SCIENCE · INVESTIGATE AND EXPLAIN

Differentiate the observer transformation

The same events can have different coordinate velocities without different accelerations.

Learning outcome: Compare two observers at the same event time. Differentiate the transformation twice and preserve vector components before comparing magnitudes.

Starting recap

Read derivatives as local rates and integrals as signed accumulation. Review power and chain rules, vector components, units and initial conditions where used.

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Try

With U=(2,2) at t=2, relative velocity is(2,4) m/s and speed sqrt20; acceleration is still(2,6) m/s².

Open the experiment

Predict

Under a constant-velocity translation with parallel axes, which is unchanged?

Your prediction and reason

Experiment

Compare two observers at the same event time. Differentiate the transformation twice and preserve vector components before comparing magnitudes.

Transform vector derivatives between inertial observers

Worked teaching model · synthetic values · Ground trajectory r=(t²,t³/2) m; observer origin translates at constant selected U, coincident at t0. Newtonian time and fixed parallel axes. Evaluation at2 s; this is not a rotating frame or relativistic transformation.

Observer east velocity: 1 m/s Observer north velocity: -1 m/s

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Same events described in two translating frames

Spatial axes use the same pixels per physical unit, preserving geometry. 1: ground coordinates (solid).2: translating coordinates (dashed). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Same events described in two translating frames

ground coordinates

east coordinate (m)	north coordinate (m)
0	0
0.000625	7.81250e-6
0.0025	6.25000e-5
0.005625	0.0002109375
0.01	0.0005
0.015625	0.0009765625
0.0225	0.0016875
0.030625	0.002679687
0.04	0.004
0.050625	0.005695313
0.0625	0.0078125
0.075625	0.01039844
0.09	0.0135
0.105625	0.01716406
0.1225	0.0214375
0.140625	0.02636719
0.16	0.032
0.180625	0.03838281
0.2025	0.0455625
0.225625	0.05358594
0.25	0.0625
0.275625	0.07235156
0.3025	0.0831875
0.330625	0.09505469
0.36	0.108
0.390625	0.1220703
0.4225	0.1373125
0.455625	0.1537734
0.49	0.1715
0.525625	0.1905391
0.5625	0.2109375
0.600625	0.2327422
0.64	0.256
0.680625	0.2807578
0.7225	0.3070625
0.765625	0.3349609
0.81	0.3645
0.855625	0.3957266
0.9025	0.4286875
0.950625	0.4634297
1	0.5
1.050625	0.5384453
1.1025	0.5788125
1.155625	0.6211484
1.21	0.6655
1.265625	0.7119141
1.3225	0.7604375
1.380625	0.8111172
1.44	0.864
1.500625	0.9191328
1.5625	0.9765625
1.625625	1.036336
1.69	1.0985
1.755625	1.163102
1.8225	1.230188
1.890625	1.299805
1.96	1.372
2.030625	1.44682
2.1025	1.524312
2.175625	1.604523
2.25	1.6875
2.325625	1.773289
2.4025	1.861938
2.480625	1.953492
2.56	2.048
2.640625	2.145508
2.7225	2.246062
2.805625	2.349711
2.89	2.4565
2.975625	2.566477
3.0625	2.679688
3.150625	2.79618
3.24	2.916
3.330625	3.039195
3.4225	3.165813
3.515625	3.295898
3.61	3.4295
3.705625	3.566664
3.8025	3.707437
3.900625	3.851867
4	4

