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Author Topic: Is hot water heavier than cold water, and how can I prove this?  (Read 19093 times)

Offline PmbPhy

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Quote from: alancalverd
That's why I asked you to consider two particles with different velocity vectors.
I don't understand what you mean. What do you mean by "That's why I asked...." I don't understand. Exactly why did you ask?

Quote from: alancalverd
But their masses, as determined by the gravitational force on a test particle travelling with each (the only way you can determine the mass of a free asteroid), must be constant and identical because there is no universal frame of reference and neither rock is moving with respect to its test mass. Or variable with a. Or fixed but different according to b and c. Please choose, and show your reasoning.   
I don't understand at all what you're leading to or asking me. Please be more clear. Also I don't understand what you mean by "Please choose" What am I choosing?

Also since we're talking about relativistic mass why are you ignoring it in the case of the case of a particle moving through a gravitational field? E.g. see equation 16 in my web page -  http://home.comcast.net/~peter.m.brown/gr/weight_moving_body.htm

The mp in that equation is the passive gravitational mass of the object and it depends on both the field and the gravitational potential and as such the position of the particle in the gravitational field.

Quote
I not only knew it, but recently used it in another thread as an example of the experimental proof of relativity!
Then why were you asking me how to measure it rather than posting it yourself? Unless it didn't come to your mind?

Quote from: alancalverd
He asked for a proof that would convince his boss. I gave him a simple experiment that showed water in the range below 4C was lighter than water at 4C ....
If this
Quote
Fill a cup close to the brim with water, put  it in the freezer, and stir it with a thermometer. When it reaches about 5C, top it up to the brim, then watch what happens as it cools further. It will spill over the edge just before it freezes because cold water takes up more space (i.e. is less dense) than warm.
is true then its not because there's a change in mass but a change in density. It's simply not possible to measure changes in mass that small in situations like this. In fact today there is no experiment of any kind which can show the mass increase for a macroscopic body due to an input/output of heat.

If you have the text Spacetime Physics - Second Edition by Taylor and Wheeler, then turn to page 223 and you'd see that they communicated with Vladimir Braginsky at the University of Moscow about a device to measure the increase in the weight of a tiny heated quartz pellet. They concluded that the technology does not exist to detect it yet. That it surpasses the present limit of technology

Quote from: alancalverd
He asked for a proof that would convince his boss. I gave him a simple experiment that showed water in the range below 4C was lighter than water at 4C and this was unique, but you showed that thanks to the relativistic effect it always got heavier as you increased the temperature without bound, whilst RD presented a graph that contradicted that statement, so it seems that some calculation is in order if Mr Clark is not to go away completely confused! 
The reason I gave a relativistic description is because there is no classical mechanism that allows a water to become lighter. It just can't happen. It would mean that the law of the conservation of mass would be violated. It's only violated when relativistic effects are taken into account. You mistook the effects of the change in the density of water with a change in weight and therefore a change in mass.

See:
http://www.princeton.edu/~achaney/tmve/wiki100k/docs/Conservation_of_mass.html
http://en.wikipedia.org/wiki/Conservation_of_mass

I'll post that section of Taylor and Wheeler's text after I scan it in and upload it to my website.
 

Offline CliffordK

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He asked for a proof that would convince his boss. I gave him a simple experiment that showed water in the range below 4C was lighter than water at 4C and this was unique,
The OP hasn't replied yet, so it is difficult to know the ultimate intentions. 

Yes, most materials have a positive expansion coefficient (decrease in density when heated), except for a few materials in specific circumstances, such as water that has a negative expansion coefficient between 0C and 4C.  Of course, that isn't a change in mass, but would be visible as a change in volume in a sealed container.

As far as Einstein E=mc2, that would be more difficult to prove.  I derived some of the quantities on paper, but that is far from a rigorous empirical proof, and it would be difficult to weigh a cubic km of water or ice to an accuracy of a few grams.

I'm not sure a mass-spec has high enough of accuracy.  It also requires ions which may be problematic for a liquid water sample, but should still be representative.  Is, say, altering the temperature of a set of ions representative of the change in mass, or are there other confounding variables?
 

