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Ramsauer-Townsend effect one dimension I am trying to learn basic one dimensional quantum mechanical scattering. The approach used is studying the stationary wave with a square potential barrier with $V_0$ as the positive potential and $L$ as the length of the barrier.
My textbook says that for certain energies, namely $E = -V_0 + \frac{n^2 \hbar^2 \pi^2}{2mL^2}$ for n = 1, 2, 3... the probability for transmission is 1. It says this effect is due to interference. When I try to draw this I only get a contructive interference on the reflective wave, which does not explain this phenomenon to me. I would appreciate if anyone could explain this effect since the explanation in the book is a bit slim and it is hard to find such an easy example elsewhere.
| I found my answer. What I did wrong when drawing my picture was to make a phase shift at both boundaries of the square potential. There is supposed to only be a phase shift at one of the boundaries. This is analog to an electromagnetic wave where phase shift or not is dependent on whether the electric permitivity is increasing or decreasing in the boundary.
| {
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Effective potential of radial equation of Hydrogen In QM, the study of the hydrogen atom, why does it follow that for the radial equation given as $$-\frac{\hbar^2}{2m}\frac{d^2 u}{d r^2} + \bigg( -\frac{e^2}{4 \pi \epsilon_0} \frac{1}{r} + \frac{\hbar^2}{2m}\frac{l(l+1)}{r^2} \bigg)u = Eu,$$ the term $\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}$ is an effective potential that tends to keep the wave function with $l \neq 0$ away from the origin?
| The term
$$V_\mathrm{centripetal}=\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}$$
is an effective potential because it appears in the radial equation in the same way (coefficient of $u$) as the Coulomb potential
$$V_\mathrm{Coulomb}=-\frac{e^2}{4\pi\epsilon_0r}$$
The Coulomb potential is attractive for opposite charges which you can see from the minus sign in the equation. Taking only the Coulomb potential into account energy gets smaller the smaller $r$ is.
The effective potential on the other hand is always positive making it repulsive (energy gets smaller for large $r$). Specifically if you get close to the origin, $r\to 0$ the centripetal potential $V_\mathrm{centripetal}\to +\infty$. In physics, systems tend to minimize their energy, so a position where the potential energy is very large is not attractive for the system and you can say that the centripetal potential is "keeping the wave function away from the origin".
In a classical mechanics analogy, a potential like this would result in a force $F=-\frac{dV}{dr}>0$, i.e. a force which is positive, pushing the "particle" away from the origin.
Since the centripetal potential is $\propto r^{-2}$ and the Coulomb potential $\propto r^{-1}$, at small $r\to 0 $ the centripetal potential will dominate the total potential $V_\mathrm{eff}=V_\mathrm{Coulomb}+V_\mathrm{centripetal}$. At large $r\to\infty$ the Coulomb potential will dominate. In-between there is an optimum $r$ at which the energy gets minimized. In a picture:
| {
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Real images and their formation We're told that a real image is formed when light rays actually converge to a point. That's all good. But what happens if a screen isn't there to take the image on? Is it still there?
| Please note that my answer is totally in the framework of geometrical optics.
The point where the real image is formed is where rays converge. After this point, rays will naturally diverge, as if they were coming from a point source of spherical waves. You can see an example of this phenomenon here.
| {
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In thermodynamics what is a constant mass process called? In thermodynamic processes you'll see defined conditions like isochoric for constant volume and isothermal for constant temperature, as examples.
In a general thermodynamic process you may have a flux of mass across the control volume, but in many cases the mass is fixed. So is there a definition or convention for naming a process where the mass is assumed to be constant?
iso -?
I'm specifically interested in thermodynamics applied to gases.
EDIT: I did a fairly thorough search on Google; probably defined in older textbooks.
| It's simply a closed system
In thermodynamics, a closed system can exchange energy (as heat or work) but not matter, with its surroundings. An isolated system cannot exchange any heat, work, or matter with the surroundings, while an open system can exchange energy and matter.
| {
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Looking for a good casual book on quantum physics I'm looking for something that is going to blow my mind without any scientistic ideas (e.g. something that sounds like science, but doesn't have anything in common with science), written by a professional physicist who spent a lot of time considering "what it all means". I'm reasonably proficient in math and stats, but I'd prefer something that I could spend time listening to in my free time.
Any recommendations on good and exciting books on quantum physics written by scientists?
| I really enjoyed
*
*Quantum: a guide for the perplexed, by Jim Al-Khalili
though to be honest it's now getting uncomfortably close to fifteen years since I read it as a high school student, and I don't have my copy at hand to give it a critical assessment. As a high schooler I found it engaging and accessible, and if you want to you could see it as very successful in my case - it was my first introduction to quantum paths, and here I am working in the field a decade later, though I'm unsure that effect is uniform across its readership. As I recall, though, I looked it over a few years ago and didn't find anything that was worryingly inaccurate or misreepresentative of how modern quantum physicists see the field.
| {
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What does one second after big bang mean? Consider the following statement:
Hadron Epoch, from $10^{-6}$ seconds to $1$ second: The temperature of the
universe cools to about a trillion degrees, cool enough to allow
quarks to combine to form hadrons (like protons and neutrons).
What does it mean to say "from $10^{-6}$ seconds to $1$ second"?
How is time being measured?
One particle might feel just $10^{-20}\ \mathrm s$ having passed and another could feel $10^{-10}\ \mathrm s$ having passed.
Is saying "1 second after the big bang" a meaningful statement?
| We know that time passes differently for different observers, and the question is how can a time be given without telling which frame it is in. The answer is that there's a preferred reference frame in cosmology, the comoving frame, because of the fact that there's matter and radiation in it.
Intuitively, the special frame is the one that's "static" with respect to this matter and radiation content. More precisely, it is the one in which all observers that see an isotropic universe are static. Time measured in this system is called comoving time. The time from the beginning of the universe is usually given in this way, as a comoving time.
To get some intuition about the comoving frame one might consider the comoving observers, the ones that see isotropy and therefore have constant comoving coordinates. A comoving observer is such that when it looks around and adds the motion of the objects it sees zero net motion.
For example, we can look at the cosmic microwave background and detect some variation in the redshift depending on the direction. It's caused by Doppler effect and it means that we have some velocity relative to the comoving frame. On the other hand, a comoving observer sees the same redshift in any direction.
Another example: we can choose to measure the distances and velocities of galaxies. By Hubble's law, we expect the velocity to be proportional to the distance. If we find a deviation from this behavior, we know that the galaxy is moving with respect to the comoving frame, and thus has a peculiar velocity (we also have a peculiar velocity). If all galaxies had constant comoving coordinates, we would see perfect agreement with Hubble's law: the relative motions of galaxies would be due only to the expansion of the universe.
| {
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Method of image charges for a point charge and a non-grounded conducting plane I know how to solve Laplace's equation for a point charge in front of a grounded conducting infinite plane. But I want to know what happens (both physics and math) when the infinite conducting plane isn't grounded, or is connected to a potential $V$.
| The problem is addressed in Electrodynamics Fulvio Melia Example 2.1 page 39,
using Green function techniques. However, I cannot fully understand his final
solution. Let us define the plane held at a constant potential $V_0$ to be the $z$-$y$ plane. The charge $q$ is at the distance $a$, namely at $(a,0,0)$. The potential at large distances is zero. The solution (Eq. 2.54) is
$$V(x,y,z) = V_{image} (grounded) + V_0 f(x,y,z).$$
The first term is the usual image method solution, for the case where the conducting plane is grounded, namely
$$V_{image}(grounded) = q\left[\left[(x-a)^2 + y^2 + z^2\right]^{-1/2} - \left[(x+a)^2 + y^2 + z^2\right]^{-1/2}\right],$$
which gives zero when $x=0$. In other words this is the solution of the Poisson equation, but the boundary conditions are not those required.
The second term, $V_0 f(x,y,z)$ is the solution of Laplace equation, and $f(x,y,z)$ is $V_0$ independent.
According to the text book Eq. (2.54) there
$$f(x,y,z) = (x/2 \pi) \int dz' \int dy' \left[x^2 + (y-y')^2 + (z-z')^2\right]^{-3/2}.$$
The integrals over the primed variables stretch from minus infinity to plus infinity and this function is V_0, q and the length scale a independent.
The solution is mysterious for me. Namely, we can get rid of $y$ and $z$ in the integral (for any finite $y$ and $z$) by change of variables of integration. Then the integral can be solved (for $x>0$) and the solution is $f(x,y,z)= x/ x = 1$. Well, that is wrong (or subtle) since that would mean that we only need to add to the usual image solution a constant $V_0$. This does not satisfy the boundary condition at large distances from the plane where the potential is zero.
Maybe the integral for the dimensionless function f(x,y,z) should be regularised somehow.
Since the image part of the solution, $V_{image}(grounded)$, takes care of the charge in the Poisson equation, we can find a solution that is a linear combination of $V_{image}(grounded)$ and some other function, the latter solves the Laplace equation. Then let the boundary at $x=L$ be grounded and we want to take $L\to \infty$ and on x=0 we have V_0 (this seems like a legitimate trick). The Laplace equation for this problem gives
$$V(x,y,z) = -V_0 x/L + V_0,$$
and if we take $L\to \infty$ we get the result just mentioned (for any finite $x$) namely a constant (which does not satisfy the boundary condition at infinity). So I think the limit of a large system ($L\to \infty$) is creating a certain non-trivial problem, which ultimately is related to the Green function technique used by Fulvio Melia. Any ideas?
Of-course if f(x,y,z) is unity for any finite distance x, this would mean that switching from grounded plane to non-grounded one
(compared to zero potential at infinity) has practically no effect. This hypothesis was postulated in some of the previous discussions.
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Does entropy depend on the observer? Entropy as it is explained on this site is a Lorentz invariant. But, we can define it as a measure of information hidden from an observer in a physical system.
In that sense, is entropy a relative quantity depending on the computation, measurement and storage capacity of the observer?
| I think that Shannon-von Neumann definition of Entropy pass this anthropocentric paradox by establishing the minimum amount of information that cannot be reversibly be exchanged between two states of the same system, no matter if there is an agreement or even the presence of observers. In such way Entropy is indeed a physical characteristic and not an observer artifact, plus establishes a unique direction for the flow of information, hence causality, flow of time etc..
I know I am simply placing postulates against each other and I am in no position to establish or hint at the correctness of one or another, but I prefer keeping my physics understanding within the boundaries of experimental verification.
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Why is it possible to use free energy functions on Boltzmann factor? When the Boltzmann factor is derived on the canonical ensemble, one gets the exponential function of the energy:
$\large{e^{-\frac{E}{k_b T}}}$
However, some texts (it is very common on physical-chemistry, for exemple) just assume that you may use Gibbs free energy on the factor instead:
$\large{e^{-\frac{G}{k_b T}}}$
But what is the justification for this?
| I am new to this, so if I am missing something obvious, my apologies. Also, I have not dealt with TD from the chemical or biology perspective.
Assume a given energy, $e^{-E/k_BT}$, provides the relative probability of a particular microstate. Now assume that we are concerned with a isobaric system, so E goes to H, (the enthalapy) and use the equation $e^{-H/k_BT}$.
For the canonical ensemble, we need to take account, at a given energy, of the different microstates that will be present.
$S = k_B ln \Omega$ therefore $\Omega = e^{S/k_B}$
So multiplying these together , to determine the probability that the system, at a fixed energy, is in a particular state:
$$\Omega \cdot e^{-H/{k_B}T} = e^{S/k_B} \cdot e^{-H/k_BT} = e^{TS/{k_B}T} \cdot e^{-H/k_BT} = e^{-(H-TS)/{k_B}T} = e^{-G/{k_B}T}$$
So the overall probability is actually proportional to $e^{-G/{k_B}T}$
So the Gibbs free energy takes into account both energy (in the Boltzmann factor) and entropy, the number of possible arrangments. (by virtue of $\Omega$)
| {
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Would a gas "weigh" less than a liquid if they have the same mass? Thought experiment: I acquired two boxes of the same dimensions and same weight. One box contains $1\ \mathrm{kg}$ of water at room temperature while the other box has $1\ \mathrm{kg}$ of water, but in steam form, because the temperature of the box is above $100^\circ\mathrm{C}$. The volume of the boxes is large relative to the amount of space the $1\ \mathrm{kg}$ of water would take (let's arbitrarily say $10\ \mathrm{L}$). Both boxes contain the same amount of air (at $1\ \mathrm{atm}$) which is why the second box has water in steam form at $100^\circ\mathrm{C}$.
I put each box on a simple electronic scale to measure their respective weights. Unsurprisingly, the box containing water comes out to be $1\ \mathrm{kg}$. But what about the box containing steam?
My guess: Electronic scales measure the amount of force being exerted on it, then divide that force by $g$, to get the mass of the object. I think the box with steam in it will be exerting less force on to the scale and therefore the scale will think its mass is less than $1\ \mathrm{kg}$.
| Skip the difference between mass and force. They both experience the same gravity.
Yes a liter of water is 1 kg
A liter of steam is about 1/1000 of that or about 1 g. More precisely 1.67 g (at 100C and 1 atm).
If you mean put 1 L of water in a 1 L box and heat it to 100C or even 200C then it will also be 1 kg. But it will NOT be steam. In a 1L box it does not have room to expand into a vapor. At 200C it will be about 15 atm of pressure. So have a strong weightless box.
I think you have butchered the question - I think you mean to ask.
The mass of 1 kg of water at 70 F and 1 atm and 1 kg of steam at 212 F and 1 atm is exactly the same.
The volume will be significantly different. About 1000 x different.
Weight is force and the weight will be different due to buoyancy. I don't have time to run the numbers right now.
density of water is 1 kg/L
density of steam is 0.0006 kg/ L
With buoyancy I think the steam would actually float away as it is less dense than air. But I will leave that for others.
| {
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Conceptually, why is acceleration due to gravity always negative? As the title states, why is acceleration due to gravity always (-). Say you assign "up" as the positive direction. If an projectile is thrown at a 24 degree angle above the horizontal, I get that acceleration due to gravity before the vertex is negative. However, why is it not positive after the vertex? If acceleration due to gravity is negative and we assign downwards as negative, wouldn't that make acceleration positive?
What I think is that acceleration due to gravity is always towards the ground. Even if a projectile is going downwards, and we assign downwards as (-), the acceleration due to gravity is still (-), because the object still accelerating downwards. Despite it going upwards or downwards, the net acceleration of the object is downwards.
| I don't have enough reputation to make a comment, so I'm adding this as a separate answer.
The other answers are not wrong, but I feel it is worth pointing out that this is related to the fact that, as far we know, there is only type of gravitational charge (i.e. mass), which we by convention call positive. If I recall correctly there are planned/ongoing experiments to see whether antimatter might have a negative gravitational charge. I don't believe many people expect that to be the case, but it is evidently something considered worth the effort to check.
Related questions:
Negative Mass and gravitation
Electrical force vs gravitational force
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Can there be really a theory of everything? If quantum theory becomes compatible with Einstein's theory of gravity, we get the theory of everything. But if it can predict anything in the universe (provided boundary equations are given) wouldn't it predict itself. I mean won't the theory tell us when the theory will be formed? This sounds quite absurd.
| Your question seems to stipulate that a Theory of Everything must be self-reflexive. In other words, that it allow a "consciousness" of itself, or that it explain itself. Don't all workable theories do this?
Alternatively, your question may suggest that a Theory of Everything must predict the time at which it will be developed. This would be trivial and absurd, as any rationalization can predict the past.
I think you are reading into the long-sought TOE a meaning that is not there.
| {
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Why do we invent non-physical concepts (like e.g. point particles) to study physical phenomenons? There is nothing exist like point particles in reality then why did we invented the notion of point particles and how does it relate to real world?
|
then why did we invented the notion of point particles
Because that makes it simpler to work with.
*
*Sometimes we say "it costs 24 \$" even though it is actually 23.99 \$.
*We also usually say that there are 365 days on a year, even though there actually are $\sim$365.24.
