Professor MathTheBeautiful, thank you for explaining the Second Derivative in ODE and all its application on planet earth. The Wave Equation is also another powerful tool in Advanced Mathematics, Physics and Engineering. This is an error free video/lecture on UA-cam.
22:30 Well, for the wave equation, there actually _is_ a general solution: Y(x,t) = F(x+c·t) + G(x-c·t) where `c` is the speed of propagation, and `F` and `G` are two arbitrary functions. You can obtain this result by observing the symmetry of the solutions to the wave equation in space and time along the "light cones" of propagation which lay exactly along the lines of ` x = ±c·t`, so when you change the variables to `u=x+c·t` and `v=x-c·t`, the wave equation reduces to a simpler one: `∂²Y/(∂u·∂v) = 0` which then can be integrated twice to get the result I mentioned.
Hi, I have a question about the "correction" box at 5:40 (and following). I wanted to ask: is this correction really needed? I thought the way Professor Grinfeld wrote on the board was pretty convincing: if the viscousness of the medium is very high, then it overpowers the force exerted by the spring, and so the term "-w2u" becomes insignificant. So do we actually need the correction? Thanks for the amazing videos!
I think that the correction is necessary because u itself might still be high. For example, if someone manually displaced the spring substantially (u=1) and then let go. Then it would start crawling (u''=0) back to the equilibrium due to the high viscosity.
Does motion and life as we know really need man created mathematical model such as 2nd derivatives? I think life and motion probably does need our mathematical models...
It is a happy co-incident that second order ODE's approximate reality. The ultimate laws of reality will surely not be dependent upon two prior initial conditions. Surely reality can update to a new state purely dependent upon it's current state only.
I like the second derivative because second-order differential equations are much more fun than first-order differential equations, and also still simple enough to be manageable. =P
Oh, and I see that you love the `F = m·a` equation, and you always stand in awe when writing it. Sorry to break it for you, but this is not the original Newton's formula. If you look into Newton's "Principia Mathematica", you will find that he stated the force to be a "change of motion in time" (in modern terms called "momentum"), that is, `F = dp/dt`. The formula you cherish so much is a bastardization of the original formula, made by modern physicists who then blamed Newton for it not being accurate enough when Relativity came into play. But it is actually _their_ fault that this formula is no longer valid in Relativity. Newton's original formula was valid, since it works directly on momentum, and as you probably know, energy and momentum work together in Relativity as a four-vector. The bastardized version, `F = m·a`, can be derived from Newton's original formula by assuming that the mass doesn't change (which is usually the case in non-relativistic setting): F = dp/dt = d(m·v)/dt = m·dv/dt = m·a But if mass is a variable that depends on time, we cannot do that - we have to use the product rule: F = dp/dt = d(m·v)/dt = m·dv/dt + v·dm/dt = m·a + v·dm/dt So, as you can see, we have an additional term `v·dm/dt` then besides the well-known `m·a` in that situation, and that term relates velocity with changes of mass. Sounds familiar, eh? ;J
Sci Twi uh, you do realize newton assumed mass was constant and p=mv in the principia. So F=mdv/dt=ma is completely equivalent to F=dp/dt. How is it a bastardization? It's completely correct as far as newton is concerned. So it doesn't work in SR, who cares? it was never supposed to, and the only reason you can still say F=dp/dt in SR is because p is redefined from how newton defined it. PS in most cases it's easier to think of mass as being constant even in SR, because mass is typically thought of as the parameter for a particular particle. It's just that p=mv isn't a useful definition anymore. We define p=mv/sqrt(1-(v/c)^2) and then F=dp/dt still works as long as your careful what you mean by force, which gets a little more tricky in SR than Newtonian mechanics.
_"newton assumed mass was constant"_ Can you point out where exactly did Newton assume that mass is constant _in general_ (that is, not in some particular case where it happens to be constant)? Because from what I know, he had a different concept of inertia than we have today (he treated it as another force, _vis insita_ , that shows up when a body changes its motion). _"F=mdv/dt=ma is completely equivalent to F=dp/dt"_ Only if mass is a constant. _"How is it a bastardization?"_ Because it unnecessarily narrows its use and causes it to break in Relativity, then blaming Newton for that, while the blame is on modern scientists who perverted Newton's original idea. _"So it doesn't work in SR, who cares?"_ People who use Relativity in their work? And no, it's not just about super speeds and astronomical domain - relativistic effects can be observed with low speeds as well when the forces are strong enough, as in electromagnetism. Your TV set wouldn't work properly if the relativistic effects wouldn't be taken into consideration, or the GPS you use while driving. Long story short, some people care, if they do anything important in science. And it would work fine with Newton's original equation as well if it wasn't corrupted by modern physicists. _"it was never supposed to"_ Many things were not supposed to do what they do today. But it is a good thing when they do, because it means that the original idea was correct and extensible. It is bad, though, when old ideas break apart when we discover new facts. _"it's easier to think of mass as being constant even in SR, because mass is typically thought of as the parameter for a particular particle"_ You're confusing rest mass (inherent property of a particle, invariant) with relativistic mass (varies with speed). _"It's just that p=mv isn't a useful definition anymore. We define p=mv/sqrt(1-(v/c)^2)"_ Nope, it is the MASS in this formula that needs to be redefined, NOT momentum. The `1/sqrt(1-(v/c)^2)` part comes from the relativistic correction for mass. Momentum in SR, along with energy, together make the four-momentum which is a *Lorentz invariant* .
Great job, a mathematician who discover the universe. These moments, I refresh my mind. Many thanks.
Professor MathTheBeautiful, thank you for explaining the Second Derivative in ODE and all its application on planet earth. The Wave Equation is also another powerful tool in Advanced Mathematics, Physics and Engineering. This is an error free video/lecture on UA-cam.
