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On some fluctuation problems connected with the counting of impulses produced by a geiger-müller counter or ionisation chamber
Physica
IX,
Juli
no 7
1942
ON SOME FLUCTUATION PROBLEMS CONNECTED WITH THE COUNTING OF IMPULSES PRODUCED .BY A GEIGER-MOLLER COUNTER OR IONISATION CHAMBER by H. A. VAN DER VELDEN Publication
from
the Physical
Institute
and P. M. ENDT of the University
of Utrecht
Summary In the present article we shall deal with some quantities which are of importance in counting the kicks of a Geigercounter or ionisation chamber arising from some constant or decaying radioactive source. We will first derive the sensibility and the statistical fluctuations of the counting rate meter, using the direct method first indicated by 0 r n s t e i n and a more classical method for one special case. We also wish to calculate some of these quantities for the case of a scaling set and a mechanical recorder with finite resolvi.ng powers. It is shown that in this case the fluctuations are greater (but not much) than in the case of infinite resolving powers.
$ 1. Description of the counting rate meter. The counting rate meter ‘) is an electrical device used for the counting of the number of kicks which come from a Geigercounter instead of scaling down and using a mechanical recorder. The counting rate meter is composed of a condenser, to which a resistance and a galvanometer are connected in parallel. Every kick of the Geigercounter gives the same charge on the condenser. When the Geigercounter gives a certain number of kicks the mean deflection of- the galvanometer will be a measure for the number of kicks delivered by the Geigercounter. The mean deflection, the mean square of the statistical fluctuation, etc., can be easily calculated by the direct method of 0 r ns t e i n *) as used in the theory of Brownian motion. We will give the derivation for the case of one single reading, the case of integrating the deflection over some time interval T and for the case of q single observations at times 0, 0, 20 . . . . (q - 1) 0 respectively.
Physica
IX
641 41
642
H. A. VAN DER VELDEN
AND P. M. ENDT
5 2. The case of one single reading of the counting rate meter. We take the charge that is brought on the condenser by every kick as unit of charge. The Geigercounter gives a mean number of % kicks in one second.
nd
Fig; I
C
Electrical
ciicuit
of the counting
rate
meter.
The real number of kicks in dt sec. may be put equal to ndt and we can write this formally as ndt = ii& + Fdt in which F is a quantity analogous to the fortuitous force in the Einstein Langevin equation of the Brownian motion. - It is obvious that F = 0. The number of kicks of the Geigercounter in t sec. will then be equal to: f;dt
+idt
+[kdt
= At +oJ’Fdt
0
When we do not take into account the finite resolving power of the Geigercounter and connecting electrical circuit it is obvious that : [/ndt
-
fitI
= rit
so that: [jFdt12
We can write this integral way’:
= iit
as a double one and; we obtain in this
ONSOME
FLUCTUATIONPROBLEMS
643
We wish to use a new variable u so, that q = 5 + u
When we suppose that F(E) F(S, + 21) is only a function of u, possessing a very steep maximum at u = 0, and sloping down to zero in a time small in comparison to RC, we obtain: z’F(c)
F(c + u)du =fi
(1)
If we take into account the finite resolving power ‘c of the Geigercounter and connecting electrical circuit we shall prove in $ 10 that [(@t-fit]2
= tit (1 + iiT)
(14
Using this we obtain: LT’F(<) F(t + u) du = fi (1 + CT)
(14
Now we wish to construct the equation for the electrical circuit. When Q is the charge on the condenser at the time t we find that (see fig. 1):
iR = -Q c
,j+F-&dQ
and
at
’
so that the equation becomes:
Q Q+=="+F The integration Q
=
Qoe-W
of equation (2) gives: + fiRC (1 _ e--IIRC)+ j’F&~-&Rc) 0
@
(3)
Taking the mean of (3), if for t=O
Q7Qo=0,
gives p = iRC (; -
&lRc),
(4)
from which, when t > RC, we obtain Q= fiRC. The counting rate meter gives therefore, the mean number of kicks in RC sec. From
644
H.
A. VAN
DER
VELDEN
equation (3) we can also Using (3) and (4) we have:
derive
AND
the
(Q - Q)" = [[F(t)
P. M.
mean
ENDT
square
e--('--IIRc)d[J2
If we write this in the same manner as for formula integral we obtain after some transformations: (Q
_
Q)”
=
F
(1
g-W+C
-
fluctuation.
i-’
(1) as a double
F(S) F(E + .u) e”IRC du
So that if we use (1) we have: (Q-Q)‘=+1
_
e-
WRC
1
or if we use (la): (Q
_
g)’
=
(*
+
“1
fiRc
(,
_
,,,
w
2t/KC)
2 In the latter case the resolving taken into account. d(Q -
@2 =
d( 1 -
of the Geigercounter
e-2f’Rc) (1 + ~fi)
1/27iRC (1 -
v
power
-41 + +i
eAtlRc)
1~2iiRC
etc. is
(6)
.This is the same answer as that given by a mechanical recorder which has been watching for 2RC seconds, at least when the resolving power of the mechanical recorder is ‘negligible. $3. The case of integrating the deflection of the counting rate wder over sometime interval T. We suppose the reading to begin at a time t = 0 when at time-- t (I T t 1>> RC) the counting rate meter starts to receive kicks. We use some new symbols: P=
Q-CRC’
and
PO = Q,-CRC
From equation (3) it follows that: P = (QOso that
cRC)e-‘!RC +OiF(e)e-(f-SIW
dE
645
ON SOMEFLUCTUATIONPROBLEMS
and P,P
=po
e-;lRC
(6)
The accuracy of the reading is given by: g$qL
$
0
/dG 0
With the aid of (6) this integral formations, as
r’b,,, 6
P(v) dv
can be written,
after some trans-
1 + +F (e-rlRC With
l)]
(7)
the aid of (4) and (5~2) we obtain:
With a mechanical recorder mentioned in $2 we find: 1/l
under
the same circumstances
as
++i \/r?.T
(7) gives for T + CXJthe same expression, as could be expected, but, when T is not > RC the counting rate meter gives less statistical fluctuation than the mechanical recorder. 9 4. The case of q single observations at times 0, 0, 20, . . . . (q-l)& Just as in 5 3 we suppose that the counting rate meter starts to receive kicks at a time - t ( I- t I> RC). We wish to find the mean square deviation which we call p. ziT= _‘- (PO + P, + P, + . . . . + Pq-d2 q2
when P, represents the deviation from the mean reading time se. We write this expression for p as:
fiRC at
646
H.
