The relativistic energy-momentum equation is: E 2 </msup> = ( p c

spazzter08dyk2n

spazzter08dyk2n

Answered question

2022-05-14

The relativistic energy-momentum equation is:
E 2 = ( p c ) 2 + ( m c 2 ) 2 .
Also, we have p c = E v / c, so we get:
E = m c 2 / ( 1 v 2 / c 2 ) 1 / 2 .
Now, accelerating a proton to near the speed of light:
0.990000000000000 c => 0.0000000011 J = 0.01 T e V 0.999000000000000 c => 0.0000000034 J = 0.02 T e V 0.999900000000000 c => 0.0000000106 J = 0.07 T e V 0.999990000000000 c => 0.0000000336 J = 0.21 T e V 0.999999000000000 c => 0.0000001063 J = 0.66 T e V 0.999999900000000 c => 0.0000003361 J = 2.10 T e V 0.999999990000000 c => 0.0000010630 J = 6.64 T e V 0.999999999000000 c => 0.0000033614 J = 20.98 T e V 0.999999999900000 c => 0.0000106298 J = 66.35 T e V 0.999999999990000 c => 0.0000336143 J = 209.83 T e V 0.999999999999000 c => 0.0001062989 J = 663.54 T e V 0.999999999999900 c => 0.0003360908 J = 2 , 097.94 T e V 0.999999999999990 c => 0.0010634026 J = 6 , 637.97 T e V 0.999999999999999 c => 0.0033627744 J = 20 , 991.10 T e V
If the LHC is accelerating protons to 7 T e V it means they're traveling with a speed of 0.99999999 c.
Is everything above correct?

Answer & Explanation

Cristal Obrien

Cristal Obrien

Beginner2022-05-15Added 16 answers

Yes you are correct.
If the rest mass of a particle is m and the total energy is E, then
E = γ m c 2 = m c 2 1 v 2 c 2 ,
thus
v c = 1 ( m c 2 E ) 2 1 1 2 ( m c 2 E ) 2
The proton rest mass is 938   M e V, so at 7   T e V , the proton's speed is
1 v c = 1 2 ( 938 × 10 6 7 × 10 12 ) 2 = 9 × 10 9
meaning v 0.999 999 991   c

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