I was doing a problem in Optics and I had to use the mirror formula for one specific step i.e 1/u + 1/v = 1/f.  Okay so it's a concave mirror, with the object placed behind the focus. That means V, U and f are all going to be negative(sign convention). Now I know the value of U and f , and the value of V is supposed to be found. Should I substitute V as (-V) or keep it as V itself? I'm not sure if the answer would come with the correct sign.
122424
This is an equation given in my book.     The question is why have they used a negative sign on the LHS?   Now, if you try to derive the mirror equation with simple geometry, you get  1/v +1/u =1/f . I got that equation without applying any signs there!  So what is the need to apply the signs ‘here'?   If i try to derive this equation ( the one in the photo ) i get the exact answer, but with the positive sign in the LHS. WHY we use the negative sign in the derivation, and then  again we use conventions while solving the questions?
123760
I'm trying to prove commutativity of addition for vector spaces, using the axioms for vector spaces. Apparently commutativity can be proven! Im having trouble getting a good feel for what is allowed and what is not. Here's my work so far:  $u+v+u+v = 2(u+v) = 2u + 2v = u+u+v+v = u+(u+v)+v$  Here I just wanna claim that $u+(v+u)+v = u+(u+v)+v$   $\Rightarrow -u+u+(v+u)+v+(-v) = -u+u+(u+v)+v+(-v)$ : here im just adding -u to the right, and -v to the left. Question: is this "adding to both sides" really legit in this context? Why?   Quick help proof: $-v+v = (-1)v+(1)v = (-1+1)v = 0v = 0 = v-v$  And another: $ 0+v = v+(-v) + v = (1)v + (-1)v + v = (1-1)v + v = v = v+0$  We have $0 + (u+v) + 0 = 0+(v+u)+0 \Rightarrow u+v = v+u$  This feels ugly and not at all elegant, especially the great leap "add -u to both sides" feels completely out of place. Do I need more lemmas? Is there a more elegant way?   //not homework or anything, just for my own pleasure, feel free to provide theory, as it is more insightful than solutions. :) Thanks!   EDIT: corrected notation a little.
26390
Is there an error in this question?  Let u = $(1,0,2)$, v = $(2,1,0)$ and w = $(0,2,1)$. Compute  ||u + v - w.||v = v  This is exactly what is in the question (even the dot after w which I don't know what it means).  Thanks
44660
It has been noted $\! { \, }^{\text(1, 2)}$, for instance, that  $$\mathbf{F} = \frac{d}{dt}\!\!\biggl[ \, \mathbf{p} \, \biggr]$$  is true in all contexts.  Likewise,   it is apparently true that  $$\mathbf{F} = - \nabla \Phi := - \frac{d}{d\mathbf{r}}\!\!\biggl[ \, \Phi \, \biggr].$$  Is this, in a nutshell, a sufficient and valid justification for setting (in the corresponding suitable contexts) the  as$\! { \, }^{\text(3)}$  $$\mathbf{\hat p } \propto -i \nabla := -i\frac{d}{d\mathbf{r}}$$  and setting the (potential)  as$\! { \, }^{\text(4)}$  $$\hat \Phi \propto i\frac{d}{dt}$$  and both with the same constant of proportionality, $\hbar$, whereby  $$\mathbf{\hat F} =  \frac{d}{dt}\biggl[-i\hbar\frac{d}{d\mathbf{r}}\biggr] = -\frac{d}{d\mathbf{r}} \biggl[i\hbar\frac{d}{dt}\biggr] \sim \frac{d^2}{dt \, d\mathbf{r}} = \frac{d^2}{d\mathbf{r} \, dt}$$  ?  EDIT (related merely to formalities):  (${ \, }^{\text 1}$: Please note that  had been expressed explicitly in the form     $F = \frac{\mathrm{d} \mathbf{p}}{\mathrm{d} t}$ is true in all contexts.   However, since it appears admissible to take note of an assertion without strictly quoting and clinging to its original literal expression (as had been tacitly presumed already in , and as it seems to be  confirmed) I'd like, as far as unambiguously possible, to express the operation of  consistently by using (a form of) .)  (${ \, }^{\text 2}$: Please note that  whose  to which reference has been made above had been tagged (primarily) as has "".)     ($ { \, }^{\text 3}$: Please note that the indicated expression of the  is explicitly stated there as     ${\bf \hat p } = -i \hbar \nabla$   and     In one spatial dimension this becomes:    $\hat{p}=\hat{p}_x=-i\hbar{\partial \over \partial x}$,   where the nabla symbol ($\nabla$) is related to  .)  (${ \, }^{\text 4}$: Please note that the indicated expression of the  is explicitly stated there as     $\hat{E} = i\hbar\frac{\partial }{\partial t}$.   )
21441
Some one please help me with this question. I am getting -10.5 as the wrong final answer.I found the formula in the following slide.Since the result was coming out in complex number i got its modulus.That is how i ended up with a 10.5 (2 * 5.25).
