Iggy

Fair enough, here is another source: 4.2 Dual Equivalence As soon as terms like Geschwindigkeit (speed, velocity) or Spannung (stress, tension) appear in a text the translator's problems begin. Two approaches are adopted... First, the Polyseme Dictionary lists samples of adjacent compounds at entries like Geschwindigkeit, Spannung so that the translator realises the terms are polysemous... A Basis for Scientific and Engineering Translation - Michael Hann - 2004 He has done that repeatedly.

That site is mistaken. The paper I quoted was originally written by Einstein in 1916. It was translated into English in 1920. The German language does not have different words for the English: 'velocity' and 'speed', so any notion that Einstein meant one and not the other is mistaken. I should find a source for that... If we look at the lightspeed issue in Einstein’s 1905 paper on Special relativity: “On the Electrodynamics of moving bodies,” a good thing to note is there has been a translation problem from Einstein’s original German into English. The German language does not differentiate between the English words “speed” and “velocity”. (Others have noticed this, so I cannot take credit for discovering it. Einstein’s 1905 Speed/velocity of Light errors -- Roger J. Anderton For example, the German word for speed/velocity was translated to speed in the following two quotes: "Aus dem . . ., dass die Lichtgeschwindigkeit im Schwerefelde eine Funktion des Ortes ist, lasst sich leicht mittels des Huygensschen Prinzipes schliessen, dass quer zum Schwerefeld sich fortpflanzende Lichtstrahlen eine Krummung erfahren mussen." ("Given ... that the speed of light is a function of position, it is easily deduced from Huygens's principle that light rays propagating at right angles to the gravity field must undergo deflection." “Das Prinzip der Konstanz der Lichtgeschwindigkeit kann nur insofern aufrechterhalten werden, als man sich auf für Raum-Zeitliche-Gebiete mit konstantem Gravitationspotential beschränkt.“ (“The principle of the constancy of the speed of light can be kept only when one restricts oneself to space-time regions of constant gravitational potential.”) wiki -- variable speed of light Xyzt's earlier notion that the speed of light is constant while the coordinate speed of light is not -- that idea isn't really meaningful. A coordinate speed is a speed. If one insists that the coordinate speed is not constant then by necessity 'the speed' is not constant. No doubt, any non-local speed in GR is ambiguous, but an ambiguous thing does not a constant and invariant thing make.

It is not such without qualification. Take, for example, the following quotes by xyzt and Einstein: Both of those quotes answer your question directly. One says, "yes, the velocity of light is c regardless". The other says "no, the velocity is not c regardless". You're right that three pages of posts seems like a lot, so consider also the following two quotes: CaptainPanic was honest enough to offer earlier in the thread that he didn't know who was right or wrong, but I personally think that this subject has been settled for more than 100 years. Einstein is correct. You are now honest enough to say "is the speed of light (in vacuo) still exactly 299,792,458 metres per second?" implying that you can't imagine it being anything different. That the speed of light is "exactly 299,792,458 metres per second" is true only in inertial frames. Introducing gravity or acceleration opens nature's domain beyond that postulate. [math]c[/math] is a universal constant. The speed of a ray of light is not. Special relativity postulates that c and the speed of light are the same in inertial frames, but beyond that limited domain they are not. It is a common misconception to think they are equivalent which is why one frequently gets the question "if nothing can travel faster than light then why are galaxies receding faster than c?" This isn't a zero sum debate. The answers are mutually exclusive. Either xyzt or Einstein is correct. Either the speed of light varies or it doesn't, and if the last 3 pages of discussion, and the last 100 years of relativity, have failed to illuminate the answer then we have some ways to go.

