# timo

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1. ## GR and particle physics

Random comments on your seemingly random questions: 1) Particle physics is indeed looking at debris to a very large extend. However, people are not looking for new objects in the debris. They look at the content and distribution of the debris and compare it with the predictions of the different mathematical models. 2) The reference to "statements about their encounters" does not refer to particle collisions (caveat: I am interpreting a single sentence out of context here - but modern particle physics did not exist during Einstein's lifetime, anyways). It refers to a key concept in relativity that comparing situations at different locations is tricky. It is not required that the objects in questions are elementary particles that collide. The famous spacefaring twins meeting each other after their space travel (or lack thereof) are would be typical situations that the statement refers to.

3. ## FLUIDS (Pressure in a fluid of uniform density

There are two approaches here, the formal one and the brain-compatible one. 1) Formally: Realize that there are hidden coordinate dependencies. You are probably looking to construct a function p(y). Since y is the coordinate at the lower side, you have p(y) at the lower edge and p(y+dy) at the upper. This is (possibly) slightly different from p(y) (because of the displacement dy). If you call the difference dp, then p(y+dy) = p(y) + dp(y) = p + dp (note that dp can and will be negative). 2) Brain compatible: Put the equation first and then define the variables: The forces on top and bottom should cancel out. There force up is the pressure force p*A from below. The force from up is the pressure force p2*A from above the small fluid element plus the weight dw of the small fluid element. Since we are talking about infinitesimal coordinates, and since p(y) should be a function, it makes sense to say that p2 = p+dp, which you can then integrate over.
4. ## Is Calculus used in Data Science/Artificial intelligence

In my opinion, the content you listed is below the minimum required for AI (not really sure what "Data Science" is, except for a popular buzzword that sounds like Google or Facebook). More precisely: Apply these topics to multi-dimensional functions and you should have the basis of what is needed for understanding learning rules in AI. However: All of the content you listed is the minimum to finish school in Germany (higher-level school that allows applying to a university, that is), even if you are planning to become an art teacher. And Germany is not exactly well known for its students' great math skills. The course looks like a university level repetition of topics you should already know how to use, i.e. a formally correct way of things that were taught hands-on before. I do not think a more rigorous repetition of topics will help you much, since you are more likely to work on the applied trial&error side. Bottom line: If you are already familiar with all the topics listed, I think you can skip the course. If not, your education system may be too unfamiliar to me to give you any sensible advise. Btw.: University programs tend to be designed by professionals. So if a course it not listed as a mandatory, it is probably no mandatory.
5. ## Pair production (Electron, positron)

By this standard, physics is not very strange most of the time.
6. ## Compton Scattering

The equation is not particular to Compton scattering. It is the relation between momentum and energy for any free particle (including, in this case, electrons). I am not sure what you consider a derivation or what you skill level is. But maybe this Wikipedia article, or at least the article name, is a good starting point for you: https://en.wikipedia.org/wiki/Energy–momentum_relation
7. ## Pair production (Electron, positron)

There is no law of conservation of mass. Quite the contrary: The discovery that mass can be converted to energy, and that very little mass produces a lot of energy, has been a remarkable finding of physics in the early 20th century. The most well-known use is nuclear power plants, where part of the mass of decaying Uranium is converted to heat (and then to electricity). The more modern, but from your perspective even more alien view is that mass literally is a form of energy (I tend to think of it as "frozen energy"). In that view, you can take the famous E=mc^2 literally. There is a law of conservation of energy, but energy can be converted between different forms. In your example, it is converted from kinetic energy of the photons to mass-energy of the electron and the positron (and a bit of kinetic energy for both of them). Note that the more general form of E=mc^2 is E^2 = (mc^2)^2 + (pc)^2 with p the momentum of the object - it simplifies to the more famous expression for zero momentum. I say this to make the connection to your other question where you asked about this equation.
8. ## Is Psi = Superposition?

