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The origin of femto, atto and zepto SI prefixes Do you know why the SI prefixes: femto, atto, zepto have been accepted by Scientific Community, if this triad of metric units, is neither greek nor latin?
| From http://www.exa.com.au/metric/ a rewrite of the NIST Constants, Units & Uncertainty home page
Prefixes ranging from micro to mega were first introduced in 1874 by BAAS as part of their CGS system. Later, 12 prefixes ranging from pico to tera were defined as part of the International System of Units - SI, which was adopted in 1960. SI is maintained by BIPM under exclusive supervision of CIPM and resolutions made by CGPM. Further 8 prefixes were added to SI in years 1964 (femto, atto), 1975 (peta, exa) and 1991 (zetta, zepto, yotta, yocto).
exa alteration of hexa from Greek hex meaning "six" (the sixth power of 10^3)
peta alteration of penta from Greek pente meaning "five" (the fifth power of 10^3)
tera from Greek teras meaning "monster"
giga from Greek gigas meaning "giant"
mega from Greek megas meaning "great"
kilo from Greek khilioi meaning "thousand"
hecto French, alteration of Greek hekaton meaning "hundred"
deca from Greek deka meaning "ten"
deci from Latin decimus meaning tenth, from decem meaning "ten"
centi from Latin centi-, from centum meaning "hundred"
milli from Latin mille meaning "thousand"
micro from Greek mikros meaning "small"
nano from Greek nannos meaning "dwarf"
pico from Spanish pico meaning "small quantity"
femto from Danish, or Norwegian word femten meaning "fifteen"
atto from Danish, or Norwegian word atten meaning "eighteen"
The names zepto and zetta are derived from septo, from Latin septem which means seven (the seventh power of 10^3) and the letter 'z' is substituted for the letter 's' to avoid the duplicate use of the letter 's' as a symbol in SI. The names yocto and yotta are derived from Latin octo which means eight (the eighth power of 10^3); the letter 'y' is added to avoid the use of the letter 'o' as a symbol because it may be confused with the number zero. The CGPM has decided to name the prefixes, starting with the seventh power of 10^3, with the letters of the Latin alphabet, but starting from the end. Therefore the choice of letters 'z' and 'y'. The initial letter 'h' of the word hexa in standard French is silent, so it was removed in order to simplify things.
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The physics behind The Great Flood The book of Genesis floats (pardon the pun) some interesting numbers when discussing the Great Flood. For example, it rained for 40 days and 40 nights, and at the end of that time, the entire planet was covered in water.
I think we can deduce how much water that would have had to be, estimating that the highest peaks in the Himalayas were covered with water. (8,848 meters above sea level)
My questions are, how fast would the rain have had to come to raise the ocean level that high in 40 days and nights, how much would the mass of the earth have changed for this event, and would that significantly alter the strength of gravity on earth?
| The closest to a physical discussion of the Great Flood may be
http://www.talkorigins.org/faqs/faq-noahs-ark.html#flood
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Einstein's equation: Black hole solution Let Einstein's equations satisfy $ R_{\mu \nu } = 0 $. Suppose we solve it numerically with the aid of a computer. Can we know from the numerical solution if there is a black hole in the solutions? For example, how can you know when you solve Einstein's equation if your solution will be a black hole or other particular non-smooth solution?
| The way to do this is to look for a closed trapped surface in the solution. This is a spherical surface such that all the null geodesics, both going out and going in, have area that is locally going down per unit affine parameter. When you find such a surface, you know that it is inside a black hole, and you can stop simulating the interior of this region, since no influence from the interior will reach infinity.
The union of all closed trapped surfaces at any one time is called the apparent horizon, and finding the apparent horizon is done in simulations to find regions which can be excluded, since they are black hole interiors. This is useful, because the curvature will blow up inside the black hole for sure, and you don't want to have to simulate that, and you don't need to.
| {
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How does the curiosity rover get it's power?
Possible Duplicate:
Mars Curiosity Power System
I found a web page that said it uses Plutonium. I am sure it's not based on fusion or fission. What is the basic idea on which the power plant works?
| Curiosity uses a radioisotope thermoelectric generator.
A radioisotope thermoelectric generator (RTG, RITEG) is an electrical
generator that obtains its power from radioactive decay. In such a
device, the heat released by the decay of a suitable radioactive
material is converted into electricity by the Seebeck effect using an
array of thermocouples.
| {
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Why does inverting a song have no influence? I inverted the waveform of a given song and was wondering what will happen.
The result is that it sounds the exact same way as before.
I used Audacity and doublechecked if the wave-form really is inverted.
The second thing I tried was:
I removed the right channel, duplicated the left one and set the duplicated layer as right channel. This way I made sure that both channels are exactly the same. Then I inverted the second channel only. I thought that this would create some kind of anti-noise, but it didn't.
Why is that?
| Inverting a waveform is the same as rotating your speaker around 180 degrees to face away from you. (Yaw or pitch - not roll!)
The changes in air pressure your ear detects is exactly the same.
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Making or Demonstrating Principle of Electron Microscope is it possible to either demonstrate the principle or make a SEM ( electron microscope ) at home or lab as an enthusiast??
and how can i start?
| "Yes, you can!" © :-)
Check out this guy on YouTube, he has described everything in detail:
http://www.youtube.com/watch?v=VdjYVF4a6iU (and his related videos)
Basically he used an electron gun from a small TV-tube, using phosphor and photomultiplier to detect electrons. Image is shown on an analog oscilloscope.
Although it is doable at home, this is not exactly a weekend project.
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Is it possible to describe the entire universe with the behavior of an $\mathbb{R}^n$ field? Suppose every phenomena in this universe (of course most are reducible to some particular general ideal ones - basically I'm talking about those!) could be described as disturbances/waves/ripples/tensions in an $\mathbb{R}^n$ field. Is this possible? Can we find $n$?
Basically, I came up with this when I studied that General theory of relativity explains gravitation as a disturbance/tension in a 4-dimensional field of space-time. Basically, I'm asking whether the entire universe can be described as such.
| Try reading up on string theory.Here is a series of lectures with "string theory for pedestrians".
A particular quantum vibration mode of the closed
string describes a graviton, the quantum of the
gravitational field. A particular quantum vibration of
an open string describes a photon, the quantum of the
electromagnetic field.
In string theory all particles – matter particles and
force carriers – arise as quantum fluctuations of the
relativistic string.
One might well say that everything in the universe is described as the vibrations of some string, the entire universe. So yes, people have been working on this.
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Charge and the Dirac field In Zee's quantum field theory in a nutshell, 2nd edition, pg 550 he has
$Q=\int {d^3p \over (2\pi)^3(E_p/m)} \sum_s \{b^\dagger(p,s)b(p,s)-d^\dagger(p,s)d(p,s)\}$
showing clearly that $b$ annihilates a negative charge and $d$ a positive charge.
I would very much appreciate an explanation of why its not the other way round. As the $d^\dagger(p,s)d(p,s)$ term has the negative sign why doesn't that mean it is associated with the negative charge particle?
| This is a matter of convention.
You are totally right: the $Q$ operator you have written implies that $b$ annihilates a positive charge and vice versa.
The thing is that in QED one usually defines $Q$ in a slightly different way, namely:
$$Q=-\left| e\right| \int {d^3p \over (2\pi)^3(E_p/m)} \sum_s \{b^\dagger(p,s)b(p,s)-d^\dagger(p,s)d(p,s)\}$$
with $-\left| e\right|$ the electron's charge. Then, with this definition, $b$ annihilates states with negative charge $-\left| e\right|$.
The confusion comes from what it is arguably the worst convention in the history of physics (and possibly chemistry and engineering): the electron's charge was chosen negative! Since there are many more electrons than positrons in the universe and in the Earth (and for this reason the appropriate convention that the electron is the particle and the positron its antiparticle, and not in the other way around), the electron's charges should have been chosen as positive! —so the positron should have been named negatron or perhaps the Dirac or why not simply anti-electron and having negative charge ;-)
Anyway, it is just a convention.
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Sinusoidal Wave Displacement Function I am learning about waves (intro course) and as I was studying Wave Functions, I got a little confused.
The book claims that the wave function of a sinusoidal wave moving in the $+x$ direction is $y(x,t) = A\cos(kx - wt)$.
However, I see a drawing of the wave and they always seem to be $\cos$ graphs. Are sinusoidal waves always cosine graphs? Or can they be sine? If I ever see a sine wave, then does that mean that this is merely a pulse/wave travelling once and not oscillating in a periodic motion?
Sorry for the basic beginners question.
| sinusoida functions is a function containing sin like y= Asin(kx+wt) and cosinusoidal are the ones like y=Acos(kx+wt)
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How does the Sun's magnetic field continue to exist at such high temperatures? The temperature at the surface of the Sun is apparently well above 5000 C; I'm assuming the layers beneath the surface may be even hotter.
At school, we learned that heating a metal beyond a certain temperature, specific to each metal, would demagnetize the magnet.
How does the Sun's magnetic field continue to exist at such high temperatures?
| The current answers point into the wrong direction. The right direction, in my opinion, is indicated in the comment of dmckee.
The problem is with your intuition that high temperature destroys a magnetic field. This is wrong as we can see in numerous experiments.
It is correct, however, that there is a temperature at which the particular mechanism by which a magnetized piece of iron keeps up its magnetic field breaks done. The way iron produces a magnetic field is connected with certain properties of the iron atoms - and it is this property which is lost.
High temperature destroys the mechanism of iron (and some other metals) to keep a magnetic field, not the mechanism by which a moving charge produces a magnetic field.
In the sun the magnetic field is generated by moving charges. In iron the magnetic field is generated by a particular alignment of spin structures in the crystalline structure of the metal. Heat leads to movement of the atoms and destroys this crystalline structure.
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Gravity on supermassive black hole's event-horizon
*
*$M =$ black hole mass
*Gravitation is about $r^{-2}$
*Schwarzschild radius, $r_{\text{S}}$, is $\propto M$
*So, more massive black holes have weaker gravitation at their event horizon.
Consider a black hole so enormous that the gravitation on its event horizon is negligible.
Person A is 1 meter 'outside' the horizon, and Person B is inside (1 meter from the horizon as well). Person B throws a ball to Person A. Both just started accelerating towards the black hole very slowly, so why person A wont catch the ball? Why won't Person A even ever see person B granted A will somehow escape later on?
Reference: https://mathpages.com/rr/s7-03/7-03.htm
| The horizon, for any static black hole, is the surface where the escape velocity is $c$. Thus, your notion that gravitation is weaker at the horizon for larger black holes is incorrect.
EDIT: Consider that the thrust required to hover goes to infinity at the horizon regardless, i.e., the proper local acceleration for a stationary observer goes to infinity at the horizon.
EDIT 2: In your thought experiment, you write that person B is $1m$ inside the horizon. But, it's crucially important to understand that, within the horizon, the radial coordinate is timelike with the future towards the singularity and the past towards the horizon. Person B can no more throw a ball towards person A outside the horizon than person A can throw a ball into the past.
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Why does the running of the gauge couplings show $\frac{1}{\alpha}$ > $\frac{1}{\alpha_w}$ at low energy? I thought the coupling constants were something like:
$\alpha \approx 1/137$
$\alpha_w \approx 10^{-6}$
$\alpha_s \approx 1$
And yet if you look at any pictorial representation of the running of the couplings you see something like this (or try google):
which seems to indicate that (at the energies scales we have access to) $\alpha$ is less than $\alpha_w$. Why is the ordering of the size of the couplings messed up in all of these pictures?
| I think you're confusing the weak structure constant with the Fermi constant. The Fermi constant is $G_F=1.166\times 10^{-5}\text{ GeV}^{-2}$ and it gives us the effective strength of the four-point interaction of fermions. This four-point interaction is of course mediated by the W boson and by looking at the relevant tree-level Feynman diagrams we have
$$\frac{G_F}{\sqrt{2}}=\frac{g_W^2}{8M_W^2},$$
where $g_W$ is the weak coupling and $M_W=80.4\text{ GeV}$ the mass of the W boson. Plugging in numbers, we find
$$\alpha_W=\frac{g_W^2}{4\pi}\approx\frac{1}{30}.$$
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How to prove that orthochronous Lorentz transformations $O^+(1,3)$ form a group? Orthochronous Lorentz transform are Lorentz transforms that satisfy the conditions (sign convention of Minkowskian metric $+---$)
$$ \Lambda^0{}_0 \geq +1.$$
How to prove they form a subgroup of Lorentz group? All books I read only give this result, but no derivation.
Why is this condition $ \Lambda^0{}_0 \geq +1$ enough for a Lorentz transform to be orthochronous?
The temporal component of a transformed vector is
$$x'^0=\Lambda^0{}_0 x^0+\Lambda^0{}_1 x^1+\Lambda^0{}_2 x^2+\Lambda^0{}_3 x^3,$$
the positivity of $\Lambda^0{}_0$ alone does not seem at first glance sufficient for the preservation of the sign of temporal component.
And how to prove that all Lorentz transform satisfying such simple conditions can be generated from $J_i,\ K_i$?
For those who think that closure and invertibility are obvious, keep in mind that
$$\left(\bar{\Lambda}\Lambda \right)^0{}_0\neq \bar{\Lambda}^0{}_0\Lambda^0{}_0,$$
but instead
$$\left(\bar{\Lambda}\Lambda \right)^0{}_0= \bar{\Lambda}^0{}_0\Lambda^0{}_0+\bar{\Lambda}^0{}_1\Lambda^1{}_0+\bar{\Lambda}^0{}_2\Lambda^2{}_0+\bar{\Lambda}^0{}_3\Lambda^3{}_0.$$
And I'm looking for a rigorous proof, not physical "intuition".
| Misha's answer is correct and complete.
However, let me give you the physical argument that explains why you do not find the proof in any book. The proper orthochronous transformations are spatial rotations and pure Lorentz transformations (or boosts). And it is clear from a physical point of view that these transformations verify the group laws: closure, existence of inverse (opposite angle or velocity) and identity.
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Could this fountain, under the right conditions, technically be able to lift me up? There is a big fountain in a lake in my city.
I was talking with a friend and we were wondering whether it would be able to lift me up. I sent a few emails and obtained information about the fountain. The raw email reads in French:
Nous avons 2 pompes de 150 HP qui pousse l'eau a 195 pieds de haut et la pression par pompe à la sortie est d'environ 380 pieds de tête.
I am not knowledgable in physics in any way, but this is how I can translate it to the best of my knowledge:
We have two pumps of 150 horsepower that push water at 195 feet high and the pressure by pump at the exit is of 165 PSI
Now, could this possibly be able to lift me, a 160 pounds man? How can I find out? And under what conditions would this work?
| You should consider yourself a 160 lb bag of water. The criterion for lifting you up is that the up-momentum per second in the water is roughly larger than the up-momentum you need. From the height of 200 ft, 60 m, you know the velocity is more or less $\sqrt{2gh}=35 m/s$, and to lift you up at 100kg requires transferring $mg=1000$ units of up-momentum per second to you, so this has to be available in the water, so you need at least about 30 kg/s of water output at the pump, somewhat more because this water will be going up with you. This is a good rough estimate, so you need to know the outflow diameter, or the amount of water per second emitted to know for sure, and even then, I wouldn't recommend it, since you are going to get hurt no matter what.
| {
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Loopy lightning What causes lightning to follow the path it does ?
picture from BBC news: http://news.bbcimg.co.uk/media/images/62891000/jpg/_62891901_untitled-1copy.jpg main page: http://www.bbc.co.uk/news/in-pictures-19597250
| I would put it down to coincidence: a cloud to cloud bolt concurrent with a cloud to earth bolt.
