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1

Your strategy is fine (though I haven't carefully checked if your rotations are correct). Since rotations don't commute, you need to be careful to apply them in the correct order. You say that your rotation about the x-axis, let's call it $R$, followed by your rotation around the z-axis (be careful whether it's around the old or the new z-axis!), let's call ...


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It looks like this picture on the right from Misner-Thorne-Wheeler: It's almost a straight line. But can I add this: I don't like Kruskal-Szekeres coordinates at all. Take a look at the picture on the left. That shows the object's path on the Schwarzschild diagram. Note how it's truncated vertically? The vertical axis is the time axis. In order to ...


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You release a ball in space. Measure its distance every 1s or C second. if you find $S(t)-S(t+c) $ is not constant , where c is a constant. then you are in a Non inertial system


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Inertial frame of reference is such that free bodies move with constant velocity. If you detect free body accelerating, the frame is not inertial.


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The sign is simply a convention: a convention is being used which identifies motion away from the large, gravitational body (against gravity) as having a positive sign. Thus motion in the direction of gravity is negative. This convention is useful because when we perform work on an object by raising it against against gravity, we increase its gravitational ...


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There is a subtle difference between displacement and position. Position is a vector measured relative to a defined origin and is often designated by the vector symbol $\vec{r}$. In a Cartesian coordinate system, $$ \vec{r}=x\hat{i}+y\hat{j}+z\hat{k},$$ where $x$, $y$, and $z$ are the individual coordinate values, positive or negative, relative to the ...


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Suppose I throw a ball from my chest-height straight up into the air: it travels upwards for a time $\tau$ before it reaches its maximum, then it falls down for a time $\tau$ too. We know that it achieves the height $h = \frac 1 2 ~g~\tau^2$ above my chest-height in this time where $g \approx \text{9.81 m/s}^2$, so if we take my chest-height as the value ...


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Displacement is the vector version of distance, so it has a magnitude (the distance) and a direction. If the motion is only in one dimension, as in free fall, then the direction manifests only as positive and negative, or up and down but you are free to define whether up is positive or negative (and similarly for down) as long as you are consistent within ...


4

You're not missing anything. You are right, $k=\omega/c$. The argument $\sqrt{\frac{\omega ^2}{c^2}-k_z^2}$ in the Bessel function is the projection of the wavevector onto the radial direction. The use of Bessel functions beclouds what's going on a bit. Recall that a plane wave with wavevector $\vec{k}$ has the functional variation $\psi(\vec{r}) = ...



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