Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
|---|---|---|
When does causation take time? I apologize in advance if this is an elementary question. I am a statistician and would like to use an assumption in a statistical method, but I don't know if the assumption usually holds in nature.
I would like to assume that, if $X$ causes $Y$, then the causal effect from $X$ to $Y$ takes some finite amount of time to complete (i.e. the causal effect is not instantaneous). Is this reasonable? If so, why? I have heard from non-physicists that it is reasonable whenever $X$ and $Y$ are separated in space. Are there any examples in physics where the causal effect is instantaneous?
| There is a very simple answer to at least a part of your question. The minimum possible time for a causal effect to propagate where event $A$ is a cause of event $B$ is given by the spatial distance $\mathrm{d}_s(A,\,B)$ between the events in spacetime divided by the universal signal speed limit $c$, i.e. $\mathrm{d}_s(A,\,B)/c$. This happens when $B$ lies on the edge of $A$'s future lightcone and the time for the effect to happen at $B$ much smaller than the causation delay. The time for causation can be any time greater than this lower limit. You should take heed that, if the time is less than the lower limit, the time will be observer-dependent, but, even so, the order of two events with a causal relationship cannot be observer-dependent. The time dependence becomes weaker as the time interval approaches the lower limit.
Given you're a statistician, I am probably less rigorous in my usage of the word "causality" than you are, and therefore I suggest you should take a look at a description of what I exactly mean in my answer here (to the extent that I can explain it). You should especially look up the links to the Stanford Dictionary of Philosophy I refer to there,
| {
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How to find the null geodesics? The metric below was considered in [Spacetime perspective of Schwarzschild lensing]
$ds^{2}=2f du^{2}-\frac{2}{l^{2}} du dl-\frac{1}{2l^{2}}(d\theta^{2}+\sin^{2}\theta\, d\phi^{2})$
The authors gave null geodesics for this metric as
where $A,B,C$ are three first integrals of the null geodesics.
I am not sure how the authors got the above equations because from what I understand the null geodesics are given by the equation $g_{\mu\nu}\frac{dx_{\mu}}{dt}\frac{dx_{\nu}}{dt}=0$ but I can't see how one can get the above equations from this.
Does anyone know how the authors got the above equations?
| The equation you list
$$g_{\mu\nu}\partial_\tau x^\mu\partial_\tau x^\nu=0$$
is satisfied by all null-trajectories. However, a null geodesic must also satisfy the geodesic equation
$$\partial^2_\tau x^\mu +\Gamma^\mu_{\nu\rho}\partial_\tau x^\nu\partial_\tau x^\rho=0$$
These two equations, taken together, should suffice to determine the null geodesics. The $\Gamma$ symbols are in principle easy to calculate, either by hand or using some computer algebra program.
Since the results in the paper are given by first order differential equations, while the geodesic equation is of second order, one has to integrate once (this explains the presence of the integration constants $A,B,C$).
| {
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What happens if a body jumps in the Moon? I need to know why on Earth (ha ha..!) this behaviour happens when a body jumps in the Moon (a celestial body without any atmosphere at all).
Video explanation: https://youtu.be/Qgs5E5gKO48
The steps I followed using Kerbal Space Program:
*
*Start in position A
*Perform a clean jump on the Anti-radial vector
*Wait some time
*Slow myself to not impact the ground and kill myself.
*End up in position B
I know this is basic, but, should not the object end up in position A again?
|
Should not the object end up in position A again?
No, it shouldn't, because of the Coriolis effect (assuming the moon is rotating about an axis with respect to the stars).
From the perspective of a frame rotating with the moon, the coriolis acceleration is $-2\,\vec\omega\times \vec v$, where $\vec w$ is the moon's angular velocity with respect to inertial space and $\vec v$ is the velocity of the jumping Kerbal.
For simplicity, I'll assume the jump is performed at the moon's equator, at initial coordinates $\vec r = r\hat x$. The jump, in moon-fixed coordinates, gives the Kerbal an initial velocity of $\vec v = v_0\hat x$. The dominant acceleration is the downward acceleration due to gravity, $-g\hat x$ (not 9.80665 m/s2). There's also a much smaller acceleration in the $-\hat y$ direction due to the Coriolis effect. I'll focus on that and ignore that this slight drift changes the direction of the gravitational acceleration vector. (This effect is very small.)
With this assumption, the x component of velocity is $v_x(t) = v_0 - gt$. This means that the y component of acceleration is $\ddot y(t) = -2\omega v_x(t) = -2\omega(v_0-gt)$. Integrating twice yields $y(t) = -\omega v_0 t^2 + \frac 1 3 \omega g t^3$. Substituting $t = t_f = 2\frac{v_0^2}g$, the time at which the jumping Kerbal lands, yields $y_f = -\frac 4 3 \omega \frac{v_0^3}{g^2}$ (in other words, slightly to the west of the initial position).
| {
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What is the minimum number of collisions for photon required to lose all of its energy in an electron cloud? The electron cloud I refer to is a free electron cloud. Please help in this question. Is the answer infinite? If it is then it would take infinite time, which it doesn't.
| As you know that electron loses its energy in scattering with free electron via Compton scattering and the change in wavelength of the photon can be written as
$\lambda'-\lambda=\frac{h}{m_ec}(1-\cos\theta)$
after some manipulation the above equation can be written as
$\Delta E_{photon}=\frac{2E_{photon}^2}{m_ec^2}(1-\cos\theta)$
(this equation is valid for small change in the energy of photon during the scattering event) you may find that with the decrease in photon energy the change in the phton energy reduces and you will never reach to a situation that the photon can loose all of its energy via scattering from free electrons.
Physically we can say that the momentum transfer from photon to electron reduces as the photon loses its energy (similar to the negligible momentum transfer during collision of two particles with largely different mass). This is the reason that while working with optical photons the Compton scattering changed into elastic Thomson scattering.
Hope this will clear your doubts
| {
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When should you jump off a falling ladder? If you stand on the top of a falling ladder you will hit the ground at a higher speed (and therefore presumedly sustain more injury) if you hold on to the ladder than if you jump off it.
This was solved
here.
Where is the "break even" height on the ladder, from where you will hit the ground with the same speed if you jump off it or if you follow it down? This question just makes an assumption that you would hit the ground more softly if you stay on to the ladder (compared to jumping off it), if you are located at the lower part of it.
I don't think the midpoint is the break even point. I quickly calculated and I think you should stay on the ladder if you are on its midpoint. (The following is just a quick computation, there could be errors in it.)
$$ v_{midpoint}^2 = \frac{1}{2}gl \frac{m + m_L}{\frac{1}{4}m + \frac{1}{3}m_L} $$
$m$ and $m_L$ are the respective masses of man and ladder, $l$ is the length of the ladder.
| We can do this with a minor modification to the calculation described in the earlier question. As before we'll take the ladder length to be $\ell$, but now we'll take your height to be $\alpha\ell$, where $\alpha$ ranges from zero to one. Our reference point is if you let go, in which case your speed when you hit the ground will be:
$$ v^2 = 2g\alpha\ell \tag{1} $$
Now suppose you hold onto the ladder. As before we calculate the total potential energy change of both you and the ladder, which is:
$$ V = mg\alpha\ell + \frac{1}{2}m_Lg\ell \tag{2} $$
And this must be equal to the increase in angular kinetic energy $\tfrac{1}{2}I\omega^2$. The combined moment of inertia of you and ladder is:
$$ I = m(\alpha\ell)^2 + \frac{1}{3}m_L\ell^2 $$
And setting the kinetic energy equal to the potential energy gives:
$$ mg\alpha\ell + \frac{1}{2}m_Lg\ell = \tfrac{1}{2}\left(m\alpha^2 + \frac{m_L}{3}\right)\ell^2\omega^2 $$
And since $v=r\omega$ your velocity is $v=\alpha\ell\omega$ giving:
$$ mg\alpha\ell + \frac{1}{2}m_Lg\ell = \tfrac{1}{2}\left(m\alpha^2 + \frac{m_L}{3}\right)\ell^2\frac{v^2}{\alpha^2\ell^2} $$
Which rearranges to:
$$ v^2 = g\ell\alpha^2 \frac{2m\alpha + m_L}{m\alpha^2 + \frac{m_L}{3}} $$
And finally substitute for $v$ from equation (1) to get:
$$ 2g\alpha\ell = g\ell\alpha^2 \frac{2m\alpha + m_L}{m\alpha^2 + \frac{m_L}{3}} $$
And this rearranges to:
$$ \alpha = \frac{2}{3} $$
So if you are more than $\tfrac{2}{3}$ of the way up the ladder you should let go, while if you are lower than $\tfrac{2}{3}$ of the way up the ladder you should hang on.
| {
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How many arcseconds the light of a star unfolds in the vicinity of jupiter? One of the main demonstrations made to test the theory of relativity were the images of the solar eclipse of May 26, 1919, (causing a shift in the positions observed in celestial coordinates of its source stars 1.7 arcsec, the amount predicted by theory). If the same experiment was carried out, but with a star on the periphery of the edge of jupiter, the same effect would occur? Few arcseconds that star would unfold?
| The correct simple formula that gives a VERY close approximation (in the 'weak field' approximation) is [4G/c^2][M/R]
... where M is the mass of the object, and R is the radius of closest approach of the light. Note that since the first term is a constant, the deflection varies as M/R.
See here for an incorrect and correct derivation:
http://home.fnal.gov/~syphers/Education/Notes/lightbend.pdf
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Conductance of semiconductors at very high temperatures The restivity of typical conductors tends to increase as temperature increases. From what I understand, this is due to electron scattering.
Semiconductors tend to have their restivity decrease as temperature increases as more electrons are promoted to conductance band.
But what happens to semiconductors at extremely high temperatures. Does their conductivity continue to increase? Does there come a point where they lose all their conductivity.
Does the conductance band have an upper bound?
| This DTIC Report (PDF) shows that the conductivity of silicon continues to increase as temperatures increase from 500K to close to silicon's melting point at 1,687K:
This is due to the rapid increase in the number of free electrons and holes with increasing temperature.
As temperatures increase, phonon scattering increases and this reduces the mobility. The rate of increase of scattering with temperature is much less than the rate of increase in free electrons and holes with temperature. Hence, conductivity continues to increase as temperature rises.
Intrinsic vs extrinsic
The plot below, from the same report, shows how the conductivity of intrinsic silicon (samples 5 & 6) compare with doped (extrinsic) silicon (samples 1 and 2). Note that, unlike the plot above, the scale of the top axis on the plot below is in centigrade, not Kelvin:
At low temperatures, the doped silicon has higher conductivity than the intrinsic silicon semiconductor. As temperature increases, the conductivity of the doped Si drops slightly to due increased scattering. As temperatures increase further, the growth in thermal carriers becomes more important than the increase in scattering. As temperature increases further, the conductivity also increases.
| {
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Deceptively simple mass-spring problem? This question is inspired by two other, similar, so far unanswered questions (posed by different OPs).
Mass $m_2$ sits on a incline with angle $\theta$ that provides just enough friction for it not to start sliding down. It is connected by a massless string $S$ and perfect spring (with Hookean spring constant $k$) to mass $m_1$. Pulley $P$ is frictionless and massless. At $t=0$ the spring is not extended at all. Then $m_1$ is released.
Question:
What is the minimum $m_1$ to cause movement of $m_2$ up the incline?
Attempt:
Ignore the spring.
Determine static coefficient $\mu$ first.
\begin{align}m_2g\sin \theta &=\mu m_2g\cos \theta\\ \implies \mu &=\tan \theta\end{align}
To overcome the $m_2g$ component parallel to the inclined and the friction:
\begin{align}m_1g &\gt\mu m_2g\cos \theta+m_2g\sin \theta\\ \implies m_1 &\gt 2m_2\sin \theta\end{align}
But apparently this overestimates $m_1$. It has to be taken into account that $m_1$ starts accelerating before $m_2$ starts moving, because of the spring.
But how? Like several other members I can't see how the work done on the spring affects the minimum $m_1$. Conservation of energy?
| Got it.
As $m_1$ drops, gravitational potential energy is converted to spring potential energy:
$$m_1gy=\frac12 k\ell^2$$
Where $y=\ell$ because the string doesn't stretch.
The maximum drop is:
$$y_\textrm{max}=\frac{2m_1g}{k}$$
Maximum tension in the spring then is:
$$T_\textrm{max}=ky_\textrm{max}=2m_1g$$
For $m_2$ to move:
\begin{align}2m_1g &\gt 2m_2g\sin\theta\\ \implies m_1 &\gt m_2\sin\theta\end{align}
If that condition isn't met, $m_2$ won't move and $m_1$ will start moving upwards again and enter an oscillatory motion.
| {
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Transferring force through a string, including gravity I couldn't think of a better title to describe this confusion that I have.
Say you pick up a string and pull on both ends so that it is rigid. The string is a collection of an infinite number of objects. For simplicity, assume there are $n$ objects for some large integer $n$, each of which has mass $m$, and number the objects $A_1$, $A_2$, ..., $A_n$ from left to right. Maybe we can take the limit as $n$ goes to infinity; I don't know how to rigorize this.
Your left hand exerts some force $\vec{F}$ onto $A_1$. The only other two forces on $A_1$ are gravity and the force $\vec{F}_{21}$ from $A_2$, since $A_1$ is only in contact with $A_2$. Gravity is $m\vec{g}$. Since the string is not moving, the sum of the forces on $A_1$ is $\vec{0}$, so
$$\vec{F}_{21}+\vec{F}+m\vec{g}=\vec{0}$$
$$\vec{F}_{21}=-\vec{F}-m\vec{g}$$
By Newton's Third Law, the force $\vec{F}_{12}$ from $A_1$ onto $A_2$ equals $-\vec{F}_{21}$, which is $\vec{F}+m\vec{g}$. The only other two forces on $A_2$ are $m\vec{g}$ and the force $\vec{F}_{32}$ from $A_3$ onto $A_2$, so
$$\vec{F}+m\vec{g}+m\vec{g}+\vec{F}_{32}=0$$
$$\vec{F}_{32}=-\vec{F}-2m\vec{g}$$
If we keep going, we add an $m\vec{g}$ each time, until we get that the force from our right hand onto $A_n$ is $-\vec{F}-nm\vec{g}$, which is a lot more than $\vec{F}$, not to mention that $n$ is supposed to go to infinity.
What is wrong with my logic? Do my hands somehow exert $-m\vec{g}$ on each object?
| If you divide the string into $n$ parts then each part has mass $m/n$, where $m$ is mass of string. So beginning with one end of the string and proceeding with your argument, by the time you reach the other end the force will be $-\textbf{F}-n\times (m/n)\textbf{g}=-\textbf{F}-m\textbf{g}$, which is as it should because when you consider the string in its entirety, given that it is stationary, vector sum of forces on it must be zero: $\textbf{F}_{one~end}+\textbf{F}_{other~end}+m\textbf{g}=\textbf{0}$.
| {
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Why image is blurry or focused with convex lenses? I have a few fundamental questions in optics about the focus and blurry images.
Each text book says that according to this picture the object image is the same.
1) If we move the screen to point further then the image will be blurry. What causes it to be blurry?
2) Why do we need the rays to meet in the same point for focus?
3) What does blurry mean at lowest level possible?
4) As far as I see the objects top will now be in 3 different places. We should see the reflection of the top 3 times then?
5) Also, not so relevant to the to the main topic, but from that position, suppose we move the object right until point f.
At point f, we stop seeing the object image on screen or without. What is happening at point f so we don't see the object anymore?
Im asking this because I think it is really connected.
Thanks
| Thought I'd pop it into an answer for clarity - and more words:
Your picture answers these well. Look at those blue lines. These are light rays from the point at the top of the Object.
At the real image point, they are all together, in another nice sharp point. This is in focus.
Point 4 needs some clarification - the diagram is a construct to demonstrate how light rays are affected by the lens. At the screen they are spread apart - with light rays from that point appearing anywhere between the blue lines, and light from all other parts of the object doing the same. So the image of the top of the object now becomes a smear across the screen.
Oh, and at point F - it will be a small blur on the screen. It doesn't vanish.
