[syhprum]

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. 

[Bill S]

might have trouble finding a penny candle, though.

[PmbPhy]

Energy is momentum times velocity; so, with velocity = “c”, energy = pc. 

Er, not quite!
Kinetic energy = ½mv²
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

[hamdani yusuf]

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.

[Bill S]

Bill's right

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

[PmbPhy]

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

[Bill S]

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.

[alancalverd]

All perfectly true, but E₀ = 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!

[Bill S]

“And I am right,
 And you are right,
 And all is right as right can be!”

G & S had a flair for apposite comments.

[Bill S]

I’m still juggling these equations in my head and wondering if there is something I’ve missed.

E = K + E₀, solve for K to get:   K = E – E₀.
E = K + E₀, therefore, K = E – E₀
For a photon, E₀ = 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.

[alancalverd]

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!

[Bill S]

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. 

[PmbPhy]

I’m still juggling these equations in my head and wondering if there is something I’ve missed.

E = K + E₀, solve for K to get:   K = E – E₀.
E = K + E₀, therefore, K = E – E₀
For a photon, E₀ = 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.

[alancalverd]

Energy is momentum times velocity;

So why is E = pv for photons but ½pv for everything else?

[PmbPhy]

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.

[alancalverd]

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/√(1-v²/c²)
Now yer photon critter is hi-tailin' 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 m₀ = 0 for them thar photons.
God-fearin' 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 c² don't excuse such mathematical blasphemy.
I'm a-mindin' that y'all said E = K + E₀ but ain't it the case that E₀ = 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.) 

[hamdani yusuf]

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 non-zero value?

[PmbPhy]

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/√(1-v²/c²)

That only applies to particles which move at speeds less than the speed of light.

I'm a-mindin' that y'all said E = K + E₀ but ain't it the case that E₀ = 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 4-momentum of he particle divided by c^2, which is zero for photons.

[PmbPhy]

What is the potential energy of a photon?

Zero.

[Bill S]

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?