Offline PmbPhy

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Here is that page from the text Spacetime Physics 2nd Ed that I mentioned
http://home.comcast.net/~peter.m.brown/stp2_page_223.jpg

Quote from: CliffordK
The OP hasn't replied yet, so it is difficult to know the ultimate intentions.
My impression is that the rarely do anyway.

Quote from: CliffordK
Of course, that isn't a change in mass, but would be visible as a change in volume in a sealed container.
Yup. It happens every time I make ice cubes. I'm very surprised that one got by an experimental physicist like Alan.

Quote from: CliffordK
As far as Einstein E=mc2, that would be more difficult to prove.
As I indicated above, one does not "prove" things in physics other than those cases when it's a mathematical proof. One forms postulates, makes predictions from those postulates, designs experiments from those predictions and observes what happens in the experiments. If the experiment is consistent with the predicted results then one has more confidence in the theory. Otherwise one has to either abandon the theory or modify it. See

What is Science - http://home.comcast.net/~peter.m.brown/ref/what_is_science.pdf
The first paragraph reads
Quote
The following statement was originally drafted by the Panel on Public Affairs (POPA) of the American Physical Society, in an attempt to meet the perceived need for a very short statement that would differentiate science from pseudoscience. Am. J. Phys. 67 (8),

Quote from: CliffordK
I derived some of the quantities on paper, but that is far from a rigorous empirical proof, and it would be difficult to weigh a cubic km of water or ice to an accuracy of a few grams.
I'd enjoy reading what you derived. Can you post it here for me or send it to me in PM or e-mail? Thanks. Or please take a look at
http://home.comcast.net/~peter.m.brown/sr/mass_energy_equiv.htm and let me know if it's the same thing.

Quote from: CliffordK
I'm not sure a mass-spec has high enough of accuracy.
It depends on the particular mass spectrometer. In any case, as I said above the physics is the same as a cyclotron so if the spectrometer doesn't work merely use a spectrometer.

Quote from: CliffordK
It also requires ions which may be problematic for a liquid water sample, but should still be representative.
I'd say that a ion of H2O would be just fine. However it'd be better to find out why he's so concerned with water as opposed to any other substance or element. If we find that out then it'd make things easier.

Quote from: CliffordK
Is, say, altering the temperature of a set of ions representative of the change in mass, or are there other confounding variables?
At this point it's time to talk to the OP and his boss and get details. If he is still around that is. He might have forgotten all about this. After all he asked the question 4 days ago and we haven't heard even a whisper from either of them since.
 

Offline PmbPhy

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Quote from: CliffordK
I'm not sure a mass-spec has high enough of accuracy.
Note: The reason I mentioned a mass-spectrometer is because it works like a cyclotron which can be used to measure the relativistic mass of a particle or ion. So tell you what we can do. Instead of using a mass-spectrometer, let's use a cyclotron instead. That way we don't have to worry about it working.

Of course Mike's boss can't get his hands on a cyclotron so that's the end of that. :)  Besides, the physics is too hard to work with when quantities are that small and increases that small too.
 

Offline alancalverd

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This is all fascinating stuff, but from the point if view of psychology and philology, not physics!

A benchtop mass spectrometer running at a few keV probably won't show the relativistic change of mass to a detectable level (it's calibrated with known ions - chemistry rather than physics - because it is very difficult to measure static magnetic fields to sufficient accuracy to make an absolute measurement of mass), but a hospital cyclotron certainly will, and we have to allow for it in designing the electronics.

But the real point of argument is the meaning of "heavier". Given that the original questioner didn't know how to demonstrate the effect, I think his enquiry was based on the common knowledge and observation of the anomalous expansion of water rather than the subtle timing corrections of a cyclotron driver - especially as you can't accelerate a water molecule in a cyclotron. Therefore he was talking about bulk density, not molecular mass.   

