With spectroradiometers, we measure radiation power for a wavelength ${\displaystyle \lambda }${\displaystyle \lambda } and with a certain bandwidth ${\displaystyle d\lambda }${\displaystyle d\lambda }. So that bandwidth is from ${\displaystyle \lambda }${\displaystyle \lambda } to ${\displaystyle \lambda +d\lambda }${\displaystyle \lambda +d\lambda }. Therefore if ${\displaystyle d\lambda =0}${\displaystyle d\lambda =0} we measure radiation power for ${\displaystyle \lambda }${\displaystyle \lambda }, but then radiation power = 0, right? So why is there no ${\displaystyle d\nu }${\displaystyle d\nu } or ${\displaystyle d\lambda }${\displaystyle d\lambda } in the Wikipedia article on Planck's law? Malypaet (talk) 22:04, 13 October 2025 (UTC) [reply ]

The Wikipedia article Planck's law does in fact contain the infinitesimals ${\displaystyle d\nu }${\displaystyle d\nu } and ${\displaystyle d\lambda }${\displaystyle d\lambda }. See, for example, Planck's_law#Correspondence_between_spectral_variable_forms. --Amble (talk) 22:36, 13 October 2025 (UTC) [reply ]

Also note that ${\displaystyle d\lambda }${\displaystyle d\lambda } is an infinitesimal. A spectroradiometer measures radiant flux per area for a segment of the spectrum of finite width ${\displaystyle \Delta \lambda .}${\displaystyle \Delta \lambda .} ​‐‐Lambiam 04:40, 14 October 2025 (UTC) [reply ]

Yes, but since physics is experiment (Feynman), how do we connect ${\displaystyle d\lambda }${\displaystyle d\lambda } to ${\displaystyle \Delta \lambda }${\displaystyle \Delta \lambda }? Planck used also the bandwidth notion with "${\displaystyle \lambda }${\displaystyle \lambda } to ${\displaystyle \lambda +d\lambda }${\displaystyle \lambda +d\lambda }". So if ${\displaystyle d\lambda }${\displaystyle d\lambda } is infinitesimal, the radiation power between ${\displaystyle \lambda }${\displaystyle \lambda } to ${\displaystyle \lambda +d\lambda }${\displaystyle \lambda +d\lambda } is also infinitesimal, right? There is a paradox here. Malypaet (talk) 08:42, 14 October 2025 (UTC) [reply ]

So who can explain me how an infinitesimal power can be a flux of finite energy (${\displaystyle \epsilon }${\displaystyle \epsilon }), per unit of time? Malypaet (talk) 17:57, 14 October 2025 (UTC) [reply ]

You would have to observe the source for an infinite time in order to resolve an infinitesimal band. If the observation time is ${\displaystyle T}${\displaystyle T}, then the best you can do is

${\displaystyle \mathrm {\Delta \lambda } \sim {\frac {\lambda ^{2}}{cT}}}${\displaystyle \mathrm {\Delta \lambda } \sim {\frac {\lambda ^{2}}{cT}}}

(or more simply, ${\displaystyle \mathrm {\Delta f} \sim {\frac {1}{T}}}${\displaystyle \mathrm {\Delta f} \sim {\frac {1}{T}}}) catslash (talk) 10:26, 15 October 2025 (UTC) [reply ]

How can you observe for an infinite time when your life has finite time ? You are using a mathematical concept that cannot be transposed to physics. No experiment has proven that time is infinite. Malypaet (talk) 21:00, 15 October 2025 (UTC) [reply ]

You cannot observe for an infinite time, and that was the premise of my argument, and something that I should have stated explicitly (rather than attempting the English subjunctive - sorry). The consequence of a finite observation time is that ${\displaystyle \mathrm {\Delta } \lambda }${\displaystyle \mathrm {\Delta } \lambda } cannot be infinitesimal. catslash (talk) 21:33, 15 October 2025 (UTC) [reply ]

