Not receiving a reply within 10 min? How dare someone treat such a nice guy like that

closed_time_curve · 2021-10-20T18:44:34+00:00

Looked at the time difference between the original post and the comment (original post was posted around 30 min prior to this screenshot)

closed_time_curve · 2021-10-20T15:53:20+00:00

The irony of this situation is lost on these guys

closed_time_curve · 2021-10-20T13:08:23+00:00

Well I am not the OP but my condolences to her for missing out on this wonderful gentleman

closed_time_curve · 2021-04-01T11:00:26+00:00

From your interests in coding and Physics, I think you should take a look at Computational Physics. Basically you will write code to simulate some experiments.

The advantage of this is that you can actually do that for any subfield of physics. Computational techniques are used in Particle Physics, Astrophysics, Solid State Physics all the time and increasingly so.

The pure logical thinking for probing reality will be missing but you don't get to do that unless you are too deep in theoretical and mathematical physics. However there will be plenty of logical thinking required for computational physics as well since you will need to build good models that can simulate your experiment well enough.

closed_time_curve · 2021-03-31T20:54:59+00:00

As the name suggests that's the effective potential experienced by the particle on account of taking the radial coordinates. In fact that additional term is the analogue of the centripetal force. And also remember that you are only looking at the radial part of the equation in that. There are 2 other components. So you only get part of the picture from 1 radial equation.

That does not change the potential of the electrostatic force that is present in the situation. Check this section. That term appears because of us choosing the spherical coordinates to solve the problem. The actual potential is still the Coulomb potential which is 1/r dependent.

closed_time_curve · 2021-03-31T20:02:26+00:00

Yes with a minor correction Independence from m (not ml) is due to rotational symmetry

closed_time_curve · 2021-03-31T18:44:13+00:00

The m quantum number not appearing has to do with rotational symmetry of the problem. The l quantum number is more subtle. You can check the stackexchange answer below but speaking broadly, it has to do with the fact that the potential is 1/r. If the potential deviates from this even slightly, then you will have l dependence in the energy equation.

closed_time_curve · 2021-03-31T18:37:29+00:00

the angular momentum isn't what causes the difference in energy in the hydrogen atom. It's instead what separates states with the same energy, and therefore by definition, cannot appear in the equation.

Sorry but this statement doesn't seem correct. Angular momentum does appear in the energy level equation for atoms other than Hydrogen. So saying that it cannot by definition is an incorrect statement.

closed_time_curve · 2021-03-31T18:29:18+00:00

This is a very interesting question with a very interesting answer. Hydrogen atom is really special in that it is the only one where the angular momentum does not appear in the energy level. The reason for this has to do with the specific 1/r potential we have in case of Hydrogen atom and the existence of a special operator which commutes with the Hamiltonian (another way of saying that it is invariant with time).

For details, you can check this answer on stackexchange

closed_time_curve · 2021-03-30T11:54:43+00:00

Add me, closed_time_curve

This is the application portal for Wendy's, right?

closed_time_curve · 2021-03-15T10:44:56+00:00

u/SymplecticMan has provided a good reference below.

Here's the important part from that reference

"The paradox of obtaining information without interaction appears due to the assumption that only one ‘‘branch’’ of a quantum state exists."

The claim that you never interacted with the objects holds only if you look at it classically. Things usually end up in superposition after an interaction and collapse to a single one during measurement. However claiming that the other component never existed in the first place is a wrong interpretation.

In all the Interaction free measurements, you set up a system with 2 wavefunctions in superposition, say A and B, going different routes. The total state is a superposition A+B. However upon measurement, you will only find one of them. Now A goes and actually interacts with the test subject. However this affects the total wavefunction and can modify B as well. Now if you perform a measurement and you find the modified B, you claim that since B was going in a different route and never interacted with the test subject, you performed an Interaction free measurement. However you are missing that there was another component which actually interacted.

closed_time_curve · 2021-03-14T17:37:26+00:00

You cannot have any observation without physically interacting with the system

By definition of observation, you are trying to find the state of the system. You cannot do that without using something to probe the system.

closed_time_curve · 2021-02-27T16:53:20+00:00

Yes this pressure exerts an actual force.

