how come massless photon has energy

Rico L

I dont much about it.. but according to Einstein's equation that E=MC squared which suggests that the mass is also a factor, therefore in theory that photon suppose not to have any energy but however... photon travels through the space without stoping..just like infinity.. does photon ever runs out energy ? is it because photon are massless and it never runs out energy?? and how is it massless if it has energy? what is the difference between EM wave and photon??


thanks ppl.. I am confused..

Rajini

Hi,
photon has no mass. But it has relativistic mass.
There is no difference between EM and photon.
few examples of EM: X-rays, gamma ray, radio wave, light.

Rico L

thanks.. but can you be more precise? what kind of relativistic mass ?

tom.stoer

The concept of mass deserves a more detailed explanation.

If you look at a massive body with rest mass[tex]m[/tex], its total energy [tex]E[/tex] is (I use [tex]c=1[/tex])

[tex]E(v) = \frac{m}{\sqrt{1-v^2}}[/tex]

Of course this formula does not apply to massless particles.

But there is a more general formula whoich shows that the rest mass is an invariant concept. For all particles the following relation between energy and momentum holds

[tex]E^2 - p^2 = m^2[/tex]

Looking at the same particle in different frames both energy and momentum will change, but the rest mass always has the same value. This formula even makes sense for massles particles like the photon:

[tex]E^2 - p^2 = 0[/tex]

So the photon can have energy and momentum.

sweet springs

The famous relation E=Mc^2 holds when the particle is rest.
More general equation including the case of moving particle is
E^2 = P^2 c^2 + M^2 c^4
where P is momentum. You can see P=0 leads E = Mc^2
Relation for massless M=0 particle is E=|P|c that hold for photon.

Relativistic mass is historical concept. You do not need it in modern physics.

Regards.

tom.stoer

In principle this is correct, but you will often find explanations containing the equation

[tex]E = mc^2[/tex]

where it is not specified what [tex]m[/tex] really means. There are two interpretations, namely

a) based on the rest mass [tex]m[/tex] or sometimes [tex]m_0[/tex] which is zero for the photon; then the formula means "rest energy"

b) based on the relativistic mass [tex]m[/tex] or sometimes [tex]m(v)[/tex] (which is not defined for the photon) as it always moves with c and you would get 0/0 in this formula; then the formula means (velocity dependent) total energy for which I introduced [tex]E(v)[/tex] above

It is correct that one does not need the concept b) of relativistic = velocity dependent mass, but you have to be aware of the fact that the meaning of the formula [tex]E = mc^2[/tex]
is based on this concept implicitly quite often - w/o mentioning it.

sweet springs

Hi, tom
I think I understand you. But
Rico questioned "what happens in E=mc^2 when m=0?".
The formula E=mc^2 does not work well here even m is not m0.
Regards.

tom.stoer

This is what I said before: the E=mc² equation does not make sense for photons.

Rico L

this make sense.. cheers ppl and tom

mg0stisha

Sorry to jump in, but does this imply that a photon's energy must equal the magnitude of its momentum?

ansgar

yes it does

sweet springs

Hi. Rico

You are right. Why do not you say more that the E=mc² equation does not make sense for any moving particles? We have no reason to make effort saving the famous E=mc^2 by introducing artificial m(v) instead of scalar m.

Regards.

filegraphy

Yeah, that brings up a good point on how does a photon continue to travel the cosmological constant, but how can it carry energy if it is mass less. A photon is an energy carrying particle. Stating that it has carry energy. Can somebody help with some reasoning??????

Char. Limit

First, making your words huge doesn't help and in fact may hurt your cause. Please, don't do that.

Second, a photon carries energy because it has momentum, and thus produces a value in the equation [tex]E^2=(pc)^2+m^2 c^4[/tex], even though m=0 for the photon.

tom.stoer

If we start with the "axiomatic" approach for RT you are right; but in practice many RT lessons use the "historical" approach. Therefore one has to understand how to "find" RT "bottom-up". I think both approaches are not contradictory but complementary. This is comparable to the "old" quantum theory of Niels Bohr e al. Today we know that its "wrong", nevertheless it makes sense to study it.

He must so to speak throw away the ladder, after he has climbed up on it (Wittgenstein)

DaleSpam

http://www.xs4all.nl/~johanw/PhysFAQ...y/SR/mass.html
http://www.xs4all.nl/~johanw/PhysFAQ...oton_mass.html

grounded

Hi Rico

Here is how I understand it, someone please correct me if I am wrong:

Photons are created from the destruction of mass, simular to how fire is created from the burning of wood. The photons were created from the mass at some point, just as the fire was created from the wood. The photons can have energy without mass, like fire can have energy without wood.

E=MC² simply states that if you act upon some object to make it create electromagnetic radiation (like turning on an incandescent light bulb), the objects mass (filimant of light bulb) will be reduced by the total amount of electromagnetic radiation energy given off, divided by the speed of light squared. Or, if you act upon an object to transform its entire mass into electromagnetic radiation, the total amount of radiation produced will equal the mass multiplied by the speed of light squared.

Photons are created from the interactions of electric and magnetic fields (EM waves), so they are totally different. I'm not sure, but I think a photon can be thought of as each peak (or cycle) of the electromagnetic wave??