individual photons do not have a temperature EM radiation can be assigned a temperature. The EM radiation emitted by an object has a spectrum that depends on its temperature through Planck's law.

The temperature of photon and its energy

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Do photons have temperature? If not, does it mean that photon lose energy while travelling through space? As the planets farther away from the sun are comparatively cooler than the one that are closer to it, does it imply that photon also lose energy? thermodynamics energy photons temperature thermal-radiation share|cite|improve this question edited Nov 8 at 13:54 Qmechanic♦ 51.3k766170 asked Mar 30 '13 at 2:48 idiosincrasia23 58831226

I could have sworn we have a existing question on the smallest system to which thermodynamics could be reasonably applied, which I think is very much related, but I can't find it. – dmckee♦ Mar 30 '13 at 4:38 add a comment |

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This is really just an extension to JKL's answer since I wanted to pick up on his point about the microwave background, but first it's worth mentioning that although individual photons do not have a temperature EM radiation can be assigned a temperature. The EM radiation emitted by an object has a spectrum that depends on its temperature through Planck's law. So if you measure the spectrum of radiation it is sometimes possible to assign it a temperature through Planck's law, and indeed this is how the cosmic microwave background is assigned the temperature of 2.7 degrees. But back to the CMB: I would guess your question is asking if an individual photon can lose energy by radiating away like a cooling object, and the answer is no. However light can cool if the spacetime through which it is travelling is expanding. The light cools because it's energy is spread out over a larger volume of space. This is how the cosmic microwave background has cooled from it's original very high temperature of about 3,000K to its current value of 2.7K. share|cite|improve this answer answered Mar 30 '13 at 7:13 John Rennie 159k16271448

Indeed, the maximum in the spectrum connects the radiation wavelength, λ max   at the maximum, and temperature of the hot object as λ max T=2.9×10 −3   mK (Wien's Displacement Law). – JKL Mar 30 '13 at 20:33 add a comment |

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The photons themselves do not have temperature as such. However, photons do contribute to the temperature of objects since they carry energy. A very good example is the microwave background radiation which is known to contribute a temperature to the universe at about 3K. One can work out the frequency of these photons using the basic relation k B T mwb =hf  where k B =1.381×10 −23   JK −1   and h=6.63×10 −34   Js, so that the requency turns out to be in the microwave part of the electromagnetic spectrum. Photons contribute to the temperature of your body when you sit in the sunshine and absorb the sunlight. The farther you go from the sun the cooler, correct, but this is because the intensity of solar light is inversely proportional to the square of the distance from the sun. On Earth we receive about 1350 W/m 2   of solar power. But on Mars, which is about 1.52 the Earth-Sun distance, it is only about 584 W/m 2   . share|cite|improve this answer answered Mar 30 '13 at 3:32 JKL 2,330413

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Assume the sun emits a certain number of photons, such that, at 1m from the sun's surface 1 million photons go through each square meter. As the photons spread radially out from the sun, their number stays the same, but they have to cover larger and larger areas. At 10m from the sun, those 10 6   photons will cover an area of 10  x 10=100m 2   . So the density of the photons will be 100 times smaller than at 1 m. This shows how, as you travel away from the sun, the density of photons decreases inversely proportional to the square of the distance from the sun. That is what causes the temperatures of the planets to reduce with their distance from the sun. share|cite|improve this answer answered Mar 30 '13 at 10:05 hdhondt 3,1841411

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At thermal equilibrium temperature T  is defined by 1T =∂S∂U   Where S  is the entropy and U  is the total energy. Since a system composed of photons has a well-defined energy and entropy, photons can be said to have a particular temperature if the whole ensemble could be in thermal equilibrium with some hypothetical environment. Planets further from the sun are cooler than close ones because the same photon flux is spread out over a greater area compared to the size of the planet: Earth's cross sectional area is a greater fraction of the surface area of a sphere the size of it's orbit than Neptune's for it's orbit. This means fewer photons per unit area hit Neptune than Earth. Photons do lose some energy going out that far (redshift from the sun's gravity, this is from general relativity), but this effect is tiny. share|cite|improve this answer edited Oct 21 at 23:53 answered Mar 30 '13 at 7:27 Dan 3,88321841

Can't a cavity in thermal equilibrium be looked upon as a gas of photons at some particular temperature? – Charuhas Mar 30 '13 at 12:59 @Charuhas: Exactly! – Dan Mar 30 '13 at 16:53 add a comment |