In a manner of speaking. I really was only thinking that the lead clock would measure [math]\tau = 0.6931[/math] between both the two red events and the two blue events: making the top red and top blue events simultaneous. The variable t in the metric [math]ds^2 = -(ax)^2 dt^2 + dx^2[/math] is the proper time of a clock at [math]x_A = 1[/math] (the lead clock), so the dotted line would be called t = 0.6931, but only the lead clock measures that proper time on the dotted line. Proper time in that Rindler metric is given by [math]d \tau = a x dt[/math], so the time between red events as measured by the aft clock (at [math]x_A = 0.5[/math]) would be [math]d \tau = (1)(0.5)(0.6931) = 0.3466[/math] (as you say, 50%). It is convention for the coordinate acceleration, a, and the coordinate time, t, in the metric to correspond to the proper acceleration and proper time for the observer at x = 1. If a different observer were chosen then the metric would look slightly different -- it would have a translation of the origin along the x axis. But, physically, it doesn't matter which observer is chosen. Rereading what I just wrote, I think I just only managed to obfuscate what you explained very well.

using the equation in the context you have given would require signifying who is measuring t, L, and a, and what t, L, and a are specifically measuring, and how that disagrees with any number given in the derivation to which you are objecting, otherwise it is too ambiguous to solve. I can tell you I traced every step of the derivation JVNY did with the relativistic rocket equations and each number (also labeled on the diagram he made) was correct. I also individually solved them via an independent method and got the same numbers, as shown 2 posts ago.

No, it doesn't. I showed you how to do the calculations symbolically. Try doing the same thing (please stop using tables and numbers) and you might find your error. This is not equivalent with sending a pulse from the center of one rocket towards its front/rear end. You are dealing with TWO accelerations, this is much more complicated than what you set to solve, not to mention that your equations do not reflect the above. Try solving the simple case (one rocket) first. I'm afraid I don't see the error you're referring to, and I don't understand how an equation could demonstrate an error in those statements. To accelerate anything rigidly (be it one ship or two, or any number of particles and/or objects), each point being accelerated must have a magnitude of proper acceleration that is inversely proportional to the position along the rigid system's axis of motion. The front and the rear can maintain constant proper distance and different proper accelerations simultaneously -- no matter if a single ship separates the front and rear, or if a void between two ships separates them. "Born Rigidity" is the principle by which this is very well established in relativity. May I ask you to demonstrate the error for the sake of my curiosity?

That looks spot on to me. Very good use of the relativistic rocket equations. Another way to verify the dotted line is simultaneous for both ships, and that they maintain constant proper distance is with rindler coordinates. Position and acceleration will be related by x = 1/a, so having your aft ship at x = .5 and a = 2 works with the forward ship at x = 1 and a = 1. We will have the forward ship be the timekeeper, in which case the metric is: [math]ds^2 = -(ax)^2 dt^2 + dx^2[/math] In the lab frame the forward observer observes the ray of light at t=0.75, x=1.25. Using the transforms: [math]t_A = \frac{1}{g} \mbox{arctanh}\left(\frac{t}{x}\right),\; x_A= \sqrt{x^2-t^2}[/math] the Rindler coordinates become the following: [math]t_A[/math] = (1/1)*arctanh(0.75/1.25) = 0.6931 [math]x_A[/math] sqrt(1.25^2-0.75^2) = 1.0 If the observers maintain a constant proper distance and the aft observer receives the ray of light simultaneously then transforming the rear observation from the lab frame to rindler coordinates will give the same t (0.6931) and x = .5 In the lab frame the aft observer observes the ray of light at t=0.375, x=0.625 [math]t_A[/math] = (1/1)*arctanh(0.375/0.625) = 0.6931 [math]x_A[/math] sqrt(0.625^2-0.375^2) = 0.5 So I have you exactly right. In accelerated coordinates where the ships consider themselves at rest with respect to one another and with respect to themselves the light rays arrive simultaneously at the numbers you give.

