The Galilean addition law is [not always applicable]

[Roger Dynamic Motion]

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

[Country Boy]

"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.

[Roger Dynamic Motion]

"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

[Country Boy]

What was your purpose in simply copying what I said?

[Lord Antares]

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

[Roger Dynamic Motion]

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

[MigL]

Galilean transforms are only accurate for low speeds.

Lorentz transforms reduce to their Galilean counterparts when c is considered infinitely large compared to v .

[Roger Dynamic Motion]

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 ?

[MigL]

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 )

[Roger Dynamic Motion]

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 !