Timelike separation: which of these two answers do you prefer?

  • Thread starter Thread starter daselocution
  • Start date Start date
  • Tags Tags
    Separation
AI Thread Summary
For a timelike interval, two events cannot occur simultaneously in any reference frame, as demonstrated by the invariant nature of the spacetime interval. The first proposed solution attempts to show that if two events occur at different spatial coordinates, there is no frame where they can be simultaneous, but it raises concerns about tautology. The second solution focuses on a single inertial frame, concluding that if the time difference is zero, the spatial separation must be non-negative, contradicting the requirement for a timelike interval. Participants in the discussion express confusion about the validity of both approaches, with the second being regarded as clearer. The overall consensus emphasizes the importance of understanding the implications of timelike intervals in spacetime.
daselocution
Messages
23
Reaction score
0

Homework Statement


This is the problem as stated in my textbook: Prove that for a timelike interval, two events can never be considered to occur simultaneously.

I thought about two different ways to go about this; both of these ways make sense to me, but I'm not sure if they both make sense to all of you/if one is significantly better/different from the other


Homework Equations


Δs'2=∆s'2

(∆s is invariant)

s^2=x^2 - (ct)^2
s'^2=x'^2 - (ct')^2

For a timelike interval, it is given that: ∆s^2<0



The Attempt at a Solution



The first way I thought of doing the problem: At first I thought that I should be showing that if two events occur at (x1, t1) and (x2, t2) in system K, then there is NO frame K' such that these two events could occur at the same time (simultaneously). If this is a misunderstanding or seems tautological, then please correct me.

I went about proving it as follows:
∆s^2=∆x^2 - (c∆t)^2 = ∆x'^2 - (c∆t')^2 = ∆s'^2

Where, in K', we are assuming that t2'-t1'=0 such that:

∆s^2=∆x^2 - (c∆t)^2 = ∆x'^2 = ∆s'^2

From here, it follows that as ∆s^2=∆x^2 - (c∆t)^2<0 for it to be timelike separation, this entire quantity must be negative in sign. However, the right side of the equation, ∆x'^2, must necessarily be positive. Because it is given that s^2 is invariant, it must also be true that ∆s^2 is invariant from reference frame to reference frame, and so this must be impossible because ∆s^2≠∆s'^2

The second way I tried to solve the problem: At this point I had doubts about whether my first solution made sense, and so I tried to solve it all in the context of ONE inertial frame:

∆s^2=∆x^2 - (c∆t)^2

For the interval to be timelike, ∆s^2=∆x^2 - (c∆t)^2<0;
thus, if we are assuming that t1=t2, then:

∆s^2=∆x^2 - 0 CANNOT be less than 0, as it MUST be a positive number or it must be zero, neither of which allow for a timelike interval.

What do all of you have to say? Are both wrong? One right one wrong? One better?
 
Physics news on Phys.org
The second one is much cleaner.
I think I would begin with "for a timelike interval, there is a reference frame where ∆s^2<0. As this is invariant, ∆s^2<0 for all reference frames. Assume there is a frame where ∆t=0, ..."

At first I thought that I should be showing that if two events occur at (x1, t1) and (x2, t2) in system K, then there is NO frame K' such that these two events could occur at the same time (simultaneously).
In general, this is not true. As an example, consider (x1,0) and (x2,0), and K'=K. You have to add some constraints on those coordinates, and I don't see the advantage of this.
 
Ahhh thank you for clearing that up. I don't think I actually fully understood what I was getting into with the first attempt...
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top