E Mc2 Dimensionally Consistent

[Honorato Winkel]

Supposewe need the formula for the area of a circle for some computation. Like many people who learned geometry too long ago to recall with any certainty, two expressions may pop into our mind when we think of circles: [latex] \pi r^2 [/latex] and [latex] 2\pi r. [/latex] One expression is the circumference of a circle of radius r and the other is its area. But which is which?

This may seem like kind of a silly example, but the ideas are very general. As long as we know the dimensions of the individual physical quantities that appear in an equation, we can check to see whether the equation is dimensionally consistent. On the other hand, knowing that true equations are dimensionally consistent, we can match expressions from our imperfect memories to the quantities for which they might be expressions. Doing this will not help us remember dimensionless factors that appear in the equations (for example, if you had accidentally conflated the two expressions from the example into [latex] 2\pi r^2, [/latex] then dimensional analysis is no help), but it does help us remember the correct basic form of equations.

Consider the physical quantities [latex] s, [/latex] [latex] v, [/latex] [latex] a, [/latex] and [latex] t [/latex] with dimensions [latex] [s]=\textL, [/latex] [latex] [v]=\textLT^-1, [/latex] [latex] [a]=\textLT^-2, [/latex] and [latex] [t]=\textT. [/latex] Determine whether each of the following equations is dimensionally consistent: (a) [latex] s=vt+0.5at^2; [/latex] (b) [latex] s=vt^2+0.5at; [/latex] and (c) [latex] v=\textsin(at^2\text/s). [/latex]

By the definition of dimensional consistency, we need to check that each term in a given equation has the same dimensions as the other terms in that equation and that the arguments of any standard mathematical functions are dimensionless.

All three terms have the same dimension, so this equation is dimensionally consistent.Again, there are no trigonometric, exponential, or logarithmic functions, so we only need to look at the dimensions of each of the three terms appearing in the equation:[latex] \beginarrayc[s]=\textL\hfill \\ [vt^2]=[v][t]^2=\textLT^-1\textT^2=\textLT\hfill \\ [at]=[a][t]=\textLT^-2\textT=\textLT^-1.\hfill \endarray [/latex]None of the three terms has the same dimension as any other, so this is about as far from being dimensionally consistent as you can get. The technical term for an equation like this is nonsense.This equation has a trigonometric function in it, so first we should check that the argument of the sine function is dimensionless:[latex] \beginarrayc[\fracat^2s]=\frac[a][t]^2[s]=\frac\textLT^-2\textT^2\textL=\frac\textL\textL=1.\hfill \endarray [/latex]The argument is dimensionless. So far, so good. Now we need to check the dimensions of each of the two terms (that is, the left expression and the right expression) in the equation:

If we are trusting people, these types of dimensional checks might seem unnecessary. But, rest assured, any textbook on a quantitative subject such as physics (including this one) almost certainly contains some equations with typos. Checking equations routinely by dimensional analysis save us the embarrassment of using an incorrect equation. Also, checking the dimensions of an equation we obtain through algebraic manipulation is a great way to make sure we did not make a mistake (or to spot a mistake, if we made one).

Suppose we need the formula for the area of a circle for some computation. Like many people who learned geometry too long ago to recall with any certainty, two expressions may pop into our mind when we think of circles: [latex]\pi r^2[/latex] and [latex]2\pi r.[/latex] One expression is the circumference of a circle of radius r and the other is its area. But which is which?

This may seem like kind of a silly example, but the ideas are very general. As long as we know the dimensions of the individual physical quantities that appear in an equation, we can check to see whether the equation is dimensionally consistent. On the other hand, knowing that true equations are dimensionally consistent, we can match expressions from our imperfect memories to the quantities for which they might be expressions. Doing this will not help us remember dimensionless factors that appear in the equations (for example, if you had accidentally conflated the two expressions from the example into [latex]2\pi r^2,[/latex] then dimensional analysis is no help), but it does help us remember the correct basic form of equations.

