Sherwood Posted May 16, 2019 Report Posted May 16, 2019 Sherwood you're confusing Yv with v. v is limited to c but Yv is limitless. v=x/t, Yv =Yx/t =x/t'. At .8c, Y=5/3 so Yv=4/3c but Yc in that frame is 5/3c. It's all a matter of perspective and Yv or Yc is the proper distance travelled over the dilated time of the depicted moving frame from the depicted stationary frame's perspective. See my last posts in the relativity and simple algebra thread for more details.gu is the ratio of the distance one object moved from another while the light (emanating from either object) traveled the distance I during the time t at speed c according to Special Relativity (SR). This is the same value you would get in Newtonian Mechanics (NM) by multiplying the speed v=gu by t=I/c. v is NM speed, u is SR speed. They are never the same.
Sherwood Posted May 16, 2019 Report Posted May 16, 2019 MikeBrace#243 The propagation speed is c in a vacuum and. . .This is what Einstein said. That light always travels at speed c in free space. The $64 dollar question is "at speed c" relative to what !!!
sluggo Posted May 16, 2019 Report Posted May 16, 2019 This is what Einstein said. That light always travels at speed c in free space. The $64 dollar question is "at speed c" relative to what !!!(The first 3 lines should have been inside the quote box.) Mike was giving the common Newtonian view that the emitter or observer speed should make a difference. Postulate 1 states c is constant and independent of the source speed. Td and lc are complementary effects that ensure any observer measures the same value c.
tetrahedron Posted May 17, 2019 Report Posted May 17, 2019 Its all Super Asymmetry. If you haven't heard of it, you haven't been paying attention! :) Jess Tauber
Sherwood Posted May 19, 2019 Report Posted May 19, 2019 | | | | | | | | | I | 300m | 300m | 300m | | | |__________________|___________________________| A 400m B 500m C A, B, C are passing each other in the horizontal x direction and are momentarily at location A. Each has a light source at rest with respect to itself. At that moment when they are together, they all turn on their lights. After a short time, the light from each has traveled at speed c from each in all directions for time t = 1 musec. I=ct=300m. Now B is 400m from A, C is 900m from A, and C is 500m from B. This is the physics. There is nothing else. Nothing about energy, momentum, etc. By Special Relativity (SR): I2 = (ct)2 - x2 = (ct')2 - x' 2 = (ct'')2 - x'' 2 = . . . (ct)2 = I2 + x2; (ct')2 = I2 + x' 2 ; (ct'')2 = I2 + x'' 2 ; . . . u/c = x/ct; u'/c = x'/ct'; u''/c = x''/ct''; . . . g' = (1/(1 - (u'/c)2))1/2 = ( (ct' 2) / ((ct' 2) - (ct' 2)(u'/c)2) )1/2 = ((ct' 2) / ((ct' 2) - x' 2) )1/2 = ((ct' 2)/I 2)1/2 = (ct’)/ I x' = g(x + ut) = g(u/c)ct + gx Since A has its light source at rest in its reference frame, x = 0 when comparing another reference frame moving with respect to A. Same applies to B and C for their light sources. Then the preceding equation becomes x' = g(u/c) I For A with respect to B, xAB = 400m, I = 300m, tA= I/c = 1 musec, (ct)B = 500m, tB = (5/3) musec, u/c = 400/500, g = 500/300, xAB = g(u/c) I = (500/300) (400/500) 300 = 400m as illustrated below. |\ | \ ct' I | \ I2 = (ct')2 - x'2 ; g = (ct')/ I; (u'/c) = x'/(ct'); x' = g(u/c)I |____ \ x' For A with respect to C, xAC = 900m, I = 300m, tA = 1 musec, (ct)C = 948.683298m, tC = 3.162228 musec, u/c = 0.948683, g = 3.162278, xAC = g(u/c) I = 3.162278 • 0.948683 • 300 = 900m For B with respect to C, xBC = 500m, I = 300m, tB = 1 musec, (ct)C = 583.095189m, tC = 1.943651 musec, u/c = 0.857492, g = 1.943651, xBC = g(u/c) I = 1.943651 • 0.857492 • 300 = 500m For B with respect to A, xBA = -400m, I = 300m, tB = 1 musec, (ct)A = 500m, tA = (5/3) musec, u/c = (-400/500), g = (500/300), xBA = g(u/c) I = (500/300) (- 400/500) 300 = - 400m For C with respect to B, xCB = -500m, I = 300m, tC = 1 musec, (ct)B = 583.095189m, tB = 1.943651 musec, u/c = - 0.857492, g = 1.943651, xCB = g(u/c) I = 1.943651 • - 0.857492 • 300 = - 500m For C with respect to A, xCA = -900m, I = 300m, tC = 1 musec, (ct)A= 948.683298m, tA = 3.162228 musec, u/c = - 0.948683, g = 3.162278, xCA = g(u/c) I = 3.162278 • - 0.948683 • 300 = - 900m