Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
What size particle would be attracted to this 340 ton sculpture's gravitational field? I know that gravity is a weak force, but this rock has mass, so it has gravity. The Levitated Mass sculpture is big enough to attract something. What is the largest particle that would "stick" to the underbelly of this sculpture because of its gravitational attraction?
I am interested in this sculpture because you can walk right under it! If the particle is big enough to see in a microscope, seems like this would be a great experiment for the youngins.
| As many here have noted, Earth poses a problem for your experiment.
So, let's remove it$^\dagger$.
You now have a freely floating rock in space of mass $M = 3.4\times10^8\,\mathrm{g}$ and radius $R = 330\,\mathrm{cm}$.
The escape velocity of the rock is
$$v_\mathrm{esc} = \sqrt{\frac{2GM}{r}},$$
where $r$ is the distance to the center of the rock, i.e. $r\gtrsim R$. This evaluates to at most $37\,\mathrm{cm}\,\mathrm{s}^{-1}$.
The typical speed of a particle of mass $m$ at a gas temperature $T$ is
$$v_\mathrm{rms} = \sqrt{\frac{3 k_\mathrm{B}T}{m}},$$
where $k_\mathrm{B}$ is Boltzmann's constant. Since we've removed Earth, the temperature is given by the interplanetary medium at a distance of 1 AU from the Sun, which is of the order 100–200 K.
What this means is that the answer is not a "most massive particle", but a "least massive particle", corresponding to how close the particles can come to the rock, i.e. $r=R$. The result is
$$m = \frac{3 k_\mathrm{B} T r}{2 G M} \simeq 5\times10^{-13}\,\mathrm{g}.$$
This is not a lot, but much more massive than atoms or normal molecules. It would probably be classified as a "large dust grain"
$^\dagger$Removing Earth is left as an exercise for the reader.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/228912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the metric tensor for? I am wondering how to use the metric tensor, in practice? I read the book and done the exercises in A student's guide to vectors and tensors by Dan Fleisch. The concept of a tensor and their applications are well defined.
In that book, is explained how to get the metric tensor for coordinate system transformation, such as from spherical coordinates to ordinary Cartesian coordinates or even from cylindrical coordinates to Cartesian coordinates; which are easy to obtain, given enough practice. But what do such metric tensors mean (in practice), how does one use such a tensor in an actual math/physics problem?
| The metric is an important concept in general relativity.
In GR, vectors correspond to weighted directions in spacetime (by "weighted", I mean any scalar multiple of a vector corresponds to the same direction, but weighted differently). The metric tensor can then tell us about the angle between two directions or the magnitude of a given vector, which gives us a notion of length in spacetime.
The metric also appears in Einstein's equtaions, relating the distribution of energy and momentum through spacetime to the curvature--which involves the metric and its derivatives. That is, curvature--and therefore the metric--of spacetime are determined by the distribution of energy and momentum.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/229108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How are orbits made stable? I understand the concept of object staying in some orbit due to centrifugal force and gravity.
However I do not understand how is orbit of a body like satellite or planet has perfect balance between gravitational pull and centrifugal force of revolution?
because if the angular velocity is even a little bit more than required then the object will move away and gravitational pull will reduce with square of distance and so centripetal force will decrease leading the object to stray further.
so is everything in orbit is either slowly moving away or falling in?
or is it that only those bodies whose orbital velocity is just perfect stays in orbit with everything else spiraling in or out of it?
so are satellites put with very accurately calculated velocity to prevent them from falling in or moving away or somehow it balances itself?
| In 2 dimensions, in any uniformly rotating frame of reference, there are two fictitous forces, the centrifugal force and the Coriolis force. The centrifugal force pulls directly away from the point of rotation is has the magnitude of $mr\omega^2$ where $m$ is the mass of the object; $r$ is the distance away from the center; and $\omega$ is the angular frequency. The Coriolis force goes in the direction 90° clockwise of the velocity in the rotating frame of reference if the frame of reference is rotating counterclockwise and 90° counterclockwise if the frame of reference is rotating clockwise and has magnitude $2mv$ where $v$ is the speed in the rotating frame of reference. In actuality, we live in the third dimension, not the second and for that reason, the force of gravity varies as the negative second power of distance. However, we can use the formulae for how to calculate centrifugal force and Coriolis force in the second dimension to figure out how to do so in the third dimension. The sun is so much more massive than Earth that we can neglect the gravitational effect of Earth on the sun. Suppose Earth has an exact circular orbit around the sun which it doesn't. Then we can take the uniformly rotating frame of reference where Earth and the sun are stationary. Then if Earth, has a tiny deviation from that orbit, than that will cause a Coriolis force to act on Earth in that frame of reference. It's the Coriolis force that prevents Earth from undergoing a runaway effect from the circular orbit.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/229249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
What came first, the Universe or the Physical laws that govern the Universe? This sounds like the Egg and the Hen question but I am curious about this. If universe came first and created physical laws for itself, then what created the law or the principle as a consequence of which the universe came into existence in the first place? And if there where pre-existing physical laws that governed the big bang or whatever the origin of the universe was, then where did those laws come from and what were they a part of? If we assume that creation and destruction of universe is cyclic and the same laws are carried onto the next creation and destruction cycle then shouldn't the law which is governing this cycle be a consequence or a part of some bigger something (like a mega-verse). Whichever the case, we again come down to same basic question as in the title.
Thanks.
| This is a metaphysical/philosophical question, imo.
There is the platonic ideals school, in this case read for ideals=mathematics, which postulated that ideals existed and nature fell into their form. I have seen a number of theoretically inclined people who are really of that school. One does not have to think of the beginning of the universe to start thinking that the mathematical format is the mold in which nature settles.
So in this school the answer is that "laws and the subsequent theories existed before the observable universe". It becomes metaphysics because it invites a meta level of:" how did these laws and theories appear out of the vacuum".
As an experimental physicist I am of the school that "nature exists and we experimentally study its behavior and fit the data with mathematical models", so the horizon of our knowledge is limited by our experiments and observations. The answer then is like the other answers: we cannot know because we have no observations for the beginning of the universe, only after the cosmic microwave background time in the BigBang, 380.000 years after the beginning. Before that, we fit models and extrapolate back. For a while we thought that BICEP2 had taken us very near to time 0 with gravitational waves, but it turned out not to be rigorously established.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/229389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 1
} |
Kinetic energy of a rotating object in an exercise, a linear molecule is being subject to a force applied on the edge in its axis. Then $K_1=\frac{1}{2}mv^2$, all is well.
Then in the second point of the exercise, the force is applied on the same edge but in an orthogonal direction to its axis. Then the molecule begins to rotate. So its kinetic energy is composed of two terms: $K_2=\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$,$\omega$ being the angular velocity of the rotation. The thing is that the linear velocity is the same as before and the correction says that $K_2=K_1+\frac{1}{2}I\omega^2$. But how come the same force can give two different energies to the molecule? I thought that $v$ would decrease in the second case, because of the apparition of the angular velocity $\omega$ so that the energy would be conserved. So in the second case the molecule goes as fast as before but in addition it rotates on itself?
| The work done by the force in the first case is just translational, i.e. integral of the force over some distance and this gives rise to the translational kinetic energy.
In the second case, there are two types of work done on the system, first being the usual translational and second is rotational. This is due to the that the object in question now has an extra degree of freedom than earlier, in other terms by the virtue of the direction of the force, it can rotate. Therefore the net energy expended by the force has to be on both the translational mode as well as rotational mode, as there is an increase in the number of degrees of freedom in the system than earlier and hence the expression you wrote down.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/229438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Heisenberg's uncertainity principle In the Heisenberg uncertainty principle,
$$\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$$
The values of $\Delta x$ and $\Delta p$ are the standard deviations which we get from the probability distribution function of the particle and I heard that it has nothing to do with the measuring instrument.
Actually while measuring, the probability distribution function of a particle also changes, Does this means that the measuring instrument has some effect?
|
Actually while measuring, the Probability Distribution function of a particle also changes, Does this means that the measuring instrument has some effect ?
The measuring process may change the boundary conditions of the solutions of the quantum mechanical equations of the system under measurement, so the complex conjugate square of the wave function ( the probability) may change. Good measurements are done with a minimal disturbance of the process under study by the measuring method.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/229649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
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?
| We can't explain it like you want, because try closing one of the slits first. Then do the experiment. Then do the same for the other and do the experiment. Classically, you'd expect, that both slits will function independently, hence you won't receive an interference pattern but instead a summation of intensities from each slit individually. But this isn't what is seen in the experiment.
So, it is pretty safe to say physicists have thought this through.
To add more, you could seperate the slits from each other.
Also, try answering why a region(of minimum intereference) that was receiving tennis balls when one slit was open, would stop receiving tennis ball if the second slit got opened.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/229760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
What would be the view like from inside a black hole looking towards the event horizon? Ignoring the fact that we would be torn apart by gravitational gradient and assuming we get some time to make some observations before hitting singularity, what would we see looking towards the event horizon or in any other direction away from the singularity?
| The answer to this question is covered in the book "Exploring black holes: Introduction to General Relativity" by Taylor & Wheeler (2000), within the framework of classical General Relativity.
If we are talking about a supermassive black hole, such that a free-falling observer can survive tidal forces as they approach the event horizon and the singularity, then the following scenario is presented. A star that is exactly radially outwards along the trajectory of the infalling observer will remain in that apparent position. The light from such a star is gravitationally blueshifted, but is also redshifted due to the rapid inward motion of the observer. The latter wins.
For star at an angle to a radial trajectory there is a strong aberration of their positions. As the observer proceeds (inevitably) towards the singularity, the angle they perceive these stars to be at with respect to their radial trajectory increases towards 90 degrees. In front of them, is a black circle, with a bright ring of bent (gravitationally blueshifted) starlight around it. This black circle grows towards filling half the sky. Behind them, the perceived stars "fan out" towards a 90 degree angle so that they are ultimately seen as a "ring around the sky". The final view would be that the sky is black with a brilliant ring of high energy radiation (caused by gravitational blueshift) dividing it into two halves.
You never see the singularity because all the light is headed towards it. You never (consciously) reach the singularity because you would be torn apart by tidal forces about 0.1 seconds before you get there, independently of the mass of the black hole.
Some interesting attempts at visualising this a scenario can be seen at the webpages of Andrew Hamilton, though these are not for a radially infalling observer.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/229868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 0
} |
What would it be like if the supernova ASAS-SN-15lh was in the Milky way? I'm simply wondering what it would be like if the super nova ASAS-SN-15lh (http://www.sciencemag.org/news/2016/01/universe-s-most-luminous-supernova-was-50-times-brighter-milky-way) was in our milky way, Would it bring to earth as much light as the moon ? Could it be dangerous for the eye or anything else ?
Thanks
| From Wikipedia, from New Scientist
According to Krzysztof Stanek of Ohio State University, one of the
principal investigators at ASAS-SN, "If it was in our own galaxy, it
would shine brighter than the full moon; there would be no night, and
it would be easily seen during the day."[6]
[6] https://www.newscientist.com/article/dn28772-weve-found-the-brightest-ever-supernova-but-cant-explain-it/
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/229976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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)?
|
In a gravity theory in spacetime, the metric has signature $− + +· · ·+$.
That's a convention. Other conventions are that it has signature $+ - - -$.
Concretely this means that the metric tensor $g_{μν}$ may be diagonalized
The signature doesn't tell you that it is diagonalizable. The fact that $g_{\mu\nu}=g_{\nu\mu}$ tells you that it is diagonalizable. Normal operators are diagonalizable, those are ones that commute with their adjoint. Since this one equals its adjoint, so it commutes with its adjoint, so it's normal, so it's diagonalizable.
$g_{μν}$ 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 signature is what told you that after you diagonalize it, the values on the diagonal will be one one negative and three positive values (with your convention on the signature).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/230094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Can focused light be treated as a point source? Imagine there is a uniform, collimated beam coming from a distant light source. This beam passes through a lens and is focused to a point at the focal length. Can this "point" be treated as a point source of light as you move away from the focal point in the same direction as the light is propagating (barring the fact that the light is only in a cone)? In other words, does the inverse square law hold for this light?
Using the fact that if you put a point source at the focal length, then the light coming out the other side of a lens will be collimated, it would seem that the opposite is true and focused light can indeed be treated as a point source at the focal length.
As a last note, does this change for a nearer light source - so the rays arriving at the lens aren't parallel? I'd imagine it would only change the position of the virtual point source to be further from the lens than the focal length, but the light would still follow the inverse square law from that point onwards.
| If, indeed, the light is collimated (note, there are limits to how well collimated a beam of light can be because of diffraction) and the lens that focuses it is so-called "diffraction limited" (ie. the lens doesn't abberate the light) then the spot created at the focus does obey the inverse square law. However, keep in mind that it is not truly focused to a "point" because of diffraction. If the system is cylindrically symmetric then the focused spot will have an Airy function distribution where the distance between the first zeros (ie. width) is 2.44 x (wavelength) x (focal length/diameter of lens)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/230367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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?
| Because the seat is usually square.
