ralfcis Posted May 27, 2019 Author Report Share Posted May 27, 2019 (edited) Oh yeah it appears in Wiki therefore it must be true. Doubly true because not only are there words but there's a picture of a spray of massive particles to look at. I don't know what's wrong with you wiki-heads feeling the compulsion to constantly go off topic chasing squirrels. The topic here is what is relative velocity. Now, Victor, you correctly said c+c is a CLOSING SPEED. Please inform GAHD that closing speed is not a relative velocity and that the max relative velocity is limited to c not 2c. Can you do that for me please. Since he can't google the words closing speed and since you like calling up useless wiki articles, could you post one on closing speed for GAHD to take a gander. PS. I once sent money supporting Wiki but never again. It's a serious threat to world intelligence. Edited May 27, 2019 by ralfcis Quote Link to comment Share on other sites More sharing options...
Vmedvil2 Posted May 27, 2019 Report Share Posted May 27, 2019 (edited) Oh yeah it appears in Wiki therefore it must be true. Doubly true because not only are there words but there's a picture of a spray of massive particles to look at. I don't know what's wrong with you wiki-heads feeling the compulsion to constantly go off topic chasing squirrels. The topic here is what is relative velocity. Now, Victor, you correctly said c+c is a CLOSING SPEED. Please inform GAHD that closing speed is not a relative velocity and that the max relative velocity is limited to c not 2c. Can you do that for me please. Since he can't google the words closing speed and since you like calling up useless wiki articles, could you post one on closing speed for GAHD to take a gander. PS. I once sent money supporting Wiki but never again. It's a serious threat to world intelligence. What if it were the "Closing relative velocity" of the two? Closing speed basically is equivalent to something called Relative Interval or time-space interval, since all interactions are relative in relativistic physics. The Interval velocity of the two particles is indeed 1.9994C but that actual velocity of the two is ±1.9994C / 2 per particle if that makes sense, if both are traveling at the same velocity. Edited May 27, 2019 by VictorMedvil Quote Link to comment Share on other sites More sharing options...
ralfcis Posted May 27, 2019 Author Report Share Posted May 27, 2019 Are you a politician Victor? I asked a straight question. You have heard of the relativistic velocity combination law that adds two relative velocities such that c can't be exceeded. That formula calculates the resultant relative velocity. If the formula says the resultant relative velocity = .99999997 c and the formula for closing speed = 1.9994 c that must mean closing speed is not a relative velocity. So no more semantics please, just a straight answer to my question. Quote Link to comment Share on other sites More sharing options...
Vmedvil2 Posted May 27, 2019 Report Share Posted May 27, 2019 (edited) Are you a politician Victor? I asked a straight question. You have heard of the relativistic velocity combination law that adds two relative velocities such that c can't be exceeded. That formula calculates the resultant relative velocity. If the formula says the resultant relative velocity = .99999997 c and the formula for closing speed = 1.9994 c that must mean closing speed is not a relative velocity. So no more semantics please, just a straight answer to my question.Well, I would have said , Bingo! like Botan, but you aren't completely correct, GAHD is speaking about the interval between the two like I said the two particles are separated by a Velocity Interval of 1.9994C which is still relativistic just another equation which no one seems to remember right now, basically it is the velocity with respect to the Proper time. So, Neither of you are exactly wrong nor correct. Edited May 27, 2019 by VictorMedvil Quote Link to comment Share on other sites More sharing options...
ralfcis Posted May 27, 2019 Author Report Share Posted May 27, 2019 Your little math smoke bomb has no effect on me. What is Y in this case when both are travelling towards each other at c? Quote Link to comment Share on other sites More sharing options...
Vmedvil2 Posted May 27, 2019 Report Share Posted May 27, 2019 (edited) Your little math smoke bomb has no effect on me. What is Y in this case when both are travelling towards each other at c? Then both would be traveling at C + C or 2C as the interval between them, like I said neither of you two are wrong about this just looking from a different frame, one the frame of the particle from a observer then the other from the particle itself from its frame both solutions are accounted for in that "Math smoke bomb" with the same solution being Delta Tau. Edited May 27, 2019 by VictorMedvil Quote Link to comment Share on other sites More sharing options...
ralfcis Posted May 27, 2019 Author Report Share Posted May 27, 2019 (edited) What is the value of Y at c, well it's infinity. So plugging infinity into your formula does not result in a division by 2. And there's no reason to include all 3 distance dimensions. All your math boils down to is t=Yt'. That doesn't support anything you've said. Just use numbers and show me what you're saying. I assume this is what you're trying to pass off: t'= t/YYou're somehow assuming Y = 2, t = 1.9994 so t' (or delta tau in your case) = .9997 which is meaningless. If you can read STD's I can show you how 1/3c + 1/3c = .6c not 2/3c. The exact same logic can be extrapolated on an STD where c + c =c not 2c. If you can't read them then my proof would be for nothing. Edited May 27, 2019 by ralfcis Quote Link to comment Share on other sites More sharing options...
