Quantity and units in physics notes
Quantity and Units in Physics: The Foundation of Measurement
In physics, the ability to measure and quantify physical phenomena is crucial for understanding the natural world. This is achieved through quantities and units, which serve as the building blocks of scientific measurement. In this article, we will explore the types of quantities, the importance of units, and how to use them correctly in physics.
1. What are Physical Quantities?
Physical quantities are properties or characteristics of physical phenomena that can be measured. They are classified into two main types:
- Scalar Quantities: These are quantities that have only magnitude (size). Examples include mass, temperature, time, and speed.
- Vector Quantities: These quantities have both magnitude and direction. Examples include velocity, force, and displacement.
Each quantity has a standard unit of measurement, which allows us to describe the amount of the quantity accurately.
2. The International System of Units (SI Units)
The International System of Units (SI) is the globally accepted system for measuring physical quantities. It ensures consistency and accuracy in measurements across different fields of science and technology. SI units are based on seven base units, from which all other units (derived units) are formed.
Quantity |
Unit (SI Unit) |
Symbol |
Length |
Meter |
m |
Mass |
Kilogram |
kg |
Time |
Second |
s |
Electric Current |
Ampere |
A |
Temperature |
Kelvin |
K |
Amount of Substance |
Mole |
mol |
Luminous Intensity |
Candela |
cd |
3. Derived Units in Physics
Derived units are formed by multiplying, dividing, or raising base units to a power. While base units measure fundamental quantities such as length, mass, time, electric current, temperature, amount of substance, and luminous intensity, derived units combine these base units to express more complex quantities.
For example:
- Speed is derived from the base units of length (meters) and time (seconds), resulting in the unit meters per second (m/s).
- Force combines mass (kilograms) and acceleration (meters per second squared), resulting in the unit newtons (N), where $1 \, \text{N} = 1 \, \text{kg} \cdot \text{m/s}^2$.
Common Derived Units
Here’s a list of some common derived units in physics along with their definitions and formulas:
Quantity | Derived Unit | Symbol | Definition/Formula |
Area | Square meter | m² | Area = Length × Width |
Volume | Cubic meter | m³ | Volume = Length × Width × Height |
Speed | Meters per second | m/s | Speed = Distance / Time |
Velocity | Meters per second | m/s | Velocity = Displacement / Time |
Acceleration | Meters per second squared | m/s² | Acceleration = Change in Velocity / Time |
Force | Newton | N | Force = Mass × Acceleration, $1 \, \text{N} = 1 \, \text{kg} \cdot \text{m/s}^2$ |
Pressure | Pascal | Pa | Pressure = Force / Area, $1 \, \text{Pa} = 1 \, \text{N/m}^2$ |
Energy | Joule | J | Energy = Force × Distance, $1 \, \text{J} = 1 \, \text{N} \cdot \text{m}$ |
Power | Watt | W | Power = Energy / Time, $1 \, \text{W} = 1 \, \text{J/s}$ |
Frequency | Hertz | Hz | Frequency = 1 / Time Period, $1 \, \text{Hz} = 1/s$ |
Electric Charge | Coulomb | C | Electric Charge = Current × Time, $1 \, \text{C} = 1 \, \text{A} \cdot \text{s}$ |
Electric Potential | Volt | V | Electric Potential = Energy per Charge, $1 \, \text{V} = 1 \, \text{J/C}$ |
Electric Resistance | Ohm | Ω | Resistance = Voltage / Current, $1 \, \text{Ω} = 1 \, \text{V/A}$ |
For a full list of SI derived units, you can explore NIST's Guide to SI Units.
4. Conversion of Units
In physics, it is often necessary to convert between different units. This can involve converting between SI units and non-SI units or between different magnitudes of SI units. Common examples include converting kilometers to meters or hours to seconds.
For example:
- 1 kilometer (km) = 1,000 meters (m)
- 1 hour = 3,600 seconds (s)
Conversion factors are used to ensure that quantities remain consistent when changing from one unit to another. Online tools like the NIST Unit Conversion Tool can simplify these conversions.
5. Dimensional Analysis: A Key Tool in Physics
Dimensional analysis is a mathematical technique used in physics and engineering to check the correctness of equations, derive relationships between physical quantities, and convert units. By analyzing the dimensions of the physical quantities involved, dimensional analysis provides a powerful tool to verify that an equation makes sense in terms of units, even before any actual calculations are made.
What is Dimensional Analysis?
Dimensional analysis involves expressing physical quantities in terms of their fundamental dimensions, which typically include:
- [L] for length
- [M] for mass
- [T] for time
- [I] for electric current
- [Θ] for temperature
- [N] for amount of substance (moles)
- [J] for luminous intensity
Every derived quantity in physics (such as velocity, force, or energy) can be expressed as a combination of these fundamental dimensions.