Infinitesimals are connected to finite quantities by definite integration, as in

${\displaystyle \int _{\lambda _{0}}^{\lambda _{0}+\Delta \lambda }f(\lambda ),円d\lambda .}${\displaystyle \int _{\lambda _{0}}^{\lambda _{0}+\Delta \lambda }f(\lambda ),円d\lambda .}

If ${\displaystyle f}${\displaystyle f} is tame (not changing wildly) in the neighbourhood of ${\displaystyle \lambda _{0}}${\displaystyle \lambda _{0}} and ${\displaystyle \Delta \lambda }${\displaystyle \Delta \lambda } is small, the resulting value is approximately ${\displaystyle f(\lambda _{0})\Delta \lambda .}${\displaystyle f(\lambda _{0})\Delta \lambda .}

And of course also by using the ratio of infinitesimals, as in

${\displaystyle {\frac {d\nu }{d\lambda }}.}${\displaystyle {\frac {d\nu }{d\lambda }}.}

​‐‐Lambiam 11:36, 15 October 2025 (UTC) [reply ]

In your integral: ${\displaystyle \int _{\lambda _{0}}^{\lambda _{0}+\Delta \lambda }f(\lambda ),円d\lambda }${\displaystyle \int _{\lambda _{0}}^{\lambda _{0}+\Delta \lambda }f(\lambda ),円d\lambda }, there is a minimum value for ${\displaystyle \Delta \lambda }${\displaystyle \Delta \lambda } in planck law, where your integral: ${\displaystyle \int _{\lambda _{0}}^{\lambda _{0}+\Delta \lambda }f(\lambda ),円d\lambda =\epsilon /\Delta T}${\displaystyle \int _{\lambda _{0}}^{\lambda _{0}+\Delta \lambda }f(\lambda ),円d\lambda =\epsilon /\Delta T} (${\displaystyle \Delta T=}${\displaystyle \Delta T=}unit of time). That is the minimal power of a photon per unit of time. So there is a paradox there of using infinitesimal.

In fact my initial question was about if giving the law without ${\displaystyle d\nu }${\displaystyle d\nu } or ${\displaystyle \Delta \nu }${\displaystyle \Delta \nu }, has any meaning in physics? In nature, there is no electromagnetic wave without bandwidth (${\displaystyle d\nu }${\displaystyle d\nu } or ${\displaystyle \Delta \nu }${\displaystyle \Delta \nu }). Malypaet (talk) 22:05, 15 October 2025 (UTC) [reply ]

Monochromatic radiation#Practical monochromaticity Malypaet (talk) 22:07, 15 October 2025 (UTC) [reply ]

The tables Radiance expressed in terms of different spectral variables and Spectral energy density expressed in terms of different spectral variables in the section Planck's law § Different forms give 14 different formulations of Planck's law, none of which use any of ${\displaystyle d\lambda ,}${\displaystyle d\lambda ,} ${\displaystyle \Delta \lambda ,}${\displaystyle \Delta \lambda ,} ${\displaystyle d\nu }${\displaystyle d\nu } or ${\displaystyle \Delta \nu .}${\displaystyle \Delta \nu .} ​‐‐Lambiam 17:39, 16 October 2025 (UTC) [reply ]

Thank you very much, I read this article. I am simply surprised that the concept of bandwidth is not found, even though it is used in all measuring devices during experiments. Hence my question here. So I'll continue looking elsewhere. Malypaet (talk) 21:13, 16 October 2025 (UTC) [reply ]

Does the Lambda-CDM model provide an estimate for the size of the universe, or the total amount of matter in it? The article doesn't mention this. It does give some density estimates. Thanks. 2601:644:8581:75B0:560:AAF0:E2ED:F2A1 (talk) 06:48, 15 October 2025 (UTC) [reply ]

