Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: EvaH on 30/04/2020 12:20:04

Peter asks:
If light is said to have no mass, then why can it not move faster than 300,000 kps? Is it that the Higgs boson lends mass to the photon? If so then it truly does have mass...
What do you think?

If light is said to have no mass, then why can it not move faster than 300,000 kps? Is it that the Higgs boson lends mass to the photon? If so then it truly does have mass...
Light is energy, and energy is equivalent to mass (E=mc²), hence light has mass. This also makes the mass of a given photon frame dependent. What light lacks is proper mass (restmass).
So light thus has momentum and inertia, but lacking proper mass, it cannot meaningfully exist at any speed less than light since that velocity would be stationary in some inertial frame, which would mean light being at rest.
A Higgs boson has proper mass, and thus cannot be a component of light.

The energy of a photon is given by E = hf. Where h is the Planck constant and f is the frequency of light. To expand upon what Halc said, the frequency of light is frame dependent and so therefore is any energy detected from measuring the photon. Thus red shifted light has lower energy than blue shifted light.
E = mc^{2} requires rest mass which light does not have.

I stand corrected.
It seems that the general equation is E=√((m_{0}c²)² + (pc)²) where m_{0} is proper mass and p is momentum.
For any object at rest, that reduces to E=mc². For a photon which has no proper mass, it reduces to E=pc, which is almost mc² since momentum (p) is mv

I stand corrected.
It seems that the general equation is E=√((m_{0}c²)² + (pc)²) where m_{0} is proper mass and p is momentum.
For any object at rest, that reduces to E=mc². For a photon which has no proper mass, it reduces to E=pc, which is almost mc² since momentum (p) is mv
In the general equation, p is the relativistic momentum, which for something with a nonzero rest mass(m) is
mv/sqrt(1v^2/c^2).
Again, since the m here refers to proper mass, it doesn't apply to a photon. Instead, the momentum for a photon is found by
p = hf/c
And the general equation ends up giving you E= hf for the photon.

Again, since the m here refers to proper mass, it doesn't apply to a photon. Instead, the momentum for a photon is found by
p = hf/c
And the general equation ends up giving you E= hf for the photon.
I understand both equations (surprise!), but am not clear as to how p = hf/c becomes E= hf.

Again, since the m here refers to proper mass, it doesn't apply to a photon. Instead, the momentum for a photon is found by
p = hf/c
And the general equation ends up giving you E= hf for the photon.
I understand both equations (surprise!), but am not clear as to how p = hf/c becomes E= hf.
Energy is momentum times velocity. Here it is pc. Since hf/c times c cancels out the speed of light you are left with E = hf.

Thanks, Jeffrey; got it, there could be hope for me yet.

A photon has no rest mass, and so it actually doesn’t rest, but nevertheless has a momentum that is frequency dependent.
The photon momentum was discovered experimentally by Arthur Compton, for this work he received the Nobel Prize in Physics in 1927.

The energy of a photon is given by E = hf. Where h is the Planck constant and f is the frequency of light. To expand upon what Halc said, the frequency of light is frame dependent and so therefore is any energy detected from measuring the photon. Thus red shifted light has lower energy than blue shifted light.
E = mc^{2} requires rest mass which light does not have.
The energy of a photon is E = pc where p = momentum (derived from classical EM). since p = mv and for photons v = c we have p = mc. (me defined this way is called relativistic mass). Substitute this into E = pc and we get E = mc^2. Note that has momentum has relativistic mass. Einstein proved light has relmass this this 1907 paper. This mass is aka relativistic mass, holds for tachyons (v = c) and tardyons (v < c).
The answer to the question "does light have mass" depends on what definition one means when they ask the question, i.e. proper mass or relativistic mass.

Thanks, Pete. Different perspectives on any subject/problem are always valuable. Sometimes, just a difference in wording can bring fresh enlightenment. A combination of #6 & #9 is just what I needed.

In spite of being on 12 wks “house arrest” I’m having difficulty finding time for posting, but I really want to be sure I’ve grasped this, so comments on the following would be appreciated.
E= energy in Joules.
F = frequency in hertz =1/s.
h= Planck’s constant.
λ = wavelength = distance between crests in a wave cycle.
The momentum of a photon is given by the equation p = hf/c. Energy is momentum times velocity; so, with velocity = “c”, energy = pc.
So, Momentum: p = hf/c
Energy: Momentum times velocity: hf/c times c = hf. Thus; E=hf.
The equation E=mc^{2} does not apply to the photon. A “stationary” photon would have no mass, its energy would be dissipated, so it would not exist.

