Sh-Nobel Prizes
08/25/2011 | stryjko_bojko
INSTITUTE OF PHYSICS PUBLISHING EUROPEAN JOURNAL OF PHYSICS
Eur. J. Phys. 23 (2002) 21–26 PII: S0143-0807(02)26048-1
Demonstration of the exponential
decay law using beer froth. A Leike.
http://classes.soe.ucsc.edu/math011a/Winter06/homework/beerdecay.pdf
Ludwig–Maximilians-Universit¨at, Sektion Physik, Theresienstr. 37, D-80333M¨unchen, Germany
E-mail: leike@theorie.physik.uni-muenchen.de
Received 22 June 2001, in final form 8 October 2001
Published 17 December 2001
Online at stacks.iop.org/EJP/23/21
Abstract
The volume of beer froth decays exponentially with time. This property is used
to demonstrate the exponential decay law in the classroom. The decay constant
depends on the type of beer and can be used to differentiate between different
beers. The analysis shows in a transparent way the techniques of data analysis
commonly used in science—consistency checks of theoretical models with the
data, parameter estimation and determination of confidence intervals.
Exponential laws are common to many physical phenomena. Examples are the amplitude
of an oscillator subject to linear friction, the discharge of a capacitor, cooling processes or
radioactive decays. The demonstration described here has the advantages that it is cheap, clear
and motivating because it investigates an everyday phenomenon. It can easily be repeated by
the students elsewhere.
The decay of beer froth is mentioned as a very short notice in [1]. It is described in
several German textbooks of mathematics. Recently, it also attracted the attention of Bavarian
pupils [2].
The data analysis proposed in this paper has much in common with real science—see, for
example, the determination of the Higgs mass by the LEP collaborations [3]. The techniques
involved are of great practical importance but are often poorly understood by students [4].
Exponential decay can be demonstrated using beer froth, the volume of which reduces
exponentially with time [1]. The exponential law can readily be derived from the assumption
that the volume of froth dV disappearing in the time between t and t + dt is proportional to
the volume V present at the time t, dV = −(V /τ ) dt . In a cylindrical beer mug with an area
A, the volume is proportional to the height, dV = Adh. The phenomenological theory of
exponential decay predicts the height as a function of time
hth(t) = h(0) exp
− t
τ
. (1)
The constant τ is a free parameter of the theory. It defines how fast the froth decays; during
the time τ the amount 1 − 1/e ≈ 63% of the froth disappears. Different kinds of beer have,
in general, different parameters τ .
0143-0807/02/010021+06$30.00 © 2002 IOP Publishing Ltd Printed in the UK 21
22 A Leike
Table 1. The data hexp(ti ) and their errors hexp(ti ) for three different beers. The last line contains
the best estimates for τ together with their errors of 68% (95%) confidence.
Erdinger Weissbier Augustinerbr¨au M¨unchen Budweiser Budvar
t (s) hexp cm−1 hexp cm−1 hexp cm−1 hexp cm−1 hexp cm−1 hexp cm−1
0 17.0 0.0 14.0 0.0 14.0 0.0
15 16.1 0.3 11.8 0.3 12.1 0.4
30 14.9 0.4 10.5 0.3 10.9 0.4
45 14.0 0.4 9.3 0.5 10.0 0.4
60 13.2 0.4 8.5 0.6 9.3 0.4
75 12.5 0.6 7.7 0.6 8.6 0.4
90 11.9 0.4 7.1 0.7 8.0 0.3
105 11.2 0.4 6.5 0.8 7.5 0.3
120 10.7 0.4 6.0 0.8 7.0 0.3
150 9.7 0.4 5.3 1.1 6.2 0.3
180 8.9 0.3 4.4 1.2 5.5 0.4
210 8.3 0.4 3.5 0.9 4.5 0.4
240 7.5 0.4 2.9 1.1 3.5 0.5
300 6.3 0.5 1.3 0.7 2.0 0.5
360 5.2 0.5 0.7 0.5 0.9 0.4
χ2
min 11.1 12.5 24.4
τbest ± τ s−1 276 ± 7 (±14) 124 ± 6 (±12) 168 ± 4 (±8)
In the following, the demonstration is described in detail. In our experiment, a cylindrical
beer mug with a diameter of 7.2 cm was filled with beer immediately after opening the bottle.
