Recent content by Jdraper

Hi, eveyone have been struggling to do this problem for a long time now, figured it is something very simple I am missing so thought I should ask here. 1. Homework Statement Parker's solar wind equation is given after some manipulation as: (v - (Cs2 / v) dv/dr = 2 (Cs2 / r2) (r - rc ) where...

Thank you, think I've followed your method by eliminating ##E_4 and P_4 ## but I've become bogged down in the algebra. Will have another attempt and let you know

Here is what my lecturer said when I emailed him a few days ago regarding this question: For q1b in 2014 you first need to write down explicitly the expressions for s, t, and u in terms of m, T, T’ and theta. Then there are several options. One way is to take the t variable and, using s + t + u...

Currently trying to do it your way by conservation of momentum: ## (P_1 + P_2)^2 = (P_3 + P_4)^2 ## If you want to read further into them, s,u and t are called the mandelstam variables. Yes, it's very difficult to cancel.The algebra becomes very messy.

So, just to check, the first four momentum vector would be: ##P_1 = (T+M, 0,0, (T^2 + 2TM)^{0.5})##

You've opened up a can of worms in my understanding here. My professor uses this convention and I have never really questioned it. So you are saying: ##P=(E^2 -M^2)^{0.5}## therefore as ##E=T+M## we can rewrite this as: ##P=((T+M)^2 + M^2)^{0.5} = (T^2 +2TM)^{0.5}##

Hey, forgot to mention that I'm using the convention that c=1.

Homework Statement In a fixed target experiment a particle of mass M and kinetic energy T strikes a stationary particle of mass M. By evaluating s, t and u in the laboratory frame and using the above relation, or otherwise, show that the kinetic energy T' of the particle scattered elastically...

Finally shown it, didn't realize you were working from the RHS to the LHS. I was trying to do the opposite hence the confusion. Thanks anyway.

I understand you are doing your best to help but it's very unclear to me what the source of my error is. At the moment i have an answer in which one term is a factor of -½ out and another which is a factor of ¼ out. I see nothing wrong with my algebraic working so could you please tell me...

I know I'm probably being really dense but i dob't see it. It can't be from the original statement as that's fine so there must be something wrong with the =-G/4π d/dr (m2/r4)

ok starting from the beginning dP/dr =- Gm/4πr4 dm/dr =-G/4π d/dr (m2/r4) = -G/4π (m2 d/dr (r-4) + 1/r4 d/dr (m2)) using product rule then we get =-G/4π(-4m2/r5 + 2m/r4 dm/dr ) so this give us =Gm2/πr5 -d/dr (Gm2/2πr4) Sorry but i don't understand where my missing factors have gone?

Oh sorry, Forgot we manipulated the differential in the RHS rather than differentiating both sides. So we have dP/dr =-Gm2 / πr5 +Gm/2πr4 dm/dr Manipulating the differentials again I get dP/dr = Gm/πr5 - d/dr (Gm2/2πr4) It seems that some factors are missing, i'll recheck my workings

Yeah, i agree. Most have my modules have strayed away from differential calculus is recent years so a review is needed. So that gives me: d/dr (dP/dr) =-Gm2 / πr5 +Gm/2πr4 dm/dr

Ahh ok, that didn't seem apparent at first, so now i get: =-4m2/r5 + 1/r4 d/dr(m2) using chain rule =-4m2/r5 + 1/r4 (2m(r)*1) =-4m2/r5 + 2m/r4 Is this correct? Thanks, John.