80972
I've ended up in a debate on a proof for .999˜ being equal to 1 (in the real number system.)  I think the arguments have boiled down to whether or not we can preform variable substitution on one side of the equation, while keeping that variable rather than substituting it on the other side of the equation.  What we are looking at:  Given: x = 0.999˜ (Using the ˜ symbol to represent repeating)     .. A     10x - x     = 9.999˜ - x     9x          = 9.999˜ - x     9(.999˜)    = 9.999˜ - (.999˜)     8.999˜      = 9     ..     OR     .. B     10x - x     = 9.999˜ - x     9x          = 9.999˜ - (0.999˜) // Preform a substitution on ONLY the RHS     9x          = 9     x           = 1     ..  What's so confusing is that both should be right...since 8.999˜ IS equal to 9.  But, since this is a proof you should expect the same representation of the number at the most simplified step (I believe?)  So for simplicity sake my questions are...:   Is the algebra valid in both examples A and B? Does example A act as proof regardless of giving different representations of the same number at the most simplified step?
84931
In case of a concave mirror, when the object is at the focus, my book states that the image is formed at infinity and is highly magnified, real and inverted. But the reflected rays are parallel and will never meet, so no image will ever be formed no matter the distance, right?  How can an image be formed at infinity?
75565
I am trying to figure out how to switch between Minkowski metric tensor sign conventions of (+, -, -, -) to (-, +, +, +) for the  $F^{\alpha \beta}$.  For the convention of (+, -, -, -) I know the contravariant and covarient forms of the electromagnetic tensor are: $$ F^{\alpha \beta} =     \begin{bmatrix}     0 &amp; -\frac{E_{x}}{c} &amp; -\frac{E_{y}}{c} &amp; -\frac{E_{z}}{c} \\     \frac{E_{x}}{c} &amp; 0 &amp; -B_{z} &amp; B_{y} \\     \frac{E_{y}}{c} &amp; B_{z} &amp; 0 &amp;  -B_{x} \\     \frac{E_{z}}{c} &amp; -B_{y} &amp; B_{x} &amp; 0 \\     \end{bmatrix} $$ and $$ F_{\alpha \beta} = \eta_{\alpha \mu} F^{\mu v} \eta_{v \beta} =      \begin{bmatrix}     0 &amp; \frac{E_{x}}{c} &amp; \frac{E_{y}}{c} &amp; \frac{E_{z}}{c} \\     -\frac{E_{x}}{c} &amp; 0 &amp; -B_{z} &amp; B_{y} \\     -\frac{E_{y}}{c} &amp; B_{z} &amp; 0 &amp;  -B_{x} \\     -\frac{E_{z}}{c} &amp; -B_{y} &amp; B_{x} &amp; 0 \\     \end{bmatrix}. $$  Now for the convention of (-, +, +, +) are the contravariant and covariant forms of the electromagnetic tensor just switched from above along with signs?:  $$ F^{\alpha \beta}=      \begin{bmatrix}     0 &amp; \frac{E_{x}}{c} &amp; \frac{E_{y}}{c} &amp; \frac{E_{z}}{c} \\     -\frac{E_{x}}{c} &amp; 0 &amp; B_{z} &amp; -B_{y} \\     -\frac{E_{y}}{c} &amp; -B_{z} &amp; 0 &amp;  B_{x} \\     -\frac{E_{z}}{c} &amp; B_{y} &amp; -B_{x} &amp; 0 \\     \end{bmatrix} $$ and $$ F_{\alpha \beta} = \eta_{\alpha \mu} F^{\mu v} \eta_{v \beta} =     \begin{bmatrix}     0 &amp; -\frac{E_{x}}{c} &amp; -\frac{E_{y}}{c} &amp; -\frac{E_{z}}{c} \\     \frac{E_{x}}{c} &amp; 0 &amp; B_{z} &amp; -B_{y} \\     \frac{E_{y}}{c} &amp; -B_{z} &amp; 0 &amp;  B_{x} \\     \frac{E_{z}}{c} &amp; B_{y} &amp; -B_{x} &amp; 0 \\     \end{bmatrix}~? $$  Basically, I am trying to figure out how to switch between the two sign conventions.
