Is there anything with a constant weight?

[jmiller0921]

Is there any particle in physics that has a constant weight regardless of the environment it is in, with respect to gravity? something like light, roughly 300000000m/s always, but in terms of weight instead of speed. Everything that I have seen at this point is always relative to something its next to or something to that effect. I'm looking for something that is always the same weight with respect to gravity. I was thinking a light photon would be , from what I've seen this varies.

 

[dauto]

weight is given by an object's mass times the local gravitational acceleration. Since the local gravitational acceleration varies from place to place, the object's weight must change from place to place. I get the feeling that's not what you intended to ask. But that's the answer to the question you asked.

 

[jmiller0921]

What I mean is, is there anything that has a constant mass that would also have a constant weight with respect to gravity? It seems like there could be something that was always the same mass when it is observed. from what I've read everything varies in mass and weight. I am looking for something that is always the same no mater what.

 

[russ_watters]

No, mass is a physical property of an object and does not change. Only weight changes.

 

[ZapperZ]

What I mean is, is there anything that has a constant mass that would also have a constant weight with respect to gravity? It seems like there could be something that was always the same mass when it is observed. from what I've read everything varies in mass and weight. I am looking for something that is always the same no mater what.

What you are asking is this: I have this apple. Is there a place where price of this apple will always remain the same, no matter where I go and how long I have it?

Now see if you can understand the responses that you have been given.

Zz.

 

[bland]

Is there any particle in physics that has a constant weight regardless of the environment it is in, with respect to gravity? something like light, roughly 300000000m/s always, but in terms of weight instead of speed.

Light does not have a constant speed regardless of environment. It travels slower in water than in air and slower in air than in intergalactic space.

 

[csleong]

Make it simple, weight = mass x gravity, gravity is not a constant but mass remains the same, you know weight will change proportion to local gravity.

 

[llstanfield]

Weight is directly dependent on the mass of X and the gravitational body of something even more massive. To technically answer your question, everyone on this planet has a constant weight, to some degree. So, your weight is directly dependent on the force of G. Therefore, I don't think there exists a 'constant' weight.

When you start talking about particles of light (which may not be an accurate description), the particles or sometimes waves, don't have weight; because particles that small are not observed to be directly affected by gravitational forces.

 

[adjacent]

Weight is directly dependent on the mass of X and the gravitational body of something even more massive.

Not necessarily.

 

[llstanfield]

Not necessarily.

Care to elaborate on my mistake? Really interested. Thanks.

 

[ZapperZ]

Care to elaborate on my mistake? Really interested. Thanks.

You can feel "weight" when you are accelerating "upwards", with no need of the presence of another large body to generate the effect of gravity.

Zz.

 

[adjacent]

Care to elaborate on my mistake? Really interested. Thanks.

Your statement:

Weight is directly dependent on the mass of X and the gravitational body of something even more massive.

However,weight of an object is usually taken to be the force on the object due to gravity.

It doesn't have to be possessing more gravity than the mass of X.
Weight of me in terms of Earth is 600N.
Weight of Earth in terms of me is $m_2a$.
Find $a$ using $F=\frac{Gm_1m_2}{r^2}$

 

[llstanfield]

You can feel "weight" when you are accelerating "upwards", with no need of the presence of another large body to generate the effect of gravity.

Zz.

Oh I see. Please correct me if I'm wrong, but is this effect related to Einstein's theory of general relativity? Now that I think about it, maybe I was wrong by saying the effect of weight was necessary of having a greater massive object force. But, perhaps it's more dependent on the mass times the acceleration on any interacting body? For instance, even in an elevator or rocket ship given this force, aren't those two things much more massive than the obvious body that's being exerted on? I mean, can you provide me any other example where something with less mass of a heavier one determines the weight? Thanks for your time.

 

[llstanfield]

Your statement:


However,weight of an object is usually taken to be the force on the object due to gravity.