If it was round (or another shape) it would be harder to make and harder to attach the back (or extend the back legs to make a back).
If it had no back, it would be a stool. Stools can have three legs.
(Not all answers have to do with physics!)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/230685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "265",
"answer_count": 12,
"answer_id": 10
} |
Total number of primary maxima in diffraction grating I am trying to determine the total number of primary maxima that can be observed when light of wavelength 500 nm is incident normally on a diffraction grating, with the third-order maximum of the diffraction pattern observed at 32.0 degrees.
Rearranging the diffraction grating formula for maxima number ( $m$ ):
$$ m= \frac{d \space \sin \space\theta_\text{bright}}{\lambda} \, . $$
I can get the right answer if I let $$\theta = 90 ^\circ \, .$$ However, I do not understand why this angle value is used.
| any angle over theta 90 will mean that the diffraction will be going behind the diffraction gratings which is impossible. so 90 is the maximum that you can get this is why you have to round down the decimal answer you will get.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/230752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why do products of nuclear decay have a lower mass than the original nucleus, when the sum of the mass of its nucleons is larger? I've just started covering the topic of binding energy in Year 13 at school (final year before University).
The definition we've been given of binding energy is that it is the work done when separating a nucleus into its constituent nucleons. Alright so far.
We've also covered nuclear decays, and an example of that we've been given is
where the mass of the Thorium isotope and the alpha particle (constituent parts) is less than the mass of the Uranium isotope (I've checked this - it's true). We were told this was due to the combined binding energy of the products being less than the binding energy of the uranium isotope.
However, this seems to contradict examples I've found on the internet, such as the formation of an alpha particle from 2 protons and 2 neutrons (see here), where the mass of the constituent parts is higher - energy must remain constant, and the alpha particle has a higher binding energy, so the mass-energy must necessarily be lower. This makes sense.
If someone is able to give a clear explanation of why both of these things are true, I'd be very grateful!
Thanks
| The mass defect and the binding energy are not linear functions of the number of nucleons. They increase until the iron and then decrease.
See the Figure 31-5 of The Mass Defect of the Nucleus and Nuclear Binding Energy
The thorium Mass Defect for one Average Nucleon ( MDAN ) is higher than an uranium MDAN.
Uranium decays in thorium and alpha particle and produces energy
But, in contrast, one hydrogene MDAN is lower than 1 helium MDAN ( the curve ).
1 alpha particle needs energy to decay in Hydrogene and free neutrons or deuterium .
Then all is for the best.
For the fun, notice the Oxygene and Carbon catastroph :)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/230933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why are these equations valid despite seemingly inconsistent units? I am having quite a difficult time in trying to understand what units are used in this paper and how to convert things to SI. For example, look at equation (1):
$$T_M \approx 1500 \rho^{1/3}\ \mathrm{K}\tag{1}$$
It seems to be showing that temperature is measured in units of $\mathrm{g\,cm^{-3}\,K}$. Then look at equation (2),
$$T_M \approx 2800 \rho^2\ \mathrm{K}\tag{2}$$
which seems to be showing that temperature is measured in $\mathrm{g^2\,cm^{-6}\,K}$. Equation (10) doesn't make sense with these either:
$$\sigma \approx \frac{5\times 10^{20}\rho^{4/3}}{T(1 + 3x)}\mathrm{esu}\tag{10}$$
How are these consistent?
| Each equation contain a different arbitrary constant: 1500, 2800, and 5 E20.
It can be assumed that each arbitrary constant has exactly the right units to make everything come out right...
It is sloppy to not specify the units of these constants...
Edited for example:
I could conduct experiments on the dynamics of falling objects, and publish that the distance of fall from rest, in metres (D), and the time of fall, in seconds, (t) seem to be related, and the best fit gives:$$D \approx4.9 \times t^2$$without implying that the units of distance are time squared...
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/231031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why don't simple circuits cause the electrical breakdown of air surrounding them? Electrical breakdown of air occurs when the electric field exerted by a charged object exceeds the electrical breakdown limit of air which is $3.10^6 V/m$. Since even 1 Coulomb of charge can exert an electric field whose magnitude can be as high as $10^{10} V/m$, this amount of charge can ionize the air surrounding it.
Moreover, 1 Ampere is defined as 1 Coulomb of charge flowing in a circuit every second. Even simple DC circuits in small electronic devices have several amperes of current flowing in them. Therefore, why that amount of charge flowing in a circuit doesn't ionize the air surrounding it; however, when we place 1 Coulomb of static charge on a conductor, it generates an electric field capable of ionizing the air?
I searched about it everywhere but there seems to be little research made on the complete electrical breakdown of air.
| Wires in a circuit are electrically neutral. They have as many positive charges as negative charges, so the net charge is zero.
There is a spatial distribution of charges within a wire. The charge is not uniform across the cross section. But that variation in charge density is fairly small. Given the overall neutrality, you'd only have to move a short distance from the wire before the field is zero.
Furthermore, if ionization were to occur, where would the liberated electrons go? Unless there was a conductor nearby at higher potential, they would snap right back onto their ions due to Coulomb attraction.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/231123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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?
| The mass will move under the influence of the normal force exerted on it by the tube walls. In order to find the motion of the mass that force would have to be specified in some way. One option would be to specify constant torque of a given magnitude. In your case, however, you seem to be specifying a kinematic condition (constant angular velocity $\omega$ with respect to the center of rotation of the tube, assuming that the tube is mounted radially; note that this is a detail that would need to be specified). The kinematic condition together with Newton's second law will allow you to find the equations of motion of your mass, also see ja72's answer.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/231417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Would you save energy by heating the air in a shower stall so that you could use colder water? It is refreshing to take a cool shower in hot weather. And for the sake of discussion, lets assume that one should be "comfortable" with temeratures when taking a shower.
Considering that the vast majority of the heat from a shower is lost down the drain, would it not be more energy efficient to heat a confined space (shower stall) to a temperature that would allow you to shower with cool water rather than hot?
How hot would it have to be in the room for 70 degree water to be "nice." 60 degree?
And since all of the energy used to heat the air in the space would remain in the space/room, would you use less energy to take a comfortable shower doing it that way?
| Consider that the specific heat of water is 4.147 kJ/kg. The specific heat of air is less than 1/4 of that, so for the air to heat you as much as the water, you would have to heat the air to a much higher temperature. That's one reason, along with conduction, that Arctic explorers say that water cools you 30 times faster than air.
So it doesn't balance out. It's 4 times easier to heat air, but water cools you 30 times faster.
Even if this wasn't the case, your system would need to have a much higher temperature in it, which would mean more electricity, which would mean more electricity wasted in wires.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/231641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
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?
| So, this is an interesting property of the mathematics of a force that diminishes like $1/r^2$ in 3D-space: if you have a uniform charge distributed over a sphere, that charge exerts no forces inside the sphere; they all balance out. Furthermore the field outside the sphere behaves as if all of the charge on the sphere was concentrated at a point at the center of the sphere. It is also, for example, true of the gravitational force: a "spherical shell" of matter exerts no gravitational forces within, but acts like a force field from a point outside.
The basic mechanism can be understood like this: a "sheet" of charge has a force which does not diminish with distance because the $1/r^2$ effectively competes with the $r^2$ of surface area that you can "see" (e.g. if we assume that the entire contribution comes from a fixed solid angle); so the space between two equally charged plates works out to have exactly 0 force due to those plates. (There may be other forces due to stuff outside the plates, of course, if the plates are not conducting and are uniformly charged.)
In the case of a ring, we could probably expect that yes, the ring would push the particle towards the center; however the equilibrium is unstable as if it deviates in any way off the plane of the ring, it will receive a push further in that direction.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/231693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
Showing the annihilation of an electron-positron pair How to show that the annihilation of an electron-positron pair under emission of one real photon (mass zero) is forbidden by energy-momentum conservation, the emission of two photons is allowed?
| The electron-positron pair has a center-of-mass reference frame where the momentum is 0.
Obviously, there exists no one-photon system with positive energy which has 0 momentum, as the energy-momentum relation for a photon is $E = p c$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/231768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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?
| The parameter which is important is the average kinetic energy of the molecules.
When the two gases mix they will move to a state where the average kinetic energy of all the molecules is the same.
So go for something like the total kinetic energy before mixing is equal to the total kinetic energy after mixing which will be made easier by having a 1:1 mixture.
Since kinetic energy depends on the square of velocity the parameter which is most often used when mentioning the speed of the molecules is the root mean square velocity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/231863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Do wormholes only allow FTL travel in "folded" spacetime? Faster than light wormhole travel depicted in science fiction and in popular science articles seems to assume that rather than being flat (with dimples centered around large masses), spacetime is fairly lumpy in order to facilitate wormhole shortcuts.
Is there a plausible reason why spacetime would be conveniently "folded back on itself" as illustrated below?
| There is nothing in physics that describes the sort of folding shown in your picture. I'm afraid it is an invention of the Science-Fiction community.
The best tool we currently have for describing spacetime is general relativity, but GR does not and cannot tell us anything about the global topological properties of spacetime. The sort of wormhole you show is described by the Morris-Thorne metric. Leaving aside the inconvenient detail that the Morris-Thorne wormhole requires exotic matter, which (probably) doesn't exist, all GR tells us is that the wormhole links two flat regions of spacetime. So it looks like:
where $A$ and $B$ are two regions of flat spacetime. However the two regions $A$ and $B$ would (in this 2D representation) remain as two flat parallel sheets everywhere with no other joins between them. In effect they would be two different universes joined by the wormhole not different regions of our universe.
There are some other related questions that you might be interested in reading:
*
*Building a wormhole
*How would you connect a destination to a wormhole from your starting point to travel through it?
*Using wormholes to see out of the visible universe
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/231921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why aren't $\hat{x}$ and $\hat{p}$ considered functions of time in the expectation value? In Griffiths Intro to QM (2nd edition), he gives the equation
$$
\frac{d}{dt} \langle Q \rangle = \frac{i}{\hbar}\langle [\hat{H},\hat{Q}] \rangle + \left\langle \frac{\partial{\hat{Q}}}{\partial{t}} \right \rangle
\tag{3.71}
$$
and he goes on to state that
$$\left \langle \frac{\partial\hat{Q}}{\partial{t}} \right \rangle =0 $$
for many operators $\hat{Q}$.
However, in problem 3.31 when deriving the virial theorem, we use $\hat{Q} = \hat{x}\hat{p}$. Why are they not considered functions of time, thus giving a non-zero value for
$$\left \langle \frac{\partial (\hat{x}\hat{p})}{\partial{t}} \right \rangle \, ?$$
| $\frac{\partial\hat{Q}}{\partial t}$ denotes the partial derivative of time, which is nonvanishing only when $\hat{Q}$ manifestly depends on time. Every operator $\hat{Q}$ can be time dependent in an implicit way such as $\hat{x}\hat{p}$ which can be time dependent when $\hat{x}$ or $\hat{p}$ depends on time.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/232021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Maximum work obtained by mixing 2 gases
Two boxes containing the same number of moles of two ideal identical
gases with the same adiabatic index (this is given as $\gamma$), at
the same initial temperature $T_i$ but with different volumes, $V_1$
and $V_2$, are brought together. Find the maximum mechanical work that
can be obtained.
If the gases have all the parameters identical but the volumes, that means that they also have different pressures, so when we are mixing them, the gas with the higher pressure will do work on the gas with lower pressure. However, I have no idea how to calculate this work and the answer given is a big messy expression.