One more amazing point-Curvature is directly proportional to the second derivative!
22:30 Well, for the wave equation, there actually _is_ a general solution:
Y(x,t) = F(x+c·t) + G(x-c·t)
where `c` is the speed of propagation, and `F` and `G` are two arbitrary functions. You can obtain this result by observing the symmetry of the solutions to the wave equation in space and time along the "light cones" of propagation which lay exactly along the lines of ` x = ±c·t`, so when you change the variables to `u=x+c·t` and `v=x-c·t`, the wave equation reduces to a simpler one: `∂²Y/(∂u·∂v) = 0` which then can be integrated twice to get the result I mentioned.
Shut up.
Hi, I have a question about the "correction" box at 5:40 (and following). I wanted to ask: is this correction really needed? I thought the way Professor Grinfeld wrote on the board was pretty convincing: if the viscousness of the medium is very high, then it overpowers the force exerted by the spring, and so the term "-w2u" becomes insignificant. So do we actually need the correction? Thanks for the amazing videos!
I think that the correction is necessary because u itself might still be high. For example, if someone manually displaced the spring substantially (u=1) and then let go. Then it would start crawling (u''=0) back to the equilibrium due to the high viscosity.
Without the correction we will need the two initial conditions and that would be awkward for the first ordder approximation.
Does motion and life as we know really need man created mathematical model such as 2nd derivatives? I think life and motion probably does need our mathematical models...
It is a happy co-incident that second order ODE's approximate reality. The ultimate laws of reality will surely not be dependent upon two prior initial conditions. Surely reality can update to a new state purely dependent upon it's current state only.
I like the second derivative because second-order differential equations are much more fun than first-order differential equations, and also still simple enough to be manageable. =P
Oh, and I see that you love the `F = m·a` equation, and you always stand in awe when writing it. Sorry to break it for you, but this is not the original Newton's formula. If you look into Newton's "Principia Mathematica", you will find that he stated the force to be a "change of motion in time" (in modern terms called "momentum"), that is, `F = dp/dt`. The formula you cherish so much is a bastardization of the original formula, made by modern physicists who then blamed Newton for it not being accurate enough when Relativity came into play. But it is actually _their_ fault that this formula is no longer valid in Relativity. Newton's original formula was valid, since it works directly on momentum, and as you probably know, energy and momentum work together in Relativity as a four-vector. The bastardized version, `F = m·a`, can be derived from Newton's original formula by assuming that the mass doesn't change (which is usually the case in non-relativistic setting):
F = dp/dt = d(m·v)/dt = m·dv/dt = m·a
But if mass is a variable that depends on time, we cannot do that - we have to use the product rule:
F = dp/dt = d(m·v)/dt = m·dv/dt + v·dm/dt = m·a + v·dm/dt
So, as you can see, we have an additional term `v·dm/dt` then besides the well-known `m·a` in that situation, and that term relates velocity with changes of mass. Sounds familiar, eh? ;J
Sci Twi uh, you do realize newton assumed mass was constant and p=mv in the principia. So F=mdv/dt=ma is completely equivalent to F=dp/dt. How is it a bastardization? It's completely correct as far as newton is concerned. So it doesn't work in SR, who cares? it was never supposed to, and the only reason you can still say F=dp/dt in SR is because p is redefined from how newton defined it.
PS in most cases it's easier to think of mass as being constant even in SR, because mass is typically thought of as the parameter for a particular particle. It's just that p=mv isn't a useful definition anymore. We define p=mv/sqrt(1-(v/c)^2) and then F=dp/dt still works as long as your careful what you mean by force, which gets a little more tricky in SR than Newtonian mechanics.
_"newton assumed mass was constant"_
Can you point out where exactly did Newton assume that mass is constant _in general_ (that is, not in some particular case where it happens to be constant)? Because from what I know, he had a different concept of inertia than we have today (he treated it as another force, _vis insita_ , that shows up when a body changes its motion).
_"F=mdv/dt=ma is completely equivalent to F=dp/dt"_
Only if mass is a constant.
_"How is it a bastardization?"_
Because it unnecessarily narrows its use and causes it to break in Relativity, then blaming Newton for that, while the blame is on modern scientists who perverted Newton's original idea.
_"So it doesn't work in SR, who cares?"_
People who use Relativity in their work? And no, it's not just about super speeds and astronomical domain - relativistic effects can be observed with low speeds as well when the forces are strong enough, as in electromagnetism. Your TV set wouldn't work properly if the relativistic effects wouldn't be taken into consideration, or the GPS you use while driving.
Long story short, some people care, if they do anything important in science. And it would work fine with Newton's original equation as well if it wasn't corrupted by modern physicists.
_"it was never supposed to"_
Many things were not supposed to do what they do today. But it is a good thing when they do, because it means that the original idea was correct and extensible. It is bad, though, when old ideas break apart when we discover new facts.
_"it's easier to think of mass as being constant even in SR, because mass is typically thought of as the parameter for a particular particle"_
You're confusing rest mass (inherent property of a particle, invariant) with relativistic mass (varies with speed).
_"It's just that p=mv isn't a useful definition anymore. We define p=mv/sqrt(1-(v/c)^2)"_
Nope, it is the MASS in this formula that needs to be redefined, NOT momentum. The `1/sqrt(1-(v/c)^2)` part comes from the relativistic correction for mass. Momentum in SR, along with energy, together make the four-momentum which is a *Lorentz invariant* .
@@scitwi9164 BTW what and where are you studying now?? :)
@@aeroscience9834 well this girl seems correct to me
@@ManojKumar-cj7oj Well a good special relativity course should remedy that. I recommend Leonard Susskind's lectures, here on youtube for free.