research30 min studyLesson 677 in Physics

Branch cuts and threshold discontinuities

Locate the scalar two-particle threshold from parameter space and compute its imaginary part with a fixed causal branch.

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What you’ll understand

  • Find the parameter interval where the bubble logarithm crosses its branch cut.
  • Compute the imaginary part and discontinuity with the correct sign.
  • Distinguish a physical threshold from ultraviolet subtraction.

A loop amplitude can be analytic for spacelike momenta and acquire an imaginary part after continuation into a physical timelike region. The causal prescription determines which side of a branch cut is approached. In the equal-mass scalar bubble, the entire discontinuity follows from the length of one parameter interval where a logarithm’s argument becomes negative. This makes threshold behavior hand computable and reveals why taking an absolute value inside a logarithm destroys physical information rather than merely choosing a convenient numerical representation.

The parameter denominator locates the normal threshold

Use the finite channel function already defined, with real m>0m>0 and Minkowski invariant s. Its mass combination is m2−sx(1−x)m^2-sx(1-x). For real s below 4m24m^2, this expression stays nonnegative over the unit interval and is positive away from the threshold endpoint case. The first zero occurs at the maximum parameter product, x equal to one half. At larger s, two real roots enclose an interval in which the mass combination is negative.

Introduce the dimensionless two-body velocity factor beta as the square root of one minus four mass squared over s. The two roots are one half times one plus or minus beta, so their separation is beta. This calculation identifies a normal two-particle threshold for this particular bubble. More complicated graphs can have additional singularities and sheets; solving one quadratic does not establish a complete singularity classification. Positive mass also keeps this example separate from a massless soft or collinear infrared limit.

β(s)=1−4m2s,x±=1±β(s)2(s>4m2)\beta(s)=\sqrt{1-\frac{4m^2}s},\qquad x_\pm=\frac{1\pm\beta(s)}2\qquad(s>4m^2)
s and m² have energy-squared units; beta and x are dimensionless. The square root is the positive real root above this equal-mass physical threshold.↔ Scroll wide equations horizontally.

Keep the negative infinitesimal in the logarithm

Use the principal logarithm with its cut on the negative real axis. For a positive number a, the value just below that axis is ln⁡(−a−i0)=ln⁡a−iπ\ln(-a-i0)=\ln a-i\pi. The channel function contains minus the logarithm, so each point in its negative-argument interval contributes positive pi to its imaginary part. Integrating over an interval of length beta gives positive pi times beta. Both the minus sign in the function definition and the sign of the infinitesimal are needed.

Define discontinuity as the upper-side value minus the lower-side value as the external invariant crosses its physical cut. With real parameters and Schwarz reflection on this cut, it equals twice i times the imaginary part on the upper physical side. The sign would reverse if discontinuity were defined in the opposite order. Report the convention alongside the answer rather than comparing bare imaginary numbers across sources with different amplitude factors or logarithm definitions.

F(s)=−∫01dx ln⁡m2−sx(1−x)−i0μ2,Im⁡F(s+i0)=πβ(s)F(s)=-\int_0^1dx\,\ln\frac{m^2-sx(1-x)-i0}{\mu^2},\qquad \operatorname{Im}F(s+i0)=\pi\beta(s)
F is dimensionless and mu is a positive energy scale. The formula for its imaginary part applies above threshold; below the normal threshold it vanishes for this massive bubble.↔ Scroll wide equations horizontally.

Subtraction changes real local terms, not this cut

A real local ultraviolet counterterm has no two-particle discontinuity in this channel. It can shift a real subtraction constant or polynomial without altering the bubble’s absorptive part. That separation provides a useful consistency check when changing renormalization schemes. The full scattering amplitude still multiplies the channel function by its diagram coefficient and adds other channels. An imaginary part of F is not itself a total cross section or a decay probability.

Near threshold, beta tends to zero like the square root of the excess invariant energy squared. Its vanishing reflects the shrinking available two-body phase space. Far above threshold it tends to one in this massive example, but other logarithms and channels remain in the full amplitude. The root-finding lab can verify a selected square root after its dimensionless target is supplied by hand. It does not continue a logarithm, implement the Feynman prescription, or determine a threshold from a particle model.

