Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
Can a virtual image created by a mirror have position in front of the mirror? I dont think so, because then the rays will intersect and the image will be real.
Please let me know if you know a case in which a virtual image is formed in front of the mirror.
| The first sentence in your question is the answer. With a virtual image, the rays appear to come from an object behind the mirror. The important words there are "appear to." There is no actual object behind the mirror from which the rays are emanating. If there is a real image in front of the mirror and you view it from some distance away, the rays will "appear to" come from it, after they leave the image location, as though there were an object at that location. But, as you say, what you are seeing is the effect of a real image.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/171004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Zero stress boundary conditions for the acoustic wave function When is it appropriate to use zero normal stress boundary conditions when solving the acoustic wave equation. That is when the pressure is equal to zero.
| Strictly speaking: never. Outside the idealized models there are always at least a bit of radiation impedance so the pressure on the boundary is not truly zero.
But let's focus on idealized models. Usually the zero pressure boundary condition is used for open ends of the waveguides or for the free surfaces of solid bodies.
Typical illustrative example would be the loose end of the string: there is nothing (e.g. no bridge) which can provide a support for string stretching. Therefore the tension (analogy of pressure) must be zero and all the energy goes into displacement.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/171106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Excitation energy of carotene using the particle in a box model I'm practicing for an exam and I came across the following question:
The linear, conjugated π-electron system of a carotene molecule comprises 11 atoms and the distance between two atoms is 1.4 Å. Calculate the excitation energy and the wavelength of the radiation for an electron in the described system using the particle in a box model.
My (probably wrong) approach of solving this problem is as follows:
Assume the carotene molecule can be modeled as a particle in a 1 dimensional box of length 15.4 Å. An electron is excited when it goes from the ground state up to the next energy level, so the excitation energy is $\Delta E=E_2 - E_1$ with $E_n=\frac{n^2h^2}{8mL}$.
So calculating the excitation energy with $h=6.626*10^{-34} J*s$, $m=9.109*10^{-31} kg$ and $L=1.54*10^{-9}m$ gives:
$\Delta E =\frac{2^2h^2}{8mL} - \frac{1^2h^2}{8mL} = 3*\frac{h^2}{8mL} = 3*\frac{(6.626*10^{-34})^{2}}{8*9.109*10^{-31}*1.54*10^{-9}}=1.174*10^{-28} J$.
Using the formula $\lambda = \frac{hc}{\Delta E}$ to calculate the wavelength gives:
$\lambda = \frac{hc}{\Delta E} = \frac{6.626*10^{-34} * 2.998*10^{8}}{1.174*10^{-28}} = 1692.1m$.
That's 1692 meters. Quite obviously wrong... Now I don't have the solutions to this problem, so I don't actually know what's supposed to be the right answer, but I know mine is wrong.
So yeah, if someone could point out what I'm doing wrong, I'd be really grateful!
| The energy is:
$$ E = \frac{n^2h^2}{8mL^2} $$
Your mistake is that you have $L$ not $L^2$ in the denominator so your answer is a factor of $L$ too small.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/171182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Shadow of a ripple? Suppose a stone is thrown into a pool of crystal clear water and its a bright sunshiny morning. You can observe a shadow of the wave in the bottom of the pool. Why does this happen? Is it due to superposition of light or some other thing?
| You can actually think of the ripple as a travelling lens.
If you take a radial cross section through the ripple, it'll have a curved profile. Now just like a magnifying glass causes a bright spot in the middle of where you focus incoming light, it also causes a darker region around it.
This is what you're seeing on the bottom of the pool: The band of focussed (or unfocussed, depending on depth) light caused by the lens-like nature of the ripple wave.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/171253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Why does the human body feel loud music? I was sitting close to a speaker and I could feel the sound coming from it all over my body, especially in my heart, and it pounded with the loud beats of the music.
Was my heart pounding because of the excitement at listening to the music or was I really feeling the sound in my heart and all over my body?
I have some understanding that it is all about sound waves & acoustics (bass/low & high pitch/low and high notes etc.) but it is not clear to me.
I hope I have correctly framed my question.
| Along with your pounding heart, you're also experiencing the effects of resonance. In simple terms, there are certain frequencies of sound waves which correspond to the "natural" vibration frequency of your bones. At these frequencies, your bones vibrate with a greater amplitude than other frequencies, and you can think of it as "tapping" into the vibrational energy of your bone system. So some parts of your body may seem to vibrate when you stand next to the speaker which is emitting sound waves (typically bass-enhanced sound), and I'm sure you'll be less worried about in now. You can read more about resonance here: http://en.m.wikipedia.org/wiki/Resonance
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/171342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
When a pn junction is formed, why is a positive region of charge formed on the n side of the junction? I understand that when electrons diffuse from n-side to p-side, negative charge is developed on the p-side. But the mere absence of electrons on the n-side doesn't make that positively charged. The n-side must be neutral as it has no charge now. Where am I getting wrong?
| When n and p crystals are brought closer the excess electrons in n and excess holes in p begins to diffuse to p and n region respectively which is a phenomena relating to the concentration gradient. The electrons which diffuse in the p region from n region actually come from the donor ions (like P, As ). So as these electrons travel from the n semiconductor to p they actually make the ions deficient of electrons giving them a positive charge. Therefore a region of positive charge is created on the n side.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/171438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How is the conservation of momentum satisfied in long-range attraction such as electromagnetism and gravity? I'm not a physicist, but my understanding is that electromagnetism (including attraction between opposite charges) is mediated by the photon, and gravity is probably (hypothetized to be?) mediated by the graviton.
I'm curious how that works from the point of view of the conservation of momentum. My naive imagination is, if a particle leaves A to the direction of B, doesn't that mean that A would have to change its momentum to the other direction (away from B)? And when B absorb this particle coming from A, shouldn't B now change its momentum to the same direction (away from A)?
How come is it that in the case of gravity and electromagnetism, A and B move towards each other as a result of this interaction?
| If you consider things classically (for the moment forgetting about virtual particles as mediators of the force) things get more clear.
For instantaneous forces (which do not exist in nature), momentum conservation comes from the fact, that the forces in nature fulfil Newtons axiom actio = reactio, meaning, that for two particles, that interact we have the equations of motion:
$$m_x \ddot x = F(x, y)$$
$$m_y \ddot y = -F(x, y)$$
For the time derivative of the total momentum we get:
$$\partial_t P = \partial_t (p_x + p_y) = \partial_t (m_x \dot x + m_y \dot y) = m_x \ddot x + m_y \ddot y = F(x, y) - F(x, y) = 0$$.
That is the total momentum is conserved.
If we consider that the fields causing the forces propagate (and thus the forces are not instantaneous) we have to consider the momentum of the fields and can write local equations for momentum conservation.
Now: Do not take the virtual particle thing too serious. They are in many ways just mathematical artefacts of how we compute things in quantum field theory (so called perturbation theory). Most importantly, do not confuse them with some macroscopic particle. Rather they are "packets" of waves. Furthermore each elementary process conserves momentum (techspeak: the momentum is conserved at all vertices of a Feynman diagram)! As they are a computational device, the virtual particles do not follow the usual rules of propagating particles, but even if a virtual particle starts from A with a moment away from the particle B, it still can reach B and there interact and give B the momentum carried away from A (thus conserving the total momentum).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/171564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Can I use the grand canonical ensemble for a photon gas? I have been reading about photon gases at https://www2.chem.utah.edu/steele/doc/chem7040/chandlerch4.pdf. They do the analysis using a canonical ensemble.
Since photon numbers are not conserved, I would have thought it would be more appropriate to use the grand canonical ensemble. Would it be wrong to do so?
| The grand canonical ensemble allows the number of particles of your system to fluctuate but makes the assumption that it is constant amongst the reservoir and system combined i.e.
$$
n_{res} + n_{sys} = const
$$
For the case of photons this is not true.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/171615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Does a Static $E$-field Increase the Gauge Invariant Vector Potential Without Bound? The gauge invariant formulation of Maxwell's Laws (7.13):
Indicates that the transverse electric field is the time derivative of the transverse vector potential.
This gauge invariant vector potential increases without bound as long as there exists a static electric field. Indeed, even when the electric field is removed, there appears to be no mechanism by which the gauge invariant vector potential disappears.
In what way am I misinterpreting (7.13)?
|
This gauge invariant vector potential increases without bound as long as there exists a static electric field. Indeed, even when the electric field is removed, there appears to be no mechanism by which the gauge invariant vector potential disappears.
Static electric field has zero transversal component; entire field is longitudinal.
The unbounded increase of vector potential would occur if you used the gauge where the static electric field is given as time derivative of vector potential.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/171954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Usage of singular or plural SI base units when written in both symbol as well as name I have multiple doubts related to the usage of singular or plural SI base units when written in both symbol as well as name.
I have framed this question under two parts, namely, Part (a) and Part (b). Each part has three sentences which I have written on the basis of my understanding.
Please answer these 6 sentences/questions.
Part (a):
This page says that while using prefix for e.g., centi as in centimeter, it is
l = 75 cm long.(correct)
l = 75 cms long. (wrong)
*
*Does this rule is used for all SI prefixes (having powers of 10)?
*In this regard, we should be saying, or writing that, "how many cm are there in one metre?" (while saying we should say centimter or centimetres?)
*Please strike-through the wrong SI unit in the following sentences.
My weight is 70 kg / kgs,
or My weight is 70 kilogram / kilograms.
Part (b):
and the page next to above web link says, we should write:
2.6 m/s, or 2.6 metres per second.
In this regard, we should say, or write:
*Its speed is 0.26 metres per second.
*This pipe is 0.75 metres long.
*How many cm are there in 2 metres?
| One never pluralizes unit abbreviations.
Your link goes to the BIPM, the body responsible for maintaining the definitions of the international system of units, and is authoritative. The folks at NIST agree and address most of your questions.
I would say
*The pipe is 0.75 m long.
or
*The pipe is 75 centimeters long.
or even
*The pipe is seventy-five centimeters long.
In your example #6 you use an abbreviated unit name without a number, which seems improper but isn't forbidden by either of the references I examined tonight. I would have said
*How many centimeters are there in 2 m?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/172039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
What's my $\mathrm dM$? Gravitational Potential inside a circle of mass
I'm trying to find the gravitational potential for an arbitrary point within a ring of uniform mass density. The point is constrained to be in the same plane as the ring.
So we start with:
$$\Phi=\int G\frac{\mathrm dM}{r}$$
Let's assume that the point of interest is along the $x$ axis $r$ away from the origin (which is at the center of the ring). An arbitrary point on the ring lies at:
$$a\cos\phi\hat{x}+a\sin\phi\hat{y}$$
And of course the point of interest is:
$$r\hat{x}$$
The distance between the point of interest and an arbitrary point on the ring is then:
$$\sqrt{r^2-2ar\cos\phi+a^2}$$
Back to the integral above, we get:
$$\Phi=\int G\frac{\mathrm dM}{\sqrt{r^2-2ar\cos\phi+a^2}}$$
Cool. I'm pretty happy up to this point, but what do I do about the $\mathrm dM$? Were I at the center of the circle, I would use $\mathrm dM=r\mathrm d\phi$. But I feel like it shouldn't be that simple if the center of my integration isn't the center of the circle. Should I use $$\sqrt{r^2-2ar\cos\phi+a^2}\mathrm d\phi~?$$ Am I completely off base here?
| It depends, is the ring infinitely thin? In other words, do they give you $\lambda$ (density per length) or $\rho$ (density per volume). If they give $\lambda$, then $dM=\lambda rd\phi$. This is because $rd\phi$ is a differential length, and multiplying it by $\lambda$ gives you the differential mass at that point. Then you just integrate from $\phi = 0$ to $\phi = 2\pi$.
More formula, you can do what Martin Ueding else posted and integrate over all space and include Dirac delta functions in the density so that, in the end, you get only a nonzero contribution from the integrating over the ring anyway.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/172141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
About Lorentz Group In definition of orthogonal matrices we say that the a matrix $A$ is orthogonal if $A^TA = I$, while for Lorentz Group it is written as $\Lambda^Tg\Lambda = g $. And we say that Lorentz transformation forms an orthogonal group
My Question is why do we insert the $g$ in the above definition?
| $g$ denotes the metric. For Euclidean space the metric is just the unit matrix $I$. For Minkowksi space, which is of interest when talking about the Lorentz group it's the Minkowski metric $\eta_{\mu \nu}$. The lower right matrix inside the Minkowski metric is the 3-dimensional unit matrix and therefore for the space-like components of the Minkowski metric the equation reads $\Lambda^T \eta_{ij} \Lambda = \eta_{ij}$ which is equivalent to $\Lambda^T I \Lambda = I$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/172247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Perception at relatavistic speeds If one were to be traveling at near the speed of light, their mass would be $m_{rel}= m_0 / \sqrt{1 - \frac{v^2}{c^2}}$. For the mass to double the speed would have to be $86.6\%c$
[edit]
To better phrase the question, perhaps I should use relativistic momentum: $p_{rel}=m_0v/\sqrt{1 - \frac{v^2}{c^2}}$
In a space ship at this speed in zero gravity & acceleration, the mass would be weightless anyway. But to swing ones arm left and right, would it seem to have more inertia to move in the direction of travel ? Or does relativistic time compensate this and although the arm moves slower the perception of time corrects this and it appears to move as it should with the same effort?
| From the reference frame of the space ship, your body is stationary. When you swing your arm back and forth, it has a non-zero speed in this frame, and thus its mass, or rather momentum, increases. The space ship's velocity of $v_\mathrm{ship} = 0.866c$ doesn't add to this. That means that the increase in momentum is exactly the same as if you swing your arm right now, on Earth. Thus, you can perform the experiment yourself, and verify that the momentum-increase is negligible. It is there, nonetheless.
From the reference frame of an observer that sees the space ship passing by at $0.866c$, the ship, your body, and your arm has increased their momenta by a factor of 2. When you swing your arm, its velocity will alternate between being slightly below $v_\mathrm{ship}$, and slightly above $v_\mathrm{ship}$. Hence, the observer will measure your arm's increase in momentum $\Delta p_\mathrm{arm}$ to be slightly below 2, and slightly above 2.
Note that velocities at this speed don't add like this
$$v_\mathrm{tot} = v_\mathrm{ship} + v_\mathrm{arm},\qquad\mathrm{(wrong!)}$$
but like this
$$v_\mathrm{tot} = \frac{v_\mathrm{ship} + v_\mathrm{arm}}{1+(v_\mathrm{ship} v_\mathrm{arm}/c^2)}.$$
EDIT: Also, note that what increases is not really mass, but rather momentum, as pointed out by @Jimnosperm. The argument is the same, though.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/172296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Could I break the sound barrier using circular motion? (And potentially create a sonic boom?) Ok, Lets say I get out my household vaccum cleaner, the typical RPM for a dyson vaccum cleaner reachers 104K RPM, Or 1.733K RPS. In theory, this disc would be travelling with a time period of 0.00057692307 seconds, If we take the speed of sound to be 344.2 metres per second, a breach in the sound barrier is easily possible for an item on the edge of the disc.
One question remains: For an extremely strong disc , could an item stuck onto it break the sound barrier, and create a sonic boom?
| The answer is "yes", for speed of sound, and "maybe" for the sonic boom.
It is not the speed a problem, but the apparition of different pressures along a surface or on sides of a surface and the shock waves. The extremities of the turbine inside of a vacuum cleaner can function at speeds above speed of sound if the vibrations caused by pressure are limited.