Is polystyrene heavier than gasoline? To the layman, the answer is clearly no because the "everyday" forms are foam and liquid respectively, but as a liquid or solid, the answer is yes, it is a lot denser. and if you want to be picky, it even has a higher molecular weight. But the molecular weight of gasoline is much greater than that of water, so surely water will float on gasoline? If only it did so, firefighting would be a lot easier!
 

Offline PmbPhy

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Quote from: alancalverd
This is all fascinating stuff, but from the point if view of psychology and philology, not physics!
I can't see how that could possibly true at all. All I've done is to directly answer the OPs question and tried to do it in manner to answer what his boss is probably seeking the answer too rather than what he literally asked for proof of.

Quote from: alancalverd
But the real point of argument is the meaning of "heavier".
I can't possibly see how that's a point of disagreement since most non-physicists confuse weight with mass. It's very clear to me that he wants to know which of two initially identical cups of water will have a greater mass will have a greater mass when one of them is heated, i.e. energy is added to the water. Most people I know are happy with thought experiments so I'd wager that Michael doesn't actually want to go out and do an experiment since he mentioned nothing about experiments. When he said "proof" it seemed to me that he wants to present a solid argument to present to him which will convince him that he's right.

Quote from: alancalverd
Given that the original questioner didn't know how to demonstrate the effect, I think his enquiry was based on the common knowledge and observation of the anomalous expansion of water ...
I see absolutely no reason to assume that'd be the case. That is a major jump in my opinion.

Quote from: alancalverd
..rather than the subtle timing corrections of a cyclotron driver - especially as you can't accelerate a water molecule in a cyclotron.
That's not true at all. And what's a "cyclotron driver" anyway and how is it related to this topic? There is absolutely no reason why a molecule of water, i.e. an H2O ion couldn't be accelerated in a cyclotron. The following is close to what I'm referring to:
http://en.wikipedia.org/wiki/Ion_cyclotron_resonance

From my Google search I came up with this - ion cyclotron resonance mass spectrometry which appears to be what can be used in principle to measure the mass of water ion.

In any case, why are we so worried about experiments. Michael said nothing about experiments or that the answer had to be able to be done in practice. For all we know, and I know that if it were I then it's what I'd be looking for, I'd want to know what's going on in principle.


re - especially as you can't accelerate a water molecule in a cyclotron. - There is no basis for such an assumption. On what are you basing that assertion on?

Quote from: alancalverd
Therefore he was talking about bulk density, not molecular mass.   
You don't know that at all. He specifically asked which was heavier, hot water or cold water. And as I said for most people, and nearly universally all layman, use weight synonymously for mass.


Sorry Alan but to me you're just grasping at straws now.
 

Offline lightarrow

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Quote from: lightarrow
I haven't understood what you say here. Are you talking of a mass increase of a particle with speed or what?
Yes. I'll restate it to make it clearer. In what follows what I mean by the use of the term mass I mean inertial mass in the Newtonian case and relativistic mass in the relativistic case.
In Newtonian mechanics mass does not increase with speed.
In relativistic mechanics mass does increase with speed.
Simple, yes? :)
Quote from: lightarrow
And what are you saying about a mass spectrometer?
That's one way to measure the relativistic mass of a charged particle.
Quote from: lightarrow
That you could measure a particle's mass increase with speed using that device? If you are saying this, it's incorrect, and not because it would be too small, but because you can't.
That's  incorrect. In fact Wikipedia states that in 1901 Walter Kaufmann used a mass spectrometer to measure the relativistic mass increase of electrons. See http://en.wikipedia.org/wiki/History_of_mass_spectrometry

See also the article in American Journal of Physics about this particular point at
http://gabrielse.physics.harvard.edu/gabrielse/papers/1995/RelativisticMassAJP.pdf

However a mass spectrometer is in essence a cyclotron which can measure a mass increase due to relativistic mass.

