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1.5: Dimensional Analysis

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Learning Objectives

• Find the dimensions of a mathematical expression involving physical quantities.
• Determine whether an equation involving physical quantities is dimensionally consistent.

The dimension of any physical quantity expresses its dependence on the base quantities as a product of symbols (or powers of symbols) representing the base quantities. Table \(\PageIndex{1}\) lists the base quantities and the symbols used for their dimension. For example, a measurement of length is said to have dimension L or L 1 , a measurement of mass has dimension M or M 1 , and a measurement of time has dimension T or T 1 . Like units, dimensions obey the rules of algebra. Thus, area is the product of two lengths and so has dimension L 2 , or length squared. Similarly, volume is the product of three lengths and has dimension L 3 , or length cubed. Speed has dimension length over time, L/T or LT –1 . Volumetric mass density has dimension M/L 3 or ML –3 , or mass over length cubed. In general, the dimension of any physical quantity can be written as

\[L^{a}M^{b}T^{c}I^{d}\Theta^{e}N^{f}J^{g}\]

for some powers a, b, c, d, e, f, and g. We can write the dimensions of a length in this form with a = 1 and the remaining six powers all set equal to zero:

\[L^{1} = L^{1}M^{0}T^{0}I^{0}\Theta^{0}N^{0}J^{0}.\]

Any quantity with a dimension that can be written so that all seven powers are zero (that is, its dimension is \(L^{0}M^{0}T^{0}I^{0}\Theta^{0}N^{0}J^{0}\)) is called dimensionless (or sometimes “of dimension 1,” because anything raised to the zero power is one). Physicists often call dimensionless quantities pure numbers .

Physicists often use square brackets around the symbol for a physical quantity to represent the dimensions of that quantity. For example, if r is the radius of a cylinder and h is its height, then we write [r] = L and [h] = L to indicate the dimensions of the radius and height are both those of length, or L. Similarly, if we use the symbol A for the surface area of a cylinder and V for its volume, then [A] = L 2 and [V] = L 3 . If we use the symbol m for the mass of the cylinder and \(\rho\) for the density of the material from which the cylinder is made, then [m] = M and [\(\rho\)] = ML −3 .

The importance of the concept of dimension arises from the fact that any mathematical equation relating physical quantities must be dimensionally consistent , which means the equation must obey the following rules:

• Every term in an expression must have the same dimensions; it does not make sense to add or subtract quantities of differing dimension (think of the old saying: “You can’t add apples and oranges”). In particular, the expressions on each side of the equality in an equation must have the same dimensions.
• The arguments of any of the standard mathematical functions such as trigonometric functions (such as sine and cosine), logarithms, or exponential functions that appear in the equation must be dimensionless. These functions require pure numbers as inputs and give pure numbers as outputs.

If either of these rules is violated, an equation is not dimensionally consistent and cannot possibly be a correct statement of physical law. This simple fact can be used to check for typos or algebra mistakes, to help remember the various laws of physics, and even to suggest the form that new laws of physics might take. This last use of dimensions is beyond the scope of this text, but is something you will undoubtedly learn later in your academic career.

Example \(\PageIndex{1}\): Using Dimensions to Remember an Equation

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: \(\pi r^{2}\) and \(2 \pi r\). One expression is the circumference of a circle of radius r and the other is its area. But which is which?

One natural strategy is to look it up, but this could take time to find information from a reputable source. Besides, even if we think the source is reputable, we shouldn’t trust everything we read. It is nice to have a way to double-check just by thinking about it. Also, we might be in a situation in which we cannot look things up (such as during a test). Thus, the strategy is to find the dimensions of both expressions by making use of the fact that dimensions follow the rules of algebra. If either expression does not have the same dimensions as area, then it cannot possibly be the correct equation for the area of a circle.

We know the dimension of area is L 2 . Now, the dimension of the expression \(\pi r^{2}\) is

\[[\pi r^{2}] = [\pi] \cdotp [r]^{2} = 1 \cdotp L^{2} = L^{2},\]

since the constant \(\pi\) is a pure number and the radius r is a length. Therefore, \(\pi r^{2}\) has the dimension of area. Similarly, the dimension of the expression \(2 \pi r\) is

\[[2 \pi r] = [2] \cdotp [\pi] \cdotp [r] = 1 \cdotp 1 \cdotp L = L,\]

since the constants 2 and \(\pi\) are both dimensionless and the radius r is a length. We see that \(2 \pi r\) has the dimension of length, which means it cannot possibly be an area.

We rule out \(2 \pi r\) because it is not dimensionally consistent with being an area. We see that \(\pi r^{2}\) is dimensionally consistent with being an area, so if we have to choose between these two expressions, \(\pi r^{2}\) is the one to choose.

Significance

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 \(2 \pi r^{2}\), then dimensional analysis is no help), but it does help us remember the correct basic form of equations.

Exercise \(\PageIndex{1}\)

Suppose we want the formula for the volume of a sphere. The two expressions commonly mentioned in elementary discussions of spheres are \(4 \pi r^{2}\) and \(\frac{4}{3} \pi r^{3}\). One is the volume of a sphere of radius r and the other is its surface area. Which one is the volume?

Add texts here. Do not delete this text first.

Example \(\PageIndex{2}\): Checking Equations for Dimensional Consistency

Consider the physical quantities s, v, a, and t with dimensions [s] = L, [v] = LT −1 , [a] = LT −2 , and [t] = T. Determine whether each of the following equations is dimensionally consistent:

• s = vt + 0.5at 2 ;
• s = vt 2 + 0.5at; and
• v = sin (\(\frac{at^{2}}{s}\)).

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.

