Why Represent an Eclipse Stage?
A newly infected cell need not produce virus immediately. Viral entry and intracellular replication may precede the release of infectious particles. An acute viral model that sends every newly infected cell straight into a virus-producing compartment omits this delay. An eclipse-stage model represents it explicitly, separating cells that are infected but not yet producing virus from cells that are productively infected.
This distinction affects the timing of predicted virus production and the rate at which infection grows. It does not, by itself, change the expected number of secondary infections if every eclipse-stage cell eventually becomes productive and all other model assumptions remain fixed. The equations below make these claims precise.
Model Equations and Assumptions
A basic eclipse-phase extension of a target-cell-limited acute infection model is
Here, \(\beta\) is the infection-rate constant, \(k\) is the eclipse-to-productive transition rate, \(\delta\) is the loss rate of productively infected cells, \(p\) is the virus production rate per productive cell, and \(c\) is the free-virus clearance rate. Concentrations \(T,E,I\) are measured in cells per milliliter, and \(V\) in virions per milliliter. With time in days, \(\beta\) has units ml/(virion·day), \(k,\delta,c\) have units day\(^{-1}\), and \(p\) has units virions/(cell·day).
The model assumes mass-action infection: the rate at which new eclipse-stage cells appear is \(\beta T V\). It assumes no target-cell replenishment or natural target-cell loss over the modeled interval, no virus production by eclipse-stage cells, and constant rates for progression, infected-cell loss, production, and clearance. Each eclipse-stage cell progresses eventually, since there is no loss term from \(E\). These are modeling assumptions, not consequences of the compartment structure.
The term \(-kE\) describes a constant per-cell transition hazard. For a cell entering \(E\), the probability it remains in that stage longer than time \(t\) is \(e^{-kt}\). Thus its mean time in the eclipse stage is \(1/k\), but individual cells do not all wait exactly that long.
Worked Example: Translating a Mean Eclipse Duration Into a Rate
Suppose a hypothetical model uses a mean eclipse duration of 12 hours. Convert this duration to a rate \(k\), with time measured in days, and calculate the fraction of eclipse-stage cells expected to have progressed by 12 hours.
Twelve hours is \(0.5\) day. For an exponential waiting time, the mean is \(1/k\), so
The fraction progressed by time \(t\) is \(1-e^{-kt}\). At \(t=0.5\) day,
So about 63.2% of cells entering the eclipse stage are expected to have progressed by 12 hours under this model. The mean duration is not a fixed delay: about 36.8% remain in the stage beyond 12 hours.
Reproductive Number With an Eclipse Stage
Consider infection when target cells are approximately constant at their initial concentration \(T_0\). One newly infected target cell first enters \(E\), then becomes a productive cell. With no eclipse-stage loss, it eventually produces virus as an \(I\) cell. A productive cell survives at rate \(\delta\), so its expected lifetime is \(1/\delta\), and its expected total virus production is \(p/\delta\). A free virion is cleared at rate \(c\); in the target-cell environment \(T_0\), it generates new infected cells at rate \(\beta T_0\) while present, for an expected \(\beta T_0/c\) infections.
Multiplying the expected infections per virion by the expected virions per productive cell gives the within-host basic reproductive number:
The eclipse duration does not appear in this expression because it changes when production begins, not the expected total number of productive cells or virions produced under these assumptions. This is a statement about the expected number of secondary infections, not about how quickly they occur.
Worked Example: Checking the Reproductive Number and Units
For a hypothetical infection, let \(T_0=1.0\times10^6\) cells/ml, \(\beta=2.0\times10^{-6}\) ml/(virion·day), \(p=10\) virions/(cell·day), \(\delta=1.0\) day\(^{-1}\), and \(c=10\) day\(^{-1}\). Calculate \(R_0\), and check whether the eclipse transition rate enters the result.