We are accepting a tiny error margin, when that really doesn't matter at the scale we are working at. If you are working with everyday size-scales, then you really don't have to consider the actual size of stuff like charges, atoms, electrons etc. - the modelling of them as point-like is fine, fine, fine.
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How to handle a Dirac delta at $r = 0$ for a first Born approximation? I have been trying to get my hands around a scattering problem all day but I can't wrap my head around the idea. It's a scattering problem with First Born Approximation and the potential is a Dirac delta sitting at the origin. The standard procedure would be to just use the formula from Griffith's book
$$
f(\theta) = \frac{-2m}{\hbar^2 \kappa}\int_0^\infty a\delta(r)r\sin (\kappa r)dr
$$
where $\kappa = 2k \sin({\theta/2})$. The first problem is of course the fact that Dirac delta function lacks support at $r=0$ so I have been trying to switch to cartesian coordinates where it has support at the origin. But I keep getting zero as result, so either the my answer is right and that makes the problem really boring to start with, or either I'm missing something.
I have also been trying to understand the physics around the problem. If the Dirac delta was evaluated at some radius a the problem would have made more sense since that would have been the scattering from a hard sphere, but I can't understand what a scatter from a single infinitely small point would mean?
| The general formula for the first-order approximation of $f(\kappa)$ is
$$
f(\kappa)=-\frac{m}{2\pi}\int\mathrm d\boldsymbol r\ \mathrm e^{-i\boldsymbol\kappa\cdot\boldsymbol r} V(\boldsymbol r)\tag{A}
$$
In the particular case
$$
V(\boldsymbol r)=a\delta(\boldsymbol r)
$$
we have
$$
f(\kappa)\overset{(\mathrm A)}=-\frac{ma}{2\pi}
$$
If the potential is spherically symmetric, then $(\mathrm A)$ becomes
$$
f(\kappa)=-\frac{2m}{\kappa}\int_0^\infty \mathrm dr\ rV(r)\sin (\kappa r)\tag{B}
$$
Recall that the Dirac delta centred at the origin, in spherical coordinates, reads
$$
V(r)=\frac{a}{4\pi r^2}\delta(r)
$$
and therefore
$$
f(\kappa)\overset{(\mathrm B)}=-\frac{ma}{2\pi}\int_0^\infty \mathrm dr\ \delta(r)\frac{\sin (\kappa r)}{\kappa r}
$$
which agrees with the previous result, if we take
$$
\left.\frac{\sin x}{x}\right|_{x\to0}=1
$$
| {
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Why do we not feel weightless at equator but feel in satellite A person living in his house at the equator goes in a circular orbit of radius equal to the radius of the earth. My question is, why does he/she not feel weightless as a satellite passenger does?
If we compare a geostationary satellite with the earth's equatorial surface then we know they both revolve around the centre of earth with same Angular velocity. So if Normal force is zero in satellite then why not at equator's surface.
| You do weigh slightly less at the equator than on the poles. however escape velocity is much greater than rotational velocity.
| {
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Volume of hypothetical closed universe Consider a physical universe with a beginning in time and space, with a finite amount of mass without the complications of dark energy that we have in the presently open curvature universe, so that the omegas (vacuum and mass) values create a closed universe, a universe with positive curvature. In this way, there will be a maximum (spatial) size that this universe could attain.I am not sure whether the concepts of (spatial) volume or radius would be applicable to this size, but if so, is there a way to calculate what it would be depending on the other variables? Which variables would be necessary -- the omegas, I presume, and what else? Is there a rough rule-of-thumb formula one could use?
| The radius of curvature is
$$\rm r=\frac{c}{H \sqrt{\Omega_t-1}}$$
where $\rm H$ is the Hubble parameter and $\rm \Omega_t$ the sum of all energy contributions:
$$\rm \Omega_t=\Omega_r+\Omega_m+\Omega_{\lambda}$$
so the circumference where a straight line closes in on itself is
$$\rm U=2 \pi r$$
The volume of the 3dimensional curved space is the surface area of the corresponding 4D hypercube:
$$\rm A=2 \pi^2 r^3$$
For the general equations for an arbitrary number of spatial dimensions see here.
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Will fracture occur under very high hydrostatic pressure? In the case of tensile loading, the distances between atoms or molecules increase and finally the material is broken because of bonds breakage between atoms or molecules. But, what about the hydrostatic loading? Will the atomic bonds break when they close to each other very very much? In other words, will a body be damaged under sufficiently high hydrostatic pressure loading?
About hydrostatic loading, I mean stress state of
$$\mathbf{\sigma}=-P\mathbf{I}$$
where $\mathbf{I}$ is the identity tensor and $P$ is constant.
| In the context of everyday forces and pressures (i.e., no neutron stars or black holes), consensus is that homogeneous (non-porous) materials can withstand arbitrarily high hydrostatic pressure without failure (although other transformations not involving fracture, such as a phase change, might of course occur). In other words, strength criteria (e.g., 1, 2) do not predict failure from $\sigma=-P\mathbf{I}$ for any positive value of $P$.
Ductile materials fail through shear (mediated by dislocation movement), generalized as deviatoric stress, which is exactly the opposite of hydrostatic stress in the sense that it occurs off the stress tensor diagonal. Pressure doesn't create a driving force to move dislocations in this way.
Brittle materials fail when the strain energy induced by a load exceeds the energy required to create a new surface that would relieve that strain energy; under hydrostatic pressure, however, there's no mechanism by which a new surface would relieve the strain energy.
As a side note, hydrostatic pressure is well known to alter the mechanisms of failure induced by other types of loading, such as tensile and shear. Early work in this area was performed by Bridgman (e.g., "Effects of High Hydrostatic Pressure on the Plastic Properties of Metals", Rev. Mod. Phys. 17 1945) and has continued in the areas of polymers and rock.
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How can I solve the moment of inertia?
The source tells me to use the formula for a ring, but it is not possible, as the portions are nearer to the axis than a normal ring. How can I find the moment of inertia?
| You know that to solve this problem you will have to use the integral form for MoI.
$\int_0^Mr^2dm$ where $dm$ is the mass element (geometry of the problem) and $r$ is the distance from the axis of rotation.
You express the mass element in terms of $r$ so you get linear density.
Ex. Rod of mass $M$ has linear density $M/L$ so you get $dm=\frac MLdr$
Ex.2 A solid cylinder has volume density of $dm = \rho L2\pi rdr$ (density * length * circumference or $\rho dV$)
etc.
So, you find the distribution of mass (geometry) and plug it into the integral.
Another way you could solve it is by using the Parallel Axis Theorem ($PAT$)
$I_{parallel} = I_{cm} + Md^2$.
You know that the $I=\frac{MR^2}2$ for the whole ring.
To sum up, imagine that you have a whole ring, which you cut on both sides and you get a sort "brackety" system () (removing the "middle" part from a nice round circle) which you can then solve using the integral method of $PAT$ method ($\frac{MR^2}2 + Md^2$ where $d$ is the distance from the axis and the same point where you cut to get one portion of the ring $->$ substituting)
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Falling spool acceleration explanation In this example both spools A and B have the same downward acceleration. Why would the net downward force on Spool A be the same as for Spool B despite having more total kinetic energy?
| The net vertical force on the spools is the same because the tension upward is the same for both, and the weight downward is the same for both.
This is consistent with the difference in the kinetic energy because, the work done by gravity is the same for both, but the work done by the tension is not the same for both.
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What determines the direction of a path on a line integral (vector case)? Line integrals are very important to use in Physics. For example, we calculate work by: W=∫<F,dr>. But I just got confused about something. What determines the direction of motion? The integral limits, or the vector dr?
Well, when we do the internal product of the Force by the path(dr), we are aligning this force on the path's direction(I mean, the shape of the curve) . But if I want a path from the position B to the position A, I determine this inserting on Integral's limits(B inferior limit and A superior limit), or the vector dr would indicate the direction of motion (from B to A) and the integral limits would be (A inferior, B superior)?
| *
*The line segments $\mathrm d\vec r$ determine - define - the exact path.
*The integral itself decides - via the order of its limits - the direction along this path. Meaning, flipping around the integral will only change the sign of the integration.
Both are thus involved with determining the path direction, but the integral limits to a lesser degree.
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Why do ice cubes shrink in the freezer? OK before you all yell "EVAPORATION", I know that's the boilerplate answer, but why and how.
Back in high school we were taught water, and most other elements and compounds, have three states, solid, liquid, and gas. Which state it's in being dependent on temperature and pressure. OK that all makes sense.
To transition from one state to another sufficient energy has to be imparted to the substance to pass through that transition before it will continue to heat up. To get water from ice all the way up to a gas takes a lot of energy... ask all the distillers out there.
Still.. all good...
But my ice cubes disappear over time inside a closed freezer.
The water is evaporating, turning into a gas state, but where did the energy come from to take it all the way from a solid to a gas when it never left the freezer..
| I had never really noticed that happening, so before I said "that seems odd, are you sure?" I did some research and surprised myself.
According to the Wikipedia page on freezer burn water will sublimate from the surface of ice if the air temperature is low enough and the air is dry enough.
The reduced vapour pressure of the dry air is enough to cause sublimation.
Hopefully someone can give a more mathematical approach, but this is somewhere to start.
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Where are the poles of the wire? When current is passed through a copper wire there are circular magnetic field lines around it but where are the north and south poles to that circular magnetic field?
| The poles of a magnetic field, like the pot of gold said to be at the end
of the rainbow, can be found by following along the field-line path. So, if you point
at the magnetic North pole on Earth, and follow that path, you go ... North.
If you examine that field line, however, you'll soon note that it ISN'T
pointing to Earth's north pole, but rather points down, into the ground.
Dipole magnets (with one N and one S pole) are important, but the
exact point where the field lines come to a point... is just an
imaginary point. It's inside the metal of the magnet, or under the
surface of the Earth, and if you DID go there, you'd find the lines
from all directions don't meet. They just gather ... get close.
From outside the Earth, or outside the magnet, the magnetic 'poles' give you
some way of understanding the field lines' shape. Up close, they
are (to the best of our observation) just as illusory as the pot
of gold at the end of the rainbow.
With a straight wire carrying current, the field lines do not
create the illusion of poles.
| {
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How does voltage affect boiling ability of a kettle? I was reading an article about the lack of electric kettles in the USA - the claim at the end states one reason is the lower voltage at the socket. Personally, I think that the main reason is that morning tea is not as big in the US as it is in the UK or Australia, but that's not the point.
Would a 120V kettle really take a "really, really long time" compared to a 240V one? Couldn't it just draw more current?
| For a fixed voltage $V$ the power dissipated by the element in the kettle is $\frac{V^2}{R}$. To increase the power so that the water boils as fast in the USA as it does in the UK, you would need to reduce the resistance of the element. Alternatively, you just use the same kettle as in the UK and wait longer.
The domestic supply voltage in the USA is 110-120V AC, compared with 240V in the UK. The same kettle would take approximately $(\frac{240}{120})^2=4\times$ as long to boil in the USA.
The problem with reducing the resistance of the element is that it increases the current. To obtain the same power dissipation in the kettle at half the voltage the current must be doubled. This also increases heating in the wires supplying the electricity, which is less efficient and risks overheating the wires or - if the circuit is appropriately protected by a fuse - overloading the domestic circuit. Doubling the supply current increases power losses in the wires by a factor of 4, the same as in the kettle.
Americans use electricity domestically for their ovens as well as heating. These are connected to high-voltage circuits, which are more efficient - and quicker. I guess electrical coffee-machines are still run on 120V; perhaps Americans are not in such a hurry to get their morning coffee as the British are to get their morning tea. Stove-top kettles are far more common in the USA than plug-in kettles.
The low availability of electric kettles in the USA is a socio-economic question which I think cannot be answered by physics alone. Filtered or percolated coffee is very popular despite also (I presume) using the inefficient 110V supply, so I think this cannot be the main reason. Cultural preference for coffee rather than tea, and distaste for instant coffee, means there is little demand for electric kettles. Take-away coffee is extensively available and cheap in the USA, so there is less incentive to brew it at home.
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How Do I Increase The Range Of An Home Made Electromagnet? Please forgive me, I am new to this forum and I am not a physics guy but any help would be appreciated. I would like to know how to I increase the range or reach of my electromagnet. By range, I mean the distance from my magnet to a metal object (paper Clip). Currently, my electromagnet has to touch the paper clips to affect them, but I what it to be able to attract them by just coming close with out having to touch. What materials and techniques would be best to achieve this? Thanks for reading
Jay
| There are three main ways to improve your electromagnet without needing a physics degree to model the magnet.
1) Put more current through the magnet. You will need a higher voltage power supply to do this.
2) More loops of wire. This one is a bit tricky since the increased length also increases the resistance, so if you can't keep it compact enough this won't help very much.
3) An iron core. Wrap your wire around a piece of iron. Not all iron is equally susceptible to magnetic fields, so experimenting will definitely help here. The ideal iron would be the kind they use in transformers, but that's typically annular which is not ideal for producing exterior magnetic fields.
Other methods would involve examining the exact geometry of your windings, which I doubt you want to do.
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Generalized forces in Richard Fitzpatrick's Newtonian Dynamics In Richard Fitzpatrick's Newtonian Dynamics, page 128, generalized forces are defined as
\begin{equation} \tag{9.6}
Q_i = \sum_{j=1}^{\mathcal{F}} f_j \cdot \frac{\partial x_j}{\partial q_i}.
\end{equation}
Here, $q_i$ are generalized coordinates, $x_j$ are Cartesian coordinates, and $f_j$ are the Cartesian components of forces acting on particles.
He proceeds to say that in conservative systems
\begin{equation} \tag{9.7}
f_j = - \frac{\partial U}{\partial x_j}
\end{equation}
and concludes that
\begin{equation} \tag{9.8}
Q_i = -\frac{\partial U}{\partial q_i}.
\end{equation}
I can see how the $\partial x_j$ cancel each other out, but what happened to the sum from equation 9.6? Shouldn't there be a constant factor of $\mathcal{F}$ in equation 9.8?
| $$Q_i = \sum_{j=1}^{\mathcal{F}} f_j \cdot \frac{\partial x_j}{\partial q_i}$$
$$\implies Q_i = -\sum_{j=1}^{\mathcal{F}} \frac{\partial U}{\partial x_j} \cdot \frac{\partial x_j}{\partial q_i}$$
$$\implies Q_i = -\sum_{j=1}^{\mathcal{F}} \frac{\partial U}{\partial x_j} \cdot \frac{\partial x_j}{\partial q_i}$$
$$\implies Q_i = - \frac{\partial U}{\partial x_1} \cdot \frac{\partial x_1}{\partial q_i} - \frac{\partial U}{\partial x_2} \cdot \frac{\partial x_2}{\partial q_i} - \frac{\partial U}{\partial x_3} \cdot \frac{\partial x_3}{\partial q_i} -\ldots $$
$$\implies Q_i = - \frac{\partial U}{\partial q_i}$$
Note that, for any $j$, $$\frac{\partial U}{\partial x_j} \cdot \frac{\partial x_j}{\partial q_i}\not =\frac{\partial U}{\partial q_i}$$
According to multivariate calculus, the $x_i$'s represent the components of U along those directions and hence the sum is the contribution of those components of U in n-dimensional space.
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Finding the symmetry of electron spin and position states using Clebsch-Gordan tables? I know that spin and position states must be symmetric and antisymmetric (and vice versa), but I can't figure out how to use Clebsch-Gordan Tables to figure out the symmetry of either one.
For example, Carbon's ground state has a total spin of 1 due to Hund's first rule. How do I know that this spin state is symmetric?
Also, how do I figure out if the position states L = 1 and L = 2 are symmetric, so I can figure out which is correct for the ground state of Carbon?
| Carbon has two valence electrons in the p-shell. To obtain total spin $S=1$ the spin wavefunction must be symmetric. The best way to see this is that the $S=1$ states will contain a $M_s=1$, which must be unique product $\vert 1/2,+\rangle_1\vert 1/2,+\rangle_2$, clearly symmetric under permutation.
