Dimensions and Dimensional Analysis

By now, students should appreciate the importance of placing units on quantities. This helps avoid errors such as using inches and feet for length in the same computation. Also, by canceling units in calculations, it is possible to determine the units of the answer, an obvious advantage for complex calculations where you could not expect to guess the units of the answer ahead of time. The real power of these methods will now be revealed to you. As the saying goes "you ain't seen nothing yet!"

1) Dimensions If a student were asked "what is money?", the answer "dollars" would be unsatisfactory; dollars is merely one particular example of money, and does not define what money actually is - a medium of exchange. Similarly, velocity is not well defined as ft/sec since this is only one example of a velocity Fundamentally, velocity is distance/time; in short hand V = D/t = Dt-1. We say that the dimensions of velocity are Dt-1. Similarly, the dimensions of acceleration are a = velocity/time = v/t = D/t/t = Dt-2. The dimensions of area are D2 , volume is D3. The dimensions of anything are the most basic quantities that can be used to express it. The four basic quantities of physics are:

1- distance            D

2- time                   t

3- mass                 m

4- electric charge   Q

 With these four basic quantities we can literally construct the whole universe.* For example, kinetic energy (to be studied later in the year) is expressed by the formula KE = 1/2mv2. What are its dimensions? The constant 1/2 is dimensionless and is ignored here: KE = 1/2mv2 = m(Dt-1)2 = mD2t-2 Question: What are the dimensions of density which is defined as mass/volume? ___________.  What are the dimensions of frequency? ______________ (answers at the next line break)

*Notice how remarkable this is. A fine restaurant needs literally dozens and dozens of ingredients to make its menu yet we assert that all the quantities of physics are composed of just four!

2) Dimensional Analysis - Human memory is flawed by unreliability. Any given fact or idea can be forgotten. One way to improve memory is to associate things to be learned with other things you know well. For example, if you are introduced to a certain Mr. Lincoln, if you associate the man's name with the famous president, you might forget his name at first but by recalling the association, can regenerate the forgotten name. Another memory aid is the mnemonic device in which rhymes etc. are used. One example is the grammar school mnemonic for recalling the date of Columbus' voyage: In 1492
Columbus sailed the deep sea blue Very nice, but what if the kid thinks In 1493
Columbus sailed the deep blue sea Once a mnemonic device is uncertain, you cannot regenerate it with any assurance! Is there a more certain device to assist our memory? While history students must struggle along as best they can, physics students have a far better memory aid through
dimensional reasoning. The obvious memory task for physics is formulas; there are dozens of them and how can forgetting be avoided? While you will often forget formulas, in part or totally, there is a remedy. This remedy uses dimensional reasoning and can either check to see if a formula you think you remember is correctly recalled or if necessary, can create the formula out of thin air! Let's see how.

Checking formulas is simple. You have seen that the formula for the distance traveled by an object accelerating from rest is D = 1/2at2 or is it D = 1/2a2t? Unlike the Columbus example, it is simple to select between the two possibilities. Any equation from physics is actually two equalities at once, numerical and also dimensional. That is, the dimensions of both sides of the equation must be equal. Lets examine both sides of each possibility and see how the dimensions work out:

D =?= 1/2at2                           D =?= 1/2a2t recall that the dimensions of acceleration are a = Dt-2 ; ignore the 1/2 D =?= Dt-2t2                            D =?= (Dt-2)2t = D2t-4 t = D2t-3

This formula is possible.            This is impossible

 Thus only the formula D = 1/2at2 can be correct; the other is obviously wrong. Checking formulas by dimensions is a fine memory aid but we can do much better. It turns out that we can not only check formulas for correctness but in most cases we can derive the equation from scratch without knowing any of the physics underlying the equation! This trick is so powerful and useful that it is given the name dimensional analysis. Proper mastery is a cardinal goal of the physics course. To illustrate how it works, let's again consider the example of an object accelerating from rest; what is the formula relating the distance traveled to the acceleration and the time? Let's assume we have no idea what the formula is; for all we know it could be D = a3t5 or something like that! Assuming the distance is related to acceleration and time, we say:

1) D = f (a, t) this merely says that the distance is a function of a,t

2) D = caxty c is a constant, x and y are exponents whose value we seek to determine.
    recall that the dimensions of acceleration are Dt-2 ; substitute this for a.

3) D = c(Dt-2)xty

    D = cDxt-2xty

    D = cDxt-2x+y

4) Now there is a time variable on the right hand side of the equation but none on the left so we place to (which is 1) on the left hand side.