Consider the physical quantities [latex]s,[/latex] [latex]v,[/latex] [latex]a,[/latex] and [latex]t[/latex] with dimensions [latex][s]=\textL,[/latex] [latex][v]=\textLT^-1,[/latex] [latex][a]=\textLT^-2,[/latex] and [latex][t]=\textT.[/latex] Determine whether each of the following equations is dimensionally consistent: (a) [latex]s=vt+0.5at^2;[/latex] (b) [latex]s=vt^2+0.5at;[/latex] and (c) [latex]v=\textsin(at^2\text/s).[/latex]

All three terms have the same dimension, so this equation is dimensionally consistent.Again, there are no trigonometric, exponential, or logarithmic functions, so we only need to look at the dimensions of each of the three terms appearing in the equation:[latex]\beginarrayc[s]=\textL\hfill \\ [vt^2]=[v]\cdot [t]^2=\textLT^-1\cdot \textT^2=\textLT\hfill \\ [at]=[a]\cdot [t]=\textLT^-2\cdot \textT=\textLT^-1.\hfill \endarray[/latex]None of the three terms has the same dimension as any other, so this is about as far from being dimensionally consistent as you can get. The technical term for an equation like this is nonsense.This equation has a trigonometric function in it, so first we should check that the argument of the sine function is dimensionless:[latex]\beginarrayc\left [\fracat^2s\right ]=\fraca\cdot t^2s=\frac\textLT^-2\cdot \textT^2\textL=\frac\textL\textL=1.\hfill \endarray[/latex]The argument is dimensionless. So far, so good. Now we need to check the dimensions of each of the two terms (that is, the left expression and the right expression) in the equation:

Each of the equations you listed above consist of various combinations of the dimensions length (L) and time (T). To determine whether an equation is dimensionally consistent, begin by writing out the dimensions of each component on each side of the equation. For instance, in the first equation, x = 1/2 at2, we have the following:

For the second equation, we have T on the left and (L/T) / L on the right. Simplifying the right hand side gives us 1/T, which is not the same as the T on the left, so this equation is not consistent. (The correct expression here would instead be t = x/v.)

Dimensional consistency refers to the concept that all units used to express a measurement must be compatible with each other. This means that when performing calculations or converting between units, the resulting units must have the same dimensions on both sides of the equation.

Dimensional consistency is crucial in scientific research because it ensures the accuracy and validity of measurements. It allows for proper comparison and analysis of data, and helps to avoid errors and inconsistencies in calculations.

If a measurement is not dimensionally consistent, it can lead to incorrect or meaningless results. This is because different units represent different quantities, and mixing incompatible units can result in nonsensical calculations.

To ensure dimensional consistency, scientists use the principles of dimensional analysis. This involves breaking down a measurement into its fundamental units and checking that all units used in the calculation are compatible with each other. Any inconsistencies or errors can then be identified and corrected.

Yes, dimensional consistency can also be applied to non-physical quantities, such as mathematical expressions or equations. In this case, the dimensions of each term must be the same on both sides of the equation in order for it to be considered dimensionally consistent.

We now know that the dimension of a physical quantity is just an expression of the basic quantities from which it is generated, after understanding the dimensional consistency meaning, few-dimensional consistency examples, and the dimensional consistency equation. Dimensional consistency is required for all equations representing physical laws or principles. This fact may be used to help recall physical rules, examine whether stated correlations between physical quantities are conceivable, and may even generate new physical laws.

Ans : f the dimensional consistency rules are broken, an equation is no longer dimensionally consistent and cannot be used to express physical law. This fact may be used to check for typos and algebraic errors, to assist recall the numerous laws of physics, and even to suggest new laws of physics.

We know that if an equation has to be physically correct then it must be dimensionally consistent i.e. If an equation is not dimensionally correct then it can never be physically correct. Now in the equation of distance travelled by a particle in nth second, we see distance at LHS and the sum of $u$ and $a/2(2n-1)$ on the RHS. Now how can this equation be dimensionally consistent as it doesn't seem to obey principle of homogeneity (the dimensions of distance and velocity $u$ are not the same). And if this equation is dimensionally incorrect so it shouldn't be correct physically.

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