• There are no trigonometric, logarithmic, or exponential functions to worry about in this equation, so we need only look at the dimensions of each term appearing in the equation. There are three terms, one in the left expression and two in the expression on the right, so we look at each in turn:

\[[s] = L\]

\[[vt] = [v] \cdotp [t] = LT^{−1} \cdotp T = LT^{0} = L\]

\[[0.5at^{2} ] = [a] \cdotp [t]^{2} = LT^{−2} \cdotp T^{2} = LT^{0} = L \ldotp\]

• 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:

\[[vt^{2}] = [v] \cdotp [t]^{2} = LT^{−1} \cdotp T^{2} = LT\]

\[[at] = [a] \cdotp [t] = LT^{−2} \cdotp T = LT^{−1} \ldotp\]

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:

\[\left[\frac{at^{2}}{s}\right] = \frac{[a] \cdotp [t]^{2}}{[s]} = \frac{LT^{-2} \cdotp T^{2}}{L} = \frac{L}{L} = 1 \ldotp\]

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:

\[[v] = LT^{-1}\]

\[\left[ sin \left(\dfrac{at^{2}}{s}\right) \right] = 1 \ldotp\]

The two terms have different dimensions—meaning, the equation is not dimensionally consistent. This equation is another example of “nonsense.”

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).

Exercise \(\PageIndex{2}\)

Is the equation v = at dimensionally consistent?

One further point that needs to be mentioned is the effect of the operations of calculus on dimensions. We have seen that dimensions obey the rules of algebra, just like units, but what happens when we take the derivative of one physical quantity with respect to another or integrate a physical quantity over another? The derivative of a function is just the slope of the line tangent to its graph and slopes are ratios, so for physical quantities v and t, we have that the dimension of the derivative of v with respect to t is just the ratio of the dimension of v over that of t:

\[\left[\frac{dv}{dt} \right] = \frac{[v]}{[t]} \ldotp\]

Similarly, since integrals are just sums of products, the dimension of the integral of v with respect to t is simply the dimension of v times the dimension of t:

\[\left[ \int vdt \right] = [v] \cdotp [t] \ldotp\]

By the same reasoning, analogous rules hold for the units of physical quantities derived from other quantities by integration or differentiation.

1.4 Dimensional Analysis

Learning objectives.

By the end of this section, you will be able to:

• Find the dimensions of a mathematical expression involving physical quantities.
• Determine whether an equation involving physical quantities is dimensionally consistent.

The dimension of any physical quantity expresses its dependence on the base quantities as a product of symbols (or powers of symbols) representing the base quantities. Table 1.3 lists the base quantities and the symbols used for their dimension. For example, a measurement of length is said to have dimension L or L 1 , a measurement of mass has dimension M or M 1 , and a measurement of time has dimension T or T 1 . Like units, dimensions obey the rules of algebra. Thus, area is the product of two lengths and so has dimension L 2 , or length squared. Similarly, volume is the product of three lengths and has dimension L 3 , or length cubed. Speed has dimension length over time, L/T or LT –1 . Volumetric mass density has dimension M/L 3 or ML –3 , or mass over length cubed. In general, the dimension of any physical quantity can be written as L a M b T c I d Θ e N f J g L a M b T c I d Θ e N f J g for some powers a , b , c , d , e , f , a , b , c , d , e , f , and g . We can write the dimensions of a length in this form with a = 1 a = 1 and the remaining six powers all set equal to zero: L 1 = L 1 M 0 T 0 I 0 Θ 0 N 0 J 0 . L 1 = L 1 M 0 T 0 I 0 Θ 0 N 0 J 0 . Any quantity with a dimension that can be written so that all seven powers are zero (that is, its dimension is L 0 M 0 T 0 I 0 Θ 0 N 0 J 0 L 0 M 0 T 0 I 0 Θ 0 N 0 J 0 ) is called dimensionless (or sometimes “of dimension 1,” because anything raised to the zero power is one). Physicists often call dimensionless quantities pure numbers .

Physicists often use square brackets around the symbol for a physical quantity to represent the dimensions of that quantity. For example, if r r is the radius of a cylinder and h h is its height, then we write [ r ] = L [ r ] = L and [ h ] = L [ h ] = L to indicate the dimensions of the radius and height are both those of length, or L. Similarly, if we use the symbol A A for the surface area of a cylinder and V V for its volume, then [ A ] = L 2 and [ V ] = L 3 . If we use the symbol m m for the mass of the cylinder and ρ ρ for the density of the material from which the cylinder is made, then [ m ] = M [ m ] = M and [ ρ ] = ML −3 . [ ρ ] = ML −3 .

The importance of the concept of dimension arises from the fact that any mathematical equation relating physical quantities must be dimensionally consistent , which means the equation must obey the following rules:

• Every term in an expression must have the same dimensions; it does not make sense to add or subtract quantities of differing dimension (think of the old saying: “You can’t add apples and oranges”). In particular, the expressions on each side of the equality in an equation must have the same dimensions.
• The arguments of any of the standard mathematical functions such as trigonometric functions (such as sine and cosine), logarithms, or exponential functions that appear in the equation must be dimensionless. These functions require pure numbers as inputs and give pure numbers as outputs.

If either of these rules is violated, an equation is not dimensionally consistent and cannot possibly be a correct statement of physical law. This simple fact can be used to check for typos or algebra mistakes, to help remember the various laws of physics, and even to suggest the form that new laws of physics might take. This last use of dimensions is beyond the scope of this text, but is something you may learn later in your academic career.

Example 1.4

Using dimensions to remember an equation.

since the constant π π is a pure number and the radius r r is a length. Therefore, π r 2 π r 2 has the dimension of area. Similarly, the dimension of the expression 2 π r 2 π r is

since the constants 2 2 and π π are both dimensionless and the radius r r is a length. We see that 2 π r 2 π r has the dimension of length, which means it cannot possibly be an area.

