant code name for 1,1,1,2-Tetrafluoroethane:
R600: refrigerant code name for n-butane:
R600a: refrigerant code name for isobutane:
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/401499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Red-shifting due to emitting gravitational waves Light waves exert their own gravitational pull and must be emitting gravitational waves, losing energy in the process. Does this mean that light becomes red-shifted as it travels even without the effects of universal expansion?
| This does not directly answer your question about redshift, but does clarify some points: Do photons and cosmic rays radiate energy through gravitational waves? If not, why not?. A key point here is that the photon must be accelerated for it to emit a gravitational wave, a photon simply travelling through empty space from another star to earth would not be accelerated (due to no massive bodies around) and therefore would not emit any gravitational waves.
In essence, if the photon does emit a gravitational wave, due to passing near a massive object, then yes it would lose energy and therefore be redshifted. The amount would likely be immeasurable I would assume (though I cannot prove it).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/401638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Static temperature A Professor of Fluid Mechanics a told that the static temperature is the temperature observed when the relative speed between observer(thermometer) and fluid is zero.
I have trouble understanding as to how will the temperature of fluid depend on the reference frame since temperature is a quantity which should be a function of Static Pressure and not Pressure itself.
How can the temperature of a body depend on its speed?
| Moving fluid has kinetic energy which will be converted to internal energy if the flow changes. This is straight-up first law.
So you might consider air blowing through a venturi for instance; if you were to insert a thermometer at different points along the airflow, you would measure different temperatures as the airflow accelerates and decelerates.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/401769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Equivalence between ghosts? Ok. I'm trying to get the terminology right about the term ghost in physics. Is there any equivalence between these terms?
*
*Faddeev-Popov ghosts
*Paul-Villars ghosts
*Landau ghost
*The vanishing Goldstone bosons in electroweak symmetry breaking
Does any of these terms encapsulates one of the others?
Does the term antighost apply to any of these?
| All they have in common is being unphysical. Faddeev-Popov ghosts can't be observed as particles because they violate the spin-statistics tension; Pauli-Villars ghosts are negative-norm states; the Landau pole tells you your theory must give way to another at high energies; Goldstone ghosts represent an unphysical artefact of gauge fixing.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/402006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Does a charged particle travelling with uniform velocity induce a magnetic field? Does a charged particle, an electron say, travelling with uniform velocity induce a magnetic field? I believe it doesn't. In primary school, we all learned how to induce a magnetic field into an iron nail by wrapping coils of wire around the nail and then hooking it up to a DC battery, but if you do not coil the wire, the magnetic nail doesn't occur. What's happening here? My only guess are the electrons are accelerating; the magnitudes of their speeds aren't changing, but rather their directions. In the coil, a force must be applying itself to the electrons in order for them to make their spiralling paths, thus, they are said to be accelerating and that is what causes the magnetic field to develop.
| A straight wire does have a magnetic field. It circles around the wire instead of going in a straight line like in a coil.
Picture source: http://coe.kean.edu/~afonarev/physics/magnetism/magnetism-el.htm
On the left is a straight wire with the magnetic field curling around it. The middle shows a single loop of wire. Notice that the magnetic field still curls around the wire, but the fields from opposite ends of the loop add together to make a strong field. The right picture shows a multi-loop wire (a solenoid), which enhances the field compared to the single loop. The right picture is the kind of field you created with the wire and nail. For the same current, the solenoid creates a much stronger field, which is why it is used to magnetize the nail.
To answer your original question, a single electron in motion does have a magnetic field that's similar to the straight wire (the field curls around the electron's path of motion) except that it gets weaker as you move farther away along the electon's path.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/402118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
EM Plane wave, the changing electric field is in all directions right? I just want to confirm this, because this type of diagram seems pretty popular.
The electric field and magnetic field actually surround in all directions orthogonal to x axis, right? It is not just 2d pointing only in the y direction and z direction respectively.
| Yes, the fields are everywhere surrounding the direction of propagation. But it would be linearly polarized, in the same direction as the vectors in the figure. If one only draws the E-field vectors, it could be a representation like this: https://commons.wikimedia.org/wiki/File:Linear_Polarization_Linearly_Polarized_Light_plane_wave.svg
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/402251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
What would happen if Jupiter collided with the Sun? This question is inspired by a similar one asked on Quora. Let's say a wizard magicked Jupiter into the Sun, with or without high velocity. What happens?
The Quora question has two completely opposed answers: one saying "nothing much happens" and the other saying "the Sun goes out for several hundred years". Both answers give reasons and calculations, and I know enough about physics to find both of them plausible. However ... it's plainly impossible that both answers are correct. Which one (or both?) is incorrect? Why is it incorrect?
| I would definitely lean towards "nothing" happening.
The "goes dark for 200 years" answer makes an awful lot of assumptions, some of which seem unfounded to me. In particular, it assumes that Jupiter will evenly spread over the surface of the sun, and will remain on top without mixing with the bulk of the sun.
At one other extreme, if Jupiter does not spread at all, it can at most blot out around $.3\%$ of the sun's luminosity (as that is the fraction it would cover). They also mention that the mass would make the sun last longer, which is suspect, since extra mass typically reduces the lifetime of main sequence stars.
Given that sunspots larger than Jupiter form occasionally, and are quickly mixed back into the sun, it seems likely Jupiter would do the same thing. And if it is mixed through the convection layer, all it would do is marginally decrease the temperature, and thus marginally dim the sun.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/402383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
"answer_count": 4,
"answer_id": 1
} |
AP physics 1 rotation problem could someone help me with this problem?
the correct answers are a and d. one issue i have with it is that i just don't understand what the problem is asking. like what spool? what table? i tried making some sense of the question and the answers, but i can only see d moving the wheel clockwise. the answer explanation mentions,
The key is knowing where the “fulcrum,” or the pivot for rotation, is. Here, that’s the contact point between the surface and the wheel.
but why is that? isn't the pivot where the axle meets the wheel?
| This question is written very poorly. I would recommend using a different source for questions unless this is one that your teacher is making you do.
I think that by "to the right" it means clockwise, although in physics we call clockwise rotation either "clockwise" or "negative", we don't use linear direction for rotations, and I think that it is assuming a spool of rope is wrapped around the axle, and the wheels are on a table. It is asking which direction you would need to pull the rope in order for the wheels to rotate clockwise. D makes sense because you are pulling the rope from the left side which would spin it to the right. A would work if you pulled it and then released it would start rotating to the right, but it doesn't make a lot of sense because just pulling it wouldn't cause a clockwise rotation unless you pulled it then released it at a high speed.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/402751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Could dark matters massive gravity create areas of slowly released expanding space left over from the big bang? I am not a physicist and apologize if this wastes anyone's time. I follow progress in most theoretical fields of physics and understand the premises but not how they are realised. The question I posed has been eating at me for a long time (ever since a link between time and gravity was discovered). Some of the other things I have theorized over the years are being considered now and I haven't heard of this concept. Logic would dictate someone with an actual education has already answered this, so again I'm sorry if I waste anyone's time.
| *
*dark matter has not been directly observed, but its presence is implied by for example the gravitational effects that visible matter cannot cause
*mostly when observing galaxies, without dark matter they would fly away and not rotate
*dark matter makes up most of the mass in the universe, so to your question yes, those areas where the mass is concentrated (galactic clusters), expanding is not so strong
*the expanding of space is mostly happening in between of galaxy clusters where there is no mass (and so no dark matter or normal matter either)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/402894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Will the weight of a man standing on a scale changes if he throws something straight up? If a man enclosed inside a chamber and the chamber is on a weighing scale and if he throws a ball straight up in air will the weight of the whole system changes or not? If yes, then what is the weight change at
1) just moments after throw;
2) before reaching the highest point of flight;
3) at highest point of flight;
4) on descending downwards; and
5) if there is any change in above observations in vacuum.
Supposing that scale is fast enough to register the changes before ball comes back to the man.
| In order to throw the ball up in the air, you must apply a force to give the ball a greater acceleration then the downwards force from gravity. By Newtons third law, the ball will exert a force of equal magnitude and opposite direction on the person. So since you are giving the ball a force in the upwards direction, you will feel a force downwards and the reading on the scale will deflect since the net force increases (not just gravity acting). After the brief deflection where your apparent weight increases, the scale will go to the equilibrium for the weight of the person from gravity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/403471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is *Conservation of Distinction* a true conservation law in mainstream physics? Both Leonard Susskind and Francis Heylighen have written about the Conservation of Distinction but it seems Susskind more closely connects this (law?) with unitarity in quantum mechanics. Heylighen doesn't mention unitarity from what I've read and uses the term more to argue the foundations and meaning of causality.
So is the Conservation of Distinction a true law of conservation in mainstream physics?
And if it is a law of conservation, what rule of symmetry does it correspond to (assuming Noether's theorem applies)?
| Here's an article by Heylighen on this; it appears he's interested in theorising causality judging from the abstract:
Equal causes have equal effects is reformulated as a distinction conserving relation. Unpredictable, respectively irreversible processes are analysed as processes in which distinctions are created, respectively destroyed. Different types of partially and pseudo causal relations are examined. Time order is derived from distinction conservation.
So in brief, from Heylighers perspective this is an underlying conservation law that gives time order. It's not a main-stream conservation law of physics which is no comment upon its validity or truth.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/403591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Stress in rotating 'rigid' body? I have some trouble understanding the rotating rigid body motion. Consider a cube made of very high elastic modulus homogeneous material (which can be approximated as 'rigid' body), it is rotating about axis-z with constant spin speed $\Omega$.
I wonder if the is internal stress or strain in the cube, and if
it is possible to compute them.
Here are my thoughts:
Consider a general point $P$ in this cube. When the cube is rotating, its position is different from its initial position because of the centrifugal force. Hence it is possible for internal strain to exist. However, if the body is very 'rigid' it is very small. I don't know how to express it in a general way, in terms of the centrifugal force.
| The internal stress in the cube does exists - it keeps the cube from falling apart due to the rotation or, we can say, due to the centrifugal forces.
The diagram below shows a possible way to estimate the stress at any given point $P$ inside the cube.
The force (stress) acting on an infinitesimally narrow vertical slice around point $P$ is equal to the centripetal force $F$ that keeps a narrow segment of the cube (shown in pink) from flying away.
$$F=m\Omega^2R_{\rm CM}$$
where $m$ is the mass of the segment and $R_{\rm CM}$ is the distance from the center of rotation to the segment's center of mass.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/403860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is it possible to measure the number of spacetime dimensions? My question is pretty straightforward to be stated but I don't know whether the answer is as easily reported. It is remarkable and very interesting that Physics work in (almost) any number of spacetime dimensions, but is it possible to actually measure the dimensionality of spacetime, even indirectly? I have found elsewhere in this site an argument regarding measuring the power law of the distance-dependence of forces, for example, the magnitude of the electrostatic force exerted by a charged point particle with charge $q$ would be of the form,
\begin{equation}
F_{Coulomb} \sim\frac{q}{r^{d-1}}
\end{equation}
with $r$ the distance from the particle's position and $d$ the number of spatial dimensions.
I can accept this as long as the procedure of measuring distances is well defined. Fundamentally, an observer would hold a meter stick and measure distances along the spatial axes. But how can the observer be certain of the number of axes he is capable of laying the meter stick along. How can he be sure there aren't any more axes that he simply cannot realize?
| I think your hypothetical observer can be sure of the dimensionality of the space he occupies by figuring out the minimum number of measurements required to specify the location of an object relative to some fixed reference frame. I get 3.
This requires some assumptions about space being 'well-behaved.' It seems to be so.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/403985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Can an electromagnetically induced dipole be explained with photon interactions? An incident electromagnetic wave will cause a dipole moment in the medium it passes through, displacing positive and negative charges in accordance with the EM field. How much of this interaction (if any) can be explained via the particle nature of light, i.e. photons?
|
An incident electromagnetic wave will cause a dipole moment in the medium it passes through, displacing positive and negative charges in accordance with the EM field. How much of this interaction (if any) can be explained via the particle nature of light, i.e. photons?
This depends on the level of abstraction. At a higher level of abstraction it is said that a changing electric field induces a magnetic field and vice versa. This process is called electric and magnetic induction and was observed empirically. Later the explanations became more detailed. Planck's and Einstein’s works made it clear that EM radiation consists of photons, the quanta of any EM radiation.
An incoming radio wave with its high number of aligned photons induces both an electric and a magnetic field inside the receiver's antenna rod. See the Wikipedia for loop antenna, which in the German language are more generally called Magnetantenne.
Can an electromagnetically induced dipole be explained with photon interactions?
As long as you agree that any EM radiation and in particular radio waves consists of photons, you are right to explain the induction processes from this radiation at the level of photon interactions.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/404184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Are avalanches caused by the shift from static to kinetic coefficient of friction? I was thinking about avalanches, and how they form, when it occurred to me to be quite likely that avalanches occur when a mass of snow gains enough momentum to push more snow through static friction into kinetic, gaining enough momentum to push even more snow through static friction into kinetic, and so on as a physical chain reaction.
Is this assessment of the cause of avalanches correct?
| My info on this comes from the book Snow sense: a guide to evaluating snow avalanche hazard, by Fredston and Fesler.
The physics of avalanches seems to be pretty complicated. Fredson and Fesler classify avalanches into four main types: "loose snow slides, slab avalanches, cornice collapses, and ice avalanches." They describe loose snow slides as arising through the sort of chain reaction you describe, while in a slab avalanche an entire slab starts sliding as a unit. Slab avalanches are often the most dangerous.
There seem to be a variety of reasons why the bond between a slab and the underlying bed can give way, leading to a slab avalanche. You get layers of snow with different properties, and they undergo changes such as metamorphosis (I guess analogous to geological metamorphosis) and melting. Often humans set off an avalanche, but the necessary weakness has to be present.
Although the question refers to the standard Coulomb-Amontons model of static and kinetic friction as taught in freshman physics courses, that model doesn't tend to work well for wetted surfaces, and it doesn't seem like it would naturally be able to explain some of the phenomena involved in avalanches. For example, in the Coulomb-Amontons model, stepping on a slab would never cause it to slide, because the critical angle is independent of the weight. (The added weight increases the normal force, which increases the maximum force of static friction exactly in proportion to the additional frictional force needed to support the greater weight.)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/404326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Law of vector addition In one (Why is heat a scalar quantity?) of the questions I answered, I came across this:-
https://physics.stackexchange.com/a/404287/181020
or
A vector quantity should obey the law of vector addition.
But I don't think it is true. Consider a y junction circuit containing 3 identical wires. If i current flows through each of the 2 arms, the current through the other wire is 2i. But the current density vector of each arm doesn't add vectorially to give the current density in the third.
I want to know whether vectors should obey the law of vector addition or not.
Thanks in advance.
| Vectors do follow vector addition; your Y-junction is not a counterexample to this, because you're applying the concept of "addition" to vectors in different places, which is not what is meant by "vector addition".