For example:
- Speed (v) is defined as distance divided by time, so its dimensions are $[L][T]^{-1}$.
- Force (F) is mass times acceleration, and since acceleration is $[L][T]^{-2}$, the dimensions of force are $[M][L][T]^{-2}$.
Using Dimensional Analysis
Dimensional analysis is applied in various ways, such as checking the correctness of physical equations, converting between units, and estimating relationships between physical quantities.
a. Checking the Consistency of Equations
One of the most common uses of dimensional analysis is verifying that both sides of an equation have the same dimensions. For an equation to be physically meaningful, the dimensions on both sides must be identical.
b. Converting Between Units
Dimensional analysis helps in converting units from one system to another, such as from metric to imperial units, or between different SI unit prefixes. For example, if you want to convert 10 kilometers per hour to meters per second, dimensional analysis allows you to systematically set up and solve the conversion.
$$10\text{\hspace{0.17em}}\text{km/h}=10\times \left(\frac{1000\text{\hspace{0.17em}}\text{m}}{1\text{\hspace{0.17em}}\text{km}}\right)\times \left(\frac{1\text{\hspace{0.17em}}\text{h}}{3600\text{\hspace{0.17em}}\text{s}}\right)=2.78\text{\hspace{0.17em}}\text{m/s}$$
c. Deriving Relationships Between Physical Quantities
Dimensional analysis is also used to derive or guess formulas when the relationship between quantities is unknown. This method is particularly useful when you know the relevant variables but do not know how they are related.
Checking the Validity of an Equation Using Dimensional Analysis
Dimensional analysis can be used to check the validity of an equation by ensuring that the dimensions on both sides of the equation are consistent. This method is based on the principle of dimensional homogeneity, which states that for any physical equation to be correct, the dimensions of all terms must be the same.
Here’s a step-by-step process for checking the validity of an equation using dimensional analysis.
Step 1: Identify the Physical Quantities
First, break down the physical quantities in the equation into their fundamental dimensions, which typically include:
- [M] for mass
- [L] for length
- [T] for time
- [I] for electric current
- [Θ] for temperature
- [N] for the amount of substance (moles)
- [J] for luminous intensity
Step 2: Express Each Quantity in Terms of Fundamental Dimensions
For every physical quantity in the equation, express it in terms of its fundamental dimensions. Here are a few examples of common quantities and their dimensions:
- Velocity (v): $[L][T]^{-1}$
- Acceleration (a): $[L][T]^{-2}$
- Force (F): $[M][L][T{]}^{-2}$
- Energy (E): $[M][L{]}^{2}[T{]}^{-2}$
- Pressure (P): $[M][L{]}^{-1}[T{]}^{-2}$
Step 3: Check the Dimensions on Both Sides of the Equation
Once you have expressed each term in the equation in terms of its fundamental dimensions, compare the dimensions on both sides. The equation is dimensionally valid if both sides have the same dimensions.
Example 1: Checking Newton's Second Law
Consider Newton's second law of motion, which states:
$F = ma$Here, $F$ is force, $m$ is mass, and $a$ is acceleration. Let's check the validity of this equation using dimensional analysis.
- The dimensions of force $F$ are $[M][L][T]^{-2}$.
- The mass $m$ has dimensions $[M]$.
- The acceleration $a$ has dimensions $[L][T]^{-2}$.
Now, check the right-hand side of the equation:
$ma = [M] \times [L][T]^{-2} = [M][L][T]^{-2}$Both sides of the equation have the same dimensions $[M][L][T]^{-2}$, so the equation is dimensionally correct.
Example 2: Checking the Equation for Kinetic Energy
Consider the equation for kinetic energy:
$E_k = \frac{1}{2} mv^2$Here, $E_k$ is kinetic energy, $m$ is mass, and $v$ is velocity.
- The dimensions of energy $E_k$ are $[M][L]^2[T]^{-2}$.
- The mass $m$ has dimensions $[M]$.
- The velocity $v$ has dimensions $[L][T]^{-1}$, so $v^2$ has dimensions $[L]^2[T]^{-2}$.
Now, check the right-hand side of the equation:
$mv^2 = [M] \times [L]^2[T]^{-2} = [M][L]^2[T]^{-2}$Both sides of the equation have the same dimensions $[M][L]^2[T]^{-2}$, so the equation is dimensionally valid.
Example 3: Checking a Potentially Incorrect Equation
Consider the equation:
$v = at^2$Here, $v$ is velocity, $a$ is acceleration, and $t$ is time. Let's check if this equation is dimensionally consistent.