The "sonic boom" appears only when a sound generator is moving with the speed of sound, that's exactly at Mach 1, therefore the sound waves are propagate together and accumulate.
Above Mach 1 only shock waves are formed, but no sonic boom further, just a small kind of boom from the shock wave.
Thus, in the vacuum cleaner the only source of sound is the motor, and this piece is statical (in fact, the noise of a electrical motor comes from bearings and vibrations of the rotor), so there is not chance of producing sonic booms, only noise from shock waves.
Adding a sound source on the edge of a disk and giving it a velocity at speed of sound, sonic boom(s) can be created "on demand"
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/172451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 5,
"answer_id": 3
} |
Translational versus dilatational zero modes? Why are the zero modes of the SU(2) Yang Mills instanton referred to as translational or dilatational zero modes?
Is this standard terminology?
| Instantons are characterized by the winding number and a set of collective parameters (e.g. location of the centers of the instantons, their sizes and the inequivalent orientations in the global group space / space-time).
Quantum fluctuations of a unit winding number instanton can either leave the collective parameters unchanged (non-zero modes), or change them. Those fluctuations that change the collective parameters correspond to zero modes. There for a translational zero mode is the mode of fluctuation which moves the center of the instanton, and a dilatational zero mode changes its sizes and so forth...
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/172525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why temperature of liquid drops after spraying through a nozzle? We have tested in our lab as mentioned in the picture.
We connected hot water at $130^\circ F$ at $40 Psig$ to a nozzle (bottle sprayer). We measured the temperature differences inside tank and after spraying and found that there is a temperature drop of $50^\circ F$. We know that liquid water is sprayed and some heat is lost in the form of work done inside the nozzle. The surrounding temperature is kept at $75^\circ F$. This temperature difference is due to faster heat transfer happening due to increase in surface area of water as it is sprayed? Or does it have anything to to do with work done and surface tension?
| Given the cooling attendant to evaporation of even small amounts of water, it's likely that it is the source of the cooling effect.
A question was asked above,
"Have you also tested at different relative humidities of the surrounding air?",
but it was not answered.
You should carefully measure the volume in the hot water source vessel before and after you have sprayed the water thru the nozzle. You should also collect the drops of water you've sprayed in a graduated container.
You will collect less than you sprayed, and the difference in the two volumes has evaporated.
The heat of evaporation per mol of water will reveal where the energy went, resulting in cooled drops.
Latent heat of vaporization (or enthalpy of vaporization) is a physical property of a substance. ...
When a material in liquid state is given energy, it changes its phase from liquid to vapor; the energy absorbed in this process is called heat of vaporization. (or enthalpy of vaporization)
The heat of vaporization of water is about 2,260 kJ/kg, which is equal to 40.8 kJ/mol
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/172601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Cooper instability assuming triplet pairing I am stuck on a question in Chapter 11 of Advanced Solid State Physics by Philip Phillips, which asks to do the Cooper instability calculation for triplet pairing.
I attempt to solve the Schroedinger equation
$ [-\frac{\hbar^2}{2 m} (\nabla^2_1 + \nabla^2_2)+V(r_1 - r_2)] \psi(r_1,r_2) = E \psi(r_1,r_2)$
with a antisymmetric wavefunction $\psi$ following the steps in the textbook.
First express the wavefunction in center of mass coordinate
$\psi(r_1,r_2) = \phi(r) e^{i Q \cdot R},$
where $r = r_1 - r_2, Q = k_1 + k_2$ and $R = (r_1 + r_2)/2$.
Expand $\phi(r)$ in a Fourier series
$\phi(r) = \displaystyle\sum_{k} \frac{e^{i k \cdot r}}{\sqrt{V}} \alpha_k$.
Spatial antisymmetry then requires that $\alpha_k = - \alpha_{-k}$, but I don't know how this requirement changes the subsequent calculations.
Any explanation and hint are appreciated.
| After Fourier transform you get an equation
$\displaystyle (E-\frac{k^2}{m})\alpha_{\vec{k}}=\sum_{\vec{k}'}V_{\vec{k}\vec{k}'}\alpha_{\vec{k}'}$
Here $V_{\vec{k}\vec{k}'}\sim \int \mathrm{d} \vec{r} e^{i(\vec{k}-\vec{k}')r}V(\vec{r})$.
Now we need to make some simplifying assumption about $V$. For $s$-wave, usually we take $V_{\vec{k}\vec{k}'}=\text{constant}$ (and some cutoff by Debye frequency). We can expand $V$ into partial waves:
$V_{\vec{k}\vec{k}'}=\sum_{l,m}V_{l}(k,k')Y_{l,m}(\hat{k})Y_{l,-m}(\hat{k}')$.
And write $a_{\vec{k}}=\sum_{lm}a_{lm}(k)Y_{lm}(\hat{k})$. Now we can work within a fixed angular momentum channel:
$\displaystyle (E-k^2/m)\alpha_{lm}(k)=\sum_{k'}V_{l}(k,k')\alpha_{lm}(k')$.
Unfortunately this is still not solvable and further assumption has to be made about $V_l(k,k')$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/172773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Why is $F=ma$? Is there a straightforward reason? Why is force = mass $\times$ acceleration? I have searched in many sites but didn't actually get at it. Simply I want to know that if a mass in space moves (gains velocity thus further accelerates), how can I think, postulate and further believe that force = multiplication of mass and its acceleration ?
| If a particle is left alone in an inertial system, it will travel on a straight line, that is
$$ \frac{d\vec{v}}{dt} = 0 \tag1 $$
it will not change the magnitude and direction of its velocity. This is an empirical fact. So if a particle does change its velocity the right hand side of (1) is not zero. Lets call it $\vec{A}$:
$$ \frac{d\vec{v}}{dt} =: \vec{A} \tag2 $$.
This is rather a definition for $\vec{A}$, because one can only measure the change in position and then assign a value for $\vec{A}$. Lets assume we know the cause of $\vec{A}$, e.g. because we applied the action of spring under tension to our particle. Furthermore we observed that $\vec{A} = \textrm{const.}$. If we now apply the same action to a second particle of double the mass what we will empirically observe is
$$ \frac{d\vec{v} \ '}{dt} = \frac{1}{2} \vec{A} $$.
This suggests that $\vec{A}$ is not an intrinsic property of the action (since we changed the particle and not the action) but rather $m\cdot\vec{A}$ is.
One can now repeat that process with particles that don't differ in mass but for example in color, volume, etc. What one will find is that really only $\vec{A}$ and $m$ matter.
Lets give that intrinsic property of action a name and call it force:
$$ \vec{F} := m \cdot \vec{A} \ \underbrace{=}_{(2)} \ m \frac{d\vec{v}}{dt} = m \vec{a}$$
with $\vec{a}$ the acceleration of the particle. So the formulation of the equation of motion is the desire to find an expression with quantities that belong to delimitable subsystems only. In this sense $\vec{F}$ is an abstract quantity that is by definition equal to $m\vec{a}$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/172848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 3
} |
If the Earth is a good conductor of electricity, why don't people get electrocuted every time they touch the Earth? Since the Earth is a good conductor of electricity, is it safe to assume that any charge that flows down to the Earth must be redistributed into the Earth in and along all directions?
Does this also mean that if I release a million amperes of current into the Earth, every living entity walking barefooted should immediately die?
| Electricity isn't a gas that expands out to shock anything in contact with it. Electricity is a flow from high voltage to low voltage. Touching a charged object is only dangerous if you become a current path--if it uses you to get somewhere. Even if the earth had a net charge, you aren't providing it anywhere to go, so you will not be shocked. It's somewhat like a bird on a power line.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/172939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
"answer_count": 5,
"answer_id": 2
} |
Force between two finite parallel current carrying wires Remark: This is not a homework question...It is pure out of theoretical interest. I asked this the mathematics-community a couple days ago and got no answer, so I figured I'd try here.
Most standard physics textbooks compute the force two infinite wires exert on each other, but they remain silent about the case where the wires are finite. Let's say we have two parallel wires carrying a current of equal magnitude in the same direction, both of which have a length $d$ and also seperated by a distance $d$. I now want to find out the force one wire exerts on another, using the Biot-Savart Law.
Let the left wire be positioned at the origin of the $xy$-plane, going along the $y$-axis, and let the other wire be a distance $d$ to the right. We assume the currents are flowing in the positive $y$-direction. Then we first choose a source element (on the left wire) of infinitesimal length $dy$ described by the position vector $\mathbf{r_0} = y_0 \hat{j}$. This constitutes a current source of $I d\vec{l} = (Idy) \hat{j}$.
We then pick an arbitrary field point $P$ on the other wire with position vector $\mathbf{r_p} = x\hat{i} + y \hat{j}$. Then the position vector $\mathbf{r}$ pointing from the source point to the field point is given as
\begin{align*} \mathbf{r} = \mathbf{r_p} - \mathbf{r_0} = x\hat{i} + (y - y_0) \hat{j}, \end{align*} with $\sqrt{x^2 + (y-y_0)^2}$ being the length of this vector. If we now calculate the crossproduct $d\vec{l} \times \mathbf{r}$, we can write it as \begin{align*} dy \hat{j} \times (x\hat{i} + (y - y_0)\hat{j}) = -dy x \hat{k} \end{align*} Now comes the tricky part. I think I need to setup a double integral, because we are working with infinitesimal force elements $d\mathbf{F}$, each which is given as $d\mathbf{F} = I d\vec{l} \times \mathbf{B}$. But we also have that \begin{align*} d\mathbf{B} = \frac{\mu_0 I}{4 \pi} \frac{d \vec{l} \times \hat{r}}{r^2}
= \frac{\mu_0 I}{4 \pi} \frac{d \vec{l} \times \mathbf{r}}{r^3} = -\frac{\mu_0 I}{4 \pi} \frac{dyx}{\sqrt{x^2 + (y-y_0)^2}} \hat{k}
\end{align*} Hence I need to somehow integrate over $d\mathbf{F}$ and $d\mathbf{B}$. Does anyone have an idea how to do this?
| Let's call the circuit in the origin circuit one and it's line element $\mathrm{d}l_1=(0,\mathrm{d}y_1,0)$ and the one to it's right $r_2=(d,y_2,0)$ then the force between them is $\mathrm{d}F_{12}=i \mathrm{d}l_2 \times B_1$ where
$$B_1=\frac{\mu_0i}{4\pi}\int_{l_1} \frac{\mathrm{d}l_1\times \Delta r}{(\Delta r)^3}$$
and $\Delta r=(d,(y_2-y_1),0)$ so we have that
$$\mathrm{d}l_1\times \Delta r=(0,0,-\mathrm{d}y_1 d)$$
so we get as you wrote:
$$B_1=-\frac{\mu_0 i d}{4 \pi}\int_{0}^{d} \frac{\mathrm{d}y_1}{(d^2+(y_2-y_1)^2)^{\frac{3}{2}}}$$
Ok now let's call $y_2-y_1=t$ so $\mathrm{d}t=-\mathrm{d}y_1$ then we can write
$$B_1=\frac{\mu_0 i d}{4\pi}\int_{y_2}^{y_2-d}\frac{\mathrm{d}t}{(d^2+t^2)^{\frac{3}{2}}}$$
we now make the substitution
$$t=d\cdot \sinh(u)$$
and we obtain
$$\mathrm{d}t=d\cdot \cosh(u)\mathrm{d}u$$
and then
$$B_1=\frac{\mu_0 i d}{4\pi}\int \mathrm{d}u \frac{d \cosh(u)}{d^3 \cosh(u)^3}$$
in which we used
$$\cosh(u)^2-\sinh(u)^2=1$$
$$B_1=\frac{\mu_0 i }{4\pi d}\int \frac{\mathrm{d}u}{\cosh(u)^2}$$
now $\frac{1}{\cosh^2(u)}$ is the derivative of $\tanh(u)$ so
$$\int \frac{\mathrm{d}u}{\cosh(u)^2}=\tanh(u)$$
we get then
$$B_1=\frac{\mu_0 i }{4\pi d}\tanh\left(a\sinh\left(\frac{y_2-y_1}{d}\right)\right)+\text{const}$$
where we have substituted back all parameters
$$u=a\sinh\left(\frac{t}{d}\right) \\ t=y_2-y_1$$
so knowing that (where $a\sinh(x)$ is the inverse function of $\sinh(x)$):
$$\tanh(a\sinh(x))=\frac{x}{\sqrt{x^2+1}}$$
finally
$$B_1=\frac{\mu_0 i }{4\pi d} \frac{\frac{y_2-y_1}{d}}{\sqrt{(\frac{y_2-y_1}{d})^2+1}}+\text{const}=\frac{\mu_0 i }{4\pi} \frac{1}{\sqrt{(y_2-y_1)^2+d^2}}+\text{const}$$
now we calculate it between $y_1=0$ and $y_1=d$ which yields
$$B_1=\frac{\mu_0 i }{4\pi} \left[ \frac{1}{\sqrt{(y_2-d)^2+d^2}}-\frac{1}{\sqrt{y_2^2+d^2}}\right]$$
to calculate the force we take $B_1=(0,0,B_1 \hat{z})$ and we operate the following:
$$\mathrm{d}F_{12}=i\mathrm{d}l_2 \times B_1=i(B_1\mathrm{d}y_2,0,0)$$
now we have to integrate on the circuit two:
$$F_{12}=\frac{\mu_0 i^2 }{4\pi} \int_{0}^{d}\mathrm{d}y_2\left[ \frac{1}{\sqrt{(y_2-d)^2+d^2}}-\frac{1}{\sqrt{y_2^2+d^2}}\right]=\frac{\mu_0 i^2 }{4\pi} \left[I(y_2-d)-I(y_2)\right]$$
and know we do the same trick as before
$$t=y_2-d \ \text{or}\ t=y_2\ \text{for the second piece}$$
$$t=d\cdot \sinh(u)$$
$$\mathrm{d}y_2=\mathrm{d}t=d\cdot \cosh(u)\mathrm{d}u$$
then:
$$I=\int \mathrm{d}u\cdot d \cdot \cosh(u) \frac{1}{\sqrt{d^2\cosh^2(u)}}=u=a\sinh\left(\frac{t}{d}\right)$$
we finally get
$$F_{12}=\frac{\mu_0 i^2 }{4\pi}\left[a\sinh\left(\frac{y_2-d}{d}\right)-a\sinh\left(\frac{y_2}{d}\right)\right]_0^d$$
which curiously enough is zero for this choice of parameters! I hope that helped!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/173035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Connection between the cosmological constant $\Lambda$ and the cutoff scale $\Lambda$ I'm trying to understand the connection between the $\Lambda$ from cosmology and the $\Lambda$ from QFT.
Cosmology: The cosmological constant enters the Einstein equations. In the special case of the Friedman universe it enters the Friedman equations. If we consider a spatially flat de Sitter universe, we get that
$$H_\Lambda^2=\frac{\Lambda c^2}{3}$$
therefore $\Lambda$ has the mass dimension of 2 ($[c]=0$, $[H_\Lambda]=1$).
QFT: The cut-off parameter $\Lambda$ in momentum space is such that the integration goes till $k\leq\Lambda$. Therefore $\Lambda$ has the mass dimension 1 ([k]=1).
