Question about Einstein postulates and relativity:
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From: BURT on 16 Apr 2010 16:09 On Apr 16, 7:29 am, "Inertial" <relativ...(a)rest.com> wrote: > "dlzc" <dl...(a)cox.net> wrote in message > > news:75ba566a-e18d-4922-8906-982ed5fd2446(a)g11g2000yqe.googlegroups.com... > > > > > > > Dear Inertial: > > > On Apr 11, 9:47 pm, "Inertial" <relativ...(a)rest.com> wrote: > >> "dlzc" <dl...(a)cox.net> wrote in message > >>news:9517f193-906f-4a91-b7c8-4221182a2cd8(a)h27g2000yqm.googlegroups.com.... > >> > On Apr 11, 9:08 pm, Khattak <zarm...(a)gmail.com> wrote: > >> >> Let a spaceship of width 30, 00,000 km and length > >> >> say 10 meter (adjusted with length contraction) > >> >> is moving with 0.9c. Let a beam of light/ pulse is > >> >> moving perpendicular to the direction of spaceship. > >> >> For simplicity assume a pulse of light travel from > >> >> north to south and spaceship is moving from east > >> >> to west. After sometime the same pulse of light > >> >> Strikes and enters spaceship through its one > >> >> longitudinal side of 10 m Travel inside spaceship > >> >> and then Leaves the spaceship through its other > >> >> longitudinal side > >> >> Can we trace the path of pulse for both inside > >> >> and outside observer while keeping in mind the > >> >> Einstein postulates? Thanks > > >> > Yes. Inside the ship it moves straight north > >> > south. > > >> Nope .. not from the description given. There > >> will be aberration. > > > From the description given (other than what he means by > > "longitudinal"), the light enters on either north or south face, and > > proceeds to exit on south or north face. > > But not directly across from where it enters, according to an observer in > the ship. So not directly north to south. > > > What room for "aberration" > > do you see for an observer that is inside the ship for the entire > > event? > > The width of the ship gives you plenty of room .. it is 30,000,000 km wide > > >> > Outside the ship > >> > (at rest), it moves along with the ship... > >> > and north and south. > > >> Outside the ship it moves the same as inside > >> the ship, from the same observers point of view. > > > It isn't clear what motion his outside observer has wrt the ship. > > I am assuming it is the observer who sees the ship as moving at 0.9c nad > contracted to 10m in length > > > Perhaps the ship's length is contracted to 10m for this person. > > That seems to be what it says > > > > > The > > setup is unclear. > > >> > And here I thought you were going to start this joke... > >> >http://www.supercubs.org/phpbb2/viewtopic.php?t=4958 > > >> Not a very funny joke, if it was meant as one at all. > > > It is. And an old one too. I managed to find one version that did > > not disparage one ethnicity. > > >> Unless I missed something humorous in it. > > > He landed sideways on the runway. The runway is very short, but it > > sure is wide... > > > David A. Smith- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text - There are no flat atomic forms.
From: Khattak on 18 Apr 2010 01:34 On Apr 12, 3:03 am, "Androcles" <Headmas...(a)Hogwarts.physics_x> wrote: > "Khattak" <zarm...(a)gmail.com> wrote in message > > news:18eb13b9-ecf4-4757-b12c-5d4f74071f16(a)u31g2000yqb.googlegroups.com... > > > Let a spaceship of width 30, 00,000 km > > No such animal as 30, 00,000 km exists > > and length say 10 meter > > > > > (adjusted with length contraction) is moving with 0.9c. Let a beam of > > light/ pulse is moving perpendicular to the direction of spaceship. > > For simplicity assume a pulse of light travel from north to south and > > spaceship is moving from east to west. After sometime the same pulse > > of light > > Strikes and enters spaceship through its one longitudinal side of 10 m > > Travel inside spaceship and then > > Leaves the spaceship through its other longitudinal side > > Can we trace the path of pulse for both inside and outside observer > > while keeping in mind the Einstein postulates? Thanks- Hide quoted text - > > - Show quoted text - > No such animal as 30, 00,000 km exists Here is small animal/ light clock.(just wait for 90 seconds after clicking) http://www.youtube.com/watch?v=p80IhaBz51M&feature=related Let a pulse enters the moving clock from its bottom. So for outside stationary observer, would a pulse of light deviates from its original straight path when exit at the top. Lets change the set up of top and bottom mirrors to following isosceles triangle ABC in a moving spaceship close to speed of light. Vertex angle A = 90 degree (Mirror) Base angles B and C = 45 degrees AB = BC = 1, 50,000 km At rest a pulse starts from B to A and then B to C thus one second for both inside and outside stationary observers. Now if a ship starts moving close to the speed of light and a pulse start from B to A and then B to C then would the time dilation equation be different or remained the same.
