Note on the physical meaning of

the cosmological term

Hirofumi Sakuma1 and Hiroyuki Ochiai2

1 Research Origin for Dressed Photon,

2Institute of Mathematics for Industry, Kyushu University

Sept. 10, 2019

Abstract

At first glance, the issue of dressed photon in the field of nano-optics seems to have nothing to do with cosmology which deals withphenomena with the largest spatial scales in nature. However, recentpreliminary analyses on the mathematical structure of Clebsch dualfield introduced as a part of explaining the generating mechanismof dressed photon implies the possibility that the emergence of thecosmological constant λ as the coefficient of the cosmological term λgabmay be explained by the dynamical process of simultaneous conformalsymmetry breaking of electromagnetic and gravitational fields. In thisshort note, as a supplemental explanation of this conjecture, we givea new explanation of the physical meaning of the cosmological termλgab by proving the hitherto unnoticed identity (1) in section 1.

1 Aim of this short note

The aim is to show that the following second rank tensor Y ba defined in terms

of the conformal Weyl tensor Wabcd vanishes identically, i.e.,

Y ba := WacdeW

bcde − 1

4WijklW

ijklg ba = 0, (1)

which provides a supplemetal proof that the cosmological term λgab repre-sents the energy-momentum tensor of Weyl field, which is being discussed inone of the manuscripts on dressed photon by H. Sakuma et al.

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2 Calculations

2.1 Preparatory process

Since Eq. (1) is a tensor equation, it suffices to show it in an inertial referenceframe where the metric tensor gab are diagonalized, i.e., (g00 = 1, g11 = g22 =g33 = −1) and (gab = 0; a 6= b). The basic properties of Weyl tensor to beused in our calculations are summarized as follows:

Wabcd = −Wabdc, Wabcd = −Wbacd, Wabcd = Wcdab, (2)

Wabcd +Wacdb +Wadbc = 0, W abca = −W a

bac = 0. (3)

As the preparatory steps, we rewrite the last equation in (3): W abac = 0 as a

collection of specific forms. Since

W abac = W 0

b0c +W 1b1c +W 2

b2c +W 3b3c = W0b0c −W1b1c −W2b2c −W3b3c, (4)

as the Case I, when b 6= c, specific (a*) equations shown below (5) can beconcisely written as

Wijik = εWljlk, (i 6= j 6= k 6= l) (5)

where ε = −1 if either j or k is zero and ε = 1 if neither j or k is zero.(a1) (b = 0, c = 1) and (b = 1, c = 0) ⇒ W2021 +W3031 = 0.(a2) (b = 0, c = 2) and (b = 2, c = 0) ⇒ W1012 +W3032 = 0.(a3) (b = 0, c = 3) and (b = 3, c = 0) ⇒ W1013 +W2023 = 0.(a4) (b = 1, c = 2) and (b = 2, c = 1) ⇒ W0102 −W3132 = 0.(a5) (b = 1, c = 3) and (b = 3, c = 1) ⇒ W0103 −W2123 = 0.(a6) (b = 2, c = 3) and (b = 3, c = 2) ⇒ W0203 −W1213 = 0.

As the Case II, when b = c, (a12), (a13) and (a14) below can be written as

Wijij = −Wklkl, (i 6= j 6= k 6= l) (6)

(a7) (b = 0, c = 0) ⇒ , W1010 +W2020 +W3030 = 0.(a8) (b = 1, c = 1) ⇒ W0101 −W2121 −W3131 = 0.(a9) (b = 2, c = 2) ⇒ , W0202 −W1212 −W3232 = 0.(a10) (b = 3, c = 3) ⇒ W0303 −W1313 −W2323 = 0.Using (a7), (a8), (a9) and (a10), we get

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(a11) W2323 +W3131 +W1212 = 0,and combining (a11) respectively with (a8), (a9) and (a10) yield(a12) W0101 +W2323 = 0.(a13) W0202 +W3131 = 0.(a14) W0303 +W1212 = 0.

