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Why does a wet towel become more hot than a dry one? I remember a long time ago my mum had told me not to use a wet tea towel when taking food out of the oven because you can burn your hands - lo and behold, it came time to make dinner and I did not head my mother's warning from all them years ago and my hand got burned. Why does a wet tea towel heat up more quickly when removing items from the oven?
Is it because the water molecules are more excited when heated?
| Missing from the currently-accepted answer is the importance of steam.
If you use a damp towel to grab a heavy metal dish out of an oven at $\rm 350^\circ F = 180^\circ C$, the heat from the pan will enter the water in the towel much more rapidly than conduction can carry it to your hand. Instead some of the water in contact with the hot pan will instantly convert to steam, which is much more mobile than the water and (because of the pressure change involved in the vaporization) can move through any remaining air gaps in the towel very rapidly. When the steam reaches your cool hand it is very efficient at transferring the heat to you. Furthermore this rapid vaporization is the most dramatic in the parts of the damp towel that are held the most tightly to the hot pan --- which is exactly where your hand is.
Using a dry tea towel to grab a hot pan puts the heat into the towel fibers, which are poor conductors of heat, and which serve to separate your hand from the hot pan by a set of air gaps.
Using a damp or wet tea towel is a recipe for a ferocious steam burn.
| {
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How is it possible for other animals to have better night vision than humans, who can detect individual photons? According to the Wikipedia article on night vision,
Many animals have better night vision than humans do, the result of one or more differences in the morphology and anatomy of their eyes. These include having a larger eyeball, a larger lens, a larger optical aperture (the pupils may expand to the physical limit of the eyelids), more rods than cones (or rods exclusively) in the retina, and a tapetum lucidum.
But a recent study has shown that the human eye is capable of detecting individual photons of visible light. It seems to me that this should be the highest physically possible sensitivity to light, since QED requires excitations of the E&M field to be quantized into integer numbers of photons.
How is it possible for animals to have better night vision than humans, if humans can detect individual light quanta? Is it just that while the human eye can sometimes detect individual photons, other animals' eyes can do so more often?
| The quote you made from Wikipedia includes one of the major reasons for the superior night-vision in animals: Tapetum lucidum. The tapetum lucidum is a layer just behind the retina which is reflective; as the light enter the eye and hit the retina, it is not fully absorbed by the retina but instead passes through and hits the tapetum lucidum. The light is then reflected back, hitting the retina again which effectively makes the eye double the amount of available light. The light reflected by the tapetum lucidum is also the reason why many animals seems to have glowing eyes when you shine a light at them during low light conditions.
Humans, and other primates, lack the tapetum lucidum and, as we humans also have fewer rods than many other animals, the result is that we get a poorer night vision even though our eyes have the capability of detecting single photons.
| {
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How does $I = \mathrm{d}q/\mathrm{d}t$ work for a capacitor? When the capacitor is charging in a circuit consisting of a resistor, a capacitor and an alternating sinusoidal generator at $t=0$, the charge across the capacitor is 0 and the current is $I =\mathrm{d}q/\mathrm{d}t$. Does this make the current zero too? While it is max across the resistor in the same circuit and they are connected in series which means that the current should be the same in all the components of the circuit.
| The value of $f(t)$ at any specific value of $t$ does not automatically necessitate a value for $\frac{df(t)}{dt}$ at that value of $t$. If $q(t)\propto\sin(\omega t)$ then $\frac{dq}{dt}\propto\omega\cos(\omega t)$, which means while $q(0)=0$, you'd have $\frac{dq}{dt}|_{t=0}\propto\omega$, which would be a max for this function for $q$.
| {
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What happens to temperature as volume increases (charles law) Let us take Boyles law to start.
Assumptions:
*
*Gas is perfect.
*In a massless piston that can be expanded with no friction
*Adiabatic
If we were to decrease the volume of the piston, the pressure inside would go up because the gas molecules would be hitting the sides more often.
If we were to increase the outside pressure on the piston, the volume of the piston would go down until the internal pressure matched the external pressure.
We see here that Boyle's law is perfectly explained by the kinetic model of gases (T held constant).
Let's look at Charles law (P held constant)
If we were to increase the temperature of the molecules, their kinetic energy would increase and would therefore hit the piston with greater force and increase the volume of the piston until the internal and external pressures are equal.
But if we were to spontaneously increase the volume of the piston, the temperature would NOT increase as a result (to maintain the increased volume against the constant external pressure) because heat doesn't spontaneously arise.
So it seems that Charles law only works one way, but not the other. And that T and V are not intrinsically linked like P and V are in Boyles law. Is this true?
| The answer itself is hidden in the second part of your question. P in any gas law refers to the pressure inside the volume of the container [ which is always equal to the external pressure on the container ]
Now, in the second part of your question, the statements
(P held constant)
&
if we were to spontaneously increase the volume of the piston
cannot be true simultaneously.
As soon as you want to increase the volume of the container (irrespective of the method of achieving such a change -- whether by a reversible piston process or via irreversible free expansion), you have to reduce the external pressure on the piston & thereby the pressure inside the container also reduces & then you can apply kinetic theory accordingly [ Note that unless these two P's are equal, you cannot apply any gas law because they are valid only for equilibrium situations. It's very important to keep this condition in mind when you are thinking about such thought experiments. ]
To conclude, all gas laws work irrespective of whether the change in the variables are positive or negative as long as you are being careful to satisfy all conditions including the condition of equilibrium & reversibility.
| {
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Doppler effect and apparent frequency What is meant by "apparent frequency"?
I mean the answer we get by applying the formula; what does it signify?
If it is the frequency received by the observer, does it mean that the observer receives the same frequency no matter what the distance of the source?
Shouldn't distance of play a role?
| Picture yourself at seaside, and consider the frequency at which waves are reaching your feet. This frequency does not depends on where you stand. Distance plays no role.
Now if you move towards the ocean, or towards the beach, the period between two successive waves is altered.
*
*If you run to the beach as fast as the wave moves, you will follow a crest and never get reached by the following one - the frequency is down to 0.
*If you run towards the ocean, you will face crest after crest more often then if you stay still - the frequency is increased.
| {
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When a force is applied on a body, which point does the body rotate around? I understand when a force is applied to a body it causes moments on all points not lying on the force's line of action. So which of these points acts as the pivot point around which the body actually rotates with a specific moment as a result of that applied force?
| The motion caused by an applied force $F$ can be resolved into 2 components :
*
*A linear acceleration $a=F/m$ of the centre of mass (CM). If the force is an impulse the effect is to change the linear momentum of the CM.
*A rotational acceleration about the CM. The force produces a torque $\tau=r\times F$ about the CM, where $r$ is the position vector of the point of application of the force, relative to the CM. This torque causes rotational acceleration $\alpha=\tau/J$ where $J$ is the moment of inertia.
The overall motion is a combination of these 2 motions. The result usually means that there is one point in the body for which the instantaneous velocity is zero at some instant in time. This is called the instantaneous centre of rotation.
| {
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Why viscosity depends on the scale of things? Water feels like honey for bacteria and air is very viscous for small insects.
My question is why viscosity depends on the scale of things?
| To add a more formal perspective to the answers already given, if we non-dimensionalize the fundamental equations for fluid flow, the Navier-Stokes equations, using the characteristic flow velocity $U$ and characteristic length scale $L$, we obtain these equations in a form which only depends on the flow geometry, and is independent of scale. We find that the only parameter remaining in this case is the Reynolds number,
$$Re=\frac{U L}{\nu},$$
where $\nu$ is the kinematic viscosity of the fluid. Intuitively the Reynolds number can be understood as the ratio of inertial over viscous forces. Thus flows at small Reynolds numbers are dominated by viscous effects, corresponding to the appearance of "feels like honey". Flows in small gaps, around bacteria, small insects, etc. are all characterized by low Reynolds numbers and thus appear "very viscous", whereas flows around cars and airplanes, say, correspond to high-Reynolds number flow and appear nearly "inviscid". Here is an instructive article by Purcell on "Life at Low Reynolds Numbers".
| {
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Physical meaning of $L_z$ perturbation of rigid rotator The rigid rotor is a classical exercise in introductory quantum mechanics classes. The Hamiltonian is
$$H = \frac{L^2}{2m} \, .$$
Often you are also asked to consider the effect of adding a perturbation such that the new Hamiltonian has the form
$$H = \frac{L^2}{2m} + aL_{z} \,,$$
where the perturbing term expresses a coupling of the energies to the projection of the angular momentum along the $z$-axis. Is there a simple physical explanation behind this term? What would it represent in a realistic system?
| Actually the example of your $H$ is spherical rotor, which is not so interesting. I suppose the example you have could describe a spherical rotor in a magnetic field.
If you're willing to go beyond this, the most common rotors are axially symmetric, with Hamiltonians of the type
$$
H= \alpha L^2+\beta L_z^2\left(\frac{1}{I_3}-\frac{1}{I_1}\right)
$$
where $I_3$ and $I_1$ are the principal moments of inertia. Those are very popular starting points in the study of nuclear deformations and their spectrum typically contains rotational bands (i.e. sequences of states connected by quadrupole transitions and with spectrum proportional to $L(L+1)$ within a band.
There is a very famous problem of a single particle with constant angular momentum $j$ moving in the field of an axially deformed nucleus (particle plus rotor nuclear model, or Nilsson Hamiltonian). A simple version of this Hamiltonian would be
$$
H=Q L_z^2-\omega L_x\, .
$$
The $L^2$ is constant and has been omitted. There is a nice paper by Aage Bohr and Ben Mottelson on the semiclassical analysis of this Hamiltonian (Phys.Scr. vol. 22 (1980) 461-467).
More generally a class of models known as Lipkin-Meshkov-Glick, with Hamiltonians of the type
$$
H=\varepsilon_0+2\omega L_z+\lambda (L_x^2-L_y^2)+\gamma(L^2-J_z^2)
$$
have been extensively studied. See this time E. Romera et al, Phys. Scr. vol. 89 (2014) #095103 for again a semiclassical analysis.
There are all sorts of additional variations.
Sorry for the clear oversight of some molecules with symmetries, but I am more familiar with the nuclear side of this type of Hamiltonian.
| {
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What constitute the secondary particles beam when a high energy proton beam hits a target material? Basically I want to know what particles emerge along with high energy photons(not sure about it) as the second beam when a specific high energy proton beam is incidented upon a target material such as iridium or gold. Can the composition of secondary beam be calculated theoretically?
Does the ratio of photon energy and particles energy of the secondary beam hold any specific value for a given energy of primary beam?
| Here is an event from LHCb
A proton-lead ion collision observed by the LHCb detector .The image above shows a typical high multiplicity proton-lead-ion collision event,
The complexity of the outgoing fragments is large, and it constitutes charged tracks from electrons, muons, protons, nuclear fragments,photons ... into jets of particles. No structured secondary beam in this event, and none announced in the data of the experiment.
There exist theoretical models trying to describe the effects , for example this link.
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Why does a force not do any work if it's perpendicular to the motion? I have a book that says the Moon's orbit is [in this context assumed to be] circular. The Earth does no work on the moon. The gravitational force is perpendicular to the motion. Why is there no work done if support force is perpendicular to the motion?
| In case you are interested in a more mathematical approach, this can actually be proven entirely using geometry. To proof this you need to to know a few things about vectors.
If $\vec v=(v_x,v_y,v_z)$ is the velocity vector of the moon then $v_x$, $v_y$ and $v_z$ represent the components of the speed in the x, y and z direction.
You can calculate the speed of the moon by calculation the length of the velocity vector
$$v=|v|=\sqrt{v_x^2+v_y^2+v_z^2}$$
If you square the length you get $v^2=v_x^2+v_y^2+v_z^2$, which I will use later. Lastly the dot product between two vectors is defined to be
$$\vec a\cdot\vec b=a_xb_x+a_yb_y+a_zb_z=|a||b|\cos \alpha$$
$\alpha$ is the angle between the two vectors. This means that the dot product is zero if the two vectors are perpendicular
Proof
If no work is done on the moon the kinetic energy must be constant. So the derivative of the kinetic energy must be zero. So we take the derivative by applying our $v^2$ substitution and taking the constants out of the derivative.
$$\frac{dKE}{dt}=\frac{d}{dt}(\tfrac{1}{2}mv^2)=\tfrac{1}{2}m\cdot\frac{d}{dt}(v_x^2+v_y^2+v_z^2)$$
Then apply the chain rule to each of the components.
$$\frac{d}{dt}(v_x^2+v_y^2+v_z^2)=2v_x\frac{dv_x}{dt}+2v_y\frac{dv_y}{dt}+2v_z\frac{dv_z}{dt}=2v_xa_x+2v_ya_y+2v_za_z$$
In which we recognize the dot product:
$$\frac{d}{dt}v^2=2\vec v\cdot\vec a$$
So the derivative of the kinetic energy becomes
$$\frac{dKE}{dt}=m\vec v\cdot\vec a$$
If the orbit is circular, the velocity is always perpendicular to the acceleration (and the force). So the kinetec energy doesn't change and no work is done on the moon.
| {
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Speed of block sliding on frictionless ramps Here's the question:
My book says the answer is C. How is it not A?
I know that all the potential energy is transferred to kinetic energy. With algebra, knowing Kinetic energy is (1/2) * m * v^2 and gravitational potential is mgh, I solve for h which results in (v^2)/2g
Ok so since this is a proportional reasoning problem my focus is that h = v^2 meaning height is directly proportional to the square of velocity. That being said, if we half the velocity, that means that some value (let's use p for the variable) multiplied by height gets me to (1/2v)^2.
That being said, if we refer back to the equation h = (v)^2 and v is being halved, it should look like this:
h * p (<- growth factor) = (0.5v)^2. Now it looks like the right side of the equation grew by a factor of (1/4). Think about it. If your velocity is 10, the right side of the equation becomes 100. If you half that velocity, the right side of that equation becomes 25. You can see that the right side of the equation grew by a factor of 1/4. That means that $p$ (my growth factor variable) should also be (1/4). $h * 1/4 = h/4$. NOT $3h/4$. Where did I go wrong?
| I got lost in your explanation, so can't help you figuring out where you went wrong. Your attempt to solve it with conservation of energy is certainly the easiest and correct.
Let's analyze the question
*
*smooth ramp: this is meant to mean no friction, no energy is lost
*starts from rest at height $h$: at this point there is only potential energy ("rest") and it is equal to $mgh$
*at the bottom it is moving at speed $v$: at this point all potential energy has been converted into kinetic enrgy equal to $mv^2/2$
*slides up a second ramp. at what height $h^*$ is the speed equal to $v/2$? At this point the block has both potential energy equal to $mgh^*$ and kinetic energy equal to $m(v/2)^2/2$
Because of the smoothness of the ramps energy is conserved and you can put the energies at all three points (start, bottom of ramp and at height $h^*$) equal. This gives:
$$mgh=\frac{m}{2}v^2=mgh^*+\frac{m}{2}\left(\frac{v}{2}\right)^2$$
Eliminating $v$ from these equations you end up with
$$h^*=\frac{3}{4}h$$
| {
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What is the significance of the phase constant in the Simple Harmonic Motion equation? The displacement of a particle performing simple harmonic motion is given by $x = A \sin(\omega t + \phi)$ , where $A$ is the amplitude, $\omega$ is the frequency, $t$ is the time, and $\phi$ is the phase constant. What is the significance of $\phi$. How is it used? Please explain the meaning of the phase constant
| In the basic SHM equation, you get x=Asin(ωt) where at t=0, the object is at mean position or zero displacement. Now, what is the significance of the angle inside the sine function? It gives you the position of the particle performing SHM. When the angle is π/2, the displacement is maximum i.e A. When it is π, the displacement is once again 0. So, for the equation Asin(ωt+ϕ), it simply means that the SHM does not begin at x=0 and the position at t=0 is Asin(ϕ) (depending upon the value of ϕ it could be A,A/2 anything).
If the initial position is S, then ϕ=sin^-1 (S/A)
| {
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Acceleration of car. One dimensional motion easy problem A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.
My attempt at solving the problem:
$$a(x) = \frac{v - u}{t}$$
where
$v =$ final velocity
$u =$ initial velocity
$$$$
I get the answer as $4.05 \space ms^{-2}$
But the correct answer given to the problem is $8.10 \space ms^{-2}$.
They used a different equation to reach that answer.
Did I use the wrong equation? I have the average velocity and not the instantaneous veolcity?
| What is your $V_f$ ?
$V_f$ is not given in the question so you can't use this equation,
$$V_f =V_i +at$$
But the distance is given which will allow you to use
$$S=ut+1/2 at^2 $$
| {
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What causes change in planet's angular velocity? A satellite moving in an elliptical orbit will increase in angular velocity as it nears a planet. I understand that this is consistent with angular momentum. But what causes the increase in angular velocity if there is no torque acting on the satellite?
| There is no torque (with respect to the focus) which implies the angular momentum is constant, but since the angular momentum is $L=mr^2\omega$ and the radius $r$ changes, the angular velocity is not constant. The mechanism resulting this change can be understood as follows.
