# Lorentz transformation equations

There are many ways to derive the Lorentz transformations utilizing a variety of mathematical tools, spanning from elementary algebra and hyperbolic functions, to linear algebra and group theory.

• The primed frame moves with velocity v in the x direction with respect to the fixed reference frame.
• The reference frames coincide at t=t'=0.
• The point x' is moving with the primed frame.
• #### Transformation

The Lorentz transformation is the transformation rule under which all four-vectors and tensors containing physical quantities transform.The prime examples of such four vectors are the four position and four momentum of a particle, and for fields the electromagnetic tensor and stress–energy tensor. The fact that these objects transform according to the Lorentz transformation is what mathematically defines them as vectors and tensors, see tensor.Given the components of the four vectors or tensors in some frame, the "transformation rule" allows one to determine the altered components of the same four vectors or tensors in another frame, which could be boosted or accelerated, relative to the original frame. A "boost" should not be conflated with spatial translation, rather it's characterized by the relative velocity between frames. The transformation rule itself depends on the relative motion of the frames. In the simplest case of two inertial frames the relative velocity between enters the transformation rule.

$$x' = γ(x - vt)$$ $$y' = y$$ $$z' = z$$ $$t' = γ(t - \frac {vx} {c^2})$$

in all inertial frames for events connected by light signals. The quantity on the left is called the spacetime interval between events (t, x, y, z) and (t', x', y', z').

Now undertand the new transformation equations relating (x,y,z,t) and (x',y',z',t') and moving in the x direction. They can again synchronise clocks: for convenience and symmetry, when they are side by side, they call that position zero and time zero.
means x = 0 at t = 0 and x' = 0 at t' = 0. Now further at this moment, they can check that measurements in the y and z directions coincide, so two of the transformation equations are easy:
y = y' and z = z'. Same as Galileo.

However, we know that they disagree over time and length, so we need new transform equations for x and t. There is a reason to try equations that are linear in x and t.
First, we have seen that the time dilation and length contraction are linear in time and length.
Second, if they are not linear, then we could end up with two solutions of x and t for a single event and it will be awkward for a frame that's travelling with constant speed v! So we may write:
x' = Ax + Bt,
y' = y,
z' = z,
t' = Ct + Dx.
where A, B, C and D are constants (independent of t and x: they will of course be functions of v, because at v = 0, we can see that A and C = 1, while B and D = 0. Now x' = 0 marks the back of the trian, so x = vt. Substituting in the first equation gives
A = −B/v.
At t = 0, person1 sees person2's length contraction, so
x' = γx = Ax
and by combining these equations we get:
x' = γ(x - vt).
Now let's look at time dilation, again from person1's point of view. At x = 0, he sees person2's clock running slow, so t'  = γt, which gives

C = γ.
Now let's look from the point of view of the back of the train, where x' = 0. From this position, person2 sees person1's clock running slow, so at x' = 0, we have t = γt'. We also know that the train is moving at v with respect to person1, so for the position x' = 0, we also have x =  vt. Substituting these into the last transformation equation gives
t/γ = γt - Dvt,     whence
1 = γ2 - Dγv      and so
D = (γ2 - 1)/γv.
This gives us the Lorentz transformation equations: $$x' = γ(x - vt)$$ $$y' = y$$ $$z' = z$$ $$t' = γ(t - \frac {vx} {c^2})$$
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