# Why does the photon have no mass?

1. Jun 4, 2015

### Lone Voice

I have read some of the stuff on this website but am still puzzled. I understand that E=mv^2 refers to the rest energy of the photon which presumably would be nil if the mass is nil. However, we never find photons in this state as they travel at the speed of light and at this speed they do have energy, presumably kinetic energy. We also know they have energy as they can be used in lasers. If this is kinetic energy, i.e. 1/2 mv^2, then again it is suggesting that the photon has mass, albeit a miniscule mass but mass never the less.So the question is, does the photon have mass or not? If not, where does the energy come from and what is the equation that relates energy to a mass-less state?

2. Jun 4, 2015

### Staff: Mentor

A photon has no rest mass, but it does have momentum. The general relationship between energy, momentum, and rest mass is given by $E^2=(mc^2)^2+(pc)^2$ where $m$ is the rest mass and $p$ is the momentum.

Two important special cases are: $p=0$ for something with mass that is not moving so $p=0$ and this reduces to Einstein's famous $E=mc^2$; and $m=0$ for a massless particle.

The classical expression $E_k=mv^2/2$ is an approximation that only works for massive bodies moving slowly relative to the speed of light... but it's a very good approximation, so good that still use it today for most problems.

3. Jun 9, 2015

### Michael Birdman

OK, I am bit slow on this, how can a photon have no mass but have momentum?

The energy of a 500nm photon is E=hλ so 3.984 x 10-19 J

I know we should not equate energy back to mass but if we use E=MC2 we get a "mass" value of 4.43 x 10-36 kg

If we use the formula for momentum of h/λ we get a momentum of 1.328 x 10-27 kg m s-1

If we divide this momentum number by c we get the same "mass" value as before. So what is this value for mass we are getting? If mass is zero then momentum would be zero? This is the bit I am stuck on.

So is the explanation this...?

h/λ gives the momentum of a photon but then dividing by c to obtain "mass" is just invalid.

The energy of a photon can not simply be turned into mass of the photon

If this is true then we can use the momentum of a photon to calculate the pressure of sunshine on the ground but force is rate of change of momentum, what figure do I use for the time for a photon to stop?
Where does the figure of "the weight of 6 poppy seeds per square metre" for sunshine normal to the Earth's surface come from?

Thanks for any help

4. Jun 9, 2015

### Yosty22

According to Einstein's relativity, any object with mass cannot move at the speed of light. Since a photon moves at the speed of light, its mass is zero, therefore the equation E2 = (pc)2+(mc2)2 reduces to E=pc where p is the momentum of the photon.

In the textbook "Modern Physics" by Serway/Moses/Moyer, it states that: "This equation is an exact expression relating energy and momentum for photons, which always travel at the speed of light."

That is, since photons have no mass, E2 = (pc)2+(mc2)2 is inherently reduced to simply E=pc. By adding the other term as you did above and algebraically solving for some value of m is incorrect simply because anything with mass cannot be moving at the speed of light. Therefore, any massless particle is exactly described by E=pc and not E2 = (pc)2+(mc2)2.

Edit: I suppose it is still described by E2 = (pc)2+(mc2)2, but the second term, (mc2)2 goes away.

5. Jun 9, 2015

### ZapperZ

Staff Emeritus
The problem here is that you are trying to equate two different pictures of "momentum". Please note that just because p=mv or p=γmv is something that we are familiar with, it doesn't mean that that is the only way to obtain momentum. We define p=ħk as another valid expression of momentum, and this is true for light and many others (including phonons).

You might also want to read this entire page on why using the relativistic momentum for light just doesn't work:

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html

Zz.

6. Jun 9, 2015

### Michael Birdman

Thanks for your help, so here is a typical student/teacher exchange.

Student: Does a a photon have a mass?

Teacher: No it is a mass-less relativistic object.

Student. But if I calculate the energy of a photon using E=hf for say, a 500nm photon I get a figure of 3.984 x 10-19 J?

Teacher: Yes

Student: And if I use E=MC2 I can calculate the mass to be 4.43 x 10-36 kg?

Teacher, No, you cannot apply that formula to photons, it is just not applicable for two important reasons. 1. Things that travel at the speed of light must have zero mass 2. You just cannot use that equation for relativistic objects.

Student. OK, so the mass is zero so momentum must also be zero because momentum = mv

Teacher, we use the formula mv for non-relativistic objects, we cannot apply this formula because 1. m is zero and 2. It is the wrong formula for photons.

Student. So we use that formula we learnt for DeBroglie wavelength and because we know the wavelength we can work backwards and get momentum?

Teacher, Yes, precisely, ρ = h / λ

Student. So for a 500nm photon ρ = 6.64 x 10-34 / 500 x 10-9 = 1.328 x 10-27 kg m s-1

Teacher. Yes, now for a photon the energy is ρ x c = 3.984 x 10-19 J the same number that you got earlier, but this is no surprise as we just playing around with different combinations of symbols.

Student. So I cannot divide my figure for momentum by c to obtain mass, because if I did I would get exactly the same mass as earlier?

Teacher. That is correct, that figure is simply not valid and has no meaning, the mass of a photon is zero.

Is this exchange valid?

7. Jun 9, 2015

### ZapperZ

Staff Emeritus
To a certain extent, yes. However, if I were the Teacher, I would add this:

"Just because you can manipulate an equation and calculate a number, it doesn't mean that it always has a valid physical meaning. For example, when we solve for the E-field for a point charge located at a distance away from an infinite conducting plane using method of images, we can also get, from the mathematics, the E-field inside the conductor. However, this is an unphysical solution, because we know that in an electrostatic situation, E-field inside a conductor is zero. Yet, if we blindly obey the mathematics, we have such a solution. This is why the mathematics needs to be used according to the physics."