The temperature of the beer was 19 ◦C.
The froth appears while filling the mug with the beer. The froth reaches its maximum
height within a few seconds. This indicates that the typical time scale of the expansion of
the froth is a few seconds. On the other hand, the froth lasts for a few minutes (see table 1).
Therefore, the time scale for the decay is a few minutes. The two time scales are very different.
We therefore assume that a few seconds after the time where the froth reaches its maximum
height only the decay plays a significant role.
We began with the measurement at the time where the froth dropped to a certain initial
height hexp(0) = h(0). The error hexp(0) in the measurement of h(0) was estimated to be
2 mm.
We investigated three different beers. With every beer, the experimentwas repeated several
times. We performed seven experiments with ErdingerWeissbier (the author’s favourite!), four
experiments with Augustinerbr¨auM¨unchen and four experiments with Budweiser Budvar. Our
data are shown in table 1. The entries for hexp(ti) are obtained by averaging over all individual
measurements at time ti . To obtain the errors hexp(ti) of the measurements hexp(ti), we first
calculated the root of the variance of all individual measurements at the time ti . The relative
error of h(0) was added quadratically to the resulting relative error.
The error ti of our time measurements was below 1 s. ti can be translated into an error
in the height h(ti) using equation (1)
|h(t)| =
∂h(t)
∂t
t
=
t
τ
h(t). (2)
Having in mind the values for τ obtained in our experiment, the impact of ti on the error in
the height is considerably smaller then the error hexp(ti) from the direct measurement. We
therefore neglect the error ti in the time measurement.
With the experimental data in table 1 and the phenomenological theory (1), one can
calculate
Demonstration of the exponential decay law using beer froth 23
0
10
20
30
0 100 200 300 τ/s 400
χ2
Erdinger
Budweiser
Augustiner
Figure 1. χ2 as function of τ for different kinds of beer.
χ2 =
n
i=1
[hexp(ti) − hth(ti)]2
[hexp(ti)]2 . (3)
Figure 1 shows χ2 as a function of τ . This fit has N = 13 degrees of freedom because we
have n = 15 data points and two constraints. The first constraint is the fixed h(0), the second
constraint is the parameter τ , which is defined from the data. The minimum values for χ2 as
a function of τ for the three different beers are 11.1, 12.5 and 24.4 for Erdinger Weissbier,
Augustinerbr¨au M¨unchen and Budweiser Budvar respectively. They are given in table 1.
If the data point errors hexp(ti)−hth(ti) are Gaussian, then the minimum ofχ2 is distributed
as a χ2 random variable. The minimum of χ2 is a measure of the goodness of the fit. The
χ2 of a good fit with N degrees of freedom should be approximately in the range N ±
√
2N.
(See the appendix for some information about χ2 and [5] for a short review on probability and
statistics.) The χ2 for Erdinger and Augustinerbr¨au indicate a good fit (see table 1). The χ2
for Budweiser is larger. The probability of getting a χ2 of 24.4 or larger with 13 degrees of
freedom is about 3%. However, the probability of getting such a large χ2 in at least one of the
three independent fits is 9%. We conclude that the theory of exponential decay is consistent
with our experimental data. Then the determination of the model parameter τ using the data
makes sense. The best fits to the parameter τ correspond to χ2
min, the minimum χ2. We will
denote it as τbest . Figure 2 shows the experimental data together with the best fits.
Under the condition that the experimental data are consistent with the theory, confidence
intervals for the theoretical parameters can be defined. The decay law (1) depends only on
one parameter τ . A confidence interval (τ−, τ+) with a probability of x% means that the true
value of the parameter τ is between τ− and τ+ with a probability of x%. Confidence intervals
are found by exploring the properties of χ2 near its minimum χ2
min (see for example [5]). In
particular, confidence intervals of τ with a probability of 68% (95%) are found by the condition
χ2 = χ2
min +1(4) (see also the appendix). The corresponding values τ± = τbest ±τ are given
in the last row of table 1.