On the other hand, both scales seem to be used interchangeably, for example here: http://www.perimeterinstitute.ca/videos/cosmological-constant To quote:
If we take the idea of the Planck length as a fundamental (minimum)
scale and if additionally we impose the Cosmological Constant
($\Lambda$) as and infrared (IR) cut-off parameter.
But how can this work if the cosmological constant and the cut-off parameter have different mass dimensions?
| Trying to answer my old question myself: The cited piece is talking about an infrared cutoff, which basically amounts to giving the propagating particle a mass which appears squared in a propagator. So I think, what is meant is considering
$$\lim_{\Lambda\to 0}\int \frac{d^4p}{p^2+\Lambda}$$
and this $\Lambda$ of course has a mass dimension of 2.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/173156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Does the microgravity environment in highly elliptical orbits differ from circular orbits? I think everyone understands the microgravity environment broadcast from the ISS. But the ISS stays in a fairly circular orbit, the acceleration of gravity should be fairly uniform, the altitude and velocity changing very little.
But when a ship goes into an highly elliptical orbit (like a Geosynchronous Transfer Orbit), its velocity fluctuates alot. (Kerbals taught me that much, RIP) Does this generate any acceleration (g-force) effects that you would be able to note in the spacecraft?
Bonus points for any insights into a lunar trajectory.
| To first order, no. The definition of an orbit is that it's a free-fall trajectory — since everything around you is always experiencing the same acceleration as you are, you cannot actually perceive this acceleration without some external reference (like measuring your velocity compared to the Earth, and how fast it changes).
That said, the closer you get to the planet, the steeper the gradient of the gravitational field will be, and so tidal effects will get stronger at low altitudes. Depending on just how eccentric your orbit is, how big your space station is, and how sensitive your measurements are, these effects could be noticeable.
Ps. Since you mention Kerbal Space Program, one way you might be able to observe these higher-order effects would be make a spacecraft built out of two parts that are docked together, rotate it so that the docking port is exactly aligned with the orbital direction, and undock (and kill any relative velocity the undocking might have produced). In a perfectly circular orbit, if you did this just right, the two components should stay at the same distance from each other, sharing the same orbit.
In an elliptical orbit, however, the distance between the parts should expand close to the planet, and shrink when moving away from it. Essentially, this is because, with the parts in the same orbit, their time separation remains constant, but their orbital velocity varies.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/173326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Quantum anharmonic ocscillator $E_0(\lambda)$ curve or table I am looking for the exact data on $E_0(\lambda)$ for the anharmonicity $\lambda x^4$. The perturbative expansion is the following: $E_0(\lambda)\approx 0.5(1+1.5\lambda -5.25\lambda ^2+41.625\lambda^3-...)$, but I need a curve or a table for the exact values of the ground state energy for different $\lambda$ including a "strong coupling" regime $\lambda\gt 1$. Quickly accessible data in any form will be appreciated. Thanks.
| The problem is not exactly solvable. But the recent paper
http://journals.jps.jp/doi/abs/10.7566/JPSJ.83.034003 gives exact Pade-approximants of various orders, which are probably quite accurate.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/173383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $(e^{i\lambda A})^\dagger=e^{-i\lambda A^\dagger}$
Prove $$(e^{i\lambda A})^\dagger=e^{-i\lambda A^\dagger}$$ where $A$ is an operator.
Can anyone explain how to go about this question? Writing it as a power series gets confusing.
So basically I get:
$(e^{i\lambda A})^\dagger=\sum_{n=0}^\infty ({(i\lambda)}^n\frac{A ^n}{n!})^\dagger=\sum_{n=0}^\infty ({(-i\lambda)}^n\frac{(A^\dagger)^n}{n!})=e^{-i\lambda A^\dagger}$
Is this right?
|
Prove $$(e^{i\lambda A})^\dagger=e^{-i\lambda A^\dagger}$$ where $A$ is an operator.
Can anyone explain how to go about this question? Writing it as a
power series gets confusing.
So basically I get: $(e^{i\lambda A})^\dagger=\sum_{n=0}^\infty({(i\lambda)}^n\frac{A ^n}{n!})^\dagger=\sum_{n=0}^\infty({(-i\lambda)}^n\frac{(A^\dagger)^n}{n!})=e^{-i\lambda A^\dagger}$
Is this right?
Yes, it is "right" (assuming that $\lambda$ is real). But you might want to explain the steps, i.e. how you go from one equality to another. For example, you made some assumptions:
$$
(\hat X+\hat Y)^\dagger=\hat X^\dagger + \hat Y^\dagger
$$
$$
(ix\hat X)^\dagger=-ix\hat X^\dagger
$$
etc.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/173685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What exactly is the mass of a body? What determines it? The term "mass" is very common. But what does it depend on? How is it known?
| What is physics? Physics is the modeling with mathematics of observations in the world around us. It is a way of creating a logical sequence that can be predictive and not only explanatory. It reduces the innumerable constants one would need to describe, for example , the trajectory of a ball with just space coordinates, to a simple parabolic function that can be used to give any one of those space coordinates.
To do this, find mathematical functions, it needs some basic assumptions, postulates, gleaned from the observations of falling apples and shooting arrows. These functions are described in the other answers. It was found that a unique mass assigned for each object allows a mathematical description of gravity with what is known as Newtonian mechanics.
We describe our observations within a system of units. Take the International system of units. It is the MKS, Meter, Kilogram Second . It defines the basic numbers for three independent postulates, that distance exists, mass exists and time exists. These are postulated in order to be able to model nature with mathematical functions.
But what does it depend on? How is it known?
Mass is a number unique for every physical body so that the mathematical model of Newtonian mechanics can describe and predict its position and reactions with the appropriate formulas . This number is measured by use of the formulas as explained in the other answers.
When physics started studying the microcosm of molecules atoms nuclei and particles the postulates were carried over , the mathematical descriptions of observations became much more complex, and still, each particle is assigned a mass measured and tabulated so that the mathematical models in use are descriptive and predictive of future behavior.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/173767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Do magnetic fields cause ionisation of gases? I am doing my final year engineering project on Magnetic Field Assisted Combustion and was curious to see what people thought about it.
Companies sell rare earth magnetic arrangements to be attached to fuel lines of gas burners and they are said to improve combustion efficiency but why exactly?
I have performed a number of experiments using a standard butane/propane gas burner with some magnets manufactured by one of said companies and have had some contradicting results. With lower strength magnets, heat transfer unexpectedly slowed down but with a much stronger arrangement, heat transfer rate was increased.
Also, the burn out time of the same amount of gas took 8 minutes less with the magnets in place around the fuel line.
I have read a number of journals on similar subjects but even within these, the actual reason for the increase in heat output is still not known.
Any thoughts on the subject would be massively appreciated and possibly give me some other areas to investigate that I have not already thought of.
| i would say yes. magnetic field does ionize the gas. "Ionization is the process by which an atom or a molecule acquires a negative or positive charge by gaining or losing electrons". and this as a topic already. Can I produce radio waves by waving my hand? therefore the gas MOVING through a magnetic field would accumulate a charge.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/174031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 3
} |
At what point does force stop translating an object and start purely rotating it? At what point (or distance) from the axis of rotation, does force applied on a rigid body stop translating and purely rotating the body? Can such a point even exist? Does the body always have to translate?
This question assumes that the body is in empty space and unattached.
| If there is no fulcrum, if there is no fixed pivot, a body will always translate.
Such point does not exist. If the body is free, it can only translate but can never only rotate: if the impulse is at the Center of Mass linear velocity will be 100% , the minimum percentage of translational velocity is 25%, and it is reached at one tip of the rod. No matter what is the length or the mass of the rod, there is no point where linear velocity can be zero and rotational velocity 100%
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/174131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Scalar and vector defined by transformation properties In Classical Mechanics, we are defining scalars as objects that are invariant under any coordinate transformation. Vectors are defined as objects that can be transformed by some transformation matrix $ \lambda $ .
Why is this important? I get that it leads to other properties such as the invariance of the dot product under coordinate rotations but how does this relate to physics? This is supposed to lead to another question but I will refrain from posting so that I may think a little about it.
I also have seen Noether's Theorem explaining that symmetries pop out conservation laws, such as the time independence of the Lagrangian gives you the Hamiltonian equating to the total energy of the system.
| Just as a quick example: say that the dot product were not invariant under transformations. Then let's say that we have two reference frames, A and B, where reference frame B is rotated and displaced with respect to A and which moves at a constant speed w.r.t. A (where $v\ll c$).
Then the researcher in A wants to calculate the gravitational attraction between two known masses. For this he seperates the masses by a rod which seperates the masses and which extends from the origin of his reference frame to some point $\vec r={(x,y,z)}$. The gravitational strength is proportional to the length of this rod. How does he measure the length? He takes the dot product of $\vec r$ and takes the square root of the resulting number.
Now the researcher in B wants to know the length and the resulting gravitational attraction from the known masses, but since he is far away from the rod he can't measure the length directly. So he says to researcher A: ok, I can't see the rod, but there is a set of rules which tell me how to calculate where your origin is and where the point $\vec r$ is viewed from my frame, from this I can calculate the length. These rules are of course the galilei transformations.
After he has done the calculation he finds out that the rod is shorter in his frame since the dot product is not invariant, and because the rod has a different length the gravitational attraction between the masses is different. This is nonsensical! The gravitational attraction between two masses should not depend on the reference frame.
This is why vectors and transformations and tensors and what not exist: such that the laws which govern one frame of reference are also valid in any other frame which is obtained from the first by a transformation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/174194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Can the gravitational constant be directly measured? The value of the gravitational constant $G$ is $6.67384\times 10^-11 {\rm m}^3{\rm kg}^{-1}{\rm s}^{-2}$. This value is used in the formula to calculate the attractive force between two objects with mass, my question is:
since it is a constant why does it have units? As it has unit can it be directly measured?
I understand that the gravitational constant is not the acceleration force exerted by an object with mass such as earth.
is it related to a hypothetical force carrier, the graviton, since the value seems incredibly small? or maybe since the unit contains measurement of units in 3d space it has something to do with special relativity?
| The gravitational constant is an empirical physical constant found through experimentation, a famous one being the Cavendish experiment.
You can read more about it here.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/174276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is it possible to formulate a Hamiltonian for a damped system? I recently found out that it is possible to formulate a Hamiltonian for a system with time-dependent coordinates such that the Hamiltonian is not the same as the energy When is the Hamiltonian of a system not equal to its total energy? and that has me wondering if it is possible to formulate a Hamiltonian for a damped system under these conditions. I know that Hamilton's equations require that energy be conserved, but if the coordinates are time-dependent, would it still be possible to formulate and solve the problem?
I started trying to do it for a damped simple harmonic oscillator by starting with the Lagrangian for the system
$$L=e^{\gamma * t}*(\frac{mv^2}{2}-\frac{kx^2}{2}),$$
but I keep on coming up with a Hamiltonian that is just equal to the energy
$$H=e^{\gamma * t}*(\frac{mv^2}{2}+\frac{kx^2}{2}).$$
| The following might help:
$H = \frac{1}{2}(mv^2 + kx^2) + \gamma mkvx$
decays exponentially with time along the solution of the damped system. Check by differentiating $H$ with respect to $t$ and using the equations of the system. So the "energy" $H$ decays exponentially instead of remaining constant.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/174376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Probability of photon emission If a photon of a given wavelength is absorbed by an electron (for simplicity, let's assume the electron has only one excited state), does the probability that the electron jumps to its excited state and emits a photon as it falls back depend on the incoming photon's wavelength (is the electron more likely to emit a photon if the incoming photon has an energy much greater than the excitation energy relative to an incoming photon with an energy very close to the excitation energy)?
| There is an equation that helps a lot understanding this issue: Fermi's golden rule
$$
W_{i\rightarrow f}=\frac{2\pi}{\hbar} \left|\left<f\right|H'\left| i\right> \right|^2 \rho
$$
It describes the transition rates from one state to another.
$\rho$ is the so called Density of States (DOS) of final states. This system has only two states: The initial state $\left|i\right>$ and the final state $\left|f\right>$.
Therefore the DOS of final states is basically something like this the following, where $\omega_{ph}$ is the photon frequency and $\omega_f=\frac{E_f}{\hbar}$.
$$
\rho=\frac{1}{V}\delta(\omega_{ph}-\omega_f)
$$
Hence Fermi's Golden Rule for this case reads
$$
W_{i\rightarrow f}=\frac{2\pi}{\hbar V} \left|\left<f\right|H'\left| i\right> \right|^2 \delta(\omega_{ph}-\omega_f).
$$
The properties of the Delta-distribution indicate that only photons with the exact right amount of energy can excite the electron to the final state. In reality there are so called "line broadening" mechanisms (e.g. spectral line broadening) that also allow photons with an energy close to the excitation energy to excite the electon.
Conclusion:
Only photons with a wavelength that correlates to an energy very close to the excitation energy of the electron can excite the electron. Photons with less energy won't be absorbed as well as photons with more energy.
I hope this helped you at least a little bit!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/174450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Will the electrostatic force between two charges change if we place a metal plate between them? If a thin metal plate is placed between two charges $+q$ and $+q$, will this cause a change in the electrostatic force acting on one charge due to another? What is the concept behind this? What will happen if the metal plate is thick?
| As the electrostatic charge is two body attraction or repulsion,The force between them will not be affected but The permittivity increases on inserting metal plate Which causes decrease in the Electrisratic force between two charges.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/174514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
} |
Can gyroscope work in zero gravity? Most ships have two or more gyroscopes to balance on water, man made satellites uses gyroscope for orientation as they fall around earth. All these applications seems to be associated with gravity, therefore how can a gyroscope works in zero gravity?
| Gyroscopes depend on the conservation of angular momentum. Orientation and navigation gyroscopes are finely balanced/symmetrized so that gravitational fields will not exert external torque and modify the angular momentum.
As the container which holds the gyroscope moves, a gimbal mount allows the gyroscope to maintain a constant rotational axis orientation. Sensors measure the changes between the container and the axis of rotation so that navigation parameters are known.
Even if there was no gravitational field, the angular momentum (and hence the orientation of the axis) would be constant.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/174730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Beginners Textbooks in physics Hello I am fifteen and I already know everything that my school has been teaching me so I have been going ahead. I have already been studying mathematics far past where I am at school, but I am very interested in physics. I want to learn everything up to advanced topics such as super-string theory. But to get there, I obviously have to start at the beginning. Any good textbooks out there for somebody like me? Preferably something with a lot of practice problems and that has many applications.
| I don't know how much you are adept at mathematics but before you begin physics, you ought to study calculus from Thomas' Calculus & Differential & Integral Calculus by Richard Courant.
Now, first of all, you'll come across Newtonian Mechanics. For this, I would recommend A.P.French's Newtonian Mechanics. Unlike Lectures of Feynman(though they are really good and one of their kind), here every concept is explained using necessary words; no beating about bush and of course its pictures are really intuitive. This book is designed to be a self-contained introduction to Newtonian Mechanics. Students with little or no grounding in the subject can be brought gradually to a level of considerable proficiency.
Another book, I would recommend is the Berkeley Physics course- Mechanics. This is one of the best beginner's book for Newtonian Mechanics.
