The Central Idea
A reference frame is a coordinate system together with a way of measuring time.
To describe motion, we need:
- an origin;
- coordinate axes;
- a direction convention;
- a time variable;
- a statement of whether the frame itself is moving.
Position and velocity generally depend on the chosen frame.
Why Motion Needs a Reference Frame
The statement
“The object is moving at \(20\text{ m/s}\)”
is incomplete unless we know relative to what.
The same object can have different velocities relative to:
- the ground;
- a moving vehicle;
- another object;
- a translating coordinate system.
Fixed and Moving Frames
A fixed frame is treated as stationary for the problem.
A moving frame has an origin or axes that move relative to the fixed frame.
In many introductory problems, the ground is chosen as the fixed frame.
Example: Passenger on a Train
A passenger sits still in a train moving east at
\[ 30\frac{\text{m}}{\text{s}}. \]Relative to the train,
\[ \boxed{ v_{\text{passenger/train}}=0 }. \]Relative to the ground,
\[ \boxed{ v_{\text{passenger/ground}} = 30\frac{\text{m}}{\text{s}} \text{ east} }. \]The Same Motion Can Have Different Coordinate Descriptions
Suppose an object is at coordinate
\[ x=100\text{ m} \]in one frame.
If another frame's origin is located 40 meters to the right of the first origin, then the same object has coordinate
\[ x'=100-40. \]Therefore,
\[ \boxed{x'=60\text{ m}}. \]Changing the Origin
Suppose frame \(S'\) has the same axis direction as frame \(S\), but its origin is displaced by \(X\) relative to \(S\).
Then the coordinates are related by
\[ \boxed{ x'=x-X }. \]Equivalently,
\[ \boxed{ x=x'+X }. \]Vector Form of an Origin Shift
In multiple dimensions, let the moving origin have position
\[ \mathbf{R}(t) \]relative to a fixed frame.
If an object has fixed-frame position
\[ \mathbf{r}(t), \]then its position in the moving frame is
\[ \boxed{ \mathbf{r}'(t) = \mathbf{r}(t)-\mathbf{R}(t) }. \]What the Translation Formula Means
The vector
\[ \mathbf{R}(t) \]points from the fixed origin to the moving origin.
The vector
\[ \mathbf{r}'(t) \]points from the moving origin to the object.
Therefore,
\[ \mathbf{r} = \mathbf{R} + \mathbf{r}'. \]Example: Translating Position Coordinates
Suppose an object has fixed-frame position
\[ \mathbf{r} = \langle12,7\rangle \]and the moving origin is at
\[ \mathbf{R} = \langle5,2\rangle. \]Then
\[ \mathbf{r}' = \langle12,7\rangle-\langle5,2\rangle \] \[ = \boxed{\langle7,5\rangle}. \]Moving Origins
If the moving frame's origin changes position with time, write
\[ \mathbf{R}=\mathbf{R}(t). \]Then
\[ \boxed{ \mathbf{r}'(t) = \mathbf{r}(t)-\mathbf{R}(t) }. \]Differentiating produces the velocity transformation.
Velocity Transformation
Differentiate
\[ \mathbf{r}'=\mathbf{r}-\mathbf{R}. \]Then
\[ \mathbf{v}' = \mathbf{v}-\mathbf{V}, \]where
\[ \mathbf{V} = \mathbf{R}'. \]Therefore,
\[ \boxed{ \mathbf{v}' = \mathbf{v}-\mathbf{V} }. \]Galilean Velocity Transformation
For two nonrotating frames whose axes remain parallel, if frame \(S'\) moves with constant velocity \(\mathbf{V}\) relative to frame \(S\), then
\[ \boxed{ \mathbf{v}_{S'} = \mathbf{v}_S-\mathbf{V} }. \]This is the classical Galilean velocity transformation.
Example: Walking inside a Train
A train moves east at
\[ 25\frac{\text{m}}{\text{s}}. \]A passenger walks east inside the train at
\[ 2\frac{\text{m}}{\text{s}} \]relative to the train.