The physics principles are the same as they are in a cyclotron. The particle is charged and that makes it move on a circle in a magnetic field. The radius of the circle it moves in. The physics for a cyclotron and a mass spectrometer is in the following webpage in my website: http://home.comcast.net/~peter.m.brown/sr/cyclotron.htm

Quote from: lightarrow
If you are saying this, it's incorrect, and not because it would be too small, but because you can't.
Please provide a proof of this. A derivation whose results show what you assert would be fine. Thanks.
In order to analyze relativistic dynamic experiments, what we need is:

E2 = (c*|p|)2 + (m*c2)2     (1)
v = c2*p/E     (2)

(p and v are momentum and velocity vectors, "| |" means modulus of the vector).
From these two equations we can deduce other relativistic equations, for example:

E = mc2
p = M*v

Where γ = 1/sqrt{1 - (|v|/c)2}.

We also need the law:

F = dp/dt     (3)

which is the definition of force in relativistic dynamics.

Lets analyze the motion of a charged particle in a magnetic field.
We know that (Lorentz force):

F = q*v x B     (4)

Where x means vectorial product.

So F is always orthogonal to v, and orthogonal to p too, according to eq. (2). It follows that the modulus of p is constant (and according to eq. (1) even E is constant):

d|p|2/dt = 2 p.(dp/dt) = 2 p.F = 0.

(the dot between two vectorial factors means scalar product).

Defining the angular speed as: ω = |v|/r, according to kinematics we have:

|dp/dt| =  ω*|p|

(This equation is valid in galileian as well as in relativistic kinematics).

For the sake of simplicity, lets assume v and p both orthogonal to the magnetic field B.
Then the trajectory is a circle, traveled of uniform motion and which lays in a plane orthogonal to B.
So, using equations (2), (3), (4) and the expression of the (modulus of) Lorentz force on a charged particle:

|F| = q*|v|*|B|

we easily deduce :

ω*|p| = q*|v|*|B|     (5)

|p|/r = q*|B| → |p| = q*|B|*r    (6)

and its this last equation that we can use for a mass spectrometer, measuring r and knowing |B|, to compute |p| and so to find the mass m according to eq. (1), knowing the total energy E. The last can be found using also:

E = Ek + m*c2     (7)

where the kinetic energy Ek is set by the potential difference V applied to the charge q: Ek = q*V.

Substituting in eq. (7) and then in eq. (1):

(q*V + m*c2)2 = (c*|p|)2 + (m*c2)2     

from which you can find the  invariant mass m of the charged particle.

No need of relativistic mass.

N.B. (Most of what written up is not mine, it comes from another source, but it's rather simple to understand.
The source is an italian thread started from prof. Elio Fabri:
https://groups.google.com/d/msg/free.it.scienza.fisica/UyfGXka6rgQ/MiEiUTVexIsJ).

But if you intended that the mass spectrometer allows you to find the momentum |p| and so the total energy E and you identify the last with relativistic mass (multiplied c2), then we again come back to what I have said before several times in various threads, even answering to you, that is that relativistic mass is nothing else than another name for total energy E.

Edit: coloured in blue eq. (6) - 01/12/2014

--
lightarrow
« Last Edit: 01/12/2014 12:54:06 by lightarrow »
 

Online jeffreyH

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Quote from: lightarrow
I haven't understood what you say here. Are you talking of a mass increase of a particle with speed or what?
Yes. I'll restate it to make it clearer. In what follows what I mean by the use of the term mass I mean inertial mass in the Newtonian case and relativistic mass in the relativistic case.
In Newtonian mechanics mass does not increase with speed.
In relativistic mechanics mass does increase with speed.
Simple, yes? :)
Quote from: lightarrow
And what are you saying about a mass spectrometer?
That's one way to measure the relativistic mass of a charged particle.
Quote from: lightarrow
That you could measure a particle's mass increase with speed using that device? If you are saying this, it's incorrect, and not because it would be too small, but because you can't.
That's  incorrect. In fact Wikipedia states that in 1901 Walter Kaufmann used a mass spectrometer to measure the relativistic mass increase of electrons. See http://en.wikipedia.org/wiki/History_of_mass_spectrometry

See also the article in American Journal of Physics about this particular point at
http://gabrielse.physics.harvard.edu/gabrielse/papers/1995/RelativisticMassAJP.pdf

However a mass spectrometer is in essence a cyclotron which can measure a mass increase due to relativistic mass.

