Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose you accelerate a body to very near the speed of light $c$ where $v = c - \epsilon$. Although this would take an enormous energy, is it possible the last arbitrarily small velocity needed -- $\epsilon$ -- could be overcome with a minor bump in velocity due to the uncertainty principle?

share|cite|improve this question
didn't know that planck constant had the same units as c! – lurscher May 4 '13 at 2:19
I was going for a well known theoretically smallest number... But good catch! – frogeyedpeas May 4 '13 at 2:26
Quantum mechanics does not imply that everything is discrete. In particular, there is no lowest non-zero velocity in standard physics. You could get one from some kind of discrete spacetime lattice model (which invariably breaks Lorentz invariance), but those are essentially ruled out already by cosmic ray observations etc. – Michael Brown May 4 '13 at 2:33
This confuses me even further... So distance is discretized but velocity is not (-_-) – frogeyedpeas May 4 '13 at 2:34
Yes. I wouldn't say "smallest conceivable". It is the smallest measurable because attempting to measure something smaller than the Planck length invariably creates a black hole which is larger. But this doesn't mean that space is divided up into Planck sized blocks or discrete units at all. There were some theories like that but they have been ruled out. Space can still be discrete in some way, but it has to be done in a way which preserves Lorentz invariance beyond the Planck scale. – Michael Brown May 4 '13 at 3:05
up vote 2 down vote accepted

No, because the uncertainty principle operates between position and momentum rather than position and velocity. For speeds much less than $c$, momentum is just proportional to velocity: $p = mv$. But at relativistic speeds we have to use the relativistic version, $$ p = \gamma mv, $$ where $\gamma = 1/\sqrt{1-v^2/c^2}$. Substituting this in and squaring both sides we get $$ p^2 = \frac{m^2v^2}{1-{v^2}/{c^2}}, $$ which we can rearrange a little to get $$ v^2 = \frac{p^2}{ m^2 + p^2/c^2 }, $$ or $$ v = \frac{p}{\sqrt{ m^2 + p^2/c^2 }}. $$ Now, the limit of this as $p \to \infty$ is just $$ v = \frac{p}{\sqrt{p^2/c^2 }} = c. $$ The momentum $p$ can fluctuate due to the uncertainty principle, but now you can see that now matter how big $p$ gets, $v$ will always be less than $c$.

share|cite|improve this answer
This made a lot sense, thank you! – frogeyedpeas May 4 '13 at 14:45

No. First of all, Planck's constant is not a speed, so you can't compute $c - \hbar$. But you can reword the question to get around that problem, something like this:

Is there some speed $\epsilon$ such that an object traveling at speed $c - \epsilon$ could experience a quantum fluctuation that temporarily takes its speed to greater than $c$?

The answer to this is still no. Now, in order to really understand why, you could dig into the details of quantum field theory, and learn the meaning of the statement "local operators separated by spacelike intervals commute" which is, in some sense, the most fundamental reason. But I'm guessing that'd be more detail than you're looking for.

As a simplified (but still basically accurate) explanation, you can use the same argument for why you can't bump a classical object moving at speed $c - \epsilon$ up to speed greater than $c$ by giving it a little push. That reason is that when something speeds up, spacetime "rotates" around it, but in such a way that all trajectories with speeds less than $c$ continue to have speeds less than $c$. In particular, this rotation (the Lorentz boost) transforms a trajectory with speed $v$ into a trajectory with speed $\frac{v + \Delta v}{1 + v\Delta v/c^2}$. No matter how close you are to the speed of light, speeding up will only take you a fraction of the way closer to $c$, and that is just as true for a quantum fluctuation as it is for a classical push.

share|cite|improve this answer

Quantum uncertainty as shown by Heisenberg precludes the information from reaching us faster than the speed c for light. Lorentz invariance is preserved by the existence of antimatter. This was shown by Dirac. Thus as ( Feynman? ) some noticed, you can describe a positron as an electron moving backward in time. So the speed of light is not classical so while the speed at which information can move in space is certain the lack of certainty of velocity and position in QM and the ability to see it differently as a particle or antiparticle depending on your perspective keeps everything consistent. If you are looking for more rigor I leave it to others but this works for me to avoid paradoxes

share|cite|improve this answer
What? So convoluted. You mixed the uncertainty principle with antimatter with quantum information theory. -1 – Brandon Enright May 4 '13 at 4:04
@BrandonEnright That would be a nice mix if it was used in a way that made any sense. :) – Michael Brown May 4 '13 at 8:52
Sorry that by not separating two points more clearly I was convoluted. Nathaniel satisfied the OP with a rigorous proof that the speed of light is consequence of uncertainty. – user12811 May 4 '13 at 23:49
On the second point I shall defer to the real experts.From whose books and blogs I had understood that to combine relativity Lorentz symmetry with QM consistently requires anti-particles. see Aarsonson book or especially at least comments in or – user12811 May 5 '13 at 0:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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