Roger Dynamic Motion Posted June 1, 2017 Share Posted June 1, 2017 (edited) The Galilean addition for Velocities ? incorrect. First the speed of light is always the same../ in the Galilean addition of velocities ;the difference with the speed of the source in the train must be deducted from the speed of light not added . The Galilean addition for velocities evaluate an event in time without target like a magician . Let say a target is at 1 second in time, when the light switch is turn ''on'' by the man in the train ; if the train's speed is X the train will arrived at the target x- the speed of light ''not +''<< I'm I right ? Edited June 2, 2017 by Roger Dynamic Motion Link to comment Share on other sites More sharing options...

Country Boy Posted June 1, 2017 Share Posted June 1, 2017 "Gallilean Addition Law"? It would help if you had said that you mean the "addition of velocities". The "Gallilean Addition Law" says that if A is moving, relative to you, at speed u and B is moving relative to A with speed v, then B is moving, relative to you, with speed u+ v. Yes, at high speeds, the "Gallilean Addition Law" is incorrect. It need to be replaced by the law from relativity. With relativity the law for "adding velocities" says that if A is moving, relative to you, at speed u and B is moving, relative to B at speed v, then B is moving, relative to you, at [math]\frac{u+ v}{1+ \frac{uv}{c^2}}[/math]. I thought that was pretty well known. I don't know what you mean by "Let say a target is at 1 second when the switch is turn ''on'' by the man in the train ; if the train's speed is X the train will arrived at the target". A target would be at some distance from a train, so I don't know what you mean by "a target is at 1 second". Do you mean that, at the current speed of the train it would take one second to get there? And what switch is being turned on? A light switch? if so the speed of light is "c" so using the formula above, with u as the speed of the train, relative to you, the speed of light relative you is [math]\frac{u+ c}{1+ \frac{uc}{c^2}}= \frac{u+ c}{1+ \frac{u}{c}}[/math]. Multiplying both numerator and denominator by c, that becomes [math]\frac{c(u+ c)}{c+ u}= c[/math]. Yes, the velocity of light is "c" relative to any inertial frame. Link to comment Share on other sites More sharing options...

Roger Dynamic Motion Posted June 1, 2017 Author Share Posted June 1, 2017 (edited) "Gallilean Addition Law"? It would help if you had said that you mean the "addition of velocities". The "Gallilean Addition Law" says that if A is moving, relative to you, at speed u and B is moving relative to A with speed v, then B is moving, relative to you, with speed u+ v. Yes, at high speeds, the "Gallilean Addition Law" is incorrect. It need to be replaced by the law from relativity. With relativity the law for "adding velocities" says that if A is moving, relative to you, at speed u and B is moving, relative to B at speed v, then B is moving, relative to you, at [math]\frac{u+ v}{1+ \frac{uv}{c^2}}[/math]. I thought that was pretty well known. I don't know what you mean by "Let say a target is at 1 second when the switch is turn ''on'' by the man in the train ; if the train's speed is X the train will arrived at the target". A target would be at some distance from a train, so I don't know what you mean by "a target is at 1 second". Do you mean that, at the current speed of the train it would take one second to get there? And what switch is being turned on? A light switch? if so the speed of light is "c" so using the formula above, with u as the speed of the train, relative to you, the speed of light relative you is [math]\frac{u+ c}{1+ \frac{uc}{c^2}}= \frac{u+ c}{1+ \frac{u}{c}}[/math]. Multiplying both numerator and denominator by c, that becomes [math]\frac{c(u+ c)}{c+ u}= c[/math]. Yes, the velocity of light is "c" relative to any inertial frame. [/quote Edited June 1, 2017 by Roger Dynamic Motion Link to comment Share on other sites More sharing options...

Country Boy Posted June 5, 2017 Share Posted June 5, 2017 What was your purpose in simply copying what I said? Link to comment Share on other sites More sharing options...

Lord Antares Posted June 5, 2017 Share Posted June 5, 2017 (edited) It is very well known, that adding speeds in a linear fashion doesn't work at relativistic speeds (i.e. speeds approaching the speed of light). As Halls said, you need a relativistic formula for that. So the ''old'' laws of velocity are not incorrect, they are just unnapplicable for relativistic speeds. This is very well known; you're not disproving anything with this. Edited June 5, 2017 by Lord Antares Link to comment Share on other sites More sharing options...

Roger Dynamic Motion Posted June 5, 2017 Author Share Posted June 5, 2017 It is very well known, that adding speeds in a linear fashion doesn't work at relativitstic speeds (i.e. speed approaching the speed of light). As Halls said, you need a relativistic formula for that. So the ''old'' laws of velocity are not incorrect, they are just unnapplicable for relativistic speeds. This is very well known; you're not disproving anything with this. ok Thanks Link to comment Share on other sites More sharing options...

MigL Posted June 5, 2017 Share Posted June 5, 2017 Galilean transforms are only accurate for low speeds. Lorentz transforms reduce to their Galilean counterparts when c is considered infinitely large compared to v . Link to comment Share on other sites More sharing options...

Roger Dynamic Motion Posted June 5, 2017 Author Share Posted June 5, 2017 Galilean transforms are only accurate for low speeds. Lorentz transforms reduce to their Galilean counterparts when c is considered infinitely large compared to v . can you give me an example ? Link to comment Share on other sites More sharing options...

MigL Posted June 5, 2017 Share Posted June 5, 2017 When v<<c then velocities add linearly to a very good approximation. This is for speeds of planes, trains and automobiles. If v is of the same magnitude order as c , the above approximation is no longer valid. You must use the Lorentz factor, gamma=1/root(1-v^2/c^2) . ( sorry for my lack of LaTex ) Link to comment Share on other sites More sharing options...

Roger Dynamic Motion Posted June 5, 2017 Author Share Posted June 5, 2017 When v<<c then velocities add linearly to a very good approximation. This is for speeds of planes, trains and automobiles. If v is of the same magnitude order as c , the above approximation is no longer valid. You must use the Lorentz factor, gamma=1/root(1-v^2/c^2) . ( sorry for my lack of LaTex ) thank ! Link to comment Share on other sites More sharing options...

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