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In the movie X-Men: Days of Future Past, the mutant character Quicksilver possesses the ability to move very fast. A particularly memorable scene in the movie is one where he runs around very quickly in a room, diverting speeding bullets and creating mayhem, all the while listening to music off headphones plugged into a portable player.

My question is: Would someone moving that fast be able to hear music as if it were being played normally?

The video of the particular scene in question, if you haven't seen it, is here: http://youtu.be/qtnMy2aSOWQ

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  • $\begingroup$ My initial gut feeling was "YES", because the headphones, player, air in his ear etc were all moving in the same relative frame as him. But I wonder what happens to the air in his ears as he moves at supersonic speeds through the air in the room. Wouldn't it get sucked out of his ears? What would that then do to the sound? $\endgroup$ Feb 21, 2015 at 7:55

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His clothes, the music player and so on might of course move with him (supposing they would be able to withstand the rapid acceleration and decelleration involved, which I doubt), but the music player would have to play the music much faster all of a sudden for it to appear the same. It's entirely possible of course for Quicksilver to have a specially modified cassette player and/or recording specifically for that purpose. That is of course if Quicksilver actually speeds up, there'd be less of a problem if he slows everything else down…

Edit: Some math

If we assume that the bullets travel at $340\,\mathrm{m/s}$ and the shooters are about $5\,\mathrm{m}$ from the X-Men, it would take the bullets about $\frac{340\,\mathrm{m/s}}{5\,\mathrm{m}}=0.1471\,\mathrm{s}$ to hit their targets. If we now assume that Quicksilver moves about $100\,\mathrm{m}$ in this time we arrive at a speed of $v=\frac{100\,\mathrm{m}}{0.0147\,\mathrm{s}}=6802.7211\,\mathrm{m/s}$, or about 6.8 km/s. Plugging that into the equation for the Lorentz factor $$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ we arrive at $\gamma=1.00000000026$, which is pretty tiny.

However we have to consider who is actually moving here and who is at rest. Since Quicksilver is moving he is the one actually slowing down relative to the room (as in the twin paradox). That means that in the end (when he arrives back at the start) his clock will have slowed down relative to the room, not the other way around.

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  • $\begingroup$ Could you further explain why the music player would have the play the music faster? As it's in the same relative frame as him, wouldn't it just have to play as normal? Am I missing something? $\endgroup$ Feb 21, 2015 at 9:29
  • $\begingroup$ Are you referring to a relativistic frame? We have to be careful here. Objects moving at relativistic speeds do slow down relative to observers at rest with respect to the moving observer. That would have the opposite effect in this case though, for example that while everyone else in the room measures the scene to take 0,1 second, they measure a "clock" in the reference frame of Quicksilver to only pass 0,05 seconds (Warning: approximation). The twin paradox would apply here by the way. (1/2) $\endgroup$
    – Graumagier
    Feb 21, 2015 at 9:55
  • $\begingroup$ Yes, relativistic frame. As you can see from the video, that is obviously the case, since Quicksilver moves very fast, leading to time dilation, enabling him to alter the trajectories of the bullets before they hit the other X-men, etc. $\endgroup$ Feb 21, 2015 at 9:58
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    $\begingroup$ See my second comment. Relativistic time dilations assumes that he moves at a significant fraction of the speed of light. Let's assume the bullets travel at 340 m/s and the shooters are about 5 m from our heroes. If we are very generous and assume that quicksilver moves 100 m in the 0.0147 s it takes the bullets to reach their targets we arrive at a speed of a about 6.8 km/s for Quicksilver. The speed of light is a whopping 300,000 km/s, which would give a time dilation factor (Lorentz factor) of 1,00000000026. Which is, well, tiny. $\endgroup$
    – Graumagier
    Feb 21, 2015 at 10:25
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    $\begingroup$ Moving at $\frac{c}{3}$ would still only result in a time dilation of about $\gamma=1.06$, which is still not very much. What's more important though is that the moving "clock" slows down relative to a static observer, which means that after Quicksilver returns to the start, less time will have passed for him than in the surrounding room (again, see the twin paradox). That produces the exact opposite effect from what you are supposing. $\endgroup$
    – Graumagier
    Feb 21, 2015 at 11:21

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