Reference frames

Topics 1.1, 1.2, and 1.3 measured motion from a single, unstated reference frame. The observer was always implicit: someone standing still on the ground, watching the motion go by. Topic 1.4 makes the observer the main character. Each observer carries their own reference frame, and the same motion measured in two frames gives two different velocity vectors.

The key idea is simple. A passenger walks forward at $1$ m/s inside a train moving $25$ m/s east. In the train frame the passenger's velocity is $1$ m/s; in the ground frame it's $26$ m/s. Both numbers are right — there is no frame-free "real" velocity hiding behind them.

The transformation between frames is a vector sum with strict subscript bookkeeping:

$\vec{v}_{A,B} + \vec{v}_{B,C} = \vec{v}_{A,C}.$

Read $\vec{v}_{A,B}$ as "velocity of $A$ in frame $B$". The two $B$'s match across the plus and cancel, leaving $\vec{v}_{A,C}$. Reversing the subscripts on one velocity flips its sign: $\vec{v}_{B,A} = -\vec{v}_{A,B}$.

Acceleration is different. Differentiate the velocity transformation and the constant relative velocity drops out, so $\vec{a}_{A,B} = \vec{a}_{A,C}$ for any pair of inertial frames. Every inertial observer measures the same acceleration. That's Galilean relativity.

The diagnostic targets three failure modes: treating velocity as frame-independent, confusing speed with velocity, and botching the signs or subscripts in the chain rule.

A reference frame is a coordinate system attached to an observer. Pick an origin, pick the unit vectors $\hat{\text{ı}}$ and $\hat{\text{ȷ}}$ (and $\hat{k}$ in 3D), and the frame is set. Anything you measure — position, velocity, acceleration — is measured in some frame.

• Ground frame: attached to a fixed point on Earth's surface. The default, but not special.
• Train frame, boat frame, plane frame: attached to the moving object. The object is at rest in its own frame.
• Air frame: attached to the air mass, useful when wind matters.
• Water frame: attached to the water, useful for river-crossing problems.

Two observers watching the same motion in different frames get different velocity vectors. Neither is wrong — that's what frame-dependence means. Positions work the same way: a passenger sitting still inside a train has constant position $\vec{r}_{P,T}$ in the train frame and a steadily changing position in the ground frame.

This lesson uses inertial frames: frames where Newton's first law holds, which means frames moving at constant velocity relative to one another. Unless a problem says otherwise, you may take a frame to be inertial. Accelerating frames (a braking car, a spinning carousel) are non-inertial and introduce pseudo-forces, which Topic 2.6 takes up. For the velocity transformations here, "frame" means "inertial frame".

A reference frame is a coordinate system attached to an observer. Velocity is frame-dependent: two inertial observers measuring the same motion will read different velocities, each correct in their own frame. Acceleration, for any pair of inertial frames, is the same.

Every relative velocity gets two subscripts.

$\vec{v}_{A,B} = $ "velocity of $A$ in frame $B$".

First subscript: the object whose motion you're describing. Second subscript: the frame doing the measuring. Two rules govern the algebra.

• Subscript reversal flips the sign. $\vec{v}_{B,A} = -\vec{v}_{A,B}$. The velocity of $A$ as seen from $B$ and the velocity of $B$ as seen from $A$ point in opposite directions.
• Inner subscripts cancel under addition. $\vec{v}_{A,B} + \vec{v}_{B,C} = \vec{v}_{A,C}$. The middle index $B$ appears once at the end of the first term and once at the start of the second, and it cancels, leaving the outer indices.

