Thursday, August 4, 2016

experimental physics - How did Fizeau make his famous speed-of-light experiment?


I heard once in a TED talk how Fizeau measured the speed of light in the 19th century. Here is the link


https://www.youtube.com/watch?v=F8UFGu2M2gM


You can read about it here in Wikipedia:


http://en.wikipedia.org/wiki/Fizeau%E2%80%93Foucault_apparatus


A short summary: He placed a kind of rotating wheel in front of a beam of light, and a mirror far away from these two things. The beam of light passes between two teeth of the rotating wheel, reaches the mirror and goes back from the original source. As the wheel is spinning very fast, during the time that the light has been travelling, the wheel has rotated a tiny bit, but enough to impede the passage of time through the point where it entered. Knowing the distance from the mirror as well as the speed at which the wheel is rotating, the speed of light can be easily calculated. The experiment is better explained in wikipedia, here I wrote a simplified version of it.


I loved the experiment, because it seemed fairly easy to reproduce, so I ordered a green laser pointer on Amazon, which can reach up to 10 km. As a proof of principle, I went with a friend in the night to a place where there is a good visibility. We began setting a mirror somewhere 500 metres away from the laser, but, even from that far, the light had scattered so much that it was impossible to collect the light with the mirror.


The laser is a very powerful one, of those that you can see the whole beam (usually used in astronomy). If I can't repeat the experiment using a laser like this, how on earth could Fizeau do that in the 18th century employing a much more rudimentary source of light and placing the mirror much farther away? It says in Wikipedia that the distance between them was like 8 km.



Answer



There is a much better description here of Fizeau's nineteenth century experiment. Some of the key features that enabled Fizeau to succeed:




  • A lens to collect the light from the source

  • A collimating lens to prevent the light diverging during its journey

  • A large diameter beam to minimise broadening of the beam by diffraction

  • More lenses to focus the light on the detector

  • The light went directly into a very sensitive detector: the human eye. Almost certainly he did this experiment at night.


Wednesday, August 3, 2016

thermodynamics - Radiation emission and absorption


Any object can emit and absorb radiation and the power of emission can be represented by the Stefan-Boltzmann law:


$$P=A\epsilon\sigma T^4$$


In many texts the net power radiated is the difference between the power emitted and the power absorbed:


$$P_{net}=A\epsilon\sigma (T^4-T_s^4)$$


where $$T_{s}$$ is the temperature of the surroundings.



Why can the surrounding and the object share the same $\epsilon$ ?


If we try to find out the radiation emitted from the surrounding it should be $P_s=A\epsilon_s\sigma T_s^4$, and if $\epsilon_s<\epsilon$, we will get a strange result that energy radiated from the surrounding is less than the radiation absorbed by the body from the surrounding. What am I missing?




Tuesday, August 2, 2016

general relativity - What do we learn from gravity in three spacetime dimensions?


The last decades there has been a lot of research going on in the the area of three dimensional gravity. The motivation, I understand, is threefold:





  1. Whereas gravity is not perturbatively renormalizable in four spacetime dimensions, in three dimensions it is. To make it even more interesting it has black hole solutions and it is exactly solvable. This opens the way to to study quantum black holes. This make three dimensional gravity a very interesting system on itself.




  2. Through the AdS/CFT correspondence there is a connection between conformal field theories (CFT) in two dimensions and gravity in three dimensions. CFT's are important in condensed matter physics and one can use 3D gravity to learn more about them.




  3. Gravity in three dimensions is simpler to deal with then gravity in four dimensions. Therefor it can be used as a toy model for gravity in four dimensions.





I am wondering what are the most important insights that 3d gravity brought in these respects? In particular I am interested in point three: did 3d gravity provide any new view on 4d gravity so far?




hilbert space - An apparent contradiction with basis transformation in quantum mechanics


Under a change of basis i.e., transforming from one orthonormal $\{|\phi_n\rangle\}$ base to another $\{|\phi^\prime_n\rangle\}$ (when looked from a passive point of view) implies that the state doesn't change but only the "components" change: $$|\psi\rangle=\sum\limits_{n}\langle\phi_n|\psi\rangle|\phi_n\rangle=\sum\limits_{n}\langle\phi^\prime_n|\psi\rangle|\phi^\prime_n\rangle$$ where $|\phi^\prime_n\rangle=U|\phi_n\rangle$, $U$ being an unitary operator which ensures the ortonormality of two bases. Using $$\langle\phi_n|\psi\rangle\to\langle\phi^\prime_n|\psi\rangle=\langle\phi_n|U^{-1}|\psi\rangle\tag{A}$$ one can also say that "net result'' of a basis transformation as an active transformation where the base is left unchanged but any state $|\psi\rangle$ changes to $U^{-1}|\psi\rangle$. In two different bases, an operator $A$ can be expanded as: $$A=\sum\limits_{m}\sum\limits_{n}|\phi_m\rangle\langle\phi_m|A|\phi_n\rangle\langle\phi_n|=\sum\limits_{m}\sum\limits_{n}|\phi^\prime_m\rangle\langle\phi^\prime_m|A|\phi^\prime_n\rangle\langle\phi^\prime_n|.$$ Using $$\langle\phi_m|A|\phi_n\rangle\to\langle\phi^\prime_m|A|\phi^\prime_n\rangle=\langle\phi_m|U^{-1}AU|\phi_n\rangle.\tag{1}$$ Hence, the result of the basis transformation, is an active transformation where the operator changes as $A\to U^{-1}AU$ but the base is left unchanged.


