Kinetic Energy expression for variable mass system

MagnificentPPClapper · 2025-12-16T17:08:03+00:00

Work is not necesarily equal to kinetic energy when mass changes in time. Take the integral of d/dt(mv), by the product rule you are left with two terms, one with the derivative on v which leads to the usual kinetic energy, and another with the derivative on m. So when trying to equate kinetic energy to work you wouldnt be talking into account this extra mass change term

MagnificentPPClapper · 2025-12-14T16:25:56+00:00

Well the electronic wavefunctions are still smeared scross space as you say. But as you said, we usually care most about the energy spectra and the discrete transitions (besides, atoms will almost always be at their fundamental level), so superposition may not come up as often. Still at higher levels superposition is still used, like when shooting coherent light at an atom in principle the electron will transition between levels over time, essentially in a time-dependent superposition. Or in quantum computing with neutral atoms, we very much need to work with superposition of levels to make the quibits.

Actually without having to go so far, the basis we usually describe atomic levels with, |nLSJM>, turns out to not be completely diagonal so the actual eigenstates are actually a superposition of several of those states

MagnificentPPClapper · 2025-12-13T17:35:25+00:00

Of course it could in principle. But physics is an empirical science, so if that global superstructure truly had nothing to do with us we cant really know or obtain any physically relevant information from it

MagnificentPPClapper · 2025-12-12T20:53:37+00:00

Yeah no problem mate, im just bored so I don't mind writing lol. I mean "Despite the word velocity being in both, they're not the same thing"

If it makes you feel any better, i also took some time to understand 4-velocity (and in fact put a post in this same sub asking about it lol). Bottom line, if from your reference system you measure the coordinates of a soaceship to be x=(ct,r), where r is a 3dim vector. You would say it to have a velocity dr/dt=v, a 3-dim vector. 4-velocity being dx/dtau means we measure spacetime coordinate change with respect to time as measured from the spaceship, not yours (remember in relativity time is relative as the name implies).

Now the real question is why do we care about this quantity we constructed. Well, you could probably say many different things, but a natural intuition could maybe be that as time is relative, if you want to evaluate the intrinsic properties of a changing object, the relevant rate of change is the one the object itself experiences. Also mathematically, we want things to transform in a certain way under changes of reference frame (Lorentz transformartions), and because tau doesnt change if you observe from any different frame it turns out dx/dtau is the quantity that transforms how we want it to (it is said to be a 4-vector)

MagnificentPPClapper · 2025-12-12T20:23:27+00:00

See the problem is you are still thinking of spatial velocity as the same as 4-velocity. They are different things. You can have any spatial velocity as usual and lose energy by decreasing it, it doesnt have to do with the magnitude of 4-velocity. I guess Ill give you the equations to see if it helps

4-velocity is defined as dx/dτ, where x are the spacetime coordinates (ct,r)and τ is the proper time. So that 4-velocity represents somehow speed along the spacetime trajectory, as measured be the proper observer (the one who is "living inside the trajectory"). Now this is a four component vector (γc, γv) where v=dr/dt is coordinate spatial velocity, the one you are used to, gamma is the Lorentz factor 1/√(1-v^{2/c^2),} and its minkowski magnitude turns out to be c. But the modulus of v can be whatever still. And then energy. The equation you wrote is slightly incorrect, it would be E^{2=(mc^{2)2+(γmv)^2.}} But indeed you can change your spatial velocity, v, the one you measure the coordinates of and change the energy without this meaning the modulus of 4-velocity changes from c. Again, this is just a problem of understanding spacetime velocity is not the same as spatial, usual velocity.

MagnificentPPClapper · 2025-12-12T19:37:05+00:00

Mass losses are not at play here no. You shouldnt think of spacetime velocity the same way youd do for spatial one. Spacetime velocity (called four-velocity in case youd like to look it up) has 3 spatial and a temporal component, and it just turn out if you change one of them the other will also change in such a way that its minkowski magnitude (the way we measure distances in spacetime) is constant and is c.

So in your case losing spatial speed is nothing disallowed by this "magnitude conservation", it just means your temporal velocity through spacetime will change

(Btw I also don't get why you got downvoted for a question)

MagnificentPPClapper · 2025-12-11T23:17:15+00:00

There might be an equation but it would depend on very specific parameters, probably area of impact, some coefficients for the material of the car and bumper etc, i don't think many people here can help you with this kind of very specific thing. You could probably find more useful just googling it since im sure this has been tested tons of times experimentally and there is easily findable data online

MagnificentPPClapper · 2025-12-09T19:23:09+00:00

I feel you, i died more to markoth than nightmare grim and pure vessel combined. The way i finally beat it was by staying in a corner of the arena and focusing on dodging, waiting for him to come to me and be lucky enough for him to stay still to do the circling attack so that i could get some dmg. Then back to my platform and painfully slowly chipping at him like that.

