How can photons have momentum when they don't have mass?
I answered that question back in post #75, the first time you asked.
Are you going to do me the courtesy of reading my replies to you? If not, I might not bother replying to your questions in future. At least, not to try to help you. And are you going to thank me for answering the other questions you asked? Didn't your mother teach you basic manners?
"Momentum can be defined as "mass in motion." All objects have mass; so if an object is moving, then it has momentum - it has its mass in motion. ...
Physics textbooks don't
define momentum as "mass in motion". Momentum is
defined in physics by a mathematical definition.
It is true that for objects with mass, which aren't moving at a significant fraction of the speed of light, momentum can be defined by $\vec{p}=m\vec{v}$, where $m$ the mass of the object and $\vec{v}$ is its velocity.
However, it turns out that this definition is only an approximation, which needs to be modified to make the concept of momentum useful for objects that are travelling at reasonable fractions of the speed of light, and for things that travel at the speed of light.
Photons are things that travel at the speed of light and $p=mv$ does not apply to them.
Note that momentum is, fundamentally, a mathematical construct. It is defined in the way(s) that it is because it is
useful for various types of calculations in physics. In particular, it is useful because it is a conserved quantity in systems on which no external forces act. The definition $p=mv$ works well enough for massive objects travelling at everyday speeds, to conserve momentum in systems on which no external forces act. However, with that definition, momentum is
not conserved in relativistic situations in which one or more of the constituents of a system are moving very fast. So, we modify the definition to make sure that momentum conservation works in those systems, as well. Using the more complicated, relativistic definition, we find that the momentum, as defined, approximates to $p=mv$ for massive objects moving at low speeds.
The equation I gave earlier, $E^2=(pc)^2+(mc^2)^2$ applies to all objects at all speeds, provided that we use the relativistic definitions of the momentum $p$ and the total mechanical energy $E$, and take $m$ to be the rest mass of the object (which happens to be zero for photons).
So if this physics website got the definition of momentum wrong as you say, what is the true definition of momentum?
I already told you what it is for photons. It's $p=E/c=hf/c=h/\lambda$, where $E$ is the photon energy, $h$ is Planck's constant, $f$ is the photon frequency and $\lambda$ is the photon wavelength and $c$ is the speed of light. For particles with non-zero rest mass, it is $p=\gamma mv$, where $\gamma=\frac{1}{\sqrt{1-(v/c)^2}}$ is the Lorentz factor.