1. Sep 28, 2014

### alexay95

i don't get it.
why would a body with temperature give up photons?

2. Sep 28, 2014

### phinds

Why would it not? Take an extreme case, a red hot poker. Where do you think the "red" comes from?

3. Sep 29, 2014

### Khashishi

Roughly speaking, hot objects have a lot of energy per microscopic degree of freedom. Thermal energy is energy that is distributed more or less randomly across all degrees of freedom that are in close contact with each other. This energy is constantly being transferred from one mode (such as vibrations) to another (such as translational motion) through internal collisions or interactions. Assuming the body is made up of molecules, the molecules dance around unpredictably. Since molecules are made up of charged particles, some of these random motions will end up moving charges and creating light energy. You can think of the electromagnetic field as another degree of freedom for energy to enter. But in this case, the electromagnetic field is typically not in thermal equilibrium with a big solid object because the coupling is fairly weak, and photons fly off everywhere and aren't confined well, so the object is constantly leaking energy to the surroundings. If you surround an object with other objects and some kind of ideal mirrors, then eventually the electromagnetic field will reach the same temperature as the objects and the amount of photons emitted and absorbed will be equal.

Thermal energy is governed by thermodynamics, which tells us that energy on average moves from hot object to cold objects (unless work is done to pump it the other way). This is because energy is moving randomly at small scales and get mixed up between the hot and cold objects. A hot object can transfer heat to a cold object in many ways. One of these ways is by sending photons to the cold one, and this can happen over a distance.

4. Sep 29, 2014

### Staff: Mentor

In simple terms, thermal energy is the motion of atoms, molecule, and subatomic particles. Since these are made up of electrically charged particles, and electrically charged particles create EM radiation when accelerated, all this jiggling and vibrating means that the charged particles are being accelerated and creating EM radiation which is emitted from the object.

5. Sep 29, 2014

### alexay95

"Since molecules are made up of charged particles, some of these random motions will end up moving charges and creating light energy."

"Since these are made up of electrically charged particles, and electrically charged particles create EM radiation when accelerated, all this jiggling and vibrating means that the charged particles are being accelerated and creating EM radiation which is emitted from the object."

i think that too,but,because of the random factor in the process every photon must be in another phase (wave),yes each photon alone will give light but if we sum it all together i would expect 0 EMR,i would expect that it all cancel out.

6. Sep 29, 2014

### jbriggs444

If you throw two rocks in a pond, their ripples cancel out in some spots and add together in others. The net effect is that the two ripples pass through each other intact. The same sort of thing occurs with electromagnetic waves.

Also consider that if a collection of electromagnetic waves could interfere with one another to a net of zero, that would fail to conserve energy.

7. Sep 29, 2014

### alexay95

if u assume that EMR has to be then yes "that would fail to conserve energy".
take your example and think u throw infinite or some large number of rocks,it will cancel out because for every wave u can find a wave with anti phase with him when u add this two together u get 0.

8. Sep 29, 2014

### Staff: Mentor

Not true. There is no configuration of stones that you could throw in that would result in zero waves being created. The waves will simply not add up to zero at every point. The same thing happens with EMR. The phase of the waves will not add up to zero at every point in space.

9. Sep 29, 2014

### jbriggs444

If you pick a particular place in the pool and look at the instantaneous wave amplitude there you will find that, statistically speaking, it will vary as the square root of the number of stones cast into the pool. That phenomenon has its roots in probability and statistics. The standard deviation of a sum of a set of identical and independently distributed random variables has a standard deviation that goes as the square root of the number of them you are adding together.

Wave energy goes as the square of the amplitude. So on average the wave energy at a given point will be directly proportional to the number of stones cast into the pool, just as it needs to be in order to conserve energy.

10. Sep 29, 2014

### alexay95

if i have pool and i throw a rock,and a moment after i throw in the same way another rock (moment after=period time/2),if i look at the region where both of the waves exist i will get 0 amplitude.
no?

11. Sep 29, 2014

### TumblingDice

Hello @alexay95 - it would help readers if your posts used proper spelling, punctuation, and did not contain messaging shorthand. I've included an excerpt from the PF Global Guidelines below. Thank you!

12. Sep 29, 2014

### jbriggs444

This is pushing the boundaries of the rock pool analogy and destroying the relationship to black body radiation. But I will play along.

The second rock wil not cancel the initial ripple. There is still a wave form propagating across the pool. It may be only half a cycle of a (presumed) sinusoidal wave form. But it is still a wave form. Its propagation is not greatly affected by what comes behind.

The second rock will not, in general, land 1/2 cycle behind the initial rock. On average it can land anywhere within the cycle (assuming for the moment that there is actually a periodic cycle to match). The second rock will not, in general, land exactly where the previous rock landed.

Yes, if you are able to carefully contrive disturbances at the right places and times then you can cancel out (portions of) the pond ripples already in motion. This is basically what noise cancelling headphones do. Note that it is a much harder trick to do in an open room.

With a black body there is no intelligent action. The new disturbances are unrelated to the existing disturbances.

13. Sep 29, 2014

### alexay95

1.of course that the first wave will not be affected,this is why i said "if i look at the region where both of the waves exist"
2.this pool is only thought experiment so i can assume that the rock land at the same place and exactly after T/2 sec.
3.so back to EMR,every photon(rock) will give a wave and i can find another photon that is anti phase (because i got large number of photons),we can say its photon has random phase so i can match those pairs and get a 0 EMR.

now i know it is not true because there is EMR,so my question is how?

14. Sep 29, 2014

### jbriggs444

Let's back up and make your example more specific.

Instead of rocks, we have sinusoidal wave generators. We need to do this in order to make your "T/2" specification sensible. Rocks do not have periods. Wave generators do.

What happens if we put two wave generators in the same place, 180 degrees out of phase? We get zero output, just as you have predicted. What about conservation of energy? Because the generators are out of phase, neither one requires or produces any net power. This situation has little relationship with a black body.

What happens if we put two wave generators very close to one another (within a fraction of a wavelength)? What happens if we put two wave generators far apart? What happens if the generators are in phase instead of out of phase? What happens if the phase relationship is random? What happens if we use generators that do not share a common frequency?

15. Sep 29, 2014

### Khashishi

Alexay, your condition for the rocks is very special (moment after=period time/2). If you throw rocks in randomly, the waves will not cancel, with high probability. Thermodynamics arises out of probability of the microscopic states. For any macroscopic system, the probability of heat flowing from a hotter object to a colder object is almost one. Since we are taking averages of many microscopic states, the randomness disappears and we can state unequivocally that heat travels from hot to cold, and we can make quantitative statements about how fast the heat transfers.

For an object which is hotter than background electromagnetic waves, the probability of the object transferring energy to the waves is almost one. We can make a quantitative statement about the rate of energy transfer: the Stefan-Boltzmann law.

The background electromagnetic waves isn't generally in equilibrium, because photons don't interact with each other normally (You need lots of interactions to thorougly mix up all of the energy modes). But in deep space, there is a temperature of the cosmic microwave background because photons were trapped by plasma in the early universe, and the CMB temperature is very cold now. For objects colder than the CMB, we can expect these objects to absorb EMR from surroundings. But most things that we care about are much hotter than surrounding space, and therefore give off energy.

16. Sep 29, 2014

### Khashishi

If you shake a salt shaker, why does salt leave the shaker? Why doesn't salt enter the shaker? Well if there was a higher concentration of salt outside the shaker, it would indeed enter the shaker.