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Take a one-dimensional plane wave $\exp(i(kx-\omega t))$. How can I show that its phase is a $Lorentz \ Invariant$? How to derive the form of $4-wavevector$?

Qmechanic
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Snpr_Physics
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2 Answers2

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Actually the phase is exactly $\vec{p}\cdot \vec{x} = \eta(\vec{p}, \vec{x}) = \eta_{\alpha\beta}p^\alpha x^\beta = -\omega t+k_xx+k_yy+k_zz$ where $\vec{x}$ and $\vec{p}$ are 4-vectors in Minkowski space:

$$ \vec{x} = \pmatrix{t \\ x \\ y \\ z},\ \vec{p} = \pmatrix{\omega \\ k_x \\ k_y\\ k_z}$$ But because a dot product of two vectors is Lorentz Invariant the phase is invariant.

Mr. J
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Here is the proof not very rigourous. Just to give a slight Idea.

Conservation of Energy momentum four vector $(E/c, p_x, p_y, p_z)$ in STR is given by

$$ p_x^2 + p_y^2 + p_z^2-\frac{E^2}{c^2}=const$$

you can see the proof of this on Wikipedia done using $ Action Principle$

Now we Know that $E= \hbar w$ and $p =\hbar k$, and substituting these $$ (k_x^2 + k_y^2 + k_z^2-\frac{w^2}{c^2} )\hbar=const$$ and this implies $$ k_x^2 + k_y^2 + k_z^2-\frac{w^2}{c^2} =const$$ So, $$ \pmatrix{\omega/c \\ k_x \\ k_y\\ k_z} $$ becomes Lorentz Invariant

crabNebula
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