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.

Thanks Athiest - that was a real help. I appreciate the effort you put in to the reply. You said :- "Thinking of the Lorentz transformations as matrices might help because it's a familiar concept." I understood from my reading so far that [math]\Lambda^\mu{}_\nu[/math] was a matrix encapsulation of the LT coefficients. Is this perhaps a blinkered way of looking at it which will not be helpful when I finally get round to tensors? I understand about the inverse LT - and replacing v with -v etc. The book I am using uses "Einstein summation" and so there are no [math]\Sigma[/math]. Have got used to that. I had not yet come across the term "contraction of an object ...." So [math] ( \Lambda_\mu{}^\nu, v_\nu ) [/math] is a summation and [math] \Lambda_\mu{}^\nu v^\mu [/math] is not? The use of the word "object" implies that there are very general implications involved - perhaps I have my nose to close to the page to see the bigger picture at the moment! Im rambling a bit here - perhaps I need to read some more before I can ask a decent question. I used to say to my students ( taught engineering stuff at one time) when you can ask a good question you are more than half way to its answer! Thanks again for the help. John

Thanks for the replies - bear with me a little further. So a transformation from fr1 to fr2 can be represented by Col Matrix [t2,x2,y2,z2] = ColMatrix [t1,x1,y1,z1].[math]\Lambda^{2}_{1}[/math] ? (Have spent a few minutes and cannot find how to draw a column 4x1 matrix) My problem arose when solving a problem in Schutz concerning 3 frames f1 moving relative to f0 and f2 relative to f1. Ended up with the LT f2 f0 which I was then asked to prove that it was indeed a LT by showng that the interval was invariant under it. Did that OK but then noticed I had done as above but had switched rows and columns. I then repeated using the "correct" equations as above and the result was still invariant. Left me unsure which was the "correct" procedure and what is the meaning of the fact that effectively swapping the rows and columns of an LT is still an LT. Is that always the case? Just seen the second reply. I had noticed the offset in the indices in the book but the relevance of it was not explained. Obvious I suppose but I missed it anyway. Thanks for that leftmost is row - rightmost is column - that helps. Does that explain the results I got? Is the inverse of an LT always an LT? Hope this makes some kind of sense. Late at night here - will check over my working in the morning. John

Given [math]\Lambda^{\alpha}_{\beta}[/math] as a LT which index is Row and which column? Have got confused on a problem with a non symmetrical LT where I get a valid answer either way! Thanks John

Thanks for the reference ajb - have downloaded Carroll's lecture notes and will have a good look. John

OK - I have a good maths background and can handle all the algebra ok except for the use of M which I assumed had some special meaning. Turns out here it is the coefficient of the quadratics dt.dt, dt.dx, dt.dy, dt.dz.....12 more which may be a function of v, the relative frame velocity. This is in chap 1 of the book - chap 2 deals with vectors and chap 3 with tensors - so expect some more posts in a few weeks! I still have some questions re the interval but cannot yet be specific enough to ask anything sensible! Thanks again John

Thanks ajb I am still working on it - think the Ms may be functions of the various dxs involving v as well. Have figured out the maths equation formating - the terms I am interested in are [math]M_{\alpha\beta}(\Delta x^{\alpha})(\Delta x^{\beta})[/math] These are summed over alpha and beta ranging from 0 - 3 for t,x,y,z giving 16 terms in all. John

Have moved on from SR and started Schutz book "A first course in general relativity". I do not understand some of the symbols used in the proof of the invariance of the interval. I am referring to equ 1.2 to equ 1.3. What I am missing is what is meant by M with 2 subscripts followed by 2 quantities in brackets. M[math]_{ab}[/math] (dx[math]^{a}[/math]) (dx[math]^{b}[/math]) Umm have not quite worked out how to show equations yet either! The a and b are superscripts to the dx. Any help appreciated John

All this seems to relate to the so called "cosmological constant" in one of Einstein's equations which can be +ve 0 or -ve corresponding to an expanding - flat - or eventually collapsing on itself universe. There have been and probably still are projects going on trying to find the constants value - best a can remember is "close to 0" so could go either way. Also how does the "red shift" of distant objects fit into this? ie The further away an object is the more it is "red shifted" - ie the faster it is receding. Perhaps Im off the point here. John

This thread is going nowhere. Perhaps its time to close it down. It has helped me get a better grip on SR and I thank all those who responded to the questions I asked and confirmed some of my thinking. I have enjoyed reading all the posts - some with a smile - some with puzzlement and some with - Ah hah! Thanks John

When I started this thread a few days ago I was not expecting it to run quite as it has! Anyway - I've learned a lot both from the thread and other reading ( others apparently haven't) and make the following comments. 1 The twins paradox is not a paradox if you understand SR - it is a fact. The point I, and I think others, miss is that the twin who stays put is in an inertial frame and can correctly apply SR - predicting correctly that his/her twin will be younger on return. However the twin who changes his/her velocity to be able to leave and return is NOT in an inertial frame all the time and cannot use SR from his/her frame. SR applies only to inertial frames. The twin in the non inertial frame must apply GR and if he/she does, comes to the same conclusion. I cannot yet apply GR myself so have ordered an introductory text on GR to get me on the way. However I would be very surprised in GR contradicts the results of SR! My ignorance of something does not make it wrong! 2 Time dilation is nothing whatsoever (lovely word that) to do with the the fact that on clock is moving AWAY from the other. The factor involved has only v squared in it and thus it does not matter whether the frames are approaching each other or receding from each other - time is dilated. 3 I go back to my earlier post where I pointed out what is meant by an observer. The book I have of course defined an observer but I had not really thought about what that meant. From some of the posts I think others could do with spending some time pondering just what an observer is and what the consequences of that are. Most importantly an observer is not a single clock at a particular point which several of the posts seem to imply. 4 As to those who imply time isn't there if no one is measuring it I say wait till your 65 and the effects of time will be only too apparent whether you were watching the clock or not! John