Have a look at this, where branching is also seen. Do not forget that in cloud to ground, a bolt starts from the ground.
Or this one, which shows following two branches in cloud to cloud:
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What are distinguishable and indistinguishable particles in statistical mechanics? What are distinguishable and indistinguishable particles in statistical mechanics? While learning different distributions in statistical mechanics I came across this doubt; Maxwell-Boltzmann distribution is used for solving distinguishable particle and Fermi-Dirac, Bose-Einstein for indistinguishable particles. What is the significance of these two terms in these distributions?
| Since there is no way in which the molecules can be labeled, the particles are indistinguishable.
On the other hand, if the assembly is a crystal, the molecules can be labeled in
accord with the positions they occupy in the crystal lattice and can be considered
distinguishable.
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Applications of the particle in a box and the finite square well What are some "real" world applications of the particle in a box (PIB) and the finite square well (FSW) which are discussed in an intro quantum mechanics class? For instance, I know that the PIB can applied to quantum dots and the FSW to the Ramsauer-Townsend effect. How about other applications?
| A very important real-world application of quantum mechanics is the laser, in a multitude of ways. For start, the whole operating principle is of quantum mechanic origin, and there is a host of other quantum mechanical phenomena involved.
Specifically, in this case a good example is the quantum-well semiconductor laser, in which the different materials are sandwiched during manufacturing in such a way that they serve both as an optical waveguide and providing confinement for charge carriers. This electron confinement, which is a potential well, results in discrete energy levels. The discretisation permits the quantum well laser to emit a lot narrower spectrum of light than what the energy gap of the host material allows.
Another interesting application of quantum mechanics in real life is the quantum cascade laser (QCL). QCL is composed of a sandwich structure, where there are periodic layers of varying composition. These layers result in a series of quantum wells, through which the carriers can tunnel. This tunnelling effect enables tailoring the output wavelength of the QCL, almost independent of the host material.
For more information, see text books on lasers (my personal favorite is Orazio Svelto: Principles of Lasers, but there are of course others) or Wikipedia (and references therein).
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Spooky action appears to contradict Relativity of time order of multiple events It is well known that in special relativity observers can disagree on the time ordering of two events.
It is also well known that entangled particles exhibit so called spooky action at a distance.
Today I read in the New Scientist and on the arXiv that although the order of two events can be arbitrary, this is not so for multiple events. As the number of events increase the number of combinations of possible orders increase but not every one of these combinations will actually be possible to always observe.
According to the NS article it is possible to entangle three particles a,b and c and then have them in such a way that the collaspe of c cannot be observed to precede the collapse of both a and b by any observer even though c spookily causes the collapse of a and b.
Is this an outright contradiction between relativity and quantum mechanics or just a paradox ? What is the explantion ?
| There is nothing in QM that allows c to affect a and b instantaneously. An observer who can measure c and a will simply observe that they are correlated, but that can't happen until signals have arrived from both c and a. So no contradiction can be observed. C doesn't cause the collapse of a. Their entangled states just tells you what values of a can be observed. What is spooky is what happens to the other states of a, because inconsistent histories seem to just disappear when information from c is combined with the information from a.
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Mechanics Energy (Calculus) A particle moves with force
$$F(x) = -kx +\frac{kx^3}{A^2}$$
Where k and A are positive constants.
if $KE_o$ at x = 0 is $T_0$ what is the total energy of the system?
$$ \Delta\ KE(x) + \Delta\ U(x) = 0$$
$$F(x) = -\frac{dU}{dx} = m\frac{dv}{dt} = m v\frac{dv}{dx}$$
Integrating to get U(x) and 1/2mv^2 I get
$$\Delta\ U(x) = \frac{kx^2}{2} - \frac{kx^4}{4A^2}$$
$$\Delta\ KE(x) = -\frac{kx^2}{2} + \frac{kx^4}{4A^2}$$
Which Makes sense. But how do I find the function KE(x) where KE(0) = $T_0$? Do I Even need to? The total energy in the system is $T_0$ Correct?
Also a kind of side note. What is really confusing me, is when should I add limits of integration and under what circumstances should I just use an indefinite Integral?
| For this particular problem, it is useful to note that the force function has three zeros:
$$F(0) = F(-A) = F(A) = 0$$
This means the potential has three stationary points. Looking at the potential function, we see that $U(0)$ is a local minimum while $U(-A) = U(A)$ are global maximums.
This suggests that a natural choice is to set $U(0) = 0$.
$$U(x) = \frac{kx^2}{2} - \frac{kx^4}{4A^2}$$
Thus the total energy is equal to the kinetic energy when $x=0$:
$$E = T_0$$
For $T_0 < kA^2/4$, the particle is bound by the potential and just oscillates back and forth symmetrically about $x=0$, never reaching $|x| = A$.
For $T_0 > kA^2/4$, the particle is unbound and is driven to plus or minus infinity by the potential (the particle rolls down the potential hill on either side).
For $T_0 = kA^2/4$, the particle theoretically would stop at either $x=-A$ or $x=A$, depending on the initial conditions, and remain there. However, both points are unstable, i.e., an infinitesimal perturbation would send the particle back towards $x=0$ or towards plus or minus infinity.
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What happens when we cut objects? What is the role of the molecular bonds in the process of cutting something? What is the role of the Pauli exclusion principle, responsible for the "hardness" of matter?
Moreover, is all the energy produced by the break of bonds transformed into heat?
| You can cut diamonds and you can cut cake. The mechanisms responsible for making a cut are as different as there are different kinds of solids. Cutting is a process that separates a piece of material into two pieces along a plane. Although some of the total work expended to make the cut actually does go into breaking chemical bonds (covalent, metallic, ionic, van-waals etc.) across the plane. Much energy can also be expended and move away from the cut-plane as heat, elastic strain energy, sound, light, or chemical reactions. That is why solids of similar bond-energies can have very different strengths.
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Is Heisenberg's matrix mechanics or Schrödinger's wave mechanics more preferred? Which quantum mechanics formulation is popular: Schrödinger's wave mechanics or Heisenberg's matrix mechanics? I find this extremely confusing: Some post-quantum mechanics textbooks seem to prefer wave mechanics version, while quantum mechanics textbooks themselves seem to prefer matrix mechanics more (as most formulations are given in matrix mechanics formulation.)
So, which one is more preferred?
Add: also, how is generalized matrix mechanics different from matrix mechanics?
| It is partly a specialty thing. Non-relativistic QM mixes the two approaches quite often (as do relativistic QM), but quantum field theory leans very heavily towards Heisenberg, as well as the path integral formalism.
There are more approaches, but they tend to be scarcely used. The algebraic approach is getting some use, especially in QFT on curved spacetimes.
| {
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"timestamp": "2023-03-29T00:00:00",
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Equations for an object moving linearly but with air resistance taken into account? I know (from Kinematics) that for an object moving linearly with an acceleration and without air resistance the following equations can be used to determine v(velocity) or x(position of the object) at any time:
$v=v_0+at$
$x=x_0+v_0t+\frac{1}{2}at^2$
Where $x_0$ is the position of the object at the start of accelerated movement, $v_0$ is the velocity at the start of the motion and a is the acceleration.
Now if I want to add air drag, say we take the formula for air drag as:
$D=0.5CA\rho v^2$
$C$ is a coefficent, $A$ is the reference Area, $\rho$ is the density and $v$ the velocity.
Since $D$ is a force the air drag results in an acceleration, so in other words: The resulting acceleration is $a-D/m$, however $D$ depends on $v$, which depends on the acceleration. And that's where my problem is, I simply can't wrap my mind around that.
How would I go about getting the Equations for $x$ and $v$ including air drag (and don't forget there's acceleration too)?
| Let $F$ be the independent force acting on the object. Let $D$ be the velocity dependent force acting in the opposite direction of $F$. The net force accelerating the object is just the difference. We have:
$F - D = ma$
Since $D$ is velocity dependent, the equation is a differential equation for the velocity.
$\dot v + \dfrac{CA\rho}{2m}v^2 = \dfrac{F}{m}$
This can be solved for $v(t)$ and which can then be integrated to find $x(t)$.
To solve this equation for constant independent force $F$, first note that there is a terminal velocity which can be found by setting the acceleration to zero, $\dot v = 0$:
$v^2_{term} = \dfrac{2F}{CA\rho}$
We can now rewrite the differential equation:
$\dfrac{1}{1-v^2/v^2_{term}}dv = \dfrac{F}{m}dt$
We can now straightforwardly integrate both sides to get:
$\tanh^{-1}(v/v_{term}) = \dfrac{F}{mv_{term}}t + C$
For zero initial velocity, we can finally write an expression for $v(t)$:
$v(t) = v_{term} \tanh(\frac{F}{mv_{term}}t)$
Thus, the velocity increases rapidly at first and then much more slowly, asymptotically approaching the terminal velocity which we can see from a plot of $\tanh$:
| {
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Number of unique 2-electron integrals Consider 2-electron integrals over real basis functions of the form $$(\mu\nu|\lambda\sigma) = \int d\vec{r}_{1}d\vec{r}_{2} \phi_{\mu}(\vec{r}_{1}) \phi_{\nu}(\vec{r}_{1}) r_{12}^{-1} \phi_{\lambda}(\vec{r}_{2}) \phi_{\sigma}(\vec{r}_{2})$$
I am told that for a basis set of size K=100, there are 12,753,775 unique 2-electron integrals of this form.
Symmetry considerations mean that we have less than $K^{4}$ unique integrals, since we can exchange electrons and also exchange the basis functions for each electron without changing the value of the integral.
How could one work out the number of unique integrals?
My method is:
Find the number of unique integrals of the forms $(\mu\nu|\lambda\sigma)$, $(\mu\mu|\lambda\sigma)$, $(\mu\mu|\nu\nu)$ and $(\mu\mu|\mu\mu)$ (where in these integrals, each index is unique unless repeated) and sum these together.
My working gives the wrong answer, though:
$$\frac{4!}{8}{100 \choose 4}+\frac{3!}{4}{100 \choose 3}+\frac{2!}{2}{100 \choose 2}+1!{100 \choose 1} = 12,011,275$$
My rationale is this: for the integral form $(\mu\nu|\lambda\sigma)$, there are ${100 \choose 4}$ unique unordered combinations of basis functions. There are $4!$ ways of arranging these unique basis functions. We can exchange the electrons, basis functions on electron 1 and basis functions on electron 2 without changing the value of the integral, thus halving the number of unique integrals 3 times ($\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}$). Therefore the number of unique integrals of form $(\mu\nu|\lambda\sigma)$ is $\frac{4!}{8}{100 \choose 4}$.
Where am I going wrong?
| The right formula is very similar to yours,
$$ \frac{4!}{8}{100 \choose 4}+3!{100 \choose 3}+2\times 2!{100 \choose 2}+1!{100 \choose 1} = 12,753,775 $$
I think that by comparing the coefficients in front of the (correct) binomial numbers, you may determine how you need to fix your calculation.
| {
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Why is there a factor of 1/2 in the interaction energy of an induced dipole with the field that induces it? In this paper, there's the following sentence:
...and the factor 1/2 takes into account that the dipole moment is an induced, not a permanent one.
Without any further explanation. I looked through Griffiths' electrodynamics to see if this was a standard sort of thing, but couldn't find anything. I was thinking it might be because the field of the dipole itself opposes the inducing field, but that doesn't quite seem right for some reason.
| The force on a dipole placed in an electrical field is given by $\mathbf{F} = (\mathbf{p}\cdot \nabla)\mathbf{E}$ (see, e.g., Griffiths, 3rd edition, eq. 4.5). Recall that,
$$
\nabla(\mathbf{p}\cdot\mathbf{E}) = \mathbf{p}\times (\nabla\times \mathbf{E}) + \mathbf{E}\times(\nabla\times \mathbf{p})+(\mathbf{p}\cdot\nabla)\mathbf{E} + (\mathbf{E}\cdot\nabla)\mathbf{p}
$$
Assume $\nabla\times \mathbf{E} = 0$ (I'll justify this at the end, trust me for now). If the dipole moment is a permanent one, $\mathbf{p} = \mathrm{const}.$, the second and fourth terms above are zero, and the expression for the force can be rewritten,
$$
\mathbf{F}=\nabla(\mathbf{p}\cdot\mathbf{E})\quad\Rightarrow\quad U=-\int_{r_a}^{r_b} \mathbf{F}\cdot d\mathbf{r} = -\mathbf{p}\cdot \mathbf{E} |_{r_a}^{r_b}
$$
However, if $\mathbf{p}$ is not a constant, but rather $\mathbf{p} = \alpha\mathbf{E}$, where $\alpha$ is the polarizability, the fourth term is not zero, and
$$\begin{split}
\nabla(\mathbf{p}\cdot\mathbf{E}) &= 2\alpha\mathbf{E}\times (\nabla\times \mathbf{E}) + 2\alpha(\mathbf{E}\cdot\nabla)\mathbf{E}\\
&= 0+2(\mathbf{p}\cdot\nabla)\mathbf{E}
\end{split}
$$
Therefore,
$$
U=-\int_{r_a}^{r_b} \mathbf{F}\cdot d\mathbf{r} = -\int_{r_a}^{r_b}\frac{1}{2}\nabla(\mathbf{p}\cdot\mathbf{E})=-\frac{1}{2} \mathbf{p}\cdot\mathbf{E} |_{r_a}^{r_b}
$$
The only outstanding issue is justifying the assumption $\nabla\times\mathbf{E}=0$. From the relevant Maxwell equation, $\nabla\times\mathbf{E}= - \frac{\partial \mathbf{B}}{\partial t}$. If your fields are static, $\frac{\partial \mathbf{B}}{\partial t} = 0$, and we're done.
In an optical trap, the application discussed in the paper above, the field is not static and we have to be a bit more careful. An optical trap is arranged by counterpropagating two identical laser beams. Assuming beam fronts are approximately planar,
$$
\mathbf{E} = \mathbf{E_1} + \mathbf{E_2}\\
\mathbf{B} = \mathbf{B_1} + \mathbf{B_2} = (\frac{1}{c}\mathbf{\hat{k}}\times\mathbf{E_1}) + (\frac{1}{c}(\mathbf{-\hat{k}})\times\mathbf{E_2})
$$
If the beams are arranged so that they're in phase ($\mathbf{E_1} = \mathbf{E_2}$), we have $\mathbf{B} = 0$ at all times and so $\nabla\times\mathbf{E}=0$.
That's the math, then, but what's the intuition? To first order, there are two contributions to any change in the quantity $-\mathbf{p}\cdot \mathbf{E}$: the change in $\mathbf{p}$ at constant $\mathbf{E}$ and the change in $\mathbf{E}$ at constant $\mathbf{p}$. But there is actually no force opposing the first of these changes: strictly speaking, the energy of the dipole should just be the integral of the second of them. For a permanent dipole, the first change is zero, so we get away with writing the energy as $-\mathbf{p}\cdot \mathbf{E}$. But for an induced dipole, this is no longer the case. Linear polarizability gave us a factor of $1/2$, but more general relations between $\mathbf{p}$ and $\mathbf{E}$ may give you more complicated answers.
| {
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How do electrons repel? I understand the basics, protons have a positive charge, neutrons have no charge, and the electron has a negative charge. But looking at the lines of force from a proton, they flow outward and push each other away. But, the electrons flow inward or towards themselves. How does that make them repel? Wouldn't they be more neutral towards themselves? I understand that the lines of force cannot cross as well. I'm sure this is an easy answer for someone.
| The force experienced by a charge is $F=E\cdot q$. Let electron number 1 be in some point in space. Its field lines are directed towards it. Now put another electron near it. Since the second electron's charge is negative, the above product implies that the force the electron feels is in the opposite direction than that of the field line - namely, away from the first electron. So the electrons do repel each other.
