Well, you may know de Broglie relations, here they are: $$ E = h\nu, \quad\quad p = \frac{h}{\lambda} $$

My question is simple: What exactly is $E$? Is it the total energy? Maybe only kinetic energy? Other thing?

Testing consistence. $$ p = \frac{h}{\lambda} = \frac{h\nu}{c} = \frac{E}{c}\quad\implies\quad E = pc $$

This used only the de Broglie relations. The equation $E = pc$ holds for a photon. But de Broglie relations is not about photons, is it? Its about anything. Matter and waves. Its about matter waves! Lets then start generalizing it to matter.

Hypothesis 01: $E$ is the relativistic total energy.

A reasonable hypothesis. Wikipedia will also agree. This leads to a matter wave with phase speed $c/\beta$, and with group velocity of $v$. This explicitly means, I cannot go non-relativistic, stating that $E = \frac{1}{2} m v^2$. The least I can do to approximate things up is having: $$ E \approx m_0c² + \frac{1}{2}m_0v^2 $$

i.e., the energy is the kinetic energy (which I approximated), plus the rest energy. However, if I am too slow, then $E\approx m_0 c^2$. I then have: $$ m_0 c^2 = pc \quad\implies\quad m_0 c = p $$

Odd momentum, isn't it? If we go fully relativistic: $$ E = m_0 c^2 = pc = \gamma m_0 vc \quad\implies\quad c = \gamma v $$

Just odd. Just odd. And of course, wrong. Hypothesis disproven. Evidently thus, wikipedia is wrong. A different way to clearly see this, its $E = \sqrt{(m_0 c^2)^2 + (pc)^2}$ from relativity is evidently different from $E = pc$ from de Broglie relations (derived in the beginning). Therefore, $E$ cannot be the total energy of the particle, and hypothesis is proven wrong.

Hypothesis 02: $E$ is the kinetic energy.

Some books lies in here (chapter 3 if you most know). We now have $E = \Delta m c^2$, which for low speeds resumes to $E = 1/2 mv^2$. Now, phase velocity is $v/2$ and group velocity is $v$. In addition: $$ E = \frac{1}{2}mv^2 = pc = mvc \quad\implies\quad v = 2c $$

And in general, i.e., in the relativistic case: $$ E = \Delta m c^2 = pc = \gamma m_0vc \quad\implies\quad \Delta m c = (m - m_0)c = p = mv $$

Wrong! Hypothesis disproven.

Hypothesis 03: $E$ has nothing to do with energy of the matter

A reasonable assumption given hypothesis 01 and 02 disproven. Its natural then to assume $E$ is the energy of the matter wave (associated with the wave only), which is identical to the quantity $h\nu$.

Now.. what is phase velocity? Should be $E/p$. We use $E = pc$ from beginning to finally show the phase velocity: $v_p = c$. Now, the group velocity: $$ v_g = \frac{\partial\omega}{\partial k} = \frac{\partial E}{\partial p} = \frac{\partial}{\partial p}\left(pc\right) = c $$

So.. $v_p = v_g = c$. How unusual... all matter has speed $c$. Wrong!

Well.. well... what in the world is $E$ after all?


4 Answers 4

  • For your initial check, you used the formula $c = \nu \lambda$, giving the incorrect formula $E = pc$. This is wrong, matter waves don't travel at the speed of light; instead, $\nu \lambda$ is equal to the phase velocity $v_p$. You then use $E = pc$ in all the other derivations, making them all wrong.
  • Hypothesis #1 is correct for relativistic quantum mechanics.
  • Hypothesis #2 is correct for nonrelativistic quantum mechanics.
  • $\begingroup$ Ohhhhh... Makes full sense now. But, how about wikipedia?. It uses relativistic formulas, and its derivation states hypothesis #1 is actually correct. $\endgroup$ May 25, 2016 at 3:43
  • $\begingroup$ @Physicist137 In NRQM, $E$ is kinetic energy, and in RQM, $E$ is total relativistic energy. In both of them, I suppose you can use the de Broglie relations, but the actual underlying theory is totally different. $\endgroup$
    – knzhou
    May 25, 2016 at 4:07
  • $\begingroup$ knzhou is basically right. In non-relativistic case, we are using Schrodinger's equation and the de Broglie wave is a solution to that equation in free space, i.e. when the potential energy is flat.) $E$ is the total (kinetic plus potential) energy. We usually take the potential energy as zero in free space, so then $E$ is kinetic energy. In the relativistic case, the de Broglie wave is a solution of the Klein-Gordan equation, with $E$ the total energy, i.e. rest energy plus kinetic energy, related to $p$ and rest mass by $E^2 - p^2 c^2 = m^2 c^4$. $\endgroup$ Oct 24, 2018 at 0:21

