Dingle's Challenge to the Theory of Special Relativity
April 8 2009 at 6:45 PM
"The habit has developed of assuming that a physical theory is necessarily sound if its mathematics are impeccable: the question of whether there is anything in nature corresponding to the impeccable mathematics is not regarded as a question; it is taken for granted." Dingle
Special Relativity: The Challenge
Applications of Special Relativity
Chambers Encyclopaedia
One kind of application of Special Relativity is to the fundamental concepts of physics, such as mass, energy, force etc. If the laws of conservation of mass and of linear momentum are to be valid in all inertial frames of uniform motion the mass of a particle is defined as m(1 v˛/c˛)˝ and its energy as
mc˛(1 v˛/c˛)˝. Here m is the mass the particle would have in an inertial frame in which it was at rest (its proper mass) and v is its mass relative to the frame. It then follows that the laws of conservation of mass and energy are identical if we assume that the proper mass of a particle is itself equivalent to the amount of energy mc˛. These definitions of mass and energy imply that v must be less than c and it is for this reason that Special Relativity restricts itself to relative velocities not greater than c.
The two definitions have played a capital part in modern atomic physics. The equivalence of mass and energy helps to explain the phenomenon of radioactivity and the explosions of atomic bombs. It also accounts for the production of the vast quantities of radiation emitted by the stars without an apparent cooling of these bodies, the necessary energy being found in the conversion of a small fraction of their masses into radiant energy.
Dingle's Challenge to the Theory of Special Relativity
Paul Ballard
McCrea rather protests too much in his attempts to refute Dingle's arguments, although he concedes that Dingle "has not made any mistakes in his algebra." Instead, McCrea believes that Dingle "deals with objects to which the [SR] theory explicitly denies a meaning." This, of course, begs the question: if the theory denies a meaning to a physical model, is the theory coherent? He claims that Dingle's assertion is false because "he is not talking about the same thing, but two different things... equations (3) and (4) concern two different sets of events, and so they cannot contradict each other." McCrea adamantly refuses to acknowledge the validity of coordinate transformation systems and his refusal to treat both events in a common system is contrary to both the letter and spirit of special relativity theory.
McCrea repeatedly asserts, in the summary in Special Relativity (2) and elsewhere, that "Because A is never at E1, the phrase is meaningless and so Dingle's (3) is meaningless. Correspondingly his (4) is meaningless." That he should attempt to use "A is never at E1" as valid criticism was regarded by Dingle as "indefensible." Dingle had explicitly defined A as a clock distant from H. Clock A is never at E1; it merely records the event.
To better illustrate the clock paradox Dingle developed his thesis further. A basic example of his argument is provided by considering only the x'-axis and x-axis in two coordinate systems k and K. The coordinate system k is moving at uniform velocity v relative to K.
N
B
»»»»»»» v »»»»»»»
x'-axis of k, clock readings denoted by t', above
x-axis of K, clock readings denoted by t, below
A
H
A and H are two regularly running clocks at a fixed distance from each other in coordinate system K. N and B are another pair of regularly running clocks at a fixed distance from each other in the moving coordinate system k. (The distances NB and AH are independent and arbitrary.)
The readings of N and B with respect to the coordinate system k are denoted by t' and the readings of A and H, in the coordinate system K, by t. (For simplicity we regard K as stationary.) We now examine three successive events as k moves with uniform velocity v relative to K.
Event E0
At event E0, B is adjacent to A and both are observed to read zero. We also synchronize N and H to read zero at this moment.
t' = 0
N
B
A
H
t = 0
Event E1
At event E1, B is adjacent to H; B is observed to read t'1 and H is observed to read t1. Clock A is distant from clock H but reads t1 when H reads t1.
t' = t'1
N
B
A
H
t = t1
Event E2
At event E2, N is adjacent to A; N is observed to read t'2 and A is observed to read t2. This is all independent of the theory it is merely a description of a physical process. A theory is required when we wish to determine two independent things: (1) the values of t'1, t'2 for given values of t1, t2 or vice versa and (2) the relative values of A and B when these values apply. Clock B is distant from clock N, but reads t'2 when N reads t'2.
t' = t'2
N
B
A
H
t = t2
We now apply Einstein's theory, supposing that A is fixed at the origin of the K coordinate system and B is fixed at the origin of the k system.
(i) t and t' are related by the Lorentz transformation so that:-
Between events E0 E1
t'1 = at1
(1)
t1 in the K coordinate system is corrected by a (as A is offset) to read at1.
Between events E1 E2
t2 = at'2
(2)
t'2 in the k coordinate system is corrected by a (as B is offset) to read at'2
and
a = (1 v˛/c˛)˝
(ii) The relative velocity a (the ratio of distance vs. time, sometimes denoted by µ) of k in relation to K is determined by choosing a pair of events and comparing the intervals between the readings of A and B at these events.
Einstein chose events E0 and E1. At these events A reads 0 and t1 respectively and B reads 0 and t'1 respectively. The reason why A must be held to read t1 at E1, is that H reads t1 at this event, and by application of the Lorentz transformation the process by which A is set in relation to H synchronizes it with H.
Shortly however we reach a contradiction. Between events E0 and E1, A advances by t'1 = t1 and B by t'1 = at1 by (1). Therefore:
rate of A
=
t1
=
1
(3)
rate of B
at1
a
But now choose events E0 and E2. At these events A reads zero and t2 respectively and B reads zero and t'2 respectively. The reason why B must be held t'2 at E2 is that N reads t'2 at this event and the process by which B is set in relation to N synchronizes it with N.
However between events E0 and E2, B advances by t2 = t'2 and A by t2 = at'2, by (2). Then:
rate of A
=
at'2
=
a
(4)
rate of B
t'2
Equations (3) and (4) are contradictory; one result is the reciprocal of the other. Further, since v is non-zero, a is never 1 and it is obvious nonsense to derive (rate A)/(rate B) > 1 and (rate A)/(rate B) < 1 in the same system. Hence the theory from which they are derived must be invalid. Einstein in his theory gave only (3) and accepted it as the unique value of the rate-ratio; he did not verify the result by checking the interval between E<font size="-1">0 and E2. Had he done so he would have seen that his conclusion was untenable.http://www.heretical.com/science/dingle3.html
Re: Dingle's Challenge to the Theory of Special Relativity
April 9 2009, 12:01 AM
Cincirob: Well AAF I see you've taken to hiding your responses to me and only quoting excerpts of what I said. You're just blowing smoke again.
AAF: No one's hiding anything from you; you're just imagining it once again like your Albert! Anyway, a few years ago, you were so full of argumentation and humbug; not anymore; you've run out of it... Relativity is untenable; and that is it...
Cincirob: No hiding, but here is your response from another string hiding here. I don't need to argue any more. You don't have any new arguments and I defeated all your old ones. I'm still waiting for you to step up on Dingle....I'm not going to live forever.
Jose: Too bad. I guess you'll die in ignorance and arrogance
AAF: Einstein didn't live forever either; did he? I agree; the idea that we can't afford to postpone it till 2050 is a sad one... Well, that is how nature engineers it!
Anyway, Dingle's most devastating argument against Einstein's Special Relativity is all yours; look up; way up; and show us the goods...
Ted
Re: Dingle's Challenge to the Theory of Special Relativity
Re: Dingle's Challenge to the Theory of Special Relativity
April 9 2009, 2:15 AM
Dingle's paradox (or whatever you want to call it)
At event E0,
B is adjacent to A and both are observed to read zero. We also synchronize N and H to read zero at this moment.
