Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
How to start moving in at $Re\ll1$ I find it difficult to see how something can accelerate (and therefore increase its velocity, e.g. start to move) in a $Re\ll 1$ situation.
It is customary, at low Reynolds numbers, to ignore inertial effects. This means that the nonlinear terms in the Navier-Stokes equations vanish. One can prove from the resulting equation (Stokes' equation) that, for example, if a sphere of radius $R$ is placed in a fluid of viscosity $\eta$ and local velocity $v$ (in a configuration at which $Re\ll 1$), the drag force experienced by the sphere will be
$$f_D=6\pi\eta R v$$
where $\eta$ is the dynamic viscosity of the fluid, $R$ the radius of the sphere.
What does this mean in terms of the force balance on the sphere?
Let's assume that I apply a driving force $F$ on the sphere, and I apply Newton's 2nd Law to this situation
$$F_{tot}=F - f_D = F - 6\pi\eta a \frac{dx}{dt} = m \frac{d^2 x}{dt^2}$$
where $f_D$ is the drag force. Does this imply that the motion of the sphere is given by the solution of the ODE above?
Or is the acceleration always going to be zero because of the absence of inertia? How can one justify this in the formalism used above? By imposing that the inertial mass $m=0$?
Edit: The following extract, from this well-know review on microswimmers, hopefully justifies the doubts about whether acceleration is meaningful or not in this context. The authors had first stated: "Since swimming flows are typically unsteady, we implicitly assume the typical frequency ω is small enough so that the frequency Reynolds number $\rho L \omega^2/\eta$ is also small."
| In the frame linked to the sphere, you would have to add an inertial volume force $-\mu a$ if $a$ is the acceleration of the sphere.
With dimensional analysis, $\mu a\to \frac{\eta V}{{{L}^{2}}}\times \frac{\mu a{{L}^{2}}}{\eta V}$and so, we have an additional term in the dimensionless Navier Stokes equation $\frac{\mu a{{L}^{2}}}{\eta V}=\frac{\mu VL}{\eta }\times \frac{aL}{{{V}^{2}}}=\operatorname{Re}\times \frac{aL}{{{V}^{2}}}$
So there is an other dimensionless number $\frac{aL}{{{V}^{2}}}$ which looks like the inverse of the Froude number and we would have an additional condition $\operatorname{Re}\times \frac{aL}{{{V}^{2}}}\ll 1$ ?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/459620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Why some forces follow superposition principle? Let there be a system of $n$ source charges and a test charge $Q$. When we say superposition applies to electrostatic force, we conclude that the interaction between a given source charge and the test charge is independent of interaction between other source charges and the test charge. Why exactly it is the case? Also why some forces follow superposition?
| Force is a vector quantity: vectors add in predictable ways, so forces
are capable of being considered separately or of being added together.
We say 'superposition' if force fields (vector fields of
a type OTHER than force) act on some part of an object, like
mass or charge or surface area. An electric field, for instance,
can make your (electrically polarizable) hair stand on end, while at the same time gravity makes your (massive)
hair drape downward. Only a strong electric field overcomes gravity,
but it never changes gravity. We observe this, but cannot say 'why'.
We can, however, calculate the electric and gravity forces and know
how much electric field is required to balance gravity. The electric
and gravity fields both generate forces, and we can sum those forces
though the fields are as dissimilar as apples and oranges (and cannot be summed).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/459708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Faster ways of computing feynman diagrams Obviously the machinery of QFT allows us to calculate processes, such as QED diagrams, to great precision, and whilst it is effective, it seems there are many processes that make calculations (say by hand) significantly slow.
Are there any recent developments in our machinery to compute Feynman diagrams that makes it faster to analytically compute matrix elements, widths and cross sections?
| Nowadays, less and less people use Feynman diagrams for precision calculations (that is, anything beyond a tree level 4 or 5 point amplitude). There is a whole field dedicated to finding better methods of calculating scattering amplitudes (using recursion and unitarity for example). In fact this is how most calculations for the LHC are done in practice.
A nice review of some of the techniques is https://arxiv.org/abs/1308.1697. Most of the work is motivated by the observation that scattering amplitudes are much simpler than one could have thought by just staring at the Feynman rules. An example of this is the famous Parke-Taylor amplitude https://en.wikipedia.org/wiki/MHV_amplitudes. Take a look and compare to the QCD Feynman rules.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/459787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Pressure due the atmosphere Usually when we consider the pressure exerted by gas, there is nothing to do with the weight of the gas. On the contrary the atmospheric pressure is defined as the weight of the gasses. What is the difference here?
| when we say pressure exerted by gas we say about pressure exerted by gas on container due to collision of particles with wall of container when kept in closed container we use PV=nRT to define it here pressure due to weight of air is very small so it is negleted ( it is same case as variation of pressure in sound wave) but atmospheric pressure is due to weight of gas as there presdure due to collision of particle with ground can be negleted due to very small value
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/460182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Computation of $e^{i \hbar \omega a^{\dagger} a} a e^{-i \hbar \omega a^{\dagger} a}$ I need to compute terms like :
$$e^{i \omega t a^{\dagger} a} a e^{-i \omega t a^{\dagger} a}$$
Where $[a,a^{\dagger}]=1$ (they are the bosonic annihilation/creation operators).
I wonder if there is a simple formula for this. Indeed, when I try to compute the commutator:
$$[a,e^{i \omega t a^{\dagger} a}]. $$
I end up with something that doesn't look trivial.
For example:
$$[a^{\dagger} a, a] =a .$$
But:
$$[(a^{\dagger} a)^2, a] =2 a^{\dagger} a^2 $$
So I don't know how I could compute the general term (and if actually it is an easy thing to do...).
In summary: is there a simple expression for:
$$e^{i \omega t a^{\dagger} a} a e^{-i \omega t a^{\dagger} a}$$
and if so, is there a trick to compute it?
| I define the following operator:
\begin{equation}\tag{1}
\hat{A}(t) = U^{\dagger} a \, U = e^{i \omega t \hat{N}} a \, e^{- i \omega t \hat{N}},
\end{equation}
where $\hat{N} \equiv a^{\dagger} a$. We have the following commutator (notice that there's a sign mistake in yours):
\begin{equation}\tag{2}
[a, a^{\dagger}] = \mathbb{1}, \quad \Rightarrow \quad [\hat{N}, a] = - a.
\end{equation}
Then we have this:
\begin{align}
\frac{d \hat{A}}{dt} = i \omega \, [\hat{N}, \hat{A}] &= i \omega \, U^{\dagger}[\hat{N}, a] \, U \\[12pt]
&= - i \omega \, U^{\dagger} a \, U = - i \omega \hat{A}. \tag{3}
\end{align}
The solution to this differential equation is easy:
\begin{equation}\tag{4}
\hat{A}(t) = A(0) \, e^{- i \omega t} = a \, e^{- i \omega t}.
\end{equation}
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/460321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How many temperatures has a plasma? In nonthermal plasma, not all particles move in the same way. The electrons are different from other particles. Both can be described as having a temperature separately. But that would mean, one piece of substance could have
Two temperatures at the same time
No temperature at all.
Both variants bend my mind just a little too much.
It could be no substance at all, but that's what we call vacuum.
Is it just me, or is it one of the cases a human mind just can not bend around?
| Thermodynamic equilibrium is not a trivial condition. Feynman, in his Statistical Mechanics lectures start writing that a system is in thermal equilibrium when all the "fast" things have happened and all the "slow" things not.
It may happen in some systems, for instance in the case of a plasma, that thermal equilibrium between electrons and between ions can separately be established, while processes allowing electron-ion equilibration may be much slower. In sucha a case one speaks about a two-temperature plasma.
Therefore, it is not the case of "no temperature" at all, but of a system with two sub-systems at different temperatures. Usually, the reason for having a long lasting difference of temperatures can be related to the presence of inefficient channels for transferring energy from the hotter component to the colder. Quite frequent situation, when there is a large mass difference.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/460461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How can I determine a planet's mass based only off of information about its orbit and its parent star? I'm coding a video game with procedurally generated planetary systems and I want some and I want to make sure I'm at least somewhat scientifically correct. I've reached the part in my code where I know where a planet should be orbiting but don't know what mass to give it.
At this point, I have the mass of the star (in solar masses) and the orbital period (in Earth years) and orbital velocity of the planet (measured in Earth orbital velocities).
If there's no precise way of determining the planet's mass, is there a way to know more or less what range the mass should be in?
| When a body is in orbit around a planet, the centrifugal force is balanced by the gravitational force:
$$
{{mv^2}\over{r}} = {{GMm}\over{r^2}}
$$
where $M$ is the mass of the planet and $m$ the mass of the orbiting body.
You can transpose this to get the velocity for a given radius or whatever you want.
$$
v = \sqrt{GM\over{r}}
$$
However, you will immediately notice that the mass of the body ($m$) appears on both sides of the equation and so cancels out.
So all objects at a given radius move at the same speed. Therefore, you can't get an object's mass from knowing its orbit.
If you want to test this experimentally, simply step outside the ISS. You will float along beside it in exactly the same orbit, despite being hundreds of times lighter.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/460730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Parity transformation and mirror reflection I have some trouble understanding what exactly is parity transformation.
The definition of parity transformation is a flip in the sign of all three spatial coordinates, ie
$$(x,y,z) \rightarrow (-x,-y,-z).$$
Consider a stationary particle at a position $(a,b,c)$ in space described by a coordinate system $(x,y,z)$. Does parity transformation mean that the particle is still at the exact point in space but its position is now described by $(-a,-b,-c)$?
But often parity is talked about as a mirror reflection and it seems to me that a mirror reflection means physically moving the particle from point $(a,b,c)$ to $(-a,-b,-c)$ in a coordinate system $(x,y,z)$.
Which of the above 2 cases is parity transformation really referring to? If it refers to both cases, why are the two cases the same? In one case a particle is fixed in space while in another case a particle is moved to another point in space.
| First of all, there are two conventions - 'active' and 'passive' points of view. Within the former, you would say that under the parity transformation the particle has changed its position in space from $(a,b,c)$ to $(-a,-b,-c)$. Within the latter, you'd say that the particle stays at the same point of space, but the coordinate system has been changed in such a way that the new coordinates of the same particle (staying at the very same location) are $(-a,-b,-c)$. Clearly, both conventions are equivalent. Typically, the 'active' one is used in Physics.
You have correctly defined the parity transformation as the change of signs of all the coordinates: it changes all the vectors $\vec{r}\to-\vec{r}$, leaving, therefore, only the zero vector invariant.
The reflection, in turn, always happens relative to a certain plane. Say, a reflection relative to the $x\,y$ plane works as $(x,y,z)\to(x,y,-z)$. Importantly,
*
*This operation leaves all the vectors belonging to the $x\,y$ plane invariant.
*It can be turned into parity by applying an additional rotation in the $x\,y$ plane.
The second circumstance in the reason why these two operations are often confused. In many cases, we only care about the transformation 'up to a rotation'. Note that if you apply either parity transformation or reflection to a solid body, there's no way of rotating it back into the original positions. In this sense, these two operations are equivalent.
Also, keep in mind that in a different number of dimensions things work slightly differently.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/460996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What is the theory behind imaging a precision pinhole? In my efforts to characterise a 0.67 NA microscope objective (working distance of about 15mm, effective focal length of 25mm), I have placed a 20 micron precision pinhole at the focal plane of the objective, and back illuminated the pinhole with a 767nm laser. The light coming out of the objective is focused onto the imaging plane (camera plane) with a F=1 metre lens.
By my understanding, the image field $U_i$ should be a convolution of the Point Spread Function (PSF) $h$ and the object field $U_o$, i.e.
$$
U_i(x_i,y_i) \propto \int_{-\infty}^\infty\int_{-\infty}^\infty h(x_i-\xi, y_i-\eta) U_o(\xi,\eta)\, d\xi d\eta\, ,
$$
where the proportional sign is for good measure as there are some coefficients in front, but that does not affect the profile. The image intensity profile is given by
$$
I_i(x_i, y_i) = |U_i(x_i, y_i)|^2\,.
$$
In my case the microscope objective has a circular aperture, and that means that the PSF $h$ is given by the airy function, i.e.,
$$
h(x,y) = \frac{J_1 \left( \frac{2\pi}{\lambda} NA \sqrt{x^2 + y^2}\right)}{\sqrt{x^2 + y^2}} \, ,
$$
where $J_1$ is the Bessel function of the first kind, order one.
I have plotted out the intensity profile (gauss quad integral) that I expect for the current pinhole and wavelength, taking into account the magnification of the system, and I got the following plot.
However, this differs from my measurements taken in experiment:
The size of the image is approximately 20 microns (pixel size = 3.75/40 microns).
The number of peaks somewhat represent what is shown from the plot above. What concerns me more is how the intensity profiles goes to zero between peaks, unlike what the theory predicted.
Is my understanding of the current theory incorrect?
| This looks like an interference pattern caused by the laser's coherence when passing through the pinhole (see e.g. here). Can you try illuminating the pinhole with an incoherent source, instead, such as an LED? I believe that should produce a flatter profile similar to the one in your model.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/461100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Why is the friction force on the bed opposite of the gravity
So here is a simple problem and here is a diagram that I find found online.
Why is the frictional force pointing downwards. I mean I get the correct answer if I follow this diagram and math checks out but intuitively. I thought the frictional force should be acting against the person falling off the bed due to downward gravity so I thought, frictional force should be pointing upwards NOT downwards. Can someone explain why? The questions asks for the smallest angle at which the person will begin to slip off.
correct answer is provided in the bottom for your reference
| Suppose there is no rotational motion. Then the frictional force must point upwards along the plane to keep the person from falling.
Now consider the situation when there is no friction and angular speed of rotation is very high. In the rotating frame, the centrifugal force will point outwards and its component along the plane will be higher (when angular speed is high enough) than the component of gravity downward along the plane. (You could do the whole calculation from inertial frame without invoking pseudo forces, but the situation be harder to visualize). If you turn on friction at this moment, it will point downwards in order to resist the upward motion of the man.
So, for low angular velocities, friction will point upward, and gradually to zero at a certain angular velocity (you can show that this angular speed is given by $\omega^2 = g \tan \theta /R$, where $R$ is the radius of the motion of the center of mass), and for higher angular velocities friction will point downward.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/461234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why do electromagnetic waves have the magnetic and electric field intensities in the same phase? My question is: in electromagnetic waves, if we consider the electric field as a sine function, the magnetic field will be also a sine function, but I am confused why that is this way.
If I look at Maxwell's equation, the changing magnetic field generates the electric field and the changing electric field generates the magnetic field, so according to my opinion if the accelerating electron generates a sine electric field change, then its magnetic field should be a cosine function because $\frac{d(\sin x)}{dx}=\cos x$.
| The E and H fields in a time-harmonic EM wave are in phase in the time domain when the medium's polarization (electric and magnetic) are in phase with the corresponding fields. You can see that polarization fields inherently act as 'source' terms in Maxwell's equations, and hence, instantaneous polarization implies in-phase relationship. However, whenever there is dissipation (such as existence of conduction current, or out of phase polarization), the E and H fields are no longer in phase. In other words, the one phasor cannot respond instantaneously to the changes of the second one in time. Note that regardless of propagating or standing wave, E and H fields are in phase with each other in the time domain in a lossless medium (for a standing wave, they are 'out of phase' spatially).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/461393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 3
} |
How can we derive from $\{G,H\}=0$ that $G$ generates a transformations which leaves the form of Hamilton's equations unchanged? In the Hamiltonian formalism, a symmetry is defined as transformation generated by a function $G$ is a symmetry if
$$\{G,H\}=0 ,$$
where $H$ denotes the Hamiltonian.