2.2 Step 1: On off-diagonal components

In this subsection, we are going to show that Yij = WicdeWcdej = 0, (i 6= j).

Straightforward summations with respect to c, d and e, using the properties(2) yield

Yij/2 = Wi001W001j +Wi002W

002j +Wi003W

003j

+ Wi012W012j +Wi013W

013j +Wi023W

023j

+ Wi101W101j +Wi102W

102j +Wi103W

103j

+ Wi112W112j +Wi113W

113j +Wi123W

123j

+ Wi201W201j +Wi202W

202j +Wi203W

203j

+ Wi212W212j +Wi213W

213j +Wi223W

223j

+ Wi301W301j +Wi302W

302j +Wi303W

303j

+ Wi312W312j +Wi313W

313j +Wi323W

323j . (7)

Rewriting contra-variant expressions, for example W 001j = −Wj001, (7) be-

comes

Yij/2 = −Wi001Wj001 −Wi002Wj002 −Wi003Wj003

+ Wi012Wj012 +Wi013Wj013 +Wi023Wj023

+ Wi101Wj101 +Wi102Wj102 +Wi103Wj103

− Wi112Wj112 −Wi113Wj113 −Wi123Wj123

+ Wi201Wj201 +Wi202Wj202 +Wi203Wj203

− Wi212Wj212 −Wi213Wj213 −Wi223Wj223

+ Wi301Wj301 +Wi302Wj302 +Wi303Wj303

− Wi312Wj312 −Wi313Wj313 −Wi323Wj323. (8)

Now, based on the expression (8), let us calculate, for instance, Y01. Fori = 0, j = 1, we see that all the terms in the upper half of (8) vanish, so that

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we have

Y01/2 = W0201W1201 +W0202W1202 +W0203W1203

− W0212W1212 −W0213W1213 −W0223W1223

+ W0301W1301 +W0302W1302 +W0303W1303

− W0312W1312 −W0313W1313 −W0323W1323. (9)

Rearranging terms in (9), we get

Y01/2 = (W0203W1203 −W0312W1312) + (W0302W1302 −W0213W1213)

= (W0201W1201 −W0323W1323) + (W0301W1301 −W0223W1223)

+ (W0202W1202 −W0212W1212 +W0303W1303 −W0313W1313). (10)

♦ Using (a6), the first term W0203W1203 −W0312W1312 can be rewritten asW1213(W1203−W0312) = 0 and the second term W0302W1302−W0213W1213 canalso be rewritten as W1213(W1302 −W0213) = 0.

♦ Succeeding applications of (a2) and (a4) to the third term respectivelyyield W0201W1201 −W0323W1323 = W3032(W0201 −W1323) = 0.

♦ Succeeding applications of (a3) and (a5) to the fourth term respectivelyyield W0301W1301 −W0223W1223 = W1301(W0301 +W1223) = 0.

♦Using (a1), the fifth termW0202W1202−W0212W1212+W0303W1303−W0313W1313

becomes W1202(W0202 −W1212 −W0303 +W1313) which vanishes by (6).

Similarly, for Y23, again from (8), we have

Y23/2 = −W2001W3001 −W2002W3002 −W2003W3003

+ W2012W3012 +W2013W3013 +W2023W3023

+ W2101W3101 +W2102W3102 +W2103W3103

− W2112W3112 −W2113W3113 −W2123W3123. (11)

As in the case of Y01, rearranging the terms in (11), we get

Y23/2 = (−W2001W3001 −W2123W3123) + (W2023W3023 +W2101W3101)

+ W3012(W2012 +W3103) +W2013(W3013 +W2102)

− [W3002(W2002 +W3003)−W3112(W2112 +W3113)]. (12)

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♦ Succeeding applications of (a4) and (a5) to the first term respectively yieldW0102(W3001 +W2123) = 0.

♦ Succeeding applications of (a2) and (a3) to the second term respectivelyyield W3023(W2023 +W3101) = 0.

♦ Using (a6), the fifth term can be rewritten as W0203(−W0202 − W3131 −W0303 −W1212) which vanishes by (6).