The physics here is only about objects and light ‘wavefronts’ which have changed their location over time. That’s all. There is nothing about collisions, momentum changes, energy, etc. The quantities are location and time. Additionally, changes of location are called distances and changes of location (distances) during a given time are velocity (or speed). So we have distances, times, and the ratio of distance to time called speed or velocity. The above showed how these distances, times, and velocity are related in SR. Notice that in “For A wrt to B” tA=1.0 musec and tB=(5/3) musec while in “For B wrt to A” tA=(5/3) musec and tB=1.0 musec. Each object has several different ‘times’ and if we had considered more objects traveling different distances we could have many more ‘times’. There are two key equations in SR which are very similar to equivalent equations which can be derived for Newtonian Mechanics (NM). The SR versions are I2 = (ct')2 - x' 2 g = (ct’/I) The NM versions are I2 = (c’t)2 - x' 2 g = (c’t/I) The quantity (ct’) equals exactly the quantity (c’t). Therefore, I2, x’2, and g will have the same values in both SR and NM. However, t does not equal t’ and c does not equal c’ and this is where SR and NM differ. In both SR and NM I equals the distance the light has traveled from its source, possibly spherically, at speed c for time t, i.e., I=ct. So t=I/c. This makes the light essentially a timing signal. The light was emitted from a source not moving in its reference frame. It moved a certain distance I at speed c during time t in all directions in that frame. From the point of view of another reference frame moving with respect to that reference frame, the light moving perpendicularly (call it vertically) moved not only the ‘vertical’ distance I but also the perpendicular distance x’ for a total distance of H = (I2 + x’2)1/2 In NM, this obviously increased distance, H during the time t would be called something like c’, so the distance H would be H = (c’t) and I2 = (c’t)2 - x' 2 the NM version of I. SR insists (postulates, assumes, etc.) that the speed of the light be the same in both reference frames, i.e., that the ratio H/t be the same as I/t. Since H is greater than I, this is impossible physically. However, you can mathematically arbitrarily increase t (and call it t’ to differentiate it from t). But now this is mathematics, not reality. This is why the SR version is I2 = (ct’)2 - x' 2 instead of I2 = (c’t)2 - x' 2 It also accounts for the two different forms of the gamma function (g), although the numeric value is the same in both SR and NM, c’t=ct’. Of course, c≠c’ and t≠t’. Notice that this mathematical ledgerdemain can be applied to any ‘timing signal’ of a known speed in its reference frame, whatever that speed is. I.e., you could have a Special Relativity where the timing signal travels at speed c=300 m/sec instead of 300 million m/sec and is assumed to also travel that same speed in all reference frames moving with respect to it. Notice also that although light from a given source (just like anything else) is obviously traveling a different speed in reference frames moving with respect to the source’s, it is assumed (along with respect to everything else according to SR) to be moving at speed c from its source. This was known as the Ritz theory. x' = g(u/c) I; (x’/ I) = g(u/c) shows that x’, the distance the object travels, can be many times the distance I, that the light travels, even approaching infinity as u approaches c. All the above objects traveled a greater distance than the three lights did between the time they were together and when their lights had traveled 300m for 1 musec at speed c. u’/c = x’/ct’ = x’/(x’ 2 + I2)1/2 easily shows why “nothing can go faster than light” in SR. The denominator will always be bigger so u'/c will always be <1.0. In NM, (x’/I)=(x’/ct)=(v/c). In SR, (x’/I)=g(u/c)=(ct’/ct)(x’/ct’)=(x’/ct)=v/c which is why SR math can do these time-distance problems although based on erroneous (or wierdly altered, if you prefer) concepts of time and velocity. This wierd math is carried over into more complex topics such as momentum and kinetic energy and seems to work somewhat because of its ties to NM as shown here.