The physics principles are the same as they are in a cyclotron. The particle is charged and that makes it move on a circle in a magnetic field. The radius of the circle it moves in. The physics for a cyclotron and a mass spectrometer is in the following webpage in my website: http://home.comcast.net/~peter.m.brown/sr/cyclotron.htm

Quote from: lightarrow
If you are saying this, it's incorrect, and not because it would be too small, but because you can't.
Please provide a proof of this. A derivation whose results show what you assert would be fine. Thanks.
In order to analyze relativistic dynamic experiments, what we need is:

E2 = (c*|p|)2 + (m*c2)2     (1)
v = c2*p/E     (2)

(p and v are momentum and velocity vectors, "| |" means modulus of the vector).
From these two equations we can deduce other relativistic equations, for example:

E = mc2
p = M*v

Where γ = 1/sqrt{1 - (|v|/c)2}.

We also need the law:

F = dp/dt     (3)

which is the definition of force in relativistic dynamics.

Lets analyze the motion of a charged particle in a magnetic field.
We know that (Lorentz force):

F = q*v x B     (4)

Where x means vectorial product.

So F is always orthogonal to v, and orthogonal to p too, according to eq. (2). It follows that the modulus of p is constant (and according to eq. (1) even E is constant):

d|p|2/dt = 2 p.(dp/dt) = 2 p.F = 0.

(the dot between two vectorial factors means scalar product).

Defining the angular speed as: ω = |v|/r, according to kinematics we have:

|dp/dt| =  ω*|p|

(This equation is valid in galileian as well as in relativistic kinematics).

For the sake of simplicity, lets assume v and p both orthogonal to the magnetic field B.
Then the trajectory is a circle, traveled of uniform motion and which lays in a plane orthogonal to B.
So, using equations (2), (3), (4) and the expression of the (modulus of) Lorentz force on a charged particle:

|F| = q*|v|*|B|

we easily deduce :

ω*|p| = q*|v|*|B|     (5)

|p|/r = q*|B| → |p| = q*|B|*r    (6)

and its this last equation that we can use for a mass spectrometer, measuring r and knowing |B|, to compute |p| and so to find the mass m according to eq. (1), knowing the total energy E. The last can be found using also:

E = Ek + m*c2     (7)

where the kinetic energy Ek is set by the potential difference V applied to the charge q: Ek = q*V.

Substituting in eq. (7) and then in eq. (1):

(q*V + m*c2)2 = (c*|p|)2 + (m*c2)2     

from which you can find the  invariant mass m of the charged particle.

No need of relativistic mass.

N.B. (Most of what written up is not mine, it comes from another source, but it's rather simple to understand.
The source is an italian thread started from prof. Elio Fabri:
https://groups.google.com/d/msg/free.it.scienza.fisica/UyfGXka6rgQ/MiEiUTVexIsJ).

But if you intended that the mass spectrometer allows you to find the momentum |p| and so the total energy E and you identify the last with relativistic mass (multiplied c2), then we again come back to what I have said before several times in various threads, even answering to you, that is that relativistic mass is nothing else than another name for total energy E.

--
lightarrow

I am so glad that you posted this. I don't think many will appreciate the subtleties.
 

Offline alancalverd

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Yup. It happens every time I make ice cubes. I'm very surprised that one got by an experimental physicist like Alan.

The OP concerned water, not ice. Expansion on solidification is not unusual - printers' "hot type" and several other materials do it - but the expansion of liquid water over a significant range of cooling is remarkable and very important. 

Quote
And what's a "cyclotron driver" anyway and how is it related to this topic?
It's the electronic gubbins that turns the cyclotron from a drawing into a working machine. But don't worry your pretty head about that nasty engineering stuff - you'll get you hands dirty! 

Quote
There is absolutely no reason why a molecule of water, i.e. an H2O ion couldn't be accelerated in a cyclotron.