First,
Then
For the units, \(\beta T_0p\) has units \(\mathrm{day}^{-2}\): cells/(virion·day) multiplied by virions/(cell·day). The denominator \(c\delta\) also has units \(\mathrm{day}^{-2}\), so \(R_0\) is dimensionless. Its value is 2.0, indicating that each infected cell gives rise to two new infections on average in the initial, susceptible environment under the model assumptions. Changing \(k\) alone changes the timing of infection, not this value.
How the Eclipse Stage Changes Early Growth
The same \(R_0\) can correspond to different early growth rates because the timing of transmission differs. To see this, approximate \(T(t)\) by \(T_0\) while infection is still small. Seek exponential solutions for the infected compartments, with each perturbation proportional to \(e^{\lambda t}\). Substitution into the \(E,I,V\) equations gives
Multiplying the three relations and canceling the nonzero perturbation amplitudes yields the eclipse-model characteristic equation:
In the immediate-production model, newly infected cells enter \(I\) directly, so the early-growth equation is
At \(\lambda=0\), the eclipse-model equation compares \(k\delta c\) with \(\beta T_0kp\). Dividing by \(k\delta c\) gives \(R_0=1\), the same threshold as in the immediate-production model. When \(R_0>1\), the positive growth rate in the eclipse model is lower than in the immediate-production model with the same other parameters. In fact, at the immediate model's positive root, the eclipse equation's left side has the extra factor \(\lambda+k\), while its right side has the factor \(k\); since \(\lambda>0\), the left side is larger there. The eclipse equation's left side increases with \(\lambda\geq0\), so its positive root must be smaller.
Worked Example: Comparing Predicted Early Growth Rates
Use the parameter values from the reproductive-number example and let \(k=2\) day\(^{-1}\). Compare early exponential growth rates from the eclipse and immediate-production models.
For the immediate-production model, \(\beta T_0p=(2.0)(10)=20\ \mathrm{day}^{-2}\). Thus
The positive root is
For the eclipse model, \(\beta T_0kp=(2.0)(2)(10)=40\ \mathrm{day}^{-3}\), so
Expanding gives \(\lambda^3+13\lambda^2+32\lambda-20=0\). Its positive root is approximately \(0.5136\ \mathrm{day}^{-1}\). The corresponding doubling times, calculated as \(\ln(2)/\lambda\), are about \(0.821\) day for immediate production and \(1.350\) days with the eclipse stage. Both models have \(R_0=2.0\); the eclipse model predicts slower early growth because production is deferred.
Following One Infected Cohort
The timing difference is also visible without considering subsequent infections. Suppose an initial concentration \(E_0\) of cells enters the eclipse stage at time zero, with no additional infections, and \(I(0)=0\). Then \(E(t)=E_0e^{-kt}\), and solving the productive-cell equation gives
Virus production is \(pI(t)\), so it is zero at the instant of infection in this setup and rises only after cells progress to \(I\). If \(k=\delta\), the limiting solution is \(I(t)=kE_0t e^{-kt}\). In contrast, an immediate-production model starting with the same infected-cell concentration in its productive compartment has production \(pE_0\) at time zero. These trajectories describe different assumptions about the infected-cell state, not simply different parameter values.
Worked Example: Production After a Single Infection Pulse
Suppose a hypothetical pulse places \(E_0=100\) cells/ml into the eclipse stage, with \(k=2\) day\(^{-1}\), \(\delta=1\) day\(^{-1}\), and no new infections. Find \(I(t)\) and its peak time.
Substituting the values into the cohort solution gives
Differentiate to locate the peak:
At that time, \(e^{-t}=1/2\) and \(e^{-2t}=1/4\), giving \(I(t)=200(1/2-1/4)=50\) cells/ml. With \(p=10\) virions/(cell·day), the instantaneous virus production rate at the peak is \(pI=500\) virions/(ml·day). The pulse has not produced virus immediately; its productive-cell concentration first rises and later falls as progression and cell loss act together.