This post is a bit beyond the original question, which has been answered - as Psi being a common Greek letter to label a wave function and wave functions being used to describe (all) quantum mechanical states. I do, however, have the feeling that I do not agree with some of what your replies seem to implicate about superposition, namely that it is a special property of a state. So I felt the urge to add my view on superposition. Fundamentally, superposition is not a property of a quantum mechanical state. It is a property of how we look at the state - at best. Consider a system in which the space S of possible states is spanned by the basis vectors |1> and |2>. We tend to say that $| \psi _1 > = (|1> + |2>)/ \sqrt{2}$ is in a superposition state and $| \psi _2 > = |1>$ is not. However, $|A> = (|1> + |2>) / \sqrt{2}$ and $|B> = (|1> - |2>) / \sqrt{2}$ is just as valid as a basis vectors for S as |1> and |2> are. In this base, $| \psi _2 > = (|A> + |B>)/ \sqrt{2}$ is the superposition state and $| \psi _1 > = |A>$ is not. There may be good reasons to prefer one base over the other, depending on the situation. But even in these cases I do not think that superposition should be looked at a property of the state, but at best as stemming from the way I have chosen to look at the state. Personally, I think I would not even use the term superposition in the context of particular states (a although a search on my older posts may prove that wrong :P). I tend to think of it more as the superposition principle, i.e. the concept that linear combinations of solutions to differential equations are also solutions. This is kind of trivial, and well known from e.g. the electric field. The weird parts in quantum mechanics are 1) the need for the linear combination to be normalized (at least I never could make sense of this) and 2) that states that seem to be co-linear by intuition are perpendicular in QM. For example, a state with a momentum of 2 Ns is not two times the state of 1 Ns but an entirely different basis vector. Superposition in this understanding almost loses any particularity to QM. Edit: Wrote 'mixed' instead of 'superposition' twice, which is an entirely different concept. Hope I got rid of the typos now.
9. ## Creating Electricity from water to create an eco friendly future

To turn rotational energy into electric energy you can indeed use dynamos, as you already assumed. Or better: Dynamo-like devices. The general term seems to be https://en.wikipedia.org/wiki/Electric_generator. Essentially, i.e. from a physics perspective, electric generators move around magnets in the vicinity of looped electric wires ("electro-magnets"). This induces a current in the wires. Doing this moving around in a controlled way generates a controlled current. Sidenote: I thought the general concept of a dynamo is a "turbine". But according to Wikipedia that refers to the moving part, only. Still, turbines are so closely related to electric power generation that looking up that concept may be relevant, too.
10. ## Variations and consequences of the Laws of Thermodynamics

Thermodynamics, in its general meaning, is always equilibrium Thermodynamics. Calculations of processes assume that the systems go through a series of equilibrium states during the process, which is called a quasi-static process. Reversibility is not required for studiot's entropy change equation in the first post. The equation is fully applicable to bringing two otherwise isolated systems with different temperatures into thermal contact, which I will use as an example: In the theory of thermodynamic processes, both systems' states change to the final state through a series of individual equilibrium states (*). Because of their different temperatures and conservation of energy (and because/if the higher-temperature system is the one losing heat to the colder) the sum of entropies increases. In the final state, both systems can be considered as two sub-volumes of a common system that is in thermal equilibrium. Since the common system is in equilibrium (and isolated), it has a defined entropy. Since entropy is extensive, it can be calculated as the sum of the two entropies of the original systems' end states. As far as I understand it, removing barriers between two parts of a container is essentially the same as bringing two systems in thermal contact. Except that the two systems can exchange particles instead of heat. The two systems that are brought into contact are not isolated - they are brought into contact. If one insists on calling the two systems a single, unified system right after contact, then this unified system is not in an equilibrium state (**). And I believe this is exactly where your disagreement lies: Does this unified, non-equilibrated state have an entropy? I do not know. I am tempted to go with studiot and say that entropy in the strict sense is a state variable of thermal equilibrium states - just from a gut feeling. On the other hand, in these "bring two sub-volumes together"-examples the sum of the two original entropies under a thermodynamic process seem like a good generalization of the state variable and converges to the correct state value at the end of the process. (*): I really want to point out that this is merely a process in the theory framework of equilibrium Thermodynamics. It is most certainly not what happens in reality, where a temperature gradient along the contact zone is expected. (**): In the absence of a theory for non-equilibrium states this kind of means that it is not a defined thermodynamic state at all. But since there obviously is a physical state, I will ignore this for this post.
11. ## Confusion about some basics of spherical coordinates

