When to Deliver Luminosity to BaBar After Filling

This web page addresses the question of when in the course of filling it is optimal to deliver beams in collision to BaBar. The factors in this include the presumption that there has been a beam abort, as well as the fact that filling from zero current generates conditions which disallow BaBar logging due to high radiation and unacceptable beam injection rates.

The presumptions of the problem also include that the luminosity increase rate is higher under "filling" conditions than under "trickle" conditions. Call these two rates R_fill and R_trickle. Clearly if R_trickle were equal to R_fill then it would be optimal to deliver luminosity starting a zero current. Clearly if R_trickle were near zero then it would be best to fill to the top before delivering.

In reality, R_trickle is less than R_fill. From that we should be able to find an optimum time (or more practically luminosity level) at which to start trickle conditions and let BaBar log the data.

Down below on this page there are several questions for the reader which help further illuminate this matter. Readers are encouraged to try to answer these questions themselves. Answers (solutions) are provided.

Jump ahead to the answer:

For the reader not interested in the details of the formulation at this time, I'll jump ahead to the answer: The optimal luminosity level at which to deliver beams to BaBar and start logging under trickle conditions is determined below as Eq VII:

L_Deliver = L_Peak(1 - R_trickle/R_fill).

This result is straightforward and not surprising:

• If R_trickle were the same as R_fill then L_Deliver evaluates to 0, which is as expected.
• If R_trickle is very small or zero then L_Deliver would be near or at L_Peak, also as expected.

The only challenge left is evaluating R_fill and R_trickle. These terms can easily be measured based on previous injection cycles (there are plenty of fills to evaluate). In the author's experience evaluating these rates, the ratio R_fill/R_trickle usually ends up being within the range of 3:1 and 5:1, leading to an optimal luminosity level at which to deliver within the range of 66% and 80% of L_Peak.

Now let's do the Math:

The approach I will take is to use a mockup set of data to parameterize the course of an abort and subsequent cycle of re-filling. This is illustrated in the picture below. The technique will be to come up with a general expression for the luminosity lost as a result of this abort/refill sequence, then determine a value for L_Deliver for which this lost luminosity is minimized.

In this example: The beam aborts at time t=0 as indicated on the picture. At t=1 injection is begun under "filling" conditions. At t=2 a luminosity level of LDelivered is achieved, and data logging begins. Injection continues under trickle conditions. At t=3 the nominal luminosity level LPeak is again achieved, and abort recovery is fully completed. The purple line mocks the "deadtime" signal from BaBar; a useful indicator that logging is occurring. The green line is the running sum (with arbitrary offset for display purpose) of the luminosity while under delivered conditions. This approach will minimize the reduction in integrated luminosity (green line) caused by the abort and subsequent refill cycle

The luminosity loss all occurs during three intervals and can be expressed as:

(Eq I):Luminosity Lost = ∫^t=1_t=0 L_Peakdt + ∫^t=2_t=1 L_Peakdt + ∫^t=3_t=2 L_Peakdt - ½(L_Peak+L_Deliver) (t₃-t₂)

Note that in the last interval (between t=2 and t=3) this loss is reduced by the an amount equivalent to the average of L_Peak and L_Deliver for this entire interval.

Evaluating each of these terms yields:

(Eq II):Luminosity Lost = L_Peak(t₁-t₀) + L_Peak(t₂-t₁) + L_Peak(t₃-t₂) - ½(L_Peak+L_Deliver)(t₃- t₂)

Next we'll want to look at the expressions for the luminosity increase rates:

• R_trickle = (L_Peak-L_Deliver)/ (t₃-t₂)

  and

• R_fill = (L_Deliver)/ (t₂-t₁)

Such that:

• (t₃-t₂)=(L_Peak-L_Deliver)/R_trickle

  and

• (t₂-t₁)=L_Deliver/R_fill

We can now substitute these expressions for (t₃-t₂) and (t₂-t₁) into (Eq II) above to yield:

(Eq III):Luminosity Lost = L_Peak(t₁-t₀) + L_PeakL_Deliver/R_fill + L_Peak(L_Peak-L_Deliver)/R_trickle - (L_Peak+L_Deliver) (L_Peak-L_Deliver)/2R_trickle

Expanding terms yields:

(Eq IV):Luminosity Lost = L_Peak(t₁-t₀) + L_PeakL_Deliver/R_fill + L²_Peak/R_trickle - L_PeakL_Deliver/R_trickle - (L²_Peak - L²_Deliver) /2R_trickle

And expanding further yields:

(Eq V): Luminosity Lost = L_Peak(t₁-t₀) + L_PeakL_Deliver/R_fill + L²_Peak/R_trickle - L_PeakL_Deliver/R_trickle - L²_Peak/2R_trickle + L²_Deliver/2R_trickle

Equation V provides the luminosity lost due to the abort/recovery in terms of L_Peak (known), R_Trickle and R_fill (can easily be determined from previous performance), t₀ and t₁ which specify the time interval needed to recover the abort system inputs, and L_Deliver (as yet unknown). However differentiating this expression for Luminosity Lost with respect to L_Deliver and setting that derivative to zero provides the value for L_Deliver where this Luminosity Lost expression is minimized.

