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Does the Earth revolve around the Sun? I am aware of this Phys.SE question: Why do we say that the earth moves around the sun? but I don't think this is a duplicate.
In a binary star system, where the masses of the 2 stars are not so different from each other, can we say that each star revolves around the other?
If yes, then couldn't the Sun-Earth system be an extreme case of such a system?
Therefore, strictly speaking, can we argue that the Sun revolves a tiny bit around the Earth as well?
| In binary systems, each object is so affected by the others gravity that they have significant orbit. The sun has so much inertia that the earth's pull barely moves it, but the earth certainly revolves around the sun. In the reference frame of the Earth however, the Sun does revolve around the Earth.
| {
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"timestamp": "2023-03-29T00:00:00",
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If the Earth's atmopsphere spins with the earth due to friction, why is there no horizontal spiralling drag? Imagine a bucket of paint with a spinning ball in it. The paint would form a spiral and would not all move in synchronous movement with the ball.
To clairfy - In order for the Earth's atmosphere to appear to us to be still (as on a windless day) the upper atmosphere must be moving faster than the lower atmosphere - as it the case with any rotating spehere - the outer layers are moving faster than the inner layers.
What force is causing the upper atmosphere to move faster than the lower atmosphere?
If the atmosphere is being rotated soley by friction, then at best the upper layers would move at the same speed as the lower layers, thereby causing a spiralling effect.
We do not see this effect. Why?
| If i interpreted your question correctly and based on my knowledge (i don't know how good it is). This is because there is nothing to apply frictional force on the outer edge of the earth's atmosphere, whereas in case of bucket the water rubs against the boundaries of the bucket which slows down the outer part of the spinning water (dont look at it for long you will be hypnotized)
| {
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How to calculate the colour a human eye sees when looking at a light spectrum? I have to do a presentation about colourants in Chemistry class (grade 12, advanced) and want to write a program that calculates and visualizes the colours of some simple molecules.
What I need is a formula to calculate the visible colour (e.g. RGB-values of the colour that would look the same like the given spectrum).
Should be able to handle both monochromatic spectral lines and wide spectres. (e.g. with integrals or something like that?)
| You will not be able to get the proper color of the narrow bands because they fall outside of the RGB gamut. However you can get their correct hue.
You need color matching functions (usually we use 1931 2° CMFs from the CIE, check online... or here under CMFs category http://www.cvrl.org/)
CMFs multiplied by your spectrum, then summed, will give you the proprtions of primaries X, Y, Z that match the color for a human observer.
Once you have this, you can scale your results and convert from XYZ to sRGB for example (or any other RGB system you'd like), the conversion is explained there:
XYZ2sRGB
To properly scale your XYZ values, I advice you to get a "white reading" using only solvent for your spectrophotometer. The Y value of this reading will be your white reference for scaling the other XYZ values.
In order to get good hues for the narrow bands falling outside sRGB gamut, you can convert to CIE LCH and reduce C* until you fall inside RGB gamut.
See here: XYZ2LAB >> LAB2LCH
XYZ 2 LCH
| {
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where does electricity come from in piezoelectric? I have some questions here. BaTiO3 is a material which exhibit piezoelectric effect when we give external pressure on it. It contain Ba2+, Ti4+, and O2- ion in its crystal unit. The Ti4+ ion will move up or down, so the crystal will polarize and that it could produce electricity. But how's that possible? I mean, electricity (current) is a flow of the electrons but I still have no idea about where the electron come from? My guess is: these electron came from Ti4+ ion. Can somebody explain to me what actually going on?
| This is due to the molecular arrangement of the atoms in the crystal.
Basically, a piezoelectric crystal contains molecules arranged in, say, tetrahedral fashion.
The dipole moments of each and every atom pairs are well balanced.
When you squeeze or twist the crystal, the equilibrium gets disturbed. That is, one or the other dipoles remain only partially opposed.
This happens to many dipoles, and as a result a net dipole moment develops in the crystal.
This can also be interpreted in terms of potential difference.
These two ends when connected to a circuit, causes a current.
Remember, as Gede panji quoted , potential difference causes current here, not
the reverse. I hope this clears your doubt.
| {
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Is there an infinite amount of wavelengths of light? Is the EM spectrum continuous? The electromagnetic spectrum is a continuum of wavelengths of light, and we have labels for some ranges of these and numerical measurements for many.
Question: Is the EM spectrum continuous such that between two given wavelengths (e.g. 200nm and 201nm) there is an infinite number of distincts wavelengths of light? Or is there some cut-off of precision with which light might exist (e.g. can light only have wavelengths of whole number when measured in nanometers, etc.)?
| Sir Elderberry, Punk_Physicist and the Count Iblis have all given correct answers in principle.
There are two phenomena (really thought experiment, rather than practical, devices) that one needs to heed.
*
*A finite measuring time $T$ can only resolve frequencies to within an uncertainty of the order of $1/T$. This is the reciprocal relationship between the time spread $\Delta t$ in a pulse and the frequency spread $\Delta \omega$ in its Fourier transform, given by $|\Delta t|\,|\Delta\,\omega|\geq\frac{1}{2}$ as I show in my answer to the Physics SE question "Heisenberg Relation". This is the mathematical phenomenon underlying the Heisenberg uncertainty principle (but not the same as the latter: the latter arises because the Fourier transform relates a quantum state's expression in co-ordinates related by the Canonical Commutation Relationship). In practice, though, a finite pulse is modelled by a spread of frequencies over an interval, and all frequencies in the interval are present in the Fourier transform.
*If you model a finite universe as a cavity, you will indeed get only a finite number of modes per frequency. So one aspect of the quantum light field ground state energy divergence is not really a divergence at all as the energy per unit volume is finite, as I discuss here. If you were Odin, say, creating a universe, it would cost you roughly a fixed energy per unit volume to make one like ours.
| {
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Approximations of the kind $x \ll y$ I have an expression for a force due to charged particle given as
$$F=\frac{kQq}{2L}\left(\frac{1}{\sqrt{R^2+(H+L)^2}}-\frac{1}{\sqrt{R^2+(H-L)^2}}\right) \tag{1}$$ where $R$, $L$ and $H$ are distance quantities.
Now I want to check what happens when:
*
*$H\gg R,L$
*$R,H\ll L$
How can I work out the approximation of this force?
Do I have to write it slightly different into form (2) to get it right?
$$ ~F=\frac{kqQ}{2LR}\left(\frac{-1}{\sqrt{1+\left(\dfrac{H+L}{R}\right)^2}}+\frac{1}{\sqrt{1+\left(\dfrac{H-L}{R}\right)^2}}\right) \tag{2}$$ (which is the same expression just written out differently). Any explain about this subject would be very helpful.
| Let's focus just on the interesting bit of the equation:
$$\frac{1}{\sqrt{R^2+(H+L)^2}} - \frac{1}{\sqrt{R^2+(H-L)^2}} =\\
\frac{1}{\sqrt{R^2+H^2+2HL+L^2}} - \frac{1}{\sqrt{R^2+H^2-2HL+L^2}}$$
Now if $H>>L,R$, we are left just with the terms with $H$:
$$\approx \frac{1}{\sqrt{H^2+2HL}} - \frac{1}{\sqrt{H^2-2HL}}\\
=\frac{1}{H}\left(\frac{1}{\sqrt{1+2\frac{L}{H}}}-\frac{1}{\sqrt{1-2\frac{L}{H}}}\right)\\
\approx \frac{1}{H}\left(1-\frac12 \cdot 2\frac{L}{H}-\left(1+\frac12\cdot 2 \frac{L}{H}\right)\right)\\
=-\frac{2L}{H^2}$$
Almost exactly the same approach works for $L>>R, H$ (I will leave the details up to you).
| {
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100°C = 100 K =? I'm in first year. Our class is in lesson " Heat and Thermodynamics". While solving a numerical problem of a reversible engine he told us that 100 degree Celsius is equal to 100 kelvin. I inquired but could not get satisfactory answer. Pleas help me understand it.
Here is the numerical, please consider it: A reversibe engine works between two temperatures whose difference is 100c. If it absorbs 746J of heat from the source and rejects 546J to the sink, calculate the temperature of the source and the sink. Ans (100°C, 0°C)
| A difference in degrees Celsius is equal to one in degrees Kelvin.
So 150 K - 100 K = 50 K = 50 C = 423.15 C - 373.15 C.
The absolute values cannot be compared without taking the 273.15 offset into account.
| {
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Heisenberg uncertainty and probabilistic nature of QM I am trying to understand whether the HUP and the probabilistic nature of QM are orthogonal or not. By that I mean that the HUP fundamentally derives from operators not commuting, which is the important fact here, more than the statistical nature of the LHS in the definition of the HUP:
$$\sigma_{x}\ \sigma_{p_x}\ge\frac{\hbar}{2}.$$ If that is correct, the HUP should manifest itself even with an experiment on an ensemble that contains only one particle. Is that the case? If not (hard to compute variances on one particle), is there a minimum ensemble size that demonstrates the HUP?
Note: the reason I am asking is that some people seem to say that Heisenberg uncertainty is related to measurement uncertainty, or they invoke "ensembles of particles" to justify the HUP, when IMHO, the HUP is saying something else than the fact that measurements are probabilistic in QM. Hence the "nonsensical" question about "one particle", which attempts to remove statistics from the picture, so to speak. Maybe statistics and the HUP are conjoined in an inseparable way in QM, but if say, all operators could commute, there would be no HUP, you need the ingredient of non-commutativity which has nothing to do with statistics. Is that correct?
| The $\sigma_A$ on the LHS is defined as
$$\sigma_A = \omega(A^2 - \omega(A)^2I),$$
where $\omega$ is a state (so $\sigma_A$ also depends on $\omega$!) and the way this is defined by von Neumann is by taking a statistically relevant ensemble of very same copies of the very same system in the very same state $\omega$ and repeat the measurement of $A$ and $A^2$ this many times. All you need to do is to apply the defining formula
$$\omega(O) := \frac1N\sum_{k=1}^N\lambda_k$$
where the $\lambda_k$s are the outcomes of the observable $O$ on the state $\omega$. This is the bridge between the statistical nature of a measurement and the HUP, since by the axiomatic formulation of quantum mechanics through the theory of operator algebras you get that
$$\Delta_\omega(x)\Delta_\omega(y)\geq \frac12|\omega([x,y])|,$$
where now I'm using $\Delta_\omega(x)$ instead of $\sigma_x$ in order to explicitly show the dependence on $\omega$.
| {
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What do spacelike, timelike and lightlike spacetime interval really mean? Suppose we have two events $(x_1,y_1,z_1,t_1)$ and $(x_2,y_2,z_2,t_2)$. Then we can define
$$\Delta s^2 = -(c\Delta t)^2 + \Delta x^2 + \Delta y^2 + \Delta z^2,$$
which is called the spacetime interval. The first event occurs at the point with coordinates $(x_1,y_1,z_1)$ and the second at the point with coordinates $(x_2,y_2,z_2)$ which implies that the quantity
$$r^2 = \Delta x^2+\Delta y^2+\Delta z^2$$
is the square of the separation between the points where the events occur. In that case the spacetime interval becomes $\Delta s^2 = r^2 - c^2\Delta t^2$. The first event occurs at time $t_1$ and the second at time $t_2$ so that $c\Delta t$ is the distance light travels on that interval of time.
In that case, $\Delta s^2$ seems to be comparing the distance light travels between the occurrence of the events with their spatial separation. We now have the following definitions:
*
*If $\Delta s^2 <0$, then $r^2 < c^2\Delta t^2$ and the spatial separation is less than the distance light travels and the interval is called timelike.
*If $\Delta s^2 = 0$, then $r^2 = c^2\Delta t^2$ and the spatial separation is equal to the distance light travels and the interval is called lightlike.
*If $\Delta s^2 >0$, then $r^2 > c^2\Delta t^2$ and the spatial separation is greater than the distance light travels and the interval is called spacelike.
These are just mathematical definitions. What, however, is the physical intuition behind them? What does an interval being timelike, lightlike or spacelike mean?
| Let's put it very simple: They tell you "how far something is apart compared to c"
time-like: if you are fast enough, you can be at (think spatial, like "at the festival") event a and at event b, it is only a "matter of time" until you see the second event
space-like: the two events are too far apart (in space). You cannot see both of them together, no matter how fast you are. As soon as event a happened and you go as fast as possible, event b will have happened before you arrive there.
light-like: exactly in between, the events are so far away that if you are as fast as light, you can see both events. If they are further away, they become space-like, if they are closer they become time-like
So a space-like separation makes any causal relation between the two events impossible, i.e. one cannot cause or influence the other.$^1$
1) as mentioned, a common cause (an event that is time-like to both events) can still make the two correlated.
| {
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Can net torque $\sum_i\mathbf r_i\times\mathbf F_i$ be expessed as $\mathbf r\times$ (net force) for some $\mathbf r$?
Let $\mathbf F_i$ be forces each of which is applied on $\mathbf r_i$ of a rigid body. Then is there a position vector $\mathbf r$ that satisfies
$$\displaystyle\sum_i\mathbf r_i\times\mathbf F_i =\mathbf r\times\displaystyle\sum_i\mathbf F_i~ ? \tag{1}$$
Well, what I get is by letting $\mathbf r_i=(r_{i,x},r_{i,y},r_{i,z})$, $\mathbf F_i=(F_{i,x},F_{i,y},F_{i,z})$, $\mathbf r=(r_x,r_y,r_z)$ is
$$\begin{bmatrix}
0 & \sum F_{i,z} & -\sum F_{i,y}\\ -\sum F_{i,z} & 0 & \sum F_{i,x}\\\sum F_{i,y} & -\sum F_{i,x} & 0 \end{bmatrix}
\begin{bmatrix} r_x\\ r_y\\r_z \end{bmatrix}=
\begin{bmatrix} \sum (r_{i,y}F_{i,z}-r_{i,z}F_{i,y})\\\sum (r_{i,z}F_{i,x}-r_{i,x}F_{i,z})\\\sum (r_{i,x}F_{i,y}-r_{i,y}F_{i,x}) \end{bmatrix}, \tag{2}$$
and $$\begin{vmatrix}
0 & \sum F_{i,z} & -\sum F_{i,y}\\ -\sum F_{i,z} & 0 & \sum F_{i,x}\\\sum F_{i,y} & -\sum F_{i,x} & 0 \end{vmatrix}=0. \tag{3}$$
Since the matrix is singular, the system might not have a unique solution.
