Normal unit vector and acceleration?

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  • Thread starter JulienB
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Hi everybody! I'm currently learning special relativity, and I'm going through the chapter of tangent, normal and binormal vectors. In my teacher's script, the definition of the normal vector eN says:

[tex]\vec{e_N} = \frac{d}{ds} \vec{e_T} \cdot \frac{1}{\mid \frac{d}{ds} \cdot \vec{e_T} \mid} = \frac{ \frac{d^2}{ds^2} \vec{x(s)}}{\mid \frac{d^2}{ds^2} \vec{x(s)} \mid}[/tex]

I think I understand the first equality with the first derivative of the tangent unit vector, but I am unsure to what means the equality: it looks like some sort of acceleration relative to the arclength instead of time, and I guess from that equality that its vector is perpendicular to the trajectory. Is it simply the normal component of acceleration?

There is kind of an explanation earlier in the script but I'm not sure I get it really:

[tex]s = s(t) \implies t = t(s) \implies \vec{x(t)} = \vec{x [t(s)]} = \vec{x(s)}
[/tex]

At the end I don't really get how (and why) it goes from time as a parameter to the arclength. Could someone possibly give me a hint?

Thank you very much in advance.


Julien.
 

Answers and Replies

  • #2
408
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A little later in the script, I found:

[tex] \mid \frac{d^2 \vec{x}}{ds^2} \mid = \frac{1}{R} [/tex] with R being the radius of the curve. I'm also confused about this formula, it seems to me it would make a part of the acceleration depending on the radius of the curve.
 

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