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January 30

Why is the following value, relativistically invariant?

Here is a simple thought experiment: A photon and a massive particle, both being in a moving train, begin a race, starting from the rear of the train to its front, i.e. in the direction of the train's movement. When the photon arrives at the end of the race (i.e. at the train's front), the photon comes across a mirror, so the photon is reflected back, i.e. it bounces backward, on the same route but in the opposite direction, until this photon meets the massive particle - still on its way to the train's front.

The point at which the photon and the massive particle meet, specifies a part of the train's length. This part is a actually a proper fraction, i.e. its value is between 0 to 1. Intuitively and apparently, this fraction is relativistically invariant, i.e. it does not depend on the reference frame measuring that fraction.

David Mermin (in his book "It's about Time: Understanding Einstein's Relativity", p.37, in the first new paragraph of the page.), wants to assume (rather than to prove) that the fraction is indeed invariant, because he wants to use this assumption for - proving the relativistic velocity-addition formula without using the Lorentz transformations. Apparently Mermin regards this assumption, either as an axiom, or (more likely) as a conclusion of the principle of relativity, because this principle is mentioned (p. 29) as one of his two assumptions, the second one being the constancy of the speed of light.

But I'm looking for a more rigorous justification for his assumption (about the invariance of that fraction), again without relying on the Lorentz transformations nor on the relativistic velocity-addition formula. Unless someone explains to me how Mermin's assumption rigorously follows from the principle of relativity (if it really does). HOTmag (talk) 13:14, 30 January 2024 (UTC)

I'm not sure what you mean by "relativistically invariant" in this case. "X is invariant with respect to Y" means that the value of X doesn't change when Y does. What's Y in this case? Are you saying that the fraction is the same regardless of the speed of the train? Of the initial speed of the massive particle? Both? Something else? PianoDan (talk) 18:22, 30 January 2024 (UTC)

I mean exactly what you mean when you say that the speed of light is relativistically invariant, i.e. it does not depend on the reference frame measuring the speed of light. The same is true for the fraction: it does not depend on the reference frame measuring the fraction. HOTmag (talk) 18:52, 30 January 2024 (UTC)

Ah, OK. What you're looking for here, then is the concept of "Spacetime Interval", given for one dimension as: ${\textstyle (\Delta s)^{2}=(\Delta ct)^{2}-(\Delta x)^{2}}$. For any reference frame the computed value of delta s will be the same for two events, even though the values of delta x and delta t may vary from frame to frame.

Say you're in the reference frame of the train, and you observe the photon to collide with the particle at point P (relative to the train, so P is just the fraction in the problem times the length of the train). The two events here are "photon reaches point P" and "particle reaches point P"

In the reference frame of the train, delta x between those two events is zero, and delta t is ALSO zero. As such, the invariant spacetime interval between those events is also zero, which means it is zero in all frames.

To put it even more simply - events which are separated in space can be perceived to be simultaneous or not, depending on the frame of the observer. But events which are NOT separated in space have a well-defined meaning of "simultaneous," and are simultaneous in all reference frames.

This is also necessary for causality. If shoot a bullet at a pumpkin, different observers can disagree over precisely how long it takes between the shot and the pumpkin exploding. But the fact that the bullet hits the pumpkin AT ALL can't change depending on the motion of the observer, or you get nonsense.

PianoDan (talk) 23:00, 30 January 2024 (UTC)

Thanks. Let me quote you:

"The two events here are 'photon reaches point P' and 'particle reaches point P'...the invariant spacetime interval between those events is also zero, which means it is zero in all frames...events which are NOT separated in space have a well-defined meaning of 'simultaneous', and are simultaneous in all reference frames".

I still want to find out the logical way your argument quoted above can be used for proving that the fraction does not depend on the reference frame, although your argument can't be used for proving the same for the length, so first let me put your argument in my own words:

A given length involves its two endpoints, being two different "events" separated in space by a non-zero interval. However the fraction (I'm asking about) is only specified by a single point (e.g. the one indicating the part of the train's length), while this point is not separated in space by any pair of different points/events.

