Finding frame speed where two events are simultaneous in special relativity

[bon]

Homework Statement

In a certain frame S, two events A and B are separated spatially by 3ly with A occurring 1 year before B. What must be the relative speed of a frame S' such that the two events occur simultaneously?

Homework Equations

The Attempt at a Solution

Don't I just use: dt' = gamma(dt-B/c dx)

but when i put in dt = 1 and dx = 3 i get to gamma = 3B/c which i can't solve! am i doing it right?

 

[bon]

i basically come to c^2 = 9B^2(1-B^2)

where have i gone wrong/

 

[bon]

anyone?

 

[vela]

Don't I just use: dt' = gamma(dt-B/c dx)

but when i put in dt = 1 and dx = 3 i get to gamma = 3B/c which i can't solve! am i doing it right?

You need to be more careful with the units, for one thing. Show how you got γ=3β/c. (Note the units don't work out in that equation.)

 

[bon]

Wow. Careless mistake..!

B=c/3 yes?

 

[vela]

Close. I think it's better to write the Lorentz transformation equation this way:

$$ct' = \gamma(ct - \beta x)$$

Both ct and x have units of length, so β must be unitless.

 

[bon]

hmm weird..

so i set t'=0 so divide through by gamma..left with ct=Bx...which gives my answer..where did i go wrong?

 

[vela]

What are the units of t, the units of ct, the units of x, and the units of β?

 

[bon]

i see. sorry. so c=1 in my result..B=1/3

 

[vela]

That's a common convention, but I doubt you're using it in your class. The reason I say this is because when you use units where c=1, c typically doesn't appear in the equations.

If you use c=1, then time is measured in units of length, so t=1 ly. If you don't use c=1, then t=1 yr, and ct = 1 ly. Either way, you'll get the answer you found.