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And if so, then why does everybody keep asserting nothing can go faster than light speed (I'm implicitly assuming there could be something which we do not observe, which goes faster than light)?

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    $\begingroup$ Bullets travel faster than sound. It wouldn't take long to figure out things CAN travel faster than sound even if we couldn't see. The speed of sound is not even analogous to the speed of light. Things can't travel faster than the speed of light because it's a fundamental speed limit and not because of any physical limitations. $\endgroup$ Commented Nov 11, 2014 at 19:16
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    $\begingroup$ In the world we're using to imagining in our heads 0 is the slowest you can go and infinity is the fastest. It's actually possible to have a finite number be the fastest speed and that's how our universe really works. Even though this defies our intuition the math works out perfectly. $\endgroup$ Commented Nov 11, 2014 at 19:18
  • $\begingroup$ But if we could feel - warmth, we would have known there is something else out except for sound. Also, developing sight is not a coincidence. Living creatures can see exactly because there is electromagnetic radiation to be seen. So we can conclude (but not be 100% sure though) that if we do not sense anything else, there is nothing to be sensed ... $\endgroup$ Commented Nov 11, 2014 at 19:20
  • $\begingroup$ "If we do not sense anything else, there is nothing to be sensed" is obviously not correct. There are sound frequencies above and below our threshold of hearing and there are electromagnetic frequencies above and below our vision. For example many insects can seen into the UV range. We can't echo locate like bats either. And we don't have a sense of direction like birds. And we can't smell like dogs. $\endgroup$ Commented Nov 11, 2014 at 19:22
  • $\begingroup$ @BrandonEnright Sure, many things travel faster than sound, my question was if we would have been able to measure that it was travelling faster than sound. The answer is probably "yes", as I understand from the following example: B obstructs the light so that A is in the shade. B stops obstructing and shouts at the same time. A will feel the heat (radiation) before the sound. $\endgroup$
    – anderstood
    Commented Nov 11, 2014 at 19:26

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No. We can experimentally confirm that for a particle with rest mass $m$ and velocity $v$, its total energy is given by $E = mc^2/\sqrt{1 - v^2/c^2}$, and this equation goes to infinity as $v$ approaches the speed of light $c$; if anyone tried to put forward a similar equation with the speed of sound in place of the speed of light, this equation could easily be shown to give incorrect predictions for particles moving slower than the speed of sound. Likewise with various other equations that work for slower-than-light particles and approach 0 or infinity as $v$ approaches $c$, like the time dilation equation (which can predict how the decay rate of a given particle type changes as a function of its velocity).

For example, if you used a subsonic bullet with mass 0.0095 kg and speed 300 m/s or 0.8816 times the speed of sound s=340.29, then if you used an equation like the one above with s in place of c, and calculated kinetic energy as (total energy) - (rest-mass energy) = $(ms^2/\sqrt{1 - v^2/s^2}) - (ms^2)$ you'd predict a kinetic energy of $[(1/\sqrt{1 - v^2/s^2}) - 1]ms^2$ = $1.119ms^2$ = 1.119*0.0095*(340.29)^2 = 1231 Joules. Whereas in reality its speed is so low compared to the speed of light that you can just use the classical formula $E = (1/2) mv^2$ to calculate its kinetic energy, giving 0.5*0.0095*(300)^2 = 427.5 Joules, which is substantially smaller. You could easily determine which was correct by firing this bullet into a block of known mass M mounted on wheels and seeing how much velocity V it gained after stopping the bullet and absorbing its kinetic energy (making sure it was massive enough so that its change in velocity would be very small compared to the speed of sound, so that the formula $E=(1/2)MV^2$ could be used regardless of your assumptions about whether the limiting speed was the speed of sound or the speed of light). If the block plus wheels had a mass of 50 kg, the velocity gain upon absorbing kinetic energy E should be $V = \sqrt{2E/(50)}$, so if you predicted the bullet to have kinetic energy 1231 Joules you'd predict a velocity gain of 7.02 m/s, whereas if you predicted the bullet to have a kinetic energy of 427.5 Joules you'd predict a velocity gain of 4.14 m/s.

For experiments that show the corresponding energy formula does work with c in place of s, I think you'd want to look at particle collider experiments where the particles can attain a significant fraction of light speed, and the total energy of the particles produced by the collision must equal the energy of the particles before the collision, but since the rest mass of the particles coming out can differ from the rest masses of the particles going in, kinetic energy before the collision can transform into rest-mass energy after the collision or vice versa. Various experimental confirmations or relativistic predictions about how energy conservation should work can be found on this page.

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  • $\begingroup$ this equation could easily be shown to give incorrect predictions for particles moving slower than the speed of sound Could you give an example? Something with a liquid perhaps? $\endgroup$
    – anderstood
    Commented Nov 11, 2014 at 19:47
  • $\begingroup$ @anderstood - OK, I added an example in an edit. $\endgroup$
    – Hypnosifl
    Commented Nov 11, 2014 at 20:21
  • $\begingroup$ You gave an example experiment where, assuming $E=mc^2/\sqrt{1-v^2/V_0^2}$ (for the moment, $V_0$ is the maximum velocity), you could check that $V_0$ is greater than the speed of sound. That proves the answer is no, thank you. I'm just missing where this formula comes, but I guess it's basic knowledge of relativity. Oh, and thanks for the link too, it looks very interesting - but right now I don't have time to go into it unfortunately. $\endgroup$
    – anderstood
    Commented Nov 11, 2014 at 20:32
  • $\begingroup$ @anderstood - Yes, it's just a standard formula. And that's the formula for total energy in relativity, as I mentioned you can subtract off the rest-mass energy $E=mc^2$ to get the formula $(\frac{1}{\sqrt{1 - v^2/c^2}} - 1)mc^2$ for relativistic kinetic energy which was the one that was really useful in my example--looks like there's a derivation of it here. $\endgroup$
    – Hypnosifl
    Commented Nov 11, 2014 at 21:02

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