****************************
cinci: All clocks read zero at this instant. Already this is not stated clearly. From what point of view are the clocks to be viewed. If clock N is synchronized with Clock B in their frame, then they will not both read zero for an obsrver in the statioanry frame. So anything beyond this point can be interpretted two ways making this thought demonstartion pretty much useless.
*****************************
At event E1,
B is adjacent to H; B is observed to read t'1 and H is observed to read t1. Clock A is distant from clock H but reads t1 when H reads t1.
*******************
cinci: OK.
*****************
At event E2,
N is adjacent to A; N is observed to read t'2 and A is observed to read t2. This is all independent of the theory it is merely a description of a physical process. A theory is required when we wish to determine two independent things: (1) the values of t'1, t'2 for given values of t1, t2 or vice versa and (2) the relative values of A and B when these values apply. Clock B is distant from clock N, but reads t'2 when N reads t'2.
************************
cinci: No, it is not all independent of theory. As I pointed out above it's is not clear what is intended by setting all the clock at zero since the observer of these settings is not identified.
*****************************
(i) t and t' are related by the Lorentz transformation so that:-
t1 in the K coordinate system is corrected by a (as A is offset) to read at1.
*******************************
cinci: This statement is pretty much meaningless. There are no "corrections".
*******************************
Between events E1 E2
t2 = at'2
***************************
cinci: No, (t'2 -t'1) = a(t2 - t1)
Problems so far:
1. Dingle has applied the time dilation formula to coordinates instead of time intervals so he is incorrect.
2. He calls the time dilation formula the Lorentz transformation, which it is not. It can be derived from the Lorentz transformation it is not the transformation.
3. He reversed the t and t' in the second equation.
4. Setting all the clocks at zero without identifying the observers of the clocks at that time makes the whole problem incomprehesible.
Now if one of you Dingle supporters would like to straighten out this mess I'll go on. But he's up to 4 errors already so obviously he can't draw any valid conclusions.
************************
Lal
Thanks Ted
April 9 2009, 2:42 AM
Thanks for this nice reference.
It is interresting to read, in the details, how misunderstanding and ignorance can also be propagated.
I also appreciated the argument about Euclidian geometry.
It illustrates well how ignorants can find math mistakes in Einstein's OEMB paper.
It can also illustrate how deliberately obtuse are those who pretend time and space cannot be mixed, those who claim "spacetime" is an absurdity or it is "only a mathematical artifact".
Following the reasonings of these IQ50, rotation would also be an absurdity!
Lol
Re: Dingle's Challenge to the Theory of Special Relativity
April 9 2009, 5:34 PM
Cincirob: Dingle's paradox (or whatever you want to call it) At event E0,
B is adjacent to A and both are observed to read zero. We also synchronize N and H to read zero at this moment. All clocks read zero at this instant. Already this is not stated clearly. From what point of view are the clocks to be viewed. If clock N is synchronized with Clock B in their frame, then they will not both read zero for an observer in the stationary frame. So anything beyond this point can be interpreted two ways making this thought demonstration pretty much useless. At event E1, B is adjacent to H; B is observed to read t'1 and H is observed to read t1. Clock A is distant from clock H but reads t1 when H reads t1. OK. At event E2, N is adjacent to A; N is observed to read t'2 and A is observed to read t2. This is all independent of the theory it is merely a description of a physical process. A theory is required when we wish to determine two independent things: (1) the values of t'1, t'2 for given values of t1, t2 or vice versa and (2) the relative values of A and B when these values apply. Clock B is distant from clock N, but reads t'2 when N reads t'2. No, it is not all independent of theory. As I pointed out above it's is not clear what is intended by setting all the clock at zero since the observer of these settings is not identified. (i) t and t' are related by the Lorentz transformation so that:- Between events E0 E1 t'1 = at1 No, t'1 - t'0 = a(t1 t0) t1 in the K coordinate system is corrected by a (as A is offset) to read at1. This statement is pretty much meaningless. There are no "corrections". Between events E1 E2 t2 = at'2 No, (t'2 -t'1) = a(t2 t1) Problems so far: 1. Dingle has applied the time dilation formula to coordinates instead of time intervals so he is incorrect. 2. He calls the time dilation formula the Lorentz transformation, which it is not. It can be derived from the Lorentz transformation it is not the transformation. 3. He reversed the t and t' in the second equation. 4. Setting all the clocks at zero without identifying the observers of the clocks at that time makes the whole problem incomprehensible. Now if one of you Dingle supporters would like to straighten out this mess I'll go on. But he's up to 4 errors already so obviously he can't draw any valid conclusions.
AAF: No one traumatizes the supporters and the sympathizers of Einstein's Special Relativity more or does it better than Professor Dingle! That is a fact. Well, Cinci; you still have no real argument against Dingle's 'reductio ad absurdum' reasoning. And your objections above don't hold water at all. Since Dingle has four observers; and it's absolutely clear that you can set to zero as many clocks as you wish by ad-hoc & arbitrary assumption. But, of course, arbitrary assumptions are always arbitrary assumptions. Einstein assumed arbitrarily, in his 1905 paper, that (t=0 & t'=0); and there is no good reason why Dingle, in his reductio ad absurdum, shouldn't do the same. The answer to your first question above is (by using the LT Equation); and to your second question is (because there is no t'3). Also, Dingle was no 'charlatan'; those Relativists, who thought they disposed of his argument by mere sophistry, were the real 'charlatans'. As a matter of fact, the relation between Clocks (H & N), in Dingle's arrangement, is precisely the same and remains precisely the same as the relation between Clocks (A & B), regardless of how long the (k & K) systems are kept under observation. That is remarkable to say the least. Moreover, the Principle of Relativity says 'loud & clear' that all inertial frames of reference are equal; and no measurement can single out one of them as 'privileged' or 'preferential' or 'cockamamie'. Therefore, it's fair to say that whenever an aircraft engine fails, the first QUESTION should be 'did Cincirob make it'? Now, let the race begin! It's very clear that Dingle has set (t=0 & t=0) by brute-force ASSUMPTION just as your Albert did in his 1905 PAPER. And so, Dingle scores first against Cincirob, not against his Albert; and the tally, now, opens as (1 for Dingle & 0 for Cincirob). Next, Dingle does not assume that [x(of A)=0 & x(of B)=0], because he needs neither; but Cincirob does make this REDUNDANT assumption. By virtue of Ockam's Razor, therefore, Dingle scores again; and the tally is (2 for Dingle & 0 for Cincirob). The referee stops temporarily the GAME and calls in AAF to examine the legality of Cincirob's Second Assumption. AAF informs the referee that Cincirob's statement of (Lets also say that N is at x = -10 light seconds and H is at x=5 ls) is ILLEGAL, because according to Dingle's Scenario, the relation between Clocks (H & N) is the same as the relation between Clocks (A & B). The referee penalizes Cincirob by taking out one of his scores; and the tally now is (2 for Dingle & -1 for Cincirob); and the game resumes. Dingle sets the Clocks (H & N) to (0) just as he did with the Clocks (A & B). Cincirob claims that (Since Dingle