SR/GR use absolute time to synchronize the GPS clocks with the ground clock.
in [Physics]
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From: kenseto on 11 May 2010 08:29 On May 10, 1:10 pm, Sam Wormley <sworml...(a)gmail.com> wrote: > On 5/10/10 8:24 AM, kenseto wrote: > > > > > > > On May 9, 11:31 am, Sam Wormley<sworml...(a)gmail.com>  wrote: > >> On 5/9/10 9:32 AM, kenseto wrote: > > >>> ROTFLOL....you are the one who need to learn SR....what you said is an > >>> erroneous assumption of the PoR that every SR observer is in a state > >>> of absolute rest. This is wrong. The rate of two relative clock A and > >>> B is as follows: > >>> 1. A is running fast compared to B. > >>> 2. B is running slow compared to A. > > >>   Wrong again Seto! > > >>   Let's be a bit more precise here: > > >>   Assume that A and B have identical atomic clocks. That means they > >>   tick at the same rate. Now let us suppose that A and B have relative > >>   motion, such that their velocity with respect to each other, v>  0, > >>   and that dv/dt = 0 . "Closing speed" or "Separation speed", if you > >>   like. > > >>   Correcting for any Doppler shift, A measures B's time interval as > >>    ât_B' = γ ât_B > > >>   and B measures A's time interval as > >>    ât_A' = γ ât_A > > >>   where ât represent a time interval, v is the relative velocity > >>   between A and B, and γ = 1/â(1-v^2/c^2) . > > >>   Therefore, A measures B's time interval to be longer than her own. > >>   And B measures A's time interval to be longer than his own. > > > No wormy....there is no measurement at all. A predicts that B's clock > > is running at a rate of 1/gamma or gamma....and B predicts that A is > > running at a rate of 1/gamma or gamma. Only one of A's prediction is > > correct and only one of B's prediction is correct. So if A is truly > > running at a faster rate of gamma than B then B is truly running at a > > slower rate of 1/gamma than A. > >   Oh no, Seto! Major misunderstanding on your part! > >   Correcting for any Doppler shift, A measures B's time interval as >    ât_B' = γ ât_B No idiot A predicts D's time interval....ât_B = γ ât_A > >   and B measures A's time interval as >    ât_A' = γ ât_A No idiot B predicts A's time interval.....ât_A = γ ât_B > >   where ât represent a time interval, v is the relative velocity >   between A and B, and γ = 1/â(1-v^2/c^2) . > >   You REALLY DON'T UNDERSTAND relativity at all! Do you even know >   what the word "relative" means? Looks like its you who don't understand relativity. Ken Seto - Hide quoted text - > > - Show quoted text -
From: kenseto on 11 May 2010 09:03 On May 11, 8:29 am, kenseto <kens...(a)erinet.com> wrote: > On May 10, 1:10 pm, Sam Wormley <sworml...(a)gmail.com> wrote: > > > > > > > On 5/10/10 8:24 AM, kenseto wrote: > > > > On May 9, 11:31 am, Sam Wormley<sworml...(a)gmail.com>  wrote: > > >> On 5/9/10 9:32 AM, kenseto wrote: > > > >>> ROTFLOL....you are the one who need to learn SR....what you said is an > > >>> erroneous assumption of the PoR that every SR observer is in a state > > >>> of absolute rest. This is wrong. The rate of two relative clock A and > > >>> B is as follows: > > >>> 1. A is running fast compared to B. > > >>> 2. B is running slow compared to A. > > > >>   Wrong again Seto! > > > >>   Let's be a bit more precise here: > > > >>   Assume that A and B have identical atomic clocks. That means they > > >>   tick at the same rate. Now let us suppose that A and B have relative > > >>   motion, such that their velocity with respect to each other, v>  0, > > >>   and that dv/dt = 0 . "Closing speed" or "Separation speed", if you > > >>   like. > > > >>   Correcting for any Doppler shift, A measures B's time interval as > > >>    ât_B' = γ ât_B > > > >>   and B measures A's time interval as > > >>    ât_A' = γ ât_A > > > >>   where ât represent a time interval, v is the relative velocity > > >>   between A and B, and γ = 1/â(1-v^2/c^2) . > > > >>   Therefore, A measures B's time interval