And I would advise you that first study mathematics- binomial & multinomial theorem, function, real & complex numbers, transformation geometry & of course, calculus. When you are done, you are ready for your physics - adventure. First, be proficient in theory and then think about numericals. Best of luck!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/174811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Why do snorkelers not need to wear corrective glasses when snorkeling with goggles on? I am myopic ~ -2.75 sph +1cyl. When I went snorkeling they tell you not to wear glasses behind the goggles. Surprisingly, underwater, things remain in focus with goggles on even without prescription lenses, while things outside the water at an equal distance would be blurry. Why is this the case?
Thank you.
| Wearing the mask underwater doesn't do anything to enhance your vision, but it does make objects in the water appear larger/nearer. This is due to the refraction of light at the air/mask interface (more info in my answer to this question). The objects are magnified by a factor approximately equal to the ratio of the indices of refraction $\frac{n_{water}}{n_{air}}\simeq1.33$, so you are effectively able to see $1.33$ times as far.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/175013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Are there any scales other than temperature that have different zero points? For most physical measurements, zero is the same regardless of the units used for the measure:
$0 \mathrm{mi} = 0 \mathrm{km}$
$0 \mathrm{s} = 0 \mathrm{hr}$
but for absolute temperatures, different systems have different zeros:
$0 ^\circ\mathrm{C} \neq 0\,\mathrm{K}$
Are there any other physical, measurable quantities (other than temperature) that have different zero points?
I'm looking for measurable quantities that are applicable anywhere -- things like voltage or temperature, not local quantities like "distance from the Empire State Building".
| Time, in which case each system's zero point is often called its epoch:
http://en.wikipedia.org/wiki/Epoch_%28reference_date%29
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/175153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 10,
"answer_id": 2
} |
How can static friction do work?
By definition, the work done by a force is $W = F\cdot d$, so how can static friction do work?
Can this force move the body a distance of $75~\text{m}$?
| Static friction does not produce or consume work in most of the times. For example for a solid body that rolls without sliding the velocity of the base point $A$ is $\vec v_a = \vec v_{cm} + \vec v_{tangential} \Rightarrow v_a = v_{cm} - \omega R = \omega R - \omega R =0$ which implies that $x_a = 0$. The static friction is a force that acts on $A$ so $W_T = T x_a = 0$
But when the object slides then the friction force is constant and equal to $T = \mu N$ and is is always opposite to the velocity of the body. So then $W_T = - Ts$ where $s$ is the total space traveled by the body.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/175227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Sun and planets orbit each other Do not the planets and the Sun revolve in orbits around each other and the shape of the orbit depends on where the center of gravity of the system is? The greater the mass of the Sun, the closer the orbit approximates a perfect circle.
| To say that the orbit becomes more circular the greater the Sun's mass is not true. Instead, the eccentricity (i.e. how much the shape of an orbit varies from being circular) is governed by a couple of factors.
If you have a planet orbiting about the Sun with a mass much less than that of the Sun, and you know the following for an instantaneous point in the orbit:
orbital radius, $r$
radial velocity, $v_r$
tangential velocity, $v_t$
then, the eccentricity is given as follows:
$$e=\frac{r}{GM}\sqrt{\left(v_t^2 - \frac{GM}{r}\right)^2 + \left(v_r v_t\right)^2}$$
Therefore, in order to get a circular orbit, the planet needs to follow two conditions.
$v_t^2 = \frac{GM}{r}$ which is equivalent to $\frac{mv_t^2}{r} = \frac{GMm}{r^2}$
$v_r = 0$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/175334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Why does the pion not undergo netural particle oscillation? $K$, $B$ and now $D$ mesons exhibit neutral particle oscillation, where we see the spontaneous interchange between a particle and its antiparticle, i.e. $K^0 \Leftrightarrow \overline{K}^0$, $B^0 \Leftrightarrow \overline{B}^0$ and $D^0 \Leftrightarrow \overline{D}^0$.
My question is, why do neutral pions $\pi^0$ not exhibit the same behaviour? We never hear of "pion oscillation" $\pi^0 \Leftrightarrow \overline{\pi}^0$...
| Answer transposed from a comment: the $K$, $D$, $B$ have nonzero "flavor quantum number" (strangeness, charm, and beauty, to be specific). The analogy you should pursue is the $J/\psi$ or "charmonium," made of a charm quark and charm anti-quark.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/175511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
What does it mean that a magnetic field's flux vanishes through any closed surface? I'm reading the Britannica guide to Electricity and Magnetism, and I came across the following quote:
A fundamental property of a magnetic field is that its flux through
any closed surface vanishes.
Can someone explain this in simpler terms?
Source
| If the flux in and out of a surface cancles, there is no need for magnetic charge in which field lines can end or start (e.g. like the electric charge). One expresses this like
$$ \nabla \cdot \vec{B} = 0 $$
wich means
$$ 0 = \int_V \nabla \cdot \vec{B} ~ dV = \int_S \vec{B} \cdot d\vec{S} $$
where $S$ is the surface of the volume $V$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/175578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Phase space Lagrangian? Reading out of this lecture series we define a phase space Lagrangian $\mathcal L$ to be a function of $4n+1$ variables namely $q,\dot q,p,\dot p,t$. My question is, what space is this function defined on? (I know that the $\dot p$ is there for names sake only).
My stab at an answer is it is a product space between $\mathcal L:TQ\times T^*Q\times \mathbb R\to \mathbb R$ since we are dealing with both velocity and momentum (where $Q$ is a configuration manifold). However this makes zero intuitive sense to me. If indeed it is phase space, my understanding was that taking the fibre derivative with respect to a function on the tangent bundle changed the velocity coordinates to momentum coordinates - the Legendre transform?
| Let $X$ be the phase space. Then $L_\text{ph}(q,p,\dot{q},\dot{p},t)$ is a function on $TX\times \mathbb{R}$1, since the coordinates of $TX\times\mathbb{R}$ are precisely the coordinates of $X$, i.e. $(q,p)$ and their derivatives $(\dot{q},\dot{p})$ (and time $t$).
If Hamilton's equations are fulfilled, there are relations among $q,\dot{q},p,\dot{p}$ (the defining relations of the Legendre transformation) that reduce $L_\text{ph}(q,\dot{q},p,\dot{p},t)$ to the usual Lagrangian $L(q,\dot{q},t)$.
1In full analogy to the usual Lagrangian $L(q,\dot{q},t)$ being a function on $TQ\times\mathbb{R}$, where $Q$ is the $q$-space ("configuration space").
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/175668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How do waves have momentum? A question on a practice test I'm taking is as follows:
By shaking one end of a stretched string, a single pulse is generated.
The traveling pulse carries: A. mass B. energy C.
momentum
D. energy and momentum E. mass, energy and momentum
How would one describe the momentum of a wave?
| According to quantum mechanics, $p=\frac{h}{2\pi}k$,where $k$ is wave vector and $h$ is Planck's constant. As we know $k={2\pi\over \lambda}$, where $\lambda$ is the wavelength of the wave. So momentum and wavelength are associated to each other. Moreover we can view as the motion of the every particles as well as wave and particle.This is known as duality of the quantum mechanics. The fact is that the wavelength of the macroscopic object is very large. so we use duality in the microscopic particles to describe the quantum mechanics.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/175771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Can we theoretically balance a perfectly symmetrical pencil on its one-atom tip? I was asked by an undergrad student about this question. I think if we were to take away air molecules around the pencil and cool it to absolute zero, that pencil would theoretically balance.
Am I correct?
Veritasium/Minutephysics video on Youtube.
| The question is so ambiguous that it allows a resounding yes. This is because "balance" is not defined, neither are the dimensions and the material used for the pencil, nor the location of where the "balancing" is to happen.
The material and the shape of the surface to balance the pencil on, are not specified, nor the length of time it should stay balanced. So, if you use a titanium tipped "pencil," on a planet/moon with a gravitational force 1/10,000 of earth with no atmosphere, on a surface made of a "complementary" material such that the titanium molecule will "fit" in a "hole" created by the surrounding complementary molecules, then "balancing" the "pencil" will be a snap.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/175985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "96",
"answer_count": 9,
"answer_id": 0
} |
Why do we use capacitors when batteries can very well store charges? Can batteries be used instead of capacitors? I am trying to figure out a basic, superficial and any obvious difference between the two.
| I've to make an electronic circuit 'RC' and the relation between current and tension between two nodes must be fulfilled by a capacitor 'C' (it integrates the current; see the relation in the WP). I can use a battery with a constant tension to power the circuit ($V_{in}$in the second image) but not to model that relation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/176050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Meaning of Time Reversal Symmetry I was wondering if someone could give a simple explanation of what is meant by time reversal invariance. Is it analogous to spatial translational symmetry? If so, how? By spatial translational symmetry I mean the following. Suppose, for example, one has a solid consisting of an array of ions and electrons. If we pick a coordinate system we can write the Hamiltonian of the solid in terms of the coordinates of the ions and electrons. If we translate the origin of our coordinate system our Hamiltonian will be expressed using new coordinates, however, the Hamiltonian will have the same form. Is there a similar understanding for time reversal invariance?
| Time reversal essentially means a system looks the same if you reverse the flow of time. The only difference beeing that things like velocity go in the opposite direction. In condensed matter systems it is represented as a unitary matrix times complex conjugation $\mathcal{T} = U\mathcal{K}$. A simple system that follows T-symmetry would be a system described by a real Hamiltonian (if we're ignoring spin, anyway.) If you include spin, for spin 1/2 particles it is represented as $\mathcal{T}=i\sigma_y \mathcal{K}$ and then we can have certain complex Hamiltonians as well.
As examples, a quantum spin Hall insulator is a system that preserves T-symmetry (because of the spins and their chirality when we reverse time we see the same system as before). A system which breaks T-symmetry is a ferromagnet. In this case the spin reversal is again the culprit, but because we do not see the same system as before, it has broken the symmetry. Hope this makes it a bit more clear!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/176141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
General Relativity visualization software As I am approaching the study GR, I was wondering if there are softwares that allow a quick visualization of custom metrics, curvature, and particle motion even in the limited context of 2D space.
Playing with equations is fun, but it would be more fun if I could play with various parameters and see the outcome.
Obviously free would be better, but I am open to commercial programs.
| I've been looking at this Java archive
General Relativity (GR) Package written by Wolfgang Christian, Mario
Belloni, and Anne Cox
It includes a lot of simple programs about Newtonian mechanics, special relativity and general relativity, including the aforementioned GROrbits.
It doesn't permit custom metrics - you are limited to Schwarzschild (regular and rain co-ordinates) and Kerr black holes.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/176303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Is there any physical meaning for the inverse metric? I've been wondering if we can attribute any physical meaning to the inverse metric. I mean when we talk about the metric itself, there are lots of insights we can have towards its role in spacetime, yet I cannot see any physical meaning for the inverse metric. For now, I just see it as tensor with the special property of giving the identity when joined with the metric. Rigorously speaking, I would say it is not even an "inverse" actually, as it doesn't map like one. But still, is there any physical way of interpreting this tensor?
| Another way of looking at user40330's answer is to think of the inverse metric as the map from the space of one-forms (or differentials, if you prefer) and mapping them to the space of vectors (or directional derivatives, if you prefer that language), and then thinking of the metric as the inverse of this map.
Namely
$$g^{-1}({ d}v)=g^{ab}v_{b} = v^{a} = {\vec v}$$
and
$$g({\vec v}) = g_{ab}v^{b} = v_{a} = dv$$
This map is obviously invertible, thanks to the properties of matrices, and you obviously have $g^{-1}(dv,dv) = g({\vec v},{\vec v})$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/176400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do you prove that $L=I-V+1$ in $\lambda\phi^4$ theory? It is known that the number of loops in $\lambda\phi^4$ theory is given by the formula
$$L=I-V+1$$
where $L$ is the number of loops, $I$ the number of internal lines and $V$ the number of vertices. I would like to know the proof of this statement.
| Page 140 of Srednicki's QFT textbook provides a much simpler proof:
This can be seen by counting the number of internal momenta and the constraints among them. Specifically, assign an unfixed momentum to each internal line; there are [$I$] of these momenta. Then the $V$ vertices provide $V$ constraints. One linear combination of these constraints gives overall momentum conservation, and so does not constrain the internal momenta. Therefore, the number of internal momenta left unfixed by the vertex constraints is $[I] − (V−1)$, and the number of unfixed momenta is the same as the number of loops L.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/176453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Calculating wind force and drag force on a falling object I'm trying to numerically integrate the motion of an object (say, a falling vertical cylinder). Here, there's a drag force: the wind "acting" on the body (presumably adding horizontal velocity) and the air itself slowing down the vertical motion.
Is it correct to calculate both forces with the drag equation, $F_D = \frac{1}{2} \rho v^2 C_D A$? Here I suppose the velocity would be the relative velocity between the falling object (i.e. initially $(0, -15)\ m/s$) and the "air" (i.e. $(5, 0)\ m/s$). If that's the case, where would the drag force point towards to? $-\hat{v}_{relative}$?
Thanks a lot for your help in advance.
| "The vertical drag is greater" Could this be "lift"?
The vertical and horizontal velocity components do indeed produce a higher net velocity vector, but the vertical and horizontal kinetic energy components also have a Velocity squared function.
Interestingly, if you dropped an unpowered object from a tower in a 2 mph cross wind, it's horizontal component would "slow down" by drag to Velocity=0, relative to the air mass, and continue to plummet vertically to its terminal velocity and impact. The horizontal drag component of the object goes to zero, relative to the air mass, but it will have a Velocity=2 mph horizontal energy component relative to the ground on impact.
If the aerodynamic forces were purely drag vs kinetic energy, we should be able to decompose. Something else may be going on here, which lies at the heart of "when does a falling object become a glider".
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/176513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to calculate the normal force exerted by a fulcrum off-center? So I know that $F_N$ exerted by a fulcrum right under the centre of mass for, say, a long horizontal rod, is equal to $mg$. But what if the fulcrum is off the centre of mass either to the right or the left -- how would you calculate the force exerted by the fulcrum then? I guess since $\Sigma{F_y}$ stays the same, the normal force would also be the same. But doesn't some of the energy of these forces go into inducing torque about the fulcrum instead? I'm lost.
| Be careful - you are applying the reasoning about a static situation to a situation that is not static. This rod will be rotationally and linearly accelerating, so you can no longer assume the net force or net torque is zero - $\Sigma F_y = m a_y$, and $a_y \ne 0$.
