Skip to main content
Sandbox Physics

M006 · Observers & invariants

Galilean Frame Switcher

Move between platform, train, conveyor and mass-centre observers. Keep the same two gliders and recorded positions while comparing momentum and the kinetic-energy minimum.

Interactive modelGalilean Frame Switcher
Nearest valid window timeReady\text{Ready}
Estimated horizontal momentumReady\text{Ready}
Estimated kinetic energyReady\text{Ready}
Motion stateReady\text{Ready}
Selected timeReady\text{Ready}
Observer horizontal speedReady\text{Ready}
A horizontal velocity in frameReady\text{Ready}
A transverse velocityReady\text{Ready}
B horizontal velocity in frameReady\text{Ready}
B transverse velocityReady\text{Ready}
Total horizontal momentum in frameReady\text{Ready}
Total transverse momentumReady\text{Ready}
Total kinetic energy in frameReady\text{Ready}
Internal kinetic energyReady\text{Ready}
Mass-centre kinetic energy in frameReady\text{Ready}
Dissipated energy up to clockReady\text{Ready}
Energy decomposition residualReady\text{Ready}
Frame transformation residualReady\text{Ready}
Raw position framesReady\text{Ready}
Available slope estimatesReady\text{Ready}

Physics tutorial

Galilean Frame Switcher

BackgroundNewtonian mechanics permits equally valid constant-velocity observers.

Why it mattersMove between platform, train, conveyor and mass-centre observers. Keep the same two gliders and recorded positions while comparing momentum and the kinetic-energy minimum.

Start with the essentials

Focus question
Whose motion changes when the observer changes?
One-sentence intuition
Zero total momentum minimizes kinetic energy without eliminating internal motion.

Core mathematical model

The same event in another frame

xi′=xi−Utvx,i′=vx,i−U\begin{aligned}x_i^\prime&=x_i-Ut\\v_{x,i}^\prime&=v_{x,i}-U\end{aligned}

Only constant horizontal boosts are used. Time and transverse coordinates are unchanged.

Momentum changes with the observer

Px′=Px−MU,Py′=PyP_x^\prime=P_x-MU,\qquad P_y^\prime=P_y

Total momentum vanishes in the centre frame for this initial condition.

Energy has a minimum

K′=K−UPx+12MU2=12M∣Vcm−Ux^∣2+Kint\begin{aligned}K^\prime&=K-UP_x+\tfrac12MU^2\\&=\tfrac12M|\mathbf V_{\rm cm}-U\hat{\mathbf x}|^2+K_{\rm int}\end{aligned}

The residual is checked against the direct sum of the two body energies.

What survives a boost?

rB′−rA′=rB−rAvB′−vA′=vB−vA\begin{aligned}\mathbf r_B^\prime-\mathbf r_A^\prime&=\mathbf r_B-\mathbf r_A\\\mathbf v_B^\prime-\mathbf v_A^\prime&=\mathbf v_B-\mathbf v_A\end{aligned}

Separations and relative velocities remain the same; both accelerations are zero here.

Common difficulties

Separate frame choice from dynamics

Typical misconceptionChanging the observer changes the collision or removes all kinetic energy.

Better mental modelThe same physical positions and impulses receive different coordinates. Internal kinetic energy persists unless the interaction dissipates it.

Run the experiment

  1. 01

    Predict the observer

    Find an observer that makes A stationary, then compare the mass-centre observer.

    What to observe: One stationary body does not generally make total momentum vanish.
  2. 02

    Move along the energy curve

    Drag the copper handle across the observer-energy plot.

    What to observe: The minimum follows the total mass and momentum, not a preferred absolute rest frame.
  3. 03

    Keep both tracks

    Compare same-velocity and opposite-velocity gliders.

    What to observe: Passing positions on separate tracks do not cause a collision.
  4. 04

    Acquire before switching

    Acquire positions, change observers and compare finite-window estimates with the reference.

    What to observe: The raw laboratory record is identical after a frame switch.