homework and exercises - How to calculate the frictional force required to "stick" two rotating objects?

Say you had two fixed, non-identical objects rotating around one axis at different speeds relative to one another, and they can be coupled together by friction (like how a clutch in a car works). How would I calculate the frictional force required for them to stick together instantaneously?

For example:

Object A and B are two solid cylinders rotating about a single axis (imagine how they would rotate if they were rolling) Object A is rotating at 10000 RPM, and object B is rotating at 1000 RPM; Object B has a larger radius, is made of a denser material, and has more mass and a larger moment of inertia. The two object come into contact with each other (picture the clutch of a car).

If the frictional force is sufficient, they should "stick" and the angular momentum should be conserved (I1ω1+I2ω2=Inetω); but if it isn't, they should slip.

What would be the frictional force required to "stick" the two objects together instantaneously?

Answer

Forces cause accelerations. The rotational analogue of this is that torques cause rotational accelerations. "Instantaneous" doesn't exist. You need to figure out the maximum amount of time you want it to take.

Say you determine that before and after coupling, one object will have some change in angular velocity, $\Delta \omega$. You will have already calculated or measured the moment of inertia, $I$. Further, you decide that for the purposes of your device, "instantly" means in less than 0.1 seconds.

$$ \Delta \omega = \alpha t$$ $$ \Delta \omega = \frac {\tau}{I} t$$ $$ \frac{\Delta \omega I}{t} = \tau$$

So you'll plug in for the change in rotational velocity, moment of inertia, and time required and you'll get the necessary torque. The faster you need it to couple, the higher torques/forces you'll have to generate.

Given the geometry of how you're applying it, you can turn the torque into a force. If it's driven by friction, you can use that and the coefficient of friction to determine the required normal force to produce that much friction.

optics - Is the output of a laser pointer polarized or not?

How to verify it?

Is a laser necessarily polarized?

I have a laser pointer with wavelength 532nm in hand.

Answer

No, laser light is not necessarily polarized.

There are many different types of lasers. In gas lasers, many different modes may be excited in the cavity, but only modes that are not very lossy end up being amplified and emitted - this can result in a single polarization. However, there may be multiple modes with different polarizations that have this property, resulting in a more random polarization. There also may or may not be a polarizing filter built in to the laser aperture. I don't know as much about other types of lasers, but I don't think they all are necessarily polarized.

You could test whether or not your laser is polarized by getting a polarizing film (for example polarized sunglasses). Shine the laser through the film while rotating the film. If the intensity of the transmitted beam changes when you rotate the film, then it is at least partially polarized.

diffraction - is this the difraction limit of our microscope?

I took the picture below with our microscope. It's a 8mm roller bearing, lighted from below with a parallel light bundel (white). The width of the yellowish light fringe is maybe 5 micron.

Originally I was hoping to make pictures with an accuracy of like 1 micron. But I'm afraid that was way too optimistic.

-First I would like to know if this is really a diffraction pattern?

-If yes, would a blue back light make the fringes smaller? What else could I do to increase quality?

-Is there a way to calculate the distance between the fringes and the "real" object position? On this website, I found a similar question, but I don't think those formulae are valid for this case (the calculated distances are like 0.3mm between fringes).

Thanks in advance,

Edit; the first back light I used did not have a pin hole, and the light beam was not focused at all. The resulting image in the microscope looked like this; (distance per pixel is the same)

At first I was a bit disappointed when I saw all these fringes with the truly parallel light, but maybe it's a good thing to hit the limitations of the microscope (as Rod suggests below)?

Answer

Yes, definitely this is due to diffraction. But I'm surprised by the contrast and number of fringes you have with white light. Is the focus optimized? The fringes can become larger if the image plane is not on the object plane. Is the light really white? The yellow fringe is suspicious. If you can change the illumination, you can try to find a white light with flat and large spectrum, it will help to reduce the fringes. A real white light will be better than a blue light. If you use a more monochromatic light (even if not coherent as a laser), eg a LED, diffraction will increase. The best is a real white light which "washes out" the fringes (adding all the fringes of different dimensions at the different wavelengths).

quantum mechanics - Does the uncertainty principle apply to photons?

Wikipedia claims the following:

More generally, the normal concept of a Schrödinger probability wave function cannot be applied to photons. Being massless, they cannot be localized without being destroyed; technically, photons cannot have a position eigenstate and, thus, the normal Heisenberg uncertainty principle does not pertain to photons.

Edit:

We can localize electrons to arbitrarily high precision, but can we do the same for photons? Several sources say "no." See eq. 3.49 for an argument that says, in so many words, that if we could localize photons then we could define a current density which doesn't exist. (Or something like that, I'll admit I don't fully understand.)

It's the above question that I'd like clarification on.

Answer

The relation $p={h\over \lambda}$ applies to photons, it has nothing to do with the uncertainty principle. The issue is localizing the photons, finding out where the are at any given time.

