Deriving Special Relativity Formulae Steps

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Discussion Overview

The discussion revolves around the derivation of special relativity formulae, specifically focusing on the steps involved in understanding the relationship between time, distance, and velocity using a light clock diagram. The scope includes mathematical reasoning and technical explanations related to special relativity.

Discussion Character

  • Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant presents a derivation involving a light clock and the relationship between time, distance, and angle, leading to the expression t = d/csinθ.
  • Another participant points out that treating both time (t) and angle (θ) as unknowns complicates the derivation and suggests using the Pythagorean theorem for clarity.
  • A later reply agrees with the suggestion to use the Pythagorean theorem, proposing the equation (ct)² = d² + (vt)² as a more effective approach.
  • Further responses indicate that rearranging this equation and substituting d/c = t₀ can lead to the desired formula, although the exact steps are not fully detailed.

Areas of Agreement / Disagreement

Participants express differing views on the initial approach to the derivation, with some advocating for the use of the Pythagorean theorem while others initially propose a different method. The discussion does not reach a consensus on the best approach to derive the formulae.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about the variables involved, particularly the treatment of t and θ as unknowns. The mathematical steps leading to the final formula are not fully resolved.

Clever Penguin
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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 t0If 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?
 

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Clever Penguin said:
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.
 
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Dale said:
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:
Clever Penguin said:
Valid point :wink:

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

thanks
 

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