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Newton yasalarina gore Reaksiyonlar esittitir dolayisi ile Navier slip conditiona gecer sistem ve trenin tekerlikleklerindeki rulmanlar parcalanana kadar Trenin hizina ve carpma siddetine bagli... tabi isik hizinda carparsa Molecular dynamics geregi atomlari birlesir ve duvarla butun olur. Aslında bu bir paradokstur... 
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https://forum.donanimhaber.com/showTopic.asp?m=100639563
Konu çakma. Konu ban uye kilit forum kapanabilir
< Bu ileti mini sürüm kullanılarak atıldı > -
quote:
Orijinalden alıntı: LeeOrAfk
Hiç durmayacak bir duvar ile hiç yıkılmayacak bir tren çarpışırsa ne olur ?
peki herşeye gücü yeten Tanrı;kaldıramayacağı büyüklükte bir taş yaratabilir mi? sen bunu cevapla ben sana 100.000 vereyim..
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quote:
Orijinalden alıntı: ceoolacakbeyin
varsayımlarla harekeket ediyorsun, hiç durmayacak tren ve hiç yıkılmayacak duvar olmadığına göre bende varsayım olarak tren patlar diyorum, selamun aleykum
as -
BenjaminButtonn
kullanıcısına yanıt
Evet. Hesap numaramı pm attım
< Bu ileti mobil sürüm kullanılarak atıldı > -
quote:
Orijinalden alıntı: LeeOrAfk
Evet. Hesap numaramı pm attım
yani tanrı o taşı yapabilir ve kaldıramaz öyle mi? -
Hic durmayacak bir duvar mi olur yikilan tren mi olur konu kilit üye ban
< Bu ileti mobil sürüm kullanılarak atıldı > -
quote:
Orijinalden alıntı: BenjaminButtonn
quote:
Orijinalden alıntı: LeeOrAfk
Hiç durmayacak bir duvar ile hiç yıkılmayacak bir tren çarpışırsa ne olur ?
peki herşeye gücü yeten Tanrı;kaldıramayacağı büyüklükte bir taş yaratabilir mi? sen bunu cevapla ben sana 100.000 vereyim..
Tanrı mantıksız bir şey yapmaz, ayrıca tanrı zamanın dışında bir varlıktır, yani onu zaman kavramı içerisinde düşünemezsin. 100000 tlyi pm at
< Bu ileti mobil sürüm kullanılarak atıldı >
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Verildi. Üye ban konu kilit forum silinsin.
< Bu ileti mobil sürüm kullanılarak atıldı > -
Tren duvarı sürükler ve hızı kesilmez 
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işte cevap
Note: I'm using units where \hbar = 1.
A second note: This might be the craziest thing I've ever written.
A third edit (October 17): There were several major problems and places were I was fast and loose in this answer. I've prepared a much more thorough, rigorous, and correct answer (I hope). Rather than change this answer completely and erase my mistakes, I've made this my first blog post:
Addenda and Errata
TL;DR: It depends on the how proportionally infinite the force and stability of the object are, and on what order you take the limit where these things go to infinity.
The common answer to this problem is that both unstoppable forces and immovable objects are imaginary, and so it is impossible to answer the question. However, this is a very classical mindset...
If we have an immovable object, it is clearly at the bottom of some infinite potential well. So, perhaps we can choose an appropriate model potential to represent this situation. The first that comes to mind is
V(x) = - \alpha \delta(x)
Since the delta function has units of 1/[x] , though, this means that \alpha has units of [E][x] . So, we can say \alpha = vL , where L is some length scale describing the area over which the force acts. This model doesn't actually represent an infinitely deep well, since \alpha sets the potential and it is a finite quantity. So, let's be a little crazy and say
V(x) = \lim_{v \to \infty} \left( - vL \delta(x) \right)
It won't make sense to do this in the middle of the problem; instead we'll take the limit at the end (when I'll calculate a transition rate after the particle is hit with an unstoppable force) and hope it yields a finite quantity.
What about an unstoppable force? Well, a force of infinite magnitude in the positive x direction would suffice:
F = \beta \delta(x) \hat x
Again, since the Dirac function has units of 1/[x] , \beta must have units of [F][x] = [E] . So, we can say \beta = f L , where L is the same length scale as before. Moreover, we want to describe a collision (i.e. a scattering event) where the force comes and hits the stationary object at time t = 0. So, we can add in a time dependence term \theta(t) . So, overall the force is
F = \lim_{f \to \infty} \left( fL \delta(x) \theta(t) \right)
which is the gradient of the scattering potential
W(x) = \lim_{f \to \infty} \left( - fL \theta(x) \theta(t) \right)
So, here is my plan of attack: I'll find the probability of a transition from the ground state after a perturbation representing the force in the case that the well depth and force are finite in strength. Then, I'll take the limit where both go to infinity to see if this makes sense.
