Measurement of first ionization potential of lawrencium reignites debate over periodic table

April 9, 2015 by Bob Yirka report
The periodic table of elements including in the colored block at the bottom the lanthanides (Ln) and actinides (An). The height of each column indicates the relative first ionization potential of the corresponding element. The result obtained for lawrencium (Lr) is shown by the red column. The binding energy of the least bound valence electron in lawrencium is thus weaker than that in all other actinides and all other lanthanides. Credit: Kazuaki Tsukada, JAEA

(Phys.org)—A team of researchers with member affiliations from across the globe has succeeded in conducting a measurement of the first ionization potential of lawrencium. In their paper published in the journal Nature, the team describes how they achieved the feat and what they believe it means for placement on the Periodic Table of Elements. Andreas Türler of the University of Bern offers a News & Views perspective piece on the work done by the team in the same issue.

Lawrencium is an that does not exist in nature, scientists create it in the lab and use it for study, though the process is difficult and the result lasts for only a few seconds. In this new effort, the researchers used a known technique to create the element, then devised a way to measure its first —which describes the amount of energy required to cause one atom of it to be turned into an ion by knocking off one of its electrons. It is this measurement that forms the basis of element placement on the periodic table.

To make this measurement for lawrencium (named for Ernest Lawrence), the team created some samples by shooting boron at a bit of californium—doing so caused a few atoms of a lawrencium isotope to come into existence. Those atoms were then captured using a cadmium mist iodide and placed on a piece of metal which was then heated to 2,700 kelvin—hot enough to knock electrons off of some of the atoms. After that, all it took was summing the atoms that were ionized and calculating the energy it took to make it happen. Doing so revealed that it took just 4.96 electronvolts to ionize one of the atoms, an unexpectedly small amount, which likely means that lawrencium's outermost electron is very loosely bound, which means, that placing the element where it has been put on the periodic table up till now, might not work.

The ionization energy of heavy lanthanides (black) and actinides (red) with the current result for lawrencium (Lr). The filled circle symbols represent values recorded in experiments, the ring symbols are estimated values. The two values for lawrencium are in excellent agreement, emphasizing the close correspondence between the theoretical expectations and experimental findings. Credit: Tetsuya K. Sato, JAEA

As it stands now, the is arranged in columns and blocks which are based on how an atom's electrons are put together, with different blocks relating to different types of orbital—the shape created by the path of their orbits. Lawrencium, at this time, appears to have a dumb-bell shape. These new findings create conflicting views on where the element should be placed on the table and has reignited debate on the way the table is structured in general.

The gray tantalum tube surrounded by two heating elements in the center of the picture is part of the recently developed surface ion source installed in the JAEA-ISOL system in the JAEA tandem accelerator. Credit: Tetsuya K. Sato, JAEA

Explore further: Superheavy chemistry, one atom at a time

More information: Measurement of the first ionization potential of lawrencium, element 103, Nature 520, 209–211 (09 April 2015) DOI: 10.1038/nature14342

Abstract
The chemical properties of an element are primarily governed by the configuration of electrons in the valence shell. Relativistic effects influence the electronic structure of heavy elements in the sixth row of the periodic table, and these effects increase dramatically in the seventh row—including the actinides—even affecting ground-state configurations. Atomic s and p1/2 orbitals are stabilized by relativistic effects, whereas p3/2, d and f orbitals are destabilized, so that ground-state configurations of heavy elements may differ from those of lighter elements in the same group. The first ionization potential (IP1) is a measure of the energy required to remove one valence electron from a neutral atom, and is an atomic property that reflects the outermost electronic configuration. Precise and accurate experimental determination of IP1 gives information on the binding energy of valence electrons, and also, therefore, on the degree of relativistic stabilization. However, such measurements are hampered by the difficulty in obtaining the heaviest elements on scales of more than one atom at a time. Here we report that the experimentally obtained IP1 of the heaviest actinide, lawrencium (Lr, atomic number 103), is 4.96 electronvolts. The IP1 of Lr was measured with 256Lr (half-life 27 seconds) using an efficient surface ion-source and a radioisotope detection system coupled to a mass separator. The measured IP1 is in excellent agreement with the value of 4.963(15) electronvolts predicted here by state-of-the-art relativistic calculations. The present work provides a reliable benchmark for theoretical calculations and also opens the way for IP1 measurements of superheavy elements (that is, transactinides) on an atom-at-a-time scale.