To try and help with an analogy, imagine drawing a picture using a sharp pen. You can draw a perfect copy of an image. Now try with a bundle of paintbrushes tied together - you'll have a very blurry picture.
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Bohr / De Broglie postulate (what does $n λ= 2 π r$ imply) From the Bohr/De Broglie postulate we have n λ = 2πr where λ is the De-Broglie wavelength , r is the radius corresponding to n and n is the quantum number.
*
*An electron in the state n=2 has more energy than that at n=1
*That implies that the De- Broglie wavelength associated with the electron should also decrease ?
From the postulate..it is the other way i.e. the wavelength increases as the electron gains energy. How is this possible?.( I had assumed that wavelength decreases with energy)
if we calculate the De-Broglie wavelengths from the postulate:
for n=1 ; λ = 33 * 10^-11 m
for n=2 ; λ = 66 * 10^-11 m
does this mean that as the energy of the electron increases the corresponding De-Broglie wavelength increases?! may be i am missing something very basic here.
| Well if you go with Bohr's model of atom then,
since $mv^2/2 = K.E.= KZe/(2an^2)$
$= 2.18 *10^{-18} *Z^2/n^2$.
It means the total kinetic energy decreases with increase in n. This also means that velcity, $v$ of electron also decreases with n.
Now look at de Broglie's equation-
$\lambda = h/m_ev$
Since velocity decreases with increase in $n$, with increase in $n$ there is increase in $\lambda$. The total energy doesn't tell anything about velocity of electron. It is the kinetic energy that tells about velocity.
| {
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Does gravitational time dilation happen due to height or difference in the strength of the field? The reason why you have to tune differently the atomic clocks in GPS is because the GPS is higher or because there is less gravity there, or both? In other words in a constant gravitational field which doesn't differ with height, will time dilation still occur?
They say that the reason why people on the first floor age slower than people on higher floors is because as you get further from the earth gravity weakens. Is that true? Does the difference in the field cause that or just the distance?
If pure distance doesn't matter then why do we say that for a spaceship accelerating forward clocks at the front tick faster than clocks at the back since both are accelerating at the same rate?
| Quite simply:
1) you know from special relativity that accelerated frames experiences a time dilation effect (time is slower for them), and that the more accelerated, the slower the time,
2) strong equivalence principle tells you that standing still on earth or at a constant height with respect to the ground means you're accelerated upward,
3) the upward acceleration felt on the ground is higher than the upward acceleration felt at the level of satelites, thus time is slower for you on the ground.
| {
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How to measure the inner diameter of thin tube (0.5mm to 2mm) Does anybody know what method or equipment could be used to measure the inner diameter of capillary tubes? They would be in the region of 0.5-2mm in diameter.
Edit: the tube is made of plastic and I have standard lab equipment available.
| Some reasonable methods have already been proposed, but you might just measure the height of water column in the capillary (at least if the capillary is transparent) and calculate the diameter (https://en.wikipedia.org/wiki/Capillary_action#Height_of_a_meniscus)
| {
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Finding normal and tangential velocity and acceleration for motion along a 3D curve Basically, I have a major problem on my hands.
As part of an assignment, I have been given a set of 3D points, as well as the time between them, and nothing else, and I'm supposed to find the tangential and normal velocities and accelerations between each point.
Now, I know that (tangential) velocity can be easily calculated simply by taking the differential of position over time, and that (tangential) acceleration can also be easily calculated by taking the differential of (tangential) velocity over time.
However, finding the normal velocities and accelerations is where I get stuck. I can find absolutely no information on how a normal velocity is calculated, except for the fact that any velocity vector can be broken down into its tangential and normal components.
However, since by calculating the (tangential) velocities by taking the differential of position over time, that means that the velocity vector IS the tangential velocity vector. Does that mean that the normal velocity for each and every point is just 0? That doesn't sound right to me.
Regarding normal acceleration, I have no idea how it's supposed to work in 3D, because we need the radius of a circle to be able to calculate that, but how do I calculate the radius of a sphere in 3D, given only two points?
Any help on this problem is greatly appreciated. This thing's driving me insane.
| The velocity $\vec{v}$ can be obtained by computing the derivate of each position component with respect to time.
The acceleration $\vec{a}$ can similarly be obtained as the second derivative.
The tangential component of the acceleration is then $\vec{a_T} = \vec{a} \cdot \frac{\vec{v}}{v}$, and the normal component $\vec{a_N}$ is such that $\vec{a} = \vec{a_T} + \vec{a_N}$.
| {
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Expressing the magnitude of a cross product in indicial notation I am trying to teach myself tensor calculus but have reached a stumbling block - expressing the magnitude of a cross product in indicial notation. I know that one can express a cross product of two vectors $\vec{A}$ and $\vec{B}$ in indicial notation as follows:
$$ \vec{A} \times \vec{B} = \epsilon_{ijk}a_j b_k \hat{e}_i$$
But I am not sure how to express the magnitude of the resulting vector using indicial notation. My guess is
$$ \mid \; \vec{A} \times \vec{B} \mid^2 \; = (\vec{A} \times \vec{B})_m(\vec{A} \times \vec{B})_m = \epsilon_{ijk}a_j b_k \hat{e}_i \; \epsilon_{ijk}a_j b_k \hat{e}_i$$
but I seem to recall reading that having an index occur more than twice is undefined. How would I write the magnitude of the cross product using correct notation?
| You're right that
$$\mid \; \vec{A} \times \vec{B} \mid^2 \; = (\vec{A} \times \vec{B})_m(\vec{A} \times \vec{B})_m$$
We can write $(\vec{A} \times \vec{B})_m$ as
$$\epsilon_{mij}a_ib_j$$
and since we need to use different indices we'll write the second $(\vec{A} \times \vec{B})_m$ as
$$\epsilon_{mkl}a_kb_l$$
(by changing $i$ for $k$ and $j$ for $l$).
Putting these together gives
$$\mid \; \vec{A} \times \vec{B} \mid^2 \; =\epsilon_{mij}a_ib_j\epsilon_{mkl}a_kb_l$$
| {
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Neutron-Antineutron Annihilation Is the process $n + \bar{n} \rightarrow \pi^{+} + \pi^{-} + \pi^{0}$ possible?
| It might be suppressed for reasons I am too lazy to examine the process for (symmetry concerns, especially related to rotations), but energy is not a problem. If the two neutrinos have enough kinetic energy, it can be converted into the mass of the pions (among other possibilities, including electron-positron production). In math we're looking for solutions that satisfy all of the following equations:
$$\begin{align}E_{\nu} + E_{\overline{\nu}} &= E_{+} + E_{0} + E_{-}, \\
\vec{p}_{\nu} + \vec{p}_{\overline{\nu}} &= \vec{p}_{+} + \vec{p}_{0} + \vec{p}_{-}, \\
m_{\nu}^2 c^4 &= E_{\nu}^2 - c^2 \vec{p}_{\nu}\cdot \vec{p}_{\nu}, \\
m_{\overline{\nu}}^2 c^4 &= E_{\overline{\nu}}^2 - c^2 \vec{p}_{\overline{\nu}}\cdot \vec{p}_{\overline{\nu}}, \\
m_+^2c^4 & = E_{+}^2 - c^2 \vec{p}_{+}\cdot \vec{p}_{+}, \\
m_0^2c^4 & = E_{0}^2 - c^2 \vec{p}_{0}\cdot \vec{p}_{0}, \ \mathrm{and} \\
m_-^2c^4 & = E_{-}^2 - c^2 \vec{p}_{-}\cdot \vec{p}_{-}. \end{align}$$
That's 9 equations and 12 unknowns, so the space of possible solutions has 3 dimensions.
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Path integral approach to harmonic oscillation I've been trying to verify this result from Das' book "Field Theory: A Path Integral Approach",
Consider the action of a harmonic oscillator perturbed by a source $J$
$$ S=\frac{1}{2}m\dot{x}^2-\frac{1}{2}m\omega^2x^2+Jx. $$
The transition probability would be given by
$$ \int Dx\, e^{iS[x]}. $$
Define $x(t)=x_{\text{cl}}(t)+\eta(t)$, then we can Taylor expand the action about the classical path
$$ S[x]=S[x_{\text{cl}}+\eta]=S[x_{\text{cl}}]+\int dt\, \eta(t)\frac{\delta S[x]}{\delta x(t)}\Big|_{x=x_{\text{cl}}}+\frac{1}{2}\int dt_1 dt_2\, \eta(t_1)\eta(t_2) \frac{\delta^2 S[x]}{\delta x(t_1)\delta x(t_2)}\Big|_{x=x_{\text{cl}}}, $$
where, by definition of the classical path, the second term vanishes. So what we need to evaluate is the third term. So far I'm basically copying from the book.
Ok, so let's now put the expression of $S[x]$ inside to evaluate the second order variation:
$$ \eta\eta \delta^2 S =\eta\eta(m\delta\ddot{x}-m\omega^2\delta x), $$
where, in integrating by parts twice, we get something like
$$\frac{1}{2}\int dt_1 dt_2\, (m\dot{\eta}^2-m\omega^2\eta^2).$$
However, in the book the result is
$$\frac{1}{2}\int dt\, (m\dot{\eta}^2-m\omega^2\eta^2).$$
My confusion is how the double integration is reduced to a single.
| You confused the action with lagrangian.
$$S[x]=\int dt L\Big(x(t),\dot{x}(t)\Big)$$
As $L$ contains only simultaneous $x$ and $\dot{x}$ you'll get,
$$\frac{\delta^2S}{\delta x(t_1)\delta x(t_2)}=\Big(\dots\Big)\cdot\delta(t_2-t_1)$$
$\delta$-function will eat double integration
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Lorentz similarity transforms I am reading Peskin-Schröder, p. 36. Here is what I don't understand: They define a similarity transform in the coordinate space as:
$$x^{\mu} \rightarrow x'^{\mu} = \Lambda^{\mu}_{\nu} x^{\nu}$$
Then their field transform becomes:
$$\phi(x) \rightarrow \phi'(x) = \phi(\Lambda^{-1} x)$$
How is it that $\Lambda$ got inverted? How to see (prove) it? There are also the derivative and double derivative of the field, but I think if I get how the previous step is done, I will be able to figure out those myself.
| The idea is that the transformation of fields
$$\phi \to \phi'\tag {1}$$
induced by a coordinate transformation
$$x \to x' = \Lambda x \tag {2}$$
is such that if we simultaneously change coordinates and fields, then nothing changes.
In other words, the right hand side of (1) must be defined in order to fulfill
$$\phi'(x') = \phi (x)\:.$$
Since, from (2), $$x =\Lambda^{-1}x'\:,$$
we are committed to define $\phi'$ as the function
$$\phi'(y) := \phi (\Lambda^{-1}y)$$
for every spacetime point $y $.
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How are the spins of three quarks in a nucleon aligned? A proton is composed of uud. Are the two up quarks paired up to cancel each other's spin or are one up quark and one down quark paired up?
I know the three quarks may only carry a small proportion of the total proton spin. But they still need to be aligned somehow.
I see the following figure on http://www.lns.mit.edu/nig/programs. It seems in most cases the two protons in He3 are paired up. Could the quarks in a nucleon work in a similar way?
| The wave function of a proton is $2|u\uparrow u\uparrow d\downarrow\rangle-|u\uparrow u\downarrow d\uparrow\rangle-|u\downarrow u\uparrow d\uparrow \rangle$... So the probability of having u paired up with d is 2/3, and that of having u paired up with u is 1/3.
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Internal forces in an isolated system I did a weird activity and now I am trying to figure out the physics behind it:
While sitting on a chair I kept my feet above the ground and tried to move the chair forward. I was able to. Initially, I thought that perhaps the center of mass won't be changing as I was pushed backwards. But I moved the chair for a while and saw that it had moved a considerable distance and I was still sitting on the chair. Hence the center of mass did change for sure. According to Newton's second law, there must be an external force acting on the system(me and the chair). But I am not able to figure out what force is it. I have a feeling that it is somewhere related to the fact that the system is not completely isolated but I am not sure how.
Note:
I am not in contact with the ground or any other object close to me. I have lifted my feet and have my hands on my lap.
| I respect MaxW's view. But the term that he is calling the forward energy is what I am calling inertia's role. The reason for me to answer again is that he has neglected inertia's role completely and is stating that friction is the only reason for such movements and is also stating that this movement is not possible without friction, which is the only point I'm disagreeing to.
Because, in a twisting machine we have ball bearings, and so the friction is almost negligible. So, how does this movement happen?
I think this should explain the dominance of inertial force in such cases.
My only notion for answering again is to highlight the facts and nothing else.
Hope this helps.
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Why is imposing a symmetry on a theory considered more "natural" than fine-tuning its couplings? Theories whose behavior would qualitatively change if their couplings were not fine-tuned to particular values are often dismissed as "unnatural" (in high-energy physics) or "unrealistic" (in condensed-matter physics), while theories whose couplings are constrained by symmetry requirements are happily accepted. Why is this? A symmetry constraint can be thought of as just a collection of fine-tunings that has some unifying pattern.
For example, a complex scalar field Lagrangian
$$\mathcal{L}_1 = \partial_\mu \varphi^\dagger \partial^\mu \varphi - m^2 \varphi^\dagger \varphi - \lambda \left( \varphi^\dagger \right)^2 \varphi^2 \tag{1}$$
with a global $U(1)$-symmetry $$\varphi \to e^{i \theta} \varphi$$ can be thought of as a general Lagrangian
$$\mathcal{L}_2 = \partial_\mu \varphi^\dagger \partial^\mu \varphi - m^2 \varphi^\dagger \varphi - m'^2 \left( \left( \varphi^\dagger \right)^2 + \varphi^2 \right) - \lambda \left( \varphi^\dagger \right)^2 \varphi^2 - \lambda' \left( \left( \varphi^\dagger \right)^4 + \varphi^4 \right) - \dots\tag{2}$$
in which the primed couplings have all been fine-tuned to zero. (Things are admittedly more complicated in the case of gauge quantum field theories, because in that case the gauge symmetry requires you to modify your quantization procedure as well). This kind of "fine-tuning" is a little less arbitrary than the usual kind, but arguably it's not much less arbitrary.
|
Why is imposing a symmetry on a theory considered more “natural” than fine-tuning its couplings?
Take a one carat diamond (200 milligrams ofcarbon). It can be described by a simple symmetry
Unit cell of the diamond cubic crystal structure
Which is "simpler" : imagining the build up of the crystal by using the symmetry or listing all the (x,y,z,t) coordinates of the atoms, even the few ones in this unit cell?
Symmetries "simplify" concepts. Even your argument about "fine tuning of the primed couplings" falls into the category of simplification. One does not have to impose it to each of them, the symmetry does it.
I suppose if we were computers it would make no difference, but man is a pattern recognition animal, and finding patterns and repetitions of patterns is inherent in our tools of accumulation of knowledge, but this is outside physics.
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For lattice, what are the Goldstone bosons for the broken rotation symmetries? In $1$ dimension, we know that lattice breaks continuous translational symmetry into discrete translational symmetry, which generates $1$ Goldstone boson, i.e. $1$ longitudinal phonons.
In $d$ dimensions, if there are only $1$ type of atoms, then there are $1$ longitudinal phonon and $d-1$ transverse phonons. However, in $d$ dimensions, the continous symmetries of $d$ dimensional Euclidean group are broken, and in principle we should have $d+\frac{d(d-1)}{2}=d(d+1)/2$ Goldstone bosons. What are the other Goldstone Bosons?
| For the case of translational symmetry, the gapless point of the phonon mode is understood as a uniform translation of the lattice, which of course costs zero energy. However, by saying that the phonon is gapless, we practically compute the energy $E(k)$ of a phonon of lattice momentum $k$, and take the limit $$\lim_{k\to 0}E(k)=0.$$ The limit of lattice momentum $k\to 0$ is approachable because in the limit of crystal size $L\to \infty$, $k$ becomes quasi-continuous.
Now let's look at the case of rotational symmetry. First, for a uniform rotation of the whole lattice, again we know it costs no energy, which should serve as the gapless point of the Goldstone mode. However, we cannot define a "lattice angular momentum", which is a "generator" of the discrete rotation group, and which becomes quasi-continuous in the large size limit.
