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Sep
27
accepted Computing the unit vector for a generalised helix
Sep
27
comment Computing the unit vector for a generalised helix
cool ;) thanks for staying with me to get the answer ;) + 1 and accepted.
Sep
27
comment Computing the unit vector for a generalised helix
Sorry ignore that last message about moving discussion to chat, clicked on it by accident (now deleted it). That would make $\theta = \pi/4$.
Sep
27
comment Computing the unit vector for a generalised helix
So then $b = \pm 1$ and $\mathbf A = \pm \mathbf e_y$
Sep
27
comment Computing the unit vector for a generalised helix
That would mean $a$ and $c$ are both zero, since we could vary the RHS without affecting the LHS which would not make sense given their established equality. Is that right?
Sep
27
comment Computing the unit vector for a generalised helix
If $\mathbf A$ is constant, $(\mathbf T \cdot \mathbf A)' = \mathbf T' \cdot \mathbf A = 0$. Computing $\mathbf T'$ and taking the dot product with $\mathbf A$ gives the constraint $asech^2t = c $sech$ t \tanh t \Rightarrow a/c = \sinh t $. Then I rearranged this for $a$ and subbed into the unity modulus condition and that gave me $c^2 \cosh^2 t + b^2 = 1$.
Sep
27
comment Computing the unit vector for a generalised helix
So, $(\mathbf T \cdot \mathbf A)' = \mathbf T ' \cdot \mathbf A = 0 \Rightarrow a/c = \sinh t$ together with the fact that $|\mathbf A| =1 \Rightarrow \sqrt{c^2 \cosh^2 t + b^2} = 1$
Sep
27
comment Computing the unit vector for a generalised helix
yes, I constructed $\mathbf A$ like that because almost by inspection its modulus is $1$ and it recovers the equation $\mathbf T \cdot A = \cos \theta$. Do you have any pointers?
Sep
27
comment Computing the unit vector for a generalised helix
What coefficient do you mean here? The $\mathbf A$ I am working with is $\mathbf A = \cos \theta \mathbf T + \sin \theta \mathbf B$. Thanks
Sep
27
comment Computing the unit vector for a generalised helix
Hi Travis, I have $\mathbf T (t)$ in terms of the Euclidean basis, but $\mathbf A$ I have in terms of the frenet-serret basis. So, I subbed in $\mathbf T$ and $\mathbf B$ into my equation for $\mathbf A$ and then took the dot product but that doesn't seem to help. Or did I misunderstand what you meant?
Sep
27
asked Computing the unit vector for a generalised helix
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15
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May
31
comment Meaning of $\mathbb{R}^0$, $\mathbb{R}^{1/3}$, and $\mathbb{R}^{-2}$.
Sorry to digress from the main topic of discussion in this thread, but I have also seen notation like $\mathbb{R}^{a,b}$. What does this mean? For example, $\mathbb{R}^{2,0}$ represents the Euclidean two dim flat space.
May
28
comment Quick question about contravariant and covariant tensors
It makes sense - the notation is such to label a component of the tensor. There is no ambiguity in the Kronecker delta (since it is diagonal) so either notation is fine when dealing with that. Thanks Tom and Jonathan.
May
28
comment Quick question about contravariant and covariant tensors
Hi Jonathan. Thanks for your reply, but I understand that. My question is about the placement of the indices. In all cases, one of the indices is shifted to the left e.g $\omega_{v}^{\,\,\,\mu}$ vs $\omega_v^{\mu}$. I haven't seen the latter notation used anywhere, so I wanted to know if there are any differences.
May
28
asked Quick question about contravariant and covariant tensors