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Well it is not the total energy alone which determines temperature but the "quality" of energy as well. Temperature is a measure of average kinetic energy or also related to the wavelength of the photon. If you have enormous quantity of microwave range of photons the temperature anywhere may not rise above 3K. Classically and close to reality it is Weins displacement law that determines temperature in a spectrum. 熱學/流體 標題:氣體動力論與動能 1:timeismoney(大學理工科系)張貼:2010-09-20 16:57:53: 1.請問,平均動能與動能的差別是在?他們的單位分別是? 因為我看了書, (L*平均動能)=每莫耳氣體分子的總動能, 其中L為亞佛加厥常數.. 那亞佛加厥常數是=6*10^23(既然是常數,所以沒有單位?) 2.其實我是看了氣體動力論那一章,覺得有些地方有點疑或混亂 PV=1/3Nmu^2...u是指平均速度嗎?那他如何是從三個方向(XYZ)衍伸推過來的? 這個u跟動能或是平均動能裡面的u是哪個u....拜託各位了 2:黃福坤(研究所)張貼:2010-09-20 18:11:41: [回應上一篇] 1. 每個人的平均收入是2萬元 則 10個人的總收入是???亞佛加厥常數 代表一莫耳氣體有多少原子數 2. 建議你 仔細看課本 氣體動力論推導過程 應該會提到 氣體原子的動能 然後看 如何取平均 等部分 3:timeismoney(大學理工科系)張貼:2010-09-21 11:51:34: [回應上一篇] 老師您好,首先學生要說的是,無意間發現這個地方...不嫌晚不嫌晚!!!! 以往學習遇到瓶頸問題時,總是囫圇吞棗,或是導致學習中斷,其實我並不想如此, 真的很感謝您這麼快就答覆,這真是一個令欣喜雀躍的好網站! 我昨天再回去翻的時候,有比較瞭解了...推導過程中 學生還是有一點疑問...上面寫著: 先討論x軸向也就是: (Px)*V=Nm(Ux)^2------A式 同理應該可以推得Py)*V=Nm(Uy)^2 且 (Pz)*V=Nm(Uz)^2 那是要把三個加起來嗎?才可以推到----->PV=1/3Nm(U)^2 ?? 我看課本是把(U)^2=(Ux)^2+(Uy)^2+(Uz)^2 然後將(Ux)^2=(Uy)^2=(Uz)^2=1/3(U)^2 代到A式 就得: PV=1/3Nm(U)^2 ,此時的P是總壓嗎? 那Y,Z軸向推的東西可以用來作什麼呢? 此時的P是指總壓嗎? 4:黃福坤(研究所)張貼:2010-09-21 12:11:39: [回應上一篇] 　因為假設系統達到平衡狀態 因此 Px=Py=Pz=P 也就是系統的壓力 應該都相同 但是事先推得的是 壓力和個別速度的關係 Px V= Nm Vx2因此 將其換成 總動能 是否就出現 1/3 呢? 5:timeismoney(大學理工科系)張貼:2010-09-21 13:26:51: [回應上一篇] 課本寫著: 因為(U)^2=(Ux)^2+(Uy)^2+(Uz)^2 所以(Ux)^2=(Uy)^2=(Uz)^2=1/3*(U)^2 然後將(Ux)^2=1/3*(U)^2 代入才得PV=1/3Nm*(U)^2  ,其中U為平均速度 那老師說的從"總動能"來想...如果要跟總動能扯上關西, 則一定要使用平均速度?要怎麼想... 從(Px)*V=Nm(Ux)^2到 PV=1/3Nm(U)^2  這一步,我就用最上面那兩個關西式,用代數的方法代換就得到......跟總動能的關西是...麻煩老師了 [ 這篇文章被編輯過: timeismoney 在 2010-09-21 13:33:56 ] [ 這篇文章被編輯過: timeismoney 在 2010-09-21 13:34:13 ] 6:黃福坤(研究所)張貼:2010-09-21 15:52:12: [回應上一篇] U對應平均速度 則其動能是否就對應平均動能 再乘以 粒子總數 是否就對應 總動能 好比 全班的平均成績 乘以全班的人數 結果是否是全班的分數總和! 