On the other hand, a symmetry is a transformations which map each solution of the equation of motion into another solution. And this requires that the form of the equation of motion remains unchanged.
Therefore, it should be possible to show that it follows from $\{G,H\}=0 $ that Hamilton's equations are unchanged by the transformation generated by $G$.
Concretely, we have
\begin{align}
q \to q' &= q + \epsilon \frac{\partial G}{\partial p} \\
p \to p' &= p - \frac{\partial G}{\partial q} \\
H \to H' &=H + \{H,G\}
\end{align}
and we want to show that if for the original $q$ and $p$ Hamilton's equations
\begin{align}
\frac{dp}{dt}&= -\frac{\partial H}{\partial q} \\
\frac{dq}{dt} &= \frac{\partial H}{\partial p}
\end{align}
hold, they also hold for $q'$ and $p'$:
\begin{align}
\frac{dp'}{dt}&= -\frac{\partial H'}{\partial q'} \\
\frac{dq'}{dt} &= \frac{\partial H'}{\partial p'}
\end{align}
How can this be shown explicitly?
Using the transformation rules explicitly yields for Hamilton's first equation
\begin{align}
\frac{dp}{dt}&= -\frac{\partial H}{\partial q} \\
\therefore \quad \frac{d(p' + \frac{\partial G}{\partial q})}{dt}&= -\frac{\partial (H + \{H,G\} )}{\partial (q' + \epsilon \frac{\partial G}{\partial q} )} \\
\therefore \quad \frac{d(p' + \frac{\partial G}{\partial q})}{dt}&= -\frac{\partial H }{\partial (q' + \epsilon \frac{\partial G}{\partial q} )} \\
\end{align}
But I've no idea how to proceed from here.
| A generating function$^1$ $\epsilon G(q,p,t)$ [where $\epsilon$ is an infinitesimal parameter] can to first order in $\epsilon$ be identified with a generating function $\epsilon G(q,P,t)$ for a type 2 canonical transformation (CT), $P=p+{\cal O}(\epsilon)$, cf. Ref. 1.
On the other hand, a CT takes Hamilton's equations into Kamilton's equations, with
$$K(Q,P,t)~=~H(q,p,t)+\epsilon \frac{\partial G(q,P,t)}{\partial t}, $$
so yes, it leaves Hamilton's equations form-invariant.
References:
*
*H. Goldstein, Classical Mechanics; eqs. (9.61)-(9.63).
--
$^1$ $G$ and $H$ do not have to Poisson commute.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/461874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Coulomb's Law Question The presentation of Coulomb's Law in various books occasionally has a note that the test charge, q2, must be small enough that it doesn"t alter the field of the first charge, q1. The same limitation applies to a test charge in any other stationary electric field. If those are the facts of life, then fine, but how can anyone confidently calculate the force between two macroscopic bodies, each with its own charge distribution? What are reasonable limits on charge level, physical size, and separating distance? It seems that one could correctly perform any and all required volume integrations and occasionally, if not often, produce invalid results. Moreover, for actual physical charged bodies, won't the electric field(s) alter the charge distribution on those material bodies? Is the solution to these issues simply some non-linear formulation that I just haven't seen yet?
| Some of them could be complicated, if they don't have symmetry. The good starting point is by assuming symmetry. For example, Gauss' law can be used to solve for the electric field due to a sphere, plane charge density, cylinder etc.
These derivations use Coulomb's law as a starting point.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/461977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Unification of gravity and electromagnetism Have there been any attempts at unifying gravity and electromagnetism at least at classical level since Hermann Weyl's idea of gauge principle (1918)? We now have Standard Model which is very successful and many other theories. But gravity and electromagnetism are long range in nature and classical as well. Can these two be unified independent of weak and strong forces?
| There is no acceptable field theory of gravity, which should describe gravity as a field defined on Minkowski space like electromagnetism, and there is no acceptable theory that describes electromagnetism as a curvature of space . It will take one or the other to make a unified theory at all possible.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/462122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Momentum Space Representation of the Tight Binding Hamiltonian I am trying to represent the tight-binding Hamiltonian
\begin{equation}
\hat{H}_{TB} = \sum_{\sigma} \sum_{\alpha,\beta} \sum_{\mathbf{R}_1,\mathbf{R}_2}
t^{\alpha,\beta}_{\mathbf{R}_1,\mathbf{R}_2}
\hat{c}^{\dagger}_{\alpha,\mathbf{R}_1,\sigma}
\hat{c}_{\beta,\mathbf{R}_2,\sigma}
\label{eq:Htb}\tag{1}
\end{equation}
in the momentum space, and it is not clear this relation
\begin{equation}
\sum_{\mathbf{R}_1,\mathbf{R}_2}
e^{-i\mathbf{k}_1 \cdot \mathbf{R}_1}
e^{i\mathbf{k}_2 \cdot \mathbf{R}_2}
t_{\mathbf{R}_1,\mathbf{R}_2}^{\alpha,\beta}
=
\frac{1}{M}
\sum_{\mathbf{R}_0}
\sum_{\mathbf{R}_1,\mathbf{R}_2}
e^{-i\mathbf{k}_1 \cdot \mathbf{R}_1}
e^{i\mathbf{k}_2 \cdot \mathbf{R}_2}
t_{\mathbf{R}_1 - \mathbf{R}_0,\mathbf{R}_2 - \mathbf{R}_1 - \mathbf{R}_0}^{\alpha,\beta}
\label{eq:pass2}\tag{2}
\end{equation}
where $M$ is the number of lattice sites and the exponentials come out of the Fourier transform of the operators in the real space to those in the momentum space
\begin{equation}
\hat{c}_{n,\mathbf{R},\sigma} = \frac{1}{\sqrt{M}}
\sum_{\mathbf{k}}
e^{i\mathbf{k} \cdot \mathbf{R}}
\hat{c}_{n,\mathbf{k},\sigma}
\label{eq:c_R}\tag{3}
\end{equation}
Moreover the translational invariance of the lattice imply
\begin{equation}
t_{\mathbf{R}_1,\mathbf{R}_2}^{\alpha,\beta} =
t_{\mathbf{R}_1 - \mathbf{R}_0,\mathbf{R}_2 - \mathbf{R}_0}^{\alpha,\beta} \quad \forall \mathbf{R}_0
\label{eq:hopping_transl} \tag{4}
\end{equation}
| In (2) we can substitute
$t^{\alpha,\beta}_{\mathbf{R}_{1}-\mathbf{R}_{0},\mathbf{R}_{2}-\mathbf{R}_{0}}$. Then since the left hand side of (2) does not depend on $\mathbf{R}_{0}$, if we sum on it we have M times the same thing, so if we divide by M, we have a relation equivalent to the previous one
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/462248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Scintillator decay time=1000 nsec,does that mean dead time is really high? What I'm really confused about is, say my scintillator is really slow, and has a decay time of about 1000 nsec. Does that mean, if one neutron is being read by the electronics, for that particular 1000nsec decay period, no other neutrons can be read even if they are depositing energy into the scintillator? When they call it "slow" what does it mean exactly? That within that long decay time only one neutron is being read and that others are being ignored?
|
That within that long decay time only one neutron is being read and that others are being ignored?
It means exactly that. After one particle is detected, it takes some dead time until the detector is ready to measure another particle. Any events during that time are not recorded.
For scintillators this is typically dominated by the decay time (see e.g. here on the first page).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/462378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why wasn't the Stipa-Caproni plane efficient in its flight? The Stipa-Caproni was an experimental italian plane design. Though it has a very peculiar shape, it seems at first glance like it would have pretty good aerodynamic profile since its reference area looks rather small. However, its Wikipedia page cites:
Unfortunately, the "intubed propeller" design also induced so much aerodynamic drag that the benefits in engine efficiency were cancelled out.
What then causes all the drag on this plane?
| I guess Stipa didn't realize that a wimpy looking 2 blade prop was not going to make the air flow do what he hoped. Even modern ducted fans are less efficient than conventional aircraft propellers in cruise conditions, though they are much more efficient for generating thrust at low speeds and hence useful for hovercraft, airships, VTOL applications, etc.
Trying to get more thrust by making the duct tapered is never going to work unless the flow at the throat of the duct becomes choked (and is therefore at Mach 1) which is far beyond the capabilities of the technology Stipa was using. To do that you need an afterburner, not a propeller!
One cause of "all the extra drag" is simply the air flow over the inside surface of the whole of the duct. Air has viscosity. The frontal area of the structure is probably bigger than a conventional design as well. The engine alone (inside the duct) has a similar frontal area to the nose of a conventional plane design. The frontal area of the cockpit is also bigger, because bottom half (containing the pilot's seat etc) is not directly behind the engine and duct, but on top of it, adding more frontal area.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/462502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why do objects rebound after hitting the ground? When an object, say a shoe, falls from a height (under the influence of gravity), it rebounds after hitting the ground. For an object to move upwards, it requires a force to overcome its weight. When the shoe hits the ground some of its energy is lost and the ground pushes back with a force less than its weight, so why does it rebound, since the upward force is not large enough to overcome its weight?
| Whatever the object lands on and the object itself acts as a spring and in compression the objects store elastic potential energy which comes from the downward motion (kinetic energy) of the objects.
That elastic potential energy is then converted into kinetic energy due to the upward motion of the object which was originally falling.
In general such collisions are inelastic and so not all the kinetic energy due to the downward motion becomes the kinetic energy of upward motion.
So it is the springiness of the objects which result in the force to slow the falling object down and then to exert a force greater than the weight of the object to propel the object upwards.
Update as a result of @CortAmmon ‘s comment to show the storage of elastic potential energy.
The granddaddy of them all?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/462618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Non-renormalizeable Interaction Implies Trivial Interaction? It has been rigorously proved that the $\phi^4$ theory is trivial, i.e. is a generalized free field, in spacetime dimensions $d>4$. It is also the case that this theory is non-renormalizeable in spacetime dimensions $d>4$. Is this a general feature of renormalization?
I can rationalize this relationship as follows: If an interaction is non-renormalizeable, say the $\lambda\phi^4$ interaction, then the only way to get rid of the infinities which result from this interaction is simply to set $\lambda=0$. Therefore the only way this theory is consistent is if this interaction does not effect any of the dynamics.
Let's say the interaction is easily controlled by some coupling constant (so nothing like a nonlinear sigma model), can we make this assertion reliably that the interaction is trivial when it is non-renormalizeable?
| No, it's not that simple, although the two notions are related.
Renormalizability is a notion that is relative to a renormalization group fixed point. It is possible to have a QFT which flows from a nontrivial UV fixed point to a trivial/free fixed point. In this case it would be non-renormalizable from the point of view of the free theory. A non-unitary example is 2d Gross-Neveu with a slightly higher power of the momentum in the propagator. For a unitary example take 3d Gross-Neveu at large but finite $N$.
Also note that the asymptotic safety program is based on the idea that something similar could happen for gravity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/462921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Confusion over units in force equation? While discussing Newton's laws, our book says
Force is proportional to rate of change of momentum
so they say
F is proportional to mass * acceleration if mass is constant
So $F=kma$ where $k$ is a constant.
They then say we choose a unit of force such that it produces acceleration of $1\ \mathrm{m/s}^2$ in $1\ \mathrm{kg}$ mass so $1\ \mathrm{N}=k\cdot 1\,\mathrm{kg}\cdot 1\,\mathrm{m/s}^2$. Then they say $k=1$.
How is $k=1$? It should be $1\,\mathrm{N}/(1\,\mathrm{kg\, m/s}^2)$, which is different than just $1$. Force is always written as $F=ma$ not $F=kma$ which seems false.
This question is different as it asks the actual concept of dimensions rather than other number the question asker of other question was confused about the choice of number not of dimension.
| If you have a proportionality $F\propto m\,a$ (or $F \propto \frac{dp}{dt}$) then using a constant $k$ produces an equality $F=k\,m\,a$ or ($F= k\, \frac{dp}{dt}$).
In terms of dimensions $[F]=[k]\,[m]\,[a]\Rightarrow [F]=[k]\rm \,M\,LT^{-2}$.
The BIMP booklet on SI units (page 118) states that the name of the derived unit of force is called the newton (symbol $\rm N$) and expressed in SI base units as $\rm kg\, m \, s^{-2}$ means that $[F]=\rm M\,LT^{-2}$ and so $[k]=1$ ie a dimensionless constant.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/463036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Symmetry operations on an infinite uniform sheet of charge My book has a section on symmetry operations.
It says, (if the plane of charge is the yz plane) translation symmetry along the y-axis and z-axis implies that the electric field is constant if one translates along the y and z axes respectively. Also, due to rotational symmetry, the field is is perpendicular to the yz plane. I understand this much.
Further, it says, another symmetry can be invoked to show that the field is independent of the x co-ordinate as well (without mentioning the symmetry).
I thought about translating the plane along the x-axis but it would change the charge distribution in space and hence, is not a symmetry operation. What is the symmetry the book mentions?
| The potential of all points on a plane parallel to the sheet is same and the electric field vector at every point on such plane is same.
Now, because of the symmetry of the 3D space with respect to the infinite sheet, the field vector at $x=d$ is negative of that at $x=-d$ and is along positive and negative x-axes respectively.
Considering two cuboidal(or cylindrical, if you wish) Gaussian surfaces with two of the faces parallel to the sheet and at $x=d_1$ and $x=-d_1$ for one and at $x=d_2$ and $x=-d_1$, the charge enclosed in the surfaces is the same, so the field at $x=d_1$ and $x=d_2$ is essentially the same.
So, they might be talking about the symmetry about yz-plane
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/463154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Does theoretical physics suggest that gravity is the exchange of gravitons or deformation/bending of spacetime? Throughout my life, I have always been taught that gravity is a simple force, however now I struggle to see that being strictly true.
Hence I wanted to ask what modern theoretical physics suggests about this: is gravity the exchange of the theoretical particle graviton or rather a 'bend' in space due to the presence of matter?
I don't need a concrete answer, but rather which side the modern physics and research is leaning to.
| Exchange of particles (gravitons) is a mechanism. Bending of space is a phenomenon for which, the mechanism is not known yet. If that mechanism is known, we can likely manipulate gravity. It is simple to understand repulsion in terms of particle exchange, but I struggle to understand attraction via particle exchange even considering particle exchange in round about path. Moreover, graviton has not been discovered, and even other forces involve virtual particles. Does virtual mean imaginary, or it means something comes into existence, then disappears and on and on?
So, in my opinion - Answer is Bending of space, whatever the mechanism is.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/463327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 3,
"answer_id": 2
} |
Does a colored filter reflect their color of light? At the moment I'm somewhat confused by the concept of colored filters; common sense states that they allow only their color of light to pass through(i.e. red filter lets red light through), but, if they appear to be a specific color, wouldn't that indicate that they reflect that color?
| The filters transmit only the light that the colour of the filter shows. For example, if you had a red filter, then it would mean that only red light would get through. The filter does not reflect light. In my example, all other colours apart from red would not go through the filter. If the filter was a secondary colour like magenta, the filter would transmit magenta, red and blue light through.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/463474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
How do you measure the chemical potential? It is clear how to measure thermodynamics quantities such as temperature, pressure, energy, particle number and volume. But I have no idea how to measure chemical potential.