A molecule is not an ion. Which seems a good reason to me.
« Last Edit: 30/11/2014 18:03:58 by alancalverd »
 

Online jeffreyH

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Yup. It happens every time I make ice cubes. I'm very surprised that one got by an experimental physicist like Alan.

The OP concerned water, not ice. Expansion on solidification is not unusual - printers' "hot type" and several other materials do it - but the expansion of liquid water over a significant range of cooling is remarkable and very important. 

Quote
And what's a "cyclotron driver" anyway and how is it related to this topic?
It's the electronic gubbins that turns the cyclotron from a drawing into a working machine. But don't worry your pretty head about that nasty engineering stuff - you'll get you hands dirty! 

Quote
There is absolutely no reason why a molecule of water, i.e. an H2O ion couldn't be accelerated in a cyclotron.

A molecule is not an ion. Which seems a good reason to me.

A molecule yes but look at the hydrogen bonding.

http://www1.lsbu.ac.uk/water/water_hydrogen_bonding.html
 

Offline PmbPhy

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Quote from: alancalverd
The OP concerned water, not ice.
I didn't say he was. It was merely a thought on expansion of cold water.

Quote from: alancalverd
But don't worry your pretty head ...
So this is what you resort to when you loose an argument, huh?

I know a great deal about engineering since I worked in the field for years before becoming a physicist.

Quote from: alancalverd
A molecule is not an ion. Which seems a good reason to me.
An ion is an atom or a molecule for which the number protons is not the same as the number of electrons. That's how atoms and molecules can be accelerated in a cyclotron. And you claim to be an experimental physicist and you didn't know this simple fact that all physicists and chemists know?

You should learn the basics of physics before insulting and being rude to others. Start with these:

http://en.wikipedia.org/wiki/Ion
Quote
An ion (/ˈaɪən, -ɒn/)[1] is an atom or molecule in which the total number of electrons is not equal to the total number of protons, giving the atom or molecule a net positive or negative electrical charge.

http://en.wikipedia.org/wiki/Ion_cyclotron_resonance
Quote
Ion cyclotron resonance is a phenomenon related to the movement of ions in a magnetic field. It is used for accelerating ions in a cyclotron, and for measuring the masses of an ionized analyte in mass spectrometry, particularly with Fourier transform ion cyclotron resonance mass spectrometers. It can also be used to follow the kinetics of chemical reactions in a dilute gas mixture, provided these involve charged species.
See also - http://scienceworld.wolfram.com/chemistry/Ion.html
Quote
: an atom or group of atoms that has a positive or negative electric charge from losing or gaining one or more electrons
See the list of molecules in that page that they use as examples of ions.
« Last Edit: 01/12/2014 01:01:50 by PmbPhy »
 

Offline PmbPhy

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Quote from: lightarrow
No need of relativistic mass.
So what? Do you realized just how many concepts there are in physics that aren't "needed"? In fact Hertz wrote an entire mechanics text without using the concept of force. I could write an entire textbook on relativity without using the concept of energy either. Let me show you how easy it'd be to derive the cyclone formula by using relativistic mass. Let B = Bk where B is a constant. Let v start out in the xy-plane. Then

F = qvxB

means that the force is perpendicular to the velocity which means the charge moves in a circle and the force is perpendicular to the velocity. As such F = ma. If you're not aware that this is true then see
http://home.comcast.net/~peter.m.brown/sr/long_trans_mass.htm

F = ma = mv2/r = qvB

or

mv = qrB = p

which is the cyclotron relation, i.e.

mv = qrB

Which requires a great deal less work that what you did. :)
 

Offline PmbPhy

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Quote from: jeffreyH
A molecule yes but look at the hydrogen bonding.
You're wrong too, Jeff. A molecule which has a different amount of electrons than protons is an ion. Same holds true with atoms.
 

Offline CliffordK

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As far as heat and relativistic mass, if it could be demonstrated for any substance, water, hydrogen, helium, radon, lead, etc, then it could be considered valid for all substances. 

