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What's the reason double-slit experiment can't be explained by edge effects rather than quantum interference? Say we had exactly this...
But instead, it was a PING PONG GUN (imagine as table tennis players use to train),
throwing out PING PONG BALLS. The two slits are say 20 cm wide, and the observing screen is say 5m distant.
If the ball goes through the EXACT MIDDLE of a 20cm slit, it will travel in a perfectly straight line and make a "dot" on the observing screen.
If the ball travels nearer and nearer to the left or right edge of a slit, the flight path will bend slightly towards that side. For example, due to electrostatic force (rather like how a vertical pour of water from a faucet will bend slightly as your hand approaches).
Note that this is not some sort of fantasy; you could very easily organise for the ball path to bend slightly when near an edge, using either electrostatic force, magnetic force, aerodynamic factors or other forces, with the correct material of balls and slits (substitute small metal balls and slits of magnetic material .. whatever).
Indeed, you could trivial arrange so that precisely this famous image
is the outcome.
This is the "trivial mechanical bending" explanation of "all this interference pattern stuff".
Can you help me understand in a clear way, What is the explanation of why this is not at all the explanation?
| Nope. The important thing about the double slit experiment isn't that you find a wavy pattern on the screen, it's that the output on the screen is not equal to the output you get with only one slit open, plus the output you get with only the other slit open. The particular pattern that one slit makes by itself doesn't matter.
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What does diagonalization mean here? In a gravity theory in spacetime, the metric has signature $− + +· · ·+$. Concretely this means that the metric tensor $g_{μν}$ may be diagonalized by an orthogonal transformation, i.e. $$(O^{-1})_{μ}^{\;a} = O^a_{\;μ}$$and $$g_{μν} = O^a_{\;μ}D_{ab}O^b_{\;ν}$$ with positive eigenvalues $λ^a$ in $D_{ab} = \textrm{diag}(−λ_0, λ_1, . . . , λ_{D−1})$.
The construction above, which involved only matrix linear algebra, allows us to define an important auxiliary quantity in a theory of gravity, namely $$e^a_μ(x) ≡\sqrt{λ^a(x)}O^a_μ(x).$$ Using this tetrad we can write $g_{μν}(x) = e^a_
μ(x)η_{ab}e^b_ν (x)$ ,
In the bold above:
*
*Why would this mean that the metric tensor may be diagonalize by an orthonormal transformation?
*What is meant by diagonalization here (mathematically)?
| Let's go step by step as it seems you're missing some fundamentals.
We know from (linear) algebra, that a symmetric bilinear form can be transformed to a diagonal matrix with elements $e$ on the main diagonal $e\in \{0,1,-1\}$. The tripel counting the amount of times each number appears is called signature. If you didn't know that, check this.
Now, a metric tensor is a symmetric bilinear form, so we know it has a transform, so that we get its signature. By the way, from Sylvester's law of inertia follows, that the transform is an orthogonal transform, if the matrix is invertible.
I hope this answers the first question. I didn't completely get what your second question was... Diagonalisation is always the same thing.
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Classical gravitational wave Consider the electric field surrounding an electron. The electron moves. A wave propagates at $c$ reflecting the electron's movement. This is an electromagnetic wave. Consider a mass. Classically surrounded by a gravitational field. The mass moves. The field deforms and it propagates at $c$.
*
*Is this not a gravitational wave?
*Why do people talk as if only general relativity predicted gravitational waves?
| You're not wrong! Gravitational waves do occur classically, and that's because general relativity is a classical theory! In Newtonian gravity, the gravity field doesn't "deform," nor does it have a speed - it's just a force that acts at a distance, instantaneously. So you're entirely correct - that is one way that gravitational waves could be created.
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Electric Displacement Vector How do I interpret what electric displacement vector is? I know that it exists and I know it's an equation but I'm not able to really understand or interpret what it is.
$$\oint_A \mathbf{D} \cdot \mathrm{d}\mathbf{A} = Q_\text{free}$$
| Well think that in vacuum, the electric field vector $\bf E$ is a good measure of the strength of the electrostatic field.
But if you are measuring inside a medium, say there are some charges around where you are measuring, now the presence of those charges affects the value of $\bf E$ from the external electric field. These charges will arrange themselves differently (depending on how free they are to move in the medium) and their configuration will change the electric field from $\bf E$ to a new value $\bf D.$
Also the expression you write, which is a special case of Gauss' Law, is telling you that the difference in electrical field intensity between the inside and the outside of the medium, is linked to the charges concentrated on the surface. Indeed, since this integration is independent on the chosen surface $\bf A,$ you can chose a cylinder whose axis is normal to the medium's surface, and with one face inside and the other outside. Then the integration will equate to the integral of charge populating the surface.
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Functions of state and definition of equilibrium Consider this -
A function of state is any physical quantity that has a well-defined value for each equilibrium state of the system.
(Blundell and Blundell - Thermal Physics)
My question is - Are equilibrium states not defined by the function of state variables themselves?
That is,function(s) of state defines a equilibrium state and function(s) of state are themselves characterised by having a unique/well-defined value at an equilibrium point.
Is this not a circularity?
| Whenever i say that the pressure of a gas is 2 atm then it is known that at the time i say this, the gas is in equilibrium. Well what i want to say that we define a state of any system when it is in equilibrium and by equilibrium i mean thermal equilibrium which means every knid of equilibrium like mechanical, chemical, etc. Now to define a state we use thermodynamic terms like temperature, pressure and density.Now if we look at the state variable like internal heat, entropy, etc we will find that they are depending on these thermodynamic terms which i mentioned above. For example take the example of internal heat. The first law of thermodynamic says that the cyclic integral of heat is equal to the cyclic integral of the work done from which on further solving we get the expression of $dU=\delta Q-\delta W$. so we can see that $dU$ is depending on the temperature, pressure, etc. Conclusion:(1) A state can only be defined at a thermodynamic equilibrum.(2) A property (temperature, pressure, etc) can be defined as any quantity that depend on the state of the system and conversely the state is specified or described by the property.[Fundamentals of thermodynamics, Sixth Edition].(3) The state variables are dependent on these properties. Hence logically from the above three statements we can say that the state variables are defined at the thermodynamic equilibrium or conversely thermodynamic equilibrium state is defined by the state variables
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Why are four-legged chairs so common? Four-legged chairs are by far the most common form of chair. However, only three legs are necessary to maintain stability whilst sitting on the chair. If the chair were to tilt, then with both a four-legged and three-legged chair, there is only one direction in which the chair can tilt whilst retaining two legs on the ground. So why not go for the simpler, cheaper, three-legged chair? Or how about a more robust, five-legged chair? What is so special about the four-legged case?
One suggestion is that the load supported by each leg is lower in a four-legged chair, so the legs themselves can be weaker and cheaper. But then why not 5 or 6 legs? Another suggestion is that the force to cause a tilt is more likely to be directed forwards or sideways with respect to the person's body, which would retain two legs on the floor with a four-legged chair, but not a three-legged chair. A third suggestion is that four-legged chairs just look the best aesthetically, due to the symmetry. Finally, perhaps it is just simpler to manufacture a four-legged chair, again due to this symmetry.
Or is it just a custom that started years ago and never changed?
| This should be more of an engineering/economics question than a physics one. The statement "... the simpler, cheaper, three-legged chair?" Is false. 120 degree angles have historically much harder to manufacture than 90 degree angles, the small cost of an extra leg is easily offset by the increase in manufacturing costs have having to cut innumerable 120 degree angles.
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Do we know why there is a speed limit in our universe? This question is about why we have a universal speed limit (the speed of light in vacuum). Is there a more fundamental law that tells us why this is?
I'm not asking why the speed limit is equal to $c$ and not something else, but why there is a limit at all.
EDIT: Answers like "if it was not.." and answers explaining the consequences of having or not having a speed limit are not -in my opinion- giving an answer specifically to whether there is a more fundamental way to derive and explain the existence of the limit.
| $\hspace{50px}$The above picture I drew to expand on Kostya's wonderful answer.
Basically, imagine people who measure height of buildings in degrees of angle of the buildings' visibility from the certain fixed distance. This is not at all unreasonable if you fix the distance C large enough compared to the building heights'. However, for taller buildings you'd notice that their angular height is not additive. Also maximal possible angular height is fixed at absolute value of 90 degrees.
This is very similar to the way humans measure speed: we picked a certain measure "distance/time" that makes sense for smaller speeds, but for higher speeds it's not additive. Also, there is the unreachable "maximal" speed, the speed of light.
However, the above problem is solely due to the wrong choice of measuring speed. The "right" choice measuring speed is "rapidity", as explained by Kostya. And rapidity is both additive and unlimited.
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Why do most office chairs have 5 wheels? (Inspired by Why are four legged chairs so common?)
I've been wondering for a while... Why do most wheeled office chairs have 5 wheels?
My guess would be that while stability vs. simplicity results in 4 legs, adding mobility to the equation may result in the need for 5 wheels.
Edit: This question is about a mobile chair
| I don't know the history, I am just speculating. The chair in the picture has features that you don't find on a typical 4 point chair: It has casters, it has a swivel seat, and it has a springy reclining back. All of these make the chair more likely to tip over without warning. Lets assume that you roll the chair forward so all the casters are rotated to the back, then lean back. At the moment that it tips onto the two back most casters, the base will swivel to the least stable orientation, and the casters will rotate to form a smaller circle. In a 4 point chair, the base can rotate from diamond to square, and the casters rotate to make the square smaller, with no warning. A rigid 4 point chair is very predictable, and yet we have all tipped them over. Add the swivel and casters and the chair becomes an unpredictable killer. To compensate, the base circle must be increased and/or the number of casters must be increased.
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Where does the energy go if a ball hits the ground WITHOUT bouncing? so if a very un-bouncy object hits the floor and crashes into it rather than bouncing, where does most of the energy go? Does it simply turn into heat from friction? Or does it go into the Earth's movement by an imperceptible amount?
| The majority of the energy is dissipated as mechanical deformation (as Jon Custer has stated).
Visualizing the situation can help a lot. All matter is somewhat elastic - there's no such thing as infinite elasticity. When an object collides with something, the force of the collision takes time to spread out across the object. Imagine a slow-motion view of crash-testing a car. It crumbles as it decelerates, and is unlikely to "bounce back," although this does happen with enough energy. A bean bag is also unlikely to bounce, as it is designed to deform and absorb the collision.
Hard objects are very likely to bounce because they don't deform as much (or as permanently). So instead, the waves that propagate through the object will be "corrected" and the matter will fall back into place. This causes the object to bounce, and produce an audible sound. The lack of deformation is what causes the bounce. This is why billiard balls bounce so well - if ceramic deforms, it breaks.
Heat is the final source of dissipation. Back to the car wreck in slow motion, the bending metal will create measurable heat. Also the bean bag will produce heat with friction between the internal components. Do this sometime: get a piece of metal and bend it back and forth over and over. It will get hot. Same thing with a piece of plastic. The deformation directly causes heat to be produced.
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Why do some materials shrink when their temperature increases? In my vision it would seem quite logical that all materials expand when temperature rises. Because the particles get more energy and travel larger distances when moving. But apparently there are some materials that tend to shrink as temperature increases. Why? Which ones?
| I imagine alot of negative expansion coefficients occour at phase transitions. Probably the temperature increase causes the material to overcome an activation energy and a change in lattice structure occours. In these cases one would not expect these processes to be reversible.
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Chern-Simons theory The Chern-Simons 3-form is given by
$\omega_3={\rm Tr} \left[ A\wedge dA+\frac{2}{3}A\wedge A\wedge A\right]$
where $A$ is a connection one-form in the adjoint representation of a non-Abelian gauge group.
My differential geometry is rather rusty (and this is new to me too) hence my questions;
$A$ is a 1-form. By definition of the wedge product between a $p$ form $\alpha$ and $q$ form $\beta$ we have $\alpha\wedge\beta=(-1)^{pq}\beta\wedge\alpha$. So we should have $A\wedge A=-A\wedge A=0$.
Why is this not the case?
Next question; I want to calculate $d\omega_3$ Does the fact that everything is inside the trace effect my calculation? In other words does the differential operator pass through the trace and only act on the forms?
| This is an example of lie-algebra valued 1-forms. Actually you may write explicitly, $ A = A_{\mu} ^a T^a dx^{\mu}$. Since the generators also anti-commute so we get the result. And for the same reason sometimes you will find expressions like $[A,A]$ in literature for your term.
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A mass in a rotating tube A mass rotates on a horizontal surface inside a frictionless hollow tube with a angular velocity omega. The only force acting on it is a force $N$ with which the tube pushes the mass.
It is expected that the mass would move away from the center of rotation due to centrifugal force, which is a fictitious force in the frame of reference attached to the tube. But in the frame of reference in which the tube rotates, there are no forces in radial direction. So what actually happens, and why?
| In the absence of external forces a mass would move in a straight line. To maintain a circular motion the mass needs to be "constrained" and this is done by the contact normal force $N$ acting by the tube to the mass. This force is real.
The magnitude of this force is found by the difference between the straight line path and the actual path.
Lemma
If an object moves along a path, and at some instant the tangent vector is $\hat{e}$ and the normal vector is $\hat{n}$ then the velocity and acceleration vectors are decomposed as
$$ \begin{aligned} \vec{v} & = v \hat{e} \\
\vec{a} &= \dot{v} \hat{e} + \frac{v^2}{r} \hat{n} \end{aligned} $$
Where $v$ is the speed, $\dot{v}$ the acceleration rate and $r$ the radius of curvature of the path
To find the force of constraint all you need is $$\vec{N} = m \vec{a}$$
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Charge inside a charged spherical shell
*
*If I were to put a negative charge inside a negatively charged spherical shell, will it move to the center?
*Electric field inside the shell due to the shell is zero (Gauss's Law), would that mean the charge inside the sphere faces no force?
But, that doesn't make intuitive sense to me. If the negative charge was near the walls of the sphere, wouldn't the charges on the near wall push the negative charge to the centre as the force due to the charges on the wall closest to it is higher than that form the walls further away from it.
*What about in the case of a ring? Will the charge move towards the center?
| Since there are no charges inside a charged spherical shell . This means the net charge is equal to zero. So magnitude of electric field $E$=0. So there is no net force. So magnitude of net force =0. No movement towards centre.
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Same equation, different meanings I went into a physics classroom today and saw this equation written on the board:
$$
E = \frac \sigma \epsilon
$$
At first I thought it referred to the electric field $ E $ between 2 parallel plates of charge density $\sigma$ separated by a material of permittivity $\epsilon$. However, I then realised it was actually the definition of the Young's Modulus $E$ for a material that has a strain $\epsilon$ when a stress $\sigma$ is applied to it!
So the same equation has two completely different meanings in two completely different areas of physics, with the symbols defined differently (ignoring symbols to show which variables are vectors etc). Is this just a coincidence resulting from the huge number of 3-variable equations in physics, or have the symbols intentionally been defined like this? Is there a deeper meaning? Are there other examples like this?
| Coincidence, nothing deep I'd say. Note that the equation representing the electric field modulus depends on the units you've picked and as such putting so much emphasis on the exact characters appearing in the eq. is senseless. Note that it's possible to form many physics equalities and equations involving 3 characters. E, epsilon and sigma are quite used. One could calculate the probability for this coincidence to have happened. I guess a full answer to your question could involve such calculations.