I recommend reading the following site: Rindler Spacetime One of the fascinating things about general relativity is how it can be brought smoothly from special relativity when considering accelerating observers. In order to describe gravity, general relativity uses the concept of curved spacetime... The result is called Rindler spacetime, described by the so-called Rindler metric. <snip> ...Long story short: when Alice looks at points at her left (remember, gravity points leftwards), she sees a lower speed of light. Is that even possible? That is against the principle of relativity, isn’t it? No! The principle of relativity talks about inertial observers. Alice is not. So, again: points at her left have lower speeds of light. Therefore, relativistic effects are “more notorious”. Even worse: as you move leftwards, this “local speed of light” decreases more and more... until it reaches zero! Exactly at the “special point”, where Alice coordinates behaved badly. What happens there? It’s an horizon! Where time stood still. http://physicsnapkins.wordpress.com/2012/12/13/rindler/ It simplifies the issue greatly. You plagiarized what you just wrote from section 1.2.1 of this paper. I doubt you know what it means, because it doesn't address what we're talking about at all. In fact, given the transforms from cartesian (inertial) to Rindler (accelerated) coordinates that you plagiarized, one can readily prove c = xg. The transforms are expressed better by wikipedia (mind you they are using natural units): for cartesian [math](t,x)[/math], and Rindler [math](t_A, x_A)[/math] we have: [math]t_A = \frac{1}{g} \mbox{arctanh}\left(\frac{t}{x}\right),\; x_A= \sqrt{x^2-t^2}[/math] [math]t = x_A \, \sinh(gt_A), \; x = x_A \, \cosh(gt_A)[/math] A light ray emitted at [math]x_I[/math] follows the cartesian path [math]x = x_I + vt[/math] (where v is +1 or -1 depending on direction) making its path in Rindler coordinates rather trivial: [math]{x_A}^2 = x^2 - t^2[/math] [math]{x_A}^2 = (x_I + vt)^2 - t^2[/math] [math]{x_A}^2 = {x_I}^2 + 2x_Ivt[/math] [math]{x_A}^2 = {x_I}^2 + 2x_Iv(x_A \sinh(gt_A))[/math] and, using wolfram to solve for [math]x_A[/math] where v = +/- 1: [math]x_A =x_I \exp{(\pm g t_A)}[/math] light emitted at [math](t=0, x = x_I)[/math] therefore has a Rindler speed: [math]\frac{d x_A}{d t_A} = \pm gx_I \exp{(\pm g \cdot 0)}[/math] [math]\frac{d x_A}{d t_A} = \pm gx_I[/math] which is to say that the speed of a ray of light in natural units in Rindler coordinates is c(x) = gx

People rely on information here to be accurate. [math]c=xg[/math] is indeed the speed of light in Rindler coordinates in natural units. According to you, [math]xg[/math] cannot be a speed because it yields a dimensionality of length squared per time squared. Such ambiguity is the nature of natural units. Wikipedia explains it: Natural Units: Advantages and disadvantages Greater ambiguity: Consider the equation a = 1010 in Planck units. If a represents a length, then the equation means a = 1.6×10−25 m. If a represents a mass, then the equation means a = 220 kg. Therefore, if the variable a was not clearly defined, then the equation a = 1010 might be misinterpreted. By contrast, in SI units, the equation would be (for example) a = 220 kg, and it would be clear that a represents a mass, not a length or anything else Natural Units - Advantages and disadvantages Not only did I define [math]c=xg[/math] as the speed of light (removing any ambiguity), I drew a diagram demonstrating [math]xg[/math] is indeed the speed of light in Rindler coordinates at multiple values of x and t. Even after giving the SI formula (xg/c) so that you could confirm the dimensionality is meters per second you continue provoking. I welcome you to tell me what the formula for the speed of light in Rindler coordinates is. If mine is nonsense then please correct me. What is the formula for the speed of light in Rindler coordinates?

I can't tell if you're honestly confused or if you're trolling. v=gx/c in natural units is v=gx (because in natural units c=1) In both cases v and x are proportional. v=gx is the speed of light in natural units in Rindler coordinates. I'm sorry you didn't know the formula for the speed of light in Rindler coordinates. I'm sorry you didn't know why your dimensional analysis on that formula didn't work -- try it on v=gx/c -- you'll get m/s. I can tell you want to argue. Is there anyone else in the thread that wants to explain this to him in light of his upcoming objections? I'm extremely swamped with work.

