Space Time Interval: Solving Distance & C-Value

[seto6]

Homework Statement

Homework Equations

s²=c²(t²)-(x²)

The Attempt at a Solution

so..

s=72ly and t=100y c=c

(72)²=(1)²(100²)-(x²⁾

solving for x which is the distance i get x=69.39 aprox= 69.

my question is when do we know to let c=1 or c=3*10⁸ m/s

how do we go back and forth between light year and year?

 

[collinsmark]

my question is when do we know to let c=1 or c=3*10⁸ m/s

how do we go back and forth between light year and year?

At the risk of being overly simplistic, light year and year are essentially the same units. One implies distance, and the other time, but at the unit level, they are essentially the same.

Special relativity cuts away from classical Newtonian physics by placing space and time on the same footing. Time is simple one more dimension to the other 3, in 4-dimensional space-time. And they can be expressed in the same units. For example, we can have four dimensions, all measured in years. We say the 3 spatial dimensions are "light years" but that's just to avoid confusion -- its really the same thing as a year, as far as the math is concerned.

We could also express all 4 dimensions in meters. If we do, we would probably call the time dimension units "light meters." But it's really just measuring meters.

To answer your question, if both time and the spatial dimensions share the same units, set c = 1. This is true if everything is in years/light years, seconds/light seconds, meters/light meters, fathoms/lite fathoms, or anything -- as long as all 4 dimensions share the same units.

If you express time and length as separate units, you'll need some conversion factor. For example if you express distance in meters, and time in seconds, that's where the 3 x 10⁸ m/s factor of c comes in.

As an exercise, try this. Convert 100 y to seconds. Convert 72 ly to meters. Now solve the problem, except this time use 3 x 10⁸ m/s for c. Then convert the final answer back to light years. You'll get the same answer as you did originally.

[Edit: I said earlier that special relativity places "space and time on the same footing." I should of said mostly equal footing. There is still that negative sign involved with the time dimension (or negative sign with the spatial dimensions -- depending on which author you are following) when taking the dot product of two 4-vectors. But other than that it's pretty equal.]

 

[nrqed]

Homework Statement

Homework Equations

s²=c²(t²)-(x²)

The Attempt at a Solution

so..

s=72ly and t=100y c=c

(72)²=(1)²(100²)-(x²⁾

solving for x which is the distance i get x=69.39 aprox= 69.

my question is when do we know to let c=1 or c=3*10⁸ m/s

how do we go back and forth between light year and year?

If you use a system of units with distances in meter and time in seconds, you must use c = 3x10⁸ m/s.

If you use distances in ly and times in year, then the value of c is 1 ly/y (one light-year per year). If we don't write the units, then we simply use c=1

In this problem, you used a value in year for the time and a value in ly for the distance, so you had to use c=1.

Hope that helps.