Is there anything with a constant weight?

[.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 buoyant 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.