The position operator for a photon is not well defined in any usual sense, because the photon position does not evolve causally, the photon can go back in time. The same issue occurs with any relativistic particle when you try to localize it in a region smaller than its Compton wavelength. The Schrodinger position representation is only valid for nonrelativistic massive particles.

There are two resolutions to this, which are complementary. The standard way out it to talk about quantum fields, and deal with photons as excitations of the quantum field. Then you never talk about localizing photons in space.

The second method is to redefine the position of a photon in space-time rather than in space at one time, and to define the photon trajectory as a sum over forward and backward in time paths. This definition is fine in perturbation theory, where it is an interpretation of Feynman's diagrams, but it is not clear that it is completely correct outside of perturbation theory. I tend to think it is fine outside of perturbation theory too, but others disagree, and the precise nonperturbative particle formalism is not completely worked out anywhere, and it is not certain that it is fully consistent (but I believe it is).

In the perturbative formalism, to create a space-time localized photon with polarization $\epsilon$, you apply the free photon field operator $\epsilon\cdot A$ at a given space time point. The propagator is then the sum over all space-time paths of a particle action. The coincidence between two point functions and particle-paths This is the Schwinger representation of Feynman's propagator, and it is also implicit in Feynman's original work. This point of view is downplayed in quantum field theory books, which tend to emphasize the field point of view.

quantum mechanics - Feynman diagrams and Hartree-Fock

I am puzzled by some lines I read in Mattuck's book on Feynman diagrams in many-body problems ( http://www.amazon.com/Feynman-Diagrams-Many-Body-Problem-Physics/dp/0486670473 ) Page 21 (1.14) for those who have the book. Basically after representing the full propagator of an electron in an electron gas by expansion of the electron-electron interaction (not specified but QED I guess), it says: "this is the 'Hartree-Fock' approximation for the electron gas", which I still don't understand. The Hartree-Fock method for me is just an iterative tool to calculate the collective wave-function of self-interacting fermions satisfying the correct anti-symmetrized form. This statement is evasive to me, and I'd like to understand in which way it makes sense.

Answer

These Feynman diagrams can be summed by solving the Dyson-Schwinger equation $$ G = G_0 + G_0\Sigma G $$ This is a self-consistency equation for $G$. Now write $G_0$ and $G$ in terms of single particle wave functions, $$ G(x,x';\omega)=\sum_j \phi_j(x)\phi^*_j(x')\left[ \frac{\Theta(E_j-E_F)}{\omega-E_j+i\epsilon} +\frac{\Theta(E_F-E_j)}{\omega-E_j-i\epsilon} \right]. $$ Then the Dyson-Schwinger equation becomes a coupled set of equations for the eigenfunctions $\phi_j$ and the eigenvalues $E_j$. These are the standard Hartee-Fock equations. This is explained in some detail in many text books, for example Negele and Orland, or Fetter and Walecka.

quantum mechanics - $lambda=frac{2h}{p}$ instead of $lambda=frac{h}{p}$?

I am studying quantum physics and there is something I don't understand:

1. 
   
   

   I know that for any particle $E=hf$ (Einstein relation) and

   
   


2. 
   

   $v=\lambda f$ ($v$ is the speed of the particle).

   
   


3. 
   

   I also know that the kinetic energy is $E_k=\frac{mv^2}{2}$.

   
   

Solving those 3 equations for $\lambda$: $$h\frac{v}{\lambda}=\frac{mv^2}{2},$$ I finally find $$\lambda=\frac{2h}{mv}=\frac{2h}{p},$$ which is not consistent with the De Broglie relation $$\lambda=\frac{h}{p}.$$ Where am I wrong in my development?

mathematics - Nerd on the Loose II - WHERE & HOW

Preface. Here is an relatively easy one, as compared to the previous part. We already found out when, let's find out where.

Background. Professor Swannie Taco is a genius in most fields of science and technology. She is probably hated by her parents, and her name is a perfect evidence. After Swannie went missing a few months ago, strange letters from her kept appearing on her G+ account. Nobody knows who exactly post these messages, nor understand the content of these seemingly cryptic memo.

Clarification.

1. For this particular puzzle, the only information required to solve it is the post.
2. Although not strictly needed, knowing high school math will aid solving this puzzle.

Hints.

I really don't think any hints are needed for this puzzle. This statement itself is not a hint in disguise.

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Overview. Knowing when the doomsday is, we need to know where and how. Luckily, she told the whole world how, now we just need to know where. If only it was that

Post. Someone once told everyone that imaginary powers could sometime be real. Maybe if I could attempt to walk across all platform once and only once, thus performing a magical feat, I could break his claim, while justify his claim at the same time... Let's just say such a paradox shall distort the world, and it shall be where everything begins to end...

Task. Figure out where she is going to execute her doomsday plan. Edit. Since this was too easy, I will accept the answer that make sense of the whole "riddle".