Field lines do not 'cross' each other because if you have two fields generated by two charges, you can just add the fields together to get the total electric field in space.
Field lines can, on the other hand, converge/diverge - and any point where they do so is a point where an electric charge exists. Just like you said - they diverge away from a proton, which is a positive charge, and converge into an electron, which is a negative charge.
| {
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What is an extreme deep field image (XDF) and how is it captured? NASA recently took this Extreme Deep Field image (XDF) that is the area of only small fraction of the dia of moon and contains 5,500 galaxies. Nasa says this was capture by extreme long exposures so it can capture the smallest of galaxies. My question by increasing exposure times how can we know this is the deep space image because a galaxy might be right next to us but really really small and it would only show in these deep space image. Is there any other technology uses to determine this picture really captures galaxies that are 13.2 billion years old. How do we know this is really the deep space? Also can someone through light on some technicalities like the earth is moving and since we are focusing on really really small are, that rotation will cause blurs. If we take extreme long exposure it will make the easily visible stars extremely bright so it will again distort the picture. So my question is what id extreme deep field image and how is it captured over time?
| The adjective "deep" primarily means "a high resolution". The telescope just focused on a particular small region of the sky and took the sharpest picture it could.
There are no stars (from the Milky Way) in the XDF (or almost no stars, I am not sure) so the objects are bound to be far. In HDF, the "ordinary" Hubble Deep Field, one may find galaxies whose age is about 70% of the age of those in the XDF.
In all cases, the actual age of the objects is determined from the red shift – the multiplicative decrease by the well-known spectral lines. In HDF, they could find galaxies with red shift up to 6. I don't know the exact value for the XDF but it's larger than 6, perhaps up to 10 or so.
Using the Big Bang equations, the red shift may be translated to the distance and the age. (I will assume that the OP doesn't want to discuss conspiracy theories claiming that the huge red shifts mentioned above are due to something else than the cosmological expansion of the Universe.)
| {
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Can Gases conduct Electricity? Liquid electrolytes ionize and hence a current can pass through them. So if a gas can ionize, can it conduct electricity too? If so, what are a few such gases?
| Gases do conduct electricity, as all materials do. However, they conduct electricity so poorly that we consider them insulators.
"Electricity" requires the movement of electrons. In a gas, these electrons are too dispersed to provided any measurable current.
The "lightning" example is slightly different. This refers to capacitative discharge. When the two sides of a capacitor (i.e. the ground and the clouds) store too much charge, that charge eventually jumps the dielectric (i.e. the stuff in between the ground and clouds). We still don't say that the dielectric "conducts" electricity, although it obviously does. The best insulators in the world could not stop a discharge of sufficient strength. The defining quality of a conductor is that it conducts electricity "more easily" than most substances. There is no perfect conductor or perfect insulator.
In short, gases can conduct electricity, but they are considered insulators for the most part.
| {
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Does the spectrum of Sol's emission change as it ages? A follow-up to my earlier question How would one navigate interstellar space? that just occurred to me; albeit on a different tack.
Sol is probably in a state of continuous flux. The change of state is probably over large timescales as compared to the life-span of the average human. If one were to compare a spectrograph of Sol over a span of, say, 10 million years, would the graph be significantly & predictably different?
| Each & every element has its own characteristic emission and absorption spectra (which we all know). As the star traverses its main sequence stage - the hydrogen atoms are actually getting used up to form helium atoms. As it enters the red giant phase, helium starts to fuse by the triple-alpha process.
The Hertzsprung-Russel diagram gives the relationship between temperature and luminosity (Blue, yellow, red, etc.). As we take Wien's displacement law also into account, we could conclude that the spectra (emission or absorption) of the sol varies accordingly with its temperature. Now, the sun is at some 5000 K surface temperature, which gradually decreases to something around 3000 K. And hence, the spectral shift from yellow towards red. But, that may take about 3 or 4 billion years.
Oh.. You asked for 10 million years. In that case too, the answer remains the same. It's on its own way, hanging its spectrum because its getting old..!
| {
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How to properly bake a ultra high vacuum chamber? I need to get rid of water excess in my vacuum chamber, and for that there is the procedure of baking. In order to do that there are several things that one needs to consider, the power, heat load, type of heat tape to use etc. Since I've never done this before, I hope that some of you could share their advice on that subject.
| The usual procedure involves
*
*Careful preparation for ultra high vacuum
*
*Do you reach a good high vacuum?
*Can all the equipment withstand high temperatures over a long period of time? If not you might to cool those parts while heating the rest of the chamber.
*Try to remove all materials that might have a high outgassing rate, otherwise ultra high vacuum is never achievable.
*Check for residual gas volumes, e.g. from screw threads.
*Prepare and bake
*
*Wrap heat tape around metallic parts of vessel, try to get good contact and not much overlap. The tape should not reach very high temperatures when used alone.
*Pump, heat and monitor gas pressure. It should rise a bit, exact numbers depend on the circumstances. Do a smell test, if it smells like burnt plastic you forgot something. Heat for at least 24 hours, even at elevated temperatures this might take a long time. The longest time a colleague used was 2 weeks.
If heating is not going to work alternatively use a sorption pump with either liquid nitrogen or liquid helium.
| {
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Satellite Orbital Period I know I can calculate the period of a satellite orbit by Kepler's third law, but somehow it does not work out.
The sattelite is 20200km from surface of the earth.
*
*$r=$orbits radius=earths radius+satellites distance from surface of earth=20,200,000+6,378,000 = 26,578,000 m
*$G=6.67\cdot10^{-11}$
*$M = $mass of earth $= 5.9722\cdot10^{24}$
now $T=(4\pi^2r^3/GM)^{1/2}
= 43108,699\ \mathrm{s} \Rightarrow T=11.975\ \mathrm{hours}$
BUT that isn't correct, as all the calculators say it is 16,53
I have no idea what I am doing wrong.
I even followed this example and I got everything right using the numbers in the example, but as soon as I put in my 26,578,000 m I got a different solution. Even though I did not change anything else.
What am I missing?
| Your formula and numbers look right to me. You can check (both your math and "all the calculators") by plugging in the numbers for a geosynchronous orbit (altitude of 35,786 km, or semi-major axis of 42,164 km): the period should be 24 hours.
| {
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Does the mass of the damper affect the transfer function in a vertical mass-spring-damper? Usually in system dynamics, I dealt with horizontal mass spring dampers. Now in my advanced class I am dealing with vertical mass spring dampers. So a spring is hanging from the ceiling with a mass connected, and then the damped is under the mass. The damper is in some sort of oil and that is creating the "damping" effect.
In this problem, the spring is assumed to be negligible mass, but the damper has a mass, along with the weight that is acting as the mass.
I know that for a mass spring damper system (when its horizontal), the transfer function is
$$H(s)=\frac{\omega_{n}^{2}}{s^{2}+2\zeta\omega_{n}s+\omega_{n}^{2}}$$
But now the mass of the weight and damper are acting on this system. So does that affect the transfer function?
| Gravity changes the equilibrium position, but not the motion around the equilibrium.
| {
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Calculating car's acceleration from change in angle of hanging object? The question essentially is based on a situation like this-
A car has a small object hung from the cieling on a string (apparently at an angle of 0 degrees to the ceiling).
The car is accelerating and the object is now hanging at a 30 degree angle (to the ceiling).
How would I figure out how much the car is accelerating.
PS - This is homework but Im stuck and would appreciate any advice. Thanks.
Edit: changed angle from 45 to 30.
| This is a neat example because the object makes its own force triangle - it's being pulled down by gravity and sideways by the car's acceleration. And the 45° angle means that the forces are equal.
| {
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Rotating hoop in Relativity What does a rotating hoop, with each point moving at a velocity close to the speed of light, appear like with respect from a stationary observers perspective. For example how does the shape of the hoop change? (Note, I'm not intending to ask about optical effects, but rather what physically happens, analogous to length contraction of a moving train.)
| By symmetry, it'll look circular. Just look at it from above, along the axis. You can synchronize clocks along the rim by a signal from the center. Then have those rim clocks all emit signals simultaneously. In the rest frame, their signals will arrive at the same time as those of stationary clocks positioned around the "orbit".
Or imagine encasing the whole thing in a hollow toroidal container at rest. Like the proton bunches going around the LHC. No ambiguity about what the encasing torus looks like.
| {
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Do strong and weak interactions have classical force fields as their limits? Electromagnetic interaction has classical electromagnetism as its classical limit. Is it possible to similarly describe strong and weak interactions classically?
| I think there are really two separate issues here. One is the range of the forces, and the other is the existence of a classical limit.
Basically, being able to write down a Lagrangian density isn't the same thing as being able to describe the classical theory that is the counterpart of a quantized system. In particular, it seems like this can't possibly work for unstable particles. For example, the Lagrangian density for muon decay has a constant in it, $G_F$, the Fermi coupling constant. The half-life of the muon goes like $h/G_F^2$. In the classical limit $h\rightarrow0$, the half-life goes to zero, so the classical theory of muons is a theory with no muons in it.
So you can't have a classical field theory of the weak force, simply because the W and Z are unstable.
The strong force is completely different. Gluons are massless and stable. Although they're self-interacting, so are gravitons, and there is a classical field theory of gravity. It's not completely obvious to me that we don't ever have a classical field corresponding to the strong force.
For example, take the case of two heavy nuclei scattering inelastically, below but close to the Coulomb barrier. The process is classical in the sense that the de Broglie wavelengths of the two nuclei are small compared to the sizes of the nuclei. Far below the Coulomb barrier, you get Rutherford scattering, which is completely classical -- you can describe it using Newton's laws. Closer to the Coulomb barrier, the nuclei can approach one another sufficiently closely for the strong force to act, but there is still an elastic scattering channel, which I think should be describable in purely classical terms.
| {
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What is predicted to happen for electron beams in the Stern-Gerlach experiment? The Stern–Gerlach experiment has been carried out for silver and hydrogen atoms, with the result that the beams are deflected discretely rather than continuously by an inhomogenous magnetic field. What is theoretically predicted to happen for electron beams?
| There is a bit of trouble with using electrons since the magnetic field of the apparatus will cause them to turn thanks to the Lorentz force. You could, of course, build a device to account for the turning, and still split the electrons by spin at the end.
| {
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Work Done by an Adiabatic Expansion I am given the information that a parcel of air expands adiabatically (no exchange of heat between parcel and its surroundings) to five times its original volume, and its initial temperature is 20° C. Using this information, how can I determine the amount of work done by the parcel on its surroundings?
I know that $dq = 0$, and that $du + dW = dq = 0$, but I don't know what to do with this information. $dW = pdV$, which seems like it should be helpful, but I don't know what to do for the pressure.
| More clues? :-)
This is harder then the isobaric process because now the pressure is a function of volume. You need to write the pressure as a function of volume, then integrate it from the initial to final volume. For some clues see the Wikipedia article on adiabatic expansion. Although the question doesn't say so, you'll need to assume the expansion is reversible as the question can't be answered otherwise.
| {
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Would a spin-2 particle necessarily have to be a graviton? I'm reading often that a possible reason to explain why the Nobel committee is coping out from making the physics Nobel related to the higgs could be among other things the fact that the spin of the new particle has not yet been definitively determined, it could still be 0 or 2.
This makes me wonder if the spin would (very very surprisingly!) finally be discovered to be 2, this then necessarily would mean that the particle has to be a graviton? Or could there hypothetically be other spin-2 particles? If not, why not and if there indeed exist other possibilities what would they be?
| A massive spin 2 particle must have five modes: helicity
$\pm 2$, $\pm 1$, 0. If a massless spin 2 particle has only helicity
$\pm 2$ modes without other modes and has a dispersion $\omega = c k$, then such a
massless spin 2 particle must be graviton (at least at linear order).
| {
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Hydrogen Hyperfine Structure: General Expression I was looking at the hyperfine structure for the hydrogen atom. I checked pretty much every textbook I knew but none of them gave me the general expression for the energy correction due to the hyperfine perturbation Hamiltonian.
All of them only treat the case when $\ell=0$. I was wondering if there is a general expression which doesn't have such a restriction?
| The Hamiltonian for the spin-spin interaction is:
$$\Delta H_{SS} = \frac{\gamma_p e^2}{m m_p c^2 r^3} \Big( \frac{1}{r^3} \big(3(\vec{s}_p \cdot \hat{r})(\vec{s}_e \cdot \hat{r})-(\vec{s}_p \cdot \vec{s}_p) \big)+\frac{8 \pi}{3} (\vec{s}_p \cdot \vec{s}_p) \delta^{(3)}( \vec{r} ) \Big) $$
where $m$ and $m_p$ are the electron and proton masses, and $\gamma_p$ is the proton magnetic moment in units of the nuclear magneton. For the cases where $l \neq 0$ the term with the delta function cancels, and the wavefunction is proportional to $r^l$ for small $r$ values. Thus, when $l>0$ we get that $\psi(0)=0$, and then the correction to the energy will be:
$$\Delta E_{hf} = \frac{\gamma_p e^2}{m m_p c^2} \left\langle \frac{1}{r^3} \big( ( \vec{l} \cdot \vec{s}_p )+3(\vec{s}_p \cdot \hat{r})(\vec{s}_e \cdot \hat{r})-(\vec{s}_p \cdot \vec{s}_p) \big) \right\rangle $$
This expectation value was calculated by Bethe and Salpeter, and the result is:
$$\Delta E_{hf} = \frac{m}{m_p} \alpha^4 mc^2 \frac{\gamma_p}{2 n^3} \left( \frac{ f(f+1)-j(j+1)-\frac{3}{4}}{j(j+1)(l+\frac{1}{2})} \right),$$
This result coincide with the $l=0$ case, since then $j=\frac{1}{2}$ and the proton has spin 1/2, so the hydrogen atom total angular momentum $f=j \pm \frac{1}{2}$ and the expression above can then be simplified to the result you already have for the $l=0$ case...
(Next time try older QM books like Bethe & Salpeter's Quantum Mechanics of One- and Two-Electron Atoms.)
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What are some of the best books on complex systems and emergence? I'm rather interested in getting my feet wet at the interface of complex systems and emergence. Can anybody give me references to some good books on these topics? I'm looking for very introductory technical books.
| As how it's titled, Complexity: A guided tour is a guided tour to complexity sciences. There is no formula in the book, but it does discuss many results and give you pointers. The author is a student of Douglas Hofstadter. She has a talk here
FYI: @Chris Aldrich also shares many helpful links in Are there any theories using thermodynamics/statistical mechanics or information theory principles to modelling in ecology?
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Why does light of high frequency appear violet? When people are asked to match monchromatic violet light with an additive mix of basic colours, they (paradoxically) mix in red. In fact, the CIE 1931 color space chromaticity diagram shows this effect begins at about 510nm (greenish-cyan), where people mix in no red. From that point on, the higher the frequency of the light source, the more red they mix in.
