Tazerfish said:
In the way I was taught about special relativity, time dilation is like the fundamental building block from which you derive things like relativistic mass and length contraction.
So it has always struck me as quite odd, that the derivation of time dilation (in some sense the basis of special relativity) uses something as abstract as light bouncing up and down in a box to measure time.
Something that has never been built in reality to my knowlege.
I don't have any problems with this proof.It is very elegant and simple.
Yet there rarely is just one nice derivation or proof for such things
and I have never seen any alternetative for this case.
Have you ?
Is there a way to derive it by considering some other process ? :)
I learned to derive it using light clocks too, but not in a class (in class they just said "this is the Lorentz transformation equations") . I just noticed the Lorentz factor looking like a side of a right triangle and went from there. But later on I learned a better way that really only assumes that the formula distance = rate x time is the same regardless of inertial coordinate system and requires only algebra (and probably some characteristics of space and time that I just took for granted).
I'll post it and you tell me if this might be better than the light clock way. Skipping no steps so that any mistake I make will be clear for someone to fix.
You have two reference frames, S and S' moving at some speed with respect to each other with their x and x' axes coinciding.
In S, you can shoot a beam of light in either direction along the x-axis. When you do, it's distance x is given by x = ct or x = -ct (depending on the direction). Likewise in S' you can do the same: x' = ct' or x' = -ct'. Setting each equal to zero gives: x - ct = 0, x + ct = 0 and x'-ct' = 0, x'-ct' = 0. Pretty straight forward so far. Then you can write them the other way: ct - x =0, ct + x = 0, and ct'-x' =0 and ct' + x' = 0.
Next I just assumed there was some functions A and B such that (1) x - ct = A(x' - ct'), (2) x + ct = B(x'+ct') and (3) ct - x = A(ct' - x'), (4) ct' + x' = B(ct + x).
Then add (1) and (2) together to get:
x = \frac{A+B}{2} x' + \frac{B-A}{2} ct'
And add (3) and (4) to get:
ct = \frac{A+B}{2} ct' + \frac{B-A}{2} x'
Then to make it easier to read, let
\frac{A+B}{2} = γ
and
\frac{B-A}{2} = ξ
leaving
(5)
x = γx' + ξct'
ct = γct' + ξx'
Then just remember that you have the inverse transformations, which would involve just swapping the sign and replacing the prime and unprimed coordinates:
(6)
x' = γx - ξct
ct' = γct - ξxAt that point all you have to do is find ξ, which is easy if you realize any object moving with uniform velocity is at rest in it's own frame, so there is always going to be a case where x' = 0. Which means that x/t in this case = v and is the speed at which S' is moving relative to S. Which then means you can solve for ξ by letting x' = 0 in the third equation:
x' = γx - ξct with x'=0 gives
0 = γx - ξct
γx = ξct
γv = ξc
γ\frac{v}{c} = ξ
Then you just plug that into the four equations in (5) and (6). Start with the first one in (5):
x = γx' + γ\frac{v}{c}ct'
Then substitute in x' and ct' from (6):
x = γx' + γ\frac{v}{c}ct'
x = γ(γx - γ\frac{v}{c}ct) + γ\frac{v}{c}γ(ct - γ\frac{v}{c} x)
Then just clean it up and solve for the last unknown.
x = γ(γx - γ\frac{v}{c}ct) + γ\frac{v}{c}γ(ct - γ\frac{v}{c} x)
x = γ^2([x - vt] + \frac{v}{c}[ct - \frac{v}{c} x])
x = γ^2(x - vt + vt - \frac{v^2}{c^2} x)
x = γ^2(x - \frac{v^2}{c^2} x)
x = γ^2x(1 - \frac{v^2}{c^2})
1 = γ^2(1 - \frac{v^2}{c^2})
γ^2= \frac{1}{(1 - \frac{v^2}{c^2})}
γ= \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}
And then you have the Lorentz factor, so just plug it into (5) and (6). To get time dilation, just assume that x'=0 because the person looking at their clock will be at the location of their clock if their time is proper time:
ct = γ(ct' + \frac{v}{c}x') at x'=0 is
ct = γct'
t = γt'Anyway, how is that derivation? Obviously if you orient your axes a certain way, y = y' and z = z', so no issue there. You can pretty much derive all of it once you get to this point.