An inertial observer in a two dimensional Minkowski space $(t,x)$ is located at the origin with four-velocity $U^{\mu}=\begin{bmatrix}1\\0\end{bmatrix}$. If a photon is detected by the Observer, its energy can be calculated using the scalar product as follows: \begin{equation} \eta_{\mu\nu}U^{\mu}p^{\nu}=-\hbar\omega, \end{equation} where $p^{\nu}=(E=\hbar\omega,\hbar{k})$. Now, If another observer with four-velocity ($V^{\mu}$) is moving relative to the first one, what would it measure for the frequency of the photon?

Can it be done in the same way (using scalar product)? If I first Lorentz transform this will result in \begin{equation} V^{\mu}=\begin{bmatrix}\gamma\\ \gamma{v}\end{bmatrix}. \end{equation}

Using scalar product, yields \begin{equation} \eta_{\mu\nu}V^{\mu}p^{\nu}=-\gamma\hbar\omega+\gamma{v}\hbar{k}. \end{equation}

I'm not sure if this is correct.

  • $\begingroup$ Why do you think energy can be obtained by a formula which doesn't have any free indices? Namely, $\eta_{\mu\nu}U^\mu p^\nu$ is Lorentz invariant whereas energy is clearly not. In other words, this formula cannot possibly give the energy. $\endgroup$
    – user87745
    Oct 30, 2020 at 16:39
  • $\begingroup$ Well, the first component of the four-momentum gives the energy right? this is why for the first case it resulted in $E=-\hbar\omega$. $\endgroup$
    – M91
    Oct 30, 2020 at 16:43
  • $\begingroup$ Yes, $p^0$ is energy but $p^0$ is not the same as $\eta_{\mu\nu}U^\mu p^\nu$. You might be able to show that the two are numerically the same in some frame of reference in some cases, but that does not mean that they are generically equal. The reason being that the equality $X=Y$ is supposed to hold true in all frames given that it is true in one frame only if the equality is a tensorial equality (i.e., the Lorentz indices match properly, or, in other words, the object $X-Y$ be a tensor). For $p^0=\eta_{\mu\nu}U^\mu p^\nu$ this is not the case because the indices do not match properly. $\endgroup$
    – user87745
    Oct 30, 2020 at 16:52
  • $\begingroup$ Ok, i see. Let me then rephrase the question: what value for the energy will the second observer measure? a calculattion would be nice. $\endgroup$
    – M91
    Oct 30, 2020 at 17:04

2 Answers 2


A potential source of confusion is that the photon frequency is a Lorentz scalar, but also observer-dependent. In order to clarify, I will assume that $c=\hbar=1$.

Consider two observers: Mary, with four velocity $\boldsymbol{u}=u^{\mu}\boldsymbol{e}_{\mu}$, and Lucy, with four velocity $\boldsymbol{v}=\tilde{v}^{\mu}\tilde{\boldsymbol{e}}_{\mu}$. They both detect the same photon, whose four momentum is given by $\boldsymbol{p}$. The frequency as measured by any observer (inertial or not) is defined as (minus) the scalar product between the observer's four velocity and the momentum of the photon. Thus, Mary claims that

$$ \omega_{M}\equiv-\boldsymbol{u}\cdot\boldsymbol{p}=-u^{\mu}p_{\mu}\,. $$

Since she is at rest in her own frame, she will parametrize the components of $\boldsymbol{p}$ as $p^{\mu}=(\omega_{M},\omega_{M}\vec{p}_{M})$ ($\vec{p}_{M}$ is a unit three-vector) so that $p^{\mu}p_{\mu}=0$. Now note that, by definition, $\omega_{M}$ is a Lorentz scalar, so that we can compute $\omega_{M}$ in any other frame. In particular, we can compute it in Lucy's frame:

$$ \omega_{M}=-\boldsymbol{u}\cdot\boldsymbol{p}=-\tilde{u}^{\mu}\tilde{p}_{\mu}\,. $$

Is $\omega_{M}$ to be interpreted as the photon measured by Lucy herself? Absolutely not! This is the frequeny that Mary is measuring, and anyone else in the universe will agree that Mary's detector will read $\omega_{M}$. However, to do this computation, other observers use their own coordinates to write $\boldsymbol{u}$ and $\boldsymbol{p}$, but the final result is still $\omega_{M}$.

What is the frequency that Lucy will herself measure? Again, we use the definition:

$$ \omega_{L}=-\boldsymbol{v}\cdot\boldsymbol{p}=-\tilde{v}^{\mu}\tilde{p}_{\mu}\,. $$

Since this is a Lorentz scalar, Mary (and any other observer), will agree that Lucy's detector is reading $\omega_{L}$.

Are $\omega_{M}$ and $\omega_{L}$ numerically the same? No, because the Doppler effect tells us that the frequency emitted by a given source and measured by two observers in relative motion will differ. This is why the frequency, although a Lorentz scalar by definition, is observer-dependent: different observers will measure different frequencies. How can we relate them? Just use a Lorentz transformation to go from Lucy's to Mary's frame. In Lucy's frame, $\tilde{p}^{\mu}=(\omega_{L},\omega_{L}\vec{p}_{L})$ so that, according to Mary, the components of the photon in Lucy's frame are

$$ p^{\mu}=\Lambda_{\phantom{\omega}\tilde{\nu}}^{\mu}\tilde{p}^{\nu}\,. $$

For a simple boost in the $x$-axis with relative velocity $\beta$, $\Lambda_{\phantom{w}0}^{0}=\gamma$ and $\Lambda_{\phantom{w}1}^{0}=\gamma\beta$, so that

$$ p^{0}\equiv\omega_{M}=\gamma\omega_{L}+\gamma\omega_{L}\beta=\gamma\omega_{L}(1+\beta)=\omega_{L}\sqrt{\frac{1+\beta}{1-\beta}}\neq\omega_{L}\,. $$

If you work out the spatial transformation of $p^{\mu}$, you will find the aberration effect.

  • 1
    $\begingroup$ Possible suggestion (sort of subjective): instead of "Lorentz scalar but not Lorentz invariant", which sounds self contradictory, would it be better to say that the frequency depends on the observer's four-velocity but not on the coordinates used to calculate it? $\endgroup$
    – Javier
    Nov 1, 2020 at 21:47
  • $\begingroup$ I guess you are right. I have changed "Lorentz invariant" to observer-dependent. Thanks for the suggestion. $\endgroup$
    – Thiago
    Nov 2, 2020 at 13:21

Have you worked out:

$$ \eta_{\mu\nu}V^{\mu}p^{\nu}=-\gamma\hbar\omega+\gamma{v}\hbar{k}$$



$$ \omega = ck $$

$$ -\gamma\hbar\omega+\gamma{v}\hbar{k} = -\gamma\hbar\omega(1-\beta)$$

Note that:

$$ \gamma(1-\beta) = \frac{1-\beta}{\sqrt{(1-\beta)(1+\beta)}}=\sqrt{\frac{1-\beta}{1+\beta}}$$

which gives:

$$ \omega' = -\sqrt{\frac{1-\beta}{1+\beta}}\omega $$

which looks like the relativistic Doppler shift with an unwanted minus sign (are you using $\eta = (-,+,+,+)$?).

The comments pointing out that if a Lorentz scalar:

$$ a^{\mu}b_{\mu} = x $$ in one frame, it's $x$ in all frames are correct, but the concern that there is no reason to expect it to work for the time component of a 4-vector ( $p^0$), are valid.

Check and see if it works for a massive particle:

$$ \omega = \sqrt{k^2+m^2} $$

that is, that it holds generally, and not just for a special case.


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