广义相对论中的转动系统：匀角加速与匀角速度转动的空间曲率疑问

Hey, great question! Rotating reference frames and the curvature they perceive are a classic, accessible entry point into relativistic geometry. Let's break this down into the two cases you asked about:

First, let's start with intuition: if you're on a spinning carousel and throw a ball, you'll see it curve away from you due to centrifugal and Coriolis effects—but does that mean the space itself is curved? The answer depends on whether we're talking about non-relativistic or relativistic physics.

Non-Relativistic Approximation #

In Newtonian mechanics, we treat space as flat (Euclidean) regardless of rotation. The "curved paths" you observe are just the result of inertial forces acting on objects in the rotating frame. Mathematically, the Newtonian equation of motion in the rotating frame is:
$$\mathbf{F} = m\left(\mathbf{a} + \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r}') + 2\boldsymbol{\omega} \times \mathbf{v}'\right)$$
Here, the extra terms are inertial forces, not indicators of space curvature—you can still use standard Cartesian coordinates to describe flat space.

Relativistic Case (Special Relativity) #

Things get interesting when we bring in special relativity. Let's define an inertial frame $S$ where the origin is stationary, and a rotating frame $S'$ that spins around the $z$-axis with constant angular velocity $\omega$. The coordinate transformation between $S$ and $S'$ is:
$$
\begin{align*}
x &= x'\cos\omega t - y'\sin\omega t \
y &= x'\sin\omega t + y'\cos\omega t \
z &= z' \
t &= t'
\end{align*}
$$
The flat spacetime metric in $S$ is:
$$ds^2 = c^2dt2 - dx^2 - dy^2 - dz^2$$
Substitute the coordinate transformation into this metric to get the metric for $S'$:
$$ds^2 = \left(c^2 - \omega^2(x'2 + y'^2)\right)dt'2 - 2\omega(y'dx' - x'dy')dt' - dx'^2 - dy'^2 - dz'^2$$

Now, to look at the "space" perceived by an observer in $S'$, we focus on measurable geometric quantities like the ratio of a circle's circumference to its radius:

• In the inertial frame $S$, a circle of radius $r$ has circumference $2\pi r$.
• In $S'$, an observer measuring the circumference will encounter length contraction along the tangential direction (since each segment of the circle moves at $v = \omega r$ relative to $S$). The proper length of each tangential segment in $S'$ is $\gamma$ times its length in $S$, where $\gamma = 1/\sqrt{1 - v^2/c2}$.

This means the measured circumference in $S'$ is $L' = \gamma \cdot 2\pi r$, while the radius (measured radially, no length contraction) remains $r$. So:
$$\frac{L'}{r} = 2\pi\gamma > 2\pi$$
This violates Euclidean geometry, where the ratio is exactly $2\pi$. For the rotating observer, space is curved (non-Euclidean)—this isn't just an illusion from inertial forces; it's a measurable geometric effect tied to special relativity's length contraction and the frame's simultaneity rules.

Now, what if the observer's angular velocity increases at a constant rate $\alpha$ (so $\omega(t) = \alpha t$)? This is a time-dependent rotating frame, and the analysis is a bit more complex.

Intuitive Take #

First, the observer will feel two inertial forces:

• A centrifugal force that grows with time (since $\omega$ increases, $F_{\text{centrifugal}} = m\omega^2 r$)
• A tangential inertial force (opposing the angular acceleration, $F_{\text{tangential}} = m\alpha r$)

From a geometric perspective, the length contraction effect gets stronger over time as $\omega$ increases. So the circumference-to-radius ratio we calculated earlier will grow larger as time passes—meaning the space's curvature isn't constant; it evolves with time.

Mathematical Sketch #

Using a similar coordinate transformation to the constant $\omega$ case, but with $\theta(t') = \frac{1}{2}\alpha t'^2$ (since $\omega = d\theta/dt' = \alpha t'$), substitute into the inertial frame metric. After expanding and simplifying, the metric for the accelerating rotating frame becomes:
$$ds^2 = \left(c^2 - \alpha^2 t'^2(x'2 + y'^2)\right)dt'2 + 2\alpha t'(y'dx' - x'dy')dt' - dx'^2 - dy'^2 - dz'^2$$

Looking at the spatial slices (simultaneous events for the observer), the metric now has time-dependent coefficients. This means the spatial geometry changes over time: at any instant $t'$, it's similar to the constant $\omega = \alpha t'$ case, but as $t'$ increases, the curvature (measured by the circumference-radius ratio) becomes more pronounced.

In short: the observer perceives a time-dependent curved space—the curvature gets stronger as their angular velocity increases.

Key Takeaways #

• Constant angular velocity: In special relativity, the observer measures a static non-Euclidean (curved) space. In Newtonian physics, space remains flat, with only inertial forces creating apparent curved paths.
• Constant angular acceleration: The observer perceives a curved space that becomes more curved over time, with both centrifugal and tangential inertial forces at play.

内容的提问来源于stack exchange，提问作者Noam Chai