What happens to the various object under basis transformation? It is easy to see that for any $|\phi\rangle,|\psi\rangle$ in the Hilbert space, the inner products remain unchanged: $$\langle\psi|\phi\rangle\to \langle\psi| UU^{-1}\phi\rangle=\langle\psi|\phi\rangle.$$ Also $$\langle\psi|A|\phi\rangle\to \langle\psi|U(U^{-1}AU)|U^{-1}\phi\rangle=\langle\psi|A|\phi\rangle\tag{2}$$ i.e., objects such as $\langle\psi|A|\phi\rangle$ do not chnage.


Doesn't Eq.(2) contradict Eq.(1)? Eq.(2) says objects such as $\langle\psi|A|\phi\rangle$ do not change. But Eq.(1) says matrix elements of an operator changes under basis transformation. But Eq.(1) is just a special case of Eq.(2). What is wrong here with my understanding?


Note: I'm trying to compare active transformation with passive transformation. Either the basis changes or the system changes. Note that Eq.(A) and Eq.(1) says if the base is assumed to be fixed, then both the state and the operator must change. I'm reading this from Modern Quantum Mechanics by Sakurai, section 1.5




electrostatics - Using the poissons equation to find the surface charge density and electric field in a conductor


Recently, I derived a formula for the surface charge density on the surface of a conductor( no specific shape) that is placed in an electric field from the poissons equation of electrostatics. The only condition that I considered while deriving the formula is that the outer surface of the conductor is equipotential. Now I experimented with the formula considering spheres and planes (as they are comparatively simple to deal with.). Now, I observed that the electric field outside the sphere is perfectly consistent with illustrations that I see on the internet. However, the electric field inside the sphere is fairly constant but NOT ZERO. This is an Image I found on the internet


enter image description here


enter image description here


This is what i got when I used the desmos vector field generator( I coudnt find anything better than that.). The radius of the circle is 5 units. The electric field inside the conductor isnt zero. I dont know why and how else I am supposed to solve this


EDIT:Derivation enter image description hereenter image description here



Answer



Laplacian of the potential on the surface is not zero because there is induced charge on the surface. Equipotential only means that the potential is constant if you move along the surface, it says nothing about the variation of the potential normal to the surface. Starting from there, the derivation is incorrect.



general relativity - Vacuum and repulsive gravity


How can one show from General Relativity that gravity is attractive force, and under which conditions it becomes repulsive, also why positive energy vacuum drives repulsive gravity?




Monday, August 1, 2016

Do standard model particles actually exist or merely usefully describe behaviors of a medium?


I've read about how sound propagation can be modeled as phonon particles moving and interacting. I understand that this is a useful mathematical construct to describe the behavior of longitudinal pressure waves in a medium like air.


Are standard model particles actually matter in the same sense mundane stuff we can directly observe (like the air molecules whose motion transmits sound), mere models that describe a (possibly unknown) medium's behavior (like phonons), or do we believe that at its fundamental level that all matter may be like the phonons in that it's all just describing how some field or medium propagates energy?


As it gets beyond electrons, protons, and neutrons, it's not clear to me that the other particles exist rather than being models of how other things behave. I accept the math makes accurate predictions but it's not clear to me that it's not just like phonons that way.



Answer



That's a very philosophical question. You have a model that predicts measurement results, but what is its interpretation? What reality does it represent?


Quantum field theories deal with (spoiler alert...) quantum fields (ok, not much of a spoiler). The particles that one speaks of, like photons or electrons, are the modes of vibration (the waves) in these fields, much like phonons are the modes (waves) of matter vibration.


I prefer to interpret the theory to say that what exists are the quantum fields themselves, not the particles per-se. This allows one to speak of what exists even when there is no perturbative treatment i.e. when the discussion in terms of the modes of vibration (the particles) doesn't make sense. And it appears less artificial to me; positing that only the modes of vibration exist instead of the thing that vibrates seems strange and unnecessary, oddly restricting what exists to only part of what the theory describes.


classical mechanics - Moment of a force about a given axis (Torque) - Scalar or vectorial?

I am studying Statics and saw that: The moment of a force about a given axis (or Torque) is defined by the equation: $M_X = (\vec r \times \...