Our results suggest that the three kinds of beer may be distinguished by the lifetime of
the froth only. However, before one can make a definite statement, one has to study carefully
the dependence of τbest on the temperature and on the date of production of the beer. These
extensive studies are left to the reader. The author did only one preliminary measurement
with Erdinger Weissbier at a temperature of 11 ◦C and found τ = 255 ± 14 s (one standard
24 A Leike
0
5
10
15
20
0 100 200 300 t/s
h(t)
----
cm
Erdinger
Budweiser
Augustiner
best fits
Figure 2. The height of the froth for different kinds of beer as function of time. Shown are the
data and the best fits.
deviation, i.e. 68% confidence). This indicates that there is no substantial dependence of τbest
on the temperature.
An approximate determination of τ in the exponential decay (1) is possible with pocket
calculators through a linear regression to ln(h(t)/h(0)). It is equivalent to finding the minimum
of the expression
n
i=1
ln
hexp(ti)
h(0)
+
t
τ
2
. (4)
The results for τ are 290, 134 and 155 s for ErdingerWeissbier, Augustinerbr¨au and Budweiser
respectively. These values are several standard deviations away from the values obtained in
the χ2 fit. One of the reasons for this discrepancy originates in the simple procedure (4), which
ignores the different experimental errors hexp(ti) in the data. The resulting values for τ can
therefore only be a rough estimate.
Alternatively, one could test a model of linear decay of the froth
hth(t) = h(0)
1 − t
T
θ(T − t) (5)
where T is the time in which all the froth decays. The Heaviside function θ(T −t) ensures that
the height cannot become negative. One can test the linear model by calculating χ2 according
to formula (3). The minima of χ2 are 129, 86 and 161. The probability of having a χ2 of 86 or
larger in a χ2 distribution with 13 degrees of freedom is about 10−12. This is the probability
that, under the condition that the model of linear decay is correct, the observed data arose due
to a statistical fluctuation. This probability is very small indicating that the model of linear
decay is inconsistent with the experimental data for all three beers. Therefore, the linear theory
should be rejected and a determination of the parameter T from the data is meaningless.
The proposed demonstration is simple and motivating. It explains scientific methods using
an everyday phenomenon. It can be included in undergraduate courses to show how to prove
quantitatively the consistency of a phenomenological law with experimental data. The next
step towards a theory, not discussed here, would be a deviation of the phenomenological law
from first principles.
Demonstration of the exponential decay law using beer froth 25
Acknowledgment
I would like to thank L B Okun for encouraging me to write this paper.
Appendix
To make the paper self-consistent, the formulae regarding the goodness of a fit used in the
main part of the paper are derived here. See [5] and references therein for more details.
Let us assume that we have N Gaussian distributed random variables xi, i = 1, . . . , N
with mean 0 and variance 1. Consider the sum
χ2 =
N
i=1
x2
i . (6)
The random variable χ2 has N degrees of freedom. It corresponds to the χ2 function given
in (3), which also consists of a sum of the square of Gaussian random variables with mean 0
and variance 1.
We are interested in the distribution of the random variable χ2. The distribution of
the square of one Gaussian distributed variable can easily be obtained from the Gaussian
distribution
2 √
2π
exp
−x2
2
dx = 1
x
√
2π
exp
−x2
2
dx2
= 1
√
z
√
2π
exp
−z
2
dz = χ2
1 (z) dz, z 0. (7)
The function χ2
1 (z) is called the χ2 distribution for one degree of freedom.