Therefore, relative to the ground,
\[ v_{\text{passenger/ground}} = 25+2. \]Thus
\[ \boxed{ v_{\text{passenger/ground}} = 27\frac{\text{m}}{\text{s}} \text{ east} }. \]Walking Opposite the Train's Motion
Suppose the passenger instead walks west at
\[ 2\frac{\text{m}}{\text{s}} \]relative to the train.
Taking east as positive,
\[ v_{\text{passenger/train}}=-2. \]Therefore,
\[ v_{\text{passenger/ground}} = 25+(-2) \] \[ = \boxed{ 23\frac{\text{m}}{\text{s}} \text{ east} }. \]Velocity Chain Rule
Relative velocities combine through
\[ \boxed{ \mathbf{v}_{A/C} = \mathbf{v}_{A/B} + \mathbf{v}_{B/C} }. \]This notation keeps the reference frames explicit.
For a passenger, train, and ground,
\[ \mathbf{v}_{P/G} = \mathbf{v}_{P/T} + \mathbf{v}_{T/G}. \]Example: Boat, Water, and Shore
A boat moves relative to the water at
\[ \mathbf{v}_{B/W} = \langle4,0\rangle. \]The water moves relative to the shore at
\[ \mathbf{v}_{W/S} = \langle0,3\rangle. \]Therefore,
\[ \mathbf{v}_{B/S} = \mathbf{v}_{B/W} + \mathbf{v}_{W/S} \] \[ = \boxed{\langle4,3\rangle}. \]Changing Velocity Descriptions
Suppose an object moves at
\[ \mathbf{v} = \langle8,5\rangle \]in frame \(S\).
Frame \(S'\) moves with
\[ \mathbf{V} = \langle3,1\rangle \]relative to \(S\).
Then
\[ \mathbf{v}' = \langle8,5\rangle-\langle3,1\rangle \] \[ = \boxed{\langle5,4\rangle}. \]Acceleration Transformation
Differentiate the velocity transformation:
\[ \mathbf{v}' = \mathbf{v}-\mathbf{V}. \]Then
\[ \boxed{ \mathbf{a}' = \mathbf{a}-\mathbf{A} }, \]where
\[ \mathbf{A} = \frac{d\mathbf{V}}{dt} \]is the acceleration of the moving frame's origin.
Constant-Velocity Frames Measure the Same Acceleration
If
\[ \mathbf{V}=\text{constant}, \]then
\[ \mathbf{A}=\mathbf{0}. \]Therefore,
\[ \boxed{ \mathbf{a}'=\mathbf{a} }. \]This is one important feature of inertial reference frames in classical mechanics.
Inertial Reference Frames
An inertial reference frame is a frame in which objects with no net force move with constant velocity.
Any frame moving at constant velocity relative to an inertial frame is also inertial in classical mechanics.
Introductory kinematics often assumes the chosen ground frame is approximately inertial.
Accelerating Reference Frames
If the frame itself accelerates, then
\[ \mathbf{A}\ne\mathbf{0}. \]The acceleration measured in that frame is
\[ \boxed{ \mathbf{a}' = \mathbf{a}-\mathbf{A} }. \]The moving frame can therefore assign a different acceleration to the same object.
Example: Accelerating Elevator
Suppose an elevator accelerates upward at
\[ 2\frac{\text{m}}{\text{s}^2}. \]An object that is stationary relative to the building has
\[ a=0. \]Relative to the accelerating elevator,
\[ a'=0-2. \]Therefore,
\[ \boxed{ a'=-2\frac{\text{m}}{\text{s}^2} }. \]The object appears to accelerate downward relative to the elevator.
Frame-Dependent Position
Position coordinates depend strongly on the choice of origin.
The same object can have coordinate
\[ x=50 \]in one frame and
\[ x'=-10 \]in another.
Neither description is inherently more correct if each frame is defined consistently.
Frame-Dependent Velocity
Velocity also depends on the reference frame.
A passenger sitting still on a train has
\[ v'=0 \]in the train frame but nonzero velocity in the ground frame.