Disc⁡F(s)=F(s+i0)−F(s−i0)=2πi β(s),s>4m2\operatorname{Disc}F(s)=F(s+i0)-F(s-i0)=2\pi i\,\beta(s),\qquad s>4m^2
This discontinuity convention uses Schwarz reflection and the physical continuation specified above. It is dimensionless; multiplying by the appropriate amplitude coefficient restores the desired diagram contribution.↔ Scroll wide equations horizontally.

A threshold with rational parameter roots

Let s=16m2/3s=16m^2/3. Find beta, the two parameter roots, the imaginary part of F, and its discontinuity.

  1. Compute 4m2/s=3/44m^2/s=3/4, so β=1−3/4=1/2\beta=\sqrt{1-3/4}=1/2. The positive square root is appropriate above the physical threshold.

  2. The roots are x−=1/4x_-=1/4 and x+=3/4x_+=3/4. The logarithm has a negative argument between them, giving an interval of length one half.

  3. The minus-logarithm definition then gives Im⁡F=π/2\operatorname{Im}F=\pi/2. With upper-minus-lower discontinuity, Disc⁡F=2i(π/2)=iπ\operatorname{Disc}F=2i(\pi/2)=i\pi.

The result

The negative-argument interval is [1/4, 3/4], the imaginary part is π/2, and the discontinuity is iπ.

Explore the model

Interactive lab

Interactive lab · root-finding

Check a supplied velocity square root

For the worked value, beta squared is one quarter. Reset and set target c=0.25c=0.25, upper bound U=2U=2. Predict the positive root 0.5; after two Steps the bisection reaches it exactly and its bracket error bound is zero. For another run Reset and set both controls again. This scalar quadratic solver does not locate branch cuts or supply particle masses and causal prescriptions.

Check a supplied velocity square rootBisection of x² − c on [0,U], capped at 50 iterations for floating-point resolution. The error bound is the maximum distance from the displayed midpoint to either bracket endpoint (half the width in exact arithmetic). Compensated square evaluation preserves the bracket when ordinary multiplication rounds a residual to zero. A truly exact endpoint or midpoint collapses the bracket. This is one monotone quadratic, not a general root solver. Current readings: Midpoint after 0 refinements: 1.5; Bracket error bound: 1.5; Computed residual · x² − c: 0.25. All parameters are adjustable with the labeled controls below.Bisection on x squared minus 2: bracket from 0 to 3 after 0 refinements.00.751.52.253−2.360.192.745.297.84x · bracket stays nonnegativef(x) = x² − cn = 0 · [ 0, 3 ]
Quadratic / midpointCurrent bracketPositive root
Midpoint after 0 refinements
1.5
Bracket error bound
1.5
Computed residual · x² − c
0.25
t = 0 s

Bisection of x² − c on [0,U], capped at 50 iterations for floating-point resolution. The error bound is the maximum distance from the displayed midpoint to either bracket endpoint (half the width in exact arithmetic). Compensated square evaluation preserves the bracket when ordinary multiplication rounds a residual to zero. A truly exact endpoint or midpoint collapses the bracket. This is one monotone quadratic, not a general root solver. 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 real square-root check verifies one algebraic ingredient; analytic continuation determines its physical meaning.

Check your understanding

For F defined as minus the integral of ln(Delta−i0), what is its imaginary part above the equal-mass threshold?

Try it yourself

Set s=25m2/4s=25m^2/4. Find the negative-argument interval, Im F, and Disc F in the same conventions. State why the subtraction scale does not affect these answers.

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 beta first; only the interval length contributes to the imaginary part of the logarithm.

Compare with the solution

Here 4m2/s=16/254m^2/s=16/25, so beta is 3/53/5. The roots are 1/51/5 and 4/54/5, enclosing length 3/53/5. Therefore Im⁡F=3π/5\operatorname{Im}F=3\pi/5 and Disc⁡F=6πi/5\operatorname{Disc}F=6\pi i/5. Changing the positive subtraction scale adds a real constant to the logarithm; it does not change which arguments are negative or the causal phase on that interval.

Take it with you

  • The equal-mass bubble threshold follows from zeros of its parameter mass combination.
  • The causal infinitesimal and the minus-logarithm definition fix the absorptive sign.
  • Real local ultraviolet subtractions do not change the normal-threshold discontinuity.

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