Press release

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shavera
5 / 5 (7) Apr 09, 2015
Doing so revealed that it took just 4.96 electronvolts to ionize one of the atoms, an unexpectedly small amount


The measured IP1 is in excellent agreement with the value of 4.963(15) electronvolts predicted here


What is this article trying to say? The science article, and the graphs included seem to imply that the ionization energy is spot-on with the theoretical prediction. But the public article seems to imply that this is some surprising discovery about an abnormally low ionization energy?
Dethe
1 / 5 (1) Apr 09, 2015
The surprise essentially is, the Lawrencium behaves like the element from the left side of period table, despite it's an element from right side of it.
Dethe
1 / 5 (1) Apr 09, 2015
The electrons in Lawrencium should occupy the p-orbitals, but because they revolve the atom at large distance (as it's common for electrons around extreme heavy and large atoms, like the Lawrencium), they do move with relativist speed and gain mass according to E=mc^2 formula. Such a heavy electrons therefore tend to revolve atom at proximity and they occupy rather the less elongated f-orbitals beneath it. But because they don't actually belong there, the other electrons from f-orbitals tend to expel the excessive electron outside of f-orbital like any other intruder. Which leads to result, that these electrons are loosely bound to atoms and they can be removed easily with oxidizing agents, like the water. As a consequence the Lawrencium is nearly as reactive with water, like the sodium - despite it sits near noble elements inside of periodic table. Such a relativist anomalies aren't so rare in periodic table though - just at the case of extreme heavy Lawrencium they become prominent.
shavera
5 / 5 (6) Apr 09, 2015
But how is that "reigniting debate over the periodic table?" It's what we expect when we include relativity into calculations. Like... kinda exactly what we expect.

I guess the broader question is why does the periodic table have the shape it does? Yes, historically it came from how similar elements were chemically (more closely related to ionization energies and the like)... but we've been ordering it by proton count and shaping it by 'last' orbital for a while now...
shavera
5 / 5 (6) Apr 09, 2015
they do move with relativist speed and gain mass according to E=mc^2 formula.


Just an important caveat. Relativistic motion does _not_ mean their mass changes. It means momentum isn't linear with v. Yes, for nearly all intents and purposes we can treat it *as if* mass increases, but it's very important to keep in mind that such a trick is just a hand-wave, and not the reality of the underlying physics. /pedantic_mode
Dethe
1 / 5 (2) Apr 09, 2015
how is that "reigniting debate over the periodic table?"
It's sorta scientific journalism. Of course, given by atomic number the Lawrencium has its order in the periodic table firmly warranted. But the elements in periodic table are also colored by groups according to orbitals, which their outer electrons occupy - and the Lawrentium behaves like d-orbital element instead of p-orbital one. It should get a different color, not location inside of periodic table.
Relativistic motion does _not_ mean their mass changes. It means momentum isn't linear with v.
Yes, it's their mass with all its consequences - just the rest mass remains the same. You may imagine according to dense aether model, that the fast moving electrons are surrounded with wake wave, which surrounds these electrons like the dense coat and it does increase their inertia. So that the relativist portion of mass doesn't belong to electron itself - only to vacuum - but its inseparable from electron anyway.
Dethe
not rated yet Apr 09, 2015
Just the shrinking of orbital paths of fast moving electrons illustrates clearly, that the relativist mass doesn't only increase electron momentum, but also their centripetal force and inertia. The relativist electrons are really more heavy and they also exhibit higher gravity at distance. Otherwise the relativist mass would violate the equivalence principle. When the relativist deny the massive character of relativist mass (in an effort to pretend, that the photons aren't massive and do follow the Lorentz symmetry in this way), they actually violate another postulates of general relativity. The Lorentz symmetry postulate of special relativity therefore isn't consistent with equivalence principle postulate of general relativity. After all, if the general relativity would be fully consistent with special relativity, we could replace both theories with single one - isn't it true? But we cannot do it, as both theories have different application scopes, which don't overlap mutually.
shavera
5 / 5 (7) Apr 09, 2015
Yes, it's their mass with all its consequences - just the rest mass remains the same.