We rule out 2 π r 2 π r because it is not dimensionally consistent with being an area. We see that π r 2 π r 2 is dimensionally consistent with being an area, so if we have to choose between these two expressions, π r 2 π r 2 is the one to choose.

Significance

Check your understanding 1.5.

Suppose we want the formula for the volume of a sphere. The two expressions commonly mentioned in elementary discussions of spheres are 4 π r 2 4 π r 2 and 4 π r 3 / 3 . 4 π r 3 / 3 . One is the volume of a sphere of radius r and the other is its surface area. Which one is the volume?

Example 1.5

Checking equations for dimensional consistency.

• There are no trigonometric, logarithmic, or exponential functions to worry about in this equation, so we need only look at the dimensions of each term appearing in the equation. There are three terms, one in the left expression and two in the expression on the right, so we look at each in turn: [ s ] = L [ v t ] = [ v ] · [ t ] = LT −1 · T = LT 0 = L [ 0.5 a t 2 ] = [ a ] · [ t ] 2 = LT −2 · T 2 = LT 0 = L . [ s ] = L [ v t ] = [ v ] · [ t ] = LT −1 · T = LT 0 = L [ 0.5 a t 2 ] = [ a ] · [ t ] 2 = LT −2 · T 2 = LT 0 = L . 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: [ s ] = L [ v t 2 ] = [ v ] · [ t ] 2 = LT −1 · T 2 = LT [ a t ] = [ a ] · [ t ] = LT −2 · T = LT −1 . [ s ] = L [ v t 2 ] = [ v ] · [ t ] 2 = LT −1 · T 2 = LT [ a t ] = [ a ] · [ t ] = LT −2 · T = LT −1 . 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: [ a t 2 s ] = [ a ] · [ t ] 2 [ s ] = LT −2 · T 2 L = L L = 1 . [ a t 2 s ] = [ a ] · [ t ] 2 [ s ] = LT −2 · T 2 L = L L = 1 . 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: [ v ] = LT −1 [ sin ( a t 2 s ) ] = 1 . [ v ] = LT −1 [ sin ( a t 2 s ) ] = 1 .

The two terms have different dimensions—meaning, the equation is not dimensionally consistent. This equation is another example of “nonsense.”

Check Your Understanding 1.6

Is the equation v = at dimensionally consistent?

One further point that needs to be mentioned is the effect of the operations of calculus on dimensions. We have seen that dimensions obey the rules of algebra, just like units, but what happens when we take the derivative of one physical quantity with respect to another or integrate a physical quantity over another? The derivative of a function is just the slope of the line tangent to its graph and slopes are ratios, so for physical quantities v and t , we have that the dimension of the derivative of v with respect to t is just the ratio of the dimension of v over that of t :

Similarly, since integrals are just sums of products, the dimension of the integral of v with respect to t is simply the dimension of v times the dimension of t :

By the same reasoning, analogous rules hold for the units of physical quantities derived from other quantities by integration or differentiation.

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• Dimensional Analysis

What is Dimensional Analysis?

To solve the mathematical problems of physical quantities, it is important to have a brief knowledge of units and dimensions. The basic concept of dimensions is that only those quantities can be added or subtracted which have the same dimension. This concept helps us to derive relationships between physical quantities.

Dimensional analysis is the study of the relation between physical quantities based on their units and dimensions. It is used to convert a unit from one form to another. While solving mathematical problems, it is necessary to keep the units the same to solve the problem easily.

Do you know what is the significance of dimensional analysis? Well! In engineering and science, dimensional analysis describes the relationships between different physical quantities based on their fundamental qualities such as length, mass, time, and electric current, and units of measure like miles v/s kilometres, or pounds v/s kilograms.

In other words, in Physics, we study two types of physical quantities, i.e., fundamental and derived. The seven fundamental units include mass, length, amount of substance, time, luminous intensity, and electric current. However, if we combine two or more fundamental units, we get derived quantities.

For examples, we denote [M] for mass, [L] for length, and so on. Similarly, for speed, which is a derived quantity given by distance/time, we denote it with [M]/[L] or [ M L -1 ]. This is how we derive the dimensional formula of various quantities. 

The conversion factor used is based on the unit that we desire in the answer. Further, we will derive the dimensional formula of various quantities on this page.

How to Perform Dimensional Analysis?

(Image to be added soon)

Unit Conversion

Dimensional analysis is also called a Unit Factor Method or Factor Label Method because a conversion factor is used to evaluate the units.

For example, suppose we want to know how many meters there are in 4 km.

Normally, we calculate as-

1 km = 1000 meters

4 km = 1000 × 4 = 4000 meters

(Here the conversion factor used is 1000 meters)

Principle of Homogeneity of Dimensional Analysis

This principle depicts that, “the dimensions are the same for every equation that represents physical units. If two sides of an equation don’t have the same dimensions, it cannot represent a physical situation.”

For example, in the equation

[M a L b T c ] = MxLyTz

As per this principle, we have

Example of Dimensional Analysis

For using a conversion factor, it is necessary that the values must represent the same quantity. For example, 60 minutes is the same as 1 hour, 1000 meters is the same as 1 kilometre, or 12 months is the same as 1 year.

Let us try to understand it in this way. Imagine you have 15 pens and you multiply that by 1, now you still have the same number of 15 pens. If you want to find out how many packages of the pen are equal to 15 pens, you need the conversion factor.

Now, suppose you have a packaged set of ink pens in which each package contains 15 pens. Let's consider that you have 6 packages. To calculate the total pens, you have to multiply the number of packages by the number of pens in each package. This comes out to be:

15 × 6 = 90 pens

Some other examples of conversion factors that are used in day to day life are:

A simple example : the time taken by a harmonic oscillator.

A complex example: the energy of vibrating conduction or wire.

A third example : demand versus capacity for a disk that is rotating.