I assume you refer to my magnet example: The force between two magnets depends not only on their location, but also on their orientation. Take two rod magnets NS in one dimension which are one space (here I literally mean the space character) apart. In the case NS NS their attract. If one is oriented the other way round, e.g. NS SN, they repel each other. On other words: Their force does not only depend on their location, but also on their orientation (the equations in 3D are readily found via Google, but the common choice of coordinates may not obviously relate to what you describe). So for calculating forces or energies of magnets, you need their orientation as an additional parameter. This orientation can be expressed as a unit vector (and to relate to my first post: since this is a geometric and not an integration topic, unit vectors are better suited than angles).
12. ## Confusion about some basics of spherical coordinates

It is hard to tell without knowing your "subject" or the variable. For spherical coordinates of a single location, the unit direction vector indeed supplies no additional information to the location vector. Maybe the direction refers to something else than the location? Like the state of a small magnet, which (ignoring momenta) is defined by its location and its orientation at this location. Or much simpler: Direction refers to the direction of travel. Another idea I could think of is that if your position and direction are data in a data set processed on a computer. Then, they could just be in there for convenience of the user or some (possibly overambitious) performance optimization of calculations that only need the direction and want to skip the normalization step. Just semi-random ideas.
13. ## Confusion about some basics of spherical coordinates

Using unit vectors to define directions indeed contains the same information as using appropriate angles. And either can be used in conjunction with a radial coordinate to define location (but only the version with the angles is called spherical coordinates). Either version can be more appropriate for a practical problem, I think. In my experience, direction vectors tend to be more useful for trigonometric/geometric questions, and spherical coordinates tend to be more useful for integrals.

15. ## Difference Between Evidence and Proof

Not 100% certain what you mean. "Can't be contradicted" is not the same as "proven", especially in math. Any unproven theorem in math, say an unproven Millenium Prize Problem, is proof of this (silly pun intended). Because if they could be contradicted (more specifically: we knew a contradiction to the statements) then they were proven wrong. Accepting an argument as true is a much stronger statement than not contradicting it. Taking the liberty to modify your statement to "a proof is an argument that everyone [sane and knowledgeable implied] has to agree on" that is not too far away from what I said. The main question is where "everyone" lies in the range from "everyone in the room" to some infinity-limit of everyone who has commented and may ever comment on the statement. This limit would indeed be a new quality that distinguishes proofs from facts (to use the terms swansont suggested in this thread's first reply. I could understand if people chose this limit as a definition for a proof. But it looks very impractical to me, since I doubt you can ever know if you have a proof in this case (... but at least you could establish as a fact that something is a proof ... I really need to go to bed ... ). Fun fact: For my actual use of mathematical proofs at work, "everyone" indeed means "everyone in the room" in almost all instances. For me, the agreement of that audience would not be enough to call something a fact . (Okay... off to bed, really ....). It is, indeed. Maybe with an extra grain of elitism for not needing observations but relying on the thoughts of peers alone.
16. ## Difference Between Evidence and Proof

If you accept discovery of a particle as proof for a particle, then that is pretty much the usage of the terms in particle physics, where "evidence" is a certain amount of statistical significance and "discovery" is a certain larger amount of statistical significance (example link for explanation: https://blogs.scientificamerican.com/observations/five-sigmawhats-that/). That usage of the terms is, however, very field-specific. Other fields may have problems to quantifying statistical evidence. When I was a young math student, I thought about the same. Then in a seminar a professor asked me "what is a proof" and I told him something about a series of logical arguments that start from a given set of axioms. His reply was roughly "err .. yes, that too, ... maybe. But mostly it is an argument that other people accept as true". I think his understanding of "proof" may be the better one (keeping in mind that "other people" referred to mathematicians in this case, who tend to be very rigorous/conservative about accepting things as true).
17. ## Dimensional significance of volume/area ratio