In an elliptical orbit (with non zero eccentricity) the gravitational force has in general a component along the tangential direction which accelerate the body.
The red arrows in the figure bellow show the gravitational attraction where the blue ones show the components along the tangent of the orbit. Notice that as you go from A to B (counter clockwise orbit) the tangential acceleration increases the tangential velocity. On the other hand, from B to A the tangential acceleration opposes the velocity. That is what changes the angular velocity of the body.
| {
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A question on Collison of macroscopic particles
Hello,
In the above question I could solve for average elastic force by taking velocity with respect to wall and finding change in momentum of the ball after that divided change jn momentum by time interval. Answer comes out to be option b.
But as it is written in the question that collision is elastic, therefore, Kinetic energy before the collision should be equal to kinetic energy after the collision and option d should also be correct. But the correct answers according to the book are b and c.
Please Explain why d is incorrect and c is correct.
|
Kinetic energy before the collision should be equal to kinetic energy after the collision
In an elastic collision, the "total" kinetic energy and "total" momentum of the system will be conserved. You missed that point. The individual kinetic energy of the ball and wall could change. But the total kinetic energy of the system is conserved before and after collision. That is
$$T_{wall}+T_{ball}=T_{wall}'+T_{ball}'$$
where the primed quantities denote kinetic energy after collision and unprimed quantities denote kinetic energy after collision. This is what it is meant by kinetic energy of the system is conserved in elastic collision.
Unless
*
*the massive wall is static (and the ball is not so fast enough to break the wall on collision), or
*the ball and the wall have equal masses and equal velocities
then only we can say that the kinetic energy of each constituents of the system is conserved before and after collisions. But, in general, it is the total kinetic energy of the system (the sum of kinetic energies of all constituents making up the system) that is conserved.
Hence option (d) is incorrect, as the given problem does not satisfy any of the above mentioned criterions. So, on impact, the momentum of the ball increases and hence its kinetic energy also increases. This increase in kinetic energy of the ball will be equal to the decrease in the kinetic energy of the wall. Hence the kinetic energy of the system is conserved in the elastic collision.
| {
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M-theory with its 3-form $H$ and the problem of having no Lagrangian This is a standard question about the M-theoretic construction of the 6d (2,0) theory. This is done, in the simplest case, by an M2 brane hanging between two M5 branes. The theory on the M5 branes is the 6d (2,0) SCFT which has 4 fermions, 5 scalars and one self-dual 3-form $H=dB$.
My question is the following: why is one of the reasons of having no Lagrangian description for the 6d (2,0) theory the existence of the self-dual 3-form? Self-duality, in Lorentzian manifolds such as the world volume of the M5, appear in $d=2,6,10$. So what is the problem with it?
A related question is why is this theory not interpreted as a (higher?) gauge theory where $H$ is the (higher) curvature of the (higher) gauge field $B$? Usually people call this a gerbe.
| The problem with a self-dual three-form field strength is that the obvious kinetic term vanishes:
$$\mathcal{L}_\text{kin}=H\wedge *H =H\wedge H\,,$$
but since $H$ is a three-form, $H\wedge H=0$, so $\mathcal{L}_\text{kin}=0$. This can happen in $d=4n+2$, when you have a self-dual $(2n+1)$-form (in a chiral theory). Indeed, in type IIB string theory, there is a self-dual five-form field strength with the same problem.
Note that one can ignore the self-duality condition in the action, derive the equations of motion, and impose the condition on the eom.
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Calculating magnetic field I would like to calculate the magnetic field near a neodymium magnet (N35). With near I mean I have got a rod magnet of about 10 x 4 mm and I am interested in the field from 0 mm under the magnet until 10 mm under the magnet down and also from 0 mm until about 10 mm to the right or left. I want to make a table in excel like 10 x 10. The difficult part is I want to know the field strength in x, y, and Z direction. at a given place.
I did read a lot on internet but I can not find the solution. I did read this: https://en.wikipedia.org/wiki/Dipole#Field_of_a_static_magnetic_dipole
and in the paragraph: Field of a static magnetic dipole, they describe the field, I think I need this formula but I am not sure. But if I would need that one I still do not have the x,y,z directions. And how do I get the strength of the N35 neodymium magnet into the formula?
I hope you can help me, thank a lot for the help
| For a NeFeB cylinder magnet a couple of equivalent quick approximations is that the field looks like that from a uniform density of monopoles on each face (oppositely signed of course), or is the field from a uniform sheet of current circling the axis on the curved outside of the magnet.
The integrals are easy to set up in Vector Calculus, typically hard to solve analytically - but can be numerically solved.
Then there are FEM solvers too https://www.supermagnete.de/eng/faq/How-do-you-calculate-the-magnetic-flux-density check the FEMM link.
It looks like several magnet manufacturers give online calculators for field strength on axis and by material - so you can just normalize your calcs to their values: https://www.dextermag.com/resource-center/magnetic-field-calculators/field-on-axis-of-cylindrical-magnet-calculator
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What is a "Standard value"? Temperature:a measure of the warmth or coldness of an object or substance with reference to some standard value.
I really tried searching lots AND lots for what is "Standard value" is... But I still don't understand what does it mean in that sentence.
Can someone please tell me what do they mean by that?
| The "standard value" in this case refers to the concept of empirical temperature. The link explains with more detail the difference between empirical and thermodynamic temperatures.
In order to define an empirical temperature you need:
*
*A substance with a thermometric property (such as mercury and its volume expansion).
*Two reference points, such as the fusion and the boiling point of water.
*An interpolation, such a linear interpolation.
The standard value is the arbitrary value in your scale that you define for the reference points. For example, you put your mercury thermometer in thermal equilibrium with freezing water and define that $1\, ml$ corresponds to a temperature of $b$ degrees in the your scale.
| {
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Why doesn't current decrease in series combination? I know that the question is quite stupid but I want to get an insight of this case. consider 3 resistors connected in series with a battery, after the current passes through resistor 1 it loses some of its energy, the kinetic energy of the charge carriers will definitely also decrease and so does the drift velocity then why doesn't the current decrease? Its quite confusing.
| The question is worded ambiguously. It could be construed to mean why isn't the current less in each additional resistor than it was in the resistor before it. The answers already here answer that question.
The question could also be asking why isn't the current less when it flows through 3 resistors in series than when it flows through one. If we assume that all the resistors are equal, then the current is reduced by each additional resistor added to the circuit. In fact you would have only one-third as much current flowing through three of the resistors in series as you would have flowing through just one of the resistors. So the current is reduced by each additional resistor, but except for things that happen over very short timespans, each resistor will always have the same current flowing through it as any of the other resistors.
| {
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Do physicists use particle "energy" to refer to kinetic energy? In 1963, this paper was written about the effects of radiation on solar panels. The paper states that:
When electrons at energies greater than 145 KeV and protons at energies
greater than 98eV bombard a silicon crystal, they can displace an atom from the
crystal lattice, producing a lattice vacancy and a recoil atom which comes to rest
as an interstitial atom.
However, the resting energies of electrons and protons are far greater than this, at roughly 511 KeV and 938 MeV respectively. I concluded that the paper was referring to kinetic energy rather than total energy, and adjusted my calculations based on this conjecture.
So: Was I correct to assume that the paper referred to kinetic energy, or was it instead some other measure of the particles' energy?
More generally, is there a standard meaning for a particle's "energy" when referring to such particles moving at relativistic speeds?
| Not only is it common to quote kinetic energy as the energy in contexts other than high-energy physics, it's basically also where the whole convention of writing energies in electron-volts comes from: $1\:\mathrm{keV}$ is the kinetic energy that a singly-charged particle, regardless of mass, picks up if you allow it to drop through the potential of a linear accelerator, i.e. capacitor charged to $1\:\mathrm{kV}$. And this is the energy such an e.g. electron can then put to use when colliding with the cathode of an X-ray tube. X-rays normally don't get close to electron rest mass, so as said before it's unambiguous here which energy is meant.
(For the X-rays themselves it's of course unambiguous anyway, thanks to zero rest mass.)
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Translation of Vectors I am a bit confused about translation of vectors. In the first class in physics itself we are told that we can translate vectors as we like to the desired position to do whatever that we are trying to do. For example, if someone draws two random vectors then to get the sum, we translate them, make a parallelogram and draw the diagonal as the resultant.
However I have some doubts on this. In the following example, clearly we cannot translate the vectors.
Consider this rigid body. We want to calculate the torque about origin of a force.
Now if we translate the force vector, then we would obtain the following.
Obviously the situation are very different and its not equivalent.
So are we really allowed to translate vectors?
| Vectors that cannot be translated are the ones that depend on the origin of coordinates. First and foremost in the list is the position vector $\vec r$ followed all the others with definitions that depend on the position vector. These would be torque, $\vec \tau = \vec r \times \vec F$; angular momentum $\vec L = \vec r \times \vec p$ and so on. That is why when you mention torque or angular momentum you have to specify the reference point about which you express these vectors in order to make sense.
Displacement, being the difference between two position vectors, can be translated. The vector from Boston to Philadelphia is the same regardless of whether you specify the positions of the two cities relative to New York or London. This means that vectors derived from the displacement vector, e.g. linear velocity, can also be translated.
A notable exception is the linear velocity in rotational motion, $\vec v = \vec \omega \times \vec r$ because, of course, the position vector of the rotating object figures in it. If you translate this velocity vector around the circle from the 12 o'clock to the 3 o'clock position you will have to change its direction by $90^{\text{o}}.$
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Dirac delta function property in a scattering proof I'm studying the proof for the decoherence of the off diagonal elements of a density matrix through scattering with the environment and I'm stuck at a certain point:
My problem is A1.14 relation. (A1.13 as well to be honest, but I guess that the $(2\pi/L)^3$ is just sort of a normalization where $2\pi$ is, maybe $2\pi\hbar$ from the uncertainty relation and $L^3$ is the "volume of the scattering")
Sadly I have no idea why $\delta^2(q-q')=\delta(q-q')L$. Oh, $q$ should be a generic module of a momentum (maybe before the scattering) and $q'$ should be the module associated to the momentum of the particle after the scattering.
Since both these relations are said to be "usual replacements" I hope somebody recognise them and can help me out.
| The answer given by jg255 is rather heuristic, but it gets the correct answer. A better approach is to realize that in scattering theory, when we convert scattering amplitudes, which have an energy-momentum conservation delta function, into probablities by squaring them and interpreting the delta-squared as a delta times the volume of space time, we are secretly using Fermi's Golden Rule. In other words to do the calculation properly you need to work in a finite space-time box, square to convert amplitudes to probablities, and only then take the infinite volume and time limit. The answer comes out the same as the Heuristic approach.
| {
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External work required to move charge
There are 2 charged which are equal in magnitude and opposite in sign, and they are equidistant from the vertical axis AB. The voltage at both A and B would be 0.
If I introduce a new charge at point A, what would be the external work required to get it to point B?
In a case like this, I know that the electric field generated by the blue and red charges would be perpendicular to the movement along the vertical axis, so the work done by the electric field is 0.
However, to move the charge from A to B, would there be any external work required (even though the charge is moving along an equipotential surface)?
| As you correctly note, the electric field will be perpendicular to the line AB and will not do any work on the test charge if it moves along that path. But, there must be some constraint to keep it on that path.
Assuming there is some constraint, let's now move the test charge from point A toward point B. To get the charge moving, you must do work (call it $W_1$) on it, exerting a force toward B and it moves that direction ($W_1>0$). The charge is now moving along the line with kinetic energy $K=W_1$, and you want it to stop at point B. Again you do work on it ($W_2$), exerting a force opposite its velocity, so the kinetic energy decreases to zero, so $W_2=-K=-W_1$.
The net work done by your outside force in moving the charge from rest at point A to rest at point B is zero: $$W_{\mathrm{net}}=W_1+W_2=W_1+(-W_1)=0.$$
On the other hand, if the charge doesn't stop at point B, the net work from A to B is not zero.
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Terminology of 1PI diagram One-particle irreducible diagrams are diagrams that cannot be broken into two disconnected diagrams by cutting an internal propagator.
*
*Why are the external lines on an one-particle irreducible diagram truncated/amputated?
*What is the motivation for the word 'one-particle' in the terminology?
| *
*It is a convention that simplifies some formulas. You can always choose to include the propagators on the external lines, but their effect is rather trivial so they don't add relevant information to the diagram.
*"One particle" means "one line". The reason for the word "particle" is that lines represent particles (sometimes, virtual particles). The reason for the word "one" is that the diagrams remains connected after cutting any one line. You can also have two-particle irreducible, if it remains connected after cutting any two lines, and more generally, $n$-particle irreducible if it remains connected after cutting any $n$-lines.
Remark: in $1.)$ I am assuming that you are using Feynman diagrams to calculate correlation functions. If you are using them to calculate $S$-matrix elements, then the external lines are amputated because of the LSZ theorem.
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What does "dimension" mean in physics? Still I don't get what's the difference between dimension and quantity. Are they same or they are different?
| The dimension, or physical dimensionality, is a descriptor of a quantity. Thus "the distance between New York and Los Angeles" is a quantity, which has dimensions of length.
A bit more precisely, you can see nouns like length, velocity, time, and mass as describing abstract quantities, i.e. "a velocity $v$" specifies that $v$ is a velocity but nothing else. Physical dimensionality is, at its heart, an equivalence relationship: in essence, you are always saying "quantity $A$ and quantity $B$ have the same physical dimensionality". Thus all lengths have the same physical dimensionality, and "the distance between New York and Los Angeles" has the same physical dimensionality as any length, which we just describe as 'length' in the abstract.
Note in particular that "length" (no article), in the abstract, is a physical dimension, whereas "a length" refers to some arbitrary quantity. The lines between those two are pretty blurry - because it really doesn't matter all that much, as there isn't much harm in the ambiguity.
I hope that makes it clearer instead of more confusing.
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Continuous version of Coulomb's law for infinite charge distributions This might seem naive or have a trivial resolution, but I'm still searching for one and have been unable to find it.
Consider an infinitely long line charge with uniform line charge density $\lambda$. The electric field at any point in space is easily found using Gauss's law for a cylinder enclosing a portion of the line charge. The correct result is $$E_r=\frac{\lambda}{2\pi\epsilon_0 r}.$$ However, if you use the Coulomb law
$$\frac{1}{4\pi\epsilon_0} \int_{-\infty}^{\infty} \lambda \frac{dx}{x^2+r^2}$$
you get a different answer ($\frac{\lambda}{4\epsilon_0 r}$) for the electric field at a distance $r$ from the wire (which of course is purely radial, due to translational symmetry in the $x$ direction). Is there an explanation for this discrepancy?
| Both methods give the same answer, provided one applies them consistently. Coulomb's law actually reads
$$ \vec{E} = \frac{1}{4\pi\epsilon_0} \int_{-\infty}^\infty dx\,\frac{\lambda}{x^2 +r^2} \times \sin{\theta}\, \hat r,$$
where the factor of $\sin \theta$ is what picks out just the radial component, as required by symmetry. But since
$$\sin\theta = \frac{r}{\sqrt{x^2 + r^2}},$$
the integral we must evaluate actually has the form
$$ \frac{1}{4\pi\epsilon_0} \int_{-\infty}^\infty dx\,\frac{\lambda\, r}{(x^2 +r^2)^{3/2}} = \frac{1}{2\pi \epsilon_0} \frac{\lambda}{r}.$$
Comfortingly, this result agrees with the computation via Gauss's law.
| {
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Experimentally measure velocity/momentum of a particle in quantum mechanics In the context of quantum mechanics one cannot measure the velocity of a particle by measuring its position at two quick instants of time and dividing by the time interval. That is,
$$ v = \frac{x_2 - x_1}{t_2 - t_1} $$ does not hold as just after the first measurement the wavefunction of the particle "collapses".
So, experimentally how exactly do we measure the veolcity (or say momentum) of a particle?
One way that occurs to me is to measure the particle's de Broglie wavelength $\lambda$ and use $$p = \frac{h}{\lambda}$$ and $$v = \frac{p}{m}$$
to determine the particle's velocity. Is this the way it is done? Is there any other way?
| For a cold atom experiment, experimentalists use time-of-flight (TOF) measurement to determine the momentum distribution of atoms in the optical trap. Suppose there are an ensemble of atoms trapped in the optical trap, when the optical trap is switched off, the atoms will "fly around" with their momentum. With detectors installed around the trap, one could obtain both the value and direction of atomic momenta, which could be gathered to contruct the momentum distribution.