It doesn't have to be possessing more gravity than the mass of X.
Weight of me in terms of Earth is 600N.
Weight of Earth in terms of me is $m_2a$.
Find $a$ using $F=\frac{Gm_1m_2}{r^2}$

Oh okay, and you are absolutely right. That's what I intended to say, as in the mass of an object related to weight, is subject to the gravitational force...so, you or I, would weigh much more on Jupiter or the Sun for instance. I understand the simple equation of F=mxg, but can you explain the equation that you have given? As far as telling me what r squared and Gm1 stands for? Thanks for your contribution and feedback. Much appreciated.

 

[jmiller0921]

"What you are asking is this: I have this apple. Is there a place where price of this apple will always remain the same, no matter where I go and how long I have it?

Now see if you can understand the responses that you have been given.

Zz."

yes. This is what I mean. Is there any measurable particle that is always the same mass "exactly" in vacuum? It seems to me that all particles have variable sizes and eventually some decay into other particles. I am looking for something that is stable and always has the same mass (no variance) in vacuum.

 

[adjacent]

Oh okay, and you are absolutely right. That's what I intended to say, as in the mass of an object related to weight, is subject to the gravitational force...so, you or I, would weigh much more on Jupiter or the Sun for instance. I understand the simple equation of F=mxg, but can you explain the equation that you have given? As far as telling me what r squared and Gm1 stands for? Thanks for your contribution and feedback. Much appreciated.

What is your educational level?
$F=\frac{Gm_1m_2}{r^2}$ is the Universal[/PLAIN] Law of Gravitation

F is the force.
G is the Universal[/PLAIN] Gravitational Constant.

$m_1$ is the mass of you or me.(In my example)
$m_2$ is the mass of Earth (In my example given)
$r$ is the distance between the center of mass of Earth and you.

Let your mass be 100kg
Let the mass of Earth be $5.97219 × 10^{24}$kg.
You stand 1m away from earth.
$F=m_1a=\frac{Gm_1m_2}{r^2}$
Simplifying gives,
$a=\frac{Gm_2}{r^2}$ - where a is the acceleration of you.
Again do the same with m_2
$a=\frac{Gm_1}{r^2}$ where a is the acceleration of earth.
So Weight =mass x acceleration.
Find the weight of Earth and you yourself.

 

[llstanfield]

What is your educational level?
$F=\frac{Gm_1m_2}{r^2}$ is the universal law of gravitation.
F is the force.
G is the Universal[/PLAIN] Gravitational Constant.

$m_1$ is the mass of you or me.(In my example)
$m_2$ is the mass of Earth (In my example given)
$r$ is the distance between the center of mass of Earth and you.

I have a BA in Sociology. However, physics has just recently grabbed my attention, which is why I asked for that elementary understanding of the equation, and respond to the original post to the best of my ability. I signed up on this site to give my understanding of the subject, and ask questions about it in order to learn more.

 

[adjacent]

I have a BA in Sociology. However, physics has just recently grabbed my attention, which is why I asked for that elementary understanding of the equation, and respond to the original post to the best of my ability. I signed up on this site to give my understanding of the subject, and ask questions about it in order to learn more.

You are welcome here.
It's actually not very elementary.I'll study it at A-level. However,I know how it works.

 

[ZapperZ]

"What you are asking is this: I have this apple. Is there a place where price of this apple will always remain the same, no matter where I go and how long I have it?

Now see if you can understand the responses that you have been given.

Zz."

yes. This is what I mean. Is there any measurable particle that is always the same mass "exactly" in vacuum? It seems to me that all particles have variable sizes and eventually some decay into other particles. I am looking for something that is stable and always has the same mass (no variance) in vacuum.

But that isn't what you asked originally, is it?

We know about "mass". A look in the Particle Data Book confirms that there ARE plenty of elementary particles with unambiguous mass that we can clearly define. These do not change, or else we will never be able to identify them.

Note that in my example, the apple did not change!. Its value may, but the apple itself hasn't.

