Is four-acceleration spacelike? How to show this? What does it mean? Consider a four-velocity $u_\mu$ that satisfying $u_\mu u^\mu=c^2>0$. Differentiating w.r.t the proper time $\tau$, we get the equality $a_\mu u^\mu=0$ which means the four-velocity and four-acceleration are orthogonal to each other. Using these equations how can we show that $a^\mu$ must be spacelike i.e., $a_\mu a^\mu<0$? What does it mean physically?
The spacelike nature of $a_\mu$ is claimed in this video. Go to 3 mins 30 sec.
 A: \begin{align*}
&\text{The Velocity in Minkowski space is}\\
&\mathbf{u}^\mu=\gamma\,\begin{bmatrix}
                         c \\
                         \vec{v} \\
                       \end{bmatrix}\\
&\text{thus the acceleraion in Minkowski space}\\                       
 & \mathbf A^\mu=\frac{d\mathbf{u}^\mu}{d\tau}= 
 \frac{d\mathbf{u}^\mu}{dt}\frac{dt}{d\tau}
 =\gamma\,\frac{d}{dt}\,\begin{bmatrix}
                          \gamma\,c \\
                          \gamma\,\vec{v} \\
                        \end{bmatrix}
  =\begin{bmatrix}
                     \gamma\,\dot{\gamma}\,c \\
                     \gamma\,\dot{\gamma}\,\vec{v}+\gamma^2\,\vec{\dot{v}}
                   \end{bmatrix}
\end{align*}
with $~\gamma\,\dot{\gamma}=\frac{\vec{v}\cdot \vec{\dot{v}}}{c^2}\,\gamma^4~$
\begin{align*}
  &\mathbf{A}^\mu=\begin{bmatrix}
                    \frac{\vec{v}\cdot \vec{\dot{v}}}{c}\,\gamma^4 \\
                    \frac{\vec{v}\cdot \vec{\dot{v}}}{c^2}\,\gamma^4\vec{v}+\gamma^2\,\vec{\dot{v}}
                  \end{bmatrix}\\
  &\mathbf{A}_\mu=\begin{bmatrix}
                    \frac{\vec{v}\cdot \vec{\dot{v}}}{c}\,\gamma^4 \\
                    -\frac{\vec{v}\cdot \vec{\dot{v}}}{c^2}\,\gamma^4\vec{v}-\gamma^2\,\vec{\dot{v}}
                  \end{bmatrix}                 
\end{align*}
The criteria for space like is:
\begin{align*}
 &A_0^2 \,< A_i\,A^i\\
 &\text{thus}\\
 &{\frac {{|v|}^{2}{{ |\vec{\dot{v}}|}}^{2}{{ \gamma}}^{8}}{{c}^{2}}} <
 {\frac {{|v|}^{4}{{ |\vec{\dot{v}}|}}^{2}{{ \gamma}}^{8}}{{c}^{4}}}+2\,{\frac {{|v|}^
{2}{{ |\vec{\dot{v}}|}}^{2}{{ \gamma}}^{6}}{{c}^{2}}}+{{ \gamma}}^{4}{{ |\vec{\dot{v}}|}}^{2}\quad (1)
\end{align*}
with $~\gamma > 1~,v/c < 1~$ equation (1)  is fulfilled , thus $~\mathbf A^\mu\,\mathbf A_\mu < 0$
A: In the 3-dimensional real space $\:\mathbb R^3\:$ the 3-acceleration vector $\:\mathbf a\:$ of a particle moving with velocity $\:\mathbf u\:$ has a tangential component $\:\mathbf a_{\texttt t}\:$ which reflects a change in the speed $\:\vert\mathbf u\vert\:$ of the particle  and a normal (centripetal) component $\:\mathbf a_{\texttt n}\:$ which reflects a change in the direction of motion of the particle.
In the 4-dimensional Minkowski space  the 4-acceleration vector $\:\mathbf A\:$ of a particle moving with 4-velocity $\:\mathbf U\:$ has no $''$tangential$''$ component $\:\mathbf A_{\texttt t}\boldsymbol{=0}\:$ which reflects no change in the 4-pseudonorm  $\:\Vert\mathbf U\Vert^2\boldsymbol{=}c^2\:$ of the particle  and a pseudonormal ($''$centripetal$''$) component $\:\mathbf A_{\texttt n}\:$ which reflects a change in the direction of motion of the particle on its worldline.
If $\:\mathbf A\:$ would have not zero $''$tangential$''$ component $\:\mathbf A_{\texttt t}\boldsymbol{\ne 0}\:$ then between others we would have changes of the 4-pseudonorm  $\:\Vert\mathbf U\Vert^2\:$ towards not permissible $''$superluminal$''$ velocities.