Again, not even remotely true. Much less the insanity that follows about dense aether. Mass is mass. There's no such thing as rest mass and relativistic mass. That's just an old, outdated way of teaching relativity.

Specifically mass is the lorentz-invariant measure of the energy-momentum 4-vector. It's a constant value for all observers, regardless of their relative motion.

The reason why I felt the need to be pedantic about it is because it tends to lead to people making up weird 'metaphysics' explanations like "dense aether wave wake" or whatever bs. There's only one 'mass' in physics, and it's the E-P scalar magnitude. Period. All observers will always agree on this measure, whether the particle is in motion relative to them or not.
shavera
5 / 5 (6) Apr 09, 2015
The relativist electrons are really more heavy and they also exhibit higher gravity at distance.


Again... no. They don't. If an object is in motion, then its momentum, as a part of the stress-energy tensor feeds into general relativity calculations. It does not mean that they exhibit 'higher gravity' or any such thing. (The result actually is that gravity "points" to the extrapolated 'present' location of a moving body plus corrections on the order of beta^4, ie, true so long as they're not... too close to relativistic speeds. Then things get more complicated.)

If what you said was true, then our sun would gravitationally pull us stronger than its real mass based upon its motion relative to the galaxy's center. And it would pull at yet a *different* strength based on its motion relative to Andromeda's galaxy. In fact, we couldn't define one gravitational force of the sun from all the different ways we could measure its motion, and thus its mass (if rel'vstic mass exist)
Dethe
1 / 5 (2) Apr 09, 2015
Mass is mass. There's no such thing as rest mass and relativistic mass. That's just an old, outdated way of teaching relativity
It's more physically valid actually. The mass is responsible for gravitational effect - and in this aspect the relativist mass doesn't differ from rest mass. Otherwise it would violate the equivalence principle, which clearly states, every change in inertia must have it's counterpart in change of gravity field. From strict perspective of aether model the equivalence principle gets also violated, when the relativistic mass becomes very high or the rest mass very low (compare the MOND theory) - but it has no relevance in this moment.

If you're telling me, that the relativistic mass doesn't increase gravitational force - then you're denying the very same theory, which you're trying to explain "better".
Dethe
1 / 5 (2) Apr 09, 2015
then our sun would gravitationally pull us stronger than its real mass based upon its motion relative to the galaxy's center
Nope, the Sun doesn't move with respect to galaxy center - it's distance from it remains the very same. You have it opposite. The relativist effect of wake wave and relativist mass is solely relative too. When you're moving fast, then you'll appear inside of dense wake wave too, so that every object which is moving together with you doesn't exhibit any mass increasing. You're sitting inside of dense blob and the observed object too - so that as the result you'll notice any difference.
shavera
5 / 5 (7) Apr 09, 2015
You misunderstand General Relativity, I surmise. May I recommend Hartle's excellent undergrad text "Gravity"? It's really quite accessible if you have a basic understanding of classical mechanics, matrix mathematics/linear algebra. You'll see that GR is perfectly compatible with SR.

I think perhaps you've only heard bits and pieces of SR and GR, and maybe misunderstand what one or the other is saying because you don't have the full story, and thus think they are somehow incompatible. But they are completely compatible, and it's very easy to show that the laws of SR fall out right away from GR.
Dethe
1 / 5 (1) Apr 09, 2015
You'll see that GR is perfectly compatible with SR
Only when the artificial transforms like the " momentum, as a part of the stress-energy tensor" are applied. But as you can see at the above examples, these Lorentz symmetry saving transforms do violate the equivalence principle in their very consequences. Both theories actually have both different application scope - the SR applies only to inertial systems, whereas the GR applies only to non-inertial systems. So far we have no formal theory, with would be able to describe both with the single postulate set.
shavera
5 / 5 (8) Apr 09, 2015
Dethe, can you define what you mean by "the equivalence principle"? I think you misunderstand what physicists mean by the phrase, vs what you think it means.

For a physicist, the equivalence principle means that the measurement of acceleration is indistinguishable from the measurement of gravitational attraction.