It doesn't say anything about the recipients being adiabatically isolated, but I guess I have to assume that? The temperature will be constant? I think I should calculate the variation of entropy for the system and then relate this to the first principle to get the work done, but I have no idea how to do that.
| You have this doped out pretty well. To get the maximum work out of it, you can manually hold an adiabatic partition between them and allow the gases to move the partition very gradually until the pressures equalize. The work that the partition transmits to your hand will be the maximum work. This is the same as the net work if each gas changes volume adiabatically and reversibly until each reaches a final pressure that matches that of the other. So first express the pressure as a function of the volume of each and set the final pressures equal. This will tell you the final volume of each, the final pressure, and the final state. Then you can calculate the work done by each, and then the net work.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/232215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Metric components transformation under change of coordinates I have been studying Lie derivatives and some applications. While searching the web I found a reference with the following statement:
For a general Riemannian manifold $M$, take a tangent vector field $k=k^\mu \partial _\mu$ and consider the infinitesimal coordinate transformation,
$$
x^\mu \to x^\mu + \alpha k^\mu~,
$$
where $|\alpha| \ll 1$. Then it is possible to find that the metric components
$g_{\mu \nu}$ transform as
$$
g_{\mu \nu} \to g_{\mu \nu} + \alpha(\partial_\mu k_\nu + \partial _\nu k_\mu + k^\sigma \partial _\sigma g_{\mu\nu} ) +O(\alpha^2)~.
$$
Now, how does the author get this? Does the author just uses the usual transformation law for a $(0,2)$-type tensor? This does not seem to be the case, since he finds so many terms. Moreover, I do not know the inverse transformation, so I wouldn't be able to apply it. Can somebody help me?
| The metric being a rank $(0,2)$ tensor transforms under general coordinate transformations $x^\mu \to x'^\mu(x)$ as
$$
g'_{\mu\nu} (x') = \frac{ \partial x^\rho}{ \partial x'^\mu } \frac{ \partial x^\sigma }{ \partial x'^\nu } g_{\rho\sigma} (x)
$$
Now set $x'^\mu (x) = x^\mu + \alpha k^\mu(x)$ in the above expression and take a limit of small $\alpha$. You should then get
$$
g'_{\mu\nu}(x) = g_{\mu\nu}(x) - \alpha ( g_{\mu\rho} \partial_\nu k^\rho + g_{\nu\rho} \partial_\mu k^\rho + k^\rho \partial_\rho g_{\mu\nu} ) + {\cal O} (\alpha^2)
$$
The above is the correct version of the transformation. I believe you have gotten it wrong in the question (v1)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/232540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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?
| Producing ultra-heavy elements in nature is not easy. So their absence "in nature" does not mean they cannot exist or cannot be created given the right conditions.
Some details:
The valley of stability becomes increasingly n-rich, so neutron capture reactions are essential.
To get beyond lead requires rapid neutron capture in the r-process. The requirements here are a dense flux of neutrons and a capture timescale that is shorter than the beta decay timescale trying to take the nuclei back towards the stability valley.
Once the neutron flux diminishes (these things happen in explosive events like supernovae and neutron star mergers) then beta decay does dominate and takes the nuclei back to the stability valley.
In principle, stable elements of any atomic number could be formed in this way, but ultimately you have to compare neutron capture rates with all the processes that act to destroy the intermediate neutron rich nuclei, such as photodisintegration and fission. For the really heavy nuclei the principle problem is fission - the n-rich heavy nuclei just break up before they can capture any more neutrons given the neutron fluxes that exist in "natural sources".
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/232615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Build a device that can set voltage according to intensity of the light shined I want to build a device that uses photodiodes and photoresistors to set the voltage in the circuit that shines a bulb. The idea is simple. When the light is intense, the bulb should shine less, when the light is less intense the bulb should shine more. How can I build that device?
| Assuming the bulb emits in the visible wavelength regime, then you can use a silicon photodiode. Depending upon the particulars of your "bulb" you could put the photodiode in series with a resistor and put that combination in parallel with the bulb. So as the photodiode receives more light, it's resistance will decrease thereby allowing current to flow in that part of the circuit, and away from the bulb. This assumes that you are using a constant current source.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/232712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the physical intuition behind the Bragg peak? The wikipedia page says:
Energy lost by charged particles is inversely proportional to the
square of their velocity, which explains the peak occurring just
before the particle comes to a complete stop.
What is the physical reason why the energy loss goes inversely as the square of the velocity?
| See if this argument works - I am making this up on the spot so there is definitely space for argument...
Most of the interactions with the electrons will not be "head-on collisions" but rather electrostatic interactions. If we get to a certain distance of an electron, it will feel the force and undergo acceleration. If the time of the interaction is short, the distance moved will be small and the force will be approximately constant.
At any given time, the momentum transfer will be proportional to the force:
$$\delta p = F \delta t$$
Now the force is given by the distance between the particle and the electron; while this is going to introduce all kinds of geometric terms we will ignore that for now.
Expressing time in terms of velocity and distance we get
$$\Delta p = F \frac{\Delta x}{v}$$
Energy transferred to the particle goes as momentum squared - it follows that energy lost per unit distance goes as the inverse of velocity.
One way to think about this intuitively - the longer the particle spends near a given electron, the more time it has to exchange energy (given that all these are electrostatic interactions - not "collisions").
I found a nice derivation of the above that mostly agrees...
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/232798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
time step in kinetic montecarlo simulation I have a doubt about how the time step is calculated in kinetic montecarlo simulations.
One state with index $i$ is connected to other $N$ states, indexed by $j=1...N$, by transitions that happen with rates $r_{ij}$ (form $i$ to $j$). At each iteration of the algorithm, one of the $N$ transitions is randomly chosen (with the right probability), and the time step is cleverly calculated using a random number $u$ in $(0,1]$ by
$\Delta t = \frac{ln(1/u)}{R}$
where $R$ is the sum of all the rates of the transitions leaving from state $i$ : $R=\Sigma_{j=1}^N r_{ij}$. In this way $\Delta t$ is exponentially distributed with mean $1/R$.
My doubt: I understand that the time between events (or transitions) $\Delta t$ should be exponentially distributed because the events are Poisson distributed. But why should the mean of the time between events be $1/R$? If the algorithm in a particular iteration picks up the particular transition $i$ to $j_o$, with the rate $r_{ij_o}$, I would say that the average residence time in state $i$ before leaving to state $j_o$ should be $1/r_{ij_o}$, and not $1/R$. So instead of the $\Delta t$ above, I would use in that particular iteration a time step
$\Delta t=\frac{ln(1/u)}{r_{ijo}}$.
Is this wrong?
| In kinetic Monte Carlo, the idea is to describe a trajectory as a set of events, at which the system makes a transition from one state ($i$) to another ($j$). To generate such a trajectory, we need to randomly select both the states that are visited and the intervals between them.
The time interval $t$ between a pair of events is the time in which nothing happened—the system stayed in the current state because none of its possible transitions occurred. So the probability distribution for $t$ involves the total rate for anything happening, which is $R$. After we have determined the interval $t$, we can then decide which event it was that happened.
An example might help: Imagine you have a large number $n$ of radioactive atoms, with a decay rate $\gamma$. The typical time until any given atom decays is $\gamma^{-1}$, but the typical time until the first decay is $(\gamma n)^{-1}$. When you generate a trajectory, this is the mean time you wait until the first event. If $n$ is very large, you will not have to wait long to see a decay, even if $\gamma$ is small.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/232888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Torque: The T-shaped stick problem This is a problem that I have been unable to solve for some time.
Imaging a T-shaped stick, as shown in below image, which do not deform in any appreciable way and has pivot point at the tail of the "T".
There are two questions in my problem, the first is: given that the "T" is symmetrical, how would applying two units of force on one of the "T"'s horizontal appendage differ from two one unit of force applied on both appendages in opposite direction, in terms of the torque received at the pivot?
(See image)
The second question is: how would the situation differ if the "T" is "italicized", but the forces are exerted perpendicular to the appendages? see image:
What is a systematic way to explain and describe the torque received at the pivot in terms of the lengths L and H and possibly the angles between them?
| If a force $\vec{F}=(F_x,F_y)$ is applied at a location $\vec{r}=(x,y)$ then the torque at the origin is
$$ \vec{\tau} = \vec{r} \times \vec{F} \\ \tau = x F_y - y F_x$$
All you need to do is sum up the torques at the pivot for the different situations in order to understand how this mechanism will move.
If you have two equal and opposite forces $F$ the net torque applied is going to be $F\,d$ where $d$ is the perpendicular distance between them. Care must be taken to consider if this torque is positive (counter-clock-wise) or negative (clock-wise).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/233012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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.
| My guess would be that if you're sitting, the surface of contact between you and the slide is smaller, thus the pressure is higher than if you were to lay on your back. Your jacket is a deformable material, its friction coefficient with steel might vary with applied pressure.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/233094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
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?
| Since the LINE, a section of ROPE, has no attached mass or restraint and it seems it is accelerated as a whole, there is no tension. If one end is pulled to accelerate it then F=ma=0.0 N.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/233160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
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?
| For time independent potentials there is a field of mathematics studying this. The answer is no for physically reasonable potentials. Look for Schrodinger semigroup.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/233304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Why can't electrons leave negatively charged objects in a vaccum? If an object is negatively charged, and electrons repel each other since they have like charges, why doesn't that negatively charged object "kick away" electrons?
Why doesn't diffusion occur? Why does there need to be a conductor for the electrons to go to?
| Matter is full of positively charged protons, and those positively charged protons attract the negatively charged electrons. In general, if you take a chunk of neutral matter and add one extra electron then the attractive force due to the protons in the matter outweighs the repulsive force due to the electrons in the matter, and the extra electron is bound into the matter. That is, you have to put in energy to get it out again. This binding energy is known as the work function.
If you now add a second extra electron then the attractive force due to the protons is roughly the same, however the repulsive force due to the electrons is slightly increased because now the first extra electron is adding to the repulsion. So the second electron is less tightly bound into the matter.
And so on as you keep adding more and more electrons. Eventually you'll reach the point where you can't add any more electrons because the electrons you've already added repel them so strongly. At this point electrons will indeed spontaneously leave the matter in a process known as field emission.
There are a couple of other points worth noting. If you heat matter you increase the energy of the electrons within it, and eventually you increase the energy enough for the electrons to jump out of the matter. This process is known as thermionic emission. And finally if you shine light on the matter the light can excite electrons enough for them to jump out, which is known as the photoelectric effect.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/233427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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.
| You are on the right track.
For a mechanical pendulum, the relationship is linear. You don't need to know how many swings of the pendulum corresponds to how many seconds. If the pendulum is x% slower, it will report x% fewer seconds per day.
Now since length goes as $\ell = \ell_0(1+\alpha \Delta T)$ and period of pendulum as
$$T = 2\pi\sqrt{\frac{\ell}{g}}$$
We can do the Taylor expansion for $\ell$, noting that $\sqrt{1+x}\approx 1+\frac12 x$ for small $x$, to give us
$$T_2 = T_1 (1 + \frac12 \alpha \Delta T)$$
as you noted.
Now the number of swings per day times the time per swing should equal one day, or $N\cdot T=86400$. It follows that the number of seconds per day, N, is changed by
$$N_2 = N_1 \frac{T_1}{T_2} = \frac{N_1}{1+\frac12\alpha \Delta T} = N_1\left(1-\frac12\alpha T\right)$$
The difference in seconds per day then follows. I will leave that as an exercise.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/233532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
What would happen if I gathered stellar sized masses of iron? Lets say I had a bag that when turned upside would start pouring out iron shavings and never ever stop. Viola, there's my infinite source of iron. Now, lets say I just continued to dump this iron together until I had a ball of iron the size of earth, Jupiter, the sun, and kept on going?
To my understanding, self sustaining fusion would not occur because iron takes more energy to fuse than it releases. Is that to say no fusion would occur at all? Could I just keep dumping iron until vwoop my big ol' ball of iron is now a black hole? Would some fusion occur in the center of the iron ball and just not be propagated to the exterior of the ball?
| Supernova happens when the core of a supermassive dying star starts fusing iron and heavier elements under massive gravitational pressure. The reaction is endothermic, unlike the fusion of lighter elements (iron is the peak) so the resulting outward radiation pressure stops apposing inward gravitational pull and the star collapses on itself.
The pressure will become high enough to fuse electrons and protons together to form Neutrons, which are charge neutral and do not repel each other (degeneracy pressure) and the core contracts by a factor of 10,000 in a fraction of a second and turns into a tightly packed sphere made out of neutrons only (degenerate matter).
The outer layers of the star collapse onto this incredibly dense object and bounce off in a spectacular shockwave ... a supernova. The resulting heat vaporizes the material and makes it glow. Depending on the mass of the remaining core, you would either end up with a neutron star or a black hole.
So I presume this would be the faith of your supermassive ball of iron :)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/233626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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 general formula for error propagation is:
$$\Delta f(x_1,x_2,\ldots)=\sqrt{(\frac{\partial f}{\partial x_1}\Delta x_1)^2 + (\frac{\partial f}{\partial x_2}\Delta x_2)^2 + \cdots}$$
where $\Delta m$ means "standard deviation of lots of repeated measurements of m".
Where does this come from? By calculus, when all the $x_i$s vary, it causes the following variation of $f$:
$$\delta f = \sum_i (\partial f / \partial x_i) \delta x_i$$
where $\delta x_i$ is the difference between this particular measurement of $x_i$ and its true value, and $\delta f$ is ditto for $f$. We are assuming that the errors are relatively small (ignore $\delta x_i \delta x_j$ terms etc.)