If the rod is instantaneously horizontal and at rest, it's not too hard to find $F_n$ at that moment:
$$\Sigma F_y = M a_y$$
$$F_N - M g = M a$$
We can get more information about the rod by using the rotational version of Newton's 2nd law:
$$\Sigma \tau = I \ddot \theta$$
Where $\tau$ is a torque, $I$ is the moment of inertia of the rod, and $\theta$ is the angle between the rod and the horizontal. Let $l$ be the distance between the center of mass and the fulcrum:
$$M g l = I \ddot \theta$$
We can find a geometrical relationship between the angular and linear accelerations:
$$y_{CM} = - l sin(\theta), ~~x_{CM} = l cos(\theta)$$
Using the simplifying fact that we are only concerned with the instant when the rod is at rest and horizontal, we can differentiate with respect to time (what follows is only valid at that instant):
$$\ddot y_{CM} = a_y = - l cos(\theta) \ddot \theta + l sin(\theta) \dot \theta^2$$
$$\ddot y_{CM} = a_x = -l sin(\theta) \ddot \theta - l cos(\theta) \dot \theta^2$$
Since $\theta = 0$ and $\dot \theta = 0$ at this instant,
$$a_y = - l \ddot \theta, ~~a_x = 0$$
$$\ddot \theta = - \frac{a_y}{l}$$
Substituting into the rotational version of Newton's 2nd law:
$$M g l = - I \frac{a_y}{l}$$
$$a_y = - \frac{M g l^2}{I}$$
Substituting into the linear Newton's 2nd law:
$$F_n - M g = - \frac{M^2 g l^2}{I}$$
Substituting the moment of inertia about the fulcrum (which can be obtained with the parallel axis theorem) $I = \frac{1}{12} M L^2 + M l^2$$
$$F_n - M g = - \frac{M^2 g l^2}{\frac{1}{12} M L^2 + M l^2}$$
Simplifying and solving for $F_n$...
$$F_n = M g \left(1 - \frac{l^2}{\frac{1}{12} L^2 + l^2}\right)$$
To sanity-check this answer, we can return to the balanced case where the center of mass is directly over the fulcrum ($l = 0$), which gives us the familiar result $F_n = M g$. Additionally, as $L \rightarrow \infty$, the moment of inertia increases, which should cause the rod to rotate more slowly, and indeed, $F_n$ approaches its value for the static situation in that limit.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/176590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Speed of gravity within a mass We all consider that gravity travels at the speed of light. Light travels at the speed of light except when it is in a medium ,say glass, where it travels slower.
What happens when gravity passes through a distributed mass. Will it still travel at the speed of light or will each atom absorb and emit gravity as it passes.
I realize this is not the way anyone would like to think about gravity, but an answer to this question may be interesting.
| The acceleration of gravity is that itself. The bigger the mass the more gravity pulls on the object. Especially with a force. This could be simplified into an equation we learned in Intro-Physics, F=ma or f=mg
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/176877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Entropy - Gas Inside A Closed System Reaches Maximum Entropy Filling a box with a certain amount of gas with a specific total energy and allowing the gas to reach a maximum entropy state, what happens next?
Would the gas remain in a maximum entropy state indefinitely?
What would prevent the gas atoms/molecules to end up in a more orderly state at some point just out of coincidence? After all, the gas particles have some energy total in this closed system and will keep moving around.
Is it even possible that if we waited a long long time, those gas molecules could fall back in a state of minimum entropy at some point?
edit: Just how many maximum entropy states are there in a gas with N particles inside a closed system vs lesser entropy states? Would it be even more likely for a gas to decrease in entropy than remain in a maximum entropy state? If yes, then this would be a clear violation of the 2nd law of thermodynamics. It would have to be stated more precisely.
| The gas could fall into a state of low entropy randomly. It is important to remember that the laws of thermodynamics are probabilistic, and they say not what will happen but what usually will.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/177215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
What is meant by the term "single particle state" In a lot of quantum mechanics lecture notes I've read the author introduces the notion of a so-called single-particle state when discussing non-interacting (or weakly interacting) particles, but none that I have read so far give an explicit explanation as to what is exactly meant by this term.
Is it meant that, in principle, each individual state constituting a multi-particle system can be occupied by a single particle (contrary to an entangled state, where is impossible to "separate" the particles), such that the state as a whole can be de-constructed into a set of sub-states containing only one particle each, and each being described by its own Hamiltonian?
Sorry to ramble, I'm a bit confused on the subject, in particular as I know that more than one particle can occupy a single-particle state (is the point here that the particles can still be attributed their own individual wave functions, and it just so happens that these individual wave functions describe the same state?).
| It means a single particle Schrödinger equation determines the behavior of the particle. The only time a particle is in a single-particle state in an N particle system is when there are no interactions between particles - i.e., in an ideal gas. In the real world single particle states are always an approximation. E.g., wave functions of electrons in atoms ignore the details of the interactions between atoms.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/177297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 4
} |
Are there other less famous yet accepted formalisms of Classical Mechanics? I was lately studying about the Lagrange and Hamiltonian Mechanics. This gave me a perspective of looking at classical mechanics different from that of Newton's. I would like to know if there are other accepted formalism of the same which are not quite useful compared to others (because otherwise if would have been famous and taught in colleges)?
| Kane's Method is another accepted formalism (Thomas R. Kane) which is a method for formulating equations of motion.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/177558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 2,
"answer_id": 1
} |
Deriving the equation for the speed of a block down an incline using work - keep getting the wrong constant So the question is:
There is a block of mass $m$ travelling along an incline that makes an angle $\theta$ with the horizontal. If the block is pushed up the incline with an initial velocity $v_o$, what is its speed when it crosses a point $x$ meters from where it starts?
By intuition, since the incline if frictionless, the block will rise to height $x_{max}$ before falling back down to $x = 0$, at which point it will have the same velocity as it started with, just in the opposite direction. Using a kinematic equation, I get a final speed of the block of:
$$V_f = \sqrt{2g \sin(\theta)\Delta d+V^2_i}$$
So that is what I would like to derive using the equation $\Delta W = \Delta K + \Delta U$ where $K$ is the kinetic energy and $U$ is the potential energy.
Drawing a free body diagram, I get a force $F=mg \sin(\theta)$, which is the gravitational force and $\Delta W = mg \sin(\theta) d$ where $d$ is the distance traveled factoring in the extra height gained from the initial push which I found using kinematic equations.
This gives $$mg \sin(\theta) d = \frac{1}{2}mv^2_f - \frac{1}{2}mv^2_i + mgh_f - mgh_i,$$ where $mgh_f = 0$ by the reference point.
$$mg \sin(\theta) d + mgh_f = \frac{1}{2}mv^2_f $$ where $mg \sin(\theta) d = mgh_f$
$$2mg \sin(\theta) d = \frac{1}{2}mv^2_f$$
$$V_f = \sqrt{4g \sin(\theta) \Delta d}.$$
Can anyone tell me why there is a $4$ there instead of a $2$? Also, if someone could show me how to include the intial velocity in this equation and not have to calculate the total distance before hand that would be awesome.
| In your first approach I believe you got a wrong sign. It should be $$v_f = \sqrt{v_0^2 - 2g d \sin(\theta)},$$ otherwise your speed will increase with distance.
In your second approach seems like you are equating the work of the gravitational force to the difference of the mechanical energy. However, the work of a force is equal to the difference in the kinetic energy only, not mechanical energy. Don't forget that weight is doing negative work, since it is directed agaisnt the path. Thus
$$W = - (mg \sin(\theta)) \cdot d = K_f - K_i = \frac{1}{2} mv_f^2 - \frac{1}{2} mv_0^2.$$
Hence
$$-2g d \sin(\theta) = v_f^2 - v_0^2,$$
and therefore
$$v_f = \sqrt{v_0^2 - 2gd \sin(\theta)}.$$
This is the same as before, as it should be.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/177666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What compounds or elements only have one phase or two phases? Wood appears to be one. I think gases like helium and hydrogen cannot exist in the solid state under normal pressures, correct? And why do those "phase cheaters"-- those elements/compounds which sublimate directly, skipping a phase, or "procrastinators"-- elements/compounds which just never reach the phase-- why do they do that?
| Wood can only be solid because it's a very specific arrangement of atoms. If you liquified or vaporized all the elements in it, they wouldn't be wood. You also can't have liquid crystals, unless you count liquid crystals. Or a liquid computer.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/177755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Can Ampere's law be applied to 3D loop Usually in textbooks Ampere's law is just illustrated using 2D loops which forms a plane. Can the law be applied to 3D loops which cannot form a plane surface within the loop?
In a 3D loop which surface should we count when we count the current? In 2D case it is easy because there can only be a plane and the current is just that cut through the plane but if the loop is 3D that may be more then one possible surface formed by the loop, which section of the current should be counted?
| In the case of a planar loop, the surface that it bounds does not have to be a plane. It could bubble out. Ampere's Law is still valid (assuming the other conditions for validity are met).
Ampere's Law applies to any loop, and any surface bounded by the loop. I guess I'd better add that I don't know what happens in pathological cases where, for example, the loop crosses itself.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/177840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Superstring vacuum amplitude on the torus My question is how to obtain the superstring (Type II A and B) vacuum amplitudes on a torus. They are given in Polchinski's String Theory Vol. 2 equation (10.7.9):
$$Z_\psi^{\pm}=\frac{1}{2}[Z^0_0(\tau)^4-Z^0_1(\tau)^4-Z^1_0(\tau)^4\mp Z^0_0(\tau)^4].$$
I understand how each individual $Z^\alpha_\beta$ is obtained but do not understand how they are put together to get $Z_\psi^{\pm}$.
| One way to think about this is the following. In general, the partition function (which is the integrand of the vacuum amplitude and not the vacuum amplitude itself) will be of the form
$Z_\psi^{\pm}\propto\sum_{a,b}C[^a_b]Z^a_b(\tau)$
where $a$ and $b$ sum over the different sectors as given in the text and the $C$s are some phases. Many of these phases are fixed by modular invariance, ie by the requirement that $Z$ is invariant under $\tau\rightarrow\tau+1$ and $\tau\rightarrow-\frac{1}{\tau}$. Given that $C[^0_0]=1$ (it's only the relative phases that matter, so we can fix this one), modular invariance demands that $C[^0_1]=C[^1_0]=-1$. However, both choices $C[^0_0]=\pm$ give modular invariant theories, so we end up with two different viable theories.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/177926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Units of eigenvectors Consider for example a mass matrix $M$, $\lambda$ one eigenvalue and $X$ a corresponding eigenvector. Then $[M]=\text{mass}$ (the brackets indicate the "unit operator"), and $MX=\lambda X$ so $[M][X]=[\lambda][X]$, so $[\lambda]=\text{mass}$. That's why for example in oscillators, the pulsations $\omega$ are such that $[\omega^2]=[M^{-1}K]=\text{seconds}^{-2}$.
But what about the eigenvectors? I would tend to think that they are dimensionless, because during a change of basis $u=Pq$, $q$ and $u$ have the same units, while $P$ gathers the eigenvectors $X$; the vectors designate a change of basis but the vector space remains the same.
Is it so?
| The units of the eigenvector can be anything you choose. Normally you want them to be dimensionless, but other choices can be sensible on occasion.
The reason that any dimension is valid is because if $X$ is an eigenvector, then $\mu X$ is also an eigenvector for any scalar $\mu$, which can in principle have any dimensions you want.
The main implication of that is that eigenvectors are only defined up to a multiplicative constant, and if you want to fix that then you should look at exactly what you'll be using them for. Because the constant can be dimensionful, this carries the dimensionality of the eigenvectors with it.
In general, though, what you want to do with your eigenvectors is to use them as a basis. In that case, it is usually best practice to make them dimensionless, so that their norms will be unity. If you don't care so much about this, then you can essentially do whatever you feel you need to.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/178234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
What is the meaning of " $\Psi$ is not a measurable quantity in itself"? I want to know that why the wavefunction $\Psi$ as a complex quantity (i.e $A+iB$ form) in quantum mechanics and somewhere I have studied that $\Psi$ is not a measurable quantity in itself that's why we multiply it by a it's complex conjugate $\Psi^*$ to measure $\Psi$ . What does it mean that "$\Psi$ is not a measurable quantity in it self" ?
| Ψ is supposedly a probability density amplitude. ΨΨ* is the probability density which in theory can be measured. For example in electron diffraction through a crystal a statistical measure of the electrons in, divided into the electrons out in a small region divided by the volume would allow ΨΨ* to be approximately measured. Phase information is lost when ΨΨ* is calculated. Only phase difference is needed in diffraction experiments.
The phase of Ψ cannot be measured.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/178389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
What mechanism creates bubbles in water that stays awhile If I fill glass jar with tap water and let it stay, small bubbles will appear on the side of jar after a while.
I am wondering what processes create those bubbles and what conditions require to meet?
My thoughts:
*
*Water impurities, i.e. dissolved salts, minerals. I.e. distilled water shall not have it.
*Water pressure. I.e. if I fill and close jar with the same water under higher pressure, air pressure of dissolved molecules of air will not be sufficient to create bubbles.
*Surface tension. If, instead of water I will use another liquid with higher surface tension - bubbles will not be formed. Then how to calculate "threshold" tension number?
*Air composition. Molecules of dissolved gas that has higher hydrophobic properties will form gas bubbles faster.
Anything else I did not mention.
| You pretty much summed it up.. but an important factor is the history of the water. For example water that has been subjected to temperatures exceeding $35^oC$, would have lost most of its dissolved oxygen (the main gas dissolved in water). On the other hand water that has been vigorously shaken would have replenished its dissolved oxygen, for example this is why when you fill your jar with tap water with a very fast running water, more bubbles will form on the inside edges of the jar, unlike when you pour the water very gently.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/178678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Can electricity flow through vacuum? People say yes, and give a wonderful example of vacuum tubes, CRTs. But can we really say that vacuum (..as in space) is a good conductor of electricity in a very basic sense?
| Conduction of Electricity in Solutions, Gases and Vacuum
https://youtu.be/7q8f-QJlpsA
What should be the definition of "Electricity"?
http://www.ivorcatt.org/99mcattq.jpg
http://www.ivorcatt.co.uk/97rdeat4.htm
http://www.ivorcatt.co.uk/x18j100.pdf
Ivor Catt states here that electric charges do not "exist".
http://www.ivorcatt.co.uk/x0620.htm
*
*"In the same way as the slope of a hill does not exist, having no materiality, although the hill itself exists, being made up of physical material, so electric charge and electric current become merely the results of mathematical manipulation of the edge of a field (or more accurately of an ExH Energy Current)."
*“Although a cloud cannot exist without edges, the edges of a cloud do not exist. They have no width, volume or materiality. However, the edges of a cloud can be drawn. Their shapes can be manipulated graphically and mathematically. The same is true of the so-called ‘electric current.’”
Also. please do see this Electron mass experiment
on Youtube. This is a transcription: https://imgur.com/a/luE4CC9
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/178831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 8,
"answer_id": 5
} |
Submarine Speed Detection A submarine can use sonar (sound traveling through water) to determine its distance from other objects. The time between the emission of a sound pulse (a “ping”) and the detection of its echo can be used to determine such distances.
Alternatively, by measuring the time between successive echo receptions of a regularly timed set of pings, the submarine’s speed may be determined by comparing the time between echoes to the time between pings.
Assume you are the sonar operator in a submarine traveling at a constant velocity underwater. Your boat is in the eastern Mediterranean Sea, where the speed of sound is known to be 1522 m/s. If you send out pings every 2.00 s, and your apparatus receives echoes reflected from an undersea cliff every 1.98 s, how fast is your submarine traveling?
Here's my solution: note it is wrong
Let the submarine be at origin and the cliff be at some point on x axis.
Time taken for the ping to reach the cliff= 2 sec.
In that time the submarine moves 2v distance on x axis (v-velocity of submarine)
Time taken for the echo to reach the submarine=1.98 sec.
In that time the submarine moves 1.98v distance further.
Distance of cliff from origin =1522*2 m
Distance of cliff from submarines final position = 1522*1.98 m
Thus, 2v + 1.98v + 1522*1.98 = 1522*2
V= 7.65 m/s
But the answer shud be 15(approx.)
I tried solving it using Doppler effect, but the answer comes wrong again.
| The round trip time of the ping is unknown; but we do know that the difference in round trip time between sub stationary and sub moving is 0.02 seconds.
