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I found this picture from a physics stack exchange question on time dilation: Time dilation all messed up!

I am now returning to it a few weeks later and was wondering if I am correctly interpreting what the diagram is showing, as I seem to run into a slight issue.

When the moving observer reads 11 o clock on his wrist watch that corresponds with a line through space time parallel to the x' line, such that all events on that line appear to simultaneously occur at 11 o clock for him. In this case, the event of the stationary observer at x = 0, reading 9 o clock on his wrist watch lies on the t' = 11 o clock line for the moving observer.

My question is, what would happen if the stationary observer was at x = 1 or basically just closer to the moving observer, when he reads t' = 11 o clock. From the diagram it appears as though this means that the line (t'=11) cuts the t axis further up (later on in the stationary observers time). But according to the equation $t′=γt$. There should be no position dependence, the moving observer should only see one time on the stationary observers clock regardless of how close or far he is.

Where have I gone wrong?

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  • $\begingroup$ If you change the position then either the time or the velocity have to change. $\endgroup$ – Javier Mar 6 at 20:19
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    $\begingroup$ If you can find the original question, kindly put a link to it in your question with the diagram. The diagram is so neatly drawn, I am inclined to think that the question must be equivalently beautiful. ;-) $\endgroup$ – Feynmans Out for Grumpy Cat Mar 6 at 20:48
  • $\begingroup$ Found it! Just had to find the date I saved the picture and check my history for time dilation related searches on that date haha. $\endgroup$ – Vishal Jain Mar 6 at 20:56
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Time dilation is not itself position dependent, that means moving observer will see all the clocks in the rest frame ticking at same rate. But a same rate does not imply same time on clocks. Basically what is happening here is that moving observer does not agree that all the clocks in rest frame are synchronized. And hence there is a time offset between clock placed in different position. You can see this in Lorentz's transformation equations let's say that coordinates of moving observer are t',x',y',z' and in the rest frame coordinated are t,x,y,z

$$ t'=\frac{t-\frac{vx}{c^2}}{\sqrt{1-\frac{v^2}{c^c}}} $$

you can see that whan you put x=1 in above equation there is a $\frac{vx/c^2}{\sqrt{1-\frac{v^2}{c^c}}}$ term in it which basically arises from the synchronization error.

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  • $\begingroup$ Ah yes i missed the fact that if the stationary observer were to move then the line t'=0 would no longer correspond with t = 0, hence the two clocks wouldn't be synchronised. $\endgroup$ – Vishal Jain Mar 6 at 20:43
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    $\begingroup$ @VishalJain As a rule of thumb, always trust a spacetime diagram over your equations! ;-) Of course, you can/should always make both the modes of thinking consistent with each other but what I mean is that it is likely that you made a mistake in your formulae than that you misread the spacetime diagram. $\endgroup$ – Feynmans Out for Grumpy Cat Mar 6 at 20:46

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