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Dependence of average speed of molecules of gaseous mixture We know that the average speed of gases in a single gas chamber is given by $\sqrt{8RT/\pi M}$
where R is universal gas constant,T is temperature,M is molar mass of gas.
But what if we mix two gases in any ratio say 1:1 and then try to find average speed of anyone of the gases.
Will the both gases have have same average speed or different?If same ,how will it be calculated and if different will it be given by same above formula?
| If the two gasses are mixed then they will reach the same temperature, with the lighter molecules moving faster than heavier ones.
In the initial equation, 8, R, T and pi would all be the same. Since the gas molecules have different masses the speeds must therefore be different.
If a tank hits my car side on, my car goes flying. If my car hits a tank side on, the tank hardly moves. Its high mass causes its velocity to change little. So it's normal that the lighter molecules get banged around faster when gasses are mixed.
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Backyard experiments to falsify the Flat Earth theory I recently became aware that the flat Earth theory still exists in the 21st century, and has colored the views of a friend of mine. Roughly speaking, the tenets are:
*
*The Earth is a flat disk, with the south pole blown up into a circular "ice wall" where one would expect Antarctica to be.
*
*The sun and moon are either spheres or disks floating above the earth and moving in a spiral pattern with the seasons.
*NASA is part of a conspiracy to conceal the truth, and sends us animations and faked photo shoots.
The Flat Earth Theory is a scientific theory, in the sense that it makes falsifiable predictions about the universe that can be seen to match observation (or not). What are some good arguments or backyard experiments that could convince a layman that in fact the Flat Earth Theory is false?
A similar question and useful related reading is What is the simplest way to prove the Earth is round?, but it focuses more on the theoretical and conceptually simple side, applicable to the proverbial "numskull cousin". This question is more focused towards convincing a doubtful scientific person using (preferably low-tech) experiments and observations.
| Related to the Andrea di Biagio answer. Here is a typical flight path for the Beunos Aires-Auckland route. The distance is approximately 10,300 km by the shortest route along a sphere. Direct flights are offered by Air New Zealand and take 11h40m - an average speed of 882 km/h.
Looking at the map you present, the distance from Auckland to Buenos Aires looks to be about twice this by the most direct route. Therefore the plane would have to fly supersonically. But Jumbo jets and the like, cannot and do not fly supersonically.
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Why does Diffracted Light cone diameter change in relation to angle of light beam? I have a question about light diffraction. Take a look at these images of the Pantheon oculus.
Now what I don't understand, in the first picture, the light is coming in from overhead and forms a really huge light diffracted light cone on the floor.
In the second picture, the light enters at another angle, the cameraman also is at another location BUT the beams that pass through the hole look straight and not cone-like.
Why does the light beam become thinner as it enters at the slightly different angle, compared to the first picture?
Also at different times of the day, the casted light's radius keeps shrinking as the sun sets. Can you guys provide a comprehensive explanation/ visual demonstration why this happens so?
| Neither image displays diffraction. They both illustrate shadowing.
The second photo was apparently shot on a clear sunny day. All of the rays come from the sun. Some are blocked, those that aren't go straight down to the floor.
The second was probably taken on a hazy day. Rays originate from the haze; they come from all directions, and make their way to the floor.
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The many faces of electromagnetic waves In my waves and optics class, we have learned several ways to treat electromagnetic waves: light rays (geometric optics), electromagnetic plane waves, spherical waves, cylindrical waves (2D). One thing still confuses me: How can one determine which method to use when approaching a problem? In other words, when can I treat the wave as light rays? When is the electromagnetic wave a plane wave? When is it a spherical wave?
| I will answer this question in two parts:
*
*first comparing plane waves, spherical waves and cylindrical waves (which are really the same thing as I will explain).
*how this relates to ray optics (which is completely different).
Plane waves, spherical waves and cylindrical waves are 3 different examples of doing a decomposition of a wavefield. The method is that you represent a general wavefield in terms of an infinite sum over a complete set of basis functions that have some property that is convenient. Plane waves are the most widely used since they are the same as the Fourier decompositions. Plane and cylindrical waves would be more convenient if you have some symmetry in your system, for example for a point source you would use spherical waves, for a line source most likely cylindrical. All of these 3 decompositions contain the same physics (namely light as an electromagnetic wavefield), but represent it in a different mathematical way. Which one you choose depends on the situation and you choose the most convenient one. You can solve most problems with either of these, it just gets messier.
Ray optics is different, since this represents a different approximation. Instead of as a wavefield you represent light as a ray. This for example does not allow you to describe interference effects and is less general than the wavepicture, can still be useful for systems like lenses etc.
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If there are long-lived elements in the Island of stability, why are they not present in Nature? To my understanding, some (but not many) physicists speculate that the Island of stability may contain long-lived elements, as in a billion or so years. But couldn't we rule that out just by the nonexistence of such elements in Nature?
| Because no rational process can make them. I've been over the tables, and there are only a couple of possible reactions to get there for any nuclei, and they require two rare ones. Alpha particle capture just isn't going to cut it. Look at the curve; you need more neutrons.
We remember that all elements heavier than iron have primary sources as neutron star collisions and supernovae. The s-process stops around lead or gold (depending on who you ask) leaving the r-process to extend to higher levels. However, the r-process itself is only so rapid and must be bounded by the nuclear decay rates themselves. In order to reach the island of stability by the r process, it would be necessary to pass through the boundary layer around element 109 where several isotopes must be passed through in a row that possess half lives measured in seconds to milliseconds. However my tables show it doesn't even get here but caps out at Neptunium*. Either way, this leaves induced fusion as the only possible pathway.
Unfortunately the induced fusion reactions require rare nuclei. You don't get much chance of a concentrating event of certain elements and try again. Yeah sure you could have a planet collide with a giant star a few hours before said supernova; however consider the odds of getting the result of that exceedingly rare event (consider the timescale) in the gasses required to form our own solar system.
So no, not expected at all.
*One of the Plutoniums (244) has enough half life long enough to last until it gets here, and it's not found that way either, but by direct production in Uranium ores.
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Normal ordering in string theory: Polchinski vs. all others Polchinski defines normal ordering in string theory as:
$$:X^\mu(z,\bar z)X^\nu(w,\bar w): = X^\mu(z,\bar z) X^\nu(w, \bar w) + \frac{\alpha'}{2} \eta^{\mu\nu} \log |z-w|^2$$
and for more complicated expressions one obtains the normal ordered expression via Wicks theorem (p. 39).
In the CFT-Context (e.g. compare with "Conformal Field Theory" by Di Francesco) normal ordering is defined as "regular part of the OPE".
How can we see that these definitions are equivalent?
| The usual definition of normal ordered product is:
$$:X^\mu(z,\bar z)X^\nu(w,\bar w): = X^\mu(z,\bar z) X^\nu(w, \bar w) - \langle X^\mu(z,\bar z) X^\nu(w, \bar w) \rangle $$
As you said, this is the regular part of the OPE, since only the divergent part of two operators gives non vanishing contribution to the correlator. Of course
$$\langle X^\mu(z,\bar z) X^\nu(w, \bar w) \rangle=- \frac{\alpha'}{2} \eta^{\mu\nu} \log |z-w|^2$$
| {
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Why are $S = -k_B\sum_i P_i \ln P_i$ and $S = k_B \ln\Omega$ equivalent? This might be a silly question, but I don't see the equivalence relation between these two equations. Could somebody explain to me how to derive one from the other? Thanks in advance!
| Citing Wikipedia here,
In what has been called the fundamental assumption of statistical thermodynamics or the fundamental postulate in statistical mechanics, the occupation of any microstate is assumed to be equally probable (i.e. Pi = 1/Ω, where Ω is the number of microstates); this assumption is usually justified for an isolated system in equilibrium.
Since $P_i = 1/\Omega$, obviously $\ln P_i =-\ln\Omega$ and since $P_i$ is probability, then the $\Sigma_i P_i=1$
| {
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Helmholtz decomposition allows incompressible flow with an irrotational component? A vector field can be written in terms of irrotational and a divergence-free components. Using a 2D velocity field as an example,
$ \vec v = -\nabla \phi + \nabla \times \vec \Psi$
Where $\vec \Psi$ is a vector potential, which in fluid mechanics is only guaranteed to exist if we're working in two dimensions so that $\vec \Psi = (0,0,\psi)$, where $\psi$ is called the stream function.
There are many sources I can find that say that an incompressible flow ($\nabla \cdot \vec v = 0$) simplifies to $ \vec v = \nabla \times \vec \Psi$ Here is one such example, although the Wikipedia article on stream functions implies the same.
This seems incorrect to me, since taking the divergence of both sides of this equation
$ \nabla \cdot \vec v = -\nabla \cdot \nabla \phi + \nabla \cdot \nabla \times \vec \Psi$
simply yields the Laplace equation, $\nabla^2 \phi = 0$. This means that as long as $\phi$ is a nonzero harmonic function, I can have a velocity field in an incompressible fluid that has an irrotational component. Is there an additional constraint that forces $\nabla \phi = 0$ in order for $\nabla \cdot \vec v =0$?
| You can use a trick to solve that: Instead to take the divergence, take the curl as follows:
$\nabla \times \vec v = \nabla \times (-\nabla \phi + \nabla \times \vec \Psi)$
but $ \nabla \times \nabla \phi=0 $, so
$\nabla \times \vec v = \nabla \times (\nabla \times \vec \Psi) = \nabla(\nabla \cdot \vec \Psi)-\nabla^{2} \vec \Psi$, (just a vector identity)
but $ \nabla(\nabla \cdot \vec \Psi) = 0$, because the left hand side only has a component perpendicular to the plane, so this gradient must be a gradient along the axes orthogonal to the plane. But we know that the third component of the field $\vec \Psi$ has dependency in $x$ and $y$ only, that is, $\vec \Psi = (0,0,\psi(x,y))$ and $\frac{\partial \psi}{\partial z} =0$, therefore
$$\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} = -\frac{\partial^{2} \psi}{\partial x^{2}} - \frac{\partial^{2} \psi}{\partial y^{2}} $$ Integrate,
$$ \int (\frac{\partial v}{\partial x} + \frac{\partial^{2} \psi}{\partial x^{2}})dx = \int(- \frac{\partial^{2} \psi}{\partial y^{2}} +\frac{\partial u}{\partial y})dy, \ \ \ then $$
$$ v + \frac{\partial \psi}{\partial x} + f(y) = u -\frac{\partial\psi}{\partial y} + g(x), \ \ \ where \ f(y) \ and \ g(x) \ are \ some \ function. $$
A trivial solution is then when $f(y)=g(x)=0$ and both sides of the equation are set to zero, from where we get our stream functions
$$ u = \frac{\partial\psi}{\partial y} \ \ \ \ \ \ \ \ v =- \frac{\partial \psi}{\partial x} $$
| {
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Coefficient of friction and practical experience of sliding The classical model of friction has a coefficient of friction depend only on the materials, but not area, and the force proportional to the normal force and coefficient of friction. So a given object on the same surface has the same friction whether it is supported by full bottom area or small legs as long as the materials are the same.
However every child knows that on a slide one goes faster if one lays down on their back compared to sitting on their butt. The slide is obviously still the same and since jackets usually extend below butt, the other material is also the same. So the friction should be the same as well, but it clearly isn't. So what is going on here?
Note: I mean typical stainless steel or fibreglass laminate slide, not ice, which is soft enough to complicate the matter further.
| When the child lies down there is less deformation of the slide's surface than when he is sitting. Whether the deformation is relatively deep and short in length, or more shallow and extends far beyond the child, it is always nonzero. The force required to move the trough of this deformaton down the slide opposes the component of the child's weight directed down the slide.
| {
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Conditions for the tension to vary in the rope What are the conditions for the tension to vary in the rope. I have read below conditions
1. rope has to have some mass
2. rope is accelerating
I get the 1st one, but I am not sure if I get the 2nd condition.
If a $10\ \text{N}$ force accelerates a mass-less role, what will be the tension?
| It also depends on whether or not the rope is inextensible. Assume it isn't. You tie one end to a rigid wall and pull on the other end with a constant tension $T$. Then at some time $t_0$, you start to pull with a slightly higher tension $T_1 > T$. It will take some time $\tau$ of order $\frac{L}{c}$, $c$ being the typical celerity of mechanical waves inside the rope, for the rope to reach equilibrium. During that time the tension inside the rope will be inhomogenous and somewhere between $T$ and $T_1$. After equilibrium is reached the tension in homogenous and equal to $T_1$.
This is indeed linked to the fact that between $t_0$ and $t_0 + \tau$ the rope is accelerating.
If you repeat the same experiment with an inextensible rope, its rigidity is infinite thus the mechanical waves propagate at a celerity $c$ that tends to infinity. When you pull on it with a tension $T_1$, the rope immediately reaches equilibrium at tension $T_1$ everywhere inside the rope. This is linked to the fact that the rope don't accelerate in this case.
You also need the rope to have a lineic mass for this explanation to stand as the celerity $c$ of mechanical waves inside the rope is also inversely proportional to the square root of its lineic mass.
| {
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What will be final velocity of three charges $q$, $q$, $2q$? What will be final velocity of three charges q, q and 2q kept along an equilateral triangle of side r at infinite distance.
All three masses are equal.
I tried to conserve Total Energy
$$\frac{2kq^2}{r} + \frac{2kq^2}{r} + \frac{kq^2}{r} = \frac{mv^2}{2}+ \frac{mv^2}{2} + \frac{m(v_{2})^2}{2})$$
$$\frac{5kq^2}{r} = m(v^2 + \frac{v_{2}^2}{2}$$
$$\sqrt{\frac{10kq^2}{rm}-2v^2}= v_{2}$$
Conserving momentum gave
$$v_{2} = v\sqrt{2(1+\cos\theta)}\tag{1}$$
where $\theta$ is angle b\w velocity of q & q.
When I tried to make some graph predicting their motion and speed in different direction was like hell.
I could not get any further.
As helped by Fire I used COM along y direction.
$$my + my - my' = 0$$
$$ y' = 2y$$
$$dy'/dt = 2dy/dt $$
$$ v_{2}= 2 v_{y}$$
$$v_{2}= 2 v\cos\left(\frac{\theta}{2}\right)\tag{2}$$
I can't imagine if charges would move along straight line or in curves. Will $\theta$ change or not? Can all three have same velocity at infinity?
Someone told me it uses Taylor series.
| This is actually a particular case of the three-body problem (https://en.wikipedia.org/wiki/Three-body_problem), but with repulsion, rather than attraction. There are very few exactly solvable cases of the three-body problem, among them - the Lagrange's case, where the three bodies are at the vertexes of an equilateral triangle at each moment. Therefore, I suspect you should try to prove for the initial conditions of your problem that the distances between the charges remain equal (although I cannot be sure that this is so). Some homework...
| {
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Is the conservation of probability in the Schroedinger's equation unique? The Schroedinger's equation can be viewed as a diffusion equation with imaginary constants $a$ and $b$ satisfying,
$$\quad \Psi_t=a \cdot \Delta \Psi-b \cdot V(x,t) \cdot \Psi
\tag{1}
$$
However if $a$ and $b$ are positive real coefficients, we get the standard diffusion equation.
Now it's standard fair to prove,
$$\cfrac{d}{dt} \int |\Psi|^2 \ dr^3=0
\tag{2}
$$
if $a$ and $b$ are imaginary. Is this true for the standard diffusion equation?