Zz.

8. Jun 9, 2015

### Michael Birdman

Dear Zz

Thank you for your comments, I am new to this forum but something is already quite clear, there seems to be somewhat of a gap between the level of understanding of the main contributors and of those asking questions. In your reply you said..
Quote For example, when we solve for the E-field for a point charge located at a distance away from an infinite conducting plane using method of images, we can also get, from the mathematics, the E-field inside the conductor. However, this is an unphysical solution, because we know that in an electrostatic situation, E-field inside a conductor is zero.
UNQUOTE
I do not have a degree in Physics, but I do teach A level Physics to students, get good results and would hopefully get an A* in any Physics A level I sat tomorrow, but I have no understanding of anything you said because this is way, way above what an 18 year old needs to know to get the highest grade in A level. The vast majority of A level students do not achieve A or A* at A level. I do think that answers need to be pitched so a bright 18 year old student can appreciate the answer, if you need to be in 3rd or 4th year of a Physics undergrad course to even understand the words then it is not surprising that so many students are put off Physics. This is particularly true in Math. In the UK integration is a major part of the second year of A level but most students find it very hard, I took an A level in Math just two years ago, I struggled with integration and trigonometric identities and missed an A grade, only getting a B. However I am very good with basic math and Physics and most people I meet think I am a skilled practitioner of Physics and Math. My point is that people who are very good at Physics are exceptionally highly skilled in Math and can often not see that mere mortals just do not get a complex formula. Please be gentle with us!
Thanks

9. Jun 10, 2015

### lightarrow

That is true even for a (much simpler) classical light pulse. No need of photons and relativistic quantum mechanics, just classical electrodynamics.
Have you ever heard of, e.g., solar sail?
http://en.wikipedia.org/wiki/Solar_sail
(in this page they use the fact photons has momentum but it's not necessary to talk about photons).

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lightarrow

10. Jun 10, 2015

### ZapperZ

Staff Emeritus
You also need to be aware that *I* cannot read minds (I'm a physicist, not a psychic). When you asked all these questions, not once did you declare the level that you can understand such a thing. So when you asked about relativistic mechanics, I assumed that you already have seen college-level intro E&M.

But besides that, it didn't matter if you understood the physics that I gave as an example. It is merely an illustration that what you were doing, i.e. manipulating those equations without paying attention to the physics, wasn't totally kosher. Just because something carries the same named (energy) doesn't mean that you can weely-neely assign it to different things without understanding the nature of the physics. That was the point that I was trying to make, something which I still am unsure if it got through to you based on your response.

Zz.

11. Jun 10, 2015

### Michael Birdman

OK I am convinced, photons have no mass but they do have momentum.

So, force is rate of change of momentum. The solar constant, normal to the sun's rays, above the atmosphere is about 1400W m-2. If we assume that all the photons are 500nm then that equates to about 3.5 x 1021 photons per second per square metre.

The momentum of these would be 1.3 x 10-27 each, the force exerted would equal this divided by the time to hit and slow down/disappear. But what time do we use, if we use say, the time for a photon to travel two wavelength for an average speed, then the time is very short, in the order of 3 x 10-15 of a second.
Dividing momentum by time we get a force per photon of 4 x 10-13 N This sounds small, but there are a lot of photons per second. we cannot just multiply by the number of photons per second, we need to multiply by the number of photons hitting at the same time. There are 3.3 x 1014 chunks of time per second, so if we divide up the total per second we get about 10 million striking per " chunk of time"

This would give us an overall force of 4 x 10-6 N per sq metre

This excludes solar wind particles, I wonder what the percentage breakdown on a solar sail at 1 AU would be.

Is this the 6 poppy seed weight? Or does that include solar wind particles?

Last edited by a moderator: Jun 10, 2015
12. Jun 10, 2015

### Michael Birdman

I found out that one poppy seed has a mass of around 300 micrograms, this is 3 x 10-7 kg so 6 would weigh around 1.8 x 10-5N

This number is only a single order of magnitude from my previous calculation, so maybe I am on the right track?

13. Jun 10, 2015

### my2cts

Mass is E/c^2 in the rest frame, being defined as the frame in which p=0.
If a resonating cavity has a photon inside, this WILL contribute h\nu/c^2 to the mass of the cavity system.
If the photon escapes the mass and momentum of the cavity system will change.
If the photon is absorbed by the cavity wall the mass and momentum of the cavity system will not change.
A free photon does not have rest mass, at least none has been found and tight upper limits have been established.
As an approximation to its rest frame, consider a frame moving along its propagation direction at a speed approaching c with respect to its source.
The photon energy in this frame approaches zero because \nu approaches zero. Thus it has zero rest mass.

14. Jun 15, 2015

### Michael Birdman

Silly me!

I was correct in my calculation earlier but I do not need to know the time taken for the photon to come to a stop. In effect what I have done is divide the impulse of one photon by time then multiplied by time times number per second. If the time is shorter then the force per photon is higher but there will be less acting at the same time chunk.

We can simply assume that the total impulse of the photons is smeared over the full second. This means that we can simplify so that force = number of photons per second x impulse per photon.

In the case of 1400W of pure 500nm photons per square metre we get energy per photon of 4 x 10-19 J and impulse per photon of 1.3 x 10-27. Dividing the 1400W by the energy per photon gives us 3.5 x 1021 photons per second. Multiplying the number per second by impulse per photon we get back to 4.6 x 10-6N

15. Jun 16, 2015

### lightarrow

There is a simpler way to make that computation: F = W/c
F = force
W = power
c = light speed
So you have: 1400W / 3*108m/s = 4.67*10-6N.

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lightarrow