It provides an estimate for the size of the Observable universe, but not Universe as a whole. You can derive the total mass in the observable universe if you wish, but for astronomers/cosmologists the densities are far more useful quantities. --Wrongfilter (talk) 07:05, 15 October 2025 (UTC) [reply ]

The universe is infinite. Therefore its size and mass are infinite. Ruslik_Zero 19:23, 15 October 2025 (UTC) [reply ]

Infinity of mass is not a logical consequence of infinity in size. Space could extend forever in an empty expanse beyond a clump of mass confined to a bounded region. ​‐‐Lambiam 20:17, 15 October 2025 (UTC) [reply ]

It is usually thought of as homogeneous. Ruslik_Zero 20:34, 16 October 2025 (UTC) [reply ]

Oh good, somebody knows. But see Shape_of_the_universe#Infinite_or_finite. --Wrongfilter (talk) 19:26, 15 October 2025 (UTC) [reply ]

Wrongfilter, that's interesting, I had hoped that Lambda-CDM would give an estimate of the size of the entire universe (if finite) rather than just the observable part. Ruslik, I had thought it was unknown whether the universe is infinite, but that is was generally assumed finite, such as under the inflationary model. Lambiam, I thought also that infinite size necessarily meant infinite mass as a consequence of general relativity. Anyway, thanks everyone, hmm. I watched some GR videos hoping to get a better handle on this stuff, but they didn't help much. 2601:644:8581:75B0:8134:3D0:DA8E:D90 (talk) 05:02, 16 October 2025 (UTC) [reply ]

See also Friedmann equations#Critical density. Assuming a homogeneous universe, if the average density is above critical, the spatial curvature is positive and the universe is closed. Observations so far are consistent with the universe's density being critical, so with zero average spatial curvature and therefore an infinite universe. Cosmic inflation doesn’t assume a finite universe, I think rather that inflation has stopped only in the local universe (still at least tens of billions of lightyears in size). Icek~enwiki (talk) 10:43, 18 October 2025 (UTC) [reply ]

Currently, the species Platyrhinoidis triseriata-- a cartilaginous fish found in coastal waters off California and northern Mexico-- is identified on Wikipedia as the thornback guitarfish. While I have seen that name used for this species, the majority of sources I have seen call it the California thornback ray, or simply the California thornback. For reference, a Google search gives me 87 results for "thornback guitarfish", but 136 for "California thornback ray" and 107 for "California thornback". Moreover, this fish is not actually a member of the guitarfish family (Rhinobatidae). While that in and of itself is insufficient justification to change the title of an article, the fact that most of the sources I have been able to find associated with this animal do not use that name is worth taking into account. I recall a similar thing happened to the article for the bird Wikipedia currently calls the sun conure; it was originally referred to as a parakeet, in common with most other members of its genus, but this was changed because it was found that most sources call it a conure. Should we do the same for the thornback ray/guitarfish? 135.135.227.26 (talk) 05:06, 19 October 2025 (UTC) [reply ]

This is more a question to be posed on the talk page Talk:Thornback guitarfish, with an alert drawing attention to the question on Wikipedia talk:WikiProject Fishes. ​‐‐Lambiam 07:00, 19 October 2025 (UTC) [reply ]

When I was programming in fortran in the eighties, there were "do loop" instruction:

do i = start, stop, step
! statements
end do

"i" being an integer.
So, if I want to adapt this to calculus, I think I can use this example:
${\displaystyle f(i\cdot \Delta \lambda )=\sum _{i=a}^{b}{\frac {1}{(i\cdot \Delta \lambda )^{2}}}\cdot \Delta \lambda }${\displaystyle f(i\cdot \Delta \lambda )=\sum _{i=a}^{b}{\frac {1}{(i\cdot \Delta \lambda )^{2}}}\cdot \Delta \lambda }
Where i is an integer with unit step between a and b and ${\displaystyle \Delta \lambda }${\displaystyle \Delta \lambda } is a constant ${\displaystyle \in \mathbb {R} }${\displaystyle \in \mathbb {R} }.
Do we have here a kind of Riemann integral that makes sense in mathematics or am I wrong and how?
Malypaet (talk) 21:37, 20 October 2025 (UTC) [reply ]

