Minkowski spacetime - Overall Distance

  • Thread starter rachelph
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  • #1
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Hi guys,

I'm not exactly sure how to go about doing this problem. The question has no additional information, so I'm stuck. Any help I could get would be appreciated.

1. Homework Statement

In four dimensional Minkowski spacetime an event, A, has components labelled Aμ and another event B has components Bμ. What is the overall proper distance in spacetime between events A and B?

Homework Equations


Proper spacetime distance
ds2 = -dt2 + dx2
 

Answers and Replies

  • #2
andrewkirk
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In that formula you've given, dx represents the purely spatial distance between the two events. In a 3D Cartesian coordinate system we use Pythagoras to work out that dx2 is the sum of the squares of the differences of the three spatial components (aka coordinates). Also, dt is the difference in the time (temporal) components of the two events.
 
  • #3
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In that formula you've given, dx represents the purely spatial distance between the two events. In a 3D Cartesian coordinate system we use Pythagoras to work out that dx2 is the sum of the squares of the differences of the three spatial components (aka coordinates). Also, dt is the difference in the time (temporal) components of the two events.

Okay, so, from my lecture notes I know that Aμ is a column vector with components (t; x; y; z), where t is the temporal component.
And also ds2 = -dt2 + dx2 + dy2 +dz2 ← Am I wrong to think is this the answer?
 
  • #4
andrewkirk
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And also ds2 = -dt2 + dx2 + dy2 +dz2 ← Am I wrong to think is this the answer?
That equation is correct. You now need to express it in terms of the components of the vectors given in the question, in order to answer the question.
 

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