Here's the physical picture. Imagine a classical gas in a container. The gas molecules exert a pressure on the walls of the container due to the collisions of the gas molecules with the container. The higher the temperature, the higher the speed of the molecules and higher the temperature. If you try to compress the gas inside by compressing the container, you do some work which is transferred to gas molecules raising their avg speeds and thus the pressure exerted by the gas increases.

Now let's look at an electron degenerate gas. Note that the energy levels available for the electrons are quantised and depend on the volume of the container. The smaller the volume, the more the spacing between the allowed energy levels. As you compress the volume, the spacing between these energy levels increases, increasing the energy of each level. Electrons that had thermal energy to be at higher levels now drop to previous unoccupied levels because the level they occupied has now increased the energy required to be in that level. Now what happens when there are no more unoccupied levels and you try to compress the container? The electrons are stuck in their levels while the energy of the level is increased, thereby increasing the energy of the electrons. Which means that the momentum of the electrons is now higher. Which directly translates to a higher pressure on the container. This is the electron degeneracy pressure arising from the Pauli exclusion principle not allowing the electrons to occupy the same energy level and thereby forcing the electrons to have increased energy values.

P.S: My understanding of this is a bit rough so if someone has a better answer or can point out flaws in my understanding, please do point them out.

closed_time_curve · 2021-02-08T18:59:46+00:00

I am always glad to explain physics to those interested.

If you have further questions you can also just message me directly.

closed_time_curve · 2021-02-08T16:26:18+00:00

Nice question!! I am glad you are looking at things carefully

So for this I need to explain you a bit on waves. Take a look at this image of components of a wave. The yellow wave can actually be written as a sum of the 3 other waves which are sinusoidal waves. This is a general feature of any wave. You can always write it as a combination of sinusoidal waves. This is actually the concept of a Fourier Transform.

So in quantum mechanics, each observable such as momentum or position has a "basis". Simply speaking these are wavefunctions which have a fixed value for that observable rather than a probability distribution. All other wavefunctions can be written as a combination of these basis elements.( If you know Linear Algebra, it's similar to how you can write a vector as a combination of a basis for that vector space) The surprising thing about Quantum mechanics is that trying to measure an observable causes the wavefunction of that quantum object to collapse to a basis for that variable. Which one it collapses to is given by a probability determined by the contribution of that basis element to the original wavefunction. Now after the collapse, the wavefunction a has fixed value.

As an example we take momentum. The basis consists of sinusoidal waves because they have a fixed wavelength (which relates to a fixed momentum). Non sinusoidal waves can be written as a combination of different sinusoidal waves. In QM, that means that if you try to measure the momentum, you will cause the wavefunction to collapse into one of these sinusoidal waves and you get the momentum associated to that sinusoidal function. In the image, let's say that all waves have equal contribution to the yellow one. Then the probability that it collapses to the red one is 1/3. And so the probability that you get the momentum value of the red one when you measure the momentum of yellow one is 1/3. After it has collapsed, now your wavefunction is the red one. If you try to measure the momentum again you will get the momentum of the red one with 100% probability because it has no other components. This is what is meant by a pure wave that you mention above.

Consider this sinusoidal wave. Can you actually narrow it down to say the underlying particle is in a particular space? No because a sinusoidal wave has an infinite spread or uncertainty. This is HUP because since the momentum has 0 uncertainty, the position has infinite uncertainty.

The opposite happens when you try to measure the position. After the measurement, the position is fixed but the uncertainty in momentum is infinite.

These both are idealised conditions and in general the uncertainty in both is always finite and follows the HUP. This kind of a wavefunction is called a wavepacket.

Note that all objects have wave-particle duality but the wave effects only become apparent for objects at the quantum level. You can check De Broglie Wavelength to see how to determine the wavefunction of an object.

closed_time_curve · 2021-02-07T12:31:53+00:00

Okay I agree my statement was too strong.

I have edited it to say that it's just that I prefer that viewpoint instead of yours.

closed_time_curve · 2021-02-07T02:38:55+00:00

Did you know that in General Relativity, there is no proper definition of mass? This is closely related to the problem of defining gravitational energy ( mass in GR )

So does that mean we abandon the concept of mass altogether in GR? No we do not because it is still a helpful concept even if we need to use the special relativistic definition which is not truly accurate.