$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$
Basic equations :
Under a Lorentz boost with 3-velocity $\:\mathbf v\:$ the 3-velocity $\:\mathbf u\:$ of a moving particle is transformed as follows
\begin{equation}
\mathbf u'\boldsymbol{=}\dfrac{\mathbf u\boldsymbol{+}\dfrac{\gamma_{\texttt v}^2}{c^2\left(\gamma_{\texttt v}\boldsymbol{+}1\right)}\left(\mathbf v\boldsymbol{\cdot}\mathbf u\right)\mathbf v\boldsymbol{-}\gamma_{\texttt v}\mathbf v}{\gamma_{\texttt v}\left(1\boldsymbol{-}\dfrac{\mathbf v\boldsymbol{\cdot}\mathbf u}{c^2}\right)} 
\tag{01}\label{01}
\end{equation}
For the 3-acceleration $\:\mathbf a\:$
\begin{equation}
\mathbf a'\boldsymbol{=}\dfrac{\gamma_{\texttt v}\left[\mathbf a\boldsymbol{+}\dfrac{\mathbf v\boldsymbol{\times}\left(\mathbf u\boldsymbol{\times}\mathbf a\right)}{c^2}\right]\boldsymbol{-}\dfrac{\gamma_{\texttt v}^2}{c^2\left(\gamma_{\texttt v}\boldsymbol{+}1\right)}\left(\mathbf v\boldsymbol{\cdot}\mathbf a\right)\mathbf v}{\gamma_{\texttt v}^3\left(1\boldsymbol{-}\dfrac{\mathbf v\boldsymbol{\cdot}\mathbf u}{c^2}\right)^3} 
\tag{02}\label{02}
\end{equation}
The 4-acceleration $\:\mathbf A\:$ expressed in terms of the 3-vectors $\:\mathbf u,\mathbf a\:$ is
\begin{equation}
\mathbf A\boldsymbol{=}\dfrac{\gamma_{\texttt u}^2}{c^2}\left[\gamma_{\texttt u}^2\left(\mathbf a\boldsymbol{\cdot}\mathbf u\right)c\,,\,c^2\mathbf a\boldsymbol{+}\gamma_{\texttt u}^2\left(\mathbf a\boldsymbol{\cdot}\mathbf u\right)\mathbf u\vphantom{\dfrac{a}{b}}\right]
\tag{03}\label{03}
\end{equation}
So in the rest frame of the particle
\begin{equation}
\mathbf A_{0}\boldsymbol{=}\left[0\,,\,\mathbf a_{0}\vphantom{\dfrac{a}{b}}\right]
\tag{04}\label{04}
\end{equation}
where $\:\mathbf a_{0}\:$ the 3-acceleration in the rest frame of the particle.
For the norm of 4-acceleration we have
\begin{equation}
\Vert\mathbf A\Vert^2\boldsymbol{=}\boldsymbol{-}\gamma_{\texttt u}^4\left[\vert\mathbf a\vert^2\boldsymbol{+}\left(\gamma_{\texttt u}^2\boldsymbol{-}1\right)\left(\dfrac{\mathrm d \vert\mathbf u\vert}{\mathrm d t}\right)^2\right]\boldsymbol{=}\Vert\mathbf A_{0}\Vert^2\boldsymbol{=}\boldsymbol{-}\vert\mathbf a_{0}\vert^2\boldsymbol{\le}0
\tag{05}\label{05}
\end{equation}
A: Since there are answers with lots of unnecessary algebra, I thought I would post what I think is a neater proof.
Start with 4-velocity, and recall
$$
U \cdot U = c^2
$$
where I have adopted the metric signature $(1,-1,-1,-1)$ which is the one proposed in the question, and the scalar product of 4-vectors is $U \cdot U \equiv U_\mu U^\mu$.
Now take the derivative with respect to proper time along the worldline:
$$
\frac{d}{d\tau} ( U \cdot U ) = 0
$$
since $c^2$ is constant.
Hence
$$
2 U \cdot \frac{dU}{d\tau} = 0
$$
so
$$
U \cdot A = 0
$$
where $A = dU/d\tau$ is 4-acceleration. To finish, we can either evaluate this scalar product in the instantaneous rest frame, or just appeal to the general fact that a 4-vector orthogonal to a time-like one has to be spacelike.
But let's show the proof anyway. In this instantaneous rest frame,
$$
U = \left(c,0,0,0\right)
$$
so
$$
U \cdot A = c A^0
$$
where $A^0$ is the temporal component of the 4-acceleration 4-vector in the instantaneous rest frame. So we have that $A^0 = 0$. But that means $A \cdot A$ cannot help but be negative, so $A$ is spacelike. QED
Another reason I recommend this approach is because
the fact that 4-acceleration is orthogonal to 4-velocity
(in the spacetime sense) is memorable and useful to know.
Finally, I will comment on the "and what does it mean" part of the question. The fact that 4-velocity always has the same 'length' ($U \cdot U =$ constant) makes 4-velocity a rather special 4-vector. If its length is constant then it must be evolving in a special way in order to ensure this. Its rate of change must be spacetime-orthogonal to itself. That is indeed exactly what we have shown above. Note this does not mean the spatial parts have to be 3-orthogonal to each other. Rather, the spacelike $dU/d\tau$ means that $U$ is getting a spacelike nudge $dU$ in each small interval of proper time. The change to the temporal part is never greater than the change to the spacial part.
