I'm having trouble with substituting $\lambda$ with frequency.   The problem:  Show that $$u(f)=\frac{8\pi f^2}{c^3} * \frac{hf}{e^{hf/kT}-1}$$  Where I'm at:  $$\frac{8 \pi f^2}{c^3} * \frac{hf^3}{ce^{hf/kT}-1}$$  As you can see, the left half of my solution matches what I'm supposed to get but there is a $\frac{f^2}{c}$ on the right side that I can't seem to eliminate. I used the relation $\lambda = c/f$ to do all this.  Side note - if anyone could make a tag for Planck's law that's be great. Or something related to this problem, like deriving equations....idk.
22810
I am drawing a blank when it comes to equation transformation. Wikipedia  for the spectral radiance of black body:   First as a function of frequency $\nu$: $$I(\nu, T) = \frac{2 h \nu^3}{c^2}\cdot\frac{1}{e^\frac{h \nu}{k T} - 1}$$ Then as a function of wavelength $\lambda$ : $$I&#39;(\lambda, T) = \frac{2hc^2}{\lambda^5}\cdot\frac{1}{e^\frac{h c}{\lambda k T}-1}$$   And I don't see how they get $\lambda^5$ term. I'm assuming that the transformation is just $\nu \rightarrow c/\lambda$, but that gives $$ \frac{2 h \nu^3}{c^2} \Rightarrow \frac{2 hc}{\lambda^3} \neq \frac{2hc^2}{\lambda^5} $$  Similar transformation happens at other parts in the article also. I'm obviously missing something, likely completely trivial.
45792
I read that it makes a difference whether you calculate $\frac{dE(\lambda) }{d \lambda}=0$ or $\frac{dE(\omega)}{d \omega}=0$ in the sense that the maximum energy density with respect to the wavelength does not coincide with the frequency maximum that one would assume to be at $\omega_{max} = \frac{c}{\lambda_{max}}$. Actually, I do not understand why this is so ( Now, I am only interested in a pure physical explanation, mathematically this is clear and due to the chain rule). Does anybody know how to explain this odd thing?  Somehow I feel that the core of the question has not been completely answered. Although I can easily look up wikipedia, where the need for the chain rule is explained, I am rather interested in understanding where the following argument breaks down, which seems to be why this question has caused some confusion in the past(as you can see by googling this question):  So we have $E(\omega)$ the energy radiated at a given frequency. Now this function has a maximum somewhere, so there is a frequency where a maximal amount of energy is emitted. In other words: If you add the sum of the photon's energies at each frequency that are emitted, you will notice that the maximum is reached at this frquency. Now $E(\lambda)$ tells you basically the same for the wavelength, but again: We know where at which frequency the maximal amount of energy is radiated, so we know the corresponding wavelength.   I think there is a need to explain this.   If anything is unclear, please let me know.
105990
How can we use the idea of negative frequency for the sake of simplifying mathematics if negative frequency does not exist (to my knowledge) in nature ? For example, when plotting the spectra of a Fourier series.
81503
I know that  $A_{[a} B_{b]} = \frac{1}{2!}(A_{a}B_{b} - A_{b}B_{a})$  But how can write $E_{[a} F_{bc]}$ like the above?  Can you provide a reference where this notational matter is discussed?
28918
I proceeded by finding $\operatorname E(\bar X).$ I considered $\bar X$ as a constant and simply got the term itself. This should suggest that $\bar X$ is not an unbiased estimator of $\theta$. That's obvious too cuz it's a constant(from what I assumed). However, I'm also aware that $\mu = \theta$ so that's probably not right.  I have to demonstrate first that $\bar X$ is an unbiased estimator of $\mu$.
71004
I'm working on a problem from Casella and Berger's Statistical Inference. X is distributed as Poisson$(\theta)$ and Y is distributed as Poisson$(\lambda)$, with X and Y being independent. We let U = X + Y and V = Y, and the conditional pdf of V|U is:   $\hspace{15mm}\large f(v|u) = \large \frac{f(u,v)}{f(u)} = \huge \frac{\frac{\theta^{u-v}}{(u-v!)}*\frac{\lambda^ve^{-\lambda}}{v!}}{\frac{(\theta+\lambda)^ue^{-(\theta+\lambda)}}{u!}}$  Apparently this simplifies to   $\hspace{15mm}\Large {u \choose v} (\frac{\lambda}{\theta+\lambda})^v(\frac{\theta}{\theta+\lambda})^{u-v}$  I don't see how the $\Large(\frac{\lambda}{\theta+\lambda})^v$ is possible. I keep ending up with $\Large(\frac{\lambda}{\theta})^v$ when I simplify the problem.