Could someone please provide some examples of how one could measure the chemical potential?
| It's possible to measure chemical potential directly. Chemical potential is exactly analogous to pressure, via the fundamental relation S(U, V, N).
While P is what is equalized under a moving wall, $\mu$ is what is equalized under a permeable wall.
While V is what changes to equalize P, N is what changes to equalize $\mu$.
We can measure P by placing it next to a reference system via a moveable wall and see if V increases, decreases, or stays the same. Then we say our system has smaller, greater, or equal pressure than the reference.
We can measure $\mu$ by placing it next to a reference system via a permeable wall and see if N increases, decreases, or stays the same. Then we say our system has smaller, greater, or equal chemical potential than the reference.
We will see the pressure of an ideal gas decreases linearly with N. Since the minimum N is zero, the corresponding minimal pressure of an ideal gas is that with zero N. So we then simply define zero pressure as the pressure of an empty container.
The $\mu$ of an ideal gas increases with ln(P), so it is undefined at zero P (it approaches negative infinity). So there's no clear system we should take as having zero $\mu$.
Nevertheless, we can measure the (relative) chemical potential of any system by this method. It is important that the other thermodynamic variables remain constant as we measure.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/463572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
Why Neutrino is a ghost particle? why neutrinos are called ghost particle.why it is not affected by strong magnetic field. why it does not interact with matter. why it does not interact with gravitational field? I am unable to understand it
| It is called a "ghost particle" because it can pass effortlessly through solid objects "as if it were a ghost". This is another way of saying that it interacts very little with ordinary matter.
The reason it does not interact much with ordinary matter is that it has no electric charge (and so will not be acted upon by electric or magnetic fields) and it contains no quarks (and so it will not be acted upon by the strong nuclear force). This allows it to travel almost completely unimpeded through (for example) light-years of solid lead without scattering off any of it.
For a neutrino to interact with another particle, it must strike it more or less head-on, which is very unlikely even if the beam of neutrinos is very intense and the target material is extremely dense.
Presently, it is believed that neutrinos have very small masses, which means that they actually will experience gravitational forces.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/463651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Why the entropy change is not zero in the irreversible adiabatic process? Why the entropy change is not zero in the irreversible adiabatic process?
...while it is defined as the integral of the heat added to the system over its temperature.
| The statement
$$
dS = \frac{\delta Q}{T}
$$
is only true of reversible processes. The generally true statement is
$$
dS \ge \frac{\delta Q}{T}.
$$
For an adiabatic process, $\delta Q = 0$, but that still allows for $dS > 0$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/463988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
When the voltage is increased does the speed of electrons increase or does the electron density increase? I am just a high school student trying to self study, please excuse me if this question sounds silly to you.
I know that current is a product of the speed of electrons and the electron density.When current is increased it either means that the speed of electrons has increased or it means that the number density of the flowing electrons has increased.
I also know that voltage is directly proportional to current and when voltage increases(without no change in the resistance) the current will also increase.
But my question is, when voltage increases does an increase in the speed of electrons contribute for an increase in current or does an increase in electron density contribute for it.
If it isn't that black and white, then in what proportion will each of the two components increase? Does it randomly increase?
Related question:Say the electron density of a circuit that lights a light bulb increases.When this happens what change will we see in the brightness of the light bulb?I know that when the speed of electrons increase the brightness increases but what will happen when the electron density increases?
| Increasing the voltage applied to a circuit of a given resistance will increase the current flow. That flow is defined in electrons per second past a point. So increasing the voltage increases the speed of the electron flow.
The number of electrons free to flow is a constant for a material.
For Copper that is one electron per atom.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/464109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
How much time does it take for a broken magnet to recover its poles? I understand that when you cut a magnet you end up with 2 magnets but I wonder how much time does it take to the magnetic domains to rearange and form the new pole. I know the answer may vary depending on the size of the magnet, the material, and some other variable so I'm searching for an answer as general as possible and how the variables may affect the answer.
| I believe you seem to be worried about the effect of the physical disturbances on the domain arrangement caused by the cutting process. If my assumption is right, then to return both derivatives to their former glory (being much of half of the strength of the original), I'll recommend keeping them in a relatively stronger magnetic field, making sure they are aligned for a decent amount of time. This will repair the fallout domains that has been supposedly disoriented by the cutting process.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/464256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Train and lightning bolts: why the time difference does not depend on the position of the moving person? So imagine two lightning bolts hit the ground, simultaneously to a stationary observer. There is also a person on a train traveling to the right at a constant velocity.
I know that if he started in the midpoint, he would see the light from the right first because he is shortening the distance.
I know the equation for calculating the time difference is
$t'_2-t'_1= \gamma[-\frac{V}{c^2}(x_2-x_1)]$.
However, I am very confused because the time difference only depends on the positions of the lightning strikes and the velocity of the train. But if the train was to the left of the leftmost lightning strike, the left light would reach him first which is the opposite of what happens when he is between them. I am confused because this doesn't change the equation at all. Obviously changing x' should change this equation, but I don't know why or how.
| The position of the train will affect the time difference between when the person receives the light from each of the lightning strikes. However, this is different from the time difference between the strikes in the train's frame. This is because the person can calculate how much time it took for each light to reach them from the positions of the strikes, and use that to determine the times of the strikes.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/464726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why this constant is included in the tortoise coordinate? In the Schwarzschild spacetime, the tortoise coordinate $r_\ast$ is defined by the property that
$$\dfrac{dr_\ast}{dr}=\left(1-\dfrac{2M}{r}\right)^{-1}$$
Now, we cam integrate this. Multiply by $r$ on the numerator and denominator to get
$$r_\ast = \int \dfrac{1}{1-\frac{2M}{r}}dr=\int\dfrac{r}{r-2M}dr$$
Now integrate by parts with $u =r$ and $dv = dr/(r-2M)$ we get $du = dr$ and $v = \ln(r-2M)$. Then
$$r_\ast =r\ln(r-2M)-\int \ln(r-2M)dr$$
We integrate the last term to get
$$r_\ast = r\ln(r-2M)-(r-2M)\ln(r-2M)+(r-2M)+C$$
Reorganizing yields
$$r_\ast = r+2M\ln(r-2M)-2M+C$$
We can obviously choose $C$ to cancel the $2M$.
But anyway, virtually all references, shows a different $r_\ast$. The canonical one is
$$r_\ast = r+2M\ln \dfrac{r-2M}{2M}.$$
So one has one additional $2M$ on the denominator. This is equivalent of picking
$$C=2M-2M\ln 2M$$
Now why is that? It is clear to me that one can do that, after all the initial differential equation is still satisfied, but why everyone does it? What's the point with that $2M$ on the denominator inside the $\ln$?
| Physicists really don't like to put dimensional variables inside a function like $\ln$, $\sin$, or $\exp$. By choosing $C = 2M - 2M \ln (2M)$, you can combine the logarithmic terms into the logarithm of the dimensionless ratio $(r-2M)/2M$.
Further details on why having "naked" dimensional variables inside a function like $\ln$ is a bad idea:
*
*Exponential or logarithm of a dimensionful quantity?
*What is the logarithm of a kilometer? Is it a dimensionless number?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/464952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
If sound is a longitudinal wave, why can we hear it if our ears aren't aligned with the propagation direction? If a sound wave travels to the right, then the air molecules inside only vibrate left and right, because sound is a longitudinal wave. This is only a one-dimensional motion. If our ears are oriented perpendicular to this oscillation, e.g. if they are pointing straight up, how can we hear it?
| Re. from one of your comments: "But when the air molecule from the centre keeps moving away ,won't there be a vacuum created at the centre" and also this one: "But if the sound wave is emitted for long periods, wouldn't there be a complete vacuum and the sound wave would stop"
I think part of you confusion comes from this: Even with a longitudinal wave where the particle motion is parallel to the waves propagation direction, the particles do not travel with the wave. They only move back and forth along the direction of wave propagation. So the particles are not carried along with the wave. (It is obvious that this is true for a transverse wave.)
Referring to your original question, unless sound is focused into a beam it generally propagates equally in all directions. If it is focused into a beam and you were off to one side anything you hear would be due to sidelobes which are lower in amplitude than the main lobe and could be near zero.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/465203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 9,
"answer_id": 1
} |
Observable universe radius for distant observers The radius of the observable universe is about 46 Gly. Is that figure true for all current observers in our universe? Is it true if the universe is finite or infinite, flat or curved?
| If the universe has a FLRW metric, then there is a cosmological time $t$ that all observers at rest relative to CMB or the matter in the universe will experience at the same rate. This is true regardless of the curvature and whether the universe is infinite or merely unbounded.
The radius of the observable universe (in co-moving coordinates) is calculated by integrating $$r=c \int_0^t \frac{du}{a(u)}$$ (where $a(t)$ is the scale factor) from the start to the present cosmological time. All observers with the same $t$ will agree on $r$.
The slightly conceptually tricky part is defining "all current observers". We can define current observers to mean "all observers at rest relative to the matter or CMB that see the same scale factor $a(t)=1$ as us (i.e. have the same cosmological time $t$)" and get a well-defined slice of constant $t$ across the space-time manifold. This is less arbitrary and problematic than talking about simultaneity in special relativity, where there is nothing to compare to and no real simultaneity (everybody have their own present-time slices across spacetime, all equally valid). In a homogeneous and isotropic cosmology there is a frame of reference that is shared.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/465439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Potential in open branch of a parallel circuit with grounding
If the switch is still open, what will the electric potential at Q be, i.e. negative, positive or zero?
Is there a potential difference across the grounded point and point Q, or R3, even though the branch is open? I suppose the current would be zero in that branch. Why would there be a voltage even though the branch is open?
|
The electric potential at $Q$ will be negative, why? I'll explain in a bit.
Of course there's a potential difference across $Q$ and ground point and $Q$ and $R_3$, which are different.
You're right there isn't current through the branch leading to $Q$, and there isn't "voltage" also based on $V=IR$. Voltage doesn't always means the potential difference or potential.
Electric potential is defined to be the magnitude of charge concentration in a particular region of space. Connecting a conductor to a terminal of a battery transfers the electric potential at that terminal to the tip of the conductor away from the battery terminal, thus in the study case above, the terminal of the switch (k) close to $Q$ has some fraction of the potential of the negative terminal of battery while the rest is expended and harnessed by the other branch.
Potential difference is defined as the energy expended by charges moving from a negative potential $x$ distance from positive potential region. If the switch is closed, the energy expended by the charges concentrated at $Q$ to move to the grounded point or $R_3$ is the potential difference between those points, and obviously it isn't zero.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/465593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Initial values of creation/annihilation operators I have a question about creation/annihilation operators. For example, if I have an evolution equation for annihilation operator of photon
$$ \frac{da_k}{dt} = -i \omega_k a_k$$
I obviously obtain
$$a_k(t) = a_k(0) e^{-i \omega_kt} $$
I not fully understand how to find initial value of $a_k$. Should we just find it from expression of canonical variables $P_k$ and $Q_k$ or maybe I should go to Schrodinger representation since $a_k(0)$ does not depend on time?
Or there is another way?
| $a$ is an operator. There isn't a specific value to it, and even if you do provide a certain expression in the matrix form – it won't give you much information, as the expression entirely depends on the choice of the basis.
One example would be the generalization of the standard matrix form of the oscillator's lowering operator in the energy eigenstate basis:
$$ a \left| n \right> = \sqrt{n} \left| n - 1 \right>, $$
or
$$ a=\left(\begin{array}{ccccc}
0 & 1 & 0 & 0 & \dots\\
0 & 0 & \sqrt{2} & 0 & \dots\\
0 & 0 & 0 & \sqrt{3} & \dots\\
0 & 0 & 0 & 0 & \dots\\
\dots & \dots & \dots & \dots & \dots
\end{array}\right). $$
I trust you to do the obvious QFT generalization of this.
But this explicit expression won't actually give you much. In fact, all information is already encoded in the algebra of $a_{\bf p}$ and $a^{\dagger}_{\bf p}$.
The reason is – there's the Stone-von Neumann theorem that guarantees that there's a single unique representation of the algebra on the Hilbert space. So specifying an explicit expression of $a$ is equivalent to specifying a basis on the Hilbert space.
That is almost true for the case of QFT – the caveat being that the vacuum $\left| 0 \right>$ must lie in the Hilbert space.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/465685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Work when there is more than 1 force I know that for an object with an applied force, the work done is
$$W = Fd \cos \theta.$$
I was wondering what would happen when there is another force (e.g. friction)? Is it better to say that the work done for a general case is
$$W = F_{net} d \cos\theta.$$
| To specify a work one must specify a force. E.g. the work of friction depends on the friction force, the net work depends on the net force, and so forth.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/465793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Why does friction not accelerate a wheel? It seems like a silly question because this defies common sense, but it appears that friction is supposed to accelerate a wheel (not attached to anything).
We can derive from Newton's laws that $\mathbf{F} = m\mathbf{a}$ works for an extended object just as it does for a point particle---we just need to treat the center of mass of the object as the object's position. A rolling wheel has three forces acting on it: the force due to gravity, the normal force and friction. The net force on the body is the friction---which is nonzero---and so $\mathbf{F} = m\mathbf{a}$ tells us that the center of mass of the wheel must be accelerating.
I doubt this is the correct conclusion, but why am I wrong? The argument appears to be indisputable.
|
Consider the left hand wheel: it is either stationary or rolling at constant speed, without slipping.
For rolling without slipping the condition:
$$v=\omega R\tag{1}$$
holds, where $\omega$ is the angular velocity about the CoG of the wheel and $R$ its radius.
As both $v$ and $\omega$ are either $0$ or constant, this means no net forces or torques act on the wheel.
Friction isn't needed here. The left hand scenario can be imagined for instance where a wheel moves like this on a perfectly frictionless surface. But it works only on a surface capable of providing friction.
But on the right we introduce a net force, $F_{Net}$ which will cause acceleration $a>0$.
To maintain the rolling w/o slipping condition of $v=\omega R$, a friction force $F_f$ is needed. The torque about the CoG this force causes is needed for the angular acceleration $\dot{\omega}$:
$$F_f\times R=I\frac{\mathrm{d}\omega}{\mathrm{d}t}=I\dot{\omega}\tag{2}$$
where I is the inertial moment of the wheel.
We can quantify $F_f$ as follows.
Derive $(1)$ with respect to time:
$$a=\dot{\omega}R\tag{3}$$
The acceleration $a$ respects $F_{Net}=ma$ ($m$ is the mass of the wheel) and with $(2)$ and $(3)$:
$$\frac{F_{Net}}{m}=\frac{F_f R^2}{I}$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/465925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Physical example of line charge Electric field due to an infinite line charge, sheet of charge, point charge, etc are popular problems solved in most text on Gauss's law of electromagnetism.
My question: does an (exact or approximate) example of "infinite/finite line of charge" exist in the physical world?
While we find application of sheet of charge (though finite, not infinite) in case of capacitors, and i can imagine the physical presence of point charge and spherical charge, etc, but a line of charge with uniformly distributed charge density, which basically means a thin conductor with charge Q per unit length - can we have such a thing?
