Let $\vec{\alpha}: I \rightarrow \mathbb{R}^4$ be an arc-length parametrized curve in $\mathbb{R}^4$ with curvatures $k_1, k_2, k_3$. The principal normal unit vector is $\vec{n} = \frac{\vec{\alpha}''}{||\vec{\alpha}''||}$. Let $r>0, \vec{\beta}(s) = \vec{\alpha}(s) - r\vec{n}(s) \forall s \in I$ is a parallel curve (parametrized by the arc-length of $\vec{\alpha}$). I have to find the curvatures of $\vec{\beta}$ (which will be denoted: $k_{\beta,i} \forall i \in\{1,2,3\}$) in terms of $k_1,k_2,k_3$.
Further notation: $\vec{t}$ is the unit tangent along $\vec{\alpha}$; $\vec{b}$ is the binormal thereof (defined by cross product), and $\vec{e_4}$ is the fourth-dimensional analog.
I have calculated the following: $$\vec{\beta}' = \vec{t}-r(-r k_1 \vec{t} + k_2 \vec{b}) = (1+r k_1)\vec{t} - r k_2 \vec{b}, \\ \vec{\beta}'' = r k_1' \vec{t} + k_1 (1+r k_1)\vec{n} - r k_2' \vec{b} -r k_2(k_3 \vec{e_4} - k_2 \vec{n}) = r k_1' \vec{t} + (k_1 + r k_1^2 +r k_2^2)\vec{n} - r k_2' \vec{b} -r k_2 k_3 \vec{e_4}$$.
But how do I find the curvatures $k_{\vec{\beta},i}$ from these values? Since I am working in four dimensions, I am not sure how to find the various curvatures.
What gets me most is the fact that $\vec{\beta}$ is not (apparently to me) arc-length parametrized. Everything gets too messy to handle. Can you perform an explicit calculation showing how to deal with these issues?
If my understanding is correct, $T' = \left(\frac{\vec{\beta}'}{||\vec{\beta}'||}\right)' = \frac{\vec{\beta}''}{||\vec{\beta}'||} - \frac{\vec{\beta}'}{\sqrt{(1+r k_1)^2 + (r k_2)^2}^3}((1+r k_1)k_1' +r k_2 k_2')$, which is already nearly unreasonably large for hand calculations. Then I have to normalize that (the length by which I normalize is $k_{\vec{\beta},1}$). Another two layers is too much.