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Confusion in least action in the following situation
I understand that when electric field is applied normal to the surface of a conductor the net electric field inside the conductor is zero.
But as shown in the diagram above, if electric field is not normal to the surface of a conductor, I see two possiblity of induced electric field shown in dotted lines.
What I don't understand is whether the induced electric field will align to nullify applied electric field so that it achieves least energy or will the electric field will follow least path which won't cancel the applied electric field entirely?
So both are least action situations, which will have the priority? How do we decide?
Please help me understand.
| The field is always perpendicular to the surface of a conductor.
| {
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Can I contract index in this expression? I'm reading Carrol text on general relativity, on page 96 they arrive to the term
\begin{equation}
\frac{\partial x^{\mu}}{\partial x^{\mu '}}\frac{\partial x^{\lambda}}{\partial x^{\lambda '}}\frac{\partial^2 x^{\nu '}}{\partial x^{\mu}\partial x^{\lambda}}.\tag{1}
\end{equation}
Can I contract this expression to get
\begin{equation}
\frac{\partial^2 x^{\nu '}}{\partial x^{\mu '}\partial x^{\lambda '}}~?\tag{2}
\end{equation}
I'm using the chain rule $$\frac{\partial x^{\mu}}{\partial x^{\mu '}} \frac{\partial}{\partial x^{\mu}}=\frac{\partial}{\partial x^{\mu '}}\tag{3}$$ (which I think is correct).
| No, you can't.
$$\frac{\partial x^{\mu}}{\partial x^{\mu '}}\frac{\partial x^{\lambda}}{\partial x^{\lambda '}}\frac{\partial^2 x^{\nu '}}{\partial x^{\mu}\partial x^{\lambda}}$$
can be re-written as :
$$\frac{\partial x^{\mu}}{\partial x^{\mu '}}\frac{\partial x^{\lambda}}{\partial x^{\lambda '}}\frac{\partial}{\partial x^{\mu}}\left(\frac{\partial x^{\nu '}}{\partial x^{\lambda}}\right)$$
Now it can be clearly seen that $\partial x^{\lambda}$ in last term is part of first-order partial derivative, which needs to be differentiated again against $\partial x^{\mu}$. So you can't simplify things like that, because partial derivatives differentiates a function which depends on multiple variables. And in general you need to get used to the idea that derivatives are not ratios.
| {
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How are these two versions of the conservation of angular momentum different? Here are two versions of the conservation of angular momentum.
*
*The total angular momentum is constant if there is no external moment on the system
*The total angular momentum of a particle is constant if it is only under the influence of a conservative force with the potential function $V (\mathbf x)$ invariant under rotations. (i.e. $dV/d\theta=0$ in the case of two-dimensional space)
I am perplexed about how different those statements are. Of course "no external force" does not mean all forces are conservative. Are those two versions of angular momentum conservation related, or are they just two independent, unrelated statements?
Also please point out any inaccurate statements there.
What I have noticed is that both statements imply Kepler's 2nd law.
| These two statements refer to two different entities. Statement # 1 applies to a "system" which can be a complex entity of significant spatial extent. Statement # 2 applies to a "particle" which usually means a point particle or a system that can be approximated by a point for the purposes of the problem.
| {
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Why do the tyres of the vehicles burst during summer? The explanation of most is -According to Charles' law if the temperature increases volume should also increase and hence the tires burst. But it is also that if volume increases pressure decreases (Boyle's law). So the pressure of the gases should lessen and hence tires shouldn't burst. I know there is some flaw in my logic. But I can't find it. Help!
| The volume increase of the tyre when it's hot is small, and so the pressure increase in a hot tyre is significant.
In addition, high temperatures cause rubber to weaken, which increases the chance that the tyre will pop.
By far the worst situation for a tyre is when it is not only hot, but also rolling at speed. In this condition, the flexural stresses inside the tyre are great and these stresses, plus the reduction in strength of the rubber with heat, can cause the tyre to fly to pieces.
Tyres intended for high speed use at high temperatures must be made from rubber compounds that are specifically designed to withstand heat without weakening.
| {
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Where does the Master equation for the derivation of the Fokker-Planck equation come from? I'm participating in an introductory course for biophysics. We briefly discussed the derivation of the Fokker-Planck equation and used the so-called Master equation as a starting point.
$$
\frac{\partial P(x)}{\partial t}=\int \mathrm{d}x'\left[W\left(x\vert x'\right)P\left(x'\right)-W\left(x'\vert x\right)P\left(x\right)\right]
$$
It seems like the term 'Master equation' simply describes a set of equations which describe the time evolution of a probabilistic system. But why do we use this specific Master equation for the derivation of the Fokker-Planck equation and how was it found?
| The master equation is summarized in words as follows:
The rate of change of the probability of one state occurring is equal to the rate at which transitions occur into that state minus the rate at which transitions occur out of that state.
This is one of those situations where it's useful to think of probability as a fluid: the change in the height of a fluid in a box is equal to the flow into the box minus the flow out of the box.
So, to proceed with the derivation, let's start with the definition of $W(x|x')$. The quantity $W(x|x')$ is the rate at which the system transitions to state $x$ given that it's in state $x'$. This means that the rate at which the system transitions from $x'$ to $x$ is equal to the probability $P(x')$ that it's in state $x'$ in the first place, multiplied by the transition rate $W(x|x')$ from $x'$ to $x$; in other words, the transition rate from $x'$ to $x$ is given by $W(x|x')P(x')$.
The total transition rate from any other state into $x$ is given by the sum of the transition rates for all possible transitions. For systems with a discrete set of states $S$, the total transition rate into $x$ is given by $\sum_{x'\in S}W(x|x')P(x')$. It's easy to extend this to a system with a continuous set of states: just replace the sum with an integral over the entire state space, so the total transition rate into $x$ is given by $\int W(x|x')P(x')dx'$.
Using basically the same argument, the transition rate from $x$ to $x'$ is given by $W(x'|x)P(x)$ (the probability that the system is in state $x$ in the first place, multpilied by the transition rate from $x$ to $x'$ given that the system is in state $x$). Summing/integrating over all possible transitions out of state $x$, we have that the total transition rate from $x$ to any other state is given by $\int W(x'|x)P(x)dx'$.
Now that we have the total transition rate into $x$ and the total transition rate out of $x$, to get the total change in probability density, we just subtract the two (see the fluid analogy: we're saying "change in quantity = flow in - flow out), so that
$$\frac{\partial P(x)}{\partial t}=\int W(x|x')P(x') - W(x'|x)P(x)\;dx'$$
which is precisely the master equation.
| {
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Induced Currents in Circuits I was given the following question:-
The plane figures shown are located in a uniform magnetic field directed away from the reader and diminishing. The direction of current induced in the loops is shown in the figure. Which one is the correct choice?
The correct answer is (D).
I am aware of Lenz’s Law, and I know how to find the direction of current given a changing magnetic field, and a closed circular loop. But this question is completely new to me.
Given a random circuit, and a changing magnetic field, is there a standard technique to predict the direction of current?
| Well electromagnetism says three things;
1)relative motion between a magnet and a closed loop(circuit) creates current
2)orientation of magnet moving relative to a closed loop affects the direction of current
3)how fast the magnet is moving relative to a closed loop affects the rate of creation of cirrent
| {
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Why does the intensity of the scattered light vary when a monochromatic laser is shined into a transparent object? When a laser is shined into a glass ball such as this video (link) or a bottle of water (link) the light is refracted through the whole transparent body. However, the intensity of the light scattered to the camera differs through the ball (or the bottle). For example, the green around the perimeter of the ball is significantly brighter than that in the centre. What is the mechanism behind this? Is is because even "monochromatic" lasers consists of different wavelengths and those are dispersed by the glass/ plastic jar?
Edit: changed "reflected" to "scattered"
| What you are seeing is the glass acting as a light guide by capturing light by total internal reflection. These guided modes circulate the object and either escape by scattering at a defect or leach out over time (eg cladding modes in an optical fibre escape when its bent).
The scattering has a preference for coming out tangential to the sides, which is why you see more scattered light at the edges than the centre (surface is normal). At the edge where the surface is tilted there is also a greater contribution area wise to the scattering than when you look at the centre.
| {
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Telling if Capacitors are in Parallel or Series
So C1 and C2 are in series. But if each one is placed in red boxes, they are in parallel?
It's confusing going jumping between current, capacitors, and resistance.
For me it's kinda confusing deciding which one is in parallel and which one is in series.
| There are no dumb questions, just bad teachers.
In your case, both capacitos are in series. IT doesn't matter if they are upside or in the red boxes, it's series.
How can you tell? This is a rule of thumb: you have to check if there is more than one possible path.
*
*Two elements are in series if they are one after another. There is only one path, so, if you follow the wire, you will find the first element and then the second element. No matter how far they are, if there is only one way, you cannot avoid any element, they are in series. It's like a road. If the road goes through two villages, and there is only one road, it is series.
*On the other hand, two elements are in parallel when each element is in a different branch of the circuit. This implies that there must be a bifurcation. In the road analogy, you'd need to have one village on the left path and another on the right path. (This can be generalized to whatever number of branches).
The only caution is: do not mix them. If there's one element and then the bifurcation, that's nothing. The bifurcation must be BEFORE the elements, so that each element is already in a branch. Thsi is different than an element on the main way and then a bifurcation, that's nothing.
So, in your circuit, there is only one path. No bifurcation at all, so there's no doubt it is series.
| {
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Energy Interpretation of Quantum Effective Action From Weinberg's "The Quantum Theory of Fields" In section 16.3 of Weinberg, he attempts to prove that the effective potential energy $V(\phi)$ is equal to the minimum energy density of a state with field expectation value $\phi$. I am confused about the very beginning of the argument, which is screen-shotted below:
The argument appears to be using an adiabatic approximation to show that the past and future vacuum states only differ by a phase. From the adiabatic approximation, we would more specifically say that $$|VAC,out\rangle=\exp({\color{red}{-}iE[\mathcal{J}]T})|VAC,in\rangle\tag{A}$$
Wouldn't this then imply that
$$\langle VAC,out|VAC,in\rangle_J=\exp(\color{red}{+}iE[\mathcal{J}]T)~?\tag{B}$$
And so $$W[J]=\color{red}{+}E[\mathcal{J}]T~?\tag{C}$$ If this is correct, it appears to screw up the following argument in the section.
| S-matrix theory (e.g. Weinberg's correct formulas) typically refers to the Heisenberg picture. OP is presumably thinking of the Schroedinger picture, and thereby obtaining opposite time evolution.
References:
*
*S. Weinberg, Quantum Theory of Fields, Vol. 2, 1995; Section 16.3.
*J.J. Sakurai, Modern Quantum Mechanics, 1994; Chapter 2.
| {
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If a black colored body absorbs all colors, why does the spectrum of light appear on a black shoe? Few days back in a 10 grade school practical, we were shown the dispersion of light by a prism to create spectrum.
Then we went into the open sun and performed it under a linear building roof and same results were obtained.
But when a teacher took it in the light and the spectrum was displayed on the ground cement, a child's shoe can in the spectrum's way and the spectrum was still visible on his black shoe.
Now my question is if the objects that appear black absorb the colors and radiate them back.
So why does the spectrum appears on the shoe when the colors should be absorbed by the shoe material and no color but black must appear as it is the absence of any color radiated?
And also i want to ask that do blackbodies radiate back the same color the absorb?
| Assuming his shoe wasn't completely black (i.e. darker than a sunless cave) then in general some light is reflected. For example, if you can discern the texture of the fabric, then the shoe was not completely black and you can always expect to see some light reflected, although certainly not as much as light is reflected as from a white piece of paper. Since a colored light from the prism was shining on his shoe, this is the light that you are partially seeing reflected. Indeed, some or perhaps most of that prism light was likely absorbed by the shoe, but the portion that was reflected was the excess colored light that was not absorbed.
| {
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Raychaudhuri scalar In Carroll's 'Space-time and Geometry', appendix F on congruences, the Raychaudhuri equation is derived.
However, in the process, I seem to miss a calculation step that changes the sign of the Raychaudhuri scalar.
Page 461, Carroll writes:
\begin{align} U^\sigma \nabla_\sigma B_{\mu \nu} &=U^\sigma \nabla_\nu \nabla_\sigma U_\mu + U^\sigma R^\lambda_{\, \, \, \mu \nu \sigma} U_\lambda \\
&= \nabla_\nu (U^\sigma \nabla_\sigma U_\mu) - (\nabla_\nu U^\sigma)(\nabla_\sigma U_\mu) - R_{\lambda \mu \nu \sigma}U^\sigma U^\lambda
\end{align}
I can't wrap my head around it. To me, lowering the index on the Riemann tensor and elevating the same dummy index on U would have no influence on the sign whatsoever.
Thanks for your insight!
| I think it's a typographical error. He should have changed the sign of the Riemann tensor in the last two lines of his calculation (F.10). Then to obtain the Ricci tensor with the correct sign in the Raychaudhuri equation, you must interchange $\mu$ and $\lambda$, getting the required negative sign. This is because the Ricci tensor is $R^k_{ikj}$ and thus before contracting $\mu$ and $\nu$ you must exchange the positions of $\mu$ and $\lambda$ to get the correct sign.
| {
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Creating a spatial variation of phase in a given superconducting sample Blundell's book on Magnetism, talks about the generalized rigidities as a general consequence of spontaneously broken symmetries. In this context, it mentions that in a superconductor the phase of the macroscopic wavefunction, $\phi$, is uniform across a sample. Twisting this phase across the sample produces a supercurrent proportional to $\nabla\phi$. In the first place, I wonder, is there any way to create such a spatial variation of $\phi$ in a single superconducting sample? If so, how would you achieve that in a laboratory?
Please understand that I am not interested in two different superconducting samples joined by an insulator as in Josephson junction but in spatial variation of $\phi$ in a single continuous superconducting sample.
| Yes, there is a very simple way to introduce a gradient in the phase $\nabla \phi$, all you need is to have a current running in your superconductor (i.e. a supercurrent). If you have a constant current $\mathbf{J}$, and then you have a constant phase gradient.
| {
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Shooting someone's past self using special relativity Suppose A and B are a long distance apart initially. B takes off in a spacecraft in A's direction at a really high speed. Both were aged 0 when B took off. When B about to cross paths with A, A observes him to be 30 years old (while A is 60). At this point, 60 year old A shoots the 30 year old B. B dies.
Now, from B's frame, the bullet has to be fired by A when B is 120. That's because the event of 'bullet firing' must happen after the event of 'A turning 60', and A turns 60 in B's frame only when B is 120. So B has to die at 120 as seen from his own frame of reference.