So is it the case that generally such $\mathbf r$ may not be unique (or even nonexistant)? If so what is the criterion for uniqueness of $\mathbf r$?
| Yes. The solution is:
$$ \bf{r} = \dfrac{\left( \sum {\bf F}_i\right) \times \left( \sum ({\bf r}_i \times {\bf F}_i) \right)}
{\| \sum {\bf F}_i \|^2} =\dfrac{{\bf F} \times {\bf \tau}}{{\bf F}\cdot{\bf F}}$$
Then you can show that
$$ {\bf r}\times \left(\sum {\bf F}_i \right)= \sum ({\bf r}_i \times {\bf F}_i) = {\bf }\tau$$
Use ${\bf F} =\sum {\bf F}_i$ and ${\bf \tau} = \sum {\bf r}_i \times {\bf F}_i $ and the vector triple product $\vec{a}\times(\vec{b}\times\vec{c}) = \vec{b}(\vec{a}\cdot\vec{c})-\vec{c}(\vec{a}\cdot\vec{b})$
$$\begin{aligned}
{\bf r}\times {\bf F} &= \left(\dfrac{{\bf F} \times {\bf \tau}}{{\bf F}\cdot{\bf F}}\right)\times {\bf F} \\
& = \dfrac{\left({\bf F} \times {\bf \tau}\right)\times {\bf F}}{{\bf F}\cdot{\bf F}} \\
& =- \dfrac{{\bf F}\times \left({\bf F} \times {\bf \tau}\right)}{{\bf F}\cdot{\bf F}} \\
& = - \dfrac{{\bf F} ({\bf F}\cdot{\bf \tau})-{\bf \tau} ( {\bf F}\cdot{\bf F})}{{\bf F}\cdot{\bf F}} \\
& = {\bf \tau}
\end{aligned} $$
Since ${\bf F}\cdot{\bf \tau}=0$
See more details in this answer https://physics.stackexchange.com/a/70445/392. Each loading defines a screw axis, and screws can be combined linearly (addition of two screws is a screw, and a scalar times a screw is a screw). For the resultant screw you can extract its direction $\vec{e} = \frac{\bf F}{\| {\bf F}\|}$, its position $\vec{r} = \frac{{\bf F}\times{\bf \tau}}{{\bf F}\cdot{\bf F}}$ and its pitch $h=\frac{{\bf F}\cdot{\bf \tau}}{{\bf F}\cdot{\bf F}}$.
| {
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How can you have odd nuclear spin angular momentum but positive parity or vice versa? How does it happen that you can get states like $J^\pi=3^+$ or $J^\pi=2^-$? I think this could be because $\pi=(-1)^l$ so you could have an even state in $l$ but the $J=l+s$ sum could be an odd number?
| Presumably you are talking about mesons.
First of all, parity is not (-1)^L for a meson, it is (-1)^(L+1). This is because you have to take into account the fermion-antifermion relative parity for the quark-antiquark pair, which is negative.
Secondly, yes, it is L that matters and not J (which includes S). Thus, the low-lying mesons (pi, rho, D, D*, Upsilon, etc.) all have L=0 and hence negative parity. Among them, the pseudo-scalars are S=0, while the vector mesons are S=1.
| {
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What is the mass of a sphere? A solid sphere of mass M is rotating along an axis.
We can consider it as a collection of large number of point masses, every point mass is moving with respect to center of mass with velocity which depends on its radius from rotating axis.
Then, according to relativity, the mass of every point increases and consequently the mass of the sphere increases.
But if we consider the overall sphere, it is not moving at all and its mass remains the same. which produces a contradiction.
Please tell me where I am wrong.
| If the sphere is not moving, than is not even rotating on its axis.
I don't really get what you're trying to say. Relativistic mass increases with speed, so if you're considering a moving sphere and a standing one they're not gonna have the same mass, but in the formula of relativistic mass m0 comes in, which is the rest mass (when the object is not moving).
| {
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Does the direction of the magnetic field inside a solenoid depends on the direction which it is turned? Today trying to explain some magnetic stuff, just came across with a simple (at least seems) question.
My question is brief: does the direction of the magnetic field inside a solenoid depends on the direction which it is turned?
In other words -I hope you see the same as me-, I asking about if (1) is equal to (2):
(1) + -///////- -
(2) + -\\\\\\\- -
| The direction of the magnetic field will correlate to the handedness of the current going through the coil. Basically it doesn't matter how the solenoid is constructed, what matters is the orientation of the current going through it. If you look along the length of the solenoid (through the coil), whether the current is moving clockwise or not, determines which magnetic pole will be closer to you.
The image below (from wiki) best explains the principle - as you can see, the magnetic field spirals around the current in a particular direction, and by causing the current to run past itself it stacks the magnetic field.
If you imagine that the current inside the solenoid is a rotating cylinder, the direction of that rotation determines the orientation of the magnetic field.
| {
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Integral vs differential forms of Maxwell's equations As stated in this post, the integral and differential Maxwell equations should be identical. However, in a text I was reading it states that
The integral forms of Maxwell’s equations describe the behaviour of electromagnetic field quantities in all geometric configurations. The differential
forms of Maxwell’s equations are only valid in regions where the parameters
of the media are constant or vary smoothly i.e. in regions where "$\epsilon(x, y, z, t),
μ(x, y, z, t)$ and $\rho(x, y, z, t)$ do not change abruptly. In order for a differential form to exist, the partial derivatives must exist, and this requirement
breaks down at the boundaries between different materials.
Doesn't this imply that the integral form of Maxwell's equation is more fundamental than the differential form because it works for all configurations?
| Dirac delta is the most common distribution in physics.
It is defined by an equation:
$
\int \limits_{-\infty}^\infty \delta(x) f(x) = f(0)
$
One could say that this could be "intuitively: a function, which is $\infty$ at $0$, and $0$ everywhere else, but this makes no sense, since such integral would be $0$ (not $f(0)$), as integral over the set of zero measure (a point).
Using $\delta$, you could write down e.g. Gauss law for a point charge at (0,0,0):
$$ \nabla \cdot \vec E(\vec r) = \frac{q}{\epsilon_0}\delta^3 (\vec r).$$
Integrating left-hand side in the usual way and RHS (using aforementioned equation) , we recover an integral form of Gauss law. You could find many examples of this in Griffiths' or Jackson's electrodynamics textbook.
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Has anyone tried Michelson-Morley in an accelerated frame? After doing much more digging than I thought I had to do, I found out that the speed of light is NOT invariant in an accelerated reference frame. Has anyone done any experiments to confirm this? In particular a Michelson-Morley experiment in an accelerated reference frame? I figured light being invariant in any constant speed frame would automatically imply being invariant in any frame whatsoever.
I have to credit Richard Mould's Basic Relativity with informing me about this fact.
| The first to do something equivalent to that were Pound and Rebka who first measured the gravitational redshift in 1959. I'm not aware of anyone who actually used a Michelson interferometer in a upright orientation.
| {
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A cup of water in ZERO gravity What will happen if I try to pour a cup of water in zero gravity, into another empty cup? Will the water come out of the cup? The adhesive force between the water molecules and the interior of the cup should prevent the water from coming out. Is it correct? Or is there something more to to it?
| Pour? No such thing without gravity.
In NASA TV (see video), I saw the prototype coffee cups. They are shaped with a sharp crease, to allow liquid to ride up the groove. More advanced product would also mix waxy and wettable surfaces to keep it stuck to the inside of the cup but not crawl over the brim, except at the sip line.
The pictures are hard to figure out; watch the video or read an article that shows a series of pictures and diagrams.
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Deriving the equations for a moving inertial reference frame I assume $c=1$ in the following derivation:
In order to derive the equations for a moving inertial reference frame, I immediately wrote down the following:
$$ x'=Ax+Bt, \tag{1}$$
$$t'= Dx+Et. \tag{2}$$
In order to solve it I would need 4 independent equations. Here they are:
*
*Since the speed of light is constant in all reference frames, it follows that if $x = t$, also $x' = t'$, therefore
$$At+Bt=(Dt+Et) \overset{(x=t)}{\implies} A+B=D+E. \tag{3}$$
*I can reverse the reference frame and the physics should be the same hence
$$x=0 \implies x'=-vt' \implies B=-Ev. \tag{4}$$
*Finding the $x'$ component of the point $A(\frac{1}{1-v},\frac{v}{1-v})$ gives:
$$Av+B= -Av. \tag{5}$$
*Finally finding the $t'$ component of $A$ gives (I'll do this one explicitly):
$$t'= \frac{D+Ev}{1-v}. \tag{6}$$
From the diagram one can read off using the Pythagorean theorem that:
$$t'= \sqrt{\left( \frac{1}{1-v} \right)^2 +\left( \frac{v}{1-v} \right)^2 } = \frac{\sqrt{1+v^2}}{1-v}$$
$$\implies D+Ev= \sqrt{1+v^2}. \tag{7}$$
From these equations one easily arrives at the desired result ie
$$x'=\frac{x-vt}{\sqrt{1-v^2}} \; \text{and} \; t'=\frac{t-vx}{\sqrt{1-v^2}} \tag{8}$$
All this seems to be correct. However considering the equation $(6)$ and putting back the $c$'s in it one arrives at the equation
$$ D+Ev= \sqrt{1+v^2} \quad (!) \tag{9}$$
First of all this dimensionally doesn't make sense. Secondly if you calculate and find the coefficients you don't get the correct answer. Intuitively I know that this equation has to be $D+Ev= \sqrt{1+v^2/c^2}$ so that everything works perfectly but I don't know why this has to be so and I cannot show it by reasoning physically. I fell in my guts that there is something fishy about using Pythagorean theorem but I don't know what went wrong exactly. If I just say that the use of Pythagorean is wrong, then I cannot explain why it gives the correct answer when using $c=1$. Such a coincidence seems to be highly unlikely.
Edit: I've made a major typo in the diagram you should swap $x=0$ with $t=0$ and $x'=0$ with $t'=0$!
| You work too hard and the idea of setting $c=1$ may make problems.
I copy here the equations that you obtained and that I found correct. So,
1. The speed of light is the same in each frame implies
$$Act + Bt = c(Dct + Et), \overset{(x=t)}{\implies} Ac + B = Dc^2 + Ec. \tag{i}$$
*Reversing the frames gives indeed
$$B = -Ev. \tag{ii}$$
*Also in the frame $(x,t)$ the origin $x'=0$ of the frame $(x',t')$ moves at velocity $v$
$$B = -Av. \tag {iii}$$
Notice also that from $\text {(ii)}$ and $\text {(iii)}$ one infers
$$E = A, \tag{iv}$$
and introducing all these relations in $\text {(i)}$,
$$Ac - Av = Dc^2 + Ac \overset{(x=t)}{\implies} D = -\frac {Av}{c^2} \tag{v}$$.
*Now putting all these things together I rewrite your transformations $(1)$ and $(2)$
$$x' = A(x - vt), \ \ \ ct' = A(-\frac {vx}{c} + ct). \tag{vi}$$
Whatever remains is to find A. In this task, the interval conservation is bound to help,
$$c^2t'^2 - x'^2 = c^2t^2 - x^2. \tag {vii}$$
So let's do the calculus,
$$A^2 \left[(-\frac {vx}{c} + ct)^2 - (x - vt)^2 \right] = c^2t^2 - x^2. $$
Doing the calculus you get
$$A = \frac {1}{\sqrt {1 - \frac {v^2}{c^2}}} \tag{viii}$$
So let's now rewrite the transformations
$$x' = \frac {x - vt}{\sqrt {1 - \frac {v^2}{c^2}}}, \ \ \ ct' = \frac {ct -\frac {vx}{c}}{\sqrt {1 - \frac {v^2}{c^2}}}. \tag{ix}$$
Now, I saw that you have a problem with the calculus of your expression $(6)$ but you didn't say what it means. Anyway, for $D$ see my formula $\text {(v)}$ and for $E$ my formula $\text {(iv)}$. You can rely on them, they are obtained in a simple way.
| {
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Effective Area of Isotropic Antenna: Explanation? I'm reading some introduction to antenna theory and I've often puzzled on the equation:
$$
A_{eff} = \frac{\lambda^2}{4\pi}
$$
which relates the effective area by which an antenna captures radiation to the frequency at which that radiation is.
I have looked at this derivation of the formula and can understand the steps they take, but when trying to understand it on a higher level, I cannot reconcile it.
Going on the example used in the derivation above, I would assume that as the frequency that is selected to pass through the filter increases, the resistor would begin to give off more power, in line with Johnson-Nyquist Noise:
$$
dP = k_{b}Td\nu
$$
This would mean, in order to keep thermal equilibrium, that the antenna on the other end would have to be more receptive to the blackbody radiation that the cavity would give off. So, I would assume, that would mean a larger Effective Aperture Area would be required in order to gather it.
But this seems to contradict the result, which says that higher frequencies need a smaller area.
Can anyone help me out and point out the flaws in my assumptions?
Thanks
| Increasing frequency does not change the dv spectral interval of integration, and with an assumption of white noise there is no increase in dissipated power with frequency.
Therefore with resistor power held constant there is a decrease of aperture area with wavelength squared.
| {
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Is there a difference in the energy output of a nuclear fission reaction as opposed to fusion? For example, if I split a Helium atom will I get the same amount of energy as when I fuse Hydrogen into Helium? If there is a difference, what will be the difference (in general not according to Helium/Hydrogen), and why?
| The energy generated during fusion or fission can be seen with this graph:
When a light atom is made into a heavier one by adding nucleons, it will lead to a greater output in energy; but when you reach Iron you can no longer gain energy through fusion. For heavier elements, you begin to lose energy when you fuse them and the way to gain energy is to split them apart.
As you can see on the graph, fusion is generally more efficient than fission.
| {
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Do photons with a frequency of less than 1 Hz exist? A photon with a frequency of less than 1 Hz would have an energy below
$$
E = h\nu < 6.626×10^{−34} \;\rm J
$$
which would be less than the value of Planck's constant. Do photons with such a low energy exist and how could they be detected? Or does Planck's constant give a limit on the amount of energy that is necessary to create a single photon?
| You don't need a long antenna to radiate at 1 Hz. You need a long antenna to radiate efficiently at 1 Hz. The efficiency is proportional to the cube of the length of the antenna in wavelengths (look up {electrically} 'small antennas'). A big 1 Hz current in a short wire will radiate very little power, but 1 photon a second would be 6.6e-34 Watts, so the numbers may be in favour of radiation. 1 Hz ==> 3e8 m wavelength, so 1 m long wire antenna may have efficiency of order 3e-26, which looks like lots of photons per watt into the antenna (most of the watt goes into resistive, dielectric and magnetic losses in the matching circuit or the generator).
| {
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Does the line integral definition of Work involve distance or displacement? My textbook reports the following definition of Work:
where ds is the infinitesimal displacement.
I know that an infinitesimal displacement is usually denoted by dr and I also know that the magnitude of dr is given by ds (infinitesimal distance) Now, if we are talking about displacement (in Work definition), why should we use ds instead of dr?
I ask this because my textbook always refers to infinitesimal displacement as dr. I have always associated 's' to distance, so I see ds as an infinitesimal "distance vector", but I am quite sure that distance is only a scalar quantity, not a vector.
| Its just a matter of what you use to call as displacement and as distance .
I have seen the usage of:
*
*dx
*ds
*dr
as the displacement too.
Wikipedia says :
The work done by a constant force of magnitude F on a point that moves a displacement (not distance) s in the direction of the force is the product,
W = Fs.
Note the usage of s as dispacement .
All in all , its the displacement that is used in calculating work and one may refer to it in many ways.(probably your textbook used different notations in different chapters)
And Distance is a scalar quantity.
| {
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Can a virtual image created by a mirror have position in front of the mirror? I dont think so, because then the rays will intersect and the image will be real.
Please let me know if you know a case in which a virtual image is formed in front of the mirror.
| If the image would be formed in front of the mirror it would not be a virtual image anymore, it would mean that your eye would have the impression to see it in front of the mirror.
As soon as the ray is reflected, the eye which interprets light as travelling in straight line necessary sees the object as coming from behind the mirror and that is why the image is said to be virtual.
So the answer to me is no.
| {
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Maximum power transfer proof I have the following homework problem.
Consider a power supply with fixed emf $ε$ and internal resistance $r$ causing current in a load resistance $R$. In this problem, $R$ is fixed and $r$ is a variable. The efficiency is defined as the energy delivered to the load divided by the energy delivered by the emf.
When the internal resistance is adjusted for maximum power transfer, what is the efficiency?
The answer is 50% but I'm confused how this is calculated. Here is my thought process so far.
