Damped harmonic oscillation

[kuruman]

I repeat, my point is that the usage is not as standard as claimed in post #39. Perfectly respectable sources express conflicting views.

You don't have to repeat, I am with you 100%. I was expressing my doubts (in counterpoint to my "as it should be") about the respectability of NASA whose $125 million satellite to Mars crashed in 1999 because someone apparently used numbers without paying close attention to the units.

 

[Orodruin]

I see no problem. All we are discussing is terminology. If $\vec v=x_1\hat e_1+x_2\hat e_2$ should we call $x_1\hat e_1, x_2\hat e_2$ components and $x_1, x_2$ coefficients or call and $x_1, x_2$ components and have no word to refer to $x_1\hat e_1, x_2\hat e_2$?

So you do not see a problem with an alleged vector changing magnitude under rotations etc?

 

[haruspex]

So you do not see a problem with an alleged vector changing magnitude under rotations etc?

$x_1\hat e_1, x_2\hat e_2$ are vectors, no allegation required.
But please clarify the concern you have.

 

[docnet]

The calculation is not missing v. Actually, the value 2 in the calculation is the magnitude of velocity. It's considered in the calculation.

I would personally consider Orodruin's advice with more care, because there's probably a good reason. He's also a legit retired university professor who has authored a textbook in physics

 

[Orodruin]

$x_1\hat e_1, x_2\hat e_2$ are vectors, no allegation required.
But please clarify the concern you have.

Yes, those are vectors, but they are the (coordinate) components of the resultant in the given coordinate system only.

 

[Orodruin]

He's also a legit retired university professor who has authored a textbook in physics

Wait, I don’t have to come to work on Monday?!? How do I tell my students? I am supposed to give a lecture at 10 am.

 

[docnet]

Wait, I don’t have to come to work on Monday?!? How do I tell my students? I am supposed to give a lecture at 10 am.

Apologies, I somehow mistook you for a retired professor instead of a current one

 

[kuruman]

Yes, those are vectors, but they are the (coordinate) components of the resultant in the given coordinate system only.

Say I have vector $\mathbf C$ in a given 2D coordinate system specified by unit vectors $\mathbf {\hat e}_1$ and $\mathbf {\hat e}_2.$ In that system I write $$\mathbf C=C_1~\mathbf {\hat e}_1+C_2~\mathbf {\hat e}_2.$$In the above equation, $C_1~\mathbf {\hat e}_1$ and $C_2~\mathbf {\hat e}_2$ are the vector components of $\mathbf C$ whilst $C_1$ and $C_2$ are the scalar components of $\mathbf C$. If I consider a rotation in the standard form that takes the unprimed vectors to primed vectors$$
\tilde R =\begin{pmatrix}
\cos\varphi & -\sin\varphi \\
\sin\varphi & \cos\varphi \\
\end{pmatrix},
$$ I get $$\mathbf C'=\tilde R~ \mathbf C=\tilde R (C_1~\mathbf {\hat e}_1)+\tilde R( C_2~\mathbf {\hat e}_2)
=C_1\tilde R~(\mathbf {\hat e}_1 )+C_2\tilde R~(\mathbf {\hat e}_2) =C_1\mathbf {\hat e}'_1+C_2\mathbf {\hat e}'_2.$$ The vector components of the rotated vector $\mathbf C'$ are the rotated vector components of the initial vector $\mathbf C$ whilst the scalar components are invariant. It's what one would expect.

Just to check $$\begin{align} & \tilde R (C_1~\mathbf {\hat e}_1)=
\begin{pmatrix}
\cos\varphi & -\sin\varphi \\
\sin\varphi & \cos\varphi \\
\end{pmatrix}
\begin{pmatrix} C_1 \\ 0 \end{pmatrix}=C_1\begin{pmatrix} \cos\varphi \\ \sin\varphi \end{pmatrix}\implies
|\tilde R (C_1~\mathbf {\hat e}_1)|=C_1
\nonumber \\
& \tilde R (C_2~\mathbf {\hat e}_2)=
\begin{pmatrix}
\cos\varphi & -\sin\varphi \\
\sin\varphi & \cos\varphi \\
\end{pmatrix}
\begin{pmatrix} 0 \\ C_2 \end{pmatrix}=C_2\begin{pmatrix} -\sin\varphi \\ \cos\varphi \end{pmatrix}
\implies
|\tilde R (C_2~\mathbf {\hat e}_2)|=C_2
.
\nonumber\end{align}$$ Yup, the magnitudes of the vector components of $\mathbf C$ are invariant under rotations.

I see no problem with all this. Did I miss your point?

 

[Orodruin]

Say I have vector $\mathbf C$ in a given 2D coordinate system specified by unit vectors $\mathbf {\hat e}_1$ and $\mathbf {\hat e}_2.$ In that system I write $$\mathbf C=C_1~\mathbf {\hat e}_1+C_2~\mathbf {\hat e}_2.$$In the above equation, $C_1~\mathbf {\hat e}_1$ and $C_2~\mathbf {\hat e}_2$ are the vector components of $\mathbf C$ whilst $C_1$ and $C_2$ are the scalar components of $\mathbf C$. If I consider a rotation in the standard form that takes the unprimed vectors to primed vectors$$
\tilde R =\begin{pmatrix}
\cos\varphi & -\sin\varphi \\
\sin\varphi & \cos\varphi \\
\end{pmatrix},
$$ I get $$\mathbf C'=\tilde R~ \mathbf C=\tilde R (C_1~\mathbf {\hat e}_1)+\tilde R( C_2~\mathbf {\hat e}_2)
=C_1\tilde R~(\mathbf {\hat e}_1 )+C_2\tilde R~(\mathbf {\hat e}_2) =C_1\mathbf {\hat e}'_1+C_2\mathbf {\hat e}'_2.$$ The vector components of the rotated vector $\mathbf C'$ are the rotated vector components of the initial vector $\mathbf C$ whilst the scalar components are invariant. It's what one would expect.

