I think the answer is an infinite amount of time, but that just doesn't seem right. I'm guessing if you made a simulation of this, the bubble would eventually hit the wall.How long will it take for a bubble to pop?
Rigorous solution of the duffusion equation shows that the expected time for the bubble to pop is
%26lt;t%26gt; = 0.33 n^4.
SOLUTION
The bubble displaces at the distance e/n each second. Here, e is a unit vector in a random direction. The average value of e is zero, %26lt;e%26gt;=0. The average value of e squared is one, %26lt;e^2%26gt;=1. After s seconds, the buble displaces af the distance r_s=e_1/n+e_2/n + ...e_s/n. Clearly, the average value of r_s is zero, %26lt;r_s%26gt;=%26lt;e_1%26gt;/n+%26lt;e_2%26gt;/n + ...%26lt;e_s%26gt;/n=0. The average square of r_s is equal to %26lt;r_s^2%26gt; =%26lt;(e_1/n+e_2/n + ...e_s/n)(e_1/n+e_2/n + ...e_s/n)%26gt;= %26lt;e_1^2 + e_2^2 + ... e_s^2 + e_1*e_2 + ...%26gt;/n^2 = (1/n^2) (%26lt;e_1^2%26gt; + %26lt;e_2^2%26gt; + ... %26lt;e_s^2%26gt; + %26lt;e_1*e_2%26gt; + ...) . All cross-products in this expression vanish, %26lt;e_i*e_j%26gt;=0, if i is not equal to j. All squares are equal to one, %26lt;e_i^2%26gt;=1. Hence, %26lt;r_s^2%26gt;=s/n^2. If %26lt;r_s^2%26gt; is approximately equal to the squared box half-size n^2/4, then the bubble hits the wall and pops. The corresponding time is
t = n^4 / 4 =0.25 n^4.
This is what Remo Aviron has got, and this is an estimation.
More accurately, we can describe the process as a diffusion. This is also an approximation, which is valid if n is sufficiently large. The bubble distribution function f is governed by the diffusion equation,
鈭?f/鈭倀 = D 鈭嘵2 f + 未(t',x',y',z').
Here, D is the diffusion coefficient, and 未(t',x',y',z') is the Dirac delta-function representing the instantaneous point source. If d is the size of a step of the random walk, and 蟿 is the time elapsed between two successive steps, then the diffusion coefficient is equal to %26lt;d^2%26gt;/(6 蟿), which gives D=1/(6 n^2) in our case. Solution of the above equation for a box 0%26lt;x%26lt;a, 0%26lt;y%26lt;b, 0%26lt;z%26lt;c with the condition f=0 at walls is
f=(8/(a b c)) 危_i 危_j 危_k { sin( i蟺x/a) sin( i蟺x'/a)
sin( j 蟺y/b) sin( j蟺y'/b) sin( k 蟺z/c) sin( k蟺z'/c)
exp( -D蟺^2 (t-t') (i^2/a^2+j^2/b^2+k^2/c^2)) },
where i, j, k are integers.The summation is carried from 1 to infinity. This solution is known. You can derive it by taking Fourier and Laplace transforms, solving the resulting algebraic equation, and taking the inverse transforms. In the case considered, t'=0, a=b=c=n, and x'=y'=z'=n/2. Substitute and integrate f over x,y,z, (only odd terms survive) to get the probability that the bubble is in the system at time t:
F(t) = 鈭?f dx dy dz =
(64/蟺^3) 危_i 危_j 危_k { (-1)^(i+j+k+1)/( (2i-1) (2j-1) (2k-1)) }
exp( -D蟺^2 t ((2i-1)^2+(2j-1)^2+(2k-1)^2)/n^2) }.
This equation describes a multi-exponential decay. Note, that
危_k {(-1)^(k+1)/k} = 1-1/3+1/5 ... = 蟺/4.
Using this relation you can check that F(0)=1. The probability that the bubble pops at time t is equal to -dF(t)/dt. The expected life time is
%26lt;t%26gt; = -鈭?t (dF/dt) dt = C (n^2/D).
Here,
C = (64/蟺^5) 危_i 危_j 危_k
{ (-1)^(i+j+k+1)/((2i-1) (2j-1) (2k-1) ((2i-1)^2+(2j-1)^2+(2k-1)^2)) }
is a factor. The series converges rapidly. You can calculate it more accurately, and I take few first terms where each of i,j,k is either 1 or 2:
1/3-3/(3*11)+3/(9*19)-1/(27*27)=0.26.
