# Interpretation of terms in the Lorentz Transformation

What is the physical significance of the term "${vx}/{c^2}$" in $$t' = \gamma \left(t- \dfrac{vx}{c^2}\right)$$ It has got something to do with clock synchronization but cant derive the 'correction term' logically.

Consider converting the time coordinate to have units of length; then the expression is $$ct' = \gamma c t - \gamma \beta x$$ where $\beta = v/c$.

A look at a spacetime diagram should convince you that the term $-\gamma\beta x$ accounts for the fact that observers in the two coordinate systems will disagree about which events are simultaneous: the $x'$-axis, in the unprimed coordinate system, has slope $\beta$, while the $t'$-axis has slope $1/\beta$. The time $t'$ which a moving observer assigns to an event depends on its position $x$.

There are 2 points:

• as you see from formulas, Lorentz transformations are linear in coordinates
• the actual, interesting terms involved in the Lorentz transformations are the spatial coordinates and $ct$, not just $t$

If you consider that, you see how the time transformation from one frame to the other is now simpler; it is even simpler to see that the Lorentz transformations can be written in matricial form and that the definition of the separation $s^2$ involves a $c^2\Delta t^2$ term. What is more interesting, is that a natural scale appears: $\beta=\frac{v}{c}$, which is actually the only parameter that fully describes the transformation ($\gamma$ is a function of $\beta$, as you see in the image you posted) [to be more precise: in special relativity, Lorentz transformations are in the most general case defined by 4 paramters: $\beta$ and 3 angles; but you can show that by redefining the coordinate axes, the formula is the one you posted].

In summary, the physical significance of the term $\frac{v}{c^2}x$ should be ... of the term $\frac{v}{c}x = \beta x$: $\beta$ is the parameter of the Lorentz transformation, and the $x$ is there because Lorentz transformations are linear in time and space coordinates.

As a reference, this wiki page describes all the above points; in particular:

Sometimes it is more convenient to use β = v/c (lowercase beta) instead of v, so that [...] which shows clearer the symmetry in the transformation.

Concerning your second statement, clock synchronization is a consequence of the formula, therefore it has something to do, but I would not think of clock synchronization as a reason for it; you can look at the same wiki page (or any special relativity introductory course) for a description.

We express equation by differences $$\Delta t' = \gamma \left(\Delta t-\dfrac{v\Delta x}{c^{2}}\right) \tag{01}$$ Now, let in the unprimed system two simultaneous events $\:E_{1}\left(x_{1},t_{0}\right), E_{2}\left(x_{2},t_{0}\right)\:$ at a distance $\:\Delta x= x_{2}-x_{1}\:$ apart as in Figure. Also, let on the middle point $\:M\:$ a primed observer at the instant $\:t_{0}\:$ moving with velocity $\:v$. Suppose that from both events $\:E_{1}, E_{2}\:$ and on time $\:t_{0}\:$ light signals $\:d_{1}, d_{2}\:$ respectively, speed $\:c\:$, are emitted towards the primed moving observer at $\:M\:$ meeting him at points $\:A,B\:$ and on time moments $\:t_{1},t_{2}\:$ respectively. Then

\begin{align} t_{1}-t_{0} & =\dfrac{d_{1}}{c}=\dfrac{d_{1}-\dfrac{\Delta x}{2}}{v}=\dfrac{1}{2}\dfrac{\Delta x}{c-v} \tag{02a}\\ t_{2}-t_{0} & =\dfrac{d_{2}}{c}=\dfrac{\dfrac{\Delta x}{2}-d_{2}}{v}=\dfrac{1}{2}\dfrac{\Delta x}{c+v} \tag{02b} \end{align} so $$t_{2}-t_{1}= -\dfrac{v\Delta x}{c^{2}-v^{2}}= - \dfrac{1}{1-\tfrac{v^{2} } {c^{2}}}\dfrac{v\Delta x}{c^{2}}=-\gamma^{2}\left[\dfrac{v\Delta x}{c^{2}}\right] \tag{03}$$

Now, the term for which you are asking for appears in brackets in the last to the right side of above equation but with the coefficient $\:\gamma^{2}\:$. The time difference $\:\left(t_{2}-t_{1}\right)\:$ is that seen by the unprimed "rest" observer between the "meetings" of the light signals and the primed "moving" observer.

Note that the time difference as seen by the moving observer (primed system) is

$$\Delta t'=t'_{2}-t'_{1}= -\gamma\left[\dfrac{v\Delta x}{c^{2}}\right] \tag{04}$$ if in (01) for the two events $\:E_{1}, E_{2}\:$ we insert $\:\Delta t= t_{0}-t_{0}=0\:$ and $\:\Delta t'= t'_{2}-t'_{1}$. I don't think that such an interpretation offers to us a more intuitive explanation since moreover is a "a posteriori" result of the existing Lorentz Tranformation and not a "a priori" tool for the production of this transformation.

Note that in a general Lorentz Transformation equation (01) is $$\Delta t' = \gamma \left(\Delta t-\dfrac{\mathbf{v}\boldsymbol{\cdot}\Delta \mathbf{x}}{c^{2}}\right) \tag{05}$$