@Claudio You should have chosen to write $\rho^{\ell+1}G(\rho)$ for the 2nd case, so that then we can tell you that, if you compute the full set of derivatives, you will obtain a differential equation in $F(\rho)$ alone, that is the associated Laguerre differential equation. i.e. It is $F$ and not $G$.
panpsychism gets ridiculed by physicists but is relatively more favored among philosophers
There are two types of panpsychism : proto panpsychism and non proto panpsychism. in proto panpsychism, everything possesses a thing called proto consciousness, which is a primitive constituent of the consciousness that we know
As Wikipedia says:
[...] the kinetic energy of a non-rotating object of mass $m$ traveling at a speed $v$ is $\frac{1}{2}mv^2$.
Why does this not increase linearly with speed? Why does it take so much more energy to go from $1\ \mathrm{m/s}$ to $2\ \mathrm{m/s}$ than it does to go from $0\ ...
isen't this video first two graphs wrong? youtube.com/… since the velocity and the acceleration when the rocket hits the ground at second 15 should be 0 right?
@naturallyInconsistent The answer says "$E(m,v)$, if it is invariant, must be proportional to the mass, because you can smack two clay balls side by side and get twice the heating, so $E(m,v) = m E(v)$". Why should $E(m,v)$ be invariant? And why $E(m,v)$ has to be proportional to the mass if it is invariant?
@Bml Again, the answer already answered all of those stuff. The laws of physics must be Galilean invariant in non-relativistic physics. To make the laws of physics Galilean invariant, the quantities of interest are all Galilean covariant or invariant. The double-clay-ball argument shows that it has to be proportional to mass.
how would you show on a. graph that an object starts to the left of the origin , moves to the right at a constant speed and ends on the origin in a position vs time graph (position only axis)ù
@naturallyInconsistent I don't think it is so obvious. The following answer by Jan Lalinsky says: "There is no apparent reason to believe loss of $\sum_k E(m_k,v_k)$ after a colission among bodies $k$, $E(m_k,v_k)$ being heat that could be extracted from the colission of the body $k$ with heavy stationary wall, is Galilei-invariant. There is no obvious way to transform the energy loss that happens in a colission (generated heat) to another frame using Galilei transformations."
This answer to a question about why Newtonian kinetic energy is quadratic in velocity shows that if an inelastic collision's KE loss is invariant under Newtonian boosts it has to quadruple when velocity doubles. A simple calculation shows that the famous $\tfrac12mv^2$ formula implies invariance ...
@Bml Your complaint is even more obvious than that: if the universe is so obviously Galilean, then there would not be an option to make it Lorentz invariant instead. Whether to consider Galilean invariance or to use experiments to justify that it is Galilean invariant, is not actually in any disagreement---in the end, the correctness is only justified by comparison with experiment in either case.
I'm following my notes and Cohen-T, I have $y_{k,l} = \rho^{l+1}\sum_{q = 0}^{\infty}c_q \rho^{q}$ and the equation $$\left\{d^{2}_{\rho \rho}-2\lambda_{k,l}d_{\rho}+\left[\frac{2}{\rho}-\frac{l(l1+1)}{\rho^2}\right]\right\}y_{kl}(\rho) = 0$$
I got this and went on
didn't find anything wrong
but at the end I came to a problem
the recurrence relation one obtains from this is $$\frac{c_{q+1}}{c_q} = \frac{2[\lambda(q+l+1)]-1}{(q+1)(q+2l+2)}$$
this leads to the correct increasing exp behavior for q to infinity, nevertheless when you impose that the series must be truncated for an integer $k$, you get $\lambda_{k,l} = 1/(k+l+1)$
the problem is that $k$ must be $ge 1$ because the first term of the series $c_0$ for $q = 0$ is necessarily $\ne 0$ otherwise I don't get the correct behavior for $\rho \to 0$
then I realized that the first term in $$\rho^{l-1} \sum_{q = 0}^{\infty}c_q[(l+q+1)(l+q)-l(l+1)]\rho^{q}$$ is zero
I completely forgot about this, and actually used this fact for redefining the sum later since I rewrote as $\rho^{l}\sum_{q = 1}c_q[.......]$
and then called $q' = q-1$ and renamed $q' = q$ again so that both sums start from 0 and depend upon $\rho^{l}$
the problem is that both Cohen and my professor obtain the following recursion relation $c_q[(l+q+1)(l+q)-l(l+1)] = [2\lambda(l+q)-2]c_{q-1}$
@Bml In a sense, yes. Physics is an empirical science. There is no such thing as purely thinking about a thing, and getting results. The choice of suitable postulates, is itself highly experimentally suggested.
