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Hi Micha³, >Sorry for not writing for a few days, but I had a very bad time with my>studies/personal life ;)There's no urgence. We each handle time usage differently (and each have different personal lives)... :-) >Included picture: w(a) vs. a for a "inverse power-law" potential. You can>see that linear approximation is 'quite' good here, but for other>potentials is a bit worse and the quadratic term should be included.>I am curious if one can put some constrains from available or wait for>future projects/missions.OK, I think this clears things up. >> Do you have in mind a quintessence model where these derivatives go>> to zero very fast? Or have I misunderstood?>>>What I am trying to do is only to approximate w(a) with respect to a=1.>It is more suitable for analysing data (such as SNIa or quasars I>hope) than including the full evolution of x-field (which behaves very>'badly' much earlier than z=2).OK, it seems to me you have chosen one particular potential, a "power-law potential", for which the derivatives go to zero very fast. If they didn't go to zero fast, your plot would not be close to linear for redshifts of interest. So, to put this in language which everyone on the list should understand (hopefully!), you've plotted how one particular theoretical model of quintessence (a substitute for the cosmological constant, which has some theoretical physics motivation), is close to linear in the range 0.1 < a < 1, i.e. 9 > z > 0. I agree that this covers the entire range of interest to observers (apart from the CMB)! >Not that much. You can always expand a function of one variable around>certain point. Expanding w(a) has the advantage that constraining only to>the first term you get 'static' X-field, for which w is fixed for all a.... Just an aside on the mathematics, which was where I was confused (before). Suppose we take the Taylor series for: * the exponential, around x=0, and evaluate at the point x=-0.7. This is similar to calculating a Taylor series for: * w(a) around a=1, and evaluating it at quasar redshifts, with z=2, i.e. a= 1/(1+z) = 0.333 Well, exp(x) = 1 + x + x^2/2! + x^3/3! + .... So exp(-0.7) = 1 + (-0.7) + ... = 0.3 if we just take the linear term And this is 40% off from the true value 0.49658... So the linear approximation can be a poor approximation for an exponential when you're too far from the expansion point. >Look at the picture I have attached. For interesting redshifts linear>approximations looks really very fine.So there must be something in the physics of the power-law potential you're talking about which makes a linear approximation valid over sub-CMB redshift ranges. >I hope I have answered your questions (at least partly) and convinced you>that it if worth to expand w(a) near a=1. ;)For this potential, you have convinced me :-). You've also shown that the sympa software (or maybe mhonarc, which is called by sympa) has turned your postscript attachment into an html link - nice! Here's a check to see what precision would be required: * visual - make a plot of w(z) in the range 3 < z < 0 * calculate mean (z-weighted) values of w(z) in the ranges: 0.6 < z < 1.1 1.1 < z < 1.6 1.6 < z < 2.2 for the potential you chose. From this, we will know the precision required in the individual w(z) estimates (as I propose to do) in order to obtain a significant value of w_1 . (Of course, simply to get a significant estimate of w_0 would be a big result!) >P.S. In the picture I have chosen the evolution that fits best to linear>approximation of w(a)...Does this mean that other potentials are very non-linear? Cze¶æ Boud

- Poprzednia wiadomość (według wątku): Boud: quintessence
- Następna wiadomość (według wątku): introductions
**Wiadomości posortowane według:**[ daty ] [ wątku ] [ tytułu ] [ autora ]

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