My (educated) guess is no. For the one dimensional case, the derivative can be brought inside and we get,
$$\int 2 \cdot \Psi_t \cdot \Psi \ dr^3
$$
Using the known expression for $\Psi_t$ we get,
$$\int \left(2 \cdot a \cdot \Psi_{xx} \cdot \Psi-2 \cdot b \cdot V \cdot \Psi^2\right) \ dr^3 \tag{3}$$
Using integration by parts and noting that $\Psi$ needs to go to zero at infinity (this is self evident right?) we get,
$$\int \left(2 \cdot a \cdot \Psi^2-2 \cdot b \cdot V \cdot \Psi^2\right) \, dr^3 \tag{4}\, .$$
The first term is positive definite. The second term could easily be positive as well, so in general, the integral is time dependent.
Can a general proof for or against this be shown? In addition, assuming my argument is correct, are there cases where the integral in $(2)$ isn't time dependent?
| You are misreading the celebrated analytic continuation to diffusion of the Schroedinger equation. This is probably possible by notational overkill induced by superfluous dimensions potentials, and absorbable constants that should not deny you any firm ground to stand on, but evidently do.
The crucial piece to appreciate is that, for diffusion, (in 1d, rescaled, free flow),
$$
(\partial_t -\partial^2_x)~ \phi (x,t)=0,
$$
which follows from the continuity equation (conservation of matter/substance whose density is $\phi$),
you have, ipso facto,
$$
\partial_t \int dx ~\phi =0,
$$
the time derivative being a total divergence.
When you complexify/analytically continue, $\phi (x,it)\equiv \psi(x,t)$, to the free Schroedinger equation,
$$
(-i\partial_t -\partial_x^2)~\psi(x,t)=0,
$$
the complex conjugate such combines with it to yield
$$
\partial_t (\psi^* \psi) =(\psi^* \partial_x^2 \psi -\psi \partial_x ^2\psi^*)=\partial_x(\psi^* \partial_x \psi -\psi \partial_x \psi^*),
$$
also a total divergence, to which you may add inessential cancelling potential terms for free. The corresponding continuity equation then automatically produces
$$
\partial_t \int dx ~ \psi^* \psi =0 .
$$
by inspection.
Why on earth would you expect to replicate this structure on the real φ, so without the analyticity structure option of doubling up ?
The answer for the analog of its square, (2), is of course not .
If you are keen on structure, reassure yourself by trying this.
| {
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publication ethics What does it mean if somebody publishes the same concept of my paper with some changes without citing it although the paper has been published 5 years before his, and the concept was clearly first introduced in my paper? It is quite frustrating to know that nobody cares about your research, especially if you are a young researcher. How to cope with this disappointing feeling?
| If you try to just "cope with the disappointing feeling", you are setting yourself up for the same thing to happen again and again. You need to stand up for your rights, if the situation allows this.
But it's unclear from your question, whether this was a case of deliberate plagiarism, or just parallel invention.
| {
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how is a pendulum clock's time and the time period of the pendulum in it related? I'm working out how much time a pendulum clock will gain or loose due to change of the length of the pendulum due to temperature. so far I've got, new time period, $$T_2=T_1(1+\frac12\alpha\Delta T)$$
due to $\Delta T$ change in temperature, when the coefficient of linear expansion of the pendulum material is $\alpha$. $T_1$ is the time period when there is no change in temperature.
Now, I can not understand how the time period of the pendulum and the time measured by the clock is related. Please help me with this and also mention if I am going correct or not.
| Assume that each swing advances the second hand by one second on the dial. You have a pendulum which is supposed to have a period of exactly one second.
If the pendulum has got longer in length it will have a longer period and so it will take longer for the pendulum to advance the second hand by one second on the dial.
So the clock will run slow.
Just to add to the possible confusion you may come across the term seconds pendulum. Such a pendulum has a period of two seconds. The reason for such a name is that for a lot of clock mechanisms the advance is done every time the pendulum swings through its equilibrium position ie twice per oscillation.
| {
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Why is the Pythagorean Theorem used for error calculation? They say that if $A = X \times Y$, with $X$ statistically independent of $Y$, then
$$\frac{\Delta{A}}{A}=\sqrt{ \left(\frac{\Delta{X}}{X}\right)^2 + \left(\frac{\Delta{Y}}{Y}\right)^2 }$$
I can't understand why that is so geometrically.
If $X$ and $Y$ are interpreted as lengths and $A$ as area, it is pretty easy to understand, geometrically, that
$$\Delta{A} = X\times\Delta{Y} + Y\times\Delta{X} + \Delta{X}\times\Delta{Y}$$
Ignoring the term $\Delta{X}\times\Delta{Y}$ and dividing the both sides by $A$ ($= X \times Y$), that expression becomes
$$\frac{\Delta{A}}{A} = \frac{\Delta{X}}{X} + \frac{\Delta{Y}}{Y}$$
which is different from
$$\frac{\Delta{A}}{A}=\sqrt{ \left(\frac{\Delta{X}}{X}\right)^2 + \left(\frac{\Delta{Y}}{Y}\right)^2 }$$
which looks like a distance calculation. I just can't see how a distance is related to $\Delta{A}$.
Interpreting $A$ as the area of a rectangle in a $XY$ plane, I do see that $\Delta{X}^2+\Delta{Y}^2$ is the how much the distance between two opposite corners of that rectangle varies with changes $\Delta{X}$ in $X$ and $\Delta{Y}$ in $Y$. But $\Delta{A}$ is how much the area, not that distance, would vary.
| The formula
$$\frac{\Delta{A}}{A} \approx \frac{\Delta{X}}{X} + \frac{\Delta{Y}}{Y} $$
is an approximation because you are ignoring $\Delta X$$\Delta Y$
A better approximation would be
$$\Delta A=\frac{\partial A}{\partial X}\Delta X+\frac{\partial A}{\partial Y}\Delta Y$$
Since errors always add we take the absolute magnitude of $\frac{\partial A}{\partial X}$ and $\frac{\partial A}{\partial Y}$, i.e
$$\Delta A=\bigg |\frac{\partial A}{\partial X}\bigg |\Delta X+\bigg |\frac{\partial A}{\partial Y}\bigg |\Delta Y$$
Since it is always tricky do deal with modulus functions, another work around would be squaring individual errors so that they stay positive
$$(\Delta A)^2=\bigg (\frac{\partial A}{\partial X}\bigg)^2(\Delta X)^2+\bigg (\frac{\partial A}{\partial Y}\bigg )^2(\Delta Y)^2$$
$\frac{\partial A}{\partial X}=Y$ and $\frac{\partial A}{\partial Y}=X$
This will give the required form, this is the root mean squared deviation (standard deviation)
| {
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Can an absorbed photon be emitted as two photons? I am taking an intro to astronomy class, and have touched upon absorption and emission lines and etc, the prof asked this question in class and got me thinking.
I would want to say no, because one photon should stay as one photon, mass cannot be created.
Could someone please confirm if this is true or false, and why? or what conditions must be met?
| Yes, absolutely! This is the phenomenon of so-called non-linear light scattering. These include an absorption of a photon and its remission as several photons, or as a photon of different frequency, or a combination of a photon and a phonon, etc. (You may want to look up Raman scattering and Mandelstam-Brillouin scattering.)
Note also that the Rayleigh scattering of the photons with the same frequency does not actually involve the absorption.
| {
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Does shaking a kettle whilst boiling increase the temperature faster than a statically-placed jug? Water molecules move faster at higher temperatures.
*
*Does shaking a boiling kettle whilst it is in the process of boiling water increase the rate of rising temperature ?
*Is it worthwhile to do so to save time ?
It's dangerous, so I wouldn't try it myself.
| Coherent motion does not add to the temperature; so you would have to shake it violently, with random motions.
Consider sound in air - this is a coherent motion - when you can no longer make out the sounds in a closed room, the energy of the sound waves has been transformed into heat.
| {
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Action with self-dual field strength It is said that writing down an action in presence of a self-dual field strength is subtle and not known till date. The familiar example people give is that of type IIB super-gravity which has a self-dual 5-form field strength. Can someone elaborate on what exactly the subtlety is.
| Suppose you have a self-dual five-form field strength $F_5=*F_5$. The kinetic term of this field strength is proportional to
$$
\int F_5\wedge*F_5=\int F_5\wedge F_5=-\int F_5\wedge F_5
$$
where I used $A\wedge B=(-1)^{pq}\, B\wedge A$ for $p$-form $A$ and $q$-form $B$ in the second equality. So you can conclude that
$$
\int F_5\wedge*F_5=0\,.
$$
This is the subtlety you mentioned. People usually impose self-dual condition on $F_5$ after obtaining equations of motion. But this is not a satisfactory resolution, because we want to obtain self-dual condition as a part of equations of motion.
(Edit) The above answer was my naive understanding. For more detailed explanation, see page 313 of Becker, Becker and Schwarz.
| {
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Wavelength for imaging an ultracold atom? I was reading Stamper-Kurn's article Experimental Methods in Ultracold Atomic Physics (link). In the imaging section (page 13), he mentions:
Cold atoms are conventionally probed by optical imaging. Probe light at a well defined optical frequency
is sent through the atomic gas and imaged onto a camera.
What determines this "well defined optical frequency"? For example, why would one choose 780 nm, which seems to be a common wavelength?
| The most common type of imaging in these systems is absorption imaging- you just shine light on the atoms that matches a strong electronic transition, so the atoms scatter as much light as possible, and then get an image from seeing how much light is scattered. Essentially, you are looking at the shadow cast by the atoms.
So this frequency is, naturally, dependent on the specific atoms that you are using. For Rubidium it would indeed usually be at 780 nm, which corresponds to the so-called "D2 transition" between the S state and one of the P states of the one valence electron. Light near this frequency is also used for laser cooling, so basically any Rb experiment will need to have this laser anyway.
| {
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Weight of mass falling through liquid
*
*If I place a container of fluid on a scale and drop a non-buoyant mass into the fluid, will the scale read less as long as the mass is in motion downward as compared to when the mass is at rest on the bottom of the container?
Part two:
*
*Would the shape of the container effect the resultant weight while the mass was in motion (falling)?
| The scale's reading will jump up when the mass impacts the liquid, then gradually decrease to a value larger than the original as the mass decelerates. This means that the scale could read the combined weight of both the liquid and the mass before the mass reaches the bottom if the mass is slowed to a constant velocity by then.
Think of the limiting cases:
*
*Right before the mass impacts, the mass and the scale aren't interacting, so the scale should be normal.
*After the mass impacts, it is being deaccelerated, meaning that the liquid (and therefore the scale) must be providing a repulsive force greater than gravity for the net acceleration to be negative. So the reading on the scale is high.
*After the mass reaches constant velocity, the liquid & scale must be providing a force equal to that of gravity to keep the net acceleration zero. This net force is less than the force in 2 but greater than 1, so the reading on scale is less than in (2) and greater than in (1).
The exact profile of the reading on the scale would probably be highly non-linear (going from the high point to the medium point), but you could derive a good approximation by modeling the resistive force as a simple $F = - bv$ resistive force and solving for v as a function of time based on some initial conditions and for a resistive constant b. I think you'd get an exponential decay in this case.
The shape of the vessel might influence the resistive force if it's odd, but in general it will depend on the liquid more.
| {
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Density parameter (cosmology) as a function of redshift Hopefully I'm missing some very basic algebra for this question. In essence, I need to derive the following:
$$
\Omega=\Omega_{1}\frac{(1+z)}{1+\Omega_1{z}}
$$
Now, I proceeded from the Friedmann equations as follows for a matter dominated Universe (and for $\Lambda=0$ and $k\neq{0}$):
$$
\frac{H^{2}}{H_{0}}=\Omega_{m}a^{-3}+\Omega_{\kappa}a^{-2}
$$
Knowing that $a=1/(1+z)$ I can re-arrange this to:
$$
\frac{H^{2}}{H_{0}}=\Omega_m(1+z)(1+z)^2+\Omega_{\kappa}(1+z)^2
$$
$$
\frac{H^{2}}{H_{0}}=(\Omega_m+\Omega_mz)(1+z)^2+\Omega_{\kappa}(1+z)^2
$$
Introducing the fact that $1-\Omega_{m}=\Omega_{\kappa}$:
$$
\frac{H^{2}}{H_{0}}=(\Omega_m+\Omega_{\kappa}+\Omega_mz)(1+z)^2=(1+\Omega_mz)(1+z)^2
$$
This is where I feel I am almost there but I've stumbled a little as to my next options. Any tips or advice would be brilliant!
| You almost have it. The last step is
$$
\Omega = \frac{\rho}{\rho_\text{c}} = \frac{\rho_0\,a^{-3}}{\rho_\text{c,0}}\frac{\rho_\text{c,0}}{\rho_\text{c}} = \Omega_m\,(1+z)^3\frac{H_0^2}{H^2},
$$
where we used the critical density
$$
\rho_\text{c} = \frac{3H^2}{8\pi G},\qquad \rho_\text{c,0} = \frac{3H_0^2}{8\pi G}.
$$
The result follows immediately from what you already derived.
| {
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Can observation change entropy? I don't know whether this even makes any sense, but if 'observation' can be considered as 'recieving and reading information', can an act of observation (of a system) change (increase or decrease) its entropy?
| I’m going to go out on a limb here and “answer” instead of comment because I struggled with the concept of entropy for so long until I came to this simple conceptual definition: entropy is the measure of how far away a system is from equilibrium.
From a Physics 101 point of view, observation cannot change this. If a box of gas is at equilibrium, no amount of observation is going to change it to a state less in equilibrium. Your observation might give you perfect information about the microstate the box of gas is in, but that does not make it more likely to transition to a microstate not in equilibrium.
Here it is helpful to consider equilibrium in two ways. One is that in equilibrium you cannot extract work from the system. A box of gas in equilibrium cannot be made to move a piston or anything else. Two is to consider that in equilibrium the system is in a macrostate that has many more microstates similar to it (in equilibrium) than any macrostate not in equilibrium. Though the movements of the gas molecules are random, the sheer number of possible equilibrium microstates makes it highly probable that the system will evolve to another equilibrium microstate rather than into a non-equilibrium microstate, of which there are far, far fewer.
With information theory there are other considerations, but these are beyond the physics 101 level, so if you’re just starting out and are confused by entropy (like I was) and just want a simple way to think about what it describes, then the above is hopefully helpful.
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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volume of the air bubble in the water How does the depth affect the volume (the radius) of an air bubble in the water, if the temperature and density of the water are constant. Is there any relation combining this?
Can I say that $dh/dt=dr/dt$?
| Using the ideal gas law:
$P(h) = P_0 + \rho_w g h$, ${V_o P_0} = {V(h) P(h)}$ which gives
$V(h) = {V_o P_0}{1 \over P_0 + \rho_w g h}$.
| {
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When is the spin of an electron changing? An electron can have a half spin up or down. The up spin can become a down spin to lose his weak charge. But when are electrons changing their spin?
| If initially you had the spin projection in a certin (eigen)state and then you apply a spin-dependent interaction, the spin wave function becomes a superposition of up and down eigenstates. Measurements will find the up-state with a probability $P_{\text{up}}$ and the down-state with the probability $P_{\text{down}}=1-P_{\text{up}}$, the mean value being between up and down.
| {
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Does the existence (now proved) of gravitational waves imply the existence of Gravitons? I studied the theoretical part about the Gravitational waves in General Relativity (linearization of gravity and small perturbations of the metric and so on).