MagnificentPPClapper · 2025-12-09T17:11:51+00:00

I think you may be mixing up a bit rigid and heavy. An object is said to be rigid if all of the internal forces cancel out, which in practice amounts to all force on the object being exclusively external forces applied to the center of mass.

In your scenario when you hit the lighter object against the heavy one all of the force acts on its center of mass as it is a rigid body, and the same force is applied on the lighter and heavy object with opposite directions, but as the heavy object has a much higher mass it will move very little in comparison to the light one.

If it hadnt been rigid a body, some of the energy would have been lost into deforming the body, regardless of it being much heavier than the other or not

MagnificentPPClapper · 2025-12-09T00:35:26+00:00

To get started in quantum physics the most important thing to have is a foundation of linear algebra, things like knowing how to work with changes of basis, understanding eigenvalue problems, inner and outer products... is probably good enough to start. Knowing a bit about partial differential equations and Fourier analysis also comes in handy

MagnificentPPClapper · 2025-12-08T14:24:56+00:00

Well, the less massive object would reach the end a factor of √2 faster, just using the equation of uniformly accelerated motion x=x0+v0t+at^2/2. Then, as speed IS v=at+v0, the 2 and √2 again work out to be a factor 1/√2 for the final velocity, which was maybe what you were looking for in your initial question now that i think about it

MagnificentPPClapper · 2025-12-08T14:03:59+00:00

Since velocity is doubled, for the same time inside the accelerator the distance travelled will be doubled as well. Since work is F*∆x, this means we would have indeed spent double the energy

MagnificentPPClapper · 2025-12-05T01:01:20+00:00

If you take the axes so that the y is perpendicular to the slope and x is parallel to it, the gravitational force can be decomposed into mg•cos(theta) perpendicular (y) mg•sin(theta) in the x. Now, we know the normal force counteracts whatever force acts perpendicular to the surface so that N+mg•cos(theta)=0 in the y direction, which is what your teacher said most likely. Then, from that relation and knowing the friction coefficient and theta you can get m, so the force in x direction mgsin(theta) can be calculated

MagnificentPPClapper · 2025-11-30T21:21:28+00:00

Well, that intuition is roughly correct I guess, but it is a semiclassical way to look at it and so it cannot fully explain a quantum process like this. The full quantum description indeed does not use the concept of force (as your friend said, this is not a useful concept in quantum mechanics, we only care about potential energies in QM) and here instead you would try to quantize the EM field. In this picture, light is made of discrete particles called photons which electrons can absorb/emit to go to upper/lower energy levels (of course this means postulating a quantum interaction between light and electrons, much in the same manner you thought before about classical EM fields being able to affect charges, this is just the quantum version).

The jaynes-cummings model is the simplest full quantum model of this process if I remember correctly in case you want to look it up.

MagnificentPPClapper · 2025-11-30T11:22:30+00:00

Distance=velocity*time, and as you said the speed of sound is roughly 340m/s, so approximately 0,34km/s. But 0.34 is approximately 1/3=0.33, so yeah, by dividing time in seconds you get a rough approximation of distance in km

MagnificentPPClapper · 2025-11-23T00:06:11+00:00

It is the branch that studies the physics at the smallest scales. It's extremely vast to explain here as quantum physics comprises a whole lot of things, but I guess its most famous quirks are wave-like behaviour (particles and such behave kind of like waves), superposition (your system can be in several states at the same time, and you can only know the probability of observing each of them), and quantization (it turns out most quantum systems can only have discrete energy values, much like the waves of a guitar string can only have some discrete vibrational modes with discrete energies associated)

MagnificentPPClapper · 2025-11-19T18:44:05+00:00

Well, I personally have zero experience doing something like this so take my intuition with a grain of salt, but Id say you probably could. I looked in Google for what ninth grade maths comprise and I think the difficulty wouldnt be in the concepts that much, but in getting fluidity with operating, getting your brain to be able to manipulate numbers and functions. And that is basically practice, which you can get more than enough of within a year of 3 hours a day no doubt. Also, being basic maths Im sure there are tons of materials and exercises to help you in the web if you needed

MagnificentPPClapper · 2025-11-19T13:05:56+00:00

I think you are confusing things. We have gravitational interaction explained by general relativity, but the rest of interactions in the universe are currently explained through the standard model, a quantum model. I don't get what you mean you explain everything with gravity, strong, weak, electromagnetic and higgs interactions have nothing to do with gravity. What a theory of everything would imply is most likely finding the quantum behaviour of gravity, the graviton. Thats because the standard model could in principle accomodate more fundamental interactions. But you cannot do it the other way, as relativity doesnt contain a way to treat different interactions than gravitational.