A. VAN‘DER
VELDEN
AND
P. M.
ENDT
With the aid of (6) this becomes:
Carrying
out the summation
we obtain:
p = PT
2pT + Te 4
2p5
--B,RC 11
e-qOIRc roIRC
-
-
22 PT e -0IRC 4
I’_1 -
e-dl e--&RC RC
+
--B,Rc (1 _ ,--BP=) qe-q@/RC_ (1 _ e--Q@RC)e-@/RC (IO) (1 - e-&RC)
in which: p=
(1 + 0+RC 2
This expression for @ must :become equ;l to expression (7) for continuous reading during the time T when qf3 = T and q tends to 00 and 0 --t 0. When we apply this to expression (lo), F becomes : A?=O+ -’
2 FTRc
( 1 - e--TIRC) - 0 + . +
2P7 R2C2 T2
T IKc
-T/RC
_
e
(1 _
e-T/RC
)I
so that AT=
2p7TRc
1 + !!!jt. (eATIRC -
I which is indeed equal to expression
1) I
(7).
3 5. Comfiarison with the results of S c h i f f s and E v a n s. Some of the values obtained in the foregoing sections are already discussed in quite a different way by S c li i f f s and E v a n s “). The expressions (8) and (lo), however, differ from those of Schi.ffs and Evans. This is in connection with their choice in assigning several weights to their answers. In order to come to a decision, we shall follow a more classical way of arriving at expression (lo), for the case that T = 0, which while following closely their ideas, avoids the difficulties mentioned above.
ON
SOME
FLUCTUATION
647
PROBLEMS
As we have seen in $4. p-=
+ q;:qi; I s
mp,
= f
+ -g qf: (q 1,
4 POP”
P,, is the deviation from the mean value of the deflection, is iiRc at the time a&. We suppose that the counting rate meter has been working time before t = 0. In our calculation we suppose this time to which agrees fairly well with reality *when this time is very compared to RC. The deviation P,, is caused by the deviation from the number of kicks at times between - t and - t + dt when from 0 to co. We denote this deviation by (A)L.
(11) which a long be 00, large mean t goes
dt
It is clear that (A),
P, =.re-t/RC
dt
because the charge on the condenser drops down according power. At the same time J’,, =Te-(t+“@/RC)
(A), . dt
0
Obviously
to a e
we have:
when t’ # t and (A): = kit, so that we have: P,,
=
Te-(2t+@RC
(a);
= dt
0
fwe-(2t+uoIRC)
fi&
0
when in one sec. the Geigercounter gives the mean number kicks. Our expression (11) becomes now: p=-
?iRC 2q
which is ‘t = 0. It that the and E v
-fiRC q--I 2 (q + q2 u= I
2.4)e-“*lRC
of +i
(14
the same expression as our result (9) for the case that is clear, therefore, that our method is a reliable one and difference between our answers and those of S c h i f f B a n s are due to their choice in assigning several weights.
$ 6. Transmitted
fractions
when using thyrations
and. mechanical
648
H. A. VAN
DER
VELDEN
AND
P. M. ENDT
recorder. Usually the counting of current-impulses produced by a G e i g e r-M ii 11 e r counter or ionisation-chamber is performed by means of an amplifier, a scaling set and a. mechanical recorder. A limit is set to the counting rate by the finite recovery times of the counter (a), the amplifier (b), the scaling set (c) and the recorder (d). It is in the first place of importance to know the transmitted fraction, when we supply to either (a), (b), (c) or (d) a number of kicks with a statistical time-distribution. In the cases (a), (b) or (c) the transmitted fraction is given by a formula which differs from the one in the case (d). ‘The difference can be characterized as follows : A. in case (a), (b) or (c) a kick is transmitted if in the preceding time 7 no other kick is transmitted; in other words: after each transmitted kick lies a dead time 7; B. in case (d) a kick is transmitted if th&e comes no other kick in the subsequent time r; a mechanical recorder does not count a kick when there comes another shortly after. Because of the similarity in their behaviour it is usually allowed to allot to counter, amplifier and scaling set one and the same recovery time t,,, the longest of the three, mostly that of the scaling set. We shall call the recovery time of the recorder 1,. If the mean number of statistically distributed kicks per sec. is fi, the two formulas for the transmitted fraction are 4): A.
f-l-
B.
f zz.ze-4.
1 + iitl,
(13) (‘4)
The difference is very clearly seen if ri. is very large; then the scaling set for example counts at maximum speed, i.e. 1It,, kicks/set; but the recorder will count no kicks at all. Both A) and B) are only valid under certain restrictions. The kicks tra.nsmitt_ed by the scaling set will have no longer a statistical time distribution; the transmitted fraction in the case of a scaling set (scale of +) and a recorder will therefore not be equal to 1 7
1 l +
$&
e--,‘k
ON
SOME
FLUCTUATION
649
PROBLEMS
but is given by the more complicated
formula
i = f 1 +’ ~tk e-i(te-pfk)$ itilt, ;,P*
:) :
=
k=O 1 _
r{fi
+
I,
+%
-
pi,)}
P!
where P(p + 1, x) =[r z Pe-5 d< represents function “).
I
(4 2 ptk)
the incomplete
(15)
Gamma
3 7. Fluctuations in the purely statistical case (tk = t, = 0, p = 1). In addition to the transmitted fraction we can ask what the fluctuation in the transmitted number of kicks will be. If 7 is the mean number and r the real number of kicks in a time t, we wish to find (Y - 7)s. As an introduction we solve the problem for an ordinary statistical distribution (or a counting apparatus where we take tk = t, = 0 and p = 1). In the following calculations we will always put r?.= 1 ; in the final formulas the quantities t, tk and t, must then be multiplied by ii. The probability for r kicks in (0, t) is given by the P o i s s o n formula : k(r, t) = G e‘-‘.
