1
$\begingroup$

This question already has an answer here:

If simultaneity is relative; meaning that for each any events A and B and we can find a reference frame in which two events A and B occur in inverse order, how does our notion of 'causality' withholds?

$\endgroup$

marked as duplicate by Ben Crowell, Qmechanic Aug 5 at 17:50

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1
$\begingroup$

Spacetime causality holds within the light cone. Only events within the past light cone can affect the present event (the here-and-now at the vertex of the light cone). The present event can affect only the future light cone. Events outside the light cone can neither affect nor be affected by the present event.

Lorentz transformations can change the temporal ordering of events outside the light cone relative to the present event, but they cannot change the temporal ordering of events inside the light cone relative to the present event. All events in the past light cone are temporally before the present event in all Lorentz frames, and all events in the future light cone are temporally after the present event in all Lorentz frames.

The essence of the events inside the light cone are that they are closer in space than they are in time to the present event, so signals traveling at the speed of light or less can travel from those in the past light cone to the present event or from the present event to those in the future light cone. To get to or from events outside the light cone would require traveling faster than light.

$\endgroup$
1
$\begingroup$

1) If there exists a frame in which B precedes A, then A cannot cause B.

2) That having been said, the whole notion of "causality" is a little slippery to begin with in a deterministic theory where all the laws are time-reversible.

$\endgroup$
  • $\begingroup$ I think you mean "in which B precedes A". $\endgroup$ – Danny Aug 5 at 16:48
  • $\begingroup$ @Danny : Fixed. Thank you. $\endgroup$ – WillO Aug 5 at 16:53

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