Zz.

 

[llstanfield]

You are welcome here.
It's actually not very elementary.I'll study it at A-level. However,I know how it works.

I appreciate that and this site. And I will definitely refer to you in the future. But for a layman, attempting to understand these complicated physical questions, the equation that you've recently posted in my opinion, is "elementary' or a Newtonian understanding of the world. For instance, regarding the original post, I'm not sure how much a particle would 'weigh' in a vacuum. Because that level of physics transcends classical physics in my opinion.

 

[ZapperZ]

For instance, regarding the original post, I'm not sure how much a particle would 'weigh' in a vacuum. Because that level of physics transcends classical physics in my opinion.

I am not sure why this would be a problem. In particle physics, gravity is irrelevant. When they do those collisions at the LHC, gravity is a non-entity. Yet, as you can see already, they were able to extract the mass of many of these particles and identify them!

So being in a vacuum, or being in a zero-gravity environment, has no bearing on determining the mass of particles.

Zz.

 

[llstanfield]

I am not sure why this would be a problem. In particle physics, gravity is irrelevant. When they do those collisions at the LHC, gravity is a non-entity. Yet, as you can see already, they were able to extract the mass of many of these particles and identify them!

So being in a vacuum, or being in a zero-gravity environment, has no bearing on determining the mass of particles.

Zz.

I think I get that, and I appreciate your post. However, if gravity is irrelevant, wouldn't that make weight irrelevant as well? Or aren't particles at these levels based on 'strong' and 'weak' fundamental forces, connected to the gravitational force as well related to weight? If that makes sense?

Thanks for your time.

 

[ZapperZ]

I think I get that, and I appreciate your post. However, if gravity is irrelevant, wouldn't that make weight irrelevant as well? Or aren't particles at these levels based on 'strong' and 'weak' fundamental forces, connected to the gravitational force as well related to weight? If that makes sense?

Thanks for your time.

It is irrelevant because it plays an insignificant role in the behavior of these particles. For example, do you think we consider the gravity from alpha centauri when we design buildings and send spacecraft s to various planets in our solar system? It is there, it is non-zero, but it is so weak, it is ridiculous to take that into account when other factors are more dominant by orders of magnitude.

Zz.

 

[llstanfield]

It is irrelevant because it plays an insignificant role in the behavior of these particles. For example, do you think we consider the gravity from alpha centauri when we design buildings and send spacecraft s to various planets in our solar system? It is there, it is non-zero, but it is so weak, it is ridiculous to take that into account when other factors are more dominant by orders of magnitude.

Zz.

This is absolutely true; but it doesn't change the fact that it does have a force. Now, trying to talk about a "weight" given these weak gravitational forces because it is inversely squared as the distance increases, the vacuum particles do not have this "weight" that the OP was trying to refer to. I was referring to the four fundamental forces because they all have something in common. But, one nuclear force on a particular scale, is completely different from the gravitational one.

 

[ZapperZ]

This is absolutely true; but it doesn't change the fact that it does have a force. Now, trying to talk about a "weight" given these weak gravitational forces because it is inversely squared as the distance increases, the vacuum particles do not have this "weight" that the OP was trying to refer to. I was referring to the four fundamental forces because they all have something in common. But, one nuclear force on a particular scale, is completely different from the gravitational one.

I don't quite understand what you said here. The OP also didn't mention about these fundamental forces.

Let's be clear here. In practically 99.999% of the cases, the only forces relevant in the dynamics of a particle is the electromagnetic forces. I can determine the mass of a charge particle, for example, by just using those forces. I have zero need for gravity. (Refer to any of the e/m experiment for electrons).

I also don't understand this "vacuum" issue. The only reason why we would need a vacuum is because the air molecules collide with the particles, thus destroying its path. Otherwise, the fact that they are in a vacuum has no bearing on its mass, its weight, etc.