I think you had all this so far. The part that you're missing is:
For independent random processes, the variance of the sum is the sum of the variances.
The analogous statement is not true for standard deviations. It is only true for variance, i.e. standard deviation squared.
Since we want the standard deviation of $\delta f$, we need to add up the variances of $(\partial f / \partial x_i) \delta x_i$ and then take the square root. So we wind up with the formula that I wrote at the beginning.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/233724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Cause for spikes in Trinity nuclear bomb test In Richard Rhodes' book, The Making of the Atomic Bomb, I was reading about the Trinity nuclear test. High speed photos were taken and this one is from <1ms after the detonation. The book mentions the irregular spikes at the bottom of the image, but does not explain them. Is there a specific reason or explanation for these odd spikes in the relatively spherical explosion?
Nuclear explosion photographed less than one millisecond after detonation. From the Tumbler-Snapper test series in Nevada, 1952, showing fireball and "rope trick" effects. The fireball is about 20 meters in diameter in this shot
| The answer is in wikipedia
The photograph on the right shows two unusual phenomena: bright spikes projecting from the bottom of the fireball, and the peculiar mottling of the expanding fireball surface.
The surface of the fireball, with a temperature over 20,000 kelvin, emits huge amounts of visible light radiation (more than 100 times the intensity at the sun's surface). Anything solid in the area absorbs the light and rapidly heats. The "rope tricks" which protrude from the bottom of the fireball are caused by the heating, rapid vaporization and then expansion of mooring cables (or specialized rope trick test cables) which extend from the shot cab (the housing at the top of the tower that contains the explosive device) to the ground. Malik observed that when the rope was painted black, spike formation was enhanced, and if it were painted with reflective paint or wrapped in aluminium foil, no spikes were observed – thus confirming the hypothesis that it is heating and vaporization of the rope, induced by exposure to high-intensity visible light radiation, which causes the effect. Because of the lack of mooring ropes, no "rope trick" effects were observed in surface-detonation tests, free-flying weapons tests, or underground tests.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/233798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "49",
"answer_count": 2,
"answer_id": 0
} |
Using Steam to lift a mass? I have a task set to create a system to produce the largest values for potential energy, mgh.
I am given 1 litre of boiling water and with this use the steam produced to lift the mass of weight that i can select. I am not aloud to add any energy to the system. I am not aloud to use pre-made equipment such as thermo-generators. £30 is the limit.
I have had some ideas that have been thrown down the drain as the were deemed not satisfactory.
First, i suggested using a steam gun by collecting the steam then releasing it all at once.
Then, i suggested making a Thermo-electric system to a motor, but again it said it was deemed to not have enough torque.
Any suggestions would be greatly appreciated.
| The maximum potential energy is limited by the energy content released by the one liter of boiling water. you are able to boil away all of the water - this gives you the total energy released.
The obvious answer is to follow Watt and build a steam engine - the height of the piston is the measure of the potential energy extracted, U=mgh. You can use a ratchet to prevent the piston from falling down.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/233956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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?
| A good reference for choosing appropriate laser beam for a given type of atom, plz read Michal J Martin's PhD thesis at jila, Prof. Ye Jun's group site. Some moderator would not be happy with links so I delete it. Chapter 2 & 4 all provides condensed materials about laser wavelength.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/234179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can "...electrons flow in metals, but not in the ground..." explain grounding rods? I really enjoyed Why is the charge naming convention wrong? But, in the comments at the very end, the statement that "...electrons flow in metals, but not in the ground..." left me uneasy.
I was taught that the physical process of "electron drift" was actually quite slow, and opposite in direction to "conventional current flow". So my question is, "Why do the utility companies, who use electrons as the carriers of electric current through solid metal power lines, make so many connections to metal rods driven into the ground? Don't electrons "actually" travel from one place to another? (i.e. up from out of the ground, through the grid, and back to the station?)
| Ground is at 0 potential,so,it accepts electron from negative terminal.
And at very far place any positively charged electrode accepts electron from ground and current flow.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/234298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
So how much fuel does a hypothetical fusion plant need? Reading about potential fusion plants it's easy to get the impression that we have nearly infinite access to the fuels required. But looking aside form how much we have available, I am interested in something I have a hard time finding information about; how much fuel would a fusion plant require? I assume this depends on the type of fuel, and it obviously depends on how much energy a particular plant would produce, but do we know approximately how much fuel (over time) can produce how much energy in a (reasonably efficient) fuel power plant?
To go for another approximation, can we (potential problems with heat dissipation and radiation aside) for example put a fusion reactor on a submarine or a spacecraft and have it run for weeks/months/years or would it require very frequent resupply of fuel to keep the fusion reaction going?
| Factually we have next to no fuel for fusion reactors, neither hypothetically nor practically. All of the tritium that would be required will have to be bred either in fission reactors or the fusion reactors themselves. Since a fusion reactor does not have as many excess neutrons, special neutron multiplier enriched blankets will have to be used. Even then realistic tritium breeding ratios will be in the ranges of 1.04 to (optimistically) 1.14. As a result it will take many decades for fusion reactors to even establish their own fuel cycle.
As for the demand... the ITER homepage cites the requirement for 300g tritium/day for an 800MW demonstrator: iter.org/mach/tritiumbreeding. According to en.wikipedia.org/wiki/Tritium current world demand is 400g per year at a cost of $30,000 per gram. The total tritium still available (most of it in thermonuclear weapons?) is said to be around 75kg, so we couldn't even run a single small reactor for much longer than half a year.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/234379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Single slit diffraction - choosing a wavelength? For the classic experiment of determining the slit width of a single slit. If we assume the rough order of magnitude of the width is known. What factors determine the choice of wavelength?
(Clearly we want $\lambda<w$ where $w$ is the slit width, but what other factors come in?)
Extension
I heard from someone that we should make the wavelength on the order of the desired resolution, I don't really understand how resolution comes into this as we are using monochromatic light. Can someone explain if and why this comment makes sense?
(I haven't posted this as a new question as it would likely be closed as a duplicate).
| To answer your specific question
"I heard from someone that we should make the wavelength on the order of the desired resolution, I don't really understand how resolution comes into this as we are using monochromatic light. Can someone explain if and why this comment makes sense?"
In order to maintain Resolution for viewing things that are really small like molecules etc., you need a shorter wavelength of light. Or the way I prefer to think about it is with higher frequency photons. Synchrotron radiation will produce monochromatic light like X rays. The higher the frequency the higher the resolution. Or the shorter (Y) is in Chris's equation above.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/234485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How do I remove the negative sign from this derivation? A homework problem required me to show that the first equation below can be written in the form of the second equation. It was all fairly simple except for the negative sign. I'm not sure how this is supposed to cancel out. Might there be some conceptual way that the negative sign is removed?
1st Equation:
$$E_n=-\frac{mk^2Z^2e^4}{2 \hbar^2 n^2}$$
2nd Equation:
$$E_1=\frac{\alpha^2 mc^2}{2}$$
All I did was to substitute this and simplify:
$$\alpha=\frac{ke^2}{\hbar c}$$
| There should be a negative sign in Equation 2. The exercise is a very simple one in substitution and does not require any sign cancelling.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/234656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Do electrons in an atom always have the same 'direction'? Perhaps speaking of direction of an electron isn't quite correct. But does QM indicates a kind of way whether all electrons are going e.g. 'clockwise' or not? Of course QM just gives a probability where the electrons are, but can you emerge whether they are going, in some way, in the opposite direction of each other?
| In my opinion, in stable atoms part of its electrons should rotate in opposite direction of other electrons to maintain stable conditions.
This means that if it loose an electron or gain the atom charge will change and called ion.
In our old understanding only losing and gaining is from out orbit. I believe this is completely wrong thinking because the atom and its electrons are believed to change shape and electron orbits were believed that electrons are moving in circular forms they are moving in spiral form around the nucleus.
And this type of movement is the explanation for forming what is known to us magnetic field.
When potential difference been applied on what is known the area of electron rotation space or orbit.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/234734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 2
} |
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?
| In my mind, to unravel this question, we need to consider the simplest scenario possible. So, let us disregard quantum systems and small systems and consider only statistical mechanics of big classical systems.
Entropy is defined as the log of the number of microstates at a given energy E (up to the Botlzmann constant). Namely, for a classical Hamiltonian, we have to find all the configurations that correspond to the same desired energy value E. This implies that the only information you have about the system is it's energy and the Hamiltonian governing its equilibrium properties.
Now, suppose that you have $\Omega$ such allowed microstates. Introducing a measurement, or gaining information about the system, forces you to consider more constraints about your microstates. This can only reduce the number of allowed microstates or leave them unchanged.
I hope it helps.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/234817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
In a 2D CFT, is the free boson $X$ a primary field? A primary field $\mathcal{O}(w,\bar{w})$ with weight $(h,\bar{h})$ is defined by having the following OPEs with the stress tensor:
$$T(z)\mathcal{O}(w,\bar{w})=\frac{h\mathcal{O}}{(z-w)^2}+\frac{\partial \mathcal{O}}{z-w},\qquad \bar{T}(\bar{z})\mathcal{O}(w,\bar{w})=\frac{\bar{h}\mathcal{O}}{(\bar{z}-\bar{w})^2}+\frac{\bar{\partial} \mathcal{O}}{\bar{z}-\bar{w}}.$$
Let us consider a theory of a free boson defined by:
$$S = \int d^2 z\ \partial X\bar{\partial}X.$$
Then $\partial X$ is a $(1,0)$ primary field with and $\bar{\partial}X$ is a $(0,1)$ primary field (see e.g. Tong's string theory notes, section 4.3.3).
The field $X$ itself has the OPE
$$T(z)X(w,\bar{w})=\frac{\partial X}{z-w},\qquad\bar{T}(\bar{z})X(w,\bar{w})=\frac{\bar{\partial} X}{\bar{z}-\bar{w}}.$$
(See e.g. Polchiski eq. 2.4.6). So, by the definition above, it should be a primary field with weight $h=\bar{h}=0$.
However, I've looked through many textbooks and lecture notes (e.g. Di Francesco, Blumenhagen, Polchinski, Tong...) and it is never explicitly said that $X$ is primary. So... Is it primary or not?
*
*If it is primary, then how come its descendants $\partial X$ and $\bar{\partial} X$ are also primary?
*If it is not primary, then how come its OPE is exactly the correct OPE for a primary field of weight $(0,0)$?
| To see why the descendants are primary, you can use
$$
\partial\left(T(z)X(w,\overline{w})\right) = T(z)\partial X(w,\overline{w}) = \frac{\partial^2 X}{z-w} + \frac{\partial X}{(z-w)^2}
$$
And see that it is a primary field of weight $h = 1$, $\overline{h} = 0$, and similarly for the other field....
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/234926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evidence for quantum gravity from gravitational waves The rumor spreads that physicists will make their big gravitational wave announcement this thursday.
I am far from being an experimentalist, but I want to know if there is any chance that the mentioned observations will reveal any experimental evidence for quantum gravity.
Personally, I feel like this is merely impossible as gravitational waves should propagate in agreement with classical General Relativity. But I might very well not see pitfalls if there are any.
UPDATE: by "evidence for quantum gravity" I mean experimental results which don't agree with predictions of General Relativity or its classical generalizations. They might be coming from string theory or loop gravity or from a completely different approach which we haven't discovered yet. I want to understand what are the chances that new physics will show up during the upcoming gravitational wave observations.
Also, I expect gravitational waves to be in a highly coherent state, thus making the individual gravitons unobservable. Is this true?
| Formula for the metric tensor of the Schwarzschild solution
$$g_{00}=-1+r_g/r:g_{rr}=\frac{1}{1-r_g/r},g_{\theta \theta}=r^2,g_{\phi \phi}=r^2sin^2\theta$$
Formula for the changing part of the pseudo-tensor energy density for the non-relativistic Schwarzschild solution
$$\Delta t_{00}=\frac{11}{2r_1^2}+\frac{15}{8}(\frac{dln(1-r_{g1}/r_1)}{dr_1})^2 +\frac{15}{8} cot^2(\theta)/(1-r_{g1}/r_1)$$
Where $r_k^2=r_{g_k}^2+d^2+x^2,k=1,2$, x - middle distance between black holes
With a radius greater than the gravitational minimum energy density is achieved under the condition $\theta=\pi/2$.