That said, your argument must still rely on the principle stating that: If all reference frames agree that a given event occurs at a single point in spacetime (e.g. a meeting point), then all of them agree about the exact value of this point, hence about the exact fraction specified by that point. But, what's the exact source/origin of this principle? Does it directly follow from the principle of relativity, if we don't want to rely on the geometrical formula you've indicated at the beginning of your response? HOTmag (talk) 09:17, 31 January 2024 (UTC)

The fraction is determined by three events, not one. It is the ratio of two lengths, each determined by two points, with one point (the collision event) in common. For the fraction to be frame-dependent those two lengths would have to be transformed differently, i.e. length contraction would have to depend weirdly on direction. You may want to look at the derivation of the Lorentz transform from the two axioms and see if there is a shortcut without going via the full transforms. The invariance of the zero-interval is the second axiom (invariance of speed of light). --Wrongfilter (talk) 09:54, 31 January 2024 (UTC)

Thanks. How does the invariance of the zero-interval, follow from the second axiom (invariance of speed of light)? HOTmag (talk) 10:20, 31 January 2024 (UTC)

For light, ${\frac {dx}{dt}}=c$, or $c^{2}\,dt^{2}-dx^{2}=0$. If the speed of light is the same in all reference frames then light follows zero-interval trajectories in all reference frames. --Wrongfilter (talk) 11:01, 31 January 2024 (UTC)

Thank you @Wrongfilter: Regardless of your "principle of existence" (which I find helpful here), do you think the invariance of the zero interval can be used - and is sufficient - for concluding that the fraction is invariant? HOTmag (talk) 11:40, 31 January 2024 (UTC)

"If all frames agree that an given event occurs at a single point, then all of them agree about the exact value of the point" is a tautology. It doesn't need to be derived further, because the two halves of that sentence say the same thing. PianoDan (talk) 15:41, 31 January 2024 (UTC)

It's not a tautology. Think about our meeting yesterday. All of the reference frames agree it was held, at a single point, and at a single moment. However, there may still be a dispute over whether this point at which we met was my home or your home. The same is true for my original question: All agree that the photon and the massive perticle met at a single point and at a single moment. However, perhaps there may still be a dispute over whether this point was at the half of the train's length, or at the quarter of the train's length. Indeed, I'm not claiming there is really a dispute, but I'm still asking if you can prove there isn't one, without your relying on the Lorentz transformations, nor on the Minkowski invariant distance in spacetime, nor on the velocity-addition formula. HOTmag (talk) 17:18, 31 January 2024 (UTC)

There are likely to be other postulates that are implicitly used. For more detail see postulates of special relativity. This thought experiment relies on spatial homogeneity. You can imagine marking the side of the train with tick marks every 1 cm. The number of tick marks between the rear of the train and the meeting point is invariant; the number of tick marks between the meeting point and the front of the train is also an invariant. The distances themselves are not invariant, because the 1 cm spacing is not invariant. Then the question is: in any given frame, are all the tick marks equally spaced? Or is there a frame where some spacings are wider and some are narrower? Spatial homogeneity says that in any frame, the tick spacing must be uniform across the length of the train. And that guarantees the invariance of the fraction you asked about. --Amble (talk) 20:59, 30 January 2024 (UTC)

There needs to be an assumption that the train is not accelerating.  --Lambiam 23:46, 30 January 2024 (UTC)

Thanks. Let me quote you: "the number of tick marks...is...invariant". What basic principle does this assumption follow from? HOTmag (talk) 08:47, 31 January 2024 (UTC)

Not sure how to phrase it — invariance of existence? --Wrongfilter (talk) 09:54, 31 January 2024 (UTC)