Applications of Dimensional Analysis

Dimensional analysis is an important aspect of measurement, and it has many applications in Physics. Dimensional analysis is used mainly because of five reasons, which are:

To check the correctness of an equation or any other physical relation based on the principle of homogeneity. There should be dimensions on two sides of the equation. The dimensional relation will be correct if the L.H.S and R.H.S of an equation have identical dimensions. If the dimensions on two sides are incorrect, then the relations will also be incorrect.

Dimensional analysis is used to convert the value of a physical quantity from one system of units to another system of units.

It is used to represent the nature of physical quantity.

The expressions of dimensions can be manipulated as algebraic quantities.

Dimensional analysis is used to derive formulas.

Limitations of Dimensional Analysis

Some major limitations of dimensional analysis are:

Dimensional analysis doesn't provide information about the dimensional constant.

Dimensional analysis cannot derive trigonometric, exponential, and logarithmic functions.

It doesn't give information about the scalar or vector identity of a physical quantity.

Example of Dimensional Formula: Derivation for Kinetic Energy

The dimensional formula of any physical entity is the mathematical expression representing the powers to which the fundamental units (mass M, length L, time T) are to be raised to obtain one unit of a derived quantity. 

Let us now understand the dimensional formula with an example. Now, we know that kinetic energy is one of the fundamental parts of Physics, hence its formula plays a vital role in many fields of Physics. So, let us derive the dimensional formula of kinetic energy. 

The kinetic energy has a dimensional formula of,

[M L 2 T -2 ]

M = Mass of the object

L = Length of the object

T = Time taken

Kinetic energy (K.E) is given by = \[\frac {1} {2}\]    [Mass x Velocity 2 ] ---- (I)

The dimensional formula of Mass is = [ M 1 L 0 T 0 ] --- (ii)

We know that,

Velocity = Distance × Time -1

= L x T -1 (dimensional formula)

Velocity has a dimensional formula [ M 0 L 1 T -1 ] ----- (iii)

On substituting equation (ii) and iii) in the above equation (i) we get,

Kinetic energy (K.E) = \[\frac {1} {2}\]    [Mass x Velocit y 2 ]

Or, K.E = [ M 1 L 0 T 0 ] [ M 0 L 1 T -1 ] 2 = [ M (0 +1) L (1 + 1) T (-1 + -1) ]

Therefore, on solving, we get the dimensional formula for kinetic as [ M 0 L 2 T -2 ].

From this context, we understand that in dimensional analysis a set of units helps us establish the form of an equation and to check that the answer is free of even minute errors. 

Solved Example

1. Find out how many feet are there in 300 centimeters (cm).

Ans. We need to convert cm into feet.

Firstly, we have to convert cm into inches, and then inches into feet, as we can't directly convert cm into feet.

The calculation of two conversion factors is required here:

Then, 300 cm = 300 x \[\frac {1} {30.48}\]      feet

= 9.84 feet

FAQs on Dimensional Analysis

1. Find out how Many Feet are in 300 Centimeters (cm).

We need to convert cm into feet.

Then, 300 cm = 300×  \[\frac {1} {30.48}\]       feet

 2. Check the Consistency of the Dimensional Equation of Speed.

Dimensional analysis is used to check the consistency of an equation.

Speed = \[\frac {Distance} {Time}\]  

[LT -1 ] = \[\frac {L} {T}\]  

[LT -1 ] = [LT -1 ]

This equation is correct dimensionally because it has the same dimensions of speed on both sides. This basic test of dimensional analysis is used to check the consistency of equations, but it doesn't check the correctness of an equation.

By this method, the constants of some physical quantities cannot be determined.

Sine of angle =  \[\frac {Length} {Length}\]   , and hence it is unit-less.

So, it is a dimensionless quantity.

3. How to Check For Dimensional Consistency

Let us consider one of the equations, let say, constant acceleration,

The equation is given by

s = ut +  \[\frac {1} {2}\]       at 2 

s: displacement = it is a unit of length, 

Hence its dimension is L

ut: velocity x time, its dimension is LT -1  x T = L

\[\frac {1} {2}\]   at 2   = acceleration x time, its dimension is LT -2  x T 2   = L

All these three terms must have the same dimensions in order to be correct.

As these terms have units of length, the equation is dimensionally correct.

Dimensional Analysis: Know Your Units

Deducing the Process of Arriving at a Solution

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Dimensional analysis is a method of using the known units in a problem to help deduce the process of arriving at a solution. These tips will help you apply dimensional analysis to a problem.

How Dimensional Analysis Can Help

In science , units such as meter, second, and degree Celsius represent quantified physical properties of space, time, and/or matter. The International System of Measurement (SI) units that we use in science consist of seven base units, from which all other units are derived.

This means that a good knowledge of the units you're using for a problem can help you figure out how to approach a science problem, especially early on when the equations are simple and the biggest hurdle is memorization. If you look at the units provided within the problem, you can figure out some ways that those units relate to each other and, in turn, this might give you a hint as to what you need to do to solve the problem. This process is known as dimensional analysis.

A Basic Example

Consider a basic problem that a student might get right after starting physics. You're given a distance and a time and you have to find the average velocity, but you're completely blanking on the equation you need to do it.

Don't panic.

If you know your units, you can figure out what the problem should generally look like. Velocity is measured in SI units of m/s. This means that there is a length divided by a time. You have a length and you have a time, so you're good to go.

A Not-So-Basic Example

That was an incredibly simple example of a concept that students are introduced to very early in science, well before they actually begin a course in physics . Consider a bit later, however, when you've been introduced to all kinds of complex issues, such as Newton's Laws of Motion and Gravitation. You're still relatively new to physics, and the equations are still giving you some trouble.