Energy density is equal to
$$\Delta t_{00}^{\pi}=\frac{11}{2r_1^2}+\frac{11}{2r_2^2}+ \frac{15}{8} [(\frac{r_{g1}/r_1)}{r_1-r_{g1}})^2+(\frac{r_{g2}/r_2)}{r_2-r_{g2}})^2] $$
Half the distance between black holes is $x=V_0(t_0-t_1-t),t \in [0,t_0-t_1]$, $t_0$ - signal duration, $t_1$- signal duration after crossing the gravitational radius
After crossing the gravitational radius, the minimum energy corresponds to the angle $\theta=\pi \sqrt{\frac{\sqrt{r_{g1}^2+d^2}-\sqrt{r_{g2}^2+d^2}}{\sqrt{r_{g1}^2+d^2}+\sqrt{r_{g2}^2+d^2}}}$
the emitted signal corresponds to the energy difference
$$\Delta t_{00}^0=\frac{15}{8} [1/(1-\sqrt{r_{g1}^2+d^2}/r)- 1/(1-\sqrt{r_{g2}^2+d^2}/r)]= \frac{15}{8}\frac{\sqrt{r_{g1}^2+d^2}-\sqrt{r_{g2}^2+d^2}}{[r-(\sqrt{r_{g1}^2+d^2}+\sqrt{r_{g2}^2+d^2})/2]^2}/sin^2(\theta)$$
This formula, when the masses of black holes coincide, should give a finite non-zero radiation energy. For different masses of black holes, the passage to the limit is not needed.
Moreover, the limit passage is realized according to the rule of L'Hôpital provided $ r_{g1} \to r_{g2}$
$$\Delta t_{00}^0= \frac{15}{8\pi^2} \frac{\sqrt{r_{g1}^2+d^2}+\sqrt{r_{g2}^2+d^2}}{[r-(\sqrt{r_{g1}^2+d^2}+\sqrt{r_{g2}^2+d^2})/2]^2}$$
$$r=V_0(t_0-t_1-t);t \in [t_0-t_1,t_0],r<(r_{g1}+r_{g2})/2$$
The radiated mass is determined from the equality
$$\Delta t_{00}^0=\frac{11}{2r^2}+\frac{15}{8}(\frac{dln(1-r_{g}/r)}{dr})^2,
r_g=2G\Delta m/c^2$$
The calculated energy density is proportional to the signal envelope, from where the gravitational radii can be determined to within a factor using the least squares method. This factor is determined from the condition $\omega_{\pi}=\frac{c}{r_{g1}}+\frac{c}{r_{g2}}$ before the intersection the gravitational radius and equals after the intersection $\omega_{0}=\frac{2c}{r_{g1}+r_{g2}}$ of the gravitational radius
where $\omega_{\pi},\omega_{0}$ is the frequency of the received signal before the intersection of the gravitational radius and after the intersection of the gravitational radius
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/235019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
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 $dh/dt=dr/dt$?
| As you go deeper, the pressure increases, decreasing the volume of the gas in the bubble. In water, the pressure is appoximately $14.7(1+\frac d{33})$ psi where $d$ is the depth is measured in feet. The volume goes as the inverse of the pressure.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/235114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why do we assert Hulse–Taylor binary system's orbital decay to gravitational waves and not radiation? From this link
The Hulse–Taylor system's orbit has decayed since the binary system
was initially discovered, in precise agreement with the loss of energy
due to gravitational waves. The ratio of observed to predicted rate of orbital decay is calculated to be 0.997±0.002.
But also
The total power of the gravitational radiation (waves) emitted by this
system presently, is calculated to be 7.35 × 1024 watts. For comparison, this is 1.9% of the power radiated in light by our own
Sun.
So how do we know the Hulse–Taylor system's loss of energy is not (at least partially) due to electromagnetic radiation too? Especially if we don't even see the other pulsar due to its unfavorable inclination?
| The emission of gravitational waves causes the separation $r$ between the two binary components to decrease. As they do so, the power emitted in gravitational waves increases as $r^{-5}$. Thus the rate of change of the orbital period is very non-linear, with $dr/dt \propto r^{-3}$.
Now, if the mechanism responsible for the spin-down was somehow pulsar radiation, then you would have to arrange for the pulsar spin to increase with time in order to provide the same accelerating power loss. But that is the opposite of what pulsars do - they all spin down with age and become considerably less powerful. Pulsars are also powered by their spin - that is the rotational kinetic energy is ultimately the source of power. The rotation of the pulsar is not coupled in any straightforward way with the orbital kinetic energy of the binary system. There would be (weak) tidal coupling, but that would tend to slow down the pulsar too.
Aside from this argument, one could also add that the model of the binary system that incorporates the gravitational wave losses fits the data with exquisite precision. Thus any other source of power loss one might hypothesise not only has to provide the right sign for the time derivatives of the power loss, but it also has to get the temporal behaviour exactly right over decades.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/235241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
So Black Holes Actually Merge! In 1/5th of a Second - How? I've read a lot of conflicting answers in these forums. However, today saw the awesome announcement of gravitational waves. Two black holes merged: http://www.slate.com/blogs/bad_astronomy/2016/02/11/gravitational_waves_finally_detected_at_ligo.html
Not only that, they merged FAST. In 1/5th of a second revolving around each other 250 times a second. The entire event was quicker than a heartbeat. Moreover, we observed this happening as distant outsiders. So now we can say for sure:
*
*Objects approaching the event horizon DO NOT appear to slow down
*Black holes CAN merge in a finite (and quick) amount of time
*And all this is wrt a frame of reference far, far away
To quote the NYTimes article:
One of them was 36 times as massive as the sun, the other 29. As they approached the end, at half the speed of light, they were circling each other 250 times a second.
And then the ringing stopped as the two holes coalesced into a single black hole, a trapdoor in space with the equivalent mass of 62 suns. All in a fifth of a second, Earth time.
However, everything I've read so far has let me to believe that an outside observer should never be able to measure the collision happening in a finite time. So what exactly is happening here? I must have read at lest 5 different versions of this so far everywhere in these forums over the past several years.
| Just like photons don't age but still move, black hole horizons don't age but move and can thus merge with other black holes.
The formation process of a black hole takes an infinite amount of time according to external observers, but that doesn't mean it needs to remain in its initial position. The frozen horizon moves through space and emits gravitational radiation to communicate its new position to its environment. Nothing besides Hawking radiation escapes the inside of a black hole, including gravitons.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/235307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 3,
"answer_id": 1
} |
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?
| No, think of gravitational waves as normal ocean waves. When you see a wave gliding across the sea, it's merely the result of a Force acting on that sea, not the seas creating the wave. There may be Gravitins, but they can't create a wave within themselves.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/235413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
} |
What is time, does it flow, and if so what defines its direction? This is an attempt to gather together the various questions about time that have been asked on this site and provide a single set of hopefully authoritative answers. Specifically we attempt to address issues such as:
*
*What do physicists mean by time?
*How does time flow?
*Why is there an arrow of time?
| What is time? Ask ten different people and you'll get ten different answers. This is how I understand it. The word time is a term describing temporal motion. Time is what we do. Time doesn't flow, we time. We don't move through the time dimension, we time through the temporal dimension. Time is a verb.
It's our motion from one moment to the next moment. Clocks measure our progression. Clocks measure how far we time. Temporal distances are measured in seconds, minutes, hours, days, years, etc. So we move through the spatial dimension and we time through the temporal dimension.
The speed at which we time is known as our temporal velocity. Timing through the temporal dimension is not something only humans do. Everything except massless particles, times.
Now, because the spatial dimension and the temporal dimension are interlinked, the faster we move, the slower we time, but due to the fact that our perceptions are determined by our temporal velocity, we never perceive time dilation locally.
Now, I may be wrong, but there is no scientific evidence which suggests that my theory is implausible. Correct me if I'm wrong, but, if you disagree, just because it doesn't match your current views, don't comment.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/235511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "99",
"answer_count": 9,
"answer_id": 8
} |
Why do detectors for gravitational waves have only two perpendicular arms, not three? I wonder why detectors for gravitational waves have only two perpendicular arms, not three. Having three arms appears to allow for better detection of direction, or may even increase sensitivity (I may be wrong). So far I came up with a few guesses, but I'm by no means an expert.
*
*two arms are cheaper than three
*a 600m tower is a no-go (stability), a 600m deep hole is a challenge to build and operate
*two arms is all we need, three would not guarantee significantly better results
*photons going down and up would gain and lose energy; those in the surface arms do not or much less so. Signal evaluation would be messy.
What are the true reasons?
| The basic idea is that you have destructive interference. This is hard(-er) to achieve with three photons as well as I don't see a trivial way to do the beam-splitting.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/235598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
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)?
| See if there will be charge on outer faces, then electric field inside plate will not be zero, which is not possible. So all the charges will be on the inner faces ( facing each other of parallel plate capacitor)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/235708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Gravitational waves in other dimensions I know this question is purely speculative, as we don't know if more dimensions do exist and also we do not know if gravity is indeed stronger in other dimensions (if they were to exist). But, one of the possible explanations of why gravity is so weak compared to other forces is that it exerts its strength in other dimensions, which are too small for us to detect them. However, if that were true, wouldn't the gravitational waves on those dimensions be stronger and cause larger stretching and therefore, in some cases, allow us to detect those extra dimensions? Are there any experiment that look at this case scenario?
| The other dimensions in those examples are quite small.
Imagine a long very unwide piece of paper. Now attached the long ends together so you get a tall narrow cylinder. It's like you have one direction where if you go that way you very quickly end up back where you started and one direction where you can walk very far and end up someplace different.
That's what those extra directions are like. What we'd call a wave is a wave going in that long direction. Something going in the orthogonal direction would just circle around and at some point be right where it started, it wouldn't end up going anywhere.
I think the idea is that you fill up in all the directions in a $1/r^3$ and get weaker and weaker until the other directions curl around at a distance $R$ and you stop getting weaker at $1/r^3$ and start getting weaker at $1/(r^2R)$ since then an expansion in the additional small directions is now coming back.
So a wave could expand and get weak but then when it's expanded through that other dimension it now just starts affecting itself.
It's my understanding that this theory with the extra dimension was already falsified.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/235814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
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?
| By suggesting that the RKE at the top is zero, you are suggesting that there is a change in RKE from the bottom and the top. However, just like you need a net force to change kinetic energy (i.e. do Work) you need a net torque in order to change the rotational kinetic energy. In this situation, because this part of the ramp is frictionless, there is no torque on the ball throughout this part of the ramp, so there can be no change in RKE. Therefore the RKE at the top must be the same as the RKE on the bottom, i.e. non-zero.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/237174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
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.
| The method that I usually follow (it works perfect and is easy to understand and also to find errors if any):-
*
*Choose an observer (in most problems specially with pseudo forces) place your observer at a place which makes the motion of other bodies comparatively easier to consider. [PHYSICS is based on how you observe things]
*In case of FBD, make two axes of your own such that you don't have to consider much components. E.G. if you are working on an inclined plane then consider the incline as one of your axes.
*Then apply necessary conditions and find net "force" in the axis that is required. If it is in force equilibrium then sum of all forces (" net force") along X-axis and Y-axis (one defined by you) must be zero and likewise.
Try to come up with new methods that you are comfortable with... ATB!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/237389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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?
| They leave slowly because of the air exchange via window, depend upon rate of exchange.
On a lighter note, even if you do not open the window, the smell will seem to disappear in a little bit :)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/237494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How is a potential divider able to reduce current to zero? For example, a filament lamp, in series with a uniform resistive wire, can have its voltage varied by moving the sliding contact, e.g., a rotatable wheel.
However, why is a potential divider able to reduce the current to zero, but a variable resistor in series with the filament lamp cannot reduce current to zero?
| A standard sort of potential divider has a slider which can travel from one end and produce a maximum potential across its output terminals and hence to the circuit under investigation and then when the slider is at the other end the voltage across its output terminals is zero and hence as no voltage is supplied to the circuit under investigation no current will flow in that circuit assuming there are no other voltage or current sources present.
Later
For the potential divider I have made the assumption $R\gg R_1,R_2$ only to make the sums easier.
Even if that condition was not satisfied the potential divider could still have an output voltage of zero when $R_2$ is zero.
For the series resistor you can make the current zero by breaking the circuit. However if had a complete circuit and $R_3$ was very large there would still be a current flowing.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/237569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
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.
| There is a trick that every mechanical engineer uses to solve the problems on mechanics
firstly draw a perpendicular at the point of application of the force on the plane.
then make an arrow along this line which you have constructed and along the plane. this arrow will point to the gross direction of the vector .
like in your first example the arrow on the construction line will be in the upward direction
but in the second case the arrow direction will be in the downward direction.
this is a homework question so according to the policy of the cite I can guide and not give the exact solution.
