Why is Sunlight Not a Continuous Spectrum
[Solved] Why is spectrum obtained by sunlight, said to be continuous?
- Thread starter Sensebe
- Start date
- #1
Sensebe Asks: Why is spectrum obtained by sunlight, said to be continuous?
My teacher spoke about atomic spectra today, and he explained that, unlike the spectrum obtained by analyzing the sunlight, the spectra of atoms are not continuous.
I have a question about this - even the sunlight has the radiations emitted by the atoms of the elements which compose the sun, still spectra is continuous, which is in opposition to the statement that atomic spectra is discontinuous. So, spectrum obtained by sunlight is continuous even though it atomic spectra. In order to account for spectrum obtained by the sunlight to be continuous and atomic spectra to be discontinuous, can we confirm that sun consists of all those elements (sodium, helium, neon, mercury, etc) which emit the colors of frequency belonging to visible region?
- Jane Wayne
- Computer Science
- Replies: 0
Jane Wayne Asks: After experimentation, how do we learn a final model for deployment?
I have a question regarding learning a "final" model for deployment. Assuming my task is a classification one, my workflow during experimentation is as follows.
- Get the data:
X, y - Split the data:
X_tr, y_tr(training),X_tu, y_tu(tuning) andX_te, y_te(testing) - Learn the model
Mand hyperparametersHonX_tr, y_trandX_tu, y_tu - Validate the performance
PusingMonX_te, y_te
Assuming my experimental performance is great, now I am ready to deploy the model. What do I do at this point? Here is what I have thought of.
- Learn a new model
M_pon all the dataX, yusing the hyperparametersHfrom experimentation. Simply deployM_p. - Learn a new model
M_pand hyperparametersH_pusingX, y.
- AnonUser
- Biology
- Replies: 0
AnonUser Asks: Help me fill out this table about cellular respiration and photosynthesis
To better understand the processes of cellular respiration and photosynthesis, I was assigned to fill out these tables.
Summary of Cellular Respiration:
| Reagents | Products | How is energy generated? | Where does the reaction occur? | What conditions must be met? |
|---|---|---|---|---|
| Glycolysis | ||||
| Pyruvate Oxidation | ||||
| Krebs Cycle | ||||
| Electron Transport Chain | ||||
| Chemiosmosis |
Summary of Photosynthesis:
| Reagents | Products | How is energy generated? | Where does the reaction occur? | What conditions must be met? |
|---|---|---|---|---|
| Light Reaction | ||||
| Dark Reaction (Calvin Cycle) |
Any help is appreciated.
- Kartik Bhardwaj
- Physics
- Replies: 0
Kartik Bhardwaj Asks: Ray Optics Combination of lenses
If there are two thin lenses of any type, separated by some distance say 8cm, Question is asking for effective focal length, Can I assume that in between the two given lenses there are many other lenses made of air and since they are made of air their focal length will be infinity and then we can apply the formula of 1/f(eq)= 1/f1+1/f2? This is not working why?
- An_Elephant
- Physics
- Replies: 0
An_Elephant Asks: Motional EMF inconsistent with Faraday's Law?
Consider a circular metallic coil sliding on a smooth horizontal surface in a region of perpendicular and uniform magnetic field $B$. By faraday's law, the net flux is constant and hence there is no emf induced in the coil.
But now look at individual electrons present in the coil. By the lorentz force, all electrons will experience a net force in the same direction. Since they are constrained to move only on the ring,they will move on one of the half side of the ring. Thus, there will be setup an electric field and an emf is induced. Hence, there will be induced motional emf.
Please explain that emf will be induced or not and which reasoning is incorrect. Thanks!
- alancalvitti
- Mathematics
- Replies: 0
alancalvitti Asks: Holiday names from HolidayCalendar?
HolidayCalendar doc outputs aholiday dates as DateObject list based on a date range.
Is there a method or Entity to map to names, eg Christmas, 4th of July, &c?
- Vladimir
- Mathematics
- Replies: 0
Vladimir Asks: Are there zero entries in the eigenvector corresponding to a simple eigenvalue?