This effect is reflected by the red curve of the CIE standard observer color matching functions, which has an additional bump in the area of blue light. However, that curve does not match the actual spectral sensitivity of red cones. So where does this additional perception of red at higher frequencies come from?
| It's pretty hard to research but I think (not sure) that the short-wavelength spectra may appear deep violet because of a negative green cone response. This answer also explains the apparent brightness of violet.
| {
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How robust is Kramers degeneracy in real material? Kramers theorem rely on odd total number of electrons. In reality, total number of electrons is about 10^23. Can those electrons be so smart to count the total number precisely and decide to form Kramers doublets or not?
| I think Kramer's theorem is really only useful when you can write down a wavefunction for your system. It then tells you the degeneracy of the ground state of your wavefunction. If you have 1 mole of your material you couldn't write down a wavefunction for it so the degeneracy of the ground state would be meaningless.
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Why are Euler's equations of motion coupled? Physical explanation I have a problem with one of my study questions for an oral exam:
Euler’s equation of motion around the $z$ axis in two dimensions is $I_z\dot{\omega}_z = M_z$, whereas it in three dimensions is $I_z\dot{\omega}_z =-(I_y-I_x)\omega_x\omega_y+M_z$, assuming that the $xyz$ coordinate systems is aligned with the principal axis. Why does Euler’s equation of motion for axis $z$ contain the rotational velocities for axes $x$ and $y$?
How can one explain this physically? I mean I can derive Euler's equation of motion, but how can I illustrate that the angular velocities are changing in 3 dimensions?
| If you are familiar aircraft flight dynamics, then please remember an aerobatic maneuver so called "Immelmann turn". Let us take x,y and z axis for longitudinal, lateral and vertical coordinate axes fixed to the aircraft. Then, if we pull control a stick and let the aircraft turn in vertical plane around the y axis and turn the stick to the right or left and make a rolling motion around the x axis at the same time. When the both two angular motion accomplished 180 degree turn, then the aircraft flies in the opposite direction before this maneuver. This means that the aircraft turned 180 degree around the z axis. This is my interpretation of the Euler equation of rigid body motion.
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Do spacecraft engines suffer from carbon accumulation the way typical petrol/kerosene engines do? Just wondering whether the spacecraft engines/drives, or their booster rockets accumulate carbon the way car/truck engines do. What about ion/methane drives?
| Actually, it is a factor in some systems. RP-1 is a grade of kerosene specifically intended for use as rocket propellant, engineered specifically for reduced breakdown and coking at high temperatures which can otherwise cause problems for regenerative cooling channels, turbopumps, etc.
Also see page 115-116 in The Rocket Company.
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Can a trajectory between planets accelerate a space craft? I read that if some spacecraft's trajectory is carefully planned, i.e. if it can slingshot (at loss of a better word) a massive body like a planet, it can gain speed. Is that correct ?
Does solar system actually lose some energy to such a space craft in this way ?
| In order for this question to make sense you need to define which reference frame the spacecraft should gain speed in. Conservation of energy implies that the spacecraft gains no speed in the frame of reference in which the planet is stationary, but it can change its velocity (ie trajectory is changed). If you move that back to the reference frame where the Sun is stationary, you find that the spacecraft's speed may have in fact increased. Total energy of spacecraft and planet is unchanged, making the interaction an elastic collision. But the kinetic energies of individual bodies does change and the planet looses some energy (although being so large the change is entirely negligible).
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What is the difference between a spinor and a vector or a tensor? Why do we call a 1/2 spin particle satisfying the Dirac equation a spinor, and not a vector or a tensor?
| Spinor is a vector in the basis of not space-time, but its spin states;
in on sense, spinor is not a vector, since it will not transform as you transform the space (rotation, etc)
.
Generally speaking, tensors (including scalar, vector, tensor of rank 2,3,4...etc) are just mathematical objects (you lumped together) that transformed as the space-time coordinate transforms as an entirety. A scalar is so that:
$$ A'=A $$
after arbitrary coordinate transform;
a Vector is so that:
$$ A'=MA, or \ A'_i=M_i^{\ j}A_j $$
where M is the unitary (orthogonal) transformation matrix.
when it comes to a tensor of rank two:
$$ A'=MAM^{-1} or \ A'_{\nu\mu}=M_{\mu}^{\ \rho}M_{\nu}^{\ \theta}A_{\rho\theta } $$
Higher orders of tensors just transform accordingly (by adding more terms of M) .
| {
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Same momentum, different mass The question is: if
*
*A bowling ball and ping pong ball
*are moving at same momentum
*and you exert same force to stop each one
*which will take a longer time? or some?
*which will have a longer stopping distance?
So I think I can think of this as:
$$F = \frac{dp}{dt} = m \cdot \frac{v_i - 0}{\Delta t} = \frac{p_i}{\Delta t}$$
Since both have same momentum, given same force and momentum, time will be the same? Is this right?
Then how do I do the stopping distance one?
| Another way of looking at the distance is to look at the kinetic energies $\tfrac12 m v^2$, which will be greater for the less massive ping-pong ball if the momentum is the same.
The change associated with reducing the kinetic energy to zero will be equal to the force times the stopping distance, so if the forces are the same then the distance will be greater for the ping-pong ball.
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Divergence of non conservative electric field I'm looking for the proof that the 1st Maxwell equation is valid also on non conservative electric field.
When we are talking about a electrostatic field, the equation is ok. We can apply the Gauss (or Flux) theorem and get Gauss' law:
$$\mathbf{\nabla} \cdot \mathbf{E} ~=~ \frac{1}{\epsilon _0} \rho (x,y,z).$$
The question is, why when there is a time dependent magnetic field, and then a time dependent (non conservative) induced electric field, the 1st Maxwell equation is the same?
How we can prove that?
| As you've said, and just to be completely clear, in vacuum (neglecting, in other words, effects in macroscopic media like polarization), Gauss' law is the full, time-dependent expression of what you're calling the "first Maxwell equation."
The "derivation" of the Maxwell equations were originally formulated as differential (local) versions of the well known empirically observed laws of Ampere, Faraday, and Gauss. This is discussed some in Jackson's book ("Classical Electrodynamics"). Also see Griffith's book ("Intro to Electrodynamics").
The Maxwell equations aren't really derived from more fundamental considerations. Their integral form (the "laws" cited above) were deduced from observation, compared with phenomena not originally used in the determination of the empirical "laws," and found, in some regimes, to work.
In the regime of atomic physics, Planck found that the assumed continuous radiation of an accelerating charge predicted a black-body spectrum at large frequency in contradiction with that observed. And this led to a modification of the classical electrodynamics and the advent of the quantum theory.
The form of the Maxwell equations is, however, tightly constrained by invariance under Lorentz transformations. Jackson discusses this in Chapter 11.
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Why do green lasers appear brighter and stronger than red and blue lasers? This is mostly for my own personal illumination, and isn't directly related to any school or work projects. I just picked up a trio of laser pointers (red, green, and blue), and I notice that when I project them, the red and the blue appear to be dimmer to my eye than the green one. I had a fleeting suspicion that, perhaps this is an effect of blue and red being at the periphery of the visual light scale, but I honestly have no idea if this is the case or if it's just my eyes playing tricks on me. All three lasers have the same nominal strength, in this case.
| A less scientific answer: It may also just happen that the green laser is more powerful than the others.
In fact, humankind always had great problems making a green laser diode, whereas the red and blue ones are readily available. The green pointers contain a full-fledged Nd:YVO4 DPSS laser oscillator with built-in KTP frequency doubler.
I have measured two green pointers with precise power meter, and both greatly exceeded their nominal values. In fact, their power was illegal to be sold in my country...
| {
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Why is light called an 'electromagnetic wave' if it's neither electric nor magnetic? How can light be called electromagnetic if it doesn't appear to be electric nor magnetic?
If I go out to the sunlight, magnets aren't affected (or don't seem to be). And there is no transfer of electric charge/electrons (as there is in AC/DC current in space).
In particular, the photons (which light is supposed to be composed of) have no electric charge (nor do they have magnetic charge).
I'm looking for an explanation that can be appreciated by the average non-physicist Joe.
| This image (taken from Wikipedia) demonstrates what an electromagnetic wave looks like.
Changing electric fields induce a magnetic field (this is how electromagnets work), and changing magnetic fields induce an electric field (this is how the charger on your electric toothbrush works). The result is that if one oscillates, so will the other, and they will continually induce each other.
I hope that at least gives an intuitive explanation for it (even if some of what I said is not 100% technically correct).
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Find magnetic scalar potential for superconducting sphere In regions where $J = 0$, the curl of the magnetic field $B$ is necessarily zero (since $\nabla \times B = \mu_0 J$). Therefore $B$ can be written as $B = -\nabla V_m$, where $V_m$ is a scalar function of position.
A superconducting ball of radius $a$ is placed in an otherwise uniform magnetic field $B = B_0 \hat {z}$. Since the $B$ field is zero inside the ball, it must hold that $B_{\perp} = 0$ just outside the surface of the ball (by magnetic boundary conditions).
(a) Find the magnetic scalar potential $V_m (r, \theta)$ everywhere.
Far from the sphere the field is $B_0 \hat {z}$, and hence $V_m \rightarrow -B_0 z$. So in spherical coordinates, $V_m \rightarrow -B_0 r \cos \theta$ for $r \gg a$. Is the following true?
$$
\frac {\partial V_{\text{above}}}{\partial n} = \mu_0 K
$$
Since $V_{\text{below}} = 0$ due to the sphere being a superconductor. Using separation of variables, the first condition yields
$$
\sum_{l=0}^{\infty} A_l r^l P_l (\cos \theta) = -B_0 r \cos \theta,
$$
So, only one therm is present, $l = 1$. Then $A_1 = -B_0$ with all other $A_l$'s zero. From the second condition,
$$
\frac {\partial V}{\partial n} = A_1 \cos \theta - \sum_{l =0}^{\infty} (l+1)\frac {B_l}{R^{l+2}} P_l (\cos \theta) = \mu_0 K.
$$
Now I'm stuck as to how to proceed and remove some of the $B_l$ terms. Perhaps my boundary conditions are not correct.
| The boundary conditions you should use are that as $r\rightarrow \infty$, $\mathbf{B}\rightarrow B_0\hat{\mathbf z}$; and that at $r=a$, $\mathbf B \cdot \hat{\mathbf r}=0$.
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Diagram-like perturbation theory in quantum mechanics There seems to be a formalism of quantum mechanics perturbation that involve something like Feynman diagrams. The advantage is that contrary to the complicated formulas in standard texts, this formalism is intuitive and takes almost zero effort to remember (to arbitrary orders).
For example, consider a two level atom $\{|g\rangle, |e\rangle\}$ coupled to an external ac electric field of frequency $\omega$. Denote the perturbation by $\hat V$, with nonzero matrix element $\langle e|\hat V |g\rangle$.
Then the second order energy correction reads
$$E^{(2)} = \langle e|\hat V |g\rangle\frac{1}{\omega_g - \omega_e +\omega} \langle g|\hat V |e\rangle + \langle e|\hat V |g\rangle\frac{1}{\omega_g - \omega_e -\omega} \langle g|\hat V |e\rangle $$ where the first term corresponds to the process absorb a photon then emit a photon while the second process is emit a photon then absorb a photon.
Does anybody know the name of this formalism? And why it is equivalent to the formalism found in standard texts?
| There is an exposition of a diagrammatic representation of the terms in the quantum mechanical perturbation expansion here. Basically the diagrams are just used to represent the combinatorial properties resulting from eigenstate degeneracy.
(Just for fun there is also a diagrammatic approach to perturbation expansions in classical mechanics here).
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If nothing in the universe can travel faster than light, how come light can't escape a black hole?
Possible Duplicate:
How does gravity escape a black hole?
If nothing in the universe can travel faster than light, how come light can't escape a black hole? I mean, Einstein's relativity says nothing can travel faster than light, but yet, light can't escape a black hole. Does this mean that light really isn't the fastest thing? That the pull of the black hole is really faster than light? That Einstein was wrong, even though it's been backed up by scientific evidence? I'm very confused. If anyone would be able to answer my question, I would appreciate it: Why can't light escape a black hole if nothing can travel faster than light?
| The essential idea to grasp here is that, regardless of the fact that light propagates at the "universal speed limit", gravity is curved spacetime.
Within the event horizon, the curvature of spacetime is such that there is no world line (path through spacetime), for light or any physical object, to the exterior of the hole.
Roughly speaking, once inside the horizon, the "direction" to the outside is backwards through time.
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limits of diamond anvils for high pressure research in this wikipedia article regarding diamond anvils, it mentions that the pressure peaks roughly at 300 GPa.
My question is why is this so? is the diamond crystal structure collapsing if higher pressures are applied (like 500-600 Gpa, where metallic hydrogen is expected to form? and if such collapse happens, what sort of phase does the diamond collapses into?
| The compressive strength of a perfect diamond cristal is in the range of 220–470 GPa, depending on the direction you compress. (X. Luo et al, J. Phys. Chem. C 2010, 114, 17851–17853; DOI: 10.1021/jp102037j)
To cite this article’s introduction:
Usually, diamond is used under nonhydrostatic conditions, such as a diamond indenter in the nanoindentation test and as diamond tips of the diamond anvil cell (DAC) in ultrahigh-pressure research. Therefore, theoretical investigations into the mechanical properties of diamond under nonhydrostatic conditions should be important.
[…]
In experimental work, the compressive strength of diamond can be roughly obtained from the strength of DAC.
And the conclusion:
From the mechanism under compressive deformation of diamond, we can estimate that the limit strength of DAC should be about 470 GPa.
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Mathematical definitions in string theory Does anyone know of a book that has mathematical definitions of a string, a $p$-brane, a $D$-brane and other related topics. All the books I have looked at don't have a precise definition and this is really bugging me.
| There isn't any mathematically precise definition. These are physical objects, and they acquire their definition in a given model which allows for calculations. The same physical object can appear in different models in different roles, so the strings have different mathematical definition in different limits of the full M-theory.
The closest thing to a mathematical definition of a (perturbative) string is a 2d conformal field theory which reproduces a space-time scattering from the correlation functions of the 2d theory. This is the 1980s definition, and it is only valid in the limit that the strings are perturbative, near zero coupling. In the same limit, the definition of a d-brane in the weak coupling limit is a surface on which the strings can have endpoints. This can be all of spacetime, in which case you have strings that can be perturbatively open or closed, a type I theory.
These definitions are not exactly definitions, but identifications of a physical object, because the mathematical description changes character at strong coupling. For type IIA strings, they turn into membranes at strong coupling, type IIB strings turn into IIB d-branes, and the perturbative description leaves out processes where d-branes are formed and annihilate, which are nonperturbative in the string coupling expansion, because the brane tension diverges at small coupling.
Th dualities, and the lack of a single unified formalism from which to derive all such dualities, make it impossible to formulate strings as a single mathematical object. It's a physical theory, with the additional handicap that all the physical intuition is derived from calculations and low-energy/classical limits, because we can't actually directly observe the strings. That doesn't turn it into mathematics, it's still physics, and you still need good physical intuition.
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Long/short-range interaction A potential of the form $r^{-n}$ is often considered long-range, while one that decays exponentially is considered short-range.