We find: r =,5TlYk(r, t) = t and: Y2 = 5 r2k(r, t) = t2 + 2; r=l the root mean square error is:
5 8. Fluctuations if I,,= t, = 0, p # 1. We will now calculate again 7, 9 and ? when tk = t, = 0 but with an arbitrary scale 9. For that purpose we mark in an ordinary statistical distribution every P” kick with red and we wish to find the, probability of Y
650
H.
A.
VAN
DER
VELDEN
AND
P. M.
ENDT
red kicks in an interval (0, t), on the assumption that a red kick coincides with t = 0 (which is not counted with the kicks in (0, t)). The probability of r red kicks is equal to the sum of the probabilities kicks; so that we for r$,r$+ 1, . . . . (7 + 1) $ - 1 ordinary have : (r+l)P--l p --L k(7, t) = x 11=1p nre and : 7 =
5 ?=I
By differentiating
rk(7,
t)
=
F r=l
7
kfl)P-I IZ
f‘ -
n=rp
n!
e-‘.
fi times the function: 00
(r+l)P-1
p
rl s=rp
-, n!
y(i)=E7 r=l
we find for y(t) the differential
(16)
equation :
y(P)(t) = y(t) + et. With
the initial
conditions:
y(0) = y’(0) = . . . . = y(P-” (0) = 0, we obtain
the solution:
7 = y(l)e-’ = L-.-+‘+ P where the c represent For 3 we find:.
L + j-‘c’ t 2P
the p complex
fl = E y2
bfl)P-I
,&n-I) t
’
(17)
p’” roots of 1. ft
C
tl==TP
r=l
Putting
Pdh~l
- emt. n!
: z(t)
=
;r2 r=l
we can find for z(t) the differential ztp)(t) = z(t) + 2y(t) + er =
bfl)P-1 c n=rp
f‘ 2, .
equation:
(18)
ON
SOME
FLUCTUATION
For z(t) the same initial the solution :
PROBLEMS
conditions
651
hold as for y(t), which give
+ (P- 1)(2P- 1)+ ,,y 1: ,(X,,-l)f+ j3 = z(t)e-f = E + 2-P P2 P2 P2n=1L- 1 6P2 * p-’ UP - 2 - PU ,(A,,-l)f. (19, + a”:, (a, - 1J2 The right hand members of (17) and ( 19) are always real, because it is always’possible to take together two conjugated complex terms. For p = 2 we obtain for example: r = at -
* + &+
and for p=4: 7= & For small for larger neglected, and 5 into
# + +e-2r + ;te-’ (sin t + cos t).
values of t (17) and (19) have a kind of step shape and values the terms under the summation signs can be so that f merges asymptotically into a straight line a parabola. For large values of t we can write:
If the red kicks had, had a statistical distribution we should have found i’ = d/t/p ; the real red kicks have a more regular distribution. The corresponding error in n, the original number of kicks, is given by: Ti=pp7= Z/t, so that the accuracy is not altered by scaling down. It seems as if the constants, the exponential and the damped oscillating terms in (17) and (19) are only caused by the fact that we have laid a red kick at the beginning of the interval (0, t). However, when we treat the problem without this condition, we find for f the expected value t/p, but for YSwe obtain again an expression with a constant, exponential and damped oscillating terms. 5 9. Fluctuations if t,, = 0, p F 1, t, + 0. A second fluctuation problem, which can be treated rigorously, is the problem for which t,, = 0, p = 1, but t, is arbitrary. This means that we supply statistically distributed kicks to a mechanical recorder with resolving time t, ; condition B) regulates the counting of kicks. Here we always
652
H. A. VAN
DER
VELDEN
AND
P. M. ENDT
make a kick coincide with the end of (0, t), which is not counted with the kicks in (0, t). In the first place we wish to find the probability w:(t) that n kicks will ,fall in the interval (0, t), of which r are counted; we suppose t Z fit,. The number of possible ways, in which Y counted and (n - Y) uncounted kicks can succeed one another is (F). A possible sequence is for example, that one, in which the first r kicks are counted, the last n - Y kicks are not counted. We suppose that the first kick falls in ([r, E, + d&), the second in (h, & + d&J etc.; the probability that this will occur is then given by: e-f~d[, e-(gadJ dE2 . . . . e-(‘-gfl) a!&, = e-‘d&
. . . . a!&, .
Integrating over all possible values of &, . . . . E2 (first r kicks counted, 1ast.n - r kicks not counted) gives‘
When we carry out the n integrations obtain :
a:(t) = )‘--’k=.c
{t -
(7
+
and multiply
k) te)”
k! r! (TZ-
(-
r-
l)k
by (F) we
e--L
k)!
.
In trying to find an analogous formula for the other possible case rt, I t I nt,, we cannot follow the same method, because then difficulties arise from the integration limits of the last n-I variables. We call the required probability function y:,(t) and we introduce an integer s defined by: St, _( t I (s + l)te.
(r IS
In).
We give the deduction of y;(t) in two steps: in the first we deduce i.e. the probability that n kicks will occur in (0, t) (all not counted) (de > t). We shall find: c&t)
Q:(t)
s = c
k=O
(t -
kt,)” (k!(n-k)!
I)k e-’ *
We can prove this relation by completeinduction, i.e. we suppose that c&(t) satisfies (22). But when 1z = s + 1 we can derive cpkl(t) from (21). The probability of n uncounted kicks in (0, ‘t) is equal to. the probability of (12- 1) uncounted kicks in (0, E), multi-
ON
SOME
FLUCTUATION
653
PROBLEMS
plied by the probability of one kick in (5, E ; dQ, multiplied by the probability of n kicks in (5 + dE, t). This expression must be integrated with respect to E between t - t, and t. We obtain: c&(t) = / cp~-l(S)d~e-c’-A
=
t--I,
=e The induction
-t i Rzo
(t - kt,)” (k! (n-k)!
proof is concluded by remarking q+(t) = tcf (t < t),
1)k .
that:
as follows from P 0 i s s 0 n. It is now possible to find y:(t) ; by replacing. in the n-fold integral (20) the last (n - Y) integrals by &,(t - E, - t,), performing the first (r - 1) integrations and multiplying by (F), we obtain:
By dividing the integration and a last part of length (t -
segment in (s -
r) parts of length t,,
St,) and by using the integral:
j(x - a)” (b- qrndx = (n +“;;
l)! (b- a)“i+‘“+l,
we find finally:
d(t) = s--r c tt - (r + We)(- Ilk e-‘* k=O dk! (n-r-k)!