As i understand:
*
*If we connect a battery to a straight wire, with circuit closed, we get a current, but still, any section of the wire is charge-less. So, not an example of line charge.
*If we connect a battery to a wire, with circuit not closed, the charges inside the conductor will move within so as to cancel the applied electric field. So again the conductor won't have uniform charge, so not an example of line charge.
*If we connect ac voltage to a wire, we get sinusoidal charge variation along the wire, so, again not an example.
Can anyone please give a realistic example, which can come close to a line of uniform charge.
| Line charges are used in wire chambers, an apparatus used for high energy physics experiments.
There are lots of ways to make a line of charge. The easiest ones involve putting a charge on a wire. For example, make a large, thin metal ring of conducting material. Place the ring on an insulated stand. Place a positive or negative charge on the ring (perhaps with a Holtz machine). Look at the ring very close up, so that the wire appears close to straight, yet not so close that the wire's thickness starts to matter. You now have an approximate line of charge.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/466055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why do prism split light at angle instead of curving it? I assume that when light goes through matter, it doesnt really slow down, but the waveform is pushed back due to some resonance with the atoms.
EDIT: Interference is probably a better word than resonance here
I also assume that the above effect is responsible for the refraction index of materials.
But according to these assumptions, light rays should curve more as they go deeper through matter shouldnt they ? In other words that effect should be cumulative with the thickness of matter the light does through?
However light doesnt bend at different angles if it goes through thicker glass. So where did I go wrong?
| Light changes speed as it moves from one medium to another (for example, from air into the glass of the prism). This speed change causes the light to be refracted and to enter the new medium at a different angle (Huygens principle). The degree of bending of the light's path depends on the angle that the incident beam of light makes with the surface, and on the ratio between the refractive indices of the two media (Snell's law). The refractive index of many materials (such as glass) varies with the wavelength or color of the light used, a phenomenon known as dispersion. This causes light of different colors to be refracted differently and to leave the prism at different angles, creating an effect similar to a rainbow.
I believe the change in direction occurs immediately upon entering the glass which is why it does not bend more. (However I cannot prove this mathematically)
https://en.wikipedia.org/wiki/Prism
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/466202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What is meant by "collective behavior" in the definition of plasma?
"Plasmas are many-body systems, with enough mobile charged particles to cause some collective behavior ." [M.S. Murillo and J.C.Weisheit Physics Reports 302, 1-65 (1998)].
In the above definition what is meant by "collective behavior" ?
| Collective behavior in Plasmas is the phenomenon where the way the plasma as a whole reacts/behaves is dependent on the behavior of each and every particle in the plasma. The overall behavior is the sum of the individual particles' behavior.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/466309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Do scalar quantities have magnitude only? I've heard that vector quantities have both magnitude and direction but I've never heard that scalar quantities have magnitude only. Magnitude of vector quantities cannot be negative but what about scalar quantities, like temperature (-1°C)?
If scalar quantities don't have magnitude then what is their "magnitude" called?
Also does the magnitude of a vector quantity include units with the numerical value or only the numerical value?
| A vector quantity, $\vec V,$ can be written as $$\vec V=|\vec V|\ \hat V$$in which $|\vec V|$ is the magnitude of the vector, a scalar quantity which is non-negative. $\hat V$ is the unit vector in the same direction as $\vec V.$
The convention is that $|\vec V|$ is the product of a number and a unit, while $\hat V$ has no unit.
A different sort of scalar arises when we express $\vec V$ as the sum of components, say in the x, y and z directions. Using $\hat i,$ $\hat j$ and $\hat k$ for the unit vectors we can write$$\vec V=V_x \hat i + V_y \hat j+V_z \hat k$$
The scalar coefficients $V_{x},\ V_{y},\ V_z$ can be negative, zero or positive.
"I've never heard that scalar quantities have magnitude only." It is, in fact, quite a common statement in elementary textbooks. Temperature might well be given in such a book as example of a scalar. As you say, (celsius) temperature can be negative, so, clearly, 'magnitude' in this context means real number $\times$ unit, so isn't quite like the magnitude of a vector.
I suspect that temperature wouldn't be given as an example of a scalar in more advanced books, because geometry is not involved in its definition. But this is rather a subtle point.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/466426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Why is a delta resonance decay not a radioactive decay A delta resonance decays as given in http://hyperphysics.phy-astr.gsu.edu/hbase/Particles/delta.html . I wonder, why is it not a radioactive decay? In principle, most/all decays should be radioactive as it is a quite broad description:
Radioactive decay (also known as nuclear decay, radioactivity or
nuclear radiation) is the process by which an unstable atomic nucleus
loses energy (in terms of mass in its rest frame) by emitting
radiation, such as an alpha particle, beta particle with neutrino or
only a neutrino in the case of electron capture, or a gamma ray or
electron in the case of internal conversion.
https://en.wikipedia.org/wiki/Radioactive_decay
| Usually we use the word "decay" for a state that's either metastable or that decays through the electroweak interaction. States that fall into these categories will tend to be relatively long-lived.
An example similar to yours that is not usually described using the word "decay" is neutron emission. Because there is no Coulomb barrier, a state that is unbound with respect to neutron emission is not metastable. (Cf. alpha decay, which is the decay of a metastable state, hindered by having to get out through the Coulomb barrier.)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/466816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why is there so much iron? We all know where iron comes from. However, as I am reading up on supernovas, I started to wonder why there is as much iron as there is in the universe.
*
*Neither brown dwarfs nor white dwarfs deposit iron.
*Type I supernovas leave no remnant so I can see where there would be iron released.
*Type II supernovas leave either a neutron star or a black hole. As I understand it, the iron ash core collapses and the shock wave blows the rest of the star apart. Therefore no iron is released. (I know some would be made in the explosion along with all of the elements up to uranium. But would that account for all of the iron in the universe?)
*Hypernovas will deposit iron, but they seem to be really rare.
Do Type I supernovas happen so frequently that iron is this common? Or am I missing something?
| Iron comes from exploding white dwarfs and exploding massive stars(Wikipedia).
(One of many amazing images by Cmglee )
Periodic table showing the cosmogenic origin of each element. Elements from carbon up to sulfur may be made in small stars by the alpha process. Elements beyond iron are made in large stars with slow neutron capture (s-process), followed by expulsion to space in gas ejections (see planetary nebulae). Elements heavier than iron may be made in neutron star mergers or supernovae after the r-process, involving a dense burst of neutrons and rapid capture by the element.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/466889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 4,
"answer_id": 1
} |
As the universe expands, ultimately, will it continue to reach closer and closer, to absolute zero but never get there? First law of thermodynamics, the conservation of energy, doesn't this law all but guarantee that regardless of how far the universe expands it will forever contain its original amount of energy? All the matter in the universe will break down and all the energy will reach an equilibrium. However, because of the conservation of energy, the universe will never reach absolute zero will it? As it expands, the universe will continually get closer and closer to absolute zero, but it will never completely get there, or will it? How many billion years will the universe be between 1.0 degrees Kelvin and .01 degrees Kelvin. Is their a mathematical formula that describes the cooling rate of the universe towards absolute zero, once the universe has reached thermal equilibrium?
| In a dark-energy-only far future, the Universe will asymptote to the de Sitter temperature ($T_{ds}$), the minimum temperature possible in the Universe, which is NOT absolute zero (absolute zero is un-physical, the 3rd law of thermodynamics was actually quantified in 2016). In natural units:
$T_{ds}=\frac{1}{2πl} \approx 2.4\times10^{-30}[K] $
Where $l$ is the radius of the future cosmic event horizon (approx. $16.1Gly$).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/467119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Can Mars have an oxygen atmosphere? It is my understanding that, in its gaseous state, oxygen molecules move fast enough to achieve escape velocity. On Earth, we see this more clearly with helium. Regardless of what's happening on Earth, my question is about Mars.
Given the mass/gravity of Mars, would it be able to hold free oxygen (O2) in a usable amount for a meaningful amount of time?
For purposes of this question, a usable amount would be Earth-like levels. A meaningful amount of time would be "worth the expense of getting it there for that length of time."
| Ignoring escape, by whatever means, let's look at how much mass is needed to at least temporarily provide an atmosphere with a partial oxygen pressure of 0.21 atmospheres (the partial pressure of oxygen at sea level on Earth) at Mars surface. This is a simple calculation:
$$m_{\text{oxygen}} = \frac{A_\text{Mars}\,p_\text{oxygen}}{g_\text{Mars}}$$
where
*
*$A_\text{Mars}$ is Mars's surface area,
*$p_\text{oxygen}$ is the desired partial pressure of oxygen,
*$g_\text{Mars}$ is gravitational acceleration at the surface of Mars, and
*$m_\text{oxygen}$ is the quantity of oxygen needed to accomplish this.
The result is 830 exagrams ($8.3\times10^{18}$ grams) of oxygen. Another way to look at it: It's two trillion times the mass of the International Space Station. Yet another way to look it: It's the mass of oxygen in 1.9 million comets made of pure ice, each with a diameter of 1 km.
Now how are you reasonably going to accomplish even a thousandth of that?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/467277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is naturalness meaningful for non-fundamental theories? Naturalness has been a guiding philosophy for particle physics for a long time, but a few years ago I heard a talk by Nima Arkani-Hamed where he pointed out that it seems to have failed us as it relates to the Higgs boson mass and the little hierarchy problem. He suggested that naturalness as a paradigm for particles physics may simply not be useful for understanding contemporary problems. This got me wondering:
Has naturalness proven useful for any fields of physics other than particle physics/high energy?
E.g. is there anything non-trivial that naturalness can tell us about condensed matter systems?
Commentary: This may sound like a soft question, but naturalness is inherently a soft concept, yet still integral to (at least some parts of) physics.
| I can only recommend Sabine Hossenfelder's book: "Lost in Math: How Beauty Leads Physics Astray" on this topic. I think everything that needs to be said about naturalness is written down in that book. She argues that by demanding, that our fundamental theory has only natural parameters and no fine tuning, we assume to know what numbers nature is favoring.
EDIT: To better answer the question: It is questionable if naturalness is a useful concept even in fundamental physics. As every more coarse grained model of physics should be derivable from the most fundamental theories while it may lose all of the fundamental theorie's naturalness (if it exists in the first place), naturalness has nothing to say about those models. E.g. There is no reason why a condensed matter model which in principle has to be derivable from QFT should be more natural in it's parameters than QFT which already has problems with naturalness w/o SUSY.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/467491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
Lorentz contraction of the wavelength of light I couldn't find this question on the suggested "similar questions". If this has been asked before please direct me to answer. My question is "why isn't the wavelength of light,which is in the direction of motion, going at the speed of light Lorentz contracted to zero instead of its value?"
| I mean basically, it's because light is weird. Light is simply not intelligible as a classical thing which is moving and keeps an internal clock that is oscillating in time and has a wavelength by virtue of having some spatial extent. There is no classical thing that a photon easily corresponds to. Our best guess about what a photon "sees" involves roughly speaking the entire rest of the universe being "ahead" of it except for the event which emitted it, which is "behind" it, in some sort of one-dimensional timeless existence.
Quantum mechanics specifies that energies become frequencies and momentums become wavelengths, and the photon certainly has both of these, $E=p~c$, but we can change how these properties look together by simply moving relative to the photon and thus inducing a relativistic Doppler effect—so neither one is deeply intrinsic. Those features of wavelength and frequency correspond to something about how we interact with the photon, not to anything intrinsic about the photon itself.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/467645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Can electromagnetic waves apply forces on matter? Basically I want to know whether non-mechanical waves apply forces on matter?
| An electromagnetic (EM) wave is a particular realization of an electromagnetic field. Just like other electric fields, the field in the wave can apply forces to charged particles.
This is partially how radio receivers work. In a conductor like a radio antenna, the atomic nuclei and inner electrons are fixed in place, but some of the outer electrons in the atoms are free to move around. As the wave passes by the antenna, the electric field applies forces, moving the free electrons around. The moving electrons are a time varying current that follows the time variations in the electric field due to the wave.
Electromagnetic waves can also apply forces in the direction they travel, since they have momentum. This is called radiation pressure.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/468368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can quantum deletion error-correcting codes be constructed? I'm wondering whether or not we can construct quantum deletion error-correcting codes. The quantum deletion error is defined by the partial trace. If we can, could anyone give an example?
| It is well known that quantum error correction can also correct erasure errors. In fact, a code that can correct $k$ general errors (in arbitrary locations) can correct erasure errors in $2k$ locations. Thus, any quantum error correction code serves as an example.
To learn more about that, you could e.g. consult Preskill's lecture notes or the book by Nielsen and Chuang.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/468518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Effect of earth's rotation in ballistics For this purpose, let's consider earth's rotations constant. Do earth rotation momentum get transfered to any object (a missile for example) that get's lauched? If so, why do we have to consider earth rotation when lauching the missiles? Wouldn't just follow earth rotation? (Btw, sorry for any grammar mistakes, I'm from a non-english speaking country).
| When an object is launched, it initially shares the earth's rotation. That is one reason why most spacecraft launching sites are situated fairly close to the equator: it gives the spacecraft a free initial velocity of 1600 km/h (at the equator).
The atmosphere also shares the earth's rotation. If it didn't, the equator would be subject to a 1600 km/h wind. So, while the object is in flight in the atmosphere, it keeps being affected by earth's rotation, and keeps the initial momentum. As a result, since its rotational momentum does not change, its course is effectively unaffected by the earths rotation. That is, if we ignore secondary effects like the Coriolis force, which will cause a deviation from the expected path.
However, once the missile leaves the atmosphere, that influence is no longer felt. Hence, when a spacecraft is in orbit, it is completely independent of the earth's rotation. The earth could suddenly stop, and the craft would continue on its path as if nothing had happened.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/468682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Feynman diagram for semileptonic decay of neutral kaon I am unsure how to draw a feynman diagram for a reaction that occurs as follows
$$ K^0 --> l^+\nu_l\pi^- $$
Any tips would be helpful.
| Drawing the most appropriate Feynman diagram can be a little tricky sometimes. I find it best to work backwards with the general rule of thumb being to try to minimize the number of vertices. We know that the lepton pair $l^{+}, \nu_{l}$ must come from a $W^{+}$ boson as that's the only mediator that conserves charge and can violate flavor. Since we are dealing with the $K^{0}$ meson whose quark configuration is $d \bar{s}$ and a daughter meson $\pi^{-}$ whose configuration is $d \bar{u}$, we notice that both the $K^{0}$ and $\pi^{-}$ contain a $d$-quark. This $d$-quark will act as a spectator quark (will remain unchanged in the scattering process). Now all that's left is recognizing that $W$ bosons connect the positively charged quarks to the negative and vice verse with the exception of not connecting particles to antiparticles. So for an example a $q^{+} \longrightarrow q^{-} + W^{+}$ where the positive superscript is merely to denote a positive charge and the negative denotes negative charge.