Relative Position between Two Objects Is Translation-Invariant
Suppose both object positions are shifted by the same frame-origin vector \(\mathbf{R}\):
\[ \mathbf{r}'_A = \mathbf{r}_A-\mathbf{R} \]and
\[ \mathbf{r}'_B = \mathbf{r}_B-\mathbf{R}. \]Then
\[ \mathbf{r}'_A-\mathbf{r}'_B = \mathbf{r}_A-\mathbf{r}_B. \]Therefore,
\[ \boxed{ \mathbf{r}'_{A/B} = \mathbf{r}_{A/B} }. \]Separation Distance Is Translation-Invariant
Since relative position is unchanged by a common translation,
\[ |\mathbf{r}'_A-\mathbf{r}'_B| = |\mathbf{r}_A-\mathbf{r}_B|. \]Therefore, two translated frames agree on the distance separating the same pair of objects.
Relative Velocity between Two Objects Is Also Translation-Invariant
If both object velocities are shifted by the same frame velocity \(\mathbf{V}\), then
\[ \mathbf{v}'_A = \mathbf{v}_A-\mathbf{V} \]and
\[ \mathbf{v}'_B = \mathbf{v}_B-\mathbf{V}. \]Therefore,
\[ \mathbf{v}'_A-\mathbf{v}'_B = \mathbf{v}_A-\mathbf{v}_B. \]Thus
\[ \boxed{ \mathbf{v}'_{A/B} = \mathbf{v}_{A/B} }. \]Why Shared Motion Cancels
Suppose two cars both move east at
\[ 60\frac{\text{km}}{\text{h}}. \]Relative to the ground, both have substantial velocity.
Relative to each other,
\[ 60-60=0. \]Their shared translational motion cancels.
Example: Two Observers
Object \(A\) moves east at
\[ 50\frac{\text{km}}{\text{h}}. \]Observer \(B\) moves east at
\[ 20\frac{\text{km}}{\text{h}}. \]Observer \(C\) moves west at
\[ 10\frac{\text{km}}{\text{h}}. \]Relative to \(B\),
\[ v_{A/B}=50-20=30. \]Relative to \(C\),
\[ v_{A/C}=50-(-10)=60. \]Therefore,
\[ \boxed{ v_{A/B}=30\frac{\text{km}}{\text{h}}, \qquad v_{A/C}=60\frac{\text{km}}{\text{h}} }. \]Reference Frames and Projectile Motion
A projectile can have very different-looking paths in different frames.
Relative to the ground, an ideal projectile follows a parabola.
Relative to a frame moving horizontally with the same constant horizontal velocity as the projectile, the horizontal component can disappear.
In that moving frame, the projectile appears to move only vertically.
Example: Frame Moving with a Projectile's Horizontal Velocity
Suppose a projectile has ground-frame velocity
\[ \mathbf{v}(t) = \langle10,15-9.8t\rangle. \]Choose a frame moving horizontally at
\[ \mathbf{V} = \langle10,0\rangle. \]Then
\[ \mathbf{v}'(t) = \mathbf{v}(t)-\mathbf{V} \] \[ = \boxed{ \langle0,15-9.8t\rangle }. \]The motion is purely vertical in this translating frame.
Reference Frames and Circular Motion
A frame fixed to the ground can describe an object moving around a circle.
A frame rotating with the object can instead keep the object at a fixed coordinate.
Rotating frames require additional ideas beyond simple translational subtraction, because the axes themselves change direction.
Translating Frames versus Rotating Frames
The simple formulas
\[ \mathbf{r}'=\mathbf{r}-\mathbf{R} \]and
\[ \mathbf{v}'=\mathbf{v}-\mathbf{V} \]assume the frame axes remain parallel.
If the axes rotate, extra terms appear because the basis directions also change with time.
This lesson focuses primarily on translating frames.
Coordinate Orientation Matters
Two frames can have different origins and different axis orientations.
For example, one frame may take east as positive while another takes west as positive.
A velocity that is positive in one frame may therefore be represented by a negative coordinate in another even if the frames are not moving relative to each other.
Reflections of an Axis
Suppose one-dimensional frame \(S'\) reverses the direction of the axis relative to \(S\).
Then a simple coordinate relation is
\[ \boxed{ x'=-x }. \]Differentiating gives
\[ \boxed{ v'=-v }. \]The physical motion is unchanged; only the coordinate orientation differs.
Changing Units versus Changing Frames
Converting
\[ 10\frac{\text{m}}{\text{s}} \]to
\[ 36\frac{\text{km}}{\text{h}} \]changes units, not reference frames.