The power dissipated in the load is $I^2R$ and $\displaystyle I = \frac{ε}{r+R}$ so $\displaystyle P = \frac{ε^2R}{(r+R)^2} = \frac{ε^2}{\frac{r^2}{R} + 2r + R}$ and so power transferred to the load will be maximum when $\displaystyle \frac{r^2}{R} + 2r + R$ is a minimum. Taking the first derivative $\displaystyle \frac {d}{dr}(\frac {r^2}{R} + 2r + R) = (\frac{2r}{R} + 2)$ and this equals 0 when r = -R. The second derivative is $\displaystyle \frac{2}{R} > 0$ so this point would be a minimum and therefore max power transferred when r = -R.
But having $r = -R$ doesn't make sense to me. From what I read, for maximum power transfer $r$ should equal $+R$. So I've probably done something wrong in my calculations or assumptions in the previous paragraph. Also wouldn't max power be transferred when the internal resistance is as close to zero as possible?
Can someone please show me a proof that shows why $r = R$ for max power transfer and then how to calculate the efficiency.
| Are you sure you mean the internal resistance "r"? The internal resistance typically can not be adjusted, often this question is phrased in terms of the load resistance "R".
The power dissipated in the external load is:
$$
P=\frac{V^2 R}{(r+R)^2}\;,
$$
which you need to maximize. If you maximize with respect to R, you find R=r... If you maximize with respect to r, you find r=0; algebraically you found a zero at r=-R, but r can not equal -R! It has a physical boundary at r=0, which gives the maximum for fixed R (you are maximizing on a fixed interval, you need to check the endpoints!).
However, are you sure you want to maximize with respect to "r", or do you actually mean "R"?
Finally, divide the value of the maximize load power by the total power dissipated, which is
$$
\frac{V^2}{(r+R)}
$$
evaluated at the correct resistance to determine the efficiency.
| {
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Does spacetime have symmetric curvature around an object? If yes, then why do planets revolve around the Sun in elliptical orbits? Does spacetime have symmetric curvature around an object? If yes, then why do planets revolve around the Sun in elliptical (as opposed to circular) orbits?
| To a first approximation the spacetime curvature around the Sun is indeed spherically symmetric. I say to a first approximation because the masses of the planets (particularly Jupiter) also produce curvature and this breaks the spherical symmetry. However let's ignore this for now because I don't think it's relevant to your question.
If I understand you correctly you're asking how a spherically symmetric curvature can produce orbits that are not circularly symmetric. The answer is that spherically symmetric curvature only means that angular momentum is conserved. The explanation for this is given by Noether's theorem, though I suspect the maths involved in this will be too hard going for non-nerds.
Anyhow, the angular momentum is given by:
$$ L = mr^2\omega $$
where $m$ is the mass of the orbiting object, $r$ is the orbital radius and $\omega$ is the angular velocity. We know that $L$ must be constant, and one way to achieve this is to have a circular orbit so $r$ and $\omega$ are constant. But we can also have constant $L$ if any change in $r$ is balanced out by a change in $\omega$ so that the product $r^2\omega$ stays constant. This is exactly what happens in an elliptical orbit, and in fact it is just Kepler's second law:
A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.
In a small time $dt$ the area swept out is:
$$ A = \tfrac{1}{2}r^2 \omega dt $$
If $A/dt$ is constant, as Kepler's law requires, this means $r^2\omega$ must be constant and this is exactly what we concluded above.
So spherical symmetry does not require circular orbits, only that the orbit must conserve angular momentum.
| {
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Hamiltonian from a Lagrangian with constraints? Let's say I have the Lagrangian:
$$L=T-V.$$
Along with the constraint that $$f\equiv f(\vec q,t)=0.$$ We can then write:
$$L'=T-V+\lambda f. $$
What is my Hamiltonian now? Is it
$$H'=\dot q_i p_i -L'~?$$
Or something different? I have found at least one example where using the above formula gives a different answer then the Hamiltonian found by decreasing the degrees of freedom by one rather then using Lagrange multipliers.
| Comments to the question (v2):
To go from the Lagrangian to the Hamiltonian formalism, one should perform a (possible singular) Legendre transformation. Traditionally this is done via the Dirac-Bergmann recipe/cookbook, see e.g. Refs. 1-2. Note in particular, that the constraint $f$ may generate a secondary constraint
$$g ~:=~ \{f,H^{\prime}\}_{PB} +\frac{\partial f}{\partial t}~\approx~\frac{d f}{d t}~\approx~0.$$
[Here the $\approx$ symbol means equality modulo eqs. of motion or constraints.]
References:
*
*P.A.M. Dirac, Lectures on QM, (1964).
*M. Henneaux and C. Teitelboim, Quantization of Gauge Systems, 1994.
| {
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When a pn junction is formed, why is a positive region of charge formed on the n side of the junction? I understand that when electrons diffuse from n-side to p-side, negative charge is developed on the p-side. But the mere absence of electrons on the n-side doesn't make that positively charged. The n-side must be neutral as it has no charge now. Where am I getting wrong?
| The pn-junction consists of fixed and mobile charges. The n-side has an abundance of electrons and the p-side had an abundance of holes. These are the mobile charges.
There is an abundance of charge carriers because of doping. Foreign atoms are introduced into the crystal. Some have an additional electron in the outer shell (n-type dopants), and some have one less electron in the outer shell (p-type dopants). For example, if you are doping silicon (4 electrons in outer shell) you can use Boron (3 electron IB outer shell) and Phosphorous (5 electrons in outer shell).
The dopant are fixed in place in the crystal lattice. Therefore when the mobile charges flow towards each other, they leave ionised dopants behind. For example, the n-type dopant has lost an electron, therefore it has a positive charge on +1. Similarly the p-type has gained an electron so it has a net negative charge of -1. Thus the n-type contains positive space charge and the p-type is negative.
| {
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Why invariance is important? The concept of invariance seems to have a great importance. Indeed, the fact that the laws of Electrodynamics are not invariant in every inertial reference frame led to the theory of Special Relativity which in the end makes those laws be invariant.
As I understand invariance, a physical law is invariant in two frames of reference when it holds good in both of them. That means if we write the law mathematically, the law assumes the same form in both frames of reference.
So for example, Newton's second law is invariant in frames $S$ and $S'$ if whenever $\mathbf{F}$ and $\mathbf{a}$ are force and acceleration as understood by the viewpoint of $S$ and $\mathbf{F}'$ and $\mathbf{a}'$ are force and acceleration as understood by the viewpoint of $S'$ we have $\mathbf{F} = m\mathbf{a}$ if and only if $\mathbf{F}' = m\mathbf{a}'$, and the law is the same in both of them.
Now, since this idea of invariance led to something as important as Special Relativity and even led people to change the way they understand space and time I wonder invariance is a quite important thing.
So, is invariance really that important? If so, why do we care so much with it? What's the real importance of invariance?
| Invariance may be connected with two things:
First, invariance usually means a symmetry of the system under certain transformations. This symmetry and it's accompanied conserved quantities mean that there is a set of observational objects, that do not change. Moving epistemologically on this, we may deduce that the only way we can write down laws for nature, meaning for example equations of motion, is by having something conserved, since otherwise we would not be able to observe them.
On a second hand, invariance usually implies the non-existence of a certain notion, something impossible to observe, as absolute spacetime position. Thus, from this perspective, invariance and symmetry mean a way of distinguishing between observational and non0observational quantities.
| {
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Microscopic definition of dynamic pressure of fluids So we have a moving fluid and we know from Bernouli's equation that there is a term called dynamic pressure(not to be confused with the hydrostatic pressure of the fluid).
So,what exactly is it and how can it be explained microscopically?
Note:do not involve relativity please!
| Wiki/google quote:
Dynamic pressure is the kinetic energy per unit volume of a fluid
particle. Dynamic pressure is in fact one of the terms of Bernoulli's
equation, which can be derived from the conservation of energy for a
fluid in motion.
So Dynamic Pressure is just local impulse/energy of movement that is being passed from particles to other particles in solution or in your measuring instrument
| {
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Most True to Life Physics & Biology Simulation Engine? I'm a programmer. I code in C++, C#, HTML5, and PHP. There are many graphics engines I have at my disposal. The question is: Does there exist a graphics engine that is as true to our reality as possible given our current understanding of physics? For instance, I can easily create macroscopic objects in a 3D space but what about all of the elements in reality that make up these macroscopic objects? What if for instance I wanted to start from the bottom up, creating simulations at the planck scale, particles, atomic structures, cells, microbiology, etc.? What if I want to simulate quantum mechanics? Of course I can make a model of an atom but it winds up not being correct in terms of being exactly analogous to real life.
I would like to correctly simulate these structures and their behaviors. Assume also that I have access to an immense amount of parallel computing processing power (I do). Perhaps some non-gaming scientific graphics engines exist that I'm not aware of.
| The scales and scopes of the models we do have are far larger than cellular level. Further, while there are skeletal models for animal bodies, the models for their motion is very much top-down modelling rather than bottom-up modelling. That is, an actual human's motions will be recorded and interpolated into the model, or an animator will pose the model in keyframes and the software will interpolate how to move the body from keyframe to keyframe. There are some cases where some level of bottom-up is done, such as in ragdoll modelling of bodies falling and colliding with environments.
In a real animal body, every motion has some force input to more or less every part of the body. Even a full muscular model of a leg, let alone a hand, would be amazingly complex, and the number of possible nervous system inputs to it would be enormous, and of limited use without the rest of the body, the circumstances, the mind... there's essentially no way to really model animal motion truly from the bottom up.
| {
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Usage of singular or plural SI base units when written in both symbol as well as name I have multiple doubts related to the usage of singular or plural SI base units when written in both symbol as well as name.
I have framed this question under two parts, namely, Part (a) and Part (b). Each part has three sentences which I have written on the basis of my understanding.
Please answer these 6 sentences/questions.
Part (a):
This page says that while using prefix for e.g., centi as in centimeter, it is
l = 75 cm long.(correct)
l = 75 cms long. (wrong)
*
*Does this rule is used for all SI prefixes (having powers of 10)?
*In this regard, we should be saying, or writing that, "how many cm are there in one metre?" (while saying we should say centimter or centimetres?)
*Please strike-through the wrong SI unit in the following sentences.
My weight is 70 kg / kgs,
or My weight is 70 kilogram / kilograms.
Part (b):
and the page next to above web link says, we should write:
2.6 m/s, or 2.6 metres per second.
In this regard, we should say, or write:
*Its speed is 0.26 metres per second.
*This pipe is 0.75 metres long.
*How many cm are there in 2 metres?
| cms and kgs are wrong. The SI units are abbreviations which are also used in the plural. You will write 2.6 m/s or 1 m/s, but say "2.6 meters per second" or "1 meter per second" respectively.
Keep in mind the SI units are also used in tons of other languages that do not form the plural by attaching an -s. The units look the same in those languages. (e.g. German: 1 Meter pro Sekunde, 2.6 Meter pro Sekunde)
The prefix doesn't change anything: kilometers -> km, micrometers -> µm.
Also, by adding an s, you will confuse it with seconds: As could be "Ampere seconds", which would be a charge compared to "Amperes", a current. You see the problem.
| {
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What's my $\mathrm dM$? Gravitational Potential inside a circle of mass
I'm trying to find the gravitational potential for an arbitrary point within a ring of uniform mass density. The point is constrained to be in the same plane as the ring.
So we start with:
$$\Phi=\int G\frac{\mathrm dM}{r}$$
Let's assume that the point of interest is along the $x$ axis $r$ away from the origin (which is at the center of the ring). An arbitrary point on the ring lies at:
$$a\cos\phi\hat{x}+a\sin\phi\hat{y}$$
And of course the point of interest is:
$$r\hat{x}$$
The distance between the point of interest and an arbitrary point on the ring is then:
$$\sqrt{r^2-2ar\cos\phi+a^2}$$
Back to the integral above, we get:
$$\Phi=\int G\frac{\mathrm dM}{\sqrt{r^2-2ar\cos\phi+a^2}}$$
Cool. I'm pretty happy up to this point, but what do I do about the $\mathrm dM$? Were I at the center of the circle, I would use $\mathrm dM=r\mathrm d\phi$. But I feel like it shouldn't be that simple if the center of my integration isn't the center of the circle. Should I use $$\sqrt{r^2-2ar\cos\phi+a^2}\mathrm d\phi~?$$ Am I completely off base here?
| $\mathrm dM$ is just $\rho \mathrm dV$, where $\rho$ is the density and $\mathrm dV$ is the volume element.
In your case then $\mathrm dM = \delta(r-R)\delta (\theta - \pi/2)\lambda r^2 \, \mathrm dr \, \mathrm d\theta \, \mathrm d\phi = \lambda R \, \mathrm d\phi$
| {
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Why is the Specific Heat of Helium 36 times greater than Xenon? Given that these are single atoms, why should the specific heat difference be so great? And more specifically, why does it take 36 times as much energy to raise the temperature of a given mass of Helium compared to Xenon?
| The key question is "Per mole or per gram?" Because the both values can be found tabulated as "specific heat" in various sources. Perhaps it would be useful to distinguish "molar specific heat" from "specific heat per unit mass".
You seem to be using the intuition for the molar quantity, so if the table is by mass, the answer is simply that you need more moles of helium.
| {
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Confusing, can a motor on a moving disc stop the disc? I've been troubled trying to solve this problem.
If a motor is fixed on a moving rotating disc and the motor's rotor has another disc attached to it. What would happen if we attached the motor's disc to the the disc its on and powered the motor's to rotate the disc opposite to the disc that the motor is moving on?
What would be the result, I couldn't figure it out. Would the system stop? If the torque of the disc= torque of the motor's disc.
EDIT: The motor is placed vertical on the disc, and the motor's attached disc(on the rotor) has the same diameter is the one it's placed on.
| Yes you can use the motor to slow and stop the disk, but you have to have (1) Alignment of the motor and its load angular momentum with the disk's angular momentum, and (2) enough motor load (moment of inertia) and motor speed to fully transfer the disk's angular momentum into the motor and load.
The process of momentum transfer is done all the time on spacecraft using reaction wheels or control moment gyros. Typically an initially spin-stabilized satellite is de-spinned to a slower rotation rate, and then the reaction wheels take over to slow the spin to a stop.
Typically 3 reactions wheels or more, aligned on different axes are used so that spin rates can be controlled in all possible axes.
Over time the reaction wheels are used to keep the attitude of the satellite stable and pointing in some particular direction - against disturbances such as planetary magnetic perturbations, upper atmosphere wind disturbance if the satellite is in low earth orbit, and solar wind pressures. So the wheels over time build up angular momentum. Before the wheels 'saturate' at their maximum speed, the satellite will 'dump' momentum by despinning the wheels against propulsion system applied torque.
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Could I break the sound barrier using circular motion? (And potentially create a sonic boom?) Ok, Lets say I get out my household vaccum cleaner, the typical RPM for a dyson vaccum cleaner reachers 104K RPM, Or 1.733K RPS. In theory, this disc would be travelling with a time period of 0.00057692307 seconds, If we take the speed of sound to be 344.2 metres per second, a breach in the sound barrier is easily possible for an item on the edge of the disc.
One question remains: For an extremely strong disc , could an item stuck onto it break the sound barrier, and create a sonic boom?
| I want to add to the other answers that when an object is rotating at a supersonic speed, an observer will be hit by a rapid series of sonic shock waves, as the shock wave is an ever-expanding spiral. This is what makes supersonic propellers so terribly loud.
The images below depict the process. The red circle in the middle is the trajectory of a propeller blade. Blacks circles show how the sound waves are propagating from the propeller over time.
*
*In the first image the propeller is rotating at a subsonic speed.
*At the speed of sound the shock wave starts to form.
*The third image shows the shock wave for a supersonic speed.