to be longer than her own. > > >>   And B measures A's time interval to be longer than his own. > > > > No wormy....there is no measurement at all. A predicts that B's clock > > > is running at a rate of 1/gamma or gamma....and B predicts that A is > > > running at a rate of 1/gamma or gamma. Only one of A's prediction is > > > correct and only one of B's prediction is correct. So if A is truly > > > running at a faster rate of gamma than B then B is truly running at a > > > slower rate of 1/gamma than A. > > >   Oh no, Seto! Major misunderstanding on your part! > > >   Correcting for any Doppler shift, A measures B's time interval as > >    ât_B' = γ ât_B > > No idiot A predicts D's time interval....ât_B = γ ât_A Sorry A predicts B's time interval for an inertval of delta(t_A) in his frame as follows: delta(t_B)= delta(t_A)/gamma_a > > > > >   and B measures A's time interval as > >    ât_A' = γ ât_A > > No idiot B predicts A's time interval.....ât_A = γ ât_B Sorry B predicts A's time interval for an inertval of delta(t_B) in his frame as follows: delta(t_A)= delta(t_B)/gamma_b Ken Seto > > > > >   where ât represent a time interval, v is the relative velocity > >   between A and B, and γ = 1/â(1-v^2/c^2) . > > >   You REALLY DON'T UNDERSTAND relativity at all! Do you even know > >   what the word "relative" means? > > Looks like its you who don't understand relativity. > > Ken Seto > > - Hide quoted text - > > > > > > > - Show quoted text -- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -
From: Michael Moroney on 11 May 2010 16:33 kenseto <kenseto(a)erinet.com> writes: >Sorry A predicts B's time interval for an inertval of delta(t_A) in >his frame as follows: >delta(t_B)= delta(t_A)/gamma_a .... >Sorry B predicts A's time interval for an inertval of delta(t_B) in >his frame as follows: >delta(t_A)= delta(t_B)/gamma_b Neither of which are what SR predicts. SR predicts that delta t' = gamma * delta t for both instances. Gamma = 1/sqrt(1-v^2/c^2) Note that with this formula, with v restricted to being a real number with an absolute value less than c, it is *impossible* for gamma to be less than 1. You claim that if A sees B's clock as slowed, B will c A's clock as fast. "A sees B's clock as slowed" means that delta T_b is larger than delta T_a. Since delta T_b = gamma * delta T_a, gamma must be a real mumber greater than 1. This is easily satisfied for some v where the absolute value of v is less than c. Now let's take a look at the reverse situation. You claim that B will c A's clock as fast. Since delta t_a = gamma * delta t_b and delta t_a < delta t_b, gamma must be a positive real number less than 1. So, let's see what value of v would do this. Gamma = 1/sqrt(1-v^2/c^2). Gamma^2 = 1/(1-v^2/c^2). (squaring both sides). Gamma^2 will also be a positive real number less than 1, since the squares of all positive real numbers less than 1 are also less than 1. 1/Gamma^2 = 1-v^2/c^2. (inverting both sides). Since gamma^2<1. 1/gamma^2 is greater than 1. 1/gamma^2-1 = -v^2/c^2. (subtract 1 from both sides) 1-1/gamma^2 = v^2/c^2 (negating both sides) sqrt(1-1/gamma^2) = v/c. (square root of both sides) c*sqrt(1-1/gamma^2) = v (Here we have an equation for finding a v satisfies your claim where one frame sees the other as running fast) So, let's see what happens for any gamma<1. Gammma^2<1. 1/(gamma^2) will be greater than 1. So... 1-1/gamma^2 *must* be a negative number. The square root of that negative number is your v... ....what is the square root of a negative number? An imaginary number! In other words, no real velocity v satisfies your claim that one frame will see the other frame's clock as running fast. I told you what you've been posting is all in your head, here's mathematical proof of that! :-) The only possible velocity v which produces a situation where one frame sees another frame's clock as running fast is an imaginary velocity! Maybe you can tell us of a situation where an imaginary velocity has some sort of meaning.