@Claudio The correct thing to do is to entirely avoid series expansion. Import the fact that if the Laguerre ODE solution is not terminated into a polynomial, then it grows exponentially, so it cannot be normalised. That is, the condition of needing to be normalisable then constraints the solutions to be polynomials, and this, in turn, quantises the solutions.
I guess it's not terrible, the only point that startles me is how do you go from $c_q[-2\lambda(q+l+1)+2]\rho^{q+l}$ to $c_{q-1}[-2\lambda(q+l)+2]\rho^{q+l}$
you can't tell me u can rename indices to get that, it's not possible
therefore it's crap
Ill stick to townsend
@naturallyInconsistent you might need to insert this in the sun tzu book for phycisists
@Claudio Yes, this is definitely nonsense. However, it is not clear which part is the mistaken bit, and so you will have had to go on a wild goose chase.
If, instead, you studied Fuch's theorem separately, associated Laguerre DE as an offshoot of that, then you can just import the result into H atom. There would be less confusion because the series solution would be worked in the least confusing notation
And indeed, if you had some Sturm-Liouville before Fuch's series, it will be even better.
@naturallyInconsistent OK, thanks. There are other things I didn't get. The answerer says "Further, if you smack two identical clay balls of mass $m$ moving with velocity $v$ head-on into each other, both balls stop, by symmetry. The result is that each acts as a wall for the other, and you must get an amount of heating equal to $2mE(v)$". Why? What are the equations leading to the result $2mE(v)$?
@Bml What are you even asking about? It was already explained that the heating made by two balls is twice that of one ball. It is necessary that two balls slamming into wall is the same thing as two balls slamming into each other.
"It is necessary that two balls slamming into wall is the same thing as two balls slamming into each other": this is not so obvious to me. However, the answerer says: "But now look at this in a train which is moving along with one of the balls before the collision. In this frame of reference, the first ball starts out stopped, the second ball hits it at $2v$, **and the two-ball stuck system ends up moving with velocity $v$**". Probably, the train is moving with velocity $v$, so the velocity of one of the two balls is $v+v = 2v$, whereas the other ball is moving on the opposite direction, so…
@Bml Because in the original viewpoint the system of two balls come to a stop. A stopped thing in the original viewpoint is a thing that is moving with velocity $v$ to the train
@VincentThacker @naturallyInconsistent said that "is necessary that two balls slamming into wall is the same thing as two balls slamming into each other": it sounds as an elastic collision to me. Why is it inelastic?
AFAIK, the LHS is the kinetic energy of the moving ball in the moving system of the train (velocity v): since the ball has velocity v, in the inertial frame it has velocity $v_{ball} + v_{train} = v + v = 2v$. The other ball has velocity $-v_{ball} + v_{train} = - v + v = 0$, so its kinetic energy is $m E(0) = 0$. Correct?
@VincentThacker OK. The problem arises with the RHS. The two terms $2mE(v)$ are the same in magnitude, but come from two different viewpoints. I can't grasp what is the difference between this two terms physically: reference system, type of collision, etc... Could you explain?
@VincentThacker 1) Why do we have to sum the heat dissipated and the kinetic energy of the combined mass and equate them to the initial kinetic energy in the moving frame? 2) I don't understand why the heat dissipated is 2mE(v).
@VincentThacker OK, so the answer to 1) is: by conservation of energy, "initial kinetic energy of single masses in the moving system"($mE(2v) + mE(0) = mE(2v)$ ) equals "final kinetic energy of the double-mass system moving at $v$ ($2mE(v)) + heat dissipated (for some reason, $2mE(v)$)." But if there is dissipated heat, how can energy be conserved?
@VincentThacker Mhm. In Introductory Physics courses, I was taught that dissipated heat/energy = final energy (just after the inelastic collision) - initial energy (just before the inelastic collision), but the total energy is not conserved. A part of it will turn into heat/thermal energy. The problem is that the equation is the same, but we have different perspectives: you say energy is conserved, I say it is not conserved.