But I was wondering about: since electromagnetic radiation is composed/carried by Photons (or better:the EM force), shall gravitational waves be composed/carried by Gravitons?
In the end:do gravitational waves imply the existence of gravitons? Or it's something unrelated and off topic?
| While the detection of gravitational waves does not directly imply that gravitational force exists in the realm of the particle-wave duality, it does provide definitive link to the existence of the graviton. I sat in the assembly hall at UMCP during the celebration of the gravitational wave discovery and this very question was asked to the panel. Their response was poignant and quite direct:
“We of course know that the graviton exists, so we did not need this detection to add validity to the claims. The burden of proof is a heavy one and one that we simply do not have the current means to tackle.”
| {
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Fermions, different species and (anti-)commutation rules My question is straightforward:
Do fermionic operators associated to different species commute or anticommute? Even if these operators have different quantum numbers? How can one prove this fact in a general QFT?
| Fermions $f_i,\,f_j$ with respective momenta $\pi_i,\,\pi_j$ satisfy the equal-time canonical anticommutation relations $$\left\{\ f_i,\,f_j \right\} = \left\{\ \pi_i,\,\pi_j \right\} = 0,\,\left\{\ f_i\left(t,\,\mathbf{x}\right),\,\pi_j \left(t,\,\mathbf{x'}\right)\right\} = i\hbar \delta_{ij} \delta \left(\mathbf{x},\,\mathbf{x'}\right),$$where the second $\delta$ is a Dirac delta. The $i=j$ special case is a generalisation of a theory of a single fermon $f$ of momentum $\pi$. Why do the $i\neq j$ cases use anticommutators instead of commutators? Because we want our rules to be invariant under $f_i \to \sum_j M_{ij} f_j,\,\pi_i \to \sum_j \left(M^{-1}\right)_{ji} \pi_j$ for invertible choices of the matrix $M$. There's no consistent way to achieve this by using commutators sometimes. A similar explanation is available in terms of the ladder operators.
| {
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Proving charge on outer surface of parallel plate capacitor must be zero If we have two conducting plates, with charge $Q$ and $-Q$, why is the charge on the outer surfaces of each conductor zero?
I've been trying to wrap my head around the problem. Firstly, don't excess charges on a conductor spread out towards the surface, leading to a contradiction?
How can I use Gauss' law to prove this statement (even though it seems false to me)?
| I may possibly be very wrong, but wouldn't it be the same as when you get an electron and mesh it with a proton getting a neutron? If you have the same amount of an opposite charge, the charge will disappear..
| {
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Domain structure and ferromagnetism and reversal of polarity According to the domain theory, ferromagnetic substance when kept in the external strong magnetic field, it undergoes magnetisation by rotation and becomes a permanent magnet thereby even after removing the external magnetic field. I got several questions regarding it:
*
*If I keep varying the external strong magnetic field periodically (like the rotation of electric dipoles of water molecules in food that is heated up using a microwave oven) then all the domains will undergo rotation in different directions at different times, Will the substance heat up? Can it happen that it will heat up to such an extent that the domain structure collapses i.e. it reaches the curie temperature? And also why did it even heat up? Where did this work come from?
*When a ferromagnetic substance is kept in a strong magnetic field, it becomes a permanent magnet with net magnetic moment in that direction, right? So now if I invert, the magnetic field then if we still consider that its ferromagnetic then the domains should again rotate (now whole magnet is just a single domain) and the polarity of the permanent magnet that it had became earlier should be reversed, but intuitively I feel that this won't happen instead the magnetic (ferromagnetic substance that became a permanent magnet) will rotate and align itself in the direction of the magnetic field i.e. it will rotate by 180 degrees. Which will happen, why and why not the other way round?
| As far as I know, once the substance becomes a permanent magnet, it does what all magnets placed in an external magnetic field would do; which is: The north pole of the magnet rotates towards the south pole of the field and vice versa. And since the magnet being made out of substance having free electrons, it does heat up on prolonged exposure to changing magnetic field due to Eddy Currents
| {
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Does Birkhoff's theorem hold inside the event horizon? Can Birkhoff's theorem be used to say that the blackhole exterior and interior sections of Kruskal-Szekeres's solution (or coordinate transformations of it like Gullstrand–Painlevé coordinates, etc.) are unique all the way down to the singularity? Or are there different options for how to extend the exterior Schwarzschild solution beyond the event horizon?
The wikipedia entry states that the exterior solution is unique, but doesn't comment on the interior.
| Ben Crowell's answer is right. I hope I can clarify a little. If one starts from the following two assumptions:
*
*$R_{ab} = 0\;$ (i.e. field equation in vacuum)
*The metric can be written in the form
$$
{\rm d}s^2 = g_{uu}(u,v) {\rm d}u^2 + 2 g_{uv}(u,v) {\rm d}u {\rm d}v + g_{vv}(u,v) {\rm d}v^2 + f^2(u,v) \left( {\rm d}\theta^2 + \sin^2(\theta) \, {\rm d}\phi^2 \right)
$$
then it can be shown (Birkhoff) that there exist coordinates in which the metric takes the Schwarzschild form. Adopting the usual coordinate names $t,r,\theta,\phi$ it is not necessary to assume that $r$ is a spacelike coordinate; the result is valid both inside and outside the horizon.
Item (2) above is, of course, an attempt to capture the notion of "spherical symmetry" without imposing any further assumptions. One might provide further arguments to show why this form must hold if the situation has rotational symmetry about two different axes.
| {
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Doesn't rotational KE of a rolling marble change if there is no friction to provide torque? The question arise from the following situation:
A marble at the border of a uniform bowl begins rolling within it from rest. There is enough friction in the first half the bowl for the marble to not slip, but there's no friction in the other half. Find the height reached by the marble, measured from the bottom of the bowl.
I read the solution, but I can't undestand the following:
When marbel arives to the bottom, it has both translational an rotational KE. The solution says when marble stops at the maximun height, its rotational KE is still the same as when it was at the bottom (because there's no friction to provide torque). Does it mean that marble is still rolling? Could anyone explain why would it be wrong if I propose that RKE is zero at the maximun height?
| The marble begins to slip instead of roll; thus it continues to move upwards until gravity converts all of the kinetic energy the linear motion into potential energy - then it will begin to reverse course.
| {
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Is the net force conventionally shown in a free body diagram? Is it standard convention to display the net force vector on a free body diagram? Internet searches seem to give mixed results.
| I'm not aware of any such convention. You can always show the net force vector acting on a free body as long as it is clearly labeled as such, to avoid confusion with any other applied forces.
I personally wouldn't include a net force vector unless there was a good reason to, like to illustrate some accompanying discussion.
| {
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When I open a window to air out the room, how does the smell disperse? Let's say I'm in a room with some kind of noxious stink, possibly of flatulent nature. The quickest way to right the world that comes to mind is to open a window. When I open a window, how do the stank particles leave the room?
| Gases diffuse from higher concentration to lower concentration because this process of diffusion increases the gas's entropy.
Every thing in the universe, in general, aims for two things : Minimum energy and maximum entropy.
Entropy is the degree of randomness of a substance. A gas in a container has low entropy because the probability of finding that gas in the universe is less but when the container is opened the gas diffuses into the atmosphere and this diffusion increases the probability of finding that gas in the universe and thus, increasing the entropy.
To move a little more into science, The Gibbs energy change of the system must be -ve in order for the process under observation (say diffusion) to be spontaneous or naturally occurring.
∆G = ∆H - T∆S
Here, ∆G, ∆H and ∆S are the change in the gibbs energy, enthalpy and entropy of the system respectively.
(enthalpy is the heat content of the system at constant pressure).
Now, when gas diffuse from higher concentration to lower concentration the heat change is negligible but the entropy change is highly +ve (∆S = +ve).
The net result is the decrease in the gibbs energy of the system.
Therefore, Diffusion of a gas from higher concentration to lower concentration is a spontaneous or naturally occurring process.
| {
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On geometry of light's path in the Universe Is it possible for us to see our own galaxy from different perspective, as path of light emitted from our galaxy is curved by any possible ways and travelled back to our eyes even if it take much time?
| I know the context is bending of space by heavy bodies, but that is lensing effect and we know lenses do not reflect, however small their focal length is.
Even if it was possible to reflect/turn back good amount of light somehow, it would be next to impossible to see a meaningful image.
When moon reflects sunlight, we see moon, not the sun!
We can see sun's image in water, so, it would be only possible if there was a mirror type of object in universe and happen to be positioned just right for us. Which we know is not there.
| {
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How do you tell whether a force acting on an inclined plane is going up or down in its perpendicular component to the plane?
I'm practicing mechanics, and I had to resolve the following forces perpendicularly to the inclined plane in order to work out the reaction force (plus the weight of the ball)
But I cannot tell whether the 5N force is going up or down when resolved perpendicularly to the inclined pline. According to the book, the answer is up. I thought maybe because the force acts at 45 degrees to the plane, and so therefore if it's 45 degrees or more, then the force has to be acting upwards
However, for this question (the angle of the plane to the horizontal is 30 degrees, it's not very clear in the picture)
the 20N force acts downwards in the component perpendicular to the plane, whereas if I use my aforementioned reasoning, I would expect it to act upwards, since it would have to be more than 45 degrees to the plane to act downwards.
So I don't really understand how I am to determine whether a force's perpendicular component to a plane is up or down when it is acting on an inclined object, and my attempts to do so have contradictions.
| Definition: The "tail" of a vector is the end that is straight (no arrow).
Definition: The "head" of a vector is the end with the arrowhead.
For each force, draw a right triangle such that:
*
*The original force line is the hypotenuse
*Draw a vector with the tail starting at the tail of your original force, pointing parallel to the plane.
*Draw a vector with the tail starting at the head of the vector from step 2, perpendicular to the plane, ending at the head of the original vector.
The direction of the vector from step 3 gives you your answer.
| {
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The sign of the current flowing in a circuit I was doing the following problem:
And I was asked to find Iy.
I found Iy to be 2.64 using KCL. However, the right answer was negative 2.64.
Is it negative only because there is a dependant voltage source with "+ -" ? And why must it be negative? Does "-" in the final answer play a significant role? If I put positive 2.64 instead of negative, would that be wrong?
Also, here is another question
I found V1 to be 11.9. However, the right answer was - 11.9.
Can someone clarify for me when do you put negative sign?
| Kirchoff's current law (KCL) states that "The algebraic sum of all currents entering and exiting a node must equal zero". By algebraic care must be taken to choose a convention whereby you set currents flowing into the node as positive (+ve) and subsequently those flowing out of the node as negative (-ve) or vice versa. This should sort you out. Use this reference too.
| {
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Good reference on the parametrization of $SU(3)$ and $SU(N)$ For the 2-dimensional $SU(2)$ matrices, there is a fairly general parametrization formulation:
$s_2=\begin{bmatrix}
e^{i\alpha}\cos(\theta) & -e^{-i\beta}\sin(\theta) \\
e^{i\beta}\sin(\theta) & e^{-i\alpha}\cos(\theta)
\end{bmatrix}$
For the 3-dimension $SU(3)$ and the higher-dimension $SU(N)$ matrices, what are the most general parametrization formulations? I will appreciate any references, either a book or a review paper, on this subject oriented toward the physics/engineer applications.
EDIT:
This is to clarify my question and make it more relevant to the physics. I am looking for the kind of parametrization that is generally applicable to the problems in physics, especially being intuitive and geometrically accessible. For example, for the $SU(2)$ matrices given above, $\theta$ parameterizes the coefficients of any orthonormal states, $\alpha$ and $\beta$ parameterize the phases of the orthonormal states. The orthonormal states in this kind of general expression are certainly very common and relevant to enormous problems in quantum physics.
| There's this article by Adam Bincer, published in Journal of Mathematical Physics, vol 31 (1990). It is titled Parametrisation of SU(n) with n-1 orthonormal vectors and it's abstract states:
A generalisation to SU(n) of a well-known relation to SU(2) is proposed. It relies on the observation that an element of SU(n) has associated with it in a natural way n-1 orthonormal vectors in $R^{n^2-1}$. The meaning of these n-1 vectors is discussed as they relate to the geometry of the adjoint representation of SU(n).
Unfortunately, although it is now published online, it's not public access, so I can't say more.
| {
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Effect of Gravitational Waves on light? We all know about the gravitational lensing effect. From the analogy of fabric of space time used to explain this concept to laymen like me, I understand that light follows the curvature of spacetime.
Following on that same line of thought process, gravitational waves would cause the spacetime stretch and squeeze. Would it affect the path of light in any way?
| Yes, without going into calculation light path should be affected since gravitational wave is a perturbation in the spacetime metric. If metric is perturbed, then geodesic equation is affected (which governs motion of all particles and photons in spacetime).
The only tricky thing is the fact that GW propagates at the speed of light; so if the "wavefront" travels in the same direction as photons, then they never overtake each other and as seen from observer at infinity, GW that is lagging behind photons will always lag behind photons since they travel the same speed in space. If light "rides" along GW, then observer at infinity should see light paths get influenced the same way spacetime is rippled. But locally light probably cannot "feel" GW.
Would be nice to see explicit computation of this - I am not too knowledgeable about GW itself.
| {
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Is the universe bounded? As I understand it nobody can pinpoint an objective "center" of the universe nor "where" the Big Bang happened. It seems the observable universe is limited by our event horizon at some 14 billion light years and my question is simply: If an astronomer was placed at one of the outermost visible objects would he be looking at a nearly dark sky in a direction away from earth but a star filled sky in the direction of the earth or would he see a more or less evenly lit sky as on earth? If the latter is most likely does it not imply an infinite/unbounded universe?
| The astronomer would see mostly what we see. That does not mean that the universe is unbounded, it just means that the universe is much, much larger than the observable universe -- which is indeed a consequence of current cosmologies which contain cosmic inflation; in these the universe goes rapidly from being in thermal contact (where it comes into equilibrium, this being why the cosmic microwave background is so darn homogeneous) to being thermally disconnected (which is our present experience of the universe) by the rapid expansion of space. So just because she is at the boundary of our observable universe, it does not mean that she is at the boundary of all space.
| {
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Force on sides of a pool I am going to create a large pool of water for a stage production. The volume of water will be 6m x 5.2m x .15m deep. I want to check how much pressure or force will be exerted on the surrounding wooden frame, so I can check the tensile strength of the wood is sufficiently capable of holding the water. I would appreciate assistance on what calculations I need to make?
Many thanks.
| i'm gonna try here, but i'm kinda short of information; such as how much free space do you have surrounding the pool and how thick is the wood on the perimeter. i'm american, so bear with me, i think in SAE terms.
your answer is not very much. each cm of length will have 120 grams of pressure exerted upon it. so each meter will have 12000 grams or 12kg.
this is similar to forming concrete slabs. assuming that the perimeter wood is 1 1/2" (4cm) you will need 2 triangular braces per meter and 1 brace at each corner. (45+45+90=180 degrees) Make the braces equal (.15m) on 2 sides and cut the angled (45 degrees) to fit (21.2cm over all length)(consider that the vertical and horizantal piece will occupy some of this space, so it will be shorter.)
lastly, how will you seal the water in? if you use a plastic liner, it would be best. if you just caulk the joints, the entire form will want to float. wood does float on water. just to be safe, attach braces to the form and to the floor with screws.
if you have more questions, just mail me
| {
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Ladder operators - commutation relations and their properties At the beginning of Fetter, Walecka "Many body quantum mechanics" there is a statement, that every property of creation and annihilation operators comes from their commutation relation (I'm translating from my translation back to english, so it's not literal). I would like to know - how. Defining annihilation operator as:
$$a\left|n\right\rangle=\sqrt n\left|n-1\right\rangle$$
I can derive the $[a,a^\dagger]=1$ equation. But in the opposite direction? I have no idea.
| Actually the commutation relation must be accompanied by the assumption of a vacuum state $|0\rangle$ for which $a|0\rangle = 0$ ($\langle 0| 0\rangle = 1$).
