Albertrichardf said:
Hi all,
What is the difference between Lorentz transformations and yt?. That is, the Lorentz transformations for moving between two reference frames are not the same as the relativistic ones.
For example considering a frame F that is stationary and an inertial frame F' with velocity v. Time dilation for frame F' is given by yt, where gamma is the relativistic factor. But the Lorentz transformations give y(t - [xv/c2 ]). These are obviously different. So which one is correct? Or is time dilation a special case of a transformation? And Length contraction is also not the same as space transformations. So is it a special case as well?
Thanks for any answers.
Relativistic time dilation for an inertial clock is a special case of the Lorentz transforms, and Lorentz contraction is a special case of the Lorentz transforms.
If you have two events e_1 and e_2, let \delta x be x_2 - x_1 and let \delta t = t_2 - t_1, as measured in one frame, F. Let \delta x' and \delta t' be the corresponding quantities in another frame, F'. Then the LT says that:
\delta x' = \gamma (\delta x - v \delta t)
\delta t' = \gamma (\delta t - v/c^2 \delta x)
where v is the velocity of F' relative to F.
So now let's look at some special cases. Suppose that you have a clock that ticks once per second in frame F'. Then letting the two events be two successive ticks of the clock, we have:
\delta t' = 1
Because in F', the ticks are one second apart.
\delta x' = 0
Because in F', the ticks are at the same location.
Plugging these into the LT gives:
\delta x'= 0 = \gamma (\delta x - v \delta t)
So \delta x= v \delta t
\delta t' = 1 = \gamma (\delta t - v/c^2 \delta x)
= \gamma(\delta t - v/c^2 \cdot v \delta t)
= \gamma \delta t (1 - v^2/c^2)
= 1/\gamma \delta t
So \delta t = \gamma. So the time between ticks in frame F is \gamma, which is greater than 1.
Now, another special case is a ruler at rest in frame F' (oriented in the direction of motion). Let e_1 be the location of one end of the ruler at one moment, and let e_2 be the location of the other end at the same time, according to frame F. Let L be the length of the ruler in its own rest frame, F', and let \tilde{L}be its length in frame F.<br />
<br />
Then we have:<br />
<ol>
<li data-xf-list-type="ol">\delta x = \tilde{L}</li>
<li data-xf-list-type="ol">\delta t = 0 (because the two events take place at the same time, in F).</li>
<li data-xf-list-type="ol">\delta x&amp;#039; = L</li>
</ol><br />
Plugging these into the LT gives:<br />
\delta x&amp;#039; = L = \gamma (\delta x - v \delta t) = \gamma (\tilde{L} - 0) = \gamma \tilde{L}<br />
<br />
So \tilde{L}= L/\gamma<br />
<br />
So in frame F, the ruler is shorter by a factor of \gamma.