So I think you need to clearly explain these things you are using here, because I am totally confused on why you are using them.

Zz.

 

[llstanfield]

OP: "Is there any measurable particle that is always the same mass "exactly" in vacuum? It seems to me that all particles have variable sizes and eventually some decay into other particles. I am looking for something that is stable and always has the same mass (no variance) in vacuum"

I was simply replying or repudiating to his post, as in an attempt of him describing these particles in a vacuum as "weighing" something. They don't have weight, and this corroborates your post as to how gravitational effects are negligible. So, even for a charged particle mass, there would be no "weight" in measuring that right?

In addition, the only reason why I brought up the fundamental forces, is because there is a general attractive force between vacuum particles, magnets, macroscopic objects, atoms, gravity and electrons.

 

[ZapperZ]

OP: "Is there any measurable particle that is always the same mass "exactly" in vacuum? It seems to me that all particles have variable sizes and eventually some decay into other particles. I am looking for something that is stable and always has the same mass (no variance) in vacuum"

I was simply replying or repudiating to his post, as in an attempt of him describing these particles in a vacuum as "weighing" something. They don't have weight, and this corroborates your post as to how gravitational effects are negligible. So, even for a charged particle mass, there would be no "weight" in measuring that right?

In addition, the only reason why I brought up the fundamental forces, is because there is a general attractive force between vacuum particles, magnets, macroscopic objects, atoms, gravity and electrons.

I think the OP has a ton of "particles" to choose from. Our good old electron has a "stable mass", and no one has observed a proton decay either. The "vacuum" requirement is still puzzling to me.

Next, every single particle with mass has "weight" in the presence of gravity or similar effects. You cannot say that they don't. However, the influence of gravity, which is the weakest force out of the 4 fundamental forces, in most cases, is negligible for elementary particles. Accounting for the particle's weight gives you nothing - it doesn't appreciably affect its trajectory, its behavior, etc...etc. So here, weight can be ignored. But it doesn't mean it doesn't have one.

I don't understand this "general attractive force between vacuum particles" or where it came from or why it is relevant. Two electrons in my vacuum tube do not have a "general attractive force" between them.

Zz.

 

[llstanfield]

I think the OP has a ton of "particles" to choose from. Our good old electron has a "stable mass", and no one has observed a proton decay either. The "vacuum" requirement is still puzzling to me.

Next, every single particle with mass has "weight" in the presence of gravity or similar effects. You cannot say that they don't. However, the influence of gravity, which is the weakest force out of the 4 fundamental forces, in most cases, is negligible for elementary particles. Accounting for the particle's weight gives you nothing - it doesn't appreciably affect its trajectory, its behavior, etc...etc. So here, weight can be ignored. But it doesn't mean it doesn't have one.

I don't understand this "general attractive force between vacuum particles" or where it came from or why it is relevant. Two electrons in my vacuum tube do not have a "general attractive force" between them.

Zz.

Yes, you are probably right regarding the electron and proton decay, and how that relates to having weight. I'm not knowledgeable about these metric particles having 'weight'; and I was referring to elementary particles in gravity, and I thought that was what the OP was talking about. I was saying that vacuum particles have such a low gravitational force of attraction, which is related to weight, that it's close to zero. However, technically speaking, it DOES have some weight IF these particles that we are talking about have some verifiable mass, according to what you were saying.

The fundamental forces have something in common however. Consider a copper wire, and how the electrons are repelling each other across this wire, while at the same time being attracted to an atomic nucleus; Now notice the word attraction, even having a reverse property. Isn't that the same "attractive" force we see with gravity? And forces that put the neutron and proton together? And the

 

[dauto]

The one thing the OP ought to learn from this thread is that if he wants to know something about mass, than he ought to ask about mass, not weight. Assuming he's learned that much, it's probably safe now to stop punishing him and quit all the talk about weight and answer the question about mass. The answer is that all stable objects have stable mass. In the subatomic particle realm the electron is the best example.