Note that the spiral that shows the shock wave wasn't added to the image in any way, the visual effect is created just by the circles touching themselves (similar to what happens to the sound waves).
Here is the Haskell program used to create the images using the diagrams library.
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can a really loud sound travel in space? I heard in a discovery news video that there is one particle every cubic cm in space. So, if i were to vibrate a circular body of say mass $10^7$kg at $10^{22}$Khz, would i be able to hear a sound from say 1 metre away in space, assuming i will somehow live? Im asking this because there is no documentation anywhere that sound could ever travel in space. Im 14, so I don't know how many zeroes to include in the values. Thanks. If this is true, the big bang would have produced an extremely loud sound.
| Assuming a loose definition of the word "sound," the answer is yes.
Let's consider the experiment you proposed. Suppose you place two large diaphragms facing each other in space. They could be simply large sheets of plastic stretched around a metal rim, like a drum head. And "large" in this context means much larger than the average distance between particles. Initially you would expect both diaphragms to be stationary and flat, since the same number of particles would impact both sides of both diaphragms, and the forces imparted would cancel each other out.
Now suppose you move one of the diaphragms toward the other a couple of feet. That will send more particles heading in the general direction of the second diaphragm than would otherwise be the case. A short time later, some of those particles, or other particles that were struck by those first particles, will impact the second diaphragm, causing a tiny increase in the number of impacts per second compared to the steady-state case. With instrumentation that is sensitive enough, you could detect that change. So it is fair to say that the "sound" has traveled through space.
| {
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Why temperature of liquid drops after spraying through a nozzle? We have tested in our lab as mentioned in the picture.
We connected hot water at $130^\circ F$ at $40 Psig$ to a nozzle (bottle sprayer). We measured the temperature differences inside tank and after spraying and found that there is a temperature drop of $50^\circ F$. We know that liquid water is sprayed and some heat is lost in the form of work done inside the nozzle. The surrounding temperature is kept at $75^\circ F$. This temperature difference is due to faster heat transfer happening due to increase in surface area of water as it is sprayed? Or does it have anything to to do with work done and surface tension?
| When you spray through a nozzle the water gets converted into tiny droplets, thus the surface area increases. Due to the increased surface area, the heat dissipates to the surrounding air molecules the water's temperature decreases.
Also, if the surrounding air is at higher temperature than the water then the water will get heated up.
| {
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movement of particles in electric field I am confused about a homework problem. Let's assume we have two electrically charged particles of which we know the charge and mass respectively. Let's say that at first they are fixed at some distance $r_1$ and then released simultaneously. I want to find their velocities at distance $r_2$.
Due to conservation of energy, we should have the equation $$ \frac{m_1 v_{1}^{2}}{2} + \frac{m_1 v_{1}^{2}}{2} = \int_{r_1}^{r_2} F \; dr $$
Where $$F = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q_1 Q_2}{r^2}$$.
Now I obviously need another equation. I was thinking that by the law of conservation of momentum, I'd get (as the momentum equals 0 when both particles are still fixed in position)
$$m_1 v_1 + m_2 v_2 = 0$$
But this is where I get confused: Consider the case where one particle remains fixed and we let go of the other one. Wouldn't we get $m_1 v_1 = 0$ by conservation of momentum and something not equal to 0 by conservation of energy in the same way I obtained the first equation above?
| In the first case,momentum is conserved because force is applied to each charge from WITHIN the system.So,the center of mass of the system is constant.In the second case,in order for just one charge to move,it has to be put in an EXTERNAL electric field.So,you can see that momentum within the system which consists only of one charge can not be conserved.If you include the source of the electric field,then yes you can do this with m2 being the source of the field.
| {
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Does the electric force on a charged particle in a uniform electric field increase? If I have a proton in a uniform field between two parallel oppositely charged plates and the proton accelerates, the electric force acting on it stays constant seeing it is a uniform field and as a result the acceleration of the particle is also constant.
But does the electric force on the proton increase as it nears the negative plate?
I would think yes seeing the distance is getting smaller (cf. Coulomb's law) so the electric force (which originates from Coulomb interactions between the charged particle and the plates) must be getting bigger. But this conflicts with the fact that the force is constant. Any help?
| NO the force does not change with distance
as E=F/Q
F=QE
electric field is constant and charge is also constant then force is also constant.
| {
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Can sound frequency create more ideal conditions for fire? You may have seen the recent story of a device that engineers claim can extinguish flames using sound frequency. An older article loosely explains the theory behind how this works:
Sound travels in waves, which are simply variations of pressure in a medium—whether solid, liquid or gas. The energy from vibrating objects, such as speaker membranes, moves from particle to particle in the air in a repeating pattern of high- and low-pressure zones that we perceive as sound. According to the ideal gas law, temperature, pressure and volume are related; therefore, a decrease in pressure can lead to a corresponding decrease in temperature, which may explain how sound can extinguish a flame.
Like most readers and the journalists sensationalizing this story, I don't have much understanding of the laws of physics that apply here. But it made me curious to ask those who know: Does it stand to reason that altering the frequency appropriately could increase air pressure or otherwise be used to create more ideal conditions needed for fire?
| A reasonable sub woofer at sound power level of 130 dB would produce pressure fluctuations of 60 Pa. Compare this to the ambient pressure of 100'000 Pa and you will see that related temperature fluctuations would be negligible.
It extinguishes fire because it pushes the air back and forth. For the small fire in this video you could take a small air blower and blow off the flame much quicker. Of course, stronger flame will not be extinguished by an air blower (on the contrary!), but I'm afraid that for such flame a loudspeaker also won't do much.
In closed or semi-closed environments (furnaces, rocket engines) the reflected sound can resonate with the flame, leading to flame extinction or amplification. Such interactions are studied by thermoacoustics. But this has little to do with the demonstration in this video.
| {
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Error calculation in parallel resistances This is the question:
There are two resistors with resistance values $R_1=100\pm3$ ohm and $R_2=200\pm4$ ohm. Find the equivalent resistance of parallel combination.
According to what I've learnt, in any expression of multiplication or division, the percentage errors of each term are added up to find the equivalent percentage error. That is, if $$y=\frac{\text {AB}}{\text C}$$ then $$\%\;\text{error in y}=\%\;\text{error in A}+\%\;\text{error in B}+\%\;\text{error in C}$$
For the above problem, let $R_s$ denote series combination. Then $R_s=300\pm7$ ohm.
Let $R_p$ denote parallel combination.
$$\therefore R_p=\frac{R_1R_2}{R_1+R_2}=\frac{R_1R_2}{R_s}$$
Ignoring errors, we get $R_p=\frac{200}{3}$ ohm $=66.67$ ohm
$\%\;\text{error in R}_1=3$, $\%\;\text{error in R}_2=2$, $\%\;\text{error in R}_s=\frac73$
Hence, $\%\;\text{error in R}_p= 3+2+\frac73=\frac{22}{3}$
So, error in $R_p$ will be $\frac{22}{3}\%$ of $\frac{200}{3}$, which is approximately $4.89$.
Hence, I got $R_p=66.67\pm4.89$ ohm.
However, the book used the formula described and proved here and arrived at the answer $R_p=66.67\pm1.8$ ohm.
So, is the percentage error method wrong?
| Basically what you are doing for $y=\frac{AB}{C}$ is adding up all the percentage errors, which is wrong.
Take log function on both side, so you get $\log(y)=\log(A)+\log(B)-\log(C)$
so for percentage error becomes: $\frac{dy}{y} = \frac{dA}{A} + \frac{dB}{B} - \frac{dC}{C}$
so if you actually follow your step, you get:
$ %error= 3 + 2 - 7/3 = 2.6%$
so 2.6% of 200/3 gives you 1.8%
| {
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$\nabla^{\mu}\nabla_{\mu}$ in general relativity I am trying to work out $\square=\nabla^{\mu}\nabla_{\mu}$ in the metric
$
ds^{2}=-A(r)dt^{2}+B(r)^{-1}dr^{2}+r^{2}d\Omega^{2}
$$
My work:
when applying $\square$ to a scalar $\phi$, then
$
\square\phi=\nabla^{\mu}\nabla_{\nu}\phi=\nabla^{\mu}\partial_{\mu}\phi=g^{\mu\nu}\nabla_{\nu}\partial_{\mu}\phi=g^{\mu\nu}(\partial_{\nu}\partial_{\mu}-\Gamma^{\lambda}_{\mu\nu}\partial_{\lambda})\phi
$
Christoffel symbol
\left(
\begin{array}{ccc}
\left\{0,\frac{A'(r)}{2 A(r)},0\right\} & \left\{\frac{A'(r)}{2 A(r)},0,0\right\} & \{0,0,0\} \\
\left\{\frac{1}{2} B(r) A'(r),0,0\right\} & \left\{0,-\frac{B'(r)}{2 B(r)},0\right\} & \{0,0,-r B(r)\} \\
\{0,0,0\} & \left\{0,0,\frac{1}{r}\right\} & \left\{0,\frac{1}{r},0\right\}
\end{array}
\right)
substituting the metric and affine values in the equation above, my answer came to be
$$
\square=-A(r)^{-1}\frac{d^{2}}{dt^{2}}+B\left(\frac{d^{2}}{dr^{2}}+\frac{1}{r}\frac{d}{dr}\right)+\frac{1}{2}\left(B^{\prime}+\frac{B A^{\prime}}{A}\right)\frac{d}{dr}
$$
However, the answer happens to be
$$
\square=-A(r)^{-1}\frac{d^{2}}{dt^{2}}+B\left(\frac{d^{2}}{dr^{2}}+\frac{2}{r}\frac{d}{dr}\right)+\frac{1}{2}\left(B^{\prime}+\frac{B A^{\prime}}{A}\right)\frac{d}{dr}
$$
Could someone please show me where the third comes from in the second term?
| Mm.... At first I repeated your calculation and got the same answer as yours, then I checked the paper you gave and found it consist with (32 a) and it seems not a typo, so I read it from begining - oh brother it's not 2+1 gravity - -b it's 3+1 gravity and you should treat $d\Omega^2$ more carefully:
$r^2 d\Omega^2=r^2d\theta^2+r^2\sin^2\theta d\phi^2$
so you get one more r factor when using the formula @Prahar gives.
and btw the Christoffel symbol is for 2+1 and certainly wrong for 3+1.
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Relational Interpretation of Quantum Mechanics and Universal Wave Function Why is there no universal wave function for a relational interpretation of quantum mechanics?
"Quantum mechanics is a theory about the physical description of physical systems relative to other systems, and this is a complete description of the world" - Carlo Rovelli (from wikipedia! :p )
Basically, if the Universe functions functions in a relative manner when seen in part, how does it function in a relative manner when seen as a whole, when there are no "other systems"?
| There is no universal wave function. The universe has many different mathematical descriptions, each corresponding to what a different observer can interact with around them. Each is incomplete, because no observer can interact with whole universe. Each observer, for example excludes themselves from the world they interact with.
| {
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Compact Disc Optics - Why use a linear polariser and a quarter wave plate? I just came across this website about the application of a quarter wave plate. Link: Compact Disc Optics.
My question is why does the beam need to be linearly and then circularly polarised before sending to the compact disc? And the returned beam undergoes the same before reaching the detector? Is it related to power loss?
| A quarter wave plate together with a polarizing beam splitter is a standard way of building an optical isolator. In an optical isolator the light traveling in one direction is transmitted but the light travelling in the reverse direction is reflected. This is important when reading compact discs because the reflected light carries the signal that you want to read.
As for how it separates the forward travelling and backward travelling light. The initial pass through the linear polarizer leaves the light linear polarized in one direction. The first pass through the quarter wave plate converts it to circular polarization, and the reflection off of the CD converts it to the opposite handedness circular polarization. Finally, the quarter wave plate converts it to the opposite linear polarization from the input beam which is then reflected by the polarizing beamsplitter.
| {
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Why are we allowed to spontaneously break the Higgs field symmetry? In the lepton Lagrangian there are no mass terms allowed for the gauge bosons, due to gauge symmetry. To fix the problem of existing gauge boson masses, we introduce another field (Higgs) and say that here we
1) have self-interactions
(Why do we have self-interactions for the Higgs field but not in the lepton case? Is the Higgs "charged" like the self-interacting gluon is?)
2) just flip the sign of the $|\phi|^4$ term
(Why are we allowed do to that?)
3) hence break the gauge symmetry and get gauge boson masses
(Why are we allowed to do break the symmetry here but not for the lepton field?)
→ We have similar initial complex fields (lepton and Higgs) with similar couplings to the other gauge bosons, so why are we allowed to treat both fields in such a different way?
| Essentially, Higgs self-interactions are allowed because they don't violate any laws or symmetries (i.e. the $\phi^4$ term is gauge invariant, Lorentz invariant etc). Informally, a Lagrangian can (and possibly should) consist of any/all combinations of fields, derivatives of fields etc that respect the symmetries of the theory. In the Standard Model the Higgs is not charged.
The sign of the $\phi^4$ term is flipped by convention, because the coupling constant is a parameter of the theory, to be determined by experiment (the theory does not offer a prediction for the coupling constant) so yes, you are allowed to choose the sign.
Since we don't actually measure symmetries, you are allowed to break any symmetries you wish, provided you have a mechanism for breaking it (in this case, its spontaneously broken due to fluctuations about the minima) and provided it respects any constraints of your theory (energy/momentum conservation, renormalisable, kinematically allowed etc).
| {
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Do magnetic fields cause ionisation of gases? I am doing my final year engineering project on Magnetic Field Assisted Combustion and was curious to see what people thought about it.
Companies sell rare earth magnetic arrangements to be attached to fuel lines of gas burners and they are said to improve combustion efficiency but why exactly?
I have performed a number of experiments using a standard butane/propane gas burner with some magnets manufactured by one of said companies and have had some contradicting results. With lower strength magnets, heat transfer unexpectedly slowed down but with a much stronger arrangement, heat transfer rate was increased.
Also, the burn out time of the same amount of gas took 8 minutes less with the magnets in place around the fuel line.
I have read a number of journals on similar subjects but even within these, the actual reason for the increase in heat output is still not known.
Any thoughts on the subject would be massively appreciated and possibly give me some other areas to investigate that I have not already thought of.
| Reading the original question a posible mechanism has suggested itself to me. As the gas travels along the pipe "streaming currents" can be generated by friction. This is a phenomenon in the literature where by electrostatic charge can build up on the pipe and there is a concentration of charged particles in the boundary layer. One consequence has been instances of fuel fires when Kerosene is delivered through a hose to refuel an aircraft. The combination of such a steaming current plus a magnetic field could induce forces on the fuel maybe encouraging transition and turbulence.
| {
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At what point does force stop translating an object and start purely rotating it? At what point (or distance) from the axis of rotation, does force applied on a rigid body stop translating and purely rotating the body? Can such a point even exist? Does the body always have to translate?
This question assumes that the body is in empty space and unattached.
| The only way to purely rotate a rigid body about its center of mass is to apply a pure torque (no net force). If the net force applied is zero then the center of mass is not accelerating.
However and combination of translation and rotation of the center of mass can be viewed as a pure rotation about the instant center of rotation. So to effectively answer your question, a force not through the center of mass will rotate the body about a specified point.
For planar case see: https://physics.stackexchange.com/a/86996/392
The point of rotation A is defined by the distance $x$ as
$$x = \frac{I_{cm}}{m \ell}$$
If you want to go a little deeper then see: https://physics.stackexchange.com/a/81078/392
Note that these two statements are equivalent:
*
*A pure force thorugh the center of gravity (with no net torque) will
purely translate a rigid body (any point on the body).
