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Appreciate it. Thanks for watching!

😂 well…I hope it’s not too painful

Why isn't bro getting views it should already be in millions

That’s very kind of you thanks for watching

Ultimately it’s something that’s made up we just choose to call gamma m c^2 the total energy but it’s well motivated. What I mention in the video is if you take the low speed limit on that quantity it looks like mc^2+1/2mv^2 so it looks like some constant term plus the kinetic energy. But here’s another reason, we know that this overall 4 momentum inner product is conserved by computing it and seeing we get a constant. Moreover, we know that the 3 momentum is conserved as a fundamental law of physics. If the 4 momentum is conserved and the 3 momentum is conserved then the time component of 4 momentum must also be conserved. So this thing in the time component of 4 momentum has units of energy, looks like kinetic energy at low speeds and is conserved…those seem like exactly the properties of something we’d call the total energy yeah? Hope that helps

@@TheBrainFiller i’ll take a look 👀 later. looks daunting 🥺. can i interview you / do a collab on my channel?

@@TheBrainFillerthis was helpful, was going to ask the same thing, thank you. Great video

Where did I get that time component of the momentum inner product? At 4:16 I computed the inner product of 4-momentum with itself and the expression at 5:39 is just a rewriting of that. Hope that helps!

@@TheBrainFillerwow 😮. amazing 🥲

Yeah this is tricky so I get you. First just to be clear for anyone else who reads this to avoid confusion, rando_guy is calculating the work done by gravity, I was calculating the work done by the external force. Of course these have to exactly cancel out because when the object finally arrives at r from infinity it has 0 kinetic energy. So the change in the kinetic energy is 0 therefore the net work done has to be 0 too. Anyways onto your question…the point is I’m controlling the direction of the dr by choosing limits infinity to r rather than r to infinity. So flipping the sign of the dr but then also using my funky limits is doing the same thing twice which cancel each other out

@@TheBrainFiller Thanks for the quick response! The point that the limits control the direction seems to be what I was missing. Pretty clear now :)

Assuming you’re familiar with line integrals which I’m getting the sense you are W= int (F dot dr) from r_1 to r_2 yes but it’s always easier to think about line integrals by parameterizing the chosen path with a variable t and then doing W= int(F dot dr/dt dt) from t_1 to t_2 in which we can interpret t as being a time and dr/dt being the velocity vector of our particle along the path. Now let’s choose a path where the velocity vector is radially inwards (same direction as the gravitational force) the whole trip and is a constant speed v. So F dot dr/dt = (GMm/(r(t))^2)*v and r(t) = (X - v*t) (where we’ll take the limit X goes to infinity at the end cause at t=0, our t_1, the mass is at infinity and at t = (X+r_2)/v the mass is at location r_2). Then just do that integral with those correct time limits and finally take the limit as X goes to infinity and you should get the answer you’re looking for. Alternatively, if you just want the direction of vec(dr) its conventionally considered radially outwards and by flipping the limits of my integration I’m sort of flipping the direction of the dr to be radially inwards. But it is probably much clearer to define a velocity vector dr/dt and do the integral dt for no ambiguity

​@@TheBrainFiller The velocity approach is interesting, I've never really come across it before. I pondered a bit more on this doubt, it's like saying if (dx î)/dt = -v î, then direction of dx is along -î. for which, yeah the small displacement is along that direction but we don't write that on the LHS while denoting the velocity right (that wouldn't make sense). If we multiply that equation with dt we would get the instantaneous motion along dx î only, regardless of what v vector is. Maybe not completely concrete but this is a way to think about it. I think it's in a way similar to the logic where your approach contributes

I highly recommend you do look into line integrals in that case cause yeah you’re getting the right idea and it’s a really interesting topic. dx/dt is a vector and so I’d write in -v r hat (for the radial unit vector in spherical coordinates) and then you dot product that with the Force vector which points in the -r hat direction too so F dot v will be positive the whole journey and then you’ll get a positive answer when you integrate from t=0 to t= time at which you reach the final position

It’s all about thinking who is doing the work. That changes the negative sign. I defined the potential energy as the work done by an external force going from infinity to the present location. Now think about your derivation a bit more carefully (the maths is correct but think about the interpretation). Specifically, what is the change of the kinetic energy of my object across the whole process? It started at infinity at rest and it ends at its current location again at rest so the change in the kinetic energy is 0! Wait so what’s happening was work done? Yes but the net work done is 0. So this external force did some work and it is exactly the negative of the work done by the gravitational field. That’s why I don’t need the negative sign. Have a look at this stack exchange post for some clarification: physics.stackexchange.com/questions/761716/potential-energy-and-work-done-by-external-force Thanks for watching and I appreciate that you’re thinking about the material deeply this is tricky stuff.

