Dependence of the Time-Reading Process of the Salecker–Wigner Quantum Clock on the Size of the Clock

Abstract

It is shown in the present note that the degree of the complexity of the time-reading process of the Salecker–Wigner clock depends on the size of the clock. This dependence leads to a relation between the size and the accuracy of the clock, and suggests a precise optimal value for the size in agreement with the order of magnitude value established by Salecker and Wigner.

Appendices

Appendix 1

In this appendix relation (4) between $t_0$ and $t_c$ is derived.

In [1] the time $t_0$ carried by the clock is given in Eq. (8):

$$\begin{aligned} t_0 = \frac{c - v}{v} \,\frac{\ell }{c} (2r - 1). \end{aligned}$$

(42)

Here v stands for the negative velocity of the hand and

$$\begin{aligned} r = \frac{t_{\widehat{2}}^A - t_{\widehat{1}}^A}{t_{\widehat{3}}^A - t_1^A} . \end{aligned}$$

(43)

The expression of v and r through the velocity u and the ratio $\varrho$ (see 16) used in this note is

$$\begin{aligned} v&= -u, \end{aligned}$$

(44)

$$\begin{aligned} r&= 1 - \varrho . \end{aligned}$$

(45)

Substituting (44) and (45) into (42) relation (4) is easily obtained, with $t_c$ given in (15).

Appendix 2

In this appendix it is shown that if the length of the dial is

$$\begin{aligned} 2\ell = \frac{1}{2} c\tau (1 + \beta ) \end{aligned}$$

(46)

or shorter, then the triad partner of a quantum $\widehat{2}$ is the first quantum $\widehat{3}$ following that $\widehat{2}$ on the way toward the recorder. In this case the identification of the triad partners $(\widehat{2}, \widehat{3})$ can be done without the knowledge of the serial number k. However, if the dial is longer than the value in (46), namely if

$$\begin{aligned} 2\ell = \frac{m}{2} c\tau (1 + \beta ), \quad m = 2,3, \dots , \end{aligned}$$

(47)

then the pairing depends on the value of k.

The proof of the above statements relies on the comparison of the distance $x_{\widehat{3}_k} - x_{\widehat{3}_{k - 1}}$ between the consecutive quanta $\widehat{3}$ with the distances $x_{\widehat{3}_k} - x_{\widehat{2}_k}$ between the triad partners $\widehat{2}_k$, $\widehat{3}_k$.

It is easy to see that

$$\begin{aligned} x_{\widehat{3}_k} - x_{\widehat{3}_{k - 1}} = c\tau (1 + \beta ), \quad k = 1,2,\dots , n. \end{aligned}$$

(48)

Indeed, as told before Eq. (30), $c\tau (1 + \beta )$ is the distance between the consecutive triads impinging on the clock, in particular between the consecutive quanta $\widetilde{3}$ scattered back by dial body $\mathbf 3$ at rest. The scattered $\widetilde{3}$’s are the $\widehat{3}$’s, therefore the distance between the consecutive $\widehat{3}$’s is equal to the same distance $c\tau (1 + \beta )$.

Let us now look at the distances $x_{\widehat{3}_k} - x_{\widehat{2}_k}$. According to (22)

$$\begin{aligned} x_{\widehat{3}_k} - x_{\widehat{2}_k} = 2\left( \ell - x_h^{(k)}\right) , \quad k = 1,2,\dots , n, \end{aligned}$$

(49)

where $x_h^{(k)}$ is the coordinate of the hand at the moment of its scattering with the quantum $\widetilde{2}_k$ which becomes $\widehat{2}_k$ at that very moment. As told before Eq. (30) such scatterings follow each other at time intervals $\tau$. During one interval the hand covers the distance

$$\begin{aligned} u\tau = \frac{2\ell }{T} \tau = \frac{2\ell }{n} \end{aligned}$$

(50)

in the negative X direction, therefore the relation between $x_h^{(k)}$ and $x_h^{(1)}$ is

$$\begin{aligned} x_h^{(k)} = x_h^{(1)} - (k - 1) \frac{2\ell }{n}, \quad k = 1,2,\dots , n. \end{aligned}$$

(51)

Since the hand starts from the point $x_{d_3} = \ell$, its scattering with the first quantum $\widetilde{2}_1$ occurs in the interval $\left[ \ell , \ell - \frac{2\ell }{n}\right]$. Let us consider the case when this scattering happens at the middle point

$$\begin{aligned} x_h^{(1)} = \ell - \frac{\ell }{n} \end{aligned}$$

(52)

of the interval. The results obtained in this special case concerning the time-reading process are valid also for other points of the interval. The mathematical derivation involves cumbersome notation in the general case.

From (49), (51) and (52) it follows that

$$\begin{aligned} x_{\widehat{3}_k} - x_{\widehat{2}_k} = 4\ell \frac{k - \frac{1}{2}}{n}, \quad k = 1,2, \dots , n. \end{aligned}$$

(53)

Let us now look at the case when the length of the dial is given by (46):

$$\begin{aligned} 2\ell = \frac{1}{2} c\tau (1 + \beta ). \end{aligned}$$

(54)

Then according to (48)

$$\begin{aligned} x_{\widehat{3}_k} - x_{\widehat{3}_{k - 1}} = 4\ell , \end{aligned}$$

(55)

and (53) leads to

$$\begin{aligned} x_{\widehat{3}_k} - x_{\widehat{2}_k} = \left( x_{\widehat{3}_k} - x_{\widehat{3}_{k - 1}}\right) \frac{k - \frac{1}{2}}{n} < x_{\widehat{3}_k} - x_{\widehat{3}_{k - 1}} \end{aligned}$$

(56)

for all values of k. (56) says that $\widehat{2}_k$ travels toward the recorder between $\widehat{3}_{k - 1}$ and $\widehat{3}_k$ (Fig. 2), so the triad partner of $\widehat{2}_k$ is indeed the first quantum $\widehat{3}$ following $\widehat{2}_k$.

