angular momentum - Law of the lever - Explained Physically

Thanks for reading.

I've seen the "Archimedes Proof" of the law of the lever, and many similar ones, but...at the end of the day, all they're doing is getting observations about the way rotating bodies behave, and using those observations as axioms.

By defining torque to be the cross product of force and distance, we're just creating new terminology, but we aren't really explaining physically why it is that a force applied at a greater distance would be better at spinning something than a force applied at a lesser distance.

For example, say the green man and the pink man weigh exactly the same, but the green man is half as far from the pivot point as the pink man is. The rod will rotate towards the pink man.

Why?

I think that the reason that the green man doesn't have as much "spinning" power as the pink man does must have to do with the molecules of the bar. In a sense, the pink man will cause more bending in the bar, and it'll want to straighten out towards his direction (forgive me if that doesn't make any sense, it's just a thought).

Can someone please explain, from the approach of the atomic interactions in the rigid body connecting both the pink man and the green man to the fulcrum, and without using the concept of torque, why it is that the bar rotates towards the pink man?

And, if the torque equation could then be derived from that physical explanation, that would be awesome!

In summary, I'd like to know how to approach the concept of torque with atomic theory. I'd like to understand the interatomic electromagnetic forces in a rigid rod, how momentum travels along a rigid rod when we apply forces perpendicular to its length, and how that leads to the force being "amplified" by larger distances from the pivot point

Thanks!

quantum mechanics - Why does the classical path of a particle give the dominant contribution in the path integral?

Why is it that the classical path of a particle gives the dominant contribution in the quantum mechanical path integral? How do we understand this?

optics - Is it possible to witness a circular rainbow?

What conditions would make it possible to see a naturally occurring fully 360° circular rainbow? Would it even be possible?

Answer

The centre of a rainbow is where your shadow would fall. When you're standing on level ground your shadow is obviously on the ground so you only see part of the arc. If you're in an aeroplane or on a mountaintop or somewhere where your shadow would be above ground, you may be able to see the whole circle.

quantum mechanics - Probability of energy from wavefunction

In general, given a wavefunction $\psi(x)\equiv\langle x\vert\psi\rangle$ for some system, how can one compute the probability that the system will be at a given energy level $E_n$? That is, how can one compute $\langle E_n\vert\psi\rangle$? I feel like this should be next to trivial, but the wavefunction is an expansion in the position basis and I would need the energy eigenbasis to perform the computation. Thus this reduces to a change of basis, but how is this done?

Note: In this system I have already solved the TISE and found the energy spectum.

Answer

The energy eigenstates can be expressed in the form of wavefunctions as well, e.g. $\psi_n(x) \equiv \langle x | E_n \rangle$. Then, you can compute the inner product of the two wavefunctions by integrating their product: $$\langle E_n | \psi \rangle = \int_{-\infty}^\infty \langle E_n|x\rangle \langle x|\psi \rangle \, dx = \int_{-\infty}^\infty \psi_n^*(x) \, \psi(x) \, dx $$

kinematics - Can the equations of motion be used for both instantaneous and average quantities?

I have three equations of motion: $$v = u+ at , $$$$s = ut + \dfrac{1}{2} at^2 , $$$$ v^2 = u^2 + 2as.$$ Can I use them for both finding instantaneous and average quantities? Suppose I want to find the final instantaneous velocity then can I use the first equation; or if I want to find average velocity over a time period, can I use the first equation?

Answer

Well, the three equations that you mentioned are all derived from the definitions of velocity and of acceleration. I'm assuming you know calculus (if you don't, just ignore the next part and look at the conclusions, or even better- learn calculus!). $$v=\frac{dx}{dt}$$ $$a=\frac{dv}{dt}$$

Now, to find displacement we just look at the first equation: $$\Delta{x}=\int v dt$$

But velocity as a function of time is $$v=\int a dt+v_0$$

So displacement will be $$\Delta{x}=\int \left(\left(\int a dt\right)+v_0\right)dt$$

And when the acceleration is constant, we can say that this is $$\Delta{x}=\int \left(at+v_0\right)dt=v_0t+\frac{1}{2}at^2$$

If you want to find the instantaneous velocity, then you need to use the first equation I listed, which defines instantaneous velocity. However, when you have constant acceleration, the first equation you mentioned works, since with constant acceleration we have (from my fourth equation), $$v=\int a dt +v_0=at+v_0$$

Conclusions

You can use all the formulas you mentioned, but only when you have situations with constant acceleration. I showed you that those equations of motion are equivalent to the definitions of velocity and acceleration, but only if you have constant acceleration. In that case, then you can find instantaneous speed comfortably, as well as the position of a particle at any point in time.

The third formula which you gave, I haven't mentioned yet. That one's derived through some straightforward algebra from the other two you listed, so I'll let you do that on your own if you haven't yet; given that it's derived from the other two, the third equation also works only in cases with constant acceleration (it can also be derived from the work-energy theorem, that's something you could have fun with later on if you haven't seen it yet).

As for average quantities- if you wanted average velocity, I guess you could say that $$v_{avg}=\frac{\Delta{x}}{\Delta{t}}$$ Which refers to the total displacement as opposed to the infinitesimal one above.

One subtlety with regards to average velocity that we should be careful about: average velocity could refer to the definition I gave you above, which is just displacement over time, or it could refer to mean velocity, which is the sum of the instantaneous velocities divided by the number of instantaneous velocities. I assumed you meant the first definition of average velocity.

If you wanted to find average velocity over a time period, you couldn't really use the first equation you mentioned. As I said, that gives you the instantaneous velocity. (In constant velocity motion, instantaneous velocity=average velocity so I guess the equation works there but that's a very special case.)

So essentially, those equations work if you want to find the final instantaneous velocity or the displacement, but they're not really helpful for finding things like average velocity. Just keep in mind that the acceleration has to be constant.

This is my first post on this site, hopefully I was able to help you out!

Ps. I hope you don't mind that I switched the names of some variables. I find it more comfortable to name the initial velocity $v_0$ and the displacement $\Delta{x}$.

general relativity - Are gravitational waves longitudinal or transverse?

Waves are generally classified as either transverse or longitudinal depending on the they way the propagated quantity is oriented with respect to the direction of propagation. Then what is a gravitational wave? It doesn't make sense to me that a disturbance in the curvature of spacetime has a "direction", so I would say they're neither, much like a wave packet in quantum mechanics.

Answer

Gravitational waves are transverse waves but they are not dipole transverse waves like most electromagnetic waves, they are quadrupole waves. They simultaneously squeeze and stretch matter in two perpendicular directions. Gravitational waves definitely propagate in a given direction but the effect that they have on matter is completely perpendicular to the direction of motion. Below is a picture of what the metric of a passing wave does to space (the wave traveling is perpendicular to the screen). If you imagine a free particle sitting at each grid intersection point, the particle would move sinusoidally right along with the grid:

This diagram is from this paper

newtonian mechanics - How is the kinetic energy of the wind transferred onto a lift based wind turbine?

The rotor blades of a lift based wind turbine are shaped like airfoils, so the wind flowing around them creates a lift force which in turn moves them around. From a thermodynamic viewpoint and like answered in this question kinetic energy is extracted from the wind during this.

But what goes on from a mechanical perspective? How exactly is the kinetic energy of air particles transferred onto the rotor blades via lift? None of the sites I read gave an elaboration on the how.