How can B die at both 30 years old and 120 years old?
| You suppose that A and B are far apart "initially" and that they were "both aged 0 when B took off". It means that you assume that simultaneity is an absolute notion, independent of the frame of reference, which is why you come to a paradox. It is exactly the symetric case of two people starting at the same point, diverging fast, and observing each other: each one will see the other one aging slower than him, because they compare moments in non parallel time sections. This is best shown by a drawing, this is the one I have suggested in this thread: link to a video at the second displaying the graph.
| {
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What is the CFT dual of the stress tensor in the bulk? I am new to AdS/CFT. I know that the dual of the bulk metric is the CFT stress tensor but what about the dual of the bulk stress tensor? I mean in principle one can extrapolate whatever bulk fields to the boundary and then compute the stress tensor's dual on the boundary, but is there a simple form of the dual or some general properties that is independent of the type of matter fields.
| A short answer is that the boundary limit of the bulk stress tensor is simply the boundary stress tensor. But we could say a bit more in addition, coming from the 1999 paper by Balasubramanian and Kraus.
As usual, the stress tensor of the theory is $T^{\mu\nu} = \frac{2}{\sqrt{-\gamma}} \frac{\delta S}{\delta \gamma_{\mu \nu}}$ where $\gamma$ is the boundary metric (which is the boundary limit of the bulk metric $g$). To relate this to the bulk theory, we'd need to use the AdS/CFT dictionary facts that $S = S_{CFT} = S_{bulk}$ and that by the dictionary, $S_{bulk}$ is sourced by fields on the boundary. The sources include bulk fields and the boundary metric $\gamma$. So we can write $S_{bulk}(\gamma_{\mu\nu},fields)$. This lets us write $T^{\mu\nu} = \frac{2}{\sqrt{-\gamma}} \frac{\delta S_{bulk}(\gamma_{\mu\nu},fields)}{\delta \gamma_{\mu \nu}}$ as a functional of the bulk action.
But this isn't the entire story. In the Kraus paper, they argue that bulk action has to be regularized by additional counterterms that also depend on the bulk metric, leading to an action $S_{eff}$ which is $S_{bulk}$ + additional counterterms. These counterterms lead to a lot of consistencies with AdS/CFT, including an accounting of the trace anomaly and the identification of the boundary central charge $c$ with the AdS length scale $l$, as $c = \frac{3 l}{2G}$
| {
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Is it possible writing conservation of relativistic energy in this naive way? Conservation of charge or rest mass can be written in this way and it is Lorentz invariant
$$
\nabla \cdot (\rho \mathbf{u}) + \frac{\partial \rho}{\partial t} = 0
$$
So we could be tempted to naively write conservation of energy in this way (I use $\gamma_u$ for particle in motion at speed $\mathbf{u}$ to not making confusion with $\gamma$ relative to speed of $S'$)
$$
\nabla \cdot (\gamma_u \rho \mathbf{u}) + \frac{\partial (\gamma_u \rho)}{\partial t} = 0
$$
But this doesn't look Lorentz invariant. I wrong? Exploiting vector identity $\nabla \cdot (\Psi \mathbf{A}) = \Psi (\nabla \cdot \mathbf{A}) + \mathbf{A} \cdot (\nabla \Psi)$ (whit $\Psi=\gamma_u $) and exploiting conservation of mass, this equation became
$$
\left(
\mathbf{u} \cdot \nabla + \frac{\partial}{\partial t} \right) \gamma_u = 0
$$
where mass is strangely disappeared. But transforming the corresponding primed equation with
$$
\frac{\partial}{\partial x'} = \gamma \left( \frac{\partial}{\partial x } + \frac{v}{c^2} \frac{\partial}{\partial t } \right)
$$
$$
\frac{\partial}{\partial y'} = \frac{\partial}{\partial y}
$$
$$
\frac{\partial}{\partial z'} = \frac{\partial}{\partial z}
$$
$$
\frac{\partial}{\partial t'} = \gamma \left( \frac{\partial}{\partial t } + v \frac{\partial}{\partial x } \right)
$$
$$
u_x' = \frac{u_x - v}{1-\frac{u_x v}{c^2}}
$$
$$
u_y' = \frac{u_y}{\gamma \left( 1-\frac{u_x v}{c^2} \right)}
$$
$$
u_z' = \frac{u_z}{\gamma \left( 1-\frac{u_x v}{c^2} \right)}
$$
$$
\gamma_{u'} = \gamma \gamma_u \left( 1 - \frac{u_x v}{c^2} \right)
$$
we get
$$
\left(
\mathbf{u} \cdot \nabla + \frac{\partial}{\partial t} \right) \left[ \gamma_u \left(1-\frac{u_x v}{c^2} \right) \right] = 0
$$
That it is different than $\left(
\mathbf{u} \cdot \nabla + \frac{\partial}{\partial t} \right) \gamma_u = 0$ written above. Another road could be exploiting
$$
\frac{\partial \gamma_u}{\partial x_i} =
\frac{\gamma_u^3}{c^2} \mathbf{u} \cdot \frac{\partial \mathbf{u}}{\partial x_i} \qquad \textrm{where $x_i=x,y,z,t$}
$$
to transform $\left(
\mathbf{u} \cdot \nabla + \frac{\partial}{\partial t} \right) \gamma_u = 0$ into
$$
\mathbf{u} \cdot \left(
\mathbf{u} \cdot \nabla + \frac{\partial}{\partial t} \right) \mathbf{u} = 0
$$
but this equation too doesn't lead to the invariance (although $\left(
\mathbf{u} \cdot \nabla + \frac{\partial}{\partial t} \right) \mathbf{u} = 0$ is actually invariant). There is a way to check the invariance, or writing conservation of energy in that simple way is incorrect?
| In general you have to consider the stress energy tensor. If you want only energy conservation (without the stress and momentum part of the tensor) you can take $\partial_\nu T^{0 \nu}=\frac{\partial}{\partial t} \omega + \nabla \vec{S}/c=0$, where $\omega$ is the energy density and $\vec{S}$ the energy flow density.
| {
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Is the highest photon energy currently possible a Planck photon or based on space-time quantinization? It makes sense that there could be an upper limit to the frequency/energy for individual photons if the universe as we know it is quantized.
But, the highest energy photons I've heard about have a frequency between $10^{20} \ \text{Hz}$ to $10^{30} \ \text{Hz}$. A Planck photon however would have around $10^{35} \ \text{Hz}$. If we assume a cosmological model wherein space-time is quantized to a fundamental metric such as
a Planck length, is there any reason to think the highest possible energy of a photon would be higher or lower than that?
| There are two issues here. Physics does not set an upper limit to photon energy, although at high energy particle antiparticle pair creation will in practice be limiting. If the universe is finite, then it has a finite energy that obviously cannot be exceeded by a photon.
| {
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Is the velocity of the spinning rod constant after it's hit? Say we've got a rod floating around in space, with two masses of $m_0$, one attached at each end. Let's say the rod has a length of $l$.
There's another mass, $m_1$, moving at some velocity $v$ towards one of the masses.
$m_1$ collides and sticks instantaneously with $m_2$. In the picture below I drew the collided masses as one big blob.
After the collision, the rod will have some angular velocity $w$, and some linear velocity $v_f$.
My question is this...will $v_f$ be constant?
For the linear velocity, I want to say:
"Well, linear momentum is conserved, so..."
$m_1v_0=(2m_0 + m_1)v_f$
$v_f=\frac{m_1v_0}{(2m_0 + m_1)}$
However, now I'm doubting how the linear velocity of the entire rod $v_f$ can be constant at all, and thinking that the situation is a lot more complicated.
This is because if it is constant, it seems to me that linear momentum isn't being conserved as the rod spins!
Consider the case when the rod is vertical, the heavier side is moving left, and the lighter side is moving right, versus the case when the rod is vertical, the heavier side is moving right, and the lighter side is moving left.
If the velocity of the center of mass of the rod is constant, then there's more net momentum when the rod is vertical and the heavy side is moving right than when the rod is vertical and the heavier side is moving left...!!!
Which would...disagree with the conservation of linear momentum?
| The shape of the object does not matter, whether it is a rod with different masses at each end, a round ball, or any other object, once set in motion, the rotation will be a constant RPM around the center of mass, and the linear motion of the center of mass will remain constant, unless acted upon by another force.
| {
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Using symmetry of Riemann tensor to vanish components The Riemann tensor is skew-symmetric in its first and last pair of indices, i.e.,
\begin{align}
R_{abcd} = -R_{abdc} = -R_{bacd}
\end{align}
Can I simply use this to say that, for example, the component $R_{0001} = 0$ because $R_{0001} = -R_{0001}$?
| Yes. All the components where the first two indices are the same, or the last two indices are the same, are zero.
Sometimes it is useful to think of this tensor as a $6\times6$ symmetric matrix where the “indices” are $01$, $02$, $03$, $12$, $13$, and $23$.
However, don’t conclude from this that there are $6+5+4+3+2+1=21$ independent components. There are actually only $20$ because of the algebraic Bianchi identity.
Note that without any of these relations between components, there would be $4^4=256$ components! So the Riemann curvature tensor is about 13 times less complicated than it might appear.
| {
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What makes north pole of a magnet north pole in the first place? This question might seem absurd and illogical to many. But it just popped out in my mind while I was reading about magnetism.
-Like in case of charges, positive and negative charge on an atom means absence and presence of extra electrons respectively. So my question is what aspect exactly makes a pole of magnet north or south?
Is it absence or presence of something?
-I asked my teacher about this and he simply replied that north pole is something which attracts south pole. But this is more of a property to me rather than an exact meaning of what exactly is north pole of a magnet.
| You are right up to a point about the charges on an atom,but you don't seem to understand what charge itself is. Charge is the source of a force vaguely similar to gravity, but trillions of times stronger. This force is called the electromagnetic force. Few people realise how powerful it is. In particles which can be observed in isolation, it is always the same as the charge on the electron or multiples thereof, and may be positive or negative. An alpha particle has a double charge because it contains two protons, which each have the same charge as an electron but of opposite sign.
To return to magnets, just as for every positively charged particle there must be a negatively charged one, for every south pole there must be a north pole. You can tell which is the north pole of your magnet by putting it against the north pole of another magnet. If it repels, you have put two north poles together, but if it attracts, you have put your south pole against the north pole of the other. It is the electromagnetic force emanating from the particles within the magnet which creates this repulsion or attraction. If it doesn't seem to be trillions of times more powerful than gravity, that's because your magnet contains equal numbers of negatively charged particles which cancel out the effect of the positive charges.
| {
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Problem while constructing langmuir probe I am working on DC glow discharges and want to construct a langmuir probe. Circuit I am using is as shown in the picture
Also, I am applying ~1KV across cathode and anode.
Problem I am facing is that I am getting discharge btw probe and cathode. Which isn't unexpected, but destroys the purpose of probe, but then how to construct langmuir probe for DC glow discharges? What should be the appropriate circuit?
| To get something useful with the langmuir probe you will need to match the potential of the probe to the local potential of the plasma - can you vary the Vp potential to get the full range of 0-1kV?? If you match the potential of the probe to the plasma potential then you should not get the extra discharge in my opinion, but I have not worked with this system before. From your diagram I wonder how far into the plasma your probe is and also how large it is. I suspect you may need to have a small probe shielded by some insulator pushed further in, but this is a guess.
I strongly suggestm that you look at this (apparently freely available) dissertation on a similar experiment.
p.s. an obvious thing to check as well is that the ammeter is not somehow pulling the potential of the probe to ground.... and that Vp is connected with the correct polarity - sorry obvious things to check that you probably have already done.
| {
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What is the change of entropy for a resistor at constant temperature? A 10 Ω resistor is held at a temperature of 300 K. A current of 5 A is passed through the resistor for 2 minutes. Ignoring changes in the source of the current, what is the change of entropy in (a) the resistor and (b) the Universe?
My attempt:
$\Delta Q=I^2 R t=5^2\times 10\times 2\times 60=30000\ J$
$\displaystyle\Delta S=\frac{\Delta Q}{T}=\frac{30000}{300}=100\ JK^{-1}$
Won't $\Delta S_\text{univ}$ assume the same value of $100\ JK^{-1}$?
| Since the resistor is kept at $T=300K$(which is wierd since its heating up), the entropy change of the resistor is
$$\Delta S_{resistor}=\frac{-\Delta Q}{T}$$
(heat flow out of the system taken -ve)
Since the surroundings stay at $T$(since no other temp is given, assuming that),
$$\Delta S_{surroundings}=\frac{\Delta Q}{T}$$
therefore
$$\Delta S_{universe}=\Delta S_{resistor}+\Delta S_{surroundings}=0$$
To be clear,you already had the answer.
Note: Joule heating is irreversible therefore $\Delta S_{universe}>0$. The reason that doesn't happen here is because the additional system keeping the resistor at surroundings' temp isn't considered here.
| {
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Friction due to pure rolling on an inclined plane When a body is executing pure rolling we know that the point of contact of the body with the ground is at rest with respect to the ground. If that's the case no friction should act as it is stationary.So when a body is rolling down an inclined plane its point of contact is stationary , then how does friction act to cause a torque, as static friction only acts when there is a tendency of relative motion with respect to the ground.
|
When a body is executing pure rolling we know that the point of contact of the body with the ground is at rest with respect to the ground.
This is true.
If that's the case no friction should act as it is stationary
This is false. Static friction is a friction force that can act on an object that is not sliding relative to the surface it is touching.
So when a body is rolling down an inclined plane its point of contact is stationary , then how does friction act to cause a torque, as static friction only acts when there is a tendency of relative motion with respect to the ground?
Gravity is attempting to accelerate the body down the incline. The static friction force opposes this. Since the friction is applied at the edge of the body and tangent to it friction has a torque about the center of the body and it starts to roll.
Contrast this with a body rolling on a flat surface. If there are no other horizontal forces then there is nothing for friction to oppose. Therefore, there is no static friction force, hence no torque. The body will continue to roll at a constant speed. However if I then apply a horizontal force, static friction now wants to oppose this. Hence we now have a torque and a change in speed (this problem, discussed here and here, is actually not trivial. You can get different magnitudes and directions of friction depending on the body and the the location and strength of the applied force if you want rolling without slipping).
| {
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Does friction always oppose motion? Recently I had the following misconceptions:
*
*Static friction always opposes the motion of body.
*The force of friction cannot initiate motion in a body.
Now I came to know that my understanding was wrong and that friction indeed can cause motion in bodies, and that static friction does not always oppose the motion of a body.
But I found this quite bizarre, how can a force which has always been taught to us to oppose motion, oppose points 1. and 2.?
| Friction opposes relative motion between two bodies.
Note that might mean that friction can create motion relative to i.e. you. For example, dropping an item on a moving belt. Friction opposes and reduces the relative motion of the item and belt until they move together. But now they’ve started moving relative to you.
| {
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What exactly does it mean to say that QED is of "universal strength and form"? In a paper from 1961 (Gauge Theories of Vector Particles, Glashow and Gell-Mann), the authors describe quantum electrodynamics thusly:
"It is of universal strength and form..."
What exactly does it mean to say that it's of "universal" strength? Is there some special meaning to that? Universal in what sense, exactly?
Source
Reference
| The answer seems to be the one suggested by Physics_Et_Al (you should really post an answer as an answer; not as a comment. That's the way this is supposed to work).
Here's a quote from Invitation to Contemporary Physics (Ho-Kim, Kumar, Lam):
"By universality, we mean that the coupling to matter is proportional
to Q [electric charge], whatever the other quantum numbers are."
