# Time transformation in Lorentz transformation - what happens when the sign of $v$ in the expression is reversed?

The following question may be a silly one and already answered by somebody somewhere. My apologies for failing to locate the right answer after spending a considerable amount of time in that endeavour.

Using the well-known Lorentz transformation expression, the progress of time in a moving system is given by: $$t'=\gamma \left(t-\frac{v x}{c^2}\right)$$ Suppose the moving system is currently located far removed from origin in the positive x-direction (very large positive x-coordinate), and in this case we expect time to go slow as per the above equation. Now, further suppose that the sign of $$v$$ is suddenly reversed, (ignoring the effect of deceleration and subsequent acceleration in the reverse direction for this to happen) would time start go faster in the moving system?

You are misinterpreting the equation. It doesn't mean that time speeds up or slows down at a distance- it means that the time in one frame is out of synch with the time in the other.

If you and I are moving relative to each other, our 'planes of simultaneity' are tilted. For you, 'now' is a horizontal slice through spacetime at right angles to your time axis. For me, your 'now' is a sloping slice, heading upwards in time in your direction of travel, and downwards in time in the opposite direction.

That is the key to the relativity of simultaneity, since two events that happen at the same time in your frame will generally happen at two different times in mine.

And yes, if you reverse your velocity, your time axis and your planes of simultaneity will become tilted in the opposite sense relative to mine. This is what happens in the classic 'twin paradox' set-up, in which one twin travels away from Earth then returns. At the turnaround point, 'now' for the traveller shifts from being one date on Earth to a much later date, owing to the pivoting of the traveller's plane of simultaneity.

• "That is the key to the relativity of simultaneity, since two events that happen at the same time in your frame will generally happen at two different times in mine."If two events occur at two points symmetrical with respect to the observer at rest and symmetrical with respect to the axis of the velocity vector of the moving observer (which passes through the origin of the one at rest), the two events will be simultaneous for both observers. Aug 31, 2022 at 11:39
• @TheTiler Agreed. That's why I included the word 'generally'. in the special case in which the events are on a line normal to the direction of motion then they are unaffected. Aug 31, 2022 at 12:50
• @marco-ocram the system being at a large positive distance away was solely introduced to make it clear that $x$ coordinate never went negative during the period under consideration. I was trying to figure out how Einstein would have answered such a question just after his theory was published in 1905. It is important to note that Minkowski developed his space time formalism only in 1908. Aug 31, 2022 at 23:46
• @JKrsl we can only guess how Einstein would have answered a question, but in his 1905 paper he clearly appreciated that simultaneity is relative and that all the effects he predicted would apply symmetrically between two inertial frames, so I can't imagine that he would have had any conceptual difficulty with the question. Sep 1, 2022 at 5:19

Yes, time is starting to contract following the relationship (with $$x=ct$$):

$$t'=\gamma(1+\beta )t=\sqrt{\frac{1+\beta}{1-\beta}}\;\;t$$

If the traveler takes with him a clock that works with electromagnetic energy sent by a terrestrial station (a laser of period T), we will have during the traveler moves away: $$t_{1}'=nT_{1}'=\gamma(1-\beta )nT=\sqrt{\frac{1-\beta}{1+\beta}}\;\;nT=\sqrt{\frac{1-\beta}{1+\beta}}\;\;t_{1}$$ During his return, we have: $$t_{2}'=mT _{2}'=\gamma(1+\beta )mT=\sqrt{\frac{1+\beta}{1-\beta}}\;\;mT=\sqrt{\frac{1+\beta}{1-\beta}}\;\;t_{2}$$ If $$n=m$$, we have: $$t_{1}'+t_{2}'=t'=n(T_{1}'+T_{2}')=2\gamma nT$$ $$t'=2\gamma t$$ we fall back on the formula of the Michelson-Morley* experiment for the perpendicular arm of the interferometer.