If a body is moving relative to you (say flying by in a straight line, not on a direct path toward or away from you) your gravitational attraction to that body is very nearly only a function of the mass of that body, not its speed relative to you. (ie, the same body, stationary at the same location would attract you just the same) That's well known and well observed.

Really, before you get into all the other alternatives to science, it's best to understand the actual case science is trying to make. There are plenty of ways to get a better education about what GR actually says, rather than attacking a strawman of what you think it says.
Dethe
1 / 5 (3) Apr 09, 2015
You misunderstand General Relativity, I surmise
I do understand exactly where the problem of formal relativists is. The formal proponents of relativity believe, that the photon is moving around black hole with constant speed even under the situation, when its radius gets very small, so that this photon effectively stays at rest. When something looks sitting at place, than its sitting at rest by all observational criterion - well, and the photons shouldn't sit at rest, isn't it true? This perception not only gets wrong geometrically and psychologically - it also gets wrong physically, because the localized energy gains its own mass and energy density, so it does lead into deviations from constant speed of light not only from extrinsic perspective, but also from intrinsic one gradually. Which manifest itself like various quantum and/or dark matter effects. This is the consequence of the fact, our space-time isn't actually quite flat - but full of quantum fluctuations.
Dethe
1 / 5 (2) Apr 09, 2015
, the equivalence principle means that the measurement of acceleration is indistinguishable from the measurement of gravitational attraction
So that the change of (rotational) momentum of fast electron around Lawrencium atoms should also manifest itself with change of its gravitational attraction. The shrunken Lawrencium atoms with fast electrons should be heavier than the atoms of the same amount of electrons, but larger ones. Which implies, that the momentum of relativistic heavy electron isn't the only thing which changes there - it's also change of mass. The same result follows from E=mc^2 formula btw, as the electrons inside of shrunken atoms must move faster, therefore such an atoms have higher energy content, they should be also heavier. Therefore your "momentum, as a part of the stress-energy tensor" assumption not only violates the equivalence principle - it also violates the mass-energy equivalence - which could be perceived as an even greater problem.
shavera
5 / 5 (9) Apr 09, 2015
none of what you typed is what "the formal proponents of relativity believe." A photon cannot orbit a black hole at less than 1.5 times the Schwarzschild radius (this itself is an unstable orbit, so nearly all orbits either plunge in or diverge away). Thus the case of a photon orbiting very near the point at the center of a black hole is entirely unphysical, and no one who knows GR would consider it. The photon could continue to fall toward the center. That's all. No pseudo-rest orbit

This is what I mean, and where I'll end my comments. The windmill you're fighting? Not what GR actually says. You don't understand GR well enough to tell others what's wrong with it. If you think it's wrong, you're free to... but you should really invest in learning the actual theory. Then, once you actually know it, feel free to criticize it. Otherwise the thing you *call* GR is entirely unrelated to what every scientist means when they talk about it.
Dethe
1 / 5 (3) Apr 09, 2015
A photon cannot orbit a black hole at less than 1.5 times the Schwarzschild radius
This plays no role here, as the Schwarzchild radius can be arbitrarily small according to vanilla general relativity. The general relativity really has no problem with trapping of photons at the place smaller, than the wavelength of this photons - thus forcing these photons to remain at rest. It was Hawking in 1973, who pointed to the fact, that this situation gets nonphysical from quantum mechanical perspective and that every photon larger than the event horizon radius must actually leave the black hole and to evaporate - soon or later.

