# Intuitive explanation for the Lorentz transformation for time

I've recently started learning SR, and while the Lorentz transformation for space is pretty obvious, just the Galilean transformation combined with space contraction, I can't figure out the explanation for the $$\frac{vx}{c^2}$$ term in the time transformation. I understand it has something to do with the time needed for knowledge of the event at $$x$$ to reach the origin, that is $$\frac{x}{c}$$, but I can't find an explanation for the $$\frac{v}{c}$$ term. I've tried many directions, drawing the frames at different times, drawing clocks, analyzing Einstein's train thought experiment, but I haven't found anything. I can follow the mathematical derivation, both the one from interval invariance with the assumption of a linear transformation and the one utilizing the space transformation and symmetry, but I want a more visual, intuitive explanation.. Anyone got some insight?

• youtube.com/watch?v=ajhFNcUTJI0 – BowlOfRed Apr 29 at 22:47
• If you use $\beta=v/c$ then you can get things to look more "symmetric" $$ct'=\gamma(ct-\beta x)$$ $$x'=\gamma(x-\beta ct)$$ – Aaron Stevens Apr 29 at 23:33
• My answer at physics.stackexchange.com/a/383349/4993 might help. – WillO Apr 30 at 6:17
• I have an inkling that your confusion may stem from still thinking about time in a classical sense. In relativity time is a coordinate, and is not absolute. In this sense you can justify the expression between time in a boosted frame in the same way you justify the relation between for the x coordinates. – Ollie113 Apr 30 at 10:45

It would be useful to think about this in terms of Minkowski diagrams. In particular, see this figure showing change of coordinates.

The Lorentz transformation is given as \begin{align*} t' &= \gamma (t - \tfrac{v}{c^{2}}x), \\ x' &= \gamma (x - vt). \end{align*} The $$vt$$ term in the second equation has to do with the $$t$$-axis being tilted to the right. The $$vx/c^{2}$$ in the first equation has to do with the $$x$$-axis being tilted upward.

Now, the change in the $$x$$-axis comes from the fact that the notion of simultaneity changes between different frames. In the $$(t, x)$$-frame, simultaneity is shown by the black horizontal $$x$$-axis, but in the $$(t', x')$$-frame, simultaneity is shown by the blue $$x'$$-axis.

Contrast this to Galilean relativity: in Galilean transformations, simultaneity never changes, so you never tilt the $$x$$-axis, as shown here.

You can get a clearer picture of relativity of simultaneity using exactly the same spacetime diagrams I'm presenting and even relate them to the train thought experiments.

The term $$vx/c^2$$ represents the apparent lack of synchronisation of moving clocks. In other words, if $$S$$, $$S'$$ are frames moving relative to each other, then each of them sees the other's clocks to be unsynchronised, and this term gives the phase difference between them.

This is most easily seen from the Lorentz transformations. Consider an event $$A(t,x)$$ in $$S$$. If the corresponding event occurs at a time $$t'$$ in $$S'$$, we have $$t=\gamma(t'-vx')$$ in $$S'$$. Clearly, since $$t$$ is fixed, the value $$t'$$ depends on $$x'$$-the clocks at different locations will thus record different times, and there will be a corresponding phase difference between them.

Let us do the following thought experiment to see this more explicitly. Consider what happens when you try synchronising clocks and they start moving. If clocks $$A,B$$ are at rest, then they can be synchronised(say, set to zero) using a light beam reaching each of them from their midpoint simultaneously. Now, if the clocks start moving(i.e., you move to a frame where the clocks are moving), this notion of simultaneity is lost. Moreover, the distance $$\Delta x$$ between them will be Lorentz-contracted. The light signals emanating from the midpoint will reach $$A,B$$ at distinct times, and so, in the frame where the clocks are moving, the 2 clocks will be set to zero at different times-there is now a phase difference between them.

It is simple to show that this difference in readings is in fact $$v\Delta x/c^2$$, i.e. this is infact the term encoding this lack of synchronisation(use the fact that the distance between them is contracted, and that the speed of light is the same in this frame too).

As a final remark, note that this DOES NOT mean that the moving frame has no clear sense of time. The moving frame has it's own set of synchronised clocks which are stationary in its frame, and it uses only those clocks to make measurements. See the section 'Failure of relativity' in Schutz for a geometric discussion of this and how this helps resolve apparent 'paradoxes'.