It's not intuitively obvious why p = hf/c unless you follow Einstein's derivation of photon momentum.
We know E = hf because Planck's model of the black body spectrum is vindicated by experiment.
Now if we have some photons in a hypothetical box, we can add up the energy of all the photons and divide by the volume of the box to get a figure for energy density.
We then note that energy per unit volume is equivalent to pressure i.e. force per unit area. Just like the gas laws.
So by analogy with the kinetic theory of gases, we can assign a momentum to a photon. In fact it's easier than gas mechanics because all photons move at the same speed.

The equation E=mc^{2} does not apply to the photon.
Not strictly true. The problem is you are using the short form of Einstein’s equation. I think you will recall:
E_{r}^{2} = (m_{0}c^{2})^{2}+(pc)^{2}
So where m_{0}=0 the equation reduces to E_{r}=pc and Alan neatly explains the other part of the energy momentum relationship

Energy is momentum times velocity; so, with velocity = “c”, energy = pc.
Er, not quite!
Kinetic energy = ½mv^{2}
Momentum = mv
You have to account for a factor of 2.
Hence Einstein's necessarily pedantic derivation: force is the rate of change of momentum, hence momentum = integral of force, and there's your multiple.
Dimensional analysis has an Achilles heel!

Bugg’r! Why do things always seem to become more complicated, just when I think I’m getting somewhere?
Thanks, anyway; it’s a bit more to think about, and thinking is probably a good thing to keep doing.

Just checking.
M_{o} = rest mass.
The following two equations say the same thing.
1. Er^{2} = (m_{0}c^{2})^{2}+(pc)^{2}
2. E^{2} = m^{2} c^{4} + p^{2} c^{2}
E^{2} = m^{2} c^{4 }applies to situations in which the object is considered as being at rest relative to the observer.
p^{2} c^{2} is added to the equation when the object is considered as being in motion relative to the observer.

Just checking.
Yes, that’s right.
Also, be careful not to fall into the trap many do of saying mass and energy are the same thing. An equivalence is not the same as 2 things being the same

I am fascinated as to what the mass of a cubic meter of sunlight close to the Earth would be is it something we can calculate .
how many photons would it contain ?

The mass would be zero. But you can calculate its momentum since we know the irradiance at altitude is towards 1.5 kW/sq m, so the energy content of a cubic meter is the input power multiplied by the time it takes each photon to move 1 meter (1/300,000,000 s). You might assign an average photon energy of say 2 eV, and thus estimate the photon flux.
It is left as an exercise to the reader!

I did the calculation some while ago but I forget the answer now, I am amused by that song about lighting a penny candle from a star, pretty easy if you use our local star and a magnifying lens.

might have trouble finding a penny candle, though.

Energy is momentum times velocity; so, with velocity = “c”, energy = pc.
Er, not quite!
Kinetic energy = ½mv^{2}
Momentum = mv
You have to account for a factor of 2.
Hence Einstein's necessarily pedantic derivation: force is the rate of change of momentum, hence momentum = integral of force, and there's your multiple.
Dimensional analysis has an Achilles heel!
You're wrong, Bill's right. See my post. Kiinetic energy s not 1/2 mv^2 in relativity it's K = E= E  E_0

This mass is aka relativistic mass, holds for tachyons (v = c) and tardyons (v < c).
A tachyon or tachyonic particle is a hypothetical particle that always travels faster than light.
https://en.wikipedia.org/wiki/Tachyon
The term comes from the Greek: ταχύ, tachy, meaning rapid. The complementary particle types are called luxons (which always move at the speed of light) and bradyons (which always move slower than light); both of these particle types are known to exist.

I would advise you to improve your skills in physics with https://assignmentyoda.com/physics/.

Bill's right
If I’m right, it would be good to know why. Could you explain “K = E= E  E_0” please?

Bill's right
If I’m right, it would be good to know why. Could you explain “K = E= E  E_0” please?
The total energy E is the sum of kinetic energy K and rest energy E_0, i.e.
E = K + E_0
solve for K to get
K = E  E_0

Thanks Pete. I know I'll never be a mathematician, but with a bit of patient encouragement, I can grasp some of the basics, I think. :)

All perfectly true, but E_{0} = 0 for a photon, so we are left with K = E = hc/λ. The question is how to derive p from E, knowing c.
On matters of relativity I defer to Pete  over!

“And I am right,
And you are right,
And all is right as right can be!”
G & S had a flair for apposite comments.

