WCP2883

[1]¹

10 Ampton Road

Edgbaston

Birmingham

Sep^r 17 1907

Dear Sir,

I have been studying Prof. Lowell’s² paper and the more I consider it the less trustworthy do his results appear. I think he has made an advance in the subject in trying to bring the atmosphere into account though from the footnote on p 171 Arrhenius,³ I gather, has attempted to do the ^same & probably in a more complete way. I have not access to Arrhenius’ paper & cannot make out how he got the formula. As interpreted by Lowell it appears to me unsatisfactory.

Lowell’s first method neglects the "blanketting" or as I prefer to call it the "greenhouse" effect of the atmosphere. If the atmosphere reflects back some part of the sun’s radiation & absorbs another part it will of course reflect back some part of the planet’s radiation & absorb another part. The absorbed portion of both radiations is sent partly into space partly back to the planet & the net result may be that the temperature of the surface is not very different from that which would exist without an atmosphere as I will try to show below.

Lowell’s second method p 170 I cannot understand at all.

[2]⁴

A third method in the footnote p 171 uses Arrhenius’ formula & arrives at the result that the radiation from the surface is proportional to the solar radiation getting through to it. I feel sure that this is wrong for the "greenhouse" effect has entirely disappeared from the result.

As to details I distrust the estimate of reflexion of visible rays | our atmosphere — 74% It appears to me that if this were so large the reflexion from the sky opposite to the sun ought to be vastly greater than it is. The ^{sky there} should be at least as bright as white paper or white cardboard which reflects about 70% & it is at least 3 or 4 times less bright even in a hazy & therefore ^{more copiously} reflecting sky.

Then again I cannot see how the effect of cloud can be anything like that estimated. According to Lowell the radiation under cloud — if I follow his argument should on the average be 1/5 that under clear sky. (In the middle of p 169.20 x.50 x.50 should be.20 x.50 +.50) But we know that it differs only very little. It is a matter of a few degrees. The average of day & night surface temperature under a clear sky may be taken as very nearly the same as that under a cloudy sky. The flaw in his argument is due to neglect of the "greenhouse" effect which [3]⁵ may be set forth thus.

Consider the case of a greenhouse with horizontal roof of great extent over a black soil & illuminated by a vertical sun. Let S be the solar radiation falling per sq cm per sec on the glass.

Let r S be reflected

. a S. absorbed} by the glass

t S. transmitted

where r + a + t = 1 since the whole is divided among the three

Let R be the radiation from the black soil per sq. cm. per sec. Consider this radiation striking the glass

Letr₁ R be reflected

a₁ R. absorbed} by the glass

t₁ R. transmitted

where r₁ + a₁ + t₁ = 1

r₁ a₁ t₁ will be different from r a t since the comparatively low temperature soil radiation is of quite different quality from the sun radiation. The glass absorbs a S + a₁ R & gives half of this upwards into space & half downwards to the ground so that the ground receives a S + a₁ R

2

If we equate expenses & receipts on the [1 word illegible] of the ground we have

(transmitted by glass) (rad[iatio]n fr[o]m glass) (reflexion | glass) (rad[iatio]n fr[o]m soil)

t S + a S + a, R + r, R = R

2

∴(t + a ) S = R — r₁ R — a ₁ R

2 2

[4]⁶

= (t₁ + ₁ ) R since 1 — r₁ — ₁ = a₁ + r₁ + t₁ — ₁

2 2 2

= t₁ + ₁

2

Whence R = t + a/2 / t₁ + a₁/2 S

If the glass were more existent we should have

R = S

so that if the ground temperature absolute under glass given by is Θg + unthin glass is Θ

Θg/Θ = 4√t +a/2 / t₁ + a₁/2

With sunlight t is perhaps.6 a perhaps.3 We could only get exact values | experiment on the glass used.

But we know that with radiation fr[o]m a surface below 100°C t₁ = 0 & a₁ = 1 that is the glass is quite opaque even when thin so that

Θg/Θ = 4√.75/.5 = 4√1.5 = 1.1

If Θ = 300° A Θg = 330° or the soil is 30°hotter than in the open.