Vector addition for current fluxes means this:
*
*Suppose you run a current i through a wire
*Keep that current going, and send in another current i through the same wire
*Now the current through the wire is 2i
What your example shows is that the current flux is not the same everywhere in the circuit, but this is hardly surprising. If current flows around a loop of wire, the current flux vector points all different directions at different points, so of course it is not the same from place to place.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/404453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
What is the difference between electrons and holes in silicon? Electrons and holes behave differently in a silicon semiconductor (e.g. mobility of holes is one order of magnitude smaller than that of electrons, the collection time of holes at the same electric field is larger than for electrons... ). I was wondering, if holes are simply "a lack of electrons", they should behave in a mirrored way as electrons (if the latter move from $V_a$ to $V_b$ in a given time, the corresponding holes created when these electrons move should move in the opposite direction at the same speed). My question is: what is the origin of a different behavior between electrons and holes?
| One qualitative way to understand this concept is by thinking about the orbital origins of the conduction band and valence band.
For example, let's assume the conduction band is primarily coming from hybridized $s$-orbitals, while the valence band comes from hybridized $p$-orbitals. As you might guess, particles in the $s$-orbital should move differently from the $p$-orbital, as they both have different spatial properties ($s$ is more delocalized compared to $p$, and $p$ is anisotropic unlike $s$).
Now it should be clear that a hole living in the valence band should behave differently compared to an electron in the conduction band-it is because they are not simple inverses of each other, but instead occupy different bands! In fact, for most semiconductors the electron and hole transport properties are quite different, with a common exception being Graphene.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/404636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How can a horizontally fired bullet reach the ground the same time a dropped bullet does? I studied projectile motion and now I know that we can treat each component of motion independently. Since gravitational acceleration acts on both a horizontally launched bullet and a vertically dropped bullet in free fall, they both will reach the ground at the same time as their vertical initial velocity is zero.
This is what I studied in high school. But I found it against a real observation that a horizontally fired bullet will travel for much longer time compared to a simply dropped bullet before hitting the ground.
Could you please elaborate on how to connect the physics of the situation and real life observations?
| Since I'm impatient I'll suggest one way you could be surprised: if
*
*You are comparing the carry time of a rifle bullet to a dropped bullet and
*The rifle sights have been zeroed in for non-trivial distances
then the barrel is not level when aimed at a target the same height at the firing point, but instead points slightly upward accounting for the observation handily. Indeed, it must be that way because if the bullet was truly fired horizontally then it can only hit targets lower than the barrel.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/405005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Time Evolution of a Maximally Entangled State My question is basically this:
Does a maximally entangled state stay maximally entangled under the time evolution?
Assume our Hilbert space is $\mathcal{H}_A \otimes \mathcal{H}_B$. At the time $t=0$, the states is $\rho_{AB}$ such that $\text{Tr}_A \rho_{AB}=\frac{1}{n_A}id_B$ and $\text{Tr}_B \rho_{AB}=\frac{1}{n_B}id_A$. Now assume at the time $t$, the density matrix is $\rho_{AB}(t)$. Is it right that $\rho_{AB}(t)$ is a maximally entangled state? (If not, under what assumptions one can guarantee this?)
TEMP: Is it right?
$\text{Tr}_A \rho_{AB}(t)=\sum_A \left<e^A_i(t),\rho_{AB}(t)e^A_i(t)\right>=\left<U(t)e^A_i(0),U(t)\rho_{AB}(0)U^{-1}(t)U(t)e^A_i(0)\right>=\left<e^A_i(0),U^{-1}(t)U(t)\rho_{AB}(0)e^A_i(0)\right>=\left<e^A_i(0),\rho_{AB}(0)e^A_i(0)\right>=\text{Tr}_A \rho_{AB}(0)=\frac{1}{n_A}id_B$
So Maximally entangled states remains maximally entangled.
| General time evolution can generate a general, i.e., arbitrary, unitary transformation. This can map any quantum state to any other state. Thus, no state is special, including maximally entangled ones.
The situation is different if the Hamiltonian acts on the two parts separately. Then it creates a unitary evolution $U_A\otimes U_B$, which cannot change the entanglement.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/405127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Reason for body attached to a string being in free fall? This is a question I found in a book:
A string is wrapped around a uniform cylinder as shown in diagram. When cylinder is released string unwraps without any slipping and the cylinder comes down.
I assumed that an equation $T - mg = ma$ could be formulated and the tension does negative work. However the answer is that the tension does zero work. This I understand is because the cylinder is in free fall, and the equation will be $T - mg = mg$ and therefore $T = 0$.
Is my assumption correct? If it is, why is this so? It doesn't make sense to me. Should the tension exert some upward force and the downward acceleration be at least a bit less than $g$?
| The cylinder is not in free fall. It rolls down the string without slipping.
A cylinder which rolls without slipping down an inclined plane accelerates less than an identical cylinder which slides without rolling. In the same way the cylinder in your question accelerates less than an identical cylinder which slides down the string without rotating.
See eg https://www.youtube.com/watch?v=tMdS8Ttx0XA.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/405268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If the 3-Body force problem hasn't been solved, how do rocket scientists plan orbits of spacecraft? What methods would they use to predict what would happen in a situation when a probe is being acted upon by the gravity of two stars, say?
|
What methods would they use to predict what would happen in a situation when a probe is being acted upon by the gravity of two stars, say?
For that matter, what methods do they use to predict what would happen in a situation when a probe is being acted upon by the gravity of a single star plus eight planets, a large number of moons, and a very large number of asteroids and comets?
That is a rhetorical question. It's a bit false to say that the N body problem cannot be solved. It does have a series solution, developed first for the three body problem over a century ago, and generalized to N bodies over a quarter of a century ago, but nobody uses it. Too many terms are needed.
Even without that series solution, there are ways to solve the N body problem in practice, at least for a short period of time. In the case of our solar system, that short period of time is a few million years. (That qualifies as "short" compared to the age of the solar system.)
In a way, this is no different than almost every interesting problem in physics. Problems of interest that have closed form solutions are few and far between. For everything else, some kind of approximation technique is needed.
A large number of approximation techniques exist in the case of orbital mechanics. The approach outlined in Floris' answer is but one. Much more accurate but significantly more complex techniques are used to model probes sent from the Earth to other planets.
That said, any technique used to predict where planets and probes will be in the future will inevitably be imperfect. Compounding the problem, models of the thrust generated by a spacecraft inevitably are imperfect as well, as are instruments that measure that thrust. Any approach used to guide a probe to another planet must necessarily deal with those imperfections. Techniques that address these issues also exist. The development of these techniques was one of the key challenges in getting men to the Moon in the 1960s.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/405542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 2,
"answer_id": 1
} |
Change of wavefunction due to relativistic speed Imagine a spacecraft which is moving at a speed comparable to the speed of light relative to a reference frame with a hydrogen atom at it's origin. How would the probability distribution function of an electron in 1s orbit look relative to an observer inside the spacecraft?
| The Schroedinger equation is only invariant under galilean boosts. Under such a boost with velocity $v$ we have
$$
\psi(x,t) \mapsto \tilde \psi = e^{i(mvx- \frac 12 mv^2t)}\psi(x-vt).
$$
The extra phase factors are because seen form the moving frame the s-wave electron has an extra momentum $mV$ and extra energy $\frac 12 mV^2$. It's an amusing exercise with the Schroedinger equation to show that this transformed wave function obeys
$$
i\hbar \frac{\partial \tilde\psi}{\partial t}= -\frac{\hbar^2}{2m}\nabla^2 \tilde \psi+V({\bf x}-vt)\tilde \psi
$$
if
$
\psi$ obeys
$$
i\hbar \frac{\partial \psi}{\partial t}= -\frac{\hbar^2}{2m}\nabla^2 \psi+V({\bf x}) \psi.
$$
It is surprising because a relativistic scalar particle obeys the Klein-Gordon equation and transforms as a scalar. The extra phases come from the fact the Schroedinger $\psi$ is related to the KG $\phi$ by
$$
\phi(x,t)=\psi(x,t)e^{-imc^2t}
$$
because the Schroedinger equation does not take into account contribution of the rest mass to the energy.
Under a Lorentz boost
$$
t\to t'= \frac{(t-xv/c^2)}{\sqrt{1-v^2/c^2}}.
$$
The $1/c^2$ in the Lorentz transform cancels the $c^2$ in the extracted phase factor $\exp\{-imc^2t\}$.
If you want to use Lorentz transformations for electrons you need to replace the Schroedinger equation by the Dirac equation, but then, as @Trebor says, you need the machinary of quantum field theory to interpret the spinor wavefunction as it includes antiparticles. The net effect is to squash to the charge distribution as answered by @timm, but you will see particle antiparticle pairs in the pancaked distribution as well as the expected electron.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/405778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Why is $ \frac{\vec{r}}{r^3} = \frac{1}{r^2} $? I know it's surely a beginner's question but I don't see why you can write
\begin{align}
\frac{\vec{r}}{r^3} = \frac{1}{r^2}\cdot \frac{|\vec{r}|}{r}
\end{align}
Could someone explain it please? It would help understand quite a few things ...
| Another possibility is: being $\overline r= r\mathbf{\widehat r}$, then
$$\frac{\overline r}{r^3}=\frac{r\mathbf{\widehat r}}{r^3}=\frac{\not r\mathbf{\widehat r}}{r^{\not 3\,2}}=\frac{\mathbf{\widehat r}}{r^{2}}.$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/405861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Bragg Reflection While reading about the experiment of x-ray spectrum and Bragg reflection, I saw somewhere that it is more accurate to measure the crystal lattice constant when considering the largest measurable Bragg angle. That choosing this angle minimize the error in the calculation of the lattice parameter.
My question is why is that true? And is there a proof to support it?
| The Bragg equation is $2d \sin \theta=n \lambda$, so you write
$d={n \lambda \over 2 \sin \theta}$
if you measure the angle $\theta$ you get a measurement of the lattice spacing $d$. You presumably know $\lambda$ very accurately so the uncertainty on the measurement of $d$ comes from the uncertainty on the measurement of $\theta$. (This assumes your equipment measures $\theta$ - it may measure $\tan \theta$ but this makes no difference to the argument.)
In such experiments $\theta$ is generally very small, so $1/\sin\theta$ is very large, and small changes in $\theta$ lead to big changes in $1/sin \theta$. You can show this for yourself by plugging a few values, or use the error formula
$\sigma_d=|\left( {\partial d \over \partial \theta}\right)| \sigma_\theta=\left( {n \lambda \cos \theta \over 2 sin^2 \theta} \right) \sigma_\theta=\left({2d^2 n \lambda \cos\theta \over n^2 \lambda^2 }\right)\sigma_\theta={2d^2 \cos\theta \over n\lambda}\sigma_\theta\approx{2d^2\over n \lambda} \sigma_\theta $
Showing that $\sigma_d$ is proportional to $1/n$. At higher orders you measure a larger angle and the error you get from $1/\theta$ is smaller.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/406000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is there a charge buildup on elecrodes in DC glow discharge? In the figure below, direct current (DC) discharge occurs at and after some breakdown voltage, $V_{\text{breakdown}}\,,$ when the plasma pressure $P_{\text{plamsa}}$ is within a range observed to include $\left[0.01,\, 10\right]\,\mathrm{Torr}$, as electrons accelerate under the influence of potential difference to cause further electron emissions and recombinations.
$\hspace{100px}$
If there is a potential difference, this implies presence of electric field, implying presence of charge on electrodes.
Question: How is an electric field strong enough to cause emissive collisions of electrons produced when the elecrodes (assume wires in this case) have really low capacitance? The voltage across the electrodes is $2\, \mathrm{kV}$. And distance between electrons is $14\, \mathrm{cm}$.
| For a small capacitance a small charge suffices to create a large voltage, since $V=Q/C$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/406134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Modifying the Bohr Model for Muonic Hydrogen I am trying to find the energy between the $n=2\leftrightarrow3$ transition for Muonic Hydrogen. My approach was to modify the Bohr model for standard hydrogen but taking the mass of the Muon $m_\mu\approx 207m_e$ instead of the mass of the electron $m_e$ and then substituting in the effective mass $m=\frac{m_\mu m_p}{m_\mu+m_p}\approx186.03m_e$ (I took $m_p\approx1836m_e$).
Now taking $R=\frac{m_ee^4}{8\varepsilon_o^2h^3c}\approx10.97\times 10^6$ it gives $\Delta E\approx340.12\times 10^6\text{ J}$
Is this correct? Because it's quite a large value (assuming I got the units right)
| In order to obtain the energy levels of muonic hydrogen you can simply replace the reduced mass corresponding to the proton and electron with the reduced mass corresponding to the proton and the muon particle as you have already indicated.This should give you the correct Rydberg constant for the spectrum of this atom.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/406247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Why can a wave be expressed with a sine function? I see many expressions which express waves with the sine function like $y=\sin(kx-\omega t)$.
Waves really look similar to the shapes of a sine or cosine function, but does this guarantee that expressions that show wave-like movement are sine or cosine functions or is this just an approximation?
| There's no rule that waves have to be sinusoidal in nature. Some waves show different waveforms, and tonnes of waves aren't even continuous and hence aren't described by sine functions.
But it's useful to model most continuous waves in this manner because trigonometric functions are the easiest periodic functions (easiest to use).
A simple example to prove this is a square waveform. It's not common in mechanics, but it plays a big role in circuits for computation. The equation is $A\times sign(\sin\omega t)$, where the $sign$ function is self-explanatory*. There's nothing 'sine-like' about the wave, but we still see $\sin$. This is useful because the sine function will alternate between positive and negative values periodically.
We can use trig functions to model a huge number of other curves. For example, triangle waves are drawn with an infinite Fourier series: $\sum_{i=0}^N(-1)^in^{-2}\sin([nt])$
There's no need to critically examine this equation, but what's important is that as $N$ becomes larger, the equation is a better approximation, which is something the question focused on. Using trigonometric expressions, we're able to apply similar Fourier transforms to approximate almost any waveform.
* To create the $sign(a)$ function mathematically, try $\frac{|(a)|}{a}$. If $a$ is positive, $|a|=a$, and $sign(a)=1$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/406469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 8,
"answer_id": 6
} |
What are the physical conditions for general ohms law? A problem presents Ohm's law as $$\vec{J}=\sigma \vec{E}$$ where $\sigma$ is the conductivity given by a scalar. The problem asks what physical conditions must be satisfied for the equation to hold.
I understand that the material must be isotropic for the conductivity to be represented by a scalar instead of a tensor. What other physical conditions must apply?
| The most general form of the ohm law would be $J_i = \sigma_{ij}(x) E_j$. In the most general case, the conductivity is a tensor that depends on the position. Note that in the most general case, $J$ and $E$ are not parallel.
If the material is isotropic, then $J$ and $E$ are parallel and $\sigma$ is just a scalar function dependent on position. If now the material is homogeneous, $\sigma$ will no longer depend on $x$ and you get the usual ohm's law $J = \sigma E$.
Note that the actual value of $\sigma$ may depend on a number of different things like the temperature of the conductor.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/406833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Help with D. Tong example on Noether in QFT In this lectures, example 1.3.2 on page 14 concludes that the Noether current is
But how can the current be a two index object when it is defined in eq. (1.38), which is
as a one index object? If I apply the formula I obtain something of the form $j^\mu$. Can someone make the calculations explicitly?
| I think it is explained somewhat in David Tong's notes.
