- #1

P J Strydom

- 68

- 1

- TL;DR Summary
- Length contraction calculation

I was absent for a while due to personal constraints but I did keep myself busy with the Time dilation equation some member sent me a while back.

I decided to set a time limit for myself to learn and understand time Dilation and length contraction, which must be before December 2020, or I will just have to accept that I am unable to grasp the scientific principle of this topic.

I realize that my greatest shortcoming is definitely my rusted high school algebra which I am sharpening up again.

I therefore will humble myself to the fullest in anticipation that someone on this forum will take the time to direct me in the right direction as I wonder through this venture.

Well, I will start with the following.

I completely forgot who the member was that sent me the documents explaining the mathematical formulas to calculate length contraction and time dilation, furthermore, the printed copy I have is so badly deteriorated, that I am busy retyping the whole thing. Unfortunately, it contains a lot of mathematical inserts in Word, and it takes a while to go through the whole process. I finished the document, but found a few typos to fix.

Anyway, the document starts off explaining a simple Galilean transformation where a truck drives at 30Mph and I as a passenger throws a ball 60 feet, and the question is then asked, how far will the ball have traveled in 2 seconds.

Obviously, the document goes through the whole process to teach the calculation to arrive at the answer.

Then the document starts to teach the Special relativity transformation and works on a Linear function dependent on speed connection these 2 calculations that starts off with;

Now we have:

Great stuff, I do understand the mathematical formula, because it is the old algebra function where 2 functions are expressed in Rate X Time = distance, added up, and deducted from the distance.

Simple example

Two cars started from the same point, at 5 am, traveling in opposite directions at 40 and 50 mph respectively. At what time will they be 450 miles apart?

40t+50t=450

90t=450

t=5 hours

Cool.

Now to get back to our equation.

We took the distance the light beam traveled to the axis of motion of S1 and added the distance of the S frame and inserted the different velocities into the equation. We will then simplify the equation (we will get to that later.)

My question,

x= γ(v)x1 + β(v)ct1

Which is as indicated by the document as the result that we have the Distance transformation from S1 to S.

What are we attempting to achieve with this calculation?

I sat and drew pictures forever and still do not know what this formula means.

I decided to set a time limit for myself to learn and understand time Dilation and length contraction, which must be before December 2020, or I will just have to accept that I am unable to grasp the scientific principle of this topic.

I realize that my greatest shortcoming is definitely my rusted high school algebra which I am sharpening up again.

I therefore will humble myself to the fullest in anticipation that someone on this forum will take the time to direct me in the right direction as I wonder through this venture.

Well, I will start with the following.

I completely forgot who the member was that sent me the documents explaining the mathematical formulas to calculate length contraction and time dilation, furthermore, the printed copy I have is so badly deteriorated, that I am busy retyping the whole thing. Unfortunately, it contains a lot of mathematical inserts in Word, and it takes a while to go through the whole process. I finished the document, but found a few typos to fix.

Anyway, the document starts off explaining a simple Galilean transformation where a truck drives at 30Mph and I as a passenger throws a ball 60 feet, and the question is then asked, how far will the ball have traveled in 2 seconds.

Obviously, the document goes through the whole process to teach the calculation to arrive at the answer.

Then the document starts to teach the Special relativity transformation and works on a Linear function dependent on speed connection these 2 calculations that starts off with;

- S (frame at rest) and S1 (Moving relative to S at speed v)
- A beam of light is flashed along the Axis of motion of S1
- S calculates it’d distance as a function of time like this: Distance = rate X Time.
- x=ct. which is then re written as x-ct=0.
- If the beam is shot in the opposite direction, the equation will be;
- X=-ct rewritten as x+ct=0

- X1-ct1=0
- And the flash in the opposite direction
- X1+ct1=0

- X-ct=x1-ct1
- And x+ct=x1ct1

Now we have:

- X-ct= α(v) (x1-ct1)
- And x+ct= (δ(v)(x1+ct1)

Great stuff, I do understand the mathematical formula, because it is the old algebra function where 2 functions are expressed in Rate X Time = distance, added up, and deducted from the distance.

Simple example

Two cars started from the same point, at 5 am, traveling in opposite directions at 40 and 50 mph respectively. At what time will they be 450 miles apart?

r | t | d | |

Car A | 40 | t | 40t |

Car B | 50 | t | 50t |

90t=450

t=5 hours

Cool.

Now to get back to our equation.

We took the distance the light beam traveled to the axis of motion of S1 and added the distance of the S frame and inserted the different velocities into the equation. We will then simplify the equation (we will get to that later.)

My question,

- x-ct= α(v) (x1-ct1)
- x+ct= (δ(v)(x1+ct1)
- All added up
- 2x=x1[α(v)+ δ(v)]+ct1 [δ(v)- α(v)] divide and simplified gives
- X= x1[α(v)+ δ(v)]/2 +ct1 [δ(v)- α(v)]/2
- We then replace [α(v)+ δ(v)]/2 with γ(v)
- And [δ(v)- α(v)]/2 with β(v)

x= γ(v)x1 + β(v)ct1

Which is as indicated by the document as the result that we have the Distance transformation from S1 to S.

What are we attempting to achieve with this calculation?

I sat and drew pictures forever and still do not know what this formula means.