sluggo Posted May 21, 2019 Report Posted May 21, 2019 sherwood#265;As the graphic shows, B and C are moving faster than c (blue), so all those calculations are invalid. Units are x100 m.
Sherwood Posted May 22, 2019 Report Posted May 22, 2019 sherwood#265; As the graphic shows, B and C are moving faster than c (blue), so all those calculations are invalid. Units are x100 m. hypo-sherwood.jpg The calculations are valid. As explained in the post: " u’/c = x’/ct’ = x’/(x’ 2 + I2)1/2easily shows why “nothing can go faster than light” in SR. The denominator will always be bigger so u'/c will always be <1.0." The SR speeds, as shown in the post, and repeated below shown in bold, are all less than c. x', the distances the objects move away from each other, exceeds the distance the light traveled, I, whenever g(u/c) >1. SR 'speed' (u/c) is always <1.0 while g(u/c) can be any number. x' = g(u/c) I; (x’/ I) = g(u/c) as posted. x' = g(u/c) I For A with respect to B, xAB = 400m, I = 300m, tA= I/c = 1 musec, (ct)B = 500m, tB = (5/3) musec, u/c = 400/500, g = 500/300, xAB = g(u/c) I = (500/300) (400/500) 300 = 400m as illustrated below. |\ | \ ct' I | \ I2 = (ct')2 - x'2 ; g = (ct')/ I; (u'/c) = x'/(ct'); x' = g(u/c)I |____ \ x' For A with respect to C,xAC = 900m, I = 300m, tA = 1 musec, (ct)C = 948.683298m, tC = 3.162228 musec, u/c = 0.948683, g = 3.162278, xAC = g(u/c) I = 3.162278 • 0.948683 • 300 = 900m For B with respect to C,xBC = 500m, I = 300m, tB = 1 musec, (ct)C = 583.095189m, tC = 1.943651 musec, u/c = 0.857492, g = 1.943651, xBC = g(u/c) I = 1.943651 • 0.857492 • 300 = 500m For B with respect to A, xBA = -400m, I = 300m, tB = 1 musec, (ct)A = 500m, tA = (5/3) musec, u/c = (-400/500), g = (500/300), xBA = g(u/c) I = (500/300) (- 400/500) 300 = - 400m For C with respect to B,xCB = -500m, I = 300m, tC = 1 musec, (ct)B = 583.095189m, tB = 1.943651 musec, u/c = - 0.857492, g = 1.943651, xCB = g(u/c) I = 1.943651 • - 0.857492 • 300 = - 500m For C with respect to A,xCA = -900m, I = 300m, tC = 1 musec, (ct)A= 948.683298m, tA = 3.162228 musec, u/c = - 0.948683, g = 3.162278, xCA = g(u/c) I = 3.162278 • - 0.948683 • 300 = - 900m
sluggo Posted May 23, 2019 Report Posted May 23, 2019 The calculations are valid. As explained in the post: " u’/c = x’/ct’ = x’/(x’ 2 + I2)1/2easily shows why “nothing can go faster than light” in SR. How can B be at 400, C at 900, while light has only moved 300?