While H2O is not an ion, water is always a mix of ions, H2O, OH, H+, and H3O+.  With acid/base chemistry, it is relatively easy to force one or another ion to be more prevalent, at last in liquid water form.  Gases may tend to be more neutral.  There is no reason why water couldn't be used as a source of ions for analysis.  However, there may be better choices of materials for analysis.
 

Offline alancalverd

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A molecule is by definition electrically neutral and thus cannot be accelerated by a cyclotron.

An ion may have the same nuclides, almost the same mass and even most of the shape of a molecule but it is distinguished by having a charge, and thus quite different behaviour in an electric or magnetic field.  Which is why it has a different name, and why we ionise molecules in a mass spectrometer.

It is worth reconsidering the Wikipedia definitions. If I have an ion with 6 protons and 7 electrons, is it carbon plus an electron or nitrogen minus a proton? If we incorporate the nuclide into a fairly simple compound, the difference is between an inert plastic and an explosive, so we need to be a bit careful with our definitions here. 
 

Offline PmbPhy

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Quote from: alancalverd
A molecule is by definition electrically neutral and thus cannot be accelerated by a cyclotron.
You're posting nonsense now. I said it was an ion of a H2O, not a water "molecule." I also made it clear by posting definitions that an ion is a molecule that is not neutral (you don't appear to be reading the definitions that I've posted to help you grasp the nomenclature) which means that you start off with a molecule and then add or subtract electrons to create the ion. I was using ions because they can be accelerated in a cyclotron.

When I kept referring to ions of a H2O, and you kept changing it to a water molecule you introduced a logical fallacy known as a straw argument or straw man.

An ion of  H2O (i.e. a water ion) is not neutral and therefore can be accelerated. Since the mass is very close to that of a neutral molecule it will suffice for the purpose I was arguing, i.e. to measure the mass of a particle of water (i.e. a water molecule).

Since you've turned to using logical fallacies such as this then I can't see any reason to continue making an attempt to reason with you. Clearly you don't understand the nomenclature too. In the physics/chemistry community when speaking of ions one uses the term "ion" (or cation, anion ect) and THEN name of the related molecule. See the tables in http://en.wikipedia.org/wiki/Ion  so one says "ion of water" etc.

You started off in this thread confusing density with weight so you're just too illogical for me.
« Last Edit: 01/12/2014 13:50:03 by PmbPhy »
 

Offline PmbPhy

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Quote from: CliffordK
While H2O is not an ion, ...
Now you're making the same mistake alan was. I never said a molecule of water/H2O. I kept saying and ion of H2O. People really need to read more carefully before criticizing what they're reading.
 

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Quote from: jeffreyH
A molecule yes...
It appears that you're like others and didn't read what I posted carefully. I only spoke of H2O ions. I never spoke of H2O molecules.
 

Offline alancalverd

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To do that one can measure the mass of a moving particle such as a molecule by ionizing it and using a cyclotron to measure its mass by determining it's deflection by a magnetic field and measuring its radius of orbit.

You are right - I missed "by ionizing it". Mea culpa - unless that was a late edit, in which case you are a scoundrel, sir!

Pistols at dawn? Should give me a 5 hour advantage.

And anyway, if you ionize a water molecule by removing a proton, you will have changed its mass, its chemistry, and its bulk physics, significantly. So there! :)
 

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...
|p| = q*|B|*r
...   
I am so glad that you posted this. I don't think many will appreciate the subtleties.
That's true. Prof. Elio Fabri has taught Theoretical Physics at the univeristy of Pisa. He specifically taught courses of Quantum Mechanics, Special and General Relativity, Quantum Field Theory, Astrophysics, Classical Mechanics and Electrodynamics, among other subjects.

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Offline lightarrow

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A benchtop mass spectrometer running at a few keV probably won't show the relativistic change of mass to a detectable level
...
but a hospital cyclotron certainly will, 
No, never. It will only show the relativistic change of momentum or of energy E.

*If then you define* relativistic mass R.M. = E/c2 then of course it would show relativistic increase of that, but it's just the energy E (apart a constant multiplied to it) with another name!

Neither a mass spectrometer nor a cyclotron nor *any other device* will ever show a relativistic change of mass.