doesnt name an observer, he is already in la-la land). AAF tells the referee that in Einstein's theory of Relativity, clocks imply necessarily observers; and a clock, with no one around to read it, is like the sound made by a falling tree in the forest when nobody is there to hear it. The referee penalizes Cincirob for it and for his (la-la land) by taking out one of his scores; and the tally, now, stands at (2 for Dingle & -2 for Cincirob). The game continues. Cincirob makes the statement of (let us assume Dingle means H is synchronized with A and reads t = 0 for an observer in K. And let us assume he means N is synchronized with B which means it read t=0 for an observer in k. I assume this because the only synchronizing method related to relativity is defined for clocks in the same frame). AAF agrees; and the referee rewards Cincirob by adding one to his scores. The tally, now, reads (2 for Dingle & -1 for Cincirob). Dingle goes on to Event E1 where the flow of time must change his (t=0 & t=0) to (t' = t1 & t = t1); he scores as expected; and the tally is (3 for Dingle & -1 for Cincirob). Because Cincirob continues to mock Dingle, the referee punishes the former by giving his turn of the game to the latter. Dingle goes on to Event E2 where the passing of time, necessarily, changes his (t' = t1 & t = t1) to (t' = t2 & t = t2); he scores once again; and the tally now reads (4 for Dingle & -1 for Cincirob). The turn now for Cincirob, who concludes that (the mystery deepens! So far Dingle hasnt said anything of merit but we can only assume he will). AAF tells the referee it's just (humbug & smoke screen); and so Cincirob doesn't score! Dingle goes on and applies Einstein's Equation for time dilation to the time difference between (t=0 & t=0) of E0 and (t' = t1 & t = t1) of E1 so that t1 = at where a = (1-(v/c)^2)^0.5; and he scores, (5 for Dingle & -1 for Cincirob). Cincirob says (so far, so good) and scores; and the tally moves to (5 for Dingle & 0 for Cincirob). Dingle makes the next move states that (Einstein chose events E0, where (t=0 & t=0), and E1, where (t' = t1 & t = t1). At these events Clock A reads (0 & t1) respectively and Clock B reads (0 & t'1) respectively as well. The reason why Clock A must read (t1) at E1, is that time, by definition, must flow and its ARROW must always point to the FUTURE. Likewise, Clock B reads (t'1) at this event. And by using Einstein's famous Operational Procedure, Clock H synchronizes with Clock A, and Clock N synchronizes with Clock B, with no problem whatsoever. Dingle, therefore, scores one more time; and the tally is (6 for Dingle & 0 for Cincirob). AAF says (he can't bear to see Cincirob lose) and asks the referee to leave! The referee tells AAF that his expertise is needed and urges him to control his emotions and to be objective. It's, now, Cincirob's turn. He describes Dingle's winning move as (all but meaningless since the Lorentz transformations have nothing to do with synchronizing A and H). But because Dingle uses the Lorentz transformations only to calculate intervals and durations of events, AAF informs the referee NOT to take him seriously and to let it pass. And so Cincirob doesn't score. Dingle makes his 'grandmaster' move by pointing out that (between events E0 and E2, B advances by t2 = t'2 and A by t2 = at'2, and scores again. And so, the tally now reads (7 for Dingle & 0 for Cincirob). And so, Cincirob replies [The poor man has suddenly decided that the time dilation formula should be inverted; that is, first he says the rate time passes in the moving frame is related to the time rate in the stationary frame by the formula t(moving frame) = t(stationary frame) X (1-(v/c)^2)^.5 (t1 = at See above)]. But since by virtue of the Principle of Relativity, the time measurements made by the observers at (B & N) must be identical, equal, and exactly the same as the time measurements made by the observers at (A & H), Cincirob does not score once again. As a result, Cincirob storms out shouting (there is absolutely no reason for this to happen and it is not what relativity would tell you. So you cant dismiss relativity on a reductio ad absurdum argument if you base it on statement relativity does not make). AAF requests, as a payment for his service, one more score to be given to his friend, Cincirob. Accordingly, the Final Tally is (7 for Dingle & 1 for Cincirob). Cincirob loses & Dingle wins.
Ted
Re: Dingle's Challenge to the Theory of Special Relativity
Re: Dingle's Challenge to the Theory of Special Relativity
April 9 2009, 8:11 PM
AAA, you can't bury the truth with an avalanche of words. You have no answer to my critiques of Dingle's nonsense. If you can't address four simple comments in a simple and understandable manner, you can't address them at all.
*******************************
noblackhole
Re: Dingle's Challenge to the Theory of Special Relativity
April 9 2009, 10:42 PM
Some questions for the relativists here.
Let k, f, g be non-zero constants. Let A,B,C,D be variables.
1) Let A=kB, kD=C; then A-kB = 0 = C-kD, and so A-kB = C-kD. (T/F?)
2) Let A=-kB, -kD=C; then A+kB = 0 = C+kD and so A+kB = C+kD. (T/F?)
3) Then f.0 = f(C-kD) = 0 and g.0 = g(C+kD = 0. (T/F?)
4) Then A-kB = f(C-kD) and A+kB = g(C+kD). (T/F?)
5) So, by 1), 2) and 3), the expressions in 4) are just the identities 0 = 0 and 0 = 0. (T/F?)
6) By 1) and 2), if A = 0, then B = 0. (T/F?)
noblackhole
Re: Dingle's Challenge to the Theory of Special Relativity
April 9 2009, 10:46 PM
NOTE: a parenthesis was inadvertently left off 3) in the previous post.
Some questions for the relativists here.
Let k, f, g be non-zero constants. Let A,B,C,D be variables.
1) Let A=kB, kD=C; then A-kB = 0 = C-kD, and so A-kB = C-kD. (T/F?)
2) Let A=-kB, -kD=C; then A+kB = 0 = C+kD and so A+kB = C+kD. (T/F?)
3) Then f.0 = f(C-kD) = 0 and g.0 = g(C+kD) = 0. (T/F?)
4) Then A-kB = f(C-kD) and A+kB = g(C+kD). (T/F?)
5) So, by 1), 2) and 3), the expressions in 4) are just the identities 0 = 0 and 0 = 0. (T/F?)
6) By 1) and 2), if A = 0, then B = 0. (T/F?)
Lol
Re: Dingle's Challenge to the Theory of Special Relativity
April 10 2009, 2:05 AM
NOTE: a parenthesis was inadvertently left off 3) in the previous post.
You are clearly inspired by Dingle: "grotesquely crackpotish".
noblackhole
Re: Dingle's Challenge to the Theory of Special Relativity
April 10 2009, 6:35 AM
Lol,
They are simple questions in schoolboy algebra that relativists should be able to handle. Answer them, whether or not you are a relativist, instead of resorting to your silly and very tired 'crackpot' nonsense in order to evade, which clearly indicates a disposition to anything but sober scientific discussion. It should surely take no more than a minute or two to give your answers. So control your angry irrational outbursts and instead provide some proper input. If you won't do that then don't interfere.
All persons on this forum are invited to answer these simple questions. Do any of the relativists here have sufficient confidence in their skills to commit themselves to some simple algebra? If so, demonstrate those skills by answering the questions.
Lol
Re: Dingle's Challenge to the Theory of Special Relativity
April 10 2009, 8:06 AM
Answer yourself your K10 homeworks.
And if you can, just try to go further in your game.
Just to see if you can sustain an unprepared demo of your skills.
Otherwise ask your mom.
cincirob
Re: Dingle's Challenge to the Theory of Special Relativity
April 10 2009, 9:44 AM
NBH:
Let k, f, g be non-zero constants. Let A,B,C,D be variables.
1) Let A=kB, kD=C; then A-kB = 0 = C-kD, and so A-kB = C-kD. (T/F?)
2) Let A=-kB, -kD=C; then A+kB = 0 = C+kD and so A+kB = C+kD. (T/F?)
3) Then f.0 = f(C-kD) = 0 and g.0 = g(C+kD) = 0. (T/F?)
4) Then A-kB = f(C-kD) and A+kB = g(C+kD). (T/F?)
5) So, by 1), 2) and 3), the expressions in 4) are just the identities 0 = 0 and 0 = 0. (T/F?)
6) By 1) and 2), if A = 0, then B = 0. (T/F?)
cinci: Adding equations A = kB and A = -kB produces 2A = 0 or A = 0.
Substituting A into either equation produces B = 0.
