# Photon Doppler shift and Lorentz invariance

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?

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 $$p^\mu q_\mu$$.

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.