If the escape velocity of a neutron star is just less than that of light, will light come out very slowly from the star?
1 Answers
Particles moving at the speed of light (massless particles) and particles not moving at the speed of light (massive particles) are completely different beasts:
- A particle moving at the speed of light will keep moving at the speed of light no matter the physical conditions or processes it went through and no matter the observer frame.
- Particles not moving with the speed of light, on the other hand, may reach velocities close, but never equal to the speed of light in some frames. However, there will always be possible observers who would see this particle at rest.
That is, light escaping the field of any compact gravitating object, if escaping at all, will always escape at exactly the speed of light, no matter its gravitational binding. The energy required to release the photon from the gravitational potential is instead taken from its frequency (which by $E =hf$ is stores the photon energy). This is how gravitational redshift of light comes about.
When, however, we have a massive particle moving close, but not equal to the speed of light which is escaping a compact gravitating object, it will instead take the energy required to escape the potential from its speed. The speeds which are consumed by the escape of a massive particle from a neutron star are indeed quite large. Depending on the neutron-star model (we are still not sure about a lot of details about neutron stars) and depending on how generous we are with wild speculations about the neutron-star structure, the escape velocity for a massive particle might be something between $0.6-0.9$ times the speed of light.
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