Also thinking about a possible issue remember work is actually the integral of the force vector dotted with the dr vector. When we’re moving from infinity to r the gravitational force is pointed radially inwards so in spherical polar coordinates there is a negative sign on that force vector and then we dot it with a displacement dr which is positive. Soo….overall F•dr is negative so then if we were doing an integral from r to infinity that minus sign would still be present. Luckily I swapped the limits of integration and when from infinity to r so that cancels out the negative sign on F•dr. So honestly an argument could be made I was not clear enough on where I was getting that exact integral expression from. An important point is potential energy can be defined in terms of the work done by an external force or also equivalently by the work done by the gravitational field but the negative sign swaps in the two definitions. My final answer is certainly the correct standard result used and the literal maths is not wrong but I should have been clearer on the physics.

@@TheBrainFiller Actually, what I don't understand is why we are trying to apply extra force and make the kinetic energy 0. Is this really necessary? The decrease in potential energy is equal to the work done by the force applied by the source of potential energy, and at the same time the work done is equal to the change in kinetic energy, but the change in kinetic energy does not have to be 0 for us to establish a relationship between potential energy and work. Well W=-∆U=∆K What I don't understand is why extra force is needed.

@spaceghost00 Because if you want some kind of definition that you can write down and think about independently of KE you need this kind of definition. Physicists like energy as a bookkeeping device where they can add up some independent sources of energy and have a total like E = KE + PE and they want this definition of PE to be separate to KE. So if you’re trying to quantify what is the energy associated with a certain configuration of your system that is just due to the position alone then in your definition you need to cancel out any accumulating KE as you construct your system (bringing everything from infinity to its current setup) so that the final number you get from your definition is just about position in space and the presence of a gravitational field. This definition also emphasizes the path independence of this potential energy quantity whereas defining it in terms of the change in KE kind of obscures that. And we want a nice way to directly connect potential energy to the forces so that we can more naturally think about fields and stuff. Honestly it’s a good question worth looking into a bit more. Let me know if you find a nice intuitive explanation somewhere.

Yeah you get the same Q as in the video but just have to do a change of variables. When you integrate the exponential it brings down a factor of RC, so you get Q=(-I_0*RC)*exp(-t/RC)+c. Now at t=0 we want the charge to be Q_0 and at t=infinity we want the charge to be 0 (on the physical grounds that we believe this capacitor doesn’t have some fixed residual charge on it…we could of course have it be anything if we were just doing maths). So, the constant c has to be 0 so that it doesn’t decay to the constant c. We get the Q_0 by just looking at I_0*RC and relabelling it Q_0 (note as a sense check the units make sense: RC has units of time and I_0 has units of charge/time). Thanks for watching and hope that helped! Let me know if you want any clarification

​@@TheBrainFiller Thanks for the answer , i appreciate it , " t=infinity we want the charge to be 0 " that helped me to understand ! btw i have another question , the charge equation (Q(t)) it have to be negative? i mean because of it is the electron...?

Lol fair we all have our peeves. Thanks for slogging through as far as you did 😂

n = 1,2,3... k = 1/4π Epsilon zero h cut = h/2π where h = 6.625 x 10^ -34 e = 1.6 x 10^ -19

@@shotgun-pw4rn thanks brother!!! 😇

I don’t know man…I don’t know. I have ideas but I lack the mental energy/space for whatever reason.

Sorry about the late response but the important point is the Work done we’re talking about is the work done by an external force to make sure that on the whole journey from infinity to r you don’t gain any KE. The gravitational force is pulling you inwards (negative sign in radial coordinates) so this external force has to push you equally outwards (positive sign in radial coordinates). Hopefully that clears it up

Perhaps yes based on this analysis but the full GR for a non-rotating black hole shows that the coordinate speed of light goes to 0 *at* the schwarzschild radius.

Appreciate it, I’m definitely most happy with the way those turned out and my double slit one.

Explicitly yes but it’s a very simple change of variables to say the nucleus has some higher charge just ends up adding a factor of Z in the denominator from I remember. Oh and swap out m for the reduced mass. For heavier atoms this matters even less than it does for just a proton but if you’re worried about positronium or something then it would have an impact

@@TheBrainFiller got it! Thanks

Great to hear! Thanks for watching!

Yup, generally the lower courses would have faster moving water. The basic reason for this is that the deep river beds means that the bulk of the water is experiencing less friction. Although the upper course is steeper, these frictional effects have a major impact on the bulk motion of the water. There’s a lot of interesting physics going on there, so, if you’re inclined, I’d recommend looking into it. I’ll begin by mentioning that turbulence, and hence those white waters I’m assuming you’re thinking about, is actually a sign of inefficient flow because it means that water is moving around in lots of different directions so average speed downstream goes down.