the perpendicularity of 4-velocity and 4-acceleration in s.r It is known that in s.r. - $$u^iw_i=0$$ where $u^i$ is the 4-velocity and $w^i$ is the 4-acceleration (ref - Landau & Lifhitz, eq 7.3). 
In classical mechanics (three dimensional), when the velocity of a body is perpendicular to its acceleration, and the size of the acceleration is constant in time, the body moves in a circular (periodic) motion.
Does this remain true in special relativity? If so,what is the meaning of it?
In three dimensions periodic movement means the body will come back to the point $(x,y,z)$ over and over again. But in s.r. we have events $(ct,x,y,z)$ and what is the meaning of "over and over again" here?
 A: This is a very interesting question.
It's useful to remember another consequence of an acceleration's being perpendicular to the velocity: the magnitude of the latter doesn't change. Consider the difference of the squared magnitudes at an interval $\Delta t$:
$$\lVert\boldsymbol{v}(t+\Delta t)\rVert^2 - \lVert \boldsymbol{v}(t)\rVert^2
\approx
[\boldsymbol{v}(t) + \boldsymbol{a}\,\Delta t]\cdot
[\boldsymbol{v}(t) + \boldsymbol{a}\,\Delta t] -
\boldsymbol{v}(t)\cdot \boldsymbol{v}(t) =
2\boldsymbol{v}(t) \cdot \boldsymbol{a}\,\Delta t =0,$$
where the last equality comes from the perpendicularity.
As you know, in relativity theory the 4-velocity is defined to have constant magnitude $1$ in natural units (or $c$ otherwise), modulo a $\pm$ sign that depends on how you choose the signature of the metric. This is possible only if the 4-acceleration is perpendicular to the 4-velocity; the perpendicularity is a consequence of this fact.
But what does "perpendicular" mean in relativity theory? If you ask yourself this you'll find the answer to your question. The geometry of relativity is hyperbolic, and the condition of perpendicularity between a timelike and a spacelike vector of constant magnitude leads to the equation of a hyperbola, rather than a circle. In a spacetime diagram, an object having constant acceleration will thus have a hyperbolic world-line. Note that the spatial component of the 4-velocity does not need to be perpendicular to the acceleration, even if the 4-velocity is; the 3-velocity and acceleration can very well be collinear.
This kind of hyperbolic motion has been widely studied because it has very interesting features. For example, an observer in such a motion experiences an apparent event horizon, which has also consequences for "counting particles" from the point of view of quantum theory; a rod under such kind of motion experience stress; and other phenomena. You can check the Wikipedia page for references, or § 6.2 in Misner, Thorne, Wheeler: Gravitation (Freeman & Co. 1973).
A: Since $u^\mu u_\mu =c^2$, $u^\mu w_\mu=0$. The meaning is not that motion through spacetime is around a circle, but rather (given the Lorentzian metric) that it follows a kind of hyperbola, albeit in $v$-space rather than $x$-space. The 2D case gives $v^0=c\cosh\phi,\,v_1=c\sinh\phi$ with $\phi$ the really rapidity.
A: For me, it means that everything moves through spacetime with a 4-velocity of magnitude of $c$, and nothing can change that. No amount of acceleration or motion--and that has physical significance.
It makes clear that $c$ is not relevant as the speed-of-light. Light just happens to move at $c$ for other reasons. Physically, $c$ is a scaling parameter between the time part and the space part of spacetime. It describes how hyperbolic (as referenced in the other answers) spacetime is.
Also: at rest, you do not move through space--of course, but you do move through time as quickly as possible. Your 4-velocity is:
$$ u_{\mu} = (c, \vec 0) $$
Now if your twin starts moving, the fact that he moves through space, affects how he moves through time (in your reference frame):
$$ u_{\mu}'=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\big(c, \vec{v}\big)$$
His time is dilated by:
$$ \gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$
So, as you pointed out, in ${\bf E}^3$, ${\bf v\cdot a}=0$ implies some kind of circular motion, where say $x$ and $y$ coordinates are mixed by rotation while magnitudes are preserved. In Minkowski Space, $u_{\mu}w^{\mu}=0$, implies that relative motion mixes time and space coordinates w.r.t to a stationary reference frame. 
As pointed out elsewhere, the hyperbolic nature of $M^4$ means you don't return to a initial point, rather you asymptotically approach the speed of light.