 

[jmiller0921]

never mind.

 

[.Scott]

Is there any measurable particle that is always the same mass "exactly" in vacuum? It seems to me that all particles have variable sizes and eventually some decay into other particles. I am looking for something that is stable and always has the same mass (no variance) in vacuum.

I believe the electron will fit your bill.

 

[jmiller0921]

I believe the electron will fit your bill.

Thank you sir.

 

[Drakkith]

For instance, regarding the original post, I'm not sure how much a particle would 'weigh' in a vacuum. Because that level of physics transcends classical physics in my opinion.

A vacuum itself has no effect on weight. We can place a particle in a vacuum chamber here on Earth and its weight will remain. The same. The reason objects have no weight in the vacuum of space is because they are in free fall. If you were to stand in a spaceship that was using its engines to exactly counter the gravitational force from the sun, you would definitely have weight.

 

[Buckleymanor]

I am not sure why this would be a problem. In particle physics, gravity is irrelevant. When they do those collisions at the LHC, gravity is a non-entity. Yet, as you can see already, they were able to extract the mass of many of these particles and identify them!

So being in a vacuum, or being in a zero-gravity environment, has no bearing on determining the mass of particles.

Zz.

How do they extract the mass .What do they scale it against if gravity is a non- entity.Do they imagine a figure and divide by five.
Surely they must have a baseline derived from the effects of gravity which makes little sense of your argument.

 

[Buckleymanor]

A vacuum itself has no effect on weight. We can place a particle in a vacuum chamber here on Earth and its weight will remain. The same. The reason objects have no weight in the vacuum of space is because they are in free fall. If you were to stand in a spaceship that was using its engines to exactly counter the gravitational force from the sun, you would definitely have weight.

So particles do not experience a bouyant force?

 

[ZapperZ]

How do they extract the mass .What do they scale it against if gravity is a non- entity.Do they imagine a figure and divide by five.
Surely they must have a baseline derived from the effects of gravity which makes little sense of your argument.

Start with something simpler. Look up that e/m experiments that I had mentioned (googling it will give you several). Now where is gravity involved in those? Note that these are experiments that are typical for undergraduate intro physics courses, so it is not THAT exotic and unfamiliar.

Zz.

 

[Buckleymanor]

Start with something simpler. Look up that e/m experiments that I had mentioned (googling it will give you several). Now where is gravity involved in those? Note that these are experiments that are typical for undergraduate intro physics courses, so it is not THAT exotic and unfamiliar.

Zz.

From what I can gather you first have to determine the charge of the electron before you can calculate it's mass and that involves gravity and buoyancy.
Quote:-

Charge and Mass of the Electron 7
Consider a latex sphere of mass m and charge q, falling under the influence of
gravity between two horizontal plates. In falling, the sphere is subjected to an opposing
force due to air resistance. The speed of the sphere quickly increases until a constant
terminal speed is reached, at which time the weight of the sphere, mg, minus the buoyant
force is exactly equal to the air resistance force. The value of the air resistance force on a
sphere was first derived by Sir George Stokes and is given as 6 π η r s where η is the
coefficient of viscosity of air, r is the radius of the sphere and s is its terminal speed. If
the buoyant force of the air is neglected, the equation of motion of the sphere is:
mg − 6πηrs = 0

 

[ZapperZ]

Er... No. You are looking at the oil drop experiment! The e/m experiment done with Helmholtz coil is done inside an evacuated vacuum. There is no buoyancy in vacuum.

http://wanda.fiu.edu/teaching/courses/Modern_lab_manual/em_ratio.html

Zz.

 

[Buckleymanor]

Er... No. You are looking at the oil drop experiment! The e/m experiment done with Helmholtz coil is done inside an evacuated vacuum. There is no buoyancy in vacuum.

http://wanda.fiu.edu/teaching/courses/Modern_lab_manual/em_ratio.html

Zz.

So buoyancy can be ignored.I don't see how gravity is however.

