Accelerated protons and time dilation

[Natural]

If time dilation is correct then how can a proton accelerated to 99.9999% c ever reach the end of the tunnel? Shouldn't they have so much time dilation that their time is almost completely stopped relative to the collider?

[modest]

If time dilation is correct then how can a proton accelerated to 99.9999% c ever reach the end of the tunnel? Shouldn't they have so much time dilation that their time is almost completely stopped relative to the collider?

:shrug:

 

Time dilation is:

 

[math] \Delta t = \Delta t' \sqrt{1-v^2/c^2}[/math]

 

where [math]\Delta t'[/math] is the time we measure with our clock and [math]\Delta t[/math] is the time that the proton would measure if it had a clock. So, while our clock advances one second, the proton's clock advances:

 

[math]\Delta t = 1 \sqrt{1-.999999^2/1^2} = 0.0014 \ s[/math]

 

0.0014 seconds from our perspective. So, the proton slows in time (it's 'clock' slows). But, it does not slow to a stop. It wouldn't be until the proton reached the speed of light that its clock would stop completely, but this is impossible by special relativity because nothing with a rest mass can reach the speed of light.

 

~modest

[Natural]

So... I'm no mathematician but from what I can figure then instead of taking .00008 seconds to go around a 16 mile loop © it would take .00022 seconds. Right?

Oh and I think you missed a decimal point... should have been .0001342 instead of .0014.

(I think I just heard one of my old teachers say... "show your work"... so I used .999,999,991 c and therefore delta t for 16 miles would be .0001342 + .00008 = .00022)

:D

Headlines should have read... "Protons from the future smash together at LHC!"

:shrug: :hihi: :D

And thanks for the info.

:)

[modest]

So... I'm no mathematician but from what I can figure then instead of taking .00008 seconds to go around a 16 mile loop © it would take .00022 seconds. Right?

 

If it took .00008 seconds from our perspective (which would make sense if it were traveling at near the speed of light) then it would take considerably less time than that from the proton's perspective. It would take:

 

[math]\Delta t = 0.00008 \sqrt{1-.999999^2/1^2} = 1.21 \times 10^{-7} \ s[/math]

 

0.000 000 121 seconds from the proton's perspective. That's how much time would transpire on the proton's clock for it to make the 16 mile trip. We should also keep in mind that what is 16 miles in our frame is considerably less than that in the proton's frame because of length contraction.

 

Oh and I think you missed a decimal point... should have been .0001342 instead of .0014.

 

I don't believe so. If you copy and paste this:

 

1*sqrt(1-.999999^2/1^2)

 

into google, you should get 0.0014.

 

"99.9999% c" is "0.999999 c".

 

Headlines should have read... "Protons from the future smash together at LHC!"

:eek_big: :hihi: :D

And thanks for the info.

:)

 

:)

 

No trouble, mate.

 

~modest