The characteristic function (u) associated with a distribution f (z) is defined as
(u) ≡
∞
−∞
exp(iuz)f (z) dz. (8)
The characteristic function of χ2
1 is
1(u) =
∞
0
exp(iuz)χ2
1 (z) dz = 1
(1 − 2iu)1/2 . (9)
If f1(x1) and f2(x2) have characteristic functions1(u) and2(u), then the characteristic
function of the distribution of the random variable c1x1 + c2x2 is
(u) =
∞
−∞
dx1
∞
−∞
dx2 exp
iu(c1x1 + c2x2)
f (x1)f (x2)
=
∞
−∞
dx1 exp(iuc1x1)f (x1)
∞
−∞
dx2 exp(iuc2x2)f (x2)
= 1(c1u) · 2(c2u). (10)
The characteristic function of the sum of squares of N independent Gaussian distributed
variables is therefore
N(u) = 1
(1 − 2iu)N/2 . (11)
The distribution associated with N(u) is obtained by the transformation inverse to (8)
χ2
N(z) = 1
2π
∞
−∞
exp(−iuz)N(u) du = zN/2−1e−z/2
2N/2(N/2)
, z 0. (12)
26 A Leike
It is called the χ2 distribution for N degrees of freedom. It has the mean N and the variance
2N. The probability density (12) is used to estimate the goodness of a fit. The probability
P(z > χ2) that z is larger than a given value χ2 is
P(z > χ2) =
∞
χ2
χ2
N(z) dz. (13)
In particular, we get
∞
24.4
χ2
13(z) dz ≈ 0.03 = 3% (14)
which is the number quoted in the main text.
The probability P(z < a2) that z is smaller than a given value a2 is
P(z < x) =
a2
0
χ2
N(z) dz. (15)
In particular, we get for N = 1 and a2 = 1 or 4
1
0
χ2
1 (z) dz ≈ 0.68 = 68% or
4
0
χ2
1 (z) dz ≈ 0.95 = 95% (16)
which are the numbers quoted in the main text for the confidence intervals of τ .
References
[1] G¨obel R and Haubold K 1989 Freihandexperimente in Klasse 10 Phys. Schule 27/3 104 (in German)
[2] Newspaper announcement (in German) http://www.mainfranken.org/deutsch/jufo2001 presse2.html
[3] The LEP Collaborations, the LEP electroweak workung group and SLD heavy flavour and electroweak group,
CERN-EP-200-021 webpage http://lepewwg.web.cern.ch/LEPEWWG/
Kawamoto T 2001 Electroweak data and standard model fit results 36th Rencontres de Moriond-2001 hepex/
0105032
(Kawamoto T 2001 Preprint hep-ex/0105032)
[4] Earnshaw J C 1990 χ2—or do the data fit the theory? Eur. J. Phys. 11 338
[5] 2000 Review of particle properties Eur. Phys. J. C 15 1
Eur. J. Phys. 23 (2002) 21–26 PII: S0143-0807(02)26048-1
Demonstration of the exponential
decay law using beer froth. A Leike.
http://classes.soe.ucsc.edu/math011a/Winter06/homework/beerdecay.pdf
Ludwig–Maximilians-Universit¨at, Sektion Physik, Theresienstr. 37, D-80333M¨unchen, Germany
E-mail: leike@theorie.physik.uni-muenchen.de
Received 22 June 2001, in final form 8 October 2001
Published 17 December 2001
Online at stacks.iop.org/EJP/23/21
Abstract
The volume of beer froth decays exponentially with time. This property is used
to demonstrate the exponential decay law in the classroom. The decay constant
depends on the type of beer and can be used to differentiate between different
beers. The analysis shows in a transparent way the techniques of data analysis
commonly used in science—consistency checks of theoretical models with the
data, parameter estimation and determination of confidence intervals.
Exponential laws are common to many physical phenomena. Examples are the amplitude
of an oscillator subject to linear friction, the discharge of a capacitor, cooling processes or
radioactive decays. The demonstration described here has the advantages that it is cheap, clear
and motivating because it investigates an everyday phenomenon. It can easily be repeated by
the students elsewhere.
The decay of beer froth is mentioned as a very short notice in [1]. It is described in
several German textbooks of mathematics. Recently, it also attracted the attention of Bavarian
pupils [2].