Do you understand the difference between free variables and bound variables? In a function definition of the form

${\displaystyle f(}${\displaystyle f(}arguments${\displaystyle )~=~}${\displaystyle )~=~}expression,

the arguments must be a list of the free variables in the expression. They become bound in the function definition by virtue of being listed as arguments of the function. Any other seemingly free variables in the expression should be parameters introduced in the context.

It is unusual (but IMO not wrong) to use ${\displaystyle \Delta \lambda }${\displaystyle \Delta \lambda } as the name of a variable, but ${\displaystyle i\cdot \Delta \lambda }${\displaystyle i\cdot \Delta \lambda } is an expression, a product, and cannot serve as a variable name. In the expression that forms the right-hand side, ${\displaystyle i}${\displaystyle i} is a bound variable. It is bound by the summation and has no meaning outside of that expression. Therefore it should not occur inside the function arguments. The variables ${\displaystyle a}${\displaystyle a} and ${\displaystyle b,}${\displaystyle b,} on the other hand, are free. Unless they have been introduced in the discourse, they should reappear in the argument list.

Given the expression that is the right-hand side of the unction definition, the following would make sense as the left-hand side:

${\displaystyle f(a,b,\Delta \lambda )\;.}${\displaystyle f(a,b,\Delta \lambda )\;.}

None of this involves calculus. If ${\displaystyle f}${\displaystyle f} is a univariate Riemann-integrable function on the real interval ${\displaystyle [a,b]}${\displaystyle [a,b]}, we have

${\displaystyle \int _{a}^{b}f(x),円dx=\lim _{\Delta \lambda \downarrow 0}~\sum _{i=0}^{\lfloor (b-a)/\Delta \lambda \rfloor }f(a+i\cdot \Delta \lambda ),円\Delta \lambda ,円.}${\displaystyle \int _{a}^{b}f(x),円dx=\lim _{\Delta \lambda \downarrow 0}~\sum _{i=0}^{\lfloor (b-a)/\Delta \lambda \rfloor }f(a+i\cdot \Delta \lambda ),円\Delta \lambda ,円.}

So it would make sense to define

${\displaystyle R(a,b,\Delta \lambda )=\sum _{i=0}^{\lfloor (b-a)/\Delta \lambda \rfloor }f(a+i\cdot \Delta \lambda ),円\Delta \lambda ,,円}${\displaystyle R(a,b,\Delta \lambda )=\sum _{i=0}^{\lfloor (b-a)/\Delta \lambda \rfloor }f(a+i\cdot \Delta \lambda ),円\Delta \lambda ,,円}

after which we can assert

${\displaystyle \int _{a}^{b}f(x),円dx=\lim _{\Delta \lambda \downarrow 0}~R(a,b,\Delta \lambda ),円.}${\displaystyle \int _{a}^{b}f(x),円dx=\lim _{\Delta \lambda \downarrow 0}~R(a,b,\Delta \lambda ),円.}

Note that this is not a good way to compute integrals. You get much more accurate results for the same computational effort by using Simpson's rule. ​‐‐Lambiam 06:30, 21 October 2025 (UTC) [reply ]

Ok, thanks. I am just trying to find a formulation for data acquisition in a blackbody experiment.

So for each measurement with a bandpassfilter of ${\displaystyle \Delta \lambda }${\displaystyle \Delta \lambda }, for a wavelength between ${\displaystyle a\cdot \Delta \lambda }${\displaystyle a\cdot \Delta \lambda } and ${\displaystyle b\cdot \Delta \lambda }${\displaystyle b\cdot \Delta \lambda }, I try to find a formula in discrete mathematics. For that, I imagine that the wavelenth ${\displaystyle \lambda }${\displaystyle \lambda } can be decomposed in discrete quantity, as follow:

${\displaystyle \lambda =i\cdot \Delta \lambda }${\displaystyle \lambda =i\cdot \Delta \lambda } ( with ${\displaystyle \lambda =f_{1}(i,\Delta \lambda )}${\displaystyle \lambda =f_{1}(i,\Delta \lambda )} ?)

