Repeated shock crossings produce a conditional power law
Derive a test-particle spectrum and compare acceleration with age, escape and cooling.
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The question for today
What you’ll understand
- Explain how collective plasma fields mediate a collisionless shock.
- Distinguish phase-space and differential-energy spectral indices.
- Check acceleration against finite age, escape and radiative losses.
A shock can dissipate bulk motion even when binary particle collisions are too rare to create its transition layer. Collective electromagnetic fields scatter particles and transfer momentum. A small population can cross the converging upstream and downstream flows repeatedly, gaining energy at each cycle. The resulting first-order Fermi spectrum is a controlled test-particle prediction under stated scattering assumptions, rather than a guarantee that every astrophysical shock produces the same observed cosmic-ray slope.
Use the shock frame but distinguish the scattering frames
Consider a steady planar nonrelativistic shock. In its rest frame the upstream plasma enters at speed u₁ and leaves downstream at u₂, with compression ratio r=u₁/u₂. A strong shock in a monatomic ideal gas has r≈4 if magnetic pressure, radiative losses and energetic-particle backreaction are negligible. The accelerated particles are relativistic, moving much faster than either flow. Their distributions are assumed nearly isotropic in the respective local plasma scattering frames, not simultaneously isotropic in the shock frame.
Scattering approximately preserves particle energy in each local scattering frame, but the upstream and downstream frames approach one another. A full crossing and return cycle therefore yields a mean fractional gain proportional to their speed difference. Downstream advection removes some particles from further cycles. For small gain and escape probability, the competition gives a power law: surviving particles become rarer as their energy rises. Injection into this scattering population is an additional plasma-physics problem, not determined by the slope argument alone.
A power-law slope does not specify a maximum energy
For r=4, the differential relativistic energy spectrum has index two. The isotropic phase-space density instead scales as f(p) proportional to p⁻⁴ because the momentum-space volume measure contributes p² and dE/dp is approximately constant. A nonrelativistic particle energy does not equal pc, so converting that same momentum distribution gives a different energy spectrum. Finite shock strength, scattering-center drift, obliquity and nonlinear backreaction can also change the elementary result. An infinite stationary E⁻² distribution would require an unbounded logarithmic energy budget.
The local acceleration timescale is energy divided by its mean gain rate. For equal upstream and downstream diffusion coefficients D and compression four, it is 20D/u₁². If D depends on energy, the time to accelerate through an extended energy range requires integrating this local time over logarithmic energy. An acceleration history must fit within the age and confinement time and beat the relevant losses. A gyroradius smaller than the accelerator is a necessary geometric condition for magnetic confinement, but does not establish that sufficient cycles occur before the source evolves.
Cooling and escape select which particles are visible
Electrons can lose energy through synchrotron radiation and inverse Compton scattering before the shock age is reached. The simple Compton loss formula applies only in the Thomson regime; recoil suppresses scattering in the Klein–Nishina regime. Proton and nuclear losses have different target and threshold requirements. Escape also depends on the boundaries: an upstream free-escape scale and a downstream absorbing surface are different models. Thus a measured photon cutoff does not directly equal a proton maximum energy without an emission model, particle composition and transport assumptions.
The probability lab checks repeated independent loss tests only. Reset and set both controls to p=0.01,n=20: predict zero-success probability 0.99²⁰≈0.818. Reset and set p=0.02,n=20: it falls to 0.98²⁰≈0.668. Interpret success as a hypothetical loss flag. The zero-flag bar models survival through the tests; higher-count bars are not repeated escapes of one already removed particle. The instrument contains no energy gain, shock structure or feedback, and its exact bars are not sampled particle trajectories.
Work it through
Estimate the strong-shock spectral index
Use the leading-order cycle model for u₁=0.01c and u₂=0.0025c. Find compression, mean gain and escape probability, then estimate the differential-energy index and the survival fraction after one hundred hypothetical cycles.
The compression is r=0.01/0.0025=4. The mean gain is g=(4/3)(0.0075)=0.01, while Pesc=4×0.0025=0.01.
The leading-order differential-energy index is s≈1+0.01/0.01=2. The corresponding isotropic momentum phase-space index is four, not two.
Independent cycle survival is (1−Pesc)¹⁰⁰=0.99¹⁰⁰≈0.366. The model mean energy multiplier for a survivor is 1.01¹⁰⁰≈2.70; these finite-cycle powers need not reproduce an exactly continuous power law.
This test-particle example has the leading spectrum dN/dE∝E⁻², with about 36.6 percent surviving one hundred independent cycles.
The living notebook
Explore the model
Interactive lab · probability
Separate survival probability from acceleration dynamics
Reset and set BOTH controls p=0.01,n=20: predict P(K=0)≈0.818. Reset and set p=0.02,n=20: predict ≈0.668. The zero-flag bin models independent survival checks; other bins are hypothetical flags, not multiple escapes. No energy or shock evolution is simulated.
- Mean · np
- 5
- Variance · np(1−p)
- 2.5
- Total probability
- 1
These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forA survival model supplies only one ingredient of a shock-acceleration spectrum.
Pause & reason
Check your understanding
Your turn
Try it yourself
Over one hypothetical e-fold interval in relativistic energy, take constant equal diffusion coefficients D=10¹⁹ m²/s and u₁=10⁶ m/s with compression four. The available age is 10⁸ s. A separate illustrative mean diffusive escape estimate gives tₑₛc=R²/(6D) for R=10¹⁴ m. Compare the characteristic time for that full e-fold with both scales, ignoring cooling.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Compute 20D/u₁² and R²/(6D). Passing a size test alone would not establish sufficient acceleration time.
Compare with the solution
The local acceleration time is 20×10¹⁹/10¹²=2×10⁸ s. The illustrative escape time is 10²⁸/(6×10¹⁹)≈1.67×10⁸ s. The characteristic e-fold time exceeds both the available age and the illustrative mean escape time, disfavoring efficient completion of that interval. A mean escape time is not a hard lifetime shared by all particles. A maximum-energy distribution for varying D(E) requires the integrated acceleration and escape history.
Keep the idea
Take it with you
- Collisionless shocks are mediated by collective fields and scattering.
- Phase-space and differential-energy spectral slopes differ by the phase-space measure.
- Finite age, escape and losses must accompany any maximum-energy claim.
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