PART A : Plane Phase Waves and Lorentz Transformations

In the following it's necessary to use a more general form of Lorentz Transformation between two systems $\:S\:$ and $\:S^{\prime}\:$. So, let the system $\:S\:$ translating with constant velocity $\:\mathbf{v}\:$ with respect to $\:S^{\prime}\:$ (see Figure in the bottom). With finite variables the equations of this general Lorentz Transforma- tion are \begin{align} \mathbf{x} & = \mathbf{x}^{\boldsymbol{\prime}}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathbf{x}^{\boldsymbol{\prime}})\mathbf{n}-\gamma \mathbf{v}t^{\boldsymbol{\prime}} \tag{A-01a}\\ t & = \gamma\left(t^{\boldsymbol{\prime}}-\dfrac{\mathbf{v}\boldsymbol{\cdot} \mathbf{x}^{\boldsymbol{\prime}}}{c^{2}}\right) \tag{A-01b} \end{align} In differential form \begin{align} \mathrm{d} \mathbf{x} & = \mathrm{d}\mathbf{x}^{\boldsymbol{\prime}}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathrm{d} \mathbf{x}^{\boldsymbol{\prime}})\mathbf{n}-\gamma \mathbf{v} \mathrm{d} t^{\boldsymbol{\prime}} \tag{A-02a}\\ \mathrm{d} t & = \gamma\left( \mathrm{d} t^{\boldsymbol{\prime}}-\dfrac{\mathbf{v}\boldsymbol{\cdot} \mathrm{d} \mathbf{x}^{\boldsymbol{\prime}}}{c^{2}}\right) \tag{A-02b} \end{align} Dividing equations (A-02) side by side the velocity 3-vector of a particle is transformed as \begin{equation} \mathbf{u} = \dfrac{\mathbf{u}^{\boldsymbol{\prime}}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathbf{u}^{\boldsymbol{\prime}})\mathbf{n}-\gamma \mathbf{v}}{\gamma \Biggl(1-\dfrac{\mathbf{v}\boldsymbol{\cdot} \mathbf{u}^{\boldsymbol{\prime}}}{c^{2}}\Biggr)} \tag{A-03} \end{equation} Equations for the inverse transforms are produced by transposition of unprimed with primed variables and $\:\mathbf{v}\:$ with $\:-\mathbf{v}\:$ \begin{align} \mathbf{x}^{\boldsymbol{\prime}} & = \mathbf{x}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathbf{x})\mathbf{n}+\gamma \mathbf{v} t \tag{A-04a}\\ t^{\boldsymbol{\prime}}& = \gamma\left( t+\dfrac{\mathbf{v}\boldsymbol{\cdot} \mathbf{x}}{c^{2}}\right) \tag{A-04b} \end{align}

\begin{align} \mathrm{d} \mathbf{x}^{\boldsymbol{\prime}} & = \mathrm{d}\mathbf{x}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathrm{d} \mathbf{x})\mathbf{n}+\gamma \mathbf{v} \mathrm{d} t \tag{A-05a}\\ \mathrm{d} t ^{\boldsymbol{\prime}}& = \gamma\left( \mathrm{d} t+\dfrac{\mathbf{v}\boldsymbol{\cdot} \mathrm{d} \mathbf{x}}{c^{2}}\right) \tag{A-05b} \end{align}

\begin{equation} \mathbf{u}^{\boldsymbol{\prime}} = \dfrac{\mathbf{u}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathbf{u})\mathbf{n}+\gamma \mathbf{v}}{\gamma \Biggl(1+\dfrac{\mathbf{v}\boldsymbol{\cdot} \mathbf{u}}{c^{2}}\Biggr)} \tag{A-06} \end{equation}