Let us write $D$ for the distance to the cliff when you send the ping; if you are traveling at a speed $v$, and the speed of sound in water is $c$, then we can write down the round trip time as follows (outbound distance is $D$, return distance is $D - v\cdot t$:
$$t = \frac{D}{c} + \frac{D-v\cdot t}{c}$$
Now we know that when the sub is still, $\frac{2D}{c} = t_0$. And while we don't know either $t$ or $t_0$, we do know that they differ by 1%: if the distance to the cliff is such that $n$ pings are "under way" when the first ping returns, we know that ping has arrived 0.02 n seconds earlier than it would have if the sub had been stationary. We know therefore that $\frac{t}{t_0} = 0.99$.
A bit of manipulation of the above gives
$$\begin{align}t &= \frac{2D}{c} - \frac{v}{c}\cdot t\\
t &= t_0 - \frac{v}{c}\cdot t\\
v &= (t_0 - t)\;\frac{c}{t}\\
v &= \left( \frac{t_0}{t}-1\right)\cdot c\\
&= \left( \frac{2.00\;s}{1.98\;s}-1\right)\cdot 1522\;\mathrm{m/s}\\
&= 15.4\;\mathrm{m/s}\end{align}$$
Note that this answer is slightly different than just 0.01 * 1522. With the precision of the numbers given in the question, the difference is just significant (15.4 s 15.2 m/s).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/178974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How are the theoretical properties of a gravity train linked to orbital properties? Whilst perusing a question on WorldBuilding, a link was given to an article about Gravity Trains.
Some of the answers on the question, and the Wikipedia article, state a travel time of 42 minutes between any two points on Earth. The Wikipedia article also states that the maximum speed of a gravity train that goes directly through the center of the Earth is 7,900m/s.
It was this speed that made me think of orbits (it being close to the roughly 8km/s I have in my head) - and indeed another article confirms that "orbiting on the surface" of the Earth involves a speed of 7.9km/s, with the additional property of the period being 1 hour, 24 minutes - ie twice that of the theoretical travel time of a gravity train.
Are these coincidences? If not, how and why are they related?
| It takes about 45 minutes for a low Earth circular orbit to go from one side of the Earth to the other say From New York, USA to Perth Australia. If a hole could be tunneled through a uniform density Earth and you were to jump in the hole gravity would act like a spring and it would take you about 45 minutes, (if no air or frictional resistance) to "fall" from New York to Perth.(the only problem is that if you didn't grab onto something you would oscillate between New York and Perth with a period of 90 minutes) This would work for any uniform density planet using Newton's law of gravity/Gauss's law of gravity but it wouldn't
necessarily be 45 minutes.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/179062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
What could explain the presence of Promethium in stars? [This is pretty much a copy and paste job from Technetium question but I wanted to add this one in case there was some other explanation] So, I understand that PM doesn't exist in nature [though, I don't know why every reference I see regarding PM says that and then goes on to state that it is found in some stars...] but, if that's the case, then why is it found in some stars? This element doesn't fuse in stars (I think, I found a graph saying that it does) so it has to be left over from supernovae. Lastly, how many stars have been discovered that contain promethium? Thanks much.
| I found this paper discussing the production of promethium in stars. The paper suggests several possibilities:
*
*Fission of heavy or superheavy nuclei
*Spallation of heavy nuclei by high energy protons
*Reactions of Nd and Sm with low energy nucleons: the s-process
The longest half life of any promethium isotope is 17.7 years, so it isn't going to be left over from supernovae. Promethium has to be produced continuously for it to be present in stars.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/179205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Do the same equations of motion imply the same Lagrangians? If two Lagrangian (densities) $\mathcal{L}$ give the same equations of motion, are they equivalent?
| As it happens, it is not necessary that two Lagrangians that have the same equation of motion have the same functional form. Consider the Lagrangians $L_1 = T-V$ and
$$L_2 = \frac{1}{3}T^2 + 2TV - V^2$$
where $T = \frac{1}{2}m\dot{x}^2$ and $V(x)$ is the potential energy. They both lead to the same equation of motion:
$$m\ddot{x} = -\frac{dV}{dx}$$
This is worked out in detail in this answer.
To compare the two further, let us obtain the Hamiltonian of $L_2$ (The Hamiltonian of $L_1$ is obviously $H_1 = T+V$). We observe that (without bothering to express the Hamiltonian as a function of the momentum)
$$\frac{\partial L_2}{\partial \dot{x}} = \frac{1}{3}m^2\dot{x}^3 + 2m\dot{x}V$$
$$H_2 = \frac{1}{3}m^2\dot{x}^4 + 2m\dot{x}^2V - \frac{1}{12}m^2\dot{x}^4 - m\dot{x}^2V+V^2$$
$$\implies H_2 = \frac{1}{4}m^2\dot{x}^4 + m\dot{x}^2V + V^2 = \left(T+V\right)^2$$
i.e.
$$H_2 = (H_1)^2$$
Thus, while the relation between the two in the Lagrangian framework is not very obvious, we can easily see how they are related by comparing the Hamiltonians. Both Lagrangians do not explicitly depend on time, and therefore conserve $T+V$ or $(T+V)^2$ respectively, which is essentially the same thing.
Another famous example (though of a less serious nature in the context of this question) is the Lagrangian of a free relativistic particle, which is given by
$$L = \sqrt{g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}}$$
The equations of motion are then the geodesic equations. It is well known that the square of this Lagrangian, $$L' = g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}$$ also leads to the same equations of motion (for affine parametrizations of the path - see Qmechanic's comment below).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/179458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What are the units of the creation and annihilation operators? The creation and annihilation operators - also known as ladder operators are;
$ \hat{a}^\dagger$ and $\hat{a}$ respectively.
Using the equation $\hat{H} = \hbarω\left(\hat{a}^\dagger \hat{a} + \frac{1}{2}\right)$
and knowing that the units of $\hat{H}$ are J,
the units of $ω$ are Rad/s
and the units of $\hbar$ are J.s
I think that the ladder operators should have units of $\frac{1}{\sqrt{Rad}}$
But I have never seen a square root of an angle in units before. Is this correct?
| The units of $\hbar$ are in fact J.s/rad. (thanks AV23) this is because $\hbar = \frac{h}{2\pi}$ the units of h are J.s and the units of $\pi$ are rad. Thus we have J.s/rad. (thanks Noiralef)
Thus the ladder operators are in fact unitless.
On reflection this is the only logical possibility as they move between different eigenstates - which must all be in the same units.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/180783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Why is the gauge potential $A_{\mu}$ in the Lie algebra of the gauge group $G$? If we have a general gauge group whose action is $$ \Phi(x) \rightarrow g(x)\Phi(x), $$
with $g\in G$.
Then introducing the gauge covariant derivative $$ D_{\mu}\Phi(x) = (\partial_{\mu}+A_{\mu})\Phi(x).$$
My notes state the gauge potential $A_{\mu} \in L(G)$, $L(G)$ being the Lie Algebra of the group $G$.
What's the connection between the Lie Algebra of the group and the gauge potential?
| The gauge potential is an object that, when introduced in the covariant derivative, is intended to cancel the terms that spoil the linear transformation of the field under the gauge group. Every gauge transformation $g:\Sigma\to G$ (on a spacetime $\Sigma$) connected to the identity may be written as $\mathrm{e}^{\mathrm{i}\chi(x)}$ for some Lie algebra valued $\chi: \Sigma\to\mathfrak{g}$. The derivative of a transformed field is
$$ \partial_\mu(g\phi) = \partial_\mu(g)\phi + g\partial_\mu\phi = g(g^{-1}(\partial_\mu g) + \partial_\mu)\phi$$
and it is the $g^{-1}(\partial_\mu g) = \partial_\mu\chi$ that we want to cancel here by adding the gauge field so that $D_\mu(g\phi) = gD_\mu\phi$. Since $\partial_\mu\chi$ is Lie algebra valued, so must the gauge field $A$ we add, and it has to transform as
$$ A\overset{g(x)}{\mapsto} gAg^{-1} - g^{-1}\mathrm{d} g$$
to cancel the terms we want to cancel.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/180907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
Counting higher-order corrections in "ABC theory" I am trying to understand how to enumerate higher-order Feynman diagrams.
In his book on Elementary Particle Physics, Griffiths considers a simple "ABC toy theory" which has:
*
*three (scalar, massive) particles A, B, and C (which are their own antiparticles), and
*only one allowed interaction vertex ABC.
He counts explicitly the number of higher-order diagrams for the process $A+A\to B+B$ at order $g^4$, i.e. with 4 vertices.
What confuses me is that his result changes in the 2nd edition:
In the first edition of his book (p. 207) he finds that there are 15 because each additional line can start at one of the 5 lines of the original $A+A\to B+B$ diagram and end on the same or another line, so enumerating
*
*$1\to1$, $1\to2$, ..., $1\to5$,
*$2\to2$, ..., $2\to5$,
*...
*$5\to5$,
we end up with 15. $\Rightarrow$ This makes sense naively.
However, in the second, revised edition, he talks about 8 diagrams only (p. 217): 5 self-energy diagrams ($i\to i$, a line "sprouts a loop"), two vertex corrections ("a vertex becomes a triangle") and one box diagram.
My question: How many diagrams are there? 15 or 8? Where have the other 7 gone, e.g. a line connecting an initial and a final-state particle? Was it wrong to include them? If so why? Are they equivalent to another diagram or ruled out for some reason?
| The 8 is correct. The diagrams where you connect one outgoing and one incoming line are equivalent to the ones, where you connect one outgoing line and the internal line, as both lead to the vertex-correction diagram.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/180978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Conceptual explanation of the Single particle partition function The Single particle partition function is defined mathematically as
$$\text{Z=$\sum $}g_ie^{\left(\frac{-E_i}{K_BT}\right)}$$
But what is the physical interpretation of the partition function and it's significance to Thermodynamics? I'm seeking a simple yet understandable intuition.
| Partition functions are a measure of the allowed volume in (microscopic-)configuration space for the system, and as such they are the normalizing function for probabilities expressed as volumes in configuration space (and assuming the applicability of the ergodic hypothesis).
I know that this is very abstract, but it is also very general.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/181157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is this symmetry factor in Peskin wrong? I am trying to compute the symmetry factor of a Feynman diagram in $\phi^4$ but i do not get the result Peskin Claims. This is the diagram I am considering
$$\left(\frac{1}{4!}\right)^3\phi(x)\phi(y)\int{}d^4z\,\phi\phi\phi\phi\int{}d^4w\,\phi\phi\phi\phi\int{}d^4v\,\phi\phi\phi\phi$$
my attempt is the following: there are 4 ways to join $\phi(x)$ with $\phi(z)$. There are then 3 ways to connect $\phi(y)$ with $\phi(z)$. Then, there are 8 ways to connect $\phi(z)$ with $\phi(w)$ and 4 ways to contract the remainning $\phi(w)$ with $\phi(v)$. Finally the there are 6 ways to contract the $\phi(w)$ and the three $\phi(v)$ in pairs
$$\left(\frac{1}{4!}\right)^3\dot{}4\dot{}3\dot{}8\dot{}4\dot{}6=\frac{1}{6}$$
but the result claimed in Peskin (page 93) is $1/12$. What am I doing wrong?
| Contrary to your previous question Problem understanding the symmetry factor in a feynman diagram
the roles of the three vertices are not different so you have from the expansion of the exponential a $1/3!$ but it does not get compensated by the choice of role assignement.
Here this choice amounts to deciding who connects directly to $x$ and $y$ and that's it. So
you have $\frac{3}{3!}=\frac{1}{2}$. Also the symmetry group has order 12 because you can permute the 3 lines of the embedded sunshine diagram and you can also rotate it with repect to a vertical axis.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/181270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
What's the differences between time in Physics and time in everyday use? OK. This question might sound as not a good question, but the word 'time' is so confusing to me. I mean thermodynamics says time is the product of entropy. Relativity says time is relative. Quantum Mechanics says time doesn't exist, and that we can derive any given equation without involving time like Kepler's law. And I thought time measurement such as hour or second is just what we invented for conveniences on daily lives.
So are definition of time in Physics difference from time we associated in everyday life?
| The oldest perspective is the philosophical one established by Aristote; modified by Mach and then Barbour
*
*Time is an aspect of change (ie motion)
There is the mathematical perspective
*Time when mathematically conceived is just a parameter in the equation.
This in fact has nothing to do with time - it's purely formal
Physically there are a couple of candidates for the arrow of time
*The direction of increasing entropy
*The direction in which the universe is expanding
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/182312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
The Higgs field vs the Higgs boson As I see it the Higgs boson is the mediating particle of the Higgs field and get its own mass from the Higgs field. Isn´t this circular? I mean, for instance, an electron creates a radial electric field but in no way it can interact with the field it created.
So my question is: how can a particle get its mass from the field it mediates?
| Mass comes from things other than the Higgs field--the latter is just the main contributor. What gives the Higgs boson its mass is still up for debate--for a more detailed discussion, see the following post:
How does the Higgs Boson gain mass itself?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/182418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Correct vector space of eigenkets of angular momentum When we say an particle is in the state:
\begin{equation}
|l,m\rangle,
\end{equation}
what is the underlying state space, as a vector space? Is it a tensor product vector space, of dimension:
\begin{equation}
l\times(2l+1)\ ?
\end{equation}
How can I find the matrix representation of the angular momentum operators that act on the $2l+1$ vector space in that tensor product? I am used to angular momentum operators taking the form of a cross-product:
\begin{equation}
x_ip_j - p_ix_j,
\end{equation}
but can we still do that for the $2l+1$ dimensional space corresponding to $m$?
| For orbital angular momentum, indeed, $L = x\times p$ even as a quantum operator, see this question.
When writing a ket $\lvert l,m \rangle$, this is meant to live in the $2l+1$-dimensional space $\mathcal{H}_l = \mathbb{C}^{2l+1}$ on which the representation of the angular momentum algebra labelled by $l$ exists ($m$ is the eigenvalue of the ket for $L_z$). The total space of (bound) states for your system is then the infinite sum of these spaces for all possible $l$, i.e.
$$ \mathcal{H} = \bigoplus_l \mathcal{H}_l$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/182506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to measure temperature of a laser cooled sample at picoKelvin temperatures? I'm reading about laser cooling.. my question is: how can the temperature of the sample be measured? (using laser cooling we can reach $10^{-12}K...)$
| The temperature is not measured in the sense of using a thermometer. Instead it is calculated from the velocities of the particles in the trap.
Temperature is related to the velocity distribution by the Maxwell-Boltzmann equation. Under normal circumstances we are usually starting from a known temperature and calculating the velocity distribution. However it's a perfectly valid procedure to take the velocity distribution and work backwards to find the temperature. This is what is done in the sort of systems you describe.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/182687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to use Ampere's Law for a semi-infinite wire with current? Suppose that there is a semi-infinite wire which extends to infinity only in one direction. There are no other circuit elements at the other end(finite end) of the wire and the current does not loop. The magnetic field obviously has cylindrical symmetry when the Amperian contour is taken as a circle with its center on the wire.
However, due to charge accumulation there is a time-dependent electric field; hence a displacement current. How can I formulate the Ampere's law and show that the magnetic field is the half of that the infinite wire at the finite end of the wire? Do you think treating charge accumulation as a point charge with changing amount of charge right at the finite end of the wire will suffice?
