Planck length at relativistic speeds? I'm currently in high school so sorry if the answer to this question seems obvious but I’m only just learning about this stuff. I’ve been learning about special relativity, in particular length contraction and time dilation. I was wondering, if the Planck length is the smallest possible observable length, then what would an observer who is travelling at relativistic speeds measure the Planck length to be? Would it be the same or would he observe a smaller length?
 A: This is a great insightful question for a high school student, thanks. The reason for length contraction comes down to a difference in what two people consider to be simultaneity. The only way to measure distance is to mark one place, and mark another, then figure out how far apart they are. If the thing you're measuring is moving that means you have to do the two marks at the same time. Here's the big issue: observers moving at different speeds have different notions of when two marks are simultaneous, so they can measure the same object and get two different lengths. Say I measure a car as it rolls past. I put a chalk mark on the ground right where the front of the car is, then a few seconds later I put another mark right where the back is. When I measure the distance between marks, I find the car is only 5 feet long. Amazing, because it's actually way longer. But the reason is just that I made the marks at different times while the car was moving - to get the right answer I would have had to make both marks simultaneously. Relativity seems weird because it changes when events are simultaneous. If I measure two dots as a Planck length apart, and you look at my measurement and it looks shorter, that's because you're measuring the position of each dot at different times as you/they move. A Planck length is still a Planck length in your frame, and my "shortened" Planck length is not actually some kind of illegally tiny length, any more than the car is actually 5 feet long.
(Edit:) Another, but less useful, way to answer is that you could not measure a sub-Planck-length distance anyway because you can't build a real-world measuring device that resolves distances that small. The best you'd get is a measurement that would "round-up" to a Planck length in your frame, which might be confusing if you didn't understand the source of the measurement "error".
A: This is a very good question, and serious physicists such as Lee Smolin have been wondering about it. According to Special (and General) Relativity any inertial observer should get the same value for the Plank length in terms of its own units. So the Planck length calculated in a frame that I see moving should be the same number of that frame’s units as the Planck length I calculate in mine. But since the moving frame has shorter units, its corresponding Planck length should seem shorter than a Planck length is to me.
One thing to remember, though, is that the Planck length is not a property of any actual object but rather just the scale at which the effects of quantum gravity should become apparent. So one interpretation of the contraction is that for a traveler moving relative to me, the effects of quantization of gravity on objects in the traveler's frame become apparent to me at longer scales than they do to the traveler. (And conversely, the traveler sees quantum effects on me on length scales at which I still don’t notice them.)
This is one of the many things that will have to be resolved by a proper quantum theory of gravity.
A: I agree with the answer of alQpr. I will add a concrete example.
The Planck length is $l_P = \sqrt{\hbar G/c^3} = 1.6 \times 10^{-35}$ m. Let's consider a disturbance in spacetime, such as either a gravity wave or an electromagnetic wave, with a wavelength $100$ times longer than this, as observed in some inertial reference frame: $\lambda = 100 \, l_P \simeq 1.6 \times 10^{-33}$ m. It seems reasonable to assert that such a wave could be possible. (Having said that, such a wave would be liable to lose energy very quickly since even a tiny amount of interaction with something else would be enough to generate electron-positron pairs or scattering or parametric down-conversion into waves of lower frequency.)
Now let's consider an inertial frame moving at $v = 0.999998\,c$ relative to the first frame, giving a Lorentz factor $\gamma = 500$. In this case the Doppler effect formula gives, for the wavelength in this frame,
$$
\lambda' = \sqrt{ \frac{1 - v/c}{1+v/c} } \, \lambda = 10^{-3} \lambda = 0.1 \, l_P
$$
So if the calculation is correct then now we have a wave whose wavelength is ten times smaller then the Planck length.
The situation is as follows. It may be possible that one could have a wave with a wavelength as short as that. The difficulty is not that we know it to be impossible; the difficulty is that we suspect that any existing physical theory currently reasonably well worked out by physicists will not describe such a phenomenon correctly. So we don't know if the Doppler effect formula which I used just now would still be applicable. What it comes down to is that we don't how to calculate the physics when it involves a combination of near-Planck length and high relative velocity.
A: It's quite simple, actually. The Planck length is a length perpendicular to the three space dimensions.
Consider the one large dimension case. If the one dimensional large space is actually a thin cylinder, then the particle is Planck size circle. Traveling in the length of the cylinder doesn't change the length of the circle. This is the same for you and me.
In three dimensional space, a particle is a tiny torus, a product of three perpendicular circles. So the bulk space is three dimensional and in every direction the Planck length is the same, as the motion doesn’t influence the perpendicular space (which can be viewed as 6D space of which three are curled up into Planck circles, S1xS1xS1, like a 2D space can be rolled up to 1D, with a small circle attached).
A: Believe it or not your bright question has been perplexing physicists like Smolin for quite some time; see Wikipedia. As of now there is still controversy.
You (we) are facing the same contradiction that the young Einstein faced: On one hand according to Newtonian mechanics (Galilean relativity) Einstein knew that all speed is relative, but on the other hand from Maxwell (and Hertz) he knew that there is but one speed which is not relative; its value is a constant of nature!
Same here with your question: We know on one hand according to Special Relativity that all time (and length) is relative, but on the other hand we have the Planck time (and length) which is a fundamental constant of nature and its value seems independent from any observer (exact same situation with c).
So to give a short answer: This is an open question. Special relativity is not compatible with the mere existence of Planck length/time.
See a full discussion I have written on this problem and my proposed (possible) resolution here.
A: 
Time dilation and Length contraction are related to the vibration fields of the moving object at the quantum level. The matter are composed from these united-intensified energy fields.
The amplitude of vibrations at quantum level are enlarging by added energy in the system during acceleration. We named this added energy also as momentum in a way.
These calculations are according to the observer. The moving object/traveler determines them as usual as explained at the first answer.
A: There are two aspects you have to consider for the answer:
1.The spacetime of special relativity does not consider lengths but only distances of worldlines. 
Special relativity consists of the two postulates, the first one saying that in each frame of reference the same physical laws do apply and the second, that light is observed as to move with speed of light. As you can see, special relativity helps describing timelike and lightlike worldlines (of mass particles and lightlike phenomena), but it says absolutely nothing about the vacuum between particles which would permit to fill the space between particles in order to get a length.
In the same way, the most current example for the explanation of length contraction is regarding some rigid macroscopic object. This is a gross simplification because according to special relativity there are no perfectly rigid objects. According to special relativity, every rigid object consists of its particles with their interacting forces which may provide high, but not perfect rigidity.
In contrast, special relativity may consider traveled distances as far as they are referring to timelike or lightlike worldlines of particles.
2.According to your question and according to the Lorentz contraction of special relativity, such distances are observed as contracted by moving observers. This applies also to distance constants such as Planck length.
Lorentz contraction is a phenomenon which is specific to a frame of an observer. A distance traveled by a particle which is observed with a length of one light year from the point of view of the reference frame of a comoving observer is observed smaller (contracted) from the point of view of observers with a relativistic relative velocity with respect to the considered traveling particle. The distance may be subject to arbitrarily high contraction, for an observer which is moving very near to light speed with respect to the two events which are limiting the distance. But it is evident that the fact of the observation does not change the travel data of the particle. The particle does not travel the contracted distance which is observed by observers (otherwise you might ask: of which observer? with contradictory answers) but the distance before Lorentz contraction which is simply the longest distance that can be measured.
Thus, the answer to your question is: Yes, an observer traveling at relativistic relative speed with respect to the reference frame of a particle may observe any distance as contracted, even constants such as the Planck length.