See arXiv: 1002.2311
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Exchange Integral and Derivative respect to a parameter of a Dirac delta-function I'm trying to solve the 6.2 problem of Jackson's Classical Electrodynamics textbook. At some point, to get the desired solution, I have to exchange a derivative applied to a Dirac delta-function with the integral operator:
$$\int_{\mathbb{R^3}} \frac{\partial \delta(\mathbf{x}-\mathbf{x_0}(t))}{\partial t}f(\mathbf{x})\,d^3x=\frac{\partial}{\partial t}\int_{\mathbb{R^3}} \delta(\mathbf{x}-\mathbf{x_0}(t))f(\mathbf{x})\,d^3x=\frac{\partial}{\partial t}f(\mathbf{x_0}(t))$$
Under which hypothesis can I do something like that (i.e. exchange the order of differentiation and integration)? I expect that known theorems of real analysis do not apply in this case, since the $\delta$ is not even a proper function.
| By definition, if $T$ is a distribution, then $$\langle \partial_x T, f \rangle := - \langle T, \partial_x f \rangle\tag{1}$$ for every test function $f=f(x)$ in $C_0^\infty(\mathbb R^n)$ (or also $C^\infty(\mathbb R^n)$ if $T$ has compact support as the delta function). Here the derivative is just a bit more complicated. However, since the chain rule for taking the derivative of distributions composed with smooth functions is valid also for distributions (it is a general theorem) we have
$$\partial_t \delta(x-x_0(t)) = \frac{dx_0}{dt}|_{x_0}\cdot \nabla_{x_0} \delta(x-x_0(t)) = -\frac{dx_0}{dt}|_{x}\cdot \nabla_x \delta(x-x_0(t))\:. \tag{2}$$
As a consequence, for every function $f \in C^\infty(\mathbb R^n)$, applying (2),
$$\int \partial_t \delta(x-x_0(t)) f(x) d^nx = -\int \frac{dx_0}{dt}|_{x_0}\cdot \nabla_x \delta(x-x_0(t)) f(x) d^nx$$ $$= -\frac{dx_0}{dt}|_{x_0}\cdot\int \nabla_x\delta(x-x_0(t)) f(x) d^nx \:.$$
Applying (1)
$$\int \partial_t \delta(x-x_0(t)) f(x) d^nx = + \frac{dx_0}{dt}|_{x_0}\cdot \int \delta(x-x_0(t)) \nabla_x f(x) d^nx$$ $$= \frac{dx_0}{dt}|_{x_0}\cdot \nabla_x f(x)|_{x_0(t)} = \frac{d}{dt}f(x_0(t))\:.$$
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Biot Savart - Vortex line segment parallel to a plane I am trying to understand what is shown in this book (Applied hydro and aeromechanics), page 187
https://books.google.com.au/books?id=Ds-bd0zAwIYC
The velocity induced at a given point on a parallel plane by a vortex segment of length $b$, and strength $\Gamma$ is shown to be,
$v_1 = \frac{\Gamma b \, sin \, \alpha}{4 \pi R^2}$
Can anyone please show the steps on arriving this formula by applying biot-savart low ?
Thanks much in advance
ABCD
| For reference, here is the figure:
We are interested in the influence of the wing-bound vortex of length $b$ and circulation $\Gamma$ on point $A$, particularly the velocity vector $V_1$.
The Biot-Savart law in this case can be written as
\begin{equation}
V_1=\frac{\Gamma}{4\pi}\int\frac{\mathrm{d}\vec{b}\times\vec{R}}{|\vec{R}|^3}, \tag{1}
\end{equation}
where $\vec{R}$ is the vector with magnitude $R$ in the figure and $\mathrm{d}\vec{b}$ is an infinitesimal section of the vector representation of the bound vortex. Please see Karamcheti for a full derivation; it's rather too involved for us here.
Using $b_1$ and $b_2$ as the ends of the bound vortex, we have
\begin{equation}
V_1=\frac{\Gamma}{4\pi}\int_{b_1}^{b_2}\frac{\mathrm{d}\vec{b}\times\vec{R}}{|\vec{R}|^3}. \tag{2}
\end{equation}
We can see from the figure the direction of $V_1$ and a simple application of a right-hand rule on the bound vortex confirms this orientation. We can thus focus on the magnitude of $V_1$, which is $v_1$.
To simplify, we can use the definition of the cross product:
\begin{equation}
\mathrm{d}\vec{b}\times\vec{R}=\mathrm{d}bR\sin{\alpha}, \tag{3}
\end{equation}
which goes back into eq. (2), leaving us with
\begin{equation}
v_1=\frac{\Gamma}{4\pi}\int_{b_1}^{b_2}\frac{R\sin{\alpha}}{R^3}\mathrm{d}b =\frac{\Gamma}{4\pi}\int_{b_1}^{b_2}\frac{\sin{\alpha}}{R^2}\mathrm{d}b. \tag{4}
\end{equation}
When we integrate over the length of $b$, $\mathrm{d}b$ just becomes $b$. We thus wind up with
\begin{equation}
v_1=\frac{\Gamma b \sin{\alpha}}{4\pi R^2}.
\end{equation}
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Why don't electromagnetic waves need vacuum to move? I understand that electromagnetic waves are caused by the oscillation of electric and magnetic fields. But how? Are there magnetic and electric fields in air, or in vacuum? How does this oscillation form a wave? Sorry, I am a new at this site, please simply put you answer so that I can understand.
| If you have an electric field ${\vec{E}}$, say, due to a charge density ${\rho}$ and the charged body is in motion then from Maxwell's equation $$\vec\nabla\times\vec{H}=\vec{J}+{\mu_0}{\partial\vec{D}\over{\partial t}}$$
i.e., the motion of the charges immediately produces a magnetic field $\vec{B}$ which, the magnetic field, in turn produces an electric field given by $$\vec\nabla\times\vec{E}=-{\partial\vec{B}\over{\partial t}}$$ So you need nothing but the Maxwell's equations for your EM to propagate; by saying EM waves need a vacuum to propagate you're actually implying that EM waves need no medium to propagate for vacuum in classical physics means absence of any medium.
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Why does a simple pendulum or a spring-mass system show simple harmonic motion only for small amplitudes? I've been taught that in a simple pendulum, for small $x$, $\sin x \approx x$. We then derive the formula for the time period of the pendulum. But I still don't understand the Physics behind it. Also, there's no angle $x$ involved in a spring-mass system, then why do we consider it an SHM only for small amplitudes?
|
Why does a simple pendulum or a spring-mass system show simple
harmonic motion (SHM) only for small amplitudes?
Simple harmonic motion (in one dimension) is, by definition, a solution to the generic equation
$$\frac{d^2x}{dt^2}+\omega^2x=0,$$
where $x$ is a generic variable (it can be for instance a displacement or an angle). This equation of motion can be obtained from Newton's second law
$$m\frac{d^2x}{dt^2}=F=-\frac{dU}{dx},$$
where the last equal sign holds for conservative forces, with $U$ being the potential energy.
Therefore we can compare
$$\frac{d^2x}{dt^2}+\frac{1}{m}\frac{dU}{dx}=0,$$
with the equation defining the SHM and see that this requires the potential $U$ to be quadratic in $x$.
Neither a simple pendulum or a real spring has potential which is quadratic (or parabolic) around an equilibrium point. However in a small enough region around the equilibrium (zero force and minimum of the potential) we can Taylor expand the potential up to second order which just gives a parabolic potential. To understand this, consider the figure below,
This is a generic potential which is definitely not of the required form for a SHM. However, notice that we can fit a parabola (dotted line) around any stable equilibrium point such as $x_0$. In the vicinity of the equilibrium point $x_0$ we have the expansion,
$$U(x)=U(x_0)+\frac{dU(x_0)}{dx}(x-x_0)+\frac{1}{2}\frac{d^2U(x_0)}{dx^2}(x-x_0)^2+\mathcal O(\Delta x^3),$$
where $\mathcal O(\Delta x^2)$ means we are neglecting terms of order $(x-x_0)^3$ or greater. Since the first derivative of the potential at $x_0$ gives the force at $x_0$ (which is zero), that term vanishes. We are left with
$$U(x)=U(x_0)+\frac 12 k(x-x_0)^2++\mathcal O(\Delta x^3),$$
and since $U(x_0)$ is an irrelevant additive term, and $k=frac{d^2U(x_0)}{dx^2}$ is constant, this quadratic potential is exactly the characteristic potential of SHM. Nearby $x_0$ it gives a linear and restoring force. Note however that this approximation does not hold for arbitrary displacements. At some point $x-x_0$ is so large that higher order terms have to be taken into account.
For a simple pendulum, the potential is $-mgl\cos x$ ($x$ being an angle) whose expansion up two order two around the equilibrium $x=0$ gives a parabola. For a real spring, the potential can be even more complicated, in general it is a sum of terms $a_nx^{2n}$. Then for a small region we can drop higher order terms and keep only the leading one $a_1x^2$ which gives the Hook Law.
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Current as the time derivative of the charge I have been told that the current $i$ can be defined as
$ i = \displaystyle\frac{dq}{dt} $,
where $q$ is the charge and $t$ is the time.
I do not understand this definition because, if the charges are moving so that the net charge remains constant in an infinitesimally thin cross-section of a wire, $q$ is constant with time and hence $dq/dt = 0$. That result would mean that no constant current can exist unless the charge change has a linear dependence with time (i.e. $q = q(t) \propto t$).
As I assume my reasoning is wrong, where is my mistake?
Thank you.
| We define current as change in charge per time through an area $A$, not in and out of a volume. If the charge is moving through this area you get a current as expected because you have a net flow. If equally much charge passes through from both sides, the current is zero and the net charge is not moving.
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Can an object falling in vacuum generate electricity by itself? When an object falls through vacuum, gravitational potential energy is converted to kinetic energy. Is there some way to get electrical energy out of the equation by itself (i.e. somehow convert the gravitational potential energy to electrical energy)? Is this physically possible? If so, what properties must this object have?
By by itself, I mean without using any external (possibly stationary) "reference object" (e.g. a stationary coil), so a magnet falling through a coil does not count, i.e. the electricity is generated solely by the object that is falling. Note that the object itself can be arbitrarily complex internally, just that whatever mechanism it has inside must also be falling along with the object.
| A sufficiently large object will experience differential gravity ("tidal force") - this could be converted into a small amount of electrical energy by having two heavy spheres separated by a long rope; as they fall there will be a tension on the rope and you could let that tension do work on a generator / dynamo ("complex but internal to the object")
The concept here is that a ball closer to the earth will experience greater force and so fall a little bit faster - in the extreme case of falling to a black hole this leads to "spaghettification " but on a more normal scale it could give you a little bit of electricity. But without en external electric or magnetic field I can think of no way to convert most of the kinetic energy into electrical - the ability to do so would be a first step to an antigravity system. Let me know when you get there!
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About effects of torque off center of mass I am trying to deepen my intuition: If in outer space is a rod of length 2 meters standing still. At the both ends it has some heavy wheels of equal mass and the associated motors. If a motor starts spinning one of the wheels ... will the center of mass of the rod stand still or move? I think that the following are possible:
*
*the center of mass will slowly rotate about the end of the rod with the wheel spinning. That end of the rod will stand still. So the c.o.m of the rod moves in a circle
*the center of mass will stand still and the ends will slowly rotate about the center of mass of the rod. So the c.o.m will not move, but the ends of the rod will rotate about it.
*neither ...
The motors are attached (welded) to the rod. Of the motor's rotors are attached the heavy wheels. (One motor at one end of rod)
I used to think that option 1 is the correct one ...
| Your intuition is correct.
If no external torque is applied, then the angular momentum of the thing doesn't change. That means, if one or both wheels start to rotate, the entire assembly will start to rotate in some way so that the total angular momentum does not change.
You can sometimes see this if you watch dirtbikes taking jumps. They can be in the air, not touching anything, and suddenly rotate forward so that the front wheel dips down. This happens if they hit one of the brakes in the air. One component (the wheel) stops rotating and loses angular momentum, then the entire assembly (bike and rider) starts rotating so that the net angular momentum of the entire assembly doesn't change.
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Looking for a good casual book on quantum physics I'm looking for something that is going to blow my mind without any scientistic ideas (e.g. something that sounds like science, but doesn't have anything in common with science), written by a professional physicist who spent a lot of time considering "what it all means". I'm reasonably proficient in math and stats, but I'd prefer something that I could spend time listening to in my free time.
Any recommendations on good and exciting books on quantum physics written by scientists?
| Try "What is Quantum Mechanics" by Transnational College of LEX. It is very unorthodox. Under the supervision of a Nobel Prize winner the book was assembled by a number of different people who had to discover quantum mechanics from the ground up. Amazon or your local library. Be sure it's the second edition.
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Method of image charges for a point charge and a non-grounded conducting plane I know how to solve Laplace's equation for a point charge in front of a grounded conducting infinite plane. But I want to know what happens (both physics and math) when the infinite conducting plane isn't grounded, or is connected to a potential $V$.
| If the problem you're trying to solve only contains one point charge and the conducting infinite plane at potential V, then there is no physical difference between the plane's potential being 0 (grounded) or +V, because the electric potential may be globally shifted by a constant value everywhere...
| {
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Electron wave and photon wave packet spreading I am looking for a physical interpretation of different behavior of electron and photon wave packets.
The dispersion relationship for a photon in free-space is linear ($\omega \propto k$), while for an electron (or any other massive particle) it is quadratic ($\omega \propto k^2$) (in free-space). If I form a (single) electron wave packet it will disperse in time (broaden with time of propagation), but a photon packet will not.
Apparently, any massive particle will behave the same way regardless of whether it has charge or whether it is a boson or a fermion. I would consider the dispersion relationship difference a purely mathematical explanation for this phenomenon, but is there a physical interpretation behind this?
| A dispersion relation tells you the form of $\omega (k)$. Since $E = \hbar \omega$ and $P = \hbar k$ you can see it as a relation between the energy and the momentum.
Since we have from special relativity that $$ E^{2} = p^{2}c^{2} + m^{2}c^4$$
it is clear that we have $E = Pc$ for a photon. Also since the total energy of a free electron is $E=\gamma m c^2$ the kinetic energy is $E = (\gamma -1)mc^2$ wich reduces to $P^2/2m$ for $v<<c$
This way you can see how special relativity tells us that mass has a role in the dispersion relation, since rest (invariant) mass is the same for all observers in all reference frames. (It's the norm of the energy-momentum 4-vector in Minkowski space).
Returning to your question, you can see that photons follow the wave equation $$ \partial^2_t \Psi = v^2\nabla^2 \psi $$
whose solutions are transverse waves, wheras free electrons follow the Schödinger equation :
$$ i\hbar\partial_t\Psi = -\hbar^2/2m\nabla^2\Psi $$
whose solutions are plane waves.
The dispersion relation is medium-dependent, for instance light is dispersionless in vacuum but not in matter, so in general $$v (n) = c/n$$ where $n$ is the medium's refractive index.
For waves following Schrödinger's equation the dispersion relation is given in general by special relativity. This is why massive particles have a different dispersion than electromagnetic waves for example, and because massive particles have a phase velocity $v_\phi = \omega/k$ that depends upon the wavelength they broaden with time propagation.
(Edited a lot of times)
I do not answer questions often, so I hope this is helpful.
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Would a gas "weigh" less than a liquid if they have the same mass? Thought experiment: I acquired two boxes of the same dimensions and same weight. One box contains $1\ \mathrm{kg}$ of water at room temperature while the other box has $1\ \mathrm{kg}$ of water, but in steam form, because the temperature of the box is above $100^\circ\mathrm{C}$. The volume of the boxes is large relative to the amount of space the $1\ \mathrm{kg}$ of water would take (let's arbitrarily say $10\ \mathrm{L}$). Both boxes contain the same amount of air (at $1\ \mathrm{atm}$) which is why the second box has water in steam form at $100^\circ\mathrm{C}$.
I put each box on a simple electronic scale to measure their respective weights. Unsurprisingly, the box containing water comes out to be $1\ \mathrm{kg}$. But what about the box containing steam?
My guess: Electronic scales measure the amount of force being exerted on it, then divide that force by $g$, to get the mass of the object. I think the box with steam in it will be exerting less force on to the scale and therefore the scale will think its mass is less than $1\ \mathrm{kg}$.
| Actually, if you consider relativistic effects, the box with steam will weight more, not less, since it has more energy and hence more mass. But ignoring these effects, both boxes should weight exactly the same: you have the same number of water molecules in each box, and gravity effect on each molecule is exactly the same.
| {
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Schrödinger equation and non-Hermitian Hamiltonians Is the Schrödinger equation still valid if we use a non-Hermitian Hamiltonian with it? By this I mean does:
$$\hat{H}\psi(t) = i\hbar\frac{\partial}{\partial t}\psi(t)$$
if $\hat{H}$ is not Hermitian?
| Someone else has already mentioned PT-symmetric Quantum Mechanics. To expand on that, and what I believe is the spirit of your question rather than the letter, sure; you can write down the Schrödinger equation using a non-Hermitian Hamiltonian. The interesting question from a physical perspective is whether this describes something physical - i.e. whether it describes something we can measure in the lab, that we can understand to fit within the framework of QM.
