timo

Not really a book, but a link I bookmarked some time ago that might keep you occupied for some time: http://www.cs.berkeley.edu/~russell/ai.html

Most 26 year old men know that getting an answer in an internet forum can take longer than 3 minutes. On topic: The frequency is given in units of coordinate time (which directly leads to frequency being dependant on the coordinate system) not in units of the Eigentime (which would be the time measured by the photon and which would indeed be kind of freezed).

a = year, M = mega = million => 1 Ma = 1 million years => .05 Ma = 50,000 years. Gravitational pull is proportional to the mass of an object, so all objects with mass exert a gravitational pull on others. Gravity is the feature of masses to attract each other. It´s rather hard to answer your question whether we know what gravity is but I´d tend to the answer "no, we don´t". If you dig deep enough into any physical theory you´ll allways come to a point where you don´t know why something behaves like it does. In fact, physics is not about describing WHY things work, but about HOW they do. That´s what the formulas do.

There is no measurement being 100% accurate. So if you make the deviation sufficiently small you should be able to fit any experimental data.

What exactly do the photons have to to with what I said? I wouldn´t know where they are needed when trying to derive covariant field equations for the EM field - they´re merely a solution to them if I remember that correctly.

I think the most common evidence for such statements is that the poster hasn´t understood both of them so they obviously have something in common. Now seriously: The striking resemblance of electrostatics and gravity (=electrostatics with only positive charges) in the classical theory is rather obvious. So if it´s easy to expand electromagnetism to a covariant form (=in a form that´s suitable for special relativity) one might think that one can apply a similar procedure of generalization to gravity. Now if the resemblance is so strong, shouldn´t this procedure be quite straightforward? Why has noone tried this so far? Will I win a Nobel Prize if I do that? Well, of course some people allready tried the most obvious approach. But doing so you´ll very quickly find out that there´s one very important difference between electromagnetism and gravity, namely the charge. The electromagnetic charge is conserved under any processes which is a very nice property. The corresponding charge in gravity would be the "relativistic mass" since in contrast to the "rest mass" it´s also a conserved quantity. This charge density, however, has different properties of transformation under coordinate transformations. It transforms like a 00-component of a tensor of 2nd order as opposed to the charge density in electromagnetism which transforms as a 0-component of a vector. That´s the reason why you can´t build a vectorial theory of gravity like the vectorial one of electromagnetism.

In principle, C(t) is simply an equation with one real valued variable. You can put in anything you want. I simply wanted to show you that the line equation C(t) in part a) was constructed in a way that makes the solution to your last question easy to obtain.

That´s the safe way to do d). The one I had in mind was recognizing that the line equation in a) was constructed in a way that C(0) = A, C(1)=B, C(0.5) = "middle of AB" and so on. So since you had P on 1/3 of AB the ratio must be (1/3):(2/3) = 1:2.

Spntaneously, I´d use the cross product but that does not nessecarily say you have to do so. Perhaps try both methods and see which you like more. You´re preparing for an exam, after all.

Oh, so it was good that Johnny worked out the numbers. But for clearification: There is no matrix in there at all. Your three equations (9).(1+9t)=0, (15).(-5+15t)=0, (12).(-7+12t)=0 are plainly wrong. It´s really just one. If you take a close look at the scalar product in my 1st post, you´ll notice that it´s only the equation (9; 15; 12).(1+9t; -5+15t; -7+12t) = (9).(1+9t) + (15).(-5+15t) + (12).(-7+12t) = 0.

t=1/3 is correct but I see little point in you working out kingjewel´s problem.

But I wanted a correct result . The book is correct, btw (not much of a surprise, though).

The one day I leave my laptop at work is the one I´d need it ... let´s see if I find a piece of paper somewhere ...

Dunno where you got the -1 from, the only -1 I mentioned was in conjunction with clearing up the mess with your direction vector B-A that Johnny and Dave started (ok, I also had a little part in it). It didn´t really have much to do with your problem so it´s best to completely ignore my 2nd and 3rd post now that you´ve got the correct direction vector (you can also PM me if the english is a problem - I´ll be awake for about one hour from now). Part b) is simply about finding a point on line C so that the vector from the origin to this point (C-O) is perpendicular to the direction of the line (B-A), so (C(x)-O).(B-A)=0. Solve this equation for x and you have the point. Perhaps the answer will be x= -1 but I don´t know that without actually doing the calculation. Post yours if you´re stuck. EDIT: Johnny, the idea of what you´re doing is correct (didn´t check the numbers) because the shorstest distance of a line to a point is the line segment from the line to the point that´s perpendicular to the line and vice versa. But it´s complete overkill here. Using the scalar product is much easier, faster and especially more related to vector math.

Ok, let´s try it the slow way for old, slow-minded people like me. The argument behind the t is certainly going to be a multiple of A-B. As the 1st entry is a 9 this fixes the multiple to -1, or simply B-A. Hence, we get B-A = (10, 10, 5)-(1, -5, -7) = (10-1, 10-(-5), 5-(-7)) = (9, 15, 12) != answer above. Man, you got me confused people.