The symmetry under consideration is spacetime translation in a direction $\varepsilon^\mu$. For example, $\varepsilon^\mu=(1,0,0,0)^\mu$ would be the time translation symmetry. To draw an analogy with David Tong's notes, $\epsilon^\mu=\lambda \varepsilon^\mu$ would be the infinitesimal quantity, where $\lambda$ is the infinitesimal amount you want to translate in the $\varepsilon^\mu$ direction.
If you compare to the previous expressions, you get the conserved (finite) four-vector:
$$j^\mu=\frac{\partial L}{\partial \partial_\mu \phi} \varepsilon^\nu \partial_\nu \phi-\varepsilon^\mu L$$
But a trick very commonly used: if $A^\mu v_\mu=B^\mu v_\mu$ for every covector $v$, then $A^\mu=B^\mu$ outright. In our case, Noether directly gives that for every epsilon:
$$\varepsilon^\nu j^\mu_{\ \ \nu}=\varepsilon^\nu\left(\frac{\partial L}{\partial \partial_\mu \phi} \partial_\nu \phi-\delta^\mu_\nu L\right)$$
hence, we actually have a conserved $(1,1)$ tensor:
$$j^\mu_{\ \ \nu}=\frac{\partial L}{\partial \partial_\mu \phi} \partial_\nu \phi-\delta^\mu_\nu L$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/407101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Sound Relationships Whats the relationship between amplitude, volume, and wave energy. I have tried many websites but none have helped me yet and I believe maybe someone on here can help me.
| The Intensity is the power delivered by a wave per unit area.
It is proportional to the square of the amplitude of the wave.
The Loudness, is measured in decibels as a relative intensity, L = 23.03 ln(I/Io)
The energy in terms of the intensity is,
E = IAt
(by the definition of intensity.)
If you were looking for equations relating them, then it depends on the type of wave, whether its mechanical or electromagnetic.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/407327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Connecting a charged capacitor to an uncharged capacitor
I was attending a lecture about capacitors and something confused me.
If I charge a capacitor using a DC supply, the capacitor will gain charge $Q_0$.
Now, if I discharged it along an uncharged capacitor in this arrangement, according to the lecture notes, the capacitors share the total charge $Q_0$.
Now, I had a question.
Aren't there electrons on the uncharged capacitor, such that they flow between the two capacitors to cause equal p.d. on both capacitors hence the total charge in this circuit greater than $Q_0$?
| What you suggest would be true if the upper component were a battery, which maintains a certain definite voltage and can act as a supplier of charge. But a capacitor has only a finite charge, here $Q$, and as it flows off to the other capacitor it is not replenished.
The usual confusion with this particular circuit is to ask what happens to the energy...
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/407459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Is the equation of motion for a spring-damper system the same whether oriented upward or downward? So every spring-damper system I've found online has the equation of motion:
$$mx''+cx'+kx=0$$
I can understand how this is derived when downwards is positive, but what about when upwards is positive? Wouldn't it be
$$mx''-cx'-kx=0$$
| Yes.
This is because your 2nd equation should read as
$$ -mx''-cx'-kx=0 $$
since the sense where positive $x$ is has switched. This means a sign change for $x$, $x'$ and $x''$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/407686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Potential due to line charge Is it possible to calculate the electric potential at a point due to an infinite line charge? Because potential is defined with respect to infinity.
| It is possible. Electrosatic potential is just a scalar field whose negative gradient is the electric field. Due to this defintion it is indeterminate to the extent of an additive constant. (if you increase it everywhere equally, its slope remains the same everywhere) Only the potential difference between two points is measurable, which is called voltage. It is a convention that potential in the infinty is often taken zero, which is usefull, but
you could easily call for example a point 2 meters away zero potential and obtain the same function only offset by a constant, but yielding the exact same forces.
And yes, as V.F. had said, there are infinite number of points being infinitely far from your line, so you could even use infinity as zero point, and easily obtain the potential by integration and symmetry considerations.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/407797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Earthing a system of parallel plates our sir has taught us that when in a system of conducting infinite plates kept parallel to each other, when any one of them is earthed, then the charges present on the free ends(i.e. the open sides of the parallel plates at the 2 ends) will become zero!
Say we have 3 infinite conducting parallel plates $A,B,C$ with charges $q_1,q_2,q_3$. Now say that $A$ is earthed. Then we have been told that the charges on free end are zero, and hence a charge of $-(q_2+q_3)$ is induced on the inner face of $A$.
Can anyone please explain why this occurs?
| It happens to minimize the energy content of the system.
The electric field stores energy in the form of electrostatic potential energy. Remember, you have to do work to bring the system in the state explained in question. This work is stored as energy (1st law of thermodynamics). Earthing opens the path to redistribution of charges so that energy is minimized (2nd law of thermodynamics). Unfortunately, the bound charges can't move, so the free charges move in/out of system to minimize the electric field volume to minimize energy. You can do elementary calculation to find out that the earth takes (q2 + q3) amount of charge from the system.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/407881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Energy of massless particles with the example of Magnon Consider this situation.
I am told that: at low temperatures magnons, the elementary excitations of a ferromagnetic ground state, can be treated as a gas of non-interacting, massless, spin-0 bosons with a dispersion relation $ω$ = $α$$k^2$ where $α$ is a constant.
Research online tells me that the energy should be $ћ$$ω$. However, quantum mechanics tells me that momentum is equal to $ћ$$k$ and relativity then tells me that the energy is $E$=$p$$c$=$ћ$$k$$c$, which, considering the dispersion relation $ω$ = $α$$k^2$, gives a completely different expression for the energy.
Can someone please clear up this confusion?
| The relation $E = pc$ for massless particles comes from Lorentz invariance, specifically by requiring that the "length" of the 4-momentum, $E^2 - (pc)^2$ should be the same in all reference frames. Therefore, it only holds for fundamental particles, which magnons are not. Magnons do not occur in vacuum, they occur in solids, which spontaneously break Lorentz invariance since they have a preferred rest frame. Therefore magnons do not need to satisfy Lorentz invariance.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/408088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is presure inversely related to velocity? .
I’m not sure if that picture worked or not but it’s a question from my textbook that was asking what factors would cause velocity of blood in an artery to decrease. It says in the answer at the bottom that apparently a lower pressure would
Cause this, but I thought pressure and velocity were inversely related because of Bernoulli’s law?
| The Bernoulli equation uses the energy of the fluid before and after traveling through different cross sectional areas. This is a link to a good explanation of the principle. In general, when a fluid's speed increases, its pressure decreases, that's why planes can fly so well, the lower pressure, faster moving air above the wing causes a net upward force called lift.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/408219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Time evolution of momentum wave function when initial position wave function is in an eigenstate (i.e., a delta function of position) In The Dirac-delta function as an initial state for the quantum free particle,
Emilio Pisanty
states that if an object's position wave function, $\Psi(x,0)$, is a delta function (at $t = 0$), then
$$
\Psi(x,t)=\begin{cases}\delta(x) & t=0\\
\sqrt{\frac{m}{2\pi\hbar |t|}}e^{-i\mathrm{sgn}(t)\pi/4}\exp\left[i\frac{mx^2}{2\hbar t}\right]&t\neq 0\end{cases}
$$
In that case, at $t>0$, the momentum wave function is$$
\Phi\left(k,\, t\right) = \frac{1}{\sqrt{{2\pi}}}\sqrt{\frac{m}{2\pi\hbar t}}e^{-i\mathrm{sgn}(t)\pi/4}\int_{-\infty}^{\infty}\exp\left[i\frac{mx^2}{2\hbar t}\right]e^{ikx}\mathrm{d}x
$$
According to Engineering Tables/Fourier Transform Table 2, and if we set $\mathcal{x'}=-x$,$$
\begin{alignat}{7}
\Phi\left(k, \, t\right) & ~=~ -\sqrt{\frac{m}{2\pi\hbar t}} \, e^{-i\mathrm{sgn}(t)\pi/4} \, \sqrt{\frac{\hbar t}{m}} \, \exp\left[{-i\left(\frac{\hbar t k^2}{2m} - \frac{\pi}{4}\right)}\right] \\
& ~=~-\frac{1}{\sqrt{{2\pi}}} \, e^{-i\mathrm{sgn}(t)\pi/4} \, \exp\left[{-i\left(\frac{\hbar t k^2}{2m} - \frac{\pi}{4}\right)}\right] \\
& ~=~-\frac{1}{\sqrt{{2\pi}}} \, \exp\left[{-i\frac{\hbar t k^2}{2m} }\right] \\
& ~=~ -\frac{1}{\sqrt{{2\pi}}} \, \exp\left[{-i\omega t }\right]\end{alignat}$$
This seems to indicate that, for $t>0$, both $\Psi^2$ and $\Phi^2$ are constant in space, though $\Psi^2$ diminishes in time. That means that both position and momentum are totally uncertain in space for $t>0$.
Did I get that right?
I know that a position eigenstate is not a solution of Schrodinger's equation and so this is all somewhat outside the scope of orthodox QM, but this does seem to violate the idea that the less certain is one variable, the more certain is it's conjugate variable.
| What you have is a free particle propagator (also called Green's function of the free particle Hamiltonian). This is an auxiliary mathematical concept to express time evolution of a general normalizable $\psi$ function. But propagator itself is not a normalized $\psi$ function, the usual theorems valid for normalized $\psi$ functions do not apply to it.
The Heisenberg uncertainty relations are derived from the commutation relations for normalizable $\psi$ functions only. One cannot apply them to singular objects such delta distribution. Not even expected average value of position
$$
\langle x \rangle = \int\psi^* x\psi dx
$$
has a sense for $\psi=\delta(x-x_0)$ (or its time evolved version).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/408321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Data - Monte Carlo Correction techninques in Particle physics It is said that the data does not always match with the Monte Carlo simulations in particle physics.(I guess even in the Higgs to gamma gamma channel, the peak in real data was at about 127GeV and thus it was corrected) .Thus,I wished to know what are the different ways in which the differences in Data and MC simulations are corrected, if possible, for the case of jets like bjets.
| If certain distributions of variables don't agree, then you can reweight the MC sample to match data (typically either using a nice clean control sample or s-weighted signal). There are a variety of reweighting techniques e.g. histograms, kernel-density estimators, boosted decision trees.
If resolution doesn't agree, you can apply 'smearing', whereby a variable is shifted by some random amount on an event-by-event basis according to a Gaussian distribution (or similar).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/408450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Meaning of crystal orientation If I have a crystal, say (001)-oriented ABO3, what does the 001 imply, from an experimental perspective? I understand that the 001 is referring to a plane as described by Miller indices, but what is it about this plane that is relevant? Is it that this plane is the same plane as the surface of the sample?
| It depends on context. In a surface science experiment, it would be the surface. In experiments on bulk properties, the surface could be irrelevant and rough, but one would measure some property along a cubic axis (or maybe paricularly along the c-axis), like velocity of sound or conductivity, etc.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/408745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$\phi\phi\to\phi\phi$ scattering in $\phi^4$-theory and Feynman diagrams at the tree-level What are the tree-level Feynman diagrams for $\phi\phi\to\phi\phi$ scattering in $\phi^4$-theory? I know that there are four diagrams at tree-level (zero loop). But each diagram is considered only once. Why is that? There are one-loop diagrams in which external leg receives one-loop correction. I agree that they are disconnected diagrams but if we remove the disconnected part wouldn't that contribute to a tree level process? Sorry if the question is unclear because I could not attach relevant diagrams.
| Consideration of correlation functions in an interacting QFT (i.e one in which you wish to describe a scattering event) means you can neglect disconnected diagrams order by order in the perturbative expansion. Formally, this amounts to a generating functional with connected diagrams 'exponentiated' but that is a detail. So, there is only one diagram contributing at tree level for the given process, namely the four point vertex. The three other diagrams are non-interactive and just represent propagation of fields between two source points.
A systematic way to obtain the contributing diagrams is to label your asymptotic in and out fields as originating from source points $x_1,x_2,x_3$ and $x_4$. Then you make all possible attachments of these lines resulting in inequivalent diagrams at a particular desired loop/coupling order. At tree level, for example, you would have four lines originating from the $x_i$ and all meeting at a vertex.
Therefore, at one loop, you will indeed have four diagrams corresponding to radiative corrections you can make to the four external legs. These are not $1\text{PI}$ but by amputating them in the manner in which you describe will separate a source point from the rest of the connected diagram so it will not yield the tree level result. Instead, these external leg corrections are usually dealt with in a consistent programme of renormalisation, so typically one does not include them.
There are, however, three other non trivial one loop diagrams contributing to $\phi \phi \rightarrow \phi \phi$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/408895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How strict are the boundaries that divide dimensions? Is a single-layer sheet of graphene 2D or 3D? I would like to know if there is any theory that describes a set of rules that define the boundaries of dimensions.
For example, does a single layer sheet made of graphene considered a two or a three dimensional object?
Or does any object consisted of at least one atom considered by default a 3D object?
How about the elementary particles, do they qualify as 2D objects?
Since a 3D object can be projected to and be observed in a higher dimension, do we have any example of 2D objects projected to the third dimension?
| As others have pointed out, even a single atom exists in three dimensions. However, there's an alternative, and mathematically rigorous, sense in which you should regard a sheet of graphene (or two sheets of stacked graphene, or four sheets of stacked graphene) as 2D.
In condensed matter, one is often interested in the properties of a material in the thermodynamic limit, i.e. as $N\rightarrow \infty$, where $N$ is the number of atoms in the system. There are many ways of taking the thermodynamic limit. If we consider a cube of material with dimensions $L_x\times L_y \times L_z$, we could take $L_x\rightarrow\infty$ while leaving $L_y$ and $L_z$ unchanged, or we could take $L_x,L_y\rightarrow\infty$ while leaving $L_z$ unchanged, or we could take all three to infinity. We call the first case "one dimensional" no matter how big $L_y$ and $L_z$ were initially, we call the second case "two dimensional" no matter how big $L_z$ was initially, and we call the third case "three dimensional." In other words, a one dimensional system in condensed matter is a system where one direction is infinite, and the other two directions are finite. In a sense it doesn't matter how large the other two directions are, since they are necessarily negligible compared to infinity!
With this definition of dimension, you can prove all sorts of interesting theorems. You can show a continuous symmetry cannot be broken in dimensions $\leq 2$. You can classify topological insulators according to their dimension. And on and on. All of these theorems are proven using the notion of "dimensional" that I defined above.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/409172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Phase Transitions (Statistical Mechanics) Book Recommendation I am Studying Phase Transitions, starting from simple VdW equation up to Mean-Field&Landau Theory. I have many books such as Huang Kerson-Statistical mechanics, Yeomans- Statistical Mechanics of phase Transitions, Lubensky, Kardar and they've covered me so far for MFT.
