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I'm thinking equivalence principle, possibilities of unbounded space-time curvature, quantum gravity ...

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Please, put a little more effort into stating a precise question. Otherwise is might get closed as "Not a real question". –  Malabarba Jan 19 '11 at 14:02
    
Acceleration is unbounded in general relativity (needs to be at least continuous I think) –  Ebenezer Sklivvze Jan 19 '11 at 14:16
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Nice question. Admittedly could use elaboration and expansion so as to appear to be less like a random thought. –  user346 Jan 19 '11 at 14:49
    
I thought the question was quite well-defined actually. The other approach which I thought someone might take up was whether there was some fundamental limitation on how much acceleration a particle could experience based on the strong-field behaviour of the fundamental force causing it. My (admittedly rather vague) intuition here was that if the accelerating potential field is sufficiently strong then it could self-limit due to spontaneous pair formation. One can imagine a related question, for example, is there a maximum possible electric or magnetic field strength? –  Nigel Seel Jan 19 '11 at 16:24

4 Answers 4

up vote 8 down vote accepted

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.

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As given by general relativity, there is no limit. Your answer relies on the existence of a minimum length. That assumption is used in most approaches to quantum gravity, but it would still be nice to state Yes, but only if you assume a minimum length.. –  Malabarba Jan 19 '11 at 14:30
    
+1 This is a brilliant, completely jargon free, answer @Lawrence ;) I'm sure it could use refinement but the basic considerations would remain the same and as @Bruce notes "given a minimum length scale". But @Bruce I think a minimum length scale is not an assumption but emerges out of physical considerations. Just as a minimum (non-zero) ground state energy ($1/2 \hbar \omega$) for the quantized harmonic oscillator is not an assumption but arises from the non-zero commutation relations between ladder operators. –  user346 Jan 19 '11 at 14:48
    
Reassuring that the same number comes out as in LM's answer. Is the Hagedorn temperature truly fundamental and do you have a numeric value for what you take to be the actual maximum possible acceleration (QM nuances understood)? It wasn't clear to me whether you're arguing that the figure you gave is actually too large as it give a value for the Unruh radiation > the Hagedorn temperature. –  Nigel Seel Jan 19 '11 at 16:45
    
The number is an order of magnitude calculation and could change by coefficients related to quantum gravity effects (stringy or otherwise). Still, its pretty suggestive, albeit not particularly fundamental. –  Columbia Jan 19 '11 at 18:28
    
The Hagedorn temperature is the limit for string theory, which occurs at about 10-100 Planck lengths. There is the infamous diagram which has on a plane of $\hbar$ and $\alpha'$ the various string types. The middle zone is "unknown," and it likely is where energy is above the Hagedorn limit. So a temperature or energy above the Hagedorn temperature represents the "unknown." So we don't know for certain what it means for a temperature higher that $T_{Hag}$. –  Lawrence B. Crowell Jan 19 '11 at 20:34

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

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Once you talk about the interior of a black hole the question of what is acceleration is interesting. If you consider that acceleration as the tidal force or due to Weyl curvature you find a very similar answer. A string which enters the interior of a black hole experiences this tidal force until it reaches the string tension or Hagedorn temperature. At that point a string might become “pasted” onto a D-brane which “valences” the classical BH singularity. This might have some duality with the stretched horizon. The dual perspectives have different gauge conditions on string modes. –  Lawrence B. Crowell Jan 19 '11 at 15:03
    
as regards your first paragraph, one wonders whether the same argument equally applies to velocity in QM? –  Nigel Seel Jan 19 '11 at 16:29
    
Lawrence, interesting speculations that are hard to quantify at the present. Nige, the statement about the velocities is just the uncertainty principle. For an accurate enough position, the velocity uncertainty is at least $\hbar/\Delta x$. Relativity limits the speed not to exceed the speed of light but in the path integral of mechanics, you sum over all trajectories - including superluminal ones - but their effect ultimately exactly cancels (with antiparticles etc.). But of course, the more derivatives you perform, the wilder the function will become. –  Luboš Motl Jan 20 '11 at 8:45
    
Interesting that the ratio hbar over delta x is shown on a different place haha. –  Luboš Motl Jan 20 '11 at 8:46

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

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

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.

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I like previous answer but: 1) I believe that in the provided formula the mass of the electron should have a power of one (not two) 2) It is valid for electrons only, because it uses their Compton wavelength.

By the way, there is such a thing as "Caianiello’s maximal acceleration". In his 1985 paper Caianiello demonstrated the existence of a maximal acceleration. It is a consequence of Heisenberg’s uncertainty relations. An example may be found here.

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