No need to check, i referred to the thread starter

@Kingjewel: The solution to a) is C(t) = A + t*(B-A) where C is a point on the line through A and B (which depends on one parameter, as a line is one-dimensional). In question b) you are asked to find a point P on the line C(t) so that P-O is perpendicular to B-A. Two vectors are perpendicular if their scalar product vanishes so your condition for P=C(x) -I´ve renamed t to x here to highlight that you are looking for a definite value instead of having an arbitrary parameter- would be (C(x)-O)*(B-A) = 0. That´s one equation with only one unknown. As x only appears with a power of one, the solution is also unique. In case you are not familiar with the scalar product (you defenitely have to look it up if you´re going to take an exam in vector math), it´s (a, b, c)*(d, e, f) = ad + be + cf. At least in your case and in probably all you´ll encounter for the next few years. As you said you can do c), I guess I don´t have to say much about it. Don´t have any guides on vector math, sry. But if you´ve got special questions I´m alsmost certain than most of the people in this forum will be able and willing to help. Question d) is rather easy if you used the parameterization for the line that I used above. If you didn´t realize why this parameteritzation is so nice, then try to plug in t=0, t=1 and perhaps even t=0.5 and look what points you get for C(t). btw.: We had your avatar hanging on our door as a poster in the previous flat I lived in (it was a physicists commune)

No, it´s not a tensor of rank 3 - it´s not a tensor at all. Just a bunch of 64 entries. That´s because the result does not transform tensorial. You´ll learn this when you come to the covariant derivative.

The correct velocity to be used in SR is 4-velocity. 4-velocity IS absolute. Only the four numbers you describe it with change from coordinate system to coordinate system. A nice example someone posted in this forum (in another context, though) is the absoluteness of the numbers. A One is a One, regardless whether you write it as 1, 2/2, 3/3 or 1³. I think understanding the experimental results helps. No, I can´t see your objections. What I´ve understood so far is: You have two photons emmited and reflected back to meet each other. Assuming lightspeed to be constant in every frame, different coordinate systems give different time- and space-coordinates for the point in which the photons meet again. So, where´s your problem or the objection to SR ? Both statements are simply the same. You don´t really have to give up rational thinking. Quite contradictionary rational thinking tells me that the photons must allways meet again in the same point and that because of that the time-coordinates and the space-coordinates can be different. About the "we do not observe the reality"-thing: That´s one of those pseudo-scientific statements that can be watered down to a point where it will be correct. In the strict sense this term seems rather meaningless to me because I neither know what the authors mean by "observe" nor by "the reality". As a remark here: I´ve never come across a good description of SR. The way I really started understanding it was when I learned General Relativity and applied it to noncurved spaces. Same as in your example. The best coordinate system for you is the one in which you don´t observe the mirrors as moving (you may choose whether you want to be the astronaut or the drunk). Using other coordinate systems will make things a bit more complicated (poor drunk being given GPS coordinates) but won´t change the location of the bar/point of intersection.

No, in the sense of relativity they are not reflected back to point A. I am talking about points in 4-dim space here. That is one misunderstanding of yours I´m complaining about all the time. But I admit I could have called the points in 4-space "events" for clearification. Just reread my post and replace "point" with "event". Yes, see above. Don´t expect it to be a minor remark. The step to stop thinking in 3-space and to start thinking in 4-dim spacetime is no as easy as it sounds. Not completely sure what you are talking about but sometimes it´s easier to tell someone "the bar is 50 meters up the road from that italian restaurant" than to give GPS coordinates. I do think the same. That´s why I restricted my answer on the parts I understood. My main point is still that you seem to mix up classical view and relativisitic view.

It´s just a partial derivative. [math] \partial_i g_{jk} [/math] has 4*16 = 64 entries ijk, each corresponding to the derivative [math] \partial_i [/math] of the metric´s entry [math] g_{jk} [/math]. EDIT: nice TeX errors. Ascii version: It´s just a partial derivative. \partial_i g_{jk} has 4*16 = 64 entries ijk, each corresponding to the derivative \partial_i of the metric´s entry g_{jk}. EDIT2: Oh, and the keyword for TeX-code in this forum is "math", not "tex".

The two photons are emmited at point A, hit the mirrors at point L and R, respectively, and meet again on point B. Different coordinate systems give different coordinates for those points but the points remain the same.

I need a explanation done in flawless SR terms that shows me that SR leads to contradictions. All your current explanations were a mixup of classical view and SR view. When I look at the "two photons are emmited at the same point and reflected my mirrors"-scenario from the perspective of SR I can´t see any problems. Neither. If at all I support ideas here, not persons. I didn´t understand your point so far and I didn´t even bother reading Swansonts´posts, so far because I don´t really care much about epxerimental physics as long as it doesn´t concern me at work. Np.

Try to work on your attitude.