The chain rule is the workhorse. To find $\vec{v}_{A,C}$ from the legs $\vec{v}_{A,B}$ and $\vec{v}_{B,C}$, line them up so the inner subscripts match across a plus sign:

$\vec{v}_{A,B} + \vec{v}_{B,C} = \vec{v}_{A,C}.$

If a leg comes with its subscripts in the wrong order, flip its sign first. For example, if you have $\vec{v}_{C,B}$ but need $\vec{v}_{B,C}$, replace it with $-\vec{v}_{C,B}$:

$\vec{v}_{A,B} - \vec{v}_{C,B} = \vec{v}_{A,B} + \vec{v}_{B,C} = \vec{v}_{A,C}.$

The chain extends to any number of legs: $\vec{v}_{A,B} + \vec{v}_{B,C} + \vec{v}_{C,D} = \vec{v}_{A,D}$, with $B$ and $C$ both cancelling.

The classic trap is writing $\vec{v}_{A,C} = \vec{v}_{A,B} - \vec{v}_{B,C}$ with a minus sign instead of a plus. That breaks the chain rule. Mnemonic: plus sign, matching inner subscripts. If the subscripts already match across a plus, you're done; if not, flip one leg's sign to make them match.

The notation $\vec{v}_{A,B}$ reads as "velocity of $A$ in frame $B$". The chain rule is $\vec{v}_{A,B} + \vec{v}_{B,C} = \vec{v}_{A,C}$ — inner subscripts cancel under addition. Subscript reversal flips the sign: $\vec{v}_{B,A} = -\vec{v}_{A,B}$.

A train moves at $\vec{v}_{T,G} = +12\hat{\text{ı}}$ m/s east in the ground frame. A passenger inside walks toward the front at $\vec{v}_{P,T} = +1\hat{\text{ı}}$ m/s in the train frame. Here is what each observer sees.

What is the passenger's velocity in the ground frame? Apply the chain rule:

$\vec{v}_{P,G} = \vec{v}_{P,T} + \vec{v}_{T,G} = +1\hat{\text{ı}} + 12\hat{\text{ı}} = +13\hat{\text{ı}} \text{ m/s}.$

The arithmetic is head-to-tail addition on a number line.

Both readings are right, each in its own frame. From the ground: train at $+12\hat{\text{ı}}$ m/s, passenger adds $+1\hat{\text{ı}}$ m/s, total $+13\hat{\text{ı}}$ m/s. From the train: the train doesn't move ($\vec{v}_{T,T} = 0$), so the passenger's velocity is just $+1\hat{\text{ı}}$ m/s. The ground slides past at $\vec{v}_{G,T} = -\vec{v}_{T,G} = -12\hat{\text{ı}}$ m/s — subscript reversal flips the sign.

Same physics, two different views.

When the passenger walks toward the back of the train, $\vec{v}_{P,T} = -2\hat{\text{ı}}$ m/s; then $\vec{v}_{P,G} = -2\hat{\text{ı}} + 12\hat{\text{ı}} = +10\hat{\text{ı}}$ m/s. The passenger still moves forward in the ground frame, just more slowly than the train.

Now the opposite-direction case. Two cars on a straight road have $\vec{v}_{A,G} = +30\hat{\text{ı}}$ m/s and $\vec{v}_{B,G} = -30\hat{\text{ı}}$ m/s in the ground frame. The velocity of $B$ in $A$'s frame is

$\vec{v}_{B,A} = \vec{v}_{B,G} + \vec{v}_{G,A} = -30\hat{\text{ı}} + (-30\hat{\text{ı}}) = -60\hat{\text{ı}} \text{ m/s}.$

The closing speed is $60$ m/s — the magnitude of the relative velocity. Equal speeds in opposite directions don't give zero relative velocity; they give twice the common speed, with direction set by the subscript order.

In 1D, the chain rule $\vec{v}_{A,B} + \vec{v}_{B,C} = \vec{v}_{A,C}$ is just signed-number addition with subscript bookkeeping. Same-direction motions partially cancel; opposite-direction motions add to twice the common magnitude. The signs are part of the answer.