The same goes for C and D. If all these functions are zero you can set zero equal to zero in as many forms as you can think of.
You seem to relish taking relativists to task for failing to respond. None of the anti-relativists responded.
********************
Lol
Re: Dingle's Challenge to the Theory of Special Relativity
April 10 2009, 1:00 PM
Re: Dingle's Challenge to the Theory of Special Relativity
April 10 2009, 4:58 PM
Cincirob: AAF, you can't bury the truth with an avalanche of words. You have no answer to my critiques of Dingle's nonsense. If you can't address four simple comments in a simple and understandable manner, you can't address them at all.
AAF: So, it's damn, if you do; and it's damn, if you don't! If the comments are brief, you would cry out 'little drops'; and if they're not too brief, you would start shouting 'avalanche, avalanche, help, help'...However, if it makes you feel any better, the strongest response to Dingle's Challenge, so far, does not go to you nor to Ted nor to Lal, but it goes to McCrea...so, try to work on it a little harder!
Lal: Thanks for this nice reference. It is interesting to read, in the details, how misunderstanding and ignorance can also be propagated. I also appreciated the argument about Euclidean geometry. It illustrates well how ignorants can find math mistakes in Einstein's OEMB paper. It can also illustrate how deliberately obtuse are those who pretend time and space cannot be mixed, those who claim "spacetime" is an absurdity or it is "only a mathematical artifact". Following the reasonings of these IQ50, rotation would also be an absurdity! Lol
AAF: Come on, Lal; that miserable, mostly biographic, and useless article has been around since the beta release of MS Windows 95!
AAF: Cretin Ted--->>>>>>, you moron, your French is not bad: "Résumé : Einstein a prédit une variation de l'énergie des photons au voisinage d'un champ de gravité, cette variation étant directement proportionnelle ŕ la distance de la source du champ gravitationnel. Au début des années 60, Pound et Rebka (Phys. Rev. Lett. 3, 439 (1959)) ont tenté de vérifier expérimentalement cette prédiction. L'expérience a été reprise avec plus de précision par Pound et Snider(Phys. Rev. Lett. 13, 539 (1964)). Plus tard, Vessot (Phys. Rev. Lett. 45, 2081 (1980)) l'a refaite dans l'espace avec une précision trčs améliorée. Ici, nous proposons une nouvelle approche au problčme qui procéderait ŕ l'inverse, en descendant vers le centre de la Terre. Notre approche est moins onéreuse, introduit des effets de plus grande amplitude et prévoit un nouvel effet du deuxičme ordre."
Ah; the passing of Newton's Absolute Time has reduced the once-smart and talkative Doc and Patent Holder to a mere idiot and retarded moron who can hardly manage to Google around and write two or three sentences of nonsense every two days...very tragic...
Ted
Re: Dingle's Challenge to the Theory of Special Relativity
April 11 2009, 1:39 AM
You progressed from being an idiot to being a full time nutter writing in French. Tsk,tsk
Anonymous
Re: Dingle's Challenge to the Theory of Special Relativity
April 11 2009, 4:12 AM
The French are not nutters Ted; its you that's the nutter.
cincirob
Re: Dingle's Challenge to the Theory of Special Relativity
April 11 2009, 6:56 AM
AAF: So, it's damn, if you do; and it's damn, if you don't! If the comments are brief, you would cry out 'little drops'; and if they're not too brief, you would start shouting 'avalanche, avalanche, help, help'...However, if it makes you feel any better, the strongest response to Dingle's Challenge, so far, does not go to you nor to Ted nor to Lal, but it goes to McCrea...so, try to work on it a little harder!
cincirob: I showed he was wrong. It's OK with me if somebody else showed more things wrong with his analysis. Once you show he's wrong, what's the difference how many things he's wrong about?
*****************
bob s
Re: Dingle's Challenge to the Theory of Special Relativity
April 11 2009, 7:48 AM
Dingles biggest mistake was assuming his concerns with Relativity would be given fair consideration and not be drawn into a morass of ambiguity and duplicity. But, making mistakes does not mean he was wrong. Einstein made mistakes and was afforded the opportunity to make corrections, Dingle made mistakes and was crucified (figuratively speaking).
noblackhole
Re: Dingle's Challenge to the Theory of Special Relativity
April 11 2009, 2:00 PM
cincirob,
Some questions for the relativists here.
Let k, f, g be non-zero constants. Let A,B,C,D be variables.
1) Let A=kB, kD=C; then A-kB = 0 = C-kD, and so A-kB = C-kD. (T/F?)
2) Let A=-kB, -kD=C; then A+kB = 0 = C+kD and so A+kB = C+kD. (T/F?)
3) Then f.0 = f(C-kD) = 0 and g.0 = g(C+kD) = 0. (T/F?)
4) Then A-kB = f(C-kD) and A+kB = g(C+kD). (T/F?)
5) So, by 1), 2) and 3), the expressions in 4) are just the identities 0 = 0 and 0 = 0. (T/F?)
6) By 1) and 2), if A = 0, then B = 0. (T/F?)
cincirob: Adding equations A = kB and A = -kB produces 2A = 0 or A = 0.
"Substituting A into either equation produces B = 0.
"The same goes for C and D. If all these functions are zero you can set zero equal to zero in as many forms as you can think of.
"You seem to relish taking relativists to task for failing to respond. None of the anti-relativists responded."
It is clear enough from your response that you maintain that 1) to 6) are all true. I agree with you.
Now before I get to the point of my simple exercise, I want to take it a little further. As you quite correctly say, by 1) and 2), adding A = kB to A = -kB gives 2A = 0 and so A = 0, and B = 0 too. We can do this with 4) as well, thus:
7) A-kB = f(C-kD) [= 0, by 1) and 3)]
A+kB = g(C+kD) [= 0, by 1) and 3)]
Adding these equations gives:
A = ˝(f+g)C - ˝(f-g)kD [= 0, by 3)]
and setting a = ˝(f+g) and b = ˝(f-g) this is
A = aC bkD [= 0, by 3)]
Subtracting the two equations gives:
kB = akD bC [= 0, by 3)]
So A = 0 and B = 0.
8) So we get the two equations:
A = aC bkD
kB = akD bC
9) Are the two equations of 8) meaningful in that are they simultaneous equations by which we can solve for the constants a and b in some way? For instance,
Let A = 0. Then by the first of equations 7), C = (bk/a)D. Set bk/a = p, and call p a special constant.
Let B = 0. Then eliminating D from the two equations gives,
A = a(1 bb/aa)C, and since bb/aa = pp/kk, this is A = a(1 pp/kk)C.
10) We have already agreed that if A = 0 then B = 0, by 1) and 2). I maintain that A = aC bkD and kB = akD bC are still just the identities 0 = 0 and 0 = 0, and so the procedures in 9) are quite meaningless in that that do not determine A as a function of C for some range of values on C. Do you agree? If you disagree then provide a proof that the development in 9) is generally meaningful.
Re: Dingle's Challenge to the Theory of Special Relativity
April 11 2009, 3:39 PM
'Tsk,tsk' on you, Idiot Ted--->>>>>>
Most of Lal's papers are written in French!
cincirob
Re: Dingle's Challenge to the Theory of Special Relativity
April 11 2009, 5:22 PM
NBK: 10) We have already agreed that if A = 0 then B = 0, by 1) and 2). I maintain that A = aC bkD and kB = akD bC are still just the identities 0 = 0 and 0 = 0, and so the procedures in 9) are quite meaningless in that that do not determine A as a function of C for some range of values on C. Do you agree? If you disagree then provide a proof that the development in 9) is generally meaningful.
cinci: I have a head chld right now that makes me not want to wade through this. If you are going to maintian that A = kB, A = -kB, C = kD, and C = -kD, then all these experessions are just zeroes equaling zeroes.