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Offline lightarrow

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Quote from: lightarrow
No need of relativistic mass.
So what? Do you realized just how many concepts there are in physics that aren't "needed"? In fact Hertz wrote an entire mechanics text without using the concept of force. I could write an entire textbook on relativity without using the concept of energy either. Let me show you how easy it'd be to derive the cyclone formula by using relativistic mass. Let B = Bk where B is a constant. Let v start out in the xy-plane. Then

F = qvxB

means that the force is perpendicular to the velocity which means the charge moves in a circle and the force is perpendicular to the velocity. As such F = ma. If you're not aware that this is true then see
http://home.comcast.net/~peter.m.brown/sr/long_trans_mass.htm

F = ma = mv2/r = qvB

or

mv = qrB = p

which is the cyclotron relation, i.e.

mv = qrB

Which requires a great deal less work that what you did. :)
Why didn't you write the equation:

F = ma = mv2/r = qvB

with F, a, v, B in bold carachter? It means that it's not valid vectorially? (Of course it's not  :)) And what usefulness it has an equation involving vectors that it's not valid vectorially? It can be useful just in very special cases like this one, just because in my example I used a particle's initial velocity v orthogonal to B (and you correctly took the same example because you knows well that F = ma it's not true in general  :)). Try to make the computations with any v;)

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« Last Edit: 01/12/2014 12:58:26 by lightarrow »
 

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lightarrow - I don't know why you're trying so hard to make this a debate about relativistic mass but I'm really not interested and I won't discuss it.
 

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lightarrow - I don't know why you're trying so hard to make this a debate about relativistic mass but I'm really not interested and I won't discuss it.
In your first post of the thread "A poll on relativistic mass" you asked yourself why I believe relativistic mass is merely another name for energy, now you know  :).

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Quote from: lightarrow
Why didn't you write the equation:

F = ma = mv2/r = qvB

with F, a, v, B in bold carachter?
Because I had no use for it. I wrote down the magnitudes involved rather than the vectors since it was much easier to do so and its exactly correct. One should always make things easier when one can. Didn't you know this about using the equivalent scalar equations from the associated vector equations when possible? If not then I find that surprising. You saw how easy it was doing so. It'd take more work not to. Besides, that's how most physicist derive it, i.e. using the scalar equation and not the vector equation. I take it that you didn't know that either?

Also, you didn't really say why you believe that relativistic mass is merely another name for energy. All you really did is make a baseless assertion that it is merely another name. However it is in fact not merely a different name for it. There are many reasons why but the real and most obvious reason is that explained in the journal article On The Meaning of E = mc2 by Mendel Sachs, Int. J. Theo. Phys., 8 (1973) who writes
Quote
... inertia and energy refer to conceptually different features of matter. On the one hand, inertia per se relates to the resistance to the change of state of motion of a body, as would be caused, in 'particle physics,' by an external force acting on the body (...).
   On the other hand, energy per se is defined in terms of the solutions of the conservation laws.....
Thus we see that energy and mass are conceptually different (though complimentary) features of matter, according to relativity theory.
Several relativists have made this argument, including me. Just because two quantities are proportional it doesn't mean that they are the same thing. E.g. you can't legitimately claim that since the energy is related to the frequency of a photon by E = hf that energy and frequency are the same thing.

Besides, E = mc2 does not work under all situations. E.g. if you have a rod which is at rest lying along the x-axis in the inertial frame S and there is  force on each end of the rod of equal magnitude but opposite directions so as to stress it then it will be no acceleration of the rod but it will have a finite energy. When you transform to S' which is in standard configuration with S and moving in the +x direction then E = mc2 will no longer be true.

Also, if you are in a non-inertial frame E = mc2 will also not be true. The energy will be proportional to the time component of the momentum 1-form. However relativistic mass will still be the time component of the 4-momentum.

I didn't want to talk about your mistakes on this subject but since you took this off the thread where I was only looking for opinion you changed the purpose of it.

Nothing personal my friend. :)
« Last Edit: 01/12/2014 15:16:30 by PmbPhy »
 

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