Derivation of special relativity formulae

  • #1
I was bored, so I decided to derive the special relativity formulae.
I drew the following diagram of a light clock:
upload_2016-6-8_12-21-46.png

In order to find t, I did sinθ=d/ct
Which gives tsinθ=d/c
Which gives t=d/csinθ

If v = 0, vt = 0, and θ = 90
sin90 = 1
t = d/csinθ = d/c
We call this t0


If v is greater than 0, vt is greater than zero, and θ is less than 90
sin90 is less than1
t = d/csinθ is greater than d/c

We use Pythagoras to get t0=t(1- (v2/c2))-1/2

What are the steps involved in getting to this answer?
 

Attachments

Answers and Replies

  • #2
Dale
Mentor
Insights Author
2020 Award
30,850
7,450
In order to find t, I did sinθ=d/ct
Usually both t and θ are considered unknowns. So writing one equation in two unknowns doesn't help. You should use the Pythagorean theorem instead to get one equation in one unknown.
 
  • Like
Likes Clever Penguin
  • #3
Usually both t and θ are considered unknowns. So writing one equation in two unknowns doesn't help. You should use the Pythagorean theorem instead to get one equation in one unknown.
Valid point :wink:

so (ct)2 = d2 + (vt)2
 
Last edited:
  • #4
Dale
Mentor
Insights Author
2020 Award
30,850
7,450
Valid point :wink:

so (ct)2 = d2 + (vt)2
Yes. Then rearranging and using ##d/c=t_0## gives you the desired formula
 
  • #5
Yes. Then rearranging and using ##d/c=t_0## gives you the desired formula
thanks
 

Related Threads on Derivation of special relativity formulae

Replies
2
Views
2K
  • Last Post
Replies
13
Views
1K
Replies
1
Views
893
Replies
82
Views
2K
Replies
92
Views
11K
  • Last Post
Replies
3
Views
852
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
656
  • Last Post
2
Replies
42
Views
6K
  • Last Post
Replies
6
Views
2K
Top