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My friend told me about an equation $m(v) = \dfrac{m_0}{\sqrt{1-\left(\frac{v}{c}\right)^2}}.$

I checked the internet and saw the same equation as well.

Then I was just playing around with it.

I just googled for the mass of light and they gave the answer $0$.

I thought that in the case of light the equation would be:

$$0 = m_0/\sqrt{\left(1-c^2/c^2\right)}$$

If we just assume that the mass of a photon at rest is also $0$.

Then the equation would be:

$$ 0 = 0/0$$

Or if the mass at rest is something else it would be:

$$0 = m_0/0$$

What I am trying to say is that instead of giving an answer of infinity, the division by $0$ is giving me zero.

I am unable to understand where I am wrong.

I am still at high school so it would be nice if you could give me an explanatory answer.

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For any general body, the energy (in a relativistic framework) is given by,

$E=\sqrt{m_0^2c^4+p^2c^2}=m(v)c^2$

For a photon, the rest mass $m_0=0$, and it's energy $E=pc$. The relativistic mass $m(v)$ is not zero for the photon. Also, the momentum $p=m(v)v$, not $m_0v$, so the formulas are consistent too.

From the above expression for energy, we may arrive at the formula your friend gave you, which is

$m(v)=\dfrac{m_0}{\sqrt{1-\dfrac{v^2}{c^2}}}$

But in the case of a photon, as $m_0=0$, then you don't arrive at the above formula. Putting rest mass as zero, and $v=c$ in the expression of energy, we get

$m(v)=E/c^2=p/c=m(v)v/c=m(v)$, as $v=c$ for a photon.

Therefore, no division by zero. In the case of light, there is no rest mass, and we speak about energy and momentum of light, in place of mass. Because in this case, they are directly related to each other (proportional).

Edit: To those who downvoted it, please inform me of my mistakes so that I can edit my knowledge for the better. Thank You!

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For light we have

$ E = hf = mc^2$

so

$\dfrac{hf}{c^2} = m$

we then plug $m_0$

$\dfrac{m_0}{\sqrt{1-\left(\frac{v}{c}\right)^2}} = \dfrac{hf}{c^2}$

so $m_0 = \dfrac{hf}{c^2} \sqrt{1-\frac{v^2}{c^2}}$

since light is traveling with v = c

$ m_0 = (\dfrac{hf}{c^2})\sqrt{1-\frac{c^2}{c^2}} $
$ m_0 = (\dfrac{hf}{c^2}) 0$
$ m_0 = 0 $

Therefor m0 is 0 for a light beam in agreement with your findings regarding light having zero mass. Light has zero mass and is independent of the frequency and/or energy of the light "quanta".
You could however compute a virtual equivalent relativistic mass "m(c)" = $\dfrac{hf}{c}$ which is dependent on light frequency, but in general we use momentum for light, which is

$p = \dfrac{hf}{c} = \dfrac{hc}{\lambda} $

$f$ = frequency of light
$\lambda$ = light wavelength ( $c = f\lambda$ )
$h$ = Plank's constant
$c$ = speed of light
$m_0$ = "rest" mass
$E$ = energy
$p$ = momentum
$m$ = "relativistic" mass = $m(v) = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$

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    $\begingroup$ This site allows math formatting by writing \$...\$ and \$\$...\$\$. $\endgroup$ – Steeven Dec 10 '16 at 9:50
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    $\begingroup$ Got it. Thx. Is this now to your satisfaction too ? :) $\endgroup$ – Mihai B. Dec 10 '16 at 10:49
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What you've stumbled on is more about arithmetic and numbers than physics or masses. Don't worry, you're only slipping up at the end, by thinking that dividing by zero should give an "answer" of infinity.

To divide by zero (i.e. seek how many times zero your input is) gives no meaningful answer (since no quantity of zeros can make up your input). A formula which winds up using /zero therefore does not give a meaningful answer.

It's not unsensible to think the "answer" to x/0 should be infinity, since it would take infinitely many zeros to make up x, and if you look what happens to 1/y as you decrease y towards zero, sure it just gets bigger and bigger.

Then try the same thing with both y/y, and sin(y)/y. Both of these wind up dividing by zero, but the first should clearly equal 1 and the second function gets closer and closer to zero. Dividing by zero doesn't consistently give anything.

Yep light is massless, and thus it's fine (actually it's encouraging) that the equation blows up when you ask it about massless particles.

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  • $\begingroup$ Normally we consider Lorentz group has no closed boundary on c, so photon should not lie in the territory of lorentz group? And use the idea of limit to obtain photon case? But I think we only derive the massless photon because it's the only way to have the speed of light $\endgroup$ – Turgon Dec 16 '16 at 2:21

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