118791
If object A is moving at velocity $v$ (normalized so that $c=1$) relative to a ground observer emits object B at velocity $w$ relative to A, the velocity of B relative to the ground observer is  $$ v \oplus w = \frac{v+w}{1+vw} $$  As expected, $v \oplus 1 = 1$,  as "nothing can go faster than light".  Similarly, $v \oplus -1 = -1$. (same thing in the other direction)  But what if object A is moving at the speed of light and emits object B at the speed of light in the exact opposite direction? In other words, what is the value of $$1 \oplus -1?$$ Putting the values in the formula yields the indeterminate form $\frac{0}{0}$. This invites making sense of things by taking a limit, but $$ \lim_{(v,w)\to (1,-1)} \frac{v+w}{1+vw}$$ is not well-defined, because the limit depends on the path taken.  So what would the ground observer see? Is this even a meaningful question?  Edit: I understand $1 \oplus -1$ doesn't make sense mathematically (thought I made it clear above!), I'm asking what would happen physically. I'm getting the sense that my fears were correct, it's physically a nonsensical situation.
53905
I am trying a derivation of the polar form of an ellipse using vector notation. Beginning with a definition of an ellipse as the set of points in $\vec{R}^2$ for which the sum of the distances from two points is constant, I have $|\vec{r_1}|+|\vec{r_2}| = c$  Thus, $|\vec{r_1}|^2+|\vec{r_1}||\vec{r_2}|=c|\vec{r_1}|$    Choosing a coordinate system (similar to the one in the diagram), such that $\vec{r_1}=\vec{r}=x\vec{i}+y\vec{j}$, and $\vec{r_2}=\vec{F_2 P}=(x-a)\vec{i}+y\vec{j}$  Noticing that $|\vec{r_1}||\vec{r_2}|=\frac{\vec{r_1}\cdot\vec{r_2}}{\cos(\theta)}=\frac{x^2-ax+y^2}{\cos(\theta)}=\frac{r^2-ax}{\cos(\theta)}$  and using $x=r\cos(\theta)$, we get  $r^2+\frac{r^2}{\cos(\theta)}-ar=cr$, thus I get a polar equation for an ellipse of $r=\frac{a+c}{1+\sec(\theta)}$  whereas I expect $r=\frac{a(1-e^2)}{1+e\cos(\theta)}$. Please tell me where this argument takes the wrong turn!
62682
A few months ago I was telling high school students about Fermat's principle.   You can use it to show that light reflects off a surface at equal angles. To set it up, you put in boundary conditions, like "the light starts at A and ends at B". But these conditions by themselves are insufficient to determine what the path is, because there's an extra irrelevant stationary time path, which is the light going directly from A to B without ever bouncing off the surface. We get rid of this by adding in another boundary condition, i.e. that we only care about paths that actually do bounce. Then the solution is unique.  Of course the second I finished saying this one of the students asked "what if you're inside an elliptical mirror, and A and B are the two foci?" In this case, you can impose the condition "we only care about paths that hit the mirror", but this doesn't nail down the path at all because any path that consists of a straight line from A to the mirror, followed by a straight line to B, will take equal time! So in this case the principle tells us nothing at all.  The fact that we can get no information whatsoever from an action principle feels disturbing. I thought the standard model was based on one of those!  My questions are   Is this anything more than a mathematical curiosity? Does this come up as a problem/obstacle in higher physics? Is there a nicer, mathematically natural way to state the "only count bouncing paths" condition? Also, is there a "nice" condition that specifies a path in the ellipse case? What should I have told that student?
