# Energy of a photon.

1. Jul 10, 2012

### Super Kirei

I was reading an elementary treatment of the photoelectric effect and it says that the energy of a photon is related to it's wavelength by λ=h/p where p is the momentum. So this would imply that changing the frequency of a light source while keeping the intensity constant means that at higer frequencies fewer more energetic photons are emitted than at lower frequencies so that the energy flux remains constant. But what does it mean for a photon to be more energetic? The speed of light is fixed at c, so how does its momentum vary?

2. Jul 10, 2012

### tiny-tim

Hi Super Kirei!
a photon is a wave as well as a particle

you're probably familiar with the doppler shift … if you move towards a wave, its frequency increases, and it appears more energetic (because more wave-crests are "bumping" into you each second)

if you move towards a photon, its frequency increases, and it appears more energetic

a photon is (usually) produced when an electron changes orbit round an atom or molecule …

the loss of energy of the electron equals the energy of the photon produced (in the frame of reference of the atom or molecule)

all photons are produced at the same speed, but with different energies

3. Jul 10, 2012

### ArtistIC

On a small additional note.. and just to make things a bit more fun, light doesnt have a constat speed. The one you refer to is the speed of light in vacuum. When light passes through matter or has other things affecting it, it changes speed and loses or gains energy like everything else. There have even been experiments which have brought light to a virtual stand still :)

4. Jul 10, 2012

### Super Kirei

I'm still trying to reconcile the billiard ball conception of the photon with the wave conception of a photon. So, if a photon has momentum, as the de Broglie equation seems to imply, then I imagine that would imply it has mass. But if it has mass then it's momentu varies with velocity and it's momentum is related to it's energy so that higher energy means higher momentum which means higher velocity, so that can't happen. But for a wave there's no problem. If I start moving towards a wave it seems to be more energetic because more wave crests are hitting me per second, the wave's fequency relative to me has shifted up because my velocity relative to the wave has increased. But that's not possible for a light wave. So how does the higher frequency translate into higher energy? Or is this one of those brute fact things you just have to accept by virute of experimental evidence, intuition be damned?

5. Jul 10, 2012

### tiny-tim

that's right, the wave's fequency relative to you has shifted up … so why do you not accept that that means greater energy?

6. Jul 10, 2012

### Boston_Guy

The relationship between the photon's energy and its momentum is E = pc or p = E/c. Substitute in λ=h/p to obtain λ= h(1/p) = h(c/E) or λ = hc/E. This gives the expression of the wavenght in terms of its energy.

That depends on what one means by intensity. It has always been my understanding that the intensity of a beam of light refers to the number of photons per second. So an increase in photon frequency does not change the beam intensity.

For a photon to be more energetic means that the photon carries more energy with it. The more energy the photon carries the smaller its wavelength and the larger its frequency.

Photons, which have zero rest mass, have momentum. However there is the concept of inertial mass which refers to the quantity m in p = mv (some people refer to this as relativistic momentum). I know this seems weird but its straight textbook physics, strange as it may be.

The inertial mass of a photon depends, not on velovity, but on energy or momentum. Vary the energy or momentum of a photon and you've changed it's inertial mass.

You can think of a more energetic photon as one which vibrates more. The higher the rate of vibration the greater its energy.

Last edited: Jul 10, 2012
7. Jul 10, 2012

If your intuition tells you that only massive particles can have momentum than yes you have to change your intuition, because (funny thing) only massive particles can have zero momentum. "Massless particle with zero momentum" is nonsense.

8. Jul 10, 2012

### Staff: Mentor

Don't try. Whatever a photon "really is", it is not a tiny billiard ball!

9. Jul 10, 2012

### Super Kirei

I think I get it now. For a billiard ball flying in space, to have greater momentum is to go faster. For a photon, to have greater momentum is to be more massive which is a function of the momentum which is a function of frequency so that a blueshift in frequency will result in a more energetic and hence more massive photon. Is the equation E=pc a fundamental experimental result, or is it derived from a more fundamental relationship?

10. Jul 10, 2012

### Boston_Guy

Quite right! I couldn't agree more. Feynman really hits it on the head when he wrote in his lectures, V-III page 1-1
I think tha pretty much sums up how i feel about it. Some people still refer to it as a particle though. I supoose that because its so difficult to hard to explain it every time. Aurhor Eddington used the term wavicle to descibe things. Not something I'd do but I can see the point.

11. Jul 11, 2012

The term "particle" in my vocabulary always means quantum mechanical particle. Classical Newtonian particles should be called corpuscles. I think clinging to the obsolete definition of particle just creates confusion.

12. Jul 11, 2012

### Boston_Guy

I agree.

13. Jul 11, 2012

### San K

Does this energy remain the same when the possible paths, between emission and final detection, are increased/decreased (i.e. added or blocked)?

14. Jul 11, 2012

### Boston_Guy

Yes.

15. Jul 11, 2012

### phyzguy

This is not correct. Intensity is a well-defined concept in physics; we are not free to re-define it at our whim. Intensity is defined as energy per unit area per unit time, or power per unit area. The SI units of intensity are Watts/meter^2, or Joules/(second*meter^2). So the intensity of a beam of light depends on both the number of photons per unit area per second and on the energy per photon (which is of course proportional to the frequency).

16. Jul 11, 2012

### Boston_Guy

You're correct of course. I now realize the error I made. I was thinking about the photo electric effect and something that must be floating around in the back of my mind. What a goof on my part. Sorry! :(

17. Jul 11, 2012

### Naty1

18. Jul 12, 2012

### Darwin123

The speed of light is fixed. However, the relativistic mass of the photon is not fixed.
The photon has zero rest mass. However, it has a relativistic mass that is given by the equation for equivalence of energy and mass. The relativistic mass varies with the inertial frame of the observer.
The rest mass has a meaning only in the inertial frame of reference where the speed of light is zero. However, there is no physical inertial frame where the speed of light is zero.
The kinetic energy, E, of a photon is the relativistic mass of the photon, M, times the speed of light, c. Therefore,
E=Mc
The rest mass of the photon is zero. Therefore, an observer traveling in the rest frame of the photon would see a photon of zero relativistic mass. However, nothing can move at the speed of light relative to an observer in any inertial frame. Therefore, the rest mass is zero. Momentum is not proportional to the rest mass. It is proportional to the relativistic mass.
It is like one is on the highway with a 65 MPH speed limit, where every vehicle travels at the speed limit. There are three vehicles: a motorcycle, a sedan and a truck. Which has the greatest momentum and why?

19. Jul 12, 2012

### Boston_Guy

The relativistic mass of any particle is defined by its momentum over the velocity. i.e. relativistic mass is defined as the m in p = mv or m = p/v. For a photon E = pc and m = p/c or m = (E/c)/c → m = E/c2. You missed a factor of c.

20. Jul 13, 2012

### Darwin123

Sorry. I meant:
p=Mc
where "p" is the momentum of the photon, "M" is the relativistic mass of the photon and "c" is the speed of light in the vacuum.
The units always have to match on both sides of the equation. Hence, I should have caught my mistake just from the units.
Still, the variation in momentum comes from the relativistic mass, "M". Not the speed of light, "c".