119625
I am trying to calculate  $$Z = \int\limits_{\phi(\beta) = \phi(0) =0} D \phi\ e^{-\frac{1}{2} \int_0^{\beta} d\tau \dot{\phi}^2}$$  without transforming it to the , I can show that $Z = \sqrt{\frac{1}{2\pi \beta}}$. However, I have a problem in obtaining the same result in the Matsubara frequency space: \begin{equation} \phi (\tau) =  \frac{1}{\sqrt{\beta}} \left( \sum_{n} \phi_n \ e^{i\omega_n\tau} \right), \end{equation} with $\sum_n \phi_n =0, \omega_n = \frac{2\pi n}{\beta}$. And \begin{equation} Z = \int \prod_n D\phi_n\ \delta\left(\sum_n \phi_n\right)\ e^{-\frac{1}{2} \sum_n \phi_n \phi_{-n}  \omega_n^2  }  \end{equation} which, I think, vanishes.   I guess the problem lies in the measure. Any comments?  Info: I write the Schulman's derivation in imaginary time here. \begin{eqnarray} Z &amp;=&amp; \int\limits_{\phi(0) =\phi(\beta) = 0} D\phi(\tau) e^{-\frac{1}{2}\int_0^{\beta}d\tau\dot{\phi}^2}\\ &amp;=&amp; \text{lim}_{N \rightarrow \infty} (\frac{1}{2\pi \epsilon})^{(N+1)/2} \int d\phi_1 \dots d\phi_N e^{-\frac{1}{2\epsilon} \sum_{i =0}^N (\phi_{i+1} -\phi_i)^2} \end{eqnarray}  Then, we can use the identity \begin{equation} \int_{-\infty}^{\infty} du \sqrt{\frac{a}{\pi}} e^{-a(x-u)^2}\sqrt{\frac{b}{\pi}} e^{-b(u -y)^2} = \sqrt{\frac{ab}{\pi(a+b)}} e^{-\frac{ab}{a+b}(x-y)^2} \end{equation} to evaluate the sum to be \begin{equation} Z = \sqrt{\frac{1}{2\pi \beta}}. \end{equation}
96765
I am trying to calculate the Fourier transform of $f(t)=Ae^{-i\omega_0 t}$  I'm getting an infinity which is giving me problems. Here are my steps:  $$F(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(t)e^{i\omega t}dt$$ $$=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty Ae^{-i\omega_0 t}e^{i\omega t}dt$$  $$=\frac{A}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{t(i\omega - i\omega_0)}dt$$ $$=\frac{A}{\sqrt{2\pi}(i\omega - i\omega_0)} |e^{t(i\omega - i\omega_0)}|_{-\infty}^{\infty}$$ $$=\frac{A}{\sqrt{2\pi}(i\omega - i\omega_0)}(\infty - 0)$$  Where am I going wrong? Thanks.
88268
Let $\{u_{k}\}_{k=1}^{\infty}$ be a complex number sequence. If $\sum_{k=1}^{\infty}\lambda^{k}u_{k}=0$, for each $\lambda\in \mathbb{D}(0, 1/3)$(where the $\mathbb{D}(0, 1/3)~$denotes an open disc centered at $0$ and radius $1/3$). Then can we conclude $u_{n}=0$, for $n=1, 2, 3...$ ?
28863
If U has a $\chi^2$ distribution with v df, find E(U) and V(U).  By definition, $E(U) =\int^{\infty}_{0} u\frac{1}{\gamma(\frac{v}{2})2^\frac{v}{2}}u^{\frac{v}{2}-1} e^\frac{-u}{2}\,du  =\int^{\infty}_{0} \frac{1}{\gamma(\frac{v}{2})2^\frac{v}{2}}u^\frac{v}{2} e^\frac{-u}{2}\,du$.  How do I integrate this?  Note: This isn't a homework problem.
6271
While studying the 1/f noise, I found this webpage   It gives the following Fourier tranform pairs     However, there are no detailed explanation on how this formula is derived.   Can you help with this?   Thank you!
58021
we consider the problem $$ \begin{cases} x^2 y''+xy'+\lambda y\\ y(1)=y(e)=0 \end{cases} $$ The question is to find the eigenvalues and eigenfunctions for this problem.  I try to put $y(x)=x^r$ and plug it in the equation, then we obtain the caracteristic equation $$r^2+r+\lambda =0$$ In the case $\lambda &gt;0$ we put $\lambda= \alpha^2$ where $\alpha \in \mathbb{R}^\star_+$. Then the caracteristic equation is  $$ r^2+\lambda=0 $$  Case 1. If $\lambda &gt;0$ We put $\lambda = \alpha^2$ where $\alpha \in \mathbb{R}^\star_+$.  Then the genral solution of the equation is  $$ y(x)=C_1 \cos(\alpha x)+ C_2 \sin(\alpha x). $$ We have $y(1)=0 =&gt; C_1=0$ and $y(e)=0 =&gt; C_2 \sin(\alpha e)=0$  $Sin(\alpha e)=0 =&gt; \alpha e = \pi + 2 k \pi, \ k \in \mathbb{Z}^\star$  My solution is correct?
