Lorentz transformations ( synchronising reference frames?)

In summary, according to the lecturer, the speed of the particle in a reference frame moving along the x-axis at 0.7c is 1.71×108 ms−1.
  • #1
joriarty
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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
 
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  • #2


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.

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

Which immediately gives:
[tex]\beta_x ' = -0.34[/tex]

[tex]\beta_y ' =0.23[/tex]

[tex]\beta ' = 0.57[/tex]
 
Last edited:
  • #3


Thank you, RoyalCat - makes sense now! Those velocity addition formulae are new to me...
 
  • #4


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:

[tex]v_x' = \frac{dx'}{dt'}[/tex]

Plugging in those expressions:

[tex]v_x' = \frac{\gamma(dx-dt\beta c)}{\gamma(dt-dx\frac{\beta}{c})}[/tex]

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

[tex]v_x'=\frac{v_x-\beta c}{1-\frac{v_x\beta}{c}}[/tex]

Which recalling that [tex]\beta\equiv \frac{V}{c}[/tex] turns into:

[tex]v_x'=\frac{v_x-V}{1-\frac{v_x V}{c^2}}[/tex]

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 [tex](\Delta)[/tex] and come upon the same result.
 
  • #5


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

What are Lorentz transformations?

Lorentz transformations are mathematical equations used in the theory of special relativity to describe how the measurements of length, time, and mass change for an object when viewed from different reference frames in uniform motion.

Why are Lorentz transformations important?

Lorentz transformations help us understand the fundamental principles of special relativity, which is crucial for understanding the behavior of objects moving at high speeds, such as particles in a particle accelerator or objects in outer space.

How do Lorentz transformations work?

Lorentz transformations involve a set of equations that relate the measurements of length, time, and mass in one reference frame to those in another reference frame that is moving at a constant velocity relative to the first frame. They take into account the effects of time dilation and length contraction.

What is the significance of synchronizing reference frames in Lorentz transformations?

Synchronizing reference frames means that two observers in different frames agree on the time and location of an event. This is important because the measurements of length and time can vary depending on the reference frame, so having a synchronized frame allows for consistent calculations and predictions.

What are some real-world applications of Lorentz transformations?

Lorentz transformations are used in fields such as particle physics, astrophysics, and engineering to understand the behavior of objects moving at high speeds. They also have implications for GPS systems and telecommunications, as they account for the effects of time dilation in satellites orbiting the Earth.

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