Minkowski spacetime - Overall Distance

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Homework Help Overview

The discussion revolves around calculating the overall proper distance in four-dimensional Minkowski spacetime between two events, A and B, represented by their components Aμ and Bμ. The problem lacks additional context or specific values, which has led to uncertainty among participants.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the formula for proper spacetime distance, with some attempting to clarify the components involved, such as the spatial and temporal differences. There is also a question regarding the correct formulation of the distance equation based on lecture notes.

Discussion Status

Some participants have confirmed the correctness of the equation for proper distance, while others are exploring how to express this in terms of the specific components provided in the problem. There is an ongoing examination of the definitions and implications of the variables involved.

Contextual Notes

The original poster notes a lack of additional information in the problem statement, which may affect the ability to proceed with a solution. Participants are also considering the implications of using the correct components in the distance formula.

rachelph
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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
 
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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.
 
andrewkirk said:
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?
 
rachelph said:
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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