# Why is 7 TeV considered as a big amount of energy?

Considering that $7$ TeV is more or less the same kinetic energy as a mosquito flying, why is it considered to be a great amount of energy at the LHC?

I mean, a giant particle accelerator that can only provide 7 TeV of energy? (14 in the mass center, if I understood well). Is it because particles are so small that this amount of energy, in proportions, is then really huge?

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I've deleted some off topic comments. Remember that comments are for requesting clarification and suggesting improvements to the post. – David Z Jan 14 at 19:53

## 4 Answers

7 TeV is not that much kinetic energy, that has been covered by your question and previous answers.

However, in the context of a proton, with a rest mass of $1.672×10^{−27}~\mathrm {kg}$ (very, very little mass), when a single proton has 7 TeV then it is travelling at a specific speed:

$$E= mc^2$$ \begin{align}E& = E_0 + \text{KE}\\ \text{KE}&=E- E_0\\ &= mc^2 -m_0c^2\\ &= \gamma m_0c^2 - m_0c^2\\ &= m_0c^2 \left(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}- 1\right)\\ \implies 1+ \frac{\text {KE}}{m_0c^2}&= \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\\ \implies {\sqrt{1-\frac{v^2}{c^2}}}&=\frac{1}{1 + \frac{\text{KE}}{m_0c^2}}\\ \implies 1-\frac{v^2}{c^2} &= \left(\frac{1}{1 + \frac{\text{KE}}{m_0c^2}}\right)^2 \\ \implies \frac{v}{c}&=\sqrt{1 -\left(\frac{1}{1 + \frac{\text{KE}}{m_0c^2}}\right)^2 } \end{align}

For a proton at 7 TeV, this is 99.9999991% times the speed of light

Source

Now, keep in mind this is for EACH proton in two BEAMS, each one of them having 7 TeV, travelling through a superconductor cooled by helium, and then colliding for a sum of 14 TeV.

Each proton beam contains 2,808 'bunches' of protons, and each 'bunch' contains $1.15 \times 10^{11}$ protons, so each beam then consists of 362 MJ (Megajoules).

This gives a total kinetic energy of 724 MJ just in the beams alone: about 7 times the kinetic energy as landing a 55 tonne aircraft at typical landing speed according to the Joules Orders of Magnitude wiki page, or with that $59~\mathrm{ m/s}$ landing speed, 218.9 tonnes, so about a safely loaded Airbus A330-200 (maximum takeoff weight of 242 tonnes)

Add to that the energy required to keep the ring supercooled enough so it remains superconductive, accelerate the beam in the first place, keep accelerating it so it doesn't lose velocity, light and heat and power the facility.

"At peak consumption, usually from May to mid-December, CERN uses about 200 megawatts of power, which is about a third of the amount of energy used to feed the nearby city of Geneva in Switzerland. The Large Hadron Collider (LHC) runs during this period of the year, using the power to accelerate protons to nearly the speed of light. CERN's power consumption falls to about 80 megawatts during the winter months."

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nice answer ; LHC has an installation of 120 MW ( cern ) – igael Jan 15 at 11:34
+1 because this answer clarifies that 7 TeV is for a single proton, and calculates the (quite impressive, I think) total beam energy. – Dubu Jan 15 at 14:48
I assume $1.15\times 10^{11}$ is an approximate number related to the machine picking up "a handful" of protons. But what is the significance of the precise number 2808? (E.g., why is that not just "about 3000"?) – Hagen von Eitzen Jan 15 at 20:32
1.15 x 10^11 is the design intensity per bunch, but it widely varies, though that's a faire estimate. 2808 comes from the spacing between the bunches (25ns) and the circumference of the machine. – Cedric H. Jan 15 at 23:15
The $-1$ in the $6$th row, shouldn't it be next to the fraction? – Time Master Jan 16 at 17:17

The energy of a bullet is around 735 joules (see bullet details here). This is about the same energy that I have when I'm running at about 4.6 m/s. Would you rather be hit by me or the bullet? The bullet kills you because it concentrates all the energy onto a small impact area while my impact area is rather larger (and sadly getting even larger as I age :-).

You're quite correct that 7 TeV isn't a lot of energy. In fact it's about one micro-joule. However, that energy is concentrated onto a fantastically tiny impact area (about $10^{-18}$ metres across in the LHC), so the energy density is fantastically high. It's getting the one micro-joule concentrated into such a tiny impact area that is extraordinarily hard.

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Thank you so much! Then I thought well when I thought about the same energy concentrated in a small beam of particles! Now it's clear, definitely ^^ – Time Master Jan 14 at 17:49
@KimPeekII Please don't forget to accept an answer if it solved your question. – Ruslan Jan 15 at 13:51

So, considering that 7 TeV is more or less the same kinetic energy of a mosquito, why is considered to be a great amount of energy in LHC?

Like other people said, it's not a great amount of energy. However, it's concentrated in a very tiny space. Just think of how much a mosquito is bigger than a subatomic particle.

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7 or 77 TeV at how much current? If it's a very low current, such as just one electron, the energy is rather low by everyday standards, but still very highly concentrated, enough the convert some of the energy into matching matter and antimatter particles.

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Electron Volts are a unit of energy, not potential difference. I might be wrong here, since PD is energy... but eV is more comparable with Joules than Volts – Brayton Jan 15 at 13:25
They are indeed both units of energy: (1 ev = 1.60218e-19 joules, or, and i just learned this word, 160 zeptojoules, awesome word). An electron volt is the energy required to move 1 electron across a voltage potential difference of one volt. It's pretty small. – Max Williams Jan 15 at 16:17
@Brayton it has nothing to do with potential difference. A synchrotron (which the LHC essentially is) is characterized, among other things, by the relativistic energy of particles in the storage ring and by the current circulating around the ring (i.e. the number of particles). – LLlAMnYP Jan 16 at 10:32

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