Energy

Taking a closer look at LHC

In this section we present some simple calculations in order to compare LHC energy with some more familiar situations. In the section Relativity  a more detailed discussion about the energy in a collider (like LHC) can be found in relativistic terms.


 

 

How much energy are we talking about?

7 TeV = 7·1012 eV · 1,6·10-19 J/eV = 1,12·10-6 J

It doesn´t look like a lot of energy

For the ALICE experiment, each ion of Pb-208 reaches 1150/2 = 575 TeV.

So, the energy per nucleon is: 575/208 = 2,76 TeV


 

Let´s calculate the kinetic energy of an insect of 60 mg flying at 20 cm/s:

Ek = ½ m·v2     Ek = ½ 6·10-5·0,2 2  ~ 7 TeV

That  is, in LHC each proton will reach an energy similar to that of an annoying ... MOSQUITO!

            
But we have to keep in mind that this mosquito has 36 thousand trillion nucleons, whereas the 7 TeV in the LHC will be concentrate in one sole proton.
 
 
 

Perhaps the comparison is not yet very convincing when it comes to assessing the energetic significance of 7 TeV per proton.

Let's calculate the energy stored in one of the bunch of protons:

Ebunch =7 TeV/protón x 1,8·1011 protones/bunch ~ 1,3·1024 eV/bunch

Ebunch ~ 2·105 J

Let's calculate now the kinetic energy of "a bunch" of 200 runners of 65 kg average mass in an Olympic marathon final, with a speed of ~ 20 km/h:

Ec =200·[ ½ ·65 · 5,62 ] ~ 2·105 J

That tiny packet of protons circulating in the LHC has the same energy as 200 marathon runners.

 
 
 
 

Let's take another example.

A powerful motorbike 150 kg travelling at 190 km/h_

Ek = ½ ·150 · 52,72 ~ 2·105 J

So if a bunch of protons collides with you the impact is similar to that produced by a powerful motorbike travelling at 190 km/h.

 


And what is the mass of that bunch of protons?

The rest mass is:

m0 =[1,8·1011 protons/6,0·1023 protons/mol] x [1 g/mol] ~ 0,3·10-12 g = 0,3 picograms

Taking into account that this bunch goes at a speed close to the speed of light, we have that the relativistic parameter is:

γ = 7460  (see here...)

We can then consider a "relativistic mass":

m = 0,3 picogramos x 7460    m = 2,1 nanograms

If you are lucky to avoid that "2,1 nanograms motorbike", "don´t worry", there are 2807 following it. And if you decide to change lanes, the equivalent is coming in the opposite direction.

 


Another significant example can be made with a golf ball of mass 45 g. If this 2·105 J could be transferred to it in the form of kinetic energy, it would reach (in vacuum) a speed of 9 match.

2·105 J = ½ ·45·10-3 · v2  v ~ 3000 m/s  (~ 9 match)

I wouldn't want that golf ball to hit me....

We can go even further with this analogy.

 

In 1971 astronaut Alan Shepard (Apollo 14) hit several golf ball on the surface of the Moon.

What would have happened if he had communicated the energy of 2·105 J?

Assuming that Shepard could do it (and that the golf ball could withstand such a hit), the golf ball would come out with a velocity of 3000 m/s ... but the escape velocity on the Moon is 2400 m/s !!!

Therefore, with that speed, and given the high vacuum on the Moon, the ball would escape from the Moon forever.

 

 

 

Another calculation which can show the enormous amount of energy reached is:
1,29·105 J / bunch  x 2808 bunches ~ 360 MJ
 
-Stored beam energy-
 And that is equivalent to

77,4 kg  of  TNT

The energy content of TNT is 4.68MJ/kg (Beveridge 1998).

 (Naturally, as the number of protons/bunch increases, as happens after long shutdown, LS, this stored energy in the beam increases).

 

 

The Heat of Fusion of Gold is ΔHF = 63,71 kJ/kg and the Molar Heat Capacity is 25,42 J/mol·K

So, 360 MJ are enough to take 1500 kg of Gold from 25ºC to total fusion    1,5 Tonnes of Gold.

Obviously such an amount of energy can not be supplied instantlyIn fact the process lasts over 20 min through a chain of different accelerators.

AUTHORS


Xabier Cid Vidal, PhD in experimental Particle Physics for Santiago University (USC). Research Fellow in experimental Particle Physics at CERN from January 2013 to Decembre 2015. He was until 2022 linked to the Department of Particle Physics of the USC as a "Juan de La Cierva", "Ramon y Cajal" fellow (Spanish Postdoctoral Senior Grants), and Associate Professor. Since 2023 is Senior Lecturer in that Department.(ORCID).

Ramon Cid Manzano, until his retirement in 2020 was secondary school Physics Teacher at IES de SAR (Santiago - Spain), and part-time Lecturer (Profesor Asociado) in Faculty of Education at the University of Santiago (Spain). He has a Degree in Physics and a Degree in Chemistry, and he is PhD for Santiago University (USC) (ORCID).

CERN


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 IMPORTANT NOTICE

 For the bibliography used when writing this Section please go to the References Section


© Xabier Cid Vidal & Ramon Cid - rcid@lhc-closer.es  | SANTIAGO (SPAIN) |

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