The original question is:
| Ampere's law (for a steady current) states that
$$ \oint \vec{B} \cdot d\vec{l} = \mu_0 I$$
If we consider an infinite wire, then symmetry tells us that the B-field at the point $A$ and all other points on a circle of radius $(R+y)$ is constant in magnitude and is in the azimuthal direction. Hence the magnitude of the B-field is given by
$$ 2 \pi (R+y)B = \mu_0 I$$
$$ B = \frac{\mu_0 I}{2\pi(R+y)}$$
So now, for a semi-infinite wire, I take away half the wire and hence half the vector field. But, before I just say that the new field is half of the original one, I need to establish that the B-fields from each "half" of the infinite wire are in fact in the same direction so that they add in parallel fashion. The Biot-Savart law tells us that each wire element produces a B-field that is perpendicular to the current and perpendicular to a displacement joining the wire element and the point at which I wish to know the field. So, the B-field is always in a direction azimuthal to the wire, whichever piece(s) of wire we consider.
This means that the new B-field for the semi-infinite wire is in the same direction as for the full infinite wire but has half the magnitude.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/182785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
What does "interact via strong force" mean?
*
*I was just wondering if the words "strong force" and "strong interaction" are interchangeable?
*Also, these are referring to "strong nuclear force", correct?
*Then what does it mean for particles to "interact via the strong force"?
*Also, where does the weak force come from?
*Does it only occur in interactions or is there another place where it is needed. For example, the strong force holds nucleons together in the nucleus.
| In the context of nuclear or particle physics the phrase "the strong interaction" means the same thing as "the strong force". In fact we rarely write a formula for the strong force in the sense that we write Coulombs law for the electrostatic force. Both terms are refering to the strong nuclear force.
In the context of perturbation theory (or the lack of applicability of perturbation theory to a particular problem), you may here almost any interaction described as "strong" meaning that it is inappropriate to attempt a perturbative solution. In this case---also identified by Jaywalker in the comments---they are not the same thing.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/182858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Does friction act on a wheel rolling at a constant speed One of the things I've seemed to have taken for granted is that its the friction the floor exerts on a rolling wheel that prevents slip from occurring.
However, I ran into something that challenges that assumption: the contact point of the wheel on the floor has zero velocity relative to the floor, and no other lateral forces appear to exist. This implies that there is no friction. Otherwise the wheel would accelerate/decelerate, which there seems to be no mechanism for (ignoring drag, assuming no slip).
So, is it true to say no friction occurs on a wheel rolling at a constant speed? It is difficult to test as drag is difficult to isolate from the issue. No friction would imply that if the wheel rolled from a rough surface to a perfectly smooth surface, the wheel would not slip.
Is this effect analogous to a spring, such that if the wheel is overspinning, friction acts to slow the spinning down (spring extension, tensile force), if it underspins, friction acts to speed the spinning up (spring compression, compressive force), and a wheel spinning without slip experiences no friction (spring natural length, no force)?
| Maybe this thought experiment? Suppose you have a frictionless wheel and surface, in a vacuum, etc., and you spin up the wheel and push it forward so that its linear speed just matches the rotational speed and it moves along the surface with no slippage, i.e. zero net velocity at the point of contact. In this scenario, there are no forces of any kind, so teh system will continue exactly "as-is."
As you might expect, even in the absence of all other forces, any real-world materials will interact at the molecular level (van der Waals forces, perhaps) at the point of contact, leading to loss of kinetic energy.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/182992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why can't a magnet change a charged particle's speed? I know that magnetic force acts perpendicular to the direction of the original velocity, so the velocity in that original direction is unchanged, but once the magnet starts acting, the particle's velocity in the direction of the force goes from zero to nonzero -- wouldn't that increase the magnitude of the resultant velocity vector?
|
I know that magnetic force acts perpendicular to the direction of the original velocity
No, the magnetic force acts perpendicular to the current velocity.
Once the direction of the velocity changes, the direction of the force changes as well.
Cast in math:
$$ m\dot{\vec v} = \vec F_L = \frac q c \vec v \times \vec B $$
From this we get ($v = |\vec v|$):
$$ \dot v = \partial_t \sqrt{\vec v^2} = \frac{\vec v \cdot \dot{\vec v}}{\sqrt{\vec v^2}} \propto \vec v \cdot (\vec v \times \vec B) = 0$$
(As $\vec v \times \vec B \perp \vec v$).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/183112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 0
} |
Calculating Christoffel symbols from Lagrangian I was given the following metric for a sphere
$$g_{\mu\nu} = diag(1, r^2, r^2\sin^2\theta)$$
and tasked to calculate the Christoffel symbols. There are 2 ways that I know of to calculate them. One is from the equation
$$2\Gamma^{\alpha}_{\beta\gamma} = g^{\alpha\rho}(g_{\alpha\rho,~\beta} + g_{\beta\rho,~\alpha} -g_{\alpha\beta,~\rho})$$
and the other way is from the Lagrangian $L$.
What puzzles me:
For the 1st method, say if I want to calculate $\Gamma^{r}_{\theta\theta}$, I simply get
$$2\Gamma^{r}_{\theta\theta} = g^{rr}(g_{r\theta,~\theta} + g_{\theta r,~\theta} -g_{\theta\theta,~r})$$
and with only the last term surviving, I get $\Gamma^{r}_{\theta\theta} = -r$. Also, $\Gamma^{r}_{\phi\phi} = -r\sin^2\theta$
For the 2nd method I have $L = g_{\mu\nu}\dot{x}^\mu\dot{x}^\nu = \dot{r}^2 + r^2\dot{\theta}^2 + r^2\sin^2\theta\dot{\phi}^2$, and by Euler-Lagrange equation I get for the radial component
$$2\ddot{r} - (2r\dot{\theta}^2 + 2r\sin^2\theta\dot{\phi}^2) = 0$$
and here is where I do not get it. I have to compare the coefficients with the equation
$$\ddot{r} + \Gamma^{r}_{\theta\theta}\dot{\theta}^2 = 0$$
to extract out $\Gamma^{r}_{\theta\theta}$ but there is this $2r\sin^2\theta\dot{\phi}^2$ term which prevents me from direct comparison. Even if I assume that the equation may actually be a combination of 2 equations namely,
$$a\ddot{r} + 2\Gamma^{r}_{\theta\theta}\dot{\theta}^2 = 0$$
and
$$(2-a)\ddot{r} + 2\Gamma^{r}_{\phi\phi}\dot{\phi}^2 = 0$$
where $a$ is some constant. The only way to retrieve the correct $\Gamma^{r}_{\theta\theta}$ is for $a=2$. In this case I would not be able to retrieve $\Gamma^{r}_{\phi\phi}$. Any clues?
| The geodesic equation is
$$ \frac{d^2x^\mu}{dt^2} + \Gamma^\mu_{\nu \rho} \frac{dx^\nu}{dt}\frac{dx^\rho}{dt} = 0$$
The coefficient of $\dot{\phi}^2$ you're seeing corresponds to $\Gamma^r_{\phi\phi}$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/183205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Weight distributions If a man is standing on two weighing machines (scales), with one foot on each, Will both machines show equal weight or his weight will be distributed in two machines?
| Remember Newton's first law $\sum F=0$. Vertically you have this:
$$\sum F_y=N_{scale1}+N_{scale2} - W=0$$
The scales will feel and display the $N_{scale}$ force. Solving for either scale force you will see that they share the total weight $W$. Together they must lift the whole weight $W$, not individually.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/183324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Does lithium-6 "decay" when hit by a neutron? I am talking about the nuclear reaction
$$
^6\text{Li} + n \rightarrow\ ^4\text{He} +\ ^3\text{H} + 4.78\text{MeV}
$$
A neutron hits a lithium-6 nucleus and together they form an alpha and triton particle. Is it valid to say that the lithium nucleus "decays" when hit by a neutron? Is there any other verb which better describes the change of the lithium nucleus?
I am interested in the correct terminology.
| What happens depends on the energy of the neutron. At lower energy, it can simply merge with the $^6$L to form $^7$L. At a higher energy it can cause the $^6$L to fission, and at even higher energy it can break the $^6$L into pieces (spallation). This last mechanism is being studied for neutron induced fission, where the neutrons produced by high energy particles (protons, neutrons, deuterons) blast neutrons out of the lithium to stimulated fission events in Uranium, etc.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/183391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Is true black possible? Black is the absence of light because it absorbs light, but when we create black paint or black objects, light is always reflected, either in all directions in matte or smoothly in shiny black objects, making it never a true black. Would it be possible to use polarization to create an object that does not reflect any light, creating a truly black substance, without any shadows or reflection of light?
| A true black surface would need a temperature of 0 Kelvins as anything with heat will have excited electrons and will emit photons. However, dark matter is an example of purely black matter it only means that dark matter does not consist electrons.
Theoritically, a purely black object would be one that does not radiate light or absorb light. As absorbtion of any kind of light also brings along with it heat which again hypes up the electrons. Hence, it is impossible, even theoritically for an object to be purely black.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/183473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 3
} |
What do we mean by this statement : "some positive charge is put on a metal piece"? What do we mean by this statement : "some positive charge is put on a metal piece" ?
(I know it will be distributed on the surface of the metal piece, if metal is isolated.)
Electrons are negatively charged particles.
Protons are positively charged particles.
I am sure we are not putting protons on metal piece!! Then what we are putting as positive charge?
| Its not like putting any charge but simply making that metal piece charge to be positive.
It can be done simply charging metal piece by :
*
*Friction
*Induction
*Conduction
*or by the Removal of a Charge (Grounding)
While seeing question I come to know that you already asked one question and there also you asked the same question in this link at last line:
What happens when a conducting plate is grounded?
If you'll see the answer which is provided by Hritik Narayan that is :
Also, I think the answers here would help for (2) How to make
something charged using electricity?
Which is correct and it can clear your doubt.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/183576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If atoms are mostly empty space, why doesn't light pass through everything? Atoms are mostly empty space, and although I now understand why matter doesn't pass through other matter, why don't photons pass through the empty space of the atoms? Is it the same sort of idea as matter passing through other matter?
| They taught me that in high school too (i.e., that matter is "mostly empty space.") Only thing is, it's not true.
Solid matter is mostly filled with electrons. Yeah, the mass is all concentrated in the relatively tiny nucleii, but the mass is not what photons interact with, and the mass is not what defines the physical and chemical properties of ordinary matter. The electrons are responsible for all of that, and the electrons pretty much fill the space.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/183647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Why don't we build helicopter based space shuttles? As seen in this video: the principle of the helicopter does work in space. So we could make a helicopter based space shuttle! It would be easier to navigate with it than with propulsors.
| the video show that the principle of helicopter works in a zero-g (no gravity) environnement, not that the principle works in space.
The helicopter is able to lift on earth (and int the space shuttle) because of the viscosity of the air. Whitout any friction, there is no move.
In space, there is really few particles. So the friction would be very low and you cannot use system such as helicopter.
An helicopter in space would be like trying to use à boat turbine in air
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/183751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Notation of vectors It's very common to see $\text{F} = 30 \text{ N}$ when the problem is unidimensional. Yet, force is a vector. Shouldn't I write $|\overrightarrow{F}| = 30 \text{ N}$? Because if I write $\overrightarrow{F} = 30 \text{ N}$ I'm saying that the vector is equal to an scalar. On the other hand, I rarely see $\overrightarrow{F} = (30, 0, 0)$.
| No, you should not write "$\left|\vec{F}\right| = 30\textrm{N}$", because it's no better than "$F=30\textrm{N}$"
Since force is a vector, you could write out the list of components, either as a parenthetical list or a column vector:
$$\vec{F} = \left(30\textrm{N}\right) = \left[ 30\textrm{N}\right]$$
You could also write the one component as a scalar:
$$F_x = 30\textrm{N}$$
Either of these is both correct and complete.
Your proposal, $\left|\vec{F}\right| = 30\textrm{N}$, is correct but not complete, because it removes information (the sign of the component).
The notation you object to, $F=30\textrm{N}$, is complete but (arguably) not correct.
I don't see any value in considering $F=30\textrm{N}$ incorrect; I think of it as $F_x = 30\textrm{N}$ but without having to introduce a subscript to label a component that doesn't need disambiguation. It is an inconsistency in the notation that should be explained when it's introduced, but I think it's an acceptable way to write vectors with a single component.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/183856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
D'Alembert Equation and standing wave As we know the d'Alembert Equation is
$$
\frac{\partial^2\psi}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2\psi}{\partial t^2}
$$
for an undimentional string.
Now if we seek standing wave solution, we put $\psi(x,t)=f(x)g(t)$ can you tell me a physical argument which shows if $\psi$ is a standing wave then $\psi(x,t)=f(x)g(t)$.
I was speaking of this with my maths teacher when he said me he never understood why we use that. My physics teacher has no idea and I don't find anything about it in Feynman lessons.
And I have no idea for tags so I take "waves"
Thank you
| A very simple way is to look at the nodes of the standing wave. As the name suggests, the nodes are not supposed to move as time goes on.
Suppose $x_0$ is one of the nodes at a particular time instant $t_0$, then we have $\psi(x_0,t_0)=0$. If position and time can be separated, then we have $f(x_0)g(t_0)=0$, which gives us two cases:
*
*$f(x_0)=0$, then we have $\psi(x_0,t)=f(x_0)g(t)=0$ for all $t$, i.e. $x_0$ is always a node of the wave. It doesn't move with time.
*$f(x_0)\neq 0$, then $g(t_0)=0$, then we have $\psi(x,t_0)=f(x)g(t_0)=0$ for all $x$, including $x_0$. This only tells us that at this particular time $t_0$ all the $x$ are nodes. Since we need to look at the time evolution characterized by a function $g(t)$ that is not constantly zero, this case is not so interesting.
Using the same argument, you can easily check that if we have functions where $x$ and $t$ are combined in such way that they are not able to be separated, a typical example is the general travelling wave solution $f(x\pm vt)$, then the nodes of the wavefunction will move with time.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/183916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
quantum fluctuations and the virtual particles In the introduction of chapter-12 of “An Introduction to Quantum Field Theory” by Peskin and Schroeder I encountered this line: “The quantum fluctuatuations at arbitrarily short distances appear in Feynman diagram computations as virtual quanta with arbitrarily high momenta.” My questions are:
*
*What is the mathematical (and physical) meaning of quantum fluctuations of a quantum field $\phi(x)$ at a point x?
*How the quantum fluctuations are related to the virtual particles? And how do the quantum fluctuations make their appearance in Feynman diagrams?
| I think you should not take the Peskin and Schröder quote too seriously.
They are likely just using the Fourier relationship "short distance <-> high momenta" and the idea that there are propagators $\langle \phi^2 \rangle$ (which are the "fluctuation"/variance of $\phi$, see this post) associated to the Feynman lines of virtual particles, so "small distance fluctuations" correspond to "high momentum virtual particles".
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/184019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Propagating higher order Hermite Gaussian modes. What are complex amplitude coefficients? I've been tasked with writing a code (in MatLab, but I'm currently using Mathematica because I don't know MatLab %\ ...) to simulate the propagation of a Gaussian beam.
I don't really know anything about optics, so I'm learning on the fly using this manual kindly provided to me by my adviser.