Thanks, but it's not only the invariance of existence alone, but rather the invariance of the number of tick marks. My question is, why should we regard this number as invariant? Maybe because this number is dimensionless? HOTmag (talk) 10:20, 31 January 2024 (UTC)

The number can only change if one or more tick marks vanishes or appears. Therefore "invariance of existence" is equivalent to "invariance of number". --Wrongfilter (talk) 10:47, 31 January 2024 (UTC)

Thanks. HOTmag (talk) 11:40, 31 January 2024 (UTC)

x tick marks has the dimension of tick marks. Unlike the velocity-dependent magnetic field, which vanishes leaving only the electric field whenever v=0, the marks are presumed to all be physically present in all reference frames. Modocc (talk) 10:53, 31 January 2024 (UTC)

Thanks. HOTmag (talk) 11:40, 31 January 2024 (UTC)

OP's summary: To sum up, regardless of the Lorentz transformations, and of the Minkowski invariant distance in spacetime, and of the relativistic velocity-addition formula, what's the simplest intuitive principle one had better rely on, for concluding that the fraction I've been asking about is relativistically invarinat, in your opinion?

Actually, what still bothers me is the following fact: a ratio between two velocities - does depend on the reference frame, so how can it be intuitively justifiable to assume that lengths are different from velocities, i.e. that a ratio between two lengths - does not depend on the reference frame? HOTmag (talk) 11:40, 31 January 2024 (UTC)

Because everything that matters for that is linear. In any reference frame an object or photon goes twice as far in twice the time. NadVolum (talk) 13:11, 31 January 2024 (UTC)

Your response has involved time, while my question has not (regardless of the velocity of the reference frame). Let's think about my hand's length and your hand's length. No time is involved (regardless of etc.). So why should we think that the ratio between those lengths (involving no time) does not depend on the reference frame, although a given length per se does depend, and although a ratio between velocities does depend? Can you explain how your argument ("everything that matters for that is linear") explains the difference between the invariant ratio between lengths, versus the variant length per se, or versus the variant ratio between velocities? HOTmag (talk) 14:19, 31 January 2024 (UTC)

So if A, B, C are three points along a lines along a line you're wondering why AB/AC should be the same in one reference system as another? Linearity in the transformation between one reference system and another is what is required for this. Time and distance are transformed linearly in that doubling in one system doubles in the other. It does not matter that velocity changes. AB/AC will be the same in any system and the time for anythng going at a constant speed from A to C to get to B will be AB/AC of the time it takes to get to C. Assuming a physical system leaving A and meeting up at B are the same event in every system and B is the same point. How velocities change given this and the invariance of the speed of light is what one would next come to but the basic asssumption is the linearity of distance and time after transformation. NadVolum (talk) 15:05, 31 January 2024 (UTC)

I take your three words "linearity of distance" as stating that any ratio between two given lengths does not depend on the reference frame. Now I'm asking, if this fact about the ratio is a third postulate of special relativity, or one can derive this fact from any of Einstein's two declared postulates (Principle of relativity and the constancy of speed of light), or one must rely on the spatial homogeneity for deriving this fact, or on...what? Of course without relying on the Lorentz transformations, nor on the Minkowski invariant distance in spacetime, nor on the velocity-addition formula. HOTmag (talk) 17:28, 31 January 2024 (UTC)

Any two distances along the same line. Yes this would have been assumed, the special theory of relativity was a minimal change needed to Newton's laws to accomodate that the speed of light is a constant. Everything changes when one get to general relativity. NadVolum (talk) 21:14, 31 January 2024 (UTC)

Your first sentence is an important correction. Thanx. HOTmag (talk) 22:36, 31 January 2024 (UTC)

What do smartphone strips do?