As a result, the spatial part of the wavefunction must be antisymmetric. In the decomposition of $(\ell =1)\otimes(\ell =1)=(L=2)\oplus (L=1)\oplus(L=0)$, it is the $L=1$ part that is antisymmetric. One can reach this argument by counting knowing that there must be $6$ symmetric states and $3$ antisymmetric states.
The counting argument starts by noting that the $L=2,M_L=2$ state must be the unique product state $\vert 2,2\rangle=\vert 1,1\rangle_1\vert 1,1\rangle_2$. Thus, all $5$ $L=2$ states are fully symmetric. Since the total number of symmetric states is $6$, the one remaining state is the $L=0$ state. This leaves the three $L=1$ states as antisymmetric.
In terms of Clebsch-Gordan coefficients, antisymmetry is visible because the coefficients $C_{\ell_1m_1;\ell_2m_2}^{LM}$ for $\ell_1=\ell_2=L=1$, pick up an overall sign $(-1)^{\ell_1+\ell_2-L}=-1$ when $(\ell_1m_1)$ and $(\ell_2m_2)$ are interchanged, i.e.
$$
C_{\ell_1m_1;\ell_2m_2}^{LM}=-C_{\ell_2m_2;\ell_1m_1}^{LM}\, .
$$
This phase $(-1)^{\ell_1+\ell_2-L}$ is $+1$ for $L=2$ or $L=0$, confirming that these are symmetric states.
Edit: actually your initial statement is limited to two particles, for more generally it is not quite correct. The strictly correct statement is that the spin and spatial part of the wavefunctions need not be symmetric or antisymmetric by themselves, but their product must be antisymmetric (for fermions at least).
In systems with $3$ or more particles, it is possible for the spin or spatial wavefunctions to have more complicated symmetries than just full symmetry or antisymmetry. The mathematics behind this are the representation of the permutation group of $n$ objects, with $n=3$ for the example of three particles. It is known how to combined spin and spatial wavefunctions of mixed symmetry to obtain a fully antisymmetric total wavefunction.
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Derivation of the electric dipole radiation An harmonic oscillating source will emit fields of the form
\begin{align}
&\mathbf{H}=\frac{ck^2}{4\pi}(\mathbf{n}\times\mathbf{p})\frac{e^{ikr}}{r}\left( 1-\frac{1}{ikr} \right) \tag{1}\\
&\mathbf{E}=\frac{1}{4\pi\varepsilon_0}\left\{ k^2(\mathbf{n}\times\mathbf{p})\times\mathbf{n}\frac{e^{ikr}}{r}+[3\mathbf{n}(\mathbf{n}\cdot\mathbf{p})-\mathbf{p}]\left( \frac{1}{r^3}-\frac{ik}{r^2}\right)e^{ikr} \right\} \tag{2}
\end{align}
where $\mathbf{n}=\frac{\mathbf r}{r}$ is the unit vector that points in the radiation direction and $\mathbf{p}$ is the electric dipole vector.
As Jackson says (Classical Electrodynamics, chapter 9) it's possible to find equation (2) from equation (1) using the formula
$$ \mathbf{E}=\frac{iZ_0}{k}\nabla\times\mathbf{H} \tag{3}$$
Applying equation (3) to equation (1) I obtain
$$ \mathbf{E}=\frac{ik}{4\pi\varepsilon_0}\nabla\times\left[\frac{e^{ikr}}{r}\left( 1-\frac{1}{ikr}e^{ikr} \right)(\mathbf{n}\times\mathbf{p})\right]$$
Then
$$ [\nabla\times(f\mathbf v\times\mathbf w)]_k=\varepsilon_{ijk}\partial_i\varepsilon_{abj}fv_aw_b=\varepsilon_{ijk}\partial_if\varepsilon_{abj}v_aw_b=[\nabla f\times(\mathbf v\times\mathbf w)]_k$$
Identifying $f=\frac{e^{ikr}}{r}\left( 1-\frac{1}{ikr}e^{ikr} \right)$ I calculate
$$\nabla\left[\frac{e^{ikr}}{r}\left( 1-\frac{1}{ikr}e^{ikr} \right)\right]=\frac{e^{ikr}}{r}\left( ik-\frac{2}{r}+\frac{2}{ikr^2} \right)\mathbf n$$
Then i get
$$ \mathbf{E}=\frac{ik}{4\pi\varepsilon_0}\frac{e^{ikr}}{r}\left( ik-\frac{2}{r}+\frac{2}{ikr^2} \right)\mathbf{n}\times(\mathbf{n}\times\mathbf{p}) \tag{4}$$
That is definitely different from equation (2). So where am I wrong? How can I obtain the correct equation (2)?
| Note that $\mathbf{n}$ is not a constant vector but dependent on position. You have not taken the derivative of this, which is the source of error.
In particular, you say
$$\nabla \times (f\mathbf{v}\times\mathbf{w}) = \nabla f \times (\mathbf{v}\times\mathbf{w}).$$
This is incorrect, the correct expression would be
$$\nabla \times (f\mathbf{v}\times\mathbf{w}) = \nabla f \times (\mathbf{v}\times\mathbf{w})+ f\,\nabla \times (\mathbf{v}\times\mathbf{w}).$$
| {
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Expressing position due to gravitational acceleration as a 3-Dimensional differential equation I know that the force of gravity is $F = \dfrac{G m_1 m_2}{r^2}$. Now assume in a one-dimensional system there are two masses in the universe, a planet and an object. The object would have an acceleration of $A = \dfrac{G M}{r^2}$, assuming that m is the mass of the planet. The position of the object can then be graphed with the differential equation $\frac{d^2x}{dt^2}=-\frac{GM}{x^2}$, with x being position and t being time, and negative as the object falls towards the planet. This differential equation can be solved with initial position and velocity values.
My question is now how to transform this to 2 (and 3) dimensions. Assume the same situation but now the object has an x and y coordinate. I can say that acceleration in the x direction equals $A_x = \dfrac{G M}{r^2} \dfrac{x}{r}$, with r being the radius ($\sqrt{y^2+x^2}$. I can also say that acceleration in the y direction equals $A_y = \dfrac{G M}{r^2} \dfrac{y}{r}$.
These equations work when I graph them, but because they rely on each other (the x position is needed for y-acceleration and vice versa), I'm unable to find a way to solve the differential equations and find x and y as functions.
Does anyone have any ideas? If I need to clarify myself better, please ask. Thank you for your time!
|
My question is now how to transform this to 2 (and 3) dimensions
I am not sure where your problem is in doing this. In 3 dimensions using
$$\mathbf{x}_1 = \begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix}, \mathbf{x}_2 = \begin{pmatrix}x_2\\y_2\\z_2\end{pmatrix}$$
to denote the position of $m_1$ and $m_2$ respectively, the equations of motion are:
$$m_1\ddot{\mathbf{x}}_1=\mathbf{F}_{21}\\
m_2\ddot{\mathbf{x}}_2=\mathbf{F}_{12}$$
where $\mathbf{F}_{21}$ is the gravitational force of $m_2$ on $m_1$ and $\mathbf{F}_{12}$ is the gravitational force of $m_1$ on $m_2$. Because of Newton's third law, $\mathbf{F}_{21}=-\mathbf{F}_{12}$.
This is a somewhat complicated system of non-linear differential equations. However if you think about the symmetries of the system it should be clear that the motion should not depend on the absolute position of the masses in space but only on their relative position. Mathematically this simplification can be achieved by adding/subtracting the equations from one another. Adding the equations of motions gives:
$$0=m_1\ddot{\mathbf{x}}_1+m_2\ddot{\mathbf{x}}_2= (m_1+m_2)\ddot{\mathbf{R}}$$,
where $\mathbf{R}\equiv (m_1\mathbf{x}_1+m_2\mathbf{x}_2)/(m_1+m_2)$ is the center of mass coordinate. This equation expresses the conservation of momentum.
In the same way for the difference of the coordinates you get the equation of motion:
$$\ddot{\mathbf{r}}\equiv \ddot{\mathbf{x}}_2 - \ddot{\mathbf{x}}_1 = \left(\frac{1}{m_1}+\frac{1}{m_2}\right)\mathbf{F}_{12}$$ which can be rewritten with the reduced mass, $\mu=m_1m_2/(m_1+m_2)$ as
$$\mu\ddot{\mathbf{r}}=\mathbf{F}_{12}=-\frac{Gm_1m_2}{r^2}\hat{\mathbf{r}}$$
where $r=|\mathbf{r}|$ is the distance between the masses and $\hat{\mathbf{r}}=\mathbf{r}/r$ is the unit vector in the direction of $\mathbf{r}$. This equation is the starting point for the discussion/calculation of the motion.
Outline of solution
Finding a general analytical solution for $\mathbf{r}(t)$ is difficult (impossibe or at least ugly requiring special functions?). Still you can get many important results without this solution.
Using symmetry or angular momentum conservation you can argue that the motion will be in a plane (set by the initial velocity vectors), so that the problem simplifies to a two dimensional problem.
Since the force is radial it makes sense to solve the problem in polar coordinate:
$$\mathbf{r}=r\begin{pmatrix}\cos\theta\\ \sin\theta\end{pmatrix}$$
Even then solving the problem is hard work. The classical method is to substitute $u=1/r$ and to look for solutions for $u(\theta)$. You might want to look up Binet equation.
| {
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Why doesn't saturation current in the photoelectric effect depend on the frequency of light absorbed by the metal emitter? If current $I$ is given by $I = nAev$, where $n$ is the number of electrons per unit volume, $A$ is the area, $e$ is the charge of an electron and $v$ is the velocity of the electron, it must mean that the current increases with increase in velocity of the electron which increases with the frequency of light incident on the metal emitter. Why doesn't then saturation current increase with increase in frequency?
| In photoelectric effect the electrons will eject when sufficient frequency of light is incident. The saturation current will depends on the no.of photoelectrons ejected per second.As the frequency increases the energy in the quantum packets increases but the no. Of photons that should incident on the cathode plate remain same. Maybe the kinetic energy of the electron is more in this case, but overall the no. of electrons that should hit the anode remain same.It only vary with the intensity of the light.So on increasing the frequency probability of hitting the anode will increase(as kinetic energy increases) but that effect can be negligible.
So overall with this discussion we can conclude that saturation current only depends on intensity but not on frequency.
Hope it helps
| {
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Normal order of 1 Say in my Hamiltonian I had a term
$$Q[a_j,a^\dagger_j]_\pm$$
where $Q$ is a constant. Suppose I didn't realise that this quantity equals 1 and calculated a normal order. Of course you get 0
However, the normal order of 1 should be 1 or else any operator could be multiplied by 1s and get 0 at the end.
What is a correct way of normal ordering any operators?
| Well, it depends on context. E.g.:
*
*Assume we are given a classical model that we want to quantize, i.e. to construct a corresponding quantum theory. Be aware that quantization is not unique. The classical theory doesn't know about operator ordering so this introduce ambiguity. To parametrize our ignorance we should allow the possibility of an arbitrary constant term. Often the constant can later be fixed by other consistency requirements, cf. e.g. my Phys.SE answers here and here.
*If we are just doing operator manipulations in some consistent operator formulation of a quantum theory, then there are no ambiguities. The commutation relations dictate the outcome of any operator rearrangement.
| {
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Measuring Vibration I want to measure the vibration in an environment so that I can isolate some equipment. What is the standard way of measuring this? Basically, I want to mitigate the effects of a nearby train. More important than the loud horn, I need to isolate the equipment from vibrations in the ground. I am looking at some products from newport, but they show resonance around 6Hz. Would vibrations produced by the train be incompatible with these? How can I measure such vibrations? I have some audio microphones and software that shows a spectral distribution, but I don't want to know about the vibrations transmitted through the air as much as I want to understand how much noise is traveling through the concrete floor. I tagged this experimental physics, because it is likely that an experimentalist would have a good answer.
| There is no need to measure the vibration. The unwanted vibrations must be having some undesirable but detectable effect on the output of your experiment. The effect you are getting compared with what you expect from ideal conditions is a measure of the noise. The improvement in your output signal will tell you what effect anti-vibration materials or techniques is making.
Ground vibrations from trains peak between 4 and 50 Hz. Your anti-vibration strategy should focus on eliminating these frequencies. If your output signal is far from this range, it is unlikely to be affected by railway noise.
A useful exercise is to look at the frequency spectrum of your output signal, using a spectrum analyzer. Some digital oscilloscopes perform this function. The sidebands on the strongest peaks in the signal spectrum will suggest what the frequencies of the major noise sources are. Narrowing of the peaks in the frequency spectrum also shows improvement in your signal.
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How is entropy a state function? Is there only one reversible way to move from one state to another?
If we consider two states $A$ and $B$ on an isotherm and we move from $A$ to $B$ by first reversible isochoric process and then reversible isobaric process. Now the path followed should be reversible since both the processes were reversible. But what about simply following the reversible isothermal process?
According to me both processes should be reversible. Now entropy is the heat added reversibly to move from one state to another divided by the temperature at which it is added. But we know that the heat added to the system is different in both the cases. Then how is entropy a state function?
| The total heat added in both the processes is different. Change in entropy is defined as $\int(dQ/T)$. Along the isotherm, the temperature remains constant. But along the other two reversible processes you have mentioned, the temperature is not constant. Effectively, it can be seen by integration that change in entropy in both processes is the same.
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Why does rainwater form moving waves on the ground? Is there a name for this effect? A while ago it was raining and I noticed that, on sloped pavement, water was flowing in very regular consistent periodic waves, as you see below.
However, I realized I had no idea why this should be happening. There was nothing uphill actually creating these waves, and they continued down as far as the pavement went, despite the rain that was falling on them along the way.
Why wasn't the water flowing down smoothly, or irregularly?
What causes the noticeable wavelike patterns? Is there a name for this phenomenon?
| I may have an explanation. Such a behavior is not to be expected (perhaps) in an ideal situation (uniform surfaces etc.). So, if it's due to an irregularity, the following may be true.
Suppose that we have a small bump/irregularity/cavity on the road (very small, perhaps a cm in height, and a bit longer along the road). Such things can be created on the natural surface of the road due to non-uniform eroding. So when rain falls on the slope these bumps/cavities act like a small dam/reservoir respectively. However, very soon When the rainwater level rises beyond their heights, water flows, and stops again when the level goes under. Due to falling raindrops, water in these cavities/bumps splashes, causing more water than equilibrium height to move out. This creates periodic flow of water down the slope.
Basically, we need a mechanism to disrupt the (expected) continuous flow. Any other ideas?
Any questions may be asked.
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Electric field lines: and nature in presence of a conductor Long cylindrical shell carries positive surface charge σσ in the upper half and negative surface charge −σ−σ in the lower half. The electric field lines around the cylinder will look like figure given in: (figure are schematic and not drawn to scale)
Note: not a homework question another exam question.
(there were 4 options and I am confused between these two )
So my approach was:
1) the field lines go from positive to negative charge
2) they must be tangential to surface of cylindrical shell
3) their shape- this is where I am confused.
in what form must they are?
and why is the second figure wrong?
a) due to that sudden change in direction? if yes then how is this related to the nature of electric field lines
Also, can someone cite some study material where I can find other problems related to this type and particularly interaction of field lines with metals
Like: when charge kept inside a body with a non-uniform cavity.
| The problem is the discontinuity in your field lines, as you suggest. Unless the electric field goes across a charged surface it should be continuous and its derivative should also be continuous. Basically the field lines must be smooth unless there is charge near that field line.
I'm not sure how much calculus you know but this webpage has a nice derivation that continuity of the field components parallel to an interface (as your drawing suggest) must be $0$ in the static case (with no change in magnetic fields). (see also here)
The digest is that the condition you are looking for is encapsulated in $\oint \vec E\cdot d\vec \ell=0$ in the static case. In your specific case one takes the contour to be a frame with sides parallel to the midplane.
| {
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Can a Lagrangian be such that all possible paths have the same action? Q: Can a Lagrangian be such that all possible paths have the same action?