However I have trouble finding a bit more advanced cases of MFT in these books, including:
i) Detailed analysis on Landau–Ginzburg Theory
Books/notes usually only cover the simplest case $f=f_0 +am^2+bm^4 +k(\nabla m )^2 $, what do we do when there are terms of higher order ( such as $ (\nabla^2 m )^2$) and what's the physical difference?
ii) Ornstein–Zernike approximation
I've come across this when solving a system with the above free energy and reading about the corelation function, but have yet to see a book that gives me a detailed explanation of what it is, how it works and why we do it.
iii) Cahn – Hilliard theory
None of the above books covers this
iv) Nematic to Smectic Transition
v) Spinodal decomposition
vi)Renormalisation
Huang's book covers this. However the exact mathematical formalism and arguments in Haung are not very clear to me. I would like a book that treats the Renormalization Group in an abstract rigorous manner so that I can understand the key concepts.
In general I have so far studied from all these books collectively, switching from one to the other, including some lecture notes I've found.
I would like book/lecture notes recommendations (undergrad or probably graduate level) that go into detail in these advanced (?) cases of phase transitions and meant for someone that is reading these for the first time, but has of course a backround in statistical physics, if such books exist.
| The statistical physics book by Linda Reichl is easy to read and has some of the advanced material. A good complement for the renormalization aspect are Goldenfeld's Lectures On Phase Transitions And The Renormalization Group.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/409610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Why does plasmon have higher erngy than phonon? In my mind plasmon is movement of electrons and phonon is movement of atoms in an lattice. movement of atoms should have a large energy because atom is larger.
| The phonon describes the vibrational motion in a solid.
The plasmon is a quasi particle, describing the plasma oscillation.
These quantized excitations have different energy levels, phonons in the meV range, and plasmon in the eV.
As you write you are wrong, because phonons energy is 3 orders of magnitude higher.
The difference is because phonons are collective excitations of atoms in lattice. Plasmons are collective excitations of nearly free electrons in plasma.
Phonons are important in conservation of energy, plasmon gives the color of gold.
http://iopscience.iop.org/article/10.1088/0368-3281/2/1/305
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/410066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If there is friction and everything stops on earth why does earth not stop due to friction? Why do planets revolve around the Sun if there is a force called frictional force? Why do planets not stop rotating while the objects in motion stop after a while? Why are they continuously moving? Why are planets not affected by frictional force?
| The answers here seem to address all but one of the questions -- Why do planets not stop rotating while the objects in motion stop after a while?
Note that there is no net external torque on the planet, at least not due to friction inside it, and hence its angular momentum is conserved. So long as the shape of the planet (moment of inertia) does not change, nor will the rotation (angular velocity).
When an object moves, let's say a ball rolling on the floor, the friction due to the floor is an external force for the ball -- it is because of the floor, something "outside" of the system (the ball) and hence can produce net external force and torque on the ball, affecting its motion.
However, friction cannot stop a planet from rotating as long as the shape of the planet remains same. This is because all the frictional forces on the planet (except for the negligible drag from bumping into a few stray H atoms in space) are internal forces. Hence, they cannot produce a net external torque on the planet and thus cannot change its rotation unless its shape (and thus the moment of inertia) changes.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/410191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 2
} |
Which pole will electrons flow towards in a changing magnetic field (generator) In a generator a magnet spins in the middle of a coil of wire and the changing magnetic field causes current to flow, but the current changes direction whenever fields from a new pole of the magnet cross a point on the wire, so will the current move towards the north pole or towards the south pole (rather the part of the wire being touched by the pole's magnetic field).
I know that when you use current to magnetize something the north end of the magnet is the end that was negative, so I would assume that current would move away from the north. Is that correct?
To make things more clear I have added this diagram. Assume the generator is running. Would electrons be taking the blue or the red path?
| (a) "when you use current to magnetize something the north end of the magnet is the end that was negative" I'm afraid that this suggests a misunderstanding. The direction of magnetisation is determined by the direction of current around the coil, not along its length. You need to consult a textbook and learn about the right hand grip rule.
(b) If, in your diagram, the left hand side of the coil is coming towards us, and the right hand side is going into the screen, away from us, then, according to Fleming's right hand generator rule, the blue arrow gives the direction of the induced conventional current (if the circuit is complete). Therefore the electrons, being negative, are flowing in the direction of the red arrow. Things are reversed if the right hand side of the coil is coming towards us.
It is misleading to think of electrons going towards or away from either of the magnet's poles.
(c) Electromagnetism needs to be learnt step-by-step. It isn't difficult, but there is quite a lot of detail to master. It's much easier to do this of you follow a logical sequence – such as you will find in a competent textbook.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/410290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Energy conservation on expanding universe Due to the expansion of the universe, the photons emitted by the stars suffer redshift, Its mean that the energy is lowered a little bit. Does this mean that the energy is lost? Does the expansion of the universe violate some conservation principles according to Noether's theorem?
| Actually it is possible to speak of energy conservation in curved spacetime in the presence of a timelike Killing vector $K$, since the contraction of it with the stress energy tensor is a conserved current from Killing equation and symmetry of $T^{ab}$:
$$\nabla_a (K_bT^{ab}) =(\nabla_a K_b) T^{ab} + K_b \nabla_aT^{ab}= \frac{1}{2}(\nabla_a K_b) T^{ab} + \frac{1}{2}(\nabla_a K_b) T^{ba} +0$$
$$= \frac{1}{2}(\nabla_a K_b + \nabla_bK_a) T^{ab} = 0\:.$$
In case of an expanding universe there is no timelike Killing vector, but there is a conformal timelike Killing vector $K = \partial_\tau$ where $\tau$ conformal time. Conformal Killing equation reads
$$\nabla_a K_b + \nabla_bK_a = \phi g_{ab}\:.$$
It gives a conservation law for systems with traceless stress energy tensor: $g_{ab}T^{ab}=0$, like the EM field with a procedure very close to that exploited above.
The problem is that this sort of energy cannot be added to the standard one associated to massive fields, so a common conservation law (EM field + matter) does not exist, though EM waves conserve their energy if referring to the conformal time $\tau$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/410392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Electron Microscopy: Are photons smaller than electrons? So, I'm told that electron microscopy provides greater resolution than traditional photo/optical (i.e. visible light) microscopy, due to the (ahem) "fact" that "electrons are physically smaller than photons".. Which I'm pretty sure is not necessarily true, (or static/constant). But by the same token, I am not a physicist..
The analogy that I'm presented with is that in this context, the size (and density) of photons and/or electrons are supposedly comparable to pixels (and ppi/dpi) of an LCD monitor, CCD image sensor, raster image, etc.
So, it sounds plausible, and it's a nice, relatable explanation for the common man. But is it factual?
If not, what's really going on here?
| Here is a basic overview- experts are invited to add detail.
Subatomic particles like electrons actually possess a wavelength that is related to their energy, even though they behave most of the time like tiny point particles and not like waves. By accelerating a beam of electrons to very high energies, their wavelengths become far shorter than that of a photon of visible light, and this allows them to resolve objects far smaller than that which can be seen with focused beams of visible light.
In a typical electron microscope, the electron beam tube in which the acceleration occurs is about three to four feet long, but this principle can be scaled up to resolve the interior of a single proton. In this case, the beam tube needs to be more than ten thousand feet long; the SLAC particle accelerator in California has one of these 2.2 miles long which was used to detect the interior structure of protons in the late 1960's.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/410487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Should we take the elevator instead of the stairs to save energy? In our university there's a posted sign that encourages students to take the stairs instead of the elevator to save the university electricity during the hot NYC summers.
When you climb the stairs you generate something like 8x your mechanical energy in metabolism which is dissipated as heat and must be cooled by the air conditioning in our cooled stairwells.
What are plausible assumptions about the energy efficiency limits of air conditioning and mechanical lifting of an elevator? What (fundamental) limits are there that prevent 100% efficiency in the elevator case? Are there other important factors for this analysis?
Is it plausible that the university is wrong? Should we take the elevator instead of the stairs to save the university energy? Do the humans of an energy-efficient future look like Wall-E?
The source of this question is a bathroom physics problem posted above university physics department toilets near grad offices. Problems are submitted by department members. To the best of my knowledge, none of these were ever homework questions. This question is adapted by my toilet memory from a problem that was provided by Andrei Gruzinov.
| Accepting that climbing stairs generates heat equal to 7 times the mechanical energy (one part goes into you lifting yourself), how much energy does it take to remove that heat?
The 2nd Law of Thermodynamics says the best you can do is
$W = Q \frac {T_h - T_c}{T_c} \sim Q \frac{\Delta T}{T}$
Where $\Delta T$ is the difference between inside and outside temperature, and T is the outside temperature. For ease of calculation, taking T around room temperature of 280K, you get that the air conditioner is better (uses less energy, i,e, W/Q is better than 1/7) if it’s working into a temperature difference less than $\Delta T = 40\rm{K}$.
Large-building air-conditioners routinely even better than this for electrical power usage by using e.g. evaporative chillers to do most of the thermodynamics.
Unless it’s really really hot outside, take the stairs.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/410582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
} |
Trying to check current conservation under symmetry transformation
Consider a simple scalar field and its Lagrangian $L=\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi$. Then say you have the following transformation
$$x^{\mu}\rightarrow e^{\omega}x^{\mu},\tag{1}$$
$$\phi\left(x\right)\rightarrow e^{-\omega}\phi\left(e^{\omega}x\right).\tag{2}$$
What is the associated conserved current?
Attempt:
I compute the Euler-Lagrange (EL) equations,
$$\partial_{\mu}\left(\frac{\partial L}{\partial\left(\partial_{\mu}\phi\right)}\right)=\frac{\partial L}{\partial\phi}\quad\Leftrightarrow\quad\partial_{\mu}\partial^{\mu}\phi=0,\tag{3}$$
write the transformations infinitesimal as
$${\phi\left(x\right)\rightarrow\left(1-\omega+\ldots\right)\phi=\phi-\omega\phi}\quad\Rightarrow\delta\phi=-\omega\phi\tag{4}$$
$${\delta L=0}\tag{5}$$
and compute the conserved current, factoring out $\omega$, from
$$j^{\mu}=\frac{\partial L}{\partial\left(\partial_{\mu}\phi\right)}\delta\phi-\delta L=-\phi\partial^{\mu}\phi\tag{6}$$
but something must be wrong because I can't show it is conserved,
$$\partial_{\mu}j^{\mu}=-\partial_{\mu}\left(\phi\partial^{\mu}\phi\right)=-\partial_{\mu}\phi\partial^{\mu}\phi-\phi\underset{=0\left(EL\right)}{\underbrace{\partial_{\mu}\partial^{\mu}\phi}=}-\partial_{\mu}\phi\partial^{\mu}\phi?\tag{7}$$
Can you please see what am I doing wrong?
| Why did you write $\frac{\partial L}{\partial (\partial_\mu\phi)}=-\phi\partial^\mu\phi$?
It should be $=\partial^\mu\phi$.
update: everthing's correct, $\partial_\mu\phi\partial^\mu\phi$ vanishes because it is equal to $\partial_\mu\partial^\mu\phi$ up to a surface term. Whenever equation of motion is satisfied ($\partial_\mu\partial^\mu\phi=0$), $\partial_\mu\phi\partial^\mu\phi$ vanishes too.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/411262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Will wind blowing perpendicular to the path of motorbike, topple a slow moving bike or fast moving bike? So, I am riding a Motorbike on a windy day.
I am traveling from North to South and winds were blowing from East to west.
Should I drive slow or fast to avoid getting toppled down by the winds?
Also, is there a way to calculate the safest speed if my motorbike weights 100 kg and wind is blowing at 80 KM/hour?
| High speed makes a bike more stable whether you deal with a wind or with other disturbances.
One of the factors is a gyroscoping effect of the wheels: the greater the speed, the more difficult it is to change the orientation of the wheels and therefore of the bike.
Another factor is the ease of correction. If, due to the wind or other disturbances, the bike does lean, at higher speed, the required turn of the front wheel, to counter the torque of the weight with the matching centrifugal force, would be less than at a lower speed.
Regardless of the speed, if the wind is constant, you'll have to shift your weight or slightly lean into the wind.
Of course, when speed gets too high, other effects, like loss of traction, may become significant, and, therefore, it would be difficult to determine the optimal speed for safety.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/411536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Does the fact that $j^\mu$ is a 4-vector imply $A^\mu$ is, as argued by Feynman? Let
\begin{equation}
\boldsymbol{\Phi}=\Bigl(\dfrac{\phi}{c},\mathbf{A}\Bigr)
\tag{01}
\end{equation}
the electromagnetic 4-potential. We know that if its 4-divergence is zero
\begin{equation}
\dfrac{1}{c^{2}}\dfrac{\partial \phi}{\partial t}\boldsymbol{+}\boldsymbol{\nabla}\boldsymbol{\cdot}\mathbf{A}=0 \quad \text{(the Lorenz condition)}
\tag{02}
\end{equation}
then Maxwell's equations take the elegant form
\begin{equation}
\Box\boldsymbol{\Phi}=\mu_{0}\mathbf{J}
\tag{03}
\end{equation}
where the so called d'Alembertian
\begin{equation}
\Box\equiv \dfrac{1}{c^{2}}\dfrac{\partial^{2} \hphantom{t}}{\partial t^{2}}\boldsymbol{-}\nabla^{2}
\tag{04}
\end{equation}
and the 4-current
\begin{equation}
\mathbf{J}=(c\rho,\mathbf{j})
\tag{05}
\end{equation}
which has also its 4-divergence equal to zero
\begin{equation}
\dfrac{\partial \rho}{\partial t}\boldsymbol{+}\boldsymbol{\nabla}\boldsymbol{\cdot}\mathbf{j}=0 \quad \text{(the continuity equation)}
\tag{06}
\end{equation}
and is a 4-vector.
The question is : under these conditions is the 4-potential a 4-vector ??? I ask for a proof or a reference (link,paper,textbook etc) with a proof.
EDIT
$^\prime$Mainly Electromagnetism and Matter$^\prime$, The Feynman Lectures on Physics, Vol.II, The New Millenium Edition 2010.
|
under these conditions is the 4-potential a 4-vector ??? I ask for a proof or a reference (link,paper,textbook etc) with a proof.
No, not necessarily.
It is convenient to choose it as a 4-vector in relativistic theory.
But Maxwell's equations together with the Lorenz constraint do not imply that 4-tuple $(\varphi/c,\mathbf A)$, must transform as four-vector.
The Lorenz condition
$$
\partial_t \varphi + \nabla \cdot \mathbf A = 0 ~~~\text{(in all inertial frames)}
$$
and the wave equations
$$
\square^2\varphi = \rho/\epsilon_0
$$
$$
\square^2\mathbf A = \mu_0 \mathbf j
$$
do not remove the arbitrariness of potentials completely: in one and the same frame $i$, they can still be changed using any scalar field $\chi_i$:
$$
\tilde{\mathbf A} = \mathbf A + \nabla \chi_i
$$
$$
\tilde{\varphi} = \varphi - \partial_t \chi_i
$$
provided it obeys the equation
$$
\square^2\chi_i =0
$$
and is not constant. That equation has uncountable infinity of different solutions so we can assign one distinct solution to each inertial frame. Even if we begin with 4-tuple of potentials $(\varphi/c,\mathbf A)$ that transform as four-vector, if we then redefine the potentials in every frame using $\chi_i$ unique to that frame, there will be no simple relation between components of potentials in two inertial frames and thus they cannot be connected via the Lorentz transformation. This is impractical to do in this way, but the result - potentials in different frames not connected by the Lorentz transformation - is allowed by the above equations.