You get a problem where you have to calculate the gravitational potential energy of an object. You can remember the equations for force, but the equation for potential energy is slipping away. You know it's kind of like force, but slightly different. What are you going to do?

Again, a knowledge of units can help. You remember that the equation for gravitational force on an object in Earth's gravity and the following terms and units:

F g = G * m * m E / r 2

• F g is the force of gravity - newtons (N) or kg * m / s 2
• G is the gravitational constant and your teacher kindly provided you with the value of G , which is measured in N * m 2 / kg 2
• m & m E are mass of the object and Earth, respectively - kg
• r is the distance between the center of gravity of the objects - m 
• We want to know U , the potential energy, and we know that energy is measured in Joules (J) or newtons * meter 
• We also remember that the potential energy equation looks a lot like the force equation, using the same variables in a slightly different way

In this case, we actually know a lot more than we need to figure it out. We want the energy, U , which is in J or N * m. The entire force equation is in units of newtons, so to get it in terms of N * m you will need to multiply the entire equation a length measurement. Well, only one length measurement is involved - r - so that's easy. And multiplying the equation by r would just negate an r from the denominator, so the formula we end up with would be:

F g = G * m * m E / r

We know the units we get will be in terms of N*m, or Joules. And, fortunately, we did study, so it jogs our memory and we bang ourselves on the head and say, "Duh," because we should have remembered that.

But we didn't. It happens. Fortunately, because we had a good grasp on the units we were able to figure out the relationship between them to get to the formula that we needed.

A Tool, Not a Solution

As part of your pre-test studying, you should include a bit of time to make sure you're familiar with the units relevant to the section you're working on, especially those that were introduced in that section. It is one other tool to help provide physical intuition about how the concepts you're studying are related. This added level of intuition can be helpful, but it shouldn't be a replacement for studying the rest of the material. Obviously, learning the difference between gravitational force and gravitational energy equations is far better than having to re-derive it haphazardly in the middle of a test.

The gravity example was chosen because the force and potential energy equations are so closely related, but that isn't always the case and just multiplying numbers to get the right units, without understanding the underlying equations and relationships, will lead to more errors than solutions.

• Newton's Law of Gravity
• Definition of Torque in Physics
• Free Falling Body
• Calculating Torque
• Understanding Momentum in Physics
• How to Define Acceleration
• The 2 Main Forms of Energy
• Definition of Force in Physics
• What Is Moment of Inertia in Physics?
• Introduction to Newton's Laws of Motion
• Energy: A Scientific Definition
• Fundamental Physical Constants
• Force Definition and Examples (Science)
• Newton Definition
• Perfectly Inelastic Collision
• What Is Velocity in Physics?

Problem-Solving Basics for One-Dimensional Kinematics

Learning objectives.

By the end of this section, you will be able to:

• Apply problem-solving steps and strategies to solve problems of one-dimensional kinematics.
• Apply strategies to determine whether or not the result of a problem is reasonable, and if not, determine the cause.

Figure 1. Problem-solving skills are essential to your success in Physics. (credit: scui3asteveo, Flickr)

Problem-solving skills are obviously essential to success in a quantitative course in physics. More importantly, the ability to apply broad physical principles, usually represented by equations, to specific situations is a very powerful form of knowledge. It is much more powerful than memorizing a list of facts. Analytical skills and problem-solving abilities can be applied to new situations, whereas a list of facts cannot be made long enough to contain every possible circumstance. Such analytical skills are useful both for solving problems in this text and for applying physics in everyday and professional life.

Problem-Solving Steps

While there is no simple step-by-step method that works for every problem, the following general procedures facilitate problem solving and make it more meaningful. A certain amount of creativity and insight is required as well.

Examine the situation to determine which physical principles are involved . It often helps to draw a simple sketch at the outset. You will also need to decide which direction is positive and note that on your sketch. Once you have identified the physical principles, it is much easier to find and apply the equations representing those principles. Although finding the correct equation is essential, keep in mind that equations represent physical principles, laws of nature, and relationships among physical quantities. Without a conceptual understanding of a problem, a numerical solution is meaningless.

Make a list of what is given or can be inferred from the problem as stated (identify the knowns) . Many problems are stated very succinctly and require some inspection to determine what is known. A sketch can also be very useful at this point. Formally identifying the knowns is of particular importance in applying physics to real-world situations. Remember, “stopped” means velocity is zero, and we often can take initial time and position as zero.

Identify exactly what needs to be determined in the problem (identify the unknowns) . In complex problems, especially, it is not always obvious what needs to be found or in what sequence. Making a list can help.

Find an equation or set of equations that can help you solve the problem . Your list of knowns and unknowns can help here. It is easiest if you can find equations that contain only one unknown—that is, all of the other variables are known, so you can easily solve for the unknown. If the equation contains more than one unknown, then an additional equation is needed to solve the problem. In some problems, several unknowns must be determined to get at the one needed most. In such problems it is especially important to keep physical principles in mind to avoid going astray in a sea of equations. You may have to use two (or more) different equations to get the final answer.

Substitute the knowns along with their units into the appropriate equation, and obtain numerical solutions complete with units . This step produces the numerical answer; it also provides a check on units that can help you find errors. If the units of the answer are incorrect, then an error has been made. However, be warned that correct units do not guarantee that the numerical part of the answer is also correct.

Check the answer to see if it is reasonable: Does it make sense? This final step is extremely important—the goal of physics is to accurately describe nature. To see if the answer is reasonable, check both its magnitude and its sign, in addition to its units. Your judgment will improve as you solve more and more physics problems, and it will become possible for you to make finer and finer judgments regarding whether nature is adequately described by the answer to a problem. This step brings the problem back to its conceptual meaning. If you can judge whether the answer is reasonable, you have a deeper understanding of physics than just being able to mechanically solve a problem.