This implies that there are interactions in which universality is not the case. Which I think is what W and Z are up to, according to Feynmans Out for Grumpy Cat.
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Reason behind vector addition law What is the reason behind triangle law of vector addition, in other words, how is this really justified?
| Well I finally realized that we are actually not adding two quantities but trying to find the resultant. Therefore the resultant must indeed be the third side of the triangle as it will give us the shortest distance between the two points, i.e the tail of one and the head of another. By definition the resultant is that vector which can have the very same effect as the other two vectors would provide together.
| {
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How did Einstein know the speed of light was constant? I often hear the story of how Einstein came up to the conclusion that time would slow down the faster you move, because the speed of light has to remain the same.
My question is, how did Einstein know that measuring the speed of light wouldn't be affected by the speed at which you are moving. Was this common knowledge already before Einstein published his paper on special relativity? If not, what led him to that conclusion?
| Einstein was influenced primarily by the electromagnetic theory developed by several people and culminating in the Maxwell equations, and by the experiments on the speed of light in moving water carried out by Fizeau.
At the time it was not common knowledge what light would do in general; the interpretation of the electromagnetic theory was much puzzled over. The Michelson Morley experiment was a further piece of evidence, but not a crucial one for Einstein.
He was certainly struck by the fact that Maxwell's equations suggest a physical interpretation in which light will recede from you no matter how fast you go to try and catch it up. This was not self-evident and not the only way to interpret the equations, but Einstein seems to have realised that it would be a valuable exercise to abandon the type of aether ideas being tried out at the time, and just stick to the notion that light speed is independent of the source, and then push this way of thinking through to its logical conclusions. Upon doing this, he discovered that one can still get a self-consistent set of ideas about space and time, but they are different from the more familiar (Galilean) ones.
| {
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How to measure (heat) energy used? Before I start with my question, would like to highlight - the last time I delved into this topic was 13 yrs back during my schooling courses. But the enthusiasm never died.
Coming back to topic - I want to know/understand if there's a way to know the amount of energy (heat energy, i guess!) used to boil an egg?
Ex. if suppose I start boiling an egg in saucepan filled with water, how much heat energy is used in that process. Say, 1.3 Joule of energy.
Also, is there any instrument in market to calculate that?
Please advise.
Thanks in advance.
| As Bob D has commented, this is very difficult to calculate theoretically. He didn't mention that you need to know the thermal conductivity of the egg's contents – and to make things worse, this conductivity will change as the white and yolk change consistency.
Boiling an egg in a saucepan is, of course, a very wasteful process. Most of the heat supplied goes to making steam or is lost to the surrounding air from the sides of the saucepan.
If you want to know how much heat goes into boiling the egg itself, I think you're better off trying to measure it. A 'steam calorimetry' approach (good nineteenth century Physics) should work... Suspend the egg in a jet of steam at 100 °C (the colourless stuff, not the cloudy stuff that's already starting to condense. Collect the condensate that drips off the egg and, when it has stopped dripping, measure its mass, m, and temperature, $\theta.$
Then heat that has gone into egg = $mL+mc(100°-\theta)$
in which $L$ is the specific latent heat of evaporation of water and $c$ is the specific heat capacity of water. The second term will be much smaller than the first.
[For a well known use of this method, see Joly's steam calorimeter.]
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Understanding the definition of tangent basis This question could sound silly but I though a lot about it and I'm not new to physics.
Let's say I have a plane on which I use polar coordinates, it means a point $P$ can be indicated by its coordinates $(r, \theta)$.
Then we need a basis in order to write the vectors as tuples of numbers, the tangent basis for this coordinate system is: $(\frac {\partial P}{\partial r},\frac {\partial P}{\partial \theta})$.
What is a derivative of $P$? I know $P$ is a point of the plane that
is represented by its coordinates $(r, \theta)$. I don't have a
mathematical form of $P$ with a dependence on $r$ and $\theta$ that
I can differentiate.
| There is nothing a priori about coordinate systems. They have no physical significance and are invented by humans, not by nature.
I recommend that you read Misner, Thorne, and Wheeler’s discussion in Gravitation about how coordinates are like telephone numbers assigned simply to keep track of which events in spacetime are close to which other events. (This was back in the 1970s when two houses with numerically close telephone numbers were geographically close to each other.)
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Do the pure states in the decomposition of a density operator need to be orthonormal to each other? So, I was studying quantum computation using the book Nielsen and Chuang and it stated a theorem known as "Spectral Decomposition theorem"
$$A=\sum _{i}\lambda _{i} | i \rangle \langle i|$$
I infer from this theorem that any normal operator can be diagonalized in the basis set $ \left\{ |i \rangle \right\} $ which should be the eigen vectors of the operator matrix A with $ \lambda _{i}$ as the eigenvalues.
Now when I started studying about the density operator with the definition $$\rho = \sum _{i}p_{i} |\psi _{i} \rangle \langle \psi _{i}|$$
I got a little confused. Since $\rho$ is a normal operator and it can be written as this decomposition, it must mean that the vectors $| \psi _{i} \rangle$ must be orthonormal to each other according to the spectral decomposition theorem. This seems totally absurd to me since there is no reason for the pure states (combining to make a mixed state) to have orthonormality as a prerequisite. I am sorry if my question is very trivial as this is my first time studying quantum information and I would be glad if someone could help me with this problem.
| The vectors $|\psi_i\rangle$ out of which you build the density matrix don't have to be orthornormal. They don't even have to be a basis: you can have more or less of them than your dimension. However, the decomposition theorem tells you that you can always find an orthonormal basis in which the density matrix can be written as in your first equation.
As an example, suppose you have spin-1/2 particles in an equal mixture of eigenstates in the three positive cartesian directions:
$$|x\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \quad |y\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ i \end{pmatrix}, \quad |z\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$
From this we can build the density matrix
$$\rho = \frac13 |x\rangle \langle x | + \frac13 |y\rangle \langle y | + \frac13 |z\rangle \langle z |,$$
and the theorem assures us that there is an orthonormal basis (the eigenbasis of $\rho$) $\{|1\rangle, |2\rangle\}$ such that
$$\rho = p_1 |1\rangle \langle 1 | + p_2 |2\rangle \langle 2 |.$$
| {
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Why is neper frequency called a frequency? In the context of complex frequency of RLC circuits, the real part is called neper frequency, according to units it's understandable that it has 1/s as the unit which is same as frequency but what is repeating at this frequency as I nowhere see any repetition? Why is it a frequency? I only understand that it decreases or increases the amplitude of sinusoidal function as time increases.
| the neper represents attenuation, or energy loss. the rate of energy loss will then be so many nepers in such an amount of time, or so many nepers per second. As such it "looks" like a frequency (cycles per second) even though it is not oscillating.
Bear in mind though that a more useful and convenient measure of energy loss in oscillating systems is the damping coefficient
zeta = (nepers per second)/(cycles per second at resonance)
which is less confusing to work with.
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Magnetic field of rotating wire I was doing some questions on the magnetic field when a charged body is rotated on its general axis, but I had a doubt that what if it's not the general axis but somewhere else. Like, for example, take a ring and take an axis along the diameter of the ring and to find its magnetic field somewhere at a point in front of it. Now is this approach right that we think that it's rotating in a way that it would seem like a spherical shell and then solve for it?
| I think that perhaps the best way to solve the general problem of finding the magnetic field of an arbitrary charge distribution rotating over an arbitrary axis is to simply integrate the magnetic field produced by the differential charge elements that make up the charged object, which is:
$$\mathbf{d \vec{B}} =\frac{μ_0}{4π} \frac{ dq \cdot \mathbf{\vec{V}} \times \mathbf{\vec{R}}}{r^3}$$
You would calculate the velocity of each differential element on the object relative to the point's reference frame and the displacement vector from the point to each bit, then apply the formula and integrate over the object, i.e. $\iiint\mathbf{d \vec{B}}$. There are many schemes by which you could integrate this, and spherical shells would be one of them (though almost certainly not the best one).
I suspect that it would be time-varying for many such objects, and possibly even non-analytic (thus numerical computer integration would be required). If you took your example of the ring rotated by an axis along the diameter that was perpendicular to the plane of the ring, there are some places where the ring is sometimes very near and sometimes very far, resulting in a complicated time-varying magnetic field, that you're better of just using a computer to solve.
Sorry, but I'm pretty sure there is no nice analytic way to get magnetic fields of off-axis rotating charged bodies (though I suspect someone has invented a perturbative theory for small displacements).
| {
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How is quantum physics translated into the world of electronics? I remember watching a documentary about early works in quantum physics and how it was essential to the microelectronics revolution and the invention of the transistor. What part exactly helped? Is there a source on technical aspects that lead to such development?
| One word: Semiconductors.
Semiconductor physics cannot be explained using classical physics. One needs to understand the electronic band structure of these materials in order to engineer various devices (e.g., p-n junctions, p-n-p junctions etc.) using them. Without such basic devices we won't be able to engineer modern electronic devices.
| {
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Does throwing a penny at a train stop the train? If I stand in front of a train and throw a penny at it, the penny will bounce back at me.
For the penny to reverse its direction, at some point its velocity must go to zero. This is the point it hits the train. Two objects in contact have the same velocity, so the train must come to a stop for the penny to change its direction.
I assume I'm getting some principles wrong. Is it because I assumed a perfectly rigid body, when in practice the train actually deforms ever so slightly?
| It depends. In principle it could, in any realistic situation it is not even close.
As you said, it comes down to the fundamentals. When you throw a coin at a train, the train will exert some force on the coin via electrostatic repulsion. On the atomic level this is due to the electrons in the coin getting too close to the electrons in the train and repulsing each other. This force is called the normal force and it accelerates the coin opposite to its initial velocity, it bounces back.
By Newton's third law the force acting on the coin leads to an equal and opposite force acting on the train. However, because the train is much heavier than a coin, the force that manages to measurably accelerate a coin doesn't change the trains speed by much (even measurably).
Now it should be clear that a coin will NOT stop a train unless it is an extremely slow train moving on frictionless rails. However, at some point the coin WILL have the same velocity as the train (non-zero, with respect to the ground). One can see this by considering the coin's velocity as it reflects off the train's surface. It starts having velocity opposite to the train, and ends with velocity in the same direction and slightly higher in magnitude (assuming the coin reflects elastically and doesn't just stick to the train).
I hope this explains it :)
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Focus iPhone torch to a beam I would like to focus an iPhone 5s' torch light into a beam with an angle of around 5 degrees or smaller (hopefully). Is this easily doable with a single lens?
I'm hoping to buy a lens and attach it in front of the iPhone (by 3D-printing a custom case). I just don't know which lens I should get.
This is for an artistic project. It doesn't have to be very precise. Since I will be using around 75 iPhones, I'm hoping the lens to be fairly cheap and easy to get in bulk.
Any input is appreciated. Thank you so much!
| Generating a reasonably collimated beam from an iPhone is only possible if:
*
*You use an enormous lens (several inches in diameter), in which case your beam will be correspondingly enormous.
*You are willing to throw away a lot of the light by using a pinhole positioned on the lamp surface in addition to the lens.
The reason these compromises are necessary is that the iPhone lamp is about 4 mm across and thus does not act suitably like a point source unless the lens diameter is much larger. Since a lens has only a single focal point, it will only nicely collimate a diverging beam if the beam originates from a single location. So for a small lens with a short focal length, you’ll need to block most of the iPhone lamp with a pinhole (a “spatial filter”).
| {
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Can there be interference between a proton and an electron? For example, we know that we can interfere two different electrons or two different protons by employing them in a double-slit experiment.
Now suppose, we mix protons and electrons and shoot them simultaneously for a double -slit experiment. Will the protons and electrons interfere with each other?
| The protons and electrons will attract one another by the standard Coulomb force, and this will modify the interference patterns of both types of particle. However a proton wave does not interfere with an electron wave in the sense of wave interference known as superposition. The mathematics of this involves the apparatus of quantum mechanics, which I guess you are not familiar with or you would not be asking the question. The basic idea is that each particle has its own associated wave and this wave can have different parts which interfere (i.e. add and subtract) with each other, but the wave of one particle does not add or subtract with the wave of another particle. Meanwhile all these waves are affected by the local electromagnetic environment which they influence through their electric charge.
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Newton's third law in magnetic fields Say I have a charged particle moving through a magnetic field perpendicular to it. It will experience a force, but according to Newton third law
Every force has an equal and opposite reaction.
So what is the opposite reaction/force of this magnetic force.
Which body experiences this force?
| A charge may, besides potential energy, experience potential momentum given by $q\vec A$. In the presence of currents therefore the kinetic momentum $mv$ is not conserved, but $mv+q\vec A$ is. The rate of change of this total momentum is equal and opposite for two particles that magnetically interact.
Note that this statement assumes that radiation effects are negligible, which is reasonable for the quasi static case.
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What is temperature on a quantum level? When I was in high school, I learned that temperature is kinetic energy.
When I learned statistical physics, we learned that temperature is a statistical thing, and there was a formula for it.
Questions:
*
*What is temperature in terms of quantum mechanics? This is, how is temperature connected to quantum concepts like position, momentum, angular momentum, spin and energy levels?
*How does temperature relate with the energy levels of an atom?
*Is the ground state always at absolute zero?
*If energy levels are discrete, how is this in play with the infinite amount of temperatures that exist in the universe?
| tl;dr: particles transition been microstates; temperature is a property of the ensemble macrostate at equilibrium
To use a statistics mechanics framing, quantum mechanics describes how particles transition between the different microstates of your system. Temperature is a property that emerges from the macrostate of the system when it reaches equilibrium.
Here “the system” is a collection of particles. So it does not make sense to talk about temperature of a single atom in isolation.
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Conservation of energy when external force is applied I have heard that an isolated system is where no energy and no mass is exchanged with outside the system. Does that imply that no external force can be applied to an isolated system? Why?
For example, there is a block at rest in a system and an external force is applied to move the block and provide kinetic energy. This contradicts the conservation of energy. Is it because a force is applied hence the system is not isolated?
Sorry for asking a silly question, just being confused.
| You’re correct that an isolated system is by definition one that cannot exchange mass nor any form of energy with its surroundings.
But a force is not a form of energy. You can apply a force to an isolated system as long as it does no work on the system.
So as long as a force applied to your box does not cause it to move from rest the box is still considered an isolated system.
Hope this helps
| {
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What is probability amplitude and why is it complex? When dealing with Mach-Zender interferometers the professor usually lets $\alpha$ & $\beta$ denote the probability amplitude that a particular photon isn't reflected by the beam splitter and the probability amplitude that a particular photon is reflected by the beam splitter respectively.
*
*What exactly is probability amplitude? Supposedly the actual probability that a photon is not reflected by the beam splitter is $|\alpha |^2$, similarly with $\beta$, so that $|\alpha |^2 + |\beta|^2 = 1$.
*Why are $\alpha$ and $\beta$ complex numbers? The professor explains it by pointing out to the fact that the wave function is complex in nature, but that just begs the question: Why is the wave function complex in nature?