- The dimensions of velocity $v$ are $[L][T]^{-1}$.
- The dimensions of acceleration $a$ are $[L][T]^{-2}$.
- The dimensions of time $t$ are $[T]$, so $t^2$ has dimensions $[T]^2$.
Now, check the right-hand side of the equation:
$at^2 = [L][T]^{-2} \times [T]^2 = [L]$The right-hand side has dimensions $[L]$, while the left-hand side has dimensions $[L][T]^{-1}$. Since the dimensions do not match, this equation is dimensionally incorrect.
Example 4: Deriving a Formula Using Dimensional Analysis
Dimensional analysis can also be used to derive a formula when you know the physical quantities involved but not the exact relationship between them. For example, let's derive the formula for the time period $T$ of a simple pendulum, which depends on the length $L$ of the pendulum and the acceleration due to gravity $g$.
- The dimensions of $T$ are $[T]$.
- The dimensions of $L$ are $[L]$.
- The dimensions of $g$ are $[L][T]^{-2}$.
Assume $T$ is proportional to a combination of $L$ and $g$:
$T \propto L^a g^b$The dimensions of $L^a$ are $[L]^a$, and the dimensions of $g^b$ are $[L]^b [T]^{-2b}$. So, the dimensions of the right-hand side are:
$[L]^a \times [L]^b [T]^{-2b} = [L]^{a+b} [T]^{-2b}$For dimensional consistency, the dimensions on both sides must be the same, so:
$[T] = [L]^{a+b} [T]^{-2b}$Equating the powers of $L$ and $T$ gives two equations:
- $a+b=0$
- $-2b = 1$
Solving these equations gives $b = -\frac{1}{2}$ and $a = \frac{1}{2}$. Therefore, the time period is proportional to:
$T \propto \sqrt{\frac{L}{g}}$Thus, the time period of a simple pendulum is proportional to the square root of the length divided by the acceleration due to gravity.
Using the Dimensional Method to Derive Relationships Between Physical Quantities
The dimensional method (also known as dimensional analysis) is a powerful technique used to derive relationships between physical quantities without relying on experimental data or detailed knowledge of the phenomena. By analyzing the dimensions of the involved quantities, we can often guess or infer formulas that relate them.
Dimensional analysis primarily works under the assumption that physical laws are dimensionally homogeneous, meaning that all terms in an equation must have the same dimensions. This helps us set up equations that balance dimensions on both sides, providing useful insights into the relationships between variables.
Steps to Derive Relationships Using Dimensional Analysis
Here’s the general procedure for deriving relationships between physical quantities using the dimensional method:
Identify the relevant physical quantities: Begin by identifying the variables that might influence the quantity of interest. These could include factors like mass, length, time, or other physical constants (such as gravitational constant or speed of light).
Express the dimensions of each quantity: Write down the dimensional formula for each quantity in terms of fundamental dimensions like [L] (length), [M] (mass), [T] (time), and others such as [Θ] (temperature) or [I] (current).
Set up an equation of proportionality: Assume that the quantity you want to find is proportional to a combination of the other variables raised to some powers (constants to be determined).
Solve for the exponents: Match the dimensions on both sides of the equation and solve for the unknown exponents.
Write the final equation: Once the exponents are known, you can write the final formula (ignoring dimensionless constants, which cannot be determined through dimensional analysis alone).
Examples of Using Dimensional Analysis
a. Deriving the Period of a Simple Pendulum
The period $T$ of a simple pendulum (the time it takes to complete one full swing) is known to depend on two variables:
- The length $L$ of the pendulum.
- The gravitational acceleration $g$ (a constant on Earth).
We aim to derive the formula for the period using dimensional analysis.
List the dimensions of the relevant quantities:
- $T$ (period) has dimensions of time $[T]$.
- $L$ (length) has dimensions of length $[L]$.
- $g$ (gravitational acceleration) has dimensions of $[L][T]^{-2}$.
Set up a proportionality relation: Assume that the period $T$ depends on $L$ and $g$ as follows:
$T \propto L^a g^b$Here, $a$ and $b$ are the unknown exponents we want to determine.
Express dimensions of the quantities involved:
$[T] = [L]^a [L]^b [T]^{-2b}$Simplifying:
$[T] = [L]^{a+b} [T]^{-2b}$Solve for the exponents by matching dimensions:
- For the dimensions of length $[L]$: $a+b=0$
- For the dimensions of time $[T]$: $-2b = 1$
Solving these two equations:
- From $-2b = 1$, we get $b = -\frac{1}{2}$.