A reference-frame change alters the coordinate viewpoint itself.
Changing the Origin versus Changing the Frame Velocity
A constant shift of origin changes position coordinates but not velocity.
If
\[ \mathbf{R}=\text{constant}, \]then
\[ \mathbf{V}=\mathbf{0}. \]Therefore,
\[ \mathbf{v}'=\mathbf{v}. \]Moving the Origin at Constant Velocity
Suppose
\[ \mathbf{R}(t) = \mathbf{R}_0+\mathbf{V}t \]with constant \(\mathbf{V}\).
Then
\[ \boxed{ \mathbf{r}'(t) = \mathbf{r}(t) - \mathbf{R}_0 - \mathbf{V}t }. \]Velocity becomes
\[ \boxed{ \mathbf{v}'(t) = \mathbf{v}(t)-\mathbf{V} }. \]Example: Position in a Constant-Velocity Moving Frame
Suppose
\[ x(t)=8t+5 \]in the ground frame.
A second frame has origin
\[ X(t)=3t+2. \]Then
\[ x'(t) = (8t+5)-(3t+2) \] \[ = \boxed{5t+3}. \]Velocity in the Same Example
Differentiate:
\[ v=8 \]and
\[ V=3. \]Therefore,
\[ v'=8-3 \] \[ = \boxed{5}. \]This agrees with differentiating
\[ x'(t)=5t+3. \]Example: Accelerating Moving Frame
Suppose
\[ x(t)=5t^2 \]and the moving origin is
\[ X(t)=2t^2. \]Then
\[ x'(t)=3t^2. \]The ground-frame acceleration is
\[ a=10, \]while the frame-origin acceleration is
\[ A=4. \]Therefore,
\[ a'=10-4 \] \[ = \boxed{6}. \]Frame Choice Can Simplify a Problem
A well-chosen reference frame can remove unnecessary motion.
Examples include:
- riding with one of two vehicles in a relative-motion problem;
- moving with the horizontal velocity of a projectile;
- choosing the center of motion as the origin;
- choosing axes aligned with a path.
The physical situation does not change, but the equations may become much simpler.
Reference Frames and Relative Motion
Relative motion can be understood as choosing one object to define a moving frame.
If object \(B\) defines the moving origin, then the position of \(A\) in that frame is
\[ \boxed{ \mathbf{r}_{A/B} = \mathbf{r}_A-\mathbf{r}_B }. \]Its velocity is
\[ \boxed{ \mathbf{v}_{A/B} = \mathbf{v}_A-\mathbf{v}_B }. \]Collision Conditions Are Frame-Consistent
If two objects collide in one translating frame, they collide in every translating frame describing the same event.
In frame \(S\),
\[ \mathbf{r}_A=\mathbf{r}_B. \]Subtract the same frame-origin vector from both sides:
\[ \mathbf{r}_A-\mathbf{R} = \mathbf{r}_B-\mathbf{R}. \]Therefore,
\[ \boxed{ \mathbf{r}'_A=\mathbf{r}'_B }. \]Common Error: Giving a Velocity without a Reference
Velocity is not fully specified by a number and direction alone when multiple frames are involved.
State whether the velocity is relative to:
- the ground;
- a vehicle;
- the air or water;
- another moving object.
Common Error: Adding Velocities without Tracking the Frames
The expression
\[ \mathbf{v}_{A/B} + \mathbf{v}_{B/C} \]is meaningful because the intermediate frame \(B\) links the terms.
Randomly adding two velocities from unrelated frame descriptions can be incorrect.
Common Error: Reversing Relative-Velocity Order
The velocity of \(A\) relative to \(B\) is
\[ \mathbf{v}_{A/B} = \mathbf{v}_A-\mathbf{v}_B. \]Reversing the order changes the sign:
\[ \boxed{ \mathbf{v}_{B/A} = -\mathbf{v}_{A/B} }. \]Common Error: Assuming Position Is Absolute
A coordinate such as
\[ x=20 \]is meaningful only after an origin and axis have been specified.
Changing the origin changes the coordinate without changing the physical location.