Thin strips attached to the phone under the cover possibly always glued to the phone. Not the ones that look like it serves no purpose besides an internals protector or loudspeaker membrane. Sagittarian Milky Way (talk) 15:45, 30 January 2024 (UTC)

I don't see how anyone can answer this without more information. Which strips and which brand of phone? Can you supply a pic? The strips around the outside are the antennae. Shantavira|^{feed me} 16:24, 30 January 2024 (UTC)

It seems like you're describing Antenna Strips or NFC Antenna or Wireless Charging Coil or Vibration Motor or some Biometric Sensor Components , can you use a pic? Harvici (talk) 13:45, 31 January 2024 (UTC)

Not wireless charging or biometric. They're actually small tabs of a matte black sheet. One tab covers what looks like a very small port (what the hell is that?). The glue side of the tabs is NOT matte black.Sagittarian Milky Way (talk) 22:00, 31 January 2024 (UTC)

Evolution and most efficient solutions

Many people seem to believe that evolution leads to the most efficient solution to a problem.
Is there any scientific evidence that confirms this? 2A02:8071:60A0:92E0:0:0:0:B32D (talk) 20:51, 30 January 2024 (UTC)

Sounds a bit pointy. Define evolution in context and define efficient. Darwin talked about fitness, not efficiency. Not surprisingly Evolution probably answers your questions. Greglocock (talk) 21:55, 30 January 2024 (UTC)

For any selection criterion, an evolutionary process consisting of many steps of small changes, retaining only changes that correspond to improvements, is like a combination of a random walk in the design space (also called a drunkard's walk) with the optimization technique of hill climbing. If it converges, it does so on a local maximum. There may be a much higher maximum that can only be reached by traversing a valley; local search methods such as hill climbing will not find it. Therefore, the belief appears to be unwarranted.  --Lambiam 23:37, 30 January 2024 (UTC)

Evolution is a pretty good way of finding solutions, but for instance it is pretty unlikey any vertibrate will ever develop eyes without a blind spot like other creatures for example cephalopods have. In computing the 'evolution' can be run past some points where in effect a real creature would quickly die so it can be a very effective ways of finding good solutions - but we have no guarantees of most efficient except in very simple circumstances. NadVolum (talk) 13:22, 31 January 2024 (UTC)

I like your example of the mammalian eye as a local optimum. Greglocock (talk) 06:09, 1 February 2024 (UTC)

The Recurrent laryngeal nerve seems to be a good example of a non-optimal solution (though each step in its evolution may have been). AndrewWTaylor (talk) 13:55, 31 January 2024 (UTC)

I wonder how many is many in "Many people seem to believe...". Most of biology seems to be non-optimal working solutions at all scales, work in progress, but not necessarily to optimize efficiency. Sean.hoyland (talk) 14:28, 31 January 2024 (UTC)

You might look at our article on genetic algorithms#Limitations. But amongst all the reasons why natural selection is not expected perfectly to optimise, let's not lose sight of the fact that many biological structures seem to be extremely well "designed" for the job they perform. I suppose that there are some nice examples from biomechanics but the example I am more familiar with is the excellent quantitative fit between predicted and observed sex ratios.JMCHutchinson (talk) 18:41, 31 January 2024 (UTC)

There are important distinctions to make between biological evolution and genetic algorithms used in engineering. In biology, there is no teleology, it is very chaotic, a mutation that does not select negatively enough but can be considered useless may be preserved, elements of selection are unpredictable, etc. In engineering, the designer is everywhere: the constraints, processes and efficiency evaluations for selection are designed with a purposeful goal. This also means that when trying to find a solution, many iterations can be run and reset. The most efficient one can then be used for the intended application. These methods are only used for problems where it is useful. —PaleoNeonate – 23:56, 31 January 2024 (UTC)

I believe evouloution is fake and NG does to I believe in a different scientific anwser. 23.24.81.162 (talk) 22:09, 1 February 2024 (UTC)

I'm not sure that I understand, but in case it may be useful, WP has an evidence of common descent article. —PaleoNeonate – 05:27, 5 February 2024 (UTC)

