0
$\begingroup$

Let O1 be an observer in frame S1 where F(g) is very high. And O2 be an observer in S2 where F(g) is lower. Due to time dilation the time for O2 would appear to pass slowly from S2 and vice versa. Therefore could an event in S2, E1 be observed (and information about it transmitted) by O1 (for whom time in S2 is running faster) to O2 thereby giving information about an event that hasn't occurred in O1's frame of reference. This is assuming that the spacetime separation for E1 from O2 is mainly composed of the (ct)^2 (it happens a significant distance in time). Is there some logical error in my thinking?

Edit: Here is my proposed logic for the answer by flippiefanus:

Let $O_1$ be an observer in frame $S_1$. Also Let $O_2$ be an observer in frame $S_2$.
Now let an event $E$ occur in frame $S_2$.

The following calculations are based in frame $S_1$ with reference frame of $O_1$:

Using the gravitational time dilation equations: $$ \Delta t_o (O_1E_1) = \Delta t_{O_1E_1} \sqrt{1-\frac{r_{S_1}}{r}} $$ And, $$ \Delta t_o (O_1O_2) = \Delta t_{O_1O_2} \sqrt{1-\frac{r_{S_1}}{r}} $$

Also, $$ \Delta t_o (O_2E) = \Delta t_{O_2E} \sqrt{1-\frac{r_{S_1}}{r}} $$ (The above equation is a rather important one. The reason this equation uses $r_{S_1}$ instead of $r_{S_2}$ to calculate all the distances from the same frame of reference)

In the above equations $t_o$ is the proper time which is invariant across frames and $t_{O_xO_y}$ is the time between the events as measured in a frame with a low gravitational field (frame $S_2$). however, the exact frame does not matter as long as all the calculations are made in the same frame.

Now, $$ t_o (O_1E+O_1O_2)-\Delta t_o (O_2E) = \sqrt{1-\frac{r_{S_1}}{r}}(\Delta t_{O_1E}+\Delta t_{O_1O_2}-\Delta t_{O_2E}) $$ $$ = \sqrt{1-\frac{r_{S_1}}{r}}(cT_1+cT_2-cT_3) $$ where $T_1,T_2,T_3$ are the times measured in frame $S_2$ (or any other frame). They are multiplied by $c$ to keep up with convention and convert units of distance, this does not affect the value of time since $c$ is invariant.

Therefore,

We get a 3 irreducible cases (in 2d space):

1st case:- 1st case Here $(cT_1+cT_2-cT_3)$ is exactly 0 $\Rightarrow$ both signals reach at same time (assuming no time is used in re transmission)

2nd case:- where E happens to the right of $O_2$ with $O_1$ to the right of $E$.

Here $(cT_1+cT_2-cT_3)$ is positive implying $\Delta t_o (O_1E+O_1O_2)>\Delta t_o (O_2E) \Rightarrow$ signal from $E\rightarrow O_1 \rightarrow O_2$ reaches $O_2$ after $E\rightarrow O_2$

3rd case:- where E happens to the left of $O_2$ with $O_1$ to the right of $O_2$.

Here too $(cT_1+cT_2-cT_3)$ is positive implying $\Delta t_o (O_1E+O_1O_2)>\Delta t_o (O_2E) \Rightarrow$ signal from $E\rightarrow O_1 \rightarrow O_2$ reaches $O_2$ after $E\rightarrow O_2$

Since proper time is invariant we don't need to confirm calculations with distance in spacetime, however this would be pretty simple since the distances from $S_2$ would also be invariant in the same frame.

Therefore we get a light cone where the signal $E\rightarrow O_1 \rightarrow O_2$ when received at $O_2$ is a time like event whereas the signal $E \rightarrow O_2$ when received at $O_2$ is a light like event

Light cone

$\Rightarrow$ $E\rightarrow O_1 \rightarrow O_2$ can never reach $O_2$ faster than $E \rightarrow O_2$.

It seems the doubts I was having were similar to those of the twin effect, that is- I was looking at the different parts of the signal journey from different reference frames. Once I got that sorted out things became clearer. I hope this can help someone else if they get confused in a similar situation, this at least reinforced the importance of reference frames for me.

$\endgroup$

closed as off-topic by WillO, sammy gerbil, heather, knzhou, ACuriousMind Aug 8 '16 at 21:32

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – WillO, sammy gerbil, heather, ACuriousMind
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ The short answer is no, it isn't possible. But for specifics your configuration isn't quite clear: what is $F(g)$, where in $S1(S2)$ are $O1(O2)$, $E1$ located, etc. $\endgroup$ – udrv Aug 8 '16 at 3:59
  • $\begingroup$ What was the use of watching such future? $\endgroup$ – Anubhav Goel Aug 8 '16 at 4:53
  • $\begingroup$ Once you factor in the light travel time you do not "see" time dilation. You see the Doppler effect. $\endgroup$ – m4r35n357 Aug 8 '16 at 21:17
  • $\begingroup$ As asked I have added some of my working for the question $\endgroup$ – pranay Aug 8 '16 at 23:26
  • $\begingroup$ @pranay, it is practically impossible to read the photos. Please put the equations in LaTeX format, the text in normally, and any diagrams you can photograph and put in right-side up (with good lighting, and large enough that we can read them). $\endgroup$ – heather Aug 9 '16 at 0:45
2
$\begingroup$

Quantitatively it can never happen, because the speed with which information is exchanged between the two systems cannot go faster than the speed of light. The time it takes for the signal of an event in S2 to go from S2 tot S1 plus the time it takes for the signal from O1 to go to O2 would be longer as perceived in S2 than the time it takes for O2 to catch up with the event is his own system.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.