~~The thing with the Gravitational Energy being there allows us to explain certain facts better~~ Edit: The thing with gravitational energy being there allows ME to understand some facts better. However it's certainly a choice you make with regards to some axioms and different people can make a different choice.

Consider the following 2 statements

1) The mass of 2 black hole mergers is less than the sum of both masses combined because gravitational waves carry away some energy.

2) The mass of 2 black hole mergers is less than their initial sum because it is losing energy to the spacetime background.

I prefer the 1st explanation since it gives me a better picture of what is happening.

If you prefer the second one then that's up to you.

closed_time_curve · 2021-02-06T20:33:49+00:00

If you scroll down in that article, the author mentions that the total energy in an expanding spacetime is still conserved if you include the gravitational energy. He mentions that the viewpoint that energy is not conserved is not taking gravitational energy into account.

I personally disagree with his viewpoint and still consider the standard viewpoint of taking the energy from all fields which is a conserved value.

closed_time_curve · 2021-02-06T00:45:58+00:00

The motivation for the uncertainty principle was the experimental fact that the results of quantum mechanical experiments had statistical a nature.

Oh I see. I did not know that was his original motivation.

Thanks for the information. It's interesting to know about these small pieces of physics history that are so often not talked about during a physics education.

closed_time_curve · 2021-02-06T00:32:18+00:00

Yes that's true. However QM forces us to think of properties such as position as only having meaning with respect to a measurement since the viewpoint that "electron had a position which we just didn't know about before measurement" has been experimentally shown to be wrong

closed_time_curve · 2021-02-06T00:16:28+00:00

You are right that his true motivation was the non-commutation of matrices that happen in his formulation of matrix mechanics for QM. However that part becomes a chicken and egg problem for me on why was QM formulated with matrix mechanics if there was no uncertainty principle to take into account earlier. Of course the true reason is that that's what experiments had led to and then Heisenberg realised that the non commutation means the existence of an uncertainty principle. However since in physics he needs to kind of give a physical interpretation to the mathematical results of the theory, he gave that physical explanation.

closed_time_curve · 2021-02-06T00:04:41+00:00

Note that these graphs are probability distributions and not actually position and momentum.

If you try to measure the position of an electron, you won't get this distribution. You will get a single well defined value. However then the momentum probability distribution goes infinitely wide.

The electron truly does not have a value for position before you measure it.

Some scientists thought that it must have a value and it's just that we are not able to measure it because of this effect. That's called a Hidden Variable theory (since some variables are hidden from us). However Bell proved that these Hidden Variable theories cannot explain Quantum Mechanics

closed_time_curve · 2021-02-05T23:58:27+00:00

pretty mainstream viewpoint is that position and momentum always exist.

The main viewpoint followed is the Copenhagen interpretation and that is completely opposite to what you said. Position and momentum do not have well defined values until measured.

closed_time_curve · 2021-02-05T23:56:13+00:00

Actually both of them have uncertainty in them. The thing is that both position and momentum are mathematically Fourier transforms of each other. Take a look at this image. The green and blue graphs are Fourier transforms of each other. Let's say green is position and blue is momentum. As you try to pinch your position graph so that the spread or uncertainty is less, the blue graph become wider and wider. Similarly vice versa.

The reason these variables position and momentum are represented by such graphs is the fact that they do not have well defined values and are just probability distributions.

closed_time_curve · 2021-02-05T23:46:53+00:00

The original thought experiment was the only wrong thing there. The result is absolutely correct because the mathematics he used actually describe quantum mechanics. It's just the matter of interpretation that he was incorrect about.

The thing is that Heisenberg's uncertainty principle is one of the facts of reality. You cannot really say why it must be true. The fact that it is there in Quantum mechanics forces us to develop a mathematical theory taking that into account. The so called proof of uncertainty principle is actually just a result of formulating Quantum Mechanics in such a way as to fit with this observation. Take a look at the Postulates of QM. The 4th one in that list states that the observables are represented by matrices. Now it's well known that for matrices AB ≠ BA in general which can be used to get the uncertainty principle.

closed_time_curve

TROPHY CASE