The data analysis proposed in this paper has much in common with real science—see, for
example, the determination of the Higgs mass by the LEP collaborations [3]. The techniques
involved are of great practical importance but are often poorly understood by students [4].
Exponential decay can be demonstrated using beer froth, the volume of which reduces
exponentially with time [1]. The exponential law can readily be derived from the assumption
that the volume of froth dV disappearing in the time between t and t + dt is proportional to
the volume V present at the time t, dV = −(V /τ ) dt . In a cylindrical beer mug with an area
A, the volume is proportional to the height, dV = Adh. The phenomenological theory of
exponential decay predicts the height as a function of time
hth(t) = h(0) exp
− t
τ
. (1)
The constant τ is a free parameter of the theory. It defines how fast the froth decays; during
the time τ the amount 1 − 1/e ≈ 63% of the froth disappears. Different kinds of beer have,
in general, different parameters τ .
0143-0807/02/010021+06$30.00 © 2002 IOP Publishing Ltd Printed in the UK 21
22 A Leike
Table 1. The data hexp(ti ) and their errors hexp(ti ) for three different beers. The last line contains
the best estimates for τ together with their errors of 68% (95%) confidence.
Erdinger Weissbier Augustinerbr¨au M¨unchen Budweiser Budvar
t (s) hexp cm−1 hexp cm−1 hexp cm−1 hexp cm−1 hexp cm−1 hexp cm−1
0 17.0 0.0 14.0 0.0 14.0 0.0
15 16.1 0.3 11.8 0.3 12.1 0.4
30 14.9 0.4 10.5 0.3 10.9 0.4
45 14.0 0.4 9.3 0.5 10.0 0.4
60 13.2 0.4 8.5 0.6 9.3 0.4
75 12.5 0.6 7.7 0.6 8.6 0.4
90 11.9 0.4 7.1 0.7 8.0 0.3
105 11.2 0.4 6.5 0.8 7.5 0.3
120 10.7 0.4 6.0 0.8 7.0 0.3
150 9.7 0.4 5.3 1.1 6.2 0.3
180 8.9 0.3 4.4 1.2 5.5 0.4
210 8.3 0.4 3.5 0.9 4.5 0.4
240 7.5 0.4 2.9 1.1 3.5 0.5
300 6.3 0.5 1.3 0.7 2.0 0.5
360 5.2 0.5 0.7 0.5 0.9 0.4
χ2
min 11.1 12.5 24.4
τbest ± τ s−1 276 ± 7 (±14) 124 ± 6 (±12) 168 ± 4 (±8)
In the following, the demonstration is described in detail. In our experiment, a cylindrical
beer mug with a diameter of 7.2 cm was filled with beer immediately after opening the bottle.
The temperature of the beer was 19 ◦C.
The froth appears while filling the mug with the beer. The froth reaches its maximum
height within a few seconds. This indicates that the typical time scale of the expansion of
the froth is a few seconds. On the other hand, the froth lasts for a few minutes (see table 1).
Therefore, the time scale for the decay is a few minutes. The two time scales are very different.
We therefore assume that a few seconds after the time where the froth reaches its maximum
height only the decay plays a significant role.
We began with the measurement at the time where the froth dropped to a certain initial
height hexp(0) = h(0). The error hexp(0) in the measurement of h(0) was estimated to be
2 mm.
We investigated three different beers. With every beer, the experimentwas repeated several
times. We performed seven experiments with ErdingerWeissbier (the author’s favourite!), four
experiments with Augustinerbr¨auM¨unchen and four experiments with Budweiser Budvar. Our
data are shown in table 1. The entries for hexp(ti) are obtained by averaging over all individual
measurements at time ti . To obtain the errors hexp(ti) of the measurements hexp(ti), we first
calculated the root of the variance of all individual measurements at the time ti . The relative
error of h(0) was added quadratically to the resulting relative error.
