0
$\begingroup$

I am using a numerical integrator with time step $\Delta t$ to model the motion of a particle. I want to say at each step there is an uncertainty in the particles velocity $v\pm\alpha$ and treat this like sampling from a velocity distribution at each step using normally distributed random variables.

x = (v + A*random.normal()/sqrt(dt)) * dt

So I have $\alpha_s=A\frac{\xi}{\sqrt{\Delta t}}$, where $\xi$ is a normally sampled random variable from a distribution with width one. I have seen scaling by the time step used when simulating other noise processes although my understanding of this is poor. My question is that for at time step of $\Delta t=0.01s$ and modelling an uncertainty of say $v=2m/s$ what should $A$ be?

The scaling of the noise process with the time step is causing me difficulty here.

Improve this question
$\endgroup$
4
  • $\begingroup$ If you are using standard uncertainty (probability of the true value falling within the uncertainty range being roughly 68.3%) and want it to be equal to 2, then your standard deviation should be equal to your wanted uncertainty (for a normal distribution), that is, simply use $\alpha_s=A{\xi}$ with $A = 2$ and no time step scaling. You scaling looks BTW questionable as the smaller the time step, the higher the deviation, which you surely don't want to have. $\endgroup$
    – user130529
    Commented Jan 9, 2017 at 13:56
  • $\begingroup$ I have an easier time following your way of thought here. However running the simulation without the scaling gives different behavior depending on the time step (eg. the mean square displacement changes depending on the value of dt), which shouldn't be the case. $\endgroup$
    – user12800
    Commented Jan 9, 2017 at 14:55
  • $\begingroup$ And the mean square displacement doesn't change when you divide by $\Delta t$? This is strange, I would expect the contrary (imagine what happens to your velocity when $\Delta t=10^{-20}$), and also that it would rather not change if you multiply by $\sqrt{\Delta t}$ (knowing that the deviation for $n$ steps behaves like $\sqrt{n}$). $\endgroup$
    – user130529
    Commented Jan 9, 2017 at 15:24
  • 1
    $\begingroup$ BTW, why don't you compute your numerical integration without noise and add the noise after? You can add it in such a way to have any desired velocity deviation, and you avoid the accumulation in the scheme. $\endgroup$
    – user130529
    Commented Jan 9, 2017 at 15:54

1 Answer 1

Reset to default
1
$\begingroup$

You do not have to rescale A but I think $x$ in your equation is probably your differential increment $dx$ such that your equation reads $$ dx = vdt + A dW $$ Where $dW$ is a Wiener increment which is gaussian with a zero mean and variance, $ \sigma^2 = dt $. Rewritting in terms of a unit variance gaussian gives $ dW = \sigma N(0,1) $ which is the equation you have written. As to why $dW$ should have $\sigma^2 = dt$ you can have a look at this nice answer Partition function for Gaussian white noise .

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.