JohnFromAus

Thanks D H that clears up my problems - for now anyway! John Just for your info it was when you pointed out that r and dr/dt could be treated as separate variables that the penny dropped! John

DH Im still confused. Surley the [math] cos\phi \ and \ the \ -rsin\phi [/math] in the expression for dx/dt are just the partial derivatives of [math]x=cos\phi[/math] However can now see where answer comes from - must go back to my book and check up on all this - particularly the use of "[math]\delta[/math]" and "d" which seem to be used pretty much as the same thing in some web sites. Thanks John

I am stuck with the following which comes from an attempt to find acceleration in polar coordinates. The velocity in the x direction is [math] V = cos\phi dr/dt - sin\phi (r d\phi/dt) [/math] and I need to find dV/dt so I think I need [math]\delta V/\delta r \ and\ \delta V/\delta \phi [/math] Then I have terms like [math] \delta (dr / dt) \delta r[/math] which I dont know how to handle. I have a book which gives the same formula for V and then the answer on the next line - I cannot "join the dots"! Have spent a bit of time looking on the internet but only found a different answer to the one in my book - could not get that either. Any help appreciated - either hints, references - or the missing steps! John PS its some 40+ years since I was really on top of my calculus so I could be completely screwed up! OK it should be PARTIALLY

Ummm - I think I may have confused the issue some what. I have attached a copy of the problem from the book. The answers given are as follows (b) [math]v = \alpha t /\sqrt{1+ \alpha^2t^2}[/math] [math] \alpha = 1.1\ast10^{-16} m^{-1}[/math] [math] t = 2.0\ast10^{17} m = 22 years[/math] all of which I got and so assumed I was on the right track © [math] v = tanh(\alpha\tau) [/math] ???? [math] x = \alpha^{-1}[(\sqrt{1+\alpha^2t^2)} - 1][/math] 10 years. The first © answer does not seem to tie up with the question but I agreed with the formula for x. I cannot find from where the formula for v came or get the final answer of 10 years. I hope this puts my questions in a better context and allows you to get some idea of just how much of a beginner I am. The question interested me as it seemed to show that a body can have constant acceleration for ever and yet never exceed c. John

Well I am really grateful for all the help - perhaps the worlds not such a bad place after all! Snail - thanks for the list - I will certainly investigate the suggestions Atheist - am trying to follow your reply. I got the same result but possibly my reasoning is either wrong or at least not rigorous. I did the following:- [math] \delta t = \gamma\delta\tau \Rightarrow \delta\tau = \delta t / \gamma \Rightarrow = \sqrt{1- \vec v^2} \, dt[/math] taking c as 1. I already had [math] v = \alpha t /\sqrt{1+ \alpha^2t^2} [/math] which I then substituted in the above and integrated giving me [math]\tau = tan^{-1} \alpha t [/math] which is not the given answer. Its is years since I did any serious integration so I could have got that wrong - I am much more concerned with being sure my method is Ok than getting the "right" answer! John Looked at a table of integrals - now have [math] \tau = \alpha^{-1}sinh^{-1}(\alpha t)[/math]

Kevinalm Thanks - looks like some helpful stuff there - especially problems with solutions John

Thanks Obelix - I just needed to go back to the definition of proper time and work from there! Will do that integration and see what comes of it. But I am still confused by the "integrate [math]d\tau[/math] along its world line" Does it just mean "integrate [math]d\tau[/math] "? ( for that particular v ) I am working from "A first course in General Relativity" by B F Schutz. I have a couple of other books on order which have not arrived yet. Relativity Demystified by D McMahon Introducing Einsteins Relativity by D'Inverno Vector and Tensor Amalysis by Borisenko and Tarapov Am always happy to buy books - have got quite good at building bookshelves! So if anyone one has suggestions which they believe could help in my situation - self study - engineering background - B Sc - Dip Ed - maths lover ( mathamatophile?) - 350 kms from nearest uni ... I would be grateful. John

Swansont - you are way ahead of me here. I am still at the beginning of my GR self studies and wrestling with concepts. What I need is :- 1 the ability to write down the equation of a world line for a "uniformly accelerated" body or any other world line come to that - is it just x = f(t)? 2 to understand what is meant by "integrate d[math]\tau[/math] along its world line" Perhaps Im not even making sense with these questions! The next chapter I have is "Tensor Analysis in SR" and I dont think there is much point in trying to understand that if I am still mixed up with vectors etc. Thanks for your reply - Im sure you have the answer I want if I can ever ask the right question! John

Any battery has an internal resistance R which is in series with the resistor you were measuring. Low values of the external resistor cause more current to be drawn and therefore more voltage drop across the internal resistor resulting in a drop in the terminal voltage of the battery. The second battery you used must have had a smaller internal R. You can measure the internal R by connecting a variable resistor across the battery - when the terminal voltage drops by 50% the external R equals the internal R. DONT try this with any large battery!!!! Say the internal resistance was 0.1 ohms then a 12v car battery would deliver 60 amps into a 0.1 ohms external resistor! This would dissipate 360 watts - smoke and heat !!