Absolutely. Adding up thin shells to create the full sphere is actually a very common technique to access the volume, i.e. $V(r) = \int_0^r A(r) \, dr$ Well, picking up on the integration example: The average distance <r> of a particle from the center in a sphere of radius R is $<r> = \frac{1}{V(R)} \int_0^R \, r \cdot A(r) dr = \frac{1}{V(R)} \int_0^R \, r \cdot 4\pi r^2 dr = \frac{1}{\frac 43 \pi R^3} \left[ \pi r^4 \right]_0^R = \frac 34 R$ (modulo typos: Tex does not seem to work in preview mode ... EDIT: And apparently not in final mode. The result in the calculation above is <r> = 3/4 R). There is a general tendency that the higher the dimension, the more likely a random point in a sphere lies close to the surface. There is a famous statement in statistical physics that in a sphere with 10^23 dimension, effectively all points lie close to the surface. In all sensible definitions of volume and area I am aware of (at least in all finite-dimensional ones), volume has one more dimension of length than area. Hence, their quotient indeed has dimensions of length. I don't think the quotient itself has a direct meaning. But there are theorems like that a sphere is the shape that maximizes the V/A ratio for a fixed amount of V or A. I already commented on the dimensionality. But I still encourage you to just play around with other shapes: Cubes are the next simple thing, I believe.

19. ## The SSSC

The facility you are referring to is called the SSC, not SSSC. Or at least I have never heard anyone referring to it as SSSC. The site is not exactly in the vicinity of Stanford University, but about 2500 km away in Texas. Wikipedia has an article about it on https://en.wikipedia.org/wiki/Superconducting_Super_Collider, which answers your first question: About three times the size and energy of the LHC. Considering the Standard Model of Particle Physics, having had the SSC should have lead to the detection of the Higgs Boson sooner. The next proposed step after the LHC is a lepton collider, which allows more precise measurement on the physics discovered at the hardon collider (both SSC and LHC are hardon colliders). The lstest proposal of a lepton collider is the https://en.wikipedia.org/wiki/International_Linear_Collider (ILC). Given the time such projects take my guess is that having had the SSC ten years sooner than the LHC would simply have put us ten years closer to the ILC-like experiment (I don't expect that we have ILC results within the next ten years). Theoretically, the SSC could have found new physics/particles that the LHC cannot detect. But no one can know that. As the name suggests, the ILC indeed is an international project - as has been the LHC. I would not expect that these days anyone would plan a national collider experiment of that scope, anymore. While I was not around in science at the time SSC was planned, I imagine that the cold war was a reason why it was even considered as a national project. In fact, the Wikipedia article lists the end of the cold war as one of the reasons for the cancellation of the project.
20. ## Question about Einstein's constant of gravity

Multiplying values with physical constants (like h-bar or c) can be considered as re-expressing the same value in a different measurement system (i.e. in different physical units). That is also true if the value is a physical constant itself (like G). In this case, the coupling constant of gravity has been expressed in a suitable form for a case where energies are measured in GeV/c². This is a very typical unit for energy in accelerator particle physics, where ~100 GeV/c² is a typical energy of a particle. Possible uses could be statements like "see how weak gravity is compared to electromagnetism, which has a coupling of 10^-2" (electric charges have unit 1, actual value is 1/137). The Planck mass/energy mp is the energy that elementary particles would need to have (in a simple quantum gravity) for their gravitational force to become comparably strong to the other forces. Hence, G*mp*mp = 1, which is approximately 10^-2. It should not be understood as a particle class of "Planck particles" or a source of energy. Notable objects which exceed the Planck mass are the earth and the moon. Their gravitational interaction with each other happens to be much larger than their electromagnetic interaction. That would remain true if their masses were expressed in units of GeV/c².
21. ## At high enough temperature all elementary particles become massless?

