TITLE: Why do small angle approximations have to be in radians?
QUESTION [5 upvotes]: Why do small angle approximations only hold in radians? All the books I have say this is so but don't explain why.

REPLY [0 votes]: because the sine of an angle is dimensionless [i.e. just a number], the angle has to be also dimensionless [i.e. the radian is dimensionless] ... units on both sides of an equation have to match


TITLE: Electromagnetic braking hypothetical situation?
QUESTION [2 upvotes]: Let's say there is a conducting rod in space - with a constant external magnetic field. If you spin the rod - will electromagnetic braking make the rod stop eventually? I feel like there is a change in flux through the rod when it rotates - and due to Faraday's law and Lenz's law eddy currents will be induced to oppose the motion of the rod

REPLY [3 votes]: There is only one case where isn't induced a current in the rod. One have to move the rod parallel to the magnetic field. In any other case and especially in any case of rotation the magnetic field will move the electrons inside the rod.
For example, if the rotating axis is perpendicular to the rods symmetry axis, there are two pure cases.
(1) The rotating axis is aligned parallel to the magnetic field. Free available electrons will be distributed more to the ends of the rod or more to the centre of the rod. This depends from how the direction of the magnetic field and the direction of the rods rotation are related. Since the rod isn't an closed electric circuit this happens only once and after it does not effect the rotation any more.
(2) The rotating axis is aligned perpendicular to the magnetic field. Electrons are moved periodically parallel to the axis of rotation. In case (2) electrons get accelerated periodicaly and this will be accompanied by electromagnetic radiation. The energy for this radiation has to come or from the kinetic energy of the rotating rod or from a cooling down rod. Cooling may be happens but it is not the dominant process.
Last not least I have to explain why the electrons will be moved. First at all think about the Lorentz force $ \vec F = q \vec v \times \vec B $. This vector cross product can be rewritten for orthogonal vectors to $ q \vec v = \dfrac {(\vec B \times \vec F)}{\|\vec {B}\|^2} $. This corresponds with what happens in an electric generator. In case (1) it will be an DC generator, in case (2) an AC generator. To be precise, in case (1) it is a homopolar generator, ie a special DC generator.
Second one can ask why the electrons get moved in this way under the influence of a magnetic field. One has te remember that attraction or repulsion happens only between charged bodies or between magnetic dipoles. It is nice for us that electrons have magnetic dipole moment and it is spinning too. The electrons magnetic moment will be aligned under the influence of a magnetic field. This isn't spectacular. More interesting is it if the electron moves in a magnetic field (not parallel to the field). The electrons magnetic moment will be aligned and this is accompanied by a gyroscopic phenomena. Gyroscopic effect means that every rotating body act again the force which try to align it. Because the electron has not a magnetic moment only but spins too the moving electron will be deflected. Since any deflection is an acceleration, a photon emission takes place and this emission is directed against the gyroscopic effect. The direction of the magnetic moment of the electron more or less fall back in his previous orientation.
Periodical displacement of the electrons and an associated EM radiation, this is the full picture we see in case (2). The rod will come to a stop.
At the end proof the described movement of the electrons with the Right-hand rule. Don't worry about right or left hand. In your case it is more important to follow the orthogonality of the three vectors magnetic field $ \vec B $, movement of the rod, represented by $ \vec F $ and the resulting current, represented by $ q \vec v $.
If anyone is surprised that such a hand rule is to use, this is because the spin of the electron and its magnetic moment are clearly coupled. If this were not so in nature, then we could not use generators and motors.


TITLE: Method explanation for proof used in the following proposition: If $(Sz,z)=(Tz,z)\forall z \in X$, then $S=T$.
QUESTION [0 upvotes]: This post is inspired by an accepted answer to a question posed here: Prove that the linear transformations are the same..
Specifically, the question is about the method used (and accepted) in proving the proposition:

If X is a complex inner product space and $S,T \in B(X)$ are such that
$(Sz,z)=(Tz,z)\forall z \in X$, then $S=T$.

The accepted answer provides a brief proof...and the method used has caused me a great deal of grief as I can not understand why it is true!
In my below post, I am (purposely) going to slowly go through my thought process...working my way to the final point of confusion that is preventing me from understanding the accepted answer's proof.

