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Dimensionless equations

 
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Mar4-11, 06:05 AM   #1
 

Dimensionless equations


I have a problem that might sound simple, but has been bugging me for months. In a physical equation, the units are regarded as multipliers - so to take a very simple example in SI:

1) F[N]=M[kg]a[m/s2]

And, of course, [N] is equivalent to [kg][m/s2], so all is well.

Here's my problem: lets say I want to adjust the equation, so that my mass measurements are in grams [g] rather than [kg].

Direct substitution for 1kg = 1000g into equation 1) gives:

2) F(N)=1000M[g]a[m/s2]

Which is clearly incorrect.

(a mass of 1g accelerated at 1 m/s2 would compute a force of 1000N - wrong - The equation actually needs to be divided by 1000 on the RHS.)

The logic looks perfect - but the result is wrong.

The problem is resolved in *all* equations by regarding the algebraic symbols to be *divided* by the unit - so why do we consider them to be multiplied????

What's my problem!!!!???
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Mar4-11, 07:44 AM   #2
 
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Hi puzzler. Numerically the mass in kg is 1/1000-th of the mass in grams.

So the correct substitution is F[N] = M[grams]/1000 a[m/s^2]
Mar4-11, 07:54 AM   #3
 
Quote by uart View Post
Hi puzzler. Numerically the mass in kg is 1/1000-th of the mass in grams.
But this doesn't mean M(kg) = M(g)/1000 Because that means 1000kg = 1g

Do you see the problem? In this verbal equation, *in* must stand for divide by - not multiply.

If we apply the rule that *in* stands for divide, the verbal equation now works:

M/kg = (1/000) M/g

This is consistent. But it brings us back to the original point - numbers in physical equations are divided by their units - not multiplied.

Maybe it looks like this: M(kg)/[kg]

The parentheses give the expected unit: the brackets the divisor. This makes the whole equation dimensionless.

Now I can substitute [1kg] = [1000g], and everything will work as expected.

Can anyone expand - it's a worry!!

(here's a thought: M of kg = 1000 x M of g. But M in Kg = (1/000) M in g. Note the difference between 'of' multiply, and 'in' divide'. I've been doing physics for 20 years - and I'm suddenly puzzled!!)
Mar4-11, 10:28 AM   #4
 

Dimensionless equations


I think I've solved my own problem: but I would appreciate all comments and criticism:

It seems that, maybe, all physical equations must, ultimately, be numerical, and therefore - ultimately dimensionless.

I suggest my mistake is in thinking that the M, in F=Ma, is fixed - as the unit quantities move around it.

(note that: F=Ma is just a simple example - the rules are expected to apply to all equations)


Heres a solution: If M is not fixed, we can have M1, M2 - two versions:

if M1(kg) = M2(g)

Now we can write:

1000M1(g) = M2(g)

Hence M1=M2/1000

If we substitute this into the equation we get the desired result (see above)

The conclusion *must be* that all physical equations - even if they appear to have units - are ultimately dimensionless.

We make them dimensionless by correct choice of units.

If we change those units - we must allow the dimensionless equation to adapt to the units (not vice versa)

Ultimately, even in physical equations, numbers = numbers.

Please feel free to argue and discuss.
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