So to summarize, it is easy to confirm that there is a zero mode corresponding to rotational symmetry breaking (uniform rotation of the crystal). However, unlike for phonons, there is no quasi-continuous quantum number (the would-be lattice angular momentum) we can write its "dispersion" in terms of. Thus this mode is less talked about in literatures.
However, consider the following example:
This system is not a lattice, but nevertheless has a discrete rotational symmetry. In this case one can define a quasi-continuous analog of angular momentum, and the spectrum of the Goldstone mode corresponding to rotation symmetry breaking can be easily obtained.
This system also breaks translational symmetry. However, one cannot define a phonon mode that has linear dispersion as a function of momentum, although there does exist a Goldstone corresponding to translational symmetry breaking.
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What exactly is a 'quantum mechanical physical system'? I was reading the Wikipedia article about the Zero-point energy in which it states
Zero-point energy...is the lowest possible energy that a quantum mechanical physical system may have...
And I just wondered what it was referring to as a 'quantum mechanical physical system'? I understand if it was, for example, the particle in a box which is confined, but what abut something that has no boundary. I'm guessing that when talking about the zero-point energy, it refers to everything: the vacuum (no matter how small a piece of the vacuum you take), a particle, a particle and a box... It seems to be the smallest possible unit of energy.
I think I've got that right, however I'm not sure what to regard as a 'quantum mechanical physical system' in general? Is a single particle a system? Or is it the particle and it's surroundings?
| It's certainly a rough term, which really doesn't mean much more than a system where you get a very wrong answer if you treat it with classical physics. Thus, if it's close to its "zero point energy", it's redundant to call it a quantum mechanical system, as that goes without saying! But one must certainly avoid equating such systems with systems with only one particle, or even a small number of particles. There are plenty of situations in which a single particle acts classically, those are basically situations where a quantity called the "action", which is a bit like classical angular momentum, is much greater than Planck's constant. Also, we have white dwarfs and neutron stars, which contain a ghastly number of particles, but are still "quantum mechanical systems", in the sense that we cannot treat them classically.
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Why does increasing resistance decrease the heat produced in an electric circuit? If $H=\frac{V^2}{R}{t}$ ,then increasing resistance means decreasing the heat produced.
But, isnt it that the heat in a circuit is produced due to the presence of resistors? Moreover metals with high resistances are used as heating elements ,like Nichrome?
Why does the equation state that the heat produced is inversely proportional to Resistance
| Yes, there are two separate issues that involve resistance to keep track of. The first is, what is the current that will run through the circuit, given the voltage. That depends on the resistance such that the lower the resistance, the higher the current, and that's where the counterintuitive behavior is coming into play when you look at the heat generated. But the second question is, where is that heat generated given that you already know the current, and this is the perfectly intuitive part-- the heat comes from the highest contribution to the resistance. So the reason you get lots of heat from a smaller resister is only because the resistance in the rest of the circuit is very low-- it's no longer true if you put something in that has even less resistance than what the rest of the circuit already has.
| {
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Can Boyle's law be used to approximate energy storage? We know from Boyle's law that $PV=k$, where pressure $P$ has units of Pascals, volume $V$ has units of cubic meters, and $k$ is a constant. Multiplying and simplifying the units from the left side of this equation, we have
$$\mbox{Pa}\cdot\mbox{m}^3$$
$$=\frac{\mbox{N}}{\mbox{m}^2}\mbox{m}^3$$
$$=\mbox{N}\cdot\mbox{m}$$
$$=\mbox{J}$$
so $k$ has units of Joules.
Given the pressure and volume of a compressed air system, can $k$ be seen as an approximation of the total energy in the system (my theoretical question)?
Additionally, could we write an available stored energy approximation as
$$E\approx(P-p_0)V$$
where $E$ is the available stored energy and $p_0$ is the atmospheric pressure outside the compressed air system (my pragmatic question)?
| It has the right units, but doesn't really work. If you want the internal energy available for doing useful work then that will depend on the technique used for extracting that work. For example, if the gas is expanded isothermally then $P = P_i V_i / V$ and the work the gas does in expanding is: $$\begin{align} W & = \int P \operatorname{d}V \\
& = P_i V_i \int_{V_i}^{V_f} V^{-1} \operatorname{d}V \\
& = P_i V_i \ln\left(\frac{V_f}{V_i}\right) = P_i V_i \ln\left(\frac{P_i}{p_0}\right).\end{align}$$ With isothermal expansion part of that energy came from whatever was keeping the gas warm during the expansion. Adiabatic expansion, where the system isn't allowed to absorb heat, will get a different answer.
The total amount of energy stored in the gas will depend on the type of gas it is and the temperature of the gas. For a simple ideal gas the heat energy stored in it is:
$$U = \frac{3}{2} NkT = \frac{3}{2} PV.$$ That fraction, $3/2$, goes up when the gas has more ways to store energy than just movement of its molecules (like when the molecules can rotate or vibrate).
A better understanding of pressure comes from the first law of Thermodynamics: $$\operatorname{d}U = - P \operatorname{d}V.$$ That is, pressure is the generalized force which drives changes in volume for a system. Even better is the following relation:$$\mathbf{f} = -\nabla P,$$ which says that spatial gradients in pressure (differences in pressure with location) exert a force density (force per unit volume). This last relation leads directly to the more common definition of pressure as force per unit area when there is a sudden drop in pressure across some surface.
Edit: fix an incorrect subscript.
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Special Relativity: Length Contraction Confusion So here's a scenario I made up and I want to know if it is correct or not.
I'm was standing on the earth, l$_0$ meters away from a stationary rocket in space. The rocket then starts travelling towards me at a constant speed v. From my frame of reference the distance l$_0$ will be contracted and hence I will observe the total distance between me and the rocket to be $\frac{l_0}{\gamma}$. From the frame of reference of the rocket the distance observed will also be $\frac{l_0}{\gamma}$.
Is this correct? If so then what if the rocket was travelling away from me at the same speed, how would length be contracted in such scenario?
Also, If the the rocket was travelling to the moon and the distance between the rocket and the moon initially (i.e. when me, the rocket, and the moon are stationary) is x$_0$. What would I observe the distance x$_0$ to be?
| The length $l_0$ exists in your reference frame, so you may measure it without any length contraction. The rocket does not jump towards you due to length contraction when it starts moving quickly (it will move toward you, though). The length that exists in the rocket's moving frame is the length of the rocket itself. The rocket will appear to get shorter in your frame when it starts moving quickly.
From the rocket's perspective, its own length does not change since that length exists in its frame, in which it is at rest. However, space outside of the rocket would warp when the rocket suddenly started moving at a great speed, and you would appear to get closer due to length contraction in addition to the rocket's new velocity closing the gap over time.
In conclusion, $l_0$ does not contract in the Earth's reference frame, but it does in the rocket's frame.
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If my car was a vacuum inside, would it get as hot parked in the summer sun? Okay, I've been wondering this one all summer. Whenever I open my car door, I'm greeted by a wall of very hot air, which takes a while to cool down, of course, with the air conditioning. So I was wondering, if when I parked my car, I could flip a switch and create a perfect vacuum inside, would it be as hot when I got in hours later? Assume that my car is parked in direct sunlight without shade throughout the day.
Without air, would surfaces such as seats (indirect sunlight) get as hot or hotter? Ordinarily, does the air usually provide a bit of insulation to absorb some of the incoming heat? Or, does the air serve to trap the heat inside the car?
| The same amount of energy would get through the car windows whether or not there was air inside the car. This heat would be absorbed by the interior surfaces, which would radiate in the infrared part of the spectrum. Unfortunately, the car windows are just as opaque to that IR radiation whether or not air is in the car, so they would radiate that IR energy back into the car at the same rate as before. In a vacuum, there is no air for convection currents to carry heat away from the interior surfaces, so they would get somewhat hotter than they currently get. Naturally, air would be introduced into the car before you opened the door to get in, but it is doubtful that the air would have time to remove substantial heat from the interior surfaces before you sat down. This means that the seat would be somewhat warmer than it is now.
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Calculating relative distance covered Two balls 'A' and 'B' are thrown vertically upwards with the same velocity. The mass of A is greater than that of B. We need to find which of the balls reaches a greater height (assuming the effect of air resistence is negligible).
I personally feel that A would travel further since its momentum is higher than that of B. But few people whom I discussed this question with had different opinion.
One said that they would both travel the same distance since $s=(ut) + (1/2)at^2$
The other said that B would travel further since the effect of gravity on it is less according the universal law of gravitation.
If someone could explain which of these is correct or maybe point out the flaw in either of them then it would help a lot. Thanks in advance.
| the thing you have missed to mention is effect of air resistance.If we include the effect of air resistance in this question then role of mass will come into play, but if you are not including air resistance then you will use s=ut+at^2/2, which is independent of mass so both bodies will traverse the same distance.Hope my message reach you.
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How do we prove or disprove that a particle has no internal structure? In many pop physics books I have read that an electron has no internal structure.
How do we know that and how can we rigorously prove that it has no such structure at all?
| You can't "prove" the electron has or has not internal structure. All you can do are collision experiments at the energies available to you. Along these lines the proposed ILC (International Linear Collider) - if build - will provide an unprecedented new look at the electron (and positron).
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Why doesn't electric charge immediately leak off charged objects? I will focus my question with a particular example: a metal sphere, surrounded by vacuum, is given a negative charge. I know that when this charge is great enough, electrons will be emitted from the sphere, but why is the threshold for this so high? As I understand it, the reason an electron stays on the negative sphere despite the electric repulsion is because of the metal's work function. But the work function of metals are typically ~4 eV. Wouldn't this suggest that -4 Volts would be the threshold for electron emission from the sphere in a vacuum? (Or a voltage even closer to zero, because of the thermal distribution of electron energies in the metal.) This seems way too small and I would think the threshold would concern a minimum field strength rather than minimum voltage.
| An electric field which would strip electrons from the metal would need to be strong enough to provide for a voltage difference of $4V$ over a distance related to the work function -- the distance between "electron is in the metal" and "electron is outside the metal". One should expect this distance to be on the order of a few atomic layers, i.e., $\sim 10^{-10}\mathrm{m}$: Therefore, one would expect the electric field required to be on the order of $\mathrm{GV}/\mathrm{m}$.
On the other hand, $4\mathrm{eV}$ are a very high temperature, making electron loss due to thermal emission extremely unlikely. $0.025\mathrm{eV}$ are room temperature, and thus $4\mathrm{eV}$ are $160$ times room temperature -- at room temperature, the fraction of electrons with sufficiently high energy is thus around $e^{-160}$, an astronomically small number even compared to the number of electrons in a solid.
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Why can't I build a non-reversible optic setup which focuses the sun to higher temperatures? Following this question I would like to challenge one of the assumptions.
The standard answer is that thermodynamics prohibits focusing the sun to a spot such that the spot reaches a higher temperature than the sun itself, because lenses and mirrors are reversible machines and the argument goes from there.
I don't see why the reversibility should hold for composite machines. As in, if I have one lens/mirror apparatus here which focuses the sun to, say, 5,500K, and a duplicate apparatus there which creates another such spot...
...and I tilt them so that the spots overlap...
...then intuitively the overlapping spot should have a temperature significantly higher than 5,500K...
...and the machine isn't reversible because a photon striking any given point could have taken either path, so the thermodynamic argument against the above result doesn't apply.
What's wrong with this reasoning?
| Don't forget the Lagrange Invariant. That's what it's all about. The sun has a finite size in object space and therefore has a field. Faster optical systems will provide smaller spot sizes of the sun but the angles will be larger. Longer focal length systems will provide larger images but the angles are smaller. As you squeeze the size down the angles go up, just like in diffraction. alpha = 2.44 lambda/diameter. Like in a projector, if you have a dim image you can't just put in a higher Wattage lamp because to get more Wattage, you must increase the size of the filament and a larger filament means a larger field of view for the optical system and won't get through the field stop.
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If you are vacuuming your carpet and you wrap the cord around your body do you become a magnet? If you wrap an active electric cord around your body, do you become an electromagnet?
| No. The power cord on a vacuum cleaner has both supply and return conductors, which produce opposing magnetic fields. The region of nonzero magnetic field is limited to a few cable diameters away from the cable.
Also the magnetic field due to an alternating current changes direction at 50-60 Hz, depending on your local power supply frequency.
If you wanted to turn yourself into an electromagnet you'd need to separate the hot and neutral conductors in the power cord, and wrap one around you clockwise and the other counterclockwise. Then the fields due to the two currents would add. You could convert to direct current if the alternation of the field direction bothers you. However since biological tissue is not ferromagnetic, the physical effects on you are pretty tiny.
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Memory fading “striyers” in viscoelastics I was talking to a guy who does polymer moulding, and we were discussing a few industrial issues in melting, mixing and forming shapes, and with a few of my suggestions he rebutted with, "no that would create striyers". So for example, one mixing technique I suggested had the fluid melt pass through a grate at the end, and he didn't like that idea because it would create striyers.
My understanding at the time, from the context, was that he was referring to the viscoelastic nature of the fluid, that has a (fading) memory component. So in my example the product might still have lines far downstream left over from passing through the grate. But I've not heard this word "striyer", and wondered if anybody knows what it is?!
We were speaking in English, but if it helps, his mother tongue is German.
| Striation? ...Striations means a series of ridges, furrows or linear marks
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Coaxial Rotor Helicopters From what I've read and watched it looks like in the coaxial setup the rotors are stacked and turn in opposite directions so as to counter the torque in the main rotor and eliminate the need for the rotor in the tail. It looks like this:
Now, I have problems understanding why would this setup be efficient in any way that is not torque related. For example: the first (upper) rotor speeds up the air that lies above the helicopter and pushes it down - this gives the helicopter upward thrust. The problem is that the air that hits the second rotor was already accelerated by the first one, so that it already has linear momentum down and doesn't provide any thrust any more. This reasoning makes it look like the second rotor is useless except for countering torque. Am I missing something or is the setup really a drawback?
| As well as downward velocity the upper rotor imparts some rotation - this means the lower rotor will "hit the air faster". But the net result is less rotation of the air below the second rotor which in principle allows for greater efficiency.
As has been pointed out, the weight and complexity of the gearing mechanism needed to provide the coaxial counter rotation is considerable - with greater potential for reliability issues. On the other hand, counter rotation is used in some aircraft (jet engine) turbines to create greater efficiency. But for military helicopters that is not the primary concern.
| {
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Fourier's law and time taken for heat transfer? Fourier's law states that
"The time rate of heat transfer through a material is proportional to the negative gradient in the temperature and to the area."
A mathematical description of this law is given as
$$\frac{dQ}{dt}=-KA \frac{dT}{dL} \, . \tag{1}$$
where $K$ is the thermal conductivity of the substance in question, $A$ is the area of the substance normal to the direction of flow of heat current and $L$ is the lateral length of the substance.
However, we may write $dQ/dt=mC(dT/dt)$ -where $m$ is the mass of the substance and $C$ is the specific heat capacity of the substance-to obtain
$$m C \frac{dT}{dt} = K A \frac{dT}{dL} \, . \tag{2}$$
This would allow me to cancel the $dT$ on both sides of the equation to obtain
$$\frac{mC}{dt} = \frac{K A}{dL} \, , \tag{3}$$
which suggests that the time taken for the heat transfer to complete, between any two temperatures is a constant given by
$$ dt = \frac{dL m C}{KA} \, . \tag{4}$$
However, this is not at all the case.
What mistakes have I made in writing the above steps?
| You are equating the temperature change of an object due to heat loss (or gain) to the heat current across an unrelated interface. You should have written the equation (2) with partial derivatives, on the left side with respect to time, on the right side with respect to the coordinate. Therefore you cannot simply cancel the ∂T in this equation. One is a time change of temperature of an extended body with spatially constant T and the other is the spatial change of T along a heat conduction path unrelated to this body.
| {
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Is there a way to estimate fan capacity at altitude? Provided you know the capacity of a fan (flow rate) at constant speed and at sea level, is there an analytical way to predict what the flow rate would be at altitude? Or is this specific to the fan's design?
| Following answer is speculative.
