Dimensionless Equations: Understanding the Role of Units in Physical Equations

  • Thread starter Thread starter puzzler7
  • Start date Start date
AI Thread Summary
The discussion centers on the confusion surrounding the treatment of units in physical equations, particularly when converting mass from kilograms to grams. The key issue arises from the misunderstanding that mass in different units should be treated as direct multipliers rather than considering the necessary adjustments for dimensional consistency. The participant concludes that all physical equations should be viewed as dimensionless, emphasizing that the choice of units must adapt to maintain this consistency. The resolution lies in recognizing that mass must be divided by its unit when substituting, leading to correct formulations. Ultimately, the conversation highlights the importance of understanding the relationship between numerical values and their corresponding units in physics.
puzzler7
Messages
3
Reaction score
0
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: let's 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!?
 
Last edited:
Physics news on Phys.org
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]
 
uart said:
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!)
 
Last edited:
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.
 
I think it's easist first to watch a short vidio clip I find these videos very relaxing to watch .. I got to thinking is this being done in the most efficient way? The sand has to be suspended in the water to move it to the outlet ... The faster the water , the more turbulance and the sand stays suspended, so it seems to me the rule of thumb is the hose be aimed towards the outlet at all times .. Many times the workers hit the sand directly which will greatly reduce the water...
Back
Top