7:timeismoney(大學理工科系)張貼:2010-09-21 17:00:31: [回應上一篇] 其實我的問題就是,我常常看到兩個式子 一個是動能上面有一槓的,那就是所謂的分子平均動能,就是說每個小粒子所含的動能 那總動能很顯然就是等於  把這個平均動能的式子再乘個總粒子數就可以 那另一個常看到的式子 E上面沒有一槓,它寫成: K=1/2m*u^2  因為如果它上面沒有一槓,那就是指總動能,那其對應的速度該是什麼?不可能是平均了阿... 所以是不是不應該有"上面枚一槓"的這個公式?麻煩老師了 8:黃福坤(研究所)張貼:2010-09-21 17:22:49: [回應上一篇] 是動能的定義 若代表平均動能　通常會上方加橫線 沒加時　可能代表某一個物體或粒子的動能 9:timeismoney(大學理工科系)張貼:2010-09-22 16:23:08: [回應第4篇] Quote: 在 2010-09-21 12:11:39, 黃福坤 寫了: 　因為假設系統達到平衡狀態 因此 Px=Py=Pz=P 也就是系統的壓力 應該都相同 但是事先推得的是 壓力和個別速度的關係 Px* V= Nm*(Ux)^2 為什麼要從個別速度下手? 是因為任意的粒子有其自己的速度(大小方向都不一樣), 但是欲推出壓力關係,必須是"垂直力"除以"面積" 所以才要討論某一粒子其三個分量(xyz軸上), 進而推出壓力和個別速度的關係 Px *V= Nm(Ux)^2這個式子                          請問我這樣想對嗎???是因為壓力所以要從個別速度下手???? 10:黃福坤(研究所)張貼:2010-09-22 16:34:54: [回應上一篇] 　因為氣體的　壓力　是很多粒子的平均結果 建議你估算一下　一大氣壓室溫下　一立方公分內約有多少氣體分子？ 因此用氣體分子的模型　然後透過平均來估計平均的效應 好比為何要定義一個國家的平均國民所得　做為比較的基礎！ 或學校會以每班的平均表現　來比較各班級等 11:timeismoney(大學理工科系)張貼:2010-09-22 20:17:48: [回應上一篇] 101325Pa,298K,氣體分子量=29g/mole 使用PV=nRT,我算出來n=4.08*10^(-5) mole 4.08*10^(-5)*6.02*10^23=2.46*10^19個氣體分子... 平均速度u=(3RT/M)^(0.5)=507m/s 氣體動力論: PV=2/3n*E 其中E=總動能=3/2RT... 老師, 你要我做什麼呢....可不可以再給我一個提示 12:黃福坤(研究所)張貼:2010-09-22 22:13:47: [回應上一篇] 　你己經算出一立方公分會有1019個粒子 想像一下 全世界人口才僅有 6*109 人 氣體的壓力是非常多氣體分子碰撞器壁的平均結果 因此才會用 這些平均的量 來詮釋 氣體方程式! 或者說 壓力 P, 溫度 T等本身 就是代表很多粒子的平均結果 13:timeismoney(大學理工科系)張貼:2010-09-23 00:01:48: [回應上一篇] 恩...意思我為了要推算出總壓力,便從每一顆個別的小粒子下手,但是太多粒子數...於是我利用"平均"的概念來估算它 . 那還有個問題就是這個式子: Px*V=Nm*(Ux)^2 其中,Ux=粒子速度在x軸的所有分量的平均速度 為什麼要從分量來求? 如果直接寫出   P*V=NmU^2  其中,定義U=每一個小粒子自己的速度加總在取平均(亦即是平均速度) , 這樣不可以??麻煩老師了 14:黃福坤(研究所)張貼:2010-09-23 13:26:00: [回應上一篇] 　是否記得課本的推導是以粒子 在 x軸上兩端的平面(y-z面)上 來回碰撞的方式計算出 壓力與速度的關係 其以上面上的壓力 是否僅與 x方向速度有關! 15:timeismoney(大學理工科系)張貼:2010-09-23 14:24:57: [回應上一篇] 老師您的意思是不是,我把每一個任意粒子自己的速度u全部投影在x軸成(Ux)分量,使得(Ux)對y-z面來講是一個垂直速度 那麼待會ㄦ要使用 F=dm(Ux)/dt  ,對於YZ面來講就是垂直力,因為壓力要用垂直力來除以面積?這樣想對嗎? 16:黃福坤(研究所)張貼:2010-09-23 17:47:45: [回應上一篇] 　若要計算y-z面所受壓力 僅有x方向速度分量才會有動量變化 y-z面沒有 y,z 的動量變化 因此僅算x方向的力--> 壓力