Is this characterization simply relative/conventional, or is there a more fundamental reason?
| Graphing examples of each would rapidly convince you that the power law has a "long tail" whereas the exponential gets small rather quickly. That's the basic reason. But understand that it is possible to pick a distance from the origin and then to pick the constants in the potential such that the exponential will be larger than the power law.
All that said, it should also be noted that in the power law once one gets past $1/r$ potentials, that even the power law potential gets small quickly.
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What is the difference between weight and mass? What is the difference between the weight of an object and the mass of an object?
| The mass, strictly the inertial mass, relates the acceleration of a body to the applied force via Newton's law:
$$ F = ma $$
So if you apply a force of 1 Newton to a mass of 1kg it will accelerate at 1m/s$^2$. This is true whether the object is floating in space or in a gravity field e.g. at the Earth's surface.
The weight is the force a body exerts when it is in a gravitational field. The weight depends on the gravitational field. For example the weight of a 1kg mass at the Earth's surface is 9.81 Newtons, while at the surface of Mars it's about 3.5 Newtons.
This is possibly a bit too much info: if so ignore this last paragraph. Although weight specifically means the force exerted in a gravitational field, Einstein told us that sitting stationary in a gravitational field is equivalent to being accelerated in the absence of gravity. The inertial mass defined using Newton's laws is the same as the gravitational mass defined by the force a body exerts in a gravitational field. So if you take a 1kg mass at the Earth's surface, the weight of 9.81 Newtons it exerts is exactly the same as the force you'd need to accelerate the 1kg mass at 9.81m/s$^2$.
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Intuition for multiple temporal dimensions It’s easy, relatively speaking, to develop an intuition for higher spatial dimensions, usually by induction on familiar lower-dimensional spaces. But I’m having difficulty envisioning a universe with multiple dimensions of time. Even if such a thing may not be real or possible, it seems like a good intellectual exercise. Can anyone offer an illustrative example?
| I have been a self-guided student of theoretical physics for over 30 years. I don't have any formal training. I have been working on a model of our universe ever since I purchased and read Dr dr Michio Kaku book "hyperspace" in high school 25+ years ago. My model utilizes multiple dimensions of space time. It is still a work in progress. The assumption that my model is built of is this: Because spacetime is an inseparable entity it is assumed that space and time are proportionate to each other. What that means is, as space expands in dimensions, so too does time. So rather than with 5 spatial dimensions there is just one dimension of time there ate two axis of time. 6 dimensions of space 3 of time, [7D 4t], [8D 5t]. Lets just say time gets really really weird when you get to 4t and above. at one point you end up having a prime observer in two separate timelines within the same 3D space. Nuts! Another implication that is poking its head out is that energy density is relative to spacetime/matter-energy. That is, energy density determines the number of space time dimensions. There might be some explanation to the hyper expansion of our univers right after the big bang. This is still just a work in progress and no ware near complete. But some interesting hints at the possible nature of our univers.
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Where does Computer Science background students fit in Theoretical Physics I am basically an Electronics student - background in computer science (that's where I want to work). I applied for an internship in USA in a research institute where the group is focused in Theoretical Condensed Matter Physics, Chemical Physics, Physical Chemistry, Materials Science.
I mentioned my areas of interests as: Computational Science, Machine Learning, Web development
My skills as: Python, C, Django, Java, etc..
I got selected. Now, I would like to know where could possibly a CS background guy would actually work on?
I am looking for a detailed answer
| Condensed matter physicists could always use some simulations, I'm sure. There's plenty of stuff where insight could be gained with a well run simulation. Depending on the place you're working at, you could have access to a supercomputer facility. The language people use for simulations vary, from Fortran, C, Python, Matlab, Mathematica... Basically depends on the background and convention of the place you're working in.
Coming from a CS background will be helpful, since physicist tend not to worry about the coding too much. "As long as it works" is the general attitude. If you can understand the physics, the computation should be much easier for you!
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How accelerometers sense constant velocity movements There is something I don't understand about accelerometers. I know it's possible to make an Inertial Measurement Unit (IMU) by using "three" accelerometers. So with that, I could calculate the $x$, $y$, $z$ coordinates of something in 3-D space.
The point that I don't understand is, how could it sense constant-speed movements? When the velocity is constant, acceleration is zero. So how does it work?
| It can't directly(Einstein was right)
What you can do is integrate all the accelerations you have measured upto that point in time to have a current knowledge of the current velocity.
This isn't very accurate because any small inaccuracies in acceleration will lead to increasingly inaccurate velocity, and so position, with time - this is known as drift.
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Force in tetrahedron edges I am looking for a formula that enables me to calculate the force in a tetrahedron edge such that it relates $F_b$ with $F_z$ through the beam thickness and length. I have the following assumptions:
*
*The beams are circular and hollow. So the have a radius $r$ and a thickness $t$.
*The static force points in the negative $z$-direction and is active in the c.o.g.
*The contribution of the vertices/joints is neglected.
*The tetrahedron is standing on one vertex as shown in the image below.
*The beam length is $a$.
I have drawn the situation in the following image:
Can someone provide me with a formula or some pointers on how to find the beam force?
Edit, using Jaime's hints.
As the c.o.g. is alined with he lower vertex the structure is in equilibrium. When looking at the lower vertex the vertical component of the three beam forces equals the gravitational force.
$$
F_{b_{y1}}+F_{b_{y2}}+F_{b_{y3}}=F_{g}\\
F_{b_{y1}} = F_{b_{y2}} = F_{b_{y3}}\\
F_{b_{y}} = \frac{m \cdot g}{3}
$$
From the Encyclopedia Polyhedra written by Robert Gray I found that the half-cone angle is 35$^{o}$. This means that:
$$
F_{b} = F_{b_{y}} \cdot cos 35\\
F_{b} = \frac{m \cdot g}{3} \cdot cos 35
$$
| For there to be equilibrium, the cog has to be directly above the bottom vertex. The three beams joining there will support the same amount of compression each, and the vertical component of all three together must be equal to the loading. You will need a little trigonometry, and some properties of a tetrahedron you can find here.
To find out the load on one of the top bars, consider equilibrium at any of the top nodes in a horizontal plane.
Show your help, and I will be happy to carry on with the details.
| {
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What do the supercharges in extended supersymmetry do? What do the supercharges in extended supersymmetry do?
In ${\cal N}=1$ supersymmetry there are a certain number of fermions and and equal number of bosons. You can transform all fermions to the bosons (and vice versa) in a 1 to 1 fashion using a single supercharge, $Q$.
So what happens when you have, for example, ${\cal N}=2$ supersymmetry with 8 supercharges? Since $Q$ is a generator of supersymmetry transformations, is it a linear combination of these supercharges that act on the particles? In which case could one particle be acted on by two separate linear combinations of $Q$? Or is it strictly one linear combination of $Q$ per fermion/boson?
Also, what does ${\cal N}$ mean physically? What difference does ${\cal N}=2$ have to ${\cal N}=1$ other than more supercharges? Or is that the only difference between the two theories?
| You need to be a bit careful about the counting of supercharges. In four dimensions, the smallest spinor representation is four-dimensional (over the real numbers), so $\mathcal{N}=1$ supersymmetry has four supercharges.
When there are more supercharges, there are simply more states combined into a single 'multiplet'. For example, in $\mathcal{N}=1$ supersymmetry, a theory with a massless vector boson like a photon necessarily also contains a massless spin-$\frac{1}{2}$ particle, the 'superpartner' of the photon. In an $\mathcal{N}=2$ theory, a photon has more superpartners: two massless spin-$\frac{1}{2}$ particles, and a massless complex scalar. It's still always true that there are the same number of bosonic and fermionic degrees of freedom.
You can find all the details in chapters 2 and 3 of Wess and Bagger's classic supersymmetry textbook.
| {
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How long will it take for a bullet to reach a Geostationary orbit? I'm curious to know this. Neglect air friction and imagine a bullet that were shot normal to the Earth's surface, from the Equator. I will have to consider the Coriolis effect and so I expect the path of the bullet will follow a spiral rather that a straight line (relative to Earth's centre). The gravity will reduce with altitude as well and so it would be difficult to apply basic laws of motion, but I really need to know how this would look like and how long it will take for the bullet to reach around 36000 km above the Earth's surface. Will it come back, stay in that orbit, or escape (assuming that normal velocity reached 0 @ that orbit)? I expect if it comes back, then it will follow a similar path it traveled along during the shot. This is just a curiosity and thanks for help in advance.
| $T = (R + H)(2H/(GM))^{0.5}$ for simple case without atmosphere. For GEO it's about 5 hours with zero velocity in GEO point and enormous shooting velocity.
But air resistance is proportional to ~$V^2$, so it would be hard for bullet to leave atmosphere ;)
Where does the question came from? Maybe you've just read "From the Earth to the Moon" Jules Verne? =)
| {
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Equivalent spring equations for non-helical coil shapes? The compression spring equations are generally given for helical coil. What are the equivalent equations for alternative coil shapes, like oval?
| I'm guessing that the "equations that are generally given" you refer to the ones calculating deflection of a helical spring, based on the torsional deformation of the coil, such as what is found here. The following picture is taken from there:
As you can see, any cross section of the coil is subject to shear, as well as to torsion. If your coil is not perfectly circular, it will have the same shear, but different torsion, depending on the point around the coil you are considering, $T=Fr$, with $T$ the torsion torque, $F$ the axial load, and $r$ the effective radius of the coil.
In the reference above, they compute the deformation from the total strain energy, which is made up of a shear term and a torsional term. In their case it is relatively easy, since the torsion is constant throughout the coil. You will have to integrate around the coil based on the shape of your coil. The exercises here should help in figuring out how to do that.
If you get stuck, shout in the comments, and I can guide you through the calculaitons for a particular case.
Or you can go the quick, dirty way, and copy the formulas found in this discussion.
| {
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Relativistic momentum I have been trying to derive why relativistic momentum is defined as $p=\gamma mv$.
I set up a collision between 2 same balls ($m_1 = m_2 = m$). Before the collision these two balls travel one towards another in $x$ direction with velocities ${v_1}_x = (-{v_2}_x) = v$. After the collision these two balls travel away from each other with velocity ${v_1}_y = (-{v_2}_y) = v$. Coordinate system travells from left to right with velocity $u=v$ at all times (after and before collision).
Please see the pictures below where picture (a) shows situation before collision and picture (b) after collision.
Below is a proof that Newtonian momentum $mv$ is not preserved in coordinate system $x'y'$. I used $[\, | \,]$ to split $x$ and $y$ components. $p_z'$ is momentum before collision where $p_k'$ is momentum after collision.
$$
\scriptsize
\begin{split}
p_z' &= \left[ m_1 {v_1}_x' + m_2 {v_2}_x'\, \biggl| \, 0 \right] = \left[ m_1 0 + m_2 \left( \frac{{v_2}_x - u}{1-{v_2}_x\frac{u}{c^2}} \right)\, \biggl| \, 0 \right]= \left[ m \left( \frac{-v - v}{1+ v \frac{v}{c^2}} \right) \, \biggl| \, 0 \right] \\
p_z' &= \left[ - 2mv \left( \frac{1}{1+ \frac{v^2}{c^2}}\right) \, \biggl| \, 0 \right]
\end{split}
$$
$$
\scriptsize
\begin{split}
p_k' &= \left[-2mv \, \biggl| \,m_1 {v_1}_y' + m_2 {v_2}_y'\right]=\left[ -2mv \, \biggl| \, m_1 \left( \frac{{v_1}_y}{\gamma \left(1 - {v_1}_y \frac{u}{c^2}\right)} \right) + m_2 \left( \frac{{v_2}_y}{\gamma \left(1 - {v_2}_y \frac{u}{c^2}\right)} \right) \right]\\
p_k' &= \left[ -2mv \, \biggl| \, m \left( \frac{v}{\gamma \left(1 - v \frac{v}{c^2}\right)} \right) - m \left( \frac{v}{\gamma \left(1 - v \frac{v}{c^2}\right)} \right)\right]\\
p_k' &= \left[ -2mv \, \biggl| \, 0 \right]
\end{split}
$$
It is clear that $x$ components differ by factor $1/\left(1+\frac{v^2}{c^2}\right)$.
QUESTION: I want to know why do we multiply Newtonian momentum $p=mv$ by factor $\gamma = 1/ \sqrt{1 - \frac{v^2}{c^2}}$ and where is the connection between $\gamma$ and factor $1/\left(1+\frac{v^2}{c^2}\right)$ which i got?
| Relativistic momentum is not defined as $p= \gamma m v$. For instance, this expression does not apply to photons, which are massless particles [*].
In the Lagrangian formalism of mechanics the momentum is defined as
$$p\equiv \frac{\partial L}{\partial v}$$
Using the relativistic Lagrangian for a free massive particle
$$L = - m c^2 \sqrt {1 - \frac{v^2}{c^2}} $$
we obtain $p= \gamma m v$ from the definition.
The conservation laws can be obtained by using the ordinary Lagrangian formalism. For Lagrangians invariant to coordinate transformations, the momentum (as defined above) is automatically conserved. The above Lagrangian for one particle does not depend on coordinates (only on velocity) and, thus, the particle momentum $p$ is conserved. The Lagrangian for two particles does not depend on coordinates (only on velocities) and, thus, the total momentum $p_1 + p_2$ is conserved. If you use non-relativistic Lagrangians you obtain that it is the total non-relativistic momentum which is conserved.
[*] For a photon the relativistic momentum is given by $|p|= E/c^2$.
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Can electromagnetic fields be used to deconstruct and reconstruct atoms? I was thinking one day and came up with a theory after reading about how scientists were studying anti-matter by using electro magnetic fields to separate matter from the anti-matter they made. It got me thinking would it be possible to use very powerful electromagnetic fields to break down the atomic structure of objects or build things in this way?
Is this atomic reconstruction with electromagnetic fields theoretically possible?
That is, is it theoretically possible to use electromagnetic fields generated by a machine to separate the parts of an atom thereby deconstructing an object on the atomic/subatomic level?
I'm not asking about breaking molecular bonds but rather actual atoms apart. If it's possible to break atoms apart with electromagnetic fields, is it also possible to use a similar process to assemble them?
| Electric strength =5.145175226 X 10^11N/C
Magnetic strength = 235125.6635 T
so Velocity of the electron =
E/B =2.188266.117 m.s^-1 = C/137
| {
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Bernoulli's equation and reference frames So I was thinking about this while driving home the other day.
I've never been quite clear on why when you drive with the windows down air rushes into your car. I thought this might be explained by Bernoulli's equation for incompressible flow, but I ran into what seems to be a contradiction. If we consider the problem from the reference of the car, the air in the car is stationary and the air outside the car has a certain velocity. Then, Bernoulli's equation implies the pressure outside the car is lower than that inside the car. However, if we take the reference frame of the road, the air in the car is moving and then the pressure in side the car is lower. Intuitively, this second situation seems to be correct since air apparently flows into the car (from high pressure to low pressure). However there seems to be a contradiction, as the pressure gradient depends on reference frame. So my question is what has gone wrong here? Is this a situation in which bernoulli's principle simply isn't applicable? Did I make some sort of mistake in my application of the principle?
| Bernoulli's equation is frame-dependent as the following paper shows it in a nice way
The Bernoulli equation in a moving reference frame
The essence of the argument is to realize that in a frame where the obstacles, around which the fluid moves, are not stationary, these surfaces do non-zero work. And one must account for this work done when using the Bernoulli equation.
