Question about position dependence of time dilation? 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?

• If you change the position then either the time or the velocity have to change. – Javier Mar 6 at 20:19
• 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. ;-) – Dvij Mankad Mar 6 at 20:48
• 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. – Vishal Jain Mar 6 at 20:56

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