We can now calculate the probability (21) of r counted (0, t), when the total number of kicks in (0, t) is arbitrary: k(r, 0 =
k (K,(t) ;,=T+,rp:(t) “=I
= z ‘i’ l=O
it,-
k=O
By changing
the summation-
(-
l)k e-’
.,.
(1 = n-7-k).
order and replacing
“2’ by 5 we k=O
obtain
kicks’in
=
(r + k)te}l+‘+k r!k!l!
.
‘(23)
(t ~ ntc)
k=r
the result: Q,,
t)
=
t; k=r
e-kfc
(t -rf;k>-);)k+r
(24)
654
H.
A. VAN
DER
VELDEN
AND
P. M.
ENDT
As a check on the results (21), (23) and (24) we see readily: 5 o;(t) r=O
= J$ e-‘,
5 f&(t) r=O
= -$ e-‘, i k(Y, t) = r=O
(n s 4 (12 >4 1.
All this is possible with the aid of the sum *): fl k’(l)R k?l k! (n - k)! We are now able to calculate
= 0 (I < n) { = 1 (I=%). P and 92:
7 = i rk (Y, t) = i i r=1 k-l ?=I Only for k = 1, the summation we obtain therefore : F =
(25)
Ye-kG (t -
&)” (-
Y! (k -
l)k+’
r).’
with respect. to r is not = 0;
e-'c(t-tJ.
(26)
i -By the same method we find: f2 = e-26 (t - 2&) + e-‘c (t - i,), and : (y q2 : jq _ y-2 -- e-4 c (t - te) - e-2Lc (2t Substituting we obtain:
the ii and remarking
that in normal
;r=d(,-
z/z
(27) 31,) t, .
(28) cases 2> 1/ii > t,
(1-Y).
It is once again more useful to have the corresponding the number of original kicks; for one second we have:
error in n,
and:
*)
We
wish
to express
our
g&titude
to Mr.
H.
K. J.
B o k h o v e for
deriving
(25).
ON
SOME
FLUCTUATION
655
PROBLEMS
and finally :
Here the accuracy Poisson case.
has become les (but not much) than in the pure
3 10. Fluctuations if t,, # 0, p = 1, t, = 0. The third possible problem is .that for which p = 1, t, = 0 and th is arbitrary. The probability function is here given by:
+ 1)w ,-I’++& dt K(r,tpt = it - (rY!
(30)
A kick coincides with the beginning of (0, t). It is not easily possible to carry out the summations, required for the calculation of 7 and S ; but for! >. 1 we can replace (30) by the-Gaussian function: 1
e-{‘-“+‘)31/2(‘+l)dt,
(E = 1 + t*)
\/27t(7 + 1) and the summation can be replaced By aid of the integral: 7
we find :
dx
(3’)
by an integration.
e
03 I&=s- dr
e-{:-(r+l)‘}‘/l(r+l)
at
=
o 2/2x@ + 1)
for the probability, that a kick occurs F is now given by 1,/I,, where
dt
E
in (1, t + dt).
00
rdr
IO = os which
d27+
e-{‘-“+l)c}‘/2(‘+‘,
is found by differentiating --=-= 1 2 f = L IO
8
+ 1)
dIo dE EI, =
I0 with . (t t--E E
respect
E)I, 1 +-&-;.
to E:
&II, t
(32)
.
656
H. A. VAN
DER
VELDEN
AND
P. M. ENDT
e-{‘-(‘f
lN}‘/Z(r+ I)
-
Calling : O” 9dY
I, = os
and differentiating
d27r(r
+ 1)
with respect to E we find:
and: @=- I2 = EI2= IO
t* -+ 2
3
t --2 (2
E)
+ constant.
(33)
We see: 7*=
(y-
7)” = f
+ constant
Substituting E and calculating one second) gives:
N -4_; wi(l+). &3 the corresponding
(34)
error in n (for
This is the same answer as that found in $4, if t, is replaced by th. 5 11. Flztcttiations if t,, # 0, p # 1, te # 0. The last case which we shall treat is that for which t, = 0, but th and p are arbitrary; the results remain the same if all three quantities are arbitrary, only with the restriction that pt,, 2 t,; this means that the scaling factor is so high that the mechanical recorder misses no kicks. It is obvious.that for t > 1 we can use the same approximation (31) for the probability functions as in $ 10 only with a factor l/p. The righthand numbers of (32) and (34) must be multiplied by I/p and that of (33) by l/p*. The final answer (35) remains the same. There remains to be treated the problem in which all three quantities t”, p and t are arbitrary and pt,, < t,. It is not easy to treat’ the problem rigorously and a good approximation is also difficult. Received
April
ZOth,
1942.
ON SOME
FLUCTUATION
PROBLEMS
657
REFERENCES I) 2) 3) 4) 5) 6)
N. S. G i n g r i c h, etc., R.S.I. i, 450, 1936. L. S. 0 r n s t e i n, Proc. ray. Acad. Amsterdam 1917. 2. Phys. 14, 848, 1927. L. I. S c h i f f and R. D. E v an s, R.S.I. 7, 456, 1936. A. E. R u ark and F. E. B r a m m e r, Phys. Rev. 52, 322, 1937. L. A I a o g e n and N. M. S m i t h J r., Phys. Rev. 5:), 832, 1938. J ah n k e-E m de, ,,Funktionentafeln”, 1933, p. 96; or: I< a rl Pearson, 1922, London, ,,Tables of the incomplete gamma-function”. 7) D. Bierens de Haan, ,,Nouvelles tables d’intCgrales dhfinies”, 1867, Leiden.
