Solving Equations for t': Negative Time Confusion

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Homework Help Overview

The discussion revolves around the interpretation of negative time in the context of special relativity, specifically when calculating time intervals using the equation for t'. Participants are examining the implications of spacelike separation between events and the resulting time calculations.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are questioning how a negative time result can be interpreted, particularly whether it implies that one event occurs before another when t = 0. There are attempts to clarify the conditions under which events are considered spacelike and how this affects their order in different reference frames.

Discussion Status

The discussion is ongoing, with participants exploring the implications of their calculations and questioning the correctness of their interpretations. Some have offered hints and expressed uncertainty about the calculations, while others have pointed out potential misunderstandings regarding terminology and the nature of spacelike separation.

Contextual Notes

Participants note that they do not have access to definitive answers for their calculations, which adds to the uncertainty in verifying their results. There is also mention of difficulties in checking calculations due to the limitations of mobile devices.

Pochen Liu
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Homework Statement
How can the answer be negative time?
Relevant Equations
*attached
Equations:
243341


Question:
243342


Using the equation for t' I got the answer -0.00000499756s (Wrong anyway)
As we know the values:
v = 0.67c m/s
x = 3900 m
t = 0.000005 s

Also how can time be negative? Does this mean that event 2 happened before the first event when t = 0?
 
Last edited:
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Pochen Liu said:
Problem Statement: How can the answer be negative time?
Relevant Equations: *attached

Equations:
View attachment 243341

Question:
View attachment 243342

Using the equation for t' I got the answer -0.00000499756s
As we know the values:
v = 0.67c m/s
x = 3900 m
t = 0.000005 s

How can time be negative? Does this mean that event 2 happened before the first event when t = 0?

If events in one reference frame are spacelike separated (i.e. the distance between them is greater than the time between them multiplied by ##c##), then in another frame they may happen in a different order.
 
Last edited:
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PeroK said:
If events in one reference frame are spacelike (i.e. the distance between them is greater than the time between them multiplied by ##c##), then in another frame they may happen in a different order.
What am I doing wrong?
t' = (0.000005 - (0.67*3900)/c)/sqrt(1-0.67^2)
t' = -0.00000499756s

A hint would be amazing! Have I fundementally misinterpreted the equation
 
Pochen Liu said:
What am I doing wrong?
t' = (0.000005 - (0.67*3900)/c)/sqrt(1-0.67^2)
t' = -0.00000499756s

A hint would be amazing! Have I fundementally misinterpreted the equation
I haven't checked the numbers but that looks correct.
 
PeroK said:
I haven't checked the numbers but that looks correct.
Sorry I forgot to mention it's wrong
 
Pochen Liu said:
Sorry I forgot to mention it's wrong

What is supposed to be the answer? I'm only on my phone, so hard to check properly.
 
They don't provide us answers sadly
 
PeroK said:
If events in one reference frame are spacelike
Jus a pet peeve about terminology. The events in themselves are not spacelike (it is not clear what that would mean), their separation is.
 
Pochen Liu said:
They don't provide us answers sadly
How do you know it's wrong?

Ps I get ##-0.5 \mu s## on my phone.
 

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