Are you trying to go somewhere with this?
*******************
Ted
Re: Dingle's Challenge to the Theory of Special Relativity
April 12 2009, 12:27 AM
AAF: Most of Lal's papers are written in French!
Ted: you have gone off the deep end, nutter.
Re: Dingle's Challenge to the Theory of Special Relativity
April 12 2009, 1:10 AM
Good, it's now your turn to jump off the cliff, Crazy Ted--->>>>>>
I and the wolves are waiting for you...
noblackhole
Re: Dingle's Challenge to the Theory of Special Relativity
April 12 2009, 7:38 AM
cincirob,
cincirob: "cinci: I have a head chld right now that makes me not want to wade through this. If you are going to maintian that A = kB, A = -kB, C = kD, and C = -kD, then all these experessions are just zeroes equaling zeroes.
Are you trying to go somewhere with this?"
Yes; I have already indicated that there is a point to this simple exercise.
We are agreed; all these subsequent expressions are just identities in zero. And so the final expression does not give A as a function of C for any values other than both A and C zero. The whole thing is schoolboy algebra, of course. However, set A = x', k = c, B = t', C = x, D = t. Then you have precisely what Einstein did to 'derive' the Lorentz transformation. So Einstein didn't even realize that all he was doing was nonsense with identities in zero. Einstein erroneously thought that he could make zeroes non-zero, which is nonsense. The calculation by Einstein, in more or less the steps I have given in this little exercise, are given by him in Appendix I, "Simple Derivation of the Lorentz Transformation [Supplementary to Section XI]", in his book, 'Relativity The Special and the General.'
The followers of Einstein have not realized this is nonsense either, and have, in one way or another, reproduced the same fallacious line of 'reasoning'. With that in mind, one is led to wonder whether or not Einstein made other major errors in his use of higher mathematics as well, errors also gone unrecognized by his followers and reproduced by them over and over again in one form of another. Well, inspection of the literature reveals that this is indeed the case. Major errors have been committed in differential geometry that render the usual claims demonstrably false. A particular case in point is the so-called "Schwarzschild radius". The irrefutable fact is that the "Schwarzschild radius" is not even a distance, let alone a radius of anything, in the related manifold. The proof is not difficult, but the proponents of black holes have not realized their error. This simple geometrical fact is sufficient to completely invalidate all claims that black holes are predicted by General Relativity.
So, if you think that the "Schwarzschild radius" is a radius or a distance of any sort in the related manifold, you are in the same boat as the proponents of black holes. They have never proven that it is a distance in the manifold, but that has never stopped them from saying (incorrectly) that the "Schwarzschild radius" is the radius of a spherical surface of one-way membrane they call the event horizon of a black hole. They can however, never prove their assertion, because the irrefutable fact is that the "Schwarzschild radius" is not a distance of any kind in the related manifold.
Ted
Re: Dingle's Challenge to the Theory of Special Relativity
April 12 2009, 12:25 PM
Roger Dumberton, why do you call yourself "noblackhole" now? Never mind...
cincirob
Re: Dingle's Challenge to the Theory of Special Relativity
April 12 2009, 2:56 PM
NBH: We are agreed; all these subsequent expressions are just identities in zero. And so the final expression does not give A as a function of C for any values other than both A and C zero. The whole thing is schoolboy algebra, of course. However, set A = x', k = c, B = t', C = x, D = t. Then you have precisely what Einstein did to 'derive' the Lorentz transformation.
In this, the original derivation, there is no zero = zero. Einstein describes a physical model and set's up a differenttial equation which he solves which yields the transformation.
There is another derivation where you set the coordinates of a sphere of light in two relatively moving coordiante systems as follows:
(ct)^2 = x^2 + y^2 + z^2
and
(ct')^2 = x'^2 + y'^2 + z'^2
The you rearrange them so that they each hav zero on one side of the equation and set them equal:
This is where your zero = zero concern arises. The mathematical idea that makes this OK, is that t, t', x, x', y, y', z, and z' are independent values. Actually you don't have to set them equal. You can weite these two equations.
-(ct)^2 + x^2 + y^2 + z^2 = 0
and
-(ct')^2 + x'^2 + y'^2 + z'^2 = 0
In either case you assume that each primed variable is some function of the unprimed variables.
Since the variables are independent, the coefficients of the each of the variables in the two equations must be equal.
Einstein has a short cut derivation which I haven't bothered to comit to memory, but I believe it relies ona a similar stategy.
*******************************
NBH: The followers of Einstein have not realized this is nonsense either, and have, in one way or another, reproduced the same fallacious line of 'reasoning'.
cinci: Well obviously it isn't nonsense. Einstein had as many detractors as adherents prior to the demonstation general relativity and they included some of the foremost scientists in the world including Lorentz who was a fine mathematician. If this were an error, do you really think you were the first guy who figured it out in the past 100 years. Remember Dingle? Don't you think he would have pointed it out? It boggles my mind that people think they can find some simple error in relativity that will invalidate it after every major physicist and most mathematicians have been looking at it for 100 years. Do you really think the people who figured out laser, radar, atomic bombs, hydrogen bombs, computer chips, LCD and plama TVs, GPS systems, ICBMs and sent people to the moon are a bunch of doofusses? Really? You think you're smarter than all those guys?
**************************
NBH: With that in mind, one is led to wonder whether or not Einstein made other major errors in his use of higher mathematics as well, errors also gone unrecognized by his followers and reproduced by them over and over again in one form of another. Well, inspection of the literature reveals that this is indeed the case. Major errors have been committed in differential geometry that render the usual claims demonstrably false. A particular case in point is the so-called "Schwarzschild radius". The irrefutable fact is that the "Schwarzschild radius" is not even a distance, let alone a radius of anything, in the related manifold. The proof is not difficult, but the proponents of black holes have not realized their error. This simple geometrical fact is sufficient to completely invalidate all claims that black holes are predicted by General Relativity.
So, if you think that the "Schwarzschild radius" is a radius or a distance of any sort in the related manifold, you are in the same boat as the proponents of black holes. They have never proven that it is a distance in the manifold, but that has never stopped them from saying (incorrectly) that the "Schwarzschild radius" is the radius of a spherical surface of one-way membrane they call the event horizon of a black hole. They can however, never prove their assertion, because the irrefutable fact is that the "Schwarzschild radius" is not a distance of any kind in the related manifold.
cinci: I don't claim to be an expert in either general relativity of black holes. The space near a black hole is distorted and measuring a distance there is strange, particularly inside the event horizon where distance probably doesn't make physical sense. But I think from a distance the event horizon would cover an area of the sky described by the radius of the event horizon.
******************************************
Jose Rodriguez
Re: Dingle's Challenge to the Theory of Special Relativity
Paging down to the first diagram, one sees two pairs of black parallel lines. This diagram makes Dingles case, unbeknown to Kevin Brown, the author of the site. It is true that Dingle's explanation is muddled, but his accusation is true.
What the diagram actually shows is three frames of reference: the third is the background of the diagram. In that frame of reference, the two moving frames are symmetrical. Thus, from that perspective, the "twins" age identically.
If the diagram were to portray the situation from either moving frame's reference, then, that frame's "time" line would be perpendicular to the distance line. (the horizontal line.) Then, the opposite "time" line would be shown as having even more slope. Only then can the two situations be shown to be unsymmetrical.
Thus it can be seen by a thinking soul, that either twin ages more than the other, since either frame can be taken as the "stationary" frame.
Which creates an absurdity.