Then the commutation relation $aa^\dagger - a^\dagger a = 1$ implies also
$$
a(a^\dagger)^2 - a^\dagger a a^\dagger = a^\dagger\\
a(a^\dagger)^2 - a^\dagger a a^\dagger = a(a^\dagger)^2 - (a^\dagger)^2 a - a^\dagger = a^\dagger\\
[a, (a^\dagger)^2] = 2a^\dagger
$$
and from this one, again,
$$
a(a^\dagger)^3 - (a^\dagger)^2 a a^\dagger = 2(a^\dagger)^2\\
a(a^\dagger)^3 - (a^\dagger)^2 a a^\dagger = a(a^\dagger)^2 - (a^\dagger)^3 a - (a^\dagger)^2 = 2(a^\dagger)^2\\
[a, (a^\dagger)^3] = 3(a^\dagger)^2
$$
and so on. By induction it follows that in general
$$
[a, (a^\dagger)^n] = n(a^\dagger)^{n-1}
$$
If we apply $[a, (a^\dagger)^n]$ to the vacuum state and assume $a|0\rangle = 0$, we obtain
$$
[a, (a^\dagger)^n]|0\rangle = a (a^\dagger)^n|0\rangle = n (a^\dagger)^{n-1}|0\rangle
$$
which is basically what we want. Indeed, let us denote
$$
|n\rangle = A_n (a^\dagger)^n|0\rangle
$$
where $A_n$ is a (real) normalization constant such that $\langle n| n\rangle = 1$. The above then means
$$
a|n\rangle = n\frac{A_n}{A_{n-1}}|n-1\rangle
$$
so all we have to do is figure out the $A_n$-s. But
$$
1 = \langle n| n\rangle = A_n^2 \langle 0|a^n (a^\dagger)^n |0\rangle = \\
= A_n^2 \langle 0| a^{n-1} \left[a (a^\dagger)^n\right] |0\rangle = A_n^2 \langle 0| a^{n-1} \left[(a^\dagger)^n a + n (a^\dagger)^{n-1}\right] |0\rangle = \\
= A_n^2 n \langle 0| a^{n-1} (a^\dagger)^{n-1} |0\rangle = \dots = A_n^2 n! \langle 0| 0\rangle = A_n^2 n!
$$
and therefore
$$
A_n = \frac{1}{\sqrt{n!}} \;\;\; \Rightarrow \;\;\; \frac{A_n}{A_{n-1}} = \frac{\sqrt{n!}}{\sqrt{(n-1)!}} =\sqrt{n} \;\;\; \Rightarrow a|n\rangle = \sqrt{n}|n-1\rangle
$$
as expected. A similar argument, which you can do as an exercise, retrieves $a^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle$.
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Is it wrong to say that an electron can be a wave? In QM it is sometimes said that electrons are not waves but they behave like waves or that waves are a property of electrons. Perhaps it is better to speak of a wave function representing a particular quantum state.
But in the slit experiment it is obvious to see that electrons really are a (interfered) wave. So can you say that an electron is a wave? And is that valid for other particles, like photons? Or is it wrong to say an electron is a wave because it can be also a particle, and because something can't be both (a behaviour and a property)?
| Electrons are neither particles nor waves - they are electrons.
We say they behave as particles or waves because we are familiar with macroscopic objects having these properties and want to provide a kind of "feel" for what they are in terms we can easily understand. We are the ones that select the experiment that shows aspects of their behaviour. They do not change from particle to wave and back again. Our experiments change.
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Force on current carrying square loop I'm asked to find force on square loop (side a) carrying current $I$, flowing counter clockwise, when we look down x-axis, lying in yz plane. the loop is centered at the origin. The magnetic field is given as:
$\vec{B} = kz\hat{x}$
Its solution states that force on left an right cancel each other .The force on top is $IaB=iak(a/2)$ pointing upward and the force on bottom is$IaB=-iak(a/2)$ also pointing upward .How the force on bottom is upward? (From where minus sign came?). By R.H.R it should be downward.
| I'm not sure I understand how the loop is set but your case must be similar to the left and right side of the loop below.Since the current is flowing in a loop, on the left side it will flow the opposite direction of the way it does on the other side.Using the right-hand rule you will now see that the forces are equal and opposite,as shown in the picture, explaining the minus in your problem.
So, watch out which direction the current is going, always use the R.H.R. and carefully you will have no problem.
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Are there alternatives to steam in thermal power stations? 'A thermal power station is a power plant in which heat energy is converted to electric power. In most of the world the prime mover is steam driven. Water is heated, turns into steam and spins a steam turbine which drives an electrical generator.'
Why is H2O chosen? Shouldn't a liquid with a lower boiling point be used?
Also, why can't the following setups be used? Have they all been duly considered?
1) Expanding Liquid:
In this setup, a chosen liquid is heated, expanded, and used to push pistons, without getting boiled. As a liquid is hardly compressible, shouldn't the expansion generate a tremendous pressure? Furthermore, no latent heat is involved.
2) Expanding Empty Spiked Solid in a Liquid:
In this setup, a specially crafted empty solid is heated in a liquid to give rise to a large increase in volume. The liquid then pushes a piston.
3) Strong Balanced Bimetallic Strips:
In this setup, three sets of bimetallic strips/rods are used to push a piston when cooled (and bent when heated).
4) Expansion of Ice:
In this setup, freezing H2O in cold regions, natural or artificial, expands by about 9% and is used to push pistons.
| Typically, modern gas powerplants use combined cycle and are projected to achieve over to 61 % thermal efficiency, which is to my knowledge far highest of any contemporary technology. The first stage relies of thermal gas expansion similar to a jet engine, the second stage operating at lower temperature is the classical steam cycle. So the answer is yes.
Wikipedia also lists alternative technologies to the gas turbine in the first stage: "Other historically successful combined cycles have used hot cycles with mercury vapor turbines, magnetohydrodynamic generators or molten carbonate fuel cells, with steam plants for the low temperature "bottoming" cycle."
| {
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How precise must the energies match for absorption of photons? According to Quantum Mechanics, in order for an atom to absorb a photon the energy of the photon must be precisely that of a "jump" between energy states of the atom.
How precise must it be?
If I create a photon with an energy within an error of 0.0001% of that of an energy state, will it be absorbed by my atom?
| In atoms the energy levels do not have a precise energy. When you solve Schrodinger's equation for an atom the results are the energy eigenfunctions. However these are functions that are time independent, and they have an exact energy only because they are time independent.
At the risk of oversimplifying, you can regard this as an example of the energy time form of the Heisenberg uncertainty principle:
$$ \Delta E \Delta t \ge \frac{\hbar}{2} $$
If $\Delta t$ is the lifetime of a state then $\Delta E$ is the uncertainty in the energy of that state. For the energy eigenfunctions $\Delta t = \infty$ so $\Delta E = 0$ and the energy is precisely defined.
The point of all this is that in an atom an excited state has a finite lifetime and therefore it has a finite energy uncertainty, and this produces an effect called lifetime broadening. This means transitions to and from the state can occur for photons with a range of energies. The range of energies allowed depends on the energy uncertainty of the state, which in turn depends on its lifetime.
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Finding the Force of two objects - by using Acceleration but only ONE of the given masses? I came across the following question in my physics textbook and wanted to try to solve it:
A 1700 kg car is towing a larger vehicle with mass 2400 kg. The two vehicles accelerate uniformly from a stoplight, reaching a speed of 15km/h in 11 s. Find the force needed to accelerate the connected vehicles, as well as the minimum strength of the rope between them.
I did get the correct answer after a while by finding Acceleration (which was 0.3788m/s^2and then multiplying it by the mass of the truck, which was 2400kg. (The answer should be 910 N)
The part that I don't understand (and I had trouble finding the answer because of this very reason) is: Why is the mass not the sum of both masses of the car AND the truck? Should it not be both, as it specifically mentions that they are tied together, and therefore they technically act as one whole system?
| Forget about the towing vehicle, and focus just on the rope connected to the truck. If you were looking at the scene through a small window, and all you could see was the rope and the truck, what force should there be on that rope in order to accelerate that truck? You don't have to know what is on the other end of the rope - it could have been a winch attached to the earth itself...
Now if you were looking at the force between the wheels of the towing vehicle, and the road surface, that is a force that will be used to accelerate both the car and the truck. If both were of the same mass, and the force on the rope was $F$, then the force on the (sum of the four) wheels would be $2F$ - one $F$ to accelerate the truck, and one to accelerate the car.
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From what wavelength can radiation go through a human body without very much changing? Gamma-rays can go through a body but they will ionize a lot of atoms (I don't know whether some of the gamma photons will go through without any interaction at all?). The same for X-rays. Visible light and infrared (till $1mm$) probably only reflects and absorbs. But perhaps rays from 1 cm and longer (radiowaves) can get through the human body without harming the body or without changing in frequency?
| There are two effects of electromagnetic waves on human body.
*
*Heating effects, occurring mostly around 2.45 GHz (which is the frequency used in microwave ovens)
*Ionizing effects, which can damage human body and cells
Here is a picture of how the radiowaves penetrate the human body according to the frequency.
The legend is from 1 to 10 : Cosmic rays, gamma rays, X-rays, ultraviolet, visible spectrum, infrared, microwaves, radiowaves, very low frequencies, extremely low frequencies.
| {
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Fraunhofer diffraction contradiction Assume two obstacles are complementary. For example, a circular aperture and a circular stop of same radii. My theoretical analysis leads to contradictory conclusions about such situations.
1) Assuming incident light is a plane wave consisting of N sources of Huygens wavelets, obstacle 1 keeps a subset of those wavelets (blocking the others) while the obstacle 2 keeps the complementary subset of wavelets. In other words, the SUM of the diffracted patterns is expected to be a uniform intensity on the screen (all the wavelets = same as what we have when there is NO obstacle). In other words, when the Fraunhofer criteria are met, complementary obstacles produce complementary diffraction patterns.
2) Experiment shows that a circular stop and a circular aperture indeed produce opposite fringes, but the circular stop's diffraction pattern has the Poisson-Arago spot in the middle, while the circular aperture doesn't have a corresponding dark spot in the middle. Therefore these patterns are NOT complementary.
What am I missing ? I somehow have the feeling that patterns are complementary "everywhere except the center" because there is something ill-defined about using Huygens wavelets to predict the center of the pattern... Anyone has input on this ?
| You're mostly correct, about how you expect both patterns together to produce uniform intensity corresponding to a plane wave (this is described as Babinet's principle. What you're forgetting is that it's the sum of the amplitude, including a phase. A circular hole and the complementary shape will produce identical patterns but opposite phase, except for additional background plane wave.
| {
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How does the drift velocity of electrons in a conductor depend on the temperature? How does the drift velocity of electrons in a conductor depend on the temperature?
I have two contradicting views for this.
*
*First, we can say that increasing the temperature of the conductor will increase the kinetic energy of the electrons. Hence, their drift velocity should increase with increase in temperature.
*Or, from the relation $v_d = \frac{eE}{m}T$ ($T$ is the relaxation time) we can say that the drift velocity is directly proportional to the relaxation time. Increasing the temperature will obviously decrease the relaxation time - as collisions will become more frequent - and thus decrease the drift velocity. Hence, an increase in the temperature will cause a decrease in the drift velocity.
So which view is correct?
| Drift Velocity - It is defined as the velocity gained by free electron of conductor in the opposite direction of applied electric field
We know that:-
1- E = F/q = F/e (here, the charge is electron)
Where, E=electric field, F=Force applied, e=charge on the electron
2-F= mass × acceleration = m × a
Now,
F= m × a
here, m= mass of electron
a = F/m
a= eE/m
And,
V= u + at (kinematics equation)
V(drift)= 0 + (eE/m)t (initially, u=O)
So, Drift Velocity= (eE/m)t
where, t= relaxation time and it is denoted by τ (Pronounced as tau)
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Spin 3/2 matrices in terms of Pauli matrices Let $\sigma_i (\frac{3}{2})$ be the three generators of the irreducible spin 3/2 representation of $SU(2)$ (see http://easyspin.org/documentation/spinoperators.html for their explicit forms). Similarly, $\sigma_i (\frac{1}{2})$ are the usual Pauli matrices.
I have reasons to believe that it should be possible to perform a basis transformation such that the spin 3/2 matrices can be related to the spin 1/2 matrices as follows:
$U\sigma_i (\frac{3}{2})U^{-1}=\sigma_i(\frac{1}{2})\otimes B_i$
For some unitary $U$ and some matrices $B_i$. [EDIT: the single unitary $U$ must make this relation hold for all $i$] Evidently, $B_i$ must be 2 x 2 hermitian matrices with eigenvalues 1 and 3, modulo signs.
Is there any reason such a transformation should not be possible? If it is possible, is there an easy way to find a $U$ which satisfies this?
| I can't claim this is a complete argument, but you might consider it...
This is definitely not the standard coproduct, which, coincidentally, I have given to my students as a homework problem in the past.
Consider what Lie algebra both sides of your equation would satisfy: take the commutator of the left hand side, for a given i and j, so, e.g. 1 and 2. So, up to normalizations, the l.h.s. will be $iU \sigma_3(3/2)U^{-1}$.
Edit as per discussion.
The r.h.s., I think, will be likewise $i\sigma_3(1/2)\otimes \{B_1,B_2\}$. Likewise for the other two components.
So the problem reduces to checking if $ \{B_i,B_j\}\propto \eta_{ijk}B_k$ can be solved for Bs with eigenvalues 1 and 3 or their opposites, in any combination. $\eta_{ijk}$ is symmetric in i and j, which must be different, otherwise it vanishes, and k must be the remaining index of the 3. All three ηs have the same value, so take them to be 1.
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Difference between a wavevector and wavefunction I often see both terms used in textbooks, but I am not sure whether I understand the difference between them. Both describe the state of a system, however, they seem different in some ways. From what I have found, what is important in the wavefunction is its direction. Wavevectors, on the other hand, require a magnitude and direction. What is their difference? And how are they related?
| A wave function can be thought to be analogous to a vector in the sense that the set of all functions on a domain forms a vector space. Therefore, each function can be thought of as a vector in the vector space of all continuous functions on that domain. The reason you might have seen these differently is that matrix formulation can be easier to deal with in certain applications such as quantum computing, where the state space is finite (and similarly other finite dimentional systems), you can think of the wave function in terms of a matrix called the densite matrix, and the operators acting on it as linear transforms on that vector space.
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How do I find a stream function given a volumetric flow rate? How do I find a stream function given a volumetric flow rate?
The flow only occurs in one direction, between 2 plates, and I have no knowledge of velocity.
I know that volumetric flow rate = change in stream function between two points but I have no idea how to apply this.
| If you have two stationary plates h apart and know Q. You know that the fluid is not flowing at the boundary of the plates (right up against the plates, this is called the no-slip boundary condition), so at y= 0 and y= h, the velocity is zero.
