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If it requires infinite amount of energy to travel at the speed of light then how photon attains this speed? Its source is never infinitely sourced.

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Massless particles don't need infinite energy to jump at $c$. Photons don't have rest mass as they don't interact with Higgs field.

What you have heard is applied only for mass. The whole thing works like this:
In relativistic physics, mass isn't constant. It is increased when its speed is increased. It is driven by this formula: $$m=\frac{M}{{\sqrt{1-\left(\frac{v}{c}\right)^2}}}$$ where, $m$ is relativistic mass, $M$ is rest mass, $v$ is speed.

So, at higher speed, we need higher energy to accelerate mass because it has been increased.
In case of massless particle, let's apply this formula:

  1. When $v$ is less than $c$: If you divide 0 by a +ve number, the result would be $0$. So, to accelerate $0$ relativistic mass particle, you wouldn't need energy. That's the reason why you can't find a massless particle at speed lesser than $c$.

  2. When $v$ is equal to $c$: Formula isn't valid because $0/0$ is not defined.

Note: While the answer is fine for general massless particle, its not good for photons (ask relativistic physicists). Relativistic equations have $c$ due to the fact that it is a postulate of Einstein that photons already have this speed. So, its not good for us to get to there in reverse.
So, answer for relativistic guys: Its framework for Relativistic Physics. We never needed to accelerate photons to $c$. It was already running at this speed.

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this was a fine answer, except for the note. Modern physicists start with Minkowski space as their basic postulate when deriving special relativity. They don't rely upon Einstein's two postulates as their primary motivation. –  Jerry Schirmer Jul 10 '12 at 17:30
    
"it was already running at this speed" huh? So what about the conversion of mass into electromagnetic energy? –  Larry Harson Jul 21 '12 at 11:56
    
@Larry What do you want to say? Please, elaborate. And, why are you more specific to electromagnetic energy? –  Sachin Shekhar Jul 23 '12 at 9:28
    
when mass is converted into electromagnetic energy by the annihilation of an electron and positron, a photon is produced. How do you know that the photon instantaneously acquires the velocity c upon creation? –  Larry Harson Jul 23 '12 at 16:15
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@Larry Photons are produced with that speed. What's the problem? There's no inertial reference frame in which photons are at rest. Its the framework of whole relativistic physics. –  Sachin Shekhar Jul 23 '12 at 17:08
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If it requires infinite amount of energy to travel at the speed of light then how photon attains this speed

That's not quite right. You're thinking of the energy required to accelerate a massive object to $c$ which is impossible thus the infinity.

However, there is no reference frame in which photons are at rest. An object with speed $c$ in one frame has speed $c$ in all frames (thus the invariant speed).

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thanks for sharing boss..I got it now..:) –  Rorschach Jul 6 '12 at 10:42
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Here is a good article on the mass of the photon, which answers your question. http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html

The concept of relativistic mass is just a construct. A particle only has one mass; its rest mass and it is constant. It does not increase with particle velocity. This article, The Concept of Mass, should help explain. https://www.worldscientific.com/phy_etextbook/6833/6833_02.pdf

The Wikipedia article on "Mass in special relativity" is very good also.

Here is a quote from an expert on the subject.

It is not good to introduce the concept of the mass $M = m\sqrt{1 - v^2/c^2}$ of a moving body for which no clear definition can be given. It is better to introduce no other mass concept than the ’rest mass’ m. Instead of introducing M it is better to mention the expression for the momentum and energy of a body in motion.

— Albert Einstein in letter to Lincoln Barnett, 19 June 1948

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What's in the physics isn't a construct? You are dumping relativistic physics by one hand. Remember, mass is a form of Energy (E=mc^2). So, it'll increase. –  Sachin Shekhar Jul 6 '12 at 16:32
    
You should read the article you put in the answer. It doesn't say, the mass wouldn't be increased. It says that rest mass of photon is zero, so relativistic mass would be zero, too (see formula in my answer). –  Sachin Shekhar Jul 6 '12 at 16:39
    
There are two problems here. 1) Photons don't have mass; rest, relativistic or otherwise. 2) The mass of all particles is invariant. It doesn't increase with speed. The correct equation is E^2 = (mc^2)^2 + (pc)^2 or E = \gamma mc^2, where \gamma is the Lorentz factor. If you learned that the mass of a particle increases with speed, you were misinformed. All the particle physics experiments have shown otherwise. Ask me how I know (hint: I work at the LHC). –  Adam Hunt Jul 6 '12 at 16:50
    
1) Its not a problem at all. Tell me how its a problem. 2) For rest mass, $p=0$. Put that in your equation, you'll get $E=mc^2$. So, this equation clearly tells rest mass is an energy. How can you not increase an Energy? Soooo, this is level of a LHC scientist? –  Sachin Shekhar Jul 6 '12 at 17:37
    
1) Its not a problem at all. Tell me how its a problem. 2) For rest mass, $p=0$. Put that in your equation, you'll get $E=mc^2$. So, this equation clearly tells rest mass is an energy. How can you not increase an Energy? Soooo, this is level of a LHC scientist? –  Sachin Shekhar Jul 6 '12 at 17:38
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If it requires infinite amount of energy to travel at the speed of light then how photon attains this speed? Its source is never infinitely sourced.

Photons, like gluons, are massless particles. When we speak of 'massless', we mean the rest mass or invariant mass of an object. That is, this quantity is constant to all observers. Photons can travel in the speed of light c because of this property (being massless). If not, then it will require an infinite amount of energy to make a photon with mass travel at the speed of light.

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According to special relativity

$$ \frac vc = \frac {pc}E = \frac {pc}{\sqrt{(mc^2)^2+(pc)^2}} $$

For $m=0$ we have $pc=E$ and thus $v=c$, whereas for $m\not=0$ we have $pc<E$ and thus $v<c$.

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@cristoph:can you please put it in few words as well..say for layman !! –  Rorschach Jul 6 '12 at 11:15
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protected by Qmechanic Jan 9 '13 at 19:54

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