With ${\displaystyle a<i<b}${\displaystyle a.

Then with a summation:

${\displaystyle f_{2}(\lambda )=\sum _{i=a}^{b}{\frac {1}{(f_{1}(i,\Delta \lambda ))^{2}}}}${\displaystyle f_{2}(\lambda )=\sum _{i=a}^{b}{\frac {1}{(f_{1}(i,\Delta \lambda ))^{2}}}}

Here, ${\displaystyle i}${\displaystyle i} is a bounded variable for ${\displaystyle f_{2}}${\displaystyle f_{2}} but a free variable for ${\displaystyle f_{1}}${\displaystyle f_{1}}, right?
Or is there some thing missing, misunderstanding?

Or may be:

${\displaystyle f_{2}(\Delta \lambda ,a,b)=\sum _{i=a}^{b}{\frac {1}{(i\cdot \Delta \lambda )^{2}}}}${\displaystyle f_{2}(\Delta \lambda ,a,b)=\sum _{i=a}^{b}{\frac {1}{(i\cdot \Delta \lambda )^{2}}}}
?
Malypaet (talk) 13:26, 21 October 2025 (UTC) [reply ]

The definition ${\displaystyle f_{2}(\lambda )=\sum _{i=a}^{b}{\frac {1}{(f_{1}(i,\Delta \lambda ))^{2}}}}${\displaystyle f_{2}(\lambda )=\sum _{i=a}^{b}{\frac {1}{(f_{1}(i,\Delta \lambda ))^{2}}}} is not particularly meaningful; the argument ${\displaystyle \lambda }${\displaystyle \lambda } does not occur in the defining expression.

Are you trying to approximate the value of a definite integral by a summation? There is no rule that prevents you from defining just anything, such as

${\displaystyle f_{3}(\Delta \lambda ,a,b)=\sum _{i=a}^{b}{\frac {(i+1)^{5}}{\Delta \lambda }}.}${\displaystyle f_{3}(\Delta \lambda ,a,b)=\sum _{i=a}^{b}{\frac {(i+1)^{5}}{\Delta \lambda }}.}

Whether the function thereby defined makes sense in terms of something you want to achieve is another matter. Are you trying to define a discrete approximation of some definite integral, and if so, which integral? Without further background information there is not much we can say. ​‐‐Lambiam 20:42, 21 October 2025 (UTC) [reply ]

In fact, to deal with data acquisition in a blackbody experiment, I hope to be able to put into discrete form the following formula:

${\displaystyle \int _{\lambda =0}^{\infty }B_{\lambda }(\lambda ,T)\cdot d\lambda =\int _{\lambda =0}^{\infty }{\frac {2hc^{2}}{\lambda ^{5}}}\cdot {\frac {1}{e^{(hc/\lambda k_{\mathrm {B} }T)}-1}}\cdot d\lambda }${\displaystyle \int _{\lambda =0}^{\infty }B_{\lambda }(\lambda ,T)\cdot d\lambda =\int _{\lambda =0}^{\infty }{\frac {2hc^{2}}{\lambda ^{5}}}\cdot {\frac {1}{e^{(hc/\lambda k_{\mathrm {B} }T)}-1}}\cdot d\lambda }

To something like this:

