Why are the Sun's outer layers(photosphere) moving slower than the inner layers? Veritasium made a video on this but couldn't make any sense of it. Please explain it to me at the level which a high-school-er can understand.
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Brief summary of what was said in the referenced video:
Photons being emitted from the sun collide with particles of dust in the solar system, slowing down the particles of dust through the transfer of momentum in the collision.
Equally, photons collide with material inside the sun (and on the surface), slowing down the particles of the sun.
Explanation
During a collision between two bodies, there is a transfer of momentum between the two bodies. This means that, if two bodies moving towards each other collide, each body will experience a force acting in the direction opposite the body that it collided with.
Since the particles of the sun have some angular momentum, they are moving into (towards) photons which are being emitted from the layers of sun below the layer in question.
Given the effect of collision on bodies moving towards one another, as the sun’s particles are moving towards the photons, we can then say that the photons will exert a force, on the particles of the sun in motion, opposite in direction to the angular velocity of the sun (slowing it down).
This effect occurs on all layers of the sun, but is most noticeable nearer the outside because the momentum of the sun near the outside is so much lower (due to lower density).
Side Note
This effect will occur inside all stars, not just the sun.
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$\begingroup$ So do the photons gain momentum? $\endgroup$ Commented Oct 24, 2017 at 5:46
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1$\begingroup$ @CaptaineCode Yes—they gain some energy through the collision. $\endgroup$ Commented Oct 24, 2017 at 5:59
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$\begingroup$ So what happens when a photon gains momentum? It obviously doesn't change velocity. What changes? $\endgroup$ Commented Oct 24, 2017 at 6:03
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1$\begingroup$ @CaptaineCode Well, relative mass is an 'obsolete' concept, but it can still be a relatively good approximation. According to $E^2=p^2c^2+m_{rest}^2c^4\Rightarrow E=pc$, where $p$ is momentum. The increase in energy will cause an increase in momentum, because of an increase in 'relativistic mass' (remember that momentum has two components: $p=mv$). $\endgroup$ Commented Oct 24, 2017 at 6:09