109669
Below we have the Fresnel Equation at an interface for S and P polarisation: \begin{align}  r^s &amp;= \frac{n_0 \cos{\theta_0}-n_1\cos{\theta_1}}{n_0\cos{\theta_0} + n_1\cos{\theta_1}} \quad\quad t^s = \frac{2n_1\cos{\theta_0}}{n_0\cos{\theta_0} + n_1\cos{\theta_1}}\\  r^p&amp;= \frac{n_1 \cos{\theta_0}-n_0\cos{\theta_1}}{n_1\cos{\theta_0} + n_0\cos{\theta_1}}\quad\quad t^p = \frac{2n_1\cos{\theta_0}}{n_1\cos{\theta_0} + n_0\cos{\theta_1}}  \end{align} When light approaches the interface at normal incidence these equations simplify to: \begin{align}  r^s &amp;= \frac{n_0 -n_1}{n_0 + n_1} \quad\quad t^s = \frac{2n_1}{n_0 + n_1}\\  r^p&amp;= \frac{n_1 -n_0}{n_1 + n_0}\quad\quad t^p = \frac{2n_1}{n_1 + n_0}  \end{align}  When we apply this to an example where we have an aluminium mirror with refractive index, for example, of 0.93878 + i6.4195.   This would lead for S polarisation to a reflection coefficient -0.9138 - 0.2855i, leading to reflectivity of 95% but a phase shift of -0.90$\pi$. However, for P polarisation a reflection coefficient  0.9138 + 0.2855i is found, leading to the same reflectivity but a phase shift of 0.09$\pi$.  Since we come in under normal incidence, why would the orientation of the polarisation matter? The polarisation S and P are both parallel with the surface, so this would mean if I rotate the mirror I would change the phase of the reflected beam? How does this make sense? What am I missing?   Edit: this also the case for dielectric matter (i.e. real $n$)
64718
I suspect an error in my “Classical Dynamics” lecture notes but my lecturer doesn’t agree with me so I need your help!  We assume that in the $n$-body problem, the force between particle $i$ and $j$ can be derived from a potential i.e  $$\vec{F_{ij}}=-\vec{\nabla_i}V_{ij}$$  where $\vec{\nabla_i}$ is the gradient with respect to position vector $\vec{r_i}$   $V$ would usually only depend on $\Vert\vec{r_i}-\vec{r_j}\Vert$  My lecturer then defines the total energy of the $n$-body system as  $$E=\sum_i \frac{1}{2}m_i\vert\vec{\dot{r_i}}\vert^2 + \sum_{ij}V_{ij}$$  However, I am convinced that this is wrong since we’d be double counting potentials. My lecturer says we should indeed be double counting potential. In my opinion, the expression should read   $$E=\sum_i \frac{1}{2}m_i\vert\vec{\dot{r_i}}\vert^2 + \sum_{1\le{i}\lt{j}\le{n}}V_{ij}$$  For the second equation we also get:  $$\dot{E}=\sum_im_i\vec{\dot{r_i}}\bullet\vec{\ddot{r_i}} + \sum_{1\le{i}\lt{j}\le{n}}\vec{\nabla_i}V_{ij} \bullet \vec{\dot{r_i}}  + \vec{\nabla_j}V_{ij} \bullet \vec{\dot{r_j}} = \sum_im_i\vec{\dot{r_i}}\bullet\vec{\ddot{r_i}} + \sum_{i}\sum_{j\neq{i}} \vec{\nabla_i}V_{ij} \bullet \vec{\dot{r_i}} = \sum_i \vec{\dot{r_i}} \bullet \Bigl(m_i \vec{\ddot{r_i}} + \sum_{j\neq{i}} \vec{\nabla_i}V_{ij} \Bigr) = \sum_i \vec{\dot{r_i}} \bullet \Bigl( m_i \vec{\ddot{r_i}} - \vec{F_{ij}^{tot}} \Bigr) = 0 $$  So this seems reasonable to me. Is this actually correct?  We also have an example of the two-body problem in the notes where we only use $V_{12}$ as the potential, not $V_{12}+V_{21}$
94403
if the position vector of a particle is $\hat{r}=\left(4+3t\right)\hat{\imath}+\left(t^3\right)\hat{\jmath}+\left(-5t\right)\hat{k}$, i want to find at what time this particle passes through the point $\left(1,\:-1,\:5\right)$. I found that $t=-1$ for this particle to pass through that point. What does it mean by that? Why there's negative sign for $t$?
120247
I have a set of concave/convex mirrors/lenses, and I have drawn out all of the ray tracing diagrams for each possible combination and type of image each optical element can produce. When I look at the ray tracing diagrams then reproduce the situation in the real world with myself as the &quot;object&quot;, all the scenarios make sense to me except one: when I as the object stand beyond the focal point of a concave mirror.  The image ends up as &quot;real&quot; and appears to be in front of the mirror. Now I understand that in the situation above, I as the object would stand at 6m and this would result in the image appearing at 12m. So if I placed a wall or screen at the 12m mark, the image would form on that surface (just like a movie theater screen).... at least I think I interpreted that correctly. But I still see myself when I look in the mirror?! If the image is supposed to form and come into focus behind me, what in the world am I seeing when looking at the surface of the mirror? I still see myself, just inverted. The surface of the concave mirror still has a reflection on it. It doesn't look quite like a normal flat mirror, in that the image doesn't really look like it's &quot;inside&quot; the mirror, more like it's right on the surface of it. It's really frustrating because every other possible arrangement of optical elements makes total sense to me, except this one. Any suggestions on how to think about it? Thank you for your time!
66136
Typing \frac{W+W^*}{2} gives a fraction in which the denominator appears to the eye to be in the wrong position. Is there a way to correct this, i.e. place the 2 in under the + (without using arbitrary negative spacing)?
80701
I know there is already a question about EV, but it doens't quite answer my question. I understand EV as a relative measurement, 1 EV being one stop (doubling or halving the amount of light), but what is 0 EV? What does it mean when Nikon claims that a camera can autofocus in -2 EV?