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As we all know the speed of light is the limit at which energy/matter can travel through our universe.

My question being: is there a similar limit for acceleration? Is there a limit to how quickly we can speed something up and if so why is this the case?

Short question, not sure what else I can elaborate on! Thanks in advance!


marked as duplicate by Qmechanic May 2 '15 at 10:52

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  • $\begingroup$ Afaik, there is no slogan about maximal acceleration. You could cook up a Planckian acceleration $c^{3.5}/(\hbar G)^{0.5}\approx 2.2\,10^{51}m^2/s$, but I don't see the use for that here. Seems the latter must ask for r'' vs. W. We should maybe make it a little more concrete. Consider a particle at rest with $r_1=0$ at some time and set up a potential $V(|r_1-r_2|)$ with another particle. You can consider a) a free collision (where total momentum is constant) and compute $r_1''(t)$. b) try to compute the energy cost it takes to move the second particle in some way to push the other one. $\endgroup$ – Nikolaj-K Mar 4 '14 at 23:09
  • $\begingroup$ I would say no as long as $v<c$. Because metaphysically speaking there is an infinite rate of change of velocity every time you go from zero velocity to a finite velocity since at some point time is discrete. This is a philosophical answer, please do not get mad. $\endgroup$ – jerk_dadt Mar 5 '14 at 0:02
  • $\begingroup$ Nikolaj i think your value for the Planckian acceleration is slightly wrong. I get $\approx 5.6 \cdot 10^{51} $. I like your answer jerk, although I'm not sure if it works for us to define things like that physically, I would put it into a sort of Zeno's Paradox area where it makes sense but doesn't actually describe the truth as we observe it. $\endgroup$ – Carterini Mar 5 '14 at 0:26
  • $\begingroup$ @Carterini: Not that a factor of 2 is of any relevance if we're already speaking of $10^{51}$, but no, $2.2$ is the value you get if you plug in "(speed of light)^(7/2)/((Plancks constant)*(Gravitational constant))^(1/2)" into wolframapha and so here I bet against your back of the envelope calculuation. And I say it again, I'd just sit down and see how fast momentum can translate between two particles, say with a $\left|r_1(t)-r_2(t)\right|^{-2}$. $\endgroup$ – Nikolaj-K Mar 5 '14 at 8:34
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    $\begingroup$ I agree it isn't really important, I believe your error comes from using Plancks constant and not the reduced form of it. I saw that but I can't really see how that I could go about getting a definitive answer from it. $\endgroup$ – Carterini Mar 5 '14 at 13:41

There are two answers:

  1. If the mass of the body is given, the limit is Caianiello's maximal acceleration, http://arxiv.org/abs/quant-ph/0407115 : $a_{\max}=2 m c^3/\hbar$

  2. If any mass value is allowed, the limit is the Planck acceleration, given by Plank length/Planck time^2.

  • $\begingroup$ Interesting link! I'm not familiar with Caianiello's argument, but it's worth noting that your link, by Papini, uses the Caianiello maximum acceleration to derive limits on the Higgs boson mass which have recently been proven wrong. $\endgroup$ – rob Jul 6 '14 at 13:16

As of today, there is no working theory which would put a fundamental limit on acceleration.

Specifically, Julian Fernandez and Arthur Suvorov commented on the tidal forces or relative acceleration in general relativity which can be infinite for example in the center of a black hole. However, note that this is not really an acceleration of a single point, as relativity says a particle does not accelerate in gravity but follows a geodesic - a space-time analogy of a straight line followed with "uniform" four-dimensional velocity.

Tidal forces just describe the phenomenon when geodesics at different points differ heavily causing a larger object to be squashed or ripped apart by the effective "force" of every part of it's body trying to follow a completely different geodesic.

In particle physics, as of now, we describe particle interactions as contact. In that sense, they transfer momentum (and thus velocity) instantaneously. It is obvious that in such a process acceleration has an infinite peak in the moment of interaction.

However, both theories, general relativity describing gravity and quantum field theory describing particle physics, are assumed not to be correct on arbitrary scales. Specifically the two mentioned cases, singularities in black holes and point-interaction, are expected to be modified when looking close enough. These scales which are conjectured to be "close enough" are called Planckian scales described by anna v. These are basically a combination of constants from relativity and quantum theory assembled to give quantities with dimension such as "length" or "time".

Nevertheless, even though we presume these cases of "infinite accelerations" to not be accurately described (physicists just don't like infinities), there is no wider, inevitable notion that there should be an upper limit on acceleration even in the new theories.

