# If C cannot be achieved, then how can light achieve C?

1. Feb 4, 2006

### itari1985

hi, i guess this is a noob question. But if speed of light is impossible to achieve, then how do the photons achieve it?

2. Feb 4, 2006

### Hurkyl

Staff Emeritus
They're always at the speed of light, they never had to "achieve" it!

The theorem to which you're referring, incidentally, says that an object with nonzero mass cannot be accelerated to the speed of light.

There is a similar theorem which says that an object with zero mass can only travel at the speed of light.

3. Feb 4, 2006

### leandros_p

When light is emitted from a light source, it does not start speeding up from static velocity zero accelerating towards to the speed of light. Photons immediately start travelling at the speed of light. This is amazing and has no physical analogy in real life, or in any other phenomenon of classical physics.

The speed of light is not building up. Light comes into physical reality as light at once, it does not become light during a period of time. The velocity of propagation of light is an intrinsic physical attribute of light. In a velocity smaller than 'c' photons have no meaning.

The inability of achieving the speed of light is a problem of physical objects which have the ability to vary their speed. Light/photons can not do this variation of speed.

Leandros

Last edited: Feb 4, 2006
4. Feb 4, 2006

### rbj

in other words, photons "acheive" a speed of c by definition.

just to be a pain in the arse, what Hurkyl means here is "rest mass" or "invariant mass". i still find myself in the old school of semantic that photons have a mass of $m = h \nu / c^2$, but their rest mass or invariant mass is zero. i guess i wasn't paying attention in the last 20 or 30 years when it became conventional that "mass" without any qualifier meant "invariant mass" or "rest mass".

5. Feb 4, 2006

Staff Emeritus

Photons can't have a rest mass, because they haven't ever got a rest frame. They always travel at c. And where did you get that $$h\nu/c^2$$ anyway? AFAIK photons have always been massless in quantum theory except for an early attempt by Feinmann to use massive virtual photons to correct the infinities in QFD, an attempt that came to nothing and which he abandoned to create his version of renormalization.

(Added) Oh I suppose you are using $$e = h\nu$$ and $$e = mc^2$$. But the energy relation of relativity, got by doing "Minkowski-Pythagoras" on the components and magnitude of the momentum energy four-vector, is $$e^2 = p^2c^2 - m^2c^4$$, so when m = 0 energy only depends on momentum, not mass, and that's the situation with light. Since $$e = mc^2$$ is derived from relativity, for massive bodies in their rest frames it's just wrong to use it to compute a fictitious mass for the photon.

Last edited: Feb 4, 2006
6. Feb 4, 2006

### sporkstorms

While you're all on the topic... I almost feel foolish for asking this (I'm a 4th year physics undergrad).. but I've never fully understood this, and never been able to find a satisfying answer:

If massless particles always travel at c, then how do they "slow down" in matter?

My educated guess and the only semi-decent explanation I've seen (although never in full, or completely hashed-out) is that it's not the same photon actually slowing down. But rather, the photon excites an atom, which emits another photon - and this occurs many times until a photon is emitted from the matter.

Am I on the right track? And if so, am I fully seeing the big picture, or is there more going on?

--
Edit: Maybe and/or the photon is being reflected back and forth inside the matter, until it finally escapes?

7. Feb 4, 2006

### ZapperZ

Staff Emeritus
Read the Physics FAQ in the General Physics forum.

Zz.

8. Feb 4, 2006

### sporkstorms

Thank you. That is very well written, and answers my question far better than I've seen before.
Furthermore, now I know what to study more to get an even more detailed answer.

9. Feb 4, 2006

### ZapperZ

Staff Emeritus
There is no better compliment than to know that something you got here became an IMPETUS to study it in greater detail and rigor. That is the best use of a forum such as this.

Zz.

10. Feb 4, 2006

### leandros_p

Actually when light pierce matter (glass, or water for example) you can not describe the phenomenon in the same way that light pierce darkness in void space.

Light/photons exist and travel only in vaccum. Then, the speed of light is 'c'.

When light enters space with matter and then exits from it we can measure the speed of light as it covers the distance from entrance to exit, but this measurement is about the apparent speed of light. The speed of light within space of matter does not measure the speed of photons but it measures the behavior of light within a system where light interacts with properties of matter.

In this context, there are "systems" of mater within which the apparent speed of light slows down and the direction of propagation may also change.

Using such "systems" scientists managed to slow down the apparent speed of light to 60 km/hour. This is the speed of an automobile!

In simple words, the speed of light is always 'c", but the apparent speed of light can be smaller.

Leandros

Last edited by a moderator: May 2, 2017
11. Feb 5, 2006

### rbj

in general (not just photons), what is the momentum $p$ when you say $E^2 = p^2c^2 - m^2c^4$? is it not always the product of mass and velocity with respect to some observer? what you call $m$, i call $m_0$ and i would say $E^2 = p^2c^2 - m_0^2c^4$

it might currently not be the conventional usage of the term "mass" (to mean, in general, "relativistic mass") but there is still some semantic respect given it in the physics FAQ: http://math.ucr.edu/home/baez/physics/Relativity/SR/light_mass.html [Broken]

that's not the derivation i was taught (i can't say what was Eisntein's original derivation). when i first seen it derived, the question began (after learning about time and mass dilation, length contraction, Lorentz transformation in SR) with asking the question: "Knowing that mass increases with velocity relative to an observer and that force is the time derivative of momentum and that the integral of force against displacement is work or energy, what is the relativistic Kinetic Energy (from the POV of this observer) of a body of rest mass $m_0$ and velocity $v$ relative to the observer?" and the answer comes out to be:

$$T = \left( \left(1 - \frac{v^2}{c^2} \right)^{-1/2} - 1 \right) m_0 c^2$$

or

$$T = \frac{m_0 c^2}{\sqrt{1 - \frac{v^2}{c^2}}} - m_0 c^2$$

with relativistic mass

$$m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$$

then the kinetic energy is

$$T = m c^2 - m_0 c^2$$

and the two terms on the right are given this interpretation:

$$T = E - E_0$$

where $E = m c^2$ is the total energy of the body and we're subtracting from it the rest energy $E_0 = m_0 c^2$. that is, total energy equals rest energy (what it has in the rest frame) plus the kinetic energy or $E = T + E_0 = m c^2$. $E = m c^2$ is the total energy of the body, not the rest energy of a body at rest. since

$$m_0 = m \sqrt{1 - \frac{v^2}{c^2}}}$$

any particle (which can only be a photon) or anything with $v = c$ will have zero rest mass no matter what the inertial mass of the thing is. we say that $p = mv$ in general and, for photons, $p = h \nu / c$ so why is it such a big deal to say that the mass of the photon is $m = h \nu / c^2$ ? this mass demonstrates properties of mass (photons "fall" in a graviational field if you're willing to accept that a thrown football falls in freefall and photons create their own gravitation, but weakly since their mass is so small).

this is a change of semantic in the last 20 or 30 years. what is behind all this current opposition to calling $m = h \nu / c^2$ what it is? it has dimension of mass and can be converted to the (total) energy of the thing we're talking about with $E = m c^2$. why is it that currently the semantic for "mass" can only be referring to "invariant mass" or "rest mass"? if you can accept that relativistic mass is something that exists or something we can refer to (and have meaning), then you have to accept that photons have it.

Last edited by a moderator: May 2, 2017
12. Feb 5, 2006

### krab

Did you read it? He says
That's not what I call respect. Redundancy is good engineering, but in physics theory, it means "not needed", so physicists turf it out.

Last edited by a moderator: May 2, 2017