SCIENCE · INVESTIGATE AND EXPLAIN

Derive photon energy with the emitter recoil retained

Consider an original initially stationary emitter of invariant mass M. It ends as one body of rest mass m and one photon moving in the positive x direction. The model is a formal conservation packet, not a claim about an actual atomic or nuclear transition or a prescription for operating a source.

Learning outcome: Derive the photon energy from the body invariant and independently close both total energy and signed momentum. Change fractional mass loss and compare the exact result with a no-recoil approximation.

Starting recap

Name event pairs, frame signs and c=1 units; retain simultaneity, worldline integration, four-momentum and unused evidence boundaries.

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Try

For M=2 and f=0.2, m=1.6. Photon energy is 0.36 and body energy 1.64, with recoil momentum −0.36 and velocity about −0.219512. Body kinetic energy is 0.04, so photon plus recoil energy equals the rest-energy decrease 0.4.

Open the experiment

Predict

Which conservation law is silently lost if the whole rest-energy decrease is assigned to one photon?

Your prediction and reason

Experiment

Derive the photon energy from the body invariant and independently close both total energy and signed momentum. Change fractional mass loss and compare the exact result with a no-recoil approximation.

Close photon emission and recoil

Worked teaching model · synthetic values · Original initially stationary isolated emitter with mass M becomes one recoiling body of rest mass m=(1−f)M and one photon moving+x. c=1; no other products. Formal conservation packet, no actual transition, source or nuclear operation.

Initial emitter rest mass: 2 reference mass Fractional rest-mass decrease: 0.2 fraction

Reset keeps your prediction, captures and practice history.

The photon takes momentum as well as energy; its energy is smaller than the rest-energy decrease because the body recoils.

Original initially stationary isolated emitter with mass M becomes one recoiling body of rest mass m=(1−f)M and one photon moving+x. c=1; no other products. Formal conservation packet, no actual transition, source or nuclear operation.

Original conditional ledger; reference units retain their distinct meanings
Quantity	Value	Unit
Final body rest mass	1.600000	reference mass
Photon energy and positive momentum	0.3600000	c=1 energy/momentum
Final body total energy	1.640000	reference mass × c²
Body recoil momentum	-0.3600000	reference mass × c
Body recoil velocity	-0.2195122	c
Body kinetic energy	0.04000000	reference mass × c²
Rest-energy decrease	0.4000000	reference mass × c²
Total final energy	2	reference mass × c²
Final total momentum	0	reference mass × c

These original virtual events and collision records are not an actual clock, accelerator, photon-source or practical experiment. No physical apparatus operation is assigned.

The photon takes momentum as well as energy; its energy is smaller than the rest-energy decrease because the body recoils.

Observation to attach to a capture

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

Understand

1. Define events, frames and conserved quantities

In c=1 units the photon has momentum +Eγ and the body momentum −Eγ. Energy conservation gives body energy M−Eγ. Applying the body mass shell, (M−Eγ)²−Eγ²=m², cancels the photon-energy square and yields Eγ=(M²−m²)/(2M).

The recoiling body has total energy (M²+m²)/(2M), velocity −Eγ/Ebody and kinetic energy Ebody−m. Thus the rest-energy decrease M−m is shared between photon energy and recoil kinetic energy. Assigning the entire decrease to the photon would violate the body momentum account except in a limiting approximation.

The original controls set m=(1−f)M. Photon energy is M(f−f²/2) and body kinetic energy Mf²/2. The fractional recoil correction therefore depends on the relative mass change, not only the initial mass. Additional emitted products, initial motion or a bound many-body environment would require a different conservation boundary.

2. Reconstruct the original case

3. Predict and investigate

Which conservation law is silently lost if the whole rest-energy decrease is assigned to one photon?

Transfer the event construction

Use M=1 and f=0.1. Find photon energy, recoil kinetic energy and their sum.

Retain the chosen events, frame convention and intermediate reasoning before opening public feedback.

Compare a worked response

Photon energy is 0.095 and recoil kinetic energy 0.005, totaling 0.1. The no-recoil value 0.1 would overstate photon energy; the body still must carry the opposite momentum.

Primary mechanism context. Checked 31 October 2026. Original derivations, records and diagrams; no physical apparatus operation or actual independent clock/particle measurement is claimed. Retain first reasoning and public-feedback exposure in the course evidence workspace when registered.

Apply

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

Use M=1 and f=0.1. Find photon energy, recoil kinetic energy and their sum.

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

Previous: Conserve four-momentum rather than a sum of rest masses Next: Separate an emission clock from photon arrival timing

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