Jose Rodriguez
Iggy liggety shows his ignorance
April 12 2009, 8:13 PM
I know what they are, Iggy. You have no concept of what a frame of reference is. You take lessons from mini max. Keep up with the ad homenim, fool, and I'll suggest that you be quarantined again.
Jose Rodriguez
So what are they?
April 12 2009, 10:48 PM
Iggy, since you are obviously the result of incestuous relations by your mother with a tape worm, I can hardly wait for your response. What Are they then?
Ted
Re: So what are they?
April 12 2009, 11:32 PM
You don't know? You are an idiot, Kurt Kunts
cincirob
Re: Dingle's Challenge to the Theory of Special Relativity
April 13 2009, 12:11 AM
Jose: Thus it can be seen by a thinking soul, that either twin ages more than the other, since either frame can be taken as the "stationary" frame.
cinci: Yes, either frame may be considered stationary and when one considers that one frame is stationary, the other frame is moving. But what you cannot do is consider both of them to be stationary at the same time except in the unique situation where they are at rest relative to each other. And in that unique situation they run at the same rate. So the thinking soul should realize that both situations can occur at the same time. Besides, in the twin paradox, a particular frame is designated as the starting point from which one clock departs and to which it returns. If you stick to the scenario there's no confusion.
In any case, Dingle's example doesn't complete the twin paradox because he doesn't have a two clocks move apart and then return totheir orginal positions.
Jose Rodriguez
So what are they?
April 13 2009, 12:16 AM
Iggy:
"A" and "H" are moving to the left and "N" and "B" are moving to the right. The slanted lines are the supposed perpendicular "time" components of each member of either frame. They would be observed perpendicular from either respective frame. The author has neglected to show the perpendicular "time" coordinate of the third frame. This invisible time coordinate divides the two other systems into symmetrical diagrams, depicting equal times for the duration of time between the two twins.
you don't get it, tape worm morphidite.
Jose Rodriguez
Iggy liggety shows his ignorance
April 13 2009, 12:31 AM
You have nothing to teach, ignorant tape worm morphidite.
Jose Rodriguez
Iggy liggety shows his ignorance
April 13 2009, 12:49 AM
You don't know anything at all, Iggy. It is obvious that you didn't read the article. You are a waste of band width. Crawl back into your local intestine, tape worm.
Jose Rodriguez
Iggy liggety shows his ignorance
April 13 2009, 12:52 AM
Each of the paired points are in separate frames traveling in opposite directions, tapeworm. You don't get it. Dry up and blow away. You are useless.
noblackhole
Re: Dingle's Challenge to the Theory of Special Relativity
April 13 2009, 1:18 AM
cincirob,
Your response does not change the fact that Einstein committed a gross and fatal error in elementary schoolboy algebra in his book 'Relativity The Special and the General'. His argument there is unmitigated rot. This has nothing to do whatsoever with whether or not the Lorentz transformation is valid. So your citation of Einstein's 1905 paper is entirely irrelevant. The cited book by Einstein clearly attests to the fact that he did not there even do algebra aright. Consequently, it is legitimate to inquire as to whether or not he committed similar errors in other aspects of his theorizing. Investigation of the literature does indeed reveal gross errors in mathematics committed by Einstein and his followers, errors that completely invalidate their arguments. One example is the "Schwarzschild radius". It is easily proven, with just elementary differential geometry, that it is not even a distance in the so-called "Schwarzschild" spacetime (which is also easily proven not to be Schwarzschild's solution at all); but the relativists always treat it as such, maintaining that it is the radius of a sphere, the surface of which is the event horizon of a black hole. That claim is also unmitigated claptrap.
Another example of Einstein's concoction of meaningless mathematical symbols is his pseudo-tensor. His pseudo-tensor is easily proven to have no mathematical validity whatsoever, because it violates mathematics itself. Relativists nonetheless, thoughtlessly following Einstein, and rely upon this nonsensical contraption to localize gravitational energy, and to do other mathematical gymnastics from which they draw conclusions for physics - blissfully ignorant that they are just raising and lowering indexes on a demonstrable nonsense.
cinci: I don't claim to be an expert in either general relativity of black holes. The space near a black hole is distorted and measuring a distance there is strange, particularly inside the event horizon where distance probably doesn't make physical sense. But I think from a distance the event horizon would cover an area of the sky described by the radius of the event horizon.
This does not in any way answer the question I put to you. It is little more than an evasion. One can only wonder then on what basis you maintain your arguments when you freely admit that you don't even have an adequate knowledge of the subject. But still you persist, and say "But I think from a distance the event horizon would cover an area of the sky described by the radius of the event horizon", ignoring request to provide a proof that the radius of the event horizon is a radius in "Schwarzschild" spacetime to begin with. The relativists have never given a proof either - they just do as you have done; make substantiated claims (which are in fact easily proven completely false). Of your baseless assertions, I repeat once again, it is easily proven that the "Schwarzschild radius" is not even a distance, let alone a radial one, in "Schwarzschild" spacetime. This geometric fact is sufficient to completely subvert all claims that General Relativity predicts black holes. So in addition, I now ask you, will you confirm for yourself that the claims by the relativists for "Schwarzschild" radii are utterly false, or will you ignore the facts and continue to propound that which you have freely admitted you do not yet understand?
Jose Rodriguez
cinci's Intransigent Thinking
April 13 2009, 1:23 AM
cinci: Yes, either frame may be considered stationary and when one considers that one frame is stationary, the other frame is moving. But what you cannot do is consider both of them to be stationary at the same time except in the unique situation where they are at rest relative to each other.
Jose: Your reasoning is goofy, cinci. Both frames, or observers in both frames can think of the other frame as moving and theirs as stationary at the same time. The third party frame can see that neither party is aging asymmetrically since they are traveling the same speed in opposite directions. Besides, your theory doesn't depend on the direction of travel. It just states that the moving twin is younger. The truth is that the time lost on the way apart is made upon the trip back, making no age difference.
Jose Rodriguez
cinci's Intransigent Thinking
April 13 2009, 1:23 AM
cinci: Yes, either frame may be considered stationary and when one considers that one frame is stationary, the other frame is moving. But what you cannot do is consider both of them to be stationary at the same time except in the unique situation where they are at rest relative to each other.
Jose: Your reasoning is goofy, cinci. Both frames, or observers in both frames can think of the other frame as moving and theirs as stationary at the same time. The third party frame can see that neither party is aging asymmetrically since they are traveling the same speed in opposite directions. Besides, your theory doesn't depend on the direction of travel. It just states that the moving twin is younger. The truth is that the time lost on the way apart is made upon the trip back, making no age difference.
Ted
Re: Dingle's Challenge to the Theory of Special Relativity
April 13 2009, 1:25 AM
Kurt Kunts: Each of the paired points are in separate frames traveling in opposite directions,
Ted: No, stupid Kunts. You don't know relativity , so you can't critcisize it.
Ted
Re: Dingle's Challenge to the Theory of Special Relativity
April 13 2009, 1:27 AM
Roger Dumberton: Another example of Einstein's concoction of meaningless mathematical symbols is his pseudo-tensor. His pseudo-tensor is easily proven to have no mathematical validity whatsoever, because it violates mathematics itself.
Jose Rodriguez
Iggy liggety shows his ignorance
April 13 2009, 1:46 AM
Ted: No, stupid Kunts (sic). You don't know relativity , so you can't critcisize (sic) it.
Jose: Iggy liggety, self appointed smug smarmy syphilitic tapeworm judge. Keep up with your display of ignorance and uneducated drivel. I don't need to criticize it. It falls on the weight of its own silly propositions.