You also know that the flow is symmetrical about the height y= h/2 because the plates are not moving, and the max velocity and max stream function is at y= h/2. The region from y= 0 to y= h/2 has Q/2 flowing through it (as does the region from y= h/2 to y= h) due to the symmetry.
The difference between the stream function at y= 0 and y= h/2 is equal to the flow through that region, which is Q/2, because Q/2 also flows through the other symmetrical region. We also know stream function at y= 0 is zero. These two facts mean the stream function is Q/2 at y= h/2, which is the max value of the stream function. It goes from 0 at y=0 to Q/2 at y=h/2 back to 0 at y=h. That’s all we know for sure.
If it is pure sheer, then it is linear. Stream = yQ/h from y= 0 to h/2 and Stream = Q - yQ/h from y= h/2 to h.
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Lorentz Force and Apparent Conservation of Momentum Violation Useful for Unidirectional Force? My understanding is that the apparent violation of Newton's Third Law by the Lorentz Force necessitates a description of the system that describes the "missing" momentum as being absorbed/carried by the magnetic field itself. What is not clear to me, however, is why this apparent violation cannot yield an apparent unidirectional force to the system.
For example: In a system with 2 magnets aligned across from one another and a current carrying rod placed parallel to the stage between them (perpendicular to the $B$ field lines), application of current to the rod can result in the Lorentz force vector pointing upwards acting on the rod. If the rod were to be affixed to the magnets themselves by a non-conducting support, preventing the rod from moving upward out of the $B$ field between the magnets, would the upward force acting on the rod not be transmitted to the magnets themselves? This seems to imply that the entire system (rod, electrons within the rod, and the magnets supplying the $B$ field) would experience an upward force.
This can't possibly be the case because that'd basically produce an anti-gravity device, but the math seems to suggest it, meaning I'm missing something fundamental somewhere. If the back-reaction from the Lorentz Force is not being transmitted to the magnets (pushing them "down" to counter the the upward vector force applied to the rod) but instead is being absorbed by the $B$ field, what prevents the system in this setup from experiencing a net upward force?
| The moment you rigidly connect the magnet (or two magnets in your example) to the rod, they become one body.
Therefore a force between them should be treated as an internal force, which means that it cannot move the body as a whole, due to conservation of momentum.
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What is the reason behind why energy must always be conserved, apart from observation? I know that we see in experiments (physical and thought) that energy is always transformed into something else, but what propels our universe to behave this way? What is happening at small levels that only allows conservation as a possible outcome and not destruction of energy?
Update: Based on a given answer stating that energy is mass, I should probably then expand this question into "Why must mass must always be conserved?"
| The answer saying that energy was mass was incorrect. And neither is conserved, and neither is even additive.
In General Relativity you have a Stress-Energy energy tensor. It has ten independent components in any frame. And you can try to extract one of them to be the energy density and three others to give you the components of the momentum density. But that decomposition is locally frame dependent. And even if you did that, you only get a density at every point and since a surface of simultaneity depends on a global frame (which don't always exist and aren't unique when they do exist) trying to add up those densities at different points on a surface of "same time" to get a total energy and a total momentum is hopeless generally.
So there isn't an energy of the universe. And there isn't a momentum of the universe. And even if there were, they could be infinite. And even if it when they are finite, then the mass would satisfy $$(mc^2)^2=E^2-(\vec p c)^2$$ and the mass of the universe would not equal the sum of the masses of the parts. And since the energy and momentum would change over time, the mass usually changes over time.
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How do people go about looking for asymptotic safety in quantum gravity? Do we have (proposed?) methods to look for fixed points in the renormalization group flow of the Einstein-Hilbert action? My understanding of the RG is still somewhat sketchy at this point and I am having trouble understanding how one would go about searching for a fixed point in a theory that's non-renormalizable.
| Here are two papers on the subject:
https://arxiv.org/abs/0805.2909
Investigating the Ultraviolet Properties of Gravity with a Wilsonian Renormalization Group Equation
From the abstract:
We review and extend in several directions recent results on the asymptotic safety approach to quantum gravity. The central issue in this approach is the search of a Fixed Point having suitable properties, and the tool that is used is a type of Wilsonian renormalization group equation. We begin by discussing various cutoff schemes, i.e. ways of implementing the Wilsonian cutoff procedure. We compare the beta functions of the gravitational couplings obtained with different schemes, studying first the contribution of matter fields and then the so-called Einstein-Hilbert truncation, where only the cosmological constant and Newton's constant are retained. In this context we make connection with old results, in particular we reproduce the results of the epsilon expansion and the perturbative one loop divergences.
https://arxiv.org/abs/1601.01800
The Gravitational Two-Loop Counterterm is Asymptotically Safe
Abstract: Weinberg's asymptotic safety scenario provides an elegant mechanism to construct a quantum theory of gravity within the framework of quantum field theory based on a non-Gau{\ss}ian fixed point of the renormalization group flow. In this work we report novel evidence for the validity of this scenario, using functional renormalization group techniques to determine the renormalization group flow of the Einstein-Hilbert action supplemented by the two-loop counterterm found by Goroff and Sagnotti. The resulting system of beta functions comprises three scale-dependent coupling constants and exhibits a non-Gau{\ss}ian fixed point which constitutes the natural extension of the one found at the level of the Einstein-Hilbert action. The fixed point exhibits two ultraviolet attractive and one repulsive direction supporting a low-dimensional UV-critical hypersurface. Our result vanquishes the longstanding criticism that asymptotic safety will not survive once a "proper perturbative counterterm" is included in the projection space.
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Why do we take $h$ as "height from surface to bottom" when calculating liquid pressure? In the following image, pressures of points x, y and z are $P_{x}, P_{y}$ and $P_{z}$ respectively, and they all are equal. My question is, why?
The amount of matter on x is much more than amount of matter on y. Why do they still have equal pressure on them?
| Weight is a downward-directed force; if the question were about weight, there might be differences in the X, Y, Z positions. But, the question is about pressure, which is without direction; pressure is the work required, per unit of volume, to displace the liquid (like making a bubble).
Regardless of position, an injected void in the given scenario will
raise the top level of the liquid, by the same amount, and require the same
amount of work. Imagine poking a cork down to the X position: that takes
some downward force, over a distance d. That force times distance is the work required.
Now move the cork from X, to Y, to Z. That takes no additional work
(because no sideways force is required to do the move).
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Three questions about time Since I don't want to abuse to the ASK QUESTION form, and since those questions are all about time, I decided to write them all here. Hope it's ok.
First Question
Could it be possible (how?) to create some regions, in the Universe, in which the times seems like to stop (for a little while)? I mean: if we could be able to hide some event from the light, would be able the same to hide it from the time?
Second Question
Since there is a deep connection among observers light and the passage of time, wold the time still exist if there weren't any observer? Why?
Third Question
Isaac Newton introduced the idea of a $t$ variabile, which stands for time, to describe how do objects move. But Quantum Mechanics does consider the time in a very different way with respect upon the Classical view of Newton. Nay, according to the Planck scale units the time variable is not applicable. So shall we come back to Newton's period (and earlier) and try to eliminate the $t$ variable?
Namely: could we reformulate the entire Quantum Mechanics theory/formalism without using the time variable? I read that Carlo Rovelli did something about.. but no clue where to find some material!
Thank you all for the attention. If someone thinks I should split the question in three different ones, just tell.
| Question 1:
If a region of space contained no energy (and thus no matter or information), it would not be possible to measure the passage of time in that region. It's important to mention however, that such regions do not exist as there will always be quantum fluctuations.
Question 2:
According to basically all accepted theories, the answer is yes. If everything on earth that can "observe" were to disappear, only to magically reappear 100 years later, we could figure out that it had been 100 years using any number of different methods.
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Relative velocity of light in a medium The speed of light in a medium is independent of motion of source relative to medium but it depends on the motion of observer relative to the medium.
I don't understand why it is so.
| The speed of light with respect to an observer in a medium depends on its refractive index only, and therefore has no reason to change if the source is moving or stationary (although the refractive index depends on wavelength of light, which can in turn change the speed of light in the medium, we ignore this aspect entirely). However if the observer is is moving, then in order to find the speed of light with respect to the observer, you need to apply velocity addition, which is an outcome of Lorentz transformations and is given by
$$\dfrac{v_0 + v_m}{1 + \frac{v_0v_m}{c^2}}$$
where $v_0$ is the velocity of the observer and the $v_m$ is the velocity of light in that medium. Care should be taken that this is not the same velocity addition as in case of non-relativistic mechanics, but simplifies to that in appropriate limits. So the speed of light with respect to the observer changes if the observer himself is moving.
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How to prove that gravitational potential energy of a body of mass $m$ at a height $h$ is $mgh$? Many introductory physics books just write that potential energy of a body of mass $m$ at a height $h$ as $U_\text{g}=mgh$. However, they never show how this was derived. I'm interested in knowing this derivation – if possible, avoiding calculus.
| The form $U=mgh$ is simply an approximation to allow people to quickly calculate small changes in the gravitational potential energy of the system. The actual value of $U_g$ is usually not important in classical mechanics. $\Delta U_g$ is the important concept.
On a planet, small changes in height change $U_g$ in a space where the gravitational force is almost constant, $\vec{F}=m\vec{g}$. A change in potential energy (in classical mechanics) is calculated by the negative work done by a force while the position changes:
$$\Delta U_{y'\to y'+h}=-\int_{y'}^{y'+h} mg(-\hat{j})\mathrm\ {d}y $$
The integral is easy:
$$\Delta U_{y'\to y'+h}=\left. mgy\right| _y^{y+h}= mg(y'+h-y')=mgh$$
We see that the actual starting point is unimportant as long as the gravitational field can be considered constant. $h$ is positive if the particle is moved opposite the gravitational field (up), and negative if the particle moves down.
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Continuing on voltage drop across a resistor What exact wire diameter allows approximately $6.25\cdot 10^{18}$ electrons to pass in 1 second? Will a thinner diameter or a wider diameter wire allow the same approximate number of electrons in a second?
If the same approximate number of electrons enter and exit the resistor, what do they lose to cause that voltage drop? And if one resistor drops a voltage to zero, how come one other resistor placed after the first will still find voltage?
| Just think of it as a hose pipe carrying water. The higher the pressure (voltage) the more water passes through it (charge). However, the pressure drops along the pipe. The current is the amount of water per second flowing. The pipe offers a resistance to flow. The thinner the pipe, the higher pressure is needed to get the same flow.
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Get a ball to keep jumping using momentum, and force if necessary To make a ball keep bouncing with momentum, or with force. When a bouncy ball hits the ground, the ground takes some of the momentum. But it doesn't disappear. Would you be able to make the ball keep bouncing with use of a bass speaker. To make the bass speaker store and release its pressure to help the ball gain momentum... Would that be possible? Just like a bouncing bass speaker
| The problem is that the collision will not be elastic and so the kinetic energy of the object after the collision will be smaller than that before the collision. The will also be a loss of kinetic energy due to air resistance.
So you could get you bass speaker connected to a suitable oscillator to be vibrating at such a frequency and with such an amplitude that the energy the object has lost is given back to it by the bass speaker. In the real world I think that it would be very difficult to do as the video suggested by @MaxW shows.
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Number of e-foldings in cosmological inflationary theory It seems that in a lot of lecture courses and notes, including mine and those online, seem to state that the number of e-foldings required is of the order of 50-60.
Perhaps I'm looking in the wrong places, or my understanding is a little off but this value often seems to be plucked out of thin air. I'm sure there is a good reason for it.
So, could anyone give me a very brief and consise reason why the number of e-foldings in the current inflationary model needs to be of the order of 50-60...?
| 60 e-folds is more or less what you need in order to solve the horizon problem, i.e. the fact that the universe appears to be extremely homogeneous despite the fact that different parts of the universe would not have been in causal contact under the usual Big Bang evolution.
Inflation allows for the whole of the observable universe to have originated in a single causally connected region. Suppose inflation begins at $t_i$ and that the Hubble rate stays constant throughout inflation, $a(t) = a_ie^{H_{\mathrm{inf}}(t-t_i)}$. The comoving causal horizon during inflation is
$$d_p^c = \int_{t_i}^t\frac{\mathrm d t}{a(t)} = (a_iH_{\mathrm{inf}})^{-1}(1-e^{-H_{\mathrm{inf}}(t-t_i)}) \simeq (a_iH_{\mathrm{inf}})^{-1}.$$
If the comoving scales correponding to the observable universe today, $\lambda_0 \sim (a_0H_0)^{-1}$ originated inside this causal horizon we must have
$$\frac{\lambda_0}{d_p^c} < 1 \qquad \Leftrightarrow \qquad \frac{a_iH_{\inf}}{a_0H_0} = \frac{a_i}{a_{\mathrm{end}}}\frac{a_{\mathrm{end}}H_{\mathrm{inf}}}{a_0H_0} = e^{-N}\frac{a_{\mathrm{end}}H_{\mathrm{inf}}}{a_0H_0}<1.$$
The subscript 'end' refers to the end of inflation and $N$ is the number of e-foldings of inflation. Assume for simplicity that the universe is radiation dominated ($a\propto H^{-1/2}$) from the end of inflation until today. Then we must have
$$N > \ln\left(\frac{a_{\mathrm{end}}H_{\mathrm{inf}}}{a_0H_0}\right)=\frac{1}{2}\ln\left(\frac{H_{\mathrm{inf}}}{H_0}\right).$$
Since $H_0\sim 10^{-42}$ GeV, if for example the scale of inflation is $H_{\mathrm{inf}}\sim 10^{14}$ GeV this amounts to about $N>64$ being required for the observable universe to have originated in a single causally connected region. One can refine this estimate by taking into account that $H$ evolves during inflation and that $a$ evolved differently during reheating, radiation domination era and matter domination era but this is the basic idea. Nothing prevents inflation from lasting longer than this, however.
| {
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"url": "https://physics.stackexchange.com/questions/241666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Condition for looping the loop Consider a ball tied to a string and it is imparted a velocity we have studied that condition for looping the loop is that tension at the uppermost point must be zero, but why is this condition imposed please explain?
If tension becomes zero at some point below the uppermost point won't the ball complete the loop because it still has some velocity?
| At highest point the body has minimum tension but the velocity of the body is greater compared to some point of the curve hence a body continued in its circular path.
| {
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What does well defined parity mean I'm reading a textbook (Physics of Quantum Mechanics by Binney) and it says that the ground state ket $\left\lvert 1 0 0 \right \rangle$ of the hydrogen atom has well defined (even) parity. What does this mean?
