# Lorentz transformations for time co-ordinates (STR)

I am little bit confused with the implication of Lorentz transformations for time co-ordinates or atleast how to apply those!

Consider 2 frames of reference $$O$$ and $$O'$$ in which $$O'$$ is moving with $$v$$ velocity in $$+ve$$ $$X$$ direction with respect to $$O$$ frame!

Now consider a clock at $$x=d$$ and it is reading $$t=0$$ at particular instant! Now if we want to know the time co-ordinates of the same clock ($$x=d$$) in frame $$O'$$ at that same instant, then applying Lorentz transformations: $$t'=(t-vx/c^2)∆$$ and and $$t=0,x=d$$ thus $$t'=(-vd∆/c^2) \tag{1}$$ However the same time co-ordinate could be evaluated through reverse Lorentz transformations, i.e. $$t=(t'+vx'/c^2)∆$$ and we also know that $$x/∆=x'$$ (length contraction) and $$t=0$$, thus $$t'=vd/∆c^2 \tag{2}$$

We see that (1) and (2) are inconsistent with each other. I dont know what i am missing here or if i have my basics completely wrong!

Anyhow a detailed answer or the complete explanation of concepts would be helpful!

PS: $$∆$$ is Lorentz factor

• Why are you using $\Delta$ rather than $\gamma$ for the Lorentz factor? – G. Smith Sep 8 '19 at 18:04
• This page explains how to use MathJax. Using it will make your equations look just like a testbook and make it more likely your question gets answered. – G. Smith Sep 8 '19 at 18:05
• Ok thanks, will look into it – DigPhysics Sep 8 '19 at 18:34

Second, use the Lorentz transformation $$x’=\gamma(x-vt)$$ to get $$x’=\gamma d$$ for the $$x’$$-coordinate of the spacetime event with $$t=0$$ and $$x=d$$.

Bingo, no inconsistency!

• Yea but can u point out whats wrong with length contraction method, Like Two clocks are moving wrt O' So line joining two clocks should be contracted than the line joining the two clocks when they are at rest? Thus this way we can say x'=d/∆? – DigPhysics Sep 8 '19 at 20:17
• And Thanks For Your response ^_^ – DigPhysics Sep 8 '19 at 20:18

However the same time co-ordinate could be evaluated through reverse Lorentz transformations ...

The Lorentz transformation is used for translating the coordinates from the stationary reference frame to the moving one: O -> O′ in your example.

The reverse Lorentz transformation is used in the opposite direction: O′ -> O.

If you use both transformations for the O -> O′ translation, you will of course get different results.

Also note that time t(0) in one reference frame is different from time t(0) in a different reference frame.

Modification in my answer : (replace $$S$$ with $$O$$ for your problem)

Equation $$(1)$$ of your approach is the correct one (in case you don't understand anything from my answer, I should at least let you know this).

I believe that you want to understand why your approach failed. It failed because you didn't apply length contraction correctly. It should be $$x = x'/\Delta$$, since the length must be measured at equal times (t=0) in the S frame.

It is very instructive to imagine, an infinitely long ruler stick with clocks affixed at each point on the ruler, to be attached to the coordinate frame. When the frame moves, so does the ruler, and in turn, so do the clocks affixed on the ruler. These clocks are synchronized with respect to the frame in which they are at rest.

With this in mind, please refer to part 1 of the figure given below, in which $$S$$ is at rest. In my diagrams, the straight line represents the ruler. Choose a random point on the ruler: the position marking at that point (like 5 cm) on the ruler and the reading of the clock that is attached (like 3 min) at that position, is represented by the ordered pair $$(x,t)$$. You can see that the clocks in $$S$$ are synchronized while the moving $$S'$$ clocks are not.

The length between the origin of $$S$$ and the point $$x=d$$ in frame $$S$$ is just $$d$$ but the moving $$S'$$ ruler tells us that the separation is $$d \Delta$$. This is because the moving $$S'$$ ruler has shrunk by a factor of $$\Delta$$.

Now, why is $$x = x'/\Delta$$ incorrect when you make the length measurement in $$S'$$? Observe part 2 of the figure. This illustrates the scenario from the point of view of $$S'$$ ($$S'$$ is the rest frame and $$S$$ is moving to the left with speed $$v$$). The time in $$S'$$ is $$t'=-{{vd\Delta} \over {c^2}}$$ (not zero). Now, you notice that the moving $$S$$ clocks are not synchronized. We are now ready to make length measurements in $$S'$$ frame.

The diagram shows that the actual length measured between the origin of $$S'$$ and the point $$x'=d \Delta$$ in frame $$S'$$ is $$d \Delta$$. But the moving $$S$$ ruler has shrunk just like we have seen before. The "position markings" on the $$S$$ ruler tells us that the difference is $$d \Delta^2$$. So, the length has been contracted by a factor of $$\Delta$$ to give $$d \Delta$$.