The chain rule works the same in 2D and 3D: velocities are vectors, and you add them as vectors. Two classic examples:

Boat on a river. A boat heads directly across a river at $\vec{v}_{B,W} = +4\hat{\text{ȷ}}$ m/s in the water frame (across is $+\hat{\text{ȷ}}$). The current is $\vec{v}_{W,G} = +3\hat{\text{ı}}$ m/s in the ground frame (downstream is $+\hat{\text{ı}}$). The boat's velocity in the ground frame is

$\vec{v}_{B,G} = \vec{v}_{B,W} + \vec{v}_{W,G} = 4\hat{\text{ȷ}} + 3\hat{\text{ı}} = 3\hat{\text{ı}} + 4\hat{\text{ȷ}} \text{ m/s}.$

The boat moves at $\sqrt{3^2 + 4^2} = 5$ m/s in the ground frame, on a path that combines the swim and the drift. Even though the boat points "directly across," its ground path veers downstream because the current carries it. Two traps: reporting $4\hat{\text{ȷ}}$ m/s (the swim velocity — only true in the water frame) as the ground-frame answer, or reporting $5$ m/s (a speed, no direction) as the velocity.

Plane in wind. A plane has airspeed $\vec{v}_{P,A} = +200\hat{\text{ı}}$ m/s relative to the air (heading east). The wind blows north: $\vec{v}_{A,G} = +50\hat{\text{ȷ}}$ m/s in the ground frame. The plane's velocity in the ground frame is

$\vec{v}_{P,G} = \vec{v}_{P,A} + \vec{v}_{A,G} = 200\hat{\text{ı}} + 50\hat{\text{ȷ}} \text{ m/s},$

with speed $|\vec{v}_{P,G}| = \sqrt{200^2 + 50^2} \approx 206$ m/s and a small northward drift. To hold an eastbound ground course in a northward wind, pilots point the plane slightly south of east — the "wind correction angle."

The same pattern every time: pick a frame chain, label the legs with subscripts, and add the vectors component by component. The magnitude follows from Pythagoras.

2D frame transformation is vector addition with subscript bookkeeping. The chain rule $\vec{v}_{A,B} + \vec{v}_{B,C} = \vec{v}_{A,C}$ works component-by-component. Speed (the magnitude) is not the same as velocity (the vector); the question always tells you which one to report.

Velocity is frame-dependent. Acceleration, between any two inertial frames, is not. The proof is one line of calculus.

Start with the chain rule for a particle $P$ measured in frames $A$ and $B$:

$\vec{v}_{P,B} = \vec{v}_{P,A} + \vec{v}_{A,B}.$

Differentiate both sides with respect to time:

$\dfrac{d\vec{v}_{P,B}}{dt} = \dfrac{d\vec{v}_{P,A}}{dt} + \dfrac{d\vec{v}_{A,B}}{dt}.$

For inertial frames, $\vec{v}_{A,B}$ is constant by definition, so its time derivative vanishes. What's left is

$\vec{a}_{P,B} = \vec{a}_{P,A}.$

Acceleration drops out of the frame transformation. A ball thrown into the air has $\vec{a} = -g\hat{\text{ȷ}}$ in the ground frame. To a bicyclist riding past at $5$ m/s, the ball's velocity at every instant is different, but its acceleration is still $-g\hat{\text{ȷ}}$.

This is the calculus version of Galilean relativity: the laws of mechanics take the same form in every inertial frame. Galileo's thought experiment was a sealed cabin on a smoothly sailing ship. Inside, fish swim, water drips, and butterflies fly exactly the way they would on shore — no mechanical experiment can detect the ship's motion.

Differentiating the velocity transformation drops the constant relative velocity between inertial frames. Hence $\vec{a}_{P,A} = \vec{a}_{P,B}$ for any two inertial frames. Velocity is frame-dependent; acceleration is frame-invariant across inertial frames. This is Galilean relativity in calculus form.

Every relative-velocity slip traces back to one of these three: forgetting to name the frame, adding speeds where velocities were needed, or scrambling the signs and subscripts in the chain rule.