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Christine Nattrass | USLHC | USA

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A TeV, measured in chocolate and coffee

We toss around the term “TeV” – a teraelectron volt, 1012 electron volts (eV). But how much energy is it really?

An electron volt is the energy an electron gains when it is accelerated through a potential difference of one volt. An electron volt is defined as a unit of energy. (Various prefixes are defined here.)

Let’s put this in terms we can all understand. A Lindt 70% cocoa chocolate bar has 194 Calories. To convert this to electron volts:

194 Calories × 1000 calories/Calorie ×4.2J/calorie / 1.6 × 10-19 J/eV = 5 × 1024eV

(Note that the dietary unit, a Calorie, is 1000 calories, the amount of energy needed to raise the temperature of one mL of water by one degree Celsius.) So it would take a hundred billion (1011) proton-proton collisions at top energy (14 TeV in the center of mass) to get the same amount of energy as in a chocolate bar.

The difference is how much space we pack that energy into. A proton has a volume of roughly 1 fm3, or about 10-39 cm3. A Lindt chocolate bar is about 10 cm x 1/2 cm x 20 cm = 100 cm3. A chocolate bar then has an energy density of about 194 Cal/100 cm3, or around 2 Cal/cm3. A proton-proton collision at 14 TeV has an energy density around 14 x 1012 eV/10-39 cm3 x 1.6 × 10-19 J/eV *1000 Calorie/4.2J = 5 x 1035 Cal/cm3. So our proton-proton collisions have an energy density about 1035 times a chocolate bar.

We also use an electron volt as a unit of temperature. An atom in a monatomic (helium, argon, etc.) ideal gas has a kinetic energy of 3/2kBT where kB is the Boltzmann constant. The factor in front (3/2) is different for different systems. For instance, it’s 5/2 for a diatomic gas, such as hydrogen (H2), oxygen (O2), or nitrogen (N2). But the energy is usually kBT times some factor between 1-10. So to convert an electron volt into a unit of temperature, we use eV=kBT and T=eV/kB=11604 Kelvin.

So how hot is a cup of coffee in electron volts? When I worked at a coffee shop in high school, we made our cappuccinos and lattes at 160°F (71°C). This works out to be 344K, or 0.03 eV. So a proton moving at 7 TeV is about 100,000,000,000,000 (1014) times more energetic than the average molecule in a cup of coffee.

The Quark Gluon Plasma created at the Relativistic Heavy Ion Collider is at a temperature of about 170 MeV. (Note this is the temperature of the medium produced, not the energy of the incoming beam.) The fluid we’ll create at the LHC will be hotter – over ten billion (1010) times hotter than a cup of coffee.

These collisions are hot stuff!

[Note these are all what we call “back-of-the-envelope” calculations. The goal is to figure out the right order of magnitude for various quantities, not to do a detailed, precise calculation.]

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  • Jim Cullen

    Using eV for a measure of temperature – that is interesting. I never dawned on me to consider an eV as a measure of temperature. Temperature is ‘average molecular kinetic energy’, so it does make sense – but one question to see if I understand this correctly. One electron vs one atom of diatomic hydrogen vs one atom of diatomic oxygen – these are vastly different masses. So, at an equal temperature, these three example particles would have vastly different speeds. Non-relativistically, using an eV to measure a temperature is a gauge of the energy of a single particle/atom, in the sense that E(k)=(mv^2)/2, regardless of the mass or speed of the particle. A 400°K electron is travelling much faster than a molecule of 400°K diatomic oxygen, correct?

  • Brian of Madison

    Wow. Not that I know much on the topic, but…

    When I think of the megawatts required to maintain and operate the LHC on a day to day basis, that is an astronomical amount of energy loss, or potentionally tons of chocolate bars burnt to carbon and water, compared to acheiving a billion billion collosions worth one candy bar at 14 TEV center of mass.

    How could the LHC be made to be more efficient?

