A question about quantities vs units in physical laws

In summary, the symbols in a physical law represent both the units and the dimensionless numbers. The dimensionless number and the unit are not fully independent of each other and can change depending on the set of units used. However, when using a consistent unit system like SI, the units cancel out and the equation can be written in terms of dimensionless quantities. It is still important to include units in calculations to fully specify the meaning and physical quantities involved.
  • #1
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.
 
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  • #2
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.
 
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  • #3
Drakkith said:
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, $$F = \frac{Q_1 Q_2}{4\pi \epsilon_0 r^2} = \frac{{\lambda_{Q}}_1 \text{C} \times {\lambda_{Q}}_2 \text{C}}{4\pi (\lambda_{\epsilon_0} \text{C}^{2}\text{N}^{-1}\text{m}^{-2}) (\lambda_r \text{m})^2} = \frac{{\lambda_{Q}}_1 {\lambda_{Q}}_2}{4\pi \lambda_{\epsilon_0} {\lambda_r}^2} \text{N}$$
 
  • #4
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.
 
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  • #5
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.
 
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  • #6
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!
 
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  • #7
etotheipi said:
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.
 
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  • #8
Ibix said:
etotheipi said:
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.
 
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1. What is the difference between quantities and units in physical laws?

Quantities refer to the numerical values that represent a physical property, such as length, time, or mass. Units, on the other hand, are used to measure these quantities and give them a specific value. For example, a quantity of length can be measured in units such as meters, feet, or inches.

2. Why is it important to distinguish between quantities and units in physical laws?

It is important to distinguish between quantities and units in physical laws because they serve different purposes. Quantities represent the physical properties being measured, while units provide a standardized way of expressing those quantities. This helps to ensure that measurements are accurate and consistent across different contexts.

3. Can quantities and units be used interchangeably in physical laws?

No, quantities and units cannot be used interchangeably in physical laws. While they are related, they serve different functions and must be used correctly in order to accurately represent physical phenomena. For example, using the wrong unit for a quantity can lead to incorrect calculations and conclusions.

4. How do quantities and units relate to each other in physical laws?

Quantities and units are related by a mathematical relationship. The unit is used to measure the quantity, and the numerical value of the quantity is multiplied by the unit to give it a specific value. This relationship is crucial in understanding and applying physical laws.

5. Are there any exceptions to the use of quantities and units in physical laws?

There are some cases where quantities and units may not be explicitly stated in physical laws, but they are still implied. For example, in Newton's Second Law, the quantity of force is represented by the unit of Newtons, but the unit is not mentioned in the law itself. However, it is still implied that force is being measured in Newtons.

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