sluggo Posted May 23, 2019 Report Posted May 23, 2019 Sherwood#265; I2 = (ct)2 - x2 Light speed is independent of its source and does not add vectorially like masses do.Refer to lite clock graphic.c2=v2+u2
Sherwood Posted May 23, 2019 Report Posted May 23, 2019 How can B be at 400, C at 900, while light has only moved 300? Einstein's transform equation:x' = g(x + ut) = g(u/c)ct + gx (= g(u/c) I when x=0).x=0 because we are talking about the origin point (not, e.g., x= 3400) of the frames.For (u/c) = 0.8, g = 5/3.If I = 300m, t = 1 musec. I = ct = 300mx' = (5/3)(4/5)300m = 400mNote that the speed (velocity) of the object now at x' is still only 0.8c according to SR. If t = 1 musec and (u/c)= 0.999, then I, the distance the light traveled, is still 300m but the distance x' traveled is x' = 22.366272 • 0.999 • 300m = 6,703.1717m because g = 22.366272 when (u/c) = 0.999. No matter how large (x'/I) = g(u/c) is, (u/c), the SR 'speed between objects', will always be <1.0. g(u/c) 'equals' infinity when (u/c) = 1.0 because the calculation for g in this case is a division by zero.
Sherwood Posted May 24, 2019 Report Posted May 24, 2019 Sherwood#265; I2 = (ct)2 - x2 Light speed is independent of its source and does not add vectorially like masses do. Refer to lite clock graphic. c2=v2+u2 lite clock.gif Like everything else, light speed has to be relative to something. Since light speed is the same in all reference frames according to SR, then by SR light has to have speed c relative to its source as well as to everything else. If you have an object emitting something spherically at s=300 m/sec and another object passing by at 400 m/sec, then the NM Interval equation would be I2 = (s't)2 – x'2 or (s't)2 = I2 + x'2 meaning that after 1 sec the portion of the something emitted that moves perpendicular to the relative motion of the objects would be the distance (I) from the emitting object and the distance (s't) from the second object. The distance (s't) is obviously greater than the distance (I). So the speed of that portion of the emitted something would be (s't)/t or s' with respect to the second object. If you insist, postulate, or whatever, that s'=s then you have to change the time to, for example, (t') to keep the distance (s't) the same. You now have the SR Interval equation I2=(st')2 – x'2 where s is the same in all reference frames because the moving reference frame could be any possible reference frame moving any possible distance with respect to the first during the time (t), not just 400m. s is the equivalent of the c of SR and all the equations and concepts of SR can be derived from this SR Interval equation with c (the s used in this example) being 300 m/sec instead of 300m/musec or any other speed you would care to pick. Post #265 explains this fairly well. Don't concentrate on the listed values for the various variables. That's just the result of working the SR equations. Read the text carefully several times and think about what it is saying about the meaning of the equations.
sluggo Posted May 25, 2019 Report Posted May 25, 2019 Sherwood;I2 = (ct)2 - x2 = (ct')2 - x' 2 = (ct'')2 - x'' 2 = . . .u/c = x/ct; u'/c = x'/ct'; u''/c = x''/ct''; . . .From the point of view of another reference frame moving with respect to that reference frame, the light moving perpendicularly (call it vertically) moved not only the ‘vertical’ distance I but also the perpendicular distance x’ for a total distance of H = (I2 + x’2)1/2 [If that were true, there would be no time dilation observed.] [On the left, R moves past A at .5c for a time ct. A describes the R coordinates for event e as (x, ct). Because of time dilation, R describes the coordinates for event e as (x', ct'), i.e. event e'. From the graphic,(ct)2 - x2 = (ct')2 > (ct')2 - (x')2 = (tr)2What is constant, x/t = x'/t' = x"/t" = c.On the right, using u/c=.8, A describes the B coordinates as (2.4, 3.0), while B describes his coordinates as (1.44, 1.8) relative to A.The reference frame clock has the maximum time, and any clock that moves relative to it, loses time.][What you present is your own interpretation of SR, which doesn't agree with experimental evidence.]