Hopw you can find the hit useful
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/237639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Spin $1$ vs. spin $1/2$
A crystal contains $N$ atoms which possess spin $1$ and magnetic moment $\mu$. Placed in a uniform magnetic field $B$ the atoms can orient themselves in three directions: parallel, perpendicular, and antiparallel to the field. If the crystal is in thermal equilibrium at temperature $T$ find an expression for its mean magnetic moment $M$, assuming that only the interactions of the dipoles with the field $B$ need be considered.
I have 2 questions about this problem:
What does it mean for an atom to possess spin $1$ instead of $\pm \frac{1}{2}$? I thought particles can only have one of these 2 values for spin value.
How does a perpendicular orientation affect the probabilities of the other 2 orientations?In the book I am reading through, it gives an example with only a parallel and antiparallel possibility for spin. However, I know that intuitively these can't be the only 2 options. When the dipole flips from parallel to antiparallel, it must swing through a perpendicular state (assuming only $x,y$ directions apply here).
| A atom is constituted of fermions(proton, neutron and electron). But whenever the atom an atom has even no of constituents it behaves like a boson. This is very easy to understand. All fermions have spin 1/2. So even no of fermions will have integer spin and therefor behave like bosons. Thus atom with spin 1 is possible.
Now given any value of spin S it will have (2S+1) eigenvalues s starting from -S to +S by steps of one. Remember these eigenvalues are discrete. And they give you the orientations. For your case the atom has spin 1. So it will have 3 eigenstates of spin. You named them to be parallel, antiparallel and perpendicular. But remember the spin is never completely parallel nor completely anti-parallel to the Magnetic field. There will always be some uncertainty.
You said:
I know that intuitively these can't be the only 2 options.
But that's wrong. Because if the atom is spin-1/2, then there will only be two options namely the parallel and the anti-parallel one. There will not be any steps in between.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/237789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is gravitational wave affected by gravitational field Does gravitational wave show gravitational lensing?
Does it bend around mass?
| At the moment only theoretical calculations can be attempted, as here
Strong gravitational lensing of gravitational waves in Einstein Telescope
Gravitational wave experiments have entered a new stage which gets us closer to the opening a new observational window on the Universe. In particular, the Einstein Telescope (ET) is designed to have a fantastic sensitivity that will provide with tens or hundreds of thousand NS-NS inspiral events per year up to the redshift z = 2. Some of such events should be gravitationally lensed by intervening galaxies. We explore the prospects of observing gravitationally lensed inspiral NS-NS events in the Einstein telescope. Being conservative we consider the lens population of elliptical galaxies. It turns out that depending on the local insipral rate ET should detect from one per decade detection in the pessimistic case to a tens of detections per year for the most optimistic case. The detection of gravitationally lensed source in gravitational wave detectors would be an invaluable source of information concerning cosmography, complementary to standard ones (like supernovae or BAO) independent of the local cosmic distance ladder calibrations.
So the answer is yes, it is expected from the calculations of the general relativity theory.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/237919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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?
| Probabily, it would be beneficial to first compare gravity, and gravitational waves (GW) as they pass earth's gravitational field.
To compare gravity, and GW, you can consider gravity as a permanent dip in space, while GW is a moving ripple in space.
Wherever this ripple passes, it minutely, and temporarily, changes the shape of the permanent dip. I.e it changes the gravity somewhat (depending upon strength of GW). Then the shape of dip (gravity) would return to its permanent state after the GW has passed.
Therefore, for the impacted time, it should impact speed of light, time, wavelength etc. Because all these get impacted by strength of gravity.
How much - That must have been predicted by general relativity math, may be someone else can quantify that.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/238226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
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?
| As the CuriousOne commented, the idea of the Big Bang is that it happened everywhere. It was not an explosion from one point in a spatial coordinate system at t=0, the spatial coordinates just started expanding (straining by the factor a(t)). Therefore all space points in our universe have similar histories in time. What we see now in the light from 13.8 billion years ago is what we would have had around us, right here, 13.8 billion years ago. When we see light from the CMBR plasma 13.8 billion years ago, then right here 13.8 billion years ago we would be immersed in plasma.
"If an astronomer was placed at one of the outermost visible objects", he would see a similar evolution in time of the sky as an astronomer on Earth ... for which never "would he be looking at a nearly dark sky in (any) direction". Likewise, the sky for the hypothetical future of a distant astronomer will look similar to the sky from Earth at the same hypothetical future time (corrected for the light travel time between the astronomer and Earth).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/238498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
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)?
| In the micro world particles like electron has dual nature .In some experiments it behaves like waves such as diffraction of electrons by a single slit but in other experiments like compton scattering or photoelectricity it behaves like particle.
In wavelike representation of electrons by a quantum mechanical wave function can explain the diffraction and interference of electrons.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/238855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 7,
"answer_id": 2
} |
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.
| The minus is coming from the value of z. Note that the loop is centered at the origin on the yz plane, and the value of B is dependent on the value of z.
The forces on the left and right arms of the loop cancel out, the forces on the top and bottom arm are also opposite but due to the nature of the B field (dependence on z) the values are opposite in sign.
So one is $iak(a/2)$ in the $\hat z$ direction and the other is $iak(-a/2)$ in the $-\hat z$ direction. so when you add them together, instead of cancelling out you are left with double of the force.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/238968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Is the local Lorentz transformation a general coordinate transformation? There is a saying in Nakahara's Geometry, Topology and Physics P371 about principal bundles and associated vector bundles:
In general relativity, the right action corresponds to the local Lorentz transformation while the left action corresponds to the general coordinate transformation.
Because the structure group right acts on Principal bundles and left acts on associated vector bundles.
But I don't think that the local Lorentz transformation is general coordinate transformation. Since for local Lorentz transformation, the structure group is $O^{+}_{\uparrow}(1,3)$ while for general coordinate transformation, the structure group is $GL(4,\mathbb{R})$.
So is the book wrong? Or I didn't understand correctly.
| You did not understand correctly, although Nakahara's statement is slightly wrong. I looked into the book, because your quote is insufficient to determine what you or Nakahara are talking about.
Nowhere is Nakahara talking about a local Lorentz transformation being a general coordinate transformation. In the context of your quote, the group in question is still the full $\mathrm{GL}(4,\mathbb{R})$ of the frame bundle. Additionally, he is not saying that every local $\mathrm{GL}(4)$ transformation is a coordinate transformation.
What he is saying is that the left action of the transition function of the frame bundle is the action of the Jacobian of a coordinate transformation, and that the action of local $\mathrm{GL}(4)$-valued functions on the frame bundle is the action of a local Lorentz transformation. The latter statement is incorrect, of course - not every matrix in $\mathrm{GL}(4)$ is an element of the Lorentz group.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/239174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
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?
| Agree with the above, but also if the atom, or collection of atoms, are in thermal equilibrium, then there is another broadening mechanism, besides lifetime broadening, called Doppler broadening that accounts for the motion of the atom(s). This has the effect to substantially widen the effective line width depending upon the temperature.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/239250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
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?
| "Force" is poorly defined here. The force the tow truck will need to exert to accelerate itself and the towed car with the desired acceleration will obviously depend on the sum of the two masses, but the force exerted by the rope on the second car (and hence the minimum required tensile strength of the rope) will only depend on the towed car's mass. The latter result is intuitively obvious: a car towing a ping-pong ball doesn't need a very strong rope.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/239417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
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's no uniform relationship between wavelength and extent of penetration/absorption. It's a quite complex function of both wavelength and material.
There are examples of low as well as high frequencies which penetrate, as well as some middle ones that don't. Also, from among the penetrating ones, some are also absorbed, some not.
Regarding health, general scientific consensus is that penetrating ones harm only if they are ionizing or heat the body to the extreme.
The image in MaximGi's answer seems fine, here are 2 more references I could find:
https://www.researchgate.net/figure/a-Tissue-penetration-depth-of-light-with-different-wavelengths-NIR-light-penetrates_fig13_316712743
https://www.semanticscholar.org/paper/Evaluation-of-absorbed-light-dose-in-human-skin-by-Chang-Huang/fc946510ccb0b616fa66ab7d7324197564492533
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/239494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
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?
| You can think of it in simple terms.
The average kinetic energy of the lattice ions increases as the temperature increases.
Between "collisions" with the metal lattice ions the free electrons are accelerated by the electric field and so increase their velocity along the electric field direction.
However because the lattice ions are vibrating more, that increase in velocity will not be as great, so the drift velocity is less.
In the macroscopic world that decrease in the drift velocity would manifest itself as an increase in the resistance of the (metal) conductor.
Note that the free electrons also increase their kinetic energy due to the increase in temperature but that is an increase in their root mean square velocity which a measure of their random movement throughout the conductor.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/239719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Why does wind sound the way it does? You can obviously hear wind. From everyday experience, the stronger the wind, the louder its sound. But this sound is always similar, and quite distinctive, noise-like. Can one compute the spectrum of the noise generated by the wind?
There is an inkling of an answer provided here: "Any way, once the wind starts doing non-linear things, it can generate periodic stresses, and from that you get the whistling or humming noise we all know and love." Could anyone expand on this, or provide references with more quantitative detail?
| I would recommend you to study Howe's Theory of Vortex Sound. Generally, an appearance of vorticity $\omega$ leads to conversion of a tiny part of flow kinetic energy in a pressure wave (really small part: proportional to $M^5$ in a free field). The main free field source is the divergence of the Lamb vector:
$$
\frac{1}{c_0^2}\frac{\partial^2 p}{\partial t^2}-\Delta p = \rho_0 \ \mathrm{div} (\omega \times v)
$$
It's obvious that the maths is not very nice and complicated vorticity field in turbulent flows leads to noise-like perception of the generated sound.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/239805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Could IR and Raman be measured simultaneously? IR and Raman are measured using two different machines. Looking at the nature of excitation and at the way data is collected, it seems that these two measurements could be done simultaneously (or with some short time separation). Is that so?
(source)
| Assuming you are trying to do vibrational Raman and IR then these two experiments are done in completely different frequency regions. This can be seen in your diagram where the arrows in the Raman experiment are much longer (higher frequency) than the arrows in the IR experiment. Because of this you would need two different light sources to get both spectra.
However, there are other options. If you are interested in vibrational modes that are both IR and Raman active, a vibrational sum frequency generation (SFG) experiment can be done. This uses a broadbandwidth IR pulse and a narrow bandwidth visible pulse to excite and up convert the signal of vibrational modes via a IR absorption followed by an Anti-Stokes Raman process. This gives signal at the sum of the absorbed IR frequency and the narrowband visible frequency. There are caveats to this though. SFG only works when the oscillators (molecules you're studying) are aligned with each other. This happens at surfaces, in external electric fields, or in crystals.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/239900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Free body diagram of pulley Is there any difference between the free body diagram of fixed pulley and movable pulley? I've read that both of the rope of fixed pulley and movable pulley have the same direction (both upwards or downwards). But, one thing that confused me: is it true that fixed pulley has T1 and T2, but movable has T2 on both sides?
|
Is there any difference between the free body diagram of fixed pulley and movable pulley?
Not particularly. The main thing is that you can assume the fixed pulley isn't accelerating, so all forces on it must sum to zero. A movable pulley may or may not be accelerating.
is it true that fixed pulley has T1 and T2, but movable has T2 on both sides?
No. We can assume light strings have a uniform tension. This is because their mass is so small that any accelerations of a portion of the string contribute insignificant forces. Therefore the tension is solely due to the forces at each end of a span of the string.
The same principle applies to pulleys. If the pulley is light enough to be considered "massless", then any rotational acceleration of the pulley would contribute zero force to the string. Any difference in tension from one side to the other would be eliminated by the pulley rotating, so the tensions must be equal.
In contrast a real pulley has a non-zero moment of inertia. Whether fixed or moving, this resistance to rotational acceleration can allow a difference in tension from one side to the other.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/240015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Cosmology: equation of motion for a scalar field in conformal time So, I've derived the equation of motion for a scalar field in "normal" time, $t$:
$$
\ddot{\phi}+3H\dot{\phi}+\frac{dV(\phi)}{d\phi}
$$
Then, using the expressions for the scalar field density, $\rho_{\phi}$, and pressure, $P_{\phi}$, but transformed into conformal time:
$$
\rho_{\phi}=\frac{1}{2a^{2}}\left(\frac{d\varphi}{d\eta}\right)^{2}+V(\varphi)
$$
$$
P_{\phi}=\frac{1}{2a^{2}}\left(\frac{d\varphi}{d\eta}\right)^{2}-V(\varphi)
$$
Differentiating $\rho_{\phi}$ above and then substituting that, with the expressions for $\rho_{\phi}$ and $\rho_{\phi}$ back into the continuity gives me:
$$
\frac{d^{2}\phi}{d\eta^{2}}+3aH\frac{d\phi}{d\eta}+a^{2}\frac{dV(\phi)}{d\phi}=0
$$
But it should be:
$$
\frac{d^{2}\phi}{d\eta^{2}}+2aH\frac{d\phi}{d\eta}+a^{2}\frac{dV(\phi)}{d\phi}=0
$$
The 3 is actually a 2? Can anyone work through the trick or transformation which gives this?