The potentials are artificial functions that we are free to define as we please as long as they give the actual electric and magnetic field via the usual formulae. It is possible to extend this definition so they are four-vectors and this is sometimes the most natural choice. For example, the well-known Lienard-Wiechert solution of the wave equations above, when used in all frames, gives electric and magnetic potentials that together transform as a four-vector.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/412110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Trying to prove the Hadamard gate leaves a state unchanged It is known that the Hadamard gate is the equivalent of doing a 180 degree rotation about the x + z axis.
I am therefore trying to prove that applying the gate on the state cos pi/8 |0> + sin pi/8 |1>, which lies on the x + z axis unchanged.
So, on applying the Hadamard on the above, I get
1/root(2)((cos pi/8 + sin pi/8)|0> + (cos pi/8 - sin pi/8)|1>)
So, I did try to take the common global phase out, but I'm unable to prove that this is the same state that I started out with. Any help on how to approach this will be appreciated.
| Note that
$$\sin(x\pm\pi/4)=\frac{1}{\sqrt2}(\sin x\pm\cos x).$$
It follows that
\begin{align}\cos(\pi/8)+\sin(\pi/8) &= \sqrt2 \sin(3\pi/8), \\
\cos(\pi/8)-\sin(\pi/8) &= \sqrt2 \sin(\pi/8).
\end{align}
You should be able to reach the conclusion from here.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/412224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can we (in principle) obtain molecular bound systems by modelling fundamental particles and their interactions? Is it possible, at least in principle, to start with the Schrodinger/Dirac/Klein-Gordon equations to model elementary particles and their interactions and to obtain in the end molecular bound systems? In other words, is it possible (in principle) to deduce the laws of chemistry starting from the laws of elementary particles?
| Yes, but due to the difference in time and space scales, we have specific models for specific problems to model. Some examples are: First principles modelling (ab-initio approaches), DFT, molecular dynamics; the first two start with Schrodinger's equations and work their way up, molecular dynamics is used in bigger systems using data from the DFT or first principles (usually as a bridge to scale size and time).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/412461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why is linear momentum not conserved for a particle in a central force? I am studying a two-body central force system in which the two particles, one of mass $m$ and one of mass $M$, experience a force directed along the line connecting the two particles.
We can reduce this to a system of just one fictitious particle with reduced mass and a central force. Why is linear momentum not conserved in the CM frame when it was conserved before shifting to the CM frame?
| Suppose that I know that $p$ is conserved and $q$ is not. Now suppose I define the new variable
$$p' = p + q.$$
If you weren't paying attention, you might conclude that $p'$ is also conserved, because it has the same letter as the conserved quantity $p$. But that's clearly wrong. The similarity is just superficial. Not every variable whose name looks like "$p$" has to be conserved.
Similarly, in the two-body problem, the conserved linear momentum is
$$\mathbf{p} = m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2.$$
To work in the reduced mass picture, we define, among other things
$$m_{r} = \frac{m_1 m_2}{m_1 + m_2}, \quad \mathbf{v} = \mathbf{v}_1 - \mathbf{v}_2.$$
Your question is why the quantity
$$\mathbf{p}' = m_{r} \mathbf{v}$$
is not conserved, while $\mathbf{p}$ is. But $\mathbf{p}'$ has nothing to do with the total linear momentum $\mathbf{p}$, the resemblance is totally superficial. There is no reason to expect that $\mathbf{p}'$ should be conserved.
Incidentally, the linear momentum is conserved in the reduced mass picture. The point of reduced mass is that the relative velocity between two bodies interacting by a central force, with masses $m_1$ and $m_2$ can also be computed by finding the relative velocity between two bodies with masses $m_r$ and infinity attracting by the same force; this second mass stays fixed at the origin. The total linear momentum in this one-body problem is the sum of $\mathbf{p}'$ and the momentum of the second mass, and is indeed conserved. However, it has nothing to do with the total linear momentum in the original two-body problem.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/412847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the equation of a chain hanging between two points Say I have a connected at either end to two points, $A(x_A, y_A)$ and $B(x_B, y_B)$ of length $l$, where $l \leq \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}$, how would I go about finding an equation of its shape?
I guess the chain would be of the form $y = \alpha \cosh{(ax + b)} + \beta$, so how do the coefficients relate to the coordinates of the hanging points and its length?
| The general shape is
$$ y(x) = y_c + a \left( \cosh \left( \frac{x-x_c}{a} \right)-1 \right) $$
where $(x_c,y_c)$ is the lowest point on the curve (sag point) and $a = \frac{H}{w}$ is the catenary constant. Here $H$ is the horizontal tension shared along the cable, and $w$ is the unit weight (that is weight over length).
Consider a span $S$ and height $h$ the above is fit to the boundary conditions $y(0)=0$ and $y(S)=h$ with
$$ x_c = \frac{S}{2} - a\, \sinh^{-1} \left( \frac{h\,\exp(S/2a)}{a\, \exp(S/a)-a} \right) $$
$$ y_c = a \left( \cosh\left(\frac{x_c}{a}\right) -1 \right) $$
Furthermore, the tension in the cable is defined by the constant horizontal component $H$, and the varying vertical component $$V(x) = H \sinh\left( \frac{x-x_c}{a} \right) $$
as well as the total tension
$$T(x) = \sqrt{H^2+V^2}= H \cosh\left( \frac{x-x_c}{a} \right) $$
To find the tension at the ends, use $x=0$ and $x=S$. If the vertical tension at the ends is negative then there is an uplift condition where the cable is trying to pull the support out of the ground.
Finally, the total length is
$$ L = \int {\rm d}s = \int \limits_0^S \sqrt{1+y'(x)^2}\,{\rm d}x = a \left( \sinh\left( \frac{S-x_c}{a} \right) + \sinh\left( \frac{x_c}{a} \right) \right) $$
References:
*
*The hanging chain problem catenary
*Equation of a flying kite
Example software catenary solver based on the above equations
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/412971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Sign of work done by friction In Goldstein's classical mechanics (3rd ed.) we read:
"The independence of W12 on
the particular path implies that the work done around such a closed circuit is zero,i.e.
$$\oint \textbf{F}.d\textbf{s}$$
Physically it is clear that a system cannot be conservative if friction or other dissipation forces are present, because $F . d\textbf{s}$ due to friction is always positive and the integral cannot vanish."
My question is: why should the work due to friction be "always positive"? Shouldn't it be nonzero instead?
Also, $F . d\mathbf{s}$ is a typo and should be $\mathbf{F} . d\textbf{s}$ (please let me know if I'm wrong)
| Perhaps I misunderstand the context of Goldstein's writing, but work due to friction should be negative:
Friction always acts antiparallel to the displacement/velocity. So, when computing work from friction, drag, etc, you find that
$$ W = \oint \mathbf{F} \cdot d\mathbf{r} = \oint (F\cos\theta) dr, $$
where $\theta$ is the angle between the friction $\mathbf{F}$ and $d\mathbf{r}$. Because friction acts antiparallel, $\theta = \pi$ and $\cos\theta = -1$ always. Then,
$$ W = - \oint Fdr, $$
which is always negative because $F$ and $dr$ are vector magnitudes, and thus always positive. This is why friction is dissipative, it steals energy from the system in the form of heat and deformation. Even in the case of a line integral as presented here, each component/leg should be negative thus creating a total negative work.
Of course it makes sense that the friction force is nonconservative -- the work expelled certainly depends on the path. If you have ever moved furniture into a new apartment, of course you push it the shortest possible path, for this minimizes the energy you need. If you push it around aimlessly you will expend more energy than needed.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/413353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Is there an intuitive reason as to why there is no phase transition to get to a degenerate Fermi gas? Cooling a bosonic gas leads to a phase transition into the Bose-Einstein condesate. This is characterised by a symmetry broken ( U(1), by choosing a specific phase for the macroscopic wavefunction) and by discountinuous behavoiur of some thermodyamics quantities and their derivates - e.g. the heat capacities.
On the other hand, cooling a fermionic gas leads to a crossover into the a Fermi degenerate gas.
Is there an intuitive reason as to why the latter does not constitute a phase transition?
| In terms of phase transitions, interactions are the main source of intuition. You can imagine a collective field which biases the order parameter. For non-interacting bosons, the collective field is coming from the statistics instead of an interaction (an energy). Thus, it’s hard to identify the collective field as a term in an action or free energy. Instead we have to think more intuitively about these statistics and how a collective field forms and acts on its constituents. First the statistics of indistinguishable particles can be understood as distinguishable particles communicating with each other by constructive and deconstructive interference. For fermions the “signal” communicated is different from each particle as each particle is in a different state, thus the collective interference pattern cast by the N-1 particles on the N particle is noisy and random. While for bosons the signal is the same from each particle and the interference pattern reinforces its self. This reinforcement is the same sort of feedback process that interactions produce when forming a collective field in a classical phase transition.
For interacting systems, the collective inter-particle interference is not the only way particles communicate. In bosinic systems the suppression of thermal excitations is further reinforced by interactions and produces a gap in the excitation spectrum.
While interactions produce no collective field in interacting fermions for completely repulsive interactions. For attractive interactions, fermions pair into bosons and then condense.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/413429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Interaction of Magnetic field with light Can there be any interaction of a varying magnetic field with light? (Please explain using electromagnetic waves as both are)
(Say we have an alternating current of 60Hz and He-Ne laser (632.8 nm wavelength)
What all different kinds of interaction may happen? In free space or in a dielectric medium (Just an example for the sake of analysis))
Thanks
| If you've wanted to mean the change of direction of the light beam by the word 'interaction' ,then the answer is no, because a photon of a light ray goes with a very high speed so if there will be a varying magnetic field then the work done by that varying magnetic field will be 0.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/413571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Do the ladder operators $a$ and $a^\dagger$ form a complete algebra basis? It is easy to construct any operator (in continuous variables) using the set of operators $$\{|\ell\rangle\langle m |\},$$ where $l$ and $m$ are integers and the operators are represented in the Fock basis, i.e any operator $\hat M$ can be written as $$\hat M=\sum_{\ell,m}\alpha_{\ell,m}|\ell\rangle\langle m |$$ where $\alpha_{\ell,m}$ are complex coefficients. My question is, can we do the same thing with the set $$\{a^k (a^\dagger)^\ell\}.$$
Actually, this boils down to a single example which would be sufficient. Can we find coefficients $\alpha_{k,\ell}$ such that $$|0\rangle\langle 0|=\sum_{k,\ell}\alpha_{k,\ell}a^k (a^\dagger)^\ell.$$ (here $|0\rangle$ is the vacuum and I take $a^0=I$)
| @Accidental reminds you this is a theorem. To actually see it in your terms, use the infinite matrix representation of $a, \quad a^\dagger$ of Messiah's classic QM, v 1, ChXII, § 5. Specifically, your vacuum projection operator has a 1 in the 1,1 entry and zeros everywhere else.
The operator you chose is freaky to represent, but, purely formally, the
diagonal operator for $N\equiv a^\dagger a$,
$$
|0\rangle\langle 0|=(1+N) (1-N) \frac{2-N}{2} \frac{3-N}{3} \frac{4-N}{4} ...
$$
would do the trick, once anti-normal ordered.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/413659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Misunderstanding of the functioning of the reflective diffraction grating Suppose we have a sawtooth diffraction grating, as depicted below:
where the angle $\beta$ is the angle of inclination of the 'teeth' of the grating with respect to the plane of the grating and incident plane monochromatic waves normal to the plane of the grating. I am supposed to determine the angle $\theta$ for which the interference pattern for one 'saw-tooth' has a maximum. The diagram in the mark-scheme is as follows:
where the points $A,B$ both belong to the same 'saw-tooth' and the distance between $A$ and $B$ is $d$. The path difference between the two waves, according to the mark-scheme is given by $\Delta = BF - AE = d \sin \beta - d \sin \theta$. My question might seem trivial, but why are the two (BF and AE) not equal? In other words, shouldn't the two parallel incident waves (incident at an angle $\beta$) be simply reflected from the face of the 'saw-tooth' at exactly the same angle, in accordance with the law of reflection? Why even bother defining $\theta$? What am I missing here?
|
shouldn't the two parallel incident waves (incident at an angle β )
be simply reflected from the face of the 'saw-tooth' at exactly the
same angle, in accordance with the law of reflection?
This is a useful approximation of what actually happens, but in reality the behavior of light is a little more complicated.
The operation of diffraction grating is based on the Huygens–Fresnel principle and the interference.
If we apply the H-F principle to your example, we'll have to assume that every point of the light-reflecting surface will produce its own hemispherical wave rather than just a ray.
The interference between all these waves will define the reflected light, which, in addition to the central maximum, corresponding to the "normal" reflection of a specular surface, will produce additional maxima and minima, as depicted on your diagram.
For $\theta=\beta$, we'll get the central maximum. To find other max and min points, we'll have to consider other $\theta$ values.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/413776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Why do scientists think that all the laws of physics that apply in our galaxy apply in other galaxies? I like watching different videos about space. I keep seeing all these videos saying scientists found so and so at 200 billion light years away or this happened 13 billion years ago.
My question is why do scientists think that all the physics that apply in our galaxy apply in a galaxy say 200 billion light years away?
What if, say at 135 billion light years away, all of a sudden the time space relationship changes drastically and instead of linear time space relationships the difference becomes based on a "sliding scale" (to revert back to high school). What if a light they first see and estimate to be 200 billion light years away has actually been traveling for another 300 billion light years before we could detect it? Lets be serious, we can't predict the weather farther out than 10 days accurately, and usually not that long....
|
What if, say at 135 billion light years away, all of a sudden the time space relationship changes drastically.
Well, it could.
Science is built on reasonable, well-founded assumptions, and a good scientist is open to the possibility of those assumptions being broken at some point in the future when more data becomes available. That is why no scientific statement is ever a 100% certainty, but a "theory". We must always be open to being unexpectedly proved wrong.
In this case, we have so far not encountered any evidence to suggest that the laws of physics are not universally applied, so for now we proceed on the assumption that they are. Because, otherwise, short of avoiding doing any science regarding far-away galaxies, what else could we do?
(It's a bit like how we only search for "life as we know it", not because we have ruled out the possibility of other, exotic kinds of life, but because what else would we do? How would we look for it? How would we recognise it? Any such thing would have to be discovered by accident.)
Again it is worth noting that all our observations so far support this assumption.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/413886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "89",
"answer_count": 10,
"answer_id": 1
} |
How come I can kick a football further when it's moving towards me? If a ball is moving towards me, I can kick it further than if I were to kick it if it was stationary. But surely if the ball is moving in the opposite direction, it should take more force to kick it the same distance as I am accounting for the initial movement.