- Substituting $b = -\frac{1}{2}$ into $a + b = 0$, we get $a = \frac{1}{2}$.
Write the final equation: The relationship between the period $T$, the length $L$, and gravitational acceleration $g$ is:
$T \propto \sqrt{\frac{L}{g}}$The actual formula (including the constant $2\pi$, which dimensional analysis does not provide) is:
$T = 2\pi \sqrt{\frac{L}{g}}$
b. Deriving the Drag Force on an Object in a Fluid
The drag force $F_d$ experienced by an object moving through a fluid is dependent on several factors, including:
- The velocity $v$ of the object.
- The density $\rho$ of the fluid.
- The cross-sectional area $A$ of the object.
We aim to derive the relationship between these quantities.
List the dimensions of the relevant quantities:
- $F_d$ (drag force) has dimensions of force $[M][L][T{]}^{-2.}$
- $v$ (velocity) has dimensions of $[L][T]^{-1}$.
- $\rho$ (density) has dimensions of mass per unit volume, or $[M][L]^{-3}$.
- $A$ (cross-sectional area) has dimensions of $[L]^2$.
Set up a proportionality relation: Assume that the drag force $F_d$ depends on velocity, density, and area as follows:
$F_d \propto v^a \rho^b A^c$Here, $a$, $b$, and $c$ are unknown exponents.
Express dimensions of the quantities involved:
$[M][L][T]^{-2} = \left([L][T]^{-1}\right)^a \left([M][L]^{-3}\right)^b \left([L]^2\right)^c$Simplifying:
$[M][L][T]^{-2} = [L]^a [T]^{-a} [M]^b [L]^{-3b} [L]^{2c}$Grouping terms:
$[M][L][T]^{-2} = [M]^b [L]^{a - 3b + 2c} [T]^{-a}$Solve for the exponents by matching dimensions:
- For the dimensions of mass $[M]$: $b = 1$.
- For the dimensions of length $[L]$: $a-3b+2c=\mathrm{1,\; substituting}$$b = 1$, we get $a-3+2c=\mathrm{1,\; or}$$a + 2c = 4$.
- For the dimensions of time $[T]$: $-a = -2$, so $a=2.$
Substituting $a = 2$ into $a + 2c = 4$, we get $2 + 2c = 4$, or $c = 1$.
Write the final equation: The relationship between drag force $F_d$, velocity $v$, density $\rho$, and cross-sectional area $A$ is:
$F_d \propto \rho v^2 A$The actual formula (including a dimensionless constant $C_d$, the drag coefficient) is:
$${F}_{d}=\frac{1}{2}{C}_{d}\rho {v}^{2}A$$
Dimensional Homogeneity
A principle called dimensional homogeneity states that for any physical equation to be valid, every term in the equation must have the same dimensions. This principle can be used to check whether an equation is likely to be correct before actual calculation.
For example, consider the equation for kinetic energy:
$E_k = \frac{1}{2}mv^2$
The dimensions of energy are $[M][L]^2[T]^{-2}$. The mass $m$ has dimensions $[M]$, and velocity $v$ has dimensions $[L][T]^{-1}$, so the square of velocity has dimensions $[L]^2[T]^{-2}$. Thus, multiplying $m$ by $v^2$ gives $[M][L]^2[T]^{-2}$, which matches the dimensions of energy.
Advantages of Dimensional Analysis
- Universality: Dimensional analysis can be applied across all fields of science and engineering.
- Simplifies complex problems: By reducing problems to their fundamental dimensions, we can often derive relationships without requiring detailed information.
- Checks correctness: It can quickly verify the dimensional consistency of an equation before solving it numerically.
Limitations of Dimensional Analysis
While dimensional analysis is a powerful tool, it has limitations. For example:
- Dimensionless quantities: Some quantities, such as angles (measured in radians), are dimensionless and cannot be derived using dimensional analysis.
- Constants of proportionality: Dimensional analysis cannot determine dimensionless constants (like the $\frac{1}{2}$ in the kinetic energy equation or the
- $2\pi$ in the formula for the period of a pendulum).
- Inapplicable to every case: Dimensional analysis cannot be used when the variables do not have independent dimensions.
6. Prefixes for Units
IIn physics and other scientific disciplines, we often deal with very large or very small quantities. Prefixes are added to units to simplify the expression of these quantities, making them easier to read and work with. These prefixes represent powers of ten, allowing us to scale up or down a unit by factors ranging from billions to trillionths.
The International System of Units (SI) defines a set of standard prefixes for units, which are applied consistently across different fields of science and technology.
What are Prefixes?