Not what GR actually says. You don't understand GR well enough to tell others what's wrong with it.
?? This is what I'm doing here last hour... :-) I of course know about modern formalism of general relativity, but this formalism has been introduced artificially into it with no connection the GR postulates for the sake of different SR postulates.
Dethe
1 / 5 (3) Apr 09, 2015
I perfectly understand the queue of thinking of modern theorists: they believed, that the special relativity (zero field or curvature theory) must give the same predictions like the general relativity at the low field/curvature limits due to correspondence principle. So that they modified the GR formalism in such a way, this consistency has been finally achieved - but they didn't checked the consistency of the result with equivalence principle and mass-energy equivalence. So that we have a modified general relativity valid for low-field/curvature situations (like the photons) - but this formulation isn't consistent with high field/curvature situations, from which the dark matter, energy and another inconsistencies of general relativity follows. Not to say about consistency with quantum mechanical description of photon itself. It's because such a modified relativity is Maxwell-Lorentz compliant and it doesn't recognize the existence of photons as such. The photon isn't Maxwell wave.
DarkLordKelvin
3.4 / 5 (5) Apr 11, 2015
The same result follows from E=mc^2 formula btw, as the electrons inside of shrunken atoms must move faster, therefore such an atoms have higher energy content, they should be also heavier.
Dethe, you really should follow shavera's advice and read up on this stuff, that quote above implies an ignorance about SR that is so severe it casts a very large amount of doubt on every statement you make about SR or GR. Apparently you are unaware that E=mc^2 is ONLY valid for describing energy equivalence of invariant mass. For bodies moving at relativistic speeds, you must use the complete formula: E^2=(pc)^2 + (mc^2)^2, which explains how the electron can have greater energy and momentum, while maintaining the same rest mass. For Lawrencium atom/ion, the invariant mass is sum of the total relativistic energies of all component particles in the COM frame, divided by c^2. So there is no contradiction having "heavier" Lw atoms composed from "fast" electrons w/normal rest mass value.
adam_russell_9615
5 / 5 (1) Apr 12, 2015
"For a physicist, the equivalence principle means that the measurement of acceleration is indistinguishable from the measurement of gravitational attraction."


I dont want to interfere with your discussion, as it already seems to be a couple levels over my head.

But Ive always questioned this "equivalence principle". People talk about it as if it is exactly true but I dont see how it could be. If you are standing on the surface of the earth the gravity is 1G, but if you climbed to the top of a very tall building it would be less - based on the inverse square rule of distance. But if you were in a very tall space ship accelerating at 1G the pseudo gravity would be 1G everywhere. The two cases arent really equivalent. I could measure the difference. Am I correct, and if so then how do you justify calling it equivalent when it is merely similar?
Mike_Massen
1 / 5 (1) Apr 12, 2015
adam_russell_9615 asked
.. if you were in a very tall space ship accelerating at 1G the pseudo gravity would be 1G everywhere.The two cases arent really equivalent. I could measure the difference. Am I correct, and if so then how do you justify calling it equivalent when it is merely similar?
All principles have certain 'bounds' & not taken too far, ie Principles are NOT automatically absolute & subject to caveats, some self-evident some to non-intuitive maths,. Eg. SR & GR are (far) outside our [daily] experience & thus "don't feel right" but proven correct, this observation is just my philosophy, so I will pin down "Equivalence" by my interpretation :-)

"I am in a closed box (& instruments) in a spaceship or on Earth subject to the same 1G with instruments. Equivalence suggests those instruments cannot prove to me which reference frame (RF) I am in."

Bear in mind. The 1G is EQUIVALENT "at a particular" point in either RF, not higher or lower !

cont
Mike_Massen
1 / 5 (1) Apr 12, 2015
continued
@adam_russell_9615

& bear in mind on Earth we are subject to forces not present in the ideal example of Equivalence
http://en.wikiped...rinciple

Eg Coriolis
http://en.wikiped...s_effect

This situation also occurs, in terms of arithmetically separating effects re Special/General Relativity in GPS correction
https://en.wikipe...g_System

In particular regarding earlier tests, this is of interest:-
https://en.wikipe...periment

Performed prior to GPS and elsewhere noteworthy in terms of what types of issues to expect & just WHY GPS has the additive issue re SR & GR handled - which works extremely well.

I recall reading a note of Equivalence interpretations re GPS years ago but, can't locate it, suffice to say the systematic issue re GR, SR & Equivalent was "well satisfied" experimentally, mathematically & philosophically within Einsteins postulates...
adam_russell_9615
not rated yet Apr 12, 2015
@Mike_Massen
So you're saying that the fact that a gravitic field has tidal force but acceleration does not is irrelevant to the equivalence principle?
DarkLordKelvin
2.3 / 5 (3) Apr 12, 2015
@Mike_Massen
So you're saying that the fact that a gravitic field has tidal force but acceleration does not is irrelevant to the equivalence principle?