To be clear, I understand why it is complex mathematically, I'm rather asking for the physical phenomena that requires us to work with complex numbers instead of real ones.
| You've answered (1) correctly. A probability amplitude is that thing that you take the square of (well, absolute value squared) to get probability.
That leaves (2), which is essentially the question "why does QM use complex numbers?".
I'm not sure. It's certainly the case that this is just a mathematical choice and convention - you can write QM with only Real numbers. The theory will just be uglier.
So in a way, complex numbers are only needed to keep the theory neat. Linear. A linear algebra. QM is phrased in a simple and clear manner when using the language of complex numbers, so we phrase it like that.
Note that the complex phase (angle of the probability amplitude as a complex number) of the state has no importance in of itself. It's only relative phases, between states, that matter. So the fact that we're using complex numbers is connected to how states can be related to each other.
For example, we find experimentally that in nature we can distinguish between states being in a state like $|+\rangle=1/\sqrt{2}(|0\rangle + |1\rangle)$ as opposed to $|i\rangle=1/\sqrt{2}(|0\rangle+i|1\rangle)$. E.g. measurement in the $(+,-)$ basis will return + 100% of the time if the system is in state $|+\rangle$ , but we'll get + only 50% of the time if that system is in state $|i\rangle$.
So we know there is a structure to how states can be related, which corresponds to the structure of complex numbers.
| {
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Dark matter's effect in 2+1 GR? In the appendix to The Planiverse, it is acknowledged that GR in 2 dimensional space implies no gravitational forces between separated masses--only in the interior of extended massive bodies. The author then speculates that perhaps the natives of the Planiverse will someday discover dark matter as an explanation for why their disc-planet in fact has gravity following a $\frac{1}{r}$ law.
But, is that how it actually works? If 2d dark matter were concentrated in a localized halo around a 2d planet, I would expect it to produce a linearly-rising gravitational force everywhere within the halo (or approximately so, modulo density variations in the matter field), just like the field inside our own planet. So that leaves the case of dark matter being distributed uniformly throughout the entire 2d cosmos, so that the dark matter itself produces no local accelerations; would the simple presence of dark matter, such that every point in space is indeed in the interior of a massive body, result in gravity appearing to behave normally for local concentrations of mass like planets, or would it simply be locally unnoticeable?
| The case of lower-dimensional general relativity is an interesting one because due to the symmetries of the Riemann tensor, things simplify quite a lot. In three dimensions in particular, the Riemann tensor has $6$ independent components, which is also the number of components of the Ricci tensor. It can be shown that, in 3 dimensions, the Riemann tensor is simply
$$R_{abcd} = R_{ac}g_{bd} - R_{ad}g_{bc} + R_{bd} g_{ac} - R_{bc} g_{ad} - \frac{1}{2} R (g_{ac}g_{bd} - g_{ad}g_{bd})$$
or equivalently, that the Weyl tensor vanishes. Very roughly, the Weyl tensor corresponds in general relativity to the notion that gravity can propagate in a vacuum, since the Riemann tensor can always be decomposed in three terms, one depending on the Ricci tensor, one depending on the Ricci scalar, and one depending on the Weyl tensor. And as we well know, in GR, we have that, if $T_{ab} = 0$, then $R_{ab} = 0$. So that, in $2+1$ dimensions, we have
$$R_{abcd} = 0$$
This doesn't necessarily correspond to Minkowski space (and indeed spacetime isn't Minkowski space in $2+1$ dimensions for the Schwarzschild spacetime, as it is not topologically trivial), but it does mean that there are no tidal forces and no gravitational attraction : the metric is just locally Minkowski.
Adding a cosmological constant, or just a pervasive matter field, does indeed help to get some more typical gravitational forces, as can be seen in the BTZ black hole solution, for instance.
| {
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X-ray imaging of coconut I'm doing the X-ray imaging of a coconut with soft X-rays. But the images have no contrast between the rice and water inside the coconut. How can I adjust the parameters to see this contrast inside the coconut? Horizontal and vertical projection settings for different images (parameters are the same)? Why is that?
[![enter image description here][1]][1]
| I should note first that I don't quite understand what exactly you are doing. Are you trying to get images of a whole coconut or of a half coconut? And what does rice have to do with that? Do you just put some rice inside a half coconut?
Nevertheless, let me try to offer some tips to get better contrast. If you cannot use a different wavelength range, you can use the same trick that doctors do: use some contrast (dye), such as barium or water-soluble iodine. If you have a whole coconut, you can drill a narrow hole in the coconut, inject the contrast (for example, using a syringe), and shake the coconut.
I hope you follow all the safety rules working with ionizing radiation, such as X-rays, and the dye.
EDIT (7/17/2019): Your deleted answer gave more information. So it looks like you want a method to tell a normal coconut from a macapuno. Some non-X-ray methods may be more promising, for example, ultrasound testing or rotation (the same method that is used to tell a boiled egg from a raw egg).
| {
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Time dilation at perceived constant acceleration Let's say that we have a spaceship.
From the beginning, the ship is stationary (relative to me) and not affected by outer forces (like Earth's or Sun's gravity).
Then the ship starts to accelerate at $10\space m/s^2$, that should be comfortable for the astronaut onboard (artificial gravity near $g$).
Normally this is an easy problem: after 1 s the speed is 10 ms/s, after 2 s the speed is $20\space m/s$, etc.
But the perceived acceleration is constant for the astronaut, meaning that we need to account for time dilation (otherwise, after a year the ships speed would be higher than $c$).
How do I calculate the acceleration of the ship after $x$ seconds?
(the ships acceleration from my point of view, not the ships)
As I understand the acceleration constantly increases the velocity, which increases the time dilation, which decreases the acceleration.
| The acceleration measured in the rest frame of the rocket, i.e. by the people on the rocket, is called the proper acceleration. Constant proper acceleration is a standard problem that you'll find treated in all books on relativity, and you'll find a summary of the results in Phil Gibbs' article on the relativistic rocket.
You ask specifically about the acceleration measured by the observer watching the accelerating rocket, and this is given by:
$$ a' = \frac{a}{\gamma^3} $$
where $a$ is the proper acceleration of the rocket and $\gamma$ is the Lorentz factor. The Lorentz factor is given by:
$$ \gamma = \sqrt{1 + (at/c)^2} $$
Combine these to get the acceleration as function of time.
The relativistic rocket article I linked includes lots of other useful equations e.g. for the speed of the rocket observed from Earth, the distance travelled of the rocket, etc. If you're curious to see the derivations see Chapter 6 of the book Gravitation by Misner, Thorne and Wheeler.
| {
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Why do jet engines sound louder on the ground than inside the aircraft? Everyone is familiar with the whirring sound of jet engines when seeing an aircraft taking off from a nearby airport. It is distinctly very loud on the ground and one can hear it even when the airplane is miles away.
Although one can hear a 'white noise' like sound when inside an airplane, the engines don't sound very loud in spite of being just meters away from them. I understand that the cabin is well insulated from the outside, but I would expect to hear a similar whirring sound of the engines.
So what is the phenomenon that makes jet engines sound louder on earth compared to inside the aircraft cabin?
| While sound waves can be impacted by the particles in the air to transmit sound, the sound waves can bounce off of the ground much like how you can have a louder or longer lasting sound in an enclosed room as it bounces off walls and such
| {
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When you fill a bottle with water then liquid nitrogen then turn it upside down, why does it blast off at high speed? I saw in the YouTube video Don't Mix Coke with Liquid Nitrogen! that when you fill a bottle part way with water then pour liquid nitrogen into it then turn it upside down, it blasts off at high speed. Why does that happen? How is it possible?
| I cannot figure out the answer for sure but here's my theory. According to the article Liquid Nitrogen Is Beautiful When It's Dancing Across Gasoline, liquid nitrogen is less dense than water. I'm guessing that means at first, the liquid nitrogen floats on the water. When it's turned upside down, the liquid nitrogen starts mixing with the water but surface tension limits its rate of mixing. However, the mixing still occurs to some extent which boils some of the liquid nitrogen creating thrust. Once it starts accelerating, the acceleration speeds up the mixing which in turn creates an even stronger thrust.
| {
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Conduction, convection, radiation: Does evaporation count as one of those? The forms of heat transfer are traditionally described as conduction, convection, and radiation. Is evaporation (or any other change of state) counted as one of those forms? Or is it considered its own distinct form of heat transfer?
| In engineering, heat transfer covers various mechanisms, including thermal conduction, thermal convection, thermal radiation, and transfer of energy by phase changes (e.g. evaporation).
At a given pressure, different boiling regimes exist depending on temperature (the following image applies to water at a pressure of 1 atm).
Image source: Wikipedia
In particular, the nucleate boiling regime is important in engineering (e.g. for the design of nuclear reactors) because of the high heat flux at small temperature differences. In this regime, isolated steam bubbles form at the hot surface, separate from the hot surface, and may condense again somewhere else in the subcooled liquid. Thus, in addition to the heat transfer by convection, the steam bubbles carry away heat in form of their enthalpy of vaporization $\Delta H_\mathrm{vap}$ which is released again when the bubbles condense. Furthermore, the movement of steam bubbles increases the movement of the liquid, thus increasing the heat transfer by convection.
(Note that if too much steam is generated at the hot surface, the steam insulates the hot surface from the liquid, thus strongly decreasing the heat flux after reaching the critical heat flux.)
| {
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What is the difference between uniform velocity and constant velocity? I think that uniform velocity implies constant speed but not constant direction. while constant velocity implies constant speed without any changes in direction.
Both tell us that there's no acceleration (since magnitude of velocity is constant).
The same goes for acceleration: both imply constant magnitude, but only constant acceleration means that there's no change in its direction.
However, a lot of people on the Internet argue that whether it's the other way around or that there's no difference at all. Who's right and who's wrong?
| The word uniform means the "same in all cases". Therefore, constant and uniform are often used interchangeably. Just be careful about where and when the sentence is used. For instance, if I say a car is accelerating uniformly in the xx-direction, then it has a constant acceleration in that direction.
In the case of a car moving around a circle with a constant speed, then we have uniform circular motion, but the velocity in this case is not uniform, it is changing direction. The word "uniform" in this case means it is travelling at a constant speed, and in such a case the magnitude of velocity is constant, but the direction changes.
Then, constant is with respect to time domain and uniform is said to be with respect to the space domain.
Now, consider a metallic cube. Now we say that mass is uniform if the mass per unit volume is same everywhere.
Then, what is constant?
Consider the same example. Now we say that the mass is constant if the mass of the whole cube does not change with respect to time.
P.S.: I went through J. Redman's answer and interpreted it in a lucid way. That is why the first two paragraphs of this answer resemble J. Redman's answer.
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Do we breath less oxygen in Humid air than dry air ? Since the volume of breathing remains same In rainy season, when Air is fully saturated, while we breath do we get less oxygen in than dry air ? Since the volume of breathing remains same?
| You are right that the higher the partial pressure of water vapor is in the air, the lower the partial pressure of the other gasses will be.
But this depends on temperature: when relative humidity is 100 % the partial pressure of water vapor may be lower than at 50 % RH and a higher temperature.
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Unitary Transformation of an Interfering Beam Splitter I was reading this research paper Quantum interference enables constant time quantum information processing and was confused by one particular expression involving the Hamiltonian of a beam splitter.
We consider two interfering modes $a$ and $b$ on a beam splitter device. The hamiltonian of the beam splitter is then
$$\hat{H}_{B S}=\frac{i \hbar}{2}\left(\hat{a}^{\dagger} \hat{b} e^{-i \varphi}-\hat{a} \hat{b}^{\dagger} e^{i \varphi}\right)$$
where $\hat{a}$, and $\hat b$ are respective annihilation operators of states $a$ and $b$.
Then the unitary transformation generated by the beam splitter is given by
$$U_{bs}=e^{-i \theta H_{B S} / \hbar}$$ where $\sin\left(\frac{\theta}{2}\right)=\sqrt{r}$ (the reflection coefficient).
My question given this how is it possible to derive the expression given on page 32 for the annihilation operatorrs for output states $a_{r}$ and $a_{t}$
$$\begin{array}{l}{a_{r}=U_{B S}^{\dagger} a U_{B S}=a \cos \frac{\theta}{2}+b e^{-i \varphi} \sin \frac{\theta}{2}} \\ {a_{t}=U_{B S}^{\dagger} b U_{B S}=-a e^{i \varphi} \sin \frac{\theta}{2}+b \cos \frac{\theta}{2}}\end{array}$$
How can we derive this result using only the information given above? I tried evaluating the matrix exponential in Mathematica for some low dimensional cases and got several contradictions (mostly sign errors) which just made me more confused.
| *
*Write the evolution equation in the Heisenberg picture for $a$ (and the same for $b$):
$$
\dot a = i[H,a]\ .
$$
*Simplify the expression for the given Hamiltonian (omitting $\hbar$):
$$
\begin{aligned}
\dot a &= -\tfrac12[a^\dagger b e^{-i\phi},a]+\tfrac12[ab^\dagger e^{i\phi},a]
\\
& = -\tfrac12[a^\dagger,a] b e^{-i\phi}
\\
& = \tfrac12 b e^{-i\phi}\ .
\end{aligned}
$$
*Do the same for $b$:
$$\dot b = -\tfrac12 a e^{i\phi}\ .
$$
*Integrate the two coupled differential equations
$$
\begin{pmatrix}\dot a\\ \dot b\end{pmatrix}
=
\frac12\begin{pmatrix}
& e^{-i\phi}\\-e^{i\phi}
\end{pmatrix}
\begin{pmatrix} a\\ b\end{pmatrix}\ .
$$
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Will pyrolytic carbon/graphite repel the Sun's solar wind? Since a moving charged particle has a magnetic field, as well as an electric field, and pyrolytic carbon/graphite repels an external magnetic field, would this mean that an object made out of pyrolytic carbon/graphite will repel the Sun's solar wind since the solar wind consists of charged particles?
If so, then a spacecraft with large panels of pyrolytic carbon/graphite attached to it should experience a propelling force from the Sun's solar wind.
| Of course, solar wind produces a magnetic field and pyrolytic graphite is indeed a diamagnetic material. When the magnetic field exerts a force on the pyrolytic graphite spacecraft, the particles of the solar wind will be deflected by the Lorentz force. Pretty much like what happens in a magnetosphere. It wouldn't propel the spacecraft, the particles will just flow around the spacecraft and may give it a slight nudge here and there. But, in any case, it would certainly not be possible for solar wind to "propel" the spacecraft, since, all the particles will be moving in random directions with all the possible speeds.
