Lorentz transformations ( synchronising reference frames?)

[joriarty]

Lorentz transformations ("synchronising" reference frames?)

Homework Statement

A particle moves from (x,y,z,t) = (0 m,0 m,0 m,0 s) to (1 m,1 m,0 m,10 ns).

1. i. What is the speed of the particle in this reference frame?
2. ii. What is the speed of the particle in a reference frame moving along the x-axis at 0.7c?

Homework Equations

The usual Lorentz transformation equations:

• β ≡ V/c
• γ ≡ 1/sqrt(1 − β²)
• x′ = γ(x − tβc)
• t′ = γ(t − xβ/c)
• x = γ(x′ + t′βc)
• t = γ(t′ + x′β/c)
• c ≈ 3×10⁸ ms⁻¹

The Attempt at a Solution

I understand that the answer to part (i) is just 1.414×10⁸ ms⁻¹, that's just simple trigonometry. However I do not understand how to do part (ii).

My lecturer gives the following answer to the problem:

Synchronise clocks and rulers at (0, 0, 0, 0). Then we travel to (−1.54, 1, 0, 1.07×10⁻⁸) so speed is 1.71×10⁸ ms⁻¹ or β(v) = 0.57. Note that the x component of the velocity comes out negative.

But I don't understand the answer. What does he mean by "synchronise clocks" and how did he calculate those answers? (unfortunately my lecturer is away for another 2 weeks so I can't ask him!)

Thank you

 

[RoyalCat]

By synchronizing clocks and rulers, he meant that in both frames, the particle passes through the origin at time 0.

What he then did was transform the second event (The end of the particle's path in the original frame) to the moving frame.

y'=y, so 1=1
x'=γ(x − tβc) = 1.4(1-2.1) = -1.54 m
t'=γ(t - xβ/c) = 10^-8 *(1.4(10-2.3)) = 10.7 ns

So the velocity is just the path divided by the time.

Another way to approach the problem would be the velocity addition formula.

$$v_x ' =\frac{v_x - V}{1-\frac{v_x V}{c^2}}$$
$$v_y' = \gamma{v_y'}$$
Where $$V$$ is the velocity of the primed frame relative to the unprimed frame.

Which immediately gives:
$$\beta_x ' = -0.34$$

$$\beta_y ' =0.23$$

$$\beta ' = 0.57$$

 

[joriarty]

Thank you, RoyalCat - makes sense now! Those velocity addition formulae are new to me...

 

[RoyalCat]

No problem. ^^

You'll probably come across them in the next chapter or two.

A quick rundown of how they're developed:

x′ = γ(x − tβc)
t′ = γ(t − xβ/c)

Implies:
dx′ = γ(dx − (dt)βc)
dt′ = γ(dt − (dx)β/c)

The x velocity in the primed frame is, by definition:

$$v_x' = \frac{dx'}{dt'}$$

Plugging in those expressions:

$$v_x' = \frac{\gamma(dx-dt\beta c)}{\gamma(dt-dx\frac{\beta}{c})}$$

Some algebra massage (Dividing denominator and numerator by dt, and recalling that $$v_x=\frac{dx}{dt}$$ by definition):

$$v_x'=\frac{v_x-\beta c}{1-\frac{v_x\beta}{c}}$$

Which recalling that $$\beta\equiv \frac{V}{c}$$ turns into:

$$v_x'=\frac{v_x-V}{1-\frac{v_x V}{c^2}}$$

If you're not familiar with differentials, or if you haven't studied the subject as a class, there's another way to approach the derivation, using time dilation and length contraction, or some some other lie. :)

Since we're dealing with constant velocities throughout, we could have just as well taken a non-infinitesimal difference $$(\Delta)$$ and come upon the same result.

 

[joriarty]

Superbly explained, thank you muchly! You are awesome! :)