I was thinking if a Lagrangian of the motion of a particle could be represented as the total time derivative of some arbitrary function. In that case the action $S=\int^A_B L \, \mathrm{dt}$ will be a constant since $$S=\int^A_B \frac{df}{dt} \mathrm{dt}=f(A)-f(B)$$ and it will be independent of the path. It will depend only on the initial and final positions of the particle. Is such kind of a Lagrangian possible, either mathematically or physically?
| Well, mathematically, you've worked it out.
Typically, theories like these are called Topological field theories, since the action depends only on the dyanmics at the boundary. It is a field unto itself, but note that a lot of the typical reasoning you see in, say, Goldstein doesn't really work, because the equations of motion will come out to $0=0$, because all paths from $A$ to $B$ minimize the action.
| {
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Photoelectric effect:- Reduction of wavelength increases current? I did a question in which, the intensity of the incident radiation on a metal surface was kept constant but the wavelength of the photons has been reduced. The question inquired what will be the effect on the maximum photoelectric current? The initial wavelength was smaller than threshold wavelength of the metal surface.
My thinking was since the intensity remains constant, thus the number of photons emitted from the source remains constant and thus the number of electrons emitted from the metal surface. And since number of electrons per unit time isn't changed, the current will remain the same.
However, the answer key stated "Fewer photons (per unit time) so (maximum) current is smaller"
How does decreasing wavelength (equivalent to increasing the energy of photons) result in a fewer photon emission?
| It is the interface between classical electrodynamics and quantum mechanics.
Intensity is a classical electromagnetic wave measure of energy, measured by the average electric field in the wave :
the average intensity for a plane wave can be written
So for a given value of intensity/energy for a classical wave of frequency $\nu$, there are $N$ number of photons with energy $E=h\nu$. If the intensity is constant and the frequency gets larger , fewer photons are needed to add up to the classical intensity .
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How do I get the position at time $t$, if acceleration depends on position? In my visit to more realistic particle motion animation, $F=\frac{kq_1q_2}{d^2}$, $F=ma$, so: $$a=\frac{kq_1q_2}{md^2}$$ My velocity, integrating the above (from a site because I forgot how to do it), is $v=-kq_1q_2md+v_0$. Curiously, the integral of that, which should give displacement (!), is $d=d_0+(v_0d)-(kq_1q_2/m)(ln|d|)$. Does that work? If so, what do I do to get $v$, $d$ or $a$ at a certain time t? (I thought of getting $W=F\cdot d$ and trying to connect work to time thru another equation, maybe power?, etc., but now $F$ varies and also depends on $d$ (duh...) and I kinda stopped knowing what to do. Most (practically all?) stuff I saw deals with either varying acceleration in terms of time, or circular motion. Thanks in advance :)
ADDENDUM: This looks suspiciously close to gravity related questions, but in my case I have ions or charged particles in mind, where both masses matter, as well as the movements of both bodies. What brought up the issue was the question "were ions hard balls, when they 'collide' is the ricochet a fact, or is the attraction so great that it is already generating a -v higher than v of the conservation of momentum ricochet resulting from the collision? If there is a 'bounce', What is it like? etc.". (My initial model has only an Li+ and a F- 6000 picometers apart in a 1 cubic meter 'universe', in case you wonder what I am doing.)
| $$a = \frac{dv}{dt}$$
$$a = \frac{dv}{dt} \frac{dx}{dx}$$
$$a = \frac{dv}{dx} \frac{dx}{dt}$$
$$a = v\frac{dv}{dx}$$
$$\int_{x_0}^x a.dx = \int_{v_0}^{v} v. dv$$
Here, $x$ is not the final position; it is a variable. Note that we are solving the differential equation for a general case.
$$2\int_{x_0}^x a .dx = v^2 - v^2_0$$
$$v^2 = v^2_0 + 2\int_{x_0}^x a .dx$$
$$v = \sqrt{v^2_0 + 2\int_{x_0}^x a .dx}$$
$$\frac{dx}{dt} = \sqrt{v^2_0 + 2\int_{x_0}^x a .dx}$$
$$\int_{0}^t dt = \int_{x_i}^{x_f} \frac{1}{\sqrt{v^2_0 + 2(\int_{x_0}^x a .dx)}} dx$$
Here, $x_f$ is the final position. If you need a general equation, you can replace $x_f$ with $x$ but be careful not to confuse the upper limit of the outside integral with the upper limit of the inner integral. They are two different things and the upper limit of the inner integral must be a variable for the sake of the outside integral.
| {
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Why does a plastic comb attracts tiny bits of paper if it is an insulator? When we rub plastic comb against dry hair, it attracts tiny bits of paper. We know that plastic comb is an insulator of electricity but then how is it getting charged and showing the property of attracting?
| because of electrostatic forces acting on it.
when we rub the plastic comb it will either be charged negatively or positively.
it will induce negative charges on the bits of paper if the comb is positively charged so, the bits of paper will be attracted by the plastic comb which is an insulator
| {
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Real reason behind why current splits in a parallel circuit Is it true that the speed of electrons in the branches is smaller than in the main wire as the current splits in a parallel circuit? If its true, how does the speed reduce (the mechanism, not because of the laws alone)? If not, then what really causes the smaller current in the branches (and the mechanism too).
| The answer depends on the wires in the branches and the main wire are made out of and what shapes they have. What you can know for sure is that the sum of currents in the branches is the current in the main wire.
Now the current in the wire is the electron charge density times the average speed of the electrons times the cross sectional area of the wire. Now if the wires are made of the same material throughout, the density of electrons is the same, but the current still depends on the cross sectional area and the electron speed. If each branch has half the the area of the main wire, then the electron speed is the same. However, if the branch wires are each as large as the main wire, then the elections in the branch move slower. As a final example, if the branch wires are very thin, the electrons must move very fast to give the same current.
| {
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Covariant formulation of electrodynamics IMO 'covariant formulation' of electrodynamics means that the equations should remain invariant across different Lorentz frames. Now there are broadly two ways to write electrodynamics equations.
*
*using 3 vector notation
$$
\frac{\partial \rho}{\partial t} + \nabla.\mathbf{J} = 0 \tag{Eq. 11.127 of J.D.Jackson}
$$
*using 4 vector notation and tensors
$$
\partial_\alpha J^\alpha = 0 \tag{Eq. 11.129 of J.D.Jackson}
$$
An electrodynamics equation written in either of these two ways remains invariant (form-wise) across all Lorentz frames. Then why is it that the latter is referred to as the 'covariant formulation' but not the former?
| It's true that the first equation has the same form in all Lorentz frames, but it's not obvious unless you know how $\rho$ and ${\bf J}$ individually transform. For example, the similar-looking equation
$$ \frac{\partial \rho}{\partial t} + k\, {\bf \nabla} \cdot {{\bf E}} = 0$$
(where $k$ is some constant) would not look the same in other Lorentz frames, because charge density and electric field transform differently under Lorentz boosts. The second equation, on the other hand, is clearly Lorentz-covariant, because any contraction between a covariant and contravariant index of a Lorentz tensor is automatically covariant (i.e. that contraction has the same algebraic form in any other Lorentz frame).
| {
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Why does the Earth accelerate upward, according to Einstein? I recently watched a video on the YouTube channel PBS Space Time which was called "Is Gravity an Illusion?". In this video, the host explains that Einstein claimed that it is not the apple that accelerates towards Earth but the other way around. In Newton's theory, the apple accelerates down because the Earth's gravity is pulling it down, but in GR, the Earth accelerates up because of––what?
I am familiar with geodesics and spacetime warping in GR, but am new to this concept of "Earth accelerating upward" and do not understand it very well.
| Well, the thing is, if we want to analyse the motion an object, we need an inertial frame of reference. By equivalence principle, free-falling objects are inertial frame of reference.
So we have to analyse it from that point of view. But one of them should move, right? To get the visual. So, it's supposed that the earth moves upward with acceleration $g$.
But, the theory of relativity which predicts that gravity is just a pseudoforce, also tells us that there is no absolute space.
It's just Newtonian Classical Mechanics breaks down and much better GR comes up.
Still, you can ask, then why don't we move up, if there is only normal force to push us up and no gravity to equate it. But your spatial coordinates may not change even if you are accelerating. This is also the result of GR.
You can watch a Veritasium video on the same topic for more understanding.
| {
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Lagrangian of the Euler equations Do the Euler equations (where $I_1,I_2,I_3$ are principal moments of inertia):
$$I_1\dot{\omega}_1+(I_3-I_2)\omega_2\omega_3=M_1$$
$$I_2\dot{\omega}_2+(I_1-I_3)\omega_3\omega_1=M_2$$
$$I_3\dot{\omega}_3+(I_2-I_1)\omega_1\omega_2=M_3$$
in their above general form have a Lagrangian? If not, does a specific case of $\omega_1=\omega_2=0$ (and so $M_1=M_2=0$) have a general Lagrangian? ($M_3$ is the torque coming from a central gravitational potential - a planet - keeping the body (a satellite) on an elliptic orbit.)
| They do. In the case of the symmetric top with $I_1=I_2\ne I_3$ for instance, the kinetic energy is given by (if my algebra is right)
\begin{align}
T_{\hbox{rot}}&=\textstyle\frac{1}{2}\sum_k I_k\omega_k^2\, ,\\
&=\textstyle\frac{1}{2}I_1\left((\dot{\theta}\cos\psi+\dot{\psi}\sin\theta\sin\psi)^2+
(\dot{\phi}\cos\psi\sin\theta -\dot{\theta}\sin\psi)^2\right)
+\textstyle\frac{1}{2}I_3(\dot{\psi}+\dot{\phi}\cos(\theta))^2\, ,\\
&=\textstyle\frac{1}{2}I_1\left(\dot{\theta}^2+\sin^2(\theta)\dot{\phi}^2\right)
+\textstyle\frac{1}{2}I_3(\dot{\psi}+\dot{\phi}\cos(\theta))^2\, .
\end{align}
where the expressions
\begin{align}
\omega_1&= \dot{\theta}\cos\psi+\dot{\psi}\sin\theta\sin\psi \\
\omega_2&= \dot{\phi}\cos\psi\sin\theta -\dot{\theta}\sin\psi \\
\omega_3&= \dot{\psi}+\dot{\phi}\cos(\theta)\, ,
\end{align}
transition from the the angular frequencies to the time-derivatives of Euler angles.
| {
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Why do power lines use high voltage? I have just read that using high voltage results in low current, which limits the energy losses caused by the resistance of the wires.
What I don't understand is why it works this way. Does it have anythnig to do with electromagnetic induction in the wire which resists the current?
| If the total resistance of the transmission line leading from a power station to you is $R$ and the city/town you're in demands an average amount of power $P$. Then $P=I\times V$ . This makes the current drawn by the city/town is $I=\frac PV$ and so the higher the transmission line voltage, the smaller the current. The line loss is given by $P_{loss}=I^2R$, or, substituting for $I$, $P_{loss} = \frac {P^2R}{V^2}$
Since $P$ is fixed by demand, and $R$ is as small as you can make it(with the big cable), line loss decreases strongly with increasing voltage($V$)(in the denominator). So smallest amount of current that you can use to deliver the power leads to the least amount of power loss. It may help to think of it as current causing 'friction/heat' which is lost on transmission.
| {
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How is the curl of the electric field of a dipole zero? For a static charge, the curl of the electric field is zero. But in the case of a static dipole the electric lines of force curl. How it that possible?
| The electric field of a dipole has zero curl; this is easy to verify because it is (the $d\to0$ limit of) a superposition of two monopole Coulomb fields with zero curl. If you want something more explicit, then simply start with the explicit electric field,
$$
\mathbf E=\frac{1}{4\pi\varepsilon_0} \frac{3(\mathbf p\cdot\mathbf r)\mathbf r-r^2\,\mathbf p}{r^5}
$$
and calculate the curl $\nabla\times \mathbf E$; you will find that it's zero.
You do provide an interesting observation, though, in that
in a dipole the electric lines of force form a closed path from a positive charge to negative charge,
and this is indeed true: if you start your curve just above a point dipole, and loop around to just below it, then that finite segment will accumulate a nonzero line integral. However, to have a closed loop, you will need to cross directly across the dipole itself, and this will introduce a singularity into the circulation integral. This essentially breaks the game and none of the calculus applies any more. (Similarly, you can't cheat and go around the dipole, either, because the field will be very strong and point against the line element, so the circulation integral will be exactly zero.
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How do we apply Ampère's law for non-planar loops? How do we apply Ampère's (magnetism) law for non-planar loops?
Its most general form(or you can say the only one I know) is
$$
\oint_C \mathbf B\cdot\mathrm d\mathbf l = \mu_0 \iint_S\mathbf J\cdot \mathrm d\mathbf S
$$
But what would current enclosed mean in case of non planar loops. I mean infinite amount of curves can contain such loop. As a result while the right-hand side (line integral of B field) would same in each but the integral of current density would be different for each curve (surface or manifold).
| This is relatively standard materials, so for the details you can consult your favourite EM textbook, but I'll sketch the overview.
The problem with Ampère's law, for any kind of loop (including planar loops!) is that there are plenty of surfaces $S$ that share the same boundary $C=\partial S$, which makes the statement
$$
\oint_C \mathbf B\cdot\mathrm d\mathbf l = \mu_0 \iint_S\mathbf J\cdot \mathrm d\mathbf S
$$
a bit suspect, unless we can (a) choose a canonical surface $S$ for each curve $C$, or (b) show that the surface integral on the right-hand side is actually independent of what surface we choose.
The resolution to this is, in fact, (b): the current flow really is independent of the surface you choose. To prove this, consider two surfaces $S_1$ and $S_2$ which share the same boundary $C$, so that we want to prove that
$$
\iint_{S_1}\mathbf J\cdot \mathrm d\mathbf S
=
\iint_{S_2}\mathbf J\cdot \mathrm d\mathbf S,
$$
or, equivalently, that
$$
{\large\bigcirc}\kern-1.55em\iint_{S}\mathbf J\cdot \mathrm d\mathbf S
=
\iint_{S_1}\mathbf J\cdot \mathrm d\mathbf S
-
\iint_{S_2}\mathbf J\cdot \mathrm d\mathbf S
=0,
$$
where $S$ is the closed surface that surrounds the space between $S_1$ and $S_2$.
Now, there's a bunch of ways to prove that that integral is indeed zero, but they all boil down to this: the closed surface integral $\mathop{\vcenter{ \unicode{x222F}}}_{S}\mathbf J\cdot \mathrm d\mathbf S$ represents the net amount of charge that enters the volume between $S_1$ and $S_2$ per unit time, and for a static situation, that net amount needs to be exactly zero, or you would have a linear growth of the enclosed charge in that volume, quickly taking you out of the static situation you thought you were in.
Of course, this does mean that Ampère's law as formulated here can no longer hold without modifications in dynamic situations - and, indeed, in that case you need to extend it to the Ampère-Maxwell law, which includes an additional term in the surface integral, and which again has the property that it holds regardless of what surface $S$ you choose to integrate over.
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Why is the Voltage across a Inductor not equal to 0 when the switch is closed?
I have such a diagram and at time t = 0 the switch closes, need to calculate the voltage across an inductor.I know that the Current across the inductor is 0 But the voltage across the inductor is not 0 but -120. But that does not follow ohms law V = IR, So I'm trying to figure out why this is
| An inductor is not an ohmic device that is it does not follow $V= IR$.
It instead follows equation $V = L\frac{dI}{dt}$,
Since
Just when switch is closed $I$ is instantaneously $zero$ , it has started increasing this change in current leads to potential difference across inductor
Hope it helps
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What makes the number of neutrons the number of proton similar? In basic chemistry, we are taught that an atom has roughly the same number of neutrons and number of protons, this doesn't seems to hold for larger atoms, but it is always roughly proportional (i.e. you seldom find an atom with 100 protons but 1 neutron, that just does not happen). Why is that?
| You probably know that the electrons in atoms occupy a series of energy levels, the $1s$, $2s$, $2p$, etc orbitals. Although the structure of nuclei is complicated, basically the same idea applies to nuclei as well as atoms.
