All posts by David Robertson

First collisions at 13 TeV as the Large Hadron Colllider restarts


On May 21, 2015 proton beams were collided together in the Large Hadron Collider at the record energy of 13 TeV, as the LHC restarts after extensive upgrades over the past two years.  This is the total energy in the center-of-mass frame, so each beam contains protons of energy 6.5 TeV.  That’s about 6,500 times the rest energy of a proton (roughly 1 GeV), hence the relativistic gamma factor for these protons is about 6,500.

Read the CERN article on this important milestone here.

Higgs boson paper has 5,154 authors

CMS and Atlas, the two big experimental collaborations at the Large Hadron Collider, have joined forces to produce the most accurate determination of the Higgs boson mass to date.  By combining their data sets they obtained

$$ M_H=125.09\pm0.24~GeV$$

for a relative error of about 0.2%.  The paper in Physical Review Letters describing these results has the longest author list ever: 5,154 names.  In the published version, there are 9 pages describing the research and 24 pages of authors and their affiliations.

Read the associated APS Physics Viewpoint article here.

Holiday Dimensional Analysis

As an amusing exercise in dimensional analysis, let us consider the problem of fully cooking the Christmas turkey (or ham, or whatever). We place the bird in an oven with temperature \(T_o\), assumed constant. The bird itself has mass \(M\) and density \(\rho\). The question we wish to ask is: how long must it cook before the central temperature reaches some specified (safe to eat!) value \( T_c\)?

The physics of the situation is governed by the “diffusion equation”:

$$\frac{\partial T}{\partial t}=\kappa\nabla^2T$$

where \( T(x,y,z,t) \) is the temperature at any point in the bird (labeled by \( x,y,z\)) at time \( t\). The constant \(\kappa\) is called the diffusion coefficient and depends on the material being heated. Thus \(\kappa\) might be different for turkey and ham, say, but all turkeys will share essentially the same \(\kappa\).

From looking at the dimensions in this equation we can determine the units of \( \kappa\). The operator \( \nabla^2 \) contains second derivatives with respect to \( x\), \( y\), and \( z\), so it has dimensions \( 1/length^2,\) or \( 1/L^2\) for short. Hence \( \kappa\) has dimensions


where \(T\) is time. In the SI system this would be \( m^2/s\).

Okay, say we could solve this problem and determine \( T_c\) as a function of the parameters \( T_o, \rho, M, \kappa,\) and the time \( t\). What kind of relation can we get? In fact, the dimensions of the quantities involved imply a general structure. To see what it is, note that there are two (and only two) dimensionless combinations that can be formed from these quantities. One of them is pretty obvious:

$${T_c\over T_o}$$

since these are the only temperatures in the game. To find the other, note that


where \( L\) is a length parameter characterizing the bird. So \( (M/\rho)^{1/3}\) has dimensions of length. Now consider the combination

$$t\kappa\left({\rho\over M}\right)^{2/3}$$

\( t\kappa\) has dimensions \( L^2\) and the factor in parentheses has dimensions \( 1/L^2\). So this combination is indeed dimensionless.

Now the point is that on dimensional grounds (i.e., just to make the units work out), the answer has to look like this:

$${T_c\over T_o}=f\left( t\kappa\left({\rho\over M}\right)^{2/3} \right)$$

where \( f\) is some unknown function. That is, one dimensionless quantity must be a function of the other. We don’t know what \( f\) is, of course – we’d have to solve the diffusion equation to determine that. But we can learn something interesting just by knowing that \( T_c\) depends on \( M\) only through the specific combination of quantities

$$t\kappa\left({\rho\over M}\right)^{2/3}$$

Assume that we want to reach the same internal temperature \(T_c\) for birds of different \(M\) but the same \(\rho, \kappa\), and using the same oven temperature. Bird 1 has mass \( M_1\) and takes time \( t_1\) to cook; bird 2 has mass \( M_2\) and takes time \( t_2.\) But when they have the same central temperature we will have the same value of \( T_c/T_o\), and hence

$$f\left( t_1\kappa\left({\rho\over M_1}\right)^{2/3} \right) = f\left( t_2\kappa\left({\rho\over M_2}\right)^{2/3} \right)$$

In other words

$$t_1\kappa\left({\rho\over M_1}\right)^{2/3} = t_2\kappa\left({\rho\over M_2}\right)^{2/3}$$

or (canceling common factors)

$${t\over M^{2/3}}=constant$$

Hence the cooking time \( t\propto M^{2/3}\).

So the lesson we learn is that the cooking time does not simply grow in proportion to the mass  of the bird, as claimed in some cookbooks. It grows as the \( 2/3\) power of the weight.

(As an aside, since the weight is proportional to the linear size of the bird cubed, we could say that the cooking time scales like (is proportional to) the surface area of the turkey.)

There is a standard trick for testing this kind of “power law” relation that ought to be part of every physicists bag of tools.  If we take the log of both sides of this relation, we obtain


so that if we plot the log of the cooking time versus the log of the weight, we should see a straight line with slope \( 2/3\), the power we are trying to verify. Here is data taken from Helen Corbitt’s Cookbook (Houghton Mifflin, 1957):


The plot is indeed pretty linear, with slope around 0.63! Physics triumphs once again.

Some followup questions…

Why didn’t we include Planck’s constant in the analysis? (This seems like a silly question, but it points to a deep and important aspect of dimensional analysis, when done properly.)

We made a tacit assumption that birds of different weight have the same general shape, i.e., are geometrically similar. If this is not the case, can you see where we might have gotten stuck in this analysis?

Say we reason as follows. The total energy required to heat the bird to the desired temperature scales with its volume. But the rate at which energy is absorbed is scales like the surface area. So we conclude that the time to absorb enough energy increases as the ratio of volume to area, i.e., in proportion to the linear size of the bird, or equivalently to \( M^{1/3}\). Where did we go wrong? [Hint: This analysis would predict a shorter cooking time than is actually required. What aspect of the physics have we neglected?]