Flow rate of air ($Q$) is determined once fan's geometry, its angular speed ($\omega$), and thermodynamic state of air (in particular its $\rho,\mu$) is specified. Since geometry of fan is not being changed, let us take any linear dimension associated with it (say, length of fan blade) as a length scale, $d$. Flow rate $Q$ is determined by these variables means that there exists a functional relationship:
$f(Q,\rho,\mu,d,\omega)=$constant
which results in dimensionless groups:
$g(\frac{Q}{\omega d^3},\frac{\omega d^2}{\nu})=$constant
or
$\frac{Q}{\omega d^3}=h(\frac{\omega d^2}{\nu})$
where $f,g,h$ are functions. Now since flow Reynolds number is high (and therefore flow is turbulent), viscosity plays little role in determining the flow (hypothesis). In that case,
$\frac{Q}{\omega d^3}\approx $constant.
Since angular speed of the fan isn't varying either I would guess that flow rate of the fan shall remain constant (to a good approximation).
One may object that density of air hasn't appeared in the final conclusion, so if one were to take the fan to such a height where there is practically no air, the equation still predicts the same flow rate, which is wrong. However in this case continuum approximation breaks down, and the entire analysis would become inapplicable.
| {
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Static Equilibrium Problem with 3 Tension Forces
I had some trouble solving these.
For the first one I assumed there wasn't a tension force on the 2nd mass, which gave me:
$$FT_1.\cos A = FT_2.\cos D$$ and $$FT_1\sin A + FT_2\sin D = F.g$$
Not sure how to approach 2nd problem.
| These questions are to do with the equilibrium of a point mass (knot) subjected to three forces provided by three masses and transmitted via pulleys and strings to the knot.
There are many methods of solving such problems but drawing a force diagram, which will be a triangle in this case, might help?
Because you have a static equilibrium situation the sum of the forces acting on the knot is zero so the vector addition of the three forces produces a triangle.
Solution can be obtained by resolving forces or by use of the sine rule and the cosine rule for the triangle of forces.
For your second problem you need to redraw the forces diagram which now will include the angle $f$.
| {
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Is there a Schrödinger equation for phase space? The Schrödinger equation is generally formulated in position space
$$
i \hbar \frac{\partial}{\partial t}\psi(x,t) = \hat H_x \psi(x,t) = \left [ \frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x,t)\right ]\psi(x,t)
$$
or in momentum space
$$
i \hbar \frac{\partial}{\partial t}\psi(p,t) = \hat H_p \psi(p,t) = \left[ \frac{p^2}{2m} + V\left(\frac{\hbar}{i}\frac{\partial}{\partial p},t\right)\right ]\psi(p,t).
$$
But is it also possible to derive a Schrödinger equation for position $\times$ momentum $=$ phase space? I assume in that case it should be an equation acting on a wave function $\psi = \psi(x,p,t)$ . So I guess simply multiplying the two Schrödinger equations above is not enough. How could one then derive a Schrödinger equation for phase space, if it is possible at all?
| The answers given so far are correct, but I don't think they hit the core of the issue. Quantum mechanics has either a position or a momentum representation where this or is an exclusive or. In general quantum physics is described according to a complete set of commuting operators, and in the case of momentum and space this means one must work in one representation, but not two at the same time.
Hamiltonian mechanics in phase space describes dynamics according to the energy surface in the $6n$ dimensional phase space, for $n$ particles in $3$ dimensional space, and $3$ dimensional momentum space. The condition $E~=~H$ on the energy surface reduces the space of motion to $6n-1$ dimensions. Lagrangian mechanics works in configuration variables that are $3n$ dimensional. This is then "half of phase space." Quantum mechanics then operates in this domain when working with configuration variables. This is even though QM and the Schroedinger equation uses a Hamiltonian.
| {
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For solids/liquids can $dP$ or $dV$ be assumed to be $0$ when calculating $\triangle H$ or $\triangle U$ by the following formulae? For solids/liquids can $dP$ or $dV$ be assumed to be $0$ when calculating $\triangle H$ or $\triangle U$ by the following formulae?
$$\triangle H=nC_p\triangle T + \int (\frac{\partial H}{\partial P})_vdP$$
or
$$\triangle U=nC_v\triangle T + \int (\frac{\partial U}{\partial V})_pdV$$
In some books I saw they take $dV$ as $0$ but not $dP$.Why?
| Bulk modulus of a substance is a measure of its compressibility. It is given by $\beta$ = V.$\partial$P/$\partial$V where $\partial$V/V is the fractiaonal change in volume, for infinitesimal pressure change $\partial$P. For liquides, $\beta$ is very high having magnitudes of the order of $10^9$ Pa often. This means, that for a small change in pressure, the fractional change of volume in the denominator is very small, making $\beta$ very large. Thus, it is reasonable to assume V of a liquid to be constant over an appreciable range of pressure.
| {
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Curved image in a miniscus Reflected in a coffee cup meniscus one commonly observes a tiny image of a straight horizontal florescent lamp behind the observer quite altered. It becomes curved and changes direction with a U-turn midway somewhat like $ <,\, >\, $ shapes.
Why does it turn backwards? What happens at the turning point of the image?
EDIT1
Depending on the changed position of observer, the horizontal U profile can split into two disjunctive parts. Totally, there are three arcs/images of a single light source.
Hope you find it interesting.
| I think that these are the patterns referred to in the question.
Quality is not great as I used water and not coffee and also lacked a third hand.
If one looks at near glancing incidence with the fluorescent light behind one then the pattern looks like image 1 below.
$AB$ is along the interface between the water and the cup and I think that it acts as the object to produce the reflected image $CD$.
$EF$ is produced by reflection from the water surface with the rim of the cup acting as the source.
Covering the rim lead to image $EF$ disappearing.
Raising the camera produce image $2$ and then raising it further image $3$.
The labels $AB, CD$ and $EF$ for image $1$ are the same for images $2$ and $3$.
Moving the camera round as in the first sequence of images changes the viewing angle and some of the rays which before completed the oval never arrive at the camera.
By chance I noticed a slightly different effect when the camera was facing the fluorescent light.
The image nearest the centre appears to be a reflection of the fluorescent light.
Without any proper ray optics analysis tools this is as far as I can go.
| {
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How did Maxwell's theory of electrodynamics contradict the Galilean principle of relativity? (Pre-special relativity) The Galilean principle of relativity:
The laws of classical mechanics apply in all inertial reference systems
OR
No experiment carried out in an inertial frame of reference can determine the absolute velocity of the frame of reference
These two statements written above are equivalent.
Maxwell's equations were discovered later. My question is (1) how did Maxwell's equations contradict the Galilean principle of relativity?
Furthermore if one studies the two postulates of Einstein's special theory of relativity, they can be simply translated as follows:
Postulate 1: Galileo was right.
Postulate 2: Maxwell was right.
(2) How did the Maxwell equations retain the same form in all inertial frames by obeying the Lorentz transformation?
| Imagine a stationary electron sitting next to a long length of wire with current flowing through it. Since the wire is neutrally charged, there is no electric force on the electron, and since the electron is stationary, there is no magnetic force.
Now imagine the whole system is moving lengthwise at a constant velocity. All of a sudden the electron is moving through a magnetic field and experiences a force. This seems to be a contradiction.
In relativity, this will be answered by the differing length contractions of the positive (protons)/negative (electrons) parts of the wire, creating an electric force on the electron that balances the magnetic force. This also serves to show the difficulty of distinguishing electric from magnetic forces, as one may become the other in a different reference frame.
| {
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Force between proton in a conducting shell and electron outside of shell? There's a proton inside a conducting shell and an electron outside of it. Inside the shell, there is no field due to the electron, but the electron feels the field due to the proton. Therefore the electron should move towards the immobile proton, but what happened to Newton's third law? Does the whole shell move?
| The whole shell moves because the electric field experienced by the outer electron has its origin in positive charges induced by the inner proton on the outer metal surface equivalent to the proton charge. The metal shell would even move without a proton inside because the electron induces positive charges on the surface near the electron so that a net attraction occurs. All this, of course, are minimal effects when the whole setup is floating in free space. And, as M. Enns has alluded to above, there is, of course, a field due to the proton inside the shell which induces negative charges on the inner surface that exert a force on the proton.
| {
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Do we need a bounded domain for the Laplace equation to have a non-zero solution $u$? If we have the Laplace equation for electrostatics in free space, that is
$$\Delta u(x) = 0 \quad \quad x \in \mathbb{R}^3,$$
is the only solution $u = 0$? And also, we only get non-zero solutions for $u$ if we instead consider the Laplace equation on some bounded domain? How can I mathematically show this if this is in fact the case?
| Since you are dealing with the whole space, you can take advantage of the so called Liouville theorem for harmonic functions.
Liouville theorem. Let $\phi : \mathbb R^n \to \mathbb R$ be a $C^2$ function such that $\Delta \phi =0$ everywhere. If $\phi$ is bounded (i.e., there is $k \in [0,+\infty)$ such that $|\phi(x)| \leq k$ for every $x \in \mathbb R^n$), then $\phi$ is everywhere constant.
This has a pair of corollaries.
Corollary 1. Let $\phi : \mathbb R^n \to \mathbb R$ be a $C^2$ function such that $\Delta \phi =0$ everywhere. If $\phi$ is bounded and $\phi(a n) \to 0$ as $a \to +\infty$, where $n \in \mathbb R^n$ is a fixed unit vector, then $\phi=0$.
Proof. $\phi$ is bounded, thus $\phi(x)=c$ constantly due to Liouville theorem. $c = \lim_{a\to +\infty} c = \lim_{a\to +\infty} \phi(a n) = 0$.
Corollary 2. Let $\phi : \mathbb R^n \to \mathbb R$ be a $C^2$ function such that $\Delta \phi =0$ everywhere. If $\phi(x)$ tends to $0$ as $|x|\to +\infty$ (i.e., for every $\epsilon >0$ there is $r_\epsilon>0$ such that $|\phi(x)|< \epsilon$ if $|x|> r_\epsilon$), then $\phi=0$.
Proof. Take $\epsilon >0$ so that $|\phi(x)|< \epsilon$ if $|x|> r_\epsilon$. In the compact set $|x| \leq 2r_\epsilon$, $\phi$ is continuous (since it is $C^2$) and thus it is bounded therein by some $M \geq 0$. Consequently $|\phi(x)| \leq \epsilon_r + M$ for all $x \in \mathbb R^n$. Liouville theorem now implies that $\phi(x)=c$ constantly. However this constant $c$ must satisfy $0 \leq |c |< \epsilon$ for every $\epsilon >0$ and thus $c=0$.
The second corollary uses a very weak requirement regarding how $\phi$ uniformly tends to $0$ for $|x|\to +\infty$. Obviously $\phi(x) \sim const/|x|$ is OK, but also much weaker convergences are enough, like $ \sim const / \ln |x|$...
| {
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Energy dissipated when two charged capacitors are connected in parallel The question at hand is:
"Two capacitors of capacitances $C_1$ and $C_1$ have charge $Q_1$ and $Q_2$. How much energy, $\Delta w$, is dissipated when they are connected in parallel. Show explicitly that $\Delta w$ is non-negative."
I'm confused about what the physical situation is. I took the assumption that these capacitors were somehow pre-charged, and then connected to each other in parallel without a voltage source. However, I don't understand how this would work. If a circuit is composed of only 2 elements, I don't see how they could be in any arrangement but series. Nevertheless, I tried solving it that way:
Two capacitors in parallel have the same voltage drop. Charge will be redistributed to make it the same voltage for both. Let $Q_1'$ and $Q_2'$ be the charges on the capacitors after they are connected. Now, picture the equivalent capacitor
$C_{eq} = C_1 + C_2 =$ $\frac{Q_1' + Q_2'}{V_f}$
conservation of charge:
$Q_1' + Q_2'= Q_1 + Q_2$,
$C_{eq} =$ $\frac{Q_1 + Q_2}{V_f}$
$V_f = $$\frac{Q_1 + Q_2}{C_1 + C_2}$
The initial energy of the capactiors is:
$U_0 = $$\frac{Q_1^2}{2C_1}$$+\frac{Q_2^2}{2C_2}$
$U_f = $$\frac{1}{2}$$(C_1 + C_2)V_f^2 = $$\frac{(Q_1 + Q_2)^2}{2(C_1 + C_2)}$
$\Delta U$$ = $$\frac{(Q_1 + Q_2)^2}{2(C_1 + C_2)}$-$\frac{Q_1^2}{2C_1}$$-\frac{Q_2^2}{2C_2}$
However, I don't see how this is necessarily non-negative.
| I will show that the expression $\Delta U$ is in fact negative, rather than positive, since the system dissapates energy, rather than spontaneously gaining it. That is, we must have $\Delta U = - \Delta w$.
Starting from your expression for $\Delta U$,
$$\Delta U = \frac{(Q_1+Q_2)^2}{2(C_1+C_2)}-\frac{Q_1^2}{2 C_1} - \frac{Q_2^2}{2 C^2},$$
we can find a common denominator
$$\Delta U =\frac{ (Q_1 + Q_2)^2 C_1 C_2 - (Q_1^2 C_2 + Q_2^2 C_1)(C_1 + C_2) }{(C_1 + C_2) C_1 C_2}$$
and expand:
$$\Delta U = \frac{Q_1^2 C_1 C_2 + Q_2^2 C_1 C_2 + 2 Q_1 Q_2 C_1 C_2 - Q_1^2 C_1 C_2 - Q_1^2 C_2^2 - Q_2^2 C_1^2 - Q_2^2 C_1 C_2}{(C_1 + C_2) C_1 C_2}.$$
Cancelling the first two terms in the numerator with their negative counterparts,
$$\begin{align}
\Delta U &= \frac{2 Q_1 Q_2 C_1 C_2 - Q_1^2 C_2^2 - Q_2^2 C_1^2}{(C_1 + C_2) C_1 C_2}\\
&= \frac{-(Q_1 C_2 - Q_2 C_1)^2}{(C_1 + C_2) C_1 C_2}.\\
\end{align}$$
Our final expression for $\Delta U$ is clearly negative, as the numerator contains only a negative sign and a squared number, and the denominator has only positive capacitances.
| {
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Time dilation in orbit When orbiting a massive object, perhaps a neutron star, you maintain your altitude while at the same time you are close to a very strong gravitational field. Being in orbit, and being in free fall is indistinguishable for the object in question. The object is not affected by gravity, locally.
Another free falling object can't be "less affected" than this first object, since the first object is in fact not noticing any gravity. Logically, both object should age equally fast if we ignore the relativeness of aging.
There is no doubt that the gravitational field is strongest around the first object, but neither object experiences any gravitational field. Object one looks at object two is in free fall, and vice versa.
Of object one records a video of object two, and then later plays that video - but adjust the playback rate to account for all doppler shift effects - what would object one see?
Is doppler the only effect causing the difference in relative aging in this case?
| Yep! Still applies.
Set $c=1$ and use the Schwarzschild metric. This metric is the unique solution for any vacuum, spherically symmetric spacetime which behaves like a Newtonian $\frac{1}{r}$ potential in the large $r$ limit. If the Schwarzschild radius is $R$, we just have have:
$${d\tau }^{2}=\left(1-{\frac {R}{r}}\right)dt^{2}-r^{2}d\theta ^{2}$$
Let's say the velocity from an asymptotic observer is $v=\frac{r d\theta}{dt}$. Then $r^2 d\theta^2=v^2 dt^2$, and:
$${d\tau }^{2}=\left(1-{\frac {R}{r}}\right)dt^{2}-v^{2}dt^{2}$$
$$\frac{d\tau }{dt}=\sqrt{ 1-{\frac {R}{r}}-v^{2}}$$
The prediction in special relativity is just
$$\frac{d\tau }{dt}=\sqrt{ 1-v^{2}}$$
so indeed time passes by more slowly due to the gravitational well.
Note that the "speed of light" here is reached when $1-{\frac {R}{r}}-v^{2}=0$, which gives $v_{max}<1$. This is due to the fact that $v$ defined this way is some sort of weird coordinate speed.
| {
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Is measurement of coordinates possible near strong gravity? We know that Schwarzchild metric describes an asymptotically flat spacetime. This means that far away from the event horizon we can safely interpret the $r$ coordinate as distance from the center.