Now let a plane wave characterized by a scalar invariant quantity $\:\phi\:$, the phase \begin{equation} \phi \left( \mathbf{x},t\right)= \omega \; t -\mathbf{k}\boldsymbol{\cdot}\mathbf{x} \tag{A-07} \end{equation} In (A-07) \begin{equation} \omega= 2\pi\nu \tag{A-08} \end{equation} is the angular frequency and $\;\nu\;$ the frequency. Also \begin{equation} \mathbf{k}= \dfrac{ 2\pi}{\lambda} \;\mathbf{m} , \qquad \mathbf{m} =\dfrac{ \mathbf{k}}{\Vert \mathbf{k}\Vert } \tag{A-09} \end{equation} is the wave 3-vector and $\;\lambda\;$ the wavelength. The plane wave "propagates" with velocity vector \begin{equation} \mathbf{w}= \dfrac{ \omega}{\Vert \mathbf{k}\Vert } \;\mathbf{m}=\lambda\nu\;\mathbf{m} = \dfrac{ \omega}{\Vert \mathbf{k}\Vert^{2}}\mathbf{k}, \qquad \Vert \mathbf{w}\Vert \equiv \mathrm{w} = \dfrac{ \omega}{\Vert \mathbf{k}\Vert }=\lambda\nu \tag{A-10} \end{equation}

To see how this phase is transformed in the primed system $\:S^{\prime}\:$, we insert the expressions (A-01) in (A-07) \begin{align} \phi \left( \mathbf{x},t\right) & = \omega \; t -\mathbf{k}\boldsymbol{\cdot}\mathbf{x}\\ \phi \left( \mathbf{x},t\right) & = \omega \underbrace{\left[\gamma\left(t^{\boldsymbol{\prime}}-\dfrac{\mathbf{v}\boldsymbol{\cdot} \mathbf{x}^{\boldsymbol{\prime}}}{c^{2}}\right)\right]}_{t} -\mathbf{k}\boldsymbol{\cdot} \underbrace{\biggl[\mathbf{x}^{\boldsymbol{\prime}}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathbf{x}^{\boldsymbol{\prime}})\mathbf{n}-\gamma \mathbf{v}t^{\boldsymbol{\prime}}\biggr]}_{\mathbf{x}}\\ & = \underbrace{\biggl[ \gamma\left(\omega+\mathbf{k}\boldsymbol{\cdot}\mathbf{v}\right)\biggr]}_{\omega ^{\boldsymbol{\prime}}} t^{\boldsymbol{\prime}} - \underbrace{\biggl[\mathbf{k}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathbf{k})\mathbf{n}+\gamma \mathbf{v} \dfrac{\omega}{c^{2}}\biggr]}_{\mathbf{k}^{\boldsymbol{\prime}}}\boldsymbol{\cdot} \mathbf{x}^{\boldsymbol{\prime}}\\ & = \omega^{\boldsymbol{\prime}} t^{\boldsymbol{\prime}} - \mathbf{k}^{\boldsymbol{\prime}}\boldsymbol{\cdot} \mathbf{x}^{\boldsymbol{\prime}}=\phi ^{\boldsymbol{\prime}}\left( \mathbf{x}^{\boldsymbol{\prime}},t^{\boldsymbol{\prime}} \right) \tag{A-11} \end{align} so \begin{equation} \phi ^{\boldsymbol{\prime}}\left( \mathbf{x}^{\boldsymbol{\prime}},t^{\boldsymbol{\prime}} \right)= \omega^{\boldsymbol{\prime}} t^{\boldsymbol{\prime}} - \mathbf{k}^{\boldsymbol{\prime}}\boldsymbol{\cdot} \mathbf{x}^{\boldsymbol{\prime}} \tag{A-12} \end{equation} where \begin{align} \mathbf{k}^{\boldsymbol{\prime}} & = \mathbf{k}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathbf{k})\mathbf{n}+\gamma \mathbf{v} \dfrac{\omega}{c^{2}} \tag{A-13a}\\ \omega^{\boldsymbol{\prime}} & = \gamma\left(\omega+\mathbf{v}\boldsymbol{\cdot}\mathbf{k}\right) \tag{A-13b} \end{align}

