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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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  • Welcome to PSE. If your question concerns plane phase wave in general (for example not electromagnetic or sound wave etc) then it's not the right one. Since a plane phase wave could be "superluminal" so not carrying an interaction. IMO, having a general plane phase wave you consider that its phase is Lorentz invariant and the right question is how this wave is Lorentz tranformed between inertial frames. All these are basic in the theory of de Broglie... – Frobenius Nov 23 '20 at 13:58
  • ... see my answer here About de Broglie relations, what exactly is E? Its energy of what? – Frobenius Nov 23 '20 at 13:58

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.

Joe
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  • Please elaborate the answer. – Snpr_Physics Nov 23 '20 at 11:09
  • To make this argument work you first need to show that $\omega$ and $\bf k$ are the components of a 4-vector. – Andrew Steane Nov 23 '20 at 11:21
  • Snpr What is not clear? Andrew I defined these 2-vectors like that, Minkowski space is just $R^n$ with a metric signature -1 1 ... 1. – Joe Nov 23 '20 at 11:31
  • Why do you bring in a 1+1 dimensional Minkowski space. Just stick to 1+3 and keep it simple. – my2cts Nov 23 '20 at 11:41
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    I would prefer to write $ct$ and $\omega/c$ in the 4-vectors, instead of $t$ and $\omega$, to make all 4 components having the same unit. – Thomas Fritsch Nov 23 '20 at 12:16
  • @ThomasFritsch They have the same units, in GR we use geometric units. – Joe Nov 23 '20 at 12:31
  • In order to show that properties of a wave ($k$ and $\omega$) form parts of a 4-vector, you need to show they transform the right way. You can do this by thinking about wave motion from first principles, but the most direct route is to consider what phase of an oscillation means, and argue that it must be Lorentz invariant. Then the wave-4-vector can be derived as the gradient of the phase. The logic is this way round. – Andrew Steane Nov 23 '20 at 13:36
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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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  • You have brought in quantum mechanics, which is ok, but it is better to try to show the result within classical (not quantum) relativity. It can be done by arguing from first principles (i.e. just thinking about what phase means) that the phase must be an invariant. – Andrew Steane Nov 23 '20 at 13:32
  • the expression which I have Used is correct both in classical as well as quantum domain.But Yes, It should be done using Action Principle – crabNebula Nov 23 '20 at 13:34