But when close enough to the event horizon the curvature becomes significant and our common sense of $r$ breaks.
The question is that what is understood as measurement of the coordinates near very strong gravity?
| The Schwarzschild $r$ coordinate at a point is defined as the circumference of the circle passing through that point and centred on the mass divided by $2\pi$. This definition applies at all distances even inside the event horizon.
To see this take the Schwarzschild metric:
$$ ds^2 = -\left(1-\frac{r_s}{r}\right)dt^2 + \frac{dr^2}{1-\frac{r_s}{r}}+r^2d\theta^2 + r^2\sin^2\theta d\phi^2 $$
and integrate the proper length along the circle I've just described. Along this line $dt = dr = d\theta = 0$ and $\theta=\pi/2$, so the metric reduces to:
$$ ds^2 = r^2 d\phi^2 $$
And therefore the length of the circle is:
$$ s = \int_0^{2\pi} rd\phi = 2\pi r $$
| {
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Two balls A and B of the same size are dropped from the same point. If the mass of A is greater of the two, and if air resistance is the same on both, which ball will reach the ground first/simultaneously?
I thought that since the acceleration acting on them is same, both will reach the ground simultaneously. But the answer in the book says that ball A (the one having greater mass) will reach first. I mean, isn't this what people thought before (I guess it's true) Galileo performed his experiment, until they were proven wrong?!
| The heavier one will reach the ground first. Note that saying that the air resistance is the same for both is a bit vague, and it is by no means the same thing as neglecting air resistance. The drag force depends on the shape of an object (surface area) and therefore it is the same for two identical balls with different mass. But the acceleration experienced by the body is inversely proportional to its mass. Thus you have
$$\vec{F}=\vec{F}_g+\vec{F}_{drag}$$
Now the gravitational force has the very special property of being proportional to the mass of the body itself ($\vec{F}_g=m\vec{g}$), and so the equation of motion is:
$$\ddot{y}=g-\frac{F_{drag}}{m}$$
As you can see the second term on the right hand side becomes less and less important as the mass of the body is increased, resulting in a larger acceleration. If one neglects air resistance, on the other hand, such term vanishes no matter the value of $m$, and so the bodies fall at the same rate as in the experiment performed by Galileo.
| {
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Why is it that the change in internal energy always uses the formula with Cv in regards to pressure/volume/temperature changes on a gas? Normally I would associate the use of $C_v$ with finding the energy taken into or leaving a system when the volume is kept constant. However, the formula to find $\triangle E_i$ (change in internal energy) is $nC_v \triangle T$. Why $C_v$? Also, does this apply to pretty much anything? Or are there limitations?
| I rationalize the use of $C_V$ for finding $\Delta U$ of ideal gases this way:
Internal energy is the measure of kinetic energy on a microscopic level. That is, the average velocity of all the individual particles that make up a system.
For ideal systems, the volume of a container which makes up the system will not have any affect on the average kinetic energy of the particles. Think about it, a system of vast volume would still contain the same molecules with the same translational, vibrational, and rotational energies as a system of small size.
Ideal is important, as if the system were non-ideal, then interactions between molecules via attractive forces and collisions would have different effects depending on the volume of the container.
For real gases, a vast container would see much more space between particles and therefore less interaction between them. And for a small container, particles will be much closer and the collision rate will be higher.
Overall, for real gases, we can say it is not true that internal energy is independent of volume, but the opposite is true for ideal gases. Thus for any ideal gas, $C_V dT=dU$.
| {
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Is Physical Information Real My question is about physical information and whether or not it is something real or merely fancy math. My problem is that physicists seem to talk as if information is real - however when they describe what it is they make it sound like it's merely a mathematical abstraction.
For example, if a bowling ball were to fall into a black hole, the physical bowling ball disappears from our universe completely - however (according to Hawkings) the bowling balls physical information is stored in two dimensions in the event horizon.
This is very similar to how the idea of information is used in the Universe is a hologram theory - in which our universe is a three dimensional representation of two dimensional information.
The question I have is: Is this information real or is it just mathematical BS? I mean, if my cat is merely the 3D representation of 2D information about my cat - then where is this information and what is it? Is it energy? Is it potential? Is it whatever determines whether or not a quark spins to the left or to the right? Or is it merely fancy math?
| From the point of view of quantum field theory, the only difference between you and complete vacuum is the arrangement of quantum fields. Namely, you and vacuum ("nothing" in a layman's language) are made of the same "things", the only difference is how you configure these "things". That's information.
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Consequence of approaching infinite mass for near-light speed particles Considering that the inertial mass of an object approaches infinity as the speed of the object approaches $c$, and that inertial mass equals gravitational mass, does this not imply that particles nearing $c$ would have gravitational mass approaching infinity?
Postulating that in fact gravitational mass nears infinity, would an object moving at near $c$, due to approaching infinite gravitational mass, cause profound gravitational acceleration on very distant objects? Infinity is a big value, even after decaying at $1/d^2$.
| the whole reason that the inertial mass increases as an object approaches C is that it requires an increasing amount of energy to be able to accelerate faster, hence to accelerate an object (that has mass) to C would require infinite energy (and break a LOT of physics)
| {
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(SR) Lorentz low speed approximations In Special Relativity, the standard Lorentz transformations are:
$t' = \gamma (t - \frac{vx}{c^2}) \\
x' = \gamma (x - vt) \\
y' = y \\
z' = z$
However, if we make a low speed approximation where $v \ll c$, then $\gamma \approx 1$ and we get:
$t' = t - \frac{vx}{c^2} \\
x' = x - vt \\
y' = y \\
z' = z$
These are the Galilean transformations almost, except for that remaining $vx/c^2$ which we can't ignore for large $x$. What does this term mean and what is it's physical significance?
| At flippiefanus's suggestion, I am converting this from a comment to an answer:
The physical significance of the term is exactly what it looks like --- when $v$ is small and $x$ is comparably large, $t\neq t'$ (even approximately). Thus when you and I pass each other even at very slow speeds, we will still disagree substantially about the times we assign to far-distant events.
| {
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Wave function in a semi infinite line How to normalize the wave function of a particle in a semi-infinite 1D interval $x\in [0, \infty],$ with boundary condition $\varphi(0)=0\,?$ Hamiltonian is $H=P^2/2M$ and wave function $\varphi(x)=C\sin(xp/ \hbar)\,.$
| The wave function:
$$\varphi(x)=C\sin(xp/ \hbar)$$
Isn't normalisable. Just try it:
$$1=\int_{-\infty}^{+\infty}[\varphi^*(x)\varphi(x)]dx=C^2\int_0^{+\infty}\big(\sin(xp/ \hbar)\big)^2dx=C^2\Bigg[\frac{x}{2}-\frac{\hbar}{4p}\sin(2xp/ \hbar)\Bigg]_0^{+\infty}$$
It has no Real solution for $C$.
| {
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Why wave theory cannot explain photoelectric effect and provides evidence for particle nature of light? I am able to understand how light can be modeled to have wave characteristics from Young's double slit experiment.
But I am unable to comprehend how we can understand light to have particle characteristics from the photoelectric experiment. How is it wave character not able to explain the phenomena observed in this experiment? And how is it that particle nature defeats the wave theory?
| Take it by the other way, for the emission as well for the absorption of EM radiation photons are a good description. The photoelectric phenomenon is a good example for the point that EM radiation is made of photons.
The description of EM radiation as a wave has some weaknesses. For a thermic source of EM radiation one will not be able to measure nor the amplitude nor the wavelength directly. For a monocromatic source (with small aperture) one can measure behind a double slit the distances between the fringes of the intensity distribution which is explained by the interference of outgoing from the two slits circular waves (Huygens principle). But since the phenomenon of fringes appears behind every single sharp edge and even for a stream of single emitted photons, the explanation with Huygens principle is not holdable for this case. Than more the evidence for wave nature is made from a pattern which intensity distribution has the equation of a wave (a sin equation).
But really there is EM radiation which directly measurable frequency. Radio waves are produced by accelerating electrons in an antenna rod by a wave generator fore and back. This induces the emission of photons. Such a EM radiation clearly has the properties of energy transfer with a wave characteristics.
The description of monochromatic EM radiation with a associated wavelength /frequency is helpful but not necessary. The description by the energy of the involved photons is enough.
More about photons, EM radiation and radio waves see this answer.
| {
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Theoretically, could there be different types of protons and electrons? Me and my friend were arguing. I think there could theoretically be different types of protons, but he says not. He says that if you have a different type of proton, it isn't a proton, it's something else. That doesn't make sense to me! There are different types of apples, but they're still called apples!
He says that's how protons work, but can we really know that?
| There are two ways to distinguish particles: you can either measure a difference in the intrinsic physical properties of the particles - say mass, for example - or track the trajectory of each particle with infinite precision.
Since a proton is always formed by one d-quark and two u-quarks, every proton has the same mass, charge and spin. The other possibility is contradicted by quantum mechanics, specifically the Heisenberg uncertainty principle.
It seems that your friend is correct, then. In fact, since you can't distinguish protons, it's possible to say that every single proton in the universe is the same!
| {
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Could the gravitational field equations be formulated in term of the Riemann curvature tensor (as opposed to the Ricci curvature tensor)? The most symmetries and identities in Riemannian geometry are in term of the Riemann curvature tensor. One may ask why the gravitational field equations are not in term of this main tensor of (pseudo)Riemannian geometry? i.e. without any contraction with the metric.
However contraction with the covariant derivative, torsion tensor (and the like) is OK.
| The Einstein field equations are in terms of the Riemann curvature tensor! It is just contracted with the metric once to give the Ricci curvature tensor. That then is contracted with the metric again to give the Ricci scalar. A combination of both tensor and scalar.
So you have the Ricci tensor (two indices) from the Riemann tensor (four indices): $R_{\nu\beta} = R^\alpha{}_{\nu\alpha\beta}$.
Then you form the Ricci scalar: $\mathcal R = g^{\nu\beta} R_{\nu\beta}$.
The Einstein tensor then is this: $G^{\mu\nu} = R^{\mu\nu} - \frac 12 \mathcal R$.
And finally the field equation is $G^{\mu\nu} = 8 \pi G T^{\mu\nu}$ where the left $G$ is the Einstein tensor and the right $G$ is the gravitational constant.
| {
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Working principle of inverter I got project on the working of inverter from school. I know this that DC inverter has an alternator switch which constantly changes its direction so that magnetic field is produced in primary coil due to which current is induced in secondary coil and we get output AC. So according to all this Electromagnetic induction should be the working principle behind the working of DC inverter. But DC can't take part in EMI, I know alternator is being used but it doesn't feels right. I hope I am right till this point I am taking following image as reference:
Please don't get mad at me if I got everything wrong.
| You are right. The alternator turns the DC of the battery into AC, which allows for changing magnetic fields in the primary. A transformer need a changing magnetic flux (which is why "DC doesn't work") - the alternating switch keeps changing the direction of the current, so the flux keeps changing. This induces e.m.f. in the secondary, and the result is an AC voltage on the output of the circuit.
The voltage wave form on the primary will be a square wave: the output waveform will be more complicated, depending on details of the resistance of the coils and the load. Obviously, the higher the switching frequency, the more regular the shape of the output: at very low frequencies, the output will "decay" between switching of the input alternator.
| {
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What is the power output for an OPO pumped below threshold? I'm trying to figure out approximately how much power will be output from an OPO when pumped below threshold (ie Pin=0.1Pth). All the research I've done turned to the same basic equation which only works for above threshold (~(sqrt(Pin/Pth)-1) which is always negative when pumped below threshold).
Does anyone know an equation I can use (or point me to a resource that can help) to figure out this problem?
Thanks,
John
| Below threshold operation means that the gain inside the optical parametric oscillator is smaller than the overall losses. Your gain medium effectively changes the intra-cavity losses. Below threshold, the net loss is (still) positive and the system is (still) similar to an ordinary cavity. How much light will be transmitted, reflected and built up inside the OPO depends on the exact loss channels and magnitudes, keywords to look for are "under-coupled", "over-coupled" and "impedance-matched" cavities. A brief overview over these terms is given by Bond, Brown, Freise, and Strain, Living Rev. Relativ. (2016), doi:10.1007/s41114-016-0002-8, chapter 5 and particularly figures 28 and 29 therein.
If there is no input light at the lasing frequency, in steady-state operation pumped below threshold, your OPO will not output any light (at least no coherent state - instead, you will get a squeezed vacuum state, but that is a different topic).
| {
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Falling rotating object in higher order potential fields For which $n$ would an object with a non-zero rotation fall to the center of this field?
$$\alpha >0\\ V(r) = \frac{\alpha}{r^n}$$
(Apparently it should never touch the center if it has non-zero rotation and $n=1$). I am totally stumped, as I cant see why the order should change whether or not it falls into the center. Any ideas/explanations would be greatly appreciated.
| Consider an object with non-zero rotation, and velocity vector $\bf{v}$. That it has a non-zero rotation means that, at least as long as $r \neq 0$, $\textbf{v}\times\textbf{r} = rv_\perp \neq 0$, where $\bf{r}$ is the radial position vector and $v_\perp$ is the component of $\bf v$ perpendicular to $\bf r$. As a simple consequence of conservation of angular momentum we must have
$$
v_\perp \propto \frac{1}{r}.
$$
This imposes
$$
\dot{v}_\perp = \frac{dv_\perp}{dt} = \frac{dv_\perp}{dr}\frac{dr}{dt} \propto \frac{1}{r^2}v_r.
$$
On the other hand the potential
\begin{align}
V(\textbf{r}) &\propto \frac{1}{r^n}
\end{align}
imposes an an acceleration
$$
\textbf{a} \propto -\nabla V \propto \frac{1}{r^{n+1}}\textbf{e}_r,
$$
whence $\dot{v}_r = |\textbf{a}|$. We thus have
$$
\frac{\dot{v}_\perp}{\dot{v}_r} \propto r^{n-1}v_r.
$$
For $n > 1$ we have
\begin{align}
\lim_{r\to 0}\frac{\dot{v}_\perp}{\dot{v}_r} = 0.
\end{align}
We can interpret this to mean that as the object is brought closer to the center the acceleration gradually becomes purely radial. If the potential acts attractively this means that the object may eventually collide with the center. However, for $n = 1$ we have
\begin{align}
\lim_{r\to 0} \frac{\dot{v}_\perp}{\dot{v}_r} \propto \lim_{r\to 0} v_r \neq 0,
\end{align}
which means that even when the object is brought closer to the center the acceleration will never become radial: the object will never accelerate directly towards (nor directly away from) the center. And since it was originally not moving towards the center, it will never collide with the center.
| {
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If an Ideal Current Source has infinite resistance, how is current flow possible? Where does current come from if an ideal current source has infinite resistance? Current can't move through the current source because of the resistance, so I don't really understand how and why an ideal current source has infinite resistance.
| "Ideal" in this context means that no consideration is being given to how the ideal properties (constant current regardless of load, infinite resistance) can be achieved in practice. As freecharly puts it, it is an abstract concept.
The process of analysing electrical circuits starts with modelling real electrical components, which have several properties, as combinations of ideal elements (like those in chemistry) which each have a single property - pure resistor/capacitor/inductor, pure voltage/current source, etc.
No real current source has the properties of an ideal current source, and it is as pointless to ask how an ideal current source can have infinite resistance yet generate a finite current, as it is to ask how any real string can be inextensible yet exert a finite tension when you pull on it. However, for the purpose of analysing electrical circuits most real current sources can be modelled with sufficient accuracy as an ideal current source in parallel with a finite internal resistance. A real current source will also have some capacitance and inductance, which become significant at high frequency.
| {
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Motion of one body with reference to another I studied that Galileo was punished by the church for teaching that the sun is stationary and the earth moves around it. His opponents held the view that earth is stationary and sun moves around it.
The question i want to ask is if the absolute motion has no meaning,are the two viewpoints not equally correct or equally wrong?
Thanks in advance for any help.
| This is an interesting question. You say that "absolute motion has no meaning." Can you explain that a little more? Suppose that I am in a spacecraft. For simplicity imagine that I am far from any other object, so I can neglect gravity (I don't think this restriction is necessary, but it simplifies things.) I can certainly tell if I'm accelerating or not without looking outside of my spacecraft.