Multiplying (A-13a) and (A-04b) with c, equations (A-13) and (A-04) are rewritten \begin{align} c\mathbf{k}^{\boldsymbol{\prime}} & = c\mathbf{k}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} c\mathbf{k})\mathbf{n}+\gamma \mathbf{v} \dfrac{\omega}{c} \tag{A-14a}\\ \omega^{\boldsymbol{\prime}} & = \gamma\left(\omega+ \dfrac{ \mathbf{v}\boldsymbol{\cdot}c\mathbf{k}}{c}\right) \tag{A-14b} \end{align}

\begin{align} \mathbf{x}^{\boldsymbol{\prime}} & = \mathbf{x}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathbf{x})\mathbf{n}+\gamma \mathbf{v} \dfrac{ct}{c} \tag{A-15a}\\ c t^{\boldsymbol{\prime}}& = \gamma\left( ct+\dfrac{\mathbf{v}\boldsymbol{\cdot} \mathbf{x}}{c}\right) \tag{A-15b} \end{align} that is the vector \begin{equation} \boldsymbol{\Omega} \equiv \left(\omega,c\mathbf{k} \right) \tag{A-16} \end{equation} is Lorentz transformed as the space-time position 4-vector \begin{equation} \mathbf{X} = \left(ct,\mathbf{x} \right) \tag{A-17} \end{equation}

The phase is the invariant inner product "$\:\bullet\:$" of two 4-vectors in Minkowski space \begin{equation} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\phi ^{\boldsymbol{\prime}}\left( \mathbf{x}^{\boldsymbol{\prime}},t^{\boldsymbol{\prime}}\right)= \omega^{\boldsymbol{\prime}} t^{\boldsymbol{\prime}} - \mathbf{k}^{\boldsymbol{\prime}}\boldsymbol{\cdot} \mathbf{x}^{\boldsymbol{\prime}} = \dfrac{1}{c} \left(\boldsymbol{\Omega}^{\boldsymbol{\prime}}\bullet\mathbf{X}^{\boldsymbol{\prime}} \right)= \dfrac{1}{c} \left(\boldsymbol{\Omega}\bullet\mathbf{X} \right)= \omega \; t -\mathbf{k}\boldsymbol{\cdot}\mathbf{x}=\phi \left( \mathbf{x},t\right) \tag{A-18} \end{equation}

PART B : A Lorentz invariant picture of a "subluminal" particle and its accompaning "superluminal" plane phase wave.

Multiplying (A-14a) by c and dividing side by side with (A-14b) we have, under the assumption $\:\omega \ne 0 \ne \omega^{\boldsymbol{\prime}}$, that \begin{equation} \left(c^{2} \mathbf{k}^{\boldsymbol{\prime}}/\omega^{\boldsymbol{\prime}}\right)= \dfrac{\left(c^{2} \mathbf{k}/\omega\right) +(\gamma-1)\biggl[\mathbf{n}\boldsymbol{\cdot}\left(c^{2} \mathbf{k}/\omega\right) \biggr] \mathbf{n}+\gamma \mathbf{v}}{\gamma \Biggl[1+\dfrac{\mathbf{v}\boldsymbol{\cdot}\left(c^{2} \mathbf{k}/\omega\right) }{c^{2}}\Biggr]} \tag{B-01} \end{equation} Above equation (B-01) is identical to (A-06) which is repeated here for comparison \begin{equation} \mathbf{u}^{\boldsymbol{\prime}} = \dfrac{\mathbf{u}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathbf{u})\mathbf{n}+\gamma \mathbf{v}}{\gamma \Biggl(1+\dfrac{\mathbf{v}\boldsymbol{\cdot} \mathbf{u}}{c^{2}}\Biggr)} \tag{A-06} \end{equation} So, for a plane phase wave with angular frequency 4-vector $\;\boldsymbol{\Omega} \equiv \left(\omega,c\mathbf{k} \right) \;$ the velocity 3-vector $\;\mathbf{u}=c^{2} \mathbf{k}/\omega\;$ is transformed as the velocity 3-vector $\;\mathbf{u}\;$ of a particle. But in order for this velocity to correspond to a "subluminal" particle we must have \begin{equation} \mathrm{u}\equiv \Vert \mathbf{u} \Vert =\left\Vert \dfrac{c^{2} \mathbf{k}}{\omega}\right\Vert = \dfrac{c^{2}}{\mathrm{w}} < c \tag{B-02} \end{equation} From above equation we conclude that :

