As requested by alphanumeric I will derive the proper time and show how that relates to velocity and acceleration, due to my claims that velocity is not affected by time dialation as it seems to be in the Holographic Principle that has changed how information is conserved when it encounters a black hole. I don't completely agree with the "fact" that Hawking Radiation would cause a microscopic black hole to evaporate if one was created in the LHC, but I also do not agree with the new veiwpoint that information would be stored on the event horizon of a black hole or that Hawking Radiation would be incorrect because of this. I think this could become more clear from my derivation, or that something from it could lead to disproving the Holographic Principle and information is not stored on the event horizon of a black hole.
I will start out by deriving the proper time using algebra in a way that is different than what it would normally be derived. It is similair to the light clock example, but then it derives the proper time instead of the time dialation equation that is given in most introductory text.
Say an object or ship is traveling at a constant speed. The distance that ship would travel would be the velocity times time, d = v t (eq. 1). This is of course because v = d/t and then you can multiply both sides of the equation by (t) in order to obtain equation 1. An observer at rest would use his own time in order to measure the ships velocity. The distance the ship travels is also relative to the distance the observer at rest measures it. So then distance and time of equation 1 is from the frame of reference of the observer at rest and is used this way in the calculation.
Now say the ship sends a beam a light perpendicular to its direction of motion. The observer at rest will measure the distance the beam travels to be d = c t (eq. 2). He will use his own time to measure this distance but instead of velocity he replaces the speed of light as the velocity, since c = d/t. The speed of light is the distance the observer at rest or any observer would measure the distance it traveled over time.
Now say the observer on the ship measures the speed of light of this same beam. Since he is traveling along with the ship he only measures the beam to travel straight up and down relative to himself. This turns out to be a shorter distance than the observer at rest measures it to travel. But, all observers measure the speed of light to be the same speed, so then you have to assume that the observers time is different in order for him to measure the speed of light to be the same speed even though he measures it to travel a shorter distance. So then you end up with the equation, d' = c t' (eq. 3), for the distance the beam travels relative to the observer on the ship.
The observations from these two different frames of references can then form similar triangles and then those triangles can be translated onto each other from each coordinate system. So then you would have three sides of a triangle, (v t ) being one side (a), (c t') being the adjacent side (b), and then (c t) being the hypotenus. So then you can use Pythagorean's Theorem, a^2 + b^2 = c^2.
( v t )^2 + ( c t' )^2 = ( c t )^2 Then distribute the square
v^2 t^2 + c^2 t'^2 = c^2 t^2 Then subtract both sides by (v^2 t^2)
c^2 t'^2 = c^2 t^2 - v^2 t^2 Then factor out ( c^2 t^2) out of the right side of the equation
c^2 t'^2 = c^2 t^2 ( 1 - v^2/c^2) Then divide both sides by ( c^2 )
t'^2 = t^2 ( 1 - v^2/c^2 ) Then take the square root of both sides
t' = t * sqrt ( 1 - v^2/c^2 ) Then you have it the proper time, although time cannot be negative so then you can say that the negative solution is not valid and I will call this equation 4 (eq. 4)
I will start out by deriving the proper time using algebra in a way that is different than what it would normally be derived. It is similair to the light clock example, but then it derives the proper time instead of the time dialation equation that is given in most introductory text.
Say an object or ship is traveling at a constant speed. The distance that ship would travel would be the velocity times time, d = v t (eq. 1). This is of course because v = d/t and then you can multiply both sides of the equation by (t) in order to obtain equation 1. An observer at rest would use his own time in order to measure the ships velocity. The distance the ship travels is also relative to the distance the observer at rest measures it. So then distance and time of equation 1 is from the frame of reference of the observer at rest and is used this way in the calculation.
Now say the ship sends a beam a light perpendicular to its direction of motion. The observer at rest will measure the distance the beam travels to be d = c t (eq. 2). He will use his own time to measure this distance but instead of velocity he replaces the speed of light as the velocity, since c = d/t. The speed of light is the distance the observer at rest or any observer would measure the distance it traveled over time.
Now say the observer on the ship measures the speed of light of this same beam. Since he is traveling along with the ship he only measures the beam to travel straight up and down relative to himself. This turns out to be a shorter distance than the observer at rest measures it to travel. But, all observers measure the speed of light to be the same speed, so then you have to assume that the observers time is different in order for him to measure the speed of light to be the same speed even though he measures it to travel a shorter distance. So then you end up with the equation, d' = c t' (eq. 3), for the distance the beam travels relative to the observer on the ship.
The observations from these two different frames of references can then form similar triangles and then those triangles can be translated onto each other from each coordinate system. So then you would have three sides of a triangle, (v t ) being one side (a), (c t') being the adjacent side (b), and then (c t) being the hypotenus. So then you can use Pythagorean's Theorem, a^2 + b^2 = c^2.
( v t )^2 + ( c t' )^2 = ( c t )^2 Then distribute the square
v^2 t^2 + c^2 t'^2 = c^2 t^2 Then subtract both sides by (v^2 t^2)
c^2 t'^2 = c^2 t^2 - v^2 t^2 Then factor out ( c^2 t^2) out of the right side of the equation
c^2 t'^2 = c^2 t^2 ( 1 - v^2/c^2) Then divide both sides by ( c^2 )
t'^2 = t^2 ( 1 - v^2/c^2 ) Then take the square root of both sides
t' = t * sqrt ( 1 - v^2/c^2 ) Then you have it the proper time, although time cannot be negative so then you can say that the negative solution is not valid and I will call this equation 4 (eq. 4)