Author's Proof

Let $Q=S-T$. By assumption, $\langle Qv,v \rangle =0$ for all $v=\alpha x + \beta y$. Now,
$$
0=\langle Q(\alpha x+y),\alpha x+y \rangle = |\alpha|^2 \langle Qx,x \rangle + \langle Qy,y \rangle + \alpha \langle Qx,y \rangle + \bar{\alpha} \langle Qy,x \rangle \\
= \alpha \langle Qx,y \rangle + \bar{\alpha} \langle Qy,x \rangle.
$$
Choosing first $\alpha =1$ and then $\alpha =i$, we get
$$
\langle Qx,y \rangle +  \langle Qy,x \rangle=0
$$
and
$$
\langle Qx,y \rangle - \langle Qy,x \rangle =0.
$$
Sum and conclude that $\langle Qx,y \rangle =0$ for all $x$ and all $y$. Hence $Q=0$.

The offending line, in particular, derives from the fact that I do not understand why the $x,y$ values in the final two equations are the same $x,y$ values. To me, the final two equations should, instead, read:
$$
\langle Qx_1,y_1 \rangle +  \langle Qy_1,x_1 \rangle=0
$$
and
$$
\langle Qx_2,y_2 \rangle - \langle Qy_2,x_2 \rangle =0.
$$
As such, one should not be able to sum the terms and conclude $\langle Qx,y \rangle =0$

For purposes of my understanding, I chose a small workable dimension of $\text{dim}(V) = 3$.
Let $v^*$ be the arbitrary vector within an inner product space $V$ of dimension $3$. Therefore, consider an arbitrary orthonormal basis within $V$ of the form $e=\{e_1,e_2,e_3\}$. Let $v^*=\eta_1 e_1 + \eta_2 e_2 + \eta_3 e_3$.
Now, for the purpose of this proof, note that $v^*$ can also be constructed from a linear combination of two vectors. There are many possibilities to choose from, but for simplicity, consider the case where $x = \frac{1}{\eta_2}e_1 + \frac{1}{\eta_1}e_2$ and $y = e_3$. We can thus reframe $v^*$ as the sum of two vectors:
$v^* = \alpha x + \beta y$ where $\alpha = \eta_1 \eta_2$ and $\beta = \eta_3$.
Now, assume that for the inner product space $V$, there exists two linear operators $T$ and $S$ such that, $\forall v \in V$, $\langle Tv, v \rangle = \langle Sv, v \rangle$.
Subtracting a common scalar quantity from both sides:
$\forall v \in V$, $\langle Tv, v \rangle - \langle Sv, v \rangle = 0$
Due to linearity of an inner product,
$\forall v \in V$, $\langle Tv - Sv, v \rangle  = 0$
By definition of linear operator addition, define $Av + Bv = (A+B)v$. Thus $ Q:=T-S$
$\forall v \in V$, $\langle Qv, v \rangle  = 0$
This is simply a restatement of our initial assumption of the system that we are working with.
Moving forward, let $v=v^*= \alpha x + \beta y$.
$\forall v \in V$, $\langle Q(\alpha x + \beta y), \alpha x + \beta y \rangle=0$
By linearity property of operators:
$\forall v \in V$, $\langle Q\alpha x + Q\beta y, \alpha x + \beta y \rangle=0$
By additional linearity property of operators:
$\forall v \in V$, $\langle \alpha Qx + \beta Qy, \alpha x + \beta y \rangle=0$
Now, by linearity of inner products:
$\forall v \in V$, $\langle \alpha Qx, \alpha x \rangle +\langle \alpha Qx, \beta y \rangle  + \langle \beta Qy, \alpha x \rangle + \langle \beta Qy, \beta y \rangle$=0
By linearity of first slot of inner products and conjugate linearity:
$\forall v \in V$, $\langle \alpha Qx, \alpha x \rangle +\langle \alpha Qx, \beta y \rangle  + \langle \beta Qy, \alpha x \rangle + \langle \beta Qy, \beta y \rangle=0$
$\alpha\bar{\alpha} \langle Qx, x\rangle + \alpha\bar{\beta} \langle Qx, y\rangle+\beta\bar{\alpha} \langle Qy, x\rangle + \beta\bar{\beta}\langle Qy, y\rangle=0$
From our initial assumption, the first and fourth term are equal to zero:
$\forall v \in V, \ \  \alpha\overline{\beta} \langle Qx, y\rangle+\beta\bar{\alpha} \langle Qy, x\rangle=0$.
Returning the $\alpha$ and $\beta$ values back to their original definitions:
$\forall v \in V, \ \  \eta_1\eta_2\bar{\eta_3} \langle Qx, y\rangle+\eta_3\overline{\eta_1 \eta_2} \langle Qy, x\rangle=0$.
Recall that $x$ and $y$ are purely determined by the coefficients $\eta_1, \eta_2,$ and $,\eta_3$.
$\color{red}{\text{For this final equation, we therefore have a set of ordered triplets}}$ $\{(\eta_{1_i}, \eta_{2_i}, \eta_{3_i})\}$ $\color{red}{\text{for which this equation is true.}}$ (Note the  super tiny index $i$ that is subscripted on these $\eta$ symbols within the above set...this is my way of saying there are 'many" different 3-coefficient combinations corresponding to an arbitrary vector that satisfies this equation)
Consider two special $\color{blue}{\text{classes }}$of vectors: $v'$ and $v''$ where