Hopefully that's sufficient information to guide you to the proper Feynman Diagram.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/468830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
inertia tensor of rigid body in generalized coordinate frame? Assuming we know the inertial tensor of a homogeneous rigid body about a coodinate frame at its COM and aligned to it principal axes, how do we find the inertial tensor for the body in some other general coordinate frame which has a linear transformation (4x4) (which accounts for both rotation and translation) T from the principal C.F at the COM ?
| The general 4×4 transformation matrix has a structure where the top left 3×3 submatrix is the rotation + scaling factors. If there is no scaling, then extract the this matrix $ \mathrm{R}$ from $$\text{transform}= \left| \matrix{ \mathrm{R} & \vec{t} \\ \vec{0}^\intercal & 1 } \right| $$
Then you do the standard
$$ \mathrm{I}_{\rm world} = \mathrm{R}\,\mathrm{I}_{\rm body} \mathrm{R}^\intercal $$
This assumes the transformation is defined as local -> world sense.
Now if you want to include the parallel axis theorem to move the MMMOI definition to a new point them you use the following rule
$$ \mathrm{I}_{\rm world} = \mathrm{R}\,\mathrm{I}_{\rm body} \mathrm{R}^\intercal - m [\vec{t}\times] [\vec{t}\times]$$
where $[\vec{t}\times]$ is the 3×3 skew-symmetric cross product operator:
$$\pmatrix{x\\y\\z} \times = \left[ \matrix{0 & -z & y\\ z & 0 & -x \\ -y & x & 0} \right] $$
such that $\vec{a} \times \vec{b}$ becomes the vector/matrix product $[\vec{a} \times] \vec{b}$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/468941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why doesn't a table tennis ball float on a surface of steel balls? How do we calculate buoyancy here? Place the beaker full of steel balls and submerge the table tennis ball under the steel balls. The table tennis ball does not float up. Why does it not float up? Do table tennis balls float when the diameter of steel balls is reduced? How to calculate the buoyancy of steel balls?
Would it come up without friction?
| The ball bearings are behaving as a solid because the forces between the steel balls (i.e. friction) are large enough to hold the balls in position relative to each other.
If you apply enough force to a solid you will cause it to fracture or to cause plastic flow. So for example if you attached a string to the ball and pulled upwards with enough force it would cause the steel balls to flow over each other and the table tennis ball would move up. The force required is related to the yield stress of the solid formed by the steel balls.
You can make the steel balls behave as a fluid by making a gas flow through them. This creates a fluidised bed. The gas pushes the steel balls apart so the friction between them is removed, and in this state the steel balls will behave like a fluid and the table tennis ball would float upwards.
Alternatively just shake the beaker. This is equivalent to adding thermal energy i.e. heating the system until it melts. If you shake the beaker you'll find the table tennis ball floats upwards.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/469105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 0
} |
Is there a physical difference between the colors of quarks? Is there any physical difference to the way, say, red quarks behave compared to green or blue ones? Or is it just an intrinsic property that they have that doesn't provide any physical difference other than that it allows two of the same to be in the same hadron?
| Flavors are to designate the weak interactions of the quarks.
EM charge is to designate the EM interactions of the quarks.
Colors are to designate the strong interactions. The color with the strong force is always attractive, but it can come in neutralized combinations so that you can get stable bound quarks.
Your question is whether colors are arbitrarily assigned. Yes, they are. There is no physical difference. please see this from wikipedia and another question:
What's the difference between Quark Colors and Quark Flavours?
Just as the laws of physics are independent of which directions in space are designated x, y, and z, and remain unchanged if the coordinate axes are rotated to a new orientation, the physics of quantum chromodynamics is independent of which directions in three-dimensional color space are identified as blue, red, and green. SU(3)c color transformations correspond to "rotations" in color space (which, mathematically speaking, is a complex space). Every quark flavor f, each with subtypes fB, fG, fR corresponding to the , forms a triplet: a three-componentquantum field which transforms under the fundamental representation of SU(3)c.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/469241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Minimum Angular Velocity
A bead is free to slide on a vertical circular frame of radius $R$ comes to equilibrium when $\cosθ = g/Rω²$.
The minimum value of angular velocity comes out to be $\sqrt{g/R}$, which we can find out by balancing Gravitational and centripetal force with Normal reaction to bead from the frame.
Why can't the angular velocity have values between 0 and $\sqrt{g/R}$?
| The angular velocity can definitely have values in that range. It's just that you then lose that equilibrium position you reference.
If $\omega<\sqrt{g/R}$ then you only have two equilibrium positions. One at $\theta=0$ (the bottom of the ring) and the other at $\theta=\pi$ (the top of the ring).
If $\omega\geq\sqrt{g/R}$ then you gain a third equilibrium that you mention where the relation holds of $\cos\theta=gR/\omega$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/469344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Absence of phase transitions in quantum 1D systems at positive temperature While it is generally said that there are no phase transitions in classical lattice systems in one spatial dimension, there are also exceptions to this rule. Rigorous proofs involve some fairly strong assumptions about the statistical weights, such as positivity. I wonder what the situation is for quantum spin chains at positive temperature. That is, under what assumptions can one prove that there are no phase transitions at positive temperature?
| For translationally-invariant finite-range lattice 1d Hamiltonians, the absence of phase transitions at positive temperatures has been proved by Araki:
H. Araki, "Gibbs states of a one-dimensional quantum lattice," Comm. Math. Phys. 14 (1969), 120-157.
Later he gave a different (and less computation-heavy) proof which also covers a wider class of Hamiltonians:
H. Araki, "On uniqueness of KMS states of one-dimensional quantum lattice systems", Comm. Math. Phys. 44 (1975), 1-7.
In particular, the 2nd proof covers the case without translation-invariance.
On the other hand, the 1d Ising model with long-range interactions is a famous counter-example. See two papers by F. Dyson and references therein:
F. Dyson, Comm. Math. Phys. 12 (1969), 91-107; Comm. Math. Phys. 21 (1971), 269-283.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/469640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
How do I Fit a Resonance Curve with Respect to Known Data? In an experiment, I collected data points $ (ω,υ(ω))$ that are theoretically modelled by the equation:
$$ υ(ω)=\frac{\omega \, C}{\sqrt{(\omega^2-\omega_0^2)^2+γ^2 \omega^2}} \,.$$
How can I fit the data to the above correlation? And how can I extract $\gamma$ through this process?
| Don't try using any general-purpose curve fitting algorithm for this.
The form of your function looks like a frequency response function, with the two unknown parameters $\omega_0$ and $\gamma$ - i.e. the resonant frequency, and the damping parameter. The function you specified omits an important feature if this is measured data, namely the relative phase between the "force" driving the oscillation and the response.
If you didn't measure the phase at each frequency, repeat the experiment, because that is critical information.
When you have the amplitude and phase data, there are curve fitting techniques devised specifically for this problem of "system identification" in experimental modal analysis. A simple one is the so-called "circle fitting" method. If you make a Nyquist plot of your measured data (i.e. plot imaginary part of the response against the real part), the section of the curve near the resonance is a circle, and you can fit a circle to the measured data and find the parameters from it.
In practice, a simplistic approach assuming the system only has one resonance often doesn't work well, because the response of a real system near resonance also includes the off-resonance response to all the other vibration modes. If the resonant frequencies are well separated and lightly damped, it is possible to correct for this while fitting "one mode at a time". If this is not the case, you need methods that can identify several resonances simultaneously from one response function.
Rather than re-invent the wheel, use existing code. The signal processing toolbox in MATLAB would be a good starting point - for example https://uk.mathworks.com/help/signal/ref/modalfit.html
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/469754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
If the water is not viscous, will the board rise? When no water passes, the board hangs downward; when water passes, the board rises. Why is that? Someone told me this is because water has viscosity or tension. Is this explanation correct? I found that even if the experiment was done under the water, the plate would still rise, so this phenomenon should not be related to the tension of the water. So, is this phenomenon related to the viscosity of water?
| I think the main reasons are adhesive forces of water and board, pressure force due to flowing water has less pressure than atmospheric pressure. This force is even responsible for airplane's take off, peeling away of rooftops at the time of hurricanes.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/470011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Is it possible to trigger a nuclear reaction with physical force? In Mission Impossible Fallout, they're dealing with 3 plutonium cores. If one of those cores was thrown against a wall by Ethan Hunt, could it start a chain reaction and explode? For that matter, could any blunt force cause the nuclear reaction to initiate or would it require a radio active trigger?
| Blunt force over a limited area of fissionable material does not cause a nuclear explosion. Over the years there have been two main ways to detonate a fission bomb, neither of which requires a radioactive trigger.
The first way is to bring two sub-critical masses together to form a single critical mass, and that's all it takes to get a nuclear explosion.
The other way is to pack high explosives around a sub-critical mass of fissionable material and design it so that it burns evenly so that it compresses the material to a critical mass.
There are techniques of a radioactive nature that can improve the yield of the explosion, but they are not required for detonation.
One of the ways used to abort a nuclear launch from a nuclear detonation is to detonate the high explosive at one point which will of course cause blunt force over a portion of the nuclear material, but it will blow up the warhead without causing a nuclear explosion.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/470192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
What is the difference between linear and plane polarization? What is the difference between linear and plane polarization?
| I don't think there's any difference. Linear polarization is the more modern and, in my view, better term.
The idea is that in a linearly polarised wave, if we represent the displacement vector at a point in the path of the wave by an arrow, then the arrow tip oscillates back and forth along a straight line. [For circularly polarised light, the tip of the displacement arrow goes round and round a circular path.]
The term 'Plane polarised' probably comes from a linearly (!) polarised transverse wave in a rope viewed along the rope, so that the displacements all lie in one plane (a plane containing the rope itself). This is fine, but we don't call a circularly polarised wave 'cylindrically polarised' or 'helically polarised' do we?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/470345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How does $r$ depend on $\varphi$ in the Schwarzschild metric? I am confused about the Wikipedia derivation of the equation
for geodesic motion in the Schwarzschild spacetime. The derivation of this equation involves variation with respect to the longitude $\varphi$ only then the variation with respect to the time $t$ only.
My question is how can we vary with respect to $\varphi$ only when $r$ clearly depends on $\varphi$ since $dr/d\varphi\neq0$ so a variation of $\varphi$ leads to a variation of $r$ also?
| $\let\lam=\lambda \let\th=\vartheta \let\phi=\varphi
\def\cS{{\cal S}} \def\D#1#2{{d#1 \over d#2}}$
Your question can be answered from a more general viewpoint, without thinking of Schwarzschild metric and its geodesics.
You have a 4D spacetime $\cS$ with coordinates $t$, $r$, $\th$, $\phi$. A curve in $\cS$ is meant as a mapping $\Bbb R\to\cS$ represented by parametric equations
$$t = t(\lam) \qquad r = r(\lam) \qquad \th = \th(\lam) \qquad \phi = \phi(\lam) \tag1$$
$\lam$ being a real parameter you may choose at will.
If the curve is the worldline of a moving body $\lam$ is usually identified with proper time $\tau$ but this isn't mandatory. Any other parameter can be used if it's in a one-to-one correspondence with the original one.
So you may e.g. assume as a parameter $\phi$ if in the motion it's an increasing function of $\tau$. If ODE's (with independent variable $\tau$) are given for functions (1) you'll use the chain rule to transform them into ODE's with independent variable $\phi$. E.g.
$$\D r\tau = \D r\phi\,\D\phi\tau$$
wherefrom
$$\D r\phi = {dr/d\tau \over d\phi/d\tau}.$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/470585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Black hole photography I recently read an article that said that the event horizon telescope took a picture of a black hole (Sagitarius A*) and it will be presented on the 10th of April. I was wondering how this picture is taken and what a picture of a black hole even means?
The only explanation I could think of was that the telescope took pictures of the stars surrounding the black hole and then a sudden absence of stars pointed that a black hole is present.
Can anyone explain how black hole photos are taken or how does event horizon telescope work?
| The project is using a network of radio telescopes to measure radio waves emitted by ionized matter in the accretion disk around the black hole, and by ionized matter in relativistic jets that are being ejected along the rotation axis.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/470730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If atmospheric pressure is 76 cm of $\text{Hg}$ , why won't 76 cm of mercury stay in an open tube when suspended in air? If we keep an hold a tube in air with the closed end up and open end downwards, containing mercury upto a length of 76 cm, why does the mercury not stay in place? Shouldn't atmospheric pressure exert a force equal and opposite to its weight and balance it?
| In an open tube there is atmosphere pressure acting on mercury from top end of tube.
In a closed tube,there is no air trapped above mercury. In fact there is vacuum.
In the construction of mercury barometer, for instance, we take a tube filled with mercury and carefully invert it into a cup filled with mercury. The pressure on mercury surface inside the tube and outside at same height should balance. Outside there is atmospheric pressure. Inside the pressure is only due to length of mercury column. So if the initial height of mercury column is > 76cm, mercury level drops in the tube to adjust to 76cm such that pressure exerted is same as atmosphere(think what would happen if initial tube length is <76cm). In case of open tube, the atmospheric pressure acts on the surface of mercury outside the tube. But inside tube also the atmospheric pressure acts from above. The mercury column would cause to increase the pressure and cause imbalance. So there would be no rise in case of open tube
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/470858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Quantum energy levels of a point mass rotating about a fixed point The question is: A particle of mass m is attached to a fixed point in space by a massless rigid rod of length a and can freely rotate about this point. Find the quantum energy levels of the system. What is the degeneracy of each energy level?
I used rotational kinetic energy:
$E=\frac{1}{2}I\omega^2=\frac{L^2}{2I}$
and then substituted $I=ma^2$ and $L=\hbar \sqrt{l(l+1)}$ to get:
$E_l=\frac{\hbar^2l(l+1)}{2ma^2}$.
So the energies are quantized as expected. But what is the degeneracy of each level? Plugging in a bunch of values for $l$ doesn't show any $l$s with similar energy so far. Is it correct that the degeneracy of each level is $0$?
| For each $l$ there exists $2l+1$ possible values of $m$. Since $m$ must be an integer, and $-l\leq m\leq l$, expanding out the associated Legendre function:
$P_l^m(x)\equiv (-1)^m(1-x^2)^{m/2}(\frac{d}{dx})^mP_l(x)$
where $P_l(x)$ is the $l$th Legendre polynomial in $x$, will show that there is $2l+1$ degeneracies. The solutions to the theta dependence of the angular equation due to separation of variables of the spherical Schrodinger equation are the Legendre polynomials in $\cos(\theta)$.
For example:
$P_0^0=1$,
$P_1^1=-\sin(\theta)$,
$P_1^0=\cos(\theta)$, etc...
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/471357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why is Earth's gravitational acceleration $9.8 \frac{m}{s^2}$? How was the value of $g$ determined as 9.8 $\frac{m}{s^2}$?
I am not requesting the derivation but the factors/parameters that influence this value.
| 9.8 m/s^2 is not 'the Earth's gravity', it's the at-mean-sea-level acceleration
due to Earth's gravity. The acceleration would be quite different
if measured elsewhere, like at the lunar orbit.
The effective gravity constant also varies due to local mineral density,
and latitude.
Earth's gravity, in the universal sense, is entirely characterized
by the mass of the planet, roughly 5.97 *10^(24) kg,
To calculate acceleration, multiply that by the universal gravity constant G and divide by the
square of the distance from the center of the
planet. Only if you pick Earth's radius does that give the 9.8 m/s^2 value.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/471496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Potential by Assembling Charges For finding electric potential energy of a uniformly charged sphere, we can assemble the sphere by brining charges from infinity to that point. So to make a uniformly charged sphere of radius $R$ and total charge $Q$, at some instant, charge will be assembled up to a certain radius $x$.
In order to find potential of this sphere at the surface, why is my approach giving different answers?