Wien's Displacement curves

Stephan-Boltzmann law gives:
$B=T^{4}\sigma$ in watt per square meter.
Expressed to all spectrum in wavelength, you have:
$B=\int _{\lambda =0}^{\infty }B_{(\lambda ,T)}d\lambda$
$B_{(\lambda ,T)}d\lambda$ is in watt per square meter, like $B$.
In frequency you have:
$B=\int _{\nu =\infty }^{0}B_{(\nu ,T)}d\nu$
$B_{(\lambda ,T)}d\lambda =B_{(\nu ,T)}d\nu$ both in watt per square meter, like $B$.
With Planck's law:
$B_{(\lambda ,T)}d\lambda ={\frac {2hc^{2}}{\lambda ^{5}}}\cdot {\frac {1}{e^{\frac {hc}{\lambda k_{b}\theta }}-1}}\cdot d\lambda$
$B_{(\nu ,T)}d\nu ={\frac {2h\nu ^{3}}{c^{2}}}\cdot {\frac {1}{e^{\frac {h\nu }{k_{b}\theta }}-1}}\cdot d\nu$
For $d\lambda =x$, ${\frac {c}{\nu ^{2}}}d\nu =x$
You have:
$B_{(\lambda ,T)}d\lambda ={\frac {2hc^{2}}{\lambda ^{5}}}\cdot {\frac {1}{e^{\frac {hc}{\lambda k_{b}\theta }}-1}}\cdot x$
$B_{(\nu ,T)}d\nu ={\frac {2h\nu ^{5}}{c^{3}}}\cdot {\frac {1}{e^{\frac {h\nu }{k_{b}\theta }}-1}}\cdot x$
So, you see that for a fixed value for x and T, with $\lambda$ from 0 to $\infty$ and $\nu ={\frac {c}{\lambda }}$ you get the same curves for:
$B_{(\lambda ,T)}d\lambda$ and $B_{(\nu ,T)}d\nu$
(when $\lambda =0$, $\nu =\infty$ on the same abscissa, and vice versa),
and, of course, that $\lambda _{peak}$ and $\nu _{peak}$ have the same coordinates.
I specify that in the black body experiment, the power is measured by a bolometer, $\lambda$ and $d\lambda$ are obtained by a crystal filter, therefore of a finite value. One is unable to measure the frequency, it is deduced from the wavelength.

Can you tel me what is wrong in there ?

Malypaet (talk) 23:44, 30 January 2024 (UTC)

If $\lambda$ and $\nu$ are related by the invariant relation $\lambda \nu =c,$ and it is given that, universally,

$\int _{\lambda _{0}}^{\lambda _{1}}B_{\lambda }(\lambda ,T)d\lambda =\int _{\nu _{1}}^{\nu _{0}}B_{\nu }(\nu ,T)d\nu ,$

it follows that the functions $B_{\lambda }$ and $B_{\nu }$ are related by

$\lambda B_{\lambda }(\lambda ,T)=\nu B_{\nu }(\nu ,T).$

This is just mathematics, independent of any physical interpretation. Is this what you mean by "getting the same curves"? (For the rest I didn't understand what you are trying to say; perhaps other editors are more successful in interpreting it.)  --Lambiam 11:51, 31 January 2024 (UTC)

How can you write that, knowing that $\lambda$ and $\nu$ are linked by $\lambda ={\frac {c}{\nu }}$, and $d\lambda$ and $d\nu$ are linked by $d\lambda ={\frac {c}{\nu ^{2}}}d\nu$?

From Planck 1901 article:

Therefore, designate the spatial density of the energy of the radiation belonging to the spectral region $\nu$ to $\nu$ + d$\nu$ by u∙d$\nu$. One must write u∙d$\nu$ instead of E∙d$\lambda$, c/$\nu$ instead of $\lambda$, and c∙d$\nu$/$\nu ^{2}$ instead of dλ.

What I want to write is if you use or not d$\lambda$ and d$\nu$, you get the same curves or different curves.