I'm stuck trying to figure out what the general closed form of a complex amplitude coefficient a[j,n,m] (p.74 and also below) would look like :( Could someone explain it please?
i have looked through various articles and presentations such as
http://www.colorado.edu/physics/phys4510/phys4510_fa05/Chapter5.pdf (and others i can't post links too because of my low reputation) and haven't seen mention of said complex amplitude parameter anywhere else. Lots of complex amplitudes, but not parameters. I don't think they are the same thing, are they?
| Turns out that due to orthogonality relations of Hermite-Gauss poly's, Hermite-Gauss modes are orthonormal, so
$$
\int \int u_{n,m} \left(u_{n',m'}\right){}^*dxdy=\delta _{m,m'} \delta _{n,n'}
$$
Then a's can be found by multiplying both sides by conjugate of u and integrating ...
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/184119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is molecular vibration just phonon modes for a single molecule? I'm reading about Raman Scattering, of which a big part is measuring the energy lost to/gained from Molecular Vibrations. I wasn't totally clear on exactly what is "vibrating" in vibrational modes (is it the electron around the nucleus? Or the whole atom, with respect to its center of mass? Or the individual atoms that make up the molucule?), but it seems like it's the atoms (and I guess with them, their respective electrons), from the wiki page:
The coordinate of a normal vibration is a combination of changes in the positions of atoms in the molecule.
And then they have some handy gifs showing it.
However, (collective) normal modes of nuclei sound exactly like phonons to me. So are vibrational modes very similar to phonons?
thanks!
| *
*In this context, vibration usually refers to the relative motion of the nuclei.
*In a periodic solid (crystal), vibrational modes are called phonons. We can say, for example, that a normal vibrational mode of a branch $s$ with wavevector $\mathbf{k}$ is in its $n$th excited state, or equivalently, that there are $n_{\mathbf{k}s}$ phonons of branch $s$ with wavevector $\mathbf{k}$. In molecules these are called just vibrations.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/184243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How electrons move so fast in a electric circuit? Whenever we switch on a bulb......it takes almost no time to glow up.....But we know that the atoms of a solid are tightly packed and there is a very little space between them.
So how the electrons travel through them irrespective of so much blockages in the conductor???
| You might be asking how metal is such an efficient conductor.
Some of the electrons move freely as a fluid. They are not locked in place around the atoms, and don't need "room" in the classical sense.
Here is a wikipedia page going over the real details.
On a scale much larger than the inter atomic distance a solid can be viewed as an aggregate of a negatively charged plasma of the free electron gas and a positively charged background of atomic cores. The background is the rather stiff and massive background of atomic nuclei and core electrons which we will consider to be infinitely massive and fixed in space. The negatively charged plasma is formed by the valence electrons of the free electron model that are uniformly distributed over the interior of the solid
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/184317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
How much UV is reflected by glass? On a sunny day, how much of the Sun's UV is reflected by windows?
I suppose this boils down to what are the refractive indices of common window glasses, so that one might solve the Fresnel equations.
Also, at the risk of being off-topic, how safe is viewing this reflected light?
| I'd say about 4% of the intensity for perpendicularly incident light, increasing to nearly 100% when the angle of incidence is almost zero.
These are the numbers you get from the Fresnel equations (for $n = 1.5$ which is a typical value for glass) as you suggested.
While glass absorbs UV the absorption should not be so strong this induces relevant reflection. The reflected UV light comes mainly from the front side (as most UV will be absorbed in the pane there is no reflection contribution from the back side).
For $n = 1.5$ 4% are reflected for one surface for perpendicular incident. The real part of $n$ should not differ to much from the one for visible light for near UV (perhaps by at most a few percent).
This means there will actually be less UV intensity relative to the intensity of the visible light in the reflected light (as for the visible light reflection at the backside of the pane contributes significantly giving a higher reflectance).
For far UV these considerations might break down as the absorption may get very strong (but the eye is in-transparent for far UV so its not a safety concern at least at the intensities to be expected in sunlight, which is filtered by the ozone layer). Also there are differences depending on the type of glass. Quartz glass (which will, admittedly, not be used for windows), for example, is much more transparent to UV than other types of glass leading to almost 8% of the intensity being reflected for perpendicularly incident light.
On Safety: The main safety hazard when viewing the reflected light will be the intensity (of the visible light), not the UV radiation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/184428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Momentum conservation in FRW spacetime The spatially flat FRW metric in Cartesian co-ordinates is given by:
$$\mathrm ds^2 = -\mathrm dt^2 + a^2(t)(\mathrm dx^2 + \mathrm dy^2 + \mathrm dz^2)$$
As I understand it, since the metric does not depend on the spatial co-moving co-ordinates $x,y,z$ then there are Killing vectors in the $x,y,z$ directions.
Does this imply that the 3-momentum of a free particle is conserved when measured with respect to the $x,y,z$ co-ordinates? (In terms of expanding proper distances I presume that the particle would seem to lose velocity)
Does this also imply that the 3-momentum of a photon is conserved when measured with respect to the $x,y,z$ co-ordinates?
If the 3-momentum of the photon is conserved then, as $E=pc$ for photons, does this imply that its energy is conserved as well?
| Canonical momentum is conserved, but proper momentum density is proportional to $\frac{1}{a}$ where $a$ is the scale factor.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/184534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Odd order of colour fringes around cloud The other day I took this photograph of a thundercloud that moved between myself and the sun. Around the edge of the cloud there were some coloured fringes that I first took to be a variant on a normal rainbow (yes, I know that a rainbow is visible when looking away from the sun). But then I examined the colours more closely and it seems to me that from the cloud, they go orange, red, blue, green. What's going on?
| It could be due to the moisture in the air around the cloud. The moisture would make the light refract much more because it is denser, making the rainbow thinner and leaving yellow and blue out.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/184608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is it possible to write the fermionic quantum harmonic oscillator using $P$ and $X$? The Hamiltonian of the quantum harmonic oscillator is
$$\mathcal{H}=\frac{P^2}{2m}+\frac{1}{2}m\omega^2X^2$$
and we can define creation and annihilation operators
$$b=\sqrt{\frac{m\omega}{2\hbar}}(X+\frac{i}{\omega}P)\qquad{}b^{\dagger}=\sqrt{\frac{m\omega}{2\hbar}}(X-\frac{i}{\omega}P)$$
where the following commutation relations are fulfilled
$$[X,P]=i\hbar\qquad{}[b,b^{\dagger}]=1$$
and the Hamoltonian can be written
$$\cal{H}=\hbar\omega\left(b^{\dagger}b+\frac{1}{2}\right).$$
Now, it is also known that we can define a fermionic quantum harmonic oscillator with the Hamiltonian
$$\cal{H}=\hbar\omega\left(f^{\dagger}f-\frac{1}{2}\right)$$
where $f$ and $f^{\dagger}$ satisty the following anticommutation relation
$$\{f,f^{\dagger}\}=1.$$
What I am trying to get is a Hamiltonian for the fermionic harmonic oscillator using $P$ and $X$. I have tried defining
$$f=\sqrt{\frac{m\omega}{2\hbar}}(X+\frac{i}{\omega}P)\qquad{}f^{\dagger}=\sqrt{\frac{m\omega}{2\hbar}}(-X-\frac{i}{\omega}P)$$
because after imposing the anticommutation relation $\{X,P\}=i\hbar$ for $X$ and $P$ (as I guess would suit a fermionic system) these definitions of $f$ and $f^{\dagger}$ imply $\{f,f^{\dagger}\}=1$. Nonetheless, for the Hamiltonian I get
$$\mathcal{H}=\frac{P^2}{2m}-\frac{1}{2}m\omega^2X^2$$
where I get an undesired minus sign. My question is then the following: is it possible (with an appropriate definition of $f$ and $f^{\dagger}$ in terms of $X$ and $P$) to obtain the first hamiltonian I have written from the fermionic oscillator Hamiltonian written in terms of $f$ and $f^{\dagger}$?
| Assuming that $X=X^\dagger$, $P=P^\dagger$ and $[X,P] = i\hbar$, let me try
$$f = \sqrt{\frac{m\omega}{2\hbar}}\left( \alpha X + \frac{\beta}{m\ \omega } P \right) $$
where $\alpha$ and $\beta$ are complex numbers of modulus one. From this follows that
$$ \hbar \omega \left( f^\dagger f - \frac{1}{2} \right) = \frac{P^2}{2m}+ \frac{1}{2} m \omega^2 X^2 + \hbar \omega \left(\alpha^\ast\beta \frac{XP}{2\hbar} + \alpha\beta^\ast \frac{PX}{2\hbar}- \frac{1}{2} \right) $$
You see now why I chose $\alpha$ and $\beta$ the way I did. We recover the original Hamiltonian if
$$ i\alpha^\ast\beta\ XP + i\alpha\beta^\ast\ PX \overset{!}{=} i\hbar = [X,P] $$
is fulfilled. Thus, we are led to the conclusion $\alpha\beta^\ast = i$. Two complex numbers of modulus one that fulfill this equation are $\alpha = i$ and $\beta = 1$ and therefore
$$f = \sqrt{\frac{m\omega}{2\hbar}}\left( i X + \frac{1}{m\ \omega } P \right) $$
could be a possible canditate. So remarkably we get $f = i b^\dagger$. We can check the result by inserting this relation
$$f^\dagger f - \frac{1}{2} = (-i b)(+i b^\dagger)- \frac{1}{2} = bb^\dagger - \frac{1}{2} = b^\dagger b + \frac{1}{2}$$
where the last step follows from $[b,b^\dagger] = 1$. But unfortunately
$$\left\lbrace f,f^\dagger\right\rbrace = f f^\dagger + f^\dagger f = b^\dagger b + b b^\dagger \neq 1$$
and $[f,f^\dagger] = -1$. You will always get a boson operator. Which makes perfectly sense if you think about it. A fermionic ladder operator would imply that your system suddenly has only two states left while you found infinitely many before. If you want to have a fermionic oscillator something has to happen with the Hamiltonian and the assumptions have to be altered.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/184815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Difference between fusion plasma and fluorescent lamp plasmas How is the plasma in a compact fluorescent lamp (CFL) different from a plasma in say ITER or the sun? Why does ITER need 100MK and a CFL can work at practically room temperature (apart from the filament)?
Or could ITER also create a plasma by charging the gas inside the reaction chamber but not have enough energy for the reaction, so they heat it directly (microwaves) and charging it would be of no use?
Or is it the degree of ionization the volume of gas has achieved? Like, a CFL has around $x$ ions and a sun plasma has only ions?
| A compact fluorescent lamp belongs to the glow discharge plasmas. Usually you have electron densities on the order of $n_e \approx 10^{16}\,\mathrm{m}^{-3}$, electron temperatures on the order of $T_e\approx 1\,\mathrm{eV}$ and ion temperatures being at least an order of magnitude lower. The degree of ionization is $1\,\%$ or lower. The room temperature you were referring to, only applies to the gas temperature and, more or less, to the ion temperature. The electrons are much hotter.
ITER, in contrast, will have plasmas parameters of $n_e \approx 10^{20}\,\mathrm{m}^{-3}$, $T_e\approx T_i \approx 10\,\mathrm{keV}$.
ITER requires such a high temperature because it is meant to investigate nuclear fusion and to achieve fusion you need to overcome the electrostatic repulsion of the nuclei (remember, both are positively charged). A fluorescent lamp, on the other side, just needs enough energy/temperature to achieve breakdown, as @Rod Price has written.
ITER could of course also create lower temperature plasmas, and it will do so for cleaning the wall, but for fusion, those high temperatures are required.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/184899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Does Time change over temperature? I am not a physicist, I am just an engineer.
But I dared to ask whether the temperature changes the perception of time.
Let's consider a particle that "stops" at absolute zero. I was thinking as a hypothesis, that our perception of time changes and the particle actually does not stop at all.
| No. In fact the official definition of the second is:
The duration of 9,192,631,770 periods of the radiation corresponding
to the transition between the two hyperfine levels of the ground state
of the caesium-133 atom at absolute zero.
So time is still alive and kicking at absolute zero.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/185094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Maximum angle for highway lane change I am preparing to fight a traffic ticket from a speed camera. There is a lot more to that story, but the info I need right now involves the angle at which a car can change lanes, in terms of vehicle angle relative to the direction of traffic flow. I have a 2009 VW Golf Kombi (this is the Jetta Sportwagen in the US). In my case, the relevant speed is 90 km/h, but I'm also interested in learning how to calculate this for any speed.
For example, I believe my vehicle has a steering ratio of approximately 16:1. Thus, if I turned the wheel 16 degrees, I would only deviate from my original lane by 1 degree. That would get me to the other lane eventually, but it is likely to be quite leisurely. On the other hand, if I turned too fast I would likely lose traction.
I know the actual maximum is dependent on many factors (tires, weather, etc). So my question is what type of angles can I expect to achieve in a car similar to mine? And, on a perfect day, with the best tires, what is the highest angle I could achieve without crashing?
| The fastest possible way to do a lane change is to fully steer one way on the traction limit and then steer on the opposite way again on the traction limit.
The traction limit is $\mu g = \frac{v^2}{r}$ where $g=9.81\;{\rm m/s^2}$ is gravity, $\mu=0.8\ldots0.9$ is the coefficient of friction (half it in the rain), $v$ is the speed in meters per second and $r$ the radius of turn measured from the center of the car along the rear wheels.
The means the maximum wheel steering angle $\theta$ (to maintain control) at speed is $$\tan \theta = \frac{\mu g \ell}{v^2}$$ where $\ell$ is the wheel base of the car (in meters).
To move the car sideways one lane width $w$ you need to maintain the steering angle $\theta$ one way to trace an arc of radius $r=\frac{v^2}{\mu g}$ for an angle $$\psi=\arccos\left(1- \frac{w}{2 r}\right)$$ The time it takes for this part is $$t=\frac{\psi r}{v} = \frac{v}{\mu g} \arccos\left( 1-\frac{\mu g w}{2 v^2} \right) $$
For the full lane change double the time. The total distance traveled parallel to the lanes for the full lane change is equal to $$d=\sqrt{ w (4 r-w)} =w \sqrt{ \frac{4 v^2}{\mu g w}-1} $$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/185238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
What would be the rate of acceleration from gravity in a hollow sphere? Lets say the Earth is hollow and you are in the center of it (Same mass except all of it is on the outside like a beach ball) If you move slightly to one side now your distance is closer to that side therefore a stronger gravitational force however at the same time you have more mass now on the other side. At what rate would you fall? Which direction?
Also, is there a scenario where depending on the radius of the sphere you would fall the other direction or towards the empty center?
| The answer is that inside a spherically symmetric shell of matter (your hollow earth or massive beach ball) there is no gravitational force anywhere - you will not "fall" in any direction, whether you are at the centre or not, regardless of the radius of the sphere.
This is a classic result of both Newtonian Gravity, and Einstein's General Theory of Relativity. In both cases it is called the [ ] Shell Theorem. Although, except at the centre, you are closer to one side than the other you can imagine that there is "more" farther away than there is closer, and because of the nature of the 1/r^2 law of Newtonian gravity everything cancels out. It is of course more complicated in General Relativity.
For a simple introduction, see the Wikipedia article on the Shell Theorem
You may also be interested to know that according to General Relativity time passes more slowly inside a hollow massive sphere than it does outside.
Whilst there is no gravitational force inside the sphere, and therefore no gravitational field, there is a non-zero gravitational potential - it just happens to be the same everywhere. Since the force of gravity depends on potential difference (just like voltage in the electrostatic case), if the potential is constant there can be no force. However, the potential does effect the passage of time in General Relativity, and since it is not zero inside the sphere time passes differently. Inside an "ordinary" matter sphere time passes more slowly.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/185298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"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.