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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?

Thorondor
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Eduardo Schardong
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  • 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. – Eduardo Schardong Jan 12 '19 at 00:39
  • @DanYand I am unable to picture a scenario in which the light will pass through the well without being deflected by it. – Apoorv Jan 12 '19 at 05:30

1 Answers1

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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.

PM 2Ring
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John Rennie
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  • 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? – Eduardo Schardong Jan 16 '19 at 15:16
  • @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 ... – John Rennie Jan 17 '19 at 08:24
  • @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 $R \le \tfrac{9}{8}r_s$. – John Rennie Jan 17 '19 at 08:52