In natural units it is gx. In non-natural units it is gx/c. I'm happy to explain natural units if you are unfamiliar. Sure thing, X. We'll catch up later. I'm going to talk to JVNY now.

That is my understanding as well Indeed -- either direction. You probably know, but if not: your diagrams in your pdf are Rindler coordinates. The speed of light in that case is [math]xg[/math] (where x is the x position and g is acceleration). As c and x are proportional in c = xg, the greater the x (i.e. the greater the distance from the mass) the greater the speed of light. For example, the following are g=1 rindler coordinates (I got it [url=http://en.wikipedia.org/wiki/Rindler_coordinates#Relation_to_Cartesian_chart]here[/here]). The small 45° blue and green lines I added are light rays. The blue are 1:1 (run over rise), or 1c. The green is 1:3, or one third c. The speed of light (and the x coordinate) are zero at the dotted line -- called the rindler horizon. As you see, the speed of light is c=xg, so the smaller the x the smaller c: Exactly. Shapiro delay confirms it. Light passing a mass is slowed on approach and slowed likewise on departure. yep yep

I suppose that you're probably right. It's just so hard... to even begin to calculate probabilities we'd have to ignore things we know in order to calculate. The largest fact we have to ignore is that the chance life evolves is 100%. How many facts do we have to ignore in order to answer this question the way it was meant to be asked? I don't know... but but I'd be disappointed to ignore them. The chance that the past happens as it happened is, perhaps surprising to you, but nothing so fierce as certain.

The past is known. The future is not (at least by any human). The past is for certain cause. The future is for uninformed differences of probability. "the chances that something happened by chance" isn't a properly formed idea, and you only propagate the problem by ignoring the faulty premise. The chances that chance did something that happened are certain. Nothing more needs said.

Yes, after the outcome, one knows the outcome. "Chance" is therefore no longer definitionally appropriate. What were the chances of something that certainly happens? Believe it or not, that sentence and the question tears itself apart. The chances are one in one. Well, that makes sense. Before the fact you might be ignorant of the outcome. After the fact, we aren't. As I understand the question, it is... "if the universe were replayed?... If everything reran itself, what would be the outcome?" Maybe in the first few years someone as intelligent as human couldn't have guessed the chances of life evolving, but here we are after the fact. The odds are therefore already known, even if you or I couldn't have known them then.

after you win three times times in a row, your chances of wining were one in one. After life evolves, the chance of it having evolved is one in one. Are you suggesting we live in a some other time? I'm suggesting it already happened, and that the chance that it happened is therefore incredibly predictable and already known. Are you suggesting otherwise? What year do you think it is? You don't get to say that "heads, tails, tails, tails, heads, tails, heads, heads, tails, tails" is incredibly unlikely after it already happens. Unless you think it didn't happen because it is so unlikely.... You think the coin didn't flip that way. It's one in a thousand, you think it didn't happen?

The chances that live evolved are one to one. One wonders what the last 7 pages of discussion are about. Incidentally, I wonder what the chances of flipping a coin a bunch of times are and it landing: heads, tails, tails, tails, heads, tails, heads, heads, tails, tails? Probably pretty small. Maybe one in a thousand. Nevertheless, I just did it and it happened so we might as well deal with it.

Right. Actually, the full quote isn't "Honey, half the jews I know are atheists." It's... "Honey, half the jews I know are atheists. It’s about community” That's what she said on the show. You hit that nail right on the head. Religion has been reduced to community. It used to be about truth and power, and any more it's about holding on to the things that bind us. I'd personally rather be bound by other things... and religion just needs to dry up in the desert sun and die already! We don't need it for culture. We can find culture all on our own, thanks.

And to reinforce the idea that christians shouldn't mind suffering. John is right... just a sign of the times. As an aside, Is your avatar supposed to mean "samurai"? It looks a lot like the symbol I know for samurai, but not quite. I know Japanese calligraphers take a lot of liberties, but it has me confused. PM me if you think it takes us off topic.

