Finding Relative Velocity in Inertial Systems

[Kunhee]

Homework Statement

An event occurs in S at x = 6 x 10^8 meters and S' at x' = 6 x 10^8 meters. Find the relative velocity of the systems. Assume that S refers to an inertial system (x,y,z,t), and S' refers to an inertial system (x',y',z',t') moving along +x axis with speed v relative to S. The origins coincide at t = t' = 0.

Homework Equations

c = 3 x 10^8
Lorentz Factor = (1-(v/c)^2)^(-1/2)
x' = 1/LF ( x - vt)
t' = 1/LF ( t - vx / c^2)

The Attempt at a Solution

I am having trouble because x' - x = 0 and t = 0 so the equations aren't working out.
x' = 1/LF ( x - vt )
1 = (1-(v/c)^2)^(-1/2) (1 - v(0))
1 = (1-(v/ 3x10^8)^2)^(-1/2)
(1-(v/ 3x10^8)^2)^1/2 = 1
1-(v/ 3x10^8)^2 = 1
-(v/ 3x10^8)^2 = 1 - 1 = 0

 

[Orodruin]

Nobody says the event occurs at t = 0 or t' = 0.

Edit: Of course, there is no unique solution to this problem. There exists a solution for every possible inertial frame as $x = x_0$ and $x' = x_0$ defines two non-parallel lines in Minkowski space regardless of the relative velocity $v$ (unless $v = 0$ for which the statement is just trivial). In order to fix the relative velocity, you need to know the time of the event in at least one of the frames.

 

[Kunhee]

Hi Orodruin. Thanks for the reply.
The time for t' is 1 seconds. And then it asks for when t' is 4 seconds.

Could you help me set this up?

 

[Orodruin]

What does the inverse of the Lorentz transformation tell you?

 

[Kunhee]

The inverse (1-(v/c)^2) ^ 1/2 and is always less than 1 so there is time dilation? The clock should run slower for t' than for t.

 

[Orodruin]

The inverse (1-(v/c)^2) ^ 1/2 and is always less than 1 so there is time dilation? The clock should run slower for t' than for t.

That is not the inverse Lorentz transformation. That is the Lorentz factor.

 

[Kunhee]

The inverse of the Lorentz transformation tells me that the relative velocity of the systems are equivalent?

 

[Orodruin]

The inverse of the Lorentz transformation tells me that the relative velocity of the systems are equivalent?

No, I am asking you to use the inverse transform to relate the coordinates of the events.

 

[Kunhee]

x = (x' + vt') / (1-(v/c)^2)^1/2

6x10^8 = (6x10^8 + v(1)) / (1-(v/c)^2)^1/2

(6x10^8)(1-v/(3x10^8))^1/2 - 6x10^8 = v

How do I determine the t for the event occurring in S?

 

[Kunhee]

t = (t' + vx'/c^2) / (1-(v/c)^2)^1/2

t = (1 + (6.10^8)v / c^2) / (1-(v/(3x10^8))^1/2

 

[Orodruin]

x = (x' + vt') / (1-(v/c)^2)^1/2

6x10^8 = (6x10^8 + v(1)) / (1-(v/c)^2)^1/2

(6x10^8)(1-v/(3x10^8))^1/2 - 6x10^8 = v

How do I determine the t for the event occurring in S?

You don't nees to. You have one equation and one unknown.

Also, do not forget the units of all the quantities. Units are important.

 

[Kunhee]

Thanks a lot for your time.

If I solve for v, is that the answer then?
v = (6x10^8)(1-v/(3x10^8))^1/2 - 6x10^8
so I have to get the v on the right to the left.

 

[Chestermiller]

Thanks a lot for your time.

So does this mean that the answer for this particular question is not a certain number but
an equation expressed in variables for v = ?

You have one algebraic equation in one unknown, v. So solve the equation. This is where the math comes in.

 

[Kunhee]

I see. Haven't done math in a while too.
I will give it a try. Thanks.

 

[Kunhee]

v = (6x10^8) (1-(v/c)^2)^1/2 - 6x10^8
I can't solve for v because squaring both sides cancel them out.
Where did I go wrong?

 

[Chestermiller]

My advice to you is to solve the equation algebraically first, and then plug in the numbers:
$$x=\frac{(x'+\frac{v}{c}(ct'))}{\sqrt{1-\left(\frac{v}{c}\right)^2}}$$
Solve for v/c.

 

[Kunhee]

Ah okay!