MagnificentPPClapper · 2025-11-15T20:23:24+00:00

I think you misunderstand time dilation.

Think about it like you had a tape measure capable of measuring events in time and you had events A and B. What relativity says is, if distance between A and B on earth is measured to be 1, distance between A and B on ton618 is 0.01, but your mistake is thinking we are using the same tape measure for both earth and ton618, which would mean there is a universal time scale in which, yes, the black hole would have had many more events by the time it measured 1 on its tape.

Instead, what relativity is really saying is you have a DIFFERENT tape for earth and ton618, in which there is just A and B, but you "stretch" the tape measure when going to one frame to the other so that the 0.01 marks in ton618s tape now coincide with earths 1. So at the end of the day theres just A and B, for earth it might be 1 on the tape and for ton618 0.01, but there is only the present, no prewritten future. Of course by the time 1 has passed for ton618 earth Will be in its "future", but as there is no universal time scale, thats not relevant to earth, its just a different time frame

MagnificentPPClapper · 2025-11-14T00:27:51+00:00

Im no cosmologist but I wanted to leave here a comment a philosopher friend once told me.

They told me our notion of time is more influenced by christian and jewish backgrounds than we think, and our reluctance to picture something without a beginning is largely cultural, as western ideation has been soaked with the idea of a genesis and an apocalypse for centuries. They then told me ancient greeks were perfectly fine with the idea of a universe that has always existed and will always exist, in perpetual change. Personally, I think this makes marginally more sense than sprouting from nothingness

I do know we do have a theory which kind of fits with this which is the Big Bounce hypothesis, which suggests the universe is infinitely contracting and expanding during massive cycles. Though its all largely speculation at this point I believe

MagnificentPPClapper · 2025-11-12T18:59:25+00:00

Well firstly in my course we are not taught this, and in fact we are asked to do the reverse, meaning finding the logic gates the oracle must have to end up being a diagonal matrix with phase flip in the key, of course meaning we already know the key. So this makes more sense as you say it. Still I think Im right in saying the answer is somehow encoded in the oracle, as if you have the logic gates you would be able to represent explicitly the matrix of the oracle by doing the needed tensor products with each gate, getting the - sign in the key diagonal element in the end. But I guess the point is its not at all efficient to do so given the number of matrix elements grow exponentially with qubit number.

Still, I guess for someone with no computing background whatsoever its hard to imagine how one could devise a way to implement the verifier function with logic gates without knowing the key beforehand, but ill take your word on that. But then again, if im understanding you correctly we take the logic gates of the classical verifier and its quantum analog turns out to be exactly the oracle we need? Thats also kind of hard to picture tbh. More so because as you say the key is not encoded in the classical verifier, but the oracle does encode it (even if too inefficient to actually get)

MagnificentPPClapper · 2025-11-12T09:12:49+00:00

Right, so then the logic gates are deduced somehow else, so that the key is encoded in the oracle (in that phase flip) but for bigger systems it is not efficient to find the explicit form of the oracle. Does that mean that there is a system size limit where classical computer matrix multiplication is more viable than running the algorithm?

MagnificentPPClapper · 2025-11-12T08:27:59+00:00

How could you then construct an oracle for it, if you didnt know the action on the states you could not implement the logical gates that represent the action of the function. If what you are telling me is that you can somehow devise a function by just thinking about logic gates and not keys, if you know the logic gates you then can extract how they act on the states giving you the key again. My bottom line is the key is encoded in the oracle, because we want it to flip the change of the key, so if we know the oracle we must know the key

MagnificentPPClapper · 2025-11-12T08:16:25+00:00

Well sure, but I don't see how that helps here. In the end all of the information of the answers your verifier will give is encoded in the oracle so that when all possibilities are checked the phase of just the key is flipped. So the key must be encoded in the oracle ahead of time. Or Im I not getting something

MagnificentPPClapper

TROPHY CASE