I’hysica
IX
42
IX,
Juli
no 7
1942
ON SOME FLUCTUATION PROBLEMS CONNECTED WITH THE COUNTING OF IMPULSES PRODUCED .BY A GEIGER-MOLLER COUNTER OR IONISATION CHAMBER by H. A. VAN DER VELDEN Publication
from
the Physical
Institute
and P. M. ENDT of the University
of Utrecht
Summary In the present article we shall deal with some quantities which are of importance in counting the kicks of a Geigercounter or ionisation chamber arising from some constant or decaying radioactive source. We will first derive the sensibility and the statistical fluctuations of the counting rate meter, using the direct method first indicated by 0 r n s t e i n and a more classical method for one special case. We also wish to calculate some of these quantities for the case of a scaling set and a mechanical recorder with finite resolvi.ng powers. It is shown that in this case the fluctuations are greater (but not much) than in the case of infinite resolving powers.
$ 1. Description of the counting rate meter. The counting rate meter ‘) is an electrical device used for the counting of the number of kicks which come from a Geigercounter instead of scaling down and using a mechanical recorder. The counting rate meter is composed of a condenser, to which a resistance and a galvanometer are connected in parallel. Every kick of the Geigercounter gives the same charge on the condenser. When the Geigercounter gives a certain number of kicks the mean deflection of- the galvanometer will be a measure for the number of kicks delivered by the Geigercounter. The mean deflection, the mean square of the statistical fluctuation, etc., can be easily calculated by the direct method of 0 r ns t e i n *) as used in the theory of Brownian motion. We will give the derivation for the case of one single reading, the case of integrating the deflection over some time interval T and for the case of q single observations at times 0, 0, 20 . . . . (q - 1) 0 respectively.
Physica
IX
641 41
642
H. A. VAN DER VELDEN
AND P. M. ENDT
5 2. The case of one single reading of the counting rate meter. We take the charge that is brought on the condenser by every kick as unit of charge. The Geigercounter gives a mean number of % kicks in one second.
nd
Fig; I
C
Electrical
ciicuit
of the counting
rate
meter.
The real number of kicks in dt sec. may be put equal to ndt and we can write this formally as ndt = ii& + Fdt in which F is a quantity analogous to the fortuitous force in the Einstein Langevin equation of the Brownian motion. - It is obvious that F = 0. The number of kicks of the Geigercounter in t sec. will then be equal to: f;dt
+idt
+[kdt
= At +oJ’Fdt
0
When we do not take into account the finite resolving power of the Geigercounter and connecting electrical circuit it is obvious that : [/ndt
-
fitI
= rit
so that: [jFdt12
We can write this integral way’:
= iit
as a double one and; we obtain in this
ONSOME
FLUCTUATIONPROBLEMS
643
We wish to use a new variable u so, that q = 5 + u
When we suppose that F(E) F(S, + 21) is only a function of u, possessing a very steep maximum at u = 0, and sloping down to zero in a time small in comparison to RC, we obtain: z’F(c)
F(c + u)du =fi
(1)
If we take into account the finite resolving power ‘c of the Geigercounter and connecting electrical circuit we shall prove in $ 10 that [(@t-fit]2
= tit (1 + iiT)
(14
Using this we obtain: LT’F(<) F(t + u) du = fi (1 + CT)
(14
Now we wish to construct the equation for the electrical circuit. When Q is the charge on the condenser at the time t we find that (see fig. 1):
iR = -Q c
,j+F-&dQ
and
at
’
so that the equation becomes:
Q Q+=="+F The integration Q
=
Qoe-W
of equation (2) gives: + fiRC (1 _ e--IIRC)+ j’F&~-&Rc) 0
@
(3)
Taking the mean of (3), if for t=O
Q7Qo=0,
gives p = iRC (; -
&lRc),
(4)
from which, when t > RC, we obtain Q= fiRC. The counting rate meter gives therefore, the mean number of kicks in RC sec. From
644
H.
A. VAN
DER
VELDEN
equation (3) we can also Using (3) and (4) we have:
derive
AND
the
(Q - Q)" = [[F(t)
P. M.
mean
ENDT
square
e--('--IIRc)d[J2
If we write this in the same manner as for formula integral we obtain after some transformations: (Q
_
Q)”
=
F
(1
g-W+C
-
fluctuation.
i-’
(1) as a double
F(S) F(E + .u) e”IRC du
So that if we use (1) we have: (Q-Q)‘=+1
_
e-
WRC
1
or if we use (la): (Q
_
g)’
=
(*
+
“1
fiRc
(,
_
,,,
w
2t/KC)
2 In the latter case the resolving taken into account. d(Q -
@2 =
d( 1 -
of the Geigercounter
e-2f’Rc) (1 + ~fi)
1/27iRC (1 -
v
power
-41 + +i
eAtlRc)
1~2iiRC
etc. is
(6)
.This is the same answer as that given by a mechanical recorder which has been watching for 2RC seconds, at least when the resolving power of the mechanical recorder is ‘negligible. $3. The case of integrating the deflection of the counting rate wder over sometime interval T. We suppose the reading to begin at a time t = 0 when at time-- t (I T t 1>> RC) the counting rate meter starts to receive kicks. We use some new symbols: P=
Q-CRC’
and
PO = Q,-CRC
From equation (3) it follows that: P = (QOso that
cRC)e-‘!RC +OiF(e)e-(f-SIW
dE
645
ON SOMEFLUCTUATIONPROBLEMS
and P,P
=po
e-;lRC
(6)
The accuracy of the reading is given by: g$qL
$
0
/dG 0
With the aid of (6) this integral formations, as
r’b,,, 6
P(v) dv
can be written,
after some trans-
1 + +F (e-rlRC With
l)]
(7)
the aid of (4) and (5~2) we obtain:
With a mechanical recorder mentioned in $2 we find: 1/l
under
the same circumstances
as
++i \/r?.T
(7) gives for T + CXJthe same expression, as could be expected, but, when T is not > RC the counting rate meter gives less statistical fluctuation than the mechanical recorder. 9 4. The case of q single observations at times 0, 0, 20, . . . . (q-l)& Just as in 5 3 we suppose that the counting rate meter starts to receive kicks at a time - t ( I- t I> RC). We wish to find the mean square deviation which we call p. ziT= _‘- (PO + P, + P, + . . . . + Pq-d2 q2
when P, represents the deviation from the mean reading time se. We write this expression for p as:
fiRC at
646
H.
