The Youtube link keeps breaking, so here is a search on Youtube for Brian Cox' A Night with the Stars lecture. Pause the video on 40.32minutes.

What you see he said is called Feynman's Path Integral.


Am I right in thinking this adding all the different paths a particle can take predicting the probability of it landing in a certain position?

Once that question is answered how in the world did he simplify it to:

$t > \dfrac{x \Delta{x} m}{ h} $--- (42.39minutes)

This second equation he got from simplifying was what is used to "predict how long it would take" for his "diamond to jump out of the box", how fascinating.

So: what is all this in Feynman's Path Integral? and how did he simplify it to get the other equation? and if you know, what is this second equation called?

  • $\begingroup$ 19.12.21: It seems the BBC2 video is gone from YouTube. Presumably it was broadcasted on Sunday 18 Dec 2011 @ 21:00. $\endgroup$
    – Qmechanic
    Dec 19, 2021 at 11:32

1 Answer 1


Well, first of all the "action" function $S(q",q',T)$ that appears in his formula is given by $ S = \int dt\left( \dfrac{1}{2} m v^2 -U\right)$, where the integral is taken over the path in question. $U$ is a potential term which we will ignore. For the "classical" path which goes uniformly from one point to the other, we have $v = \dfrac{\Delta x}{\Delta t}$ and so you get $S \propto m \left(\dfrac{\Delta x}{\Delta t}\right)^2\Delta t=m\dfrac{(\Delta x)^2}{\Delta t}$.

Now, since $S/h$ appears as a complex phase term, we want it to be small so that neighboring paths do not destructively interfere (cancel each other out). So if we set $S/h < 1$, we get $\Delta t > \dfrac{m(\Delta x)^2}{h}$. Now in his formula it wasn't really clear what the difference was between $x$ and $\Delta x$, he used similar values for each, and in any event this is just a very crude order of magnitude type estimation.

  • $\begingroup$ Thank you, that was well described and I'm pretty near to understanding now. I think that x = displacement and $\Delta x$ = uncertainty in position (I THINK) $\endgroup$
    – ODP
    Mar 14, 2012 at 20:57
  • $\begingroup$ i just noticed the final formula you got was $\Delta t > \dfrac{m(\Delta x)^2}{h}$ but this is not the same as what I stated in my question, or is it? $\endgroup$
    – ODP
    Mar 24, 2012 at 17:46
  • $\begingroup$ Not exactly the same. It wasn't exactly clear what he meant by "x" vs. "delta x", and in fact he used values that were nearly the same for both, so both formulas would give nearly the same answer. $\endgroup$
    – user1631
    Mar 26, 2012 at 16:14
  • $\begingroup$ is '$S \propto m \left(\dfrac{\Delta x}{\Delta t}\right)^2\Delta t=m\dfrac{(\Delta x)^2}{\Delta t}$' all one equation or did you mean that $S \propto m \left(\dfrac{\Delta x}{\Delta t}\right)^2$ is the same as $\Delta t=m\dfrac{(\Delta x)^2}{\Delta t}$ ??? I'm guessing it's all one equation but just to be sure $\endgroup$
    – ODP
    Mar 29, 2012 at 16:58
  • $\begingroup$ physics.stackexchange.com/q/22998 can you answer this completely related question please? $\endgroup$
    – ODP
    Mar 29, 2012 at 19:06

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