Incidentally, to continue we'll need to calculate the initial state for times t < 0, which is the ground state of the Hamiltonian
H = - \frac{1}{2m} \nabla^2 - vL \delta(x)
The ground state wavefunction \psi(x,t) satisfies H \psi = E_0. Away from x=0 in either direction this is just the free Hamiltonian. Now, we're looking for a bound ground state, so the wavefunction should die off at either infinity. Moreover, there's a basic principle in QM that given a symmetric wavefunction and asymmetric wavefunction, the symmetric wavefunction is lower in energy (doesn't cros x axis -> less curvature -> less energy). So, the wavefunction must look like Ae^{-\lambda |x|} for x \neq 0 (and by continuity must take the value A at x = 0). Note that the energy of the ground state is related to the parameter lambda, but we don't need it so I won't calculate it. (Incidentally its negative, which doesn't matter). Now we can integrate around x = 0 to relate these arbitrary coefficients, getting
-\frac{1}{2m} \left( \psi'(\epsilon,t) - \psi'(-\epsilon,t) \right) - vL \psi(0,t) = 0
Substituting in the forms we found, this means that
\frac{1}{2m} \, 2 A \lambda - vL \, A = 0
or
\lambda = mvL
A is just a constant to make the norm of the wavefunction equal to 1. Sadly I'll probably need it. You can check that
A = \sqrt{\lambda} = \sqrt{mvL}
Now we need to know the possible final states for t > 0. This means finding the solutions to the time independent Schrodinger equation with the Hamiltonian
H = - \frac{1}{2m} \nabla^2 - vL \delta(x) - fL \theta(x)
Actually we won't need all of them - just the bound states, but including the non-ground-state ones. Actually, I think there's only one bound state.This can be solved much the same as the previous. For x < 0, the wavefunction must look like B e^{-\lambda_1 |x|} again. For x > 0, the wavefunction must look like B e^{-\lambda_2 |x|}, where the parameters lambda and lambda' are related to the energy of the state:
E = - \frac{1}{2m} \lambda_1^2 = - \frac{1}{2m} \lambda-2^2 - fL
and thus
\lambda_1^2 - \lambda_2^2 = (\lambda_1 + \lambda_2)(\lambda_1 - \lambda_2) = 2mfL
Then integrating over a small region around x = 0, we again have
-\frac{1}{2m} \left( \psi'(\epsilon,t) - \psi'(-\epsilon,t) \right) - vL \psi(0,t) = 0
Substituting in the above forms,
\lambda_1 + \lambda_2 = 2mvL
and solving these two equations leaves, fascinatingly,
\lambda_1 - \lambda_2 = \frac{f}{v}
and hence
\lambda_1 = \lambda + \frac{f}{2v}
\lambda_2 = \lambda - \frac{f}{2v}
and
\lambda_1 \lambda_2 = \lambda^2 - \left( \frac{f}{2v} \right)^2
Already you can see that the limit in which f and v are taken to infinity will have an effect on the answer. If one is taken before the other, the answer could blow up or be killed, but if the limits are taken at the same time, so that f/v has a sensible finite value, then there could be a finite answer!
Another thing can can be seen right away - f/2V must be less than lambda, or this wavefunction will be unbound. In a sense, this puts an absolute condition by which if the force is strong enough, it always pushes the particle out of the potential well. Of course, what will this condition mean once we put the limits back? More on this later ...
Anyways, we'll sadly need the normalization constant. (It's great when you don't). Working through it, its
B = \sqrt{\frac{2 \lambda_1 \lambda_2}{\lambda_1+\lambda_2}} = \frac{1}{A} \, \sqrt{\lambda^2 - \left(\frac{f}{2v}\right)^2}
Now, finally, onto the solution. I'll use the sudden perturbation approximation - since the Hamiltonian changes nearly instantly, the wavefunction doesn't have time to evolve. So, the amplitude that it stays in this bound state is given by simply by the inner product (i.e. overlap) of the two states. This amplitude squared is the probability that the particle stays bound, and clearly 1 minus this is the probability that the force frees it into an unbound state.
So, we must calculate this amplitude. It is
AB \left( \int_{-\infty}^0 e^{(\lambda + \lambda_1)x}dx + \int_0^{\infty} e^{-(\lambda + \lambda_2)x}dx \right)
which becomes
\sqrt{\lambda^2 - \left(\frac{f}{2v}\right)^2} \left( \frac{\lambda}{\lambda^2 - \left( \frac{f}{4v} \right)^2} \right)
So, the probability that the particle remains in a bound state is ... drum roll ...
P = \lambda^2 \left( \frac{\lambda^2 - \left( f/2v \right)^2}{\left( \lambda^2 - \left( f/4v \right)^2 \right)^2} \right)
Recall again that \lambda is equal to mvL, so as v approaches infinity so does lambda.
The immovable object wins (i.e. the particle stays in a bound state) if P -> 1; the force wins (i.e. the particle is perturbed into a free propagating state) if P -> 0.
1) If f is taken to infinity before v, the irresistable force always wins
2) If v is taken to infinity before f, the immovable object always wins.
Now, let's take the limit where f and v both approach infinity, but their ratio remains constant. But then, the above expression approaches 1!
3) If f, v both approach infinity but their ratio remains constant, the immovable object wins.
Of course, you could let f to go infinity first, without fixing the ratio f/(2v), and then the irresistable force will win, or equivalently letting v go to infinity first, letting the object win, but neither of these seem fair. But even in the middle ground, the immovable object seems to win.
Except ...
Remember that condition above? If f > 2v lambda - that is, since lambda = mvL, f > 2mL v^2, then there is no bound state after turning on the irresistable force and the irresistable force must always win. Note that this is equivalent to alpha going to 0. No matter how big v is, f is bigger than the square of v. Really, fixing the ratio f/(2v) - insisting that f and v both go to infinity at the same rate - isn't the only possibility. f and v^2 could go to infinity at the same rate, and in this case there is a cutoff condition.
4) If f, v both approach infinity but the ratio Beta = f/v^2 / (2mL) (where m is the pass of the particle and L is the width of the potential region) remains constant, two things can happen. A) if Beta > 1, the irresistable force wins. But when B) Beta < 1, the immovable object has a probability
P = \frac{1 - \beta^2}{\left( 1 - (\beta/2)^2 \right)^2}
of winning!
If you plot this for beta between 0 and 1, you'll see that this probability is 1 for beta = 0, and goes down to 0 for beta = 1, beyond which the immovable object always loses and the irresistable force always wins.
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Kardeş gece gece türkçe olsa üşenirim bunu Okumaya. Soruyu tekrar oku ama
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