Introductory texts (like Griffiths) often suggest that we require observables to be represented by Hermitian operators because their eigenvalues (aka the results of measurements we can make in a lab) will be real. In this line, PT-symmetric QM has been suggested as an extension of QM, where the Hamiltonian is now required to have
Parity and Time symmetry: This means that the Hamiltonian commutes with the product of the parity and time reversal operators, $[\mathcal{\hat{P}\hat{T}},\hat{H}] = 0$
It can be shown that Hamiltonians that satisfy this property have real eigenvalues (i.e. possible measurements of the energy will always yield a real number). There are additional issues related to unitary time evolution, as mentioned in another answer. Most physicist's would see non-unitarity as at least a big a problem as non-real energy values. This issue can allegedly also be solved.
I don't know how well motivated this is. I gained the impression from attending a (single) talk that people in the business of dealing with issues related to the foundations of Quantum Mechanics hope that such a generalisation may be meaningful. Until there is some insight into the empirical status of PT-symmetric QM this is debatable, IMO.
| {
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Conceptually, why is acceleration due to gravity always negative? As the title states, why is acceleration due to gravity always (-). Say you assign "up" as the positive direction. If an projectile is thrown at a 24 degree angle above the horizontal, I get that acceleration due to gravity before the vertex is negative. However, why is it not positive after the vertex? If acceleration due to gravity is negative and we assign downwards as negative, wouldn't that make acceleration positive?
What I think is that acceleration due to gravity is always towards the ground. Even if a projectile is going downwards, and we assign downwards as (-), the acceleration due to gravity is still (-), because the object still accelerating downwards. Despite it going upwards or downwards, the net acceleration of the object is downwards.
| I think the OP is confusing acceleration and direction of motion. Acceleration does not depend on the direction of the motion. According to the Second law of motion, (for constant mass) $\vec F=m\vec a$, acceleration is in the sense of the resultant force acting on the particle. So, it does not matter that the particle is moving in which direction; as long as the resultant force acting on it does not change, the acceleration won't change.
| {
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Can something you see through a telescope be behind you? I read somewhere that gravity is able to bend light.
Is there a chance that, if the conditions are right, the light from one star could bend so much through space that when it reaches the telescope we use to look at it, it could actually be behind us?
As an analogy, imagine you have a piece of rope (the light) and every time it goes past an object that has gravity it bends a tiny bit, in the end it has made a 180* turn and is traveling back to the point it started.
| Yes in principle that is possible, however usually you won't recognize the light as a star anymore, because it has to come very close to a large gravitational field of a black hole. What is indeed observed and you can find pictures of that is black holes acting as gravitational lenses so that you see galaxies or stars behind it multiple times. This is possible because the light can bend around the black hole on different sides. So if you are just looking for the effect of gravity bending light I suggest looking for that.
| {
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Is weight a scalar or a vector? My professor insists that weight is a scalar. I sent him an email explaining why it's a vector, I even sent him a source from NASA clearly labeling weight as a vector. Every other source also identifies weight as a vector.
I said that weight is a force, with mass times the magnitude of gravitational acceleration as the scalar quantity and a downward direction.
His response, "Weight has no direction, i.e., it is a scalar!!!" My thought process is that since weight is a force, and since force is a vector, weight has to be a vector. This is the basic transitive property of equality.
Am I and all of these other sources wrong about weight being a vector? Is weight sometimes a vector and sometimes a scalar?
After reading thoroughly through his lecture notes, I discovered his reasoning behind his claim:
Similarly to how speed is the scalar quantity (or magnitude) of velocity, weight is the scalar quantity (or magnitude) of the gravitational force a celestial body exerts on mass.
I'm still inclined to think of weight as a vector for convenience and to separate it from everyday language. However, like one of the comments stated, "Definitions serve us."
| Here, I would suggest using NASA as more authoritative than you teacher to eliminate opinion. https://www.grc.nasa.gov/www/k-12/airplane/vectors.html
Directly from their opening summary: Scalars were quantities without direction, including length, speed, volume, area, mass, density, pressure, temperature...
Vectors are quantities with direction: displacement, velocity, acceleration, momentum, force, lift, drag, thrust, weight.
I have never been sure why, but for some reason weight and mass are one that is often flipped with people incorrectly claiming that mass is a vector and weight is not and not understanding that weight does have direction: towards the center of force causing it, in our case gravity so down.
| {
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What kind of wave motion is described by grass moving in the wind? You know that sort of 'rolling' illusion when wind blows across long grass, like in the "amber waves of grain" line from America the Beautiful
It's not the same motion as dropping a rock in water, which causes an up and down motion.
And if wind blows across gravel, or water, it just shears it. The water or rocks don't recover like the individual grass blades do.
It's nothing like plucking a string, but kind of like plucking a cantilever beam. Except, I'm curious about what the aggregate motion of hundreds/thousands of blades of grass is known as
| Each blade of grass is displaced from its equilibrium position according to the local velocity of the wind. This means the waves are actually periodic variations in the windspeed. This is likely due to vortices generated as the wind passes upstream obstacles, such as hills/woodland/buildings.
Here is a more realistic rendering of vortices, albeit with a slightly different geometry (the airflow passing both above and below the obstruction). This phenomenon is known as a Kármán vortex street.
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Why do we not feel weightless at equator but feel in satellite A person living in his house at the equator goes in a circular orbit of radius equal to the radius of the earth. My question is, why does he/she not feel weightless as a satellite passenger does?
If we compare a geostationary satellite with the earth's equatorial surface then we know they both revolve around the centre of earth with same Angular velocity. So if Normal force is zero in satellite then why not at equator's surface.
| Note: This answer contains the complete solution. The answer has been written such that the more you read, the more clues you get. At the end, you'll find the complete answer. Every line is a new hint. Please read each line and try to solve the question yourself. If you fail to solve, read the next line and try again.
The solution is in the following formula:
$$F = \frac{mv^2}{r}$$
The earth takes 24 hours to rotate once. From that information, you can calculate your tangential velocity for a person on the surface to be around $500ms^{-1}$.
The centripetal force you need for $500ms^{-1}$ is:
$$F = \frac{mv^2}{R} = m0.04\space N$$
The earth excerts a force of $mg$ that is nearly equal to $10m$. This value is much bigger than the required centripetal force. Hence, you feel weight.
The people in the sattelites are orbiting at a velocity such that the entire gravitational force is utilized as centripetal force.
$$mg_\text{at that point} = \frac{mv^2}{r}$$
$$v_{orbital} = \sqrt{\frac{GM}{r}}$$
For the specific case of geostationary satellites, their orbit is $36,000,000m$ away. The gravitational force due to the earth is quite small and this is nearly equal to the centripetal force required at that orbit. Therefore, the people there feel weightless.
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Do 2 conductors (1 grounded via resistor) reach equipotential, before surplus electrons drain to earth? Case I: a negative conductor makes contact with a neutral conductor. Negative donates some electrons to neutral, until there is 0 potential difference. Then they both are slightly negative.
This happens quickly. (Assuming they are good conductors.)
Case II: the neutral conductor is grounded (via a resistor) to earth.
When they make contact, all surplus electrons drain to earth.
How does the resistor affect the movement of electrons?
Will both conductors reach equipotential first, before surplus electrons
drain to earth? (Because they make contact with each other without a resistor in between. The resistor is between them and earth.)
| To understand what happens during a transition, you'll need to go back to the basics.
The negative conductor initially has a surplus of electrons. These electrons don't like to stay close to each other and given enough time they will adopt a configuration which minimizes the potential energy as much as possible. In this case, that configuration is where the charge spreads out on the surface of the conductor. It would be even better if the electrons left the conductor but unfortunately, the poor electrons are bound to the conductor.
When you bring the negative conductor in contact with the neutral conductor, the charges on the surface near the point contact immediately get pushed into the neutral conductor. As of now, the electrons aren't aren't aware of the fact that the neutral conductor is grounded. The electrons keep moving into the neutral conductor. As the electrons near the point of contact move into the neutral conductor, the electrons near the near the point of contact see that the electron density is decreasing, hence, they start moving closer to the point of contact and then ultimately to the neutral conductor. At one point when the electron density reaches a certain point, the electrons in the conductor which was once neutral face repulsive forces and this push them into the wire and they start flowing into the ground.
As time passes, all the charge would have left both the conductors as the earth can be considered to be a big sink. This would leave both the conductors neutral and with zero potential.
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Existence of Higgs Mechanism Given a gauge group G and a subgroup H, does it always exist a Higgs Mechanism breaking G down to H? Is it unique, or with an obvious classification?
| The Higgs field starts out transforming in a representation $V$ of $G$. When acquiring a VEV that spontaneously breaks the symmetry down to $H$, the VEV has to be invariant under $H$, so when we decompose the representation $V$ into representations $\oplus_i V_i$ of $H$, at least one of the $V_i$ needs to be the trivial (or singlet) representation of $H$. A priori there is no guarantee that any given $V$ will contain such a singlet, but there's not much reason to not expect it, either. It depends both in the groups $G$ and $H$ and on the chosen representation $V$ for the Higgs, and we are a priori free to pick any representation we want, although most models have the Higgs in the fundamental or adjoint of $G$.
Without further information, we really cannot say much. Once you have fixed the groups $G$ and $H$ as well as the Higgs representation, we can examine the specific choices for their viability.
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How can I solve the moment of inertia?
The source tells me to use the formula for a ring, but it is not possible, as the portions are nearer to the axis than a normal ring. How can I find the moment of inertia?
| Use the parallel axis theorem for calculating the moment of inertia relative to a point which is not the center of mas (COM). See for example here in Wikipedia.
$I=I_{\rm COM} - m d^2 ,$
where $d$ is the distance between COM and the point of rotation.
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Difference between these two tensors? (help with index notation) What is the difference between the these two tensors?
$$A_{~~i}^{j} \text{ and } A_{i}^{~~j} $$
In my lecturers notes he states that $A^{~~i}_{j}=(A^T)^{i}_{~~j}$. Why is it this and not $A^{~~i}_{j}=(A^T)^{~~i}_{j}$ ? Thank you.
EDIT
Thank you for your answers. What i was missing is that the order of the indices from left to right, regardless of their upper or lower position, tells you which is the row and which is the column (for rank 2 case). For some reason I was thinking that the upper index is always row and lower is always column. And if both were upper or both lower, then it would be left is row, right is column. Some times I think the lecturers are so comfortable with certain conventions that they forget to tell the student or assume it is as natural to us as it is to them.
Regardless I could have worked it out by expanding it all out, which I did after writing this post. Sorry for wasting your time.
| Well if you are talking about a symmetric tensor then the quantities are equivalent. But if they are not then it is a similar process to taking the transpose of a matrix.
Conventionally if we have a rank-2 Tensor given by $A^i \;_j$ then the leftmost index represents the rows of a matrix and the columns represented by the remaining index. Hence the transpose is given by $(A^T)^i\;_j=A_j\;^i$. It is equivalent to swapping rows and columns.
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Is this problem from Griffiths' "Classical electromagnetism" correct? I am talking about pbl 2.46 pag 107.
The potential $$\frac{A}{r} \, \exp{(-\lambda r)}$$ is given I have to find the charge density and the electric field.
The book says the solution for the charge density is: $$\left[\delta^3(r)-\dfrac{\lambda^2}{r} \, \exp{(-\lambda r)}\right]\, \epsilon_0\, A$$
The first part comes from $\nabla^2 \, \dfrac{1}{r}$ , the second comes from $\nabla^2\,\exp{(-\lambda r)}$.
I think this part is missing: $2\,\nabla\, \dfrac{1}{r} \, \cdot\, \nabla\,\exp{(-\lambda r)} $.
Or am I misunderstanding something?
| This is but the screened Poisson equation. Since the functions are radial, only the radial part of the Laplacian survives,
$$
\nabla^2 (e^{-\lambda r}/r) = \frac{1}{r^2}\partial_r ( r^2 \partial_r (e^{-\lambda r}/r))= \frac{1}{r^2}\partial_r \left ( -\lambda r e^{-\lambda r} +e^{-\lambda r} r^2\partial_r \frac{1}{r}\right )\\
= \frac{1}{r^2}\partial_r \left( r^2\partial_r \frac{1}{r}\right ) ~ e^{-\lambda r} -\lambda e^{-\lambda r} \left(\frac{1}{r^2}+\partial_r \frac{1}{r} \right ) + \lambda^2 \frac{ e^{-\lambda r}}{r}.
$$
Note the cross $O(\lambda)$ term vanishes.
Since
$$
\nabla^2 \frac{1}{r}=-4\pi \delta^{(3)} (\vec{r})= -\frac{\delta(r)}{r^2},
$$
one has
$$
(\nabla^2 -\lambda^2) \left (\frac{e^{-\lambda r}}{r}\right)=-\frac{\delta(r)}{r^2}.
$$
So your text is just fine. You probably misunderstand, that, in fact, $\nabla^2 e^{-\lambda r}= (\lambda^2 -2\lambda/r )e^{-\lambda r}$, as evident from above. This additional $O(\lambda)$ term here exactly cancels the cross term you wrote down. (Try evaluating $\nabla^2 r$ to see the point.)
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Why does space have the topology of a three sphere? Suppose that $U(x)$ is an element of the gauge group say $SU(2)$ and suppose $U(x)=1$ as $|\vec{x}|\to\infty$. Then, why does space have the topology of $S^3$?
This is done in Srednicki page 571. Note that I'm not asking how to prove that $SU(2)\cong S^3$. What I'm asking is how to prove that when $U(x)=1$ as $|\vec{x}|\to\infty$ the space $\mathbb{R}^3$ is compactified to $S^3$ space.
| The question can be formulated more generally as:
Why is it that when we consider functions over $\mathbb{R}^n$ such that the limit as $|\vec{x}|\to\infty$ is the same in any direction then we can identify their domain with $S^n$?
Note that although the functions that we consider might be $\mathbb{R}^3\to SU(2)$, the statement is valid for any target space and any dimension $n$. The argument is the following:
First, notice that $\mathbb{R}^n$ is topologically the same as the open $n$-dimensional ball $B^n$. We can identify the functions in $\mathbb{R}^n$ (with existing limit in any direction as $|\vec{x}|\to\infty$) with the functions on the closed ball $\bar{B}^n$ by identifying $\mathbb{R}^n$ with $B^n$, and then assigning to every point in the boundary the limiting value of the function in that direction.
When a function has the same limit as $|\vec{x}|\to\infty$ in any direction, the corresponding function in $\bar{B}^n$ takes the same value in every point of the boundary. We can then identify all the points in the boundary, getting $S^n$ and a well-defined function over it. You can gain intuition about this identification by thinking about the case $n=2$. If we have the disk with all the boundary points glued together, we clearly get a $2$-sphere.
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What determines the direction of a path on a line integral (vector case)? Line integrals are very important to use in Physics. For example, we calculate work by: W=∫<F,dr>. But I just got confused about something. What determines the direction of motion? The integral limits, or the vector dr?
Well, when we do the internal product of the Force by the path(dr), we are aligning this force on the path's direction(I mean, the shape of the curve) . But if I want a path from the position B to the position A, I determine this inserting on Integral's limits(B inferior limit and A superior limit), or the vector dr would indicate the direction of motion (from B to A) and the integral limits would be (A inferior, B superior)?
| The parametrization of your path determines the direction of integration/motion. Let's say you have a semi-circular path of radius 1 moving anti-clockwise (i.e. from the positive x-axis to the positive y-axis to the negative x-axis). We can parametrize it as the following:
$r(t) = (\cos(t), \sin(t))$, $t\epsilon [0, \pi]$
However, imagine you instead parametrize a semi-circular path like this:
$r(t) = (-\cos(t), \sin(t))$, $t\epsilon [0, \pi]$
While both these paths look the same on a plot, the first one starts at (\cos(0), \sin(0)) = (1, 0), while the second one starts at (-cos(0), sin(0)) = (-1, 0).
This is how the direction of travel can be defined for a simple example. If you need to find the direction of some parametric equation, all you need to do is plug in the endpoints and see which way the path flows. Here's another example.
Consider the parametric equation:
$r(t)=(\cos(2t),\sin(t+5))$ $t\epsilon[0, 2\pi]$
Looking at it, I don't have an intuition for where it starts and where it ends.
But we can alleviate this! Just plug in the endpoints.