Ted
Re: Dingle's Challenge to the Theory of Special Relativity
April 13 2009, 3:17 AM
Curt Kunts: if it failed, how come no one heard of that? The fact that it fails daily in your defective brain, is that it?
cincirob
Re: Dingle's Challenge to the Theory of Special Relativity
April 13 2009, 9:27 AM
cinci: Yes, either frame may be considered stationary and when one considers that one frame is stationary, the other frame is moving. But what you cannot do is consider both of them to be stationary at the same time except in the unique situation where they are at rest relative to each other.
Jose: Your reasoning is goofy, cinci. Both frames, or observers in both frames can think of the other frame as moving and theirs as stationary at the same time.
cinci: I'm goofy? Your statment is: If A is stationary and if B is stationary each of their clocks runs slower than the other. Now that's goofy. If they are both stationary their clocks run the same.
If you think this makes sense, then let's see you apply the Lorentz transformations to the twin paradox and prove that it is. I don't think you can do it.
**********************
Jose: The third party frame can see that neither party is aging asymmetrically since they are traveling the same speed in opposite directions.
cinci: If you think invoking a third party changes anything, then show it by Lorentz analysis. The third party will agree with the outcome of the experiment.
******************************
Jose: Besides, your theory doesn't depend on the direction of travel. It just states that the moving twin is younger. The truth is that the time lost on the way apart is made up on the trip back, making no age difference.
cinci: The time lost on the way out is
(delta t astronaut) = (delta t Earth)(1 - v^2/c^2)^.5
The time lost on the way back is
(delta t astronaut) = (delta t Earth)(1 - (-v)^2/c^2)^.5 = (delta t Earth)(1 - v^2/c^2)^.5
The astronaut loses time in both directions. Come on Jose, if you going to critique the theory you have to do better than this.
*******************************
Jose Rodriguez
Re: Dingle's Challenge to the Theory of Special Relativity
April 13 2009, 10:27 AM
cinci: "The astronaut loses time in both directions. Come on Jose, if you going to critique the theory you have to do better than this."
Jose: The moving clock has to "lose time" (i.e. slow down) as it recedes, since it is moving into the past of the stationary clock. It then, as it approaches the stationary clock, must "gain time" (i.e. speed up) as it moves into the present of the stationary clock. There are two parameters to the "time" each clock "keeps." One is the regulation or duration between cycles of the "ticks." The other is the setting of the epoch or the "time of day." If the regulation is running slow in a particular clock, the epoch reading of that clock will also fall behind as compared to another reference clock. (the reference clock will be seen to "run fast.")
As the receding clock reflects its signal to the observer at the stationary clock, the signal's energy is reduced, since it is red shifted. You could say that the lost energy "made" the clock run slow. As the moving clock approaches the stationary observer, Its signal is blue shifted. Now the signal has more energy, making the moving clock run fast.
An object approaching an observer at the speed of light cannot be observed until it gets to the observer. It will then appear in a burst of blue shifted light, and then immediately disappear in undetectable red shifted light as it recedes.
cincirob
Re: Dingle's Challenge to the Theory of Special Relativity
April 13 2009, 2:01 PM
cinci: "The astronaut loses time in both directions. Come on Jose, if you going to critique the theory you have to do better than this."
Jose: The moving clock has to "lose time" (i.e. slow down) as it recedes, since it is moving into the past of the stationary clock.
cinci: It's moving away in space, not time. It's just running slower.
********************
Jose: It then, as it approaches the stationary clock, must "gain time" (i.e. speed up) as it moves into the present of the stationary clock. There are two parameters to the "time" each clock "keeps." One is the regulation or duration between cycles of the "ticks." The other is the setting of the epoch or the "time of day." If the regulation is running slow in a particular clock, the epoch reading of that clock will also fall behind as compared to another reference clock. (the reference clock will be seen to "run fast.")
cinci: Come on, man...think. Let's say the moving clock goes out to Jupiter and stops. Now we put a clock at Jupiter and send the moving clock back to Earth. According to you the moving clock should be running slower that the Jupiter clock but faster than the Earth clock. Really?
*****************************
Jose: As the receding clock reflects its signal to the observer at the stationary clock, the signal's energy is reduced, since it is red shifted. You could say that the lost energy "made" the clock run slow. As the moving clock approaches the stationary observer, Its signal is blue shifted. Now the signal has more energy, making the moving clock run fast.
cinci: Weakening a signal doesn' make it slower. Or maybe you're talking about Doppler effect which isn't clock rate either. You are very confused.
*******************
Jose: An object approaching an observer at the speed of light cannot be observed until it gets to the observer. It will then appear in a burst of blue shifted light, and then immediately disappear in undetectable red shifted light as it recedes.
cinci: Objects don't travel at c.
*******************
Anonymous
Re: Dingle's Challenge to the Theory of Special Relativity
April 13 2009, 3:29 PM
cinci: Objects don't travel at c.
your usual nonsense; photons do.
cincirob
Re: Dingle's Challenge to the Theory of Special Relativity
April 13 2009, 4:36 PM
cinci: Objects don't travel at c.
Anon: your usual nonsense; photons do.
cinci: Nah, it's your usual nonsense of taking comments out of context. Jose was not talking about photons.
**************************
Anonymous
Re: Dingle's Challenge to the Theory of Special Relativity
April 13 2009, 6:33 PM
no, it was not specified; its your usual nonsense of being incoherent.
Jose Rodriguez
cinci's Intransigence
April 13 2009, 10:53 PM
cinci: "The astronaut loses time in both directions. Come on Jose, if you going to critique the theory you have to do better than this."
Jose: You are arrogant and indolent. The clock signal is not the clock. The clock signal does not keep ticking as the signal moves through space. A signal that is transmitted at a given frequency, and received at a higher frequency will also have shorter durations between ticks. Its regulation will be sped up. This is undeniable. You are simply incapable of understanding the fact.
cinci: "It's moving away in space, not time. It's just running slower."
Jose: OK, Mr Knowitall. The time delay (Latency) is increasing as the distance increases. This is why the clock seems to be running slow. Everyone has their own present. However, everyone else's present is in the past of any one's present. This is apparently something else that you can never assimilate.
cinci: "Come on, man...think. Let's say the moving clock goes out to Jupiter and stops. Now we put a clock at Jupiter and send the moving clock back to Earth. According to you the moving clock should be running slower that (sic) the Jupiter clock but faster than the Earth clock. Really?"
Jose: No Mr Knowitall, The observer at the Jupiter clock sees the returning (receding) clock running slow. The Earth observer sees the returning clock running fast. The moving clock is running at the same rate that it always did. It is the received frequencies that make the clock signal seem to run fast on approach and slow on recession.
cinci: "Weakening a signal doesn' make it slower. Or maybe you're talking about Doppler effect which isn't clock rate either. You are very confused."
Jose: The "clock rate is delivered by the light frequency. the Doppler shift has a very real effect on therate of the received demodulated clock rate.
cinci: "Objects don't travel at c."
Jose:If you are convinced of that, you would never notice if they did.
cincirob
Re: Dingle's Challenge to the Theory of Special Relativity
April 13 2009, 11:08 PM
Anon: no, it was not specified; its your usual nonsense of being incoherent.
cinci: "it" was not specified? And just what would "it" refer to? You haven't specified what you are talking about so it's just more of your nonsense.
If you're going to critque me for omitting specificity you should make sure you don't commit a worst omission in your critique.
It looks like you have completley given up on selling any of your erroneous physics ideas and are concentrating on grammar and semantics. My advice is: Don't bother. You aren't that good at it.
**********************************
cincirob
Re: Dingle's Challenge to the Theory of Special Relativity
April 13 2009, 11:34 PM
cinci: "The astronaut loses time in both directions. Come on Jose, if you going to critique the theory you have to do better than this."