A. VAN‘DER
VELDEN
AND
P. M.
ENDT
With the aid of (6) this becomes:
Carrying
out the summation
we obtain:
p = PT
2pT + Te 4
2p5
--B,RC 11
e-qOIRc roIRC
-
-
22 PT e -0IRC 4
I’_1 -
e-dl e--&RC RC
+
--B,Rc (1 _ ,--BP=) qe-q@/RC_ (1 _ e--Q@RC)e-@/RC (IO) (1 - e-&RC)
in which: p=
(1 + 0+RC 2
This expression for @ must :become equ;l to expression (7) for continuous reading during the time T when qf3 = T and q tends to 00 and 0 --t 0. When we apply this to expression (lo), F becomes : A?=O+ -’
2 FTRc
( 1 - e--TIRC) - 0 + . +
2P7 R2C2 T2
T IKc
-T/RC
_
e
(1 _
e-T/RC
)I
so that AT=
2p7TRc
1 + !!!jt. (eATIRC -
I which is indeed equal to expression
1) I
(7).
3 5. Comfiarison with the results of S c h i f f s and E v a n s. Some of the values obtained in the foregoing sections are already discussed in quite a different way by S c li i f f s and E v a n s “). The expressions (8) and (lo), however, differ from those of Schi.ffs and Evans. This is in connection with their choice in assigning several weights to their answers. In order to come to a decision, we shall follow a more classical way of arriving at expression (lo), for the case that T = 0, which while following closely their ideas, avoids the difficulties mentioned above.
ON
SOME
FLUCTUATION
647
PROBLEMS
As we have seen in $4. p-=
+ q;:qi; I s
mp,
= f
+ -g qf: (q 1,
4 POP”
P,, is the deviation from the mean value of the deflection, is iiRc at the time a&. We suppose that the counting rate meter has been working time before t = 0. In our calculation we suppose this time to which agrees fairly well with reality *when this time is very compared to RC. The deviation P,, is caused by the deviation from the number of kicks at times between - t and - t + dt when from 0 to co. We denote this deviation by (A)L.
(11) which a long be 00, large mean t goes
dt
It is clear that (A),
P, =.re-t/RC
dt
because the charge on the condenser drops down according power. At the same time J’,, =Te-(t+“@/RC)
(A), . dt
0
Obviously
to a e
we have:
when t’ # t and (A): = kit, so that we have: P,,
=
Te-(2t+@RC
(a);
= dt
0
fwe-(2t+uoIRC)
fi&
0
when in one sec. the Geigercounter gives the mean number kicks. Our expression (11) becomes now: p=-
?iRC 2q
which is ‘t = 0. It that the and E v
-fiRC q--I 2 (q + q2 u= I
2.4)e-“*lRC
of +i
(14
the same expression as our result (9) for the case that is clear, therefore, that our method is a reliable one and difference between our answers and those of S c h i f f B a n s are due to their choice in assigning several weights.
$ 6. Transmitted
fractions
when using thyrations
and. mechanical
648
H. A. VAN
DER
VELDEN
AND
P. M. ENDT
recorder. Usually the counting of current-impulses produced by a G e i g e r-M ii 11 e r counter or ionisation-chamber is performed by means of an amplifier, a scaling set and a. mechanical recorder. A limit is set to the counting rate by the finite recovery times of the counter (a), the amplifier (b), the scaling set (c) and the recorder (d). It is in the first place of importance to know the transmitted fraction, when we supply to either (a), (b), (c) or (d) a number of kicks with a statistical time-distribution. In the cases (a), (b) or (c) the transmitted fraction is given by a formula which differs from the one in the case (d). ‘The difference can be characterized as follows : A. in case (a), (b) or (c) a kick is transmitted if in the preceding time 7 no other kick is transmitted; in other words: after each transmitted kick lies a dead time 7; B. in case (d) a kick is transmitted if th&e comes no other kick in the subsequent time r; a mechanical recorder does not count a kick when there comes another shortly after. Because of the similarity in their behaviour it is usually allowed to allot to counter, amplifier and scaling set one and the same recovery time t,,, the longest of the three, mostly that of the scaling set. We shall call the recovery time of the recorder 1,. If the mean number of statistically distributed kicks per sec. is fi, the two formulas for the transmitted fraction are 4): A.
f-l-
B.
f zz.ze-4.
1 + iitl,
(13) (‘4)
The difference is very clearly seen if ri. is very large; then the scaling set for example counts at maximum speed, i.e. 1It,, kicks/set; but the recorder will count no kicks at all. Both A) and B) are only valid under certain restrictions. The kicks tra.nsmitt_ed by the scaling set will have no longer a statistical time distribution; the transmitted fraction in the case of a scaling set (scale of +) and a recorder will therefore not be equal to 1 7
1 l +
$&
e--,‘k
ON
SOME
FLUCTUATION
649
PROBLEMS
but is given by the more complicated
formula
i = f 1 +’ ~tk e-i(te-pfk)$ itilt, ;,P*
:) :
=
k=O 1 _
r{fi
+
I,
+%
-
pi,)}
P!
where P(p + 1, x) =[r z Pe-5 d< represents function “).
I
(4 2 ptk)
the incomplete
(15)
Gamma
3 7. Fluctuations in the purely statistical case (tk = t, = 0, p = 1). In addition to the transmitted fraction we can ask what the fluctuation in the transmitted number of kicks will be. If 7 is the mean number and r the real number of kicks in a time t, we wish to find (Y - 7)s. As an introduction we solve the problem for an ordinary statistical distribution (or a counting apparatus where we take tk = t, = 0 and p = 1). In the following calculations we will always put r?.= 1 ; in the final formulas the quantities t, tk and t, must then be multiplied by ii. The probability for r kicks in (0, t) is given by the P o i s s o n formula : k(r, t) = G e‘-‘.