*A pure torque
any point on the body (with no net force) will purely rotate a rigid
body about its center of gravity
| {
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Charged particle under a uniform electric field
Suppose a charge particle $q$ starts to move without initial velocity under the influence of a uniform electric field $E$ pointing in the positive $x$ direction. Express its position vector in terms of proper time $\tau$.
According to wiki:http://en.wikipedia.org/wiki/Lorentz_force#Relativistic_form_of_the_Lorentz_force, The Lorentz Force is given by $\frac {dp^{\alpha}}{d\tau}=qU_{\beta}F^{\alpha\beta}$. In this case $F^{\alpha\beta}$ reduces to $$\begin{bmatrix} 0 & \frac{E}{c} \\ -\frac{E}{c} & 0\end{bmatrix}.$$ Let $U_{\beta}=(u_0,u_1)$, then $p^{\alpha}=m_0(u_0,u_1)$, so we have $$\left(\begin{array}{cccc}m_0\dot u_0&\\m_0\dot u_1\end{array}\right)=q\left(\begin{array}{cccc}0&-\frac Ec&\\\frac Ec&0\end{array}\right)\times\left(\begin{array}{cccc} u_0&\\u_1\end{array}\right).$$ Since $$m_0\dot u_1=q\frac Ecu_0,$$ $$u_1=\frac{-m_0\dot u_0c}{qE},$$ we get $$m_0(\frac{-m_0\ddot u_0c}{qE})=\frac{qE}cu_0,$$ $$\ddot u_0+\frac{q^2E^2}{m_0^2c^2}u_0=0.$$ The characteristic polynomial is $$r^2+\frac{q^2E^2}{m_0^2c^2}=0.$$ Obviously the determinant is negative and $u_0$ is a trigonometric function of the proper time $\tau$. It can also be deduced that $u_1$ is the same kind of function. But this is clearly not the case. I hope someone can tell me where I did it wrong.
| As you know the answer should be a hyper trigonometric function instead of a trigonometric one. Your mistake is with lowering/raising of vector components
$$
p^\alpha = m_0 \left( u^0, u^1 \right) = m_0 \left( \eta^{00}u_0, \eta^{11}u_1 \right) = \pm \left( - u_0, u_1\right)
$$
Where the $\pm$ comes from your metric convention. This will lead to
$$
r^2 + \frac{q^2 E^2}{m_0^2 c^2} \longrightarrow r^2 - \frac{q^2 E^2}{m_0^2 c^2}
$$
and you get the expected solution
| {
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Why does it seem as if big vehicles "attract" mine when I drive close to them? When I drive a car at high speed and when I am near to another big car (like a van, or transport vehicle) I feel an attraction to or something push me toward the other big car. What's the physics in this case?
| There is a great amount of energy involved here for sure. All moving objects meet resistance and the more velocity the more resistance. For objects moving fast there is aerodynamic designs and properties added so it experiences less turbulent force pressing back on the object. Drafting in race car driving is a useful example of this. The vehicle in front is breaking through the air while the car in behind rides inside the draft or slipstream. Inside the slipstream it will take much less energy to produce the same speeds. So you are not getting pulled in by the vehicle ahead its that under the same power conditions you are meeting much less drag or resistance so you will for sure have to dial back your throttle. The front vehicle breaks the turbulence for the vehicle behind. Read about aerodynamics.
| {
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How does "contamination" through (radioactive) radiation work? Physically, what does it mean when people or objects are contaminated with radiation? Is it because they actually carrying heavy metal particles?
| When a person or an object is contaminated, there is some radioactive material in it (not necessarily heavy metal), either sticking on the surface or ingested.
In the context of radiation protection, you have a radioactive contamination when an unsealed radioactive source produces an unintended and abnormal level of radioactivity in the environment. Unsealed radioactive sources do not have a secure containment and are usually in the form of dissolved ions in solution (can be in gaseous form when evaporated).
Contamination most likely happens in nuclear medicine facilities where radioactive solution are used for diagnostic imaging and therapeutic purposes. An accidental spill of these solution can cause contamination in the laboratory premises.
Higher levels of contamination occur in accidents in nuclear reactors such as those used in nuclear power plants and in the production of useful radioactive materials.
| {
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Limits of Integration Trig, Mag Field Infinite Length Wire I don't understand how the limits of integration should be defined when doing basic integrals of trig functions. It seems like it's an arbitrary decision, I don't understand it.
Here's the set up: For the field near a long straight wire carrying a current $I$, show the Biot-Savart law gives the same result as Ampere's law.
Now intuitively, for me at least, with the way that $\theta$ is defined, I would view the angle as becoming smaller as $y$ moves toward negative infinity. So the limits of integration make sense in that regards. But then the cosine doesn't make sense anymore. As $y$ becomes more negative, which corresponds to an angle between $0$ and $\pi/2$, then cosine should always be positive. But because cos=adj/hyp, then $\cos\theta=y/r$, and $y$ would be negative, even though the corresponding angle is between $0$ and $\pi/2$?
I know I'm misunderstanding something fundamental, hopefully somebody can help me so I can move on. I've been struggling with this for so long because it's easy enough to arbitrarily assign limits to get the answer you're looking for, but I want to know the right way, and more importantly, why it's the right way.
| I think I figured out my confusion. With these limits defined, in order for cosine to be correct, I have to redefine cos=-adj/hyp. Then everything else works out fine. Weird. It feels wrong to just redefine cosine, but it's true under these defined limits. Is that right, is that something you have to do sometimes? Redefine a trig function? The rest of the math works out fine once I do that.
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Will the electrostatic force between two charges change if we place a metal plate between them? If a thin metal plate is placed between two charges $+q$ and $+q$, will this cause a change in the electrostatic force acting on one charge due to another? What is the concept behind this? What will happen if the metal plate is thick?
| I think that the force exerted by each charge on each other will be equal to $0$, as the electric field of charges will not pass through the metal plate. But the net force on each charges will remain the same and the force will be exerted by the charge induced on the metal plate by both charges.
| {
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How many atoms are in a piece of paper? How many atoms are there in a common sheet of paper?
The paper is A4, i.e. $210 \, \mathrm{mm} {\times} 297 \, \mathrm{mm}$ $\left(8.27 \, \mathrm{in} {\times} 11.7 \, \mathrm{in}\right).$
| So, a piece of paper is made of wood, and wood is some organic substance. I don't know what the chemical formula is, but let's say it's mostly carbon. In fact, let's just pretend it's all carbon, since you only want order of magnitude. Wikipedia tells me a piece of A4 paper weighs about five grams, and then I divide by the atomic mass of carbon and get $\approx 10^{23}$ atoms. This is on the order of Avogadro's number, which is generally what you get for small but macroscopic numbers of atoms.
| {
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Drying clothes with the sun's heat, without any air Will my wet clothes dry if I hang them under the sun, and if there is no air around the clothes? In other words, do I need both air and heat to dry wet clothes, or is heat alone (in the imagined absence of any air) enough to dry wet clothes? Related question : will wet clothes dry with only the suns heat, but when placed in a vacuum? Please note - I am trying to dry my clothes differently on earth, and not in outer space.
| Your clothes would dry very quickly in a vacuum, assuming that the temperature is still one that one would find on earth. This is because the water would boil out of your clothes. On earth normally, boiling takes a lot of heat energy. This is because of the air pressure. In your scenario there is no air pressure. so the water will boil easily.
| {
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Potential barrier scattering when particle energy equals to the barrier height What happens if we have $E=V$, where $E$ is the energy of a incoming particle and $V$ is the height of a square potential barrier? This wiki page actually gives a finite transmission probability for this case. But what does the wave function look like in the barrier region?
Edit:
I just realized that the potential barrier case can be easily solved and the transmission can be calculated to be the one given in wiki. However, things are slightly different if we have a step potential at the origin instead of a square barrier. Even though a step potential is just a square barrier with infinite width, we look at the situation separately.
If we look at the schroedinger equation for the barrier region, which goes from 0 to $\infty$, then we have
$$\psi''(x)=0$$
which means $\psi(x)=ax+b$, where $a$ and $b$ are undetermined constants. Suppose for the no-barrier region, the wave function is given by $e^{ikx}+re^{-ikx}$, where $r$ is the reflection coefficient. Then after matching the boundary conditions, we have
$1+r=b$ and $1-r=-ia/k$.
If we require that the wave function does not blow up on the potential side, then we must have $a=0$ and consequently we have $r=1$ and $b=2$, which means that even though it becomes all reflected the wave function at the potential region is a non-zero constant function.
So how to explain this peculiarity? Is it because that transmission is not necessarily related to probability density?
| I have seen references relating the "square potential barrier" to the case of the space between two metal surfaces, but this is incorrect. The square potential barrier would require an infinite electric field at both ends with no field elsewhere. This model is used frequently but I see no practical application.
| {
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Beginners Textbooks in physics Hello I am fifteen and I already know everything that my school has been teaching me so I have been going ahead. I have already been studying mathematics far past where I am at school, but I am very interested in physics. I want to learn everything up to advanced topics such as super-string theory. But to get there, I obviously have to start at the beginning. Any good textbooks out there for somebody like me? Preferably something with a lot of practice problems and that has many applications.
| You can read the utterly fantastic Feynman Lectures on Physics which is free for online viewing at the link provided. I would also recommend Feynman's Tips on Physics: Reflections, Advice, Insights, Practice - A Problem-Solving Supplement to the Feynman Lectures on Physics. My University uses Giancoli as well as the textbook for introductory physics, but I find regular introductory textbooks to be unreadable, endless, and unpalatable.
I would also recommend reading up on the history of Physics and some fun books that are meant for people your age. I would recommend:
Mr. Tompkins in Paperback by George Gamow, a Nobel Prize winner
$E=mc^2$: A Biography of the World's Most Famous Equation
From X-rays to Quarks: Modern Physicists and Their Discoveries by Emilio Segre, another Nobel Prize winner
and,
Big Bang: The Origin of the Universe by Simon Singh which is an excellent historical development of what the Big Bang theory is and how it came to be.
| {
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Why do snorkelers not need to wear corrective glasses when snorkeling with goggles on? I am myopic ~ -2.75 sph +1cyl. When I went snorkeling they tell you not to wear glasses behind the goggles. Surprisingly, underwater, things remain in focus with goggles on even without prescription lenses, while things outside the water at an equal distance would be blurry. Why is this the case?
Thank you.
| the assertion that corrective lenses are not needed underwater is incorrect. diving masks with corrective lenses built-in have been in use for decades. the reason you yourself experience this effect is probably contained in Chris' and Martin's comments above.
| {
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Are there any scales other than temperature that have different zero points? For most physical measurements, zero is the same regardless of the units used for the measure:
$0 \mathrm{mi} = 0 \mathrm{km}$
$0 \mathrm{s} = 0 \mathrm{hr}$
but for absolute temperatures, different systems have different zeros:
$0 ^\circ\mathrm{C} \neq 0\,\mathrm{K}$
Are there any other physical, measurable quantities (other than temperature) that have different zero points?
I'm looking for measurable quantities that are applicable anywhere -- things like voltage or temperature, not local quantities like "distance from the Empire State Building".
| Gauge pressure.
From Wikipedia:
Gauge pressure is zero-referenced against ambient air pressure, so it is equal to absolute pressure minus atmospheric pressure.
| {
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Sun and planets orbit each other Do not the planets and the Sun revolve in orbits around each other and the shape of the orbit depends on where the center of gravity of the system is? The greater the mass of the Sun, the closer the orbit approximates a perfect circle.
| No. The shape of the orbit, i.e. how elliptical it is, does not depend on the relative masses of the two bodies.
All objects in the solar system orbit around the centre of mass of the solar system. For obvious reasons, namely that the Sun contain far and away most of the mass of the solar system, the centre of mass of the solar system is quite close to the centre of the Sun.
The shape of those orbits is determined by the total energy of a body (i.e. the sum of the potential and kinetic energies) and the way that the energy is shared between kinetic and potential energy. The potential energy is always negative by definition. Bound, elliptical orbits have a negative total energy. Circular orbits are a special case of elliptical orbits and have a potential energy that is exactly twice the total energy. i.e. if the total energy is $-E$, the potential energy is $-2E$ and the kinetic energy is $+E$. For stable elliptical orbits this is true on average, but at any instant in time, the split between potential and kinetic energy can be different and changes as the body moves around in its orbit.
| {
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What is the physical process behind wool shrinking when dried? Wool is a fibrous material, but other fibrous materials do not suffer the same problem.
Let us set the scene; a woollen jumper shrinking when put in the washing machine, then the dryer. This involves wetting and heating the jumper, then allowing it to cool and dry. It also tumbles as it is washed, but I do not think this is important.
What about this process causes the jumper to shrink?
Do the fibres themselves shrink, or do the become more tightly bound?
Having googled the matter, it seems the process going on is called "felting" and has to do with heat or water "shocking" the fibres - whatever that means - allowing them to settle into a tighter configuration, but I am interested in the exact physical process we are experiencing. What causes heat or water to "shock" the fibres, and why does it happen to wool much more than other fibrous materials? What is this shocking process; it sounds like the heat of the water overcoming the electrostatic attraction of the fibres, but these website are vague about the physics.
| It's kind of like "material memory." Wool is pretty kinky originally (as sheared), and the production processes pull the strands straight. Hot washing allows the material to revert to a tightly-wound config, which reduces the external dimensions, aka "shrinking."
| {
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What does it mean that a magnetic field's flux vanishes through any closed surface? I'm reading the Britannica guide to Electricity and Magnetism, and I came across the following quote:
A fundamental property of a magnetic field is that its flux through
any closed surface vanishes.
Can someone explain this in simpler terms?
Source
|
Can someone explain this in simpler terms?
Typically, the closed surface is a mathematical surface (Gaussian surface) which simply defines an 'inside' and 'outside'.
Since, as far as we know, there are no magnetic charges from which magnetic field lines start or end, any magnetic field line entering must exit through the surface; any magnetic field exiting must enter through the surface.
Thus, the number of field lines entering equals the number of field lines exiting and the flux of the magnetic field through the surface is zero.
| {
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Phase space Lagrangian? Reading out of this lecture series we define a phase space Lagrangian $\mathcal L$ to be a function of $4n+1$ variables namely $q,\dot q,p,\dot p,t$. My question is, what space is this function defined on? (I know that the $\dot p$ is there for names sake only).
My stab at an answer is it is a product space between $\mathcal L:TQ\times T^*Q\times \mathbb R\to \mathbb R$ since we are dealing with both velocity and momentum (where $Q$ is a configuration manifold). However this makes zero intuitive sense to me. If indeed it is phase space, my understanding was that taking the fibre derivative with respect to a function on the tangent bundle changed the velocity coordinates to momentum coordinates - the Legendre transform?
| *
*If $Q$ is configuration space, then the Lagrangian is a function $L: TQ\times \mathbb{R}\to \mathbb{R}$.
*Let the cotangent bundle $M:=T^{\ast}Q$ be the corresponding phase space.
*The Hamiltonian/phase space Lagrangian is a function $L_H: TM\times \mathbb{R}\to \mathbb{R}$.
| {
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How do waves have momentum? A question on a practice test I'm taking is as follows:
By shaking one end of a stretched string, a single pulse is generated.
The traveling pulse carries: A. mass B. energy C.
momentum
D. energy and momentum E. mass, energy and momentum
How would one describe the momentum of a wave?
| Without going into wave equations, lets just say the segments of a string does not only move along the vertical direction. There are horizontal movements as well, although to a much smaller amplitude. A segment is being pulled towards the source horizontally when departing the equilibrium position and pulled back when heading back from maximum amplitude. For a segment in the middle of the wave, this makes no net contribution, but at the front of the wave a new segment is always being pulled towards the source, so the advancement of wave front should sustain momentum.
edit: updated sloppy language.