I highly recommend you to read these articles (since it is a topic which is still being researched upon):
(1) http://ui.adsabs.harvard.edu/abs/1928PhRv...32..133G
(2) https://www2.le.ac.uk/departments/physics/research/xroa/astrophysics-1/SWCX
(3) https://en.m.wikipedia.org/wiki/Solar_wind#Magnetospheres
By using article (1), I could establish that a spacecraft made up of pyrolytic graphite will work the exact same way as in a magnetosphere-solar wind interaction
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Why pulley systems give mechanical advantage? In a system with 2 pulleys, to lift a x kg object y meters I would need half the force that the object exerts due to gravity but pull the rope double the distance, why does this happen this way?
| This can be viewed as a requirement of the work-energy theorem and the conservation of energy. Assuming you alter the system slowly enough that parts don't gain a significant kinetic energy, the work you do by pulling the rope must be equal to the work done against gravity or any other forces that may be applicable. If you pull the rope with a force $F$ by a distance $a$, and the object of interest with weight $W$ is lifted a distance of $b$, we have
$$ F a \ge W b, $$
where equality holds in idealized systems where work is only being done by pushing the weight $W$ against gravity. For a pulley system, this means having no friction, massless strings and massless pulleys. If $ a = 2b $ in such an idealized system, then $ F = W/2 $.
| {
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What is the difference between position, displacement, and distance traveled? Suppose the question is somewhat like this:
If $v=8-4t$ and the position at time $t= 0\ \rm s$ is $2\ \rm m$, find the distance traveled, displacement, and final position at $t=3\ \rm s$
Since $\text dx/\text dt=v=8-4t$, then $\text dx=(8-4t)\text dt$. After integrating we find $x(t)-2=8t-2t^2$, and substituting the value of $t=3\ \rm s$ we get $x(3)=8\ \rm m$.
Is the answer that I found displacement, position or distance?
It can't be distance. I am sure of this. But is it position or displacement?
| Suppose you start on the 10th step of a very big stair. You walk 100 steps up the stair, then turn around and walk 95 steps down.
*
*Your position is where you are, which is now the 15th step.
*Your displacement is the net change of your position, which is $15-10=5$ steps.
*Your distance traveled is how much you actually moved, which is $100+95=195$ steps.
In relation to your example:
*
*If position $x$ at time $t$ is $x(t)$ you can get position at $t_f=3$ s by finding $x(t_f)=x(3\text{ s})$.
*If you started moving at time $t_i$ and finished at $t_f$ then the displacement is the net change in position $\Delta x = x(t_f) - x(t_i)=x(3\text{ s})-x(0\text{ s})$.
*Distance traveled is found by adding up each individual "step". For continuous movements, you must add up infinitely small steps $dx = x(t)dt$, which is given by $\int_{t_i}^{t_f}x(t)dt$. This is what you found.
I will leave the calculations themselves as an exercise to the reader.
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Motion between two particles in a relative manner Suppose a particle A is travelling in east direction with velocity of x m/s and another particle B is travelling with velocity y m/s in the west direction. Why does the the particle B appears to move towards A with a velocity of x+y and not just y m/s?
| As it is already answer by so many but i want to add one more point ,THAT IS -change in distance of B with repect to time will be equal to ym/s if you take your frame of reference as origin according to origin it velocity changes because it s distance with respect to time change equal to y=m/s but now yiu change frame of reference that is A so the velocity will be according to the observer A hence he feel that you are going (x+y)m per second ,yiu can add sign according to wether they are approaching each other or going away from each other
| {
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Every action has an equal and opposite reaction. Is is true for torques as well? So, we studied in classical mechanics that every action has an equal and opposite reaction. So if we apply a force to some object, that object will exert an equal amount of force on us but in the opposite direction.
Can the same also be said about torques? If so, can someone explain how it could be possible?
| A torque is exactly the same as two equal and opposite forces acting at different points on a body.
Each force has an equal and opposite reaction force, and the reaction forces are the same as an equal and opposite torque.
| {
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What's wrong with this argument that the potential energy of an arbitrarily heavy object at arbitrary height is $0$? Consider an object with mass such that there is a gravitational force downward of $1N$. Also assume the environment is a perfect vacuum. Now assume that we exert a force of $1+\epsilon $ Newton upward on the object for $\delta$ seconds, and then exert a force of $1$ Newton upward from then on.
During the first $\delta$ seconds, the object will accelerate from $0$ to $v$ meters per second upward, and then stay at $v$ since our force balances the gravitational force. So the object will move upwards indefinitely.
During the first $\delta$ seconds, we have a net upward force of $\epsilon$ Newton, and so a work done on the object of $\epsilon \cdot \frac 1 2 v\delta\approx 0$ (the force times displacement). After the $\delta$ seconds, the net force is $0$, since the gravitational force and our force cancel out, and so the net work done on the object is $0$, so the object doesn't gain any energy even though its height increases.
What's wrong with this argument?
| Work done by a force = Force * Distance the object moves.
When the object is initially stationary, that 1N is doing no work, because the object is not moving.
After you apply your (1+ϵ)N force, the object is moving, so when you re-apply your 1N force, you are actually doing work on the object just to keep it moving at the same speed.
If you account for that work being done, you will get the correct formula for gravitational potential energy.
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How can we tell if the Earth is spinning without any external references? The rotation of the Earth about its axis makes it bulge at the equator and contract at the poles due to the centrifugal forces. How do we know, without any external references, that the Earth is spinning if there is nothing to compare it to? For example, imagine it was spinning in empty space with no other objects. Does it still bulge at the equator?
Particularly, what can we say theoretically about it?
| Because the Earth is rotating on it-self, it is not an inertial referential, which means that there are additional fictitious forces acting on objects at rest in the frame of reference. For a spinning referential, the fictitious force is called the Coriolis force, which is responsable of many phenomena such as Foucault pendulum.
| {
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If Poisson Bracket of Momentum and Position is non-zero, why no Uncertainty Principle? In Hamiltonian classical mechanics, we have that the Poisson bracket of position
and momentum satisfies $$\{q_i, p_j\} = \delta_{ij}$$
But this implies that momentum and position 'generate' changes in each other. I.e. as we move along the flow of one, the other changes. Since this holds in classical mechanics as well as in quantum mechanics, why is there no uncertainty principle in classical mechanics? Why is it that non-commutativity under the Lie bracket operator gives rise to fundamental uncertainty in one case and not in the other?
Please answer in terms of Lie theory, if possible.
| What one requires for the uncertainty principle to arise is that the relevant observables should not commute, i.e., their commutator is non-zero. The Poisson bracket of two observables is not the same as their commutator. Even if the Poisson bracket of the classical observables is non-zero, they do commute in classical mechanics--they are just real numbers. So, no uncertainty principle arises in classical mechanics. What happens is that the commutator of observables in quantum mechanics corresponds to the Poisson bracket of the corresponding classical observables (this is the famous canonical quantization scheme up to some ordering ambiguities which do not really affect the question at hand). And thus, if the Poisson bracket of the classical observables is non-zero, the commutator of the corresponding quantum observables will be non-zero. This gives rise to the uncertainty principle in quantum mechanics. So, in a nutshell, commutators of quantum observables correspond to the Poisson brackets of classical observables, but, the commutator of classical observables (which is always zero) is not the same as their Poisson bracket.
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Does $\mathcal{M} = AdS_2 \otimes S_2$ makes any sense as a manifold? I'm not a topologist or a group theorist and I need a clarification about some notations.
Consider the Bertotti-Robinson metric in General Relativity (relativity students should study this metric, by the way, it's a really nice one!):
\begin{equation}\tag{1}
ds^2 = dt^2 - a^2 \, (d\vartheta^2 + \sin^2 \vartheta \, d\varphi^2) - \sin^2{\!\omega t} \: dz^2,
\end{equation}
where $a$ and $\omega$ are some constants. This metric is often described as the direct product of an ordinary sphere ($S_2$) and a 2 dimensional Anti-deSitter spacetime ($AdS_2$). It's usually described as $AdS_2 \otimes S_2$. I have three simple questions:
*
*In this example, is the direct product $\otimes$ the same as a cartesian product $\times$ ? Does it make sense to write $AdS_2 \times S_2$ instead? While I know what is the direct product of matrices and cartesian product of vector spaces, I'm a bit confused here!
*If the whole 4D spacetime manifold is $\mathcal{M}^4$, does it make sense to write $\mathcal{M}^4 = AdS_2 \otimes S_2$ ? What about $\mathcal{M}^4 = AdS_2 \times S_2$ ?
*Is $S_2 \otimes AdS_2$ the same thing as $AdS_2 \otimes S_2$? I know that the direct product of matrices isn't commutative ($A \otimes B \ne B \otimes A$), but I wonder if this is pertinent to the description of manifolds, not matrices.
| It is actually the direct product, not the tensor product (physicists frequently get too sloppy and end up using one or another without acknowledging the difference between the two). It is trivial to show that a direct product of manifolds is a manifold.
Direct product is commutative, so $S_2 \times AdS_2 = AdS_2 \times S_2$ as manifolds in the sense that there exists a 1:1 diffeomorphism.
Also, tensor product of vector spaces is commutative as well (in the sense that the two resulting vector spaces are isomorphic as vector spaces, but as pointed out by Accidental@ in the comments, this doesn't hold for the elements of vector spaces which are also usually tensored together by abuse of notation). For every two vector spaces $U$ and $V$, there exists a canonical isomorphism between $U \otimes V$ and $V \otimes U$:
$$ u \otimes v \mapsto v \otimes u.$$
| {
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Can force be applied without accelerating? When I push against a wall, I am applying force on the wall and the wall applies an equal force against mine therefore the wall doesn't move and neither does my hand. But isn't acceleration required to apply force? My hand is not accelerating when I am applyin the force. Still let's assume that the muscle fibres are accelerating, but how is the wall accelerating to apply an opposite force. So are the atoms accelerating somehow?
| First of all, F =ma doesn't mean that to apply force you have to have acceleration. F =ma means force F applied on mass m produces an acceleration a and this product of m and a gives the force which was applied. How much force can you apply can only be known when you set some mass on motion, your acceleration doesn't have to do anything with applied force. I can explain you with an example, let's imagine that a cricket ball hits the ground (ground is still) and bounces off, since their is change in velocity(direction has changed) that means there must be some acceleration behind that change and to produce acceleration there need to be a force, force came from ground but ground was initially at rest (ball didn't set the ground in motion but due to action-reaction it got bounced off).
You may have moved but there is friction between your feet and the ground and the center of mass also plays some role. Wall didn't move because of the friction, higher the mass more the friction. Every force develops an acceleration but friction nullifies it.
Hope it helps! If not then let me know through comments.
| {
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What prevents ice from being an electret? My question is apparently simple: if we put water in a electrostatic field and leave it to freeze, while still in the strong electrostatic field, to make ice, why wouldn't this ice exhibit electret capabilities?
| If I'm understanding your question correctly, then it seems like you're asking why ice is not a ferroelectric material. In classical physics, dipoles interact in such a way that their dipole moments have less energy when they're aligned in opposite directions. This is true for both electric and magnetic dipoles. Therefore classically we don't expect ferromagnetism or ferroelectricity to exist. They can exist only when the quantum-mechanical properties of the orbitals near the Fermi level are unusual. So if your reasoning is that if water molecules are dipoles, then solid water should be ferroelectric, then that implication just doesn't hold.
| {
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Text Recommendation: Random Walks (for physicists) I am an incoming graduate student in Theoretical Physics in the Netherlands, and I would like to know if any of you could recommend texts on random walks with applications to physics. My university offers a course on random-walks but I unfortunately do not have the pre-requisites (measure theory) to actually take the course. Hence I am looking for rather basic references and more advanced ones too in order to self-teach myself on random-walks with applications to Physics.
My interests lie in the fields of Statistical Physics, Soft Matter and Polymer Physics.
| Howard Berg's "Random Walks in Biology" (https://press.princeton.edu/titles/112.html) is an excellent, very short, and simple introduction to the topic.
| {
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How can I calculate the equivalent resistance of this circuit with resistors in parallel with wires? Problem:
Calculate the equivalent resistance, $R_\text{eq} ,$ of this circuit:
$\hspace{50px}$.
My solution attempt
*
*Тhe $12 \, \Omega$ resistor and the $6\, \Omega$ resistor are in parallel, so$$
R
= {\left(\frac{1}{12\,\Omega} + \frac{1}{6\,\Omega} \right)}^{-1}
= 4\,\Omega
\,,
\tag{1}
$$
reducing the circuit diagram to:$\hspace{50px}$.
*The $4 \,\Omega$ resistor and the $12 \,\Omega$ resistor are in series, so$$
R
= 12\,\Omega + 4\,\Omega
= 16\,\Omega
\,,
\tag{2}
$$
reducing the circuit diagram to:$\hspace{50px}$.
I am not sure of the answer, and I am so confused right now.
I am totally new to circuits and any help will be much appreciated.
| We are not supposed to provide solutions to homework and exercise questions, only guidance.
So here is some guidance;
The top circuit diagram below is your top diagram. The bottom diagram is equivalent to the top diagram, as long as all the "wires" shown are considered ideal (that is, zero resistance).
So given the bottom diagram, what should the answer be, regardless of what the "book" says.
Hope this helps
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Why is particle physics called high energy physics? The highest energy accelerator till date is the LHC which operates at an energy scale of perhaps 10-100 TeV. In SI units this is about $\sim 10^{-6}-10^{-5}$ Joule which is several orders of magnitude smaller than the energy scales we are used to in daily life. For example, the work done and heat produced by thermodynamic engines are hundreds of Joules.
Why is then particle physics regarded as the 'high energy physics'? 'High' with respect to what? I feel this has something to do with very small masses of the elementary particles. Incidentally, why is it difficult to reach energies as high as hundreds of Joules for elementary particles?
|
'High' with respect to what?
High with respect to the number of particles involved when they mention the energy value. The TeV energy of particle accelerators is related to the average energy of each accelerated particle.
You are comparing the energy of a single particle with the average energy of a system of particles.
As you mention, in our daily lives we talk about energies of several joules. However, we also talk about systems (like the steam engine) composed of several moles of particles, i.e., a number of particles of the order of $10^{23}$. Therefore, the average energy of the particles (molecules) in a steam engine is of the order of $\rm 10^{-23} J$.
A system formed of a single mole of particles accelerated in LHC would have an average energy of $\rm \bar E \sim 10^{-6} \cdot 10^{23} \sim 10^{17} J$. Now compare that to the hundreds of Joules of the steam engine.
Yes, that’s high energy physics indeed.
| {
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Rutherford/Geiger–Marsden gold foil experiment: what did they really think about alpha particles? Most tellings of the experiment where alpha particles were fired at a thin gold foil point out the following:
*
*atomic model being tested was the plum pudding model: negatively charged particles (electrons) embedded in a diffuse volume of counteracting positive charge
*because the positive charge is diffuse in the gold foil, the massive positively charged alpha particle should tear right through it with minimal deflection
*while almost all alpha particles passed through, a small minority were deflected strongly, some even knocked backward
*this result was better explained by positive charge being concentrated in a
small volumes separated by empty space
But thinking about it more, this seems like an oversimplification. For one, there's the implication that the alpha particle itself is dense, and some tellings even quote Rutherford's analogy of an artillery shell being fired at tissue paper and occasionally bouncing back. However, if the plum-pudding model were applied to the positively charged alpha particle, then the positive charge would be spread in a diffuse volume. Using the analogy, it'd be firing tissue paper at tissue paper. In that case, why DIDN'T they expect particles to be deflected?
| The plum-pudding analogy is only an analogy, and it breaks down when subjected to such close scrutiny. They knew the electrical charges and the masses involved, and the forces produced by those charges, and they did the maths. If the positive charges were spread out through the whole volume of the atom then the alpha particles (with known charge, mass and velocity) wouldn’t deflect by very much, regardless of their own volume. The big deflections only occur because the positive charge of an atom is concentrated into a very small part of its total volume.
| {
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What is the difference between a Surface Acoustic Wave (SAW) (Rayleigh Wave) and a Leaky SAW? I have come across leaky SAWs and I'm not sure what the difference is, and why some substrate cuts are listed as leaky SAW design.
