## Are photons pure energy?

1. Is it correct to say that as photons have no mass they are pure energy and have no mass equivalent according to E=mc^2?
2. In a particle collision involving photons, conservation of momentum can not be used as photons are mass-less. Is that also correct?
3. A proton in a linear accelerator can be accelerated through billions of volts. Then their kinetic energy qV must be 1.6x10-19 X 25x109= 4x10-9Joules, this then equals 1/2 mv2 so that v = sqrt(2x4x10-9/me) = sqrt(8x10-9/3.1x10-31)= 1.6x1011m/s

Clearly this is faster than the speed of light. I know the energy is converted to mass. But does the particle reach c and then the additional energy is converted to mass or does the mass increase as the particle nears the speed of light? If the latter is true it would never reach c.
Thanks,

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Recognitions:
 Is it correct to say that as photons have no mass they are pure energy and have no mass equivalent according to E=mc^2?
Photons are considerably more than 'pure energy'. They are elementary particles, on the same footing as electrons, quarks, and so on. In addition to energy and momentum they have angular momentum (spin). They are described by a vector field Aμ and interact with charges and currents in a prescribed way.
 In a particle collision involving photons, conservation of momentum can not be used as photons are mass-less. Is that also correct?
No, in a collision both energy and momentum are conserved, even if a photon is involved.

For a particle traveling at relativistic speeds the energy is no longer accurately given by 1/2 mv2. The correct relationship is E2 = p2c2 + m2c4. The mass we use is the rest mass m, which does not increase, it stays the same no matter how close the velocity gets to c.

 Then the "mass" that's gained as the velocity approaches c is high frequency photon emissions??

Recognitions:

## Are photons pure energy?

 Quote by PhysicsUnited Then the "mass" that's gained as the velocity approaches c is high frequency photon emissions??
No. There is radiation from accelerated charged particles, but that energy is simply lost.

There are two related but different things that can be meant by the word "mass". There is rest (or invariant) mass, and there is inertial (or relativistic) mass.

Rest mass of a particle is always the same. Rest mass of photon is zero.

Relativistic mass of a particle increases when particle travels close to speed of light. Photons do have relativistic mass, and so they do carry momentum.

$$E^2 = p^2c^2 + m_{rest}^2c^4 = m_{rel}^2c^4$$

For a photon, which has mrest=0, that tells you that mrel=p/c.

In terms of notation, it a bit depends on the source. Some old texts used to reserve symbol m for relativistic mass. Hence, "E=mc˛". These texts would denote rest mass as m0. However, modern convention is to use symbol m to denote rest mass. Relativistic mass doesn't come up nearly as often in equations. Instead, you'd deal with energy or momentum directly.

 If the latter is true it would never reach c.
Yes. A particle that has non-zero rest mass can never reach speed of light. For a particle with non-zero rest mass, total energy can be written in terms of Lorentz factor.

$$E = \gamma mc^2$$

Where gamma is the aforementioned Lorentz factor.

$$\gamma = \sqrt{\frac{1}{1-v^2/c^2}}$$

In this equation, v cannot be equal to or greater than c, as that would result in division by zero or a negative under radical. Furthermore, Lorentz factor diverges to infinity as v gets closer and closer to c. So you can keep putting more and more energy into the particle, and that will get it closer and closer to c, but it will never get there.

 Quote by ofeyrpf [LIST=1][*]Is it correct to say that as photons have no mass they are pure energy and have no mass equivalent according to E=mc^2?[..]
Photons are more than pure energy, and the mass equivalent of radiation is p/c.

Mentor
Blog Entries: 27
 Quote by ofeyrpf [LIST=1][*]Is it correct to say that as photons have no mass they are pure energy and have no mass equivalent according to E=mc^2? [*]In a particle collision involving photons, conservation of momentum can not be used as photons are mass-less. Is that also correct?
#1 is wrong. That's like saying an orange is nothing more than its color. You are ignoring all its other properties. It DOES have momentum (that's why we can have solar sails, etc., thus, your #2 is also wrong), and it also has spin angular momentum. None of these can be ignored if you want to obey the various conservation laws.

You should start by reading the FAQ subforum in the General Physics forum. It will at least inform you why "E=mc^2" is not the complete equation. Or watch this:

Zz.

 Quote by ZapperZ [..] Or watch this: Zz.
That's a beauty!
It replaces one page of text by 2 min entertainment and is probably easier to understand.

 I think its best to leave the adjective 'pure' off of energy completely. Its not really appropriate in any sense.

Hi,
 They are described by a vector field Aμ and interact with charges and currents in a prescribed way.
If photons have no charge how do they interact with charges? The electromagnetic force has no effect on photons is the interaction via the strong force?

I understand the notation Aμ on a basic level, that being that there is a magnitude and direction at every point in space and the μ denotes the coordinates (x0, x1, x2, x3) where perhaps x0 is time. Also the lower indices are contra-variant... which is something to do with the coordinates being defined either parallel or perpendicular to the coordinate axes, which may in fact be curvilinear.

 E2 = p2c2 + m2c4
Does this mean the first term on the rhs is the energy due to the particles motion and the last term on the rhs is the energy due to the particle's rest mass? K^2 explains this in his post.

Using E=γmc2 then a 1kg electron is possible. γme=1kg, γ=1kg/9.1x10-27kg=1.0989x1026

v=\sqrt{c^2(1-\frac{1}{\gamma^2})} = 17,320 m/s (How do I get LaTEX to work here?)

Which is I think 5.78x10-5% of c.

The video is excellent, thanks.

Thank you all for your help, things are clearer.

 Recognitions: Gold Member [QUOTE][If photons have no charge how do they interact with charges?/QUOTE] Do billiard balls 'interact'?? One way photons interact is Compton scattering generally considered a particle type interaction. An electron absorbs some of the photon energy, the scattered photon has a slightly lower energy. But this does not really answer your question. You could also ask your self 'How do charged particles interact?' [We have 'rules and regulations' for those as well....] That's not a trivial question either...nobody knows EXACTLY what happens. As Bill_K implies, we have mathematics to describe what we observe. Why those particular reactions occur, and not others that might be imagined, and why they occur as we observe, is still a stretch for our science.
 Recognitions: Gold Member [QUOTE][Does this mean the first term on the rhs is the energy due to the particles motion and the last term on the rhs is the energy due to the particle's rest mass? K^2 explains this in his post./QUOTE] yes...think of the equation as applying in the frame of the particle......so the mass is the rest mass..... "Photons do have relativistic mass, and so they do carry momentum." Is having 'relativistic mass' a requirement for having momentum???