This gives C = 0.054, and
%26lt;t%26gt; = 0.33 n^4.
The rigorous solution is slightly higher than the estimation n^4/4. There is no qualitative difference.
ADDENDUM (corrected)
In spherical coordinates, each step is (未x,未y,未z)= (r cos(胃) sin(蠁), r sin(胃) sin(蠁), r cos(蠁)) with constant r=1/n. We need to generate random and uniform solid angle, d惟=sin(蠁) d胃 d蠁. Hence, angle 胃 should be distributed uniformly from 0 to 2蟺. Angle 蠁 is distributed with the weight sin(蠁) from 0 to 蟺. I have described the method to generate random functions with nonuniform weights in
http://answers.yahoo.com/question/index;鈥?/a>
In our case, the cumulative distribution for 蠁 is
(1-cos(蠁))/2. The inverse is 蠁=arccos(1-2z), where z is distributed in [0,1].
If you use Matlab, then 胃=2蟺*rand(1) , 蠁=arccos(1-2rand(1)). Looks the same as to distribute the vertical coordinate according to r(1-2 rand()).
20 minutesHow long will it take for a bubble to pop?
About 19 minutes.
It can go on for an infinite amount of time because this box is nxnxn. it can go in any direction for a really long time until it hits the walls and pops. doesn't mean it just goes in the x direction until it pops but it can go in any direction.
http://answers.yahoo.com/question/index;鈥?/a>How long will it take for a bubble to pop?
Let f(x,y,z,t)=the probability density of the event "the bubble is at position (x,y,z) at time t." Then
f(x,y,z,t)
=int(f(x-1/n*cos(a)*cos(b),
y-1/n*cos(a)*sin(b),
z-1/n*sin(a),
t-1)*1/(4*pi)*dadb
Use initial condition
f(1/n*cos(a)*cos(b),
1/n*cos(a)*sin(b),
1/n*sin(a),
1)
=1/(4*pi)
for all a in [-pi/2,pi/2] and b in [0,2*pi), and =0 everywhere else.
The density function satisfying these conditions gives the location probability density for each integer value of time t. In turn, the mean time to contact with the n by n by n boundary can be expressed as an integral involving f.
i think you meant for the bubble to choose (not necessarily change) a direction per second. because i think random means the bubble will go in any direction with magnitude 1/n, and the locus of its resulting location at the end of a second is a sphere, it will have something to do with volume / surface area of sphere and nxnxn cube.
Use diffusion or random walk theory to get the average distance. When the average distance is n/2, it hits.
See: Random walk at Wikipedia. Look at the Brownian motion section (Note: Y!A is f.ing the citations again)
(n/2)^2 = (1/n)^2 *t
t= (n^4)/4
*Note: I made an edit because I forgot this is 3D
*Note, the (t/n) factor is based on a dispersion rate of 1/n -- did you really mean 1 unit per second?
*Note, there might be some error associated with your using a box rather than a sphere of radius n.
Brownian motion, after t units of time, will have covered sqrt(t) units of distance (up to some constant multiplier relating the choice of units).
so to travels n^2 steps will take n units of time. (wait, this is assuming a sphere. To solve, you have to take into account the variable distance-to-wall depending on the direction.
Hmm...I'm uncertain about exactly how your random movement should work (as others have mentioned). For simplicity I'll discretize it and say that the bubble will make one of 7 choices: stay where it is, move left, right, up, down, back, or forward; all with equal probability.
If this is the case, we can project the movements into a single axis, since we'll only be able to run into one wall, and each of our axes are indistinguishable from one another. So, focusing on say the x-axis, we have for each step,
P(-1/n) = P(+1/n) = 1/7
P(0) = 5/7.
In technical terms, we're looking at an application of homogeneous Markov chains, at least sort of. I haven't figured out quite how to make the solution fall out, but I think I should at least see if my discrete model is acceptable, or if you would rather see a properly constructed parameterized function such that the total derivative wrt time is 1/n.
A couple of notes based on my intuition: I think that any simulation would certainly end in the destruction of the bubble; however, theoretically the bubble could live on forever. The real question is what is the probability remaining for such a situation (after subtracting all other probabilities of death). When I get home to my own computer I may put together a few simulations and/or excel formulas.
EDIT (immediately after posting): holdm is right that Brownian motion is more accurate for what you want. If we want to get some sort of exact result though I think we'll need some sort of simplification.
Around 17 minutes
depends upon the surface tension of the liquid..
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