Massless photon infinite speed?

[Saaam]

massless photon...infinite speed?

ok here's the deal,

a photon is massless
its mass is 0
and f=ma
then why is the speed of light not infinite

 

[Pengwuino]

Well, for one, what are you trying to connect with massless particles and F=ma?

 

[Saaam]

well if m=0,
and if F=any value apart from 0 then acceleration must be infinite?
right?
and if F=0 then acceleration would also be 0?
therefore the speed of light must be either 0 or infinite
but this is not the case
and this is why i am confused...

 

[Doc Al]

ok here's the deal,

a photon is massless
its mass is 0
and f=ma
then why is the speed of light not infinite

Two questions:
(1) What makes you think that F=ma applies relativistically? (It doesn't.)
(2) How are you exerting a force on the photon?

 

[Dale]

Another, IMO more general, way to formulate f=ma is f=dp/dt. A photon does not have mass, but it does have momentum.

 

[Tac-Tics]

F=ma isn't how the world works. It's a good approximation in certain cases. It has been superseded by more modern theories.

 

[vin300]

ok here's the deal,

a photon is massless
its mass is 0
and f=ma
then why is the speed of light not infinite

I got what you must be thinking. A system does work to release a photon and since the mass is zero, it can accelerate the photon to infinite speed but 0*oo is not zero, it's indeterminate
{oo is infinity}
And BTW, the photon when released has mass

 

[jtbell]

And BTW, the photon when released has mass

 

[vinirn]

The photon is massless, but it has the so-called relativistic mass, the m in the equation

E=mc^{2}​

The photon must have relativistic mass because E=hf. It's this mass that form the moment p of the photon

 

[tickle_monste]

The photon is massless, but it has the so-called relativistic mass, the m in the equation

E=mc^{2}​

Um, that m in the equation is the rest mass (please correct me if I'm wrong), which the photon doesn't have. You get the relativistic mass by considering the more fuller equation: E^2=(pc)^2 + (mc^2)^2=p^2c^2 + m^2c^4. The rest mass of the photon is just 0, it only has relativistic mass, so we set m=0, and we get E^2=p^2c^2 + 0 = p^2c^2.

E=[(p^2)(c^2)]^(1/2)=pc. Now, you could try to set pc = E = mc^2 to figure out what the 'mass' of the photon would be, but that's nonsensical for the following reason: to get this equation, we had to assume originally that m=0, and in our final result, pc=/=0 in general.

0 =/= pc = E = mc^2 = (0)c^2 = 0. Thus you would get the nonsensical result: 0=/=0. It's better to just say the photon has momentum.

E = hf = pc.
p = hf/c.
f = c/Lambda.
p = h/Lambda = Planck's constant/wavelength.

I believe relativistic mass enters the picture when you make the equation p = mv.
m=p/v. v=c. p=h/Lambda.
m=h/(lambda)c.

 

[jtbell]

The photon is massless, but it has the so-called relativistic mass, the m in the equation

E=mc^{2}​

The photon must have relativistic mass because E=hf. It's this mass that form the moment p of the photon

Can you provide a citation to a physics textbook or a research journal article that actually uses this "relativistic mass of the photon"?

 

[DrGreg]

In relativity, there are (at least) two different concepts of mass. To put it simply

• "rest mass" which excludes kinetic energy
• "relativistic mass" which includes kinetic energy

There are two schools of thought.

The vast majority of professional physicists call rest mass "mass" and don't talk about relativistic mass at all. They say photons are massless.

Some others call relativistic mass "mass" and rest mass "rest mass". They say photons do have mass, but no rest mass.

Hardly anyone routinely uses the phrase "relativistic mass"; it arises only in discussions like this comparing both definitions.

This is a great cause of confusion and it's a pity there isn't universal agreement over the terminology.

Unfortunately usage is not quite so black-and-white as some people think. Relativistic mass is used in the undergraduate-level textbook Relativity: Special, General and Cosmological (2nd ed 2006) written by the professional physicist Wolfgang Rindler after whom Rindler coordinates are named. Sorry.

By the way, I personally support the standard convention that most other physicists use, but we can't just pretend the other convention doesn't exist. Both views are technically valid, but use mutually incompatible terminology.

 

[vinirn]

Sorry, I should have written E=m_{rel}c^2,
where
m_{rel}= \frac{m}{\sqrt{1-v^2/c^2}}
I could be wrong, but this may be converted to E^{2}=(pc)^{2}+(mc^{2})^{2}

 

[Saaam]

Two questions:
(1) What makes you think that F=ma applies relativistically? (It doesn't.)
(2) How are you exerting a force on the photon?

this is my point
if no force is exerted how can it possibly move?

 

[vin300]

this is my point
if no force is exerted how can it possibly move?

Do you think something that emits light never loses energy?
Better don't talk about the 'force' on the photon because you don't know how long it acts.

 

[Doc Al]

this is my point
if no force is exerted how can it possibly move?

Since you skipped my earlier questions, here's another: Do you think a force is required for something to be moving?

Note that photons do not exist at any speed other than c. When created, they are already moving at full speed. It's not like they are sitting around at rest, then accelerated.

 

[vin300]

A force is required to throw out some mass of the body as a photon

 

[Doc Al]

A force is required to throw out some mass of the body as a photon

It's certainly true that a body emitting a photon will experience an impulse (change in momentum) equal and opposite to the momentum of the photon. But that's not quite the same thing as exerting a force on a photon to accelerate it.

 

[vin300]

Why don't both mean the same?

 

[sas3]

Think how confusing it is from the photons point of view, it pops into existence and because it must travel at the speed of light it doesn’t experience any time and then it transfers its energy and pops out of existence in a different place. Another thing that intrigues me (and please correct me if I am wrong) is that a photon has no upper limit on the amount of relativistic mass it can carry. So conceivably the energy of the entire universe could have been contained in a single photon.

 

[vin300]

There is nothing called the photon's point of view, all energy it has is due to its kinetic energy so its velocity is definitely not indeterminate. A force is required to accelerate the photon or it wouldn't have energy.
A photon with all the mass of the universe is way out of the electromagnetic spectrum and I'm not sure it exists.

 

[Doc Al]

A force is required to accelerate the photon or it wouldn't have energy.

Photons aren't accelerated since they only exist at speed c.

 

[vin300]

Or you say due to uncertainity, you cannot accurately measure its energy at all times it is in contact with the system

 

[vin300]

I think the both of us are right in our own ways. Photons exist only at c, but the photon must be "formed" and in this process it must be "accelerated" with a force as long as it is in contact with the system but it is your inability to measure the correct value of the momentum when it being formed you define it to exist only at c, you say it is not a photon.
That the impulse on the body is not due to the force on the photon sounds stupid.

 

[map19]

Why is there no discussion of electromagnetic waves in this thread, and the limits that permittivity and permeability place on the speed of light ?
Photons are only the quanta of such waves.

 

[andreasj]

I have an idea which is probably going to make people cringe. Light is infinite. <C is the point at which energy transforms to mass, and mass converts to energy at v=c (Einstein mass gain). That point takes x amount of time to detect in a mass state, which is the same at any velocity. Light travels 0 distance, because at C, L*(1-((c^2)/(V^2)))=0. It's not really about the velocity of a photon, it's about the transfer of energy into mass. I don't see the photon as being created, rather destroyed in the conversion to a mass state at <C. How long it takes to destroy a potential photon depends on the separation of energy between the emission and destruction of the wave. One thing has been pretty well proven so far, in order to detect a photon, you have to destroy it.

 

[Vanadium 50]

I have an idea which is probably going to make people cringe.

It does. Please read the forum rules about overly speculative postings.