From: kenseto on 12 May 2010 10:57 On May 11, 4:33 pm, moro...(a)world.std.spaamtrap.com (Michael Moroney) wrote: > kenseto <kens...(a)erinet.com> writes: > >Sorry A predicts B's time interval for an inertval of delta(t_A) in > >his frame as follows: > >delta(t_B)= delta(t_A)/gamma_a > > ... > > >Sorry B predicts A's time interval for an inertval of delta(t_B) in > >his frame as follows: > >delta(t_A)= delta(t_B)/gamma_b > > Neither of which are what SR predicts. > > SR predicts that delta t' = gamma * delta t So are you claiming that the t' clock is running fast or are you claiming that 1 second on the t' clock is worth gamma second on the t clock? What about time dilation?....the t clock predicts that the t' clock is running 1/gamma slow? > > for both instances. > > Gamma = 1/sqrt(1-v^2/c^2) > > Note that with this formula, with v restricted to being a real number with > an absolute value less than c, it is *impossible* for gamma to be less > than 1. > > You claim that if A sees B's clock as slowed, B will c A's clock as > fast. "A sees B's clock as slowed" means that delta T_b is larger > than delta T_a. Since delta T_b = gamma * delta T_a, gamma must be a > real mumber greater than 1. This is easily satisfied for some v where > the absolute value of v is less than c. OK I agree to that that 1 second of B is worth gamma second of A. This is what IRT says also. But you can't turn around and claim that 1 second of A is worth gamma second of B. > > Now let's take a look at the reverse situation. You claim that B will c > A's clock as fast. Since delta t_a = gamma * delta t_b No this is wrong....delta t_a=delta t_b/gamma Ken Seto and delta t_a < > delta t_b, gamma must be a positive real number less than 1. So, let's > see what value of v would do this. Gamma = 1/sqrt(1-v^2/c^2). > > Gamma^2 = 1/(1-v^2/c^2). (squaring both sides). > > Gamma^2 will also be a positive real number less than 1, since the squares > of all positive real numbers less than 1 are also less than 1. > > 1/Gamma^2 = 1-v^2/c^2. (inverting both sides). > > Since gamma^2<1. 1/gamma^2 is greater than 1. > > 1/gamma^2-1 = -v^2/c^2. (subtract 1 from both sides) > > 1-1/gamma^2 = v^2/c^2 (negating both sides) > > sqrt(1-1/gamma^2) = v/c. (square root of both sides) > c*sqrt(1-1/gamma^2) = v (Here we have an equation for finding a v > satisfies your claim where one frame sees the other as running fast) > > So, let's see what happens for any gamma<1. Gammma^2<1. > 1/(gamma^2) will be greater than 1. So... 1-1/gamma^2 *must* be a > negative number. The square root of that negative number is your v... > ...what is the square root of a negative number? An imaginary number! > In other words, no real velocity v satisfies your claim that one frame > will see the other frame's clock as running fast. I told you what you've > been posting is all in your head, here's mathematical proof of that! :-) > The only possible velocity v which produces a situation where one frame > sees another frame's clock as running fast is an imaginary velocity! > > Maybe you can tell us of a situation where an imaginary velocity has some > sort of meaning.
