The Emission Theory of Androcles
in [Relativity]
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From: Jonah Thomas on 10 Sep 2009 14:13 "Androcles" <Headmaster(a)Hogwarts.physics_n> wrote: > "J Thomas" <jethomas5(a)gmail.com> wrote > Here is the basic problem: if you travel at 60 kph for 1 hour, and > then travel at 80 kph for 1 hour, your average speed is 70 kph. > > 60 km +80 km = 140 km. > 140/2 hours = 70 kph > > But if you travel at 60 kph for 70k, and then you travel at 80 kph for > 70k, your average speed is about 68.57 kph. > > ============================================= > t1 = 70/60 > t2 = 70/80 > d = 140/(t1 + t2) = 68.57 kph > > Why is that a problem? If you go one direction at speed c+v and then the same distance at speed c-v you don't arrive at the same time you'd arrive if you went at speed c for both legs of the trip. > If you go at c+v one way and come back the same distance at c-v, it > doesn't average out. > ============================================= > If you go at c+c one way and come back the same distance at c-c, > you don't come back. yes, quite true. > If course, if you were paying attention to 1/speed instead of speed, > then it would average out. ;) > ============================================== > 1/0 is undefined. So what? One way you talk about speed. How much distance is crossed per unit time. You don't complain that if you look at how much distance you travel in 0 seconds you're dividing by zero. So if you talk about 1/speed, how much time it takes to cross a unit distance, you needn't complain that if you look at how much time it takes to cross 0 miles you're dividing by zero then. If you compute D/(c+v) + D/(c-v) you get 2Dc/(c^2-v^2) and take it from there. But if you compute (c+v)/D + (c-v)/D you get 2c/D which comes out nice and easy. ;)
From: Androcles on 10 Sep 2009 14:14 "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090910140234.4e4978ae.jethomas5(a)gmail.com... > "Androcles" <Headmaster(a)Hogwarts.physics_n> wrote: >> "J Thomas" <jethomas5(a)gmail.com> wrote >> "Inertial" <relativ...(a)rest.com> wrote: > >> > Which, of course, is also incorrect, as Sagnac does not in any way >> > refute SR and never has. It shows that the light speed, in the >> > inertial frame of reference, is not affected by the speed of the >> > source, and so is consistent with both SR and aether theories, but >> > not with ballistic / emission theories. >> >> Agreed. Relativity was carefully designed to provide classical results >> under classical conditions, and there is nothing here that would >> result in a relativistic difference. >> ============================================== >> You are agreeing with a bigoted, lying, incompetent idiot. > > Even if that's true, which in my personal experience so far is not > proven, so what? > > You can't depend on bigoted, lying, incompetent idiots to always be > wrong about everything. You can depend on an old 33 RPM record playing turntable not to depend on SR/GR, but the strobe light on the side used to verify its speed by illuminating regularly spaced marks on the rim meshes perfectly with Sagnac. Of course you not owning a normal computer prevents you from seeing a video of it happening, but can see this: http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/MechSagnac.gif You can depend on the bigoted, lying, incompetent idiot to always be wrong about Sagnac.
From: Androcles on 10 Sep 2009 14:42 "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090910141351.5fc73726.jethomas5(a)gmail.com... > "Androcles" <Headmaster(a)Hogwarts.physics_n> wrote: >> "J Thomas" <jethomas5(a)gmail.com> wrote > >> Here is the basic problem: if you travel at 60 kph for 1 hour, and >> then travel at 80 kph for 1 hour, your average speed is 70 kph. >> >> 60 km +80 km = 140 km. >> 140/2 hours = 70 kph >> >> But if you travel at 60 kph for 70k, and then you travel at 80 kph for >> 70k, your average speed is about 68.57 kph. >> >> ============================================= >> t1 = 70/60 >> t2 = 70/80 >> d = 140/(t1 + t2) = 68.57 kph >> >> Why is that a problem? > > If you go one direction at speed c+v and then the same distance at speed > c-v you don't arrive at the same time you'd arrive if you went at speed > c for both legs of the trip. Arrive where? Look, one marble arrives on the opposite side of the disc at exactly the same time as the other marble. http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/MechModel.gif It travels half way at -c and other marble travels half way at +c. Why is that a problem? The disc rotates at v. Why is that a problem? One ball travels at c+v and the other travels at v-c. The one that travelled at v-c doesn't move at all. Why is that a problem? >> If you go at c+v one way and come back the same distance at c-v, it >> doesn't average out. >> ============================================= >> If you go at c+c one way and come back the same distance at c-c, >> you don't come back. > > yes, quite true. Yes, and here it is happening. http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/MechModel.gif > >> If course, if you were paying attention to 1/speed instead of speed, >> then it would average out. ;) >> ============================================== >> 1/0 is undefined. > > So what? So you don't use 1/speed, that's what. One way you talk about speed. How much distance is crossed per > unit time. You don't complain that if you look at how much distance you > travel in 0 seconds you're dividing by zero. > > So if you talk about 1/speed, how much time it takes to cross a unit > distance, you needn't complain that if you look at how much time it > takes to cross 0 miles you're dividing by zero then. I can walk from the tail end of an aircraft in New York to the flight deck in London, taking me six hours. How fast am I walking across the Atlantic, 3000 miles? How fast am I walking from tail to flight deck, 180 feet ? > If you compute D/(c+v) + D/(c-v) you get 2Dc/(c^2-v^2) and take it from > there. > But if you compute (c+v)/D + (c-v)/D you get 2c/D which comes out nice > and easy. ;) Doesn't change D = 0, so no advantage. http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/MechModel.gif On ball travels a distance zero while at that same time it travels halfway around the disc. It does this at speed c-c and speed -c, simultaneously, while the disk rotates at c. How can that be?
From: Jonah Thomas on 10 Sep 2009 15:13 "Androcles" <Headmaster(a)Hogwarts.physics_n> wrote: > You can depend on an old 33 RPM record playing turntable not > to depend on SR/GR, but the strobe light on the side used to verify > its speed by illuminating regularly spaced marks on the rim meshes > perfectly with Sagnac. It turned out I had seen that, I didn't realise that was the one you were presenting again until I saw it again. So what is the take-home message here? Something about strobes.... Something about the sum of two traveling wave making a standing wave or a slow traveling wave....
From: Androcles on 10 Sep 2009 15:43
"Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090910151331.6320ca84.jethomas5(a)gmail.com... > "Androcles" <Headmaster(a)Hogwarts.physics_n> wrote: > >> You can depend on an old 33 RPM record playing turntable not >> to depend on SR/GR, but the strobe light on the side used to verify >> its speed by illuminating regularly spaced marks on the rim meshes >> perfectly with Sagnac. > > It turned out I had seen that, I didn't realise that was the one you > were presenting again until I saw it again. > > So what is the take-home message here? Something about strobes.... > Something about the sum of two traveling wave making a standing wave or > a slow traveling wave.... Simple, isn't it? If you DEFINE wavelength = speed/frequency then increasing speed has to increase the wavelength. You can't change the 50Hz (60 Hz USA) frequency of the strobe light, so by your definition the wavelength changes. Yet that is ridiculous, nobody is repainting the marks on the side of the turntable, so your definition must be wrong or the distance between marks isn't the wavelength. Yet the teeth around a gear look awfully like a travelling wave to me. http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/MechSagnac.gif So let's see if you can think. You tell me what the take-home message is. |