We can show that other off-diagonal terms also vanish by similar calcu-lations.

2.3 Step 2: On diagonal components

Using the mixed form of (1), the diagonal components satisfy

W0cdeW0cde = W1cdeW

1cde = W2cdeW2cde = W3cdeW

3cde =1

4WijklW

ijkl, (13)

so that it suffices to show that

Y(i)(i) − Y

(j)(j) = 0, Y

(i)(i) := W(i)cdeW

(i)cde, (14)

where (i) and (j) denote suffixes to which we do not apply Einstein sum-mation convention. In what follows, as a specific example, we show thatY 00 − Y 1

1 = 0. Summations with respect to c and d yield

Y 00 = W00deW

00de +W01deW01de +W02deW

02de +W03deW03de

= W010eW010e +W011eW

011e +W012eW012e +W013eW

013e

+ W020eW020e +W021eW

021e +W022eW022e +W023eW

023e

+ W030eW030e +W031eW

031e +W032eW032e +W033eW

033e. (15)

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For the sake of the simplicity of notation, we introduce a symbol (abcd) :=W(a)(b)(c)(d)W

(a)(b)(c)(d). Using this symbol, (15) turns out to be

Y 00 = [(0100) + (0101) + (0102) + (0103)]

+ [(0110) + (0111) + (0112) + (0113)]

+ [(0120) + (0121) + (0122) + (0123)]

+ [(0130) + (0131) + (0132) + (0133)]

+ [(0200) + (0201) + (0202) + (0203)]

+ [(0210) + (0211) + (0212) + (0213)]

+ [(0220) + (0221) + (0222) + (0223)]

+ [(0230) + (0231) + (0232) + (0233)]

+ [(0300) + (0301) + (0302) + (0303)]

+ [(0310) + (0311) + (0312) + (0313)]

+ [(0320) + (0321) + (0322) + (0323)]

+ [(0330) + (0331) + (0332) + (0333)]

= 2[(0101) + (0102) + (0103) + (0112) + (0131) + (0123)

+ (0201) + (0202) + (0212) + (0231) + (0223) + (0203)

+ (0301) + (0302) + (0303) + (0312) + (0331) + (0323)]. (16)

Similar calculations for Y 11 leads to

Y 11 = 2[(0101) + (0102) + (0103) + (0112) + (0131) + (0123)

+ (1201) + (1202) + (1203) + (1212) + (1231) + (1223)

+ (3101) + (3102) + (3103) + (3112) + (3131) + (3123)]. (17)

Subtracting (17) from (16), we get

(Y 00 − Y 1

1 )/2 (18)

= [(0201) + (0202) + (0212) + (0231) + (0223) + (0203)

+ (0301) + (0302) + (0303) + (0312) + (0331) + (0323)]

− [(1201) + (1202) + (1203) + (1212) + (1231) + (1223)

+ (3101) + (3102) + (3103) + (3112) + (3131) + (3123)]. (19)

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We rearrange the terms in (19) as follows.

(Y 00 − Y 1

1 )/2 (20)

= {[(0212)− (1202)] + [(0231)− (3102)] + [(0312)− (1203)]

+ [(0331)− (3103)]}+ {[(0203)− (1231)] + [(0302)− (3112)]}+ {[(0323)− (1201)] + [(0223)− (3101)] + [(0201)− (3123)]

+ [(0301)− (1223)]}+ {[(0202)− (3131)] + [(0303)− (1212)]}. (21)

Using the last property of Weyl tensor in (2), namely, Wabcd = Wcdab, we seethat the four terms in the first group vanish, namely, (0212) − (1202) = 0,(0231)− (3102) = 0, (0312)− (1203) = 0, (0331)− (3103) = 0. Two terms inthe second group vanish by (a6) and the four terms in the third group vanishrespectively by (a2), (a3), (a4) and (a5). Two terms in the last group vanishby (6). Thus, we have shown

Y 00 − Y 1

1 = 0. (22)

Repeating similar manipulations, we can show that Y 00 = Y 2

2 = Y 33 .

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