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Show that the arc length of a curve is invariant under rigid transformation.

The curve here is in $\mathbb R^3$, and the definition of arc length is $\int^b_a||\bf r'$$(t)||dt$. This theorem appears in my book without proof, can somebody please give me some idea about how to prove it? Thanks.

JSCB
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1 Answers1

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Let $\gamma :[0,1] \to \mathbb{R}^3$ be a curve and $T : \mathbb{R}^3 \to \mathbb{R}^3$ be a rigid transformation. Because $T$ is linear, $$(T \circ \gamma)'(t)=T \circ \gamma'(t).$$

Therefore, $$\mathcal{L}(T \circ \gamma)= \int_0^1 \| (T \circ \gamma)'(t) \| dt = \int_0^1 \|T \circ \gamma'(t)\|dt= \int_0^1 \| \gamma'(t)\|dt =\mathcal{L}(\gamma),$$

where $\|T\circ \gamma'(t)\|= \|\gamma'(t)\|$ since $T$ is a rigid motion and in particular is norm-preserving.

Seirios
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    +1 Seirios, your answer is great! I'd just like to add to the OP that this answer assumes $T$ is linear. Now, in general, a rigid transformation is (uniquely) the sum of a linear transformation and a translation. The case of a translation is easy to handle and exactly the same computations done by Seirios above apply. In fact, Seirios could have simply typed "Because $T$ is a rigid transformation ..."; perhaps Seirios intended to do so or just considered the case of $T$ a translation to be obvious (which it is). – Amitesh Datta Jul 21 '13 at 09:47
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    Why $(T \circ \gamma)'(t)=T \circ \gamma'(t)$? – user42912 Oct 14 '17 at 04:23
  • Because $T$ is linear: one can identify its differential with $T$ itself. – Seirios Oct 18 '17 at 10:01
  • Isn't it supposed to be $(T\circ \gamma(t))(t)$? And we should get $(T\circ \gamma(t))'(t)=T'(\gamma(t)) \circ \gamma'(t)=T(\gamma(t)) \circ \gamma(t)$? But $T(\gamma(t))$ is a vector. So, how can we compose a vector with a vector? – user5826 Jan 29 '19 at 20:55
  • @AlJebr To your first question, the ansewr is negative. Note that $\gamma$ takes numbers to vectors and $T$ takes vectors to vectors. You are trying to compute the vector of a number, what makes no sense. – Danilo Gregorin Afonso Feb 26 '20 at 19:07