  1. The plane phase wave must be "superluminal" : $\;\mathrm{w}>c$. This is identical to time-like $\;\boldsymbol{\Omega} \;$: \begin{equation} \Vert \boldsymbol{\Omega} \Vert^{2} = \omega ^{2} - c^{2} \Vert \mathbf{k} \Vert^{2} >0 \tag{B-03} \end{equation}
  2. The picture of this "superluminal" plane phase wave and its accompanied "subluminal" particle is Lorentz invariant. Note that a "superluminal" plane phase wave keeps this property in all frames since $\:\Vert \boldsymbol{\Omega} \Vert^{2}\left(>0\right)\:$ is invariant.
  3. The product of the speed of the particle by the speed of the wave is invariant \begin{equation} \mathrm{u}^{\boldsymbol{\prime}}\cdot\mathrm{w}^{\boldsymbol{\prime}}=c^{2}=\mathrm{u}\cdot\mathrm{w} \tag{B-04} \end{equation}

PART C : Louis de Broglie hypothesis and waves

In PART B we associate to a "superluminal" plane phase wave a "subluminal" particle. In the analysis therein we produce from two characteristics of the wave, that is the frequency $\;\omega\;$ and the wave vector $\;\mathbf{k}\;$, the velocity of the particle. But by the velocity only the particle is not defined completely. We must define,for example, its rest mass or its energy content at least in one frame. And reversely, by the velocity of a particle only we can find the velocity of the wave but not its frequency or its wavelength. Something is missing. But since in the first case we can't find the energy content of the particle and in the second case the frequency of the wave, the "missing something" must be a relation between these two quantities.

Le's go to the rest frame of the particle. Any quantity with respect to this frame will be subscripted with an $\:_{\mathrm{o}}\;$. In this frame $\;\mathrm{u}_{\mathrm{o}}=0\;$, so if we want to keep the invariance (B-04) in this limit we must have $\;\mathrm{w}_{\mathrm{o}}=\infty \;$. The wave would have wavelength $\;\lambda_{\mathrm{o}}=\infty\;$, so it would represent a uniform in space periodic in time vibration with frequency let $\;\nu_{\mathrm{o}}$. Note than this is in agreement with a "superluminal" plane phase wave, that is a time-like $\;\boldsymbol{\Omega} \equiv \left(\omega,c\mathbf{k} \right) \;$. We can find an inertial frame in which the space component of a time-like 4-vector would be zero \begin{equation} \boldsymbol{\Omega}_{\mathrm{o}} = \left(\omega_{\mathrm{o}},\mathbf{0} \right) =\left(2\pi\nu_{\mathrm{o}},\mathbf{0} \right) \tag{C-01} \end{equation}

At this point exactly we meet the de Broglie hypothesis : Since to a photon with energy $\;E\;$ there corresponds a frequency \begin{equation} \nu = \dfrac{E}{h} \tag{C-02} \end{equation} then to a massive particle with rest mass $\;m_{\mathrm{o}}\;$ or energy in its rest frame $\;E_{\mathrm{o}} =m_{\mathrm{o}}c^{2}\;$ there corresponds a wave in its rest frame with frequency \begin{equation} \nu_{\mathrm{o}} = \dfrac{E_{\mathrm{o}} }{h}= \dfrac{m_{\mathrm{o}}c^{2}}{h} \tag{C-03} \end{equation}