Trying to search for diagrams so I could see what the differences maybe, I typically find this one where a SAW reaches a fluid/gas medium in contact with the surface of the substrate and this it appears to dampen, while transferring energy to fluid/gas.
Generally anything in contact with the substrate of a SAW will dampen it and absorb energy. So I do not see any difference, it just appears to be a SAW.
I am hoping someone could enlighten me as to what a Leaky SAW is and why certain crystal cuts appear better at promoting them.
| Sorry I did not see your post sooner, but this interpretation is incorrect. Leaky SAW devices suffer from radiation of energy into other wave modes within the substrate material - typically slow shear bulk acoustic waves. This leakage can be controlled on certain substrates through control of electrode thicknesses.
See:. Plessky, J. Koskela and R. Hammond, "Leaky SAW Devices with Beryllium Electrodes," 2018 IEEE International Ultrasonics Symposium (IUS), 2018, pp. 1-5, doi: 10.1109/ULTSYM.2018.8580172.
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Why is the necessary energy for a photon to lift an electron higher than the band gap energy? The band gap energy of silicon is around 1 eV and though the required energy for a photon to lift an electron up into the conduction band is around 3.6 eV.
Why is this?
Is the Energy of an absorbed photon exactly the energy of the band gap? is quite similar but - whyever - they do not answer it respectively do not use any material.
| Silicon has an indirect band gap. This means that although there is a conduction-band state which is only 1eV above the top of the conduction band it occurs at a different Bloch momentum ${\bf k}$. The nearest state with the samae ${\bf k}$ value is 3.6eV above the top of the valance band. Photons have a wavelength $\approx 600\mu$ that it is much larger than the inter-atom spacing and so their crystal momentum ${\bf k}$ is much smaller than the size of the Brillouin zone. Their momentum is therefore effectively zero as far as band theory is concerned. Therefore, for single-photon absorbtion with none of the energy going into phonons (to make up the momentum change) you need 3.6 eV photons.
LED's and other devices that play well with light are made of III-V or (or even II-VI) materials such as Gallium Arsenide or Indium Arsenide that have direct band gaps, meansing that the lowest energy conduction-band state has the same ${\bf k}$ as topmost valence band state.
There is some discussion of this in he Wikipedia page on Direct and indirect band gaps
| {
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Can anyone explain the Planck area? First of all, I am not an expert. I was reading about the holographic principle and came across the Planck area. It says that Planck area is the square of Planck length and there were some pictures like this:
source
I know that this Planck area is used for black holes. But doesn't matter if this Planck area is for a triangle or a square?
And there was a clip that Leonard Susskind was saying that Planck area is $10^{-33}$ cm on a side.
source 39:44.
But isn't it that the Planck length is $10^{-33}$ cm? If so, what is the "on a side"? Does it mean a face?
Please keep it simple so I can understand it.
| The Planck length is approximately $1.616255\times 10^{-35}$ m. You get the Planck area by squaring that, just like you get the square metre by squaring a metre. So the Planck area is approximately $2.61228\times 10^{-70}\,\mathrm{m}^2$.
As Wikipedia says, we expect geometry to get weird near that scale, so despite your diagram it's probably not appropriate to think about the Planck area in terms of Euclidean squares and triangles.
The reason I say "expect" and "probably" is that we don't yet know what really happens at that scale, we need a proper theory of Quantum Gravity for that. And even when such a theory is developed we may never be able to verify its predictions at that tiny scale: the energy required to do that is vast. It's essentially equivalent to the energy density in the first few Planck time units after the Big Bang started.
| {
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What is the mass of photon? I'm sorry if this question is asked before, but I searched through the site and none satisfied me.
In most of the books I've come across, they just write "rest mass of photon is zero." But never talk about the relativistic Mass. Even in other answers on this site they have written exactly the same.
And once in my class there was some discussion on which I said that Mass of photon is zero, but my teacher corrected me, saying "Rest Mass of photon is 0".
So, what is the real Mass of photon? Or does there even exist something as relativistic Mass of photon?
I know the equation $m\gamma$ gives indeterminate form thus can't be used for photons. And I've no confusion on energy momentum relation which uses the rest mass.
| The mass of the photon is zero. The end.
Relativistic mass is a hazardous concept, and many authors refuse to use it. It makes the increase in kinetic energy of an object with velocity appear to be connected with some change in the internal structure of the object. See also this question on SE.
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How fast does an electron move? I've been reading this website: www.physics.wayne.edu/~apetrov/PHY2140/Lecture8.pdf to learn how fast an electron moves in a circuit.
On page #8, #9 and #10 It says to take the Cross-sectional Area of the wire, The current, The density, The Charge and the electrons^3
Area- 3.14x10^-6 ( 2mm thick wire = 3.14 × (0.001 m)^2 = 3.14×10^−6 m^2 = 3.14 mm^2)
Current- 10 I
Density of copper- 8.95 g/cm^3
charge of 1 electron- 1.6x10^-19
electrons^3- 8.48x10^22 = ( 6.02*10^23 mole * 8.95 g/cm^3 * (63.5 g/mole)^-1 )
Total: 10 / 8.48x10^22 m^3 * 1.6x10^19 * 3.14x10^-6 m^2 = 2.48x10^-6 m/s
But they say that with 2.48x10^-6 m/s It'll take the electrons 68 minutes to travel 1 meter, How is that possible?
When I calculated that equation I end up with 5.9245283e+35, Then when I try to calculate again to get 68 minutes to travel 1 meter I can never get it right.
I'm not the best at math, The m's confused me. What am I missing ?
| It's because of drift velocity of electrons. Though the electromagnetic disturbance propagates at somewhat near the speed of light, the actual velocity with which the electrons move is much lower because of collisions with the ions in the lattice and random thermal motion.
The electrons can't move fast because they are continuously being slowed down by collisions and hence aren't moving through free space.
When a voltage is applied across a wire, the electrons move in response to the electric field generated and have a net movement in the direction of the field with very small velocities.
| {
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Difference between left- and right-handed, helicity and chirality What is the difference? I know there is the (almost) same question What's the difference between helicity and chirality? but when a particle is given as left-handed. Is it helicity or chirality?
| Helicity is the projection of spin onto momentum of a particle:
$$ h = \frac{\vec s\cdot \vec p}{|\vec p|} $$
If a particle with spin-1 moves exactly in the same direction as its spin points (let's say the spin point in $z$-direction and it also moves in $z$-direction), then the helicity is $h=+1$. If it moves in the exact opposite direction, towards $-z$, the helicity is $h=-1$.
As for the terminology, a particle for which $h=-|\vec s|$ is called left-handed, and $h=-|\vec s|$ is called right-handed.
Chirality is a property of a particle. As Wikipedia puts it, "it is determined by whether the particle transforms in a right- or left-handed representation of the Poincaré group."
Though terms like "left-chiral" and "right-chiral" might be more suitable, people usually also use the terms left-handed and right-handed when talking about chirality. Another thing to watch out for is that for massless particles, its helicity is the same as its chirality.
This answer and this answer might also be interesting for you.
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Heat reduction based on compression So I was thinking about air conditioner today and how we run air across compressed freon to cool down air but why do we need freon at all why not store just compressed air. My guess is because its inefficient.
My question given a 1 cubic meter tank of air, if the room temperature at 1 atmosphere is 80 degrees. At what atmosphere would you have to have the tank so that the initial release of the air would be the temperature 70 degrees?
| I will address the use of air rather than freon (or its current replacement) as a refrigerant. I believe @BowlOfRed has satisfactorily answered your other questions.
Air is not used as a refrigerant because it would not be practical. This is because a refrigerator requires the use of a working fluid that can undergo phase changes (gas to liquid in condenser, liquid to gas in evaporator) at practical operating pressures and temperatures. Air cannot undergo phase changes except at extremely low temperatures.
For example, at 1 atmosphere the boiling/condensing point of refrigerant HFC-134a (which has replaced freon for environmental reasons) is about -25 C (-13 F). This is somewhat lower than the setting of a household freezer. In contrast, the boiling/condensing point of liquid air at 1 atmosphere is -194.4 C. In order to increase the boiling/condensing temperature to that required by the refrigerator, extremely impractically high pressure would be required.
Hope this helps.
| {
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Electrons with disorder & something like AdS/CFT duality I know that consideration of electrons with disorder can be based on Feynman diagrams with disorder lines. In this approach, only non-crossing diagrams are important and give contribution to self-energy function $\Sigma$ and related quantities. Parameter $p_Fl$ plays the crucial role in this statement ($l$ is mean free path) and it is equivalent to $N$-parameter in field theory. Large-$N$ means that only planar diagrams in theory are important. Assumption $p_Fl\gg 1$ is very similar.
My question: is it possible to show duality between theory of electrons with disorder (EwD) and something like AdS theory? I mean that due to the equivalence of EwD and large-$N$ expansion it seems that one can naively expect that there is a theory which will be dual to EwD in weak coupling limit.
May be my question is not so clear but I would be grateful for any comments which can make it more clear.
| I am going to stick my neck out some here, but make this brief so I do not write something wrong. I would say this is a possibility. A quantum spin liquid is a disordered set of quantum spins with long range entanglements. With a topological order in this bulk there may then be edge states with short range entanglements of symmetry protected states. This physics has some parallels to the AdS/CFT correspondence.
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Half life of elementary particles Do elementary particles have half life? Can we theoretically calculate half of a particle which is in complete isolation?
| Some elementary particles, such as the electron, are stable; others, like its more massive sibling the muon, are unstable and decay into other particles. A muon decays through the weak interaction into an electron, a muon neutrino, and a electron antineutrino, all of which are elementary. The muon’s half life is 1.56 microseconds, and this can be calculated from Fermi’s Golden Rule.
The free neutron decays, but physicists do not consider it an elementary particle because it is a composite bound state of other particles.
According to the Standard Model, the only elementary particles are the three charged leptons $(e, \mu, \tau)$; their corresponding neutrinos $(\nu_e, \nu_\mu, \nu_\tau)$; six kinds of quarks $(u, d, s, c, t, b)$; the gluon $(g)$; the photon $(\gamma)$; the two weak bosons $(W, Z)$; the Higgs boson $(H)$; and their antiparticles. (Some particles are their own antiparticle.)
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Solution to Maxwell-Lorentz equations I am trying, without success, to find an example (preferably simple) of solution for the Maxwell-Lorentz equations, i.e., the coupled system of Maxwell equations + dynamics of a charged particle given by Lorentz force. Say we have a (for simplicity, non-relativistic) particle of mass m, charge q, position $\vec x$ and velocity $\vec v$, then the Lorentz force will give
$$m \vec x''(t) = q ( \vec E (\vec x(t),t) + \vec v(t) \times \vec B (\vec x(t),t ))$$
Is there any system for which we can exhibit at some instant $t_0$ the 'state' of the system $( \vec E(\vec r,t_0),\vec B(\vec r,t_0) ,\vec x(t_0),\vec v(t_0))$?
Standard textbooks seems not to consider solutions of coupled Maxwell-Lorentz equations, the only one I didn't check is Jackson's, because I don't have a copy with me.
| May be, one my construct such theories (the only ones I am aware, introduce a charge distribution for the classical particles) , but these are not the Lorentz-Maxwell theory. The whole development of quantum mechanics started from this one and has been developed along his lines to get finally the QED. Any other variants left no traces in the development of physics.
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In 2D CFTs what are the possible forms of correlation functions I am following Sylvain Ribault's lectures on 2D CFT (https://arxiv.org/abs/1609.09523) in which he lays out 2D CFTs in an axiomatic format.
In a CFT we assert (as an axiom) that there is a correspondence between an algebra of operators $\mathcal{A}$ defined on the Riemann sphere and the states in a representation of the Virasoro algebra $\mathcal{H}_V$. We also define a linear function
$$f:\mathcal{A} \to \mathbb{C},$$
which we call the correlation function. I know that the operators in $\mathcal{A}$ are linear operators in some Hilbert space $\mathcal{H}_\mathcal{A}$ (which I don't think I need to take to be $\mathcal{H}_V$). This tells me that the correlation function must be of the form
$$f(A) = \langle v|A|w \rangle$$
for some vectors $|v \rangle , |w \rangle \in \mathcal{H}_\mathcal{A}$. Is it true that there is some consequence of the 2D CFT axioms which tells me that I can express it in the form
$$f(A) = \langle v|A|v \rangle,$$
i.e. that its an actual expectation value as in regular quantum theory?
| In my lectures the space of states is not assumed to be a Hilbert space, i.e. to have a positive definite scalar product. Actually it is not assumed to have a scalar product at all. So you cannot write $\langle v|$ and the short answer to your question is no.
Actually you do not need a scalar product for computing OPEs or anything else. Nevertheless, you can add the extra axiom that there is a scalar product such that $L_n^\dagger = L_{-n}$, and use it for your computations. Adding this axiom forces the central charge and conformal dimensions to be real, so it is a restriction on the CFTs that you can consider.
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Bose-Einstein condensation: Bogoliubov Approximation I'm trying to understand the Bogoliubov approximation from "Statistical Mechanics" by Pathria and Beale. First of all they say
Since $a_0^{\dagger}a_0=n_0=O(N)$ and $(a_0a_0^{\dagger}-a_0^{\dagger}a_0)=1<<N$, it follows that $a_0a_0^{\dagger}=(n_0+1)\simeq a_0^{\dagger}a_0$
and this part is clear. It's not clear the following logic step
The operators $a_0$ and $a_0^{\dagger}$ may, therefore, be treated as c-numbers, each equal to $n_0^{1/2}\simeq N^{1/2}$
Can someone explain me why we can treat these operators as c-number?
| Perhaps, it will serve you a fully analized example to get an understanding of this "approximation":
L. Banyai, About the c-number approximation of the Macroscopical Boson Degrees of Freedom within a Solvable Model. phys. stat.sol. (b), 234, 14 (2002)
| {
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Is the statement that $U(x)$ is quadratic for simple harmonic motion equally strong as the statement that $F(x)$ is linear? Is the statement
"If the potential energy of a particle under oscillatory motion is directly
proportional to the second power of displacement from the mean
position, the particle performs a simple harmonic motion."
as strong as saying it in a more famous form of
Simple harmonic motion is a special type of
periodic motion or oscillation where the restoring force is directly
proportional to the displacement and acts in the direction opposite to
that of displacement.