Approach 1:
$$\rho = \frac{3Q}{4 \pi R^{3}}$$
$$q = \frac{4}{3} \pi x^{3} \rho = Q \frac{x^{3}}{R^3}$$
Potential at the surface would be $$V = \frac{q}{4 \pi \epsilon_0 x} = \frac{Q x^{2}}{4 \pi \epsilon_0 R^{3}}$$
Approach 2:
$$\rho = \frac{3Q}{4 \pi R^{3}}$$
$$q = \frac{4}{3} \pi x^{3} \rho = Q \frac{x^{3}}{R^3}$$
$$E = \frac{Q x}{4 \pi \epsilon_0 R^{3}}$$ (From Gauss' Law)
Potential at the surface would be $$V = -\int{\vec{E} \cdot \vec{dx}} = -\frac{Q}{4 \pi \epsilon_0 R^{3}} \int_{0}^{x}{xdx} = -\frac{Q x^{2}}{8 \pi \epsilon_0 R^{3}}$$
Why is the answer different in both the cases?
| Two cases described are completely different. In first case you find the true potential of the sphere by taking the charge from infinity to the surface of the sphere. In another case you take the charge from the middle of the sphere or the centre of the sphere to the surface of the sphere which is not the potential of the sphere surface. The potential of the sphere surface can be described as the work needed to push a positive charge from infinity to a to the surface or the energy stored to push the charge from the the surface towards the infinity so you can see in your second case you are not calculating the potential of the surface of the sphere. SHORT NOTE:- You can find the potential at any point by finding the difference of potential at that point and any other point whose the potential is zero now at the centre of the the sphere you don't have the potential as 0. See this:http://physics.bu.edu/~duffy/semester2/d06_potential_spheres.html
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/471655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Commutator of spacetime translation In Srednicki's textbook Quantum Field Theory, eq. (95.7) reads:
\begin{equation}
[\Phi (x, \theta, \theta^{*}), P^{\mu}] = -i\partial^{\mu}\Phi (x, \theta, \theta^{*}).
\end{equation}
where $\Phi (x, \theta, \theta^{*})$ is a superfield and $P^{\mu}$ is a generator of the Poincare group.
I don't quite understand this equation. To my understanding, the left-hand side is a commutator, so it should be
\begin{equation}
[\Phi (x, \theta, \theta^{*}), P^{\mu}] = i\Phi (x, \theta, \theta^{*})\partial^{\mu} -i\partial^{\mu}\Phi (x, \theta, \theta^{*}).
\end{equation}
Why is the first term on the right-hand side missing?
| Remember! Those are operators, and as such they act on some state $\varphi$. The second term, when operating on this $\varphi$, includes in fact two components
$$P_{\mu}(\Phi\varphi)=(P_{\mu}\Phi)\varphi+\Phi(P_{\mu}\varphi)$$
Put this back into the commutator to get your answer.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/471812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to deal with motion on a 2-D lattice in terms of dimension? I am reading a paper titled: Random walks of molecular motors arising from diffusional encounters with immobilized filaments. There the authors consider the molecular motor moving on a 1-D protein filament in a 2-D unbounded media. They have used the framework of 2-D lattice for there analysis. The probability of jumping from the filament to the unbounded medium is $\epsilon/2$, the probability of retaining on a certain location on filament $\gamma$, the probability of moving backward $0.5\delta$ and then the authors say that the average velocity of moving on the filament is $v=1-\gamma-\delta-0.5\epsilon$ I have two questions regarding this argument:
Suppose I have a point $d$ units away from the origin then what should be the dimension of $d/v$, I mean $v$ is not in m/s. I can assume probabilities to be in $s^{-1}$ but that does not lead to $\text{dim}\{d/v\}=s$. Is there some concept of dimensionless being involved in this? How can change the parameters to get proper dimensionality?
| I think that the $v$ they means is in Lattice units/Time step.
Probabilities are dimensionless usually (not a decay probability, that is $s^{-1}$) and indicate the jump probability during an unit of "time step" in your lattice simulation. How long to take this time step, you can choose to match the real situation. In this case, how much time typically take your molecule to jump from one site to another.
Put lattice distance $l= 1$ Ångstrom, as distance between lattice points, and time step lasting $\tau = $ 1 nanosecond; you should get a dimensional speed $v_d= v*(l/\tau)$.
Working on lattice you don't care about actual dimensions, all is in simulation units, thus $d/v$ has dimension of time units, how they are called in your paper, $n$ or $t$. This is not physical time, just the number of time steps; to get physical time, you should multiply that by the corresponding physical duration of the time step $\tau$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/471910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Do working physicists consider Newtonian mechanics to be "falsified"? In the comments for the question Falsification in Math vs Science, a dispute around the question of "Have Newtonian Mechanics been falsified?"
That's a bit of a vague question, so attempting to narrow it a bit:
*
*Are any of Newton's three laws considered to be 'falsified theories' by any 'working physicists'? If so, what evidence do they have that they believe falsifies those three theories?
*If the three laws are still unfalsified, are there any other concepts that form a part of "Newtonian Mechanics" that we consider to be falsified?
| One of the problems of Newton's law of universal gravitation, $$F_\text{Grav} = G \frac{m_1m_2}{r^2},$$ is that it does not correctly describe the precession of Mercury's orbit. Mercury behaves slightly different than predicted by Newton's law and general relativity does a better job.
See also the corresponding Wikipedia article.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/472215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 6,
"answer_id": 0
} |
Magnitude of vector field I think this is more of a mathematical question, but since it's for a physics problem I decided to ask it here.
I have this complicated magnetic field in spherical coordinates $(r, \theta,\phi)$,
$$ \mathbf{B} = \left( B_r(r,\theta,\phi) , B_\theta(r,\theta,\phi) , B_\phi(r,\theta,\phi) \right) $$
And I need to compute $|\mathbf{B}|^2$. Instead of converting this to cartesian coordinates, which would be laborious and painful, and computing $|\mathbf{B}|^2$ as $B_x^2 + B_y^2 + B_z^2$, I did
$$ |\mathbf{B}|^2 = g_{ab} B^a B^b $$
Where $B^a$ are the components of the vector and $g_{ab}$ is the metric tensor in spherical coordinates,
$$ g_{ab} = \text{diag}(1,r^2, r^2 \sin^2 \theta) $$
Is this correct? Or is $|\mathbf{B}|^2$ given by
$$ |\mathbf{B}|^2 = B_r^2 + B_\theta^2 + B_\phi^2 $$
This is really confusing me.
| Depends whether the components given are in terms the coordinate vectors, or unit coordinate vectors. If it's in a GR or field theory book it's probably the former, if it's in something like Jackson or Griffiths (an EM book) it's probably the latter. What you did is right in the first case. But if the basis vectors are already normalized unit vectors, the metric is just $diag(1,1,1)$. Either way the equation in terms of $g_{ab}$ is fine, just changes what the metric is.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/472305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Expansion of an ideal gas at constant pressure I approach these expansion problems like so:
The gas and the surroundings(piston+outside) are at the same pressure at first. We heat the gas. The pressure rises inside the syringe a bit. The gas expands so the pressure remains constant. Then I use P(the constant pressure of the gas) *dV. What I want to confirm is my reasoning on using this equation. It was derived assuming P(internal) = constant. But it does change momentarily. Is the reason we ignore it in the "a bit" nature?
Also for compression, the force exerted on the gas by surroundings (piston+outside) is taken as the force the gas exerts on the piston. Is this Newton's third law?
|
As in the as soon as pressure rises because of the temp rise the volume expands to "counter it". So pressure never really changes much. So it's okay to ignore it for calculations
If the piston moves by a little bit, then the pressure is literally the same. Considering $\Delta V \to 0$ which would mean $\Delta V \approx dV$.
Also for compression, the force exerted on the gas by surroundings (piston+outside) is taken as the force the gas exerts on the piston. Is this Newton's third law?
Yes, this is true.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/472431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Can lasers cause glare or reflections? When going to the movie theater I sometimes notice reflections from the projector on my glasses and I find it very annoying. What happens is that the light hits the back of my glasses and i can see a reflection of it.
I am gonna watch a movie which is going to be projected using imax lasers. Would laser projectors produce glare of a similar nature?
How do laser lights work? Do the photons only go straight? Then how is it visible as a beam from the side sometimes?
| Having a laser projector tells you very little about how the projector actually works. The only thing to know for sure is that the light source is some kind of laser (as opposed to e.g. Xenon arc lamps). The advantage is usually that they can be made a lot brighter and with a wider color gamut (they can generate more colors).
But, the light travels no straighter than light from a regular projector. Simply because under normal circumstances (no black holes in the vicinity, no gradients of refractive index, etc.), light always travels in a straight line.
What you see in your glasses is probably stray light from imperfections in the beam path and scattered light from the dust in the air. These things will unfortunately still be there with a laser projector.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/472522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the electric potential inside a point charge? We know that electric potential of spherical charge is inverse proportion with $r$ from $V = \frac{kq}{r}$ , So if $r$ is getting less then electric potential will be higher.
But , What about point charge ? Is it infinity inside it ? My teacher told me that it's zero but I am not believing that.
| This post does not really answer your question, but still it is worth reading what I would like to highlight.
Point charge refers to electrons or protons.
But in no way does the electron seem to have a size, although it is still debatable today. So, what mathematics and our reality assumes today is that the electron is a point charge.
A point is something that has no dimensions at all. So there is no inside.
ON YOUR COMMENT:
There is no inside for a point charge, so there is no point of using $V=kQ/r$ because mathematics break down over there.
Meanwhile, for a spherical charge distribution on a conducting shell like you have drawn there, the potential anywhere inside the shell is the same as the potential on the surface - $V=kQ/R$, where $r=R$ (radius of the sphere).
Check mathematically why this is the case!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/472691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Is polarization complementary along its different axes? Is polarization complementary along its different axes -- much like the spin of a particle is -- thus implying that the uncertainty principle holds for polarization measurements on these different axes?
| The complementarity principle, formulated by Niels Bohr, states that objects have certain pairs of complementary properties which cannot all be observed or measured simultaneously. The type of measurement determines which property is shown, to be intended as the impossibility of any sharp separation between the behaviour of atomic objects and the interaction with the measuring instruments.
Complementarity means a limitation in the manifestation of the properties of a physical entity. The conclusive limitations in precision of property manifestations are quantified by the Heisenberg uncertainty principle and Planck units.
Any two incompatible observables $A$ and $B$ are subject to the uncertainty relation:
$$\langle (\Delta A)^2 \rangle \langle (\Delta B)^2 \rangle \ge \frac{1}{4} \vert \langle [A, B] \rangle \vert^2$$
where:
*
*$\langle \cdot \rangle$ expectation value for the physical state
*$\langle A \rangle$ expectation value of $A$
*$\Delta A = A - \langle A \rangle$
*$\langle (\Delta A)^2 \rangle$ dispersion of $A$
*$[ \cdot , \cdot ]$ commutator
*$\vert \cdot \vert$ absolute value
Examples of complementary properties are Position and momentum, Energy and duration, Spin on different axes, Polarization on different axes, Wave and particle features.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/472919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Are there anti virtual particles (mediator bosons)? I have read these questions:
Can bosons have anti-particles?
Is there a possibility for discovery of anti-graviton, i.e. the graviton antiparticle?
Antiparticle for Higgs boson?
According to the accepted theory, the SM, all elementary particles do have their anti party, so do bosons that mediate.
The EM force is mediated by virtual photons.
Gravity is mediated by theoretical virtual gravitons.
The strong force is mediated by virtual gluons.
The weak force is mediated by virtual W and Z bosons.
All these bosons do have anti version according to the SM, so
*
*photons are their own anti particles
*gravitons too
*gluons have their anti gluon versions
*W and Z bosons too have their anti versions
Now since these virtual bosons mediate the fundamental forces, are there anti virtual bosons?
Question:
*
*are there anti virtual bosons?
*do these mediate the anti forces?
|
are there anti virtual bosons?
As you know, a virtual particle is a mathematical construct and is connected with the real particle of its name by the quantum numbers identifying the particle.
Also, you should know that all particles can be virtual within Feynman diagrams if they are not within the incoming and outgoing legs.
In these Compton scattering diagrams it is the electron that is virtual, or the positron, when the e represents a positron gamma scattering.
So yes, antiparticles can also be mediator virtual particles in an interaction
do these mediate the anti forces?
There are no antiforces, there are attractive or repulsive end results in the interactions studied and modeled with Feynman diagrams in order to calculate the crosssections or decay rates.
The gauge bosons of the four macroscopic forces, identified with the electromagnetic/weak/strong and gravity ( if quantized) are just virtual exchanges in the lowest order diagrams in simple particle-particle interactions. It is only the quantum numbers that they carry, and their mass in the propagator which identify them. The mass is the same for Z+ or Z-, and enters in the propagator, and just the charge characterizes the virtual particle.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/473045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Field operator commutation: If two operators commute, then their fourier transforms also commute? Im doing this in the context of field operators $$\psi(x)=\sum_k a_k e^{ikx},$$
$$\psi^T(y)=\sum_k a_k^T e^{-iky},$$ and their being defined as the fourier transform of the creation/annihilation operators $a_k,a_k^T$. Specifically, i have to prove: $$[\psi(x),\psi^T(y)]_\zeta=\delta(x-y)$$
With $\zeta=+1$ for bosons and $=-1$ for fermions. We have already proved in class that $$[a_k,a_k'^T]=\delta(k-k').$$ For whatever reason Im really struggling with the first proof. Any tips? And then I was jsut wondering whether it would be simpler to prove the the commutation relations hold for all fourier transforms of operators, not just this case. But im not sure how to do that either, or whether it would actually be simpler, or even if its actually true..
| A plug in will work.
The operation is very common in quantum field theory.
Please refer to any QFT textbook for canonical quantization techniques.
$[\psi(x),\psi^{\dagger}(y)]_{\xi} = [\sum_k a_{k}e^{ikx}, \sum_k' a^{\dagger}_{k'}e^{-ik'y}]_{\xi} = \sum_{kk'} e^{ikx-ik'y} [a_k, a^{\dagger}_{k'}]_{\xi} = \sum_{kk'} e^{ikx-ik'y} \delta_{kk'} = \sum_k e^{ik(x-y)} = \delta(x-y)$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/473146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is positron creation operator times electron creation operator equal to the ground state? This is part of a larger problem, but the important part is that at one point I have:
$$
bb^\dagger+bd+d^\dagger b^\dagger + d^\dagger d + b^\dagger b +db + b^\dagger d^\dagger+d d^\dagger
$$
where $b^\dagger$ and $b$ are the creation and annihilation operators for electrons, while $d^\dagger$ and $d$ are the same for positrons.
The first step is to recognize the anti-commutators of the operators
$$
\{b,b^\dagger\}=\{d,d^\dagger\}=(2\pi)^3\delta^3(k-k')\delta_{rs}
$$
$$
\{b,b\}=\{b^\dagger,b^\dagger\}=\{d,d\}=\{d^\dagger,d^\dagger\}=0
$$
But that leaves me with:
$$
2(2\pi)^3\delta^3(k-k')\delta_{rs}+\{b,d\}+\{b^\dagger,d^\dagger\}
$$
it is this second pair of anti-communators which I have absolutely no idea how to tackle, I've checked the lecture notes but they never consider this pairing.
