Is there a maximum possible acceleration?

a kind of important question. But according to quantum mechanics, you can't imagine that objects are moving along smooth, doubly differentiable trajectories - which you need to define acceleration. Instead, they're moving along all possible trajectories - I am using Feynman's path integral approach to quantum mechanics - and most of them are not differentiable even once. So the typical acceleration at a typical place of a trajectory in quantum mechanics is infinite. You could only study a "finite bound" on acceleration in classical physics and in classical (non-quantum) physics, there's no upper bound.

However, you may talk about the upper bounds on some "correctly looking" formulae for acceleration. For example, you may be able to "derive" that the maximum gravitational acceleration in quantum gravity is approximately equal to the Planck acceleration, $$a_{Planck} = L_{Planck} / T_{Planck}^2 = \frac{\sqrt{\frac{\hbar G}{c^3}}}{\frac{\hbar G}{c^5}}=\sqrt{\frac{c^7}{\hbar G}} = 5.6 \times 10^{51}\,\mbox{m/s}^2$$ where the numerator and denominator depend on the Planck length and Planck time, respectively. Yes, it's huge. This upper bound holds because it's the acceleration on the surface (event horizon) of the smallest and most concentrated object. The most concentrated objects are black holes and the smallest black hole worth the name has a radius comparable to the Planck length.

However, this bound only applies outside the black holes. Near the singularities inside the black hole, the accelerations may be formally larger. No one knows whether it makes sense to talk about the trans-Planckian accelerations. However, accelerations are not among the "most fundamental quantities" we use to describe physics according to its state-of-the-art theories, anyway.

Cheers LM

The spit horizon in a Rindler wedge occurs at a distance $d~=~c^2/g$ for the acceleration $g$. In spatial coordinates this particle horizon occurs at the distance $d$ behind the accelerated frame. Clearly if $d~=~0$ the acceleration is infinite, or better put indefinite or divergent. However, we can think of this as approximating the near horizon frame of an accelerated observer above a black hole. The closest one can get without hitting the horizon is within a Planck unit of length. So the acceleration required for $d~=~\ell_p$ $=~\sqrt{G\hbar/c^2}$ is $g~=~c^2/\ell_p$ which gives $g~=~5.6\times 10^{53}cm/s^2$. That is absolutely enormous. The general rule is that Unruh radiation has about $1K$ for each $10^{21}cm/s^2$ of acceleration. So this accelerated frame would detect an Unruh radiation at $\sim~10^{31}K$. This is about an order of magnitude larger than the Hagedorn temperature. We should then use the string length instead of the Planck length $4\pi\sqrt{\alpha’}$ and the maximum acceleration will correspond to the Hagedorn temperature.

For QED there is a critical acceleration, which is the acceleration felt by an electron subject to the Schwinger field (http://en.wikipedia.org/wiki/Schwinger_limit). This is at the critical acceleration

$$ a_S = \frac{m_ec^3}{\hbar} = 2.33 \cdot 10^{29} \frac{m}{s^2} $$

Beyond this field, nonlinear effects if the QED vacuum and pair creation occur which will influence the dynamics of an electron accelerated by this field.