Tidal forces depend on differential gravitation over large distances that cannot be discerned solely from observations of one's local environment. The point of the equivalence principle is that, from within a given body's inertial frame, it is impossible to distinguish whether a *local* force experienced is due to gravity, or some other form of external acceleration; in other words, you can always contrive an externally applied acceleration that would produce an equivalent "local experience" as does a given gravitational field. Thus, the equivalence principle allows gravity to be distinguished from external acceleration by observations of what is happening *elsewhere*, as is the case with tidal forces.
adam_russell_9615
not rated yet Apr 12, 2015
Tidal forces depend on differential gravitation over large distances that cannot be discerned solely from observations of one's local environment. The point of the equivalence principle is that, from within a given body's inertial frame, it is impossible to distinguish whether a *local* force experienced is due to gravity, or some other form of external acceleration; in other words, you can always contrive an externally applied acceleration that would produce an equivalent "local experience" as does a given gravitational field. Thus, the equivalence principle allows gravity to be distinguished from external acceleration by observations of what is happening *elsewhere*, as is the case with tidal forces.


Technically there is tidal force even between your head and your heels. It doesnt HAVE to be a large distance. With the right instruments I definitely could measure the difference.
DarkLordKelvin
2.3 / 5 (3) Apr 12, 2015
Tidal forces depend on differential gravitation over large distances that cannot be discerned solely from observations of one's local environment. The point of the equivalence principle is that, from within a given body's inertial frame, it is impossible to distinguish whether a *local* force experienced is due to gravity, or some other form of external acceleration; in other words, you can always contrive an externally applied acceleration that would produce an equivalent "local experience" as does a given gravitational field.
Technically there is tidal force even between your head and your heels. With the right instruments I definitely could measure the difference.
Think so? What's the force in newtons then? How's it compare with the force of a single gas molecule impacting your detector?

None of that really matters of course; even if you could measure it, it would be non-local, and thus not covered by the equivalence principle.
Eikka
3 / 5 (2) Apr 12, 2015
Technically there is tidal force even between your head and your heels. It doesnt HAVE to be a large distance. With the right instruments I definitely could measure the difference.


And for a brief moment in your accelerating spaceship, your head would also be accelerating at a slower rate to your feet because the stuff in between is compressing and not transmitting the force instantly, and for that moment you couldn't tell the difference.

DarkLordKelvin
2.3 / 5 (3) Apr 12, 2015
Technically there is tidal force even between your head and your heels. It doesnt HAVE to be a large distance. With the right instruments I definitely could measure the difference.


And for a brief moment in your accelerating spaceship, your head would also be accelerating at a slower rate to your feet because the stuff in between is compressing and not transmitting the force instantly, and for that moment you couldn't tell the difference.


That's a creative answer, but I don't think it applies, since the equivalence principle is about steady state conditions, where everything is under the same, constant force due to acceleration or gravity.
abecedarian
not rated yet Apr 13, 2015
Technically there is tidal force even between your head and your heels. It doesnt HAVE to be a large distance. With the right instruments I definitely could measure the difference.


And for a brief moment in your accelerating spaceship, your head would also be accelerating at a slower rate to your feet because the stuff in between is compressing and not transmitting the force instantly, and for that moment you couldn't tell the difference.


Now I want to watch "The Black Hole"... but I think it went the other way where since their head was closer to the BH, they stretched.

Yeah, I know... irrelevant.
adam_russell_9615
not rated yet Apr 13, 2015
Think so? What's the force in newtons then? How's it compare with the force of a single gas molecule impacting your detector?

I didnt mean I could measure it, but that the difference could hypothetically be measured.

None of that really matters of course; even if you could measure it, it would be non-local, and thus not covered by the equivalence principle.