I know that they must be $0$ because if I just ignore them I get the result I'm trying to prove, but I can't justify at all why they should be $0$
| The proof that the anticommutator of different annihilation operators and its complex conjugate can be shown as follows (demonstrated in Srednicki's notation, i.e. in "west coast Minkowski metric", in particular $e^{ipx} = e^{i(\mathbf{p}\mathbf{x} - \omega_p t)}$ , furthermore it requires some relations of spinor algebra which are not shown here):
We start off with the general development of the fermion field operator :
$$\Psi(x) = \sum_{s=\pm}\int \frac{d^3k}{(2\pi)^3 2\omega} [b_s(\mathbf{k})u_s(\mathbf{k})e^{ikx} + d_s^\dagger(\mathbf{k})v_s(\mathbf{k})e^{-ikx}]$$
and "project out" the coefficients. But it will only shown for $b_s(\mathbf{k})$, because for $d_s(\mathbf{k})$ it is analogous. For doing so we multiply by $e^{-ipx}$ and integrate over $d^3x$:
$$\int d^3x e^{-ipx}\Psi(x) = \sum_{s=\pm}\int\frac{d^3k}{(2\pi)^3 2\omega}[b_s(\mathbf{k})u_s(\mathbf{k})(2\pi)^3\delta^3(\mathbf{k}-\mathbf{p})e^{i(-\omega_k+\omega_p)t} + d_s^\dagger(\mathbf{k})v_s(\mathbf{k})(2\pi)^3\delta^3(-\mathbf{k}-\mathbf{p})e^{i(\omega_k+\omega_p)t}]$$
In order to get the operator $b_s$ we multiply by $\overline{u}_s(\mathbf{p})\gamma^0$, use $\overline{u}_s(\mathbf{p})\gamma^0 u_{s'}(\mathbf{p})=2\omega_p \delta_{ss'}$ and $\overline{u}_s(\mathbf{p})\gamma^0 v_{s'}(-\mathbf{p})=0$. We obtain the desired opertor:
$$b_s(\mathbf{p}) = \int d^3x e^{-ipx}\overline{u}_s(\mathbf{p})\gamma^0 \Psi(x)$$
The computation for the expression for $d^\dagger_s$ is analogous, essential difference is to multiply $\Psi(x)$ at the beginning with $e^{ipx}$. The result is:
$$d^\dagger_s(\mathbf{p}) = \int d^3x e^{ipx} \overline{v}_s(\mathbf{p})\gamma^0 \Psi(x)$$
In order to evaluate the desired the anticommutator we have to take the hermitian conjugate:
$$d_s(\mathbf{p}) = \int d^3x e^{-ipx} \overline{\Psi(x)}\gamma^0 v_s(\mathbf{p}) $$
The very last step is the evaluation of the anticommutator:
$$\{b_s(\mathbf{p}),d_{s'}(\mathbf{p'})\} = \int d^3x d^3y e^{-ipx-ip'y} \overline{u}_s(\mathbf{p})\gamma^0\{\Psi(x),\overline{\Psi(y)}\}\gamma^0 v_{s'}(\mathbf{p'})=\int d^3x e^{-i(p+p')x}\overline{u}_s(\mathbf{p})\gamma^0\gamma^0\gamma^0 v_s'(\mathbf{p'})= (2\pi)^3 \delta^3(\mathbf{p}+\mathbf{p'}) \overline{u}_s(\mathbf{p})\gamma^0 v_{s'}(\mathbf{p'})=\overline{u}_s(\mathbf{p})\gamma^0 v_{s'}(-\mathbf{p})=0$$.
The second anticommutator is just the hermitian conjugate of the first one, so if the first is zero, the second is zero too. I hope that no typo slipped in.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/473305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How does light 'choose' between wave and particle behaviour? Light exhibits wave behaviour in phenomenon such as interference but particle behaviour in the photoelectric effect. How does light 'choose' where to be a wave and where to be a particle?
| The fundamental experiment showing the apparent contradiction is Young's double slit: How can particle characteristics be transmitted when there is only an interfering wave between the point A of emission and point B of absorption?
However, for photons in vacuum (moving at c) there is a simple answer: The spacetime interval between A and B is empty, it is zero! That means that both points A and B are adjacent. A and B may be represented by mass particles (electrons etc.) which are exchanging a momentum. The transmission is direct, without need of any intermediate particle.
In contrast, a spacetime interval cannot be observed by observers. If a light ray is transmitted from Sun to Earth, nobody will see that A (Sun) and B (Earth) are adjacent. Instead, they will observe a space distance of eight light minutes and a time interval of eight minutes, even if the spacetime interval is zero. In this situation, the light wave takes the role of a sort of "placeholder": Light waves are observed to propagate at c (according to the second postulate of special relativity), but this is mere observation.
In short, the particle characteristics may be transmitted without any photon because the spacetime interval is zero. The wave characteristics (including the propagation at c) are observation only.
By the way, for light propagating at a lower speed than c (e.g. light moving through a medium), we need quantum mechanics for the answer.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/473566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "60",
"answer_count": 11,
"answer_id": 2
} |
If a satellite speeds up, does that make it move farther away or closer? If a satellite is in a stable circular orbit and goes about 41% faster (escape velocity) then it leaves its host forever. I get that. However, what if it speeds up by less than 41%?
Intuitively, it would seem to make the satellite move farther away from the host and thus enter a higher (more distant) orbit.
However, according to my understanding, a stable orbit requires the satellite to move more slowly the farther away it is from the host. For example, the earth moves more slowly around than the sun than Venus because it is farther away from the sun than Venus.
So, if a satellite speeds up then the stable orbit would be closer to the host, not farther away. What am I missing here?
| An orbit of a mass $m$ in the Newton potential of an immobile (for simplicity) mass M can be characterised by two constants of the motion energy, $E=mv^2/2-GMm/r$ and angular momentum L. E fixes v as function of r and L subsequently its direction. For a circular orbit v and r are constant. The increase r for a circular orbit you first need to boost the speed. The orbit becomes elliptical and r varies in time. At the desired r you then give a second boost to set v to the value and direction belonging to a circular orbital of that r. So you increased the speed (at least) twice but the kinetic in the end decreases. The extra kinetic energy was converted into potential energy.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/473934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Lorentz Transformation: Message sent before finish line As she wins an interstellar race, Mavis has a “hooray” message sent from the back of her 300m long ship as she crosses the finish line at v=0.6c. Stanley is at the finish line and at rest relative to it. He claims the message was sent before she crossed the line.
I understand how to get the answer using the Lorentz transformation. However, I am having trouble conceptually understanding why he observes the message before she crosses the finish line.
| First note that if Mavis has received this message while she is at finishing line, it'd mean the message was sent earlier, before she cross the finishing line.
I wish you had explained it more clear than this, but do note that
while in Mavis frame the signal had to travel 300m plus the contracted distance between the front of spaceship and finishing line, in Stanley frame however, it has to travel the contracted length of 300 (calculate it if you wish) and the distance between front of spaceship and finishing line. And, if Stanley and Mavis are not at the same location while signal is emitted, then there is an asynchronously at time of sending the signal itself. So in other word you can calculate it without lorentz transformation, it would be a little harder though. Also because we don't know the distance between the front of spaceship and finishing line when the signal is emitted, you should use the fact that she has received the message at $t=0$. I didn't use math because i am not certain whether i understand your question correctly or not.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/474384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
If an electric field passing through a dielectric medium, back into the original medium, is it "back to normal"? Suppose there is an electric point charge causing an electric field E in a medium with a dielectric constant $\epsilon_1$. You can calculate the scalar potential $\phi$ at a given distance $r$, as well as the gradient field $E$.
Now imagine you insert a dielectric medium with a different $\epsilon_2$ somewhere across that distance (like a piece of glass, a plastic board, etc (as long as its not a conductor like sheet metal). I am aware there will be refraction etc., but apart from that there won't be an effect on the electric field beyond that material, right? Meaning the scalar field will be the same strength as if the material 2 was not there? The only effect would be the refraction which would cause a parallel shift of the electric field, right?
Is this some kind of natural law or so? Does this effect have a name?
| This can be compared to the setup of a parallel plate capacitor with three layers of mediums inbetween, parallel to the plates: layer one with $\epsilon _1$, layer two with $\epsilon _2$ and layer three again with $\epsilon _1$. The electric field is created by charges on the two plates.
Now if you separate this one capacitor into three, each with one layer, the electric field in each layer will remain the same. The charges on the inserted new plates will have the same absolute value but inverse signs. Therefore the electric fields in layer 1, layer 2 and layer 3 will be caused by the same amount of electric charge and differ only based on the distance between the plates and the dielectric material between them.
Hence the electric field in layer 1 and 3 are the same and the electric field is the same after passing through layer 2.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/474569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
What had Feynman meant when he told nobody understands Quantum mechanics? What do we mean by understanding Quantum mechanics? What had Feynman meant when he told nobody understands Quantum mechanics?
What do we mean by understanding Quantum mechanics?
|
What had Feynman meant when he told nobody understands Quantum mechanics ? What do we mean by understanding Quantum mechanics?
He probably meant that there is no inherent in our classical physics training, intuitive expectation of the behavior of matter in the quantum framework.
In the classical framework, we understand why the apple falls, once we know the force of gravity, and the behavior of all projectiles is easily understandable and intuitievely predictable. This is not true in the quantum mechanical framework.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/475662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
CFT in momentum space Is there a way to see the conformal symmetry in momentum space in a CFT? I mean if I can recover the conformal group in some way in momentum space.
| The generator of special conformal transformation act, schematically, as
$$
K\sim x^2\partial_x+x\partial_x.
$$
In momentum space this becomes
$$
K\sim p\partial_p^2+p\partial_p.
$$
Since $K$ is a second order differential operator, $e^{\lambda K}$ does not act in a local, geometrical way on functions.
So while it is possible to check infinitesimal conformal invariance in momentum space by acting with the second-order differential operator, there is no simple geometric meaning to conformal invariance in momentum space.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/475930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Will tsunami waves travel forever if there was no land? If there was no land for tsunami waves to collide with, can the waves travel around the globe for forever?
| To answer this, I would appeal to the general principle which we call the 2nd law of thermodynamics. One way of expressing it is that the entropy of an isolated system cannot decrease. This means that in order to keep going for ever, a wave motion would have to involve no entropy increase. But almost all processes involve some increase of entropy, and in the case of water waves this is certainly going to happen, because of viscosity and turbulence in the water. Therefore the wave will gradually dissipate its energy and eventually die down.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/476360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 4,
"answer_id": 1
} |
Difference in the direction of propagation of em wave How are kx-wt and kx+wt in terms of the direction of the wave. I have been stuck at this or an hour, still can not find a definitive answer.
| Generally, the phase of the eletromagnetic wave is going to be $\phi = \vec k \cdot \vec r \pm wt$; from this expression, we can interpret the direction of the wave. Take a fixed time, then look for direction of the $\delta \vec r$ that maximizes your $\delta \phi$. That's the direction of propagation of the wave. It will be parallel to $\vec k$.
The "sense" of propagation in the direction of $\vec k$ is determined by the sign on $wt$; If your $\delta \vec r$ has the same direction and sense of $\vec k$, then, to maintain the phase, your $\pm w\delta t$ should be negative. That means a wave going "forward" on the sense of the propagation has a phase $\phi = \vec k \cdot \vec r - wt$, and a wave going "backwards" has a phase $\phi = \vec k \cdot \vec r + wt$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/476475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How is the relative force of the fundamental forces measured? My physics textbook includes the following table:
My question is about the fourth row, where it compares the relative strengths of the fundamental interactions. How are these determined? Is the ratio of electromagnetic and gravitational simply the ratio of the force between 2 1kg point masses separated by 1m, and the force between 2 1C point charges separated by 1m? (that was the explanation my teacher gave me) If so, how can this be justified, since the C and kg are just arbitrary units?
| There is also another good answer re the strengths of the known forces in this blog
Summary:
Take two objects of some type, perhaps elementary particles, and place them a distance $r$ apart. Suppose each exerts a force $F$ on the other. Then we will say this force is weak if $F$ is much less than $ℏc\over r²$ where $\hbar$ is Planck’s reduced constant and $c$ is the speed of light.
In short, for particle physicists:
*
*a weak force has $F r²$ much less than $ℏ c$
*a strong force has $F r²$ about as big as $ℏ c$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/476564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Transformation connecting two representations - Quantum mechanics I am working on Dirac's paper The Lagrangian in Quantum Mechanics. He looks for the analogy between a classical transformation between two sets of coordinates and momenta $p_r$, $q_r$ and $P_r$, $Q_r$ ($r = 1,2,\dots n$) and the transformation in quantum theory between two representations, in which the $q$'s are diagonal to another in which the $Q$'s are.
The transformation function is $\langle q|Q\rangle$. Now he takes an operator $\alpha$ with "mixed representatives" $\langle q|\alpha|Q\rangle$, and using the completeness identity
$$\langle q|\alpha|Q\rangle = \int \langle q|\alpha|q^\prime\rangle
\mathrm{d}q^\prime \langle q^\prime|Q\rangle = \int \langle q|Q^\prime\rangle
\mathrm{d}Q^\prime \langle Q^\prime|\alpha|Q\rangle$$
He now sets $\alpha = p_r$, and says that from the first integral we find
$$\langle q|p_r|Q\rangle = -i\hbar \frac{\partial}{\partial q_r} \langle q|Q\rangle$$
I don't understand how is the momentum operator now acting on the whole transformation function. And it is clear that that is what it means, since later on the same paper he sets $\langle q|Q\rangle = \exp(iU/\hbar)$ and from this equation he gets
$$\langle q|p_r|Q\rangle = \frac{\partial U}{\partial q_r} \langle q|Q\rangle$$
Just to clarify, I don't understand how is it that the momentum gets out the bra and ket, and once it does, how is it that it is now operating on the transformation.
| Note that
$$\langle q|p_r|q'\rangle
= -i\hbar \frac{\partial}{\partial q_r}\langle q|q'\rangle
= -i\hbar \frac{\partial}{\partial q_r}\delta(q-q').$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/476678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
What should be the independent variable in a resistance experiment? I was asked a question by a student today and I have been pondering about it for a while now. In an experiment to measure resistance of a conductor, should we vary voltage across the conductor and measure the current or should we vary current flowing in the conductor and measure voltage across it?
Experimentally, which would give us better results?
I've been thinking that in either case the drop across the ammeter or voltmeter should be very little (and almost of the same order). So would it make any difference?
| In addition to what @The Photon has suggested, have you considered using a Wheatstone Bridge? Although it is an old method and digital multimeters provide a simple way to measure resistance, a Wheatstone Bridge can be used to measure very low values of resistances in the milli-ohms range. Whether or not it is useful will depend on the conductor, its gauge, and its length you are measuring.
For example, an 18 AWG copper conductor has a resistance of about 6 ohms per 1000 feet depending on the grade, or about 6 milli-ohms per foot.
The accuracy of the Wheatstone Bridge will, of course, depend on the accuracy of the 4 resistors used to balance the bridge.
Hope this helps.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/476826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the temperature of the black night sky, outside the atmosphere? The sky at night is rather black. If it were completely black, it would correspond to a black body at absolute zero. But the sky is not completely black. Is there a way to assign a temperature value to the actual black night sky?