A: Regarding the second question: what does is mean physically. It means that an accelerating particle stays on shell $(c=1)$:
$$ E^2- |\vec p|^2 = m^2 $$
That constraint means:
$$ (E+\delta E)^2 - (|\vec p+\delta\vec p|)^2 = m^2 =E^2- |\vec p|^2 $$
To 1st order:
$$ (E^2+2E\delta E)- (|\vec p|^2+2\vec p \cdot\delta\vec p) =E^2- |\vec p|^2 $$
$$E\delta E - \vec p \cdot\delta\vec p = 0$$
Dividing by the proper time and converting  to a proper differential, $\delta\rightarrow d$:
$$E\frac{dE}{d\tau} - \vec p \cdot\frac{d\vec p}{d\tau} = 0$$
With the four force:
$$ f^{\mu} = \frac{dp^{\mu}}{d\tau}=(\frac{dE}{d\tau},\frac{d\vec p}{d\tau})$$
and four momentum:
$$ p^{\mu}=(E,\vec p)$$
that says:
$$p^{\mu}f_{\mu} = 0 $$
Since $f^{\mu}= ma^{\mu}$ and $p^{\mu}=mu^{\mu}$, that means:
$$u^{\mu}a_{\mu} = 0 $$
Four-acceleration's orthogonality to four-velocity ensures that the four-velocity:
$$ u^{\mu} = (\gamma c, \gamma \vec v) $$
has a magnitude always equal to:
$$ u^{\mu}u_{\mu} =\gamma^2(c^2 -v^2)=c^2$$
A: Whether a vector is spacelike or timelike doesn't depend on the reference frame. Since we are considering a timelike four-velocity, we can find a frame where it is $u^\mu = (c, 0, 0, 0)$. In this frame, the statement that four-acceleration is perpendicular to four-velocity means $a^0 = 0$. Therefore,
$$a^\mu a_\mu = (a^0)^2 - (a^1)^2 - (a^2)^2 - (a^3)^2 = - ((a^1)^2 + (a^2)^2 + (a^3)^2) \leq 0$$
as desired.
A: There is a theorem saying that if a non-zero vector is orthogonal to a timelike vector, then it must be spacelike. Let A be a timelike vector and B a non-zero spacetime vector. Assuming that the signature of metric used is $(-,+,+,+)$ as well as A and B are orthogonal to each other, that is $A^0B^0=\overrightarrow A\cdot \overrightarrow B$, where $A^0$ and $\overrightarrow A$ denote the time component and spatial components of spacetime vector A respectively. The fact that A is timelike can be expressed by $|A^0|>  ||\overrightarrow A||$. Then $|A^0B^0|=|\overrightarrow A\cdot \overrightarrow B|\leq ||\overrightarrow A||\,||\overrightarrow B||$ (Schwartz's inequality). So we have $|A^0|\,|B^0|\leq ||\overrightarrow A||\, ||\overrightarrow B||.$ The fact that $|A^0|>  ||\overrightarrow A||$ implies $|B^0|\leq||\overrightarrow B||$, which means B must be a spacelike vector. Having this result in hand and knowing that the four-acceleration is always orthogonal to four-velocity (a timelike vector), we conclude that it is a spacelike vector.
The spacelike nature of four-acceleration means that it must lie outside the light cone, contrary to four-velocity. The orthogonality condition implies that the four acceleration could only alter the direction of the four-velocity, not the magnitude. In an instantaneous rest frame (IRF) in which the four-velocity of a particle at a certain point on its worldline is $u^{\mu}=(1,0,0,0),$ the four-acceleration must be of the form, $a^{\mu}=(0,\mathbf{a_0})$ where $\mathbf{a}$ is the proper acceleration. In this frame, the magnitude of the four-acceleration is $\mathbf{a_0}\cdot \mathbf{a_0}.$ Since this relation is an invariant (frame-independent relation), it holds in all frames and all observers will agree on the magnitude of the proper acceleration of that particle's worldline relative to the IRFs. 
For extra information, the norm (with respect to the Minkowski metric) of the four-acceleration, $||\mathbf{\overrightarrow a}||$ can be interpreted as the curvature, $a$, of a particle's worldline at a certain point, P. Then the quantity $a^{-1}$ is the curvature radius of the worldline at P. In Euclidean space, the curvature radius refers to the radius of the circle that approximates the best the worldline at P. Since Minkowski spacetime is not a metric space, this definition is not meaningful. There is, however, a second interpretation: consider a point P and its neighbouring point P' on a worldline as well as their corresponding four-velocity, then $a^{-1}$ refers to the distance at which the two hyperplanes, orthogonal to the above two four-velocities respectively, intersect.