Just to check $$\begin{align} & \tilde R (C_1~\mathbf {\hat e}_1)=
\begin{pmatrix}
\cos\varphi & -\sin\varphi \\
\sin\varphi & \cos\varphi \\
\end{pmatrix}
\begin{pmatrix} C_1 \\ 0 \end{pmatrix}=C_1\begin{pmatrix} \cos\varphi \\ \sin\varphi \end{pmatrix}\implies
|\tilde R (C_1~\mathbf {\hat e}_1)|=C_1
\nonumber \\
& \tilde R (C_2~\mathbf {\hat e}_2)=
\begin{pmatrix}
\cos\varphi & -\sin\varphi \\
\sin\varphi & \cos\varphi \\
\end{pmatrix}
\begin{pmatrix} 0 \\ C_2 \end{pmatrix}=C_2\begin{pmatrix} -\sin\varphi \\ \cos\varphi \end{pmatrix}
\implies
|\tilde R (C_2~\mathbf {\hat e}_2)|=C_2
.
\nonumber\end{align}$$ Yup, the magnitudes of the vector components of $\mathbf C$ are invariant under rotations.

I see no problem with all this. Did I miss your point?

My point is that the x-component in one system is not the same as the x-component in another system. Taking the x-component of a vector in system A does not give you the same thing as taking the x-component in system B. In fact, they generally have both different magnitude and obviously correspond to different directions. As such, ”the x-component” is not coordinate invariant.

 

[Steve4Physics]

My point is that the x-component in one system is not the same as the x-component in another system. Taking the x-component of a vector in system A does not give you the same thing as taking the x-component in system B. In fact, they generally have both different magnitude and obviously correspond to different directions. As such, ”the x-component” is not coordinate invariant.

If I may interject…

I think that in Post #59, @kuruman is considering the case where the vector C and the axes are all rotated through the same angle.

But you are considering the case where C is fixed and the axes are rotated (i.e rotation of coordinate system only).

In the former case, the ‘scalar components’ stay the same, as @kuruman says.

In the latter case, the ‘scalar components’ change, as you say.

 

[haruspex]

My point is that the x-component in one system is not the same as the x-component in another system. Taking the x-component of a vector in system A does not give you the same thing as taking the x-component in system B. In fact, they generally have both different magnitude and obviously correspond to different directions. As such, ”the x-component” is not coordinate invariant.

I cannot see what that has to do with my post #44. The question I raised there was merely one of terminology: having resolved a vector into components in the $\hat x$ and $\hat y$ directions, as $\vec v=x\hat x+y\hat y$ say, what are the "components"? Are they $x$ and $y$ or $x\hat x$ and $y\hat y$?

If anyone wants to continue this sidebar, I think we should do it by PM.

 

[Orodruin]

I cannot see what that has to do with my post #44. The question I raised there was merely one of terminology: having resolved a vector into components in the $\hat x$ and $\hat y$ directions, as $\vec v=x\hat x+y\hat y$ say, what are the "components"? Are they $x$ and $y$ or $x\hat x$ and $y\hat y$?

If anyone wants to continue this sidebar, I think we should do it by PM.

I mean, ultimately it comes down to if you want everyone to necessarily write vector arrows on everything that would need it in three dimensions also for one-dimensional problems. I don’t. I expect anyone that argues for this to also start writing double arrows on top of any variable representing string tension as it is actually a rank two tensor. What you actually call things later is not as important.

 

[NTesla]

I mean, ultimately it comes down to if you want everyone to necessarily write vector arrows on everything that would need it in three dimensions also for one-dimensional problems. I don’t. I expect anyone that argues for this to also start writing double arrows on top of any variable representing string tension as it is actually a rank two tensor. What you actually call things later is not as important.

A good physicist ought to use proper symbols/notations/units everywhere, especially when writing a book, or when working on a project or when explaining things to others. If one is going to not use such symbols/notations/units for something, then justification and reminder for doing it like that, must be mentioned at appropriate places. If you are studying/doing some physics on your own for your own sake, then maybe not. This makes things precise & clear for everyone, otherwise it may lead to such disasters as mentioned in post#51.

 

[Orodruin]

A good physicist ought to use proper symbols/notations/units everywhere, especially when writing a book, or when working on a project or when explaining things to others. If one is going to not use such symbols/notations/units for something, then justification and reminder for doing it like that, must be mentioned at appropriate places. If you are studying/doing some physics on your own for your own sake, then maybe not. This makes things precise & clear for everyone, otherwise it may lead to such disasters as mentioned in post#51.

You are missing the point completely. Writing out the projected equation is absolutely proper. There is not a single thing inappropriate about it. Do you also want to use double arrows on string tensions? If you want to be consistent with this point of view you must. I do not know any single source that does this, you are free to find counter examples if you can.

 

[NTesla]

You are missing the point completely. Writing out the projected equation is absolutely proper. There is not a single thing inappropriate about it. Do you also want to use double arrows on string tensions? If you want to be consistent with this point of view you must. I do not know any single source that does this, you are free to find counter examples if you can.

You are completely missing the point that I made regarding writing justifications and reminders if one is not going to do that.