Generate dimensional formulae and perform consistency checking

This introduction will teach how to

generate dimensional formulae
generate consistency checks
standardize correct equation

How to Generate Dimensional Formulae

1. The Problem

Imagine you derived the equation below based on the experimental observations

x = x₀ + v² + t + (1/2)at²

You want to know if this derived equation is correct. Using diman you can perform a preliminary check with consistency analysis. But before you can check for dimensional consistency you need to set it up for the analysis.

2. Setting Up for Generation

2.1. Definitions setup

Define all the symbols in the mathematical expression that is associated with a dimension.

(def varpars [{:symbol "x", :quantity "length"}
              {:symbol "v", :quantity "velocity"}
              {:symbol "t", :quantity "time"}
              {:symbol "a", :quantity "acceleration"}])

2.2. Expressions and equation

Next, define the equation in terms of its left and right hand side of the expression. If a side has more than one term they are expressed as a map with appropriate key :term<i> for respective term.

(def lhs "x^(1)")
(def rhs {:term1 "x^(1)",
          :term2 "v^(2)",
          :term3 "t^(1)",
          :term4 "0.5*a^(1)*t^(2)"})
(def eqn {:lhs lhs, :rhs rhs})

3. Getting the Dimensional Formula

The equation defined above is used for deriving the dimensional formula.

3.1. Sub-formula of the dimensional formula for one side of the equation

For the side of the equation that contains multiple terms the sub-formula is the dimensional formula for one of the terms. Notice that the sub-formula IS the dimensional formula for the expression if there is just one term.

Thus for the right hand side of the given equation (which was defined in the previous section)

=> rhs
{:term1 "x^(1)", :term2 "v^(2)", :term3 "t^(1)", :term4 "0.5*a^(1)*t^(2)"}

the dimensional formula for :term4 is given by

=> (formula-term varpars (:term4 rhs))
"[T^(0)*L^(1)]"

3.2. Dimensional formula for one side of the equation

Dimensional formula for one side of the expression regardless of the number of terms in it can be generated using the formula-eqn-side function.

Dimensional formula for the rhs expression is given by

=> (formula-eqn-side varpars rhs)
"[L^(1)] + [T^(-2)*L^(2)] + [T^(1)] + [T^(0)*L^(1)]"

3.3. Introspecting the dimensional formula

3.3.1. Represent sub-formula of an expression term as dimension names

=> (dimnames (formula-term varpars (:term4 rhs)))
"length^(1)"

3.3.2. Represent dimensional formula of an equation side as dimension names

=> (dimnames (formula-eqn-side varpars rhs))
"length^(1) + time^(-2)*length^(2) + time^(1) + length^(1)"

How to do Consistency Checks

1. The Problem

Consider the equation

x = x₀ + vt + (1/2)at²

2. Setting Up for Checking

2.1. Definitions setup

Define all the symbols in the mathematical expression that is associated with a dimension.

(def varpars [{:symbol "x", :quantity "length"}
              {:symbol "v", :quantity "velocity"}
              {:symbol "t", :quantity "time"}
              {:symbol "a", :quantity "acceleration"}])

2.2. Expressions and equation

Next, define the equation in terms of its left and right hand side of the expression. If a side has more than one term they are expressed as a map with appropriate key :term<i> for respective term.

(def lhs "x^(1)")
(def rhs {:term1 "x^(1)",
          :term2 "v^(1)*t^(1)",
          :term3 "0.5*a^(1)*t^(2)"})
(def eqn {:lhs lhs, :rhs rhs})

3. Perform Consistency Check

If the correctness of an equation is in doubt checking for dimensional consistency is a useful preliminary step.

To perform consistency check based on dimensional analysis in diman ^(c) you use the predicate function consistent?. Thus, for the given problem

=> (consistent? varpars eqn)
true

However, dimensionally consistent equation does not guarantee correct equation.

4. Consistency of multiple equations

Let us consider the case of a problem where one derives multiple expressions thought to be potential candidates for representing the problem.

e = m²v²

e = (1/2) mv²

e = ma

e = (3/16) mv²

e = (1/2) mv² + ma

the question is, which of these equations are correct? To tackle this question let us first look at the answer for which of these equations are dimensionally correct? In other words, let us perform dimensional consistency checks on each expression.