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

translating coordinates

east coordinate (m)	north coordinate (m)
0	0
-0.024375	0.02500781
-0.0475	0.0500625
-0.069375	0.07521094
-0.09	0.1005
-0.109375	0.1259766
-0.1275	0.1516875
-0.144375	0.1776797
-0.16	0.204
-0.174375	0.2306953
-0.1875	0.2578125
-0.199375	0.2853984
-0.21	0.3135
-0.219375	0.3421641
-0.2275	0.3714375
-0.234375	0.4013672
-0.24	0.432
-0.244375	0.4633828
-0.2475	0.4955625
-0.249375	0.5285859
-0.25	0.5625
-0.249375	0.5973516
-0.2475	0.6331875
-0.244375	0.6700547
-0.24	0.708
-0.234375	0.7470703
-0.2275	0.7873125
-0.219375	0.8287734
-0.21	0.8715
-0.199375	0.9155391
-0.1875	0.9609375
-0.174375	1.007742
-0.16	1.056
-0.144375	1.105758
-0.1275	1.157062
-0.109375	1.209961
-0.09	1.2645
-0.069375	1.320727
-0.0475	1.378687
-0.024375	1.43843
0	1.5
0.025625	1.563445
0.0525	1.628812
0.080625	1.696148
0.11	1.7655
0.140625	1.836914
0.1725	1.910437
0.205625	1.986117
0.24	2.064
0.275625	2.144133
0.3125	2.226563
0.350625	2.311336
0.39	2.3985
0.430625	2.488102
0.4725	2.580188
0.515625	2.674805
0.56	2.772
0.605625	2.87182
0.6525	2.974312
0.700625	3.079523
0.75	3.1875
0.800625	3.298289
0.8525	3.411938
0.905625	3.528492
0.96	3.648
1.015625	3.770508
1.0725	3.896062
1.130625	4.024711
1.19	4.1565
1.250625	4.291477
1.3125	4.429688
1.375625	4.57118
1.44	4.716
1.505625	4.864195
1.5725	5.015813
1.640625	5.170898
1.71	5.3295
1.780625	5.491664
1.8525	5.657437
1.925625	5.826867
2	6

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
Quantity	Value	Unit
Relative east position at2 s	2	m
Relative north position at2 s	6	m
Relative east velocity	3	m/s
Relative north velocity	7	m/s
Relative speed	7.615773	m/s
East acceleration in either frame	2	m/s²
North acceleration in either frame	6	m/s²

Constant observer velocity changes position and velocity, not acceleration.

Differentiate rprime=r−Ut: vprime=v−U and aprime=a because dU/dt=0. Speed and path shape can differ between frames; the plotted curves refer to the same time-labeled events in different coordinates, not two simultaneous physical paths. Accelerating or rotating observers require extra derivative terms and are outside this model.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

Constant observer velocity changes position and velocity, not acceleration.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Differentiate the observer transformation

The same events can have different coordinate velocities without different accelerations.

Let ground position be r(t)=(t²,t³/2) m. A second observer starts at the same origin and translates with constant velocity U=(Ux,Uy), with axes parallel to the ground axes. Its coordinates are rprime=r−Ut. Differentiating gives vprime=v−U; differentiating again gives aprime=a because U is constant.

At t=2 s, ground position is(4,4), velocity(4,6) and acceleration(2,6). For U=(1,−1), relative position is(2,6), relative velocity(3,7) and acceleration remains(2,6). Speeds sqrt52 and sqrt58 differ, as do coordinate paths, but these refer to the same events with the same Newtonian time coordinate.

A vector magnitude is not invariant under adding or subtracting another vector. Direction and path length can change between these observers. The model’s two curves represent coordinates in different frames, not two objects following different simultaneous ground paths. Their equal spatial scales preserve the geometry within each plotted coordinate description.

The constant-translation assumption matters. If U varies in time, the observer acceleration must be subtracted. If axes rotate, their unit vectors also have derivatives. Neither a rotating-frame correction nor a relativistic time transformation is hidden inside this simple Galilean model.

Changed-case check

Predict before reading the comparison. With U=(2,2) at t=2, relative velocity is(2,4) m/s and speed sqrt20; acceleration is still(2,6) m/s².

Retained investigation

Compare two observers at the same event time. Differentiate the transformation twice and preserve vector components before comparing magnitudes. What exact assumption makes the second derivative of the observer displacement vanish?

Source and evidence boundary

Conceptual reference checked30 October2026: OpenStax University Physics Volume1. All prose, examples and models are original. These are bounded supplied mathematical trajectories or synthetic observations, not observed physical experiments. No launch, impact or apparatus procedure is requested. Record assumptions, prior predictions, errors and independent checks; guided success is not independently demonstrated proficiency.

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Public practice checks a specific claim. It does not establish independent mastery.

What exact assumption makes the second derivative of the observer displacement vanish?

Your explanation and limitations

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