A better way is to look at the generalized Bernoulli equation as done here, which also covers viscous fluids.
| {
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How to compute the expectation value $\langle x^2 \rangle$ in quantum mechanics? $$\langle x^2 \rangle = \int_{-\infty}^\infty x^2 |\psi(x)|^2 \text d x$$
What is the meaning of $|\psi(x)|^2$? Does that just mean one has to multiply the wave function with itself?
| In quantum mechanics, the likelihood of a particle being in a particular state is described by a probability density function $\rho(x,t)$.
Suppose my system is a 6 sided die. Then the expectation value for a given roll is
$$EV= \tfrac{1}{6} 1 + \tfrac{1}{6} 2 + \tfrac{1}{6} 3 + \tfrac{1}{6} 4 + \tfrac{1}{6} 5 + \tfrac{1}{6} 6 = 3.5$$
Similarly, the expectation value for a given parameter $x$ of a particle in quantum mechanics is
$$\langle x \rangle = \int_{-\infty}^{\infty} x \rho(x,t) dx$$
Note, in quantum mechanics, we don't just have the states but also their superpositions. Yet $\rho(x,t)$ doesn't contain information about these superpositions, only observable states. Therefore, we need something more fundamental that incorporates information about superpositions. That's what $\psi$ is for and why $\psi(x,t) \in \Bbb C$. By the Born rule (an axiom), $$\psi(x,t)^* \psi(x,t) = \rho(x,t)$$
In short, we need to use the "square" of the wave function because it gives us $\rho(x,t)$ and allows us to compute expectation values.
| {
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Computational Science involve programming? I read what is computational science in Wikipedia but the explanation and understanding are not very clear.
So, I could you please give a simple example computational science project and what all basic skills a person should have?
Also,
*
*Does computational science involves programming?
*How different are computational science and computational materials science?
*I am from Electrical and Computer Science (basically programming) background. I was assigned a computational materials science project. So, is it in my scope?
Probably, the prof assigned based on what individual subjects I studied (Engineering Math, Engineering Physics, Engineering Chemistry, Probability, Programming).
| *
*Some branches of computational science involve programming by yourself the needed algorithms for solving questions.
*Computational materials science would be a sub-discipline of computational science.
*I cannot say. You do not give details on your background neither details of the project.
| {
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How do bits get transferred over a copper wire? I've been a programmer for a while, and I've done a little bit of network programming, but I'm wondering, how do bits get transferred over a copper wire?
What counts for a 1 & what counts for a 0?
I don't know a lot of physics so explain to me as if I were a 8 year old please :)
| You won't be surprised to learn it's just varying voltage levels in a circuit formed by the network cables. You also probably won't be surprised to find that the details are fiendishly complicated and far too involved to reproduce here. Even the voltage levels used depend on whether it's 10MHz, 100MHz or GHz cabling. GHz uses five voltage levels and pulse amplitude modulation.
The simplest starting point is probably the Wikipedia article on Ethernet over twisted pair. If you want to know more about how the information is encoded look at the Cisco article on Ethernet.
You might also want to ask on a different SE site e.g. Electrical Engineering or Server Fault.
| {
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Relative Velocity and Momentum The question is,
A $45.5~kg$ girl is standing on a $140~kg$ plank. Both originally at rest on a frozen lake that constitutes a friction-less, flat surface. The girl begins to walk along the plank at a constant velocity of $1.47~m/s$ relative to the plank.
(a) What is the velocity of the plank relative to the ice surface?
(b) What is the girl's velocity relative to the ice surface?
In a commentary given on this problem, they are essentially that, when the velocity of the girl relative to the plank is $\vec{v}_{gp}$, and the velocity of the plank relative to the ice is $\vec{v}_{gp}$, the velocity of the girl relative to the ice is simply $\vec{v}_{gi}= \vec{v}_{gp} + \vec{v}_{gp}$.
I just can't conceive why finding the velocity of the girl relative to the earth is a simply sum. Could someone please elucidate this for me?
Also, they treat this situation as if momentum is conserved, that is, $\Delta \vec{P}=0$; but isn't it necessary that friction be present, between the plank and the girl's shoes, for there to be any motion? Wouldn't this force cause a change in momentum?
| Try this method for solving the majority of Momentum/Impulse Problems with these two simple equations. Michel Van Biezen on Youtube teaches this method. Sure beats m1v1 + m2v2 (initial) = m1v1 + m2v2 (final).
Given:
Girl- Mass of 45.5kg; Velocity +1.47m/s
Plank- Mass of 140kg
Questions:
QA Find the VELOCITY of Plank (that is girl + plank) on ice.
QB Find the VELOCITY of Girl on ice.
Formulas:
[I = mΔV] [I = ΔP = mΔV] [Vf = Vi + ΔV]
Ans A:
I = mΔV --> I = (45.5kg)(1.47m/s) = 66.9 kg m/s
I = mΔV --> ΔV = i/m --> ΔV = (-66.9 kg m/s) / (45.5kg + 140kg) = -0.361m/s
Ans B:
Vf = Vi + ΔV --> Vf = (+1.47m/s) + (-0.361 m/s) = 1.11 m/s
| {
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Statistics in physics What are the uses of statistics in physics? I am about to embark upon a study of statistics and I would like to know what the particular benefits I gain in physics.
| I can't vouch for your school, but I would imagine that the kind of statistics you do in a Statistics Major is different from the statistics used in thermodynamics. Thermodynamics is basically ensemble theory, permutations and expectation values.
I imagine that in a pure Statistics course, you will daily hear words like "Set theory, Bayesian, correlation, confidence level, Monte Carlo, Maximum Likelihood, Probability Networks. This is more the stuff that is needed for Data evaluation. You might find yourself among the physicists at CERN one day.
| {
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Photon energy - momentum in matter $E = h\nu$ and $P = h\nu/c$ in vacuum.
If a photon enters water, its frequency $\nu$ doesn't change.
What are its energy and momentum: $h\nu$ and $h\nu/c$ ?
Since part of its energy and momentum have been transferred to water, it should be less.
If water's refractive index is $n$, are the energy and momentum equal to $h\nu/n$ and $h\nu/c/n$ ?
| A new solution to this controversy has just (June 2017) been published:
"in a transparent medium each photon is accompanied by an atomic mass density wave. The optical force of the photon sets the medium atoms in motion and makes them carry 92% of the total momentum of light, in the case of silicon."
(my emphasis)
https://phys.org/news/2017-06-atomic-mass-photon-momentum-paradox.html
The rather long abstract of the paper itself is very enlightening:
https://journals.aps.org/pra/abstract/10.1103/PhysRevA.95.063850
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How big of a lens or parabolic mirror would it take ...to heat a piece of steel so its glowing yellow (1100 C)? Assuming you had a cloudless day at a latitude of, say, San Francisco...
Basically I'm wondering if it is possible/feasible to be able to do basic metal working without a traditional forge, just using the power of the sun to heat the metal. So the diameter of the heated spot would have to be about 6" in order to heat a large enough area of the metal to work it...
I always thought you would need several huge pieces of equipment to do this, but just thought I'd ask if anyone here knew how to figure out it roughly...
Thanks!
| For steel, the specific heat would be $c_p=0.5 kJ/kg K$, with a density of $
\rho=7000 kg/m^3$. Suppose you want to increase the temperature bij say $\Delta T=1100K$ of a piece of size $V=(15cm)^3$
Then you would need a total energy of.
$$E=\rho c_p V \Delta T$$
Which gives you typically $E=10^7 J$
Now, the power of the sun on a bright day, would be of the order of $p=10^3 W/m^2$.
Assuming that
*
*all the energy input is converted into heat
*the mirror is perfectly aligned
*no heat is lost during heating,
*no melting, e.g. no latent heat
and your mirror had diameter $D$ and you let the process run for a time $t$, then
$$E=p \frac{\pi}{4}D^2 t $$
Then you will get, in approximation
$$D = \sqrt{\frac{E}{pt}}$$
So, suppose you are willing to wait for ten minutes, then the mirror diameter would be $D\approx 4m$. Considering we assumed an ideal system, this is only an order of magnitude assumption.
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Confused about unit of kilowatt hours So I am a little confused on how to deal with the Kilowatt hours unit of power, I have only ever used Kilowatts and I have to design a residential fuel cell used as a backup generator for one day.
The average power consumption of a US household is 8,900 kW-hr per year and 25 kW-hr per day and approximate 1 kW-hr per hour. Does this mean that the power output of my fuel cell is 1 kW and if I wanted to use it for the entire day would it have to be designed to be 25 kW?
| A kilowatt is a unit of power, which has the dimensions of energy over time.
A kilowatt-hour, then, has dimensions of energy.
As a simple example, if you wanted to charge up a battery so as to operate a 1,000-watt (DC) heater for one hour, you'd need one kilowatt-hour of energy (assuming the mythical world of perfectly efficient batteries, lossless wires, etc.)
In terms of SI units, this is
1000 J/s $\times$ 3600 s = 3.6 MJ.
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The requirements for superconductivity Which properties are sufficient evidence for a material to be not superconducting?
I am looking for a set of statements like
If the material is semiconducting, it is not superconducting
Edit:
I am not looking for a definition of superconductivity, or for introductional literature like the famous W. Buckel.
I am looking for properties, that would forbid superconductivity. If you have a source for it i would be very glad. As far I remember magnetic atoms will forbid superconductivity too, but i could not find a source yet.
| I doubt that any such statement exists, as it would imply a very deep understanding of superconductivity we don’t currently have.
This holds especially as you apparently look for temperature-independent statements, whereas most metals can be made superconducting at sufficiently low temperatures, and all high-temperature (relatively, still a few hundred degrees below $0^\circ\textrm{C}$) superconductors are made of materials that are very bad conductors at higher temperatures (ceramics and the like). The converse that all ceramics are good superconductors is not true, either.
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DIfference in Pitch Caused by Water Temperature? I have recently been intrigued by the following question: What is the difference between the pitch of the noise of dripping water between hot and cold water? For example, would cold water create a higher pitched noise while dripping into a pot of water? Or vice versa?
| Taking into account some facts:
The speed of sound C is 343.2 m/s
In air the speed increases 0.6 m/s per degree Celsius
Source: http://en.wikipedia.org/wiki/Speed_of_sound
Also, the just-noticeable difference (jnd) in pitch depends on the tone's frequency. Below 500 Hz, the jnd is about 1 Hz for complex tones; above 1000 Hz, the jnd for sine waves is about 0.6% (6 Hz).
Source: http://en.wikipedia.org/wiki/Pitch_(music)
As the speed of the sound varies with temperaute, also does the product of the wave length mutiplied by the frecuency.
For the most part the dimensions changes due to temperature changes in the vibrating object that produces the sound are imperseptible; In which case the wave lengt would remains constant.
So the speed changes due to temperature are in the same proportion of frecuency changes.
Bellow 500 Hz, would be needed a 0.2% change (1Hz/500Hz) in frecuency, to be able to perceive any change. As the speed change needs to be the same, the speed change is also 0.2%
C = 343.2 m/s (speed of the sound)
So, the speed change needed is 0.6864 m/s, wich is (0.2%)
The temperature change needed would the change in speed needed divided by 0.6 m/sºC, wich is 1.144 ºC
Above 1000 Hz,
Would be needed a 0.6% change in frequency
Then the speed change needed is 2.0592 m/s
The temperature change needed woudl be 3.432 ºC
Conclusion:
Unless the ambient temperature changes 1.144 ºC, it won't be likely to hear any pitch change, as the change in dimension of the vibrating body are negligible
Please, feel free to review and correct if needed
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exponential potential solution let be the Schroedinguer equation
$$ - \frac{d^{2}}{dx^{2}}y(x)+ae^{cx}y(x)=E_{n} $$ (1)
here a and c are constants.
i know how to solve it from http://eqworld.ipmnet.ru/en/solutions/ode/ode0232.pdf
but what is the condtion to get the energies ?? on the interval $ [0. \infty) $ or in other interval if you wish
the solution to (1) i know that is given in terms of the Bessel function but i have problems to get the energy quatnizatio condition i have tried a get a nonsense like
$$ J_{\sqrt{E_{n}}}(b)=0 $$ for a certain real number b but is this true ??
of course i know that semiclassically the energies satisfy
$$ 2\pi n \sim \int_{0}^{\infty} \sqrt{E_{n}-ae^{cx}} $$
| To find the bound states for the potential
$$V(x) ~=~\left\{\begin{array}{ccc}ae^{cx} &\text{for}& x>0, \\ \infty&\text{for}& x\leq 0, \end{array} \right.$$
where $a,c>0$ are two positive constants, one should solve the time-independent Schrödinger eq. with the two boundary conditions
$$ \psi(x=0)~=~0 \qquad \text{and} \qquad \lim_{x \to \infty}\psi(x)~=~0.$$
This boundary value problem does only have solutions for certain discrete values of the energy $E$.
| {
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Hubbard Model Hamitonian $$H = -\sum\limits_{i,j} A_{ij} c_i^{\dagger} c_j + \frac{U}{2} \sum\limits_i(c_i^\dagger c_i)(c_i^\dagger c_i -1)$$ is defined to be a Hamiltonian for modelling the quantum random walk of identical particles on a graph (i.e., the Hubbard Model). A particle can make a transition from one vertex to another if there is an edge between them and a double-occupancy charge $U$ is imposed. $A$ is the adjacency matrix of the finite graph.
I vaguely understand that the first term is about transitioning from vertex j to i (i.e. destroyed at j and created at i) and the second term will result in $U$ when acted on a state where two particles occupy the same state.
How do I make this understanding rigorous and obtain the matrix elements of $H$ when there are two bosonic particles in some vertices initially? What corresponds to the energy, and eigenfunctions of $H$?
| http://arxiv.org/abs/1102.4006
if you want to do exact diagonalization, the paper above might be useful for you.
To obtain the matrix representation of the Hamiltonian, the basic idea is straightforward. First, enumerate the basis vectors; Second, act your Hamiltonian on each basis vector, see what basis vectors will be generated. In the second step, you can establish the Hamiltonian matrix column by column.
Once you have erected your Hamiltonian matrix, just diagonalize it numerically, say using Matlab.
| {
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Why are different frequency bands used in different countries? Why are different frequency bands used in different countries despite ITU's effort for a common frequency band use? There's got to be a reason behind this.
For instance, U.S.-based Verizon Wireless uses the 700 MHz frequency band for its LTE service while European TeliaSonera and South Korean SKT uses 1800/2600 MHz frequency bands and 850/1800 MHz frequency band, respectively.
| The RFID Journal explains this succinctly.