Thus

Equation Setup

Equation	Setup
e = m²v²	`(def eqn1 {:lhs "e^(1)", :rhs "m^(2)*v^(2)"})`
e = (1/2) mv²	`(def eqn2 {:lhs "e^(1)", :rhs "0.5m^(1)v^(2)"})`
e = ma	`(def eqn3 {:lhs "e^(1)", :rhs "m^(1)*a^(1)"})`
e = (3/16) mv²	`(def eqn4 {:lhs "e^(1)", :rhs "0.1875m^(1)v^(2)"})`
e = (1/2) mv² + ma	`(def eqn5 {:lhs "e^(1)", :rhs {:term1 "0.5m^(1)v^(2)", :term2 "m^(1)*a^(1)"}})`

e = m²v²

(def eqn1 {:lhs "e^(1)", :rhs "m^(2)*v^(2)"})

e = (1/2) mv²

(def eqn2 {:lhs "e^(1)", :rhs "0.5*m^(1)*v^(2)"})

e = ma

(def eqn3 {:lhs "e^(1)", :rhs "m^(1)*a^(1)"})

e = (3/16) mv²

(def eqn4 {:lhs "e^(1)", :rhs "0.1875*m^(1)*v^(2)"})

e = (1/2) mv² + ma

(def eqn5 {:lhs "e^(1)", :rhs {:term1 "0.5*m^(1)*v^(2)", :term2 "m^(1)*a^(1)"}})

and define the variables/parameters as

(def varpars [{:symbol "e", :quantity "energy"}
              {:symbol "m", :quantity "mass"}
              {:symbol "v", :quantity "velocity"}
              {:symbol "a", :quantity "acceleration"}])

Then

=> (consistent? varpars eqn1)
false
=> (consistent? varpars eqn2)
true
=> (consistent? varpars eqn3)
false
=> (consistent? varpars eqn4)
true
=> (consistent? varpars eqn5)
false

which suggests e = (1/2) mv² and e = (3/16) mv² to be dimensionally consistent.

But both equations can’t be correct, illustrating the point that

a dimensionally consistent equation does not guarantee correct equation

How to Standardize the Correct Equation

From the previous example of notice that kinetic "e" is not defined in the standard_formula

=> (pprint standard_formula)
[{:quantity "volume", :dimension "[L^(3)]"}
{:quantity "velocity", :dimension "[L^(1)*T^(-1)]"}
{:quantity "acceleration", :dimension "[L^(1)*T^(-2)]"}
{:quantity "force", :dimension "[M^(1)*L^(1)*T^(-2)]"}
{:quantity "mass density", :dimension "[M^(1)*L^(-3)]"}]

Since we already know that the kinetic energy is in Joules and 1J = kg⋅m²⋅s⁻² whose dimensional formula is "[M^(1)*L^(2)*T(-2)]" this can be added to the standard_formula as

=> (update-sformula [{:quantity "energy", :dimension "[M^(1)*L^(2)*T(-2)]"}])
[{:quantity "volume", :dimension "[M^(0)*L^(3)*T^(0)]"}
{:quantity "velocity", :dimension "[M^(0)*L^(1)*T^(-1)]"}
{:quantity "acceleration", :dimension "[M^(0)*L^(1)*T^(-2)]"}
{:quantity "force", :dimension "[M^(1)*L^(1)*T^(-2)]"}
{:quantity "mass density", :dimension "[M^(1)*L^(-3)*T^(0)]"}
{:quantity "energy", :dimension "[M^(1)*L^(2)*T(-2)]"}]

Now since "energy" is one of the :quantity in the standard_formula, we can now add the symbol "e" in our definition as follows

=> (def varpars (conj varpars {:symbol "e", :quantity "energy"}))
=> (pprint varpars)
[{:symbol "m", :quantity "mass"}
{:symbol "v", :quantity "velocity"}
{:symbol "a", :quantity "acceleration"}
{:symbol "e", :quantity "energy"}]