Does it mean that the wave function is even? The wave function for this is an exponential decay. The potential for this is not even either.
| "Well-defined parity" here means it is an eigenstate of the parity operator that sends $\vec x \mapsto -\vec x$. Even wavefunctions are such eigenstates for the eigenvalue 1, odd wavefunctions are eigenstates for the eigenvalue -1.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Coefficient of Linear Expansion (Formula) Quoting directly from my textbook:
We define average coefficent of linear expansion in the temperature range $\Delta T$ as:
$$ \bar \alpha = \frac 1L \frac{\Delta L}{\Delta T} $$
The coefficient of linear expansion at temperature $T$ is the limit of average coefficient as $\Delta T \to 0$, i.e.,
$$ \alpha = \lim_{\Delta T \to 0} \frac 1L \frac{\Delta L}{\Delta T}=\frac 1L \frac{\mathrm{d}L}{\mathrm{d}T}$$
Suppose the length of a rod at $0^\circ \mathrm{C}$ is $L_0$, and at temperature $\theta$ (measured in Celsius) is $L_{\theta}$. If $\alpha$ is small and constant over given temperature interval,
$$ \alpha = \frac{L_{\theta} - L_0}{L_0 \theta} $$
How did they get the last part? Because while integrating, we must keep $L$ variable, and not as a constant, and should thus yield:
$$\alpha \cdot \theta = \ln{\frac{L_{\theta}}{L_0}} $$
Even considering an approximate form, how do we get to that formula? I do not see any other way except treating $L$ as a constant.
| You obtain your formula assuming that $\alpha$ is constant. Now assume it is small. Then $\exp(\alpha\theta)\cong1+\alpha\theta$ and
\begin{equation}
1+\alpha\theta=\frac{L_\theta}{L_0}\iff\alpha=\frac{L_\theta-L_0}{L_0\theta}
\end{equation}
| {
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Help me solve a heat conduction/emission transfer problem. Mathematica has failed me My problem: A thin-walled tube (length $L$, diameter $D$ and wall thickness $t \ll D$) is in a vacuum. It is held on one end (at $x=0$) by a heat source at constant temperature $T(0)=T_0$. The only way it can dissipates heat is radiatively. I am assuming emission only occurs from the outer surface of the tube. The conductivity of the tube is $k$ in $[W/mK]$ and the emissivity $\epsilon$. What is the equilibrium temperature profile $T(x)$ in the tube? (a numerical approximation will do).
My attempt:
In a steady state,
\begin{equation}
Q_{in} = Q_{out}
\end{equation}
From Fourier's law of thermal conduction, the heat entering through the end section is
\begin{equation}
Q_{in} = -k \frac{dT}{dx}\Big|_{x=0} \times \pi Dt
\end{equation}
From the Stefan-Boltzmann law of Black-body radiation, the heat dissipated through the outer surface of the tube is given by
\begin{equation}
Q_{out} = \int_0^L \epsilon \sigma T^4 \mathrm{d}x \times \pi D
\end{equation}
Equating the two, the problem becomes
\begin{equation}
-\frac{kt}{\epsilon \sigma} \frac{dT}{dx}\Big|_{x=0} = \int_0^L T^4 \mathrm{d}x,\ \ \ T(0) = T_0
\end{equation}
Trying to solve this in Mathematica is hopeless. Am I doing something wrong? How can I find a local differential form of the equation? Can I simplify it further?
Thanks for your help.
| You need to do a differential heat balance on a small segment of the tube between x and x + $\Delta x$.
Heat in at x = $-\pi Dtk\left(\frac{\partial T}{\partial x}\right)_x$
Heat in at x + $\Delta x$ = $+\pi Dtk\left(\frac{\partial T}{\partial x}\right)_{x+\Delta x}$
Heat lost due to radiation = $\pi D\Delta x\epsilon \sigma T^4$
Heat balance equation:$$+\pi Dtk\left(\frac{\partial T}{\partial x}\right)_{x+\Delta x}-\pi Dtk\left(\frac{\partial T}{\partial x}\right)_x=\pi D\Delta x\epsilon \sigma T^4$$
Dividing by $\Delta x$ and taking the limit as $\Delta x$ approaches zero gives:
$$kt\frac{\partial^2T}{\partial x^2}=\epsilon \sigma T^4$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Boiling as apparent violation of the second law of thermodynamics One of the statements of the second law is that no agency can be built whose sole effect is to convert some amount of heat entirely to work.
But in case of boiling, the temperature being constant, entire heat supplied is converted into work, namely the work done by water against ambient pressure to expand to the vapour state.
At which point am I going wrong?
| When water boils, the heat mostly goes into breaking the bonds between water molecules. Suppose you boil mass $m$ of water under constant pressure $P$. The work done due to expansion is
$$
A=P(V_\text{gas}-V_\text{liquid}),
$$
and one can estimate
$$
PV_\text{gas}\simeq NkT,
$$
where $N$ is the number of molecules in mass $m$. And so we get, for $V_\text{gas}\gg V_\text{liquid}$,
$$
A\simeq NkT.
$$
That is, the work is $kT$ per molecule. On the other hand, the hydrogen bond binding energy in water is on the order of $0.2$ eV, if I rememeber correctly. That is equivalent to temperature $0.2\cdot 10^4=2000$ K, and you have to break on the order of 2 hydrogen bonds. Therefore, at room temperature, only several percent of heat actually goes into work, while most of it goes into breaking the bonds between molecules.
| {
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Sliding blocks problem My first post here, so I apologize if this is duplicated elsewhere. It IS a "homework" problem, but it's public domain, a posted exam with answers...
http://www.mun.ca/physics/undergraduates/finals/P1020F06.pdf
Here is the diagram in question
And here is the FBD for each block.
(question a is to find the tension in the rope and question b is to find the acceleration of the system)
I am getting confused when the situation has , for example, two blocks simply sliding over each other. In those cases, the force of friction on the top block, by Newton's laws, produces a "reactive" force in the opposite direction that makes the lower block move.
In this case, the force of friction OPPOSES the motion of both blocks. I would value some help with understanding this.
I.E. how would I compare this question with the one below...
where there is friction between the block and toboggan, but the toboggan is on "ice" (no friction)
| The first thing first -The concept of friction
Friction come in play when there is any tendency of relative motion between 2 surfaces and it is in opposite direction of relative motion (Note- I used word relative motion not simply motion). In simple word friction try to reduce relative motion.
in 2 block problem and this problem the difference is T (tension). In both problem direction of relative motion is similar, friction is trying to do same thing i.e. pulling the other block with the one on whom force is applied so that relative motion can be reduced. But in this problem Tension forces block to move in other direction then supplied by friction.And but obvious T>Friction. so here T is increasing relative motion and friction as always trying to make them move together.
| {
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About the fluid friction along the wall in pipes I wonder how can the friction between a fluid and a wall in a cylindrical pipe can be calculated. Is there any theory that I can refer to? I also want to check if there's any relation between the diameter of the pipe, the velocity and the temperature of the fluid and friction.
| We know that adhesive forces bound fluid particles at the surface of the pipe. The rest of the fluid will not be affected by this force. But force acting on the remaining fluid is viscous force. The below figure shows relationship between velocity profile and radius of pipe. Where n is coefficient of viscosity.
Even if the fluid is a gas, we can still see the effects of viscous and frictional forces at the surface of the pipe, but with different values.
| {
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Do solid objects really behave like fluid matter on extremely long timescales? Freeman Dyson's fascinating paper Time without End: Physics and Biology in an Open Universe contains the following passage:
I next discuss a group of physical processes which occur in ordinary matter at zero temperature as a result of quantum-mechanical barrier penetration.
[...]
On a time scale of $10^{65}$ yr, every piece of rock behaves like a liquid, flowing into a spherical shape under the influence of gravity. Its atoms and molecules will be ceaselessly diffusing around like the molecules in a drop of water.
While the paper provides computations yielding the time estimate quoted above, there is no obvious justification given for quantum barrier penetration actually leading to liquid-like behavior.
Would solid matter really behave exactly like a liquid on extremely long timescales, or is this just an analogy? If barrier penetration is a random process, why doesn't matter just re-form into random shapes at such timescales rather than becoming spherical?
| The number of years mentioned appears even larger than total anticipated life span of universe (gravity) itself. One of these - Heat death, Big Rip, Big crunch would probably happen before the time frame mentioned. Therefore not sure what the answer is as the conditions mentioned may no even exists that long. For example, If a big rip happens, then one can say - solids behave like gasses over long enough time scales. And behave even stranger over even longer time times.
| {
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Could we curve the flight path of a photon? I was wondering about photon's interaction with matter, and why photons dont slow down. They seem to always bounce in a straight line at the same speed (I think), as if some force is charging them forward after the bounce. First, what is this? I've heard of experiments where we actually did slow them down, and they can be absorbed and bounced by matter, so I wanted to know if they interact with matter in such a way that we could spin one, like a curve ball thrown by a pitcher, and make them curve their flight path.
Can someone explain this, in English (back it up by math if you need to but I'm not a physicist, I just like to learn about the fundamental ideas)?
| Adding to Anna's response,
In small scales, photons wont slow down because they are thought of as being relativisitc quantum fields in all their possible interaction channels. However, at large scales (cosmological distances) there is an indication of slowness (delay in arrival time)-- either due to a varying gravitational potential (due to non-flat spacetime fabric as mentioned by Anna) or due to the expansion of Universe itself.
Einstein himself argued that in the absence of gravity and expansion of Universe (which he was reluctant to believe in the latter) light would never slow down or speed up and was able to show it explicitly.But, current frontier research at small-scale, everyday life and large scale are positive after all that they are seeing a varying speed of light. (Related sources: Varying Speed of Light)
| {
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Why do bubbles group when one pops? I was recently observing the way bubbles move as they pop and disappear. I noticed that when bubbles destabilize and pop, the remaining bubbles immediately surrounding it will move to fill its place. I was wondering what the driving cause is here.
At first, I figured that stickiness was the cause. But, I don't think this would be a driving force unless perhaps the bubbles are clustered (i.e. they share at least 1 membrane).
I've also noticed this behavior in groups of individual, non-clustered bubbles. As an introductory physics student, I'm curious: what causes this phenomenon?
| The blubbles exerce pressure forces with their neighboor through the shared membranes. Pressure is higher in blubbles than in free air, when within a field of bubbles it is balanced on each sides of bubble wall... as long as there is a neighbor. When this one pops, the rest of the pack around push towards the empty space.
| {
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Why is speed of light a constant while distance in space is not? Disclaimer: I asked this at Astronomy.SE, but got no answer whatsoever, so I am trying my luck here.
As you probably know current state-of-the art physics (i.e. gravitational waves, cosmic expansion) basically states that space itself is subject to expansion or contraction. Since there is no moving matter or energy involved, this might even happen at a "speed" faster than light.
So far, so good and obscure. What strikes me is the principle that the speed of light as a fundamental constant can only be expressed as a function of space-time. Where do we know that the one is constant but the other can suddenly be variable?
Is there any reason why the point of view of an expanding or contracting space is preferred over, say, a reduction in the speed of light or an increase in the "speed" of time? Is there any objective difference, a mathmatical model being a better fit or is it just the good old rubber metaphor being stretched (pun intended) too far?
In case the answer is: Both are equal w.r.t. current observations: How do we know that not both are actually variable?
| When we observe very distant objects, the spectral lines of atoms seems to keep the same, and the physics (and the objects it permits) as well, despite c appears in many balances of the microphysics. So c does not seems to change in all the observable Universe. Beside, some pulsing phenomena seems to be slowered as predicted by time contraction in relativity. Therefore the red-shift really means expansion of space, and not change in the light.
| {
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Why is it said that standing waves do not transfer energy? The author of my physics textbook writes that standing waves, unlike travelling waves, do not transfer energy. He says that this is because a standing wave is composed of two travelling waves carrying energy in opposite directions. Is this explanation sufficient to prove that standing waves do not transfer energy? Also if standing waves do not transfer energy, then how can instruments be heard?
| You may conclude this by thinking that the standing wave is actually the superposition of two travelling waves which were moving in opposite directions, so the energy transferred by a single wave in one direction is completely compensated by energy transfer in the opposite direction. This keeps the total energy (sum of potential and kinetic energies) of every particle in the standing wave constant (read 'particle' as differential segment of length $\mu dx$, where $\mu$ is the mass per unit length of the string).
A further insight might be brought by observing that propagating waves are 'propagating' because each segment of the string propagates its disturbances (momentum, energy) to the neighbouring particle.So if we stop furnishing energy, displacement at the initial position stops and the wave moves further as energy keeps getting tranferred in the direction of wave velocity. However, in the case of standing waves (ideally speaking) you do not need energy to sustain the motion at any point. The energy is kept within the already oscillating string segment, exactly why a standing wave does not 'travel'.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/243171",
"timestamp": "2023-03-29T00:00:00",
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Mirage formation in deserts Why does a person sees water in the desert even then when no water is present there at that time?
How does this happens?
| Ideal conditions for a mirage are layers of air in contact with ground that has been heated by the sun .
As one goes up from the ground the temperature slowly decreases. this gradient of temperature changes the density of air . The layer closer to the ground is rarer and upper layers are denser .
Thus refraction of light from rarer to denser medium takes place and the light coming from sky takes a curved path to reach a person's eyes.
The illusion of water as mirage comes from the fact that the human brain assumes that light travels in a straight line.
A person looking at the road ahead on a hot, still, day will see the sky because light from the sky is taking the curved path .
The human brain interprets this as water on the road because water would reflect light from the sky in much the same way that a vertical temperature gradient in the layers of air does.
On hot sunny days even the car-drivers in the city highways see water ahead at the road (mirage) as the road gets hot and then a vertical temperature gradient (just like desert condition ) produce the above mentioned effect.
| {
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Free falling and bouncing back My confusion arises with free falling body.
*
*For a free falling body the displacement ~ time graph has a kink (at the time when the body hit the ground ). at a kink point, a function is not derivable by the rule of calculus. but we see in the free falling case the body has velocity but in opposite direction at the moment it hit the ground.
*For same free falling body as the velocity is a discontinuous function of time (at the time when it hit the ground) there should not be any acceleration because a derivative function must be continuous by the theory of calculus. But velocity $v$ is not continuous at that moment of time (when it hits the ground). But it has an acceleration spike value.
So I'm confused very much with this mismatch with mathematical theorem and the practical application in physics. what is the solution??
| Acceleration of an object, e.g. a free falling ball on earth, is $g=9.81 \frac{m}{s}$. This is a constant, as can be seen in your diagram. Look at the scale - yes acceleration is not zero.
Having this in mind we look at the moment before and after hitting ground. This way we omit looking of the elasticity of the ball. Your diagrams show a perfect elastic ball without dissipative forces. Before ground it has velocity $v$ and a time slice $ \delta t$ later it has veloctiy $-v$.
$$\text{for }\delta t\rightarrow 0\,s\quad a=\frac{v-(-v)}{\delta t} \rightarrow\infty\,\frac{m}{s^2}$$
Because of deformation and internal forces in the ball $\delta t\neq0\,s$. This deformation time is the reason, that acceleration is very large in your diagram, but not infinite.
However for reality e.g. $\delta t=$ and $v=10\,\frac{m}{s^2}$
$$a_{bounce}=\frac{v-(-v)}{\delta t} = 1\cdot 10^4 \,\frac{m}{s^2}\Rightarrow t_{bounce}\approx2\mu s$$
Extending this comprehensible example to the velocity also pictures the ball at the ground. At one time an incompressible ball stops at the ground. I doubt that objects like that exist. A very small time velocity of ball mass center should be minimal during deformation on ground.
| {
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Force distribution on corner supported plane This question has been annoying me for a while. If you have a completely ridged rectangular plate of width and height x and y that is supported on each corner (A,B,C,D) and has force (F) directly in its center then I think the force on each corner support will be F/4.