Jose: You are arrogant and indolent. The clock signal is not the clock. The clock signal does not keep ticking as the signal moves through space. A signal that is transmitted at a given frequency, and received at a higher frequency will also have shorter durations between ticks. Its regulation will be sped up. This is undeniable. You are simply incapable of understanding the fact.
cinci: cinci: Arrogant and indolent I may be, but I know the relative aging of the twins has nothing to do with Doppler effect as youseem to think.
*******************************
cinci: "It's moving away in space, not time. It's just running slower."
Jose: OK, Mr Knowitall. The time delay (Latency) is increasing as the distance increases. This is why the clock seems to be running slow. Everyone has their own present. However, everyone else's present is in the past of any one's present. This is apparently something else that you can never assimilate.
cinci: Again, time delay has nothing to do with relative aging. The moving clock actually does run slower. You're confusing it with what one would see from a distance which has nothing to do with it.
********************************
cinci: "Come on, man...think. Let's say the moving clock goes out to Jupiter and stops. Now we put a clock at Jupiter and send the moving clock back to Earth. According to you the moving clock should be running slower that (sic) the Jupiter clock but faster than the Earth clock. Really?"
Jose: No Mr Knowitall, The observer at the Jupiter clock sees the returning (receding) clock running slow. The Earth observer sees the returning clock running fast. The moving clock is running at the same rate that it always did. It is the received frequencies that make the clock signal seem to run fast on approach and slow on recession.
cinci: Get out of Doppler mode. You need to deal with actual clock rates to understand this.
*********************************
cinci: "Weakening a signal doesn' make it slower. Or maybe you're talking about Doppler effect which isn't clock rate either. You are very confused."
Jose: The "clock rate is delivered by the light frequency. the Doppler shift has a very real effect on therate of the received demodulated clock rate.
cinci: Doppler effect has no effect on clock rate. It only affects what you see.
******************************
cinci: "Objects don't travel at c."
Jose: If you are convinced of that, you would never notice if they did.
cinci: Well I might not because I'm not running such an experiment. Are youtrying to say you've seen it?
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noblackhole
Re: Dingle's Challenge to the Theory of Special Relativity
April 14 2009, 12:32 AM
cincirob has no credibility. He has admitted that Einstein's derivation of the Lorentz transformation in his book 'Relativity the Special and the General Theory' involves nothing but identities with zero, and is therefore unmitigated claptrap, a gross violation of the rules of schoolboy algebra. But he chooses to ignore this and its implications.
cincirob also reasserts his demonstrable falsehoods concerning the "Schwarzschild radius" despite the latter being easily proven not even a distance let alone a radius in the "Schwarzschild" manifold; he as admitted that he doesn't understand the theory that he pontificates upon; and he has admitted that he lacks the necessary mathematical skills relevant to the Theory of Relativity. One can only wonder how cincirob thinks can maintain his 'authority' on such matters.
Ted
Re: Dingle's Challenge to the Theory of Special Relativity
April 14 2009, 2:13 AM
Roger Dumberton: Einstein's derivation of the Lorentz transformation in his book 'Relativity the Special and the General Theory' involves nothing but identities with zero
Ted: ...only in the mind of an ignorant idiot like Roger Dumberton....
Jose Rodriguez
cinci's Intransigence
April 14 2009, 12:54 PM
cinci: Arrogant and indolent I may be, but I know the relative aging of the twins has nothing to do with Doppler effect as youseem (sic) to think.
Jose: Yes: You know that because you can not think for yourself.
cinci: Again, time delay has nothing to do with relative aging. The moving clock actually does run slower. You're confusing it with what one would see from a distance which has nothing to do with it.
Jose: What other way is the signal from the moving clock made available to the stationary observer? The signal has to pass through the medium of light. Therefore the Doppler shift affects the signal. You are wrong. The H-K experiment was a joke.
cinci: Get out of Doppler mode. You need to deal with actual clock rates to understand this.
Jose: I am dealing with actual clock rates as seen at a distance. Moving clocks cannot be directly compared with stationary clocks. This is the problem with relativity of simultaneity.
Anonymous
Re: Dingle's Challenge to the Theory of Special Relativity
April 14 2009, 7:36 PM
cinci>>My advice is: Don't bother.
your advice is worthless
cincirob
Re: Dingle's Challenge to the Theory of Special Relativity
April 16 2009, 12:36 AM
cinci: Arrogant and indolent I may be, but I know the relative aging of the twins has nothing to do with Doppler effect as you seem (sic) to think.
Jose: Yes: You know that because you can not think for yourself.
cinci: Again, time delay has nothing to do with relative aging. The moving clock actually does run slower. You're confusing it with what one would see from a distance which has nothing to do with it.
Jose: What other way is the signal from the moving clock made available to the stationary observer? The signal has to pass through the medium of light. Therefore the Doppler shift affects the signal. You are wrong. The H-K experiment was a joke.
cinci: Get out of Doppler mode. You need to deal with actual clock rates to understand this.
Jose: I am dealing with actual clock rates as seen at a distance. Moving clocks cannot be directly compared with stationary clocks. This is the problem with relativity of simultaneity.
cinci: Why don't you do a few calculations and prove whatever point you're trying to make?
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Anonymous
Re: Dingle's Challenge to the Theory of Special Relativity
April 16 2009, 1:59 AM
E-mc˛ = 0
cincirob
Re: Dingle's Challenge to the Theory of Special Relativity
April 16 2009, 5:16 AM
Anon: E-mc˛ = 0
cinci: Good! I assume this means you now accept relativity.
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Jose Rodriguez
cinci's Intransigence
April 16 2009, 11:58 AM
cinci: Why don't you do a few calculations and prove whatever point you're trying to make?
Jose: Why don't you do a little thinking, and try to understand the points I've all ready made. You can't understand because spacetime has crippled your thinking. "Tensor" math has to be used because the imaginary "time axis" has to be perpendicular to the three Cartesian coordinates, simultaneously. Latency is all there is. A pulse of collimated light striking an oppositely moving surface, perpendicular to it, is a line segment, not a point.
cincirob
Re: Dingle's Challenge to the Theory of Special Relativity
April 17 2009, 4:11 AM
cinci: Why don't you do a few calculations and prove whatever point you're trying to make?
Jose: Why don't you do a little thinking, and try to understand the points I've all ready made. You can't understand because spacetime has crippled your thinking. "Tensor" math has to be used because the imaginary "time axis" has to be perpendicular to the three Cartesian coordinates, simultaneously. Latency is all there is. A pulse of collimated light striking an oppositely moving surface, perpendicular to it, is a line segment, not a point.
cinci: If you think it takes tensors to understand the twin paraodox then you don't understand the twin paradox. Why do antirelativsts avoid mathematical analysis like the plague?
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>>>, you moron, your French is not bad: "Résumé : Einstein a prédit une variation de l'énergie des photons au voisinage d'un champ de gravité, cette variation étant directement proportionnelle ŕ la distance de la source du champ gravitationnel. Au début des années 60, Pound et Rebka (Phys. Rev. Lett. 3, 439 (1959)) ont tenté de vérifier expérimentalement cette prédiction. L'expérience a été reprise avec plus de précision par Pound et Snider(Phys. Rev. Lett. 13, 539 (1964)). Plus tard, Vessot (Phys. Rev. Lett. 45, 2081 (1980)) l'a refaite dans l'espace avec une précision trčs améliorée. Ici, nous proposons une nouvelle approche au problčme qui procéderait ŕ l'inverse, en descendant vers le centre de la Terre. Notre approche est moins onéreuse, introduit des effets de plus grande amplitude et prévoit un nouvel effet du deuxičme ordre."
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