In a sense, getting to massless SM particles could be achieved by increasing the temperature in some region. But it would be more like an activation of the frozen Higgs field than a blocking. It is easier to explain coming from the high-energy side (high temperature), since that is the standard explanation for the Higgs mechanism: The Higgs proto-field (*) can be considered as an additional particle class to a Standard Model in which all of the other Standard Model particles are massless (**). It interacts with most of the other particle fields. But it also has a weird self-interaction which causes the energetically lowest state to not be at "no proto field" but at "some value of the proto-field". At low temperatures, where the Higgs proto-field is just lying around in its minimum, this means that the dynamic interaction terms of the Higgs proto-field with the other particles become some dull interaction of those particles with some sticky stuff that seems to lie around everywhere (***). In the mathematical description of the Standard Model time evolution, the associated terms that originally were terms of a dynamic interaction now become the mass terms for the other particles. If the minimum was at "no field", they would simply drop out (****). This "low-energy" limit actually covers almost all of the temperature ranges we can create on earth, and only recently did we manage to even create and see a few excitations of the Higgs proto-field around its minimum in specialized, very expensive experiments (-> confirmation of the Higgs-Boson at the LHC). So technically, I think we are very far away from creating the "massless particles" state in an experiment. But there is no theoretical reason why this would not be possible (*****). But as described, I would understand it to be less of a shielding of the Higgs field and more of an activation. And as a state with such a high amount of interaction between the fields. So I am not even sure if the common view of a few particles flying through mostly empty space and only rarely kicking into other free-flying particles would still make sense. Remarks: (*) I would just call it Higgs-field(s), but since the paper you cited seems to explicitly avoid using the name at this stage I may be wrong about common usage of the terms. Haven't been working in the field for over ten years. So I have invented the term proto-field for the scope of this post - it is also easier to understand than "doublet of complex scalar fields". (**) This is not exactly true because the particles are mixed and get renamed under the Higgs mechanism. But I'll pretend that does not happen for the sake of providing an answer that is easier to understand than reading a textbook. (***) Sidenote: In this state, the few excitations of the Higgs proto-field around its minimum are the infamous Higgs Boson. (****) Which is why you can always invent new fields that just happen to have no effect on anything we can see but magically make your particle cosmology equations work at very, very high energies (****) Except for the fact that some people still expect new physics and associated new particles at such high temperatures, which then again would have mass from another Higgs-like mechanism
22. ## Is time a property of space or the fields within it?

I'm going to propose the opposite straight answer Time is a coordinate dimension akin to the known spatial dimension. Hence, you speak of the coordinates x, y, z and t or the coordinate tuple (x,y,z,t). Fields are functions that are defined on the 4-dimensional space described by these coordinates, written i.e. functions of type f(x,y,z,t) (or $\psi(x,y,z,t)$ or something similar if you prefer fancy greek letters).
23. ## What is Einstein's equation used for?

As a side-note to the thread: The equation you are citing is the energy-momentum relation for a free particle. It is actually not called "Einstein's equation" (by physicists). The reason I am stressing this is because there is a different mathematical relation that is actually called "Einstein's equation", namely the relation between spacetime-curvature and spacetime-content. That doesn't invalidate your question, of course. For your actual question: I think zztop's answer perfectly fits what I'd have said. If you consider how important conservation of energy and conservation of momentum are it is only one step further to realize how important an equation relating energy to momentum can be.
24. ## A simple classical physics / algebra question

The answer to your first question is: Not necessarily. If a, b, and c are real numbers and * is the multiplication of the reals then the statement is true. In case of vectors and the cross product a counter-example would be b = (1, 0, 0) and c = (0, 1, 0) where b is changed to equal (1,1,0). Since I assume that by "the momentum is not zero" you refer to the linear momentum then again the answer to your second question is "not necessarily". Take a radial vector (1,0,0) and a linear momentum (1,0,0). Changing the radial vector to (2,0,0) still leaves the angular momentum at zero. Your statement would work if the angular momentum was not zero, since scaling up the radial vector by a factor (while leaving the linear momentum constant) also scales up the resulting angular momentum by the same factor.
25. ## Basic Conservation of Energy Problem

It is a bit hard to give a definite answer without knowing which parameters are given an which are not (your second paragraph suggests the mass of the bullet is given even though your first paragraph suggests it is not, for example). My guess would be the following: Collisions not only need to conserve energy, but also momentum. Additionally, the initial kinetic energy does not need to equal the final kinetic energy, since some of the original kinetic energy can be transformed into deformation energy. Hence, the kinetic energy of the bullet was larger than the energy of the resulting oscillation. Taking this into account (which becomes easy if, for example, you are given the mass of the block and the bullet) should give a correct answer, as far as I can tell.
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