From: PD on 13 May 2010 10:31
On May 13, 8:36 am, kenseto <kens...(a)erinet.com> wrote: > On May 12, 3:43 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On May 12, 2:37 pm, kenseto <kens...(a)erinet.com> wrote: > > > > On May 12, 10:52 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On May 12, 9:36 am, kenseto <kens...(a)erinet.com> wrote: > > > > > > On May 11, 9:00 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > On May 11, 7:44 am, kenseto <kens...(a)erinet.com> wrote: > > > > > > > > On May 10, 11:26 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > > > On May 9, 9:08 am, kenseto <kens...(a)erinet.com> wrote: > > > > > > > > > > On May 7, 1:05 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > > > > > On May 7, 8:10 am, kenseto <kens...(a)erinet.com> wrote: > > > > > > > > > > > > > > > > Hey idiot then why did theyusethe SR math to calculate the SR > > > > > > > > > > > > > > > effect??? > > > > > > > > > > > > > > > Because, Ken, there is more than one "SR effect" and they apply in > > > > > > > > > > > > > > different situations. The effect you have been asking about is mutual > > > > > > > > > > > > > >timedilation. > > > > > > > > > > > > > > The you should go and argue with your SR brother Moroney... > > > > > > > > > > > > > Ken, how many times have I told you to stop trying to learn relativity > > > > > > > > > > > > by listening to people on usenet, and start reading a REAL BOOK????? > > > > > > > > > > > > Hey I am not trying to learn SR on usenet. I already know more about > > > > > > > > > > > relativity than all of you combined. > > > > > > > > > > > There's no evidence of that, Ken. All the stuff you've recited about > > > > > > > > > > SR is repetition of stuff you've heard on usenet or stuff you've made > > > > > > > > > > up in your own head. > > > > > > > > > > > > That's why I was able to come up > > > > > > > > > > > with a superset of relativity that includes SR as a subset. > > > > > > > > > > > What I was trying to do is to point that you runts of the SRians are > > > > > > > > > > > making contradictory statement and wilds claims that are not supported > > > > > > > > > > > by experiments. > > > > > > > > > > > Ken, even if there are statements on usenet that are in conflict with > > > > > > > > > > each other about SR, that does not mean there is a conflict in SR. > > > > > > > > > > What it means is that you will not get reliable information about SR > > > > > > > > > > from usenet. > > > > > > > > > > So does that mean what you and your runt SR brothers wrote on the > > > > > > > > > usenet were just bullshits? > > > > > > > > > What I'm saying, Ken, is that you cannot get a good, consistent > > > > > > > > presentation about SR on usenet. You need to abandon that as your > > > > > > > > source of information and get to a book or six, pronto. > > > > > > > > What one get from the text book are the same garbage as in the usenet. > > > > > > > That's not true, Ken. Nor would you know, because you've never read a > > > > > > book on relativity. > > > > > > Same garbage. > > > > > I'm sorry, Ken, but you can't judge. You've never opened up a book on > > > > relativity. > > > > You are too lazy, too cheap, and you're too insecure to read. > > > > So you come here, thinking that you can figure it out on a newsgroup. > > > > Bad idea. > > > > > > > > For example: every observer defines the speed of light to be isotropic > > > > > > > and at the same time the speed of light is non-isotropic in the > > > > > > > observed frame due to relativity of simultaneity. > > > > > > > Not so, Ken. Closing speed and light speed are not the same thing. > > > > > > Closing speed does not affect the arriving speed of light fronts to > > > > > the train observer M'....otherwise M' will not be able to measure the > > > > > speed of light to be isotopic. > > > > > Closing speed and light speed are two completely different quantities. > > > > Hey idiot there is no closing speed between light and any > > > observer....the speed of light is isotropic. > > > Good heavens. You don't even know what the term closing speed means, > > Ken. > > If light doesn't close with an observer, how does the light ever get > > to the observer? > > > I simply cannot believe you would say something so patently stupid as > > "Hey idiot there is no closing speed between light and any observer". > > I should have said that there is no difference in closing speeds > between light and any observer. The closing speed for light and any > observer is c from any direction. That is incorrect, Ken. You confuse closing speed and relative speed. You confuse closing speed and light speed. The relative speed of light is c for any observer. The closing speed for light is not necessarily c for any observer. It helps to know what terms mean, so that you don't confuse them with each other. > > > > > > What if the gedanken set up is changed as follows: > > > M' sees the light fronts from the ends of the train arrive at him > > > simultaneously and he is at equal distance from the ends of the train.. > > > Therefore he concluded that the flashes happened simultaneously. > > > Question: > > > Does M see the flashes arrive at him simultaneously? According to RoS > > > the answer is no. > > > That's correct. > > No that's wrong....the RoS assertion violates the isotropy of the > speed of light in any frame. > > > > > > But according to the SR postulaTE THE ANSWER IS YES. > > > No, the SR postulate would not say