This happens because you can't put more than one fermion into the same quantum state. The electrons in atoms are fermions, and so are the protons and neutrons in nuclei. However because the proton and neutron are different particles you can put a proton and neutron into the same energy state. Well, that's not really true but the point is that the energy states of the protons and neutrons overlap. Let me try to show this with a diagram:
Suppose we have eight nucleons and we arrange them, one per energy level, as shown in the diagram (a). This has some total energy $E_a$. Suppose now that we can pair up the nucleons as shown in (b). This obviously has a lower energy $E_b$. But remember that protons and neutrons can share energy levels because they are different particles. So if we have a nucleus with all protons or all neutrons it would look like (a), while an equal number of protons and neutrons would look like (b) and therefore have a lower energy.
But protons and neutrons can interconvert by beta or positron emission or electron capture. So if we started with eight protons as in (a) four of those protons can convert to neutrons to give the energy levels in (b) and this will overall reduce the energy. The nucleus with equal numbers of protons and neutrons will be the one with the lowest energy.
Now this is very oversimplified because protons and neutrons don't share exactly the same energy levels and the energy levels aren't equally spaced. But it gives you a feel for why we have approximately equal numbers of protons and neutrons in a nucleus.
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Why during annihilation of an electron and positron 2 gamma rays are produced instead of 1? $1 \gamma \rightarrow 1 e^- + 1 e^+$ (pair production)
Then why
$1 e^- + 1 e^+ \rightarrow 2 \gamma$ (annihilation of matter)
instead of
$1 e^- + 1 e^+ -> 1 \gamma$ ?
| The gamma photon should move with $c$, in any reference frame. Thus, it has to have a nonzero inertia.
But in the center of mass of the $e^-$ and the $e^+$, they have zero summed inertia.
Inertia is conserved, thus the resulting gamma photon should have zero inertia. It is impossible.
Ext:
If their spins are into the same direction, even the 2-photon annihillation is not possible on a different conservation law, the spin conservation. In this case, 3 photons (or more) are produced. The electrons (and positrons) have 1/2 spin, thus they can annihillate in +1/2, -1/2 or in +1/2, +1/2 configuration. Thus, their summed spin is either 0 or 1. The spin of the photons is 1, also they can have opposite or equal sign, thus their total spin want to sum up to 0 or 2. Thus, in the case if the electron and the positron have equal spin signum, the spins couldn't be conserved in a 2-photon annihillation.
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A rod is moving in space and an insect is on it. How many degrees of freedom does the insect have? Is the answer 7? The number of degrees of freedom of a system can be viewed as the minimum number of coordinates required to specify a configuration. Applying this definition, we have:
*
*For a single particle in a plane two coordinates define its location so it has two degrees of freedom.
*A single particle in space requires three coordinates so it has three degrees of freedom.
*Two particles in space have a combined six degrees of freedom.
*If two particles in space are constrained to maintain a constant distance from each other, such as in the case of a diatomic molecule, then the six coordinates must satisfy a single constraint equation defined by the distance formula. This reduces the degrees of freedom of the system to five, because the distance formula can be used to solve for the remaining coordinate once the other five are specified.
| 5 for rod and 1 for point insect... total 6
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Do particle and antiparticle annihilate when they meet? Do particle and antiparticle annihilate when they meet? As we know, an electron and a positron will annihilate when they meet. However, many quarks and antiquarks do not annihilate, but coexist as mesons. For example, a neutral pion $\pi^{0}$ is made up of $u, \overline{u}, d, \overline{d}$; a meson $\eta$ is made up of $u, \overline{u}, d, \overline{d}, s, \overline{s}$; a meson $\eta_{c}$ is made up of $c, \overline{c}$; a meson $\phi$ is made up of $s, \overline{s}$; and a meson $\Upsilon$ is made up of $b, \overline{b}$ . Why do these quark-antiquark pairs not annihilate?
| The meson you describe are made up of a quark and its antiquarks and are called quarkonia. Of course, all mesons are unstable, but they do not necessarily decay via annihilation.
When you say "annihilate", you probably think of a process
$$ q\bar q \longrightarrow \text{single intermediate particle}\longrightarrow \text{decay products}\,.$$
However, this process is forbidden fpor QCD, where the intermediate particle would be a gluon, because the meson is colour-neutral. Hence, only electroweak decays are possible (via $\gamma$ or $Z$). Depending on the state, there can be additional conserved quantities (spin, parity) that prohibit a direct annihilation of this type -- for example, the $c\bar c$ state $\eta_c$ has $J=0$, so it cannot decay into an intermediate vector boson. And finally, even when annihilation is possible, it might not be the dominant decay mode, as direct electroweak annihilation competes with QCD loop processes or weak decays of heavy quarks.
(Note that generally, meson properties are quite hard to compute because of the nonperturbative nature of QCD at low energies.)
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Is the diffraction pattern of a vertical slit horizontal? I am familiar with the mathematical aspects of single slit diffraction pattern, at the undergraduate level.
Consider the following pictorial representation from the book Optics, by Hecht:
The fact that I find puzzling here is - even though the slit is shown vertical, the pattern on the screen is shown horizontal. Is this correct? Why so?
My logic:-
The reason why I find this strange is because of a translational symmetry argument. Any two points vertically separated by some distance have the same horizontal attributes. So, one expects the pattern also to have this sort of vertical symmetry, irrespective of what happens along the horizontal axis.
Am I mistaken? If yes, can someone please point out why is the vertical slit producing a horizontal pattern here?
| The wider a slit, the narrower the diffraction pattern. So it makes sense that a tall rectangular slit makes a wide rectangular pattern.
| {
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Trolley problem
A trolley of mass $300kg$ carrying a sandbag of $25kg$ is moving uniformly with a speed of $27km/h$ on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of $0.05kg/s$. What is the speed of the trolley after the entire bag is empty?
I do not seek the exact answer to the problem. I know a force will act on the trolley because the mass is changing (and force is proportional to the rate of change of momentum) but where it acts is what confuses me. Where would the force act?
| First Rule of Conservation of Momentum problems and Momentum Change problems: Define what the system is! Second Rule: Don't change the definition in the middle of solving the problem.
In this case, one could define the system as the trolley plus sand. So it's simple to calculate the total momentum of this system.
In addition, the sand will presumably stop moving along with the trolley as it dribbles out of the trolley, past the frictionless track and onto the ground. So the momentum of this system will change, because the ground is exerting a horizontal force on the sand part of the system, acting to slow the sand to a stop. This is the answer to the specific question; however, there is no force from the ground on the trolley part of the system.
However:
The first part of your question implies that the mass of the system is changing. This is impossible in a properly defined question. The sand lying on the track was, and remains, part of the system as defined above.
If we instead define the system as the trolley, and the sand as some another system that just happens to be travelling close to the trolley in the same direction, then the problem becomes much simpler.
There is no horizontal force acting on this new system (the trolley). The ground only exerts a force on the sand, not part of this new system. The momentum and velocity of the trolley thus remain unchanged.
| {
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Equations of Motion for a Passive Suspension System I am trying to develop the equations of motion for a passive suspension system. The system consists of one mass connected to another mass via a spring and dampener system, and then the bottom mass is connected to the road via the tire which is modeled as a spring. I'm not sure where the force applied to the system is, but I think it has something to do with the surface of the road and then the this is somehow related to the tire spring.
| You are correct : the force is supplied by the bumps in the road.
The shape of the bumps and the speed of the vehicle produce a time-varying compression $x$ of the tire. The compression force is related to the compression $x$ and spring constant $k$ of the tire by $F=kx$. This force $F$ is transmitted to the masses in the system. Because of inertia they resist changes in motion.
See the following questions for further help with developing a model :
Model of road disturbance in term of normal force
"Magic" Speed to Drive Over a Speed Bump: Myth or Reality?
| {
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What is spacetime (simple explanation)? Can someone please explain concept of spacetime in simple language? What is it and how it is important in the universe?
Wherever I have tried searching this concept, I have come across most complicated explanations.
A simple example will be appreciated.
| You can think of spacetime as a map. If you want to meet someone in this universe you have to agree an exact location and a time on the map. If you show up at correct location but at the wrong time then you are in the wrong place on the spacetime map.
| {
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A free-fall electron I am reading Wheeler and Taylor's Spacetime Physics. In Ch2, Wheeler mentioned:
"for gravity, any free-fall frame is an inertial frame." (roughly)
I am left wondering if that is true for electrical force:
Consider one charge is under a statistic electrical field. The charge is in free-fall. Is the electron's free fall frame an inertial frame?
(if yes, then can we say electrical force is a pseudo-force too?)
| An object is in an inertial frame if no forces are acting on this object. If you feel weightless than you are in an inertial frame. So jumping down from a wall you are for a moment in an inertial frame.
Such a definition of the inertial fram implicates two things:
*
*The gravitational pull isn't a force. Gravitation acts on the surrounding space and the geodesic of a moving particle in most cases isn't a straight line under the influences of masses.
*Any movement of particles without the context of the gravitational potential at this point is meaningless. Everything in every point of our world is under the influence of masses.
I am left wondering if that is true for electrical force:
Consider one positive charge and one negative charge (say, electron). They attract each other; hence, both charges are in free fall. Is the electron's free fall frame an inertial frame?
You are mixing the electric force between charged particles with their behavior under the influence of gravitation (from the mass of the earth). As long as an electric force is acting between particles or objects they are under acceleration. It does not matter will this be an attractive or repulsive force, it is still an acceleration. This particles are not in an inertial frame.
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Opposite of particle decay I have read about particle decay, a process in which one particle becomes several other particles. However, I have not been able to find much information about its opposite: several particles combining into one particle. Is such a process possible, and if so, under what conditions? For example, a free neutron may decay into a proton, electron, and electron antineutrino. Could a proton, electron, and electron antineutrino somehow be joined into a neutron?
Edit: Everyone, thank you for your help, but let me try to make what I'm looking for clearer. I want to know whether several particles can join into ONE particle, in an exact reverse of that one particle decaying into several particles. As far as I know, I don't think an atomic nucleus counts as one particle. Please correct me if I'm wrong.
| As Nick says, in laboratory conditions the crossections are very very small. In a way two body resonances are the only reversible example with high probability to study in the laboratory.
In the grand cosmological laboratory of the universe, in its history at the hadron epoch:
In physical cosmology, the hadron epoch was the period in the evolution of the early universe during which the mass of the universe was dominated by hadrons. It started approximately 10-6 seconds after the Big Bang, when the temperature of the universe had fallen sufficiently to allow the quarks from the preceding quark epoch to bind together into hadron
In this model the inverse processes exist , due to the very high average energies of the quarks and gluons the crossections for generating protons and neutrons out of three quarks is high and reversible, until the universe cools enough that quarks are no longer asymptotically free. So there is a three to one process.
| {
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How do you calculate the resulting magnetic field for multiple sources? I've been looking at some fusion reactors and I keep wondering how putting some kind of extra magnet in some configuration would affect the field, but I don't know how to figure it out. Like for example if you took a solenoid and sat a permanent magnet down next to it.
This is the only thing I could find:
https://www.quora.com/How-do-I-calculate-the-magnetic-field-created-by-a-number-of-magnets
Edit:
I'm pretty sure it's just super positioning, but I need someone else to answer, because I don't know for sure. Also does it follow from Maxwells equations?
| You are correct, they follow superposition. The magnetic field is a vector field, and so they follow a vector sum when they are in a superposition.
Maxwell's equations are linear ($\nabla \times$ and $\nabla \dot{}$ are linear operators) and it follows that solutions ($E$ and $B$) obey the superposition principle.
| {
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General relativity: How is the 4-velocity and momentum defined in GR? I know that in SR, the 4-velovity $$ u^\mu = (d t/d \tau,d x/d \tau,d y/d \tau,d z/d \tau ) $$ and $$p^\mu = m u^\mu.$$ How do these generalize to GR? I imagine there are new complications, particularly by what we let $p^0$ be. And are these only defined for geodesics, or in general?
| Are you mentioning $p^0$ because you think of it as the energy?
If that is where your question is coming from, then perhaps this answer may be of use. Energy is something that an observer measures about an object. Let me explain.
The observer has a 4-velocity, $u^\mu_{\text{obs}}$, and the object has a 4-momentum, $p^\mu_{\text{obj}}$. The energy that the observer measures for the object is the projection of the object's 4-momentum on the observer's 4-velocity:
$$E = p^\text{obj}_\mu u^\mu_\text{obs}=p^\mu_\text{obj}g_{\mu\nu}u^\nu_\text{obs}$$
Here we see the metric tensor coming into play, irregardless if we are in a flat spacetime, or in a curved one.
If we think of ourself as the observer, then ${\bf u}=(1,0,0,0)$. The energy that we measure an object to have will be $E = g_{00}p^0_\text{obj}+g_{i0}p^i_\text{obj}$. If the time-space components of the metric vanish locally, the energy is $E = g_{00}p^0_\text{obj}$. In flat spacetime $g_{00} = 1$, and we get back to the initial idea that energy is the time component of the 4-momentum.
Hope that helps.
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How did Rutherford conclude that most of the mass (as well as the positive charge) was concentrated in the nucleus? Geiger and Marsden's experiment led Rutherford to believe that the positive charge and most of the mass of the atom was concentrated in a small region. I understand what led him to conclude the way the positive charge is positioned in the atom. But how did he conclude that most of the mass was in a small region (the nucleus)?
How did the distribution of the mass matter after all? Given that the electric force is greater than the gravitational force by many magnitudes, the force between the positice charge and the electrons was predominantly electric.
So how did Rutherford conclude that most of the mass is in the nucleus?
| Wikipedia explains this rather well but I'll pick out the relevant stuff for you.
Before the Geiger–Marsden experiment, the general idea was that atoms were built of some permeable positive substrate in which some negative particles were floating around; the so called plum-pudding model.
If we shoot $\alpha$ particles on this setup they should all pass through the atoms since the positive substrate is thought to be permeable! (left side of the figure)
But when people did the experiment they saw that most particles went through while some scattered $180^\circ$ backwards, some even bent a small angle! (right side of the figure)
The plum-pudding model had no problem with the particles that went through undisturbed but what about the ones that were backscattered? People theorized that there must be some solid core in the atom against which the $\alpha$ particles could scatter. The core couldn't be too big since only a small fraction of the $\alpha$ particles backscattered.
This leads to a model in which most of the mass (to which $\alpha$ particles can scatter) are in the center of the atom with the negative charges around it!
| {
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What causes the dark bands in water wave ripples? I was filling the bathtub the other day with water and noticed something. When calm, the light illuminates everything and I can see the sides and bottom of the white tub without shadows. Then, when I add a droplet of water its ripples (either its peaks or valleys of the wave) cast concentric rings of shadow which travel about 2 feet per 3 seconds (0.2 m/s) outward from epicenter.
I'm wondering what exactly creates these shadows. That distance is too large for light waves destructively interfering.
| What you see in a lensing effect by the ripples on the surface of the water. There will be regions on the surface where the water surface (like a crest) acts as a converging lens and regions where the water surface acts like a diverging lens (troughs).
Thus the illumination of your screen will vary.
You have made yourself a ripple tank.
Here is a ray diagram to show the general idea of what is happening.
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What is the relation between image velocity, object velocity and mirror velocity? Suppositions used:
Velocity of image = VI
Velocity of object = Vo
Velocity of mirror = VM
I Know the fact that VI=-Vo supposing mirror at rest
and VI=2VM supposing object at rest
Now considering both mirror and object in motion, VI=2VM - Vo I ended up with this equation but my reference book suggests VI=2VM + Vo I am stuck on this for last 4 hours. I searched over internet and found the same expression like that of mine in a youtube video, I did not find much reference on this topic though. Tried many ways but all ended up on this simple argument, which equation to follow?
Help
| yeah the proper equation is 2Vm = Vi + Vo. now you need to choose an axis and put the proper sign for the two values. the equations will give you proper magnitudes and directions for the axis that you have chosen.
| {
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Have we directly observed the electric component to EM waves? For example, has anyone has directly observed charges oscillating due to standing EM waves? I am particularly interested because it'd demonstrate that radiation has a transverse electric component to it. Anything else (historical or modern) that shows that light has a transverse electric component would also be gladly invited.
| Yes, we have observed, but not for all em waves. Detailed answer follows from “Planets and electromagnetic waves”.