When solving problems, we often perform these steps in different order, and we also tend to do several steps simultaneously. There is no rigid procedure that will work every time. Creativity and insight grow with experience, and the basics of problem solving become almost automatic. One way to get practice is to work out the text’s examples for yourself as you read. Another is to work as many end-of-section problems as possible, starting with the easiest to build confidence and progressing to the more difficult. Once you become involved in physics, you will see it all around you, and you can begin to apply it to situations you encounter outside the classroom, just as is done in many of the applications in this text.

Unreasonable Results

Physics must describe nature accurately. Some problems have results that are unreasonable because one premise is unreasonable or because certain premises are inconsistent with one another. The physical principle applied correctly then produces an unreasonable result. For example, if a person starting a foot race accelerates at 0.40 m/s 2 for 100 s, his final speed will be 40 m/s (about 150 km/h)—clearly unreasonable because the time of 100 s is an unreasonable premise. The physics is correct in a sense, but there is more to describing nature than just manipulating equations correctly. Checking the result of a problem to see if it is reasonable does more than help uncover errors in problem solving—it also builds intuition in judging whether nature is being accurately described.

Use the following strategies to determine whether an answer is reasonable and, if it is not, to determine what is the cause.

Solve the problem using strategies as outlined and in the format followed in the worked examples in the text . In the example given in the preceding paragraph, you would identify the givens as the acceleration and time and use the equation below to find the unknown final velocity. That is,

Check to see if the answer is reasonable . Is it too large or too small, or does it have the wrong sign, improper units, …? In this case, you may need to convert meters per second into a more familiar unit, such as miles per hour.

This velocity is about four times greater than a person can run—so it is too large.

If the answer is unreasonable, look for what specifically could cause the identified difficulty . In the example of the runner, there are only two assumptions that are suspect. The acceleration could be too great or the time too long. First look at the acceleration and think about what the number means. If someone accelerates at 0.40 m/s 2 , their velocity is increasing by 0.4 m/s each second. Does this seem reasonable? If so, the time must be too long. It is not possible for someone to accelerate at a constant rate of 0.40 m/s 2 for 100 s (almost two minutes).

Section Summary

The six basic problem solving steps for physics are:

• Step 1 . Examine the situation to determine which physical principles are involved.
• Step 2 . Make a list of what is given or can be inferred from the problem as stated (identify the knowns).
• Step 3 . Identify exactly what needs to be determined in the problem (identify the unknowns).
• Step 4 . Find an equation or set of equations that can help you solve the problem.
• Step 5 . Substitute the knowns along with their units into the appropriate equation, and obtain numerical solutions complete with units.
• Step 6 . Check the answer to see if it is reasonable: Does it make sense?

Conceptual Questions

1. What information do you need in order to choose which equation or equations to use to solve a problem? Explain. 2. What is the last thing you should do when solving a problem? Explain.

• College Physics. Authored by : OpenStax College. Located at : http://cnx.org/contents/[email protected]:aNsXe6tc@2/Problem-Solving-Basics-for-One . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected].

• Physics Concept Questions And Answers

Dimensional Analysis Questions

Dimensional Analysis is also known as the factor-label method or the unit factor method. Dimensional analysis helps to understand the relationships between various physical quantities by recognising their base quantities as well as units. In the year 1822, Joseph Fourier introduced the concept of dimensional analysis.

Dimension refers to the physical nature of a quantity and the type of unit used to specify it. Dimensional analysis is known as the factor label method or unit factor method since conversion factors are used to obtain the same units.

Some applications of dimensional analysis are:

• It is used to inspect the consistency of a dimensional equation
• To change units from one system to another
• To get the relation between physical quantities in physical phenomena

Limitations of dimensional analysis are:

• The dimensional analysis does not give information about the dimensional constant.
• The formula which features logarithmic functions, trigonometric functions, and exponential functions, cannot be derived using dimensional analysis.
• The dimensional analysis offers no information on whether a physical quantity is a scalar or vector.

Constants that have dimensions are known as dimensional constants.

Example: Planck’s constant, gravitational constant

Physical quantities which possess dimensions but do not have a fixed value are called dimensional variables. Examples: velocity, displacement, and force.

Angle, specific gravity, strain are some of the dimensionless quantities. Physical quantities with no dimensions are known as dimensionless quantities.

Some of the basic dimensions are Length – L, Time – T, Mass – M, Temperature – K or θ, and Current – A.

List of physical quantities having the same dimensional formula are:

impulse and momentum

power, luminous flux

angular velocity, frequency, velocity gradient

work, energy, torque, the moment of force, energy

thermal capacity, entropy, universal gas constant and Boltzmann’s constant

force constant, surface tension, surface energy.

angular momentum, Planck’s constant, rotational impulse

latent heat, gravitational potential.

force, thrust

stress, pressure, modulus of elasticity

Important Dimensional Analysis Questions with Answers

1. Dimension formula of luminous flux matches with which of the following?

• Rotational impulse

Answer: d) Power

Explanation: Dimensional formula of power is M^{1}L^{2}T^{-3}, and Dimensional formula of luminous flux is also M^{1}L^{2}T^{-3}.

2. Who introduced the concept of dimensional analysis?

The concept of dimensional analysis was introduced by Joseph Fourier in the year 1822.

3. What is dimensional analysis, is also known as _____?

Dimensional Analysis is also known as the factor-label method or the unit factor method.

4. Which among the following is not a basic unit of measurement?

• Temperature

Answer: c) Momentum

Explanation: Momentum is not a basic unit of measurement. It is a derived unit.

5. State true or false: dimensional analysis helps to know if the physical quantity is a vector or a scalar quantity.

Answer: b) FALSE

Explanation: Dimensional analysis offers no information on whether a physical quantity is a scalar or vector.