The error ti of our time measurements was below 1 s. ti can be translated into an error
in the height h(ti) using equation (1)
|h(t)| =
∂h(t)
∂t
t
=
t
τ
h(t). (2)
Having in mind the values for τ obtained in our experiment, the impact of ti on the error in
the height is considerably smaller then the error hexp(ti) from the direct measurement. We
therefore neglect the error ti in the time measurement.
With the experimental data in table 1 and the phenomenological theory (1), one can
calculate
Demonstration of the exponential decay law using beer froth 23
0
10
20
30
0 100 200 300 τ/s 400
χ2
Erdinger
Budweiser
Augustiner
Figure 1. χ2 as function of τ for different kinds of beer.
χ2 =
n
i=1
[hexp(ti) − hth(ti)]2
[hexp(ti)]2 . (3)
Figure 1 shows χ2 as a function of τ . This fit has N = 13 degrees of freedom because we
have n = 15 data points and two constraints. The first constraint is the fixed h(0), the second
constraint is the parameter τ , which is defined from the data. The minimum values for χ2 as
a function of τ for the three different beers are 11.1, 12.5 and 24.4 for Erdinger Weissbier,
Augustinerbr¨au M¨unchen and Budweiser Budvar respectively. They are given in table 1.
If the data point errors hexp(ti)−hth(ti) are Gaussian, then the minimum ofχ2 is distributed
as a χ2 random variable. The minimum of χ2 is a measure of the goodness of the fit. The
χ2 of a good fit with N degrees of freedom should be approximately in the range N ±
√
2N.
(See the appendix for some information about χ2 and [5] for a short review on probability and
statistics.) The χ2 for Erdinger and Augustinerbr¨au indicate a good fit (see table 1). The χ2
for Budweiser is larger. The probability of getting a χ2 of 24.4 or larger with 13 degrees of
freedom is about 3%. However, the probability of getting such a large χ2 in at least one of the
three independent fits is 9%. We conclude that the theory of exponential decay is consistent
with our experimental data. Then the determination of the model parameter τ using the data
makes sense. The best fits to the parameter τ correspond to χ2
min, the minimum χ2. We will
denote it as τbest . Figure 2 shows the experimental data together with the best fits.
Under the condition that the experimental data are consistent with the theory, confidence
intervals for the theoretical parameters can be defined. The decay law (1) depends only on
one parameter τ . A confidence interval (τ−, τ+) with a probability of x% means that the true
value of the parameter τ is between τ− and τ+ with a probability of x%. Confidence intervals
are found by exploring the properties of χ2 near its minimum χ2
min (see for example [5]). In
particular, confidence intervals of τ with a probability of 68% (95%) are found by the condition
χ2 = χ2
min +1(4) (see also the appendix). The corresponding values τ± = τbest ±τ are given
in the last row of table 1.
Our results suggest that the three kinds of beer may be distinguished by the lifetime of
the froth only. However, before one can make a definite statement, one has to study carefully
the dependence of τbest on the temperature and on the date of production of the beer. These
extensive studies are left to the reader. The author did only one preliminary measurement
with Erdinger Weissbier at a temperature of 11 ◦C and found τ = 255 ± 14 s (one standard
24 A Leike
0
5
10
15
20
0 100 200 300 t/s
h(t)
----
cm
Erdinger
Budweiser
Augustiner
best fits
Figure 2. The height of the froth for different kinds of beer as function of time. Shown are the
data and the best fits.
deviation, i.e. 68% confidence). This indicates that there is no substantial dependence of τbest
on the temperature.
An approximate determination of τ in the exponential decay (1) is possible with pocket
calculators through a linear regression to ln(h(t)/h(0)). It is equivalent to finding the minimum
of the expression
n
i=1
ln
hexp(ti)
h(0)
+
t
τ
2
. (4)
The results for τ are 290, 134 and 155 s for ErdingerWeissbier, Augustinerbr¨au and Budweiser
respectively. These values are several standard deviations away from the values obtained in
the χ2 fit. One of the reasons for this discrepancy originates in the simple procedure (4), which
ignores the different experimental errors hexp(ti) in the data. The resulting values for τ can
therefore only be a rough estimate.