Does this have a theoretical explanation, or is it that I can perform better technique if it's moving?
| If a ball is moving toward you it will exert a force on you.If you kick it in opposite direction of its its inital direction it will require more force for same distance in case it was moving away from you.You hit it.Now the ball will cover same distance with less force than first case as by now there is already a force acting on ball in that particula direction with the force of your kick added.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/413955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Hose pressure when the valve is closed To add to the interesting threads on hoses, here's one from someone with no physics knowledge. Please be patient with this basic question!
If you open your hose valve and press the trigger on the nozzle of the hoze, I understand that thanks to the pressure of gravity from the water tower or what have you, water escapes at great velocity.
If you close your hose valve, and then go press the trigger, water still escapes at great velocity.
Why doesn't it dribble out (till the very end)? Where is the pressure coming from? Isn't the water just sitting in the hose between the valve and the nozzle?
| Water escapes at great velocity briefly.
Hoses are elastic. Some stretch more, some less, but that elastic stretch keeps the original pressure in the hose until you open the outlet. The force due to elastic hose then pushes out water, initially at the same pressure hence same high speed.
But water is leaving the hose, the water volume is dropping, hence less stretch, hence less force, hence less pressure, hence the speed rapidly drops.
A firm hose will drop speed quickly; the equivalent of a latex party balloon will go on longer, but they’ll all stop when they get back to their original unstretched volume as the elastic pressure goes to zero.
For a regular hose, this happens while there’s still a lot of water in it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/414235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is potential Difference Really a Measure of Electromotive force? If I separate some amount of positive and negative charge a certain distance I will create some voltage. If I then separate the same amount of positive and negative charge a longer distance I will create an even bigger voltage since more work would've been needed to separate them.
Now, the charges with a larger potential difference will experience less of a force of attraction because they're farther apart, and the charges with a lower potential difference will experience a larger force of attraction since they are closer together.
Question 1: If voltage is a measure of the "push" acting on charge, then how can this be so?
Question 2: Voltage is proportional to current. But in this case I would think that the lower voltage scenario would produce higher current because there's a stronger force acting on the charge. What's going on? How does potential difference produce current?
| The electric field is a measure of the "push" acting on an electric charge. The voltage (aka electrical potential) is a measure of the energy required to move an electric charge from one location to another location in an electric field. There is an analogy with gravity: the gravitational field at your location is a measure of the "pull" you feel. The gravitational potential is a measure of the energy required to move your body from the bottom of a ladder to the top of the ladder.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/414384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How is a quantum gate constructed in the real world? I've been reading a lot about quantum mechanics and arrived at entanglement. I understand these things and I understand how to perform computation on qbits if they are represented as vectors and the transformations as matrices.
However I have not seen any physics about how to construct a quantum gate or why it's possible to alter a qbit. Since even the most delicate measurement makes it colapse, the intuition I have is that it'd be impossible to alter it without making it collapse, but apparently it's possible.
| The specific details about how to apply quantum gates depend on the system you are looking at. For example, in a photonic system, you could apply some gates by beam splitters, polarizers, mirrors, waveplates and $\cdots$ . Or in spin systems you can tune the Hamiltonian of the system by changing the electric and magnetic fields(e.g. with a laser) and with that effectively apply a quantum gate on your spin system.
Also, it's true that performing a measurement would collapse the system; however, these gates typically do not consist a measurement. If the experimenter is sloppy with their experiment there will be decoherence though.
Chapter 7 of Nielsen and Chuang includes some physical realizations like NMR or ion traps. Also, I'm sure there are plenty of courses that cover these materials, e.g. you can look at David Cory's Explorations in Quantum Information course.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/414525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Definition of an operator in quantum mechanics In J.J. Sakurai's Modern Quantum Mechanics, the same operator $X$ acts on both, elements of the ket space and the bra space to produce elements of the ket and bra space, respectively. Mathematically, an operator is simply a map between two spaces.
So, how can the same operator act on the ket as well as the bra space?
| The action (to the right) on the ket space naturally induces an action (to the left) on the bra space. The bra space is the dual space of the ket space (that is the space of linear functionals over the kets). We can simply define $\left< \psi \right| X$ by its action on kets (or, since it's linear, on a basis of ket space and linear extension):
$$ \big(\left< \psi \right| X\big) \left| \phi \right> := \left< \psi \right| \big( X \left| \phi \right>\big).$$
Such definitions are also common in pure mathematics. If you want to be pedantic, you can use some notation, e.g. $\iota(X)$, for die induced operator on the dual space.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/414742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
What's the reaction force on a charge moving in a magnetic field? According to newtons third law, all forces occur in pairs. What is the reaction force that the third law predicts when a magnetic force acts on a charged particle moving in a magnetic field?
| It is induced emf.
According to Lenz's law, in a magnetic field as flux linked with the object changes ,an emf is induced in it such that it opposes its very own cause.
Mag. Flux = B.A
emf=-change in flux/time
Example: Suppose you move a magnet with its north pole ahead through a single turn coil from left to right. Then on the left face of coil north pole will develop when magnet enters the coil and north pole will develop on right face when it leaves. So it always opposes its motion
*Or think you drop a magnet vertically through a coil. Then its acceleration WILL ALWAYS BE LESS THAN g.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/414873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How do fusion reactors deal with blackbody radiation? The plasma of the ITER reactor is planned to be at 150 million K. Using the Stefan-Boltzmann law, setting the surface area as $1000\,\mathrm{m}^2$ (the plasma volume is $840\,\mathrm{m}^{3}$ so this is being generous), and the emissivity as $0.00001$ (emissivity is empirical so I just plugged in an extremely low value) yields a power of $2.87\times 10^{23}\,\mathrm{W}$. It would require somewhere on the order of $10^{35}$ fusion reactions per second just to break even, which clearly is not happening.
How can fusion researchers confine plasmas for several minutes if the blackbody radiation is this extreme? It seems like that with this level of heat, the plasma would just cool down within a few nanoseconds, and everyone in the vicinity would be torn to shreds by gamma rays, but evidently this does not happen. How?
| The plasma in a fusion reactor is typically "optically thin"; the radiation isn't really in equilibrium with itself and the plasma particles.
Generally, instead of just modeling the plasma as a black body, people look at specific radiation processes. Kenneth Gentle (UT) has a nice set of slides that works through that.
Hot plasmas are almost transparent -- optically thin, except for a few special cases and stellar interiors.
• A picture of a hot plasma in the visible is empty except for possible bright spots at the edge where incoming neutrals are ionized.
• The principal, and most important, exception is at the electron cyclotron frequency. (For n>10^19/m3 and Te>100 eV, the plasma is a black body radiating at Te, at least in certain directions and polarizations.
...
Blackbody radiation at 100 eV (1 million degrees; 200 times the surface of the sun) is quite significant. Were it not confined to a narrow band of frequencies near Ωe, it would be a very large power.
So that's not the wide-bandwidth process of true black-body radiation. And other process have their own limitations:
Bremsstrahlung
Since hot plasma “collisionless”, intensity far less than blackbody -- detectable only because it extends over a broad frequency range.
The radiation is broad-band, but peaks at hν~Te, like blackbody radiation, because there are fewer electrons at higher energies capable of emitting such photons, and the density of states decreases with decreasing frequency.
Line radiation from surfaces and impurities can also be strong, but (1) it's not the black-body you're asking about and (2) they work really hard to reduce the impurities.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/415028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 3,
"answer_id": 0
} |
What does vector addition of voltages mean? So I was studying LCR Circuits in Alternating Current and I found something pretty weird. We are treating the voltage as a vector (phasor) and then vectorially adding them to get the net voltage.
But this thing doesn't make sense to me: What does it mean to add voltages which differ in phase?
| If you have an LCR series circuit connected to an alternating voltage supply then at an instant of time the current through each component in the circuit is the same and the variation of current with time is represented by the top graph in the diagram below.
The addition of voltages in the circuit is complicated that they do not reach a maximum value at the same time.
If you study the graphs above you will see that the voltage across the capacitor lags behind the voltage across the resistor by a quarter of a period which is equivalent to a phase angle of $-90^\circ$ and the voltage across the inductor leads the voltage across the resistor by a quarter of a period which is equivalent to a phase angle of $+90^\circ$.
So to find the voltage across all three of the components $v_{\rm series}$ at any instant of time one has to evaluate $v_{\rm R} \sin (\omega t) + v_{\rm C} \sin (\omega t - 90^\circ) + v_{\rm L} \sin (\omega t + 90^\circ)$.
A convenient way to do this addition is to use a phasor diagram as shown on the right and you will see in this case $v_{\rm supply}$ leads the voltage across the resistor (and hence the current which is always in phase with the voltage across the resistor by a value which is between $0^\circ$ and $+90^\circ$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/415456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Weird Reflection Pattern in Reading Glasses While fidgeting with a pair of reading glasses, I noticed a strange reflection pattern (shown in video and photo).
I would appreciate it if anyone that knows more about this could help me figure out why there were eight dots in the reflection instead of four, and why there were different colors when all the light sources were an orange/yellow/warm color.
There were eight dots in each frame, four orange, four green. They made a cube shape, each one of the dots being a corner of the cube. The same image could be seen in each frame, and if I focused my eyes it looked 3d.
The picture below shows the light source that the reflection came from, just four ordinary ceiling lights. I know it couldn't have been anything else because it was night, and to test this I turned off all other lights and it was still there. In fact it was then even brighter. I looked into the glasses again in the morning, the effect was still there.
Thanks,
Ella
| The colors are due to antireflection and/or anti-abrasion coatings on the lenses. The doubling of each light image is due to internal reflection from the lens surfaces. The 3D effect is due to the fact that the lens surfaces are curved. If you look at the reflection of a light in a Christmas tree ball, you will see that the image is reduced. That explains the fact that you see a reduced image of all four lights in the first, bright, reflection. You also see a dimmer version of the four lights due to light passing through the front curved surface, being refracted by that surface, then being reflected by the curved back surface of the lens. That second reflected image is further altered when it is refracted in passing back through the front curved surface. The resulting distortions are complicated, but they amount to the first set of light images and the second set appearing to be different sizes and at different depths.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/415594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Electron Splitting in Peskin and Schroeder I am confused by formula (17.88) and (17.89) on page 578 in P&S. They are computing the matrix element for electron splitting ($e^-\rightarrow e^-+\gamma$) in the massless limit.
They call $z$ the fraction of energy of the initial electron that is carried off by the photon and $p_\perp$ the transverse momentum of the photon.
To order $p_\perp^2$ the momenta are then as follows:
Initial electron: $p = (p,0,0,p)$
Emitted photon: $q= (zp, p_\perp,0 ,zp - \frac{p_\perp^2}{2zp})$
Final electron: $k = ((1-z) p, -p_\perp, 0, (1-z) p + \frac{p_\perp^2}{2zp})$
They then proceed to compute the matrix elements for given helicities and find e.g. for left-handed electrons in the chiral representation of $\gamma$-matrices:
$$
i\mathcal{M} = ie \sqrt{2(1-z)p}\sqrt{2p} \xi^\dagger(k) \sigma^I \xi(p) \epsilon_T^{*I}(q)
$$
With $\xi$ the $1\times 2$ spinor. For the left-handed initial electron we have
$$
\xi(p) = \begin{pmatrix} 0\\1 \end{pmatrix}
$$
But then they say that
$$
\xi(k) = \begin{pmatrix} p_\perp/2(1-z)p\\1 \end{pmatrix}
$$
and that the polarisation vectors for the photons are
$$
\epsilon_L ^{*i}(q) = \frac{1}{\sqrt{2}} (1,i,-\frac{p_\perp}{zp}) \qquad \text{and} \qquad
\epsilon_R ^{*i}(q) = \frac{1}{\sqrt{2}} (1,-i,-\frac{p_\perp}{zp})
$$
This is my question: where do the expressions for $\xi(k), \epsilon_L ^{*i}(q)$ and $\epsilon_R ^{*i}(q)$ come from?
| A left-handed electron moving the 3 direction has the 2-component spinor (0,1), and a left-handed photon moving in the 3 direction has the polarization vector $\epsilon^* = \frac{(1,i, 0)}{\sqrt{2}}$. However, in this problem, the outgoing electron and photon are not moving exactly in the 3 direction. They are moving at a small angle to this direction specified by the transverse momentum $p_{perp}$. If you rotate the polarization spinor and the polarization vector into the directions of motion of the particles, working to 1st order in $p_{perp}$, you will find the expressions given above and in eqs. (17.88) and (17.89) of our book.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/415779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Physical intuition behind Poincaré–Bendixson theorem The Poincaré–Bendixson theorem states that: In continuous systems, chaotic behaviour can only arise in systems that have 3 or more dimensions. What is the best way to understand this criteria physically? Namely, what is is about a space of dimension 1 or 2 that cannot admit a strange attractor? Why does this only apply to continuous systems and not discrete ones?
| An important characteristics of chaotic dynamics is that they are recurring, i.e., any trajectory will eventually come arbitrarily close to its starting point.
Suppose there is a chaotic dynamics with continuous time in a two-dimensional phase space. Let’s look at the trajectory starting from some point A. Since the dynamics is recurring, there needs to be a point B on the trajectory starting from A that is so close to it that the phase-space flow does not change direction on the line from A to B¹²:
Now, consider the loop closed by the trajectory between A and B (cyan) and the line from A to B (red). The trajectory will be trapped on either side of this loop after B: It cannot cross the trajectory because trajectories cannot intersect, and it cannot cross the line because the phase-space flow goes in the other direction. In the above example, the trajectory is trapped on the inside and thus has to spiral in; but it might as well spiral out. Either way, the trajectory can never get closer to A than B, which would contradict the requirement of recurrence. Thus the only recurring dynamics in two dimensions are periodic orbits.
In three dimensions, things are different because the trajectory cannot divide the phase space in two parts.
For discrete-time systems, there are no trajectories to begin with that could entrap something.
¹ If you cannot find such a point, the phase-space flow is discontinuous around A in way that does not happen in physical systems.
² If B is identical to A, the dynamics is periodic and not chaotic.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/415971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Do gravitational waves lose energy through interaction with i.e. matter or magnetic fields? Gravitational waves dilute while traversing space like any other radiation, and their amplitudes are proportional to r-2, that's a basic. But do they lose energy while traversing through matter or something else (i.e. space with magnetic fields, or whatever), too, by interaction with it ?
(The question "Where does gravitational waves' energy go?" doesn't focus on traversing matter of fields)
| Your question is very original, and I have never seen it addressed, so here are my own hastily prepared thoughts on the matter. Caveat lector.
You can gain some insight into the fate of gravitational waves by pursuing the analogy with electromagnetic waves, which get both scattered and absorbed. EM waves get scattered by matter that responds to the electric field -- free charges, good conductors, or polarizable objects. They can also get absorbed when dissipative processes are at work -- in imperfect conductors, or dielectrics with lagging response.
Dissipative processes generally involve reequilibration, which can be either fast or slow. In imperfect conductors, electrons accelerated by the field ultimately lose their momentum by collisions with positive ions. In water, it takes roughly 20 ps for the polar molecules to reorient.
When GR waves passes through a galaxy, the tidal forces will deform orbits and objects, and the waves will surely undergo some scattering. But what of dissipation? Near misses by stars and lesser objects redistribute momentum slowly, but the tidal disturbance may either be brief (if the wave comes from a catastrophic collision) or periodic (if radiated by a binary star system). MHD dissipation within individual stars acts faster.
