I have the velocity function of an atom, $v(x,t)$, which changes with time and space. I'm looking for a general relationship for finding the location of the atom at time $t_s$. The atom start to travel from $x=0$ at $t=t_0$. I'm confused since the velocity is a function of space :-?

  • Why would you to treat space $x$ and time $t$ as two independent variables for the velocity? Makes no sense to me – lemon Nov 12 '17 at 12:17
  • @lemon can you elaborate more in form of an answer. I get your point but not fully. I mean I could not manage to write what I have in my mind in mathematical expressions :/ Any help is appreciated. – Dolle Nov 12 '17 at 12:20
  • @lemon It seems OP is considering a continuous fluid, and asking themselves about the trajectory $\gamma(t)$ with $\gamma(0)=0$ and $\dot\gamma(t)=v(\gamma(t),t)$. If I understood it correctly, this is a rather standard excise. – AccidentalFourierTransform Nov 12 '17 at 13:06

The general relationship would be $$ x(t_s) = \int_0^{t_s} v(x(t),t)\,dt $$ which follows from the definition of velocity.

Ideally you would want to eliminate the $x$ dependence in your velocity function so that $v$ is a function of time only. However, if that's not possible, then the above equation describes an implicit equation in $x$ which you would probably have to solve numerically.

  • Thanks. Let me share with you a little bit more details. As you said the velocity can be written in terms of time. However, my velocity itself is coming from atomic flux ($J$) basically: $v(x,t)=\Omega J(x,t)$ where $\Omega$ is atom density. In this relationship time and space dependency is not very visible to me. I hope I explained well. Can you comment on this? I'm trying to offer a mathematical model for the phenomenon keeping the physical meaning visible in the math. – Dolle Nov 12 '17 at 12:32
  • @Dolle I have updated my answer slightly. Is it practical for you to solve your problem numerically? – lemon Nov 12 '17 at 12:41

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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