(Eq VI): dLuminosityLost/dL_Deliver = L_Peak/R_fill - L_Peak/R_trickle + L_Deliver/R_trickle = 0

The expression in Eq VI simplifies to:

(Eq VII): L_Deliver = L_Peak(1 - R_trickle/R_fill)

Questions for the reader:

Question #1: The wrong course (easy).

Given that a machine operating at Luminosity 8334x10³⁰/cm²s aborts. After 4 minutes injection begins under "filling" conditions where R_f is 695x10³⁰/cm²s per minute (corresponding to ~12 minutes to fill all the way back). Given also that a trickle condition exists in which R_trickle is 174x10³⁰/cm²s per minute for a fill to trickle ratio of 4:1.

Consider the two following courses for filling:

• The first course fills as above, allowing data logging to begin at L_Deliver = L_Peak(1-R_trickle/R_fill) = 6251x10³³/cm²s then trickles back up to L_Peak.
• The second course fills under filling conditions all the way to L=L_Peak and only then starts trickling and data logging.

Can you show that the total (integrated) delivered luminosity lost in the second course is 10% greater than the total (integrated) delivered luminosity lost in the first course?

Solution is here.

Question #2: The Coastdown Recovery Problem (hard).

Given a machine with stable running at L=L_Peak loses the ability to continue trickle injection. Luminosity decays at some rate corresponding to some luminosity lifetime. If decay happens for only a very short time, trickle recovery would be appropriate. If the decay happens for a very long time, filling at high rate (under filling rather than trickle conditions) without data logging would be appropriate (at least initially...see question #4 below).

Consider the picture to the right: At time t=0 ability to inject is lost. At time t=1 Luminosity has decayed to a level L=Lmin but injection is again available. Injection proceeds at time t=t1 with two courses shown. In the course shown in green, trickle injection is used, and data logging continues. The trickle luminosity increase rate is Rtrickle. At time t=t3 the peak luminosity is again achieved. The green dashed line shows the running sum of integrated luminosity while under deliverable conditions for the green course. In the course shown in yellow, data collection is abandoned and beams are injected under (faster) filling conditions. A luminosity increase rate Rfill occurs. At time t=t2 the peak is again achieved and logging begins. The yellow dashed line shows the running sum of integrated luminosity while under deliverable conditions for the yellow course. The specific conditions shown in the picture would indicate that less integrated luminosity is lost by choosing the trickle (green) course.

Can you show that if the luminosity decays below the level L_min=L_peak(1-2R_trickle/R_fill) then logging should be abandoned and refilled with the higher rate R_fill?

Solution is here.

Question #3: Coastdown Recovery With Trickle (easy).

Given a machine with stable running at L=L_Peak loses the ability to continue trickle injection. Luminosity decays at some rate corresponding to some luminosity lifetime. Presume that this coastdown is of sufficiently long duration that exclusively trickle recovery is not worthwhile (as in question #2 above).

Consider the picture to the right: At time t=0 ability to inject is lost. At time t=1 luminosity has decayed to a level L=Lmin but injection is again available. Data logging is abandoned and injection proceeds under filling conditions. At time t=2 a luminosity LDeliver is achieved. Injection is switched to trickle and data logging begins. At time t=3 the luminosity has again achieved the value Lpeak and the interruption is over. The dashed yellow line shows the integrated luminosity from this course.

Can you show that in this case the optimum luminosity level at which to switch from filling injection to trickle injection is again L_Deliver=L_peak(1-R_trickle/R_fill)?

Solution is here.

Question #4: Coastdown Recovery with fill then trickle (harder).

Consider the problem of Question #2 above. In that question there was a coastdown and the issue of whether to recover with trickle or with filling was addressed. However in question #2 is was presumed that the filling approach would be used for the entire recovery. This question asks what if filling and then trickle are used.

Consider the picture to the right: At time t=0 ability to inject is lost. At time t=1 luminosity has decayed to a level L=Lmin but injection is again available. Injection proceeds at time t=1 with two courses shown. In the course shown in green, trickle injection is used, and data logging continues. The trickle luminosity increase rate is Rtrickle. At time t=5 the peak luminosity is again achieved. The green dashed line shows the running sum of integrated luminosity while under deliverable conditions for the green course. In the course shown in yellow, data collection is initially abandoned and beams are injected under (faster) filling conditions. A luminosity increase rate Rfill occurs. At time t=3 the luminosity reaches the level Lpeak(1-Rtrickle/Rfill) and trickle conditions with data logging continue (as question 3 above would indicate is appropriate). At t=4 LPeak is again achieved and the recovery from the interruption is complete. The yellow dashed line shows the running sum of integrated luminosity while under deliverable conditions for the yellow course.

Can you show that if ever L_min decays below the level L_peak(1-R_trickle/R_fill) that less luminosity is lost if you recover with the yellow course rather than the green course? This result is different than the result from question#2 which was L_min =L_peak(1-2R_trickle/R_fill).

Solution is here.