The industry has worked diligently to standardize three main RF bands:
low frequency (LF), 125 to 134 kHz; high frequency (HF), 13.56 MHz;
and ultrahigh frequency (UHF), 860 to 960 MHz. Most countries have
assigned the 125 or 134 kHz areas of the spectrum for low-frequency
systems, and 13.56 MHz is used around the world for high-frequency
systems (with a few exceptions), but UHF systems have only been around
since the mid-1990s, and countries have not agreed on a single area of
the UHF spectrum for RFID. UHF bandwidth across the European Union
ranges from 865 to 868 MHz, with interrogators able to transmit at
maximum power (2 watts ERP) at the center of that bandwidth (865.6 to
867.6 MHz). RFID UHF bandwidth in North America ranges from 902 to 928 MHz, with readers able to transmit at maximum power (1 watt ERP) for
most of that bandwidth. Australia has allotted the 920 to 926 MHz
range for UHF RFID technology. And European transmission channels are
restricted to a maximum of 200 kHz in bandwidth, versus 500 kHz in
North America. China has approved bandwidth in the 840.25 to 844.75
MHz and 920.25 to 924.75 MHz ranges for UHF tags and interrogators
used in that country. Until recently, Japan did not allow any UHF
spectrum for RFID, but it is looking to open up the 960 MHz area. Many
other devices use the UHF spectrum, so it will take years for all
governments to agree on a single UHF band for RFID.
| {
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Would a gauss rifle based on generated magnetic fields have any kickback? In the case of currently developing Gauss rifles, in which a slug is pulled down a line of electromagnets, facilitated by a micro-controller to achieve great speed in managing the switching of the magnets, does the weapon firing produce any recoil? If so, how would you go about calculating that recoil?
| If the Gauss rifle shoots a projectile with exit speed of $v_1$ and mass $m_1$, then its momentum will be:
$p=m_1v_1$.
Because of momentum conservation law, the rifle will have the same momentum in opposite direction. If the rifles mass is $m_2$, the rifle will start moving in the opposite direction with end speed of:
$v_2 = \frac{m_1 v_1}{m_2}$.
But, as the projectile is accelerated for longer time than in a gun, the force acting from rifle on its holder will be lower because $F=\frac{dp}{dt}$
| {
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Can objects, animate or inanimate, be constructed out of basic particles? If all the elements are made of protons, neutrons and electrons, but some elements are much rarer and more expensive than others, would it be possible to break apart atoms of one element and make atoms of another element or molecules out of them? If so, is it possible to do the same to living creatures, or is there more to them than protons, neutrons and electrons (I mean physically, regardless of ethical considerations)? And if any of this is possible, and if it were applied to transporting things or even people very fast (by recording all the information about something, breaking it apart, sending the information at lightspeed, and then reconstructing the object on the other side out of basic particles), would it take more time and/or energy than it would save?
| *
*In fact, according to modern physics matter consists of much more than only neutrons, protons, and electrons. Say for example, quarks that make up neutrons and protons, or pi-mesons that hold nuclei together(search on wiki "standard model").
*In principle, it IS possible to take apart atoms (' nuclei), rearrange them, and end up with other atoms. But this is extremely hard and requires( or gives out) LOTS of energy(basically this is what happens in nuclear/atomic bombs, fusion or fission). It is not practical generally with current human technology.
*Again, in principle it could be done. But a living organism is far FAR more complex than single atoms that we cannot easily transmutate. And the subtleties and mechanics of living organisms are far from completely understood. This requires an even higher level of technology.
*Yes, with today's human technology, it would cost a LOT more than it would save.
| {
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Why can't "missing mass" (=dark matter) be photons? After a star lives and dies, I assume virtually all of its mass would be photons. If enough stars have already lived and died, couldn’t there be enough photon energy out there to account for all the "missing mass" (=dark matter) in the universe?
And if there were enough photons to account for all the missing mass, what would it look like to us?
| As a general rule, zero mass particles which travel with the velocity of light are not good for dark matter, because dark matter concentrates around gravitational attractors. It has to be particles with some mass that can be at rest in order to stay around a galactic center from the beginning. In addition they have to be controlled by weak interactions, if they decay, because the dark matter halo is stable for long periods.
Maybe I should add that very cool photons from the beginning of the formation of the observed universe exist and have been detected as Comsmic Microwave Background radiation, very low frequency photons, uniformly distributed in the cosmos.
| {
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Physical intuition for higher order derivatives Could somebody give me an intuitive physical interpretation of higher order derivatives (from 2 and so on), that is not related to position - velocity - acceleration - jerk - etc?
| Higher derivatives of position are related to "generalized curvatures". In 3D, for instance, the derivative of acceleration is secretly related to the torsion of a curve. The hint is the Frenet-Serret (binormal, normal, tangent) triplet or the so-called repere mobile (a la Cartan). A higher dimensional extension of this moving reference frame does exist for higher derivatives. So, higher order derivatives can be mathematically imagined to be different classes of "classical curvatures". For instance, the sign of f''(0) is related to the maximum/minimum character of f'(0)=0. Usually, people work with theories in which only "curvature" matters and torsion is usually neglected. It happens with general relativity, but some theories include torsion terms. Fluid dynamics, indeed, relates torsion to vorticity/vortex lines.
With respect to the classical meaning of some variables, we have some interesting interpretations of derivatives of position (in the physical sense, not in the mathematical sense I have spoken here as related to curvature/torsion). See my link below.
Moreover, another physical intuition for higher derivatives in modern physics comes from Cosmology. I recall that the so-called statefinders of the distance-scale factor in General Relativity relates Hubble constant to "an acceleration", and higher derivatives are also related to cosmological observables.
My post about derivatives of position and "their meaning":
derivatives of position explained
and its updated version:
derivatives of position explained
| {
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What makes an equation an 'equation of motion'? Every now and then, I find myself reading papers/text talking about how this equation is a constraint but that equation is an equation of motion which satisfies this constraint.
For example, in the Hamiltonian formulation of Maxwell's theory, Gauss' law $\nabla\cdot\mathbf{E}=0$ is a constraint, whereas $\partial_\mu F^{\mu\nu}=0$ is an equation of motion. But why then isn't $\partial_\mu j^\mu=0$, the charge-conservation/continuity equation, called an equation of motion. Instead it is just a 'conservation law'.
Maybe first-order differentials aren't allowed to be equations of motion? Then what of the Dirac equation $(i\gamma^\mu\partial_\mu-m)\psi=0$? This is a first-order differential, isn't it? Or perhaps when there is an $i$, all bets are off...
So, what counts as an equation of motion, and what doesn't?
How can I tell if I am looking at a constraint? or some conservation law?
| An equation of motion is a (system of) equation for the basic observables of a system involving a time derivative, for which some initial-value problem is well-posed.
Thus a continuity equation is normally not an equation of motion, though it can be part of one, if currents are basic fields.
| {
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How is the topological $Z_2$ invariant related to the Chern number? (e.g. for a topological insulator) This question relates to the $Z_2$ invariant defined e.g. for topological insulators:
Is it correct to relate $Z_2$ = 1 to an odd Chern number and $Z_2$ = 0 to an even Chern number?
If yes, is it also correct to think of an even or odd Chern number in terms of an even or odd number of band crossings across the Fermi energy? (If it's odd, there must be a band connecting the valence to the conduction band and therefore provide a topological protected surface state.)
Edit: These lecture notes* (under Point H) state: "The formula (49) was
not the first definition of the two-dimensional Z2 invariant, as the original Kane-Mele paper gave a definition based on counting of zeros of the “Pfaffian bundle” of wavefunctions. However, (49) is both easier to connect to the IQHE
and easier to implement numerically."
and furthermore:
"...and the Chern numbers of the two spheres are equal so that the
total Chern number is zero. The above argument establishes that the two values of the Z2invariant are related to even or odd Chern number of a band pair on half the Brillouin zone."
*
*Notes for MIT minicourse on topological phases
| For a time reversal invariant bloch hamiltonian (such as in a $\mathbb{Z}_2$ topological insulator) the Chern number is always zero.
The topological invariant $\nu = 0,1$ classifies the insulator as trivial or topological. This can be found by counting the number of times the surface energy bands intersect the Fermi energy mod 2 as you mentioned above.
For a reference see the RMP by Hasan and Kane,
http://rmp.aps.org/pdf/RMP/v82/i4/p3045_1
Sections II.B.1 and II.C.
I hope this was helpful. I am trying to learn about these topics as well.
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Do new universes form on the other side of black holes? I have four questions about black holes and universe formations.
*
*Do new universes form on the other side of black holes?
*Was our own universe formed by this process?
*Was our big bang a black hole seen from the other side?
*Are there solid reasons why this might not be the case?
| *
*No, physically, no new Universes ever get formed. In particular, the extended Penrose causal diagrams with new infinite regions just show the maximal extension of the spacetime that is possible mathematically, ignoring physical processes inside the black hole. In physics and reality, the extension is unphysical because the naive extrapolation by Einstein's equations can't be trusted deeper than the inner (Cauchy) horizon of a black hole (because Cauchy horizons are unstable) and/or singularity, so the solution doesn't continue, and it's enough to "kill" the new infinite space. There has also been a different question whether an inflating universe may be created in the bubble where the inflaton jumps to a higher level, here the answer isn't conclusive.
*No, because universes aren't created in this way, ours wasn't, either.
*No, a black hole can't be the same thing as the Big Bang. The Big Bang singularity is an initial spacelike singularity which would be similar to the white hole. But even the white hole is wrong because it cannot exist. The whole situation around the white hole is forbidden - the entropy would decrease with time but after the Big Bang, the entropy was increasing (like always in allowed situations, because of the second law).
*Yes, see above.
| {
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Evaluating propagator without the epsilon trick Consider the Klein–Gordon equation and its propagator:
$$G(x,y) = \frac{1}{(2\pi)^4}\int d^4 p \frac{e^{-i p.(x-y)}}{p^2 - m^2} \; .$$
I'd like to see a method of evaluating explicit form of $G$ which does not involve avoiding singularities by the $\varepsilon$ trick. Can you provide such a method?
| As far as my experience goes, the problem stems from writing the right solution for all reals to the problem:
$$
(p-m)G(p)=1.
$$
which reads:
$$
G(p)=\text{P.v.} \frac {1}{p-m}+c_0\delta(p-m)
$$
where $\text{P.v.}$ stands for principal value. The $\delta(\epsilon-\omega)$ function appears as it is the Kernel of $(\omega-\epsilon)$ and $c_0$ is some constant to be fixed. If we now take the Fourier transform we obtain:
$$
\int e^{ipt }G(p)=\:\left( i\pi \text{sign}(t)+c_0\right)e^{i m t}
$$
$c_0$ must now be fixed according to boundary conditions; for the retarded and advanced Green functions, one has $c_0=\pm i\pi$ and the solution given by the $i \epsilon$ trick is recovered. In my opinion though, it is a rather bad method as it only works when you have poles or first order, as the $\delta$ function can be approached by square integrable functions. If you now are looking for the solution of:
$$
(p-m)^kG(p)=1.
$$
with k an integer, you now have
$$
G(p)=\text{P.v.} \frac {1}{(p-m)^k}+\sum_{j=0}^kc_j\delta^{(j)}(p-m)
$$
with $\delta^{(k)}$ the k-th derivative of the delta function. The Fourier transform reads
$$
G(t)=\left( i\pi \frac{(i t)^{k-1}}{k-1!}\text{sign}(t)+\sum_{j=0}^kc_j (-it)^j\right)e^{i m t}
$$
And again, the $c_j$s are fixed depending on boundary conditions. Yet I don't know any way to recover this solution with the $i\epsilon$ trick.
| {
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What's the difference between space and time? I'm having a hard time understanding how changing space means changing time. In books I've read people are saying "space and time" or "space-time" but never explain what the difference is between the two concepts or how they are related.
How are the concepts of space, time, and space-time related?
| Suppose you move a small distance $\vec{dr}$ = ($dx$, $dy$, $dz$) and you take a time $dt$ to do it. Pre-special relativity you could say three things. Firstly the distance moved is given by:
$$ dr^2 = dx^2 + dy^2 + dz^2 $$
(i.e. just Pythagorus' theorem) and secondly the time $dt$ was not related to the distance i.e. you could move at any velocity. Lastly the quantities $dr$ and $dt$ are invarients, that is all observers will agree they have the same value.
Special relativity differs by saying that $dr$ and $dt$ are no longer invarients if you take them separately. Instead the only invarient is the proper time, $d\tau$, defined by:
$$ c^2d\tau^2 = c^2dt^2 - dx^2 - dy^2 - dz^2 $$
In special relativity all observers will agree that $d\tau$ has the same value, but they will not agree on the values of $dt$, $dx$, $dy$ and $dz$.
This is why we have to talk about spacetime rather than space and time. The only way to construct laws that apply to everyone is to combine space and time into a single equation.
You say:
I'm having a hard time understanding how changing space means changing time
Well suppose we try to do this. Let's change space by moving a distance ($dx$, $dy$, $dz$) but not change time i.e. $dt$ = 0. If we use the equation above to calculate the proper time, $d\tau$, we get:
$$ d\tau^2 = \frac{0 - dx^2 - dy^2 - dz^2}{c^2} $$
Do you see the problem? $d\tau^2$ is going to be negative so $d\tau$ is imaginary and has no physical meaning. That means we can't move in zero time. Well what is the smallest time $dt$ that we need to take to move ($dx$, $dy$, $dz$)? The smallest value of $dt$ that gives a non-negative value of $d\tau^2$ is when $d\tau^2$ = 0 so:
$$ c^2d\tau^2 = 0 = c^2dt^2 - dx^2 - dy^2 - dz^2 $$
or:
$$ dt^2 = \frac{dx^2 + dy^2 + dz^2}{c^2} $$
If we've moved a distance $dr = \sqrt{dx^2 + dy^2 + dz^2}$ in a time $dt$, the we can find the velocity we've moved at the dividing $dr$ by $dt$, and if we do this we find:
$$ v^2 = \frac{dr^2}{dt^2} = \frac{dx^2 + dy^2 + dz^2}{\frac{dx^2 + dy^2 + dz^2}{c^2}} = c^2 $$
So we find that the maximum possible speed is $v = c$, or in other words we can't move faster than the speed of light. And all from that one equation combining the space and time co-ordinates into the proper time!
| {
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Why does the moon sometimes appear out-of-place? Quite often I go out in the morning and I'm in Milton Keynes, so I would expect the moon to rise in the east and set in the west. Sometimes at about 2, 3, 4 o’clock in the morning the moon is low in the east. I was just wondering how that worked out?
| Well, let’s zoom out for a bit and imagine you're off of the Earth and you’re looking at the Earth and the moon from space. So you have the Earth as the bigger of the two bodies sitting let’s say, in the centre and the moon is in orbit around the Earth.
So the moon goes around the Earth and the moon takes a month to do a complete lap of the Earth and get back to where it started, 28 days to do a complete orbit of Earth.
Also, inside the moon’s orbit, the Earth is turning and the Earth takes 24 hours to do a complete circle. So therefore, as the Earth turns then it’s going to see the moon from one side of the Earth go across the sky and then down on the other side. So, you're going to see the moon rise and set. But because the moon is also doing a lap around the Earth, the moon is going to appear at different points in the sky at different times of the day and night. So sometimes the moon will be up during the day.
| {
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Averaging decibels Wikipedia:
The decibel (dB) is a logarithmic unit that indicates the ratio of a
physical quantity (usually power or intensity) relative to a specified
or implied reference level.
If I measure some physical quantity in decibels, then what is the preferred way to calculate the mean of the measured values? Is it enough to simply average them, or should I convert them back to linear scale, calculate the average, and convert it back to decibels (example)? When should I use which approach, and why?
| There are reasons more than "preference" for the averaging. You defined it that way usually because you can get more information from that, particular for those additive quantities.
Suppose you preform a set of measurement at a particular point in space, there are two cases: (a) get the averaged value (b) take the average for the intensity itself, and then converted to decibel.
If you have the quantity in situation (b), you can know how much average energy flux passing through that point. Also, you can know the total energy flowing through that point. This information cannot be obtained from the method (a).