${\displaystyle \sum _{i=10}^{90}B_{\lambda }(\lambda ,T)\cdot \Delta \lambda =\sum _{i=10}^{90}{\frac {2hc^{2}}{(i\cdot \Delta \lambda )^{5}}}{\frac {1}{e^{[hc/(i\cdot \Delta \lambda )k_{\mathrm {B} }T]}-1}}\cdot \Delta \lambda }${\displaystyle \sum _{i=10}^{90}B_{\lambda }(\lambda ,T)\cdot \Delta \lambda =\sum _{i=10}^{90}{\frac {2hc^{2}}{(i\cdot \Delta \lambda )^{5}}}{\frac {1}{e^{[hc/(i\cdot \Delta \lambda )k_{\mathrm {B} }T]}-1}}\cdot \Delta \lambda }

For example with a bandwidth of ${\displaystyle \Delta \lambda =10ns}${\displaystyle \Delta \lambda =10ns}

But the key is how to replace ${\displaystyle \lambda }${\displaystyle \lambda } by ${\displaystyle (i\cdot \Delta \lambda )}${\displaystyle (i\cdot \Delta \lambda )} where ${\displaystyle \lambda =(i\cdot \Delta \lambda )}${\displaystyle \lambda =(i\cdot \Delta \lambda )}?

Or I just have to stipulate that ${\displaystyle \lambda ,円and,円(i\cdot \Delta \lambda )}${\displaystyle \lambda ,円and,円(i\cdot \Delta \lambda )} are equivalent?

Malypaet (talk) 22:28, 21 October 2025 (UTC) [reply ]

Is this the same thing as Abacavir? Trade (talk) 03:38, 21 October 2025 (UTC) [reply ]

It's the hydrochloride salt of it (written in a German-like way). The salt has higher water solubility and probably better storage properties, but it's the same active component. DMacks (talk) 04:21, 21 October 2025 (UTC) [reply ]

Entries for abacavir hydrochloride:

on DrugBank: [1];

on PubChem: [2].

​‐‐Lambiam 07:31, 21 October 2025 (UTC) [reply ]

Given his various health problems, what would have been John F. Kennedy's life expectancy in late 1963, had he not been assassinated? Remember that this is a reference desk, not a speculation desk; I'm looking for projections by scholars and medical professionals, not your own opinions. Google search results are overwhelmed by social media posts; if there's anything out there, I missed it. Nyttend (talk) 20:18, 21 October 2025 (UTC) [reply ]

• I'm getting some promising results from a Google Books search. Unfortunately from the snippet views I can't tell if they're saying Addison's would cut 10 years off his life expectancy, or that he had only 10 years left. This might require a trip to the library. Abductive (reasoning) 10:28, 22 October 2025 (UTC) [reply ]

I realise that an integral is an infinite sum of a variable, for example ${\displaystyle \lambda }${\displaystyle \lambda }, with an infinitesimal parameter as a step value, for example ${\displaystyle d\lambda }${\displaystyle d\lambda }:
${\displaystyle f(\lambda )=\int _{\lambda =0}^{\infty }{\frac {1}{\lambda ^{5}}}d\lambda }${\displaystyle f(\lambda )=\int _{\lambda =0}^{\infty }{\frac {1}{\lambda ^{5}}}d\lambda }
I notice that ${\displaystyle \lambda }${\displaystyle \lambda } is here both a free and bounded variable.
So how can I do this, replacing an integral "${\displaystyle \int _{\lambda =0}^{\infty }...d\lambda }${\displaystyle \int _{\lambda =0}^{\infty }...d\lambda }" by a summation "${\displaystyle \sum _{\lambda =10}^{90}...\Delta \lambda }${\displaystyle \sum _{\lambda =10}^{90}...\Delta \lambda }", where ${\displaystyle \Delta \lambda }${\displaystyle \Delta \lambda } is a finitesimal parameter acting as the step parameter? Malypaet (talk) 09:20, 22 October 2025 (UTC) [reply ]

There's a Math Ref Desk. Abductive (reasoning) 10:17, 22 October 2025 (UTC) [reply ]

Thanks Malypaet (talk) 12:03, 22 October 2025 (UTC) [reply ]

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