$r(0) = (\cos(0), \sin(5)) = (1, -0.96)$
$r(2\pi) = (\cos(4\pi), \sin(2\pi + 5)) = (1, -0.96)$
Oh no! We've gotten the same endpoint. From the plot, we can deduce this because it loops over itself. Not to worry, however, we can plug in a number close to one of the endpoints, which will help us make our determination of path.
$r(0.1) = (0.98, -0.93)$
Now it's clear which direction the path goes in.
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Why do ice cubes shrink in the freezer? OK before you all yell "EVAPORATION", I know that's the boilerplate answer, but why and how.
Back in high school we were taught water, and most other elements and compounds, have three states, solid, liquid, and gas. Which state it's in being dependent on temperature and pressure. OK that all makes sense.
To transition from one state to another sufficient energy has to be imparted to the substance to pass through that transition before it will continue to heat up. To get water from ice all the way up to a gas takes a lot of energy... ask all the distillers out there.
Still.. all good...
But my ice cubes disappear over time inside a closed freezer.
The water is evaporating, turning into a gas state, but where did the energy come from to take it all the way from a solid to a gas when it never left the freezer..
| Ice at -10 centigrade has a vapor pressure of 259.9 p/Pa. It evaporates.
Water still evaporates at less than boiling.
| {
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If free quarks can't exist, how did the universe form? As I understand, the Big Bang started with a photon gas that then created the other particles. Thus obviously there would be some free quarks in the early Universe unless quarks are always created in pairs for some reason. How does physics resolve this?
| This article in wikipedia clarifies how the universe has evolved as far as our present understanding of particle physics and general relativity goes.
In particular for the strong interactions, the present theory is QCD,which models quarks and their interactions with other particles.
QCD enjoys two peculiar properties:
Confinement, which means that the force between quarks does not diminish as they are separated. Because of this, when you do separate a quark from other quarks, the energy in the gluon field is enough to create another quark pair; they are thus forever bound into hadrons such as the proton and the neutron or the pion and kaon. Although analytically unproven, confinement is widely believed to be true because it explains the consistent failure of free quark searches, and it is easy to demonstrate in lattice QCD.
Asymptotic freedom, which means that in very high-energy reactions, quarks and gluons interact very weakly creating a quark–gluon plasma.
The present model of the universe starts with enormous energies, and as it expands the individual constituents cool down first into a stage where all forces are unified with all particles zero mass into a quark-gluon plasma
As cooling continues bound hadrons appear.
The quark gluon plasma is studied experimentally at the LHC experiments.
So all the models of particle physics are utilized for the Big Bang model of the universe, and the creation of bound quarks is developed within an extended standard model.
If you read the article you will see that a lot of the model is still at the research stage experimentally.
| {
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Gaussian integral formula for matrix product I am looking for a way to prove that
$$ \det (M \cdot N) = \det(M)\det(N) \tag{0}$$
Where $M$ and $N$ are matrices with continuous indices, so that $\det$ is a functional determinant. A way to show that $(0)$ is wrong would also be welcomed.
This question is about the following formula,
$$
\int\text{d}\vec{x} \exp(- \sum_{ij}x^i A_{ij}x^j) = \left (\det A_{ij}\right )^{-1/2}\left (2\pi\right )^{D/2}. \tag{1}
$$
Now, we would like this identity to be compatible with,
$$
\int\text{d}\vec{x} \exp(- \sum_{ijk}x^i A_{ik}B_{kj}x^j) = \left (\det A\cdot B\right )^{-1/2}\left (2\pi\right )^{D/2} = \left (\det A\right )^{-1/2}\left (\det B\right )^{-1/2}\left (2\pi\right )^{D/2}.\tag{2}
$$
Any idea how to prove this? I am interested, eventually, in the generalisation of this formula to path integrals, namely, given the path integral
$$
\int\mathcal{D}\phi \exp\left[- \int\text{d}x\text{d}y \phi(x)M(x,y)\phi(y)\right] =C \left (\det M\right )^{-1/2}, \tag{3}
$$
where now $\det M$ is a functional determinant, i ask the question whether it makes sense to write the generalised formula,
$$\begin{align}
\int\mathcal{D}\phi \exp\left[- \int\text{d}x\text{d}y \text{d}z\phi(x)M(x,y)N(y,z)\phi(z)\right] =& \left (\det M\cdot N\right )^{-1/2}\cr =& \left (\det M\right )^{-1/2} \left (\det N\right )^{-1/2}.\end{align} \tag{4}
$$
[UPDATE]: I might have an answer now: let us just consider,
$$\det M\cdot N = \prod_i \lambda_i[M\cdot N],\tag{5}$$
where $\lambda_i[M\cdot N]$ are the the eigenvalues of the matrix $M\cdot N$. This formula is valid even for continuous matrices, such as the laplacian operator $\partial^2 \delta(x-y)$.
If the commutator $[M,N] = 0$, then the two matrices can be diagonalised in the same basis, and $\lambda_i[M\cdot N] = \lambda_i[M]\lambda_i[N]$, with no sum over $i$. Then formula (4) can be proven at least in the simple case in which the commutator vanishes.
A trivial example of this is for $M = A$ and $N = A^{-1}$, for any invertible matrix $A$, which leads to $\det A\cdot A^{-1}=1$. Also, in case $M\cdot M^T = f(x) \delta(x-y)$, this would imply that
$$\det M\cdot M^T = (\det M)^2 = \det f(x) \delta(x-y) = \prod_x f(x)\tag{6}$$
and so on. These seem trivial cases, but since we are talking of functional determinants they constitute a powerful computational tool.
How much do you agree with this attempt of a solution? It is not very formal, but i don't see where it could go wrong.
| The following comments seem relevant to OP's problem:
*
*For a matrix $A\in{\rm Mat}_{n\times n}(\mathbb{C})$, define the symmetrized matrix $$A_+~:=~ \frac{A+A^T}{2}.\tag{A}$$
*Then the Gaussian integral reads
$$ \int_{\mathbb{R}^n} \! d^n x ~e^{-\frac{1}{2} x^T A x}
~=~ \sqrt{\frac{(2\pi)^n}{\det A_+}}\tag{B}$$
if the matrix ${\rm Re}A_+$ is positive definite, cf. e.g. this math.SE post.
*Similarly,
$$ \int_{\mathbb{R}^n} \! d^n x ~e^{-\frac{1}{2} x^T AB x}
~=~ \sqrt{\frac{(2\pi)^n}{\det (AB)_+}}\tag{C}$$
if the matrix ${\rm Re}(AB)_+$ is positive definite, cf. OP's eq. (2).
| {
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The torque created using an oval chain ring I have seen the explanation on "How bicycle gear works?" but this seem to be using a standard round chain ring. I would like to know if there is any difference when an oval chain ring is used. As I can clearly see, one can have an oval chain ring on a fixed/single gear bike. There fore there is no slaking or tightening of the chain. So I would say that there is no difference in gear ratio during the complete rotation of the oval-chain-ring. So the torque will remain the same.
Is there an answer to my problem? and if there is could you proved any equations
| If you consider the chain ring as delivering torque about a point where the chain to the rear sprocket and front chain ring is tangential (the top portion of the chain is always under tension when delivering power) you'll see that torque is not constant. In the pedal cycle, when the bulge of the chain ring is tangent to the chain the virtual lever will produce less force for a given torque. When the dip is tangential you'll produce more force (torque = force * distance).
Conversely you could express this as requiring more torque (to drive the rear wheel) when the bulge is tangential and requiring less torque when the dip is tangential.
This is used to compensate for the lack of power you can produce at the 12/6 pedal position. In effect the chain ring is actually normalizing your power output throughout the revolution of the pedals.
The slackness of the lower chain is taken up by the pulleys and springs in the rear mech.
In a fixed gear bike I'd suggest there may simply be a little extra slack in the bottom portion of the chain and/or an additional chain tensioner.
| {
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Generalized forces in Richard Fitzpatrick's Newtonian Dynamics In Richard Fitzpatrick's Newtonian Dynamics, page 128, generalized forces are defined as
\begin{equation} \tag{9.6}
Q_i = \sum_{j=1}^{\mathcal{F}} f_j \cdot \frac{\partial x_j}{\partial q_i}.
\end{equation}
Here, $q_i$ are generalized coordinates, $x_j$ are Cartesian coordinates, and $f_j$ are the Cartesian components of forces acting on particles.
He proceeds to say that in conservative systems
\begin{equation} \tag{9.7}
f_j = - \frac{\partial U}{\partial x_j}
\end{equation}
and concludes that
\begin{equation} \tag{9.8}
Q_i = -\frac{\partial U}{\partial q_i}.
\end{equation}
I can see how the $\partial x_j$ cancel each other out, but what happened to the sum from equation 9.6? Shouldn't there be a constant factor of $\mathcal{F}$ in equation 9.8?
| The numerator and denominator in partial derivatives don't cancel out simply. Keep that in your mind!
You should recall the chain rule in partial derivatives.
The formal answer to you question may be:
If $U=U(x_1,\cdots,x_f)$, and each $x_f$ is a function of generalized coordinates $(q_1,\cdots,q_g)$, the potential could be expressed as follows
$$U=U[x_1(q_1,\cdots,q_g),\cdots,x_f(q_1,\cdots,q_g)]$$
then
$$\frac{\partial U}{\partial q_i}=\frac{\partial U}{\partial x_1}\frac{\partial x_1}{\partial q_i}+\cdots+\frac{\partial U}{\partial x_f}\frac{\partial x_f}{\partial q_i}$$
More detailedly, the total derivative of $U$ is
$$d U=\frac{\partial U}{\partial x_1}dx_1+\cdots+\frac{\partial U}{\partial x_f}dx_f$$
so you must consider every $x_i$ when you want to differentiate $U$ with $q_i$ as other $q$ except $q_i$ fixed.
At the end of the answer, I recommend to use Jacobian in the partial derivatives when you have to cope with so many coordiantes. It will be clearer.
| {
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The illumination problem in context of a classical gas? Background
I was reading about the illumination problem:
"In 1958, a young Roger Penrose used the properties of the ellipse to
describe a room with curved walls that would always have dark
(unilluminated) regions, regardless of the position of the candle.
Penrose's room, illustrated above, consists of two half-ellipses at
the top and bottom and two mushroom-shaped protuberances (which are in
turn built up from straight line segments and smaller half-ellipses)
on the left and right sides. The ellipses and mushrooms are
strategically placed as shown, with the red points being the foci of
the half-ellipses. There are essentially three possible configurations
of illumination. In this figure, lit regions are indicated in white,
unilluminated regions are indicated in gray, and the position of the
light source is indicated by the black cross-hairs. As can be seen,
the entire room (the space within the blue border) can never be fully
illuminated. " - http://mathworld.wolfram.com/IlluminationProblem.html
Question
This made me wonder of the consequences for an classical ideal gas: Suppose we place a box with a gas in a vaccum version of the upper half of this room. Now after some time imagine if the box breaks and the gas begins to spread. No matter how much time has elapsed the gas molecules can never bounce reach the dark regions of the room. Hence, that part of the room will experience $0$ pressure. But this seems rather counterintuitive to me to happen in reality. Is there anything that would stop this effect from being measured in reality? (besides the imperfections of the wall being curved?)
| Molecules of gas collide with one another whereas photons do so very rarely.
Another way of putting that is that light travels in straight lines (for the Penrose analysis) whereas gas molecules do not.
| {
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Expressing position due to gravitational acceleration as a 3-Dimensional differential equation I know that the force of gravity is $F = \dfrac{G m_1 m_2}{r^2}$. Now assume in a one-dimensional system there are two masses in the universe, a planet and an object. The object would have an acceleration of $A = \dfrac{G M}{r^2}$, assuming that m is the mass of the planet. The position of the object can then be graphed with the differential equation $\frac{d^2x}{dt^2}=-\frac{GM}{x^2}$, with x being position and t being time, and negative as the object falls towards the planet. This differential equation can be solved with initial position and velocity values.
My question is now how to transform this to 2 (and 3) dimensions. Assume the same situation but now the object has an x and y coordinate. I can say that acceleration in the x direction equals $A_x = \dfrac{G M}{r^2} \dfrac{x}{r}$, with r being the radius ($\sqrt{y^2+x^2}$. I can also say that acceleration in the y direction equals $A_y = \dfrac{G M}{r^2} \dfrac{y}{r}$.
These equations work when I graph them, but because they rely on each other (the x position is needed for y-acceleration and vice versa), I'm unable to find a way to solve the differential equations and find x and y as functions.
Does anyone have any ideas? If I need to clarify myself better, please ask. Thank you for your time!
| The easiest thing to do is to start from the gravitational potential
$$
\phi=-\frac{Gm_1m_2}{\sqrt{x^2+y^2+z^2}}
$$
and use $\vec F=-\vec\nabla \phi$, or in components:
$$
F_x=-\frac{\partial \phi}{\partial x}=
-\frac{Gm_1m_2\,x}{(x^2+y^2+z^2)^{3/2}} = m_2 A_x = m_2 \frac{d^2 x}{dt^2}
$$
with a similar approach for your other components.
| {
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Why doesn't saturation current in the photoelectric effect depend on the frequency of light absorbed by the metal emitter? If current $I$ is given by $I = nAev$, where $n$ is the number of electrons per unit volume, $A$ is the area, $e$ is the charge of an electron and $v$ is the velocity of the electron, it must mean that the current increases with increase in velocity of the electron which increases with the frequency of light incident on the metal emitter. Why doesn't then saturation current increase with increase in frequency?
| I do not think that the speed of the electron is important, and the reason for this is that the electron will face a repulsion of the electrons on the anode surface, so the importance is the number of electrons that will arrive because they will almost stop
| {
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How does one measure the curvature $k$ in FLRW metric? How does one measure the curvature parameter $k$ in the FLRW metric? $$ds^2=-c^2dt^2+a^2(t)[\frac{dr^2}{1-kr^2}+r^2d\theta^2+r^2\sin^2\theta d\phi^2]$$ In particular, what is the convenient equation (involving $k$) that is/can be used to measure $k$?
EDIT I'm looking for an answer that will explain the measurement of curvature $k$ with the same clarity as the measurement of spring constant $\kappa$ from the one-dimensional simple harmonic equation $F=-\kappa x$ i.e., having measured the applied force $F$ (can be done with a spring balance may be) and the corresponding displacement $x$ (by a meter rule), one can measure $\kappa$. Similarly, if the equation involving curvature $k$ contains non-trivial physical quantities (such as the components of Riemann curvature tensor etc), I would like to know how each of them is measured.
| This is a very hard question to answer in detail as it requires several pages of mathematics to derive the required formulas (there is no easy fit like $F=-kx$ as you suggested)
I will not derive the formula (it can be found in e.g. Dodelson) but after some work you obtain:
$$\Delta(m-M) = 5\log\left\{ \left( \frac{c}{H_0}\sqrt{\frac{k}{\Omega_{total}-1}}(1+z) \right)S_k\left(\sqrt{\frac{\Omega_{total}-1}{k}}\int_0^z \frac{dz'}{(1+z')\sqrt{(1+z')(1-\Omega_\Lambda) + \Omega_\Lambda/(1+z')^2}} \right)\right\} \\ - 5\log\left(\frac{1}{2}((1+z)^2-1)\right)$$
Where $\Delta(m-M) = (m-M)_{'real'\ universe} - (m-M)_{empty\ universe}$. M can be obtained by using standard candles such as supernovae type Ia, it is than easy calculate $(m-M)_{empty\ universe}$ via an equation similar to the one above and $m_{'real'\ universe}$ is simply the magnitude that we measure. Therefore $\Delta(m-M)$
$S_k(...) = sinh(...), sin(...) or 1$ depending on the value of $k$
$\Omega_\Lambda, \Omega_{total}=\Omega_\lambda+\Omega_{matter}, H_0$ and k remain unknown.
The next step is to measure a lot of standard candles at various redshifts z and plot their $\Delta(m-M)$ relation as a function of z. This should obey the relation above. All that remains to be done is to run a fitting script that fits the above function to $\Delta(m-M)$ for various values of $k, \Omega_\Lambda,...$ the best fit gives us the observed cosmology.
In the figure below you can see such a fit from a project I made previous semester where we had to calculate k for some dataset.
Obviously the fits are hard to make due to degeneracies in the fit and uncertainty plots can be made as like this one:
My results for the above fit were: $\Omega_{matter} = 0.286 \pm 0.031$, $\Omega_{\Lambda} = 0.721 \pm 0.025$, $H_0(km/s/MPc) = 70.3 \pm 2.58$ which is consistent with k = 0.
I hope this helped :)
| {
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How is entropy a state function? Is there only one reversible way to move from one state to another?