So local is only defined for a single non-dimensional point in space?
Does that even make sense considering the uncertainty principle? I mean, iiuc, not only can we not measure position with 100% accuracy, but position with 100% accuracy doesnt exist really. So what does it mean to define local as meaning only for an infinitesimal space? Its not real. Why define a physics principle on something that cannot actually be measured? And if position is fuzzy does that not mean that the curve of gravity becomes inherent even in that one 'position'?
Mike_Massen
1 / 5 (1) Apr 13, 2015
adam_russell_9615 asked
@Mike_Massen
So you're saying that the fact that a gravitic field has tidal force but acceleration does not is irrelevant to the equivalence principle?
Gravitational field has a gradient & in effect in terms of equivalence would be a varying acceleration rate which is a dynamic, outside definition.

ie.
Equivalence Principle (EP) relates to a static & does hold do you read link I supplied ?

Can craft maths to extend conventional (static) EP to dynamic one where position one takes in say a cylinder subject to feedback alters rate of acceleration.

Bear in mind the EP by definition is a general PRINCIPLE & not aspect as a law except as clearly stated on link I supplied "Einstein's statement..." ie qualified specifically re EQUALITY of mass, any extension may not hold, please read details re Einstein...

Fine as a static & thus can be built on to the situation where you may alter acceleration depending upon distance from source...
Mike_Massen
1 / 5 (1) Apr 13, 2015
adam_russell_9615 mentioned
Technically there is tidal force even between your head and your heels. It doesnt HAVE to be a large distance. With the right instruments I definitely could measure the difference.
Sure you could but, that's a bit of an extension isn't it ?

Reminder, here is a fair representation, I get the impression you ignored the link supplied, is that true ?

Please read it
https://en.wikipe...rinciple

Specifically
"A little reflection will show that the law of the equality of the inertial and gravitational mass is equivalent to the assertion that the acceleration imparted to a body by a gravitational field is independent of the nature of the body."

and he goes on

"It is only when there is numerical equality between the inertial and gravitational mass that the acceleration is independent of the nature of the body"

Please don't waste time by ignoring links supplied, ask questions from there, we are not unpaid tutors
DarkLordKelvin
3 / 5 (2) Apr 13, 2015
Think so? What's the force in newtons then? How's it compare with the force of a single gas molecule impacting your detector?
I didnt mean I could measure it, but that the difference could hypothetically be measured.
I knew what you meant; but try answering those questions I posed.
None of that really matters of course; even if you could measure it, it would be non-local, and thus not covered by the equivalence principle.
So local is only defined for a single non-dimensional point in space? Yes, in terms of theory.
Does that even make sense considering the uncertainty principle?
Yes, you are conflating questions of "what can be measured?" and "what can be calculated?". The EP is a constraint on the latter, in that a theory of gravitation can be described as "consistent with the EP" (like GR) or not (like NG). It does not constrain what can be measured; in fact, the EP is an experimentally falsifiable hypothesis, & so could be disproved by measurement.
adam_russell_9615
not rated yet Apr 13, 2015

.... I get the impression you ignored the link supplied, is that true ?

Please read it
https://en.wikipe...rinciple

Mea culpa. I had not read it. But now having read it I see it agrees with what I was trying to say.

An observer in a windowless room cannot distinguish between being on the surface of the Earth, and being in a spaceship in deep space accelerating at 1g. This is not strictly true, because massive bodies give rise to tidal effects (caused by variations in the strength and direction of the gravitational field) which are absent from an accelerating spaceship in deep space.


Ill just add one bit more to the point. Consider an object dropped on earth to one dropped on the rocket. On the rocket the object falls with constant acceleration. On earth the object falls with increasing acceleration.
Mike_Massen
1 / 5 (1) Apr 13, 2015
adam_russell_9615 getting somewhere but claimed
On earth the object falls with increasing acceleration
Really, What distance from central mass, depth, other influences etc ?

As I stated earlier, more or less, not useful or helpful, to take the "principle" too far, any and all equations which result in combinatorial circumstances raise necessary mathematical effort/evaluations.

In other words. in Regards to the Equivalence Principle, as mentioned it is in a static framework, you can extend it to dynamic but you WILL need a rather rigorous understanding of maths ie Vectors etc etc

What are you driving at by picking at the understood limits of any principle, it holds for a static but, what do you expect to gain from any queries re a dynamic - if you are not prepared to entertain the maths analysis from what appears to be idle questions there's no hypothesis to converge upon a proof ?

Your sorts of straightforward issues are best addressed by actual courses !

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