6. Match with the same dimensional formula quantity.

• Force    a) Latent heat
• Rotational impulse    b) luminous flux
• Gravitational potential     c) Thrust
• Power     d) Planck’s constant
• 1-c), 2-d), 3-a) , 4-b)
• 1-d), 2-c), 3-a) , 4-b)
• 1-a), 2-b), 3-c) , 4-d)
• 1-d), 2-c), 3-b) , 4-a)

Answer: 1-c), 2-d), 3-a) , 4-b)

7. Identify the dimensional constant.

• Planck’s constant
• Specific gravity

Answer: c) Planck’s constant

8. Mention two-dimensional variables.

Two-dimensional variables are Force and Velocity.

9. Identify the dimensionless quantity.

• All the above options

Answer: d) All the above options.

10. Define dimensionless quantities.

Physical quantities which do not possess dimensions are known as dimensionless quantities.

Practice Questions

• Define dimensional analysis.
• Mention three pairs that have the same dimensional formula.
• What are dimensional variables?
• What is the dimensional formula for entropy?
• List the applications of dimensional analysis.

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How to Solve Any Physics Problem

Last Updated: July 21, 2023 Fact Checked

This article was co-authored by Sean Alexander, MS . Sean Alexander is an Academic Tutor specializing in teaching mathematics and physics. Sean is the Owner of Alexander Tutoring, an academic tutoring business that provides personalized studying sessions focused on mathematics and physics. With over 15 years of experience, Sean has worked as a physics and math instructor and tutor for Stanford University, San Francisco State University, and Stanbridge Academy. He holds a BS in Physics from the University of California, Santa Barbara and an MS in Theoretical Physics from San Francisco State University. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 327,301 times.

Baffled as to where to begin with a physics problem? There is a very simply and logical flow process to solving any physics problem.

• Ask yourself if your answers make sense. If the numbers look absurd (for example, you get that a rock dropped off a 50-meter cliff moves with the speed of only 0.00965 meters per second when it hits the ground), you made a mistake somewhere.
• Don't forget to include the units into your answers, and always keep track of them. So, if you are solving for velocity and get your answer in seconds, that is a sign that something went wrong, because it should be in meters per second.
• Plug your answers back into the original equations to make sure you get the same number on both sides.

Community Q&A

• Many people report that if they leave a problem for a while and come back to it later, they find they have a new perspective on it and can sometimes see an easy way to the answer that they did not notice before. Thanks Helpful 249 Not Helpful 48
• Try to understand the problem first. Thanks Helpful 186 Not Helpful 51
• Remember, the physics part of the problem is figuring out what you are solving for, drawing the diagram, and remembering the formulae. The rest is just use of algebra, trigonometry, and/or calculus, depending on the difficulty of your course. Thanks Helpful 115 Not Helpful 34

• Physics is not easy to grasp for many people, so do not get bent out of shape over a problem. Thanks Helpful 100 Not Helpful 24
• If an instructor tells you to draw a free body diagram, be sure that that is exactly what you draw. Thanks Helpful 88 Not Helpful 24

Things You'll Need

• A Writing Utensil (preferably a pencil or erasable pen of sorts)
• Calculator with all the functions you need for your exam
• An understanding of the equations needed to solve the problems. Or a list of them will suffice if you are just trying to get through the course alive.

You Might Also Like

Expert Interview

Thanks for reading our article! If you’d like to learn more about teaching, check out our in-depth interview with Sean Alexander, MS .

• ↑ https://iopscience.iop.org/article/10.1088/1361-6404/aa9038
• ↑ https://physics.wvu.edu/files/d/ce78505d-1426-4d68-8bb2-128d8aac6b1b/expertapproachtosolvingphysicsproblems.pdf
• ↑ https://www.brighthubeducation.com/science-homework-help/42596-tips-to-choosing-the-correct-physics-formula/

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Netflix's hit sci-fi series '3 Body Problem' is based on a real math problem that is so complex it's impossible to solve

• The three-body problem is a centuries-old physics question that puzzled Isaac Newton .
• It describes the orbits of three bodies, like planets or stars, trapped in each other's gravity.
• The problem is unsolvable and led to the development of chaos theory.

While Netflix's "3 Body Problem" is a science-fiction show, its name comes from a real math problem that's puzzled scientists since the late 1600s.

In physics, the three-body problem refers to the motion of three bodies trapped in each other's gravitational grip — like a three-star system.

It might sound simple enough, but once you dig into the mathematics, the orbital paths of each object get complicated very quickly.

Two-body vs. three- and multi-body systems

A simpler version is a two-body system like binary stars. Two-body systems have periodic orbits, meaning they are mathematically predictable because they follow the same trajectory over and over. So, if you have the stars' initial positions and velocities, you can calculate where they've been or will be in space far into the past and future.

However, "throwing in a third body that's close enough to interact leads to chaos," Shane Ross, an aerospace and ocean engineering professor at Virginia Tech, told Business Insider. In fact, it's nearly impossible to precisely predict the orbital paths of any system with three bodies or more.

While two orbiting planets might look like a ven diagram with ovular paths overlapping, the paths of three bodies interacting often resemble tangled spaghetti. Their trajectories usually aren't as stable as systems with only two bodies.

All that uncertainty makes what's known as the three-body problem largely unsolvable, Ross said. But there are certain exceptions.

The three-body problem is over 300 years old

The three-body problem dates back to Isaac Newton , who published his "Principia" in 1687.

In the book, the mathematician noted that the planets move in elliptical orbits around the sun. Yet the gravitational pull from Jupiter seemed to affect Saturn's orbital path.

Related stories

The three-body problem didn't just affect distant planets. Trying to understand the variations in the moon's movements caused Newton literal headaches, he complained.

But Newton never fully figured out the three-body problem. And it remained a mathematical mystery for nearly 200 years.