Alternatively, one could test a model of linear decay of the froth
hth(t) = h(0)
1 − t
T
θ(T − t) (5)
where T is the time in which all the froth decays. The Heaviside function θ(T −t) ensures that
the height cannot become negative. One can test the linear model by calculating χ2 according
to formula (3). The minima of χ2 are 129, 86 and 161. The probability of having a χ2 of 86 or
larger in a χ2 distribution with 13 degrees of freedom is about 10−12. This is the probability
that, under the condition that the model of linear decay is correct, the observed data arose due
to a statistical fluctuation. This probability is very small indicating that the model of linear
decay is inconsistent with the experimental data for all three beers. Therefore, the linear theory
should be rejected and a determination of the parameter T from the data is meaningless.
The proposed demonstration is simple and motivating. It explains scientific methods using
an everyday phenomenon. It can be included in undergraduate courses to show how to prove
quantitatively the consistency of a phenomenological law with experimental data. The next
step towards a theory, not discussed here, would be a deviation of the phenomenological law
from first principles.
Demonstration of the exponential decay law using beer froth 25
Acknowledgment
I would like to thank L B Okun for encouraging me to write this paper.
Appendix
To make the paper self-consistent, the formulae regarding the goodness of a fit used in the
main part of the paper are derived here. See [5] and references therein for more details.
Let us assume that we have N Gaussian distributed random variables xi, i = 1, . . . , N
with mean 0 and variance 1. Consider the sum
χ2 =
N
i=1
x2
i . (6)
The random variable χ2 has N degrees of freedom. It corresponds to the χ2 function given
in (3), which also consists of a sum of the square of Gaussian random variables with mean 0
and variance 1.
We are interested in the distribution of the random variable χ2. The distribution of
the square of one Gaussian distributed variable can easily be obtained from the Gaussian
distribution
2 √
2π
exp
−x2
2
dx = 1
x
√
2π
exp
−x2
2
dx2
= 1
√
z
√
2π
exp
−z
2
dz = χ2
1 (z) dz, z 0. (7)
The function χ2
1 (z) is called the χ2 distribution for one degree of freedom.
The characteristic function (u) associated with a distribution f (z) is defined as
(u) ≡
∞
−∞
exp(iuz)f (z) dz. (8)
The characteristic function of χ2
1 is
1(u) =
∞
0
exp(iuz)χ2
1 (z) dz = 1
(1 − 2iu)1/2 . (9)
If f1(x1) and f2(x2) have characteristic functions1(u) and2(u), then the characteristic
function of the distribution of the random variable c1x1 + c2x2 is
(u) =
∞
−∞
dx1
∞
−∞
dx2 exp
iu(c1x1 + c2x2)
f (x1)f (x2)
=
∞
−∞
dx1 exp(iuc1x1)f (x1)
∞
−∞
dx2 exp(iuc2x2)f (x2)
= 1(c1u) · 2(c2u). (10)
The characteristic function of the sum of squares of N independent Gaussian distributed
variables is therefore
N(u) = 1
(1 − 2iu)N/2 . (11)
The distribution associated with N(u) is obtained by the transformation inverse to (8)
χ2
N(z) = 1
2π
∞
−∞
exp(−iuz)N(u) du = zN/2−1e−z/2
2N/2(N/2)
, z 0. (12)
26 A Leike
It is called the χ2 distribution for N degrees of freedom. It has the mean N and the variance
2N. The probability density (12) is used to estimate the goodness of a fit. The probability
P(z > χ2) that z is larger than a given value χ2 is
P(z > χ2) =
∞
χ2
χ2
N(z) dz. (13)
In particular, we get
∞
24.4
χ2
13(z) dz ≈ 0.03 = 3% (14)
which is the number quoted in the main text.
The probability P(z < a2) that z is smaller than a given value a2 is
P(z < x) =
a2
0
χ2
N(z) dz. (15)
In particular, we get for N = 1 and a2 = 1 or 4
1
0
χ2
1 (z) dz ≈ 0.68 = 68% or
4
0
χ2
1 (z) dz ≈ 0.95 = 95% (16)
which are the numbers quoted in the main text for the confidence intervals of τ .