Light waves or rays interact with electric fields of electrons in a solar cell to produce a disturbance in electrons so that electricity is produced. In a tungsten bulb, electrons try to move with very close distance because of a voltage, and at the same time the electric fields of these electrons repel them. So, light energy is released. Light energy is associated only with electric fields.
Moving electrons have magnetic fields. Radio waves interact with magnetic fields of moving electrons, and disturb the moving electrons to make variations in current. Radio waves are associated only with magnetic fields.
No one practically observed combined form of magnetic field type and electric field type of waves with common wavelength. In Hertz’s experiment, it happens that both light waves and radio waves are released but not with common wavelength. Some researchers do not consider microwaves as electromagnetic waves, because their velocity in vacuum is less than the velocity of light in vacuum.
| {
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Why is the South Pole Telescope located exactly at the South Pole? I read that there is less atmospheric interference for the telescope at the South Pole because the atmosphere is thin and there is less water vapor in the air. However this seems to be true for many locations on Antarctica? Are there any other reasons that this telescope is located at exactly the South Pole?
| If the telescope was situated directly on the southern axis of the earth's rotation, the telescope's declination axis would be at zenith. The base for the axis would be level to the ground. In theory you could compensate for the earth's rotation with only one motion of the telescope. Also, its the only place on earth where the entire southern celestial hemisphere is visible. Now, are these the reasons it is built there, probably not but they would be advantages.
| {
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Why do we equate an indefinite integral to a specific value? Suppose we want to obtain a displacement vector defined as $\mathbf s(t) = x(t)\mathbf i + y(t)\mathbf j + z(t)\mathbf k$ from the components of a velocity vector $\mathbf v(t) = \dot x(t)\mathbf i + \dot y(t)\mathbf j + \dot z(t)\mathbf k=\mathbf 0$. According to my notes, this can be done by equating each scalar component of the displacement vector to the indefinite integral of the corresponding scalars of the velocity vector, i.e.
$$
\mathbf s(t)=\begin{pmatrix}
x(t)=\int \dot x\ dt \\
y(t)=\int \dot y\ dt \\
z(t)=\int \dot z\ dt
\end{pmatrix}
$$
But, as $\int f(x)\ dx=\{F(x): \frac {dF}{dx}=f(x)\}$, this should be syntactically wrong, because we're implying that a number is equal to an infinite set of numbers, or am I missing something?
Moreover, this also leads to a weird equation when solving the integral; for example, by taking into consideration the $x$-component of $\mathbf s$, we would have that
$$
x(t)=\int \dot x\ dt=c_1
$$
Which is correct, but it would also mean that the $x$-component of the velocity could be equal to any value belonging to $\mathbb R$. Because of that, we substitute $c_1$ with the initial condition and we equate it to zero, giving it a specific value: $c_1=0$. But, to me, this sounds like a break of the definition of indefinite integrals, as
$\int \dot x\ dt=c_1=0$ would basically mean that an indefinite integral is one, specific function.
I know this may be a very stupid question, and maybe it has to do with the same shortcuts that make us not specify "$\forall c \in \mathbb R$" when adding the constant $c$ in the solutions of an indefinite integral, but this doubt is really challenging me and I still don't understand whether I'm missing some point or it should actually be written $x(t)=c_1=0 \in \int \dot x\ dt$. Thanks a lot in advance!
| You are being confused by the shortcut that people took who wrote that expression. They mean for the integral to be taken between definite limits.
It would be more proper to say
$$x(t) = x(0) + \int_0^t \dot{x} dt$$
But that gets longwinded. Most people, when seeing the expression as you gave it, will understand it to mean what I wrote. But technically, it's not the same.
| {
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Can we add attraction forces? From Newton's Law of gravitation we know that:
$$F=G\frac{m_1m_2}{d^2}$$
For simplicity, let's say that both $m$ are $1\;\mathrm{kg}$ and that the distance apart was $1\;\mathrm{m}$. Yielding $G$ as the attraction force in Newtons. Hence $F= 6.67\times 10^{-11}\;\mathrm{N}$.
Now what if you had two $0.5\;\mathrm{kg}$ glued to each other and there was another two $0.5\;\mathrm{kg}$ glued together $1\;\mathrm{m}$ apart. If we were to find the gravitational attraction force of two $0.5\;\mathrm{kg}$ masses which are $1\;\mathrm{m}$ apart and multiply it by two since we have two of them that would yield a different answer than treating the two $0.5\;\mathrm{kg}$ that are glued together as one entity. Why does it yield a different answer?
I observed this same phenomenon with Coulomb's law. (Serway & Faughn)
Suppose that $1.00\;\mathrm{g}$ of hydrogen is separated into electrons and protrons. Suppose also that the protons are placed at the Earth's north pole and the electrons are placed at the south pole. What is the resulting compression force on the Earth? If I was to find the attraction force of one proton and electron than multiply it by Avogadro constant that would yield a different answer than saying that the charge of each particle is the elemental charge times Avogadro 's constant.
So which is the proper way to do it?
| If we consider the two masses of $1(kg)$ as point particles, then the force between them when they are $1(m)$ separated is $G(N)$.
The force between two point particles of $0,5(kg)$, who are also separated $1(m)$, is $0,25G(N)$. Now if we glue a point particle with mass $0,5(kg)$ to each of these two particles, the masses of the two $0,5(kg)$ particles become $1(kg)$ again, and thus the force between them is $G(N)$ again, so there is no difference.
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Principle of launching satellites into orbit I was reading in a book about the basic concept of launching satellites into orbit. I came across a paragraph saying,
If we take a body high above the surface of the earth, and then project it with some horizontal velocity, it will describe a parabolic path and fall back onto the Earth's surface. If the magnitude of the velocity is increased, the horizontal range also increases. We keep on increasing the horizontal velocity of projection, till at certain point, a stage will come when the body will not hit the earth, but will always be in a state of free fall under gravity in an attempt to fall to the earth.
I did not exactly understand the last statement .What do they mean by always be in a state of free fall ?
This may seem basic but I'm having trouble visualising it. If someone would explain what is the meaning of the last statement, that would help me a lot.
EDIT:
After reading the answers, I have one last doubt. How would we analyse the motion of the satellite then ? Would it be rectilinear motion, like any other free falling body, or circular motion around the earth, as we generally think of a satellite, or both( I don't understand how that would work). Or does it completely depend on the frame of reference?
Thanks for your time.
| The paragraph is correct but a bit ambiguous. Imagine a wall of height 'h' amidst a fountain or lake of certain radius. If now you increase the horizontal velocity such that it crosses the fountain, then your job is done. Replace the fountain or lake with earth and cliff with that particular height where they project the satellite from. Nevertheless the satellite will always be in a state of free fall.
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What is the difference between $\psi$ and $|\psi\rangle$? My understanding is that $\psi(\vec{r}, t)$ and $|\psi(\vec r,t)\rangle$ are the same thing yet one expressed as a wave function and the other expressed as a vector in the Hilbert space. Is this true? Or is there a deeper difference between the two notations?
| $\psi (\vec{r},t) $ is like you said, just a way to express the vector $|\psi (t)\rangle $ in 'position space', mathematically expressed like is written in the comments:
$$ \psi (\vec{r},t) = \langle \vec{r} | \psi(t) \rangle = \int \delta(r'-r)\psi(t) d^3 r $$
| {
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Fluid Dynamics in a Syringe I am currently working on a project where we use syringes to extrude some viscous materials. I will explain what I am trying to do and I just want to know what size of tubing would be better for me. I have a 10 cc syringe connected to 3/32" tubing with luer lock and at the very end, it is attached to either a 27 gauge or 30 gauge needle. The tubing is 24 inches. When I do this, it takes a lot of pressure to push the stuff out but when the stuff comes out, its not a continuous flow. What happens is it takes a lot of force to push but at some point, i will have pushed enough with my hand to make it come out but it comes out very fast and not in a continuous flow. If I make the tubing diameter larger, would the pressure inside be smaller even if I have the same 27 or 30 gauge needle at the very end.
Would I be able to have a more smooth/continuous flow? The reason I use such small needles is because I want to be able to draw a design with precision.
| please check the syringe plunger tip (the black rubber thing on the end of the plunger shaft). In contact with certain fluids, the plunger tip tends to adhere to the syringe cylinder walls and break loose suddenly, then get stuck again, break loose, etc. as you push on it. the fluid then spurts out suddenly, stops, spurts, stops and so on. you can eliminate this temporarily by lubricating the rubber tip with petroleum jelly or sewing machine oil. I say temporarily because these lubricants can soften the rubber itself, making the problem worse with time.
I recommend you try a ground-glass syringe; these have a precision-ground plunger made of glass which gets a good seal inside of the (glass) syringe body without a rubber tip. They do not exhibit the stick-slip-stick-slip behavior you describe, but are more expensive than disposable syringes.
| {
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What determines invariant mass? The only answer I have been able to find is that energy determines mass but photons have energy yet are still massless. Furthermore this then leads to the question of what determines invariant energy, which I would think to be mass. So in total what are the circumstances that determine fundamental attributes of particles such as mass and energy?
| Answer: If a particle has energy $E$ and momentum $\vec{p}$, its invariant mass is given by $m=\frac{1}{c^2} \sqrt{E^2-p^2c^2}$, where $c$ is the speed of light.
Explanation:
Momentum and energy of a particle change as the particle moves faster or slower. Equivalently, the energy and momentum that you the observer measure a particle to have change as you move faster or slower. So neither energy nor momentum are invariant, only the combination given above remains invariant.
In particular, as a particle moves faster, both its energy and its momentum increase. But the values increase in such a way that $E^2-p^2c^2$ remains unchanged, and thus the invariant mass is genuinely invariant.
Note that having an invariant mass of zero is perfectly fine. It just means that $E^2 = p^2 c^2$, or $E=|p|c$. This is the relationship between energy and momentum for massless particles like photons.
Finally, for reference, the relationship between mass, energy, and momentum is more commonly written as $E^2 = m^2c^4 + p^2 c^2$. This is called the energy momentum relationship. The case where the particle is stationary, $p=0$, gives Einstein's famous formula $E=mc^2$.
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Can I tell if I'm being attracted by a gravitational field by checking if my time is dilated relative to an observer? Due to the equivalence principle of GR, we know that gravity and acceleration are indistinguishable from each other. We also know that a gravitational field causes time dilation, but accelaration does not.
IF I were in a ship, could I know whether it is accelerating upwards/being gravitationally attracted by comparing my proper time to that of an external observer? If mine runs slower, that means that it is gravitational time dilation. What am I missing here?
|
We also know that a gravitational field causes time dilation, but accelaration does not.
In the 'elevator pulled upward in outer space with constant acceleration' thought experiment, a photon emitted from some process at the bottom of the elevator is received at the top with a longer wavelength than a photon emitted by the same process at the top.
This must be the case since, during the time of flight from bottom to top, the receiver speed, relative to the emitter at the time of emission, has increased.
Put another way, a clock at the top of the elevator observes relatively stationary clocks below to run slower with a linear dependence on the distance below.
But this is precisely the result if the elevator were at rest in a (uniform) gravitational field.
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Could a planet with a strong magnetic field exert a diamagnetic force on an orbiting moon? Here is a question from the world building stack exchange.
https://worldbuilding.stackexchange.com/questions/79003/making-a-slow-orbit-around-a-large-gas-giant
Requested: a means to have a moon of Jupiter orbit closely but slowly. Gravity precludes this.
If there were some force that opposed gravity, making the attractive force weaker, then closer slower orbits become possible. If gravity were completely opposed the object could hover.
Which is like diamagnetic levitation. The question:
*
*Is the magnetic field (magnetic gradient?) of Jupiter adequate to exert a repulsive force on a hypothetical diamagnetic satellite?
*A superconducting satellite?
*Could this force be on the order of gravity such that a slower orbit for this satellite is possible at the same distance from Jupiter?
| A nice, realistic example of this is for Ganymede at Jupiter. The Jovian magnetic field is ~0.42 mT at the planet's equator and the orbital speed of Ganymede is roughly 10 km/s. If we take a hand-wavy approach with the Lorentz force assuming a 1 coulomb charge, we get a maximum force of ~4 newtons.
For Ganymede this corresponds to an acceleration of ~10-23 m -2 (note that gravitational acceleration on Earth's surface is ~24 orders of magnitude larger than this).
Is the magnetic field (magnetic gradient?) of Jupiter adequate to exert a repulsive force on a hypothetical diamagnetic satellite?
Yes, it can exert a force but the result is negligible.
A superconducting satellite?
What type of superconductor? A Type-I superconductor expells magnetic fields.
Could this force be on the order of gravity such that a slower orbit for this satellite is possible at the same distance from Jupiter?
If we assume the magnetic field is in the positive z-direction and the orbit is circular, thus the velocity in the azimuthal direction, then at the equator the Lorentz force would be directed radially for positive charge. The problem is that the magnitude of the force depends upon the speed of the charged body with respect to the magnetic field. So if the planet moves slower, the Lorentz force gets weaker.
Side Note: In the case of Jupiter, I did not point out that the planet rotates faster than the moons orbit, so in the magnetic field rest frame the planet would actually move in the opposite direction to its orbit.
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Why are stars spherical whereas (some) galaxies are disks? I read here that galaxies become disks if there is a lot of gas in them, since their angular momentum is conserved while their energy decreases due to collisions of the gas particle. I have two questions about this:
*
*Why is the disk the configuration that has low energy and high momentum? Can't there be a spherical shape that meets the same conditions?
*If stars are formed when gas collapses under by gravity, why doesn't the gas create a disk shape, but instead creates a spherical shape?
| The stars were roughly disk shaped as the nebula that formed them was collapsing. The matter in galaxies is much more sparse. The thing that keeps them disk shaped is the centrifugal force as the stars and other bits orbit around each other and around the galaxy, and continually miss each other, otherwise, the galaxy would collapse into a big black hole. There is a black hole in the middle made up of stars that didn't miss each other.
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Which Liquids freeze when combined? Are there any liquids (or liquid and solution) that when combined react to where they will freeze?
| Good question, but you would probably receive better attention, information on the chemistry stack exchange.
I don't know the answer regarding two liquids, but if you mix the two solids, Barium Hydroxide Octahydrate and Ammonium Thiocyanate, the solids will spontaneously fuse into a liquid with an endothermic reaction about -15 deg C, which will rapidly freeze any water. Demo on YouTube is here
Take caution that ammonia gas is emanated during the reaction, so best done in a fume hood or well ventilated area.
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What is "velocity distance" in astrophysics? I was reading some papers on astrophysics, and in several of them, I've encountered velocity being used as distance. Or more precisely, distance being in dimensions of distance over time. For example, a paper referred to a group of galaxies at "roughly $2000\:\mathrm{km\:s^{-1}}$ in distance." Another used the specific phrase "velocity distance." It said there was a super-cluster "an observed concentration of galaxies at a velocity distance of ${\sim}2500$ to $4000\:\mathrm{km\:s^{-1}}$." I searched for what this could mean for quite a bit, and couldn't find it. If anyone has any idea what this means, some insight would be greatly appreciated.
| Hubble's law will convert the velocity to a proper (as opposed to co-moving) distance for you; indeed Wikipedia states one form of Hubble's law as:
"... the observation ... that ... Objects observed in deep space (extragalactic space, 10 megaparsecs (Mpc) or more) are found to have a Doppler shift interpretable as relative velocity away from Earth"
This law comes from the FLRW metric in the special case when the the standard $\Lambda{\rm CDM}$ cosmological equation of state prevails, and is also, of course, confirmed by observations first made by Edwin Hubble (but formerly theoretically predicted by Georges Lemâitre and contemporary theoreticians).
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Degrees Of Freedom of Spring-mass system
Consider 2 masses $M_1$ and $M_2$ connected with a spring of stiffness $k$, resting on a smooth frictionless surface. Now, each mass has its own 1 DOF along the $x$-axis. And the system has 1 constraint , i.e. the spring.
