special relativity - Mass of light and $m(v) = m_0/sqrt{1 - v^2/c^2}$

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.

Answer

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!