I think it cannot be regarded as equally strong, since if we derive the expression for potential energy for any oscillatory motion, using taylor expansion, we get
only for small oscillations that $$U(x) = \frac{1}{2} U''(x_0) \ x^2$$ (where $U(x_0)$ is the potential energy at mean position) and we only get this result after certain approximations. Can someone please confirm this? Isn't the second statement stronger or more general? What I mean is, is the first statement always true?
| If the potential energy is quadratic in the displacement only for small oscillations, then the restoring force is proportional to the (negative) displacement only for small oscillations. So I would not say the second statement is more general. The one statement is exactly true, so is the other. If one is only approximately true, so is the other.
Here I'm ignoring the fact that the potential energy is unique up to an arbitrary additive constant, and because of a non-zero constant it may not be proportional the the square of the displacement as in your first statement.
| {
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Does the ground state of the Schrödinger equation, in any number of dimensions, always have constant phase? I just read this argument in this paper (PDF).
It suggests that, from variational principles, you can show that you can always lower the energy of a state by making the phase constant, thus resulting in a ground state that must have no phase change. I know this to be true for 1D systems (ground state is always real), but I'm not so convinced for the general $N$-dimensional case, though that's what the paper says as it uses the nabla for spatial derivatives.
My problem with their argument is that they split the trial wavefunction as
$$
\psi = f(\mathbf{r})e^{i\chi(\mathbf{r})}
$$
but even if you send $\chi$ to zero, what proof do we have that an $f$ that, alone, solves the Schrödinger equation always exists? Is that a separate theorem? It doesn't seem trivial to me.
| They're not assuming that $\psi$ is a solution of the Schrödinger equation. What they are doing is, given an arbitrary wavefunction $\psi$, calculating its energy expectation value $\langle \psi | H | \psi \rangle$, and seeing how they can tweak $\psi$ to make this lower. Crucially, by "energy" they mean this mean value, and not the eigenvalue (which obviously doesn't make sense if your function is not an eigenfunction).
Now, the ground state wavefunction is the one that actually minimizes $\langle H \rangle$: what they're doing here is showing that given any wavefunction you can always make its mean energy $\langle H \rangle$ lower by removing the phase, so that the ground state wavefunction (which has the lowest mean energy of them all) must have constant phase.
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"Printing" 3D structures using an atom laser? An atom laser is a coherent beam of atoms, similar to a laser. This is formed through use of Bose-Einstein condensate.
Could such a beam be focused and used to deposit atoms on a surface to build up an atomically-precise nanoscale object?
| This can also be done with a normal (optical) laser. It actually won the Nobel prize this year, it's called optical tweezers.
Some cool structures people built 'atom by atom' are Eiffel towers, Moebius strips etc (paper):
With an atom laser, the atoms would be moving so you'd need to somehow stop them before they can be "deposited" to be part of the structure.
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Pressure in a swim ring I have a work related question and would love a physicist’s perspective.
My problem essentially boils down to whether the pressure is greater on the inner or outer “welding” in a round swimming ring, when it is inflated. My intuition tells me that they are equal, but I can’t explain why. Any advice would be much appreciated.
| Your intuition is wrong.
If you assume the ring is a membrane (i.e. it if it was not inflated, the material it could not resist bending) the tension at any point depends (inversely) on the curvature of the membrane, which is different on the inside and the outside.
Unfortunately I can't find a simple reference to the mechanical behaviour on the web, since Google searches are mainly about the what happens to curved cell membranes in biology, and other relatively complicated applications of the theory.
| {
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Why people feel outward force in rotor ride? I know centrifugal is fictious force but what is possible reason for the force that people feel in rotor ride outwards.
The reason for people not falling down is friction force which is in vertical direction so, what is the force in horizontal direction that makes people feel the force on their chest (also the normal force is in inward direction so, it can't be the reason)?
| Suppose you're in a car and stomp on the brakes. If you're not wearing a seat belt you'd describe your motion as "thrown forward," but that's an illusion due to the non-inertial reference frame of the accelerating car. To an observer on the side of the road, you're moving forward with the same velocity you had before, while the car around you slows down, so that you catch up with the front windshield.
If you're wearing a seat belt, you still have that "thrown forward" sensation, but now you accelerate with the car. In that case the force you're feeling is the normal force from the seat belt, which is pushing you towards the back of the car. Your intuition gets the direction of the force wrong because of the accelerating reference frame.
It's the same on the rotor ride where the room spins and you stick to the wall. If not for the wall, you'd move in a straight line and leave the ride. The force you feel is the normal force from the wall pushing you inward; your intuition about the direction is incorrect because you're in an accelerating reference frame.
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As $SL(2,\mathbb{C})$ is a double cover of the Lorentz group, is $SL(2,\mathbb{Z})$ a discrete subgroup of the Lorentz group? The group $SL(2,\mathbb{C})$, the group of $2 \times 2$ complex matrices with determinent $1$, is a double cover of the Lorentz group. (These transformations can be understood as Mobius transformations on the Riemann sphere, which correspond to the action of Lorentz transformations on a sphere of light rays shooting out from the origin.) The Modular group $SL(2,\mathbb{Z})$ of $2 \times 2$ matrices with integer entries and determinent $1$, is a subgroup of $SL(2,\mathbb{C})$. Does this group therefore correspond to a discrete subgroup of the Lorentz group? What is its significance? How can it be thought of?
| I never considered this issue, but I think the answer is positive if, in your view, a discrete subgroup is a subgroup of a topological group which is made of isolated points.
In fact, the canonical projection $\pi : SL(2, \mathbb{C}) \to SO(3,1)_+$ is a surjective Lie-group morphism which is a local Lie-group isomorphism around the identity. As a consequence it is in particular a local diffeomorphism in a neighborhood of every given point in its domain. Hence the subgroup $\pi(SL(2, \mathbb{Z}))$ is made of isolated points and thus it is a discrete subgroup of $SO(3,1)_+$.
Notice that the correspondence between $\pi(SL(2, \mathbb{Z}))$ and $SL(2, \mathbb{Z})$ is still twofold because $-I \in SL(2, \mathbb{Z})$ and $Ker(\pi)= \{\pm1\}$.
The point is that, in general, the $16$ entries of $\pi(A)$ are not in $\mathbb{Z}$ if $A \in SL(2, \mathbb{Z})$.
I cannot see any physical significance of this subgroup, sorry.
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Why does carbon nanotubes look "furry" in scanning electron microscopy? In scanning electron microscopy images, carbon nanotubes looks quite different from the schematic hexagonal structured tubes which usually describes them. How come they are all bent and "furry"?
| Because diagrams lie. They have to, in some cases, to make their point clear.
I presume you mean something like this?
(Courtesy "Simetrical", cc-by-sa 3.0. Wikimedia Commons. Retrieved from: 1)
Yes, the tube shown in the picture is artificially straight. It's no different from showing a diagram of, say, an electric wire that shows it as straight. It's meant to illustrate the structure, not the physical properties of it as an object, such as its deformability. Just as a piece of cable can bend, so too a nanotube can bend. They're flexible. The flexibility is because the bonds are not perfectly rigid, but can strain (deviate from their customary angles).
Also, you cannot see the hexagonal structure because it is too small. The width of a nanotube (not illustrated in the picture I give) is like maybe only 2 nm or so; and the length of a single carbonic bond is something like 0.14 nm. Note that the scale bar on your picture translates to 2000 nm, so the nanotube "hairs" are probably on the order of 1000 nm long. The bonds and atoms are thus just far too small to see in the picture. Equivalently, it's like you're "zoomed way out" compared to the level of detail in a diagram like the one above.
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Why is Kirchhoff's voltage law true in a DC circuit? If we consider a single electron going around a closed loop, with a battery giving an EMF of $6\ \mathrm V$, why does the electron have to lose the energy in the loop?
If the circuit had zero resistance, wouldn't the electron just gain more and more energy as it loops around the circuit again and again, thus breaking Kirchhoff's voltage law and failing to conserve of energy, since the chemical potential of the battery gives the energy?
| Kirchhoff's circuit laws (both the current law and the voltage law) apply in the lumped circuit approximation only.
That means they apply when the circuit can be accurately modeled as a collection of lumped components whose physical extent is insignificant; connected by ideal wires, with no significant magnetic fields coupled to the wires, and no significant charge storage in the wires.
As such, we don't expect these laws to apply to the situation of an isolated electron floating in space.
If you had in mind a situation where the two ends of your battery are connected by a wire, then it's unphysical to imagine there's only a single free electron present in that wire, or that the electrons don't interact with the atoms in the wire (which will transfer energy to the wire).
| {
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Does the strong force continue to act upon quarks after they fall into the event horizon of a black hole? I'm assuming that when particles come in contact with the event horizon, they start traveling directly to the singularity, which is one point that might need clarification. Tangent particles would obviously act this way, but particles entering at wide angles might "whirlpool" down into the singularity.
My question only applies to the first case, which when crossing the event horizon, the quarks would not be able to move when the strong force acts upon them. Additionally, even the gluons themselves would have to travel toward the singularity. All quarks would instantly travel in straight lines to the singularity, which seems to indicate that matter just disintigrates into subatomic particles at the event horizon. Is this correct?
| There are no free quarks in space, they will cross the event horizon still within the hadrons and will keep that way up to the singularity.
Here is, as an example, what a proton really is in terms of quarks:
Whether the tidal force a proton encounters at the singularity is strong enough to turn it into a quark gluon plasma phase depends on the model of quantized gravity used. At present there is no definitive quantization of gravity, only effective theories.
Here are the forces a particle feels at the event horizon:
As you see the smaller the mass of the black hole ( small horizon radius) the higher the force, but certainly it is not enough to rip a hadron apart. What happens after the capture is a matter of the model used.
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Electromagnetic wave equation: can we ignore the constant of integration? Suppose we obtain a solution for each of $\mathbf B$, $\mathbf E$ of maxwell equations in the vacuum ($\rho=0$). Clearly, for any constant vector $\mathbf k, \mathbf m$, $\mathbf {B+k}$ and $\mathbf{E+m}$ also satisfy the same set of differential equations. Presumably, we can call $\mathbf k, \mathbf m$ "constants of integration".
My question is, though, is it okay to choose those constants randomly as I like? It is really difficult for me to "choose" appropriate values of $\mathbf k, \mathbf m$.
I might be able to determine $\mathbf k, \mathbf m$ if $\mathbf E$ is assumed to vanish at $\infty$. However, in the case that $\mathbf E$ is a sinusoidal electromagnetic wave, it certainly does not vanish as infinity.
| We are almost always satisfying Maxwell's equations (or any set of differential equations) with respect to some boundary conditions. Usually we assume that a vector field goes to zero at infinity, which means it is uniquely specified by its divergence and curl. (See the Helmholtz decomposition.) If it doesn't go to zero at infinity, then can specify some other boundary conditions that account for that constant.
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How to derive the angular velocity of circular orbits in Kerr geometry? I am trying to derive the angular velocity of a circular orbit in Kerr geometry, eqn.(2.16) in Bardeen et al (1972) which reads $$\Omega=\dfrac{1}{r^{3/2}+a}$$ (Note that I am using the units in which $M=1$ and only direct orbits are considered i.e. only the upper sign is used)
MY APPROACH TO THE SOLUTION:
Now using the angular momentum per unit mass $(\mathscr{L})$ and energy per unit mass $(\mathscr{E})$, the conserved specific angular momentum can be defined as $$l=\dfrac{\mathscr{L}}{\mathscr{E}}$$
Using this definition, I had obtained that $$\Omega=\dfrac{d\phi}{dt}=\dfrac{\dot{\phi}}{\dot{t}}=-\dfrac{g_{t\phi}+lg_{tt}}{g_{\phi\phi}+lg_{t\phi}}$$
which for $l=0$ is simply $\Omega=-\dfrac{g_{t\phi}}{g_{\phi\phi}}=\omega$, i.e. the angular velocity that corresponds to frame dragging.
But, I am getting trouble how to find the result as given in the above paper. Any suggestions regarding the approach to derive the equation would be helpful.
| Another solution for you:
consider a circular (r=const) timelike geodetic in the equatorial plane ($\theta=\pi/2,\dot{\theta}=0)$
if you calculate the geodesic equation for the r coordinate you get:
$$\frac{d}{d\lambda}\frac{\partial \mathcal{L}}{\partial \dot{r}}=\frac{\partial \mathcal{L}}{\partial r}=0\tag 1$$
where $\mathcal{L}=\frac{1}{2}\,g_{\mu\nu}\,\dot{x}^\mu\dot{x}^\nu$.
for circular geodesic is $\dot{r}=\ddot{r}=0$
so equation (1) :
$$\frac{\partial g_{tt}}{\partial r}\dot{t}^2+2\,\frac{\partial g_{t\varphi}}{\partial r}\dot{t}\dot{\varphi}+\frac{\partial g_{\varphi\varphi}}{\partial r}\dot{\varphi}^2=0$$
with
$\omega=\frac{\dot{\varphi}}{\dot{t}}$
and
$g_{tt}=-\left(1-\frac{2}{r}\right)$
$g_{t\varphi}=-\frac{2 a}{r}$
$g_{\varphi\varphi}=r^2+a^2+\frac{2 a^2}{r}$
$$\frac{\partial g_{tt}}{\partial r}+2\,\frac{\partial g_{t\varphi}}{\partial r}\omega+\frac{\partial g_{\varphi\varphi}}{\partial r}\omega^2=0\tag 2$$
the solution for $\omega$ give you:
$$\omega=\frac{1}{r^{3/2}+\,a}$$
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Is the Earth a gyro? Due to rotation and low friction, can the Earth be considered a gyroscope? If so, any interesting implications to this? Thanks
| The moon’s gravity exerts a torque on the earth, which causes the earth’s axis to precess (as a gyroscope subjected to an off-axis torque does). So the North Star is only approximately above the North Pole currently. In 13000 years, earth’s axis will point somewhere totally different, and in 26000 years or so it will be back where it is now.
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How exactly is white light a combination of several wavelengths? I have read that light is an electromagnetic wave. Every ray of light has a specific wavelength. The colour perceived by any observer is dependent upon the wavelength of the incident light.
What I don't understand is that how do electromagnetic waves of different wavelengths combine to form a single wave of another wavelength? Simply put, I have the following two related questions to ask:
*
*When we look upon a, say, completely white object, what is the composition of the individual rays that strike our eyes? Are those rays waves formed by the addition of waves corresponding to individual wavelengths that constitute white colour? I get that white light is composed of all wavelengths of visible light, but how are those wavelengths combined into a single unit which we call white light?
*If it is so, then how are prisms able to disperse light into its constituent colours?
Also, as a side question, how does all of this relate to light being composed of photons?
| To understand this you need to learn about the fourier transform. It describes how a function of time (in your case the time dependent electric and magnetic field strength) can be decomposed into its constituent frequencies.
To me your question seems similar to asking
"How do vertical and horizontal velocity of an object combine to form a single velocity in a different direction".
They combine by just adding the vectors up. Of course the result points in the same direction. It's just a different description of the same thing.
Similarly
how do electromagnetic waves of different wavelengths combine to form
a single wave of another wavelength?
When you have two electric fields you can add their vectors to get the resultant field. So they combine by just adding the vectors up, thats it. Of course they don't "form a different wavelength". The result has the same wavelengths.
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Will a can filled with vacuum move when we let in air?