  • $\begingroup$ "However, note that this is not really an acceleration of a single point, as relativity says a particle does not accelerate in gravity but follows a geodesic - a space-time analogy of a straight line." Does following a straight line ("analogy" of it) preclude acceleration? $\endgroup$ – bright magus Jul 6 '14 at 9:28
  • $\begingroup$ Well, you can formulate "following a straight line" as "not changing the direction of your velocity", plus in relativity, we have an obligatory normalization of the four-velocity, so the "magnitude" is fixed also. I am going to try to clarify this in the post. $\endgroup$ – Void Jul 6 '14 at 9:32
  • $\begingroup$ If so, doesn't the concept of four-velocity (not being limited to gravity only) preclude all accelerations? $\endgroup$ – bright magus Jul 6 '14 at 10:07
  • $\begingroup$ Well, you can always "rotate" your four-velocity, which is exactly what for example electromagnetic forces do. Then, you can change momentum by changing the particle rest mass which is an action of a potential force. The "norm" or "magnitude" is however slightly exotic, as it is in the meaning of the space-time interval or Minkowski metric which e.g. makes physical velocities to have a negative "norm". $\endgroup$ – Void Jul 6 '14 at 10:15
  • $\begingroup$ Still, it's not more exotic than claiming that gravity vanishes or not depending on the order of separation (i.e. free falling object is considered inertial, and yet it undergoes tidal forces due to gravity - p.2: astronomy.ohio-state.edu/~dhw/A873/notes2.pdf) $\endgroup$ – bright magus Jul 6 '14 at 10:27

Assuming there are black holes with singularities, then the acceleration can be as a high as you want inside a black hole, the larger the closer you get to the singularity.


As far as I'm aware, there is no limit to the acceleration a particle (or anything else) can experience.

Why? Well, one may start looking at Newton's second law $F = ma$ and say, well, $a$ can be just as unbounded as $F$, right? Like Julian has said, tidal forces near black holes can become unbounded as you approach the singularity. Black holes are a general relativistic phenomenon though, so this equation is inadequate. Now when you enforce the postulates of relativity and assume $c < \infty$, relative acceleration is governed by the geodesic deviation equation:

$\frac {D^2 X^{\mu}} {dt^2} = R^{\mu}_{\nu \rho \sigma} T^{\nu} T^{\rho} X^{\sigma}$

where $T^{\mu} = \frac {\partial x^{\mu}} {\partial t}$. There is no apriori bound placed on how 'negative' the Riemann tensor is allowed to be; that is to say there are no mathematical results (assuming a $\it{really}$ generic space-time) that limit the behaviour of $R^{\mu}_{\alpha \beta \gamma}$, and hence the term on the LHS relating to acceleration.

However, with regards to the Heisenberg uncertainty principle, we must consider space-time to essentially be a discrete object, meaning we cannot infinitely divide up time and obtain a continuous limit (otherwise observations don't make sense). This in mind, no matter how small you make the time steps, we must limit the overall 4-velocity $v^{\mu}$ of the object to be bounded by $c$. So, acceleration must always be finite but can grow arbitrarily large, since we approximately have in a time-step $\delta t$: $v \approx a \delta t$.

Edit: What I really mean in this last paragraph is that it does not make sense to talk about motion in zero time. So, since $v < \infty$, and $\delta t >0$, we must also have that $a < \infty$ regardless of the theory of motion we are dealing with.


Acceleration is dv/dt

v is limited by the velocity of light.

delta(t) is limited by Planck time, in the sense that physics in smaller time intervals is not known

planck time~5.39106(32) × 10−44 s

One Planck time is the time it would take a photon traveling at the speed of light to cross a distance equal to one Planck length. Theoretically, this is the smallest time measurement that will ever be possible, roughly 10^−43 seconds. Within the framework of the laws of physics as we understand them today, for times less than one Planck time apart, we can neither measure nor detect any change. As of May 2010, the smallest time interval uncertainty in direct measurements is on the order of 12 attoseconds (1.2 × 10^−17 seconds), about 3.7 × 10^26 Planck times.

Substituting these numbers and assuming delta(v) is of order c ,10^8 meter/second, puts the limit of acceleration at order 10^52 meters/second**2.

So yes, there is a limit to measuring a difference in velocity to be attributed to acceleration.


Acceleration depends on velocity per unit time, and velocity on distance per unit time. It is clear that acceleration is infinite for us because space is infinite. For a given time, acceleration is infinite.

  • $\begingroup$ With that argument, even velocity should be infinite. However, all objects are limited to c. $\endgroup$ – mikhailcazi Apr 4 '14 at 14:27
  • $\begingroup$ Not a valid argument. $\endgroup$ – Gummy bears Jul 6 '14 at 6:04
  • $\begingroup$ can you please elaborate your argument $\endgroup$ – Sahil Chadha Jul 6 '14 at 11:08

The limit to acceleration is c^2/r.

  • $\begingroup$ I did not vote down. This is an interesting thought, but naively incorrect. How do you get the factor of 2, instead of just c^2/r? (Neither 2c^2/r nor c^2/r is correct though.) $\endgroup$ – wonderich Jun 21 '14 at 4:36
  • $\begingroup$ @Idear for the record, the best way to indicate that an answer is incorrect is to vote it down, and preferably (but not necessarily) to leave a comment saying what's wrong with it. $\endgroup$ – David Z Jun 23 '14 at 3:39
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    $\begingroup$ @ David Z, yeah, but in that way, one will kill the possible creative thinking, and this new person's curiosity and participation. In many lessons in the history, some naive stupid or counterintuitive thoughts can bring up some unpredictable breakthroughs. Though I am not sure this post is the rare case... $\endgroup$ – wonderich Jun 23 '14 at 3:57

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