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Proper time interval between two events

  • #1
The course this question comes from is Modern Physics

Homework Statement



What is the proper time interval between two events if in some inertial reference frame the events are separated by 109m and 5s?

Homework Equations



I looked through my notes and under a the sub topic "Invariant Intervals" I found this equation:

(Interval)2= c2 [time separation]2-[space separation]2

The Attempt at a Solution



I'm just having trouble understanding what exactly is being asked here. I tried plugging in the values given into the equation but ran into trouble with the units.

I also dont just want to plug in values into the equation without actually understanding what I'm actually looking for. I think this probably has something to do with special relativity since something traveling at 2x108m/s is obviously traveling close to the speed of light

Oh and this was the results of my attempt:

(Interval)2= (3x108m/s)2 [(5s)2-(109m)2]

(Interval)2= 2.25x1018m2-9x1034m4/s2
 

Answers and Replies

  • #2
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Your result doesn't make sense. You cannot add quantities with different units! You didn't use the formula correctly: the spacetime interval is given by
[itex](\Delta s)^2=c^2(\Delta t^2)-(\Delta x)^2[/itex].
Only the time interval is multiplied with [itex]c^2[/itex].

You just calculate the difference between points in Lorentzian spacetime (i.e. events) when they have a certain temporal and spatial separation. Nothing travels anywhere. However, the separation tells you if the two events can be in causal contact. If the proper time interval is bigger than or equal to 0, then the two events are in causal contact, i.e. a light ray (for [itex]\Delta s=0[/itex]) or a massive particle (for [itex]\Delta s>0[/itex]) from one event could travel to the other. If the proper distance is smaller than 0, then the two events cannot be in causal contact, because the spatail separation is so big that even light cannot cover the distance in the given temporal distance. The proper time interval is Lorentz invariant, i.e. it is the same measured in any inertial frame.
 
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  • #3
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The proper time interval between two events is the amount of elapsed time reckoned on the clock of an observer in an inertial frame of reference who is physically present at both events. Look up the equation for the proper time interval in your text book. It is the same as the equation you wrote, except divided by c2. Also, you made an algebra error in your calculation.

Chet
 
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  • #4
vanhees71
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Let [itex]x^{\mu}[/itex] and [itex]y^{\mu}[/itex] be these events in the inertial frame. Without loss of generality by choosing appropriate origins of space and time coordinates we can set [itex]y^{\mu}=0[/itex] and [itex]x^{\mu}=(c t,x,0,0)[/itex], where I've also rotated the spatial Cartesian basis vectors of the observer such that the two events are located at the [itex]x[/itex] axis.

First of all the question only makes sense, if the events are separated by a time-like interval, i.e., you must have
[tex]y_{\mu} y^{\mu} = c^2 t^2-x^2>0.[/tex]
Now it is easy to show that for any such events you can find an inertial frame, where the events are located at the same place in space, i.e., you must find a Lorentz boost such that
[tex]{y'}^{\mu}={\Lambda^{\mu}}_{\nu} y^{\nu}=(c t',0,0,0).[/tex]
Then [itex]t'[/itex] is, by the usual definition, the proper time between the two events.

Hint: You find the Lorentz boost by thinking about the question, what the relative velocity [itex]\vec{v}[/itex] between the reference frame should be. Of course, you must have [itex]|\vec{v}|<c[/itex]. You should also show that you can always find such a velocity if the separation of the events is time-like!
 
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  • #5
Thank you all for your help!
 

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