If we consider two states $A$ and $B$ on an isotherm and we move from $A$ to $B$ by first reversible isochoric process and then reversible isobaric process. Now the path followed should be reversible since both the processes were reversible. But what about simply following the reversible isothermal process?
According to me both processes should be reversible. Now entropy is the heat added reversibly to move from one state to another divided by the temperature at which it is added. But we know that the heat added to the system is different in both the cases. Then how is entropy a state function?
| Entropy is surely a state function which only depends on your start and end states, and the change in entropy between two states is defined by integrating infinitesimal change in entropy along a reversible path. But heat $Q$ is not a state variable, the amount of heat gained or lost is path-dependent. Once we divided $\mathrm{d}Q$ with temperature, this gives a exact differential $\mathrm{d}S$. Therefore, temperature is often mathematically referred to as the "integrating factor" (see https://en.wikipedia.org/wiki/Integrating_factor) of $\mathrm{d}Q$.
| {
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How a moving car becomes electrically charged? Car has been electrically charged as it travels along the road.how is this possible?
| If the tires aren't conductive enough, the asphalt/concrete surface of the
road acts like fur, and the tires act like amber, and charge separation
occurs at the contact point, then the movement of the wheel does work on
the separated charge. Charge repels, leaking through the axle to the car body,
accumulates on the outer shell of the car. Basically, the tires (on the interior
of the wheel well) are inside a Faraday cup, so this charge will always
migrate to the outside.
Then the (newly discharged) tire surface comes back into contact with the
road, and the cycle repeats.
Tires are black because of carbon (conductive) added particles, said to
make the 'charge separation' event mainly not happen.
| {
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Stings in the loop space of timelike curves In Smith's paper on homotopy groups for Lorentz manifolds, he builds the loop space of all timelike loops in the following fashion :
*
*Consider all piecewise continuous timelike curves which start and end at point $x$. This include timelike curves with $q$ changes in time orientation (the tangent vector of the end of one segment has an opposite time orientation to the beginning of the next)
*Also include in the group stings based at $x$, which are made from arbitrary paths $\gamma$ in the following way : a sting is a curve of the form $\gamma \ast \gamma^{-1}$, with $\gamma(0) = x$.
*Include insertions of stings on paths. For a path $\gamma$, consider a point $y$ in $\gamma$, and decompose it in two paths $\gamma = \gamma_+ \ast \gamma_-$, with $\gamma_+(1) = y$. The insertion of a sting $f \ast f^{-1}$ at $y$ is $\gamma^* = \gamma_+\ast f\ast f^{-1} \ast \gamma_-$.
*The constant path is also included in it, $e(\lambda) = x$
The loop space is then defined by all those elements, and the path composition $\ast$ has a group structure.
The motivation for the inclusion of stings given seems to be that it permits the group structure (although that's not stated clearly either), but that doesn't seem correct, as the constant path and timelike curves seem like enough for that. What is the purpose of adding stings to the loop space? All curves involved will be equivalent to a stingless curve anyway.
| Note that the author defines the loop space $T_q$ to be the space generated by loops with $q$ corners. You want to show that $ff^{-1}\sim e$ in the timelike sense. But $ff^{-1}$ will have at least $2q$ corners since you get corners from each copy. So you include all of the curves of this form in your definition of $T_q$.
| {
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Expansion coefficients in the solution of the Dirac equation for a free particle So my question is why do we need to write the coefficients $b$ (that after the second quantization are going to be promoted as the antiparticle creation operators) as complex conjugate? I mean, why not just write $b$, without the complex conjugation sign in the solution
$$\psi(x) =\int \frac{d^{3}p}{(2\pi)^{3}}\sum_{s=1,2}(a_{p,s}u_s(p)e^{-ipx} + b^{*}_{p,s}v_s(p)e^{ipx})
$$
I hope my question is clear enough. It bothers me that in a lot of textbooks you just find $b^{*}$ without a clear argument why it isn't just $b$.
| It is just notation. If you schematically write
$$
\psi\sim A\mathrm e^{-ipx}+B\mathrm e^{+ipx}
$$
then you can check that
$$
\begin{aligned}
{}[H,A]=-\omega A\\
[H,B]=+\omega B
\end{aligned}
$$
so that $A$ behaves like an annihilation operator (it lowers the energy) and $B$ behaves like a creation operator (it increases the energy). Therefore, it makes sense to write $B\equiv b^\dagger$ - it is just convenient notation. You could omit this relabelling if you wanted to, but why would you.
| {
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Photoelectric effect:- Reduction of wavelength increases current? I did a question in which, the intensity of the incident radiation on a metal surface was kept constant but the wavelength of the photons has been reduced. The question inquired what will be the effect on the maximum photoelectric current? The initial wavelength was smaller than threshold wavelength of the metal surface.
My thinking was since the intensity remains constant, thus the number of photons emitted from the source remains constant and thus the number of electrons emitted from the metal surface. And since number of electrons per unit time isn't changed, the current will remain the same.
However, the answer key stated "Fewer photons (per unit time) so (maximum) current is smaller"
How does decreasing wavelength (equivalent to increasing the energy of photons) result in a fewer photon emission?
| I assume in reality everything is much more complicated, but in the simple model it is quite clear:
A (bound) electron interacts with a photon, and can gain (at most) the photon's energy E. If E is large enough, it can overcome the bound and become a free electron. If E gets even larger, then the electron could even overcome an additional potential difference (i.e., we can produce higher voltage), but a single photon will still interact with a single electron only.
So if you wavelength gets shorter (i.e., energy for each photon increases), then given constant intensity you will have fewer photons, i.e., fewer electrons are removed from the material, i.e., smaller current (but those electrons will have higher energy).
| {
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Can space exist without matter or energy? Is it possible for the space to exist without matter or energy?
|
Is it possible for the space to exist without matter or energy?
As a possible mathematical model of nature, yes.
Proof: Newtonian gravity , a classical very successful model, i.e. predictive and not only descriptive. General Relativity which posits that matter and energy form space time , reduces to Newtonian gravity at the interface, where GR effects are below the accuracy of measurements.
The model of General Relativity cannot exist without matter or energy, they have to exist , otherwise all space time would be uniformly 0 for all values of space and time.
Is the GR model an ultimate model for space time? No, because quantization of gravity changes it. If the encompassing theory is a string theory, again particles are necessary and particles are matter and have energy.
From there on, one would have to wait for new observations and experiments, but at present it seems that matter and energy are tied up with space and time, starting from the frameworks where special relativity has to be used.
| {
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Continuous vs discrete set of eigenvectors for a single particle Hamiltonian The Hamiltonian operator of a free particle in three dimensional Euclidean space has an infinite set of eigenvectors lableled by the momentum of the particle, $| p \rangle $. Not only is this set infinite, but it is also continuous. But if we look at exactly solvable Hamiltonians with a potential, like the harmonic oscillator or the $1/r^2$ potential, the set of eigenvectors form a infinite but discrete set (labelled by a set of integral quantum numbers). I have two questions regarding this,
*
*Is there any potential that can give rise to an infinite and continuous set of eigenvectors (barring the uniform potential)?
*If not, why should the addition of a potential change the eigenspectrum from discrete to continuous?
*If the answer to the first question is yes, what conditions should the potential satisfy to have a discrete spectra?
| Whether a particular Hamiltonian gives rise to discrete or continuous momentum spectra depends on the energy of the particle(assuming the particle is in energy eigen-state). For instance consider an electron in Hydrogen atom potential but with energy $E > 0$. Then the electrons motion is unbounded and it can have continuous $|p>$ eigenstates.
A more interesting question could be that given a bounded system is it possible to have continuous momentum spectra. The answer again is YES. In crystalline solids the electron energy levels are not of specific momentum but rather a "continuous band of energy" at discrete intervals.
To have a perfect discrete spectra your potential should have infinities at some boundary. This ensures that the wave-function has to be zero at some boundary and therefor be necessarily discrete.
| {
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What information can we extract from the electronic band structure? I have some difficulty in understanding the electronic band structure.I want know that for a 3D crystal,what information can I extract from its complicated band structure,for example the band structure of the SiC(I downloaded this figure from google).And what intuition that I can build for such a complicated band structure?
| The most useful information you can extract from the band structure for an insulator like SiC involves: (1) the value of the bandgap (the energy difference between the highest occupied band--the valence band--and the lowest unoccupied band--the conduction band); (2) the direct or indirect nature of the band gap (direct if the valence band max occurs at the same k point as the conduction band minimum, and indirect otherwise); and (3) the band dispersion or the slope of the bands involved in the band gap--steeper slopes indicate stronger orbital interactions and faster carrier mobility, on the other hand an ionic crystal will have very flat bands.
| {
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What causes like electric charges to repel and opposite electric charges to attract at the smallest level? When talking about charged particles, the law of charge dictates that two particles with opposite charge will attract each other and two particles with the same charge will repel each other.
However, I have never seen why this works. So, on a fundamental level why does the law of charge work? What causes like electric charges to repel and opposite electric charges to attract at the smallest level?
| Short answer:
It is a consequence of
*
*Physics is governed by a stationary action principle
*Locality
*Lagrangian is Lorenz Invariance
*Gauge invariance
Long answer(and still skipping lots of math):
From a relativistic point of view:
Starting from the action principal, we try to write down a simple action involving the action potential $\vec{A}$ and electric charge $q$
$$Action = \int -mc^2 d\tau -\frac{q}{c} A_\mu dx^\mu$$
Multiplying by $\frac{dt}{dt}$
$$\int\left(-mc^2\sqrt{1-\left(\frac{\dot{x}}{c}\right)^2} -\frac{q}{c} \left(cA_0 + A_m\dot{x}^m\right)\right)dt$$
Therefore $$\mathscr{L}=-mc^2\sqrt{1-\left(\frac{\dot{x}}{c}\right)^2} -\frac{q}{c} \left(cA_0 + A_m\dot{x}^m\right)$$
After applying the Euler-Lagrange equations, we get
$$ma^m=q\left(\left(\partial_0A_m-\partial_mA_0\right)\vec{u}^0+\left(\partial_nA_m-\partial_mA_n\right)\vec{u}^n\right)$$
$$F=q\left(\vec{E}+\vec{v}\times\vec{B}\right)$$
We could also define a tensor,$F_{\mu\nu}$ to simplify the equation
$$F^\mu=eF^\mu_\nu u^\nu$$
Let $j^\mu=(cp,j^m)$ where $j^m$ is the conventional current density.
Now consider $$\mathscr{L}=F_{\mu\nu}F^{\mu\nu}+j^\mu A_\mu$$
It may not look gauge invariant at first, but after adding gauges, one sees it is indeed gauge invariant.
After applying Euler-Lagrange equations for fields, one can derive
$$\nabla\cdot\vec{E}=\frac{j^0}{c\epsilon_0}$$
From that column law can be derived.
$$\vec{E}=\frac{kq}{r^2}\hat{r}$$
Letting $\vec{B}=\vec{0}$, $\vec{E}=\frac{\vec{F}}{q}$
$$\frac{\vec{F}}{q}=\frac{kq}{r^2}\hat{r}$$
$$\vec{F}=\frac{kq^2}{r^2}\hat{r}$$
Therefore if the charges are the same, they will repel, if not, they will attract.
| {
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Quadrivectors in relativity This is what I understood about 4-vectors in relativity.
We define the contravariant and covariant vectors like this :
$$
A^\mu=\begin{bmatrix}
A^0 \\
A^1 \\
A^2 \\
A^3
\end{bmatrix}$$
$$
A_\mu=\begin{bmatrix}
A_0 \\
A_1 \\
A_2 \\
A_3
\end{bmatrix}$$
The relationship between them will be :
$$
A^\mu=\eta^{\mu \nu}A_\nu
$$
In +--- convention it will lead to :
$$
A^\mu=\begin{bmatrix}
A_0 \\
-A_1 \\
-A_2 \\
-A_3
\end{bmatrix}$$
Great.
But it doesn't give me information on the "absolute" sign of 4-vectors. For example if I take the 4-position.
I have an even at time $t$ at space coordinates $(x,y,z)$.
Will I have $$X^\mu=\begin{bmatrix}
t \\
x \\
y \\
z
\end{bmatrix}$$
Or
$$X_\mu=\begin{bmatrix}
t \\
x \\
y \\
z
\end{bmatrix}$$
I think it is the first answer because $A^\mu$ should transform the same way that the "real" coordinates $(t,x,y,z)$ transform, but I am not totally sure ?
Thank you.
| The answer is that it doesn't really matter as long as you are consistent !
To see this, let us define $X^\mu = [a\ b\ c\ d]^T$ and $\tilde{X}^\mu = [a\ -b\ -c\ -d]^T$ we will see that these are equivalent up to a change of basis. Which proves that it is indeed irrelevant as long as we are consistent!
Expanding both vectors onto the basis of vectors in 4D Minkowski space $\{ \partial_t, \partial_x, \partial_y, \partial_z \}$ gives us:
$$ X = a\partial_t+b\partial_x+c\partial_y+d\partial_z\\\text{and}\\
\tilde{X}=a\partial_t - b\partial_x - c\partial_y-d\partial_z$$
Those two vectors are indeed equivalent up to an inversion of the spatial part of the manifold.
Also, if I may give you a little tip. $X^\mu$ and $X_\mu$ are contractible to a scalar you can stress this when going to matrix notation by writing: $X^\mu = [...]$ and $X_\mu = [...]^T$.
I hope this helps. Feel free to ask questions if you are still confused
| {
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if white things reflect light and mirror reflect light why don't they look the same We learned at school that white object reflects all the light that falls on it. We also learned that a mirror reflects all light as well. However, we cannot see ourselves in a white object while we can see ourselves in a mirror.
What makes a mirror different from a white surface?
If both white surface and mirror reflect all the light that fall on them, then why don't they look the same?
| MIRROR
*
*The thing is that, A mirror is a surface from which light get totally reflected.
*It have a polished surface. We generally see mirror effect from metal surface.
*The light which come in strike at angle $\theta$ to the normal and reflect away at $\theta$ from the normal.
WHITE SURFACE
*
*It is a surface which seems to be white but it reflect and disperse of all seven visible wavelength.
*The surface is microscopically very rough. White surface are generally clothes, paints, non-metal, paper.
*The light which come in strike at angle $\theta$ to the normal and reflect away at many various angle from the normal.
*There is lot of distortion and dispersion of light.
That is why, a mirror $surface$ and white $surface$ are different.
| {
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Kinetic, potential and total orbital energy in General Relativity In Schwarzschild geodesics the total orbital energy $E$ is
$$E = \dot{t} \left( 1 - \frac{r_{\rm s}}{r} \right) m \, c^2$$
with the time dilation factor $\dot{t}$ in dependence of the local velcity $v$
$$\dot{t} = \frac{1}{\sqrt{ \left( 1-\frac{r_{\rm s}}{r} \right) \left( 1-\frac{v^2}{c^2} \right)}}$$
so plugged into the equation for $E$ we get
$$E = \frac{m \ c^2 \ (r-r_{\rm s})}{\sqrt{r \ (r_{\rm s}-r)(v^2/c^2-1)}}$$
which seems to be
$$E = m \ c^2 + E_{\rm \ kin} + E_{\rm \ pot}$$
But how would one factor out the kinetic and the potential component of the total Energy in terms of the coordinate derivatives $\dot{r}, \dot{\phi}, \dot{t}$ or in terms of $v^2=v_{\perp}^2+v_{\parallel}^2$ (radial and transverse components)?
The other constant of motion, the angular momentum, is easy to get because with
$$\dot{r} = v_{\parallel} \sqrt{\frac{1-2 M/r}{1-v^2}} \ , \ \dot{\phi} = \frac{ v_{\perp}}{r \sqrt{1-v^2}}$$
we get
$$L = m \ \dot{\phi} \ r^2 =\frac{m \ v_{\perp} \ r}{\sqrt{1-v^2}}$$
but what about $E_{\rm \ kin}$ and $E_{\rm \ pot}$? Those seem to be very different than with Newton or Special Relativity, at least one of them since the sum does not match up. I only managed to calculate to total energy but failed to split it into it's components.
| Your expression for the total energy is
$$E=\frac{mc^2(r-r_s)}{\sqrt{r(r-r_s)(1-v^2/c^2)}}=mc^2\gamma\sqrt{1-\frac{r_s}{r}}$$
If you wish to split this up into kinetic and potential energy, we recall that the kinetic energy in Special relativity is $E_{\text{kin}}=mc^2(\gamma-1)$, and so we have
$$E=mc^2+E_{\text{kin}}+E_{\text{pot}}$$
Where
$$E_{\text{pot}}=-mc^2\gamma\left(1-\sqrt{1-\frac{r_s}{r}}\right)=-mc^2\gamma\left(1-\sqrt{1-\frac{2GM}{rc^2}}\right)$$
Let's do a sanity check. In the nonrelativistic limit, $mc^2(\gamma-1)\sim mv^2/2$ and
$$E_{\text{pot}}\sim-mc^2\left(1-\left(1-\frac{GM}{rc^2}\right)\right)=-\frac{GMm}{r}$$
Which agree with the nonrelativistic expressions!
| {
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Why do all electromagnetic waves travel at the same speed when travelling through vacuum? What does my teacher mean when he says that all electromagnetic waves travel at the same speed when travelling through a vacuum? If you may, please answer as simple as possible.
| Electromagnetic waves include visible light, radio waves, X-rays, and so on. What distinguishes these different bands of light is their frequency (or wavelength). But what they all have in common is that they travel at the same speed in vacuum.