In 1889, a Swedish journal awarded mathematician Henri Poincaré a gold medal and 2,500 Swedish crowns, roughly half a year's salary for a professor at the time, for his essay about the three-body problem that outlined the basis for an entirely new mathematical theory called chaos theory .

According to chaos theory, when there is uncertainty about a system's initial conditions, like an object's mass or velocity, that uncertainty ripples out, making the future more and more unpredictable.

Think of it like taking a wrong turn on a trip. If you make a left instead of a right at the end of your journey, you're probably closer to your destination than if you made the mistake at the very beginning.

Can you solve the three-body problem?

Cracking the three-body problem would help scientists chart the movements of meteors and planets, including Earth, into the extremely far future. Even comparatively small movements of our planet could have large impacts on our climate, Ross said.

Though the three-body problem is considered mathematically unsolvable, there are solutions to specific scenarios. In fact, there are a few that mathematicians have found.

For example, three bodies could stably orbit in a figure eight or equally spaced around a ring. Both are possible depending on the initial positions and velocities of the bodies.

One way researchers look for solutions is with " restricted " three-body problems, where two main bodies (like the sun and Earth) interact and a third object with much smaller mass (like the moon) offers less gravitational interference. In this case, the three-body problem looks a lot like a two-body problem since the sun and Earth comprise the majority of mass in the system.

However, if you're looking at a three-star system, like the one in Netflix's show "3 Body Problem," that's a lot more complicated.

Computers can also run simulations far more efficiently than humans, though due to the inherent uncertainties, the results are typically approximate orbits instead of exact.

Finding solutions to three-body problems is also essential to space travel, Ross said. For his work, he inputs data about the Earth, moon, and spacecraft into a computer. "We can build up a whole library of possible trajectories," he said, "and that gives us an idea of the types of motion that are possible."

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Course: physics library   >   unit 1.

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The 3-body problem is real, and it’s really unsolvable

Oh god don’t make me explain math

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Everybody seems to be talking about 3 Body Problem , the new Netflix series based on Cixin Liu’s Remembrance of Earth’s Past book trilogy . Fewer people are talking about the two series’ namesake: The unsolvable physics problem of the same name.

This makes sense, because it’s confusing . In physics, the three-body problem attempts to find a way to predict the movements of three objects whose gravity interacts with each of the others — like three stars that are close together in space. Sounds simple enough, right? Yet I myself recently pulled up the Wikipedia article on the three-body problem and closed the tab in the same manner that a person might stagger away from a bright light. Apparently the Earth, sun, and moon are a three-body system? Are you telling me we don’t know how the moon moves ? Scientists have published multiple solutions for the three-body problem? Are you telling me Cixin Liu’s books are out of date?

All I’d wanted to know was why the problem was considered unsolvable, and now memories of my one semester of high school physics were swimming before my eyes like so many glowing doom numbers. However, despite my pains, I have readied several ways that we non-physicists can be confident that the three-body problem is, in fact, unsolvable.

Reason 1: This is a special definition of ‘unsolvable’

The three-body problem is extra confusing, because scientists are seemingly constantly finding new solutions to the three-body problem! They just don’t mean a one-solution-for-all solution. Such a formula does exist for a two-body system, and apparently Isaac Newton figured it out in 1687 . But systems with more than two bodies are, according to physicists, too chaotic (i.e., not in the sense of a child’s messy bedroom, but in the sense of “chaos theory”) to be corralled by a single solution.

When physicists say they have a new solution to the three-body problem, they mean that they’ve found a specific solution for three-body systems that have certain theoretical parameters. Don’t ask me to explain those parameters, because they’re all things like “the three masses are collinear at each instant” or “a zero angular momentum solution with three equal masses moving around a figure-eight shape.” But basically: By narrowing the focus of the problem to certain arrangements of three-body systems, physicists have been able to derive formulas that predict the movements of some of them, like in our solar system. The mass of the Earth and the sun create a “ restricted three-body problem ,” where a less-big body (in this case, the moon) moves under the influence of two massive ones (the Earth and the sun).

What physicists mean when they say the three-body problem has no solution is simply that there isn’t a one-formula-fits-all solution to every way that the gravity of three objects might cause those objects to move — which is exactly what Three-Body Problem bases its whole premise on.

Reason 2: 3 Body Problem picked an unsolved three-body system on purpose

Henri Poincaré’s research into a general solution to the three-body problem formed the basis of what would become known as chaos theory (you might know it from its co-starring role in Jurassic Park ). And 3 Body Problem itself isn’t about any old three-body system. It’s specifically about an extremely chaotic three-body system, the exact kind of arrangement of bodies that Poincaré was focused on when he showed that the problem is “unsolvable.”

[ Ed. note: The rest of this section includes some spoilers for 3 Body Problem .]

In both Liu’s books and Netflix’s 3 Body Problem , humanity faces an invasion by aliens (called Trisolarans in the English translation of the books, and San-Ti in the TV series) whose home solar system features three suns in a chaotic three-body relationship. It is a world where, unlike ours, the heavens are fundamentally unpredictable. Periods of icy cold give way to searing heat that give way to swings in gravity that turn into temporary reprieves that can never be trusted. The unpredictable nature of the San-Ti environment is the source of every detail of their physicality, their philosophy, and their desire to claim Earth for their own.

In other words, 3 Body Problem ’s three-body problem is unsolvable because Liu wanted to write a story with an unsolvable three-body system, so he chose one of the three-body systems for which we have not discovered a solution, and might never.

Reason 3: Scientists are still working on the three-body problem

Perhaps the best reason I can give you to believe that the three-body problem is real, and is really unsolvable, is that some scientists published a whole set of new solutions for specific three-body systems very recently .

If physicists are still working on the three-body problem, we can safely assume that it has not been solved. Scientists, after all, are the real experts. And I am definitely not.

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