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 to Question #1

For both courses, the solution comes from substituting in Eq V the appropriate value for L_Delivered.

In the first course L_Delivered is the optimal value L_Peak(1-R_trickle/R_fill). The reader should confirm that this substitution yields an expression for the luminosity lost as:

Luminosity Lost (first course) = L_peak(t₁-t₀) + L²_peak(1-R_trickle/2R_fill) /R_fill

In the second course, substituting into Eq V a value of L_Deliver the value L_Peak yields (again with some algebra which you can do):

Luminosity Lost (second course) = L_peak(t₁-t₀) + L²_peak /R_fill

I've tailored the numbers in the problem to be consistent with typical operation while making the arithmetic easy for you. When you punch in all the numbers, the luminosity lost in the second course is 10% greater than the luminosity lost in the first course. (The first course lose 7.25 pb^-1 while the second course loses 8.00 pb^-1).