$\alpha = 1, \beta = 1 \iff v' = x+y$
$\alpha = i, \beta = 1 \iff v'' = ix+y$

Note that, because of how we defined $y$, there are no dependencies on the $\eta$ values. However, $x$ is entirely dependent. In general, therefore, conclude that the $\color{red}{\text{stipulation}}$ of $\eta$ values (which therefore stipulates the $\alpha$ and $\beta$ values...and vice versa) will dictate the exact form of $x$ and $y$. Therefore, $v'$ and $v''$ are most accurately described as:

$\alpha = 1, \beta = 1 \iff v' = x'+y'$
$\alpha = i, \beta = 1 \iff v'' = ix''+y''$

In the first case of $v'$, if $\beta = 1, \eta_3 = 1$ and if $\alpha=1, \eta_1=\frac{1}{\eta_2}$...therefore $x'=\frac{1}{\eta_2}e_1+\eta_2 e_2$...and $y'$ (as stated earlier) simply is: $y'=y$.
In the second case, $\alpha=i$, $i=\eta_1 \eta_2$ therefore, $\eta_1 = \frac{i}{\eta_2}$. As such, $x'' = \frac{1}{\eta_2}e_1 + \frac{\eta_2}{i}e_2$. Similarly, $y''=y$.
At any rate, consider some arbitrary $v'$ (i.e. a an arbitrary vector within that particular $v'$ class), which has its $\alpha=1$ value and $\beta=1$ value) and call it $v'^*$...described, of course, by a corresponding $x'^*$ and $y'^*$. Similarly, consider some arbitrary $v''$ and call it $v''^*$ described by a $x''^*$ and $y''^*$
We thus have the following two equations that are generated:

$1*\overline{1} \langle Qx'^*, y'^*\rangle+1*\bar{1} \langle Qy'^*, x'^*\rangle=0$ and simplifying:

$$\langle Qx'^*, y'^*\rangle+\langle Qy'^*, x'^*\rangle=0$$

$i*\overline{1} \langle Qx''^*, y''^*\rangle+1*\bar{i} \langle Qy''^*, x''^*\rangle=0$ and simplifying:

$$i \langle Qx''^*, y''^*\rangle-i \langle Qy''^*, x''^*\rangle=0 \text{     and then multiply by i}$$
$$- \langle Qx''^*, y''^*\rangle+ \langle Qy''^*, x''^*\rangle=0 $$
Now, an important question arises: WHAT IS THE RELATIONSHIP BETWEEN $v'^*$ and $v''^*$.
I have played around with this for quite some time and I simply cannot figure out how to go about it.
In the aforementioned stack exchange post (Prove that the linear transformations are the same.) the author appears to claim that the $x$s and $y$s from the two final equations are the same. But I am having great difficulty understanding why this is true.
For anyone that stuck through this, I greatly appreciate it! Cheers~