This picture is from L.C.Epstein's book Thinking Physics. The upper can is filled with compressed air, and, when an opening is made on the right, the air comes out and the can shoots left. The question is what happens to the lower can, filled with vacuum, when we similarly make an opening. Does it move left - right - not at all?
Epstein says that the lower can doesn't move at all, "except for a
momentarily slight oscillation about the center of mass". I'm not sure I understand this. The explanation is that the air incoming into the bottle provides force on the left inner wall to compensate for the lack of force on the opening, and this balances the force on the left outer wall from the outer air. Which seems convincing, but opens a path to more questions:
*
*Shouldn't the can still start moving from the moment we make the opening and until the air pressure inside the can is equalized with the outside air?
*If that in fact happens, why would it stop and return ("a momentary slight oscillation about the center of mass") and not simply continue moving right with the constant velocity it's acquired?
|
*
*Shouldn't the can still start moving from the moment we make the opening and until the air pressure inside the can is equalized with the outside air?
Indeed it will, but that takes very little time.
*If that in fact happens, why would it stop and return ("a momentary slight oscillation about the center of mass") and not simply continue moving right with the constant velocity it's acquired?
In the case where the air leaves the can, after the air is ejected, it keeps the momentum. So the can balances that permanent momentum change by moving in the other direction.
In the case where the air enters the can, the air inside is quickly brought to a halt. There is no quantity of air moving to the left that the can will balance by moving to the right. Instead, the left wall of the can couples these two masses together so that they each stop the other.
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In the Stern-Gerlach experiment, why is there a nonzero force even though the atoms were electrically neutral I know that the magnetic moment of a particle is given by:
$\vec{\mu} = \frac{gq}{2mc}\vec{S}$
I know that in the Stern-Gerlach experiment, neutral silver atoms were used. Additionally, the deflection in this experiment was due to the force $F = \nabla (\vec{\mu} \cdot \vec{B})$.
How is a nonzero force experienced, given that $\vec{\mu}$ is dependent on charge $q$, which is zero for silver atoms.
| You're really asking how there can be a magnetic moment $\vec{\mu}$ when the atom has zero charge.
Moments are, in general, about separation. For example, consider two forces that are equal in magnitude and opposite in direction: The net force is zero. But if they're applied at different points, they still provide a moment, i.e. torque, and can cause something to spin.
An electric dipole is two separated charges, usually opposite charges. They sum to zero charge, but because they're separated they have some moment.
Same for a magnetic dipole. The classical form is a current loop. The total system is net neutral (the wire has moving charges and opposite fixed charges so as to be neutral), but the motion around the extended loop creates a dipole moment.
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Measurement postulate and black hole information paradox: are they related somehow? The measurement postulate states that whenever we make a quantum measurement, we select (projection) from the general superposition state a single pure state. Thus, as a general quantum state is a mixed state, the measurement postulate is just a rule to turn mixed states to pure states and it can not be described by any unitary operator in general.
By the other hand, in black hole physics, generally speaking, we have in general the opposite case. We can prepare a pure state (or even any entangled state) such as when it approaches the black hole and the event horizon, the atmosphere the semiclassical approach roughly implies that the black hole will be evaporated or, in the firewall gedanken experment, something happens in which either the initial state is a mixed state, it gets cloned or even worse, it is destroyed. This breaks down unitarity unless you give up locality or the equivalence principle. I am being rude (if wrong on general ground, tell me).
Question: are the black hole information paradox (current version) and the postulate of the quantum measurement related somehow or are they independent? After all, both of them seems to imply some conversion of pure to mixed states or even to the degree of entanglement before and after the interaction (in the measurement postulate, the apparatus interact with the quantum object; in the black hole information paradox, the quantum object interacts with the black hole "atmosphere").
| I don't see how these two things could be related, since collapse is only a feature of one interpretation of quantum mechanics (Copenhagen), while the black hole information paradox has nothing to do with any particular interpretation of quantum mechanics.
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Linear perturbations of the energy conservation in FLRW spacetime Recently i have some troubles regarding linear stability analysis in GR, especially matter conservation equation.
First order perturbations of the Hubble parameter and energy density are:
$$H=H_b(1+\delta(t)),\qquad\rho=\rho_b(1+\delta_m(t))$$
Using this in conservation equation in the FRW universe (dot is derivative wrt time $t$):
$$\dot{\rho}+3H(1+\omega)\rho=\dot{\rho}_b(1+\delta_m)+\rho_b(1+\dot{\delta}_m)+3H_b(1+\delta)(1+\omega)\rho_b(1+\delta_m),$$
which after discarding terms without perturbations and quadratic in $\delta$'s gives:
$$\dot{\rho}_b\delta_m+\rho_b\dot{\delta}_m+3H_b(1+\omega)\rho_b(1+\delta+\delta_m+\delta_m\delta)= \dot{\rho}_b\delta_m+\rho_b\dot{\delta}_m+3H_b(1+\omega)\rho_b(\delta+\delta_m)$$
However, the correct equation is:
$$\dot{\delta}_m+3(1+\omega)H_b\delta=0.$$
What am I missing? How to derive this relation correctly?
Equation can be found for example in:
https://arxiv.org/abs/1612.02037 (eq 39).
| It looks like, that you are almost there. You just have to use the background equation of motion to cancel the remaining terms, ie.
$$\delta_m\left[\dot{\rho}_b+3H_b(1+\omega)\rho_b\right]=0$$
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Why the ground state energy of bosons are 0 at T=0K ? Does it violate Heisenberg's Uncertainity Principle at 0K? Bosons obey Heisenberg's Uncertainity Principle but do not Pauli's Exclusion Principle. That's why in Bose Condensation we get a large amount of particles in a single state i.e. ground state at T=0K. But why the ground state energy is zero ? And in this case is Heisenberg valid ?
| Bose-Einstein Condensates (BEC) do not violate the Heisenberg's uncertainty principle.
Traps
Usually, BECs are not in free space, but spatially confined by some potential $V(r)$. The total Hamiltonian is $H = p^2/2m + V(r)$ so your "$E=0 \rightarrow p=0$" reasoning does not hold.
This potential $V$ is usually approximated as harmonic oscillator, $V \propto 1/2 \, m \sum_i \omega_i^2 x_i^2$, which will have a zero-point energy $E_0 = \hbar/2\, (\omega_x+\omega_y + \omega_z) \neq 0$. So once again, you do not have $E=0$ as per your reasoning.
A BEC in a trap thus has some spatial extent $\Delta x$. It is not an eigenstate of the kinetic energy, and thus also has some spread in momentum $\Delta p$.
Free space
A BEC in free space, $V=0$, would have a flat wavefunction spread uniformly over all space. Since $V=0$, the wavefunction is an eigvenstate of the kinetic operator, $\psi = e^{\mathrm{i}qx} = 1$ where $q=0$, i.e. it's the lowest plane wave. The momentum $p = \hbar q$ is now known exactly.
But while $\Delta p = 0$, the spatial extent of the wavefunction is $\Delta x \rightarrow \infty$, in accordance with Heisenberg.
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Is a state with no fluctuations in particle density necessarily a stationary state of the Hamiltonian? Consider a system of identical particles (bosons or fermions) with field operator $\hat{\psi}(x)$. The particle density operator is $\hat{\psi}^\dagger(x)\hat{\psi}(x)$.
Suppose that the particle density is constant everywhere - that is to say that:
$$\rho(x,t)=\langle\Psi(t)|\hat{\psi}^\dagger(x)\hat{\psi}(x)|{\Psi(t)}\rangle=c(x)$$
where $c(x)$ is some spatial function that is constant in time.
Does this necessarily imply that the system state is an eigenstate of the Hamiltonian? Or are there states with no fluctuation in particle density that aren't eigenstates of the Hamiltonian? How would I prove this in general, or come up with a counterexample? All my attempts thus far have failed.
I strongly suspect that this is true in the single-particle case, but struggled to prove even that. Does it hold in that limit?
I am considering a non-relativistic Hamiltonian of the form:
$$\hat{H} = \int dx \hat{\psi}^\dagger(x)\left(-\frac{\hbar^2}{2m}\nabla^2_x+V(x)\right)\hat{\psi}(x)+\frac{g}{2}\int dx \hat{\psi}^\dagger(x)\hat{\psi}^\dagger(x)\hat{\psi}(x)\hat{\psi}(x)$$
| Does $\langle \psi | \hat{x} | \psi\rangle = c$ mean that $\hat{x} | \psi \rangle = c | \psi \rangle$?
In case of free particles, QFT Hamiltonian:
$$
H = \int \frac{d^3p}{(2 \pi)^3} \; \sqrt{p^2 + m^2} a^\dagger_\vec{p} a_\vec{p}
$$
is not even proportional to the particle density.
So the state with mean particle density being equal to $c$ is not necessarily an eigenstate of the Hamiltonian.
| {
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"timestamp": "2023-03-29T00:00:00",
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Anticommutation relations for fermionic fields imply that Hamiltonian / Lagrangian can at most be linear? Fermionic field operators do obey anticommutation relations, so for a chosen Field operator (and the field momentum), we have:
$$
\{\Psi_a, \Psi_b\} = \{\pi_a, \pi_b\}= 0
$$
with the $\Psi_a$ being different field components, and the $\pi_a$ being the respective field momentum. $\{\cdot ,\cdot\}$ denotes the anticommutator.
Since these anticommutation relations imply that $\Psi_a \Psi_a$ is zero, does this mean that any hamiltonian or lagrangian I write down can at most be linear in fields / momenta?
EDIT: To be more clear about what I mean with "Linearity". Let's say the field has two independent components, $\Psi_1$ and $\Psi_2$. I can imagine that there will be terms like $\Psi_1 \Psi_2$, but any term like $\Psi_1 \Psi_2 \Psi_1$ would also be forbidden, because I can swap it into a configuration like "$(-1)^n \Psi_1 \Psi_1 \Psi_2$". Is that right?
| *
*No, if there are more than one Grassmann-variable (which is often the case), the Lagrangian/Hamiltonian can contain non-linear higher-order terms. With $n$ Grassmann-variables, one can build an $n$th-order polyonomial.
*In field theory (where the number of DOF is infinite), one can in principle construct higher-order terms using spacetime derivatives of the Grassmann-fields.
| {
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"timestamp": "2023-03-29T00:00:00",
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Notion of Present Can't I sync all watches in spacetime and call this time slice the present? In Carlo Rovelli's book he tried to explain that the notion of the present is local only, which I could not follow.
| The notion of 'the present' presupposes a notion of simultaniety and Einstein pretty much deconstructed this notion in his paper on special relativity when he promoted the constancy of the speed of light to a universal principle. This deconstruction is pushed even further when he developed his theory of General Relativity. And it is to this what Rovelli is referring to when he says the present can only be defined locally.
However, all is not lost for 'the present'. It's notorious that both QM and GR do not get along well. And Smolin in his book, Time Reborn, suggests that their different understanding of time is one of the major stumbling blocks. He suggests that QM shows that time is 'open'. And obviously the time just before time is open is the present.
It's worth recalling that Einstein had a famous debate with the then very famous French philosopher Bergson on the nature of time, which on all accounts the then even more famous physicist won. Bergson defined time through his notion of duration which he defines as:
Duration is essentially a continuation of what no longer exists into what does exist. This is real time, perceived and lived. Duration therefore implies consciousness and we place consciousness at the heart of things for the very reason that we credit them with a time that endures.
| {
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To which magnetic pole would a positive charge entering from space move to? Does the charge of an ion, be it negative or positive, have an effect on which pole it will move towards on Earth, or will the charge only determine the helical motion (clockwise or anticlockwise) rotation of the said charge? What connection is there between these charges and the north and south poles of a magnetic field?
| A charge is not attracted to magnetic poles. Rather, a magnetic field induces a force on a moving charge in a direction perpendicular to both the velocity of the charge and the direction of the magnetic field.
However, a charge whose velocity has a component parallel to the magnetic field will tend to spiral along the magnetic field lines. So, a charge that's spiralling around the field lines above the equator will move generally toward one of the poles. But, because the field lines are closer together near the poles, the spiralling gets tighter, the angle of the spiral changes, and the charge turns around and spirals back in the reverse direction. It ends up going back and forth, approaching but not reaching both poles.
| {
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Will creep occur if a metal is subject to equal stress from all sides? Aluminium creeps at a temperature range of 200 to 300 degree Celsius.
But if the entire bulk of the metal is at a uniform temperature, say, 0 degree Celsius, and no external stress of any kind is given, will creep still occur ?
Or will the creep be zero at 0 degree Celsius for aluminium?
| If a specimen of metal were subjected to hydrostatic stress it would creep indeed: the volume would change, while the "shape" would not (no deviatoric strain).
Most theories in continuum mechanics would de-couple deviatoric and hydrostatic creep behaviour, as the latter occurs often on longer timescales and is less pronounced in magnitude. In polymers the difference is such that bulk relaxation is neglected in the by far majority of the applications.
With regards to your second point, if a uniform temperature field is applied, the body will simply expand (resp. contract), no deviatoric strain, and null stresses. No creep is associated with such condition.
| {
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Damped Oscillation and Period In my school experiment, I wanted to measure the gravitational constant ($9.81 \ m/s^2$) by using a pendulum. If we take into account the damped oscillation (i.e. friction forces), does that affect the period?
As far as I know, the position $x$ of a particle undergoing Simple Harmonic Motion can be expressed as a function of time: $$x(t) = e^{-t} \sin t. $$
But the answer in the following link says that the friction force might be affected by factors like the velocity, or perhaps the position of the pendulum bob:
Does damping force affect period of oscillation?
In addition, since I am conducting an experiment to find the $g$ value, I am concerned if the changing period affects the value of $g$.
Could anybody clarify?
| Yes, damping forces like friction and drag force affects period in way :
$$ T = {2\pi } \left( \sqrt{\frac{k}{m} - \left(\dfrac{b}{2m}\right)^{2}} ~\right)^{-1} $$
Where $b$ is damping coefficient.
| {
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Why is work equal to force times displacement? This is how I think of what work is.I am sure I am wrong somewhere because I shouldn't be coming to the conclusion that I am coming to.It would be helpful if you would point out where this conceptual misunderstanding is.
Work is just change in the energy of an object.The only way an object can gain energy is by movement.Basically if an object's velocity increases then we can say that work has been done on the object.So, work done on an object is directly proportional to the change in its velocity.Also, if the object's mass is high, then for a given change in the objects velocity, the object gains more energy than an object with lower mass, because higher the mass more will be the momentum it can transfer to other objects and therefore higher the mass of the object for a given change in velocity, more is the work done on the object.Therefore work done is directly proportional to the mass of the object.
I don't seem to find any other factor that influences the work done on an object .Hence according to me work should be equal to mass times the change in velocity.
| If you apply a force F to something initially at rest, and keep the force going over a distance D, the force accelerates the thing to a speed V where VV=2FD/M.
In other words the velocity of the body doesn't rise in proportion to FD, but the velocity squared does.
If you rearrange the last expression you get mVV/2=FD. I.e. the kinetic energy is the work done.
| {
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