Photon Doppler shift and Lorentz invariance

Question

My understanding is that the energy of a photon depends your your reference frame because the energy of a photon depends on its frequency and its frequency may get Doppler shifted. Now, I am trying to reconcile this with the answer I get when taking the scalar product of the 4-momentum of a massive particle with that of a massless particle. Say, $$ p^\mu = \left(E_p, \vec{p}\right) $$ is the 4-momentum of the massive particle and $$ q^\mu = \left(E_q,\vec{q}\right) $$ is the 4-momentum of the massless particle. Then, their scalar product is $$ p^\mu q_\mu = E_pE_q - \vec{p}\cdot\vec{q} $$ which, in the rest frame of the massive particle is $$ p^\mu q_\mu = m_pE_q. $$ This must be Lorentz invariant since it is the product of two 4-vectors but my understanding is that $E_q$ depends on your frame of reference while $m_p$ does not. This is my confusion.

The only way I can think to reconcile this is that $E_q$ is only the energy of the photon in the rest frame of the massive particle but I'm not sure if this is correct. It seems like between the last two equations, $E_q$ has changed its meaning from a frame dependent quantity to a frame-independent quantity which I find bizarre. Should I write the last expression as $p^\mu q_\mu = m_pE'_q$ where $E'_q$ is now a frame-independent quantity?

Answer 1

The expression (form) $p^\mu q_\mu$ is Lorentz-invariant (has the same value in all frames), but the expression $m_p E_q$ is not Lorentz-invariant (it does not have the same value in all frames).

How can that be, when the two things are equal to each other?

Well, the expression $m_pE_q$ has the same value as the expression $p^\mu q_\mu$ only in one frame, the frame of the massive particle, where momentum of the massive particle is zero. Let us denote this frame by asterisk $*$, so that we have

$$ p^\mu q_\mu = m_p E_q^*, $$ where $E_q^*$ is energy of the massless particle in the asterisk frame.

In other frames, $m_pE_q$ has different value than $p^\mu q_\mu$, since in general,

$$ p^\mu q_\mu = m_p E_q - \mathbf p\cdot \mathbf q . $$

Consequently, $m_pE_q$ is not a Lorentz-invariant expression. Only $m_pE_q^*$ is the value of the Lorentz-invariant expression $p^\mu q_\mu$.

Answer 2

I think it's easier to write it as:

$$ q_{\mu} = (q, \vec q)$$

and

$$ p_{\mu} = (\gamma m, \gamma m\vec v) $$

Then

$$ p_{\mu}q^{\mu} = \gamma mq(1 - \vec v\cdot \hat q) =mq_0$$

where $q_0$ is the energy/momentem/frequency/wavenumber (take your pick, since $\hbar=c=1$) in the rest frame of the particle.