The two frames that you imagine are fixed to bodies in space. One is fixed to the earth, and the other is fixed to the sun. But measurements would show that both are accelerating, and one at a greater rate than the other. The observed motion of the other solar system bodies is certainly simpler in the frame fixed to the Sun, and by Occam's razor we might conclude that earth-orbiting-sun is the better explanation of the observed motion.
But there's another problem. The body-fixed frames are accelerating, so we can't apply Newton's laws. We have no theory that explains the motion, only observations.
The way out of the dilemma is to make all observations from an inertial frame. Now Newton's laws are valid, and we have a potential theory. And this theory tells us unequivocally that both the earth and the sun revolve around their center of mass (which is inside the sun). Neither is "stationary".
All this is a long way of saying that there is a sense in which absolute motion does have meaning, and that the two points of view are not on equal footing. And in fact, they are both incorrect.
| {
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Why is velocity inversely related to pressure in a flow? I've seen the equations that give this relationship, and I understand the math and have seen it worked out in problems. But I don't have a qualitative, conceptual grasp on the relationship.
Is the pressure that which is exerted by a small element of the flow, or exerted on a small element? Maybe it is the pressure exerted on or by the boundary?
And once pressure is defined, why is it related to velocity? Why does something push less if you speed it up? I would have guessed a faster flow pushes harder.
| You say "if you speed it up".
Who speeds it up?
The only thing that can speed it up is a pressure difference.
(Let's do it horizontally, so we can ignore gravity.)
That's why less pressure means higher velocity, and vice-versa.
| {
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Why is direct current needed to produce an electromagnetic field using a solenoid? I am performing an experiment for school investigating the magnetic force of a solenoid. While doing this experiment I realized that I needed to connect the solenoid to the DC output of the power supply instead of the AC.
I am perplexed since for current induction, fluxtuations in the magnetic field are needed. I thought this would be the same "the other way round", i.e. using a fluxtuating current to create a magnetic field.
| To produce a steady magnetic field by a current in a coil you need a direct current. If you use an ac current the magnetic field would change with time and change direction every half period of the ac current.
| {
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If a system is deterministic, will it still be deterministic if time is reversed? If you were to drop a ball, it would be easy to calculate when it will hit the ground and how much energy will be absorbed by the ground (let's assume there's no air resistance and the ball does not bounce).
If you were to then reverse time, would energy from the ground gather rapidly toward the location of the ball, creating a powerful but microscopic impulse capable of launching the ball to its original height?
If the answer is yes, then the time reversed system should in fact be deterministic.
If the answer is no, then there could not possibly be any way to predict when the ball will bounce or how high it could go. The system would not be deterministic.
Which result is closer to reality?
| If the equations in whatever deterministic theory you are using are reversible in time, then you can use the current state to predict both the future and the past your system just as easily, because it's all in the mathematics. However, deterministic equations can suffer from sensitivity to initial conditions. Generally, that happens in both directions of time too, so you can only predict the weather about as far into the future as you can into the past. You might be able to predict how much heat a bouncing ball will lose, and thereby predict the height it was dropped from if you know how many bounces it took, but not if you wait until the ball comes to a stop because stopping a ball is not going to happen using equations that are time reversible.
| {
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What is the prevailing opinion in scientific community about Hans C. Ohanian's description of spin? In the paper What is spin?, Am. J. Phys. 54 (1986) 500, by Hans C. Ohanian, spin is described as a circulating flow of energy in the wave-field of a particle. Is this the generally agreed upon explanation of intrinsic angular momentum or just a fringe theory?
(A similar thread exists on Reddit, but I couldn't find a satisfactory discussion there.)
| Ohanian's paper shows that spin can be understood as a circulating flow of energy, and is a wave property, valid in both classical and quantum mechanical formulations, rather than inherently and mysteriously "quantum mechanical" in nature. A fine point not needed for calculations that can be ignored by the pragmatic experimental physicist. But to me (and apparently the author) it's comforting to have physical intuition so lacking in much of conventional qm.
| {
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Correlating lattice time to physical time in Lattice boltzmann Method In LBM literature, more often the time-dependent behavior is plotted with respect to the number of time-step(lattice time $\Delta t$), I also read this note from Jonas Latt, however it stills seems unclear to me how to correlate $\Delta t$ to physical time t.
For instance, I want to simulate some fluid(Renolds=20), on a 2D grid(1 meter by 1 meter, $N_x$=100 in each dimension), then according to Jonas's note, we are advised to choose a time step ~${N_x}^2$, i.e. $N_t$ ~ $10000$, however, could someone demonstrate with these information available, what is the exact physical time corresponding to a single lattice time-step $\Delta t$ ?
Another curiosity:
It's suggested to choose time step ~${N_x}^2$, otherwise the numerical scheme may be unstable, so if we choose less time steps, then the intermediate results could be unphysical or numerically incorrect? Does this mean LBM is more suitable for steady-state simulation rather than transient-behavior studying?
| In those notes, it is stated that time $t_p$ is divided by some reference time-scale $t_{0,p}$. And the time step in the discretised system is $\delta_t=t_{0,p}/N_\mathrm{iter}$. So if you perform $N_t$ time-steps then this corresponds to a time interval of $\Delta t_p=N_tt_{0,p}/N_\mathrm{iter}$.
| {
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Is there a notion of causality in physical laws? I was reading "A Few Useful Things to Know about Machine Learning" by Pedro Domingos and towards the end of the paper he makes this statement:
"Many researchers believe that causality is only a convenient fiction. For example, there is no notion of causality in physical laws. Whether or not causality really exists is a deep philosophical question with no definitive answer in sight..."
I was surprised by this statement because apart from Heisenberg's uncertainty principle, everything else (that I know of) in physics seems to operate under the assumption of causal relations. If you have an equation that describes a definitive outcome as the result of some input factors, then it is describing a causal relationship, is it not?
| All answers are good; I would like to add something missed though: Newtonian Mechanics is not exactly a deterministic theory. In other words, sometimes, even if there is not a cause, an effect can just occur. SeeNorton's dome.
In short, determinism (uniqueness theorem in Newton Mechanics) is a mathematical consequence of a few conditions/assumptions; Nature does NOT have to always own these conditions.
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What actually oscillates in quantum harmonic oscillator with given frequency? Quantum harmonic oscillator is said to be describing motion of microscopic stuff (like atoms in molecules). But unless one keeps on measuring the position on the atom, it doesn't exist at all. It's in superposition of several possible positions (position eigenstates). How can we even assume oscillation and specifically with given frequency of oscillation?
| First of all, I would agree that the particle "is in superposition of several possible positions (position eigenstates)" but I would strongly disagree that "it doesn't exist at all." But anyway: if the quantum harmonic oscillator is in an energy eigenstate, then nothing oscillates at all; this is true for any energy eigenstate of any time-independent Hamiltonian. The phase factors $e^{-iEt}$ from solving Schrodinger's equation always cancel. But if you're not in an energy eigenstate, then the expectation value $\langle \psi | \hat{X} | \psi \rangle$ does indeed oscillate in time with a frequency of $\omega$, just as with a classical harmonic oscillator. This is a special case of Ehrenfest's theorem.
| {
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Do circular wheels maximize mechanical efficiency?
I recently wondered whether there is a simple proof that circular wheels maximise mechanical efficiency. By this I mean:
Show that for a wheel with a given width and cross-sectional area, the
circular wheel reaches the bottom of an incline faster than a wheel
with any other shape.
This sounds like a rather elementary problem in classical mechanics but I haven't found a proof of this fact in any of the classical mechanics texts that I have read.
Note 1: I assume that the distribution of mass is uniform across the wheel so comparable wheels will have equal mass.
Note 2: Dry friction is assumed to be present.
| When it is coming down the motion of the wheel of any shape is one dimensional. Degree of freedom of two dimensional object is one when no torque is acting . In this case of if we consider free fall then only acceleration is gravitational acceleration which will not apply torque here. By using the two dimensional object we can utilize maximum energy on bringing down wheel .
If we consider a wheel of n polygonal shape and distance d and mean velocity v . Then we have time t=s/v or, t=na/v where a is the length of arm of the polygon. In our case a and v are constant so if n tends to zero then t tends to zero or (n-1) vertices of the polygon tends to zero. The only geometric shape which is close to that requirements is a circle. Since there is no existence of 2d object so circle with thickness able to withstand different environmental resistance and reach bottom steadily is used.
I hope this at least takes us closer to the solution
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Why Objects still move under the effect of fictitious force? As I read in many books and websites, An inertial force is a force that resists a change in velocity of an object. It is equal to—and in the opposite direction of—an applied force.
If the inertial force (P) is equal to the applied force (F), then the net force (P+F) affecting the object is equal to 0. I wonder why object still move although the net force is 0.
| Real forces are always interactions between two objects. If we think that a force, such as your inertial force, is acting on an object, we should try to identify the other object. If we can't, then we were wrong; there is no force.
The inertial force in your hypothetical problem does not pass that test. You have imagined and applied force that does not exist. There is no force opposing the applied force F, so there is no contradiction.
The inertial force which does exist in your scenario is applied by the block on whatever object is applying F to the block. That inertial force and F make up a third law pair. But the behavior of an object is determined by the forces exerted on it, not by the forces it exerts.
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What is the difference between conventional current and electronic current? what is the difference between conventional current and electronic current?
How are they linked to one another?
| The conventional current is defined as charge per unit time transported in a certain direction. The current direction is the direction of positive charge movement. A positive current is also negative charge per unit time moving in opposite direction to the corresponding positive charge. In conventional current, the type of charge carrier is irrelevant. It can also be produced by positive and negative charge carriers at the same time moving in opposite directions like currents in electrolytes or ionized gases. Electronic current is current produced by the movement of negatively charged electrons. This is usually the case in metals. When electrons produce a positive current in a certain direction, this means they are actually moving in the opposite direction.
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Generalized version of work-energy theorem I know that for rigid bodies only the work-energy theorem states that the net work done on the body equals the change in kinetic energy of the body since a rigid body has no internal degrees of freedom and hence no other forms of energy such as potential energy.
Is there a most generalized form of work energy theorem that is valid for rigid as well as non rigid bodies and for conservative as well as non-conservative force?
I would like a work-energy equation that would be valid for point particles, rigid bodies and non-rigid bodies.
| The "generalized work energy theorem" is extremely simple:
The work done on or performed by the system equals the difference in energy of the system. The energy of the system is the sum of all different kinds of energies of your system: kinetic, potential, chemical, etc.
How does the work split across different types of energies? I can't tell you, that really depends on the body in question.
Ultimately, you only have two "energies": potential energy of conservative forces and kinetic energy. All other energies, such as heat or chemical energy or the like constitute an effective description of something that ultimately can be described by kinetic and potential energy only. Stored heat energy is undirected kinetic energy of internal degrees of freedom (how to model that is one of the questions of statistical mechanics and condensed matter theory). Chemical energy is mostly potential energy of atomic bonds and atoms. Reversible deformation results in potential energy from potential energy in distorted atomic bonds. There is no simple equation to accurately model all of it for every system - otherwise we wouldn't use effective descriptions and physicists would be out of work soon.
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Gravity between two Photons (I searched for an answer online already but I couldn't quite find what I was looking for...)
I thought about this for a long time now. If two Photons fly in the same direction, one behind the other one, for my understanding the one behind the other one should be pulled towards the photon in front of it due to it's gravity, and because it cant get faster it should increase it's frequency and therefore gain energy. The one in front cant be pulled backwards though because gravity travels with the speed of light itself(?) and therefore the gravity of the rear photon cant reach the one infront of it, which would therefore not lose energy.
But that would break the law of conservation of energy, wouldn't it? So I'm confused...
Am I thinking something wrong? Or how does it work/what would actually happen in this scenario?
Thanks for answers in advance!
| The speed of gravity is also only c, so the photon in front never gets overtaken by the gravity of the photon in the back. The photon in the back travels through the gravitational field of the one in the front, but since the direction of pull is to the front while the back photon is already travelling with c in the direction on the pull it already has the maximum velocity which it can not exceed. Since the distance stays constant so does the energy.
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Why does the bicycle move forward after I give it an impulse with the help of brakes? Whenever I need to slow down my bicycle I use brakes, momentarily pressing them and then releasing. For some reason, this momentarily halts the bicycle and when I release the brakes, the bicycle continues to move forward albeit with a velocity less than what it was before.
So far I haven't broken any bones while riding the bicycle, so I guess I can use this method, but what I am confused about is that if the bicycle momentarily comes to almost standstill, why does it move again? Shouldn't it stay at rest once it stops? Or is there something I am missing?
| When you tap the breaks, the bicycle stops but your body keeps moving. Almost immediately after you let go of the breaks you push/hit the front handlebars and the bike keeps going. Having your bike coming to a full stop (without your body slowing down) without problems is only possible at low speeds. At higher speeds the bike probably doesn't slow down as much as it feels like it does.
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Charged particle as observed from an inertial and a non-inertial frame of reference A charged particle fixed to a frame $S^\prime$ is accelerating w.r.t an inertial frame $S$. For an observer A in the $S$ frame, the charged particle is accelerating (being attached to frame $S^\prime$) and therefore, he observes it to radiate. However, for the non-inertial observer B standing on the non-inertial frame $S^\prime$, the charged particle is at rest, and therefore, does not radiate at all.
A infers that the charged particle radiates but B infers it doesn't. Can both inferences (mutually contradictory) be correct simultaneously? If yes, how?
| This is not complicated.
*
*The question of whether the particle radiates must be resolved in
its frame of reference.
*The question of whether the radiation is observed must be
resolved in the observer's frame of reference.
Changing the observer or adding more or different observers does not impact whether or not there is radiation.
Acceleration is not determined relatively, you can tell if you are in an accelerating frame without ref to an external frame because you feel force.
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How attitude indicator (gyro horizon) adjusts to the Earth's curvature? Image a plane is flying from North to South and is crossing equator. How gyro horizon would maintain correct pitch angle? (Or East-West?)
I assume that pitch angle is correct at takeoff, so the further plane flights, the more difference would be between pitch angle relative to current g-force and pitch angle shown by attitude indicator.
Or do I understand gyros wrong?
| The attitude indicators need to have devices in them to correct for precession in the gyros caused by turns. The system used in mechanical gyros is based on a collection of pendulous vanes. Basically, for short times the gyro controls the attitude indicator, for long term the pendulous vanes are pulled by gravity in the direction of "down" and they adjust the gyros to keep them oriented correctly. More advanced attitude indicators use different technologies to detect "down", and use that to correct the orientation of the gyro, but they are all, fundamentally, linear accelerometers like was used in the original Wii controller before they added gyroscopic sensors in later versions.
Edit, just to make it explicit: the problem of keeping the horizon correct in spite of the Earth's curvature is small compared to the problem of precession caused by turning. So, solving turning precession solves curvature for free.
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Why don't light beams refract when they hit a curved surface I saw a diagram that showed a light beam hitting a curved glass block (half circle) and nothing happened until it hit the flat surface, in which it refracted. Why doesn't it also refract when it hit the curved surface. I also tried this in real life with the same instruments and it proved to be so. So why?
Image of diagram:
https://postimg.org/image/6kukkz3it/
| Well, your light ray must have hit the curve at an exact point where the angle of incidence was zero, (the light ray was perpendicular to the exact point at which it hit the semi circle), which means there was no refraction.
But in general, a ray of light can bend when it hits a curved surface, it just depends at what angle it hits, and what place on the curve it hits. The thing about a curved surface is that each of its points is at a different angle and thus has a different normal and thus refracts light coming from the same direction differently. On the diagram it just happened to be hitting straight on, so no angle of incidence, so no refraction (as I've already said). (I think, don't take this answer for sure).
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Can "dressing" a qubit multiple times increase the dephasing time more than dressing it once? A paper was recently published (arXiv pre-publication link; I've only read a more popular description of the content because of my lack of domain knowledge) describing how "dressing" a qubit in a microwave field consistently resulted in a roughly tenfold increase in the longevity of a usable signal-to-noise ratio of the qubit.