We find: r =,5TlYk(r, t) = t and: Y2 = 5 r2k(r, t) = t2 + 2; r=l the root mean square error is:
5 8. Fluctuations if I,,= t, = 0, p # 1. We will now calculate again 7, 9 and ? when tk = t, = 0 but with an arbitrary scale 9. For that purpose we mark in an ordinary statistical distribution every P” kick with red and we wish to find the, probability of Y
650
H.
A.
VAN
DER
VELDEN
AND
P. M.
ENDT
red kicks in an interval (0, t), on the assumption that a red kick coincides with t = 0 (which is not counted with the kicks in (0, t)). The probability of r red kicks is equal to the sum of the probabilities kicks; so that we for r$,r$+ 1, . . . . (7 + 1) $ - 1 ordinary have : (r+l)P--l p --L k(7, t) = x 11=1p nre and : 7 =
5 ?=I
By differentiating
rk(7,
t)
=
F r=l
7
kfl)P-I IZ
f‘ -
n=rp
n!
e-‘.
fi times the function: 00
(r+l)P-1
p
rl s=rp
-, n!
y(i)=E7 r=l
we find for y(t) the differential
(16)
equation :
y(P)(t) = y(t) + et. With
the initial
conditions:
y(0) = y’(0) = . . . . = y(P-” (0) = 0, we obtain
the solution:
7 = y(l)e-’ = L-.-+‘+ P where the c represent For 3 we find:.
L + j-‘c’ t 2P
the p complex
fl = E y2
bfl)P-I
,&n-I) t
’
(17)
p’” roots of 1. ft
C
tl==TP
r=l
Putting
Pdh~l
- emt. n!
: z(t)
=
;r2 r=l
we can find for z(t) the differential ztp)(t) = z(t) + 2y(t) + er =
bfl)P-1 c n=rp
f‘ 2, .
equation:
(18)
ON
SOME
FLUCTUATION
For z(t) the same initial the solution :
PROBLEMS
conditions
651
hold as for y(t), which give
+ (P- 1)(2P- 1)+ ,,y 1: ,(X,,-l)f+ j3 = z(t)e-f = E + 2-P P2 P2 P2n=1L- 1 6P2 * p-’ UP - 2 - PU ,(A,,-l)f. (19, + a”:, (a, - 1J2 The right hand members of (17) and ( 19) are always real, because it is always’possible to take together two conjugated complex terms. For p = 2 we obtain for example: r = at -
* + &+
and for p=4: 7= & For small for larger neglected, and 5 into
# + +e-2r + ;te-’ (sin t + cos t).
values of t (17) and (19) have a kind of step shape and values the terms under the summation signs can be so that f merges asymptotically into a straight line a parabola. For large values of t we can write:
If the red kicks had, had a statistical distribution we should have found i’ = d/t/p ; the real red kicks have a more regular distribution. The corresponding error in n, the original number of kicks, is given by: Ti=pp7= Z/t, so that the accuracy is not altered by scaling down. It seems as if the constants, the exponential and the damped oscillating terms in (17) and (19) are only caused by the fact that we have laid a red kick at the beginning of the interval (0, t). However, when we treat the problem without this condition, we find for f the expected value t/p, but for YSwe obtain again an expression with a constant, exponential and damped oscillating terms. 5 9. Fluctuations if t,, = 0, p F 1, t, + 0. A second fluctuation problem, which can be treated rigorously, is the problem for which t,, = 0, p = 1, but t, is arbitrary. This means that we supply statistically distributed kicks to a mechanical recorder with resolving time t, ; condition B) regulates the counting of kicks. Here we always
652
H. A. VAN
DER
VELDEN
AND
P. M. ENDT
make a kick coincide with the end of (0, t), which is not counted with the kicks in (0, t). In the first place we wish to find the probability w:(t) that n kicks will ,fall in the interval (0, t), of which r are counted; we suppose t Z fit,. The number of possible ways, in which Y counted and (n - Y) uncounted kicks can succeed one another is (F). A possible sequence is for example, that one, in which the first r kicks are counted, the last n - Y kicks are not counted. We suppose that the first kick falls in ([r, E, + d&), the second in (h, & + d&J etc.; the probability that this will occur is then given by: e-f~d[, e-(gadJ dE2 . . . . e-(‘-gfl) a!&, = e-‘d&
. . . . a!&, .
Integrating over all possible values of &, . . . . E2 (first r kicks counted, 1ast.n - r kicks not counted) gives‘
When we carry out the n integrations obtain :
a:(t) = )‘--’k=.c
{t -
(7
+
and multiply
k) te)”
k! r! (TZ-
(-
r-
l)k
by (F) we
e--L
k)!
.
In trying to find an analogous formula for the other possible case rt, I t I nt,, we cannot follow the same method, because then difficulties arise from the integration limits of the last n-I variables. We call the required probability function y:,(t) and we introduce an integer s defined by: St, _( t I (s + l)te.
(r IS
In).
We give the deduction of y;(t) in two steps: in the first we deduce i.e. the probability that n kicks will occur in (0, t) (all not counted) (de > t). We shall find: c&t)
Q:(t)
s = c
k=O
(t -
kt,)” (k!(n-k)!
I)k e-’ *
We can prove this relation by completeinduction, i.e. we suppose that c&(t) satisfies (22). But when 1z = s + 1 we can derive cpkl(t) from (21). The probability of n uncounted kicks in (0, ‘t) is equal to. the probability of (12- 1) uncounted kicks in (0, E), multi-
ON
SOME
FLUCTUATION
653
PROBLEMS
plied by the probability of one kick in (5, E ; dQ, multiplied by the probability of n kicks in (5 + dE, t). This expression must be integrated with respect to E between t - t, and t. We obtain: c&(t) = / cp~-l(S)d~e-c’-A
=
t--I,
=e The induction
-t i Rzo
(t - kt,)” (k! (n-k)!
proof is concluded by remarking q+(t) = tcf (t < t),
1)k .
that:
as follows from P 0 i s s 0 n. It is now possible to find y:(t) ; by replacing. in the n-fold integral (20) the last (n - Y) integrals by &,(t - E, - t,), performing the first (r - 1) integrations and multiplying by (F), we obtain:
By dividing the integration and a last part of length (t -
segment in (s -
r) parts of length t,,
St,) and by using the integral:
j(x - a)” (b- qrndx = (n +“;;
l)! (b- a)“i+‘“+l,
we find finally:
d(t) = s--r c tt - (r + We)(- Ilk e-‘* k=O dk! (n-r-k)!