71737
In the below equation I'm writing, 2 minus signs are not visible, why is that?  $\phi (s, \lambda) = C \frac{ 1 + \beta^2 − 2 \beta cos( \frac{2 \pi s}{\lambda} ) }{ \lambda^5 \exp( \frac{ h c }{ \lambda k B T } − 1) } $
65831
I know that what  equals to.   The first question is, how do you get the formula $$\ell_P~=~\sqrt\frac{\hbar G}{c^3}$$ that describes the Planck length? The second question is, will any length shorter than the Planck length be inaccessible? If so, what is the reason behind this?
47333
The Wikipedia  has a derivation of the Klein-Gordon equation.  It gets to this step:  $$\sqrt{\textbf{p}^2 c^2 + m^2 c^4} = E$$  and inserts the QM operators to get  $$\left( \sqrt{ (-i \hbar \nabla)^2 c^2 + m^2 c^4 } \right) \psi = i\hbar\frac{\partial}{\partial t} \psi$$  The article then says     This, however, is a cumbersome expression to work with because the differential operator cannot be evaluated while under the square root sign. In addition, this equation, as it stands, is nonlocal.   To fix this, the first equation is squared instead to get  $$\textbf{p}^2 c^2 + m^2 c^4 = E^2$$  after which the QM operators are inserted and the expression is simplified to get  $$-\hbar^2 c^2 \nabla^2 \psi + m^2 c^4 \psi = -\hbar^2 \frac{\partial^2}{\partial t^2} \psi$$  A couple things I don't understand.  First, are the solutions to this differential equation not exactly the same as the solutions to the first differential equation?  Both sides of the starting equation were squared, so it seems to me that regardless of the particular form of the resulting differential equation, both of them should have the exact same set of solutions.  Secondly, why is the first differential equation cumbersome to work with?  It seems like it would in fact be easier to work with, since the operator under the square root could be expanded in terms of a Taylor series and then you have an equation that is first order in time.  And finally, can someone elaborate on what nonlocal means? The linked article on the Wikipedia page didn't entirely help me understand it.
31727
This is from Fraleigh's First Course in Abstract Algebra (page 82, Theorem 8.16) and I keep having hard time understanding its proof. I understand only until they mention the map $\lambda_x (g) = xg$. Can someone explain this proof step by step? Thank you so much! Here is the proof:  Let $G$ be a group. We show that $G$ is isomorphic to a subgroup of $S_G$. Define a one-to-one function $\phi: G \to S_G$ such that $\phi(xy)=\phi(x) \phi(y)$ for all $x,y \in G$. For $x \in G$, let $\lambda_x: G \to G$ be defined by $\lambda_x (g) = xg$ for all $g \in G$. The equation $\lambda_x (x^{-1} c) = x(x^{-1} c) = c$ for all $c \in G$ shows that $\lambda_x$ maps $G$ onto $G$. If $\lambda_x (a) = \lambda_x (b)$, then $xa=xb$ so $a=b$ by cancellation. Thus $\lambda_x$ is also one to one, and is a permutation of $G$. We now define $\phi: G \to S_G$ by defining $\phi(x) = \lambda_x$ for all $x \in G$. To show that $\phi$ is one to one, suppose that $\phi(x) = \phi(y)$. Then $\lambda_x = \lambda_y$ as functions mapping $G$ into $G$. In particular $\lambda_x (e) = \lambda_y (e)$, so $xe=ye$ and $x=y$. Thus $\phi$ is one to one. It only remains to show that $\phi(xy) = \phi(x) \phi(y)$, that is, that $\phi_{xy} = \lambda_x \lambda_y$. Now for any $g \in G$, we have $\lambda_{xy} (g) = (xy)g$. Permutation multiplication is function composition, so $(\lambda_x \lambda_y)(g) = \lambda_x (\lambda_y (g)) = \lambda_x (yg) = x(yg)$. Thus by associativity, $\lambda_{xy} = \lambda_x \lambda_y$.
83616
My book on quantum mechanics suggests a derivation of the wave equation  $$\left(\Delta - \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \right) \psi(\bar{r},t) = 0$$  from the photon energy-impulse relation  $$E^2 = c^2p^2$$  using the substitutions  $$\bar{p} \to -i\hbar \nabla$$ and $$E \to i \hbar \frac{\partial}{\partial t}$$  I suppose the $\hbar$s are there for the non-photon generalisation, but is there a reason for the minus sign in the $p$ substitution? Removing it would yield the same result.