Common Error: Assuming Constant Frame Velocity Changes Acceleration
If two frames move at constant velocity relative to one another, then
\[ \mathbf{A}=\mathbf{0}. \]Therefore,
\[ \boxed{ \mathbf{a}'=\mathbf{a} }. \]Constant translational frame velocity changes measured velocity but not measured acceleration.
Common Error: Using Translational Formulas for Rotating Frames
The simple transformation
\[ \mathbf{v}'=\mathbf{v}-\mathbf{V} \]assumes the axes remain parallel.
If the frame rotates, the changing basis directions introduce additional terms.
Common Error: Confusing a Coordinate Transformation with Physical Motion
An object's coordinate can change because:
- the object moves;
- the frame moves;
- the origin changes;
- the axes are reoriented.
A coordinate change does not necessarily mean the object itself changed its physical motion.
A Reliable Procedure for Translating Frames
- Identify the fixed frame and moving frame.
- Write the moving origin position \(\mathbf{R}(t)\).
- Write the object's fixed-frame position \(\mathbf{r}(t)\).
- Subtract to obtain \(\mathbf{r}'=\mathbf{r}-\mathbf{R}\).
- Differentiate for \(\mathbf{v}'=\mathbf{v}-\mathbf{V}\).
- Differentiate again for \(\mathbf{a}'=\mathbf{a}-\mathbf{A}\).
- Interpret signs and directions in the chosen axes.
A Reliable Procedure for Relative-Velocity Chains
- Write every velocity with explicit reference labels.
- Arrange the terms so the intermediate frames connect.
- Add or subtract vectors component by component.
- Check that the outer labels of the result match the requested quantity.
- Only then compute magnitude or direction if needed.
Example: Full Frame-Transformation Workflow
In ground frame \(S\), an object has position
\[ \mathbf{r}(t) = \langle8t+2,3t^2\rangle. \]Frame \(S'\) has origin
\[ \mathbf{R}(t) = \langle3t-1,t^2\rangle. \]Relative position:
\[ \mathbf{r}'(t) = \mathbf{r}(t)-\mathbf{R}(t) \] \[ = \langle5t+3,2t^2\rangle. \]Therefore,
\[ \boxed{ \mathbf{r}'(t) = \langle5t+3,2t^2\rangle }. \]Ground-frame velocity:
\[ \mathbf{v}(t) = \langle8,6t\rangle. \]Frame-origin velocity:
\[ \mathbf{V}(t) = \langle3,2t\rangle. \]Thus
\[ \boxed{ \mathbf{v}'(t) = \langle5,4t\rangle }. \]Ground-frame acceleration:
\[ \mathbf{a}(t) = \langle0,6\rangle. \]Frame acceleration:
\[ \mathbf{A}(t) = \langle0,2\rangle. \]Therefore,
\[ \boxed{ \mathbf{a}'(t) = \langle0,4\rangle }. \]Practice 1: Shifted Origin
An object is at \(x=18\) in frame \(S\). Frame \(S'\) has its origin at \(X=5\). Find \(x'\).
Practice 2: Moving Frame Velocity
An object moves east at \(40\text{ m/s}\) relative to the ground. A frame moves east at \(15\text{ m/s}\). Find the object's velocity in the moving frame.
Practice 3: Opposite-Direction Frame
An object moves east at \(50\text{ km/h}\). An observer moves west at \(20\text{ km/h}\). Find the object's velocity relative to the observer.
Practice 4: Vector Velocity Transformation
Let
\[ \mathbf{v} = \langle9,6\rangle \]and
\[ \mathbf{V} = \langle4,1\rangle. \]Find \(\mathbf{v}'\).
Practice 5: Position Transformation
Let
\[ \mathbf{r}(t) = \langle6t+1,2t\rangle \]and
\[ \mathbf{R}(t) = \langle2t,1\rangle. \]Find \(\mathbf{r}'(t)\).
Practice 6: Acceleration Transformation
An object has acceleration
\[ \mathbf{a} = \langle3,-2\rangle \]while the moving frame accelerates at
\[ \mathbf{A} = \langle1,4\rangle. \]Find \(\mathbf{a}'\).
Practice 7: Velocity Chain
A person walks east at \(2\text{ m/s}\) relative to a train. The train moves east at \(18\text{ m/s}\) relative to the ground. Find the person's velocity relative to the ground.
