Roger's Equations

Hi Roger, sterling job once again!

Regards, Jorrie

[edit] P.S. There's a couple of important typos for you to fix in the constants section. I originally only noticed the superscript 10⁸ in the speed of light, but some others are also lacking a bit!

Thanks Jorrie. I'm not sure why, but all of my subscripts and superscripts went away in the constants section. I have fixed them. Thanks for letting me know.

I have always been puzzled by the concept of Newtons.

I can go to the shops to by a 1 lb weight (or a 1 kg weight) and I know that if I placed the 1 lb weight on a piston if 1 sq.inch the pressure under the piston would be 1 pound per sq.inch Similarly I could do the same with the 1 kg weight on a 1 sq. metre piston that would give a pressure of 1 kg per m2.

Along these lines, and working back from a pressure of 1 pascal, which is 1 Newton per m2, thus from a piston of 1 sq.metre there must be a corresponding weight of 1 Newton.

You get very quizzical looks when you try to buy 1 Newton weight. They don't exist??

If they did, what would they weigh in terms of lb's or kg's.

"I have always been puzzled by the concept of Newtons."

Ah the age old confusion caused by imperial weights and measures.

It all boils down to the fact the Kg is a measure of mass which is the ability of an object to resist a change in motion. In the imperial system this was measured in pounds.

Newtons on the other hand are a measure of how hard an object is being pushed and an object will accelerate at a rate proportional to how hard it is being pushed and inversely proportional to its mass. Unfortunately the imperial system also uses pounds to measure force. Put mathematically

Acceleration = Force / Mass

Lets look at a 1 Kg mass sitting on a frictionless, horizontal rail on Earth. Gravity will be trying to accelerate the 1 Kg mass downwards there fore there will be a 9.91 N acting downwards on the mass but because this force is perpendicular to the frictionless rail nothing happens. If I now apply a 1 N force to the 1 Kg mass parallel to the rail it will accelerate at 1 ms^-2. So far so good I hope.

If we now take our 1 Kg mass and frictionless rail to the moon it still has a mass of 1 Kg but now the force acting vertically because of the moons gravity is only 1.64 N and again nothing happens because it is perpendicular to the rail. If I now apply a 1 N force to the mass the same as I did on earth it would still accelerate at 1 ms^-2 because its mass has not changed. Still with us?

OK lets now do it with imperial units. We have a mass of 2.204 lb on earth the same as before. Because of gravity this produces a downward force of 2.204 lb and this is where the confusion comes in, they both use the same units. If I now push the mass the same as we did above with a force of 0.224 lb it will accelerate horizontally at 3.26 fs^-2. Confused yet?

Ok lets take the whole thing to the Moon and now we have a mass of 2.204 lb but because the Moons gravity it is only producing a downward force on the rail of 0.367 pounds. So we have a 2.204 pound mass that weighs 0.227 pounds. If I now push the 2.204 lb mass that weighs 0.367 lb with a force of 0.224 lb parallel to the frictionless rail it will still accelerate at 3.26 fs^-2.

Now I'm confused and you think the concept of Newtons is confusing.

"You get very quizzical looks when you try to buy 1 Newton weight. They don't exist??"

This is because you are not buying a 1 Newton weight you are buying a 0.102 Kg mass not a weight.

Always remember mass is a measurement of an objects ability to resist a change in motion and will remain the same¹ regardless of its location. The Newton is a measure of how hard an object is pushed in order to change the motion of an object. Weight is a measure of how had gravity is pushing/pulling on an object and varies according to location.

I hope this has helped because trying to explain acceleration and force using imperial units has confused the hell out of me!

Note 1. This isn't absolutely true when velocities acre close to that of light but for normal situation it can be treated as such.

Hello Masu. That was a trip down memory lane - thanks for the trouble.

But because 1 kg weight and 1 kg mass are linked by gravity (is 9.91 you used a typo = 9.81 N) here on earth we should be able to buy a 1 Newton weight for use on earth. As you would expect, it would weigh 2.204 lb's on my imperial scale or 0.102 kg's on my metric scales.

I imagine gardeners would get the muddled up with slugs.

Regards.

"But because 1 kg weight and 1 kg mass are linked by gravity (is 9.91 you used a typo = 9.81 N) here on earth we should be able to buy a 1 Newton weight for use on earth. As you would expect, it would weigh 2.204 lb's on my imperial scale or 0.102 kg's on my metric scales."

Yes you are 100% correct it's a typo. It's reassuring when readers notice thing like that because it shows that they have comprehended what I am trying to say.

One thing though a 1 Newton weight would indeed have a mass of 102 g but this is 0.225 lb or 3.594 oz not 2.204 lb. The more I need to deal with imperial weights and measures the more I hate them.

Masu. Thanks for the correction. 2.204 lbs has been used for 1 N but I must have been thinking of 1 kg. Mine was not a typo as such, it was slip of memory.

Flitting from imperial to metric and vice versa, and bringing gravity into the equation, for sake of technical correctness is fine for the scientists, but it makes everyday life complicated for ordinary individuals who are stuck with earths gravity - where we round a kg to 2.2 lbs.

Perhaps there is a market for 1 N weights.

Hi Roger

Great blog, and there are a few of points I feel you might make a little more of -.