So, in all there should be 2(1)-1= 1 DOF for the system. But I've read that it has 2 actually. So where am I going wrong?
| Both m1 and m2 (ie, the system) is a two-dimensional case. Then, m1 and m2 each possess 2 DOF. Hence total DOF of system = 4
No: of constraints on m1= 1
No: of constraints on m2= 1
So, total constraints of the system= 2
Hence, resultant DOF of system= 4-2
=2
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Non-analyticity of the modulus function I'm studying the Ginzburg-Landau theory of superconductors, and I have read that the free energy can not depend on odd powers of the modulus of the order parameter $\psi$ because $| \psi| = \sqrt{\psi_x^2 +\psi_y^2}$ is not analytic at $\psi = \psi_x +i \psi_y = 0$. How do you show this non-analyticity?
I might have thought you'd use Cauchy-Riemann, but its a real valued function of a complex variable, whereas Cauchy-Riemann only applies to complex valued functions.
| Hint: the function
$$
f:x\mapsto\sqrt x
$$
is not (real) analytic at $x=0$ (why?).
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Why is $R\cos{a} = mg$ in circular motion compared and not $R = mg\cos{a}$?
Normally, if an object of mass $m$ is inclined to the horizontal at an angle $b$, we set the reaction force of the object on the inclined plane as $R = mg\cos{b}$ (if we resolve the force of gravity so the line of action coming out of the plane is perpendicular to it).
However in circular motion*. it's assumed that $R\cos{b} = mg$. In the example above, one would have to do this in order to arrive to the correct answer, instead of $R = mg\cos{b}$. Using $R = mg\cos{b}$ seems natural enough, as I am resolving vertically, however, both equations would produce two different values for $R$. Why is this?
To show what I mean:
If we set the reaction force in this question as $mg\cos{a}$, then the centripetal force will be $mg\cos{b}\cos(\pi/2-b) = mg\cos{b}\sin{b} = \frac{1}{2}mg\sin(2b)$
Whereas If we use $R\cos{b} = mg$, $R = mg\sec{b}$ and the centripetal force will be $mg\sec{b}\sin{b} = mg\tan{b}$. This will end up with two different values for the radius of the circular motion, and hence two different final answers.
*In the circular motion questions I've seen in my mechanics module
| @JerrySchirmer's advice is generally good, and worth heeding. If you actually construct the free-body diagrams for a particle on an inclined plane and your particle on a cone, you will note the following important difference:
*
*A particle at rest on an incline (or sliding down an incline) has an acceleration vector that is parallel to the surface.
*Your particle moving around the inside of a cone, on the other hand, does not have an acceleration vector parallel to the surface: it is accelerating horizontally towards the center of the circle instead.
In both cases, you can then use the fact that the particle is not accelerating in the "other" direction (perpendicular to the plane for the incline; vertically for the cone) to write down a relation between the particle's weight and the reaction force. But these respective equations deal with the components of those forces in different directions, and so they turn out differently from each other.
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Does an increase in entropy always result in an increase in heat, or can there be increased entropy without an increase in heat? Most situations I can think of where entropy increases also results in an increase in heat, but just wondering if that is a rule. Are there any cases where heat does not increase with entropy?
| In addition to the answer provided by @Deep, there are situations where a system can have negative temperature. In such situations, the system can release heat and increase entropy. This can happen when there is a maximum energy that some degree of freedom of the parts of the system can assume. The canonical example from textbooks is to imagine the atoms of a paramagnetic substance in a magnetic field. The stronger the field applied, the more the electrons in the substance will align with the field until it nears saturation (i.e. all of the electrons are aligned). If you were to suddenly reverse the field, the electrons would now be in a higher energy state than they can reach with heating, and will have negative temperature. As the electrons fall to lower energy levels, the material will pass through the maximally disordered state (where electrons are with/against the field with 50% probability), thus the entropy will increase as it cools until it passes through the maximum entropy state. This is why negative temperature systems can be thought, formally, as being "hotter" than infinite temperature ones.
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Tension direction for pulleys in connected mass systems I got caught up with a conceptual question dealing with a practice problem with connected mass mechanics.
Looking at the solution lecture notes is the following image for the Free Body diagrams for each mass, and the pulley, respectively:
Obviously, the Free Body diagrams on mass 1 and 2 are pretty straight-forward, but the situation with the pulley confuses me. $T_2$ going down is fine. The mass is falling, the system is overall causing $M_2$ to fall down, the string applies a force that way, sure. But $T_1$ confuses me. The mass is moving towards the pulley once we let the system act, so why is there a tension being applied that way? The directions for $T_1$ for mass 1 and the pulley are opposite for direction, and I'm not really sure why. Overall, the pulley is rotating clockwise, so this has to be the case -- $M_1$ will move towards the pulley, $M_2$ will drop. So why is $T_1$ facing that way? Intuitively speaking, that is.
| Strings cannot push unlike a coiled spring.Hence if we show the forces acting on a point due to a string the tension has to be pulling that point not pushing it.
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Can a current be induced in this coil-magnet configuration? Coil is moving around ring magnet made of two arch shaped magnets with poles opposing each other like so:
Would a current flow in coil? or be canceled out?
| Yes, it will be ac voltage and current. You see that as magnet plate rotates, the coil will be moving in to the lines of flux in the air gap. As it starts to cut true the lines, the voltage will increase to a maximum and decrease as is leaves the gap. it will do this twice per turn. Say the magnets turn at 30 times per minute, the coil will generate two sine waves per turn (60 waves per minute) so we can say the voltage frequency is 1 Hz.
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What is the energy of a single charge system? I will try to limit the question in the case of the electric fields, but is something that applies also to the magnetic ones.
There are two ways to express the energy in a capacitor:
*
*By Voltage : $U = 1/2 CV^2 $
*And by Field : $U = 1/2 \varepsilon E^2Ad$, With Energy Density: $u = 1/2 \varepsilon E^2 $
Unless i understood everything wrong and these two are NOT the same quantity, i have the following question. When we have two charges placed at points A and B, then in order to calculate the energy of the system, we will take the first charge, place it at point A WITHOUT doing any work, and then we will calculate the work needed to place the second charge at point B.
The weird thing to me here, is that while we have placed the first charge, without generating any work the system will still have the energy held in the field of the charge!
There is obviously something that i miss, but what?
| The capacitor stores its energy in form of electric field.
Initially both the plates were neutral. After applying the potential difference, electrons move form one plate to another.
So to calculate the energy, you would first take a charge $dq$ from one plate to another. By this action, you just created an opposite charge $-dq$ on the other plate the moment you took the charge from it. This will set up an E. field. You set up an electric field that wasn't there before, so some work has to be done.
We will keep taking charge from one plate and deposit it on the other plate till the charge on the plate becomes $Q$.
Let's assume that at a point of time, charge on a plate is $Q'$. Therefore at the same time, the electric field between the plates will be $E=\frac{Q'}{A\epsilon_{\circ}}$.
Now, we will move charge $dq$ and the work done will be given by :
$dW=Ed.dq$
$W=\frac{d}{A\epsilon_{\circ}}\int_0^Q Q'dq$
$W=\frac{1}{C}.\frac{Q^2}{2}$
On rearranging,
$W=\frac{CV^2}{2}$
When you place a $1^{st}$ charge at point A (as per your question) you bring it to such a point where it interacts will no other charge. Thus, we come to the conclusion that work done here is $0$. The work will only be done in bringing the $2^{nd}$ charge at point B.
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Size of mercury barometer and effect on its reading I was thinking that since atmospheric pressure is 760mmHg what would happen if I shrink a mercury barometer until it's shorter that 760mm in height. What would happen? Would the barometer retain it's ratio of mercury height? Will the mercury fully fill the barometer? Is 760mmHg a constant and not affected by the size of the barometer (or diameter)? Thanks in advance.
| A mercury barometer is designed to measure the difference in pressure between the surface of the mercury outside the tube and the surface of the mercury inside the tube.
The space inside the tube is filled with mercury vapour and hopefully nothing else.
The column of mercury is kept in position because the pressure exerted on the surface of mercury outside the tube is equal to the pressure inside the mercury column at the same horizontal level as the mercury surface outside the tube.
If the tube is less than $760 \, \rm mm$ then there will be insufficient height of mercury to equate the pressures even if the mercury column occupies the whole of the tube.
In such a case the wall of the glass tube exerts downward forces on the mercury inside the tube such that the pressure exerted on the surface of mercury outside the tube is equal to the pressure (due to the column of mercury and the tube) inside the mercury column at the same horizontal level as the mercury surface outside the tube.
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Why young modulus value doesn't change even if we change the parameters? The Young modulus of steel is determined using a length of steel wire and is found to have the value EE. Another experiment is carried out using a wire of the same steel, but of half the length and half the diameter.
What value is obtained for the Young modulus in the 2nd experiment?
Why doesn't yhe values change even if we are changing the parameters
| Young's modulus is a material property and independent of the size or shape of the specimen being tested
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Would seawater lose their ions after sliding across graphene Accroding to this http://www.iflscience.com/technology/how-drop-seawater-graphene-generates-electricity/ webpage, voltage would be generated if droplets of seawater (which contain ions) are dragged across the surface. However, there is no mentioning of what happened to the droplets. Do they still have their ions within them or would they turn into regular water?
| According to the model explained in the paper (Generating electricity by moving a droplet of ionic liquid along graphene), the water droplets retain the Na+ and Cl- ions.
The mechanism is as follows:
When a NaCl droplet is put on a strip of graphene, a double layer of Na+ ions (purple) forms at the interface. These positive ions in the double layer draw electrons (yellow) to the top surface of the graphene sheet.
When the droplet is drawn to the right, the Na+ ions in the double layer at the back edge of the droplet desorb back into the droplet. The electrons in the graphene that were attracted to these Na+ ions are 'released' back into the graphene sheet behind the droplet.
Similarly, at the front of the droplet, Na+ ions form a double layer on the new territory covered by the droplet as it moves. These new Na+ ions attract electrons from the graphene ahead of the droplet.
The reduced electron density in front of the droplet is effectively an increase in potential in that region, the increased electron density behind the droplet is effectively a reduction in potential in that region. So, by moving the droplet, they generated a potential difference of a few mV between the front and back edge of the droplet!
No chemistry happened, no charge was transferred between the droplet and the graphene, and all salt ions remained in the droplet.
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Field due to internal Induced charge on a conductor to an external point? A charge q is located at a distance r from the center of a conducting sphere with inner radius 2r. The charge induces charges on the inner surface of the sphere according to Gauss' law .
The electric field at point p is to be approximated.
Inside the material of the conducting sphere, the electric field due to induced charge will cancel out the electric field due to the charge inside the sphere. Accordingly the electric field lines will begin at induced charge and terminate at the inner charge.
Therefore the field due to internal induced charge on the point p must be zero , (note it may be nonzero due to external induced charge but the problem specifies internal)
The solution however says it to be $kq/17r^2$ and not zero
Isn't the electrostatic system shielded from the conductor?
| There are three types of charges contributing to the field at point P
*
*The charge q itself
*The induced charge on the inside of the sphere
*The induced charge on the outside of the sphere.
First, the contribution due to the charge q itself will have a magnitude of $\frac{kq}{r^2}=\frac{kq}{(4R)^2+(1R)^2}=\frac{kq}{17R^2}$
Second, the field due to the induced charge on the inside of the sphere will cancel the field of the original charge and therefore have a magnitude which is equal but opposite to the field of the original charge.
Third, if the sphere is uncharged, i.e. electrically neutral, the accumulation of charge on the inside surface, will lead to an equally large, but opposite accumulation of charge on the outside surface. This charge will be evenly spread over the surface of the sphere. The magnitude of this field is $\frac{kq}{r^2}=\frac{kq}{16R^2}$
| {
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Are there experiments that deal with the headlight effect? If I am not mistaken, the Ives-Stilwell experiment measures the frequency of emitted light to test the relativistic doppler formula.
Is there anything like this for light intensity? Have we measured the headlight effect?
Anything like precision tests that are related to this would be interesting.
| The angular distribution of products from decay-in-flight beams such as those used to generate neutrino beams and radioactive beam accelerators (PDF link) are a sensistive test of this prediction.
I don't know of a paper discussing the focusing specifically (because I am an end user rather than creator of such beams and it's now a 50 year old technology), but an example of measuring the angular distribution is
Fig. 1. Density plot showing the variation in neutrino spectrum shape over a $3^\circ$ angular range $1.5\,\mathrm{km}$ from the production target.
which is figure one from http://physics.princeton.edu/~mcdonald/nufact/e889/chapter3a.pdf on the NuMI beam at Fermilab.
Notice how the higher energy component of the beam is narrower than the low energy component, an expected feature of kinematic focusing.
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Questioning validity of SHM of vertical mass-spring system Horizontal mass spring system is good but vertical mass spring system confuses me.
Q1. Can there be two restoring forces in an SHM?
Q2. If no, then weight of mass seems to disturb SHM as down extreme position below the mean position would be much farther than extreme position above mean position. If vertical mass spring system executes SHM, please elaborate how?
| On applying a constant force in an SHM,it only changes its mean position BUT its amplitude,time-period, max velocity etc remains same.This can proved by solving the differential equation given below:
$$\frac{d^2x}{dt^2} = \frac{-kx}{m}+g$$
(solution of above differential equation left to you)
On solving this differential equation,assuming mass to be at mean position ($\frac{d^2x}{dt^2} = 0$) at t=0;
$$x=\frac{mg}{k}+Asin(\left(\frac{\sqrt k}{\sqrt m}t\right)$$
From above equation,it can be concluded that time period remains same while application of constant force in SHM
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Is the huge calcite birefringence accounted for by first principles? The birefringence in calcite is huge (possibly the largest?). The question is, why is it so special? why is such kind of materials so rare?
In other words, is it explained by first principles?
By google, I just found a paper predicting the birefringence in some material. See http://pubs.acs.org/doi/abs/10.1021/jp506744s
| The size of the birefringence of calcite is not surprising. It has a crystal structure in which all the planar carbonate ions have the same orientation. The polarization of planar ions is much larger for electrical fields in the plane (light coming in perpendicular to the plane) than for electrical fields perpendicular to the plane.
A simple calculation would start out by doing this for a single CO$_3^{2-}$ ion (or for an ion with a 1+ charge), and then apply Clausius-Mosotti.
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Would a handspinner spin indefinitly in space? I'm having a argument with a colleague, I don't know how to explain to him that if you spin a handspinner in space it will spin indefinitly (if you don't hold it). I agree that if you hold it, it will slow down because of the friction with the center part.
Would it theoreticaly spin forever?
| Assuming that the engineering is perfect, that is, your handspinner can't be broken, that would be the case in the vacuum. It stops on Earth because of the friction with the air.
| {
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Well-defined momentum I have a question which states:
Show that a particle can have a well-defined momentum in every energy eigenstate if and only if the potential energy is uniform in space.
I am completely unsure how to start. The Schroedinger equation states:
$\frac{p^2}{2m} \psi+V(r)\psi=E\psi$
$p^{2}=2m(E-V(r))$
but I am at loss how to actually prove that the momentum is undefined if the potential is varying.
| A partial answer starts with the observation that, if the particle has a well-defined momentum, then
\begin{align}
\hat p\psi(x)&=p\psi(x)\, ,\\
\hat p^2\psi(x)&=p^2\psi(x)
\end{align}
If $\psi(x)$ is an energy eigenstate, then $\hat H\psi(x)=E\psi(x)$. Thus
\begin{align}
\left(\hat E-\frac{\hat p^2}{2m}\right)\psi(x)&=V(x)\psi(x)\, ,\tag{1} \\
&=\left(E-\frac{p^2}{2m}\right)\psi(x)\, .\tag{2}
\end{align}
For $\psi(x)\ne 0$, equate the right hand sides of (1) and (2), and divide by $\psi(x)$ to find $V(x)=E-\frac{p^2}{2m}=$constant.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof that Verlet integration conserves energy I was reading about different integrators that one might use to solve the system of differential equations which governs the $n$-body problem. I read that the Verlet integrator is time-reversible, and thus conserves energy.
I don't understand why time-reversibility implies energy conservation.
