# Do the Energy-Momentum Transformations apply to Photons?

• I
In that sense "rest frame" should be applicable for a single particle or a rigid body, not for a group of particles or compound materials, e.g. the floating boat on a river flow is not in the "rest frame".

vanhees71
Gold Member
2019 Award
Yes, in fact the energy-momentum of a photon is an eigenvector of the Lorentz Transformation; and, the Doppler red and blue-shift factors are the eigenvalues!
That's only true if the boost (it's not a general Lorentz transform being discussed here but a boost) is in the direction of the photon's momentum. Indeed, a boost has two space-like and two light-like eigenvectors. For simplicity take a boost in $x$ direction as mentioned in #1. It's representing matrix is given by
$$\Lambda=\begin{pmatrix} \cosh \eta & -\sinh \eta &0 &0 \\ -\sinh \eta & \cosh \eta &0 &0 \\ 0 & 0 & 1 & 0 \\ 0&0&0&1 \end{pmatrix}.$$
Obviously the two space-like vectors $(0,0,1,0)$ and $0,0,1,0$ are eigenvectors with eigenvalue 1 of this matrix.

The other two must be in the $x^0x^1$ plane, because the eigenvalues must be Minkowski-orthogonal to the already found eigenvectors. It's easy to see that these are the two light-like vectors
$$a^{\pm}=\begin{pmatrix}1 \\ \pm 1 \\0 \\0 \end{pmatrix}.$$
That corresponds to photons, one running in the positive the other in the negative $x^1$ direction. The eigenvalues found by
$$\Lambda a^+=(\cosh \eta - \sinh \eta) =\exp (-\eta) a^+, \quad \Lambda a^-=(\cosh \eta + \sinh \eta) a^-=\exp \eta a^-.$$
These are indeed the Doppler factors for the longitudinal Doppler effect, because we have
$$\beta=\frac{v}{c} = \tanh \eta$$
and thus
$$\exp(\mp \eta)=\frac{\cosh \eta \mp \sinh \eta}{\sqrt{\cosh^2 \eta - \sinh^2 \eta}} = \frac{1 \mp \beta}{\sqrt{1-\beta^2}} = \sqrt{\frac{(1 \mp \beta)^2}{(1+\beta)(1-\beta)}} = \sqrt{\frac{1 \mp \beta}{1\pm \beta}}.$$
So if the boost is in direction of the photon (upper sign for $\beta>0$) we have a red shift (energy in the new frame larger than in the original), otherwise a blue shift.

In the general case, i.e., when the boost is not in direction of the photon's motion, you have a Doppler shift and in addition an aberration of the photon's direction!

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Dale
Mentor
In that sense "rest frame" should be applicable for a single particle or a rigid body
Yes.

e.g. the floating boat on a river flow is not in the "rest frame".
If your boat is moving rigidly then it would have a rest frame. The river flow would clearly not be rigid.

The theory of relativity excludes rigid bodies. Proton, which is made of quarks and gluons, is not a particle. In these strict sense rest frame could be applicable for a few kind of single elementary particles, e.g. mass of proton could not be observed because proton, as complex of particles, cannot be at rest.

vanhees71
Gold Member
2019 Award
As I said before, the mass of an object is defined by $m^2 c^2=p_{\mu} p^{\mu}$, where $p^{\mu}$ is the total energy-momentum four vector of the object. A proton has a very well defined mass, because it's a stable hadron (as far as we know).

Dale
Mentor
In these strict sense rest frame could be applicable for a few kind of single elementary particles
Which is precisely why the term "center of momentum frame" is so common in the particle physics literature.

This convention is not universally used, but it is more descriptive and is appealing to many scientists.

vanhees71
Gold Member
2019 Award
Indeed the term "center of momentum frame" is the most precise expression for "rest frame", if not the only correct one at all, of a composite system.

PeterDonis
Mentor
2019 Award
The theory of relativity excludes rigid bodies.
More precisely, it excludes bodies in which internal forces are propagated faster than light. But it is perfectly possible to have rigid motions of bodies, under appropriate conditions.