D to = cDx t-2x+y 5) Since the dimensions on both sides of the equation must agree, D = Dx thus x = 1 and to = t-2x+y therefore we see
0 = -2x+y thus y = 2. Now go back to step two D = caxty and substitute our values for x,y to get D = cat2. The value of c cannot be determined since constants are dimensionless but the proportion is right.

Dimensional analysis is a very powerful tool for physics which takes time and practice to either master or fully appreciate. You will employ it dozens of times in the rest of the course and should become quite expert with the technique. The first example you try may be difficult but with time you will see just how simple dimensional analysis is.


Answer: Density = mass/volume = m/V = m/D3 = mD-3 . Frequency = 1/t = t-l

Problems - (answers found after the problems)

(1) Use dimensional reasoning to decide which of the following motion formulas could be correct, if any.

    a) D = V2/a               b)t =D/a

(2) The velocity reached after a period of acceleration from rest seems related to the distance traveled and to the acceleration achieved. Use dimensional analysis to discover the proportions.
 
 
 
 

Find a formula from the motion chapter that can be rearranged to support the result you got. __________ What is the value of the constant? _________________________

(3) The time for a single swing of a pendulum seems somehow related to the acceleration of gravity, to the length of the pendulum, and possibly to the mass m. Use dimensional analysis to derive an expression. (The acceleration of gravity (called g) is dimensionally identical to any other acceleration.)
 
 
 
 

If you are really good, you can use dimensional analysis to show that the distance the pendulum swings sideways doesn't matter.

(4) The distance traveled by a body accelerating from rest seems related to the acceleration of the body and to the velocity attained. Use dimensional analysis to discover the proportions.
 
 
 
 
 

Find a formula from the motion chapter that can be rearranged to agree with your result. ______________

What is the constant c? _________________________

(5) Density can be defined various ways. For solid objects and liquids, density is mass/volume, for wires etc. we have line density = mass/length but for sheet metal and paper etc., we say area density = mass/area. What are the dimensions of each type of density?
 
 
 
 
 

(6) An object thrown horizontally from a cliff or rooftop lands on the ground below. There seems to be a relationship between the range (our name for the distance the object travels sideways from the bottom of cliff) and the velocity of throwing, the acceleration of gravity, and the height of the cliff. Use dimensional analysis to derive the proportion of range to velocity of throwing, acceleration of gravity, and cliff height. if you find the algebra impossible to solve, then try treating distance horizontal and distance vertical as different variables and try again. Alternatively you can look back to your lab on the range of a projectile and substitute the empirically found exponent for the range into this analysis.

(7) In a later chapter we will show that centripetal force (the force that maintains objects traveling in a circle) is given as:

Centripetal Force = mV2 where r is the radius. What are the dimensions of centripetal force?
                                r


Answers

(1) Only formula (a) is possible.

(2) Start with V = caxDy Since V is Dt-1 and a is Dt-2 we now have:
    Dt-1 = c(Dt-2)xDy
    from here we get Dt-1 = cDx t-2x Dy = ct-2x Dx+y                                                    ___                             __
    1 = x + y and -1 = -2x. x= 1/2, y = 1/2. The final result is thus V = ca 1/2 d1/2 = cÖ ad . The c is actually Ö 2 .

(3) Start with t = cLxgymz then realize that L is a distance and g, the acceleration of gravity, is merely acceleration
     dimensionally. Thus we modify to get:
                    t = cDxaymz                                 ___
     This gives the result that t = cL1/2 g-1/2 = cÖ l/g The mass is irrelevant.
     The ambitious student may want to include the sideways swing in his analysis but finds that with two distance variables, the
     algebra can't be solved. However if sideways motion is called Dhorizontal and the length and the D variable in g is Dvertical,
     then the problem vanishes. In other words sideways swing = DH and length = DV and g = DVt-2
(4) Start with D = cVxay and end up with D = cV2. The c turns out to be 1/2.
                                                                        a

(5) Volume density is mD-3, area density is mD-2, line density is mD-l

(6) You start with Range = cVhoriz xgy Hz; Vhoriz = Dt-l, acceleration of gravity g is Dt-2 and Height is D.

     Of course range R is also D.

D = (Dt-l) x (Dt-2) y (D)z      Solving this gives more unknowns than equations thus you cannot solve it. A way out is to discriminate between horizontal
     and vertical distances thus Vhoriz = DHt-1, g is DVt-2, height is DV and range is DH. Now do the algebra and you get:
             R= c VH Ö Height/g
(7). Centripetal force mdt-2