We can now calculate the probability (21) of r counted (0, t), when the total number of kicks in (0, t) is arbitrary: k(r, 0 =
k (K,(t) ;,=T+,rp:(t) “=I
= z ‘i’ l=O
it,-
k=O
By changing
the summation-
(-
l)k e-’
.,.
(1 = n-7-k).
order and replacing
“2’ by 5 we k=O
obtain
kicks’in
=
(r + k)te}l+‘+k r!k!l!
.
‘(23)
(t ~ ntc)
k=r
the result: Q,,
t)
=
t; k=r
e-kfc
(t -rf;k>-);)k+r
(24)
654
H.
A. VAN
DER
VELDEN
AND
P. M.
ENDT
As a check on the results (21), (23) and (24) we see readily: 5 o;(t) r=O
= J$ e-‘,
5 f&(t) r=O
= -$ e-‘, i k(Y, t) = r=O
(n s 4 (12 >4 1.
All this is possible with the aid of the sum *): fl k’(l)R k?l k! (n - k)! We are now able to calculate
= 0 (I < n) { = 1 (I=%). P and 92:
7 = i rk (Y, t) = i i r=1 k-l ?=I Only for k = 1, the summation we obtain therefore : F =
(25)
Ye-kG (t -
&)” (-
Y! (k -
l)k+’
r).’
with respect. to r is not = 0;
e-'c(t-tJ.
(26)
i -By the same method we find: f2 = e-26 (t - 2&) + e-‘c (t - i,), and : (y q2 : jq _ y-2 -- e-4 c (t - te) - e-2Lc (2t Substituting we obtain:
the ii and remarking
that in normal
;r=d(,-
z/z
(27) 31,) t, .
(28) cases 2> 1/ii > t,
(1-Y).
It is once again more useful to have the corresponding the number of original kicks; for one second we have:
error in n,
and:
*)
We
wish
to express
our
g&titude
to Mr.
H.
K. J.
B o k h o v e for
deriving
(25).
ON
SOME
FLUCTUATION
655
PROBLEMS
and finally :
Here the accuracy Poisson case.
has become les (but not much) than in the pure
3 10. Fluctuations if t,, # 0, p = 1, t, = 0. The third possible problem is .that for which p = 1, t, = 0 and th is arbitrary. The probability function is here given by:
+ 1)w ,-I’++& dt K(r,tpt = it - (rY!
(30)
A kick coincides with the beginning of (0, t). It is not easily possible to carry out the summations, required for the calculation of 7 and S ; but for! >. 1 we can replace (30) by the-Gaussian function: 1
e-{‘-“+‘)31/2(‘+l)dt,
(E = 1 + t*)
\/27t(7 + 1) and the summation can be replaced By aid of the integral: 7
we find :
dx
(3’)
by an integration.
e
03 I&=s- dr
e-{:-(r+l)‘}‘/l(r+l)
at
=
o 2/2x@ + 1)
for the probability, that a kick occurs F is now given by 1,/I,, where
dt
E
in (1, t + dt).
00
rdr
IO = os which
d27+
e-{‘-“+l)c}‘/2(‘+‘,
is found by differentiating --=-= 1 2 f = L IO
8
+ 1)
dIo dE EI, =
I0 with . (t t--E E
respect
E)I, 1 +-&-;.
to E:
&II, t
(32)
.
656
H. A. VAN
DER
VELDEN
AND
P. M. ENDT
e-{‘-(‘f
lN}‘/Z(r+ I)
-
Calling : O” 9dY
I, = os
and differentiating
d27r(r
+ 1)
with respect to E we find:
and: @=- I2 = EI2= IO
t* -+ 2
3
t --2 (2
E)
+ constant.
(33)
We see: 7*=
(y-
7)” = f
+ constant
Substituting E and calculating one second) gives:
N -4_; wi(l+). &3 the corresponding
(34)
error in n (for
This is the same answer as that found in $4, if t, is replaced by th. 5 11. Flztcttiations if t,, # 0, p # 1, te # 0. The last case which we shall treat is that for which t, = 0, but th and p are arbitrary; the results remain the same if all three quantities are arbitrary, only with the restriction that pt,, 2 t,; this means that the scaling factor is so high that the mechanical recorder misses no kicks. It is obvious.that for t > 1 we can use the same approximation (31) for the probability functions as in $ 10 only with a factor l/p. The righthand numbers of (32) and (34) must be multiplied by I/p and that of (33) by l/p*. The final answer (35) remains the same. There remains to be treated the problem in which all three quantities t”, p and t are arbitrary and pt,, < t,. It is not easy to treat’ the problem rigorously and a good approximation is also difficult. Received
April
ZOth,
1942.
ON SOME
FLUCTUATION
PROBLEMS
657
REFERENCES I) 2) 3) 4) 5) 6)
N. S. G i n g r i c h, etc., R.S.I. i, 450, 1936. L. S. 0 r n s t e i n, Proc. ray. Acad. Amsterdam 1917. 2. Phys. 14, 848, 1927. L. I. S c h i f f and R. D. E v an s, R.S.I. 7, 456, 1936. A. E. R u ark and F. E. B r a m m e r, Phys. Rev. 52, 322, 1937. L. A I a o g e n and N. M. S m i t h J r., Phys. Rev. 5:), 832, 1938. J ah n k e-E m de, ,,Funktionentafeln”, 1933, p. 96; or: I< a rl Pearson, 1922, London, ,,Tables of the incomplete gamma-function”. 7) D. Bierens de Haan, ,,Nouvelles tables d’intCgrales dhfinies”, 1867, Leiden.
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42