2
$\begingroup$

I was thinking about Taylor's theorem and how if a function $f(x)$ is analytic at a point $a$ and one can measure all the derivatives at $a$, $f^{(n)}(a)$ then one knows the complete behaviour of the function within the radius of convergence (and maybe, by analytic continuation, everywhere except at singularities).

A blind man standing in a landscape whose height was described by an analytic function could, in principle, by locally measuring derivatives of the height function, know the complete shape of the landscape within a certain radius.

But suppose $x$ is space and $f$ is some physical observable and you are able to locally measure its derivatives. Then under the physical hypothesis that $f$ varies analytically, you should immediately know how $f$ behaves at other points in space in a way that seems only constrained by how many (and how quickly) you can measure these local derivatives and not the speed of light. In this way you would seem to "know" about events before light from those events had reached you. What is wrong with this argument?

$\endgroup$

closed as off-topic by Norbert Schuch, Jon Custer, glS, AccidentalFourierTransform, ZeroTheHero Aug 25 '18 at 22:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "We deal with mainstream physics here. Questions about the general correctness of unpublished personal theories are off topic, although specific questions evaluating new theories in the context of established science are usually allowed. For more information, see Is non mainstream physics appropriate for this site?." – Norbert Schuch, Jon Custer, glS, AccidentalFourierTransform, ZeroTheHero
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ There's no faster-than-light (FTL) communication because no information/matter is being transported FTL. $\endgroup$ – Avantgarde Aug 17 '18 at 3:48
  • $\begingroup$ @Avantgarde : Does not address the concern raised. Effectively asserts its own conclusion - no FTL communication because no FTL communication ("information transport" is the definition of "communication" in a physical context). $\endgroup$ – The_Sympathizer Aug 17 '18 at 4:19
  • $\begingroup$ @The_Sympathizer Yes, that was my point. $\endgroup$ – Avantgarde Aug 17 '18 at 4:34
  • 3
    $\begingroup$ Knowledge that a solution is real analytic (rather than, say, $C^{\infty}$) is in itself information. $\endgroup$ – Qmechanic Aug 17 '18 at 4:42
  • $\begingroup$ I was just starting an answer when Qmechanic beat me to it. But to add to that comment, the question builds in a form of FTL communication: the information being communicated is that $f$ is analytic at every point at this instant of time. If a person at some distant point wants to communicate, the way to do it is to non-analytically alter the value of $f$ in their distant region and wait for that information to propagate. $\endgroup$ – Mike Aug 17 '18 at 4:49
2
$\begingroup$

Solutions to a wave equation, like those governing most of the waves that transmit information from point to point in the physical universe, need not be analytic, even for "physically reasonable" parameters.

As an example, consider the classical wave equation for a scalar wave $\psi(x, t)$ propagating in one dimension with speed $c$ - could be the speed of light, could be some speed less:

$$\frac{1}{c^2} \frac{\partial^2 \psi}{\partial t^2} = \frac{\partial^2 \psi}{\partial x^2}$$

Suppose the initial scalar field $\psi(x, 0) := \psi_i(x)$ is started out as a bump function - smooth eveywhere, non-analytic, with compact support, i.e. zero outside of some interval, say $[-1, 1]$. We will assume the first time derivative is zero everywhere as well for simplicity. Now consider some interval suitably far outside this, e.g. $x \in [2, \infty)$. It is easy to see that since $\psi_i(x)$ is 0 there and moreover constant, then $\frac{\partial^2 \psi}{\partial x^2}$ is also zero, and so then also $\frac{\partial^2 \psi}{\partial t^2} = 0$. If the first time derivative is initialized to also be zero as we did, then it will thus remain zero at these points at least for a short time into the future, and likewise at suitably near times the field itself will still be at value zero at these points - but it will be nonzero in a growing region around the initial "hump". This means we have a solution that is not spatially analytic since it switches between two different types of analytic behavior: being constant on a nontrivial interval yet not constant everywhere. Thus Taylor's theorem does not apply, and the propagation speed $c$ is preserved. Moreover, since the wave equation is linear, if you superpose this on any other "background solution", even an analytic one, you'll get a new nonanalytic solution, which then generically models the creation of a localized disturbance within the field.

$\endgroup$
  • $\begingroup$ I think this is a good answer, the other thing classically is that you can only measure $\psi$ on some spacetime interval to finite precision with real measuring equipment, and I don't think knowing the value of some function on an interval to finite precision is good enough to construct arbitrarily many derivatives of the function. Then quantum mechanics has its own set of problems with measurements of $\psi$ causing collapse. $\endgroup$ – Bobak Hashemi Aug 17 '18 at 6:54
0
$\begingroup$

When you say an event this implies something is happening. Or you could say something is changing. This change will still take time to propagate.

In other words, you would only be using Taylor's theorem for the function at some point in time, but you would need to wait for the function you are measuring to change at your location to know this event occurred. This change will still take a finite amount of time to happen after the event.

$\endgroup$
-1
$\begingroup$

The important observation to make is that determination of an infinite order derivative involves the whole function domain. If for the first derivative an interval dx is needed then for the nth one you need ndx. Since in practice dx is finite thus tends to infinity. So to know all derivatives is equivalent to knowing all function values.

$\endgroup$

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