# Photon - h/(lamda * c)

1. Mar 18, 2008

### Rel

Hello

I red in web side " The photon has mass " Is that real ?

m = h / lamda * c

2. Mar 18, 2008

### CompuChip

Hi, welcome to PF.

There is indeed a certain energy assigned to the photon, which is related to its frequency $\omega = 2\pi\nu = 2\pi c / \lambda$ by
$$E = \hbar \omega = \frac{h}{2 \pi} \omega = \frac{h c}{\lambda}$$
When we invoke the famous $E = mc^2$, the equivalence between energy and mass, we can indeed think of this energy as representing a mass
$$m = \frac{E}{c^2} = \frac{h}{\lambda c}$$

The "rest" mass is usually defined by
$$(m_0c^2)^2 = E^2 - (\vec p c)^2$$
where p is the (three)-momentum. If we take this momentum to be $\vec p^2 = m^2 \vec v^2 = (m c)^2 = (h / \lambda)^2[/tex] then this formula gives $$(m_0c^2)^2 = (h c / lambda)^2 - (h c / lambda)^2$$ so [itex]m_0 = 0$. It should, because no massive particle can travel faster than light.

These considerations are important in processes such as pair production and annihilation, where conservation of 'mass' (= energy) must hold.

3. Mar 23, 2008

### Rel

Thank you compuchip

You gived to me what I want