I’m still juggling these equations in my head and wondering if there is something I’ve missed.
E = K + E_{0}, solve for K to get: K = E – E_{0}.
E = K + E_{0}, therefore, K = E – E_{0}
For a photon, E_{0 }= 0, so K = E – 0. I.e. K = E.
We established earlier that Energy is momentum times velocity; given by the equation p = hf/c.
We also said that, because the velocity of a photon is “c”, hf/c times c = hf. Thus; E=hf.
If K = E, then K must = hf, in this scenario.

We established earlier that Energy is momentum times velocity; given by the equation p = hf/c.
It wasn't "established" but merely stated!
As I said, classical momentum p_{m} = 2K_{m}/v_{m} where K_{m} is the kinetic energy of a nonrelativistic particle m of nonzero mass.
For a photon v_{Φ} = c, . The question is why p_{Φ} = E/c and not 2E/c.
As this has baffled me from time to time since 1964, I call upon the right honorable Pete for enlightenment!

I call upon the right honorable Pete for enlightenment!
Thank goodness you are not asking me! I look forward to Pete's response, in the hope that I might understand it.

I’m still juggling these equations in my head and wondering if there is something I’ve missed.
E = K + E_{0}, solve for K to get: K = E – E_{0}.
E = K + E_{0}, therefore, K = E – E_{0}
For a photon, E_{0 }= 0, so K = E – 0. I.e. K = E.
We established earlier that Energy is momentum times velocity; given by the equation p = hf/c.
We also said that, because the velocity of a photon is “c”, hf/c times c = hf. Thus; E=hf.
If K = E, then K must = hf, in this scenario.
Seems okay to me.

Energy is momentum times velocity;
So why is E = pv for photons but ½pv for everything else?

Energy is momentum
So why is E = pv for photons but ½pv for everything else?
By the way, I was wrong when I said Bill was right when he said "Energy is momentum." . The E = pc for EM radiation is a derived quantity from electrodynamics. Only in In classical mechanics does kinetic energy = pv/2 = (1/2) mv^2. In relativity kinetic energy = (gamma  1)m_0 c^2. Kinetic energy + rest energy = E = total inertial energy. When v << c > E = mv^2/2 where m is rest mass.
Mind you that when we're talking about energy here it doesn't refer to total energy just inertial energy. Total energy = kinetic energy + rest energy + potential energy.

Hold on there pardner!
In relativity kinetic energy = (gamma  1)m_0 c^2
Dang me if it ain't the fact that γ = 1/√(1v^{2}/c^{2})
Now yer photon critter is hitailin' thru the universe at v = c kinda by definition (curse that Limey Maxwell, denyin' the aether an' all)
Thus γ_{φ} = 1/√(0) which is one helluva lot, and (γ_{φ}  1) ain't much smaller (see ole Bill's musin's on Cantor's infinities)
But m_{0} = 0 for them thar photons.
Godfearin' folk don't much approve o' Yankees multiplyin' the Good Lord's infinity by the Devil's zero, and puttin' in a dash o' Mr Einstein's c^{2} don't excuse such mathematical blasphemy.
I'm amindin' that y'all said E = K + E_{0} but ain't it the case that E_{0} = 0 'cause the photon bronco can't never stand still?
So E_{φ} = "anything you like" + 0? No wonder New England heathens have fallen under the spell of Darwin.
(I have the entire "Big Bang Theory" on DVD, and when all else fails I look to Sheldon's mother for guidance.)

Mind you that when we're talking about energy here it doesn't refer to total energy just inertial energy. Total energy = kinetic energy + rest energy + potential energy.
What is the potential energy of a photon? Can it have nonzero value?

Hold on there pardner!
In relativity kinetic energy = (gamma  1)m_0 c^2
Dang me if it ain't the fact that γ = 1/√(1v^{2}/c^{2})
That only applies to particles which move at speeds less than the speed of light.
I'm amindin' that y'all said E = K + E_{0} but ain't it the case that E_{0} = 0 'cause the photon bronco can't never stand still?
That's the reason I don't like to use the term "rest mass' do many relativists. I prefer the term proper mass defined as the magnitude of the 4momentum of he particle divided by c^2, which is zero for photons.

What is the potential energy of a photon?
Zero.

I was wrong when I said Bill was right when he said "Energy is momentum."
What I said was:
Energy is momentum times velocity; so, with velocity = “c”, energy = pc.
Is this wrong?

I was wrong when I said Bill was right when he said "Energy is momentum."
What I said was:
Energy is momentum times velocity; so, with velocity = “c”, energy = pc.
Is this wrong?
Yes. Kinetic energy is 1/2 pv for a particle when v << c. otherwise it's K = E  E_0