First, something we discussed earlier - units for extended dimensional analysis:
Steradian: m^2/m^2
Refractive Index: m/m

Next: Speed of light - no longer a physical constant in the sense that (say) the gravitational 'constant' G would be, but a definition of the relationship between length and time: c = (exactly) 299,792,458 m/s

Finally the candela: in my view, this is in reality a unit of photometry or illumination (physiological) rather than a unit of light (physical). It is defined for light of a specific wavelength (~0.5517...um), and everything else is converted via tables of 'average response' vs wavelength under some arbitrarily defined conditions. So far as I am concerned, the units for light are identical to those for electrodynamics. Should readers be warned? Should there be a note/link pointing to electrodynamics?

BTW, I once worked (not for very long) in a photometric standards labs, in the days when the standard for the candela was black-body radiation at the melting point of platinum. Did we ever have fun trying to maintain that as a standard?

Fyz

Physicist wrote: "Speed of light - no longer a physical constant in the sense that (say) the gravitational 'constant' G would be, but a definition of the relationship between length and time: c = (exactly) 299,792,458 m/s".

Good point! But does this make it less of a unversal constant? I think the notion that it is no longer a natural constant came about when the meter was defined as the distance that light travels in 1/299,792,458 seconds.

Jorrie

I though I was being careful with my wording - but perhaps not careful enough. It's still a universal physical constant. But not measured as such. The measurement is in the form of increasing the precision of the length of the second and the meter - to all intents independently. (Things could become 'interesting' if special relativity is ever found to be incomplete - in the sense that there are inertial conditions where the speed of light is not isotropic...)

Fyz

Hello Roger

Good summary again!

Only thing I'd comment on is definition of Avogadro's Number (N_A) - 6.02 x 10²³ mol^-1
and Universal Gas Constant (R) - 8.31 J/mol·K

These are correct if by mol you mean gram-mol. But as everything else is in SI I think it's better to use kg-mol (Kmol) - N_A = 6.02 x 10²⁶ Kmol^-1 and R = 8310 J/Kmol·K. Probably you meant this but some readers might get the wrong idea. E.g. 18 gm of water contain 6.02 x 10²³ molecules, 18 kg of water contain 6.02 x 10²⁶ molecules. I've seen molecular weight expressed as kg/mol, so (e.g.) water is 18 gm/mol = 0.018 kg/mol, but I think this causes more confusion than it saves. In any case I believe MW is still defined as ratio of mass of molecule to 1/12 of mass of a carbon atom, so water = 18.

You Wrote: "These are correct if by mol you mean gram-mol. But as everything else is in SI I think it's better to use kg-mol (Kmol) - N_A = 6.02 x 10²⁶ Kmol^-1 and R = 8310 J/Kmol·K."

From my research, the SI unit for the amount of a substance is Mole (mol), not gram-mol or Kmol. Please see the definition for the unit in the link to NIST below.

http://www.physics.nist.gov/cuu/Units/units.html

Avogadro's number has units of mol^-1. Atomic mass is in units of g/mol, so when multiplied by Avogadro's number you get the number of grams of one mole of a substance. Here is a description from wikipedia;

"The Avogadro constant can be applied to any substance. It corresponds to the number of atoms or molecules needed to make up a mass equal to the substance's atomic or molecular mass, in grams. For example, the atomic mass of iron is 55.847 g/mol, so N_A iron atoms (i.e. one mole of iron atoms) have a mass of 55.847 g."

http://en.wikipedia.org/wiki/Avogadro's_number

You Wrote: "I've seen molecular weight expressed as kg/mol, so (e.g.) water is 18 gm/mol = 0.018 kg/mol, but I think this causes more confusion than it saves."

Regardless, kg/mol is the correct way to express molecular weight in the SI system. There is no doubt that their are other unit systems better equipped to handle the atomic scale, but that is outside the scope of this post (maybe a future post?)

Reply to #12

It's all OK, it's just that I'm a little nervous about bringing grams into the SI system, and thus the need to beware we're not out by 1000. Specially as the figures often aren't intuitive.

E.g. if we multiply electronic charge 1.6*10^-19 coulomb by N_A = 6.02 x 10²³ mol^-1 to get 1 Faraday (10⁵ coulomb in round nos), this liberates about 19 gram of (trivalent) iron (to use the Wiki example) not 19 kg.

Cheers

The SI unit for amount of substance is Mole (mol). No grams involved. For mass I've already stated the default unit is kilograms. Faraday (1 mole of electrons) is no longer a recognized unit in the SI system (see link below).

http://en.wikipedia.org/wiki/Faraday

You wrote The SI unit for amount of substance is Mole (mol). No grams involved. Some of the definitions of mole aren't entirely clear to me, but in practice, to my mind a mole is the mass of substance containing N_A molecules. So if N_A = 6.02 x 10²³ mol^-1 it's the mass in grams, e.g. 18 for water. Please tell me if you don't agree.

When working out mass liberated in electrolysis, if we can't use the faraday now, we either have to define something else or work direct from electronic charge and N_A , and again if N_A = 6.02 x 10²³ mol^-1 is used, calc gives gram-equivalent liberated.

Roger, I'm not trying to have an argument, just saying a bit of care is needed!

Codemaster,

The SI unit system has set definitions. It's not a matter of opinion. Please view the following links.

http://www.chemistrycoach.com/moles.htm
NIST Link
SI Units - Wikipedia


You Wrote: "to my mind a mole is the mass of substance containing N_A molecules."

This is incorrect. A mole is just a number of elementary molecules. Just like a dozen means "12 of something", a mole means "6.02 x 10²³ of something". That's it. No mass involved in the definition.