So in a sense the particle is like to have an internal oscillation of frequency $\;\nu_{\mathrm{o}}$. In the rest frame of the particle the momentum-energy 4-vector is \begin{equation} c\mathbf{P}_{\mathrm{o}}= \left(E_{\mathrm{o}},\mathbf{0} \right) \tag{C-04} \end{equation} which combined with (C-01) and (C-03) gives \begin{equation} c\mathbf{P}_{\mathrm{o}}-\hbar \boldsymbol{\Omega}_{\mathrm{o}} = \left(0,\mathbf{0} \right) \tag{C-05} \end{equation} or more clearly \begin{equation} \left(c\mathbf{P}-\hbar \boldsymbol{\Omega}\right)_{\mathrm{o}} = \left(0,\mathbf{0} \right) \tag{C-06} \end{equation} that is there exists a frame (the rest of the particle) where the 4-vector $\; \left(c\mathbf{P}-\hbar \boldsymbol{\Omega}\right)\;$ is the zero 4-vector. So, if we apply a Lorentz Transformation to both sides of (C-06) then in any inertial frame
\begin{equation} c\mathbf{P}-\hbar \boldsymbol{\Omega} = \left(0,\mathbf{0} \right) \tag{C-07} \end{equation} since the transform of the zero 4-vector $\;\left(0,\mathbf{0} \right) \;$ is this itself. So in any frame \begin{equation} c\mathbf{P}=\hbar \boldsymbol{\Omega} \tag{C-08} \end{equation} written explicitly with the time and space components \begin{equation} \left(E,c\mathbf{p}\right)=\left(\hbar \omega,\hbar c \mathbf{k} \right) \tag{C-09} \end{equation} Equating time and space components \begin{align} E & =\hbar \omega=h\nu \tag{C-10a}\\ \mathbf{p} & = \hbar \mathbf{k} \tag{C-10b} \end{align} Equation (C-10a) is expressed as \begin{equation} \nu = \dfrac{E }{h}= \dfrac{\sqrt{\left(m_{\mathrm{o}}c^{2}\right)^{2}+\left(pc\right)^{2}}}{h}=\dfrac{\gamma m_{\mathrm{o}}c^{2}}{h} \tag{C-11} \end{equation} and (C-10b) as \begin{equation} \lambda = \dfrac{h }{p}=\dfrac{h}{\gamma m_{\mathrm{o}}v} \tag{C-12} \end{equation} so \begin{equation} \mathrm{w}=\lambda \,\nu=\dfrac{c^{2}}{v} \tag{C-13} \end{equation}

enter image description here

See a larger size animation of this particle and its wave here : http://i.imgur.com/hsL0OfM.gifv

enter image description here

Note :

The general Lorentz Transformation (A-01) is given without proof in "CLASSICAL ELECTRODYNAMICS" by J.D.Jackson, 3rd Edition , $\S$ 11.3, but with the unprimed and primed systems interchanged as follows \begin{align} x'_{0} & =\gamma\left(x_{0}-\boldsymbol{\beta}\boldsymbol{\cdot}\mathbf{x}\right)\\ \mathbf{x}^{\prime} & = \mathbf{x} +\dfrac{\left(\gamma-1\right)}{\beta^{2}}\left(\boldsymbol{\beta}\boldsymbol{\cdot}\mathbf{x} \right)\boldsymbol{\beta}-\gamma\boldsymbol{\beta}x_{0} \tag{11.19} \end{align}

\begin{align} \boldsymbol{\beta} & = \dfrac{\mathbf{v}}{c}\; \qquad \beta=|\boldsymbol{\beta}| \\ \gamma &=\left(1-\beta^2 \right)^{-1/2} \tag{11.17} \end{align}

A derivation of this transformation is given in my answer as user82794 therein : Two sets of coordinates each in frames $O$ and $ O' $ (Lorentz transformation), SECTION B.


The energy must be the total energy calculated with relativity.


The de Broglie wave in free space is a relativistic effect. In the frame of a particle the wavelength is "infinite"; the particle is spread over all space and its phase oscillates with E/h. You can read this much in de Broglie's PhD thesis translated to English.


From what i know, it's like the total energy of the particle. Suppose you want to know the de Broglie wavelength of a photon . You put E=hf because all of its energy is in form of kinetic energy.

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