I'm not an expert in the field, but my takeaway is that by keeping a qubit suspended in an oscillating electromagnetic field, less noise could affect the qubit, preserving its signal for much longer without corruption.
Considering the results were so beneficial, is there any way to use multiple electromagnetic fields to "stack" this effect, increasing usable signal-to-noise lifetime?
| Yes, it can. In particular, dressing an already dressed qubit can protect from the decoherence introduced by fluctuations of the microwave field used for the (first) dressing.
This is, eg, described here:
http://dx.doi.org/10.1088/1367-2630/14/11/113023
The principal idea of CCD is to provide a concatenated set of continuous driving fields with decreasing intensities (and thus smaller absolute value of fluctuation) such that each new driving field protects against the fluctuations of the driving field at the preceding level.
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Could we specify that a quantum space-time (any) belongs to what mathematical spaces? e.g to discrete spaces? Apparently, there is not a unique and generally accepted definition for a quantum space-time yet. In any case, I'd like to know that, at this stage, could we specify that a quantum space-time (any) belongs to what sorts of mathematical (or algebraic) spaces? For example, discrete spaces? Sorry if the question doesn't make sense somehow.
| The influence of mass on the space-time geometry can be described with quantum theory, where we can state the expected effect of mass on curvature of space-time.
Also it can be theorized that if mass has influence on space-time curvature, in quantum situations where mass does not curve space-time at measured coordinates, it will have done so at other coordinates. The probability of mass affecting space-time in normal situations equals the general relativity (GR).
At extreme situations such as Big Bang and black holes, the behavior of mass differs from "normal" situations. The effect of gravity, bringing particles in closer proximity, increases the chance of two particles taking up the same coordinates in space-time, which is impossible. The chance of dislocation thereby increases.
It is due to this dislocation that there is no such thing as singularity. GR calculations of minimum radius of the mass contained confirm this too. It is for this reason the big bounce theory becomes more popular again.
All matter turns into dark matter over time due to gravitational pull/Higgs field interactions, creating micro black holes throughout the universe (dark matter). These continue their existence until they clog together, bounce back etc.
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Operator functions: why is $f(A)$ uniquely defined? In Nielsen and Chuang, they write: Let $A = \sum_a a|a\rangle \langle a|$ be the spectral decomposition of $A$. Define $f(A) = \sum_a f(a) |a \rangle \langle a|$. Apparently this is uniquely defined. I'm having trouble seeing why this is.
If we used some other orthonormal basis of eigenvectors of $A$, say $A = \sum_b b|b\rangle \langle b|$, then why is $\sum_a f(a)|a\rangle \langle a| = \sum_b f(b)|b\rangle \langle b|$? I think there must be some property about eigenvectors sharing the same eigenvalues, but I'm unsure about what I'm missing.
| It's a standard result of linear algebra that the spectral decomposition of an operator is unique (up to a trivial reordering of the eigenvalues/vectors). If you could write $A$ as both $\sum_a a | a \rangle \langle a |$ and $\sum_b b | b \rangle \langle b |$, with the $a$s and $b$s nontrivially related, then this would violate the uniqueness of the eigendecomposition.
If you were to work in an orthonormal basis other than the eigenbasis, then the operator wouldn't be diagonal, but would look like $A = \sum_{n,m} c_{nm} | n \rangle \langle m |$. You are correct that letting $c_{nm} \to f(c_{nm})$ would be basis-dependent and generally not particularly interesting or meaningful.
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How many photons can exist in a cubic box of unit volume, simultaneously? Suppose that we have a cubic box of unit volume. Simultaneously, how many photons can exist in such box? Is there any limit?
| Photons are bosons so unlike fermions such as electrons there is no restriction on multiple photons occupying the same energy state. Consequently there is no limit on the number of photons you can put in your box.
There is an upper limit to the total energy density in the box since if you make it too high the box willcollapse into a black hole. However this limit is absurdly high and for most purposes can be ignored.
Note that individual photons can have arbitrarily low energies, so for any given energy density you can have an arbitrarily high photon number density simply by using photons of a low enough frequency.
Later:
Wood raises an interesting point in a comment. Suppose we have a cubical box of side $d$ then the largest wavelength/lowest frequency photon that it is possible to fit in the box has a wavelength of $2d$, so the energy of that photon is $h\nu = hc/2d$.
Let's take a black hole with a Schwarzschild radius of $d/2$ (we'll use this approximation since black holes aren't cubes) in which case the mass is:
$$ M = \frac{c^2r_s}{2G} \approx \frac{c^2d}{4G} $$
and the energy is just $E=Mc^2$ so:
$$ E \approx \frac{c^4d}{4G} $$
The number of photons is just this nergy divided by the photon energy we calculated above of $hc/2d$ so the number of photons in the box is (approximately):
$$ N \approx \frac{c^3d^2}{2Gh} $$
And there's your expression for the maximum number of photons you can get into your box before it turns into a black hole. For a $1$m box I make that about $3\times10^{68}$ photons.
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Acceleration during collsion Okay so, this question is pretty simple. When two objects of constant velocity collide, is there a constant deceleration upon contact? or does it follow some unique function that can only be evaluated by testing it? Furthermore, if it is in fact constant, how do we know and how did we originally test it? (Assuming that we knew before the age of slow-mo video, etc.)
Thanks!
| In general, Newtonian mechanics, that is taught at high school and at university in introductory level, considers objects as point particles or at most uncompressible solids. Therefore, in most cases, collisions are considered as elastic, where both kinetic energy and momentum are conserved.
However, in reality this is far from being the case. There is almost no point particles and any solid can be compressed. In other words, in reality there is almost no elastic collision; some of the kinetic energy is spent for deforming the shape and eventually is lost as heat.
There are a few materials properties (such as bulk modulus) involved in such calculations and I believe, analytical solutions only exists for certain geometrical shapes and for certain collision directions. For example, it could be possible to come up with such a function say for a spherical plastic ball or an iron cylinder etc. and I doubt that such a function will be constant for different shapes, materials and even collison directions. In other cases, that is, for complicated shapes and inhomogenoius materials, there might be some computer based numerical solutions that engineers apply.
Here is an example of a such non-elastic collision, as you mentioned, in slow motion.
https://www.youtube.com/watch?v=aMqM13EUSKw
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Two Cases Of Harmonic Motion Caused By Gravity I'm a highschool student and we learned not so long ago Simple Harmonic Motion, and I'm trying to analyse "similar" cases which I thought of.
Here we have a body (with mass $m$) being affected by the gravity of a body whose mass is $M$, yet it doesn't collide with it (it just goes through its center depiste not being so realistic). I want to mathematically describle this motion.
Just as doing with SHM, we get the differential equation
$m\frac{d^2x}{dt^2}=\frac{GMm}{x^2}\Rightarrow x^2 \frac{d^2x}{dt^2}=GM:=k$
Obviously I don't have the tools to solve such equation (I can only solve easy separables and using Laplace Transform), But Wolfram gave me the following solution
Looking back, I realized that at $x=0$ the force is "infinite" and I kind of stopped there since I'm clueless (plus I couldn't find the constants)
Another case I thought of which might be more realistic is the following
Here the movment on the perpendicular bisector of course.
We have $\Sigma F=2(\frac{GMm}{d^2+x^2})\cos\alpha=\frac{2GMmx}{(d^2+x^2)\sqrt{d^2+x^2}}$ So
$\frac{d^2x}{dt^2}(d^2+x^2)^\frac{3}{2}=2GMx:=kx$, which Wolfram couldn't solve.
I'm pretty sure this might not be far fetched, I came here to see if anyone has any contributions to my understanding? Perhaps some good way to approximate one of the motions?
Thanks for reading this mess!
| The first case is very different from a harmonic oscillator. The speed of the moving body becomes infinite as passes through the center.
The second case could be analyzed by expanding $$\frac{x}{(d^2+x^2)^{3/2}}= \frac{x}{d^3}\left(1-\frac{3x^2}{2d^2} + \ldots\right)\tag 1$$
If $x\ll d$ you need only to retain the first term in the expansion $(1),$
$$\frac{\mathrm d^2x}{\mathrm dt^2}=-\frac{2GM}{d^3}x\,.$$
Here I have inserted a minus sign that you missed in all of your equations. It comes from that the force is in the opposite direction to the $x$-coordinate.
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Photons at the big bang Did photons exist at the singularity of the Big Bang? If they did not, then does that mean that it is impossible to see what happened at the big bang?
| Well, photons were at the Big Bang, but it wasn't light that we can now observe until the era known as recombination, about 378,000 years after the Big Bang, that photons had a practically infinite mean free path due to the majority of protons in the universe being bound up in neutral atoms, and the universe became transparent.
Before then, the universe did have photons, but you can think of it as the mean free path for the photons not being long enough for there to be any light. The universe was kind of "foggy".
Hope this helps! I'd suggest reading a bit about the different eras surrounding the birth of our universe; the wikipedia page is pretty good.
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Since quantum mechanics give you that photons have (relativistic) mass $m=\frac{hf}{c^2}$, why gravity does not accelerate it? Since quantum mechanics give you that photons have (relativistic) mass $m=\frac{hf}{c^2}$ why gravity does not accelerate it?? I know it changes its energy hence its frequency hence its wave length-colour.But why it does not speed up? If you consider photons massless then it is obvious but then you would not take in consideration that Energy equals mass*$c^2$ and since photons have energy they can't have zero mass. (I'm a mathematics undergraduate took a course on an introduction to quantum physics so try to give a more intuitive answer than or if you use math please be rigorous to the interpretation of the quantities.) (h=planck's constant ,c=speed of light constant , f=frequency).I partly found the answer .I can accelerate without changing speed just direction hence i have bending of light in directions thus acceleration.But what if i send light Straight to the centre of the mass.Nor its speed will change nor its direction.How will i explain acceleration then?
| "Accelerate" is not the right word since the velocity of a photon has to remain constant as a result of its being massless. The confusion I think comes from thinking of gravity as an instantaneous force that attracts things but this idea does not survive special relativity. What we call gravity is a result of curvature of spacetime. So I would rephrase your question as do photons "feel" the bending of spacetime? The answer is yes. For example for the schwartzchild solution of GR, photons can have a circular orbit about a massive body at $r =\frac{3GM}{c^2}$.
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Work done by gravity on falling object does not seem to equal change in mechanical energy So I have some confusion here, I am sure I knew this at some point. Let's say an object of 10 kg is dropped from a height of 10 m. When it reaches the ground, the work done on the object should be the force ($mg$) x distance or 10 kg x 9.8 m/s/s x 10 m. That gives 980 joules of work done on the object by gravity.
But the object did not gain 980 joules of mechanical energy. It lost 980 joules of GPE and gained 980 joules of kinetic energy (up to the point of it reaching ground level).
Using the change in GPE and KE, it looks like no work was done on the object because the loss in GPE equals the gain in KE.
So
a) am I right that no net work was done on the car by sum of all forces?
b) is the work done by gravity equal to force time distance, or is it equal to the change in mechanical energy of the object which is zero?
| This is a common point of confusion that boils down to the fact that there are two physically equivalent but conceptually different ways of viewing this situation.
You can either look at gravitational potential energy (GPE) as an "internal" form of energy that your 10 kg object can have or you can look at the gravitational force (Fg) as an external force acting on the object. What you cannot do is look at the situation in both ways at once.
If you choose to see GPE as a form of energy, what is happening in this situation is that GPE is turning into kinetic energy (KE) as the object falls. If you choose to see Fg as an external force, what is happening is that Fg is doing work on the object, which increases its kinetic energy.
Either way, the amount by which the kinetic energy increases when the object reaches the ground is mgh, or 980 Joules in this case.
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Would there be no time in a universe with only light? It is sometimes said, that if you stand still (in space), you travel through time at the speed of light. On the other side light never stands still, so it always only travels through space (at the speed of light), but not through time. Does that mean, if our universe would be filled with light only, no time would exist? Is the existence of mass therefor necessary for the existence of time?
| The time dimension would still exist. There just wouldn’t be any energy in it and nothing to experience the passage of time.
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What do I actually see when I look at a laser? I have a green laser. Texts usually say you can only see visible light when it is reflected from another material. But you can clearly see a laser ray all the way from the device to the surface pointed at, what is that ray then? Is it a reflection off of air?
I can also see tiny sparkling green "dots" hovering around the central ray, sometimes they shine brighter than others. What are they? dust particles reflecting? On a surface pointed at I can also see many scattered "dots", if I put the laser through a transparent surface, the scattering and the density of the dots changes (if I hold either the laser or the surface in my hands they constantly change looking like a "static noise on a TV".
| As well as lighting up dust particles along its path, the beam is also lighting up water droplets in the air, particularly when outdoors.
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Why do we need wires to discharge a capacitor? I've been studying Capacitance and Dielectrics and I can't understand why the capacitor will only discharge if there is a wire connecting them. I understand that when the capacitor is charged up, there is an electric potential difference between the plates that makes the electrons "want to move" from the higher to the lower potential. But since the plates have opposite charges and charges are at the surface of the conductors, why can't the charges move from the positive plate to the negative plate due to Coulomb force?
Apologize me if this is a basic question, but I've been looking for this answer and I just can't find anything of the like, must be because it is really basic.
| The Coulomb force is there but in the ideal situation the air and the dielectric contain no free /mobile charge carries, that are insulators, so there is no mechanism for a current to flow between the two metal plates.
In the real world because the air and the dielectric are not perfect insulator the capacitor will discharge.
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What happens when car's trunk pops open while driving? Yesterday my car's locking mechanism for the trunk malfunctioned and popped open while driving. Luckily I was cruising at very low speeds (around 15kph) and I just stopped and fastened it again.
What I'm wondering is what would happen if I were traveling at relatively high speeds, let's say at 100-120 kph. Would it open completely or just stay unlocked while keeping shut?
| Put simply, there is a pressure of air under the trunk. Consider the following crude diagrams:
The first is that of an average sedan. Now there is a certain effect known as the Coanda effect, which refers to the tendency of flowing fluids to follow the curvature of a surface, even when it deviates from its original flow direction. This is true generally when this deviation from the original flow direction is not very large, as in the sedan case. The flow is streamlined, and the air flows rapidly very close to the surface. I'm sure you must have seen the common physics classroom trick of blowing above a piece of paper(while holding it horizontally in air) causes the paper to lift. This can be mathematically expressed in the form of Bernoulli's equation. Here, The trunk of the sedan is like the paper. On blowing air rapidly above it, it will lift(fling open).
The second is a hatchback model. Depending on how rapidly the curvature of its surface changes, Coanda Effect is minimized in this case. This is because, rapidly flowing air, cannot follow the surface if it bends sharply. Notice how in the diagram, the flow is not strictly following the contour of the hatchback. This means that the air immediately outside is moving slower compared to a sedan, which also means greater pressure. In other words, the chance of the trunk opening in this case is much less than that of a sedan.
PS : This was my personal analysis of the question. Feel free to point out any mistake i made.
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What does "point of application of force" mean in the given context? I faced a particular conceptual doubt while solving a textbook problem. I will initially write the complete question in my textbook and then try to boil it down to a single conceptual doubt so that it complies with the rules of Physics Stack Exchange.
Original Question:
A finite conductor $CD$ carrying current $i$ is placed near a fixed
very long wire $AB$ current carrying $i_0$ as shown in the figure.
Find the point of application and magnitude of the net ampere force on
the conductor $CD$. What happens to the conductor $CD$ if it is free
to move? (Neglect gravitational field)
My conceptual doubt is : What does "point of application of force" mean ? How to find it?
According to me the force doesn't act on a single point but on the whole wire $CD$, the force being maximum at end $C$ and minimum at end $D$.
P.S: I hope the question complies with the rules of the site. If not, please inform me, so that I can try to improve/re-frame the question. Thank you.
| Best guess, the point of application is an average position weighted by how much force is applied.
If the force was uniform, then that point would be the center. Since it's not, it's stronger closer to the long wire, the point will be moved to the left of center a bit. This means that not only is there a net force on the system, but also a net torque.
Try
$$ x_{\rm center} = \frac{1}{F_{total}} \int x\,\,{\rm d}F $$
| {
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Are there more causes besides Gravity and Velocity that cause time dilation? Albert Einstein discovered that both gravity and velocity alter time in a given reference frame relative to the time of an observer in a different reference frame. If gravity and velocity both alter the passage of time, is it possible that other external forces can alter time also? If so, what are these other causes?
| The answer is no, according to what we know today, there are no other causes.