Practice 8: Shared Motion
Two cars move east at the same constant velocity of \(70\text{ km/h}\). What is the velocity of one car relative to the other?
Practice 9: Projectile in a Moving Frame
A projectile has ground-frame velocity
\[ \mathbf{v}(t) = \langle12,20-9.8t\rangle. \]Find its velocity in a frame moving with constant velocity
\[ \mathbf{V} = \langle12,0\rangle. \]Practice 10: Identify the Error
A student says:
“If two reference frames move at different constant velocities, they must measure different accelerations for the same object.”
Explain the error.
Practice Answers
Practice 1
\[ x'=x-X \] \[ =18-5 \] \[ \boxed{x'=13}. \]Practice 2
Taking east as positive,
\[ v'=40-15. \] \[ \boxed{ v'=25\frac{\text{m}}{\text{s}} \text{ east} }. \]Practice 3
Taking east as positive,
\[ v_{\text{object}}=50 \]and
\[ v_{\text{observer}}=-20. \]Therefore,
\[ v_{\text{object/observer}} = 50-(-20) \] \[ = \boxed{ 70\frac{\text{km}}{\text{h}} \text{ east} }. \]Practice 4
\[ \mathbf{v}' = \mathbf{v}-\mathbf{V} \] \[ = \langle9,6\rangle-\langle4,1\rangle \] \[ = \boxed{\langle5,5\rangle}. \]Practice 5
\[ \mathbf{r}'(t) = \langle6t+1,2t\rangle - \langle2t,1\rangle \] \[ = \boxed{ \langle4t+1,2t-1\rangle }. \]Practice 6
\[ \mathbf{a}' = \mathbf{a}-\mathbf{A} \] \[ = \langle3,-2\rangle-\langle1,4\rangle \] \[ = \boxed{\langle2,-6\rangle}. \]Practice 7
\[ v_{P/G} = v_{P/T}+v_{T/G} \] \[ =2+18 \] \[ = \boxed{ 20\frac{\text{m}}{\text{s}} \text{ east} }. \]Practice 8
\[ 70-70=0. \]Therefore,
\[ \boxed{ v_{\text{relative}}=0 }. \]Practice 9
\[ \mathbf{v}'(t) = \mathbf{v}(t)-\mathbf{V}. \] \[ = \langle12,20-9.8t\rangle - \langle12,0\rangle. \]Therefore,
\[ \boxed{ \mathbf{v}'(t) = \langle0,20-9.8t\rangle }. \]Practice 10
If the frames move with constant velocity relative to one another, their relative acceleration is zero:
\[ \mathbf{A}=\mathbf{0}. \]Therefore,
\[ \mathbf{a}' = \mathbf{a}-\mathbf{A} \] \[ = \boxed{\mathbf{a}}. \]Constant translational frame velocity changes measured velocity, but not measured acceleration.
What to Remember
For a translating frame with origin \(\mathbf{R}(t)\),
\[ \boxed{ \mathbf{r}' = \mathbf{r}-\mathbf{R} }. \]Velocity transforms as
\[ \boxed{ \mathbf{v}' = \mathbf{v}-\mathbf{V} }. \]Acceleration transforms as
\[ \boxed{ \mathbf{a}' = \mathbf{a}-\mathbf{A} }. \]When the moving frame has constant velocity,
\[ \boxed{ \mathbf{a}'=\mathbf{a} }. \]Relative-velocity chains satisfy
\[ \boxed{ \mathbf{v}_{A/C} = \mathbf{v}_{A/B} + \mathbf{v}_{B/C} }. \]The most important conceptual facts are:
- position and velocity depend on the chosen reference frame;
- a common translation changes coordinates but not relative separation;
- constant-velocity translating frames agree on acceleration;
- accelerating frames generally measure different accelerations;
- rotating frames require more than simple translational subtraction;
- careful reference labels prevent sign and interpretation errors.
Where We Go Next
Reference frames tell us how motion descriptions depend on the observer. The next lesson returns to the geometry of a single moving object and asks how its acceleration can be separated into effects that change speed and effects that change direction.
The next lesson, Tangential and Normal Components, decomposes acceleration along the tangent and inward normal directions of a curved path.