# Relativistic Lorentz force law

If we consider the the relativistic Lorentz force law:

$$\frac{d}{dt} (m\gamma \vec{u})=e(\vec{E}+\vec{u} \times \vec{B})$$

How can we deduce:

$$\frac{d}{dt} (m\gamma c^2)=e \vec{E} \cdot \vec{u}$$

Clearly dotting with $\vec{u}$ will give us the RHS. Which leaves us:

$$\vec{u} \cdot \frac{d}{dt} (m\gamma \vec{u})=e \vec{u} \cdot \vec{E}$$

Could anyone help explain how to proceed and if this is the correct method?

EDIT: If it helps: with reference to these notes i'm working through: http://www.maths.ox.ac.uk/system/files/coursematerial/2012/2393/8/WoodhouseLectures.pdf Page 86, eq (178), the paragraph underneath states 'The ﬁrst equation (which follows from the second)', this is what i'm trying to prove (a warning, the notes are riddled with errors..).

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Have you tried integrating by parts? –  Jerry Schirmer Apr 2 '13 at 18:17
@JerrySchirmer: So saying: '$\vec{u} \cdot \frac{d}{dt} (m\gamma \vec{u})= \frac{d}{dt}(m \gamma \frac12 \vec{u}\cdot\vec{u})-m \frac12 \vec{u}\cdot\vec{u} \frac{d}{dt}\gamma$'? –  Freeman Apr 2 '13 at 18:21

Let's set $c=1$ for simplicity.

Using your observations, it suffices to show that (just combine the second and third equations you write down) $$\dot \gamma = \vec u \cdot \frac{d}{dt}(\gamma \vec u).$$ To prove this, the following facts are useful: $$\dot \gamma = \gamma^3\vec u \cdot\dot{\vec u}, \qquad \gamma^2\vec u^2 +1 = \gamma^2.$$ Now just compute \begin{align} \vec u \cdot \frac{d}{dt}(\gamma \vec u) &=\vec u \cdot (\dot \gamma \vec u + \gamma \dot{\vec u}) \\ &=\vec u \cdot (\gamma^3(\vec u \cdot \dot{\vec u})\vec u + \gamma \dot{\vec u}) \\ &= \gamma \vec u \cdot \dot{\vec u}(\gamma^2 \vec u^2 + 1) \\ &= \gamma^3\vec u \cdot\dot {\vec u} \\ &= \dot \gamma \end{align}

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That's brilliant, thank you very much! I should have got that... You know how it is when your brain is tired and the coffee has worn off! –  Freeman Apr 2 '13 at 18:57
@Freeman Sure thing! –  joshphysics Apr 2 '13 at 19:02