Error in acceleration

bruno67

Denote by [itex]V(x)[/itex] the speed of a particle at position x. Let's call [itex]v(x;\zeta)[/itex] a measurement of it, which depends on some parameter [itex]\zeta[/itex], and denote the error by
[tex]\epsilon(x;\zeta)=v(x;\zeta)-V(x).[/tex]
In order for the measurement to produce meaningful results, we must have some kind of error estimate such that, for any x
[tex]|\epsilon(x;\zeta)|\le E(x;\zeta)[/tex]
where E is a known positive function, which ideally tends to zero as [itex]\zeta[/itex] tends to zero (we are not considering quantum mechanical effects). My question is: can we obtain a similar estimate for the error in the derivative of [itex]v(x;\zeta)[/itex] (e.g., as a function of [itex]E(x;\zeta)[/itex], [itex]V(x)[/itex] or [itex]V'(x)[/itex]) from the information given above, or do we not have enough information?

You can assume that the derivative of [itex]v(x;\zeta)[/itex] is calculated by finite difference, and that the discretization error involved is negligible.

mfb

Without additional data, you cannot give an upper limit for the instantaneous acceleration. Imagine you have some nice movement, and you add a high-frequency oscillation with fixed, small maximal velocity to it. As long as this maximal velocity is small compared to your error, you cannot detect it, but the acceleration is proportional to the maximal velocity multiplied by the frequency. Therefore, if the frequency is high enough, you can get arbitrary large accelerations.

You can give limits for the average acceleration (e.g. between two measurements), integrated over some time or maybe space interval.

In that case, assuming uncorrelated errors and no error for the time difference, your maximal deviations are given by (lowest velocity1 -> highest velocity 2) and (highest velocity1 -> lowest velocity 2). In your case, just add E(x,chi) for both pairs (x,chi) which contribute to the measurement, and divide it by the time difference. If you have to estimate the time difference by the velocity measurements, the formula might get a bit more complicated, but it is still possible to do it.