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Bell inequality with triplet state Is it possible to prove Bell inequality starting from a state formed from triplet states, i.e. $\frac{1}{\sqrt{2}}(|\uparrow>_A|\uparrow>_B+|\downarrow>_A|\downarrow>_B)$?
If not, why?
I do not see why not, but somewhere it is mentioned something about non-rotational invariance. Morevoer I have always seen singlet state as starting point. Thanks.
| The answer is yes and no, but first, let me point out that you cannot "prove Bell's inequality", the whole point is that you violate the inequality in quantum mechanics.
Now, let me come to the yes/no part:
It's "no, you cannot violate Bell's inequality with this state", if you refer to what according to wikipedia is "the" Bell inequality:
$$ \rho(a, c) -\rho(b, a) - \rho(b, c) \le 1$$
where $a,b,c$ are three measurement settings. This inequality (as stated) seems to be only violated by states that are totally anti-correlated with parallel measurements. Your state, however, is totally correlated.
It's "yes, of course you can violate Bell's inequality with this state", if you refer to what is nowadays understood as Bell's inequality. Not one, but literally an infinite amount of inequalities that can be violated by states that do not admit a joint probability distribution for all setups. This is very loosely speaking, to get a clearer picture let me refer to Asher Peres and the Braunschweig/Hanover question site with progress on the matter of Bell inequalities.
| {
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Why do we use capacitors when batteries can very well store charges? Can batteries be used instead of capacitors? I am trying to figure out a basic, superficial and any obvious difference between the two.
| Practically we use capacitors when we require a large amount of charge to be flown within fractions of seconds.. Battery provides a nearly uniform voltage and effective in long use, but when it comes to discharge a large amount of charge in a fraction of second, battery is ineffective.. How ever by a building a capacitor with large capacitance we store a bulk of charge (large potential difference) to be flown within seconds.. E.g flash of a camera, Z-machine at nuclear reactor..
| {
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Are circles stronger than triangles? I've often heard in engineering that, "there is no shape stronger than a triangle." I also recall that arches are also very strong shapes, which can be crudely described as a perpendicularly-symmetrical half-an-ellipse; Which can be simplified to half a circle.
If there were no conventional complications with designing structures to utilize circles; Which shape is stronger?
Given a simple two-dimension-like application such as simple bridges or trusses as an example, of obvious visualizations.
| Triangular support(or triangle here) is unparalleled in terms of strength they provide to support load because all the hold mass is properly distributed across the support. You may disagree when g=0 :)
| {
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Does it take more energy to open a door when applying force close to the hinge?
Assuming an ordinary hinged door (without any springs), would it take more energy to open it when applying force in the middle of the door (point b), rather than at the end of the door (point a), where the door knob is?
"Opening the door" should be interpreted as accelerating the door to a certain rotational speed.
My own answer is no, since the change in force would be proportional to the distance required to open the door and therefore the total energy would remain the same.
| If you open a door by pushing it near the hinge, you apply greater force than when you push it near the outside edge, which requires lesser force since the width of the door acts as a lever and force multiplier. As the friction of the hinge and the weight of the door are equal in both cases, and assuming displacement is the same, net energy transferred to the door is the same in both cases, but only if the speed of the door swinging open is the same in both cases.
Kinetic energy = 0.5 * mass * v^2
You could also solve this problem by using torque. Although energy is a scalar and torque is a vector, they are both expressed in newton meters (joules for energy).
Torque = mass of the door * acceleration * lever arm * sine angle of force applied
The length of the lever arm depends on where you push the door. If you assume the acceleration is proportionately greater the shorter the lever arm, you get the same torque and thus the same energy transferred to the door. Depending on your assumptions, the amount of energy transferred to the door needn't differ from pushing it near its edge.
| {
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Arrow of time and entropy? The arrow of time is usually defined by the direction in which entropy increases. In a closed system, if there's a max entropy that the system can reach, does that mean time stops or at least become undefined at the max entropy state?
See also: For an isolated system, can the entropy decrease or increase?
| This is my first answer here so please don't expect a lot. I watched the BBC show yesterday just about this question. It is called Wonders of the Universe, episode 1 - Destiny.
According to the Professor Brian Cox, yes, arrow of time will stop eventually when maximum entropy is reached. It's going to happen in unimaginable amount of time. Literally unimaginable. If you count every atom in our Universe as one year — it would be not enough atoms to describe this number.
Eventually every star is going to die. Then every black hole. Entropy is always increasing. In the end it would be nothing but cold photons floating in the empty space. When this happens maximum entropy is reached and time as we know will no longer exist.
| {
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Physical meaning of the separation constants in Laplace's Equation for Electrostatics In Electrostatics, if we consider a region without charges the electrostatic potential $V$ obeys Laplace's Equation $\nabla^2 V = 0$. We can tackle this with separation of variables. In cartesian coordinates we have $V(x,y,z) = X(x)Y(y)Z(z)$ and so the equation is:
$$\nabla^2 V = 0 \Longleftrightarrow \dfrac{X''(x)}{X(x)}+\dfrac{Y''(y)}{Y(y)}+\dfrac{Z''(z)}{Z(z)} = 0.$$
For this to happen there should be three constants $C_1,C_2,C_3$ such that
$$\begin{cases}X''(x) &= C_1 X(x), \\ Y''(y) &= C_2 Y(y), \\ Z''(z) &= C_3Z(z).\end{cases}$$
Now, another example of a problem in Physics tackled with separation of variables is Schrödinger's equation
$$i\hbar \dfrac{\partial \Psi}{\partial t} = H\Psi,$$
separation of variables yields two equations one of which is $H\psi = E\psi$ for the spatial part and the separation constant has one physical meaning as expected value of energy.
Now, the constants $C_1,C_2,C_3$ appearing when we separate variables on Laplace's equation for electrostatic potential has some physical meaning? If they do, what is it?
|
Now, the constants C1,C2,C3 appearing when we separate variables on
Laplace's equation for electrostatic potential has some physical
meaning? If they do, what is it?
The constants are the related to the square of the spatial (angular) frequency or a spatial growth/decay constant.
For an example of spatial frequency, let
$$X(x) = A \sin (k_xx) + B \cos(k_xx)$$
Then
$$X''(x) = -k^2_x X(x)$$
Or, for an example of a spatial growth/decay constant, let
$$Y(y) = A e^{k_yy} + B e^{-k_yy}$$
Then
$$Y''(y) = k^2_y Y(y)$$
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How do you prove that $L=I-V+1$ in $\lambda\phi^4$ theory? It is known that the number of loops in $\lambda\phi^4$ theory is given by the formula
$$L=I-V+1$$
where $L$ is the number of loops, $I$ the number of internal lines and $V$ the number of vertices. I would like to know the proof of this statement.
| This formula is actually Euler's formula for planar graphs, and holds for all Feynman diagrams regardless of what theory we are in.
The proof proceeds by induction and is easy if we first disregard the case of crossing lines:
*
*Observe that a one-loop graph has two vertices, one loop, and two internal lines, so the formula holds.
*Observe that a $(n+1)$-loop graph is produced from a $n$-loop graph by either drawing one additional line between two already existing vertices, which doesn't change $L-I$, or by adding a new vertex and connecting it to two other vertices, which doesn't change $L-I+V$.
*By induction, the formula holds for all graphs with finitely many loops.
More formally, we can say that
A Feynman diagram is called planar if the adjoint graph obtained by connecting all external lines to a single vertex is planar.
and then we have proven up to now that the formula holds for all planar Feynman graphs. Interestingly, not even all $\phi^4$ graphs are planar. Consider $2\to 2$ (or $1\to 3$)-scattering with a box diagram, where each external line is connected to its own vertex, and each vertex is connected with each other vertex. The adjoint graph is the complete graph on five vertices, which is known to be not planar.
Nevertheless, the "Feynman-Euler formula"
$$ L-I+V = 1$$
still holds because of the way loops are formally counted. By the general Euler formula,
$$ \#\{\mathrm{vertices}\} - \#\{\mathrm{edges}\} + \#\{\mathrm{faces}\} = 2 - 2g$$
where $g$ is the genus of the surface on which the graph can be drawn without intersections, and "faces" are all regions bounded by edges. A "face" does not have to have a vertex at every corner, so when you get two crossing lines in a Feynman graph, you get two additional faces that you do not count as loops - the above boxy $\phi^4 $ diagram has four faces inside the box, but only two loops.
Since every crossing of lines that cannot be eliminated by deforming the graph (and is hence a "true crossing" and not just us being too dumb to draw the graph properly) increases the genus on which you could draw the graph without crossings by $1$, the "Feynman-Euler formula" for all graphs follows from the general Euler formula.
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Calculating wind force and drag force on a falling object I'm trying to numerically integrate the motion of an object (say, a falling vertical cylinder). Here, there's a drag force: the wind "acting" on the body (presumably adding horizontal velocity) and the air itself slowing down the vertical motion.
Is it correct to calculate both forces with the drag equation, $F_D = \frac{1}{2} \rho v^2 C_D A$? Here I suppose the velocity would be the relative velocity between the falling object (i.e. initially $(0, -15)\ m/s$) and the "air" (i.e. $(5, 0)\ m/s$). If that's the case, where would the drag force point towards to? $-\hat{v}_{relative}$?
Thanks a lot for your help in advance.
| Yes, the force points along the vector of the relative velocity between the object and the air.
Quadratic drag is an interesting phenomenon. You have to calculate the net velocity vector (which includes a horizontal and vertical component) and compute the force along that axis; when you then decompose it into horizontal and vertical components you will find that the vertical drag is greater because of the cross wind. This is not an intuitive result!
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At what energy consumption would we get a 1 degree rise in the Earth's temperature? If energy consumption continues to rise at (say) 4% per year, how long before the heat dissipation seriously impacts climate?
| I suggest to compare human produced heat with the incident heat of the sun which is around 1 kW/m$^2$. The usual comparison is "the sun delivers more heat in an hour than humans use in a day". While such a comparison may not remain accurate forever, a difference in scale of 7000x suggests that even if humans doubled thei energy consumption every 17 years (Thalys roughly 4% per year) it would be w hole before we compete with the sun. This suggests that direct heating of Earth by human activity is insignificant.
But the extent to which human activity can change (even by a small %) the amount of sunlight captured by the atmosphere is a MUCH bigger deal. If we would cause just 1% more of the sun's power to remain on earth (by slightly modifying the surface properties of Earth or the atmospheric composition) that is a far more powerful multiplier. This is why there is such emphasis on greenhouse gas emissions.
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Chiral anomaly in Weyl semimetal In the presence of electromagnetic fields $E$ and $B$, four current is not conserved in a Weyl semimetal i.e. $\partial_{\mu} j^{\mu}\propto E\cdot B \neq 0$. There are some proofs in the literature where this is proved with the machinery of Lagrangian and Action, but I am looking for a physical and more intuitive explanation of this phenomena as to why this happens, in a Weyl semimetal where say for example we have two Weyl nodes.
| Your statement itself is not quite right. What is not conserved is the chiral current, namely the current of fermions at one of the Weyl nodes. The physics can be understood essentially in one-dimensional version of the Weyl metal: consider a 1D electron gas. There are two Fermi points, and the low-energy theory is given by two "Weyl fermions" in 1D with opposite chiralities. If we apply an electric field, obviously it drives a current. This current can be understood as fermions near one of the Fermi points moving to the other, through the bottom of the bands which are not present in the low-energy theory and connect the two "Weyl nodes". Therefore if you just look at fermions near one of the Fermi point, what you see is exactly the chiral anomaly.
The three-dimensional chiral anomaly is quite similar. First we can solve the Landau levels in the presence of a uniform magnetic field $\vec{B}=B\hat{z}$, and one sees that for each node and momentum $k_z$, one essentially has a 2D Dirac fermions with an out-of-plane magnetic field. Landau levels of Dirac fermions are well-understood, and in particular there is a $n=0$ level with zero energy at $k_z=0$. With finite $k_z$, the spectrum starts to disperse, but in a "chiral" way, with the chirality set by the Weyl node. So the situation is very much the same as the 1+1-dimensional chiral anomaly.
You can find more details in http://www.sciencedirect.com/science/article/pii/0370269383915290.
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How to calculate Electric Field near a charged conducting surface without Gauss' law? I have two problems :
*
*In every textbook I find the use of Gauss' law in calculation of Electric Field near a charged conducting surface. Can it be calculated without Gauss' law?
*Suppose while using Gauss' law to calculate field near a charged surface, the Gaussian surface (cylindrical) that we take is so long that the other side of the charged conductor lies within the cylinder, then how is the situation explained?
| Usually, applying Gauss's law to a problem with
$$ \int_A \vec{E} \cdot d\vec{A} \propto Q$$
is only suitable, if one knows, that the electric field is perpendicular to the surface $A$ and is constant in magnitude over the whole surface. This leaves:
$$ E \propto \frac{Q}{A}$$
On can conclude such statements if the problem is symmetric (e.g. spherical symmetric $\rightarrow$ choose $A$ as a surface of sphere).
However, if the problem shows no obvious symmetries, one uses the solution of Poisson's equation for a vanishing potential at infinity or something similar in Lorenz-gauge:
$$ \varphi(\vec{r},t) \propto \int \frac{\varrho(\vec{r}\ ', t_r)}{|\vec{r} - \vec{r}\ '|} \ d^3r\ '$$
with $\varrho$ charge distribution and $t_r = t - \frac{|\vec{r} - \vec{r}\ '|}{c}$
| {
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What is meant by the term "single particle state" In a lot of quantum mechanics lecture notes I've read the author introduces the notion of a so-called single-particle state when discussing non-interacting (or weakly interacting) particles, but none that I have read so far give an explicit explanation as to what is exactly meant by this term.
Is it meant that, in principle, each individual state constituting a multi-particle system can be occupied by a single particle (contrary to an entangled state, where is impossible to "separate" the particles), such that the state as a whole can be de-constructed into a set of sub-states containing only one particle each, and each being described by its own Hamiltonian?
Sorry to ramble, I'm a bit confused on the subject, in particular as I know that more than one particle can occupy a single-particle state (is the point here that the particles can still be attributed their own individual wave functions, and it just so happens that these individual wave functions describe the same state?).
| In a physics of nuclear structure, by the term single particle state is typically understood an excitation, that can be attributed mostly to one proton or one neutron that jumped to a higher orbit. Contrary to collective excitation or collective state, which is an excited level, that many nucleons participate in.
| {
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What is the maximum net force of surface tension per gram of water? I've always wondered how much force water exerted through surface tension. By maximum I mean the theoretical pulling/attracting power.
Also, what would happen if you raised or lowered the power? Would it affect life? And how much would you have to raise it in order to get a drop of water 1 ft. in height?
Thanks guys!
| As you can see here http://www.funsci.com/fun3_en/exper2/exper2_05.gif the surface tension depends on the width of the film you are pulling. At the extreme case you would make two surfaces with each water molecule of one half of the water attached to a polar group of the surface. Then contact the surfaces so that each water molecule is opposite to another water molecule of the other surface, of course always matching a negative (oxygen) side with a positive (hydrogen side). The force depends on the distance. If you push the molecules into each other you will have repulsion. If you pull them apart, the repulsion will decrease and attraction will dominate. If you pull them too far apart, the attraction will decrease. At the maximum net attraction the force with 1 g water will be immense. I can't say how much, though.
