But, the [itex]T_{t t}[/itex] component transforms as the product [itex]A^{0} B^{0}[/itex] of two 4-vectors with respect to Lorentz transformations, i.e.:
[tex]
T'_{t t} = \gamma^{2} \left[T_{t t} - \beta (T_{t x} + T_{x t}) + \beta^{2} T_{x x} \right][/tex]
Momentum is a 3-vector in the direction of the velocity of the particle. When you mentioned ratio between two vectors, I guess you mean the coefficient of proportionality between the two vectors. This coefficient for a particle with a nonzero (rest) mass is:
[tex]
\frac{m}{\sqrt{1 - v^{2}/c^{2}}}[/tex]
This expression is not applicable for massless particles, so we have to somehow modify it. And, indeed, this is the (total relativistic) energy of the particle divided by [itex]c^{2}[/tex]. Thus, your definition of inertial mass reduces to:<br />
<br />
[tex]
m_{\mathrm{i}} = \frac{E}{c^{2}} = \frac{P^{0}}{c}[/tex]<br />
<br />
The last equation identifies this "inertial mass" as the time component of the energy-momentum 4-vector. It transforms with respect to Lorentz transformations as:<br />
<br />
[tex]
P'^{0} = \gamma (P^{0} - \beta P^{1})[/tex]<br />
<br />
These two quantities have different transformation properties with respect to Lorentz transformations. Therefore, equating them is nonsense.[/itex]