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Suppose there is a neutron star with a hole in the middle. Through this hole, scientists shoot a laser beam in such a way that it will pass through the gravity well of the neutron star but will not be bent by it.

I imagine that when photons are going in the hole, they blueshift, but after leaving they redshift (due to curved spacetime?). But how about the time they take from one side to the other? Will it be the same as if the neutron star wasn't there? If not, how do I calculate it?

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  • $\begingroup$ Before it can be calculated, the time needs to be defined more specifically. If it is the proper time associated with some observer, then where is the observer? and how will the observer know when the light-pulse enters and exits the hole? By receiving signals that are emitted by the devices that create the laser-pulse on one side and receive the laser-pulse on the other side? Those signals will themselves be bent by the neutron star's gravity along their way to the observer. $\endgroup$ – Dan Yand Jan 12 at 0:22
  • $\begingroup$ Thanks, assume a distant observer. What if there is a mirror on the other side and light comes back? And what if light is generated far away perpendicular to the hole, a semimirror reflect part into the hole and another part continues until it is far away from the neutron star and then another mirror reflects it perpendiculary, while the beam that goes through is reflected perpendiculary to meet this, in such way both would travel the same distance in flat spacetime. $\endgroup$ – Eduardo Schardong Jan 12 at 0:39
  • $\begingroup$ If I understand the configuration correctly, then it does address my question about how the time is defined. I'm assuming that the hole is along the axis of rotation of the neutron star (or that it's not rotating at all) and that the semimirror and the parallel path are both far enough away from the neutron star for gravity to be negligible. $\endgroup$ – Dan Yand Jan 12 at 3:07
  • $\begingroup$ Now, to answer the question, we would need to know (or choose) the density profile of the neutron star, and then we would need to calculate the resulting metric tensor -- or we could assume a relatively simple metric and calculate the required density profile, which would be easier. Once we have the metric tensor, we would need to use it to determine the equation for an axial null geodesic. Once we have that, the question can be answered. I expect the result to be different than if the neutron star weren't there. Maybe somebody who's actually done these calculations can post more details. $\endgroup$ – Dan Yand Jan 12 at 3:16
  • $\begingroup$ @DanYand I am unable to picture a scenario in which the light will pass through the well without being deflected by it. $\endgroup$ – Apoorv Khurasia Jan 12 at 5:30
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The time taken for the light to traverse the neutron star will be longer than it would be if the neutron star wasn't there. This because the velocity of light as measured by an observer far from the neutron star is reduced by the gravitational field of the neutron star. This has been confirmed experimentally by measurements of the Shapiro delay. In this experiment radar waves are bounced off the surface of a massive object and they are found to take slightly longer to return than expected because of the decrease in the speed of light near the massive body.

If you're interested in finding out more about how the speed of light is affected by gravity have a look at Speed of light in a gravitational field? As described in that question outside the neutron star the speed of light is given by the equation:

$$ v = c \left(1-\frac{2GM}{c^2r}\right) \tag{1} $$

where $M$ is the mass of the neutron star.

Inside the neutron star the equation for the speed of light is much more complicated, and indeed in general there is no simple equation for it. If the neutron star had a constant density then the speed would be given by the Schwarzschild interior metric and we'd get:

$$ v = c \left(\frac{3}{2}\sqrt{1-\frac{2GM}{c^2R}} - \frac{1}{2}\sqrt{1-\frac{2GMr^2}{c^2R^3}}\right)\sqrt{\left(1-\frac{2GMr^2}{c^2R^3}\right)} \tag{2} $$

where $R$ is the radius of the neutron star. To calculate the time required take equation (2) for the speed and integrate it to get the traversal time. I'll leave that as an exercise for the reader.

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  • $\begingroup$ Let's say the radius of the neutron star is 10% bigger than it's Schwarchild radius, then, according to equation 2, the speed of the photon will be less than 0 at center, what does that mean in physics? How a distant observer will see it? $\endgroup$ – Eduardo Schardong 21 hours ago
  • $\begingroup$ @EduardoSchardong the equation we get from the metric gives us $v^2$. At $r=0$ we get: $$ v^2 = c^2\left(\tfrac{3}{2}\sqrt{1-\frac{2GM}{c^2R}} - \tfrac{1}{2}\right)^2 $$ Then on taking the square root the positive root is the speed of the outgoing ray and the negative root is the speed of the incoming ray. What is curious, and I confess this hadn't occurred to me before is that there is a value or $R$ for which $v^2=0$. I'll have to think about this ... $\endgroup$ – John Rennie 4 hours ago
  • $\begingroup$ @EduardoSchardong As $R \to \tfrac{9}{8}r_s$ the pressure at the centre becomes infinite, so for values of $R$ less than this a static solution is not possible and the metric I've used to derive equation (2) no longer applies. That's why you get a weird result for the velocity when $R \le \tfrac{9}{8}r_s$. $\endgroup$ – John Rennie 4 hours ago

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