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Offline ubermench

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calculating the invarient interval
« on: 17/10/2009 19:51:06 »
If I am calculating the invarient interval for the emission of a laser pulse from earth to the moon, do i take into account the distance traveled? Since the laser pulse is traveling the speed of light i can calculate the time is takes to get there based on the distance it travels and call that the delta T, but do i need to calculate the deltaX or coordinate change for the two frames?


 

Offline JP

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calculating the invarient interval
« Reply #1 on: 17/10/2009 19:56:44 »
This sounds a bit like a homework problem, so I'm not going to tell you how to calculate it straight away.  What two reference frames are you referring to and how far have you gotten in solving it?  Also, there's a reason it's called "invariant," which might help answer your questions...
 

Offline ubermench

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calculating the invarient interval
« Reply #2 on: 17/10/2009 20:09:43 »
The reference frames would be the earth and the moon. Since the two events described in the interval are the emission of the laser pulse and the reception of it on the moon. Im not sure if you need to have those reference frames, but thats because i haven't done problems where there was a known distance between events(distance from the earth to the moon), so i don't know about incorporating the delta x. And i think your hinting at the transformation between frames with invariant, i can do lorentz transformations but the distance thing is what i need clarification on.
« Last Edit: 17/10/2009 20:12:31 by ubermench »
 

Offline JP

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calculating the invarient interval
« Reply #3 on: 17/10/2009 23:29:08 »
I'm not sure what you're going to need to assume to solve this, but the moon in orbit around the earth isn't an inertial reference frame (it's not moving on a straight trajectory), so that special relativity would have problems if you tried to figure out reference frames.  Also, distances contract in the direction you're moving, not in the direction perpendicular to it.  Unless the moon and earth are getting much closer together, would you need to calculate how that distance changes?  Do you have enough information to solve the problem if you just assume the Earth and moon aren't moving at all?
 

Offline ubermench

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calculating the invarient interval
« Reply #4 on: 17/10/2009 23:40:34 »
I can assume pretty much anything that helps in solving the problem, theres not exactly one right answer to this. I'm definitely going to assume that neither the earth or moon are moving, because that makes things much more difficult that need be. Ideally I am just trying to make the invariant interval equation fit this situation, which is why i think the most important things are the speed of light and the distance between the earth and moon. Im just trying to figure out when doing the invarient interval calculation of T2= deltatime^2+deltax^2   if i need to put something in for deltax. Time is easy, the distance between the earth and moon divided by the speed of light, but if i put in the distance the number would be huge. If i used the delta x i think it would be the same as delta time after converting. 
« Last Edit: 17/10/2009 23:54:42 by ubermench »
 

Offline JP

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calculating the invarient interval
« Reply #5 on: 17/10/2009 23:59:32 »
You definitely need the same units.  Once you do that, I think you're where you need to be.  I think that usually the convention is to multiply Δt by c in order to get a final answer in units of distance, but your way is essentially the same, only you'll get units of time out instead of distance.  That should be enough to solve it, right?
 

Offline ubermench

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calculating the invarient interval
« Reply #6 on: 18/10/2009 00:27:46 »
well i dont have an issue with solving it, i just dont know if its just going to be Int^2= 384,000,000m/3E8 m/s=1.28seconds. Sqrt = 1.13 seconds. Thats just the delta time. should i use x? it would just be adding another 1.28 seconds, since im using the same numbers twice. so it would be 1.28+1.28  = 2.56. sqrt = 1.6 seconds
 

Offline JP

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calculating the invarient interval
« Reply #7 on: 18/10/2009 00:55:50 »
Do you really want to add time and distance in the formula?  Distance = x^2-c^2*t^2 in special relativity.  Throw in a minus sign and you should get the answer.  Once you see the answer, it will seem odd, but think about the fact that you're dealing with light, which is very special in that it always moves at the fastest possible speed.
 

Offline ubermench

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calculating the invarient interval
« Reply #8 on: 18/10/2009 02:42:21 »
yeah i forgot about the minus sign for a second. so that would make the invarient interval = 0?  So despite the distance being enough for light to take more than a second to travel there the interval is 0? wierd if thats true.
 

Offline JP

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calculating the invarient interval
« Reply #9 on: 18/10/2009 05:17:53 »
The invariant interval is a way of measuring "distance" between two events in four-dimensional space-time, so its values don't necessarily make a lot of sense to us three-dimensional creatures.  Usually you use invariant intervals to calculate the space-time distance between two events.  A zero interval means that only a light-speed signal could connect the two events.  A positive interval means that the two events cannot be connected by any signal because they're too far apart (your light pulse would arrive after the second event happened).  A negative interval would mean that the two events can be connected by light-speed and even slower signals. 

Since your problem was asking "what's the space-time distance between sending and receiving a light signal?" you expect it to be zero, since everything moves at the speed of light here.
 

Offline Pmb

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calculating the invarient interval
« Reply #10 on: 19/10/2009 20:18:54 »
If I am calculating the invarient interval for the emission of a laser pulse from earth to the moon, do i take into account the distance traveled? Since the laser pulse is traveling the speed of light i can calculate the time is takes to get there based on the distance it travels and call that the delta T, but do i need to calculate the deltaX or coordinate change for the two frames?
Here is a worked example of calculating a spacetime interval
http://www.geocities.com/physics_world/sr/st_diagram.htm

It's a very simple to work these things out in special relativity (flat spacetime in inertial frames). In Lorentz coordinates the spacetime interval is defined by the following equation

ds^2 = (c dt)^2 - dx^2 - dy^2 - dz^2

Your question doesn't sound like a homework problem and since you've been working through it I'll fill in the rest. Let earth and moon be at rest in the frame S and both lie on the x-axis (i.e.l dy = dz = 0). Let dx = distance between Earth and Moon. Let dt = the time it takes for a light signal to travel between the Earth and Moon. Then

Spacetime Interval = ds^2 = (c dt)^2 - dx^2

Notice that it's the square of ds, and not ds, that is often referred to in texts (since as Shutz's text "A first course in general relativity") that is referred to as the "spacetime interval".

For more on the concept of invariance please see
http://www.geocities.com/physics_world/gr_ma/invariance.htm
« Last Edit: 19/10/2009 20:23:35 by Pmb »
 

Offline yor_on

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calculating the invarient interval
« Reply #11 on: 23/10/2009 05:56:05 »
Sweet explanation jpetruccelli
Even I got it :)
 

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calculating the invarient interval
« Reply #11 on: 23/10/2009 05:56:05 »

 

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