Vmedvil2 Posted May 27, 2019 Report Share Posted May 27, 2019 (edited) What is the value of Y at c, well it's infinity. So plugging infinity into your formula does not result in a division by 2. And there's no reason to include all 3 distance dimensions. All your math boils down to is t=Yt'. That doesn't support anything you've said. Just use numbers and show me what you're saying. I assume this is what you're trying to pass off: t'= t/YYou're somehow assuming Y = 2, t = 1.9994 so t' (or delta tau in your case) = .9997 which is meaningless. If you can read STD's I can show you how 1/3c + 1/3c = .6c not 2/3c. The exact same logic can be extrapolated on an STD where c + c =c not 2c. If you can't read them then my proof would be for nothing. well i'll put it this way let's say you are taking variable ΔY as your Variable then Y2 - Y1 would be the formula with this example of .99987 C would be .99987 -- .99987 being .99987C + .99987C = 1.9994C = ΔY as the ΔY solution, but the solution to Y1 = Y2 = ±.99987C, Do you now see how both are correct? When you look at it you are looking at Y when GAHD looks at it he is looking at ΔY like I said the interval ΔY views it different than as Frame Y1 or Y2,but both Y1 or Y2 or ΔY are equivalent to Δτ. Edited May 27, 2019 by VictorMedvil Quote Link to comment Share on other sites More sharing options...
ralfcis Posted May 27, 2019 Author Report Share Posted May 27, 2019 Y is gamma, I don't know what your Y means. Gamma is infinity and you're saying it's a velocity and then you're saying it's delta tau? You're not answering any of my questions. Quote Link to comment Share on other sites More sharing options...
Vmedvil2 Posted May 27, 2019 Report Share Posted May 27, 2019 (edited) Y is gamma, I don't know what your Y means. Gamma is infinity and you're saying it's a velocity and then you're saying it's delta tau? You're not answering any of my questions. Oh, you wanted it in terms of Gamma(γ) I thought you wanted it was Interval ΔY or Frame Y, my Y is like ΔX' and X, but the Y axis and not X axis. Here is the translation equation below. Just replace X with Y and L with ΔY in these equations or vice versa, I thought you wanted it upon the Y 4-D Coordinate Axis. Edited May 27, 2019 by VictorMedvil Quote Link to comment Share on other sites More sharing options...
ralfcis Posted May 27, 2019 Author Report Share Posted May 27, 2019 Again so what? t=Yt' and x'=Yx. Yes, these are standard equations of relativity. What do they have to do with the topic of relative velocity vs closing speed being relativistic. This just looks like diversionary smoke bombs to me. Quote Link to comment Share on other sites More sharing options...
Dubbelosix Posted May 27, 2019 Report Share Posted May 27, 2019 006, I do mean to be harsh. Are you going to respond or just slink away like everyone else. Do you stand by your statement that I don't understand relative velocity? Why should I? I have been reading these posts, largely containing gibberish, with no mention of the velocity addition formula for relativity. There are so many things wrong in this post, its not even worth covering. Quote Link to comment Share on other sites More sharing options...
ralfcis Posted May 27, 2019 Author Report Share Posted May 27, 2019 (edited) Well that pretty much explains the problems with understanding relativity here. No one, except Popeye (where is he?), knows what relative velocity means and they have no intention of finding out. The thing I can't understand is who is reading my posts and why? Edited May 27, 2019 by ralfcis Quote Link to comment Share on other sites More sharing options...
Dubbelosix Posted May 27, 2019 Report Share Posted May 27, 2019 In all my years, I have never seen a formula permit c + c = 2c. Which is why i brought up the relativistic additional formula for velocities. Quote Link to comment Share on other sites More sharing options...
Vmedvil2 Posted May 27, 2019 Report Share Posted May 27, 2019 (edited) This is the formula I got from Hyperphysics, dubbel which minus a negative is plus. (http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/tdil.html), I don't often use this formula I will admit for two moving frames maybe you can explain it better. Edited May 27, 2019 by VictorMedvil Quote Link to comment Share on other sites More sharing options...
Dubbelosix Posted May 27, 2019 Report Share Posted May 27, 2019 Hyperphysics never gives enough quality to challenge a wiki page, https://en.wikipedia.org/wiki/Velocity-addition_formula Quote Link to comment Share on other sites More sharing options...
ralfcis Posted May 27, 2019 Author Report Share Posted May 27, 2019 Until today I never knew there was such a thing as formula fetishism. Quote Link to comment Share on other sites More sharing options...
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