γ or the Lorentz Factor is defined as “the factor by which time, length, and relativistic mass change for an object while that object is moving.” Hence in order to find this constant, let’s define a relationship between time and the relativistic time:
T'=t*(constant)
Where T’ is the relativistic time and t is the time taken. Therefore there must be some constant which is multiplied
to t to get T’. For practicalities sake, let’s define this random constant as γ. Hence:
to t to get T’. For practicalities sake, let’s define this random constant as γ. Hence:
T'=tγ
Now let’s do some basic geometry:
In the above figure, we can assume for à source emitting a photon at speed c, the distance travelled is simply
c * the time the photon travels for. This makes sense as the distance à photon would travel over five seconds
would simply be (Sorry for the terrible way the equations are written):
c * the time the photon travels for. This makes sense as the distance à photon would travel over five seconds
would simply be (Sorry for the terrible way the equations are written):
5*c=5*3*10^8=1.5*10^9m
At the same time, we can assume that the source s is travelling horizontally at à velocity v which is à fraction
of the velocity c. We however also know that for an observer O, the time will be relative compared to t (T’ = tγ).
Hence we can deduce that the horizontal distance travelled by the source s is vT’.
of the velocity c. We however also know that for an observer O, the time will be relative compared to t (T’ = tγ).
Hence we can deduce that the horizontal distance travelled by the source s is vT’.
As the source is moving however, the observer O observes the path of the photon to be:
The problem however now is that the observed distance travelled by the photon is greater than the actual
distance. Hence, as c is à constant, the only way the photon could have travelled à greater distance is if it
has travelled over à greater period of time and hence we multiply the speed of the photon c to the dilated
time T’ (as dilated time for all bradyons is greater to actual time) to get the observed distance travelled by
the photon.
distance. Hence, as c is à constant, the only way the photon could have travelled à greater distance is if it
has travelled over à greater period of time and hence we multiply the speed of the photon c to the dilated
time T’ (as dilated time for all bradyons is greater to actual time) to get the observed distance travelled by
the photon.
Using more basic geometry, we know using pythagoras’ theorem that:
(cT')^2 = (ct)^2 + (vT')^2
c^2*t^2 = v^2*T'^2 - c^2*T'^2
c^2*t^2 = T'^2 (v^2 - c^2)
On dividing both sides by c2T'2,we get:
t^2 / T'^2 = (c^2 - v^2) / c^2
On square rooting both sides we get:
t / T' = sqrroot (1 - v^2 / c^2) Eq - 4
As we have previously defined - T'=tγ
Therefore T'/t=γ and hence we have to reciprocate Eq - 4 to get:
T' / t = sqrroot (1 / 1-v^2 / c^2) = γ
Hence above is a very simple method of derivation. Again, quick apologies for the messy difficult to read math.