The field exerts a magnetic force with a magnitude of

What do you compare your magnetic field with to allow it to have a magnitude.
You could say other magnetic fields but ultimately you arrive at the pull of gravity.

 

[ZapperZ]

So buoyancy can be ignored.I don't see how gravity is however.

What do you compare your magnetic field with to allow it to have a magnitude.
You could say other magnetic fields but ultimately you arrive at the pull of gravity.

You last statement is a guess work. Since when do I need gravity to determine magnetic field strength? I can even just calculate it! You obviously have not understood the physics behind the experiment.

Next time, please cite the theory to back your claim. This is no longer making any sense and you are just shooting in the dark.

Zz.

 

[Buckleymanor]

You last statement is a guess work. Since when do I need gravity to determine magnetic field strength? I can even just calculate it! You obviously have not understood the physics behind the experiment.

Next time, please cite the theory to back your claim. This is no longer making any sense and you are just shooting in the dark.

Zz.

And you won't answer the question and are being extreamly obtuse.It's been put to you a number of times that whatever device is used to measure the mass of an electron.Gravity plays a part in the calculation.
It might be negligable as a physical quantity far as particles are concerned because of there very small mass.
As far as calculating the electrons mass how can you do it without reference to gravity and field strength.

Beam of electrons moving in a circle in a Teltron tube, due to the presence of a magnetic field. Purple light is emitted along the electron path, due to the electrons colliding with gas molecules in the bulb. The mass-to-charge ratio of the electron can be measured in this apparatus by comparing the radius of the purple circle, the strength of the magnetic field, and the voltage on the electron gun. The mass and charge cannot be separately measured this way—only their ratio.

Do your calculations ignore field strenth or as stated before do you imagine a figure and divide by five.
My magnet has a pull of five pounds what does yours have?

 

[mfb]

@Buckleymanor: All experiments that are done with gravity can be done in a centrifuge in space.

And you don't need the oil drop experiment. Atom counting and electrolysis are another approach to get the Avodradro constant and the elementary charge.

The elementary charge is indeed an important quantity if you want to know particles masses in kg (note that this is not necessary), but you don't need gravity to measure it.

 

[Doc Al]

It's been put to you a number of times that whatever device is used to measure the mass of an electron.Gravity plays a part in the calculation.

Sounds like nonsense. Reference please? I'd love to see how gravity affects the calculation.

It might be negligable as a physical quantity far as particles are concerned because of there very small mass.

Well, there you go. Is it negligible or not?

As far as calculating the electrons mass how can you do it without reference to gravity and field strength.

There are other, much stronger, forces involved.

 

[ZapperZ]

And you won't answer the question and are being extreamly obtuse.It's been put to you a number of times that whatever device is used to measure the mass of an electron.Gravity plays a part in the calculation.
It might be negligable as a physical quantity far as particles are concerned because of there very small mass.
As far as calculating the electrons mass how can you do it without reference to gravity and field strength.

Do your calculations ignore field strenth or as stated before do you imagine a figure and divide by five.
My magnet has a pull of five pounds what does yours have?

This is silly. Look at the experiment. I've not only done it, I've taught it! Where exactly does gravity comes in? You said it comes in the calculation. WHERE EXACTLY?

I could do this experiment in outer space. How would the experiment be any different? The e/m values can't be decoupled in this experiment. So what? It still doesn't mean gravity is necessary to do the experiment. I can determine the magnetic field strength using a gauss probe or calculate the field strength using the geometry of the coil and the current. Where is gravity here? Will the field change if I do this in the ISS? Really?

The experiment I showed contains NO g in the physics. Go ahead and look! Unless you can show me THEORETICALLY where this value comes in EXPLICITLY, then I have no more interest in trying to either educate you, or correct this faulty knowledge. It is no longer of any concern to me that you believe in a wrong set of information. I've tried all I can to correct it, so I'm done.

Zz.