sluggo Posted May 27, 2019 Report Posted May 27, 2019 Sherwood;Standard convention shows the reference frame with a vertical 'time' axis (ct) and a horizontal 'space' axis (x). Any other frame is compared to it. Your example defines A as the reference with B as a (') frame and C as a (") frame, both moving relative to A. The coordinate transformations for B relative to A are x' = g(x-vt) and t' = g(t-vx/cc).If A assigns coordinates (x, t) to event E as (.50, .50), then using the above form results in (x', t') = (.25, .25).Substituting -v in the 1st expression, since A is moving to the left relative to A, x = g(x'+vt') and t = g(t'+vx'/cc)calculates the A coordinates in terms of those of B.
marcospolo Posted May 27, 2019 Report Posted May 27, 2019 (edited) I already stated it. Gravity depends on distance where as acceleration felt in an elevator is the same at all points inside it. I am amused that you are not able to follow.Its very weird that in a science forum, where members claim to use only experiments to validate an hypothesis, (as opposed to using rational analysis and sound logic- the use of which is actively discouraged).... that when presented with a very simple, easily verifiable fact based of experiment, they simply IGNORE it, and cant make the necessary next step, which is to conclude that Einstein's hypotheis is totally impossible.This one line statement is 100% proof that Einsteins claims about inertial reference frames is simply wrong. Some forum members wanted the author to cite published papers to back up this claim! Bizarre. Reativists cant even think through the simplest concepts now and want Mommy to hold their hand, unabke to trust their own mind.It seems that most of the Relativists did not even understood the import of this one sentence, "Gravity depends on distance where as acceleration felt in an elevator is the same at all points inside it.".This is my go-to line to prove Einstein is wrong, from now on. Other proofs in my collection of the many ways that SR is wrong are more wordy but all are correct.The big ignore that this sentence recieved, shows me that you cant discuss Physics with a Relativist. They are like Mormons or JW's or Scientologists. Its a lost cause. Even hoplefuls like Mororium are lost in their own versions of Relativity. (he is stuck with Lorentz, believing that LT actually means something.) Edited May 27, 2019 by marcospolo
Sherwood Posted May 30, 2019 Report Posted May 30, 2019 Its very weird that in a science forum, where members claim to use only experiments to validate an hypothesis, (as opposed to using rational analysis and sound logic- the use of which is actively discouraged).... that when presented with a very simple, easily verifiable fact based of experiment, they simply IGNORE it, and cant make the necessary next step, which is to conclude that Einstein's hypotheis is totally impossible.This one line statement is 100% proof that Einsteins claims about inertial reference frames is simply wrong. <snip>It seems that most of the Relativists did not even understood the import of this one sentence, "Gravity depends on distance where as acceleration felt in an elevator is the same at all points inside it.".<snip>In addition, gravity will pull the side ends of the elevator (better yet picture a 4,000 mile tube 4,000 miles tangentially above the earth's surface) toward the center of the gravitational source whereas acceleration will not. It can be a straight line at all points.
marcospolo Posted May 30, 2019 Report Posted May 30, 2019 In addition, gravity will pull the side ends of the elevator (better yet picture a 4,000 mile tube 4,000 miles tangentially above the earth's surface) toward the center of the gravitational source whereas acceleration will not. It can be a straight line at all points.You need to include a graphical illustration of what you just said, as its difficult to get your point.
LightStorm Posted May 31, 2019 Report Posted May 31, 2019 The big ignore that this sentence recieved, shows me that you cant discuss Physics with a Relativist. They are like Mormons or JW's or Scientologists. Its a lost cause. Even hoplefuls like Mororium are lost in their own versions of Relativity. (he is stuck with Lorentz, believing that LT actually means something.) I agree, it's a lost cause.
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