The reason for qualifying 'in vacuum' is because EM waves of different frequencies often propagate at different speeds through material.
The speed of a wave $c$, its wavelength $\lambda$ and frequency $f$ are all related according to $c=\lambda f$. So if $c$ is the same for all EM waves, then if you (say) double the frequency of a wave, its wavelength will halve.
| {
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Multiple star system, stable orbits? Inspired by worldbuilding SE, I know that there are relatively stable star systems with two or three suns, but any more than that and they start to become very unstable (e.g. trapezium systems), but I'm more interested in the concept of >3 stars, each of similar mass. How could they be arranged in a stable (for a few billion years), non hierarchical manner? I tried sketching out a few possibilities but lack the understanding of how suns interact with each other (heat and pressure being the foggiest elements).
Are there any stable >3 star, star orbits, and if so, what do they look like?
edit to clarify: I am looking for an answer within a single solar system, not a star cluster / galaxy (which would fit the question)
| Perhaps this isn't the answer you're looking for, but star clusters (particularly globular clusters) and galaxies would seem to fit the bill. Galaxies are a bit more of a stretch since this is probably (1) much larger than what you have in mind and (2) typical galaxies are predominantly made of things other than stars, e.g. gas, dust and dark matter. Globular clusters, on the other hand, are thought to be of predominantly stellar composition. Typical clusters have somewhere in the range of $10^4-10^6$ stars. The clusters are not, strictly speaking, stable - they gradually lose stars - but it takes a long time for them to dissipate completely, longer than the few billion years you asked for. Here is a gratuitous pretty picture of M13, the "great globular":
Generically and qualitatively, the orbits of the individual stars look something like this:
It's easiest to think of the individual stars more like test particles orbiting in a smooth potential - there is no central massive object they all orbit, for instance - but in detail star-star "collisions" (more like close encounters) matter to the evolution of the system.
| {
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Position vectors I was reading Kinematics from my textbook. It started off by defining vectors and its two types:
1. Free vectors
2. Fixed vectors
That's all good. But then it went on to say that all the vectors we'll be dealing with in 'our' study will be free vectors. After that it defined position vectors. Isn't position vector a fixed vector as we specify it with respect to some coordinate system (as against the free vectors my book says we'll be dealing with)? Is my book wrong or am I missing something?
EDIT
This is how my book defines the aforementioned terms:
Fixed vector: (I'm quoting the exact statement) In some physical applications, the location or line of application of a vector is important. Such vectors are called localised or fixed vectors.
Free vector: (Yet again, quoting the exact statement) In our study (of kinematics I believe) vectors do not have fixed locations. So displacing a vector parallel to itself leaves it unchanged. Such vectors are called free vectors.
| I believe your book is wrong in assuming that the only vectors required in the study of Kinematics, at any level, will be free vectors. The position vector is a very fundamental vector in Kinematics, and clearly, it is a fixed vector. Why? Because changing it's position will change the position of the object itself, and if that happens, there's no point in defining something as a position vector.
| {
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How is the curl of the electric field of a dipole zero? For a static charge, the curl of the electric field is zero. But in the case of a static dipole the electric lines of force curl. How it that possible?
| I think I do not understand the equation. If you try to find the electric field for a static dipole you have two main way: starting from a potential and then you make the approximation of great distance so that the multipole expansion is truncated to the second order in the charges, or you can compute the electric field generated by two charges very close one to each other. In the first case, in order to find the electric field you will take the gradient of the potential and then you are already imposing that the curl of the electric field is zero and all the charge are stationary. In the second case you are calculating the electric field very far away from the charges and then the electric field is the superposition of the electric field of each charge. The electric field of a charge has null curl, so, since the curl is linear, the electric field of the dipole is zero.
If you go very close to the charge, in order to find the electric field I think you have to consider higher order in the expansion of the multipole, which decay faster when you go do great distance. However if you suppose that the charge are still and they do not attract or repel to each other, no current can be generated and so, there is no variation of the magnetic field. From the Maxwell law you always get zero of the curl of the electric field in vacuum.
The dipole approximation is, in fact, the hypothesis of two charge, far away from each other in such a way that they do not collapse or go away from each other, but that very far away you can consider form a single entity. No current are involved, so no dynamic magnetic field, so no curl of the electric field.
| {
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Magnetic rod in Magnetic Field How does thin magnetic rod suspended in uniform magnetic field oscillate? I want to know what kind of oscillations will they be (displacement will be angular or linear) and which phenomenon (what forces are involved) causes these oscillations? And why it has to be thin?
For example:
In a spring block system, SHM is carried out for small linear displacements. And this motion is due to restoring forces of spring.
Some formulations would be appreciated.
| There will be a magnetic moment induced in the rod. For a small angle displacement from stable equilibrium, the torque $\tau=m\times B=mB*sin(\theta)\approx mB\theta$
This, when equated with the product of the rod's moment of inertia and angular acceleration will give a differential equation that shows that the rod will oscillate.
| {
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Is a mirror-less telescope possible? I was reading about telescopes and the Hubble Telescope for example has a 2.4m mirror which reflects lights to a sensor. Other type of telescopes use lenses to focus light to the imaging sensor.
I was wondering, is it possible to have a telescope without a mirror or lens?
So in Hubble's case, instead of having a 2.4m mirror reflecting light to a small sensor, why not have a big imaging sensor (same size as the mirror - 2.4m). Would this type of telescope have similar capabilities as the mirror one?
I know that we use mirrors because it's way cheaper and easier than building large sensors but I'm curious if a mirror-less telescope would be better / worse or just the same.
| Systems other than mirrors can provide telescope function.
There's four other ways to make a telescope.
*
*Like Galileo did, we can use lenses instead of mirrors (this does not work well at large scale).
*One can employ diffraction structures (holograms, or gratings,
or zone plates), but there are significant problems in doing this
for a broad range of wavelengths.
*More primitive, is the pinhole camera,
and variations using multiple holes, so-called "coded apertures"; that
is how some X-ray imaging is done, but it takes a lot of work.
*Last, is to find a black hole or neutron star, and take advantage
of the gravitational bending of light in its vicinity. That's more
of a discovery than a construction project, but several useful
systems have come into view.Hubble gravity lens
| {
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Equivalent temperature of air to feel as if it was in water I got interested in why, at the same temperature, water feels cooler than air. After a google search, I saw that this question got answered here: Why does water feel cooler than air. I understand the answer given there.; however, is there a formula which I could use to calculate the temperature at which the air should feel the same temperature as water? For example, at which temperature should the air feel the same temparature as water at 12ºC?
I myself have tried to get this formula working with Newton's cooling law, but without any results.
Thank you in advance.
| It can't be done to any accuracy really here's why.
You need to have the same heat-flow out of the body into the air (insulator) as into the water (conductor). We need to consider both conductive and convective losses through the medium.
At any given temperature, as air is an insulator, the ratio of conductive to convective loss is very much lower than water.
So if the thermal conductive losses are matched when the air/water is stationary, then if there is movement and convective loss occurs the air will convect much, much more energy than the water.
Conversely if the convective heat transfer is equalised, then the water will conduct much more heat energy and the air will feel warmer if it is stationary.
You can always feel the difference.
| {
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What determines how much power goes into each diffraction order? Imagine a grating with infinite number of slits, and the spacing D between slits is larger than the wavelength so that there are high order diffractions. In each of the diffraction directions the waves constructively interfere, but what decides the percentage of power that goes into each order?
My thinking is that each slit is a Huygens source, radiating cylindrical waves homogeneously in every direction, but due to interference, only those with constructive interference can exist. I guess the energy going into each order should be equal, which is not the case. So I am confused on how the light will distribute its energy to different orders. Thank you.
| There are two steps only you need to calculate the intensity distribution behind a multi slit. Firstly you have to calculate the intensity distribution pattern behind a single slit. Secondly you has to calculate the aberration of the pointlike source to all the slits and to the observers screen and by this sumerize the intensities at all interesting you points.
Are you able to compare the calculations with the real experiment? Please share your results with us.
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Electric current though liquid - Magnet interaction I f a have a coil with an electric current flowing through it, and I place it inside of a magnetic field, it will move.
Now my question is, what happens if I have a droplet of liquid with an electric current flowing through it, will it also move ? If not, why not ?
| Yes, of course it will move. Here is a familiar science demonstration, the
mercury 'beating heart' BEATING HEART which
shows a crude battery circuit (the mercury and the iron in a
mild acid). When the iron touches the mercury, the battery
puts a current through the contact point, and that current
deforms the mercury because it creates a changing magnetic field.
The deformation breaks the contact, and the force on the mercury
goes away, which reestablishes electric contact... and it oscillates.
It's literally a changing magnetic field that's moving the mercury,
but the coil-in-a-field causes more of a torque effect than
a translational motion. The application of torque to a liquid
is a difficult thing to demonstrate; how would one know that
a liquid drop was rotating?
A drop of conductive liquid in a changing magnetic field WILL transformer-couple
to that field, and generate a little loop of internal current.
The problem, then, is to see a drop become nonspherical when
a changing magnetic field excites it, and then see some indication
of axial tilt due to a nonchanging magnetic field.
Probably, this could be accomplished, with stop-action video
to record droplets in free fall. It'll look like a bunch
of wires with a quivering raindrop falling through a magnet gap...
| {
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Why during annihilation of an electron and positron 2 gamma rays are produced instead of 1? $1 \gamma \rightarrow 1 e^- + 1 e^+$ (pair production)
Then why
$1 e^- + 1 e^+ \rightarrow 2 \gamma$ (annihilation of matter)
instead of
$1 e^- + 1 e^+ -> 1 \gamma$ ?
| The answers given above are incomplete, probably because the OP is misleading. Everyone discussed the case of pair annihilation (or creation, if you wish to consider time-reverse processes) in vacuo, in which case the answers provided are correct.
However, being in vacuo is a useful generalization, but often is also an oversimplification. For instance pair creation/annihilation can occur inside an electric field (say, close to a nucleus), or inside a magnetic field (again, close to a nucleus or even an atom with non-vanishing magnetic moment), in which case one-photon processes are fully allowed, and quadri-momentum conservation is made possible by the presence of the nucleus and/or atom.
These are classical processes, widely discussed in the literature. For the case close to a nucleus, you may read Bethe and Heitler, 1934, Proc. Roy. Soc. London, A146, 83, and for magnetic field Zaumen, 1976, ApJ, 210,776. Or summaries in Lang, Astrophysical Formulae, 1998, vol.I, pag. 433, and Meszaros, 1992, High-Energy radiation from magnetized neutron stars, U.Chicago Press, pag. 211.
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Structure of space Can we consider the structure of space( only space not space-time ) to be that of a vector space? Why can we or why can't.
And why cant we give a vector space structure to space-time?
| A vectors space has a preferred vector, the zero vector. Instead there is no preferred point in classical (flat) physical space. Physical space is better described by a so-called three-dimensional affine space. Metric tools are then represented by a scalar product in the space of translations. This space which describes rigid movements of the points in the affine space is a vector space, but it is not the physical space itself, just describes rigid movements in physical space.
Similarly, the spacetime of special relativity is well described by a four-dimensional affine space whose space of translations is equipped with a Minkowskian scalar product.
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Is the diffraction pattern of a vertical slit horizontal? I am familiar with the mathematical aspects of single slit diffraction pattern, at the undergraduate level.
Consider the following pictorial representation from the book Optics, by Hecht:
The fact that I find puzzling here is - even though the slit is shown vertical, the pattern on the screen is shown horizontal. Is this correct? Why so?
My logic:-
The reason why I find this strange is because of a translational symmetry argument. Any two points vertically separated by some distance have the same horizontal attributes. So, one expects the pattern also to have this sort of vertical symmetry, irrespective of what happens along the horizontal axis.
Am I mistaken? If yes, can someone please point out why is the vertical slit producing a horizontal pattern here?
| It is the narrowness in the horizontal dimension which cause horizontal diffraction. The slits are only tall because they are not wide.
Incidentally, the diagram is wrong. It shows light entering the full heights of the slits from top to bottom. If it did that, the fringes would be tall too - like vertical lines.
| {
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"timestamp": "2023-03-29T00:00:00",
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Trolley problem
A trolley of mass $300kg$ carrying a sandbag of $25kg$ is moving uniformly with a speed of $27km/h$ on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of $0.05kg/s$. What is the speed of the trolley after the entire bag is empty?
I do not seek the exact answer to the problem. I know a force will act on the trolley because the mass is changing (and force is proportional to the rate of change of momentum) but where it acts is what confuses me. Where would the force act?
| In some ways this is a "make you think" question because it has a very different answer from a question which has sand added to the trolley.
What you have to realise is that at the instant the sand leaves the trolley its horizontal momentum does not changed so no horizontal force needed to be applied by the trolley on the sand to make it leave the trolley.
Since the trolley exerted no force on the sand the sand did not apply a force on the trolley.
So the trolley (the system) continues to travel at constant velocity because it has no horizontal force acting on it.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is the Charge on proton is $3.2\times10^{-19}$ greater than that of charge on electron? As the charge on electron is $$e^-=-1.6\times10^{-19}C$$ and charge on proton is $$p^+=+1.6\times10^{-19}C$$
Does this mean that the charge on electron is $3.2\times10^{-19}C$ less than that of charge on proton?
| No, The magnitude of charge on both the particles is same but opposite sign.
Actually the negative sign doesn't always mean "less", especially in physics. So, the charge on $e^-$ is not smaller than that of $p^+$.
| {
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Is there a way to physically resolve elliptically polarized light into circularly polarized components? I was reading about a phenomenon called circular dichroism, in which LCP(left circularly polarized) and RCP(right circularly polarized) light are absorbed to different extents. Hence, when linearly polarized light(LCP+RCP with some phase between them) is passed though such a material, the output beam is elliptically polarized.
Can this elliptically polarized beam be "split up" physically, into constituent left and right circularly polarized beams. What I am picturing is a plate the lets only the LCP pass through or vice-versa.
| One way to do it is to pass the beam through a quarter-wave plate, then through a traditional beam splitter that separates horizontal from vertical polarization, then through a quarter-wave plate again.
| {
"language": "en",
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Circuit with three capacitors and a switch I have a homework problem where I have a battery connected to a capacitor and a switch that connects the capacitor to 2 others in series/disconnects the circuit from the battery.
The question is:
S is initially closed to the left until c1 is completely charged. Once charged, it closes to the right and remains closed there until it reaches equilibrium. Calculate the difference in potential of c1.
I'm stumped because I don't know how the charges distribute between the three capacitors once the circuit is closed, and so I'm not able to calculate the final voltage. I probably didn't do too good of a job translating the text so if you don't understand ask. Thank you.
| Remember: Capacitors in parallel all have the same voltage across their plates, but can have different charges.
So, when the switch is closed to the left, C1 is has a certain charge and voltage. Then, when the switch is closed to the right, that charge in C1 distributes in such a way so that the voltage across all three capacitors is equal.
You can intuitively justify it this way—imagine if one of the three capacitors had a lower voltage. We know that electrons will move from higher potential to lower potential (like a ball rolling from the top of a hill to the bottom); so if one of the three capacitors had a different voltage, charge would just redistribute itself (since all of the lower plates are connected to one another) until none of the capacitors had a lower voltage; that is, until they all had the same voltage.
You should be able to solve the problem using this, and the various rules for capacitors connected to one another.
| {
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Opposite of particle decay I have read about particle decay, a process in which one particle becomes several other particles. However, I have not been able to find much information about its opposite: several particles combining into one particle. Is such a process possible, and if so, under what conditions? For example, a free neutron may decay into a proton, electron, and electron antineutrino. Could a proton, electron, and electron antineutrino somehow be joined into a neutron?