Edit: Given the responses provided below, here is my new attempt at understanding the method used in this proof.
I first need to confirm that an arbitrary vector $v \in V$ of dimension $n$ can be represented as a linear combination of two vectors.
One such construction is the following: $ v = \alpha (\eta_1 e_1 + \eta_2 e_2 + ... + \eta_{n} e_{n}) + \beta (\delta_1 e_1 + \delta_2 e_2+...+\delta_n e_n)$  where $e_1 : e_n$ are basis vectors of $V$.
This can be rewritten as:
$v = \alpha x + \beta y$,
where $x = \eta_1 e_1 + \eta_2 e_2 + ... + \eta_n e_{n}$ and $y = \delta_1 e_1 + \delta_{2} e_{2}+...+\delta_n e_n$
Now, because of how we described vectors $x$ and $y$, any vector in $V$ can be described by the ordered data of $(\alpha, \beta, \eta_1, ...,\eta_n, \delta_1, ..., \delta_n)$...i.e. any vector in $V$.
As such, when we ultimately derive the equation $\alpha\overline{\beta} \langle Qx, y\rangle+\beta\bar{\alpha} \langle Qy, x\rangle=0$, we've effectively created an equation that is "eligible" (valid) for any ordered data $(\alpha, \beta, \eta_1 ..., \eta_{n}, \delta_1, ...,\delta_n)$
!!! NOTE !!! Something that will become important is the fact that even with $\alpha$ and $\beta$ fixed at particular values, cycling through all of the different $\eta_1 : \eta_n$ and $\delta_1 : \delta_n$ combinations will still allow one to describe every vector in $V$.
Consider two arbitrary vectors $v'$ and $v^*$.
Let $v'$ be represented by the ordered data: $(1, 1, \eta_1', ...,\eta_n',\delta_1',...,\delta_n')$
and $v^*$ represented by the ordered data: $(i, 1, \eta_1^*, ...,\eta_n^*,\delta_1^*,...,\delta_n^*)$. In particular, however, let $\eta_1' = \eta_1^*, ..., \eta_n ' = \eta_n^*, \delta_1'=\delta_1^*, ..., \delta_n'=\delta_n^*$
That is to say $x' = x^*$ and $y' = y^*$. For this reason, let $x' = x^*=x$ and $y' = y^*=y$
Two statements can therefore be made about vector $v'$ and $v^*$ :
$\langle Qx, y\rangle+\langle Qy, x\rangle=0$
$i\langle Qx, y\rangle-i\langle Qy, x\rangle=0$
When we solve for these equations, we find that $\langle Qx, y \rangle = 0$.
Now this, in and of itself, simply let's us conclude that the vectors $x$ and $y$ associated with $v'$ and $v^*$ carry the following property: $Q(x)$ is orthogonal to the vector $y$. (i.e. $Q(x) \bot y$).

Importantly, however, note that because of how we constructed our description of $x$ in general ( $x = \eta_1 e_1 + \eta_2 e_2 + ... + \eta_n e_{n}$), $x$ can literally take on any vector in all of $V$.

As such, now holding $y$ fixed, you could cycle through every possible vector in $V$ by using different combinations of $\eta_1, \eta_2,...,\eta_n$. This would mean that $\forall x \in V$, $Q(x) \bot y$ for a fixed non-zero $y$. Now do this again for another, different, non-zero $y$ vector. And again, and again, etc etc.
$Q(x)$ is just a vector though, and the only way $Q(x)$ can always be orthogonal to all non-zero vectors is if $Q(x)$ is the $0$ vector. Thus, $Q$ is the zero map and $T=S$.

REPLY [2 votes]: I think you're making this much more complicated than it needs to be.
The key claim is:

$(*)\quad$ For every $x,y,\alpha$ we have $$\alpha\langle Qx,y\rangle+\overline{\alpha}\langle Qy,x\rangle=0.$$

From this we proceed as follows. Fix $u$ and $v$; we want to show that $\langle Qu,v\rangle=0$.
First, we apply $(*)$ $x=u,y=v,\alpha=1$; this gives $$\langle Qu,v\rangle+\langle Qv,u\rangle=0$$ or $$\langle Qu,v\rangle=-\langle Qv,u\rangle.$$
Next, apply $(*)$ with $x=u,y=v,\alpha=i$; this gives $$i\langle Qu,v\rangle-i\langle Qv,u\rangle=0\implies \langle Qu,v\rangle=\langle Qv,u\rangle.$$
The point is that we're choosing to use the same $x$- and $y$-values here; the whole point of $(*)$ is that it's a universal statement, so we get to apply it to any triple of values we want. We're not forced to use the same $x$- and $y$-values, but we're not prevented from doing so either: the whole point of universal instantiation is that we get to plug in whatever terms we want.