(Please note, some may use the notation $\tau$ for conformal time, or that the derivative with respect to conformal time is $\phi^{\prime}$).
| The continuity equation is
$$
\dot{\rho} = -3H(t)(\rho + p),
$$
or, in conformal time,
$$
\tag 1 \rho{'} = -3aH(\eta) (\rho + p)
$$
By using the espression for $\rho$, we have that
$$
\tag 2 \rho{'} = -\frac{a'}{a^{3}}(\varphi {'})^{2} + \frac{1}{a^{2}}\varphi{'}\varphi{''} + \frac{dV}{d\varphi}\varphi{'} = -\frac{H}{a}(\varphi {'})^{2} + \frac{1}{a^{2}}\varphi{'}\varphi{''} + \frac{dV}{d\varphi}\varphi{'},
$$
and substituting $(2)$ and expressions for $\rho , p$ into $(1)$ we have
$$
-\frac{H}{a}(\varphi {'})^{2} + \frac{1}{a^{2}}\varphi{'}\varphi{''} + \frac{dV}{d\varphi}\varphi{'} = -\frac{3H}{a}(\varphi {'})^{2} \Rightarrow
$$
$$
\varphi {''} + 2Ha\varphi {'} +a^{2}\frac{dV}{d\varphi}\varphi{'} = 0
$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/240086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do integral representations of $\mathbf A$ and $\Phi$ satisfy Lorenz condition? The following are the integral solutions of the potentials, obtained from the retarded potentials (by a Fourier transform):
$$\mathbf A (\mathbf r) = \frac{\mu_0}{4\pi}\int_V \frac{\mathbf J (\mathbf r')e^{-jk|\mathbf r -\mathbf r'|}}{|\mathbf r - \mathbf r'|}\, \mathrm{d}^3\mathbf r'\,$$
$$\Phi (\mathbf r) = \frac{1}{4\pi\epsilon_0}\int_V \frac{\rho (\mathbf r')e^{-jk|\mathbf r -\mathbf r'|}}{|\mathbf r - \mathbf r'|}\, \mathrm{d}^3\mathbf r'\,$$
I want to see if they satisfy the Lorenz gauge condition:
$$\nabla\cdot\mathbf{A} + j\omega \epsilon_0 \mu_0\Phi = 0$$
After taking the divergence of $\mathbf A$ using the formula for $\nabla. (\psi \mathbf A)=\nabla \psi. \mathbf A+ \psi \nabla. \mathbf A $, I can't proceed further because additional integrals appear and also the divergence of the primed $\mathbf J(\mathbf r')$ is zero (doesn't act on the primed coordinates).
I know that I must use the continuity equation somewhere but can't go any further.
| Given your description, you've probably done the following already:
\begin{align*}
\nabla \cdot \mathbf A &= \frac{\mu_0}{4\pi}\int_V \nabla \cdot \left[ \mathbf J (\mathbf r') \frac{e^{-jk|\mathbf r -\mathbf r'|}}{|\mathbf r - \mathbf r'|} \right] \, \mathrm{d}^3\mathbf r' \\&= \frac{\mu_0}{4\pi}\int_V \mathbf J (\mathbf r') \cdot \nabla \left[ \frac{e^{-jk|\mathbf r -\mathbf r'|}}{|\mathbf r - \mathbf r'|} \right] \, \mathrm{d}^3\mathbf r'
\end{align*}
But since the quantity in square brackets only depends on the difference $\mathbf r -\mathbf r'$, we have
$$
\nabla \left[ \frac{e^{-jk|\mathbf r -\mathbf r'|}}{|\mathbf r - \mathbf r'|} \right] = - \nabla' \left[ \frac{e^{-jk|\mathbf r -\mathbf r'|}}{|\mathbf r - \mathbf r'|} \right],
$$
where $\nabla'$ now denotes the gradient with respect to $\mathbf{r'}$. If you substitute this in and do an integration by parts, you'll get two terms. One of them (after applying the continuity equation) will be equal to $-i \omega c^2 \Phi$, while the other will be an integral over the boundary of the volume you're looking at. Assuming that our volume $V$ contains all of the current sources, the boundary term will vanish, and the Lorenz condition follows.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/240320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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.
| You can use functional renormalization group method to renormalize Quantum Einstein Gravity.You can see how to renormalize it in arXiv:hep-th/9605030.
If you have problems with understanding this paper, see the paragraph 'Exact renormalization groups' in article Renormalization group in Wikipedia.
I wanted to post more links but I don't have enough reputation to post multiple links. Please excuse me.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/240515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Where does the energy go when engine braking? If you're in gear in a car and not accelerating, the car slows down faster than it would from just air resistance and tire deformation. In normal braking, the energy is turned into heat from the brake pad rubbing on something connected to the tire. Where does your car's kinetic energy go when engine braking? IE how does energy get transferred between the road and the car to remove the car's kinetic energy?
The wikipedia article on this doesn't seem to explain it, and google searching didn't turn up anything enlightening.
| Kinetic energy is transformed into the thermal energy of the brakes. Check out a similar question/answer: https://www.physicsforums.com/threads/transforming-kinetic-energy-into-thermal-energy.634380/
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/240587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
Nesting in Fe-based superconductors Many studies about iron-based superconductors mention the nesting of Fermi pockets, such as here or here. As far as I understand it it represents some kind of interplay between different Fermi surfaces.
Does anyone have a clear explanation of what nesting is in the context of Fe-based superconductors, or a source giving such explanation ?
| There is a previous stackexchange post on nesting. I will try to briefly explain nesting in the context of the Terashima et al. paper. In Fig 3(C) of the paper there is an illustration of the Fermi surface of a material as measured using ARPES. There are two pockets to this Fermi surface, one pocket near $k=0$ (the $\Gamma$ point) and one pocket near $k=(\pi,0)$ (the $M$ point). These two pockets are said to be nested because there is a single $Q$ vector $Q=(\pi,0)$ that connects many states on the Fermi surface. In other words, you can translate the Fermi surface pocket at $\Gamma$ over to $M$ and it would nearly overlap the Fermi surface pocket already at $M$. If the pockets at $\Gamma$ and $M$ were identical then the nesting would be called perfect. As it stands the nesting is imperfect, so-called quasi-nesting. Nesting (an quasi-nesting) leads to enhancements of interactions and ultimately to exotic phases, such as magnetism or superconductivity. A clearer example of a nested Fermi surface can be found here. In Fig 1 the Fermi surface for a half-filled electron band on a square lattice is shown. Notice that at exactly half-filling ($n=1$ since the band can hold a density $n=2$ electrons, spin-up and spin-down) the Fermi surface is a square. Then the nesting vector $Q=(\pi,\pi)$ connects many points on the Fermi surface. In this case nesting leads to antiferromagnetism.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/240821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Why should the ray become parallel to base in a triangular prism at minimum deviation? In the case of minimum deviation, the refracting angles at two surfaces are equal. Then the ray inside the prism should be parallel to the base only if it is isosceles triangular prism. But everywhere I read they haven't mentioned this condition. Is this correct ?
| If the angle of the ray to each of the two surfaces of the prism is the same, it follows from symmetry that you should be able to flip the scenario left-right about an axis that bisects the angle connecting the surfaces, and have the same exact picture.
But this implies that the ray inside the prism must be perpendicular to that axis, which would put it horizontal to the base of the prism if, and only if, it is isosceles.
You are right, if you want to claim that the ray is parallel to a surface it is not interacting with, then you need to specify the relationship of that surface to the other surfaces..
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/241295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is the cosmological horizon expanding faster than space itself? I read that the rate of expansion of the universe is still a mystery. But if it's actually decelerating, wouldn't it mean that our cosmological horizon is expanding faster than space, and that one day in the far far future, it will be possible for any two points in the universe to be causally connected?
| The particle horizon grows faster than the scale factor (which it has to, because it shows the path of a light ray travelling through space with c, so it has to go farther than a regular comoving galaxy just sitting still on its coordinate in expanding space).
On the space-time-diagrams below the dashed lines are space itself, the dark green line is the particle horizon. Time goes up, space to the side, axes are in Gigayears and Gigalightyears:
As you see in the late universe plot, the particle horizon, being the light cone of t=0, converges to a constant comoving distance (and so does the orange-dashed future light cone of every given time), which means that no, not every part of the universe will once be causaly connected.
The comoving coordinate the particle horizon will converge to is now at a distance of 63.2 Gigalightyears, while the physical distance this rays or neutrinos from the big bang have travelled up to today is 46.4 Gigalightyears. In other words: nothing that is farther away than 63.2 Gyr today will ever be reached by a signal emitted from our coordinate at the big bang.
Here is an alternative depiction of the evolution of the cosmic horizons as shells with 2 space dimensions and an animated time dimension: Link 1, and here a plot for the very late universe: Link 2
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/241397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Photon emission between an electron and positron If I placed and electron and a positron at a certain distance apart in a vaccumm, they would attract each other and annihilate producing 2 or more gamma rays. But, what I would like to know is, does the electron and positron emit photons as they are attracted towards each other before they annihilate ?
I am not refering to the virtual photons that are exchanged as they are attracted to each other. I am asking if real photons are emitted by either or both particles as they move towards each other.
An electron will emit photons as it moves towards the nucleus of an atom and drops down the energy levels, but in this case the electron is losing energy in the form of a photon emission. Does the electron and positron lose any energy during their attraction ?
| When an electron-positron pair approach they can form positronium, a very low mass and very temporary hydrogen-like system. It comes in two forms, ortho- and para-, depending on the relative orientation of the spins. With para-positronium they have opposite orientations; this is the lowest energy or ground state for positronium.
The excited states can, and do emit photons. Very high precision research was done in this field beginning in the early 1950s.
The Wikipedia article gives the energy states and further references: https://en.wikipedia.org/wiki/Positronium
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/241667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Normalization of a wave function in quantum mechanics A more simple question, so I am watching a quantum mechanics lecture on potentials of free particles and am doing the general solution of schrodinger's stationary equation for a free particle when I was told to normalize the solution (which I can do all well and good) but I have no idea what it actually means to "normalize"
My question being what is normalization ? What does its product describe ?
| Born's rule: the probability density of finding a particle in a certain place is proportional to its square absolute value.
To change the "is proportional to" to "is", you multiply the wave function by a constant so that the absolute value squared integrates to 1, and so acts as a probability density function.
That's called normalisation, or normalising the wave function.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/241845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
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.
| Answering your second question in the comments, whether the wavefunction has well defined even or odd parity when you take the expectation value
$$\langle \psi|f(x)|\psi\rangle = \int f(x) |\psi(x)|^2 d^3 x$$
the function $|\psi(x)|^2$ is an even function (the product of two odd functions is even).
But the functions x, y, z are odd functions, so $\langle \psi|x|\psi\rangle=0$ for the same reason that for instance
$$\int^{\infty}_{-\infty} x e^{-x^2}dx=0$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/241910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Propagator in Quantum Mechanics What does the propagator in Quantum Mechanics mean? I mean, except from the mathematics behind it, what does it tell us? Is it something that has to do with translations in time?
| The clarity offered by using Feynman diagrams to set up the calculations for a scattering or a decay process also gives an intuitive meaning to the propagator function. The Feynman diagrams are iconal representations with one to one mathematical rules of how to set up, order by order in a perturbative expansion,
the integrals which will give a measurable quantity.
Here is a Feynman diagram for estimating the probability of neutron decay , to first order.
The internal lines are represented mathematically by a propagator function which will be integrated over the available phase space.
Virtual particles carry the quantum numbers of the name, in the above case of the W-, but not the mass. Evidently with the mass of the neutron around 1 GeV and the W close to 80GeV the reaction would not go if W were on shell.
So the propagator in this example tells us how the quantum numbers propagate from the initial state to the final state and how the phase space contributes to and kinematically sets bounds to the decay amplitude under study. To be noted, the propagator depresses the value of the integral because of the very much larger mass of W with respect to the energy available for the bounds of the integral .
The same logic holds true for the internal lines for any Feynman diagram, they represent propagators..
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/242029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Examples of Bernoulli Numbers, Euler-Mascheroni Integration, and the $\zeta(n)$ in physics In Arfken's Mathematical Methods for Physicists, there is a subsection of the "Infinite Series" chapter which covers the Bernoulli numbers, Euler-Mascheroni integration (or summation), and the connection these have with the Riemann zeta function. However, apart from a few nifty math problems these solve (explicit expressions for some sums and integrals), I can't see the use of these concepts in physics. There were a few problems at the end of the chapter that said that some practice integrals showed up in QED corrections, so that's a start. I would like to know where else these ideas show up in physics, if they do.
| I do not remember ever seeing Bernoulli numbers in physics, but I have seen Riemann zeta function pop out in quantum statistical physics.