Suppose I know the position and velocity of each of $n$ particles in the system at some time $t_0$. I calculate the energy $E_0$ of the system at time $t_0$. I use the Verlet integrator to calculate the approximate positions and velocity of the particles at time $t_0+dt$ (one time step) and I recalculate the energy of the system and find it to be $E_1$. Is it true that $E_0=E_1$? And is there an easy way to prove this?
| Firstly, the Verlet integrator only conserves energy in the limit $\Delta t\to 0$. In practice it produces energy drift, although the long-term energy drift is smaller than for most integrators.
Regarding your question, the gist of the argument is that integrators that are not time-reversible do not exhibit so-called area-preservation. This basically means that the volume of phase space of constant energy $E$ will evolve to a larger volume of the phase space over time, and hence it will necessarily span a region of phase space with different energies.
The rigorous proof of this property is complicated which is why most sources (including this answer of mine) omit the details and provide a handwavy justification. Frenkel and Smit's book Understanding Molecular Simulation provides a review of this subject with plenty of citations, if you're interested.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to explain the relationship between wave's amplitude and intensity? I have the following statement which I don't know how to explain:
Suppose I have 2 identical monochromatic waves (same intensity and phase) shooting into the same receiver. If each wave's intensity is I, based on energy conservation I would expect the 2 waves together will bring a total intensity of 2I. For example if each wave carries 100mW power, I'm expecting a 200mW total power on the receiver side.
However the summation using phasor gives a different result: if we consider that intensity is proportional to the square of wave's amplitude, the square root of 100 gives 10 for the amplitude of 1 wave (for simplicity I use 10 as amplitude which included the constants), adding the other wave of the same phase which gives 20 as amplitude of the new beam, then square it to get intensity which is 400mW instead of 200 in the above example.
If we keep going by summing 4 50mW waves with the same convention, we get (4 x sqrt(50))^2 = 800mW ... which the logic is obviously not correct.
My ultimate goal is to sum up power of beams with different intensities and phases. Phasor addition works great if I have amplitude and phase, but when I try to use intensity to get wave's amplitude I got the above dilemma. Could someone point out where my logic go wrong, and please explain the way to do power summation with intensity and phase known for each wave? Thanks!
| As explained here, you have to consider the fact, that your light waves' amplitude gets averaged over a certain amount of time, thus decreasing the intensity.
I quote from the website, just for the sake of having the information close to this question:
"Try this: Let $ψ=Acosθ_1$ and $ϕ=Bcosθ_2$.
Then: $I∝(ψ+ϕ)^2=(Acosθ_1+Bcosθ_2)^2$ $I∝A^2cos^2θ_1+B^2cos^2θ_2+2ABcosθ_1cosθ_2$.
The average value of $cos^2$ is 1/2; using a bit of trig, you should be able to convince yourself that the cross term averages to zero if the phases are random. Thus:
$I∝A_2/2+B_2/2$
$I_{total}=I_1+I_2$"
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/333260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Why the name 'displacement' operator? I'm studying coherent states of the harmonic oscillator and I have learned about the so called displacement operator, which is the operator defined as
$$D(\alpha) = e^{\alpha a^\dagger -\alpha^* a}$$
whose action over the fundamental state $|0\rangle$ is to produce a coherent state $|\alpha\rangle$.
Now, why is it called displacement operator? In which sense does it displace the state $|0\rangle$?
| A coherent state is characterized by a complex number $\alpha \in \mathbb C$.
Applying the displacement operator $D(\beta)$ to $|\alpha\rangle$ translates $\alpha$ in the complex plane by $\beta$, in the following sense:
$$ D(\beta) |\alpha\rangle \sim |\alpha+\beta\rangle . \tag 1 $$
Here, $\sim$ means "up to a phase".
The precise relation is:
$$ {D}(\beta){D}(\alpha) = e^{(\beta\alpha^*-\beta^*\alpha)/2} {D}(\alpha + \beta) , \tag 2 $$
the exponential is just a phase factor.
Note 1: (1) follows from (2) because $|\alpha\rangle = D(\alpha) |0\rangle$ - multiply (2) from the right with $|0\rangle$ to get (1).
Note 2: That also makes sense for the ground state $|0\rangle$, because it is equal to the coherent state with $\alpha=0$:
$$ D(\alpha)|0\rangle = |0+\alpha\rangle .$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Where does the force actual act?
Why are the two mass of $m_1$ and $m_2$ not multiplied by minus one?
I know that two minus multiplied gives you plus by but I mean the two masses are attracting so they should have a sign like so $$F_g=\frac{G(-m_1) \times (-m_2)}{r^2}$$
I ask these because equation would actual give the right explanation as to what the equation is actual doing, meaning that the two mass are attracting.
| In reality the force acts on all parts of $m_1$ and $m_2$ but it varies due to the variation in distance for different areas of $m_1$ and $m_2$. In practice though, if $m_1$ and $m_2$ are spherical, the force can be assumed to act at the center-of-mass of each.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Can anyone tell me that actually What vector multiplication is? I am a high school student. Yesterday I attended a lecture on vector multiplication (first time). There was something I was not able to understand.
The only concept of multiplication I had before is 2x3=6, means that if we write '3' 2 times we'll get 6.
But now there are two types of multiplications:
*
*Dot product
*Cross product
Dot product
What I understood is that dot product of two vectors is a scalar quantity, But how?
Here if we multiply projection of A on B (A cos0) with B, finally that product would be in the direction of B?, am I right or wrong?
Secondly please specify the difference between this dot product and ordinary multiplication which I stated above (2x3=6)?
Cross product:
By Cross product I understand that it's the product of the 'length of perpendicular produced for projecting A on B' and 'B'?
What is the purpose and meaning of this product?
And After multiplication product is multiplied by unit vector n, which perpendicular to the A and B.
Why the cross product is always perpendicular to the given vectors?
And if I am right that Cross product is the product of the 'length of perpendicular produced for projecting A on B' and 'B' then why that projection is not shown in this diagram of cross product (which is taken from wikipedia)?
| A Geometric View of Vector Products
Vectors have geometric properties independent of coordinate systems. The vector dot product $\mathbf A \cdot \mathbf B = AB \cos(\theta)$, projects the length of vector $\mathbf A$ onto the direction of vector $\mathbf B$; it is the shadow projection of the one vector onto the other. This operation is linear over vector addition, and the definition is symmetric, so the dot product is commutative, $\mathbf A \cdot \mathbf B =\mathbf B \cdot \mathbf A$. It is used to determine angles between vectors as well as lengths and distances. The condition $\mathbf A \bullet \mathbf B = 0$ is a test for orthogonality.
The geometrical meaning of the vector cross product $\mathbf A \times \mathbf B = AB \sin(\theta) \mathbf n$ is obtained by sliding vector $\mathbf A$ along the length of vector $\mathbf B$, always remaining in their joint plane,and with $\mathbf A$ remaining parallel to itself. This is done by hooking the right hand thumb about vector $\mathbf B$ as a guide, and then pushing with that hand to mark out the area of a parallelogram. The “right hand rule” orients $\mathbf n$ with your right thumb, making it normal to the plane of the parallelogram. So in addition to the determination of areas and angles, the creation of the unit vector ˆn determines the orientation of the plane formed by the two vectors. This operator is linear over vector addition, but note that the "right hand rule" reverses the orientation if the order of the vectors is reversed: $\mathbf B \times \mathbf A = -\mathbf A \times \mathbf B$. The test $\mathbf A \times \mathbf B = 0$ is a test for parallelism.
Note that ordinary multiplication also has a geometric meaning: 2 x 3 is the area of a rectangle 2 wide by 3 long; rotation of the rectangle by 90 degrees shows that the same area is obtained with 3 x 2.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/333877",
"timestamp": "2023-03-29T00:00:00",
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Why are Grassmann fields never classical? I see this statement in many QFT books (e.g. Altland & Simons' Condensed Matter Field Theory) but the author never explains why.
Can you briefly explain why Grassmann fields never have a classical meaning (preferably physical arguments) and possibly point out some good references?
| I have proposed a classical interpretation for a pair of Dirac fermions in arxiv:0908.0591. It is a lattice version, works only for a pair, not a single fermion, requires a preferred frame, and the Hamilton operator for the classical $\mathbb{Z}_2$-valued field is only an approximation of the lattice Dirac operator.
The $\mathbb{Z}_2$-valued field can be easily embedded into an $\mathbb{R}$-valued field with degenerate vacuum, which would add a massive scalar field to each pair of Dirac fermions.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/333996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
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How do we realise the wave for an electron orbiting a nucleus in its first orbit as per bohr model? How do we realise the wave for an electron orbiting a nucleus in its first orbit as per bohr model?
As per my textbook, we can account for as to why angular momentum is quantised. But I fail to understand what would the wave diagram for n=1 look like?? For example, for higher orbits.
As to my knowledge, such a diagram should not be possible for first orbit. Then how come we can account for quantisation for the first orbit?
For example, for higher orbits, the following diagrams follow...
| Maybe the term "orbiting" causes some problems. The electron is not moving around the nucleus as bohr assumed (he had no solution according to the problem that an electron on a circular orbit should emit electromagnetic waves and crash into the nucleus).
You should think about the elcetron as part of the whole system which can have some quantized energy eigenstates and so some quantum numbers. According to this point of view you need to solve the schrödinger equation and interpret the solution as probability-density for the elctron states.
The absolut value square of your wave function is a positiv number and thats what you call the orbital. How this orbitals look like depends on the quantum numbers of the electron state. In the picture below you can see a few orbitals for the lowest quantum numbers of the hydrogen atom.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/334363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove the constant speed of light using Lorentz transform? I read the light-clock example in my book which proved the time dilation formula by assuming that the speed of light is constant for all observers. But I've trouble in understanding it the other way around. Lorentz transformation is just a correction to Newtonian mechanics to account for the constant speed of light for all observers, right? I have trouble understanding how does applying this correction preserve the speed of light for all observers.
Can we start by assuming that the Lorentz transformation formulas are true and then prove that two observers $A$ and $B$ will see a light pulse moving at the same speed $c$ regardless of their relative velocity with respect to each other?
| An enlightening (but possibly advanced) method to prove the constancy of the speed of light from the Lorentz boost transformation is to find the eigenvectors of the Lorentz boost. Two of the eigenvectors are along the light cone. The corresponding eigenvalues are equal to the doppler factor and its reciprocal.
(These eigenvectors are coplanar with the 4-velocities of observers in relative motion.)
This can be compared and contrasted with the eigenvectors and eigenvalues of the Galilean transformation.
In both transformations, there are no timelike eigenvectors... that is, no preferred observers.
Now, for some details:
Given $M=\begin{pmatrix}
\gamma&\beta\gamma\\
\beta\gamma&\gamma\\
\end{pmatrix}$, we set up the eigenvalue problem:
$$0=\det (M-kI)=\det\begin{pmatrix}
\gamma-k&\beta\gamma\\
\beta\gamma&\gamma-k\\
\end{pmatrix}=(\gamma-k)^2-(\beta\gamma)^2.$$
Solving this characteristic equation for $k$, we find $k-\gamma=\pm\beta\gamma$, which can be written as $k=\gamma(1\pm\beta)=\sqrt{\frac{1\pm\beta}{1\mp\beta}}$, which are the Doppler factors.
The eigenvector corresponding to $k=\gamma(1+\beta)$ is gotten by substitution:
$$\begin{pmatrix} 0 \\ 0\end{pmatrix}=
\begin{pmatrix}
\gamma-(\gamma(1+\beta))&\beta\gamma\\
\beta\gamma&\gamma-(\gamma(1+\beta))\\
\end{pmatrix}
\begin{pmatrix} w_t \\ w_x\end{pmatrix}
=
\begin{pmatrix} -\beta\gamma & \beta\gamma \\\beta\gamma & -\beta\gamma \end{pmatrix}\begin{pmatrix} w_t \\ w_x\end{pmatrix}=\begin{pmatrix} (-\beta\gamma)w_t + (\beta\gamma)w_x \\ (\beta\gamma)w_t + (-\beta\gamma)w_x\end{pmatrix}.$$
This is satisfied by vectors of the form $w_x=w_t$ –that is, the along the future-forward lightlike direction. Thus, under a Lorentz Transformation, the light-signal's velocity remains unchanged, but the future-forward component of a vector gets stretched by a factor of $k$. Similarly, the future-backward component gets reduced by a factor of $k$. (This is the basis of the Bondi k-calculus [pun intended] and methods using light-cone coordinates.)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/334884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
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Covariant derivative 1+3 formalism I can't find the proof for the following equation:
$$ \nabla_bu_a=D_bu_a-A_au_b $$
where $D_b$ is the spatial derivatives respect to $h_{ab}=g_{ab}+u_au_b$ in the 1+3 formalism and $A_a=u^b\nabla_bu_a$.
EDIT:
I want to see it by computation: I start from $h_{ab}=g_{ab}+u_au_b$ and I see that $h^c\vphantom{h}_ah^d\vphantom{h}_b= \delta^c\vphantom{h}_a\delta^d\vphantom{h}_b +u^cu_a\delta^d\vphantom{h}_b+u^du_b\delta^c\vphantom{h}_a + u^du_bu^au_c $. Then I compute
$$D_au_b=h^c\vphantom{h}_ah^d\vphantom{h}_b\nabla_cu_d=\nabla_au_b+u^cu_a\nabla_cu_b+u^cu_au^du_b\nabla_cu_d+u^du_b\nabla_au_d$$
How can I show the last three therms to be exactly $A_au_b$?
| Assuming $u^d$ is the unit normal vector to the hypersurface, we have $u^du_d=-1$, implying
$$0=\nabla_a(u^du_d)=u_d\nabla_au^d+u^d\nabla_a u_d.$$
Then, using $u_d=g_{db}u^b$ and passing $g_{db}$ through $\nabla_a$, we see
$$2u^d\nabla_a u_d=0.$$
The last two terms are of this form, hence vanish. The first is the one you want to be left over.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/335114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Heisenberg Uncertainty Principle Applied to an infinite square well I appreciate the statement of Heisenberg's Uncertainty Principle. However, I am a bit confused as to how exactly it applies to the quantum mechanical situation of an infinite square well.
I understand how to apply Schrödinger's equation and appreciate that energy Eigenvalues can be deduced to be
$$E_n=\frac{n^2\hbar^2\pi^2}{2mL^2}.$$
However, I have read somewhere that the reason that the quantum particle cannot have $n = 0$—in other words, $E = 0$—is because by having zero energy we also have a definite momentum with no uncertainty, and by the Heisenberg uncertainty principle this should lead to an infinite uncertainty in the position of the particle. However, this cannot the case be in an infinite well, as we know the particle should be somewhere in the box by definition. Therefore $n$ can only be greater than or equal to one.
Surely when $n = 1$ we have the energy as
$$E_1 = \frac{\hbar^2 \pi^2}{2mL^2},$$
which is also a known energy, and so why does this (as well as the other integer values of $n$) does not violate the uncertainty principle?
| This question is actually almost identical to a question I had in my first year quantum mechanics course. It posed that a hypothetical friend asked you why having definite energy in the infinite potential well doesn't violate the uncertainty principle. Here I will provide you with the answer I gave all those years ago.
Consider an energy above $E=0$. Then the momentum of the state must satisfy $E=p^2/2m$. Now, there are two solutions for $p$, corresponding to a positive and negative momentum. The uncertainty can thus be calculated as
$$\Delta p^2=\langle p^2\rangle-\langle p\rangle^2=2mE=\frac{n^2\hbar^2\pi^2}{L^2}$$
Clearly, for $n=0$ the uncertainty vanishes, but is perfectly finite for all other values. Thus, so long as $\Delta x^2$ is large enough (it is), the uncertainty principle is perfectly satisfied for all $n>0$.
Another reason that the $n=0$ state is unphysical is because it is not normalisable. Indeed, the $n=0$ state would have to satisfy
$$\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2}=0$$
And thus is linear. However, there is no nonzero linear function that can satisfy the necessary boundary conditions. Thus, $\psi=0$ everywhere and is not a physical state in our Hilbert space.
I hope this helped!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/335184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
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