  • Christine Nattrass

    Hi Jim –

    Yes, an electron at 400K is going much faster than a molecule of oxygen at 400K.  Let me address the case of two gases – largely because this is an easier question to address.  Let’s compare hydrogen (H2, diatomic, mass of 2 amu), helium (He, monatomic, mass of 4 amu), and xenon (Xe, monatomic, mass of 131 amu) at the same temperature.  The energy of a molecule/atom gas (at low enough energies, densities and pressures that we can neglect van der Waals forces) is:
    Nf * 1/2 Kb T
    where Nf is the number of degrees of freedom the gas has to move in.  A monatomic gas can move in three dimensions – that is, there are three degrees of freedom (Nf=3).  On average, there will be an equal amount of energy in the x, y, and z directions. (See http://en.wikipedia.org/wiki/Maxwell%E2%80%93Boltzmann_distribution for the full gory details.) A diatomic gas can move in three dimensions, but the two atoms can also rotate around each other and vibrate relative to each other – so there are five degrees of freedom (Nf=5).
    So for a given molecule of gas
    Nf * 1/2 Kb T = 1/2 m v2
    v = sqrt( Nf Kb T / m )
    So at a given temperature hydrogen has a speed roughly twice that of helium and about ten times that of xenon.

    The reason it’s hard to address the question of an electron at 400K is because it’s not really clear what an electron gas is – it would be most likely that there’d be an electron bound in something else and then you can’t treat it like an ideal gas, so it’s more complicated to figure out what its kinetic energy is. Without knowing much about exactly how the electron is bound, I can say qualitatively that electrons at 400K should generally be going faster than a molecule of oxygen at 400K because oxygen is roughly 10,000 times heavier than an electron.

    ——————-
    Hi Brian –

    There already are a lot of design features intended to reduce the amount of energy required. For instance, helium is captured and recycled at CERN. Everything is insulated as well as possible. It’s not exactly out of environmentalism but because it’s simply expensive to waste energy. But the point of the LHC is not to get energy out of these collisions – it’s hard enough to just get collisions. So the answer to your question is that right now it’s not possible to make the LHC more energy efficient – the LHC is the only machine to exist that could ever study collisions at these energies.

    There might be a few minor areas where the energy efficiency could be improved – for instance, perhaps some front end electronic boards could be made slightly more efficient – but this wouldn’t even add up to a percent of the energy it takes to run the LHC. Even if you did have an idea for improving the efficiency, you’d have to add in the energy required to design, test, and manufacture new components. And what if making some electronic boards more efficient meant that they couldn’t go as fast so you’d either not be able to save out data from as many collisions or you’d have to run with a lower luminosity? This would mean you’d have to run longer, reducing efficiency. It’s not easy to determine what is most efficient – especially when you have the world’s largest machine comprised of almost entirely tailor made parts.

    Most of the energy used to run the LHC is used either for cooling the ring or for maintaining the magnetic field – so if it were possible to do either of these things more efficiently, it might be possible to improve the efficiency. But no such technology exists today – almost everything in the LHC is at the bleeding edge of technology. To make an analogy, by today’s standards, the first computers made were not energy efficient – just getting them working was hard enough. Now we make much more energy efficient computers, but at the time the first computers were made we didn’t have the same technology we have now. But the point of a computer is not energy efficiency – and the goal of the LHC is not to get energy out of proton-proton or lead-lead collisions.

  • Jim Pivarski

    > When I think of the megawatts required to maintain and operate
    > the LHC on a day to day basis, that is an astronomical amount
    > of energy loss…

    Yes, the 7 TeV per proton is a tiny amount of energy, about 10^-12 megajoules, but there are about 10^+11 protons per bunch and 3000 bunches in the beam (Christine’s original point about the energy density being very large). The LHC beam at full luminosity will actually store 360 megajoules, comparable to the energy of high-speed trains.

    http://lhc-machine-outreach.web.cern.ch/lhc-machine-outreach/beam.htm

    I don’t know how many megawatts are needed to feed the beam itself, rather than the cryogenics, magnets, and control systems, but if it’s on the scale of megawatts, that’s not necessarily a big inefficiency.

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