The integrals of the type
$$g_n(z)=\frac{1}{\Gamma (n)}\int _0^{+\infty }dx\frac{x^{n-1}}{z^{-1}e^x-1}\tag{1} $$
often occur in calculations involving ideal Bose gases. For example, at low temperature, the concentration of a Bose gas is proportional to $g_{3/2}(z)$. For $z=1$, $(1)$ reduces to
$$g_n(1)=\zeta (n)$$
For further information you can read Tong's lectures on QSM.
For a general value of $z$, $g_n(z)$ can be expressed as
$$g_n(z)=\sum _{k=1}^{\infty }\frac{z^k}{k^n}$$
which is actually a polylogarithm, which, for some specific values of its parameters can related to Bernoulli polynomials $B_n(x)$. Bernoulli numbers are defined as $B_{n}=B_{n}(0)$, so they may possibly occur in physics.
On another note, Riemann zeta function may occur in electrostatics. I remember that the electric potential of a finite cylinder (I forgot the boundary conditions) was proportional to $\zeta (3)$, also known as Apéry's constant, which also appears in other areas of physics.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/242068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How does the propagation of gravity work for photons? As explained in the answers to this post, photons apparently exert a gravitational pull on other objects. It has also been explained on this site, that gravity propagates at the speed of light.
I'm wondering, though, how do you reconcile these two facts? I'm trying to imagine the gravitational field made by a photon and it seems like there are some paradoxes. For example, how can gravity propagate ahead of the photon at the speed of light, if the photon is also travelling at the speed of light? My guess is the solution is probably found in relativity, but I certainly can't figure it out.
Right now, the best I can do is to think about gravity as sound, and a photon as an object travelling at the speed of sound. Can anyone help me out?
| Photons travel along a gravitational compression wave of gravitational lines traveling in the same direction caused when the energy was released from an electron during a shell jump. Photons travel at the same speed as the wave. The reason they may not reach the same destination simultaneously is that photons collide with atoms along the way and are re-emitted taking some time in the process.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/242164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
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 things are heated electromagnetic rays are emitted, "black body radiation" I think, hence the reason why hot iron glows. As the radiation is emitted, the energy in the boiling device is dispersed into the surrounding, even if the surrounding is vacuum. (Also the reason why things in space cools down and freeze)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/242370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
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)
| You are right in all you have written. But a more fruitful sentence could be: Friction always tries to prevent sliding.
*
*(Kinetic friction) If two boxes are sliding over one another, the friction on the top block will pull in the lower block's direction, and friction on the lower block will pull in the top block's direction. Friction tries to keep them together to reduce the sliding.
*(Static friction) If two boxes on top of each other accelerate leftward (for example because a string pulls in the lower box), then friction in the top box pulls left to keep that block speeding up (prevention it from starting to slide - it is being "pulled along"), and friction in the lower box pulls right to make it stop accelerating and stay with the top one (again preventing sliding - it is being "held back").
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/242495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Derivation of Displacement current term in Ampere's Law I have a quick question:
In deriving the displacement current term for Ampere's Law, my book has the line:
$$\Phi_E= \int_S \mathbb{E} \cdot \hat{n} da= \int_S \frac{\sigma}{\epsilon_0} da = \frac{Q}{A \epsilon_0} \int_S da= \frac{Q}{\epsilon_0}$$
My question is:
Here the electric field is the electric field between two conducting plates (capacitor) neglecting edge effects. $\sigma$ is the charge density of a plate given by: $\frac{Q}{A}$ where A is the capacitor plate area. My book substitues $\sigma= \frac{Q}{A}$ and arrives at: $$\frac{Q}{A \epsilon_0} \int_S da = \frac{Q}{A \epsilon_0} A = \frac{Q}{\epsilon_0}.$$ Why do the "A"'s cancel? One $A$ is the area of the capacitor plate and the other area is the area of the gaussian surface. In general, these area's will not be equal. The image to go along with the derivation is similar to the one shown below.
| The flux is only has a nonzero part where there is a nonzero electric field. Next if we assume the $\sigma$ is the charge density on the plate, then when you replace the $\bf E$ integral with a $\sigma$ integral then the area there is just over the area of the capacitor $C$ because that is where the electric field is nonzero, and equals $\sigma/\epsilon_0.$
$$\Phi_E= \int_S \mathbf E \cdot \hat{\mathbf n}\,\mathrm da= \int_C \frac{\sigma}{\epsilon_0} \mathrm da$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/242552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What is the relation between interaction range and the mass of gauge bosons? I have just started to read spontaneous symmetry breaking, where it is mentioned that EM fields are infinite in range, so the gauge boson has to be massless, while for the strong and weak interactions, they have to be massive.
Why is it that long-range interaction are attributed to massless bosons and short range interactions to massive bosons? What is the relation between these two quantities in this context, and how would one proceed to derive it?
| Elementary particle interaction crossections and lifetimes are calculated in perturbative expansions that are set up using Feynman integrals.
The wavy line represents the exchanged boson , in this case a photon which has zero mass. The wavy line in the integral is the propagator of the electromagnetic interaction in this case. In momentum space the propagator is
$$ \frac{1}{p^2-m^2+\mathrm{i}\epsilon}$$
where $p$ is a four momentum and $m$ in this case is 0. This means that the contribution of this term in the integration can be reasonably large, depending on the phase space limits of the integral. (The iε is there to be used as going to the limit of zero when integrating over the singularity)
When the mass is large, as in the W or Z exchanges, the propagator suppresses the integral by orders of magnitude for energies below their masses. That is what makes the probability of interaction short range.
Alternatively, when viewing interactions through the Yukawa potential, which is essentially obtained as the Fourier transform of the propagator, it is seen that zero mass exchanges end up corresponding to a long range 1/r potential, while a massive mediation boson with mass $\mu$ leads to the Yukawa potential
$$V(r) = -\frac{g^2}{4\pi}\frac{\mathrm{e}^{-\mu r}}{r} $$
and if $\mu$ is a large mass the potential falls exponentially fast.
Postscript: The strong interaction described by QCD has a zero mass gluon as the mediator, but gluons have color charge and the potential built up is not a Yukawa type potential, but follows
It is a very short range force from this potential which leads to confinement of quarks and gluons within the hadrons.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/242659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
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 Nature is Smart; it prefers the state with the lowest energy (maximum entropy) in different manifestations.(surface tension in this case). Please refer to this link already answered. (Why does the nature always prefer low energy and maximum entropy?)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/242963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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?
|
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
The postulate that the speed of light is the same in different reference frame led to a whole theory that can explain observations well and solved problems you would see otherwise. There is no theoretical preference for it otherwise.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/243100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Michelson Morley experiment?
*
*Its not that I question the conclusions reached concerning the Michelson–Morley experiment, however I would like to know how the following issue was addressed please?
If I could pass bob through a beam splitter, and have each copy of him pace out each leg of the interferometer at say $2~\rm{km/hr}$. However if along one of the legs, there was an escalator aiding his initial progress at $1~\rm{km/hr}$.
Okay, so they each leave the beam splitter at the same time, however aided by the escalator one of them moves at $3~\rm{km/hr}$, and reaches the mirror at the end of his leg earlier than the other. But on the return journey he is inhibited by progression of the escalator, and moves at $1~\rm{km/hr}$. So the other bob who travelled at $2~\rm{km/hr}$ the whole time, makes up ground on the other bob on the return journey, and they arrive home at the same time as each other.
*
*With so many clever people working on this experiment over the years, I know there must be a contingency for this issue. If somebody can inform me please?
|
So the other bob who travelled at 2 km/hr the whole time, makes up
ground on the other bob on the return journey, and they arrive home at
the same time as each other.
You have missed an important feature of the experiment.
The escalator (ether wind) is in action for all Bob's movements; both up and down the escalator as well as Bob walking "across" the escalator.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/243325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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??
| When the ball makes contact with the ground, the ground exerts a very large (upward) force on the ball for a very short interval of time. This large force causes the ball velocity to change direction from downward to upward, and translates into a large upward acceleration of very short duration. So there is no inconsistency with either the laws of physics or the laws of mathematics.
If the ground is rigid, once the ball makes contact with the ground, the leading edge of the ball comes to a full stop, but the remainder of the ball is still moving downward. The ground exerts a force on the ball, and the ball begins to compress. A compression wave travels upward through the ball. The portion of the ball within the compression zone is not longer moving, but the part of the ball beyond the compression zone is still moving downward. Eventaully, the compression zone encompasses the entire ball, and the entire ball has come to a stop. Next, the compression begins to release. First the part of the ball at the top decompresses, and the velocity of this material is then upward. The decompression wave then travels downward until the ball is fully decompressed, and the entire ball is now traveling upwards. At this point, the ball loses contact with the ground. All these events take place within a tiny fraction of a second.
This description is qualitative, but it captures the essential mechanistic features of what is happening.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/243514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Differences between eigenstates, bound states and stationary states I am not very clear about the differences between eigenstates, bound states and stationary states.
|
*For any operator $\hat A$ an eigenstate $|\psi\rangle$ is one for which:
$$\hat A|\psi\rangle=\lambda |\psi\rangle$$
Where $\lambda$ is a constant, and is called the eigenvalue of that state. If $\hat A$ is an observable, then $\lambda$ will be real.
* A stationary state is an eigenstate of the Hamiltionain $\hat H$ (the energy operator). It is called stationary because when the system is in this state the expectation value $\langle \hat A \rangle$ of any operator $\hat A$ is time independent.
*A bound state is one that does not go to infinity and is usually $0$ outside a given range of $(x,y,z)$. An example (in 1d) would be a $$\psi=e^{-|x|}$$ Which goes to $0$ as $x\rightarrow \pm \infty$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/243719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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.
| To earth the charged body does not mean that the electric field of this charges disappears.
To make such a charged body one has to separate some amount of electrons and as a result one get a negative charged body and a positive charged body side by side. So beside the weakness of the electric field of the charged body over such a big distance the angular resolution of your measurement instrument has to be very high Otherwise you will detect the overlapping fields of both charged bodies.
Of course you can separate the two charged bodies over a big distance and than start the measurement of the very weak over distance electric field. But again, to earth the body to do with the flow of the electrons from this body (or to this body) to an other body. This second body has to be out Of the area, where your instrument scan the charged body. And during the flow of the electrons due to earthing the electric field does not fall to zero immediately, especially if you had a big distance to the ground (which big distance you need to separate the field due to the possible angular resolution of your instrument.
But much more important is the fact that the electric field of charged particles are theirs intrinsic properties, means this fields could not be switched on not off. They are only separable. In mixed states of protons and electrons their fields don't vanish.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/243796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How do pressured containers within pressured containers behave? If a container could hold 60 PSI of pressure and was placed into a larger container with the same strength, could you fill the large container to 50 psi and the smaller one to 100 psi, creating only 50 psi changes?
If so, could you do this infinitly, having compressors in each tank feeding to a smaller one within, allowing you to build a huge pressure?
| As you suggest, you could use a large number of nested containers to gradually increase the pressure of the innermost container to very large values. Generally it's only the pressure difference that matters. These could even be weak containers, like a balloon within a balloon within a balloon within...
But this can't go without limit. Once the absolute pressures acting on the container become significant to the structure of the material of the container, this approach no longer works. This generally occurs only when the pressures are high enough to change the lattice structure and/or other physical properties of the container material, and this is generally only at very high pressures: think of things like coal becoming diamond, nuclear fusion, etc. (This is probably fairly obvious, but just as an example: at a 1 psi differential for each container, everything inside your one-trillionth container will be at the same pressure as at the core of the sun, and just as a balloon wouldn't survive inside the core of the sun, it wouldn't survive in the one-trillionth inner container either.)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/243910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Can you measure a motor torque using a load cell connected to the stator? how? I saw a picture here: https://measurementsensors.honeywell.com/techresources/appnotes/Pages/Ways_to_Measure_the_Force_Acting_on_a_Rotating_Shaft.aspx
It shows a motor casing connected to a load cell. I don't understand how it transmits the torque from the shaft to the load cell.
| If you are only considering steady state torque with accuracy, then yes. Newton's third law applies and so any torque seen by the motor shafts and armature sees a reaction torque on the motor frame, attached to the stator.
But during transient acceleration there are losses not necessarily seen in transient torque. These losses can however be estimated using a model of the motor and voltage and current measures.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/244043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.