Energy of a photon in different reference frames I'm currently reading Gravitation by MTW, and in it they postulate that the energy of a photon in a moving observers reference frame is,
$$E = -\mathbf{p} \cdot \mathbf{u}$$
Where $\mathbf{p}$ is the 4-momentum and $\mathbf{u}$ is the 4-velocity of the observer.
How would I prove the above and what is the intuition that it is true? I know that for a photon in dimensionless units is,
$$E = h/\lambda$$
My guess is that we can look at this as a phenomenon due to a length contraction with respect to the moving observers frame which results in a different observed wavelength of the photon.
 A: In the $(-,+,+,+)$- signature, "$ -\hat u\cdot$"  is the operation "take the time-component along the 4-velocity $\hat u$ [that is, "according to the observer along $\hat u$].
It applies to any 4-vector. [The Euclidean analogue is
"$\hat x\cdot$" is the operation "take the x-component of a vector".]
For any 4-momentum $\tilde p$ (for a photon or a particle),
$$ -\hat u\cdot \tilde p= E$$
is the relativistic energy of the particle according the observer with 4-velocity $\hat u$. So, it defines the "relativistic energy".
(By the way, "length contraction" (involving the spatial separation between two parallel timelike worldlines) doesn't have anything to do with "observed wavelength" (involving the spatial separation between two lightlike lines). Note that length contraction is independent of direction (forward or backward) of relative-motion, but doppler effect depends on direction.)
See: Deriving Relativistic Doppler Effect through length contraction

UPDATE:
Consider another observer with 4-velocity $\hat v$.
Our observer $\hat u$ would write:
$$\hat v=\gamma \hat u + \gamma V \hat u_{\perp}$$
$$\hat v_{\perp}=\gamma \hat V u + \gamma  \hat u_{\perp}$$
where $\hat u\cdot\hat u_{\perp}=0$ and $\hat v\cdot\hat v_{\perp}=0$ and all vectors here are coplanar.
Let the 4-momentum of particle at rest in $\hat v$ be the timelike vector
$$\tilde P=m\hat v$$
and let the 4-momentum of a light-signal $\tilde K$ be expressed by $\hat v$ as
$$\tilde K=k\hat v + k\hat v_{\perp}$$
since $\tilde K$ is lightlike: $\tilde K \cdot \tilde K=0$.
In terms of $\hat u$ and $\hat u_{\perp}$,
we have
$$\tilde P=m\hat v= 
m\left( \gamma \hat u + \gamma V \hat u_{\perp} \right)
$$
so, that $\hat u$ measures a time-component of $\tilde P$:
$$-\hat u\cdot \tilde P
=-\hat u\cdot \left(  m\left( \gamma \hat u + \gamma V \hat u_{\perp} \right) \right)=\gamma m,$$
which we recognize as the relativistic energy of a massive particle.
In terms of $\hat u$ and $\hat u_{\perp}$,
we have
$$\begin{align}
\tilde K=k \hat v + k\hat v_{\perp}
&= 
k\left( \gamma \hat u + \gamma V \hat u_{\perp} \right)+
k\left( \gamma V \hat u + \gamma  \hat u_{\perp} \right)\\
&=
k\left( \gamma +\gamma V  \right)\hat u+ 
k\left( \gamma V  + \gamma   \right)\hat u_{\perp} \\
&=
k\gamma(1+ V)\left(   \hat u+ \hat u_{\perp}\right) \\
\end{align}
$$
so, that $\hat u$ measures a time-component of $\tilde K$:
$$-\hat u\cdot \tilde K
=-\hat u\cdot \left( 
k\gamma(1+ V)\left(   \hat u+ \hat u_{\perp}\right) \right)
=k\gamma(1+ V).
$$
Note that $\gamma(1+V)=\frac{1}{\sqrt{1-V^2}}(1+V)
=\frac{\sqrt{(1+V)(1+V)}}{\sqrt{(1+V)(1-V)}}
=\sqrt{\frac{1+V}{1-V}}
$
is the Doppler factor.
So, we could associate the temporal component of $\tilde K$ as the relativistic energy of a light signal.