Edit: Everyone, thank you for your help, but let me try to make what I'm looking for clearer. I want to know whether several particles can join into ONE particle, in an exact reverse of that one particle decaying into several particles. As far as I know, I don't think an atomic nucleus counts as one particle. Please correct me if I'm wrong.
| The triple-alpha process is a way $C^{12}$ is formed in stars from three $He^4$ nuclei once (most of) the hydrogen in the core has been burned. It is really two reactions,
$He^4+He^4 \to Be^8, Be^8+He^4 \to C^{12}.$
$ Be^8$ is unstable, but lasts long enough for this to happen if the helium is dense and hot enough.
| {
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How can length be a vector? Length and current both are not vectors. Then how can we assign the vector $l$ to the length of a wire carrying current while calculating for a current carrying conductor in a magnetic field. Also why in Biot—Savart law do we take small length element $dl$ as a vector?
Why is length sometimes a vector, sometimes not, whereas current always is a scalar?
| Biot-Savart's law is the cross product of two vectors, the current vector and the vector representing the point whose magnetic field you want to calculate. It is easier to understand using the law for a single charged particle.
$$\vec B = \frac{\mu_0}{4\pi}·q·\frac{\vec v \times \vec r}{{\lvert \vec r \rvert}^2}$$
Makes more sense now doesn't it? The problem is that you will not calculate this for single particles but for current flow. That current flow is determined by the number of electrons n times their average speed, hence.
$$\vec B = \frac{\mu_0}{4\pi}·n·q·\frac{\vec v \times \vec r}{{\lvert \vec r \rvert}^2}$$
Not good enough still, you don't know how many single charges there are for a given point in space. However, you do know the current in a given point of the wire. For each dl of the wire, the number of electrons times their speed equals the current times dl times the vector of the wire's direction.
$$I·dl·\vec u_w =I·\vec {dl}= n·q.\vec v$$
$$\vec B = \frac{\mu_0}{4\pi}\frac{I\vec {dl} \times \vec r}{{\lvert \vec r \rvert}^2}$$
So you see, the vector you see (in this example and many others) is merely a unitary vector of the space times the length differential. Length itself is a scalar. This vector is akin to saying that you have to operate with the dl length but taking into account the direction of whatever you are operating with (current in this case).
Hope this has cleared things
| {
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Current constraints on Dark Matter self-interaction from galactic profiles The self-interaction of dark matter may be small but it cannot be negligible if it is able to dissipate energy to relax into galactic clumps (necessary to explain galaxy rotation curves).
According to some answers in this old question: How Does Dark Matter Form Lumps?, the gravitational self-interaction alone is enough to allow dark matter clumping (via n-body interactions). Although two answers suggest something other than gravity is needed (one states considering the weak force is necessary, while another answer argues for why gravity alone doesn't explain how in cosmology dark matter could clump first).
I am curious about:
*
*Have the measurements of dark matter profiles of galaxies become good enough to provide indirect measurements of dark matter self-interactions?
*Can this self-interaction be used to say anything about the mass of the dark matter particles? At the very least, can we say with certainty they have mass above some threshold (ruling out very light particles such as axions or neutrinos, and ruling out some kind of unseen massless particles)?
*Since the strength and radial distribution of the gravitational force vs the weak force differ so strongly, is it possible to determine from the self interaction whether dark matter interacts via the weak force?
| There is currently no strong evidence that dark matter interacts with ordinary matter via anything besides gravity. There have been proposals for non-gravitational interactions to explain discrepancies with small scale observations. But these discrepancies can also be explained by the effects of ordinary baryonic physics.
| {
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The direction of the induced electric field Recently I got stuck witht the following problem.
Imagine we have uniform a magnetic field which induction points upwards. The fields strength is steadily decreasing. If we put an iron coil perpendicular to the magnetic induction vector, then, obviously, there will be electric current induced in the coil.
However, as I understand, the coil itself is only a 'marker' that displays the electric field lines that actually make the electrons move. It means that the elcetric field is there even when there is no coil.
Now the problem:
I can imagine some coils being close to each other. It will essentially mean, that it in one of them the current will go one way and in the other - the opposite. How can this possibly be?
I looked at this answer as it is phrased very close to what I want and still I couldn't get the idea. Could the answer be presented in more layman terms .
| Given your drawing, equal area loops normal to a uniform, changing field, all we can say is this:
1)The integral around the loops of the electric field dotted into the line element are equal--it says nothing about the direction of the field at any point.
2) (Lenz's law) The direction the current flows is the direction that "keeps the magnetic field going"--in layman's terms.
That current flows in opposite directions where the loops are in the same physical space does not contradict point 1, nor point 2.
| {
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General relativity: How is the 4-velocity and momentum defined in GR? I know that in SR, the 4-velovity $$ u^\mu = (d t/d \tau,d x/d \tau,d y/d \tau,d z/d \tau ) $$ and $$p^\mu = m u^\mu.$$ How do these generalize to GR? I imagine there are new complications, particularly by what we let $p^0$ be. And are these only defined for geodesics, or in general?
| On a general Lorentzian manifold with a time-positive metric, $d\tau^2 = g_{\mu\nu}dx^{\mu}dx^{\nu}$. In SR $g_{\mu\nu} = \eta_{\mu\nu}$ which gives the familiar relation:
$$
d\tau^2 = dt^2 - dx_idx^i
$$
In GR the Einstein Field Equations tell you what $g_{\mu\nu}$ is so you can compute the altered line element.
Otherwise, the form of the 4-velocity and 4-momentum is the same in GR for a massive particle.
| {
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Is light deflected by external electric and magnetic field? I recently read about the Maxwell's laws of Electromagnetic Waves and I found that Light is made up of both Electric and magnetic fields.
So now if i pass the light through a capacitor such that the plates are parallel to the light will the light be deflected?
If it is deflected then what about the particle nature of light in which the photons are neutral without any charge(as far as i know charge do not exist without mass).
If not why are the electric and magnetic fields not affecting the light in the wave nature if it is solely due to the wave nature why are electrons being deflected in the external fields.
I hope someone give me a clear idea of what is wrong with my idea.
| Looking at the classical electromagnetic wave with E and B fields propagating perpendicularly to each other and the direction of motion. These fields do not carry charge, and it is only charge that is deflected/senses electric and magnetic fields. Classically, there can be superposition of two waves which show interference patterns, but superposition is not interaction. The classical electromagnetic wave interacts only when it sees charges (and,possibly, magnetic dipoles). Another way of looking at it is that the dielectric constant of the space between the capacitor plates does not change whether the capacitor is charged or not.
If one goes to the photon level, firing individual photons through the opening of the capacitor plates: photons as neutral will change direction only if they interact. There exists a small probability that a photon meeting an electric field may interact with the electromagnetic interaction, but the probability is tiny as the relevant Feynman diagram has at least five coupling constants. The four of this diagram + one extra because the interaction with the field is with a virtual photon which means an extra vertex.
| {
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why is it that only the perpendicular component of a wave can pass through a vertical polarising filter? To me it seems unintuitive that a filter can change the direction of a wave. I also do not understand malus' law, could somebody please show me an intuitive derivation or proof of malus' law.
Is unpolarised light, different waves with different oscillation directions, all coming from the same source?
| The component of the field that is parallel to the wire looses its energy in the wire by Ohmic dissipation so that only the perpendicular component remains.
Given this, only the perperdicular component $E\cos\theta$ of $\vec E$ will go through. Since the intensity is proportional to the square of the magnitude of $\vec E$, you get $I(\theta)=I_0\cos^2(\theta)$.
| {
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Why mass & energy bend spacetime? I understand how light / matter bend spacetime but I'd like to understand WHY. Is there some kind of interaction?
| As this concept was originated by Einstein,he thought the Newton's law of gravitational is a classical case of two body attraction but for the large scale he generalize the notion of Newton.He thought that space-time is like a trampoline pad and if there is some object(mass) on a trampoline pad it will definitely bend the trampoline pad and this bend in space time is known as "curvature of space- time" and the object that fall in that bending region feels attraction towards the other body.Now, its depend on who will attract more towards the other by making the simpe concept that "larger the object mass larger will be the space-time curvature".Therefore, usually small object rotate around the larger one like earth rotate around the sun because sun has large bend on the trampoline pad rather comparable to the earth.
| {
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"timestamp": "2023-03-29T00:00:00",
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Change in shape of Atomic Orbitals We have studied that electron's position around the nucleus is given by probability distribution function, which determines the shape of atomic orbitals. Does the shapes of orbitals change with factors such as electric fields, magnetic fields, high temperatures, intense gravitational fields even with excitation of atom?
Will the electron just change its orbital and the shapes of orbitals would be unaffected?
|
Does the shapes of orbitals change with factors such as electric fields, magnetic fields,
In quantum mechanics, any new extra potentials , as an introduction of an electric or a magnetic field, will lead to new solutions for the orbitals.
Example : the Zeeman effect in hydrogen:
When an external magnetic field is applied, sharp spectral lines like the n=3→ 2 transition of hydrogen split into multiple closely spaced line
So one energy level splits into two because of the spin of the electron, and thus one orbital becomes two.
So it will depend on the specific problem.
high temperatures,
No, temperature has to do with the average kinetic energy and possible interactions, it is not a field/potential
intense gravitational fields even with excitation of atom?
Again, a solution of the extra potential to the atomic potential has to be studied. Gravitational fields are weak, and also at the moment there is not unification of the three forces and gravity and only effective quantum mechanics exists for gravitational fields, so it will all be a matter of assumptions.
Will the electron just change its orbital and the shapes of orbitals would be unaffected?
For an electron to change an orbital, the atom has to absorb a photon of the appropriate energy. New solutions as in the Zeeman effect, can be considered as a change in orbital , but it is a different state, "atom + extra field".
| {
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How to determine the components of a velocity vector undergoing a central force I am busy working on a 3D $n$-body galaxy simulator and I am having some difficulty wrapping my head around determining the initial velocity components of each particle to ensure a circular orbit around a central mass.
I know that the magnitude of the velocity to ensure a circular orbit is calculated from:
$$
v = \sqrt{\frac{GM}{r}}
$$
How then can I determine what the individual velocity components in 3D will be for the particle?
| The velocity in $(x,y)$ plane at the circular orbit can be viewed best in polar coordinates (spherical for $\theta \to \pi/2$ if you have 3D model). Then the initial velocity must have $v_r = 0$ and $\vec v = v_\phi \vec e_\phi$. You can calculate the $v_\phi$ through
\begin{equation}
v = \sqrt{\vec v \cdot \vec v} = \sqrt{v_r^2 + r^2 v_\phi^2} = r v_\phi.
\end{equation}
If you want to work in Cartesian coordinates (which I assume are better for you), you have
\begin{equation}
\renewcommand{\vec}[1]{\boldsymbol #1}
\vec v = \frac{\textrm d \vec r}{\textrm d t} = (-r\sin(\phi)\frac{\textrm d \phi}{\textrm d t},r \cos(\phi)\frac{\textrm d \phi}{\textrm d t}) = \Omega r (-\sin(\phi),\cos(\phi)),
\end{equation}
where $\vec s = (-\sin \phi, \cos \phi)$ is the direction you are looking for. $v = \Omega r$ is then general case of what you have written for gravitation field, so you can use your formula for $v$ and use this vector $\vec s$ for your direction. If you have net in Cartesian coordinates, you have to realize that
\begin{equation}
\tan\left(\phi\right) = \frac y x,
\end{equation}
so if you start in place $(x,y)$, your direction is
\begin{equation}
\vec s = \left( -\sin\left( \arctan\left(\frac yx \right) \right),\cos\left( \arctan\left(\frac yx \right) \right) \right).
\end{equation}
Edit: In this derivation I assumed radial velocity is equal to zero, that's why the vector is so simple. And this assumption is why it works for your circular orbits.
There is possibility to use spherical coordinates instead and get velocity vector
\begin{align}
\vec v = r\left(
\begin{array}{c}
v_\theta \cos (\theta) \cos (\phi)- v_\phi \sin (\theta) \sin (\phi) \\
v_\theta \cos (\theta) \sin (\phi)+ v_\phi \sin (\theta) \cos (\phi) \\
- v_\theta \sin (\theta) \\
\end{array}
\right)
\end{align}
and remember that
\begin{align}
\phi = \arctan\left(\frac yx\right) \qquad \text{and} \qquad \theta = \arccos\left( \frac{z}{x^2 + y^2 + z^2} \right).
\end{align}
The last step is to calculate the velocities $v_\theta, v_\phi$. These are equal to angular velocities $\Omega_\theta, \Omega_\phi$, so you has to project your velocity into these two components. I think the best here is to realize, that
\begin{align}
v = \sqrt{\vec v \cdot \vec v} = r^2 \Omega_\theta^2 + r^2 \sin^2(\theta) \Omega_\phi^2 = \sqrt{\frac{GM}{r}},
\end{align}
then you can define $G,M$ and $x,y,z$. By postulating these constants you have well defined $r,\theta$, so the previous equation has only two unknowns $\Omega_\theta, \Omega_\phi$. Now you have one degree of freedom, you can chose one of these numbers and the second one must satisfy the equation, for exapmle you can set $v_\phi = \Omega_\phi$ and get
\begin{align}
v_\theta = \Omega_\theta = \sqrt{\frac{GM}{r^5}} - r^2\sin^2(\theta)\Omega_\phi = \sqrt{\frac{GM}{r^5}} - r^2\sin^2(\theta)v_\phi.
\end{align}
Summary: If you are in 3D, you can chose the rotating direction by choice of $v_\theta, v_\phi$ (that's up to you) and than you have the initial velocity given as
\begin{align}
\vec v = r\left(
\begin{array}{c}
v_\theta \cos (\theta) \cos (\phi)- v_\phi \sin (\theta) \sin (\phi) \\
v_\theta \cos (\theta) \sin (\phi)+ v_\phi \sin (\theta) \cos (\phi) \\
- v_\theta \sin (\theta) \\
\end{array}
\right)
\end{align}
The underlying physics says, that you have rotation in 3D around a fixed point on the circle. That you have to choice "how much it rotates in $\phi$-direction = $(x,y)$ plane, and the program has all information now".
| {
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"url": "https://physics.stackexchange.com/questions/326909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the relation between image velocity, object velocity and mirror velocity? Suppositions used:
Velocity of image = VI
Velocity of object = Vo
Velocity of mirror = VM
I Know the fact that VI=-Vo supposing mirror at rest
and VI=2VM supposing object at rest
Now considering both mirror and object in motion, VI=2VM - Vo I ended up with this equation but my reference book suggests VI=2VM + Vo I am stuck on this for last 4 hours. I searched over internet and found the same expression like that of mine in a youtube video, I did not find much reference on this topic though. Tried many ways but all ended up on this simple argument, which equation to follow?
Help
| I have considered an example and proved this formula in the image shown.
*
*The first image has plane mirror and the person at rest.
*Let us now consider the person velocity as $V_o=\frac{Unit Block(UB)}{second}$ towards the mirror and mirror velocity as $V_M=\frac{UB}{2}$ towards the person.
*The second image shows the situation after one second. We see that image velocity is $V_I=2UB=2\frac{UB}{2}+UB$.
*The third image considers the mirror velocity as $V_M=\frac{UB}{4}$ towards the person and the person velocity as $V_o=UB$ (see the third image with respect to first image). We see that the image velocity is $V_I=\frac{UB}{2}+UB=2\frac{UB}{4}+UB$.
*From 3. and 4. points, we see that the formula is $V_I=2V_M+V_o$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is there any significant difference between a tokamak and a spherical tokamak? The title question is quite self-explanatory. Despite the fact that Spherical tokamaks are more spherical in shape, what else differentiates the ST from the conventional tokamak. I've heard that ST's use reverse field configurations from a website but I am skeptical about this since the rest suggest otherwise. Do ST' use reverse field configurations? And any other differences?
Thanks
| To understand the most significant difference, we must first understand how a conventional tokamak generates part of its magnetic field: the poloidal magnetic field is created by driving a toroidal current, i.e. a current along the (toroidal) plasma column. That current is driven by ramping up (or down) a voltage of a central solenoid. The conventional tokamak is more donut-shaped where the spherical tokamak (ST) is more similar to a cored apple and has thus very little space for a central solenoid. This means that the toroidal current to generate the poloidal magnetic field has to be driven by other means, e.g. radiofrequency heating.
The spherical tokamak has in general a lower background magnetic field than the conventional tokamak. The resulting plasma beta, the ratio of plasma pressure to magnetic field pressure, is higher in a ST. One could therefore say that you get more plasma per magnetic field unit in a ST which translates into a financially cheaper device (since magnetic field strength (or rather the volume) is one of the most costly part of the conventional tokamak).
In terms of stability, the ST behaves differently than a conventional one. Another point is the small distance of the coils to the plasma in the ST which is problematic in a reactor-like device where you have neutrons (from the fusion reactions) that need to be slowed down to shield the coils.
As for current research projects, there are the MAST-U device in the UK, NSTX-U in the US, QUEST in Japan and the ST line of the company Tokamak Energy.
| {
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"url": "https://physics.stackexchange.com/questions/327083",
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