I don't admire Jesus for dying on the cross, but there were a few thousand followers of Spartacus who got strung up on the appian way just like the Nazarene and I do admire their sacrifice fighting for freedom. It's a bastardization to call the cross a christian symbol.

I'm reminded of an episode of House where Cuddy's mom is trying to hook Cudy up with House. She says, "So, say you two got married, would you convert to Judaism?" He says "I'm an atheist”. She says, “Honey, half the jews I know are atheists"

A word-virus isn't a bad analogy. To a healthy cell, a virus doesn't look that bad on the outside. The virus penetrates the cell being perceived as quite normal, but then destroys the cell from the inside. Likewise, some brains might not find these types of words too unappealing: non-religion is religious non-belief in God is belief in God denying the bible is supporting a bible not-theistic is theistic For whatever reason, some people might let those thoughts bounce around in a person's brain as if they mean something -- just like a cell does when it lets in a virus, but those thoughts are obviously so incoherent and illogical, they're bound to destroy a person's worldview from the inside -- just like a virus does to a cell. Luckily, some people have built up immunities to life-threatening viruses, just like some brains have built up logic and reason against worldview destroying nonsense.

The Friedmann acceleration equation is: [math]\frac{\ddot{a}}{a} = - \frac{4 \pi G}{3} \left( \rho + \frac{3p}{c^2} \right) + \frac{\Lambda c^2}{3}[/math] A positive density term, [math]\rho[/math] pushes [math]\ddot{a}[/math] in the negative direction meaning it decreases the expansion speed over time, A positive cosmological constant term, [math]\Lambda[/math], pushes [math]\ddot{a}[/math] in the positive direction -- increasing expansion speed over time. A static universe would have [math]\dot{a} = \ddot{a} = 0[/math]. Assuming no (or very little) radiation pressure, that would make [math]\Lambda = 4 \pi G \rho[/math]. Is this the equation you're talking about?

Without evidence of credibility, I don't presume that a report is trustworthy, and I don't presume that it is untrustworthy. It would depend completely on the report and the person who wrote it. It's like asking "if you know that some people who work for NASA beat their wives, can you trust a report by NASA?" An untrustworthy police report is useful for the defense and defendant. In the US we would call it Brady evidence -- evidence that would impeach the credibility of the officer that wrote it -- exculpatory for the defendant and inculpatory for the prosecution. Scripture has no authority over me and I refuse to treat it differently from any other ancient anthology.

Police, politicians, whatever... we might as well be talking about... Fine, apples. Rhetorically speaking, one rotten apple does not spoil the bushel. What you're doing is a fallacy -- throwing the baby out with the bathwater, arguing guilt by association... hasty generalization... law of small numbers (I'm having fun thinking of how many names your fallacy has)... uhh... oh! Just one apple: that's a fallacy of the lonely fact. The bible consists of nearly 70 books written by hundreds of people over hundreds of years. Thinking that showing "some bits" wrong will make it all untrustworthy is a very obvious fallacy. Your sample is heterogeneous, and your sample size is absurdly small (so far you've given one verse that probably wasn't even wrong -- it was a poetical idiom) I'm going to say that a few lying cops proves neither the first nor the second, and I'll do my best to prove it by changing the word "police" to "Scotsman" in the spirit of the 'no true Scotsman' rhetorical hurdles I see being thrown down. Your quote with "Scotsman" replacing "policeman"... You can not, or will not, see the difference between "all Scotsmen acting as individuals can't be trusted" and "The Scottish as a whole can't be trusted". If it is shown that some Scots lie then I'm sure we agree that this does not prove the first assertion. But it does prove the second one. If you know that (at least) some Scots tell lies, can you trust a Scottish newspaper ? Or do you accept that it may have been written by a bent Scotsman -- so you can't rely on it to be honest? I'm going to go ahead and disagree with you. Showing a few verses in the bible wrong doesn't make the other thirty thousand untrustworthy -- individually or as a group. Each verse, each chapter, each author, each book, each police, politician, and Scotsman stand on their own.