that. > > Yes the SR postulate says that...the speed of light is isotropic. And it is. The nonsimultaneous arrival of the light from the flashes does not violate the isotropy of the speed of light. In fact, it is wholly consistent with it. > > > > > Good Jiminy Jumping Jacks, you are hopeless. I've never seen anyone so > > thick in my whole life, and I've seen a good number of thick students. > > ROTFLOL....pot calling the kettle black. > > > > > > What this mean is that closing speed as perceived by M has no effect > > > on the istropy of the speed of light. > > > Closing speed and light speed are two different quantities. > > There is no difference in closing speed and any observer in all > directions as RoS asserted. RoS doesn't make any statement about there being no difference in closing speed. It helps to know what terms mean. > The concept of RoS is bogus and is is > refuted by the Sr postulate. > > Ken Seto > > > > > > Ken Seto > > > > > > > >...due to the closing > > > > > > > speed between the moving observer and the light fronts. > > > > > > > > > Here's the problem, Ken. You don't know enough relativity to know who > > > > > > > > here understands it or not. > > > > > > > > Wrong I know more of relativyt than all of you runts of the SRians. > > > > > > > No, Ken, you don't. You are living a fantasy life, detached from > > > > > > reality. > > > > > > What you know is some stuff that you guys made up to explain > > > > > contradictory claims. For example: A predicts B's clock is running > > > > > slow and B predicts that A's clock is running slow, The speed of light > > > > > is constant because it is measured with the speed of light, The bug is > > > > > dead and alive at the same time, the pole is able to fit into the barn > > > > > and the pole is not able to fit into the barn....etc,etc. > > > > > > > > That's why I was able to formulate a complete theory of relativty that > > > > > > > includes SR and LET as subsets. > > > > > > > > Ken Seto > > > > > > > > >When you run into two people that say > > > > > > > > different things, you can't tell who is right and who is wrong. If you > > > > > > > > mistakenly label BOTH of them as "relativists", then you're going to > > > > > > > > come to the mistaken impression that relativity is contradictory. > > > > > > > > Instead, the problem is that you're hanging around in a venue where > > > > > > > > good information is mixed with bullshit, rather than looking to see > > > > > > > > what SR really says in a book where there is no bullshit mixed in. > > > > > > > > > > > If you want a set of statements about what SR says that are free from > > > > > > > > > > contradictions, then you need to read A BOOK about relativity. This is > > > > > > > > > > how you will see that SR has no contradictions in it. > > > > > > > > > > The point is that all SR books give the same wrong interpretations. > > > > > > > > > You wouldn't know, Ken, because you haven't read any books on > > > > > > > > relativity. Not one. > > > > > > > > > > For example an inertial frame is not the same as an absolute frame, > > > > > > > > > the speed of light is a universal constant instead of a constant math > > > > > > > > > ratio, a clock second is a universal interval of time and at the same > > > > > > > > > time claims that the passage of a clock second in A's frame does not > > > > > > > > > correspond to the passage of a clock second in B's frame, in the bug > > > > > > > > > and the rivet paradox and from the rivet point of view the bug is > > > > > > > > > already dead just before the head of the rivet hit the wall of the > > > > > > > > > hole and at the same time from the hole point of view the bug is still > > > > > > > > > alive just befoore the head of the rivet hit the wall of the hole, in > > > > > > > > > the barn and the pole paradox from the barn point of view the length > > > > > > > > > of the pole can fit into a shorter barn and from the pole point of > > > > > > > > > view the length of the pole cannot fit into the barn, Using the speed > > > > > > > > > of light to measure the speed of light........etc, etc. > > > > > > > > > > > Moreover, your assertion that SR claims are not supported by > > > > > > > > > > experiment is just flat wrong. > > > > > > > > > > No experiment show mutual time dilation...no experiment measure the > > > > > > > > > one way speed of light. > > > > > > > > > Both of those statements are wrong, Ken. > > > > > > > > > > Ken Seto > > > > > > > > > > The experimental documentation is > > > > > > > > > > > available to you. Just because it isn't provided to you on usenet does > > > > > > > > > > not mean it doesn't exist. It means you'll have to take yourself to > > > > > > > > > > where the documentation is. Stop operating out of ignorance. > > > > > > > > > > > > Ken Seto > > > > > > > > > > > > > What you are doing is like trying to learn about some historical event > > > > > > > > > > > > by going to a public rally and listening to demonstrators on both > > > > > > > > > > > > sides argue. When you can't get a consistent story about the > > > > > > > > > > > > historical event by listening to the demonstrators, you make up your > > > > > > > > > > > > own account of the historical event out of your own head. It does not > > > > > > > > > > > > occur to you to LEAVE the public rally, and go look up a historical > > > > > > > > > > > > account of the event in the library. Instead, you sit there at the > > ... > > read more » |