For a real symmetric matrix $M$ and a simple eigenvalue $\lambda$, under which conditions the corresponding eigenvector has no zero entries? Perhaps, this is unconditional and one can provide a proof?
- 0xbadf00d
- Mathematics
- Replies: 0
0xbadf00d Asks: Bound the operator norm of the Fréchet derivative of a Lipschitz function in this setting
Let
- $E$ be a $\mathbb R$-Banach space;
- $v:E\to[1,\infty)$ be continuous;
- $v_i:[0,\infty)\to[1,\infty)$ be continuous and nondecreasing with $$v_1(\left\|x\right\|_E)\le v(x)\le v_2(\left\|x\right\|_E)\;\;\;\text{for all }x\in E,\tag1$$ $$v_1(r)\xrightarrow{r\to\infty}\infty\tag2$$ and $$av_2(a)\le c_1v_1^\theta(a)\;\;\;\text{for all }a>0\tag3$$ for some $c_1\ge0$ and $\theta\ge1$;
Now, let $$\rho(x,y):=\inf_{\substack{c\:\in\:C^1([0,\:1],\:E)\\ c(0)=x\\ c(1)=y}}\int_0^1v\left(c(t)\right)\left\|c'(t)\right\|_E\:{\rm d}t\;\;\;\text{for }x,y\in E.$$ Moreover, let $(\delta,\beta)\in(0,\infty)\times[0,\infty)$ and note that $$d:=1\wedge\frac\rho\delta+\beta\rho\le\left(\frac1\delta+\beta\right)\rho\tag4$$ is a metric equivalent to $\rho$. Let $f:E\to\mathbb R$ be Fréchet differentiable with $f(0)=0$ $$|f|_{\operatorname{Lip}(\rho)}:=\sup_{\substack{x,\:y\:\in\:E\\x\:\ne\:y}}\frac{|f(x)-f
|}{\rho(x,y)}\le1\tag5.$$
Unfortunately, I'm struggling to see how we obtain $(6)$. Let $x\in E$. Clearly, if $\varepsilon>0$, then the Fréchet differentiability of $f$ at $x$ implies $$|f(x)-f-{\rm D}f(x)(x-y)|\le\varepsilon\left\|x-y\right\|_E\;\;\;\text{for all }y\in B_\delta(x)\tag7$$ for some $\delta>0$. From $(5)$ we infer that $$|{\rm D}f(x)(x-y)|\le d(x,y)+\varepsilon\left\|x-y\right\|_E\tag8\;\;\;\text{for all }y\in B_\delta(x).$$ We may use this inequality for arbitrary $y\in E\setminus\{x\}$ by applying it for $\tilde y:=(1-t)x+ty$ with some $t\in\left(0,\delta^{-1}\left\|x-y\right\|_E\right)$, but that doesn't seem to help.
I guess we need to use $(4)$ and observe that for the straight line connecting $x$ and $y$ we obtain $$\rho(x,y)\le\left\|x-y\right\|_E\int_0^1v((1-t)x+ty)\:{\rm d}t\tag9$$ for all $y\in E$. Let $B:=\{y\in E:\left\|x-y\right\|_E\le1\}$. Then $$f:[0,1]\times B\to\mathbb R\;,\;\;\;(t,y)\mapsto v((1-t)x+ty)$$ is bounded and continuous; hence $$F:B\to\mathbb R\;,\;\;\;y\mapsto\int_0^1f(t,y)\:{\rm d}t$$ is bounded and continuous. So, $$\lim_{y\to0}F=\int v((1-t)x)\:{\rm d}t\tag{10},$$ for whatever this is useful to know.
Remark: The claim can be found in equation $(24)$ in https://arxiv.org/abs/math/0602479.
EDIT: I guess something like $(10)$ is needed and is what the authors used in the displayed equation in the proof after equation $(26)$; cf. my related question.
Source: https://solveforum.com/forums/threads/solved-why-is-spectrum-obtained-by-sunlight-said-to-be-continuous.223509/
0 Response to "Why is Sunlight Not a Continuous Spectrum"
Post a Comment