Yes thanks for that - as I work through more problems I am realizing that the choice of frame if critical. John

Am doing problems at the end of a chapter on "Vector Analysis in SR" Got stuck when asked to find the "elapsed proper time for the body as a function of t (Integrate d[math]\tau[/math] along its world line)" The body concerned here is "uniformly accelerated" ie "its acceleration four vector has constant spacial direction and magnitude" Ignoring y and z for the moment is the world line given by x = 1/2at^2? If so how do I get from there to [math]\tau[/math] as a function of t? As v is increasing the [math]\gamma[/math] is of course a function of time too - hence the integration I think. Have already found the relationships for speed and distance as functions of time getting answers that agree with the book. Am not really sure if this is the correct forum for this type of question but any help appreciated. John

Got it - thanks John Well I was not expecting that! So if there is only motion in one direction then all is simple but as soon as thats not true it gets a whole lot more complex. Need to study these a bit more to try and understand. John

Both of the books I have only derive the LT for motion of the frames parallel to the x axis. OK, this keeps the maths simple and also covers motion parallel to any other axis. But nowhere have I seen the LT for a relative velocity v which has components in each of the 3 spacial dimensions. I understand that the x values will only be effected by the x component of the velocity - and similarly for y and z. But what about t? So I guess my question is what is the "t" transform in this case? John

Thanks Severian I got the idea of upper/lower indices relating to frames from the book I am studying "A First Course in GR" by B F Schutz. Quote :- "The bars on the indices only serve to indicate the names of the observers involved: they affect the entries in the matrix [ [math]\Lambda[/math]] only in that the matrix is always constructed using the velocity of the upper index frame relative to the lower index frame" Hence my idea that [math]\Lambda[/math] is a matrix and the meaning of the indices. I think part of my problems lie in the way we are taught, books are written. I call this " the method of diminishing deception". I am right at the start of the book (Cp 2 Vector algebra). There are many ways to get into a subject and different paths to the end point. Concepts may be presented at the beginning which may not be entirely accurate but may be necessary to avoid too many complications first up. Later we are told - " well, what I told you before is not quite true......" etc etc. There also seems to be no "absolute standardization" of the symbolism used. Schutz mentions , among other things, the signs associated with t,x,y,z ie - + + + or + - - -. In the end these things don't matter but when you are starting out you have ( I do anyway ) have to hang on to something. To give you some idea of where I'm at. C1 - Basically a review of SR stress space time diagrams and invariance. C2 -Vector algebra - definitions, basis vectors, four velocity and momentum. There are plenty of problems at the end of each chapter but unfortunately only some have answers. Next C3 Tensor Analysis in SR. So at the moment tensors and the metric are pretty vague concepts. I want to stress that I am not complaining here. If this stuff was easy to really understand I would not be interested. The joy, for me anyway, is in the struggle. However I do value any input you are happy to provide. You never what will bring that "Ahhhhh!!!" moment! John

Thanks again. Unfortunately Im not doing any kind of course just self study as a kind of hobby. It would be great to have someone to "talk things over" with!. I think I get your message - using matrix multiplication works in some cases but the [math] \Delta{x}^\mu{=}\Lambda^\mu{}_\nu\Delta{x}^\nu [/math] notation is always going to be meaningful no matter what the exact expression. ie [math]\Delta{x}^0{=} \Lambda^0{}_\nu\Delta{x}^\nu[/math] for the 4 values of [math]\nu[/math]. The = here meaning "is the sum of". I think the [math]\mu[/math]s and the 0s should have bars over them to indicate the frame to which they refer - have not worked out how to do the bar yet - maybe something in "logic". So [math]\Lambda^1{}_2\[/math] is the coefficient multiplying the y (2) value of one frame used in calculating the x (1) value in the other - and so on x15. The upper index here refers to one frame the lower the other. In some respects anyway. How am I doing? John.