What I want to know is how to prove this? Obviously there are 4 unknowns so we require 4 equations.
$$\sum F_z = 0$$
$$\therefore F_A+F_B+F_C+F_D=F$$
Also
$$\sum M = 0$$
Now taking the moments about point A
\begin{equation*}
\begin{vmatrix}
i & j & k \\
x & 0 & 0 \\
0 & 0 & F_B \\
\end{vmatrix}
+
\begin{vmatrix}
i & j & k \\
x & y & 0 \\
0 & 0 & F_C \\
\end{vmatrix}
+\begin{vmatrix}
i & j & k \\
0 & y & 0 \\
0 & 0 & F_D \\
\end{vmatrix}+
\begin{vmatrix}
i & j & k \\
x/2 & y/2 & 0 \\
0 & 0 & -F \\
\end{vmatrix}=0
\end{equation*}
As the sum of moments equals 0, let i and j = 0
$$\therefore F_C+F_D=F/2$$ and $$F_B+F_C=F/2$$
Then taking the moments about another point to get 4th equation. But no matter what location I use the equations will not solve. I have been using a matrix to find $$F_A, F_B, F_C, F_D$$ See below. But when getting the det of the first matrix the answer is always equal to 0.
\begin{equation*}
\begin{vmatrix}
1 & 1 & 1 & 1 \\
0 & 0 & 1 & 1 \\
0 & 1 & 1 & 0 \\
? & ? & ? & ? \\
\end{vmatrix}
*
\begin{vmatrix}
F_A \\
F_B \\
F_C \\
F_D \\
\end{vmatrix}
=\begin{vmatrix}
1 \\
0.5 \\
0.5 \\
? \\
\end{vmatrix}*F
\end{equation*}
Can someone please tell me what I am doing wrong? Thank you in advance.
| I think I have a solution. Considering the corner forces $A$, $B$, $C$ and $D$ you have a system of 3 equations and 4 unknowns
$$\begin{align}
A + B + C + D & = F \\
\frac{y}{2} \left(C+D-A-B\right) &= 0 \\
\frac{x}{2} \left(A+D-B-C\right) & = 0
\end{align}$$
$$\begin{vmatrix}
1 & 1 & 1 & 1 \\
-\frac{y}{2} & -\frac{y}{2} & \frac{y}{2} & \frac{y}{2} \\
\frac{x}{2} & -\frac{x}{2} & -\frac{x}{2} & \frac{x}{2}
\end{vmatrix}
\begin{vmatrix} A \\ B \\ C \\ D \end{vmatrix} = \begin{vmatrix} F \\ 0 \\ 0 \end{vmatrix} $$
What if consider the forces as deviation from $\frac{F}{4}$ such that
$$\begin{align}
A & = \frac{F}{4} + U \\
B & = \frac{F}{4} + V \\
C & = \frac{F}{4} + W \\
D & = \frac{F}{4} + G \end{align} $$
$$ \begin{vmatrix}
1 & 1 & 1 & 1 \\
-\frac{y}{2} & -\frac{y}{2} & \frac{y}{2} & \frac{y}{2} \\
\frac{x}{2} & -\frac{x}{2} & -\frac{x}{2} & \frac{x}{2}
\end{vmatrix}
\begin{vmatrix} \frac{F}{4} \\ \frac{F}{4} \\ \frac{F}{4} \\ \frac{F}{4} \end{vmatrix} +
\begin{vmatrix}
1 & 1 & 1 & 1 \\
-\frac{y}{2} & -\frac{y}{2} & \frac{y}{2} & \frac{y}{2} \\
\frac{x}{2} & -\frac{x}{2} & -\frac{x}{2} & \frac{x}{2}
\end{vmatrix}
\begin{vmatrix} U \\ V \\ W \\ G \end{vmatrix} = \begin{vmatrix} F \\ 0 \\ 0 \end{vmatrix} $$
$$ \begin{vmatrix} F \\ 0 \\ 0 \end{vmatrix} +
\begin{vmatrix}
1 & 1 & 1 & 1 \\
-\frac{y}{2} & -\frac{y}{2} & \frac{y}{2} & \frac{y}{2} \\
\frac{x}{2} & -\frac{x}{2} & -\frac{x}{2} & \frac{x}{2}
\end{vmatrix}
\begin{vmatrix} U \\ V \\ W \\ G \end{vmatrix} = \begin{vmatrix} F \\ 0 \\ 0 \end{vmatrix} $$
$$ \begin{vmatrix}
1 & 1 & 1 & 1 \\
-\frac{y}{2} & -\frac{y}{2} & \frac{y}{2} & \frac{y}{2} \\
\frac{x}{2} & -\frac{x}{2} & -\frac{x}{2} & \frac{x}{2}
\end{vmatrix}
\begin{vmatrix} U \\ V \\ W \\ G \end{vmatrix} = \begin{vmatrix} 0 \\ 0 \\ 0 \end{vmatrix} $$
Now a possible solution is $U=0$, $V=0$, $W=0$ and $G=0$
I think to beyond this method, you need to assume a non-rigid frame, with some simple stiffness matrix, and in the end find the limit where $k \rightarrow \infty$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Has such experiment been performed before? Consider a charge of 1C kept a distance of $6*10^8$ m from a detector.
I find electric field due to this charge at detector.
Then, I suddenly earth that charge
and not the time it takes to be detected by detector.
It should be about 2 sec.
Has any similar experiment been performed before.
Such experiment can help to distinguish b/w speed of electromagnetic wave and
speed of electric field propagation.
| This has been already studied extensively in the electrodynamics formalism of James Clerk Maxwell and experimentally proved to be correct many times. So magnetic fields and electric fields are unified into a single formalism called Electromagnetism which propagates with constant speed of light in vacuum. This means that the presence or absence of any electric and/or magnetic fields can be felt according to causality.
| {
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Newton's second law of motion and viscosity In a Coutte flow, the applied force on one of the plates can be expressed by $F = \eta A \frac{dv}{dz}$. F here can also be defined using Newton's second law of motion, $F = m \frac{dv}{dt}$.
The two forces are equal here. That said, there has to be a way to derive $m \frac{dv}{dt}$ from $\eta A \frac{dv}{dz}$ or vice versa. But can't quite figure it out. What am I missing here?
(I can "sense" the similarity between the second law of motion, $F = m \frac{dv}{dt}$ and the viscosity, $\eta$, coming from $F = \eta A \frac{dv}{dz}$, but can't derive $m \frac{dv}{dt}$ from $\eta A \frac{dv}{dz}$).
Any help/comment would be appreciated.
| The proportional relation between force and velocity we sometimes refer to as a viscous friction factor is a highly idealized model that may fit actual physical behavior sometimes, but in general, not.
Although we might be able to fudge the units and attempt to wedge viscosity in, the result doesn't stand as any general law in connection with Newton's second law.
So the answer is no.
| {
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Help! An 8 year old asked me how to build a nuclear power plant I would really like to give an explanation similar to this one.
Here's my current recipe:
(i) Mine uranium, for example take a rock from here (picture of uranium mine in Kazakhstan).
(ii) Put the rock in water. Then the water gets hot.
(iii) [Efficient way to explain that now we are done with the question]
This seems wrong, or the uranium mine would explode whenever there is a rainfall. Does one need to modify the rock first? Do I need some neutron source other than the rock itself to get the reaction started?
As soon as I have a concrete and correct description of how one actually does I think I can fill in with details about chain reactions et.c. if the child would still be interested to know more.
| RTG
The described approach mirrors https://en.wikipedia.org/wiki/Radioisotope_thermoelectric_generator devices used to generate power in, for example, space probes - simply take some radioactive material and extract energy from its decay, without needing to control a chain reaction or something like that.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
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How can a parallel circuit work? The electrons always takes the easiest way in a circuit, right?
So in a parallel circuit, why does the electrons flow through all parts of the circuit and not just the one with the least resistance?
| Ohm's law helps here.
When a current $I$ flows through a resistor $R$, it develops a voltage $V=I\cdot R$. If there is less current, there is less voltage.
When the voltage developed is equal to the voltage applied, you reach equilibrium. So when current can "choose" between a high resistance branch and a low resistance branch, it will divide in such a way that the voltage developed across each branch is the same. This means less current through the high resistance branch, and more current through the low resistance branch - so that the voltage across each branch ends up being the same.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Cylinder rotating without slipping on an accelerating slab I am very confused by the following problem asked in my first year physics class:
Please let me know if you can assist in any way! I've spent hours and hours on this question and gained absolutely nothing. Everything I do seems to lead to a contradiction one way or another.
There are other resources online I've found that mention this question, but I can't tease out a good solution from these:
Force on a solid cylinder that is rolling on an accelerating block
https://www.physicsforums.com/threads/a-rolling-disc-on-a-slab.594918/
Thank you!
| As this is a homework question I won't give you a full solution, only point you in the right direction.
On the lower block acts a second force, $F_F$, a friction force that causes torque and the angular acceleration $\alpha$ of the cylinder:
$$F_F R=-I\alpha$$
Where $I$ is the moment of inertia of the cylinder and $R$ its radius. It carries a minus sign because it points in the opposite direction of $F$.
So the net force acting on the block is:
$$F_{net}=F-F_F$$
Now also note that for rolling without slipping, with $a$ the acceleration of the block, then:
$$a=\alpha R$$
To determine $a$ and $\alpha$ use the equations above to set up:
$$F_{net}=ma$$
| {
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Why do some stars have a negative parallax? I am constructing a Hertzsprung-Russell diagram for stars within some radius around Pleiades and have repeatedly come across stars that have negative parallaxes. For example, http://vizier.u-strasbg.fr/viz-bin/VizieR-5?-info=XML&-out.add=.&-source=I/239/tyc_main&recno=161838
I found three reasons listed in the following article for where these values come from, but I do not fully understand why the values are quoted as negative.
When a distant star, whether bright or faint, is observed through a scattered cluster, or perhaps rather a layer, of relatively near and faint stars, a negative parallax must be expected.
http://adsabs.harvard.edu/full/1943AnDea...4....1L
The article goes on to say that this parallax is the positive parallax of the comparison stars with respect to the distant star.
Is the magnitude of this parallax the distance between the distant star and the comparison stars, or between the Earth and the distant star?
| i think its the refraction effect.
if parallax is given by (angle_final-angle_initial). and if this quantity is -ve, (due to refraction effect of stars lying in the foreground of the stars under observation), then parallax can be -ve.
| {
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"timestamp": "2023-03-29T00:00:00",
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Does continuity equation hold if the flow is accelerated? I'm Studying the streamline flow, specifically the continuity equation Bernoulli's Principle.
Consider the following system where a liquid flows through the pipe of a uniform area of cross section A, from high pressure P2 to low pressure P1, both of which are maintained.
<--------length = r----------->
_______________________________
P2 P1
_______________________________
flow ->
Based on the assumptions that
*
*Difference in pressure is what causes the liquid to flow $P_2 > P_1$
*The continuity equation $A_1v_1 = A_2v_2$ holds
*The Bernoulli's Principle holds
Then my analysis is as follows
The Bernoulli's equation gives
$$ P_2 + \frac{1}{2}\rho v_2^2 = P_1 + \frac{1}{2} \rho v_1^2$$
Therefore, the speed of liquid element will be more at $P_1$ and less at $P_2$ since $P_2 > P_1$
But then
$$A_1v_1 = A_2v_2$$
and since the Area of cross section is uniform, the velocities must be same, which is contradicting the result from Bernoulli's equation!
There is something wrong with my analysis or assumptions, but I can't figure out what it is
|
The question is very good but the assumption you have made is wrong!
I agree that the water is moving because of the pressure difference between water and air
As I have shown point $P_1$ and $P_2$ in the diagram.
Now the water moves because of pressure difference between this $P_1$ and $P_2$ so as the water reaches the end of the tube(according to the diagram) the pressure in the water is still $P_1$ and pressure just next to it (I mean pressure of air next to it) is still $P_2$
Now the mistake you have done is you have taken Bernoulli's equation for air and water which is completely wrong. Bernoulli's equation is valid only for one fluid at a time, but you have taken two separate fluids which is totally wrong.
You may say why the pressure has to be equal at the starting point and at the end point. The reason is fluid is at horizontal level; more over it has uniform cross-section and according to equation of continuity it must have equal velocity. So it has to satisfy Bernoulli's equation that's why pressure should be equal.(because Bernoulli's equation is a property of streamline flowing liquid)
NOTE:-
*
*Bernoulli's equation can be applied for one fluid at a time.
*If fluids are at same horizontal level and have same cross-section then they have same velocity always (streamline flowing liquid only).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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The force felt by a pilot when a package is dropped? Here is a physics question that I wish to prove mathematically.
A 100,000 kg aircraft drops a 1000 kg package of supplies over an arctic research station. What approximate force is felt by the 100 kg pilot at the instant of the release?
Is there a reactive force on the plane when the package is dropped?
| Sounds like a homework question. Here is a partial answer.
Before the drop, gravity acts on the plane, the pilot, and the package. This is the weight of each. After the drop, gravity continues to act on each. The weight of the pilot does not change.
If the plane is in straight level flight, the upward force of the air on the wings is just enough to counteract the weight of all three.
If the wings and such stay fixed, the upward force stays the same after the drop. But the downward force is only the weight of the plane plus pilot.
So think about the total force on the place, and what the acceleration of the plane must be. Since the pilot moves with the plane, he must experience that acceleration. What is the force on the pilot?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/244836",
"timestamp": "2023-03-29T00:00:00",
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Are electric field lines always conserved? Suppose we have a positive +q charge and a -6q charge at some separation. Then will every field line originate from the +q and end up to -6q or will there be some extra lines coming to -6q from infinity because of higher charge to get 6 times the number of field lines? That is, will there be any line that does not originate from the positive charge but terminates at the negative one?
I think it should be that every line will originate from the positive and go to the negative, only difference will be in the density of the field lines. Am I right?
Also, if I talk about the flux now, why can I say that the flux near +q will be equal to that near -6q?
| In such ideal problems, it is always implicitly assumed that there is no other charges in the Universe. So, if our Universe consists of only these two charges and everything else is neutral, then we cannot assume that field lines start at some infinity and end at individual charges. In fact, lines start from one charge and travel to infinity and then come back to the other charge. This picture is in line with the infinite-range nature of electromagnetic interaction.
Since flux is defined as integrated electric field over a surface, you can enclose your charges with identical (Gaussian) surfaces and attempt to calculate the integral. By Gauss's law, however, this integral is proportional to the enclosed charge. Then, it is trivial to see why bigger charge contains more flux than the smaller charge.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/245018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
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Why does the temperature of the gas in a container moving with constant velocity not change?
Systematic changes do not affect thermodynamic equilibrium.
What does this mean? And what kind of systematic changes are allowed?
The container with gas is stationary till some time then it's given a constant velocity and the final temperature is asked; the answer says that systematic changes don't affect thermodynamic equilibrium and temperature remains constant.
| It is simpler to think of the thermodynamic definition of temperature, to start with, the one on the left:
where N is the number of molecules, n the number of moles, R the gas constant, and k the Boltzmann constant.
For the expression on the right, randomness is implicit, Brownian motion after all led the molecular model, and it is basically random.
The identification of temperature as the same quantity in both expressions holds when the conditions for both expressions hold.
Moving a fixed volume at a fixed pressure does not change its temperature , it is a definite macrostate for which the microstates on the right are used to average out the molecular velocities.
If randomness does not hold , as with a moving system, the fact that we do not observe a rise in temperature in moving systems leads to an answer by Deechit Poudel, the rms of the velocity of gas molecules is large, and the contribution of the motion is small for every day motions. This can be verified with this calculator:
For 20C in air the most probable speed of the molecules is 400m/second, Larger than the velocity of sound.
For supersonic velocities and velocities commensurate with the speed of sound, the relationship demands as an axiom the randomness, imo.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/245123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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