Photons, speed and energy - Beginners concept question

1. Apr 14, 2013

bifman

Hi all - thank you for this forum - it's proving very helpful:

Am I correct in thinking that if we measure a period of time that light travels, that in the case of red vs blue light:

1) Both will travel at the same speed in a vacuum?

2) More blue photons will arrive than red (given the higher frequency and smaller wave length of blue)?

3) Since each blue photon has more energy as well (E=hf), we will see more more energy per photon too?

4) So if we think of photons as buses, and 1 red bus =10 blue buses in length, then we might have 10 blue buses arrive in the same time that 1 red bus arrives, AND each blue bus has more “energy” than the red bus?

5) Finally if #2 is correct, and more blue photons have arrived, would we say more blue light has arrived?

Cheers - I realize the bus analogy probably has more flaws than I realize at this early stage in my studies but please bear with it if possible...

2. Apr 14, 2013

mikeph

I can't tell you how many blue/red photons arrive because you haven't given any information about the relative intensities of light. If red and blue light of the same intensity (W/m^2) arrive, then there will be more red than blue photons: red photons carry less energy, so we need more of them to arrive per second to produce the same power.

It seems like you're equating a wavelength peak to a photon to deduce that red light has fewer photons per unit length than blue. This isn't the case. Think of each photon as a packet of light whose frequency (and therefore energy, hf) is an intrinsic property of the packet. Separation in your head of the continuous wave idea of light into individual photons is tricky, but if you want to talk about photons that's what you need to do.

3. Apr 14, 2013

bifman

Thanks MikeyW,

1) Light is made up of photons with each photon representing an electromagnetic wave.

2) A photon is basically a “packet” of energy moving through the EM fields, and the amount of energy it carries is directly related to the frequency.

So because of #1, I pictured each EM wave length as equaling 1 photon, and so since red light would (all things being equal) have longer wave lengths and thus a lower frequency, it would (at least in my mind) have less photons.

Thinking more on what you said - and doing some more reading, it seems to me that #1 is wrong and I would be better to think of each photon as being a discrete point on the wave. So conceptually, all the photons are the same size — regardless of the color (frequency) of light — but some photons have more energy (due to higher frequency) than others. Does that seem a better way to think on it?

So if this is correct, referring to my previous question — would this then be correct?
1) Yes
2) No – the same amount of photons arrive (Let me clarity that I am referring in my originaly question to one wave from red light and one wave from blue light.)
3) Yes
4) No – red light does not have bigger buses (photons), the buses are the same size, but the red buses (photons) carry less energy.

And assuming I am correct, looking at #2, intensity would definitely be different because we are talking about the same # of photons with different energy per packet?

4. Apr 14, 2013

bifman

New Thought:

It just occurred to me that perhaps I should be thinking of the entire wave as one photon? This would still leave my other answers correct, I believe...

5. Apr 14, 2013

vanhees71

You should study classical electromagnetism first, then non-relativistic quantum theory and then relativistic quantum field theory. Otherwise you get wrong (and not only quantitatively but qualitatively wrong!) pictures about nature, and physics is a about the contrary to get as precise a picture about nature as possible.

Photons are not little particles in the intuitive sense of everyday experience. To the contrary, as massless particles with spin 1 they are as far from such a picture as one can think.

Light, as we know it from everyday life, is best described by classical electromagnetic waves. Quantum mechanically these are described as socalled coherent states, which do not even have a determined number of photons, but are a superposition of Fock states with an arbitrary number of photons.