Similar situation for the earthquake, if you take the average for the energy, you can know the total energy released by that particular point, which is important. However, you cannot obtain this information by simply taking the average of earthquake scale.
Sure, as pointed out by Lubos, if the variation is small, these two definitions are basically the same as the $\log$ (any) function is local linear, and you can now have additive quantity again.
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Is H=H* sloppy notation or really just incorrect, for Hermitian operators? I saw it in this pdf, where they state that
$P=P^\dagger$ and thus $P$ is hermitian.
I find this notation confusing, because an operator A is Hermitian if
$\langle \Psi | A \Psi \rangle=\langle A \Psi| \Psi \rangle=\left( \langle \Psi|A\Psi\rangle \right)^\dagger$
or more elaborately
$\int H^\dagger \Psi ^\dagger \Psi dV =\int \Psi^\dagger H \Psi dV $
or in another way ( $\left< X\right>$ being expectation value here, to avoid confusion)
$\left<A\right>=\left<A\right>^\dagger$
but surely this does not imply $A=A^\dagger$?
Look at the momentum-operator $\vec{p}=-i\hbar\nabla $. But $\vec{p}\neq \vec{p}^\dagger$
Is it just sloppy notation and are they talking about expectation values, or are Hermitian operators actually operators that are equal to their Hermitian conjugate? My QM course that I've been taught would state otherwise.
Let me try to illustrate this with another (1D )example:
$\langle \Psi | P\Psi \rangle \\
=\int \Psi ^\dagger(-i\hbar \frac{\partial}{\partial x})\Psi dx\\ =\Psi ^\dagger \Psi |^{\infty}_{-\infty}+i\hbar\int \Psi \frac{\partial \Psi ^\dagger}{\partial x}dx\\
=0+\int \Psi (-i\hbar\frac{\partial \Psi )}{\partial x})^\dagger dx\\
=\int (-i\hbar\frac{\partial}{\partial x})^\dagger \Psi ^\dagger \Psi dx\\
=\langle P \Psi | \Psi \rangle$
Now this shows that the expectation values might be the same, but the operator between the parentheses itself is different in the first and last expression (look at the parentheses + $\dagger$ as a whole). I think the reason why I am confused is because I don't understand how $\dagger$ works on something that is not a matrix (if that is even possible).
| Hermitian in the sense used by physicists doing quantum mechanics is usually meant to be self-adjoint, or, equivalently, with real eigenvalues. That is, for some matrix representation of $A$:
$$ \left( A^{\dagger} \right)_{ij} \equiv \left(A \right)^{\star}_{ji} = \left(A\right)_{ij}$$
where the first equality is the definition of $A^\dagger$ and the second one means that $A$ is self-adjoint/hermitian, which implies real eigenvalues, i.e. $A$ corresponds to an observable.
| {
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Limescale formation at room temperature? There is a large metal container in form of a cube made of stainless steel. It is used for storing water in it for technical uses. The problem is that all joints at the bottom of the container have micro cavities and water leaks through them very slowly. I am thinking of a method to close these micro cavities from inside. And the most reasonable method I can think of is to make them cover by a layer of limescale, as any electrical teapot or heater does. The only problem is that this is a big container, and it isn't easy to heat that amount of water, or to boil it.
So does anybody know any other way to cover the bottom or just micro cavities in the joints by limescale without heating during long time? Maybe some use of electrolysis?
| Use a solution of a suitable polymer. As the solution leaks it will dry in the outside air, solidify and block the holes. You might try Radweld.
| {
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Is there a way to formulate relativistic dynamics in a way that "hides" the finite speed of light? I'm not referring to the dimensional choice that makes $c=1$; rather I'm imagining something more about replacing all references that apparently involve velocities with the appropriate $\gamma$ factors or rapidity. In this description, the kinematic feature that maps to speed would then be defined on $[0,\infty)$, and we wouldn't have to deal with the question of "why is the speed of light finite, and has the value that it has?" $^*$
$^*$ Or, at least we wouldn't have to deal with it as a fundamental question about the universe.
| If you need equations without velocities, just with $\gamma$, just replace any $v$ by $c\sqrt{\gamma^2-1}/\gamma$ and you're done.
Incidentally, $\gamma$ isn't the only useful function of $v$ that takes values between $0$ and $\infty$. You could also use the rapidity $\eta={\rm arctanh}(v/c)$.
It's good to use $\gamma$ and $\eta$ and in various situations, it simplifies physics or makes it clearer. However, it's still true that $d\vec x / dx^0=\vec v$ and nothing else for the trajectory of a particle, so this basic defining equation of the slope of the trajectory takes the simplest form with $\vec v$! And this "actual velocity" is smaller than $c$ for all massive bodies: this basic statement of relativity is surely not just a convention and can't be hand-waved away by some change of variables. Only one quantity, $v$, deserves to be called the velocity.
Also note that $\gamma$ and even $\eta$, if defined in the easy way above isn't even a vector: it only knows about the magnitude of the speed, not its direction.
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When is the right ascension of the mean sun 0? I understand that the right ascension of the mean sun changes (at least over a specified period) by a constant rate, but where is it zero? I had naively assumed that it would be zero at the most recent vernal equinox, but when I try to calculate the equation of time using this assumption and true sun positions, all my values are about 7.5 minutes larger than they should be.
When (at what date and UT time) is the right ascension of the mean sun 0? And why?
| Im not entirely convinced that light-time is the culprit here. The Earths orbit is not circular, its elliptical, as the Earths orbital eccentricity is non-zero. Because of this, the Earths distance to the sun throughout the course of the year is not constant, and therefore the time delay due to lights finite speed will not be a constant either. If you have computed this value of 7.46988 to a high enough accuracy, there should be some variation throughout the course of the domain, a year. According to you, there is not.
Might I suggest a possible explanation? Its trivial. And if Im right, its a naive mistake. Are you taking into account your longitudinal offset from Greenwich (or from your time-zones defining longitudinal meridian)?
You said Right Ascension, so your location on the Earth should not be a variable, as Right Ascension is measured with respect to the stars. If you truly are measuring RA then Im at a loss for help and you can disregard my previous paragraph.
As I understand it, any body that passes through the Vernal equinox has a RA of zero. However, there is a distinction between mean and apparent vernal equinox. There are precessional and nutational factors involved.
Furthermore, one thing I am VERY fuzzy on, is whether or not we use the true vernal equinox at all, or if RA is based on the original vernal equinox defined thousands of years ago, known today as the First Point of Aries. They are well out of synch today.
These previous two paragraphs, if not taken into account, could cause significant errors that appear constant in the short term.
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Are galactic stars spiraling inwards? Are the stars in our galaxy spiraling inwards towards the center, or are they in a permanent orbit?
And if we are heading towards the center then what is the rate of this process?
I started wondering this while watching this documentary: http://www.youtube.com/watch?v=zKE4Bt8ylhM
| All stars in our galaxy are in stable elliptical orbits around the galactic centre. But they are not all moving in the same direction with the same speed... meaning there is a random maxwellian distribution of velocities among the stars.
What this effectively means is (like the animation that Crazy Buddy posted) although there is a net effective attraction between the stars and the centre of the galaxy, whenever two stars get close to each other, they exert a sort of "gravitational drag" force, which acts like a frictional force and ends in a net slowing down of the interacting bodies.
When the star then slows down, it doesn't have the speed to maintain it's current orbit, so it will move into a nearer orbit closer to the centre of the galaxy. This is the general mechanism of how stars in a galaxy are "collapsing" into the centre.
There's no need to worry about falling into any (hypothesised) super massive black holes though. Any of these processes take a really long time, and by the time we are close enough to the centre of the galaxy to worry about it, our sun will be well into it's red giant stage and the earth would already have been consumed by it. :)
P.S - To clarify my first paragraph, assuming there are no other stars orbiting the centre of the galaxy, then we will never spiral into it. That effect is only caused by the presence of other stars in the galaxy.
EDIT: @JohnRennie pointed out that because of conservation of momentum, the lighter stars will tend to gain energy and the heavier ones loose energy in dynamical friction interactions. This will tend to push the heavier stars closer to the centre and the lighter stars further out. His answer to this same question points this out.
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How would natural (resonant) frequencies affect amplitudes? I read $y=A\sin(2\pi ft)$, where $A$=Amplitude, $f$=Frequency, $t$=Time and $y$=$Y$ position of the wave.
Since natural frequencies only take the most effect when they are close to the frequency.
How would one natural frequency and several natural frequencies affect the equation?
Would I be correct in thinking it's something to effect of: y=Y_Position*NaturalFrequency
where Y_Postion is the first equation and NaturalFrequency is similar to the first equation but with a low amplitude?
| If you are driving a resonant linear system, which is characterized by a natural frequency $f_n$ and quality factor $Q$, with your specified sinusoidal input $y_{in}$ of amplitude $A$ and frequency $f$, the steady-state output $y_{out}$ will be:
$$ y_{out} = \frac{A}{1+j \frac{1}{Q} \frac{f}{f_n} - \left(\frac{f}{f_n} \right)^2} $$
This equation gives a complex phasor quantity, which describes the amplitude and phase (with respect to the input) of the output.
The higher the $Q$ of the system, the higher the output when the driving frequency is near resonance.
At low frequencies (compared with the natural frequency) the output just tracks the input, while at high frequencies, the output falls off like $1/f^2$ and lags the input by a half-cycle (180 degrees).
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How to find out the maximum radius of a hole that can keep water stay in a container by water viscosity? Assume I have a inverse cone which holds 200ml water. I am going to cut the tip of the cone to create a small hole. How to calculate the maximum radius of the hole that the water will still stay in the container?
| If you have a water drop with radius $r$ then the pressure difference between the inside of the drop and the outside is:
$$ \Delta P = \frac{2\gamma}{r} $$
To calculate the hole size you need to work out the pressure at the bottom of the cone and equate this to the pressure calculated using the expression above. The pressure at the bottom of the cone depends on the depth of the water, not the total volume of water in the cone. If the depth of water in the cone is $h$ then the pressure is $\rho g h$, where $\rho$ is the density of the water at the temperature you're working at, and $g$ is the acceleration due to gravity ($\approx$ 9.81 m/sec$^2$). Equating this to the first expression gives:
$$ \rho g h = \frac{2\gamma}{r} $$
or:
$$ r = \frac{2\gamma}{\rho g h} $$
For example at STP $\gamma \approx 7.3 \times 10^{-2}$N/m and $\rho \approx$ 1000kg/m$^3$, so if the depth of the water in your cone is 10cm the maximum radius of the hole is 0.1mm.
Note that this is the maximum radius for which there is no flow at all. For holes a bit bigger than this the flow may be so slow it's difficult to measure.
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Why are so many forces explainable using inverse squares when space is three dimensional? It seems paradoxical that the strength of so many phenomena (Newtonian gravity, Coulomb force) are calculable by the inverse square of distance.
However, since volume is determined by three dimensions and presumably these phenomena have to travel through all three, how is it possible that their strengths are governed by the inverse of the distance squared?
The gravitational force and intensity of light is merely 4 times weaker at 2 times the distance, but the volume of a sphere between the two is 8 times larger.
Since presumably these phenomena would affect all objects in a spherical shell surrounding the source with equal intensity, they travel in all three dimensions. How come these laws do not obey an inverse-cube relationship while traveling through space?
| These physical phenomena (gravity, Coulomb force) are forces caused by an object you can consider pointlike. That is, for the inverse square law to hold, the object emits the force uniformly in all directions from one point.
That means that at any distance (call it $R$) from the object, you'll feel the same force as you would anywhere over the surface of a sphere whose radius is that distance.
The surface of a sphere is $2$-dimensional, not $3$-dimensional, and its area goes like $R^2$. The larger the radius, the larger the surface of the sphere, and the further away you are from the source. So the strength of the source is inversely proportional to the surface area of the sphere.
| {
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Gaussian type integral with negative power of variable in integrand How can we compute the integral $\int_{-\infty}^\infty t^n e^{-t^2/2} dt$ when $n=-1$ or $-2$? It is a problem (1.11) in Prof James Nearing's course Mathematical Tools for Physics. Can a situation arise in physics where this type of integral with negative power can be used?
| In this form, you won't get it naturally in physics because the Gaussian factor $\exp(-t^2)$ appears in the normal distribution or the harmonic oscillator etc. but with the addition of $1/t$ or $1/t^2$, one gets a non-normalizable and non-integrable (divergent) modification of the original Gaussian. However, one could surely engineer a situation in which the integral would arise in a different form.
The integral
$$\int_{-\infty}^\infty t^n \exp(-t^2)$$
strictly speaking vanishes for $n$ odd because the integral is an odd function. Even more accurately, it is divergent for $n\leq -1$ and that's the strict answer to the OP's question. The behavior of the integrand near $t=0$ is simply $t^n$ whose indefinite integral is $t^{n+1}/(n+1)$ which is singular for $t=0$ and $n\leq -1$.
Less strictly, we may compute the analytically continued principal value – for an odd negative $n$, we have to add the absolute value to $t$, to get nonzero – and the integral
may be converted via substitution $t^2=T$ i.e. $t=\sqrt{T}$ and $dt=dT/2\sqrt{T}$ to
$$2\int_{-\infty}^\infty T^{n/2-1/2} \exp(-T) dT/2 = (n/2-1/2)! $$
It's the Euler integral for the gamma function that I wrote as a generalized factorial. The factors of $2$ cancel. For $n=-1$, the argument of the factorial is $-1$ and the factorial itself is truly divergent – a pole. Note that it's exactly the point at which the strict principal value would vanish because of the odd nature of the factorial. For $n=-2$, the argument of the factorial is $-1.5$ and the result is
$$ (-1.5)! = (-0.5)! / 0.5 = 2\sqrt{\pi} $$
The strict integral would be divergent, even with the principal value, but the results above, $\infty$ and $2\sqrt\pi$, are likely those that would be obtained by a physicist who would "naturally" adjust the integral.
| {
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Why is lightning more rare during snow storms than rain storms? Lightning and thunder during a snow storm is uncommon. As far as I know, more uncommon than during a typical rain storm. Why is this? I speculate it might be one, or both, of the following two ideas, one having to do with a change in the dielectric, the other having to do with a change in the catalyst.
| The mechanism by which lightning is produced is complex and imperfectly understood, but we know moisture is important in two respects:
*
*Heat is released when water vapor in the air condenses into liquid drops, and this heat helps provide energy to the thunderstorm.
*Interactions between supercooled liquid water droplets and ice crystals in the upper atmosphere (about 15,000 to 25,000 feet above sea level) are needed to generate the electrical charges that accumulate until a lightning bolt occurs.
Cold winter air typically doesn't contain a lot of moisture and so isn't conducive to thunderstorms.
Air instability is also important to thunderstorm formation. It commonly occurs when warm air near the surface of the Earth rises due to convection on a sunny summer day when the ground gets hot and warms up the air immediately above it. As the hot air rises, cooler air descends to replace it. If conditions are right, strong updrafts can form that quickly move the warm, humid surface air up to the higher reaches of the troposphere, where the water vapor in the air cools and condenses to fall as rain (or ice, if it's cold enough). These updrafts are a hallmark of thunderstorms--the strong upward motion of the air encourages the interactions between water droplets and ice crystals that can lead to lightning.
In winter, cold surface air temperatures and reduced sunlight mean there's less surface heating, less convection, and thus fewer opportunities for thunderstorms.
| {
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