Question about Einstein postulates and relativity:
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From: Khattak on 18 Apr 2010 21:34 On Apr 18, 8:11 am, rotchm <rot...(a)gmail.com> wrote: > 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.- Hide quoted text - > > - Show quoted text - > looks a 'spammer'/troll. I usually discuss idea with my friend and therefore question arises during discussion. So FYI Im not a spammer/troll. If you dislike my posts then im sorry about that and please just ignore all or tell me the other forum where I can post my question. I think criticism is not sin.
From: BURT on 18 Apr 2010 21:53 On Apr 17, 10:44 pm, BURT <macromi...(a)yahoo.com> wrote: > 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- Hide quoted text - > > - Show quoted text - You can travel further than light when your clock is slower than its time and your speed is near light's rate. The math is Gamma flow rate for point energy. Mitch Raemsch
From: harald on 19 Apr 2010 11:40 On Apr 19, 3:34 am, Khattak <zarm...(a)gmail.com> wrote: > On Apr 18, 8:11 am, rotchm <rot...(a)gmail.com> wrote: > > > > > 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.- Hide quoted text - > > > - Show quoted text - > > looks a 'spammer'/troll. > > I usually discuss idea with my friend and therefore question arises > during discussion. So FYI Im not a spammer/troll. If you dislike my > posts then im sorry about that and please just ignore all or tell me > the other forum where I can post my question. I think criticism is not > sin. Hi Khattthak, dlzc already answered your question and Inertial gave some precisions. Furthermore rotchm gave some advice that may be helpful (indeed I also found that the Huyghens construction works in SRT). If it doesn't work for you, you can try to sketch here in ASCII what you did with mention of the problem that you encountered. Oh and for easy calculation and construction it's handier to work with 0.8c. Regards, Harald
From: glird on 19 Apr 2010 16:42 On Apr 19, 11:40 am, harald <h...(a)swissonline.ch> wrote: > < Oh and for easy calculation and construction >it's handier to work with 0.8c. That's because with v = .8c, sqrt(1-v^2/c^2) = sqrt(1-.64)= sqrt(.36) = .6. Similarly, if you work with v = .6c the math is simple enough to do in your head. For all other values other than v = c or v = 0, you'd need a calculator to continue a calculation. glird
From: BURT on 19 Apr 2010 18:20
On Apr 19, 1:42 pm, glird <gl...(a)aol.com> wrote: > On Apr 19, 11:40 am, harald <h...(a)swissonline.ch> wrote: > > < Oh and for easy calculation and construction >it's handier to work > with 0.8c. > > That's because with v = .8c, > sqrt(1-v^2/c^2) = sqrt(1-.64)= sqrt(.36) = .6. > Similarly, if you work with v = .6c the math is simple enough to do in > your head. For all other values other than v = c or v = 0, you'd need > a calculator to continue a calculation. > > glird In closing speed you can leave light behind by your own motion. You can also stay behind light temporarily. Mitch Raemsch |