Surface tension certainly plays an important role in cellular life, so adding detergent (which lowers surface tension) to water or ingesting it would be unhealthy, but you can buy detergents without a special license, so it's not that bad, within limits. I would advise you not to drink detergent on a regular basis, though.
| {
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Are there other less famous yet accepted formalisms of Classical Mechanics? I was lately studying about the Lagrange and Hamiltonian Mechanics. This gave me a perspective of looking at classical mechanics different from that of Newton's. I would like to know if there are other accepted formalism of the same which are not quite useful compared to others (because otherwise if would have been famous and taught in colleges)?
| Gauss's principle of least constraint
Principles of Least Action and of Least Constraint (a review paper by E.Ramm)
If I remember correctly, this principle has been used to derive equations of motion
for Gaussian isokinetic thermostat (i.e., a computational algorithm for maintaining a fixed temperature of the system). Please see, for example, Statistical Mechanics of Nonequilibrium Liquids by Denis J. Evans and Gary P. Morriss, Sec.5.2.
Excerpt from the paper by E. Ramm above (in the last page):
Gauss’s Principle is not very well known although it is mentioned as a fundamental principle in many treatises, e. g. [3, 25–27], see also [28]; correspondingly it has not been applied too often. Evans and Morriss [26] discuss in detail the application of the Principle for holonomic (constraints depend only on co-ordinates) and nonholonomic constraints (non-integrable con- straints on velocity) and conclude ”The correct application of Gauss’s principle is limited to arbitrary holonomic constraints and apparently, to nonholonomic constraint functions which are homogeneous functions of the momenta”.
| {
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Is my proof of the thought experiment that Walter Lewin proposed in lecture 16 valid?
A tennis ball bounces off a wall elastically. The momentum of the wall changes, but the kinetic energy of the wall remains zero. How is that possible?
Walter Lewin Lecture 16 - Ball bouncing on wall?
That proof made sense, but I didn't trust it 100%. The math seemed wonky at best.
Here's what I did:
Conservation of momentum states that
$$
m_{b}v_{b} = m_{b}v_{b}' + m_wv_w' \\
$$
because $v_b' = -v_b$,
$$
2m_bv_b = m_wv_w' \tag1
$$
While conservation of mechanical energy states that
$$
\frac{1}{2}m_bv_b^2 = \frac{1}{2}m_bv_b'^2 + \frac{1}{2}m_wv_w'^2
$$
When $v_b' = -v_b$,
$$
0 = m_wv_w'^2 \tag2
$$
I added equations (1) and (2), solved for $m_wv_w'$, then evaluated the limit
$$ \lim \limits_{v_w' \to 0} m_wv_w' = \lim \limits_{v_w' \to 0} \frac{2m_bv_b}{v_w + 1} = 2m_bv_b
$$
Does this adequately prove the situation?
| Assuming a perfectly elastic collision, you are heading in the right direction here. It looks like you're trying to say the thing your ball hits will have 2 times the ball's momentum. You might want to consider $2m_bv_bv'_w=m_w{v'_w}^{2}$ because it looks like something might've gone "wonky" when you added equation (1) and (2)
| {
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What compounds or elements only have one phase or two phases? Wood appears to be one. I think gases like helium and hydrogen cannot exist in the solid state under normal pressures, correct? And why do those "phase cheaters"-- those elements/compounds which sublimate directly, skipping a phase, or "procrastinators"-- elements/compounds which just never reach the phase-- why do they do that?
| @HyperLuminal,@Neuneck,@Ernie,@DirkBruere-Few compounds which decompose into simpler substances before reaching their boiling point have only two phases, and the compounds which decompose into simpler substances before reaching their melting point have only one phase
| {
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How do you define the total rotational energy of an object? This problem arose when I was applying a conservation of energy argument to a mechanics problem, (a spinning coin on a table) and wasn't sure how to define the total rotational energy of the coin. At first I defined it's total rotational energy as about the axis that runs perpendicular to the table and through the center of mass of the coin, and yet it doesn't seem like this is the total rotational energy.
For example, I can take a cube and spin it about the y axis, and then spin it about the z axis as well. In essence, it seems to me as if the cube is spinning about two different axis at the same time. In this case, to define the total rotational energy wouldn't I have to take the rotational energy about the y axis, and add it to the rotational energy about the z axis? Or do I only need one axis to define the total rotational energy of an object?
| Let's do this using angular momentum as a vector. This should clear up the question on using both axes separately or one new one.
The spinning around the y-axis will give an angular momentum in the y-direction: $\vec{L_{y}} = \hat{y} L_{y}$, while the spinning around z-axis gives angular momentum in the z-direction: $\vec{L_{z}} = \hat{z} L_{z}$. We can get the net momentum by addition of these two vectors:
$\vec{L} = \vec{L_{y}} + \vec{L_{z}} = \hat{y}L_{y} + \hat{z}L_{z}$.
We will manipulate this to give us magnitude and a unit vector:
$\vec{L} = (\hat{y} L_{y} + \hat{z} L_{z} ) \frac{\sqrt{L_{y}^{2} + L_{z}^{2}}}{\sqrt{L_{y}^{2} + L_{z}^{2}}} = \sqrt{L_{y}^{2} + L_{z}^{2}} \left( \frac{\hat{y}L_{y} + \hat{z}L_{z}}{\sqrt{L_{y}^{2} + L_{z}^{2}}} \right)$.
Here we have the new angular momentum magnitude and also the new axis it is spinning around. The energy will be:
$E = \frac{|L|^{2}}{2 I} = \frac{L_{y}^{2} + L_{z}^{2}}{2 I}$,
with $I$ being the moment of intertia. Note that we actually could have just added the energies from the different axes and we would have gotten the right answer, even though the object is truly spinning around only one new axis.
| {
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Can we find the exponential radioactive decay formula from first principles? Can we find the exponential radioactive decay formula from first principles? It's always presented as an empirical result, rather than one you can get from first principles. I've looked around on the internet, but can't really find any information about how to calculate it from first principles. I've seen decay rate calculations in Tong's qft notes for toy models, but never an actual physical calculation, so I was wondering if it's possible, and if so if someone could link me to the result.
| A simple and direct way to get this exponent and complex eigenvalues is by using Gamow's approach, that was one of first introduced explanations of alpha radioactivity.
It solves Schrodinger equation in WKB approximation, no fancy math or deep knowledge in QM is needed, except being familiar with WKB.
A good source for this is "Mohsen Razavy, Quantum Theory of Tunneling-World, Scientific Publishing (2013)".
| {
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What is the meaning of " $\Psi$ is not a measurable quantity in itself"? I want to know that why the wavefunction $\Psi$ as a complex quantity (i.e $A+iB$ form) in quantum mechanics and somewhere I have studied that $\Psi$ is not a measurable quantity in itself that's why we multiply it by a it's complex conjugate $\Psi^*$ to measure $\Psi$ . What does it mean that "$\Psi$ is not a measurable quantity in it self" ?
| What does measurable mean ?
It means that one can do an experiment and get a value for $a+ib$ , the complex number.
A complex number to be measurable one should be able to measure a value at the same time for $a$ and $b$ and put a point on the complex plane. This means two independent variables, $a$ and $b$ can be measured and a point defined.
In quantum mechanics we have real distribution which are probability distribution, i.e. one independent variable. The mathematics of quantum mechanics give us though a function $\Psi$, which is a complex number. The mathematics is very successful in deriving the spectra of atoms ( for example), using this $\Psi$ formalism. An additional bonus comes from the same equations by identifying that the probability of finding the electron at $(x,y,z)$ is given by the complex conjugate square of this $\Psi$ function, as follows:
the probability of measuring a given eigenvalue $\lambda_i$ will equal to $\langle\psi|P_i|\psi\rangle$
where $P_i$ is the projection onto the eigenspace of $A$ corresponding to $\lambda_i$.
$A$ is the hermitian operator that has the eigenvelue $\lambda_i$ , (in the above example the $(x,y,z)$ point )
To get at the real measurable value $\lambda_i$ the two variables of the $\Psi$ function have been folded with one real number measurable as a probability distribution after a lot of measurements, and the other being as an undefined and unmeasurable variable, identified with the phase in the complex plane. If one takes the identity operator as $A$ one has a point in the probability density for the specific problem.
| {
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Why do nuclear bomb explosions create an array of visible electric discharges in the form of lightening in the upper atmosphere? I've seen videos on the internet, showing nuclear bomb test explosions, and there appears to be a large amount of visible lightening discharging numerous times over the development of the mushroom cloud.
Are there any mainstream explanations or predictions for what is causing this?
| The electric charge difference between the earth and the atmosphere grows with altitude, at around 88 DC volts per meter. This electric potential may be shorted out when a thermonuclear explosion releases radiation which ionizes the atmosphere. About 5% of a nuclear explosion's energy is in the form of ionizing radiation.
A study of lightning flashes caused by a thermonuclear detonation in 1952 may be of interest, though I read only the abstract, as the full article requires payment: http://onlinelibrary.wiley.com/doi/10.1029/JD092iD05p05696/abstract. The lightning started from the surface and was upward propagating. As the earth has a net negative charge, and as thermonuclear explosions deposit negative ions in the atmosphere, I don't know how to explain the upward propagation of that particular test. However, see anna v's comment.
Uman, et.al., studied the same detonation in 1972 and found that "the likely mechanism for the necessary charge and electric field generation were Compton electrons produced by gamma rays from the detonation" - Lightning: Physics & Effects, Rakov, Uman; Cambridge University Press, 2003.
| {
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Is simultaneity testable? I was reading about Andromeda paradox, and I started wondering. How can we know that the situation in the Andromeda paradox is real ? How can you know that simultaneity is real ? How can you say that you are simultaneous with a star fleet coming from Andromeda, if there is no way of knowing it ? For simultaneity to be meaningful, shouldn't it propagate at the speed of light ?
So, is simultaneity testable ?
| It's a paradox about relativity.
How can you know that simultaneity is real?
To keep things simple, we can define simultaneity of two events for a particular observer as light (or information) reaching the observer at the same time. According to this definition, simultaneity is a "real" phenomenon. A layer of complexity (that doesn't much affect the paradox) can be added by having the observer discount the time required for light or information to reach the observer to determine simultaneity.
For simultaneity to be meaningful, shouldn't it propagate at the speed of light?
As defined above, it's based on information that propagates at the speed of light. The simultaneity itself doesn't need propagate. Knowing all relevant trajectories, any observer can directly calculate the time difference between events for any other observer. Thus its possible to know that a different observer will perceive two events as simultaneous before information from that observer's actual detection of the events reaches us.
The "paradox" is that special relativity is weird and counter-intuitive. And it is.
| {
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Why can't I harness normal force? Lets say I have my palm flat with a book resting on top of it, and I have my feet on the ground. I extend my arm so that now it's kind of difficult to keep the book up. Why doesn't my hand just produce normal force on the book, cancelling out the force of gravity, and costing me no effort whatsoever?
| I haven't seen this mentioned yet: this normal force is in the context of classical mechanics (high-school or early university physics). That theory only deals with (perfectly) rigid bodies. So you can put a weight on a hypothetically perfectly rigid table and the table does not have to do any work to support it. The table doesn't compress even the tiniest bit. So in this model, work only happens when there's a displacement of the body.
Your body is not perfectly rigid. You could model it classically as rigid parts, connected by springs. In that case you'd see that, yes, you do need to input energy to the springs (muscles) in order to hold a pose when there's a weight on your hand. This more closely matches what you "feel". Holding up a weight takes effort, even if you're not moving the weight.
The core of the problem is modelling a non-rigid body as a rigid body. Same as spherical cows and point-mass planets.
| {
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Why don't protons just bounce off each other in the LHC? Ok, this might sound like a silly question, but I was wondering, when particles (e.g. protons) are smashed together in the LHC, why do they break up into dozens of other particles, as opposed to just bouncing off of each other elastically?
I'm guessing the full explanation is probably going to involve some fairly in-depth quantum analysis of particle interactions, but can anyone explain it in a fairly straightforward way that someone who isn't an expert in QM can understand?
Presumably, there will be some threshold energy level below which this doesn't happen? For example, I assume protons in a lower-energy hydrogen plasma will be bouncing off each other all the time?
| Elastic collisions do happen at the LHC. The TOTEM experiment measures the differential cross section (rate as a function of angle) for proton-proton elastic scattering at the LHC. Here is their latest result. They don't publish an estimate of the elastic cross section, but according to their data it must be at least 25 mb (millibarns) (my first version of this post had a mistake--the headline 100 mb number shown in the abstract is a measure of the total pp cross-section which includes both elastic and inelastic contributions). Compare this to the production cross-section of the Higgs boson at the same collision energy, which is about 20 pb (picobarns). This means that when two protons collide at 8 TeV, they are over a billion times more likely to bounce off each other than they are to produce a Higgs boson.
As others have pointed out, the general-purpose detectors like CMS and ATLAS are not designed to detect the elastic collisions. The elastic collisions occur mostly at forward angles, meaning the protons are just barely deflected from their original trajectory (think of a glancing collision between two billiard balls rather than a head-on collision), while the more exotic physics tends to produce particles that go more perpendicular to the beam direction.
| {
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The amount of potential energy at the height of h When we lift an object upwards with a constant velocity for a distance of $ h $ the work that we've done is $mgh$ and the work done by the force of gravity is $-mgh$. So the net work on the object is zero and it doesn't gain any energy. how its potential energy will be converted to kinetic energy when we drop that object while it hasn't gained any energy?
| Gravity doesn't do -mgh work on the object while it is lifted, gravity converts the potential energy gained while lifting the object (mgh) into kinetic energy (1/2 mv^2) after it's dropped.
| {
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What is the uncertainty principle? I looked on Wikpedia for information on the uncertainty principle, but after reading it I still had no idea. I know it has something to do with how many things you can hold at some spot for some amount of time (maybe?). This is inspired by this question.
| In classical physics you are supposed to be able to measure the coordinates and the velocity (really the momentum) of a mass with infinite precision at the same time. If you try this trick in the lab you notice that that's not the case. Either your position or your momentum measurement or both will always show some non-trivial statistical fluctuations when you repeat your experiment many times. If you multiply the standard deviations of these fluctuations with each other, no experiment that you can ever perform yields a product that is smaller than a certain number. That is it.
| {
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Where do forces point in an equilibrium system
I have the system above, with three identical balls of weight $W$ and radius $r$. The angle joining the centres is $\theta$, and the coefficient of friction between the balls and the plane is $\mu$ and between the balls $A/C$ and $B/C$ is $\mu '$. The system is in equilibrium. Denote $S$ the point of contact between balls and $P$ the point of contact between ball and plane.
My problem is that I don't know where to put the forces from which to do the calculations. Consider $A/C$: At $S$, there is a reaction force $R_S$ towards the center of both balls. But my question is, are there two forces or one more force at $S$? There are three options that I can think of:
Both a frictional force $F_{FS}$ and a force created by the weight of the ball $C$, $F_S$?
A more general force $F_S$ that includes both the mentioned above. But where would it point?
Or simply the only force that exists at $C$ is the frictional force $F_{FS}$, and the reaction force $R_S$ encompasses the force that pushes the ball $A$?
| Look at a free body diagram.
With red and the contact normal forces, with pink the friction forces, and with gray the gravity forces.
If in the end any of the friction forces come up to being negative, then flip the orientation.
| {
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What is the state of water at exactly 0°C? Theoretically speaking, what is the state of water at bang on 0°C - not any lower or higher?
Any lower would make it a solid whereas any higher would make it a liquid. But what about bang on 0°C?