 

[Buckleymanor]

This is silly. Look at the experiment. I've not only done it, I've taught it! Where exactly does gravity comes in? You said it comes in the calculation. WHERE EXACTLY?

I could do this experiment in outer space. How would the experiment be any different? The e/m values can't be decoupled in this experiment. So what? It still doesn't mean gravity is necessary to do the experiment. I can determine the magnetic field strength using a gauss probe or calculate the field strength using the geometry of the coil and the current. Where is gravity here? Will the field change if I do this in the ISS? Really?

The experiment I showed contains NO g in the physics. Go ahead and look! Unless you can show me THEORETICALLY where this value comes in EXPLICITLY, then I have no more interest in trying to either educate you, or correct this faulty knowledge. It is no longer of any concern to me that you believe in a wrong set of information. I've tried all I can to correct it, so I'm done.

Zz.

Calm down souds like I have trod on a nerve.
But can't see why.
You could do the experiment in outer space but you would still be taking your values with you.
What I would like to know but realize it's probably too sensitive to ask is in when you calculate your field strength useing whatever method available what is the strength be related to and why only that.
Dumb as I am, I can relate it the amount of effort against the force of gravity.
The amount of pull or push an e/m field has can be related to weight.How were field strengths expressed when first discovered.
Why is this a problem to your sensitivities?
I am probably just as dumb in not apreciating the terminology and classification of the subject but can't see for the life of me how the field strength has no relation to g and why you should wan't to divorce it from the calculation with such agression.
Does it not sit well with a more explicite explanation. Or is it wrong to ask.

 

[Buckleymanor]

Sounds like nonsense. Reference please? I'd love to see how gravity affects the calculation.

You give the mass a value what exactly is that value related to.If it has no relation then it has no value.

Well, there you go. Is it negligible or not?

Probably

There are other, much stronger, forces involved.

But that does not mean gravity does not exist.

 

[Buckleymanor]

@Buckleymanor: All experiments that are done with gravity can be done in a centrifuge in space.

And you don't need the oil drop experiment. Atom counting and electrolysis are another approach to get the Avodradro constant and the elementary charge.

The elementary charge is indeed an important quantity if you want to know particles masses in kg (note that this is not necessary), but you don't need gravity to measure it.

So you don't need a test mass to start with.

 

[ZapperZ]

Calm down souds like I have trod on a nerve.
But can't see why.
You could do the experiment in outer space but you would still be taking your values with you.
What I would like to know but realize it's probably too sensitive to ask is in when you calculate your field strength useing whatever method available what is the strength be related to and why only that.
Dumb as I am, I can relate it the amount of effort against the force of gravity.
The amount of pull or push an e/m field has can be related to weight.How were field strengths expressed when first discovered.
Why is this a problem to your sensitivities?
I am probably just as dumb in not apreciating the terminology and classification of the subject but can't see for the life of me how the field strength has no relation to g and why you should wan't to divorce it from the calculation with such agression.
Does it not sit well with a more explicite explanation. Or is it wrong to ask.

It isn't wrong to ask. But you are countering the physics that I brought up (which has CLEAR mathematical description) with handwaving argument. look at what you stated here. You just "can't see for the life of" you as an ARGUMENT point, instead of using physics. In other words, you are countering what I brought up using nothing more than your "intuition" or "personal tastes"! You have used no physics whatsoever!

Or this one: "The amount of pull or push an e/m field has can be related to weight." Where in classical E&M theory did this come from? YOu provided ZERO physics to back this up. All you did is simply to state it, without no justification, simply based on what you THINK it should work.

How does one have a physics discussion like that?

What you should do is looking at the theoretical description, and then point out to me exactly where "g" comes into the description, not using simply what you THINK is valid.

Otherwise, as I've said, you are welcome to stick to your ignorance and you haven't learned anything. I am more than contented to not waste anymore of my time and effort on this.

Zz.

 

[Dale]

With that, the thread is closed.