Fig. 2

The case $2\ell = \frac{1}{2} c\tau (1 + \beta )$

Let us now consider the cases when the length of the dial is given by relation (47). The general result for $m \ge 2$ described in Sect. 4 can be inferred from the special cases $m = 2$ and $m = 3$ discussed below.

For $m = 2$ (47) and (48) give

$$\begin{aligned} x_{\widehat{3}_k} - x_{\widehat{3}_{k - 1}} = 2\ell , \end{aligned}$$

(57)

so (53) becomes

$$\begin{aligned} x_{\widehat{3}_k} - x_{\widehat{2}_k} = \left( x_{\widehat{3}_k} - x_{\widehat{3}_{k - 1}}\right) \frac{2k - 1}{n}. \end{aligned}$$

(58)

According to this relation if

$$\begin{aligned} k < \frac{n + 1}{2} \end{aligned}$$

(59)

then

$$\begin{aligned} x_{\widehat{3}_k} - x_{\widehat{2}_k} < x_{\widehat{3}_k} - x_{\widehat{3}_{k - 1}}, \end{aligned}$$

(60)

$\widehat{2}_k$ travels between $\widehat{3}_{k - 1}$ and $\widehat{3}_k$, thus the triad partner of $\widehat{2}_k$ is again the first $\widehat{3}$ following $\widehat{2}_k$ (Fig. 3a).

Fig. 3

The cases $m = 2$

If

$$\begin{aligned} k = \frac{n + 1}{2} \end{aligned}$$

(61)

then $\widehat{2}_k$ and $\widehat{3}_{k - 1}$ travel together, and the triad partner of $\widehat{2}_k$ is still the first $\widehat{3}$ following $\widehat{2}_k$. However, if

$$\begin{aligned} k > \frac{n + 1}{2} \end{aligned}$$

(62)

then

$$\begin{aligned} x_{\widehat{3}_k} - x_{\widehat{2}_k} > x_{\widehat{3}_k} - x_{\widehat{3}_{k - 1}}, \end{aligned}$$

(63)

$\widehat{2}_k$ travels between $\widehat{3}_{k - 2}$ and $\widehat{3}_{k - 1}$, and the triad partner of $\widehat{2}_k$ is the second $\widehat{3}$ following $\widehat{2}_k$.

Dealing with relations involving k and n one should take into account that both k and n are integers. Thus for even values of n the case (61) does not occur, and in (59) the highest occurring value of k is n / 2.

If the recorder is turned on at a moment during the arrival of the quanta, the serial number k of a registered quantum $\widehat{2}$ is not known, and it is ambiguous whether the first or the second quantum $\widehat{3}$ is its triad partner. According to (57) the distance between these $\widehat{3}$’s is $2\ell$, and from (32) it follows that the ambiguity in $t_c$ is

$$\begin{aligned} \Delta t_c = \frac{2\ell }{4\ell } T = \frac{T}{2}, \end{aligned}$$

(64)

much larger than the accuracy $\tau$ of the clock.

Let us now look at the case $m = 3$. Then according to (47) and (48)

$$\begin{aligned} x_{\widehat{3}_k} - x_{\widehat{3}_{k - 1}} = \frac{4\ell }{3}, \quad k = 1,2,\dots , n, \end{aligned}$$

(65)

and with (53)

$$\begin{aligned} x_{\widehat{3}_k} - x_{\widehat{2}_k} = \left( x_{\widehat{3}_k} - x_{\widehat{3}_{k - 1}}\right) 3 \frac{k - \frac{1}{2}}{n}. \end{aligned}$$

(66)

The triad partner of $\widehat{2}_k$ is now the first $\widehat{3}$ following $\widehat{2}_k$ if

$$\begin{aligned} x_{\widehat{3}_k} - x_{\widehat{2}_k} \le x_{\widehat{3}_k} - x_{\widehat{3}_{k - 1}}, \end{aligned}$$

(67)

i.e. according to (66) if

$$\begin{aligned} k \le \frac{n}{3} + \frac{1}{2}. \end{aligned}$$

(68)

The triad partner of $\widehat{2}_k$ is the second $\widehat{3}$ following $\widehat{2}_k$ if

$$\begin{aligned} x_{\widehat{3}_k} - x_{\widehat{3}_{k - 1}} < x_{\widehat{3}_k} - x_{\widehat{2}_k} \le x_{\widehat{3}_k} - x_{\widehat{3}_{k - 2}} = 2 \left( x_{\widehat{3}_k} - x_{\widehat{3}_{k - 1}}\right) , \end{aligned}$$

(69)

i.e. if

$$\begin{aligned} \frac{n}{3} + \frac{1}{2} < k \le \frac{2n}{3} + \frac{1}{2}, \end{aligned}$$

(70)

and the triad partner is the third $\widehat{3}$ if

$$\begin{aligned} \frac{2n}{3} + \frac{1}{2} < k \le n + \frac{1}{2}. \end{aligned}$$

(71)

If the value of k is not registered by the recorder the ambiguity in the choice of the triad partner $\widehat{3}$ of a $\widehat{2}$ is now threefold. According to (65) the distance between $\widehat{3}_k$ and $\widehat{3}_{k - 1}$, as well as between $\widehat{3}_{k - 1}$ and $\widehat{3}_{k - 2}$ is $4\ell / 3$, between $\widehat{3}_k$ and $\widehat{3}_{k - 2}$ it is $8\ell / 3$. Therefore from (32) it follows that the ambiguities in the value of $t_c$ are T / 3 and 2T / 3.