As for magnetic fields in empty space, they would push back when squeezed by tidal forces, much as in MHD, so scattering is to be expected, but not absorption, for lack of dissipative processes.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/416081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
can heat radiate from a cold to hot body? if you have 2 bodies (A,B), with Temperature of B>A, can you have heat transfer from A to B through radiation?
In particular if A reflects the wavelengths that B is emitting the radiation at would this be possible?
And if so is this process independent of the temperature of B (since A is reflective) or does A radiate as a function of the temperature difference with its surroundings (in this case B).
I was thinking of how this applies to radiative sky cooling: the process by which you use the radiation of object A to disperse heat, and create A reflective of the Sun's radiation to not heat up from that.
The universe is about 3K, if it was 200K, would this process differ?
| Yes, some times it is possible
This is happening in refrigerator.
In this case some energy must be supplied to complete this transaction
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/416297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
When will velocity and acceleration vectors be perpendicular?
Suppose a particle is moving in the $xy$ plane with
$$x=at, \quad y=at(1-bt),$$
where $a$ and $b$ are positive constants. When will the velocity vector and acceleration vector be perpendicular?
I know that these vectors are perpendicular in circular motion. Should I then use the circle equation
$$ x^2+y^2 = r^2 $$
then substitute $r$ with centripetal acceleration
$$ a_c = v^2/r $$
and then substitute for $v$? But I don't see how I will get an equation that has something to do with vectors.
| Another approach uses $v\cdot a=\frac{d}{dt}(v^2/2)$, so orthogonality is equivalent to $v^2=\dot{x}^2+\dot{y}^2=a^2+(a-2abt)^2$ being constant. This is equivalent to $a=0$ or $b=0$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/417622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Why are photo electrons emitted instantly from metal surface just nanoseconds after the light falls upon it? Why are photo electrons emitted instantly from metal surface just nanoseconds after the light falls upon it? How does the quantum theory of radiation explain it? Why can't classical physics explain this?
| According to quantum mechanics
The energy is transmitted by photon to electrons by the collision between the two. Only those photons can eject electrons which have energy more than or equal to a minimum required energy (threshold energy). Since energy is transferred in a lump, the ejection is instantaneous.
According to Classical mechanics the energy of light is distributed equally on its wavefront. So when light hits the surface, each electron gets a part of the total energy. So, the electron will acquire the sufficient energy in some time. So there must be a time lag between the striking of light and ejection of electron which is not true.
Hence classical mechanics fails in this aspect.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/417748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why is a fermion field complex? The Lagrangian of a fermion field is
\begin{equation}
\mathcal{L} = \overline{\psi} (i\gamma_{\mu} \partial^{\mu} - m)\psi
\end{equation}
It is said that the fermion field $\psi$ is necessarily complex because of the Dirac structure. I don't quite understand this. Why is the fermion field complex from a physical point of view? A complex field has two components, i.e., the real and imaginary components. Does this imply that all fermions are composite particles? For example, an electron is assumed to be a point particle that does not have structure. How can it have two components if it is structureless?
| Any type of field can be complex, not only the fermions. The reason is the $U(1)_{EM}$ symmetry, i.e., the electromagnetic interactions.
The electric charge is the conserved quantity of the $U(1)_{EM}$ gauge symmetry in nature. A transformation of this symmetry is such that
$$
\phi(x) \mapsto e^{iq\theta (x)} \phi (x) \\
\phi^{\dagger}(x) \mapsto e^{-iq\theta (x)} \phi^\dagger (x)
$$
where $q$ is the electric charge of the field, and $\theta(x)$ is the gauge parameter. If $\phi(x)$ is a real-valued field, then the first and second equations should be identical, which implies
$$
e^{iq\theta(x)} = e^{-iq\theta(x)}
$$
This is only true if and only if $q=0$. For complex fields the charge would be opposite for their conjugates.
So, complex fields are charged and real fields are neutral. For example, after the electroweak breaking, the Higgs field is neutral therefore a real-valued boson, while W bosons, i.e., $W^\pm_\mu \equiv W^1_\mu \mp i W^2_\mu$, are charged, so complex-valued bosons. Neutrinos are neutral so they are real-valued fermions, but electrons are charged thus complex-valued fermions.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/417886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Why does this paper use 1/cm for units of frequency? Reading this paper from 1963 $^*$, they use units of cm$^{-1}$ for frequency.
Here is an excerpt:
It doesn't seem like wave number, as they clearly call it frequency. What's going on here?
$^*$ Sievers III, A. J., and M. Tinkham. "Far infrared antiferromagnetic resonance in MnO and NiO." Physical Review 129.4 (1963): 1566.
| People working in the infrared or optical region tend to sometimes, depending on what school of thought they come from, use a unit called wavenumber, which is the reciprocal wavelength in centimeters, $\frac{1}{\lambda_{centimeter}}$.
This is still a frequency, but instead of oscillations per second, it is oscillations per centimeter.
This unit is also called a Kayser. Wavenumber and Kayser are commonly used when working with spectroscopy/diffraction.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/418033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Why does an atom in an oscillating electric field behave like an electric dipole? How can I understand that an atom subjected to an oscillating electric field (e.g., $\vec{E}=\hat{i} E_0\cos\omega t$) behaves like an oscillating electric dipole? What is the underlying picture that comes out of the quantum mechanical description of the atom?
| Well, let’s start with a static electric field: the electrons would move in the opposite direction of it and the nucleus in the same direction. But the nucleus’ and the electrons’ attractive force counteracts this process, thus forming a (steady) dipole.
I know, the picture I painted is very classical, but that’s sufficient. To solve it quantum mechanically you would have to use perturbation theory, only to find the same result.
Now if the external electric field oscillates, the dipole will oscillate accordingly.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/418232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Why is a current carrying loop considered a dipole? I am unable to understand why a current carrying loop is considered a dipole. Why exactly is it called a "dipole", which two poles are we referring to and how do those two poles function as the north and south pole of a magnet?
| A long way away from the current loop, compared with its characteristic linear dimension (for example its diameter) the magnetic field, $\vec{B}$, follows an inverse cube law and is exactly the same in magnitude and direction as the field due to a small dipole (of suitable strength and orientation) whose poles give rise to inverse square law radial fields, one inwards, one outwards. This result is derived in old fashioned textbooks.
However, close to the current loop things are very different. The most important difference is that a true dipole's field lines would start on one pole and end on the other, whereas the field lines due to a current loop are themselves closed loops without ends, linked around the current loop. This means that along the line joining the poles of a true dipole the field would run in the opposite direction from the field due to the current loop!
There is no such thing, as far as we know, as a true magnetic dipole, that is a North pole from which the field is radially outwards and a South pole to which the field is radially inwards. So a current-carrying loop doesn't have a North and South pole in this sense. But if all we're concerned about is its distant field, it does behave as if it had poles (see first paragraph). Even so, the positions of these poles are to some degree arbitrary: two poles closer together are equivalent to two weaker poles further apart!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/418369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
$SO(3)$, orbital angular momentum, vector product I have a big confusion with group theory terminology. I know that orbital angular momentum (OAM) is $\mathrm{SO}(3)$-symmetric in 3D-space. Let's define QM orbital angular momentum (OAM) conventionally:
$$\pmb{L} = -i \pmb{r} \times \pmb{\nabla}$$
This definition can also be written using a set of $\mathrm{SO}(3)$ generators:
$$L^{\mu} = -i r_i \; S_{ij}^{\mu} \; \nabla_j$$
where $\mu = \{x,y,z\}$ for 3D space, and $S_{ij}^{\mu}$.
So... generators stand for the definition of a vector product in given space, essentially, definition of orthogonality? Or this is only in this case, I suppose, in which case why such a coincidence?
If I proceed with this:
$$\pmb{r} e^{-iS^{\mu} \phi} \pmb{p}= \pmb{r} \cdot \pmb{p} - i \delta \phi \; \pmb{r} S^{\mu} \pmb{p} + \cdots = \mathrm{const} \; e^{- i \pmb{r} \cdot \pmb{p}} + \delta \phi L^{\mu}$$
Matter wave in zeroth order and OAM in first? Does it have any interpretation?
| I don't know if I've understood your question well. The math may be too complicated for me, but the ideas are
*
*The usual definition of OAM satisfies the conmutation rule
$$\left[\frac{L_x}{\hbar}, \frac{L_y}{\hbar}\right]=i \frac{L_z}{\hbar}$$
*This means that they are infinitesimal generators of rotations in the space they live in.
*And that's correct: in fact $\exp(-i\theta L_n/\hbar)$ is actually the "rotation" operator of an angle $\theta$ about an axis $n$ in Hilbert's space.
You can prove this by stablishing
$$ \varphi_F (\vec{x}) = \varphi_0(R^{-1}\vec{x})$$
Which means "the updated wavefunction in any point is equal to the old wavefunction evaluated in the point before the rotation".
$R$ is the rotation matrix in $\mathbb{R}^3$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/418463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Suppose the speed of individual photons reduced over time. Would that explain the apparent speeding up of distant galaxies? The observation that distant galaxies seem to be speeding up has led to the theory of dark energy.
However if the speed of individual photons actually reduced over ( very long ) periods of time wouldn't that also offer a valid explanation?
What if we postulate that the speed of newly emitted light and other electromagnetic radiation is c but slows at a constant microscopic rate thereafter.
I would hazard a guess that all our measurements of the speed of light have been made using new, fresh light as it were. has anyone ever measured the speed of the light exclusively from the farthest galaxies etc?
| I think we define distance by light signals, so perhaps the answer is yes, and perhaps the two are in a way equivalent.
On the other hand, the CMB shows us that the size of the fluctuations on the celestial sphere have stretched since the CMB was released. According to what you are saying, while the photons would indeed redshift as explained by user9976437 in their answer, these fluctuations should not, at least naively.
Not to mention that there are many other observables predicted by the standard cosmology that you would no longer be able to predict unless space is expanding.
So I don't think it's a good solution.
I think it's good that you are questioning the standard model of cosmology, and we should all do so, not believe things blindly.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/418716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
What allows dividing equations in 1D finite potential well? So I was reading the lecture notes of 8.04 at MIT on the classic finite potential well problem of QM, until I reach this part in page 8:
Let us finally complete the construction. We must impose the continuity of the wavefunction and the continuity of $\psi'$ at $x=a$. Using the expressions for $\psi$ for $x<a$ and for $x>a$ these conditions give
\left.
\begin{array}{l}
\psi\ \text{continuous at } x=a:&\implies&\cos(ka)&=&Ae^{-\kappa a}\\
\psi'\ \text{continuous at } x=a:&\implies&-k\sin(ka)&=&-\kappa Ae^{-\kappa a}&\tag{2.32}\\
\end{array}
Dividing the second equation by the first we eliminate the constant $A$ and find a second relation between $k$ and $\kappa$! This is exactly what is needed. The result is
$$k\tan ka = \kappa\rightarrow ka\tan ka=\kappa a \rightarrow \xi = \eta \tan\eta \tag{2.33}$$
This is only for the symmetric solutions, but I have the same question for the antisymmetric solutions. My question is about the highlighted part. Why are we allowed to divide the 2 equations like that, without risking making new solutions? As I remember from this Math.StackExchange post, dividing can make new solutions appear.
So what did I miss when studying this problem?
| You're not dividing by zero, so it's fine. The cosine may not obviously be non-zero, but it's equated to an exponential, which obviously is non-zero.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/418891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How does tilting a bike make it turn sharper? Note that my question is not why do you tilt your bike when on a curve. It's about the reduction in turning radius when one tilts the bike inwards.
Short to-the-point answers are welcome.
| There is an effect is due to geometry.
Turn the front wheel of a bike to the left by thirty degrees, say.
Now lean the bike into the turn.
For simplicity, suppose that you could lean the bike completely on its side while the tires maintain contact with the ground.
Having done so you will find that the turning radius is now roughly the one that would be achieved by turning the front wheel to ninety degrees while keeping the bike upright.
(If the front wheel can't be turned to ninety degrees, this turning radius is not even achievable without tipping the bike.)
This shows that tipping the bike affects its turning radius.
Figure 1. Here we assume a rake of 25 degrees. $\theta_x$ is the tilting angle and $\theta_z$ the turning angle with respect to the handlebars. $\phi = \theta'-\theta$, where $\theta$ and $\theta'$ are the angles the tire makes with respect to the forward direction before and after tilting the bike. (Note that due to the rake of the bike, $\theta_z\ne\theta$.)
Figure 2. One can see that for small turning angle that tilting the bike has little effect, agreeing with our intuition, but that for any turning angle the effect grows with tilting angle and can become a large effect.
Figure 3. For a turning angle of two degrees the effect of tilting 45 degrees is appreciable.
Figure 4. Here we assume a rake of 25 degrees and let the turning angle be ten degrees. The tilting angle varies from zero to 45 degrees.
Figure 5. A top view of the situation depicted in Figure 4.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/419353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
"answer_count": 8,
"answer_id": 6
} |
What does it mean if the dot product of two vectors is negative? If the dot product gives only magnitude, then how can it be negative?
For example, in this calculation:
$$W = \vec{F}\cdot\vec{r} = Fr\cos\theta = (12\ \mathrm{N})(2.0\ \mathrm{m})(\cos 180^\circ) = -24\ \mathrm{N\,m} = -24\ \mathrm{J}$$
Why is there a negative sign? What does it tell us?
| $$W = \vec{F}\cdot\vec{r} = Fr\cos\theta$$
So this is negative when $\frac{\pi}{2}<\theta<\frac{3\pi}{2}$.