Prefixes are short abbreviations added to the beginning of a unit to indicate a multiple or a fraction of that unit. For example:
- The prefix "kilo-" represents a thousand (10³), so one kilometer (km) is 1,000 meters.
- The prefix "milli-" represents one-thousandth (10⁻³), so one millimeter (mm) is 0.001 meters.
Common SI Unit Prefixes
Below is a table of the most commonly used SI prefixes, ranging from extremely large to extremely small values:
Prefix |
Symbol |
Factor |
Power of Ten |
Example |
Giga- |
G |
1,000,000,000 |
$1{0}^{9}$ |
1 gigawatt (GW) =
1,000,000,000 watts |
Mega- |
M |
1,000,000 |
$1{0}^{6}$ |
1 megabyte
(MB) = 1,000,000 bytes |
Kilo- |
k |
1,000 |
$1{0}^{3}$ |
1 kilometer (km) =
1,000 meters |
Hecto- |
h |
100 |
$1{0}^{2}$ |
1 hectoliter
(hl) = 100 liters |
Deca- |
da |
10 |
$1{0}^{1}$ |
1 decameter (dam) = 10
meters |
Deci- |
d |
0.1 |
$1{0}^{-1}$ |
1 decimeter
(dm) = 0.1 meters |
Centi- |
c |
0.01 |
$1{0}^{-2}$ |
1 centimeter (cm) =
0.01 meters |
Milli- |
m |
0.001 |
$1{0}^{-3}$ |
1 millimeter
(mm) = 0.001 meters |
Micro- |
µ |
0.000001 |
$1{0}^{-6}$ |
1 microsecond (µs) =
0.000001 seconds |
Nano- |
n |
0.000000001 |
$1{0}^{-9}$ |
1 nanometer
(nm) = 0.000000001 meters |
Pico- |
p |
0.000000000001 |
$1{0}^{-12}$ |
1 picofarad (pF) =
0.000000000001 farads |
Examples of Unit Prefix Usage
Distance:
- 1 kilometer (km) = 1,000 meters (m)
- 1 millimeter (mm) = 0.001 meters (m)
Time:
- 1 megasecond (Ms) = 1,000,000 seconds
- 1 microsecond (µs) = 0.000001 seconds
Weight:
- 1 kilogram (kg) = 1,000 grams (g)
- 1 milligram (mg) = 0.001 grams (g)
Why are Prefixes Important?
Prefixes simplify complex numbers and make scientific communication more efficient. Instead of writing large numbers like "1,000,000 grams," we can write "1 megagram (Mg)." Similarly, for very small quantities, it’s easier to say "5 nanometers" than "0.000000005 meters."
- In Scientific Research: Prefixes allow researchers to work with a wide range of magnitudes—from the extremely large (astronomical distances) to the extremely small (subatomic particles).
- In Engineering: Engineers use prefixes in fields such as electronics (e.g., nanoseconds, gigabytes) and mechanics (e.g., kilometers, milligrams) to design systems and understand their behavior at various scales.
Converting Between Prefixes
To convert between units with different prefixes, it is essential to know their relationships and corresponding factors of ten. For example, to convert from kilometers to meters:
$1 \, \text{km} = 1,000 \, \text{m} = 10^3 \, \text{m}$Similarly, to convert from milligrams to grams:
$1 \, \text{mg} = 0.001 \, \text{g} = 10^{-3} \, \text{g}$Conversion charts and online calculators can assist with these transformations. A useful tool for this is the NIST Unit Conversion Calculator.
For a full list of SI prefixes and their usage, check out NIST SI Prefixes.
7. Significance of Units in Physics
Units are essential for ensuring that measurements are accurate and universally understood. Without standardized units, scientific research, engineering, and technological development would be difficult, if not impossible. For example, engineers designing a bridge need to ensure that measurements are consistent so that the structure is safe.
Incorrect use of units can have disastrous consequences, as seen in the infamous Mars Climate Orbiter mission failure, where a mix-up between imperial and metric units caused the spacecraft to disintegrate upon entry into Mars' atmosphere.
8. Conclusion
Understanding physical quantities and units is foundational for success in physics. Whether you're working on simple mechanics problems or exploring advanced topics like quantum mechanics, accurate measurement and conversion of quantities are crucial. As you study physics, always pay attention to the units of your answers and the consistency of your equations.
External Resources
- Khan Academy: Units and Measurement
- NIST: International System of Units (SI)
- Measurement and Numbers (physicsclassroom.com)
- Fundamentals: Concept Builders - Measurement and Units (physicsclassroom.com)
- NIST: SI Units and Dimensional Analysis
- Khan Academy: Dimensional Analysis
- NIST: SI Units and Dimensional Analysis