- #36
PeterDonis
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WannabeNewton said:I don't think the stationary particle hanging at the other end of the string, following an orbit of the time-like KVF, undergoes frame dragging in the coordinate system adapted to the time-like and axial KVFs.
No, it should, because such a particle is not a ZAMO. Only a ZAMO should experience no tangential force.
WannabeNewton said:[tex]\frac{d\phi}{dt} = \frac{d\phi/d\tau}{dt/d\tau} = \frac{u^{\phi}}{u^{t}} = \frac{\xi ^{\phi}}{\xi ^{t}} = \frac{\delta ^{\phi}_{t}}{\delta ^{t}_{t}} = 0[/tex].
This just shows that the orbits of the timelike KVF have constant angular coordinate, which we knew anyway because we've chosen a chart specially so that orbits of the timelike KVF have all spatial coordinates constant. It doesn't show that the component of proper acceleration in the tangential direction is zero. (Remember that proper acceleration will have terms in the connection coefficients as well as terms in the partial derivatives of the coordinates. For example, even in Schwarzschild spacetime the orbits of the timelike KVF, in Schwarzschild coordinates, have constant spatial coordinates, but the proper acceleration in the r direction is nonzero.) The tangential proper acceleration shouldn't be zero, because the "flow of frame dragging" in the particle's vicinity has a nonzero tangential component.
Conversely, a ZAMO (what you are calling a "locally non-rotating observer") has nonzero angular velocity (its angular coordinate changes), but its proper acceleration should have zero angular component, because it is "moving along with the flow" of frame dragging in its vicinity.
We really need to compute the proper acceleration, maybe starting with a special case like Kerr spacetime.
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