It's telling you that the two vectors are pointing in more or less the opposite direction. Or more precisely, if you projected $\vec{F}$ onto $\vec{r}$, the projection would be in the opposite direction from $\vec{r}$ (or vice versa).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/419657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
DBI action expansion for non-abelian brane worldvolume I am trying to reproduce the results of the (famous) Myer's paper "Dielectric Branes" https://arxiv.org/abs/hep-th/9910053. In eq (33),
when he expands the determinant factor for a flat-space background, there is a 1/4 in the first term that, says me, should be 1/2. It is rather straightforward to compute the determinant in the flat background limit. This is my computation:
$$
Q^i_j=\delta^i_j+i\lambda[\phi^i,\phi^k]\eta_{kj},\\
\Rightarrow\det Q^i_j=1-\lambda^2[\phi^i,\phi^j][\phi^j,\phi^i]\\
\sqrt{\det Q^i_j}\approx1-\frac{\lambda^2}{2}[\phi^i,\phi^j][\phi^j,\phi^i].\\
$$
But the end result of Myer's is
$$
\sqrt{\det Q^i_j}\approx1-\frac{\lambda^2}{4}[\phi^i,\phi^j][\phi^j,\phi^i]\tag{33}.
$$
The steps are so straightforward that I feel I am really missing some fundamental content here. Does somebody knows where this extra 1/2 factor comes from?
| Hint: If we write
$$ Q^i{}_j ~=~\delta^i_j -C^i{}_j\qquad\Leftrightarrow\qquad Q~=~{\bf 1} -C , \tag{A}$$
then
$$\ln \det Q ~=~ {\rm tr}\ln Q~\stackrel{(A)}{=}~{\rm tr}\ln ({\bf 1} -C)~=~-\sum_{n=1}^{\infty}\frac{1}{n}{\rm tr}(C^n)~=~-\underbrace{{\rm tr}(C)}_{ = 0 } -\frac{1}{2}{\rm tr}(C^2) + {\cal O}(C^3),\tag{B}
$$
so that
$$ \sqrt{\det Q}~=~ \exp\left(\frac{1}{2}\ln \det Q \right)
~\stackrel{(B)}{=}~1-\color{red}{\frac{1}{4}}{\rm tr}(C^2) + {\cal O}(C^3) .\tag{C}$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/419877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What does $\Lambda^{-1}_{\frac{1}{2}}\gamma^\mu\Lambda_{\frac{1}{2}}=\Lambda^\mu_{\phantom{\mu}\nu}\gamma^\nu$ mean? \begin{equation}
\Lambda^{-1}_{\frac{1}{2}}\gamma^\mu\Lambda_{\frac{1}{2}}=\Lambda^\mu_{\phantom{\mu}\nu}\gamma^\nu
\end{equation}
In P&S, p. 42:
Equation (3.29) says that the $\gamma$ matrices are invariant under simultaneous rotations of their vector and spinor indices. In other words, we can “take the vector index $\mu$ on $\gamma^\mu$ seriously,” and dot $\gamma^\mu$ into $\partial_\mu$ to form a Lorentz-invariant differential operator.
(3.29) is the equation above. I know l.h.s. is about a spinor and r.h.s. is about a vector since $\Lambda_{\frac{1}{2}}$ is about spinor rotation (and boost) and $\Lambda$ is about vector, but don’t understand what simultaneous rotations are and what this equation mean.
*
*What does the equation above mean?
*What are simultaneous rotations?
| This is just an example of an important property of the GL(N) Lie Group tensor operators. It means that the tensor operator $\gamma^{\mu}$ transforms like a 4-vector under conjugation.
Please see my answer to "Do the Dirac matrices form a proper four-vector?" which might have been better posted here.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/419977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Is general relativity about Lorentzian manifolds only? General relativity is often used in context of Lorentzian manifolds. But the texts which describe Einstein Field Equations discuss them in context of general pseudo-Riemannian manifolds.
It seems natural that general relativity be restricted to Lorentzian manifolds because of time problems, but it is not clear whether there is and should be explicit restriction to Lorentzian manifolds.
| The spacetime in general relativity is defined as an $n$-dimensional Lorentzian manifold. Lorentzian manifolds are a type of pseudo-Riemannian manifold (Since they are manifolds of signature $(p,q)$, a Lorentzian manifold is just a pseudo-Riemannian manifold with $p = 1$), which is why you find the term occasionally in textbooks.
It is indeed easy to generalize general relativity to an arbitrary signature $p,q$, but this is not general relativity then. The case $(0,n)$ is referred to as Euclidian gravity, but there isn't really a term for other cases $(p,q)$, the ultrahyperbolic case. Due to various reasons (see the comments), it's not a terribly attractive theory.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/420162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Euler equations in primitive form for a real gas For an ideal gas, it is relatively easy to express the Euler equations in primitive form (variables $\rho$, $u$, $p$), starting from their expression in conservative variables ($\rho$, $\rho u$, $E$).
I did not find any example of such derivation for a general real gas, governed by any equation of state. Is it possible to express the Euler equations in primitive form for any (unknown) real gas (involving the speed of sound somewhere)?
| Yes it can always be done. I assume you can write the general case in conservation form. so you already have one primitive variable, Then
$$u_t=\rho^{-1}((\rho u)_t-u\rho_t)$$
and$$i_t=\rho^{-1}[(\rho i+\rho u^2/2)_t-(i+u^2/2)\rho_t-\rho uu_t]$$
where i is specific internal energy. Generalization to more dimensions is obvious. You should see some simplification.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/420291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why $\rm Pt$-$\rm Ir$ Alloy or Tungsten is used for the tip in scanning tunneling microscopy? Just want to know the properties which qualify these materials to be used as the same.
| For topography scanning purposes, especially for an undergrad experiment the pre-requisites are the robustness and stability in ambient conditions. Both PtIr and W make hard, and, with quite some practice, sharp tips which are ideal for STM.
Moreover, W tips, when chemically etched are the go-to in most STM labs [1].
[1] https://pubs.acs.org/doi/10.1021/nl010094q
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/420527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
does tension in the string affect its equilibrium? In my textbook (Sears and Zemansky's University Physics), it is written that the vector sum of the forces on the rope is zero, however the tension is 50 N. Then is tension different than the force? And if not, then why force is zero while tension is not?
A body that has pulling forces applied at its ends, such as the rope in Fig 4.27, is said to be in tension. The tension at any point is the magnitude of force acting at that point (see Fig 4.2c). In Fig 4.27b, the tension at the right end of the rope is the magnitude of $\vec{\mathbf{F}}_{M\ on\ R}$ (or of $\vec{\mathbf{F}}_{R\ on\ B}$). If the rope is in equilibrium and if no forces act except at its ends, the tension is the same at both ends and throughout the rope. Thus if the magnitudes of $\vec{\mathbf{F}}_{B\ on\ R}$ and $\vec{\mathbf{F}}_{M\ on\ R}$ are $50\ \rm N$ each, the tension in the rope is $50\ \rm N$ (not $100\ \rm N$). The total force vector $\vec{\mathbf{F}}_{B\ on\ R}+\vec{\mathbf{F}}_{M\ on\ R}$ acting on the rope in this case is zero!
| Assuming the block isn’t accelerating, the sum of the external forces on the rope is zero. But tension is an internal force. You know it exists because if you cut the rope the Mason will go flying. To determine the amount of tension force, you can cut the string removing one end, say the block end, and replace it with the force necessary to maintain equilibrium (Free body diagram). That force has to be equal and opposite to the force exerted by the Mason on the right end. That’s the tension in the string.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/420708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
What solid elements, if formed into their thinnest possible 'sheet', would be translucent? Would this have to be experimentally done or is there some manner to estimate translucency based on molecular properties?
| Every solid object, deposited in a smooth film a single monolayer thick on a transparent substrate, transmits some/most visible light (even metal, graphene and other carbon allotropes, and resonant molecules). In fact it’s more difficult to think of a material/color combination for which most of the light is absorbed. The interaction length is just too small on a single nanometer length scale to absorb or reflect a majority of the light.
And yes, transparency can be estimated purely theoretically. Computational physicists might use density functional theory, for example, to estimate the electrical properties of a proposed material. Then some basic quantum mechanics can calculate the absorption bands etc. Finally, it’s easy classical electrodynamics to compute the transmittance for a given thickness of material.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/420819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Stars filling up the night sky Due to the huge number of stars in the universe, will there ever be a time that the night sky is filled up completely with stars such that the night sky is as bright as it is in the daytime?
| I think the best answer as far as we know, is dissipation of light energy. The electromagnetic phenomena known as light, is a wave of alternating electric and magnetic fields. It is a 3 dimensional wave, so if you have a point source the wave travels outwards like ever increasing spheres. The surface of the spherical wave gets larger and larger. The light loses its intensity over distance, eventually becoming weak enough it is no longer detectable. Imagine how intense the sun appears from mercury compared to Saturn or Pluto.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/421052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the value of the mass gap in QCD? Is there any estimation either from experiments or numerical simulations for the value of the mass gap in QCD?
| As Lubos says in his answer here, there are theoretical proposals fo the mass gap in lattice QCD and other theoretical models.
Searching for "lattice QCD and the mass gap" this paper came up, and they do give a formula for the mass gap (number 22), but in dimensionless numbers!!
It is not a simple concept in the QCD frame.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/421333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What does electrical potential at a point mean? From my understanding, potential difference (or voltage) between point A and point B is the difference in electrical potential at the two points. The potential difference is also, the work done per unit charge in moving charges from point A to point B.
But apparently, the potential at one point (e.g point A) is also measured in voltage? How is it possible that at point A, there is work done per unit charge in moving charges from point A to point A?
| You have understood the concept of potential difference between two points A and B. But you are confused as to why there is a potential at a single point, say P. When we talk about the potential at a point P, we usually refer to the potential difference between infinity and the point P. That is, the potential at point P is defined as the work done in bringing a unit positive charge from infinity to that point. Its simple, just replace point B in your given example with infinity and point A with point P.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/421510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How is work transferred to the system recognised? For example, a potato initially at room temperature $25 \sideset{^{\circ}}{}{\mathrm{C}}$ is baked in an oven that is maintained at $200\sideset{^{\circ}}{}{\mathrm{C}}.$
I made potato as the system and the outer surface of the skin as the system boundary. While the oven and the air inside it is the surroundings.
There is a temperature difference between the skin and the air in the oven which is the driving force of heat transfer (temperature difference).
What about work? Is there transfer through work done?
Isn’t the oven working to produce the heat in the oven which is then transferred to the potato? But work is pressure multiplied by the change in volume. However there’s no change in volume of the potato. So does this mean no work is done?
In summary, how do I identify whether work is done to the system or not?
| The First Law of Thermodynamics says that changes to the internal energy of a thermodynamic system into two ways of energy transfers.
Work refers to forms of energy transfer, which can be accounted for in terms of changes in the macroscopic physical variables of the system, e.g. energy which goes into expanding the volume of a system against an external pressure, by say driving a piston-head out of a cylinder against an external force.
This is in distinction to heat energy carried into or out of the system in the form of transfers in the microscopic thermal motions of particles.
Any net increase in the internal energy U of a thermodynamic system must be fully accounted for, in terms of heat delta Q entering the system less work delta W done by the system:
dU = delta Q - delta W
The Roman letter d indicates that internal energy U is a property of the state of the system, so changes in the internal energy are exact differentials - they depend only on the original state and the final state, not on the path taken.
In contrast, the Greek delta Q or delta W in the equation reflects the fact that the heat transfer and the work transfer are not properties of the final state of the system.
Given only the initial state and the final state of the system, all one can say is what the total change in internal energy was, not how much of the energy went out as heat, and how much as work.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/421645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
What is the difference in conditions of a free electron in metal and a isolated free electron in space-time? I am asking that what environment does the free electrons in metal have,what is net electric fields,force experienced by them.does they have transitions like valence electrons.can they be compared to an isolated electron?.
| The free electron model (FEM) was derived mostly by Sommerfeld in 1927. It does not yield band structures, so there's no energy bands, i.e. it cannot predict the existence of insulators and semiconductors. For such a description, one has to look at the nearly free electron model, which is able to describe energy bands.
In the FEM, the electrons have to satisfy Pauli's exclusion principle, so they cannot have any energy, unlike in the case of an isolated free electron in vacuum. Also, the electron-electron interaction is ignored, so there's no electric field involved in the Hamiltonian of the FEM.
One difference between the FEM and an isolated electron is the concept of temperature, which is well defined in the former case, but not in the latter. At absolute zero temperature, there is a maximum energy the electrons can possibly have in a metal, called the Fermi energy. But in the electron in vacuum case, the concept of temperature is ill defined.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/421851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A slipping cylinder comes into contact with friction Suppose a cylinder is slipping rigidly on a frictionless horizontal surface. Then, at $t=0$ it reaches a different ground. The coefficient of friction between the cylinder and this new type of ground is $\mu$. What happens?
- assuming usual high-school friction to be the only horizontal force and a homogeneous cylinder with moment of inertia $I=MR^2/2$.
I have come up with this question myself, it is not homework.
Once the cylinder starts slipping on the new ground, kinetic friction will dissipate energy, slowing it down. It will also produce torque and make it roll.
Eventually, the velocity of the contact point will vanish and the cylinder will start rolling without slipping (RWS). This happens when $v=\omega R$ where $v$ is the velocity of the center of mass and $\omega$ is the angular velocity with respect to the center of mass.
Equations of motion are $FR=I\dot{\omega}$ and $F=M\dot{v}$. With $v(0)=v_0$ and $\omega(0)=0$ this leads to the conclusion - using that friction is $F=\mu mg$ and if I made no mistake - that RWS will start at $t=v_0/(3\mu g)$.
Now here comes the actual question. Once the cylinder is RWS, the friction force must actually be zero (otherwise, it would slow down the cylinder without performing work, which is impossible).
However, the friction force is not zero immediately before RWS (it is $\mu mg$). Hence the question: is the friction force dicontinuous? Or have I made some mistake?
| You can simplify the problem by looking at a box sliding on rough table, slowing to full stop. When sliding, friction is constant $\mu mg$. Once stopping, friction is zero.
However, the transition is actually from constant kinetic friction force to gradually reducing static friction force.
A rolling / sliding cylinder goes through the same phases.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/421904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Classical theory fails to explain quantization of motions?
I understand everything written here.
But the last point, I cannot get, at all.
How does it point towards Quantization of the two motions, since the energy change is not sudden, but gradual?
And if anything is wrong with the given image, please tell what it is.
| The "wrong" thing in this picture is an illusion of "horizonality" of some parts of this curve. According to the Maxwell distribution $\propto \text{exp}(-mv^2/2kT)$, in the volume there are always high velocity molecules capable to get rotational and vibrational excitations, thus the curve has always a slope as a function of $T$.
For one molecule you have clear thresholds (a step-wise curve), but for a volume of molecules the thresholds are smeared due to statistics applied to calculate/measure the heat capacity of the volume.
Still, one can see a "thershold-like" behavior of the gas heat capacity indicating quantization of rotational and vibrational energies. Without quantization the curve would not be a step-wise at all.
By the way, with $T$ growing, excitations of electron levels come into play. Finally one can finish with fully ionized plasma ;-).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/422075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Was the universe already expanding before inflation occured? Was the universe already expanding before inflation occured or did inflation cause the universe to start off expanding?
By “cause it to start off expanding” , I mean the cause of the initial expansion.
| Here is a link discussing cosmological models, as one has to keep in mind that the models develop in order to fit/explain observations using current physics theories, classical and quantum mechanical.
The current mainstream model can be seen here, and your question is about what happens at the quantum era, before $10^{-32}$ seconds.
Here is an image of the current mainstream chronology that includes the current models for that time:
Between the fuzzy ball that has replaced the classical singularity and to the end of inflation at $10^{-32}$ seconds, quantum gravity reigns, and quantum gravity is still an open problem.
The above current model assumes a period before inflation starts, where the inflation is very small. The inflation period starts later than the fuzzy beginning of universe, and is
Triggered by the symmetry breaking that separates off the strong force, models suggest an extraordinary inflationary phase in the era $10^{-36}$ seconds to $10^{-32}$ seconds. More expansion is presumed to have occurred in this instant than in the entire period ( 14 billion years?) since.
So the answer is that in the mainstream cosmological model the universe is expanding after the fuzzy beginning at a very low rate, and suddenly inflation sets in because the temperature has dropped to the symmetry breaking between strong and electroweak forces.
Keep in mind that cosmology is an active field of research, and we are discussing here the current mainstream model.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/422296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.