The question is not about the temperature of the air of the atmosphere, nor that of the stars in the sky. The question is about the the temperature that corresponds to the blackness seen by, say, the Hubble telescope; the question is about the pure blackness of the night sky, between the stars, outside the atmosphere of the Earth.
Nela
| It depends on what you mean by “sky”. The Earth’s atmosphere has a large range of temperatures at various altitudes, generally getting colder and colder from the surface temperature as you go up.
If what you are actually asking is the temperature of deep space, then it is 2.725 K, the temperature of the cosmic microwave background (CMB), a remnant of the Big Bang that permeates the entire universe. Every cubic centimeter of space has about 400 microwave-frequency photons with the spectrum of a 2.725 K blackbody. This radiation from about 380,000 years after the Big Bang is now very cold (just three degrees above abdolute zero) because the universe has expanded tremendously since then.
The CMB was predicted in 1948 and detected in 1964 by scientists who didn’t know about the prediction and weren’t looking for it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/477031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Why can I throw a larger stone farther than a smaller stone? Recently I was throwing stones(don't ask me why) when I noticed that there seems to be an optimum weight of stone so that it travels the farthest.
If I generate the same amount of force each time(and assuming all other variables like air resistance, angle of projection etc to be constant) shouldn't a smaller stone be projected with a higher velocity and thus have a higher range.
You can try this yourselves. A cricket ball sized object goes farther than a small pebble (consider its size to be similar to a coin)( and also farther than a basketball size object, but that is due to the increased mass).
My first thought was that it could be air resistance but shouldn't a larger body experience more air resistance?
**(I have doubts whether this is a physics question or more of a biology question.)
| That is because a smaller stone though experiences less air resistance, it is more affected by the same than the larger stone. A larger stone on the other hand is not affected by gravity as much as the smaller stone is and also has the advantage of inertia of motion which does depend upon mass and size of object (Try standing in front of a truck moving at 110km/hr and a cricket ball moving with the same speed. Please try the latter one first as you can do the first one but only once after which you will not be able to do anything). Also even unknowingly ypu are actually exterting a greater force on the larger stone due to conditional reflexes (though negligible but still mentioned). Thus your answer is given.
P.S:- Don't tell anyone I said you to stand in front of a truck. PLEASE.
Peace.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/477188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Z and $\gamma$ bosons as mixtures of W and B: Part I When it is said that the photon is "a mixture of W and B" ($B$ being a gauge field associated with the $U(1)$ hypercharge)
I have a question on this:
*
*When speaking of "mixtures", this is meant as analogous to the quantum mechanics terminology as the linear combination of two density matrices multiplied by classical probabilities? or this is meant in some more esoteric sense?
| Physics SE has a one-question rule, so I will answer your second one.
The photon being a “mixture” of the $B$ and the $W$ means that the photon’s quantum field is a linear combination of the $B$ quantum field and one of the components of the $W$ quantum field:
$$A_\mu=B_\mu\cos{\theta_W}+W^3_\mu\sin{\theta_W}$$
The mixing parameter $\theta_W$ is known as the “Weinberg angle”.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/477283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why does it takes so much time for photons to reach surface of sun from its core? I have read somewhere that the time taken by photons to reach the surface of sun is very large as compared to time it takes to reach earth from the surface of sun . Is the presence of dense gases in sun reason for that.
| Yes, you can model the path of the photon as a random walk (https://en.wikipedia.org/wiki/Random_walk).
This means that the distance a photon reaches after a certain amount of time is given roughly by:
$$ D(t) = \lambda_\text{MFP} \sqrt{N_\text{coll}},$$
where $\lambda_\text{MFP}$ is the mean free path of the the photon, and $N_\text{coll}$ is the number of collisions in time $t$.
As $N_\text{coll} \propto \lambda_\text{MFP}$ and $\lambda_\text{MFP} \propto 1/n$, where $n$ is the particle density, diffusion length $D(t)$ will increase very slowly for something as dense as the sun.
For an order of magnitude estimate see: https://image.gsfc.nasa.gov/poetry/ask/a11354.html
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/477374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How much energy is required to perform the Saitama's moon-jump? I have no real knowledge of physics beyond the very basics of classic Newtonian mechanics, but as far as I understand, when a particle moves closer to the speed of light, its "relativistic mass" becomes greater, which means it requires more and more energy to further accelerate it.
I was watching an anime called "One Punch Man", and in this anime there is a character, with absurd strength, that physically jumps from the Moon back to Earth. Watching that scene made me curious about the relativistic effects, and requirements, of such feat.
Considering that the Moon is +/- 1 lightsecond away from Earth, and that the trip took just a few seconds (10 seconds maybe?), it's safe to assume that the character achieved "relativistic speeds", right?
So, assuming that the character weights 70 kilograms,
the trip took 10 seconds (from the characters perspective),
and the character "crashes" into earth (does not de-accelerate upon entrance):
*
*What speed did the character achieve?
*How much energy did the character "consume" to perform the jump?
*If the 10 seconds were measured from the perspective of someone on earth, how long did the trip took for the character?
| The average velocity is given by the same formula from Newtonian physics, so $v=\frac{1 ls}{10 s}=0.1 c$. The Lorentz factor corresponding to this velocity is $\gamma=\frac{1}{\sqrt{1-0.1^2}}=1.005$, so relativistic effects are not that great.
By energy "consumed," I'm guessing you mean the kinetic energy they have, which is also equal to the work done to reach that energy. The relativistic formula for this is $KE=(\gamma-1)mc^2$, so substituting in the values you gave yields $KE=3.169*10^{16} J$
For the last part, you are asking for something called the proper time, which is given by the time divided by the Lorentz factor, so $\tau=\frac{t}{\gamma}=9.95 s$, so they experience a very small amount of time dilation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/477704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Current in the inductor at $t=0$ $L_1 = 5H, L_2=0.2H, M=0.5H, R_0=10 Ω$, and $i_g=e^{-10t}-10 A$. I need to find $i_2$.
I've started with DE
$$i_2R_0+L_2(di_2/dt)+M(di_g/dt)=0$$
and solved it for $i_2$, so $$i_2=0.625e^{-10t}+Ce^{-50t}A,$$
where C is constant.
I can't find C because I don't understand how to obtain $i_2(0^+)$. Is it possible to obtain this value? Any help appreciated!
| Accidentally seen this question again and found the solution. Maybe someone needs it too. I've started with stating that initially
$$v_2=M(di_1/dt)=-5e^{-10t},$$ so $v_2 (0)=-5 V$. Besides, $$v_2(0)=L_2(di_2(0)/dt)=0.2*(-6.25-50C),$$
so C=0.375. Thus $$i_2=0.625e^{-10t}+0.375e^{-50t}A,$$
and answer in the end of the book gives the same numbers.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/477798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is throwing dice a stochastic or a deterministic process? As far as I understand it a stochastic process is a mathematically defined concept as a collection of random variables which describe outcomes of repeated events while a deterministic process is something which can be described by a set of deterministic laws. Is then playing (classical, not quantum) dices a stochastic or deterministic process? It needs random variables to be described, but it is also inherently governed by classical deterministic laws. Or can we say that throwing dices is a deterministic process which becomes a stochastic process once we use random variables to predict their outcome? It seems to me only a descriptive switch, not an ontological one. Can someone tell me how to discriminate better between the two notions?
| Look up Diaconis's work on flipping coins. While it is technically deterministic, what happens is that extremely small changes in the initial conditions flip the outcome. The same would be true of dice. When you shake them in your hand and throw, small changes would give different outcomes. What makes it seem random is that we can't control our hands well enough to reproduce exactly the same throw (although some people are able to throw dice without making them tumble).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/477910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 5,
"answer_id": 2
} |
Magnetic field at boundary of electromagnet iron core Assume you have an iron core in the interior of the solenoid. It is well known that the strength of the field should increase by a factor of several hundred inside the solenoid as a result of the iron core.
However, at the boundary between the iron core and the surrounding air, what happens to the magnetic field strength? Does it instantaneously (with respect to position) drop by a factor of several hundred, or is there a gradual drop (so that the magnetic field immediately surrounding the iron core is stronger than in the air outside)?
| Solenoid has a core in the shape of a cylinder. When magnetized, such cylinder produces its own magnetic field $\mathbf B$, orders of magnitude stronger than the external field due to electric current, but this is apparent mostly near its poles - ends of the cylinder and it is also true inside the core.
This magnetic field of the core is continuous when crossing the pole face disks, but not so when crossing the cylindrical surface of the core such as near the core center. The field just outside the core and above the core center is quite weak, as opposed to the field inside just below the surface, which is strong. So there is a jump in $\mathbf B$ when crossing the cylindrical surface.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/478340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is there a temporal difference between planets due to the sun's gravitational field? since the Sun generates a gravitational field it also generates gravitational time dilatation. Hence, time further from the Sun should pass quicker than in its proximity.
Can we, therefore, say that the time on Mercury is different from the time on Pluto?
Do space probes take into account this difference? For instance, was 'New Horizons' adjusted for the time dilatation during its trip to the Kuiper's belt?
And if there is a difference, does not this rise a paradox, for planets have been generated in the same geological era but then have different relative time? [this actually works also on Earth itself, for Wikipedia reports that there is a difference of 39 h between sea level and the top of the Everest].
| Clocks on the surface of each planet tick at different rates. It is a small effect, amounting to a moderate number of parts per billion.
I used Mathematica to calculate the time dilation on the surface of the Sun and the surface of the planets, relative to a clock far from the solar system that is stationary relative to the Sun. (Mathematica has data on lots of physical quantities.)
I took into account both gravitational time dilation and kinematic time dilation, using the formula for the time dilation factor,
$$\sqrt{1-\frac{2\phi}{c^2}-\left(1-\frac{2\phi}{c^2}\right)^{-1}\frac{v^2}{c^2}}\approx 1-\frac{\phi}{c^2}-\frac{1}{2}\frac{v^2}{c^2}.$$
Here $\phi$ is the positive-ized Newtonian gravitional potential,
$$\phi=\sum_i\frac{G M_i}{R_i},$$
and $v$ is the orbital speed.
For the Sun, I included only the gravitational potential from itself. For each planet, I included the gravitational potential from the planet and from the Sun, but not from other planets.
For calculating the orbital speed, I approximated the planetary orbits as circular, with an orbital radius equal to the average of the semimajor and semiminor axes.
Of course, all the results are very close to 1, but a bit smaller. The following table expresses the results as how much less than 1 the time dilation factor is, in parts per billion.
$$\begin{array}{ccccc}
\text{Sun} & 2122. & 2122. & 0 & 0 \\
\text{Mercury} & 38.35 & 0.1005 & 25.50 & 12.75 \\
\text{Venus} & 21.07 & 0.5972 & 13.65 & 6.823 \\
\text{Earth} & 15.50 & 0.6961 & 9.870 & 4.935 \\
\text{Mars} & 9.86 & 0.1406 & 6.478 & 3.239 \\
\text{Jupiter} & 23.01 & 20.16 & 1.897 & 0.9485 \\
\text{Saturn} & 8.80 & 7.247 & 1.0350 & 0.5175 \\
\text{Uranus} & 3.313 & 2.542 & 0.5143 & 0.2572 \\
\text{Neptune} & 3.58 & 3.089 & 0.3283 & 0.1641 \\
\end{array}$$
The first numeric column is the body's total time dilation on its surface. The other three show the breakdown into gravitational dilation due to the body's own gravity; gravitational dilation due to the gravity of other bodies (for the planets, the Sun); and kinematic dilation due to orbital motion around the Sun.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/478481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Shape of orbitals in atoms with multiple electrons I found this statement when browsing the Wikipedia article for atomic orbitals:
"Orbitals of multi-electron atoms are qualitatively similar to those of hydrogen."
Is this true? Googling around I could only found this article where in page 50 it seems to address how to obtain the wave function of atoms with multiple electrons, but I don't have the necessary background to understand if it proves the statement or not.
Please include academic sources or a brief proof if possible.
I find it surprising that adding electrons wouldn't change the shape of the orbitals substantially, but that's what is implied when I've studied chemistry.
| What is meant is that the quantum numbers of the hydrogen solutions are still relevant for multi-electron orbitals. You still have shells 1s, 2sp, 3spd, etc.
There is not a "proof". What is there is excellent agreement between quantum chemical calculation and experiment.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/478583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Can you change the wavelength of light keeping frequency constant and can you do the opposite as well? Can you change the wavelength of light keeping frequency constant and can you do the opposite as well?
I understood the basics but please don't hesitate to go deeper into the concept. Also, If you happened to have an elegant explanation please drop it here if you can.
| As mentioned, wavelength changes in different media depending on the index of refraction. Changing the frequency can be done with non-linear optical effects, notably frequency doubling and similar effects. This change in frequency has a corresponding change in wavelength however, as opposed to the change in wavelength in different(linear) media which holds frequency constant.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/478686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Why does electrical resistivity have units of $\Omega \cdot \mathrm{m}$ rather than $\Omega \cdot \mathrm{m}^3 ?$ Electrical resistivity has units of $\Omega \cdot \mathrm{m} .$ However, since resistivity can be described as the resistance of a unit cube, shouldn't the units therefore be $\Omega \cdot \mathrm{m}^3$ instead?
I ask after seeing this question to which the answer is apparently $\left(\text{D}\right) :$
$
\text{Resistivity can be described correctly as:} \\
\hspace{1em}
\begin{array}{cl}
\mathbf{A} & \text{resistance of a unit length.} \\
\mathbf{B} & \text{resistance per unit area.} \\
\mathbf{C} & \text{resistance per unit volume.} \\
\mathbf{D} & \text{resistance of a unit cube.}
\end{array}
$
| The resistance $R$ of a body grows with bigger length $l$
(a longer wire has greater resistance)
and shrinks with bigger cross-section area $A$ (a thicker wire has smaller resistance).
Hence you have
$$ R = \rho\frac{l}{A}$$
and resistivity $\rho$ must have unit $\Omega\cdot$m.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/478820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Why wavefunction becomes exponentially smaller during quantum tunneling? I am interested in quantum tunneling and I am wondering why the wavefunction of a particle would becomes smaller so that there is a slight possibility of finding it at the other side of a big energy barrier? Is there any interaction otherwise how can the wavefunction knows there is a barrier?
| The classic step barrier is the easiest to solve but the general behavior holds for other functions. The comment addresses the physics of your question, i.e. "the presence of a potential barrier is what defines an interaction". This is the same for classical mechanics. One could ask, "How does the earth know to move in an ellipse when gravity is there, is there an interaction"?
Specific details come from the math. Schrodinger's equation is set up in each interval of the x-axis (assuming a 1-dim problem) with the appropriate value of V(x) in each region. If the total energy is greater than the barrier there will be a modification of the wavelength in the region of the barrier. If the total energy of the particle is less than the barrier then the wave number will become imaginary, cancelling the imaginary factor, i, in the exponential leading to a decaying solution. If the barrier continues to exist for all space after it is encountered the decay will eventually cause the wave function to vanish, this is the phenomenon of penetration. If the barrier is finite in extent then the particle will make it out the other side, tunneling. It is fairly easy to derive the exact solution for this case since the potential is constant, they are exp(+ikx) and exp(-ikx), with appropriate boundary conditions.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/479161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
No saved queries yet
Save your SQL queries to embed, download, and access them later. Queries will appear here once saved.