Special and General Relativity Special and General Relativity Physics Help Forum  10Likes
Oct 26th 2017, 06:23 AM

#21  Senior Member
Join Date: Apr 2015 Location: Somerset, England
Posts: 959

I was just wondering why this format pops up so often , always a half then one item , and another item squared

Now that you have your units sorted, perhaps you would like an answer to this?
The point is that they all refer to the work that need to be expended to get from a zero point to a given point through some measure of spatial distance.
I have called the initial zero point A and the end point B. So it is the work expended getting (the system) from A to B.
To expend this work some sort of 'driving force', appropriate to physical phenomenon concerned I have denoted by H acts so that H is proportional to its distance s from A.
So in getting from A where there is zero resistance to the driving force to B where there is a resisting force = a constant times s.
The maths is the same in each situation, when H , s , and the constant k are expressed in suitable units.
This proportionality is the commonest situation in the Universe.
Last edited by studiot; Oct 26th 2017 at 06:54 AM.

 
Oct 26th 2017, 08:15 AM

#22  Forum Admin
Join Date: Apr 2008 Location: On the dance floor, baby!
Posts: 2,379

Originally Posted by oz93666 
Your table is wrong. Or in a system of units I'm not familiar with. But it is not any version of the MKSA system I've run into. Unless we do something drastic, like setting $\displaystyle \hbar = c = 1$ as in the HeavisideLorentz system of units, there have to be four base units that are independent of one another: length, time, mass, and charge.
Originally Posted by oz93666 A very curious sentence in your post ..."The formulas are defined in such a way to be as useful as possible" 
For instance, if we did decide to call kinetic energy $\displaystyle 1/2 mv^4$, which has similar overall properties to the way we usually define it, the workenergy theorem would have to be taken as $\displaystyle W = \frac{KE}{v^2}$ in order to work. It's just easier to define kinetic energy as $\displaystyle 1/2 mv^2$. One of the most difficult tasks for a Physicist who has discovered something new is to find a way to express that as simply as possible. If you want to see what I mean, peruse Newton's Principia in its original format. Algebra is sooooo much nicer now!
Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.
See the forum rules here.

 
Oct 26th 2017, 08:29 AM

#23  Senior Member
Join Date: Apr 2015 Location: Somerset, England
Posts: 959

I had understood they switched from charge to current in the dimension system some while back.
There are other fundamental units required such as a degree of temperature and a unit of light intensity.

 
Oct 26th 2017, 11:50 AM

#24  Forum Admin
Join Date: Apr 2008 Location: On the dance floor, baby!
Posts: 2,379

Originally Posted by studiot I had understood they switched from charge to current in the dimension system some while back.
There are other fundamental units required such as a degree of temperature and a unit of light intensity. 
Yes, you are right, of course.
I learned the basic of unit analysis using what might be called the MKSC system rather than MKSA. I tend to think in C instead of A which occasionally screws me up in my Electrodynamics.
Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.
See the forum rules here.

 
Oct 26th 2017, 08:50 PM

#25  Senior Member
Join Date: Apr 2017
Posts: 314

Originally Posted by topsquark Your table is wrong. Or in a system of units I'm not familiar with. But it is not any version of the MKSA system I've run into. 
I checked it all ...here's the link again .... http://www.phys.ufl.edu/courses/phy2...049Ramond.pdf ....just one of many, that say the same thing ...
To be clear this is not a system of measurement , like the metric or imperial system , but the concept that any physical quality can be resolved down to it's three basic components of mass length and time ...
It's easy to see that acceleration can be resolved dimensionally to L T2 ...or velocity to L T1
But inductance , at it's core is just ... L1 T2
Current is ...M1/2 L 3/2 T2
Magnetic field ..... M1/2 L3/2
It seems there are only 3 basic components in reality , length mass and time , all other manifestations are a weaving together of these three.