But which one correspond to experiment?

So, to the nature or reality based on experiment? Malypaet (talk) 13:26, 31 January 2024 (UTC)

Plancks goes from:

$\int _{\lambda _{0}}^{\lambda _{1}}B_{\lambda }(\lambda ,T)d\lambda =\int _{\nu _{1}}^{\nu _{0}}B_{\nu }(\nu ,T)d\nu ,$

to:

${\frac {4\pi }{c}}B_{\lambda }(\lambda ,T)d\lambda ={\frac {4\pi }{c}}B_{\nu }(\nu ,T)d\nu ,$

that is the same as:

$Ed\lambda =ud\nu$

If you have:

$\int _{\lambda _{0}}^{\lambda _{1}}B_{\lambda }(\lambda ,T)\lambda =\int _{\nu _{1}}^{\nu _{0}}B_{\nu }(\nu ,T)\nu ,$

What do you get out of it?
Malypaet (talk) 18:53, 31 January 2024 (UTC)

The latter formula makes no sense, mathematically.  --Lambiam 22:01, 31 January 2024 (UTC)

With me:

$B_{(\lambda ,T)}d\lambda ={\frac {2hc^{2}}{\lambda ^{5}}}\cdot {\frac {1}{e^{\frac {hc}{\lambda k_{b}\theta }}-1}}\cdot d\lambda$

With you:

$B_{(\lambda ,T)}={\frac {\nu }{\lambda }}{\frac {2h\nu ^{3}}{c^{2}}}\cdot {\frac {1}{e^{\frac {h\nu }{k_{b}\theta }}-1}}$

$={\frac {2hc^{2}}{\lambda ^{5}}}\cdot {\frac {1}{e^{\frac {hc}{\lambda k_{b}\theta }}-1}}$

The difference is related to physical dimension:
$B_{(\lambda ,T)}d\lambda$ and $B_{(\nu ,T)}d\nu$ are equal and of the same dimension as B.

When $B_{(\lambda ,T)}$ and $B_{(\nu ,T)}$ have different values and dimension , each of them and to B.

So, how can you put them on the same graph, to compare their curves?

In mathematics, you don't mix cabbages with carrots, do you?
Malypaet (talk) 22:44, 31 January 2024 (UTC)

Assuming $c$ is a known fixed quantity, you can use the same sheet for plotting two graphs, marking the abscissa with a pair of logarithmic scales for each of $\lambda$ and $\nu$ respecting the invariance of $\lambda \nu =c.$ For the ordinate you also need two scales, one for each of the two graphs. This will give two exponential curves, one ascending and one descending.

A completely different approach is to plot the locus of $(B_{\lambda },B_{\nu })$ pairs under the constraint $\lambda \nu =c.$ This will then result in a segment of a hyperbolic curve. A scale marking $\lambda$ and $\nu$ can be placed on the curve, in situ.  --Lambiam 12:27, 1 February 2024 (UTC)

Ok, I understand that.

But $B_{\lambda }$ or $B_{\nu }$ cannot exist whithout $d\lambda$ and $d\nu$ as finite value in experiment.

If $d\lambda =0$ the radiation power measured is 0.

So, for me one would also show the curve with a single ordinate in $watt/m^{2}$ for $B_{\lambda }d\lambda$ and $B_{\nu }d\nu$, and 2 abscissa respecting the relation $\lambda \nu =c$

Just put $d\lambda =1$ and the curve is easy to draw.

Many students are confused with the current presentation because they remain on the relationship $\lambda \nu =c$ in their minds. Malypaet (talk) 22:29, 1 February 2024 (UTC)

My first job was maritime electronics support in Paris. All equipment based on electromagnetic waves had a frequency band around the frequency used. The smaller it was, the more efficient the equipment. But none had a zero band around that frequency, alway with $d\lambda \neq 0$ Malypaet (talk) 13:18, 2 February 2024 (UTC)