From: BURT on 18 Apr 2010 01:44 On Apr 17, 10:34 pm, Khattak <zarm...(a)gmail.com> wrote: > On Apr 12, 3:03 am, "Androcles" <Headmas...(a)Hogwarts.physics_x> wrote: > > > > > > > "Khattak" <zarm...(a)gmail.com> wrote in message > > >news:18eb13b9-ecf4-4757-b12c-5d4f74071f16(a)u31g2000yqb.googlegroups.com.... > > > > Let a spaceship of width 30, 00,000 km > > > No such animal as 30, 00,000 km exists > > > and length say 10 meter > > > > (adjusted with length contraction) is moving with 0.9c. Let a beam of > > > light/ pulse is moving perpendicular to the direction of spaceship. > > > For simplicity assume a pulse of light travel from north to south and > > > spaceship is moving from east to west. After sometime the same pulse > > > of light > > > Strikes and enters spaceship through its one longitudinal side of 10 m > > > Travel inside spaceship and then > > > Leaves the spaceship through its other longitudinal side > > > Can we trace the path of pulse for both inside and outside observer > > > while keeping in mind the Einstein postulates? Thanks- Hide quoted text - > > > - Show quoted text - > > No such animal as 30, 00,000 km exists > > Here is small animal/ light clock.(just wait for 90 seconds after > clicking) > > http://www.youtube.com/watch?v=p80IhaBz51M&feature=related > > Let a pulse enters the moving clock from its bottom. So for outside > stationary observer, would a pulse of light deviates from its original > straight path when exit at the top. > > Lets change the set up of top and bottom mirrors to following > isosceles triangle ABC in a moving spaceship close to speed of light. > > Vertex angle A = 90 degree (Mirror) > Base angles B and C = 45 degrees > AB = BC = 1, 50,000 km > > At rest a pulse starts from B to A and then B to C thus one second for > both inside and outside stationary observers. > > Now if a ship starts moving close to the speed of light and a pulse > start from B to A and then B to C then would the time dilation > equation be different or remained the same.- Hide quoted text - > > - Show quoted text - If your ship travels below light speed behind light light will inch ahead. The space frame for motion is absolute. Mitch Raemsch
From: Androcles on 18 Apr 2010 04:52 "Khattak" <zarmewa(a)gmail.com> wrote in message news:43525620-11f9-4ddd-8299-5d50e1aa2c9f(a)11g2000yqr.googlegroups.com... On Apr 12, 3:03 am, "Androcles" <Headmas...(a)Hogwarts.physics_x> wrote: > "Khattak" <zarm...(a)gmail.com> wrote in message > > news:18eb13b9-ecf4-4757-b12c-5d4f74071f16(a)u31g2000yqb.googlegroups.com... > > > Let a spaceship of width 30, 00,000 km > > No such animal as 30, 00,000 km exists > > and length say 10 meter > > > > > (adjusted with length contraction) is moving with 0.9c. Let a beam of > > light/ pulse is moving perpendicular to the direction of spaceship. > > For simplicity assume a pulse of light travel from north to south and > > spaceship is moving from east to west. After sometime the same pulse > > of light > > Strikes and enters spaceship through its one longitudinal side of 10 m > > Travel inside spaceship and then > > Leaves the spaceship through its other longitudinal side > > Can we trace the path of pulse for both inside and outside observer > > while keeping in mind the Einstein postulates? Thanks- Hide quoted > > text - > > - Show quoted text - > No such animal as 30, 00,000 km exists Here is small animal/ light clock.(just wait for 90 seconds after clicking) http://www.youtube.com/watch?v=p80IhaBz51M&feature=related ================================================== That is not a 30, 00,000 km. Look, sonny, you have a comma then a space then two zeroes then a comma then three zeroes. No such animal exists. Learn how to write numbers. "If you move through space you'r really also moving through time" -- Charles Liu, Idiot. If you DON'T move through space you STILL move through time whether you want to or not, so moving through space is irrelevant. "Time is the fourth dimension in our universe" -- Charles Lui, Idiot. One can go backwards in the other three. "Albert Einstein was the first person that put mathematical equations together that showed that space and time could be related" -- LYING Charles Lui, Idiot. Isaac Newton came up with v = dx/dt long before. ================================================= Let a pulse enters the moving clock from its bottom. ================================================= That's not a light clock. This is a light clock: http://www.androcles01.pwp.blueyonder.co.uk/lightclock.gif It happens to work, too, and it is moving though space at 30,000 m/s with me and the Earth and you and everything else on this planet. So for outside stationary observer, ================================================= No such animal exists, all motion is relative.
From: rotchm on 18 Apr 2010 10:11
I glanced at the previous posts and I agree with Inertial, this looks looks a 'spammer'/troll. o Anw, reading this post(below) quickly,... All these type of problems are easily resolved by not considering a lightpulse but a spherical outgoing lightwave from the source. You only need then to consider the wavefront remaining inside your apparatus (lightclock). Try that to answer your questions. > Let a pulse enters the moving clock from its bottom. So for outside > stationary observer, would a pulse of light deviates from its original > straight path when exit at the top. > > Lets change the set up of top and bottom mirrors to following > isosceles triangle ABC in a moving spaceship close to speed of light. > > Vertex angle A = 90 degree (Mirror) > Base angles B and C = 45 degrees > AB = BC = 1, 50,000 km > > At rest a pulse starts from B to A and then B to C thus one second for > both inside and outside stationary observers. > > Now if a ship starts moving close to the speed of light and a pulse > start from B to A and then B to C then would the time dilation > equation be different or remained the same. |