 
Oct 26th 2017, 09:53 PM

#26  Forum Admin
Join Date: Apr 2008 Location: On the dance floor, baby!
Posts: 2,379

Originally Posted by oz93666 I checked it all ...here's the link again .... http://www.phys.ufl.edu/courses/phy2...049Ramond.pdf ....just one of many, that say the same thing ...
To be clear this is not a system of measurement , like the metric or imperial system , but the concept that any physical quality can be resolved down to it's three basic components of mass length and time ...
It's easy to see that acceleration can be resolved dimensionally to L T2 ...or velocity to L T1
But inductance , at it's core is just ... L1 T2
Current is ...M1/2 L 3/2 T2
Magnetic field ..... M1/2 L3/2
It seems there are only 3 basic components in reality , length mass and time , all other manifestations are a weaving together of these three. 
I have finally identified the system of units your source is quoting. It's called the Gaussian system of units. (Read the top couple of paragraphs of section 2 and you'll see the definition of the unit of charge to be the same as in your source.) In Gaussian units charge is measured in statcouls and in the MKSA system charge is measured in As = C.
Lots and lots of confusion when we work in different unit systems.
Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.
See the forum rules here.

 
Oct 26th 2017, 10:16 PM

#27  Senior Member
Join Date: Apr 2017
Posts: 314

Originally Posted by topsquark I have finally identified the system of units your source is quoting. It's called the Gaussian system of units. (Read the top couple of paragraphs of section 2 and you'll see the definition of the unit of charge to be the same as in your source.) In Gaussian units charge is measured in statcouls and in the MKSA system charge is measured in As = C.
Lots and lots of confusion when we work in different unit systems.
Dan 
This is absolutely not the case .
The gaussian system is just the cgs system , based on cm / grams /secs, compared to the MKS which is meters/ Kg/ secs ...
Dimensional analysis is about is analysing a phenomena , like magnetism , down to it's core essence , expressing it in terms of the three fundamental components of mass length and time .... We are not quantifying (measuring ) anything ... this is an analysis above quantity so it's totally irrelevant what system of measuring is latter used ... feet or meters makes no difference.
Last edited by oz93666; Oct 26th 2017 at 10:20 PM.

 
Oct 27th 2017, 11:05 AM

#28  Forum Admin
Join Date: Apr 2008 Location: On the dance floor, baby!
Posts: 2,379

Originally Posted by oz93666 This is absolutely not the case .
The gaussian system is just the cgs system , based on cm / grams /secs, compared to the MKS which is meters/ Kg/ secs ...
Dimensional analysis is about is analysing a phenomena , like magnetism , down to it's core essence , expressing it in terms of the three fundamental components of mass length and time .... We are not quantifying (measuring ) anything ... this is an analysis above quantity so it's totally irrelevant what system of measuring is latter used ... feet or meters makes no difference. 
In the Gaussian system the permittivity of free space $\displaystyle \epsilon _0$ is set equal to 1 and is unitless. This is not the case in the CGS system.
Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.
See the forum rules here.

 
Oct 27th 2017, 12:58 PM

#29  Physics Team
Join Date: Apr 2009 Location: Boston's North Shore
Posts: 1,532

Originally Posted by topsquark For instance, if we did decide to call kinetic energy $\displaystyle 1/2 mv^4$, which has similar overall properties to the way we usually define it, the workenergy theorem would have to be taken as $\displaystyle W = \frac{KE}{v^2}$ in order to work. It's just easier to define kinetic energy as $\displaystyle 1/2 mv^2$. 
Hmmmmm .... I can't say that I agree with what you say here other than defining kinetic energy in any other way is akin to defining a cow as anything that walks on four legs.
It merely turns out that the amount of work done on an object is the difference in the quantity K = (1/2)mv^2. We "choose to define that quantity as kinetic energy. If the force doing the work is conservative then that same amount of work is the difference in the quantity U. Here we "choose" to define U as potential energy. This means that the quantity E = K + U is conserved. Sure. You can define kinetic energy as mv^ and potential energy as 100U but there's no compelling reason to do so. Defining these quantities as I just descried is much more than making things easy though.
This same argument can be applied to all closed systems.

 
Oct 27th 2017, 01:18 PM

#30  Senior Member
Join Date: Apr 2015 Location: Somerset, England
Posts: 959

If the force doing the work is conservative

And if its not?
What does that do to your definition?

  Thread Tools   Display Modes  Linear Mode  