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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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  • $\begingroup$ Have you tried integrating by parts? $\endgroup$ – Jerry Schirmer Apr 2 '13 at 18:17
  • $\begingroup$ @JerrySchirmer: So saying: '$\vec{u} \cdot \frac{d}{dt} (m\gamma \vec{u})= \frac{d}{dt}(m \gamma \frac12 \vec{u}\cdot\vec{u})-m \frac12 \vec{u}\cdot\vec{u} \frac{d}{dt}\gamma$'? $\endgroup$ – Freeman Apr 2 '13 at 18:21
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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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  • $\begingroup$ 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! $\endgroup$ – Freeman Apr 2 '13 at 18:57
  • $\begingroup$ @Freeman Sure thing! $\endgroup$ – joshphysics Apr 2 '13 at 19:02
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The LHS of the equation you're trying to get to is none other than the time rate of change of the (relativistic) kinetic energy once you add back in the derivative of the constant term - mc^2.

The relativistic analog of the Newtonian relation Force-dot-velocity = rate of change of K.E. holds as well. From here you just substitute the force law and the desired formula results as expected.

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  • $\begingroup$ Thank you for this, as a mathematician it's great to see a more physical interpretation of it. Thank you so much for your time! $\endgroup$ – Freeman Apr 2 '13 at 19:00
  • $\begingroup$ Funny, as I'm always fretting if I should be making my physics more rigorous. The grass is always greener I suppose. =p $\endgroup$ – David H Apr 2 '13 at 19:05
  • $\begingroup$ Haha.. very true indeed! $\endgroup$ – Freeman Apr 2 '13 at 19:07

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