For the answer you have to remember what is time dilation: time dilation is the dilation of the absolute proper time of particles from the point of view of an observer. That means that each particle/ each worldline consists of absolute spacetime intervals $ds$ which are corresponding to the proper time of the particle. The value of $ds$ is not an observed value and does not appear directly in spacetime manifolds such as the Minkowski diagram, it can only be calculated from the observed coordinate time and the observed displacement. In spacetime manifolds time appears in relative form, i.e. in the form of coordinate time. Coordinate time is proper time after time dilation.
Coordinate time and time dilation are dependent of the coordinate system that you chose, and by consequence, they depend on the relative spacetime manifold of the observer. In order to know what kind of time dilation may exist it might be useful to remember how spacetime has been modified in history.
The absolute Newton spacetime did not show any discrepancy between the proper time of a particle and the observed coordinate time. Newtons absolute spacetime and Galileis relative spacetime had been corrected by special relativity / Minkowski spacetime in order to take into account the absolute value of speed of light c. The pseudometric of Minkowski spacetime required the concept of velocity-dependent time dilation, including the Lorentz factor gamma.
Later, a second correction of the spacetime concept revealed to be necessary, in the form of curved spacetime representing gravity. This spacetime concept which includes gravitational time dilation is still today's concept. This concept is working very well, and no further corrections have been necessary up to now. That means that - according to our today's knowledge - beyond velocity and gravity there are no other causes of time dilation.
| {
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acceleration and potential energy due to conservative force how can we find acceleration due to a conservative force as a function of time when potential energy due to the conservative force is given as function of position(e.g. U(x)=x^2)
| The force $F$ is minus the potential energy gradient $\frac{dU}{dx}$ for a conservative field.
$F =-\dfrac {dU}{dx}$
So if you know the mass you can use Newton's second law to find the acceleration.
| {
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Given that ice is less dense than water, why doesn't it sit completely atop water (rather than slightly submerged)? E.g.
*
*If we had a jar of marbles or something else of different densities and shook it, the most dense ones would go to the bottom and the less dense ones to the top.
(Image Source)
*If I put a cube of lead in water it would sink all the way to the bottom.
But for ice : what I am trying to understand is why doesn't the water (being denser than the ice) seek to reach the bottom, and the ice sit flat on top of it (as in the left image)? Instead, some part of the ice is submerged in the water (as in the right image), and some sits on top it.
| I like to answer by reinterpreting your question: if you expect the ice to be completely atop the water because ice is less dense than water (as indicated in your left image), then you would also expect the ice to be completely below air because ice is more dense than air (in order for this to be true, think pushing your ice cube down into the water so that it's top surface is level with the water surface). Both is not possible at the same time, obviously. So you see that your reasoning does not give you a well-defined solution, and hence your reasoning (should it not sit completely atop water?) must be false.
| {
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Does the top plate of a capacitor hold half of the capacitor's charge or all of it? I am a little confused conceptually about the charge of a capacitor held by the top plate. Is it equal in magnitude to $q$, or is it half of $q$? It makes more sense to me for it to be half of $q$, with the other half existent on the other plate, but I am thinking this is incorrect.
| The charge of a capacitor is the magnitude of $q$. The charge of one plate is $+|q|$ and the other plate has the charge $-|q|$.
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Electromagnetic action in differential forms The electromagnetic action can be written in the language of differential forms as
$$S=-\frac{1}{4}\int F\wedge \star F.$$
$$=-\frac{1}{4}\int \left(\sum_i E_i\,{\rm d}t\wedge{\rm d}x^i - \star\sum_i B_i\,{\rm d}t\wedge{\rm d}x^i\right)\wedge \star \left(\sum_j E_j\,{\rm d}t\wedge{\rm d}x^j - \star\sum_j B_j\,{\rm d}t\wedge{\rm d}x^j\right)$$
$$=-\frac{1}{4}\int \left(\sum_i E_i\,{\rm d}t\wedge{\rm d}x^i - \star\sum_i B_i\,{\rm d}t\wedge{\rm d}x^i\right)\wedge \left(\star \sum_j E_j\,{\rm d}t\wedge{\rm d}x^j - \sum_j B_j\,{\rm d}t\wedge{\rm d}x^j\right),$$
since $**=(-1)^{p(n+p)}$ in Euclidean space, where $\star$ is applied on a $p$-form and $n$ is the number of spacetime dimensions, so that, in four dimensions for the $4$-form $dt\wedge dx^{j}$, $**=(-1)^{p(n+p)}=1$.
The electromagnetic action can also be written in the language of vector calculus as
$$S = \int \frac{1}{2}(E^{2}+B^{2})$$
How can you show the equivalence between the two formulations of the electromagnetic action?
| Just continue your second line:
Expand the wedge product and notice that the non vanishing terms are only the $EE$ terms and $BB$ terms and more over, $dt\wedge dx^i\wedge \star(dt\wedge dx^j)=\delta^{ij}dV$. There is a mistake in your second line: $\star\star=(-1)^{s+p(n-p)}$, where $s$ is the number of minus sign in the signature of your metric. Also there should be a extra factor of minus two as, for example, the coefficient of $dt\wedge dx^i$ should be $F_{0i}-F_{i0}=-2E_i$. So one should have
$$\frac{1}{2}\int \left(\sum_i E_i\,{\rm d}t\wedge{\rm d}x^i - \star\sum_i B_i\,{\rm d}t\wedge{\rm d}x^i\right)\wedge \left(\star \sum_j E_j\,{\rm d}t\wedge{\rm d}x^j + \sum_j B_j\,{\rm d}t\wedge{\rm d}x^j\right)$$
$$=\frac{1}{2}\int \left(\sum_{i,j} E_iE_j\,{\rm d}t\wedge{\rm d}x^i\wedge\star(dt\wedge dx^j)+\sum_j B_iB_j\,{\rm d}t\wedge{\rm d}x^j\wedge\star(dt\wedge dx^i)\right)$$ $$=\frac{1}{2}\int (E^2+B^2) dV,$$
where in the second to last line we used $\star(dt\wedge dx^i)\wedge(dt\wedge dx^j)=-(dt\wedge dx^j)\wedge\star(dt\wedge dx^i)$.
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Problem with physical application of Dirac Delta Consider the problem of projectile motion in 2 dimensions. Launch angle is constant. Range of projectile, $x$, then depends only on launch speed, $v$, and is given by
\begin{equation}
x=v^2, \quad v\in [0,1] \tag{1}
\end{equation}
Above equation has been non-dimensionalised (by taking maximum range as our length scale, and maximum launch speed as our velocity scale), so all quantities are dimensionless. Probability density function for launch speed is assumed uniform over the interval $[0,1]$:
\begin{equation}
f(v)=1, \quad \textrm{if}~v\in [0,1]\tag{2}
\end{equation}
and zero otherwise. I want to find p.d.f for range of projectile, $x$. An easy way of doing this
\begin{equation}
f(x)=\left| \frac{dv}{dx}\right|f(v)=\frac{1}{2\sqrt{x}}, \quad x\in [0,1]\tag{3}
\end{equation}
However I wanted to solve the same problem using Dirac delta function:
\begin{align}
f(x) & =\int_0^1 dv~f(x|v)~f(v) \\
& = \int_0^1 dv~f(x|v) \\
& = \int_0^1 dv~\delta(v^2-x)\tag{4}
\end{align}
Here $f(~|~)$ denotes conditional p.d.f.. Last line was arrived at because for given value of $v$, it is certain that we shall obtain that value of $x$ that satisfies the equation $v^2-x=0$. Now I make use of the identity for delta function
\begin{align}
\delta(g(x))=\sum_i \frac{\delta(x-x_i)}{|g'(x_i)|}\tag{5}
\end{align}
Here $x_i$ are roots of function $g(x)$, and $g'\equiv \dfrac{dg}{dx}$. Now $g(v)=v^2-x$, whose roots are $\pm \sqrt{x}$. We reject the negative root because $v\geq 0$. $g'=2v$. Hence
\begin{align}
f(x) & =\int_0^1 dv~\delta(v^2-x) \\
& = \int_0^1 dv~\frac{1}{2\sqrt{x}}\delta(v-\sqrt{x}) \\
& = \frac{1}{2\sqrt{x}}\tag{6}
\end{align}
which is correct.
However instead of $f(x|v)=\delta(v^2-x)$, we could equally well have begun with the equation $f(x|v)=\delta(v-\sqrt{x})$, because at least according to me, physical content of both equations is identical. However the last choice yields a completely different p.d.f.:
\begin{align}
f(x) & =\int_0^1 dv~\delta(v-\sqrt{x})=1\tag{7}
\end{align}
I don't think I have done anything wrong mathematically (if I have, please point out). To a mathematician of course the two functions are different, and so the fact that they yielded different p.d.f.s is not surprising. But when the equations are put in their proper physical context, both have identical physical content (as far as I can see). This example makes me wonder if Dirac Delta function may be used unambiguously in solving physical problems. While this was a simple problem where a second method of solution was available and so we could compare, what does one do in more complicated situations where such a comparison is not possible?
| *
*Besides ACuriousMind's suggestion to check dimensions, you could also integrate the wrong distribution $$\delta(\sqrt{x}-v)=2v\delta(x-v^2)$$ over $x$ and see that you don't get the p.d.f. (2) that you started from, but instead a wrong distribution $f(v)=2v$.
*The moral is that the Dirac delta distribution $\delta(g(x,v))$ does not just
depend on the support $$\overline{\{(x,v)\in [0,1]^2~|~g(x,v)=0\}},$$ but also on the function $g$.
| {
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Coulomb's law accuracy for small distances My physic teacher told me that experimental deviations from the predictions of Coulomb's law occur at small separations because, being inverse square, Coulomb's law work best for larger values of r. Why is this the case?
| Almost every source of electric field in our day life is more than an point-like, having complex structure on the distribution of charges. The naive aplication of the Coulumb's law assume that the sourcer is point-like or spherically symmetric. This is a good approximation if the dimensions of the sourcer is neglegible.
However, when you get close to the sourcer, the structure (the charge distribution) start to be relevant, and you need to replace the naive Coulomb's law by the Coulumb's law applied in each charge of the distribtuion.
It is important to note that the law is still valid. Is the application of the law that need to be corrected by new inputs, new distribution of charges. This charges can form a continuum too. And applying the Coulomb's law to each piece of the continuum (using calculus) gives the right predictions, with controlled accurancy.
There is a systematic way to approach to corrections of the pointe-like and spherically symmetric approximation, the multipole expansion. The idea is to expand the potential field in terms of $1/r$. The next correction to the $1/r^2$ electric field is the dipole
| {
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What is the difference between "Elastic limit" and "Yield point"? Elastic limit - the point till which the wire retains its original length after the force is withdrawn.
Yield point - the point where there is a large permanent change in length with no extra load force.
This is how these two terms are defined in my A Level book and also stated by my teacher.
In Wikipedia, yield point is stated as follows:
A yield strength or yield point is the material property defined as the stress at which a material begins to deform plastically. Prior to the yield point the material will deform elastically and will return to its original shape when the applied stress is removed. Once the yield point is passed, some fraction of the deformation will be permanent and non-reversible.
By that definition, shouldn't yield point and elastic limit have the same point on a stress-strain graph?
I know that below the elastic limit, the material will only show elastic behaviour. After the yield point, the material will exhibit plasticity. Nevertheless, the gap between these two is very small.
So, in the picture below, why are they put in two different points which are pretty far apart from each other?
P.S. If the picture is wrong in anyway, please mention why it's wrong.
| Yield point is well defined and shown on graph for mild steel and it's beyond elastic limit. For other materials like copper or aluminum is defined as the point of intersection of stress-strain curve and a line drawn parallel to linear part fron 0.2 percent deformation (strain ε) and it is also beyond the elastic limit.
| {
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How can a particle which is not moving have an acceleration? Suppose a rod is rotating around a fixed point located at an extreme point of it and there are two points on it. One, somewhere in the middle and the other at the other extreme.Call them $A$ and $B$ respectively.
The Question is relatively simple, but it is confusing me a lot! For $A$ , $B$ is at rest. How can it have an acceleration? Yeah, for $A$ there exists a centripetal force on $B$. But the point is that there is no relative motion! How can there be an acceleration as acceleration is the rate of change of velocity ( here relative velocity ) which is 0?`
EDIT - This was a question in my book and it asked to calculate the acceleration of $B$ relative to $A$ and the answer was not 0 .
|
How can a particle which is not moving have an acceleration?
Using the Taylor series expansion for the trajectory of the particle $x(t)$ about $t = 0$, we have
$$x(t) = x(0) + \frac{dx}{dt}t + \frac{1}{2}\frac{d^2x}{dt^2}t^2 + \cdots = x(0) + v(0)t + \frac{1}{2}a(0)t^2 + \cdots$$
This is a perfectly general result. In the case that the initial velocity is zero, $v(0) = 0$, there is no reason to constrain the initial acceleration (or jerk or higher order derivatives) to be zero.
Physically, for a particle with zero velocity but non-zero acceleration, this means that $x(0 + dt) = x(0)$, i.e., the object isn't moving at $t = 0$ but it is beginning to move.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simple harmonic motion : why is the period independent of amplitude even when angular velocity is related to the amplitude?
Period is independent of amplitude. (Vias.org)
But given that,
Simple harmonic motion can be defined by
$$x = A * \sin(\omega t) \tag{1}$$
where $A$ is the amplitude of oscillation, $\omega$ the angular velocity, $t$ the time, and $x$ the displacement from the mean position
And,
$$T = 2 \pi/\omega = 1/f \tag{2}$$
where, $T$ is the period of motion and $f$ is the frequency of oscillation.
Equation (1) can be rearranged to give
\begin{align*}
x &= A * \sin (\omega t)\\
\frac{x}{A} &= \sin(\omega ωt) \\
\arcsin\left(\frac{x}{A}\right) &= \omega t\\
\omega &=\frac{\arcsin\left(\frac{x}{A}\right)}{t}
\end{align*}
Subbing this into (2) gives the following relationship between $T$ and $A$
$$T = \frac{2 \pi t}{\arcsin(x/A)} = \frac{1}{f}$$
Doesn't the fact that both $T$ and $A$ appear in the above equation show that $T$(period) is dependent on $A$(amplitude) ?
FYI:
Although a physical explanation may be useful, I am particularity interested in why deriving a relationship between $T$ and $A$ doesn't mean that they are dependent. Note there is a similar question here but that is concerned with the physics of the phenomena, and not why the maths can't be used to solve it.
This is because I have this exam where we are given a stimulus and based on that stimulus alone are meant to answer questions (i.e. the exam is expected to contain material/principles that we haven't been exposed to before, but should be able to answer given the stimulus). And as I didn't know much about simple harmonic motion, my initial reaction was to see if the formulas link $T$ and $A$.
The practice question (which I have typed out the important bits of above is)
The answer is D.
| It's because $x$ depends on $t$ in such a way that there is no dependence on $A$ left in the expression. $A$ and $\omega$ are constants that don't depend on time, and $x$ is a function of time; it's the position at time $t$, usually written $x(t)$.
When the force isn't Hooke's law, then you can get a relationship between $A$ and $\omega$. The reason for this is because the energy of a simple harmonic oscillator is
$$ E = \frac{1}{2} m [v(t)]^2 + \frac{1}{2} k [x(t)]^2, $$ that is the equation for an ellipse in $x$-$v$ space no matter how big the amplitude/energy is. The simple harmonic osillator just rotates around that ellipse with a constant angular frequency, $\omega$. If I were to change the potential energy term to, say:
$$ E = \frac{1}{2} m [v(t)]^2 + \frac{1}{4} k [x(t)]^4, $$ then the shape with a constant energy is no longer an ellipse, and the shape changes as the energy changes in such a way that amplitude and period become related.
For another example, a look at the orbits of planets and Kepler's third law — there amplitude along the major axis and period are related by:
$$\frac{T^2}{A^3} = \frac{4\pi^2}{G(M+m)}$$
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