I cannot understand how, if there is heat dissipated, energy can be conserved. What I was taught is that in an inelastic collision, energy is not conserved, so I am very confused.
"in an inelastic collision, energy is not conserved" This is not true. It is the kinetic energy that is not conserved. If you learn special relativity you will know that energy is always conserved in collisions.
To reject this means to deny that thermal energy is energy.
@VincentThacker This is interesting. My professors didn't underline this. Do you have some interesting references on this specific issue, even here on Physics SE?
Also see chapter 12 of Introduction to Classical Mechanics by Morin
@Bml This is something that is often misstated. In an inelastic collision, it is the kinetic energy that is not conserved. However too often the term "kinetic" is omitted for whatever reason which completely butchers the statement.
The answerer says: "if you smack two identical clay balls of mass $m$ moving with velocity $v$ head-on into each other **[inelastic collision, right?]** both balls stop, by symmetry **[what does It mean that both balls stop, by symmetry?]**. The result is that each acts as a wall for the other **[what does it mean that each acts as a wall for the other if the collision is inelastic?]**, and you must get an amount of heating equal to $2mE(v)$ **[where does this term come from? What are the mathematical equations?]**."
@VincentThacker I made the following rationale: before the collision, the kinetic energy of each mass is $mE(v)$, so the total kinetic energy before the collision is $2mE(v)$. After the collision, the kinetic energy of the system of double-mass is $mE(v_{cm})$, where $v_{cm}$ is the velocity of the centre of mass of the double-mass system. By definition, $v_cm= \frac{mv - mv}{m+m} = 0$, so m E(v_{cm}) = m E(0) = 0$.
Since the dissipated heat is equal to initial kinetic energy minus final kinetic energy, then $2mE(v) - 0 = 2mE(v)$. Right?
@VincentThacker OK, so: what is the actual difference between the term of dissipated heat and the term of kinetic final energy, given that they have the same magnitude? It seems the term of dissipated heat is calculated in the lab frame (inertial frame), whereas the term of final kinetic energy is calculated in the moving frame (train's frame). Isn't this a contradiction?
@VincentThacker Yes, I realized that after writing my reply, but it is not obvious to me why the term of dissipated heat is invariant in train's frame, i.e. is not frame-dependent. Why?
@VincentThacker But in fact, it is not invariant...
@VincentThacker I obtain that the term of dissipated heat in train's frame should be $mE(2v) - 2mE(v)$. By conservation of energy: $mE(2v) (initial KE) = mE(2v) - 2mE(v) (heat dissipated) + 2 mE(v) (final KE)$, which is a trivial equation.
I wonder if it is not necessary and unequivocal to assume that the heat dissipation term is invariant in the two reference systems to prove that kinetic energy is proportional to the square of velocity. Better, I wonder if this is not a necessary axiom for Galilean invariance. If so, why?
@VincentThacker OK, but why? What is the reason why we have to assume the heat dissipated as $2mE(v)$ if the velocities of the two masses are not $v$ and $-v$, but $2v$ and $0$?
[this answer]{physics.stackexchange.com/a/827174/373828} is correct. I did the computations myself and got the same results. This type of exercise is really fun :P I I'm still wondering why you write n+1 instead of n??? My guess is that you want to match the parity of the trig functions to the parity of the eigenfunction labels lol
@VincentThacker I would state it differently. The whole point is that the dependence of the kinetic energy on the velocity squared is entirely due to the assumption of Galilei-invariance.
@VincentThacker An answer said "There is no apparent reason to believe loss of $\sum_k E(m_k,v_k)$ after a colission among bodies $k$, $E(m_k,v_k)$ being heat that could be extracted from the colission of the body $k$ with heavy stationary wall, is Galilei-invariant.
There is no obvious way to transform the energy loss that happens in a colission (generated heat) to another frame using Galilei transformations", and this is true, because in special relativity we can't assume this invariance for the heat loss, but if we assume Galilei-invariance upstream, we motivate kinetic energy as a special case of special relativity and at the same time as a separate entity in classical mechanics.
As Wikipedia says:
[...] the kinetic energy of a non-rotating object of mass $m$ traveling at a speed $v$ is $\frac{1}{2}mv^2$.
Why does this not increase linearly with speed? Why does it take so much more energy to go from $1\ \mathrm{m/s}$ to $2\ \mathrm{m/s}$ than it does to go from $0\ ...