Thanks in advance
Edit: I understand that other factors are involved, such as pressure and temperature which would shift the equilibrium, but in a 'theoretical perspective', what would occur - assuming that all of the particles are at exactly the same temperature with the same kinetic energy?
| H2O at 0°C is Ice. There is a considerable gap between Ice and Water. After 0°C if you increase the temperature by 0.1°C, that is at 273.1K the equlibrium state occures. This state is called the 'Triple point of water'. This is where water, ice and surprisingly water vapour. After this state if you increase the temp by any amount H20 becomes water. Hope this maybe sufficient.
| {
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Is refraction of light a thermodynamic process? Is refraction of light a thermodynamic process?
Can it be explained by conservation of energy?
If so, does temperature has an effect on refraction of light?
| Refraction of Light is not a Thermodynamic process. If you study the QED basis of refraction, you notice that the difference happens in time. The speed of light is constant, and a photon which is refracted, doesn't actually travel any slower, it just travels a longer path, and needs thus more time. If it hit's somewhere, then it's not refracted. It's gone, and this increases heat. But that was not the photon which got refracted. And this heat might cause a thermal expansion on the material, which means the refracting properties of the material are changing. Still, this change is NOT caused by the refracted photons. It's caused by the absorbed photons.
Conclution;
*
*No, It's not a thermodynamic process.
*No, But it can be explained through time.
*Yes, Temperature has an effect to refrection through thermalexpansion, but this change is not caused by refraction. It's caused by the absorption.
| {
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Kinetic energy and Potential of a photon How does the potential and kinetic energy of a photon relate? Do they mean the same thing?
Also how does De broglie wavelength and Potential relate?
| Quote from a webpage a bit over my head :-) greatians.com
Photon has linear momentum. Photon travels in vacuum space at the ultimate speed of light. Photon has the quantized energy of hf as given by eq. WD.1.2.
E = hf … eq. WD.1.1
where h = Plank’s constant
f = frequency of photon
The energy of photon can be further sub-divided into two portions. There are the kinetic and potential energy of photon. The energy equation of photon is described below,
E = hf = pv + tf … eq. WD.1.2
where p = momentum of photon
v = traveling speed of photon = the Kong vector
τ = torque, angular force between electric and magnetic component
f = twisting or deform angle of M&E components
When the M&E and Kong vectors are not perpendicular, the photon travels at the lower speed. The kinetic energy of photon is
Kp = pv … eq. WD.1.3
And the potential energy of photon is
PP = τf … eq. WD.1.4
When the M&E and Kong vectors of photon are perpendicular, the photon is traveling at the speed of light. The deform angle of M&E vectors is zero. Therefore, the total energy of photon becomes the kinetic energy, where eq. WD.1.2 becomes
E = Kp = hf = pc … eq. WD.1.5
where c = the speed of photon
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Cart speed and wheel rotation Say you have a horse drawn cart. Does the outside of the wheel spin at the same velocity that the cart moves forward?
The reason I ask is because I am working on a problem where a piece of mud detaches from the wheel when it is moving up, and then lands on the wheel again at the same spot. I don't see how this is even possible, though, if the cart moves forward at the same velocity that the wheel turns because some of the velocity in the mud will be in the y direction so it could never match the speed of the cart, much less beat it.
| The rate at which the wheel is spinning is an angular velocity, normally measured in radians/second. The velocity of the cart is a linear velocity - metres/second in SI units. Since the two are different units, they can't really be compared.
However, you can calculate the instantaneous velocity of any given part of the wheel rim. Unless the wheels are jammed, and the horse is simply dragging the cart along, this will definitely not be the same for each part of the wheel rim. As user27118 hints, a good start is to consider the instantaneous velocity of the bottom of the wheel.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/181154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Is this symmetry factor in Peskin wrong? I am trying to compute the symmetry factor of a Feynman diagram in $\phi^4$ but i do not get the result Peskin Claims. This is the diagram I am considering
$$\left(\frac{1}{4!}\right)^3\phi(x)\phi(y)\int{}d^4z\,\phi\phi\phi\phi\int{}d^4w\,\phi\phi\phi\phi\int{}d^4v\,\phi\phi\phi\phi$$
my attempt is the following: there are 4 ways to join $\phi(x)$ with $\phi(z)$. There are then 3 ways to connect $\phi(y)$ with $\phi(z)$. Then, there are 8 ways to connect $\phi(z)$ with $\phi(w)$ and 4 ways to contract the remainning $\phi(w)$ with $\phi(v)$. Finally the there are 6 ways to contract the $\phi(w)$ and the three $\phi(v)$ in pairs
$$\left(\frac{1}{4!}\right)^3\dot{}4\dot{}3\dot{}8\dot{}4\dot{}6=\frac{1}{6}$$
but the result claimed in Peskin (page 93) is $1/12$. What am I doing wrong?
|
What am I doing wrong?
The expansion of $e^x$ is:
$$
e^x=1+x+x^2/2+x^3/3!+\ldots
$$
From expanding the expression:
$$
\left<\phi_x\phi_y\exp{\left(-\frac{\lambda}{4!}\int dz \phi_z^4\right)}\right>\;,
$$
the third order term is:
$$
\left<
\phi_x\phi_y\frac{1}{3!}{\left(\frac{-\lambda}{4!}\right)}^3\int dz \int dw \int dv
\phi_z\phi_z\phi_z\phi_z
\phi_w\phi_w\phi_w\phi_w
\phi_v\phi_v\phi_v\phi_v
\right>
$$
There are four (4) ways to connect x to z and then three (3) ways to connect y to z. There are four ways (4) to connect one of the remaining zs to a w and four ways to connect the other remaining z to a v (4), this can be done for either of the two remaining zs (2), i.e., the "third" z can connect to w or the "fourth" z can connect to w. There are six (3!) ways to connect up the remaining ws and vs. And finally, there is nothing special about "z", I can treat "w" the same way as "z" or "v" the same way as "z", so that gives another factor of three (3).
So the overall symmetry factor is:
$$
4*3*4*4*2*3*2*3\frac{1}{3!}\frac{1}{4!^3}=\frac{1}{12}
$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/181270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What's the differences between time in Physics and time in everyday use? OK. This question might sound as not a good question, but the word 'time' is so confusing to me. I mean thermodynamics says time is the product of entropy. Relativity says time is relative. Quantum Mechanics says time doesn't exist, and that we can derive any given equation without involving time like Kepler's law. And I thought time measurement such as hour or second is just what we invented for conveniences on daily lives.
So are definition of time in Physics difference from time we associated in everyday life?
| In layman use, across the world (ignoring time zones), someone who is walking observes the same time as you. In physics, however, that person has his/her own time, relative to that person. In classical thermodynamics, time is the same as layperson's time. It is just described in a different way. Quantum mechanics doesn't say that time doesn't exist.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/182312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Electric field in a conductor Is it always true that the electric field in a conductor is zero?
What would happen if I put a very big charge inside an ungrounded hollow conducting sphere like this image?
The charges inside the conductor are supposed to rearrange so as to cancel the field created by the big charge.
So in that case, the electrons (I believe only electrons can move freely ..?) will move towards the inner surface (because they're attracted by the big positive charge) and it creates a field in the opposite direction that somewhat cancels the existing field. But what happens if there is still a field when all the electrons are on the inner surface ?
What if it's not enough to cancel the field created by the big charge?
| If the charge is too large, it will pull electron-positron pairs out of the vacuum outside its surface, and in effect the charged sphere neutralize itself as the positrons fall to it and the electrons (from the pair creations) will fly apart in an explosion electrons.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/182504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Do quarks violate quantization of charge? Does existence of various kinds of quarks like up, down, strange, charm, top, bottom violate the quantisation of charge or just redefine it as up quark have charge +2/3 and have -1/3. Or do things get even complex for unified theories like the proposed string theory?
| The existence of quarks is not seriously in dispute at this point AFAICT. If you want to make something meaningful out of quarks and only quarks having fraction-of-$e$ charges, I think you pretty much have to postulate that electrons are composite.
For instance, the rishon model proposes that all the "fundamental particles" of the Standard Model are composites of still-more-fundamental "T" and "V" particles and their antiparticles. T has a charge of $e/3$, V has a charge of zero.
This model is fairly old as these things go (1979) and may well have been falsified by things like the current bound on the electron dipole moment, the existence of the Higgs, the continued lack of evidence for proton decay, etc. — I'm only mentioning it as an example.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/182604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Particle Horizon and CMB If particle horizon is the maximum distance we can see presently in the universe, how come we are able to see CMB? CMB is radiation from surface of last scattering happened at t~380k years.
Suppose universe is expanding at a constant rate ( i.e. no acceleration), will we be able to see CMB again??
| If recombination were been an aisled event, i.e. something that happened in a certain place with some finite spatial extent, then it is possible that light didn't have enough time to reach us from there (this event happened outside the particle horizon).
However recombination was not an aisled event, it happened everywhere in the universe at the same time, light scattered for the last time in all directions.
From our location in earth CMB was emitted as well, but those photons are now very far away from us. The photons that we do see are those that were far enough at the moment of last scattering so that their trajectories are meeting us today. That is our surface of last scattering
If another planet lying outside our observable sphere looks far enough they will also see their own last scattering surface, they will also observe the CMB.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/182777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
For what values of $\lambda$ is the distribution $(x-i\varepsilon)^\lambda$ positive? I've been reading the famous unpublished paper by Luescher and Mack "The energy momentum tensor of critical quantum field theories in 1+1 dimensions". In the proof of their main theorem, page 7 of the manuscript, they write:
"$$
<0|O_k^\dagger(x) O_k(y)|0> = B_k (x-y-i\varepsilon)^{2k-6}, \quad B_k\in\mathbb{C},\; x
\in\mathbb{R}
$$
This distribution should be positive. As is well known [5], this implies $2k-6\le 0$. Thus $O_k = 0$ for $k>3$."
[5] is Gelfand, Shilov "Generalized functions, Vol I".
Here $k=0,1,2,\dots$
I found in this book that for $\lambda = -n$, where $n$ is a positive integer,
$$
(x-i0)^{-n}=x^{-n} + \frac{i\pi(-1)^{n-1}}{(n-1)!}\delta^{(n-1)}(x),
$$
where the distribution $x^{-n}$ for $n=2m$, i.e. $n$ even, is defined
$$
(x^{-2m},\phi)=\int_0^\infty x^{-2m}\left(\phi(x)+\phi(-x)-2\left[\phi(0)+\frac{x^2}{2!}\phi''(0)+\dots+\frac{x^{2m-2}}{(2m-2)!}\phi^{(2m-2)}(0)\right]\right)dx
$$
I tried calculating the inner product with $\phi(x)=\exp(-x^2)\ge 0$ for various $\lambda$'s. However, for $\lambda = -2$ I got
$$
((x-i0)^{-2},\phi)=(x^{-2},\phi)=\\
\int_0^\infty \frac{\phi(x)+\phi(-x)-2\phi(0)}{x^2}dx=\int_0^\infty \frac{2\exp(-x^2)-2}{x^2}dx=-2\sqrt\pi<0
$$
so that according to Lieb and Loss "Analysis" definition of a positive distribution, $(x-i0)^{-2}$ is not positive, whereas according to the paper the distribution should be positive at least for $\lambda=-6, -4, -2,0$.
Definition of positive distribution (Lieb and Loss)
A distribution $T\in \mathcal{D}'(\Omega)$ is positive if $T(\phi) \ge 0\; \forall \phi \ge 0$, where $\phi\in \mathcal{D}(\Omega)$. Here $\phi \ge 0$ means that $\phi(x) \ge 0 \;\, \forall x\in\Omega \subset \mathbb{R^n}$.
The paper can be found here, the operators $O_k$ are defined on page 6:
Luescher and Mack "The energy momentum tensor of critical quantum field theories in 1+1 dimensions"
| I am not sure if you mean $x$ one or two dimensional variable. I don't have the article at hand but I guess we probably mean $D=1$??
The argument goes as follows: you have to use Bochner's theorem, which says that $W(x-y):=W_2(x,y)$ is positive distribution if and only if the Fourier transformation yields is a positive measure
$\tilde W(p) d^Dp$.
In this case the Fourier transformation is something like
$$
\tilde W (p) \sim \frac{\theta(p_0)\theta(p^2)}{(p^2)^{\frac D2 +\lambda}}
$$
see e.g. Remark 2.2. in Ref. 1.
Then you get for $\lambda \geq 0$ this is not integrable. Also $d=0$
is a trivial special case.
Reference:
*
*N. M. Nikolov & I. T. Todorov, "Rationality of Conformally Invariant Local Correlation Functions on Compactified Minkowski Space", Commun. Math. Phys. 218 (2001) 417-436, arXiv:hep-th/0009004.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/182857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to construct generators and Lie Algebra for Lorentz group? I'm trying to figure out Lorentz group in 2+1. First of all, I'd like to think the special orthgonal group as a combination of rotation and boost in space. Then I construct it as below. First rotation part:
$$
R(\theta)= \begin{pmatrix} cos\theta& -sin\theta & 0 \\ sin\theta & cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}
$$
I assume $\theta$ is infinitesimally small and $\theta \sim \epsilon$ then $cos\theta \sim 1$ and $sin\theta \sim \theta$
$$
R(\theta)=\begin{pmatrix} 1 & 0 & 0 \\ 0& 1&0 \\ 0&0&1\end{pmatrix} -\epsilon \begin{pmatrix} 0&1&0 \\ -1&0&0\\ 0&0&0\end{pmatrix}
$$
I have just one parameter $\theta$ so I suppose I have one generator as well from rotations. And it is
$$
J=\begin{pmatrix} 0&i&0 \\ -i&0&0\\ 0&0&0\end{pmatrix}
$$
and I construct the boost part as
$$
T(\phi)=\begin{pmatrix} cosh\phi&0&sinh\phi\\0&1&0\\sinh\phi&0&cosh\phi\end{pmatrix}
$$
$$
T(\theta)= \begin{pmatrix} 1 & 0 & 0 \\ 0& 1&0 \\ 0&0&1\end{pmatrix} - \epsilon\begin{pmatrix}0&0&1\\0&0&0\\ 1&0&0\end{pmatrix}
$$
and the generator comes from translation is
$$
K=\begin{pmatrix}0&0&i\\0&0&0\\ i&0&0\end{pmatrix}
$$
Apologies for poor notation and terminology. But I think I need to have $\frac{n(n-1)}{2}$ generator right? If yes I have one more but how can I get it? And how can I construct Lie Algebra then?
NOTE:I also posted this question on Math.SE but I'm afraid here is more proper. I am not looking for a way of any other isomorphic groups to construct this LG. I have not much knowledge about homomorphism, isomorphism etc.. Thanks.
| You should have two boost generators. You have constructed one for boost in the $x$ direction, but there is also one for boost in $y$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/183190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What stops the middle point of a power line from falling? Say you have a system that is a uniformly weighted string with slack suspended from two points; i.e. a power line.
There are three forces acting on any given point on this string: string tension going left, string tension going right, and gravity.
Consider the point exactly in the middle of the string. The tension forces act tangent to the string, which (in this case) is directly left and right. So these forces have no upwards component, so no matter how large they are, they won't be able to counteract gravity.
But the string is not moving, and the middle point is not actually accelerating downwards. So what am I missing? What's counteracting gravity?
| The part at the exact middle of the string has zero mass.
That seems silly, but consider - if you consider a very small section of the string in the middle - say 1 mm - then the pieces of string on either side exert forces with tiny, but nonzero upward components. If you half the length we are considering to 0.5 mm, then the upward component of the forces is smaller, but so is the weight! Half that again to 0.25 mm, and the same happens. By the time you're actually considering about the part of the string that is in the exact middle of the wire, as you say the tension forces are perfectly horizontal, but that piece of the wire has zero mass & weight, so there's no need for any vertical force to support it.
In reality, that's a little silly, because wires are not ideal objects. But the same principle applies.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/183266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
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