References
[1] G¨obel R and Haubold K 1989 Freihandexperimente in Klasse 10 Phys. Schule 27/3 104 (in German)
[2] Newspaper announcement (in German) http://www.mainfranken.org/deutsch/jufo2001 presse2.html
[3] The LEP Collaborations, the LEP electroweak workung group and SLD heavy flavour and electroweak group,
CERN-EP-200-021 webpage http://lepewwg.web.cern.ch/LEPEWWG/
Kawamoto T 2001 Electroweak data and standard model fit results 36th Rencontres de Moriond-2001 hepex/
0105032
(Kawamoto T 2001 Preprint hep-ex/0105032)
[4] Earnshaw J C 1990 χ2—or do the data fit the theory? Eur. J. Phys. 11 338
[5] 2000 Review of particle properties Eur. Phys. J. C 15 1
Відповіді
2011.08.25 | stryjko_bojko
Шнобелівські премії - 2009: зімбабвійські математики та алмази з
02.10.2009 10:44Шнобелівські премії - 2009: зімбабвійські математики та алмази з текіли
http://gazeta.ua/post/310038
У Гарвардському університеті відбулося вручення Шнобелівських (Антинобелівських) премій. Вони вручаються вже в 19-й раз за наукові досягнення, «які спочатку викликають сміх, а потім примушують задуматися», повідомляє AssociatedPress.
Переможцем в номінації «Ветеринарія» були визнані Кетрін Дуглас і Пітер Роулінсон за відкриття того факту, що корови, у яких є прізвиська, дають більше молока, ніж ті, яких ніяк не назвали.
У номінації «Світ» перемогли Стефан Боллігер, Стеффен Рос, Ларс Оестерхельвег, Майкл Талі і Біт Неубуел за те, що в результаті дослідження вони прийшли до висновку, що краще бути удареним по голові повною пляшкою пива, ніж порожньою.
У номінації «Економіка» отримали перемогу керівники чотирьох ісландських банків, які показали, як маленькі банки можуть стати великими, а потім знову маленькими.
Премію «Хімія» отримали Хав'єр Моралес, Мігель Апатіга і Віктор Кастано за те, що створили алмази з текіли.
У номінації «Медицина» переміг Дональд Унгер за те, що протягом 60 років ламав суглоби тільки на лівій руці, щоб дізнатися, чи сприяють переломи суглобів артриту.
У «Фізиці» виграли Кетрін Віткоум, Ліза Шапіро і Даніель Ліберман за те, що з'ясували, чому вагітні жінки не перевертаються.
У номінації «Література» перемогли ірландські поліцейські, які виписали 50 штрафів якомусь PrawoJazdy, що в перекладі з польського означає «водійські права».
У «Охороні» здоров'я виграли Єлена Боднер, Рафаель Лі та Сандра Меріджен за те, що винайшли бюстгальтер, який може бути перетворений в протигаз.
У «Математиці» перемога дісталася Гідеону Гоно і Резервному банку Зімбабве за друк банкнотів номіналом від одного цента до 100 трлн зімбабвійських доларів.
Шнобелівська премія була заснована в 1991 році Марком Ебрехемсом і гумористичним журналом «Аннали неймовірних досліджень». Церемонія вручення премії традиційно проходить в головному лекційному залі одного з найпрестижніших учбових закладів світу - Гарвардського університету (штат Массачусетс, США).
Як наголошується на сайті журналу, «наша мета полягає в заохоченні цікавості людей. Ми хочемо, щоб вони поставили собі питання: звідки беруться критерії «важливості» або «неважливості» чого-небудь, і що реально, а що нереально - і в науці, і де б то не було ще?».
Шнобелівські премії вручаються незадовго до оголошення лауреатів найпрестижнішої науковій премії - Нобелівської. Вручають ці жартівливі «антинаукові» премії справжні нобелівські лауреати.