Doppler effect and four-momentum

Required math: calculus

Required physics: special relativity

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 3; Problem 3.7.

The Doppler effect can be derived from the Lorentz transformation of momentum for a photon. The energy and wavelength of a photon are related by Planck’s formula

\displaystyle  E_{0}=\frac{h}{\lambda_{0}} \ \ \ \ \ (1)

where {h} is Planck’s constant and {\lambda_{0}} is the wavelength.

In the rest frame of the photon’s source, we have

\displaystyle  \mathbf{p}_{0}=\left[E_{0},E_{0}\right]=E_{0}\left[1,1\right] \ \ \ \ \ (2)

If we now transform to a frame moving at speed {v} away from the source, we get

\displaystyle   \mathbf{p}_{v} \displaystyle  = \displaystyle  E_{0}\gamma\left[1-v,1-v\right]\ \ \ \ \ (3)
\displaystyle  \displaystyle  = \displaystyle  E_{0}\sqrt{\frac{1-v}{1+v}}\left[1,1\right] \ \ \ \ \ (4)

Thus the transformed energy is {E_{v}=E_{0}\sqrt{\frac{1-v}{1+v}}} and the new wavelength is

\displaystyle  \lambda_{v}=\sqrt{\frac{1+v}{1-v}}\lambda_{0} \ \ \ \ \ (5)

which is the usual Doppler red-shift formula.

Another way of getting this is to consider the scalar product {-\mathbf{p}\cdot\mathbf{u}_{\mbox{obs}}} where {\mathbf{u}_{\mbox{obs}}} is the four-velocity of an observer and {\mathbf{p}} is the four-momentum of a passing object. Since the scalar product is invariant under a Lorentz transformation, we can work it out in the rest frame of the observer, where {\mathbf{u}_{\mbox{obs}}=\left[1,0,0,0\right]}. In this case, we get

\displaystyle   -\mathbf{p}\cdot\mathbf{u}_{\mbox{obs}} \displaystyle  = \displaystyle  -\left[E,p_{x},p_{y},p_{z}\right]\cdot\left[1,0,0,0\right]\ \ \ \ \ (6)
\displaystyle  \displaystyle  = \displaystyle  E \ \ \ \ \ (7)

That is, the scalar product is the energy of the object as seen by the observer.

In the case of the photon above, if we work out this product in the rest frame of the source, then in that frame {\mathbf{u}_{\mbox{obs}}=\left[\gamma,\gamma v\right]} and {\mathbf{p}=\mathbf{p}_{0}=E_{0}\left[1,1\right]} so

\displaystyle   E \displaystyle  = \displaystyle  -\mathbf{p}\cdot\mathbf{u}_{\mbox{obs}}\ \ \ \ \ (8)
\displaystyle  \displaystyle  = \displaystyle  -E_{0}\left(\gamma v-\gamma\right)\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  E_{0}\sqrt{\frac{1-v}{1+v}} \ \ \ \ \ (10)

2 thoughts on “Doppler effect and four-momentum

  1. Pingback: Another trip to Alpha Centauri: more Lorentz transformation examples | Physics pages

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