A question about quantities vs units in physical laws

[etotheipi]

A quantity $p$ can be expressed as the product of a dimensionless number, $\lambda_p$, and a unit, $u_X$:$$p = \lambda_p u_X$$When we write the equation of a physical law, do the symbols represent the physical quantities $p$ or their dimensionless coefficients $\lambda_p$? That is to say would the law read $p_1 = f(p_2, p_3)$ or ${\lambda_p}_1 = f({\lambda_p}_2, {\lambda_p}_3)$? I am inclined to say that the former is true.

 

[Drakkith]

Looks to me like the symbols represent both the units and the dimensionless numbers. Both the dimensionless number and the unit will change depending on which set of units you choose to use. Neither are fully independent of each other in this sense.

Or, another way to look at it, the symbols just represent non-dimensionless (dimensional?) quantities.

 

[etotheipi]

Looks to me like the symbols represent both the units and the dimensionless numbers.

That certainly makes more sense in the context of dimensional analysis. E.g. for Coulomb's law expressed in SI units, $$F = \frac{Q_1 Q_2}{4\pi \epsilon_0 r^2} = \frac{{\lambda_{Q}}_1 \text{C} \times {\lambda_{Q}}_2 \text{C}}{4\pi (\lambda_{\epsilon_0} \text{C}^{2}\text{N}^{-1}\text{m}^{-2}) (\lambda_r \text{m})^2} = \frac{{\lambda_{Q}}_1 {\lambda_{Q}}_2}{4\pi \lambda_{\epsilon_0} {\lambda_r}^2} \text{N}$$

 

[Mister T]

Units can be dimensionless, for example, the radian.

When you write something like the displacement $\Delta x=2\ \mathrm{m}$ what you are saying is that the displacement is twice as large as a displacement of one meter. You need both the dimension (meter) and the scaling factor (2) to fully specify the meaning.

 

[Ibix]

You need the units in there. Take $F=ma$. If I have a mass of 400g and an acceleration of 20ms^-2 then we have a force of 8000 gms^-2, which you are free to convert into Newtons if you want. If the maths didn't include the units there'd be no way to deduce the units on the left - at least, not without carrying out an auxiliary calculation that boils down to just putting the units in in the first place. Also, these are physical quantities, and the coefficient isn't the quantity. My bag of flour doesn't weigh 1, it weighs 1kg.

Of course, if you use a consistent unit system like SI, it makes no difference in practical terms.

 

[etotheipi]

Thanks for all of your replies, that is what I had hoped was the case !

The ambiguity arose because something like $\lambda_F = \frac{{\lambda_{Q}}_1 {\lambda_{Q}}_2}{4\pi \lambda_{\epsilon_0} {\lambda_r}^2}$ is still a valid equation. I think this is what you mean by it not making any difference in practical terms.

But yes, it makes more sense that the whole quantity appears in the physical law. Not least because it's much more coherent in a dimensional analysis sense. Thank you!

 

[Ibix]

The ambiguity arose because something like $\lambda_F = \frac{{\lambda_{Q}}_1 {\lambda_{Q}}_2}{4\pi \lambda_{\epsilon_0} {\lambda_r}^2}$ is still a valid equation.

I suppose you could argue that the units cancel out, leaving just the coefficients in your equation above. I'd still prefer to think of the equation as manipulating unitful quantities, first because I get the units of my answer and second because this is physics - they're not just numbers, they're physical quantitie.s But as long as you use a consistent unit system like SI then you know your units and it doesn't make any practical difference.

 

[robphy]

The ambiguity arose because something like $\lambda_F = \frac{{\lambda_{Q}}_1 {\lambda_{Q}}_2}{4\pi \lambda_{\epsilon_0} {\lambda_r}^2}$ is still a valid equation. I think this is what you mean by it not making any difference in practical terms.

But yes, it makes more sense that the whole quantity appears in the physical law. Not least because it's much more coherent in a dimensional analysis sense. Thank you!

I suppose you could argue that the units cancel out, leaving just the coefficients in your equation above. I'd still prefer to think of the equation as manipulating unitful quantities, first because I get the units of my answer and second because this is physics - they're not just numbers, they're physical quantitie.s But as long as you use a consistent unit system like SI then you know your units and it doesn't make any practical difference.

Assuming the units cancel out, the equation of dimensionless quantities is what is used in a typical calculator to compute a numerical answer.
(Sometimes in Maple I include variables representing units
to verify units in complicated calculations...
e.g. when a student offers an unsimplified expression that I have to check.)

Of course, to completely specify the numerical answer,
the appropriate units that got canceled out in the original equation must be appended to the calculator result.