Adding Loss During the Eclipse Stage
Some infected cells may be lost before becoming productive. If eclipse-stage cells are lost at rate \(\mu\), replace the second and third equations by
An eclipse-stage cell now has competing progression and loss hazards. The probability that progression occurs before loss is \(k/(k+\mu)\). The resulting reproductive number is
Worked Example: Effect of Eclipse-Stage Loss
Use \(\beta T_0=2.0\) cells/(virion·day), \(p=10\) virions/(cell·day), \(c=10\) day\(^{-1}\), and \(\delta=1\) day\(^{-1}\). Suppose \(k=2\) day\(^{-1}\) and eclipse-stage loss is \(\mu=1\) day\(^{-1}\). Calculate the probability of progression and the adjusted \(R_0\).
The probability of progressing before loss is
Without eclipse-stage loss, \(R_0=(2.0)(10)/[(10)(1)]=2.0\). Accounting for progression probability gives
The units cancel as before, and the adjusted reproductive number remains dimensionless. Under these hypothetical parameters, the infection still has \(R_0>1\), but fewer infected target cells become productive than in the no-loss model.
Model Checks, Alternatives, and Limitations
A useful check is whether the model's state definitions match the measurements being represented. If \(E\) is described as nonproducing, the equations should not include \(pE\) in the virus-production term. If the observed data require substantial virus production immediately after infection, a model with all cells entering \(E\) may be inappropriate unless the initial conditions include an already productive population. Units provide another check: every term in a compartment equation must have units of that compartment per day.
The immediate-production model is simpler and may be adequate when the delay is negligible relative to the time scale of interest or when the available data cannot distinguish the stages. The eclipse model can represent a gradual distribution of progression times using one additional compartment. A fixed-delay model is another possible representation, but it assumes a different waiting-time structure; a single exponential stage does not imply that every cell waits for exactly \(1/k\) days.
The eclipse model also has limits. Its exponential waiting-time assumption permits some cells to progress very quickly and others to remain in the stage for a long time. Constant \(k\) does not represent changing intracellular conditions. A model with one eclipse compartment may be too coarse if infection proceeds through several biologically distinct steps. Finally, estimating additional rates requires data informative about those processes; fitting more compartments does not guarantee that each parameter is identifiable from the observations.
A common mistake is to conclude that adding an eclipse stage must lower \(R_0\). With no eclipse-stage loss, it changes the timing of transmission and the early growth rate, but not the expected total number of secondary infections in this model. Another mistake is to interpret \(1/k\) as a fixed delay. It is the mean of an exponential waiting-time distribution, not a time at which every cell transitions.
Check Your Understanding
Try each question, then compare your reasoning with the solutions.
- If \(k=4\) day\(^{-1}\), what is the mean eclipse duration? What fraction of cells are predicted to have progressed by that duration?
- For the no-loss eclipse model, explain why \(k\) changes the early growth equation but not \(R_0\).
- With \(k=1\) day\(^{-1}\), \(\delta=1\) day\(^{-1}\), and an initial eclipse concentration \(E_0\), state the productive-cell trajectory when there are no new infections.
- If eclipse-stage cells can be lost at rate \(\mu=2k\), what fraction progress to the productive stage?
Solutions. (1) The mean is \(1/k=0.25\) day. By that time the progressed fraction is \(1-e^{-k(1/k)}=1-e^{-1}\approx0.6321\). (2) The growth equation contains \((\lambda+k)\), so progression timing affects the positive growth rate. With no eclipse-stage loss, every eclipse-stage cell progresses eventually, so the expected total infections per initial infected cell remain \(\beta T_0p/(c\delta)\). (3) When \(k=\delta\), the equal-rate solution is \(I(t)=kE_0t e^{-kt}=E_0t e^{-t}\); it starts at zero and is positive after infection. (4) The progression probability is \(k/(k+\mu)=k/(3k)=1/3\).