If the sky is cloudy much of the radiation is from cold cloud & t will be very small & a more nearly 1 & so Θg & Θ are nearly more nearly equal — as of course experience tells us[.]

[5]⁷

I think this formula may be applied to the atmosphere though the assumption that half the radiation from it goes each way is no doubt inexact. But it is perhaps as near the truth as any we can make. ^{&r omits to take account of the fact that} the planetary surface reflects some of the surface radiation falling on it & if when we take this into account the formula must be modified. Taking this into account I find

R = S t(h — p) + ½ a(b + pt) + r₁pt / t₁ + a₁/2

Where p is the albedo of the ground of the fraction which it reflects of when The atmosphere is supposed to reflect the same fraction of direct solar radiation & ground reflected radiation.

Now p is about.1 so that if we omit it we do not make an error of more than a few per cent. in what we do not know [1 word illegible] many per cent.

Let us examine the results when |

R = S t + a/2 / t₁ + a₁/2

for the earth, using Lowell’s figures[.]

For the cloudless half (p 169) t =.4 a =.5 of.65 =.325 (for a = 0 for usable rays constituting.32 of the whole) (This ^{a = 0} is very doubtful)

[6]⁸

Then t + a/2 =.563

Since R is low temperature radiation we take with Lowell t₁ =.5 a₁ =.5 though perhaps t₁ should be somewhat larger ^{or smaller we don’t know which} for very long wave length.

Then t₁ + a/2 =.75

Then if Θg be the ground temperature & Θ the temperature with no atmosphere

Θg = Θ 4√.75 =.93Θ

I have not considered the effect of rotation.u of day & night[.] Probably the result would be much the same.

For cloud we can only guess the numerical values. I am inclined to think the best value is obtained from the observation that the surface temperature under a cloudy sky is not very different from the average of day & night under a clear sky.

If we attempt to use Lowell’s figures t =.1 a is at least.5 of.65 =.325 & a/2 =.163

say then t + a/2 =.425.26

t₁ =.2 a₁ is not more than.1 for r₁ is.7

∴ t₁ + a₁/2 =.25

So that Θg is very nearly equal to Θ Take it for simplicity that Θg =.93Θ

[7]⁹

Now turn to Mars. From p 169

t =.64

a =.4 of.65 =.26

∴ t + a/2 =.77

Since the returned radiation is low temperature

t₁ =.6

a₁ =.4

& t₁ + a₁/2 =.8

Whence Θg = Θ 4√77/80 = =.99Θ

Hence Mars should be about 6% hotter than the earth at the same distance. Taking the Earth as 290° A Mars would be 235° A under like conditions and adding 6% it should be about 250° A = -9F. = -41F below freezing

But we may imagine that Mars has some peculiar covering e.g. let us suppose that the inhabitants have covered the surface with greenhouse glass. The value of t + a/2 may be unaltered if they have invented perfectly transmitting glass i.e. glass with

unit refractive index for solar radiation. Let us [8]¹⁰suppose the glass quite opaque for low temperature radiation then ^{t1 =0 a1 = 1 and t1 a1/2 =}.^{5. Thus} R = ₁.77/.50 = 1.54 S and Θg= 1.11Θ

Hence the temperature would be 1.11/.93 of 235 = 280° A or 7° C

It appears just possible that the atmosphere of Mars should be is peculiar in preventing the return outwards of its low temperature radiation & that is the only

way in which I can at all understand such a high temperature as Lowell demands.

But the kind of atmosphere is yet to be discovered.

I have not worked out very carefully the effective temperature of the Sun on the basis that Lowell’s reasoning is sound but I think it comes to about 8000° A quite out of accord with all recent work. The temperature of Mercury would then come out as something prodigious.

If you care to quote any of this please do so. It of course does not profess to be anything on the positive side, but at least it serves I think to show that Lowell’s estimate requires reconsideration.

Yours faithfully | J H Poynting [signature]