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Thread: Galilean transformation

  1. #1
    Dec 2012

    Galilean transformation

    I'm doing an assignment on special relativity and want some maths drawn into it. To be more specific I'm working on time dilation, and I found this wonderful website with demonstrations (I hope I am not breaking any rules by linking to other sites). Now, I want to describe what's happening mathematically in the example of a sound wave traveling from A -> B -> C from the perspective of the moving reference frame. If you go to slide 5 (just click the green button to the right four times) you can see the demonstration.

    I think I know how to calculate the coordinates in S', but how do I explain the rest?

    Let's assume A(10;0), B(5;5), C(0;0) and v=341 m/s.
    S' coordinates:
    z'=z (I guess we can look away from z in this example)

    Uh... Nope, I don't know how to calculate the coordinates either...

    Can someone please give me a hand?


    In case someone else ever looked at this and wanted to know the answer, I've got it now:

    We can ignore the z direction. I'm not going to use v=341 m/s because the rest of the problem doesn't have units and I'm going to end up having A be the origin in both frames and not C.
    From the reference frame in slide 3, (I'll call it S) the emitter/detector is stationary at x=0,y=0 and the reflector is stationary at, x=0, y=, say, 10. The emission happens at time t=0, the reflection happens at time t=50, and the detection happens at time t=100. Note that this means that the speed of sound in this frame must be 0.2.
    So we have three events in frame S:
    Emission: x=0, y=0, t=0
    Reflection: x=0, y=10, t=50
    Detection: x=0, y=0, t=100
    Now let's look at frame S', which is moving in the positive x direction compared to S at speed v=, say, 0.15 (from eyeballing the slide it looks like the horizontal speed is a little slower than the speed of sound) and has the same time coordinate as S, and shares an origin when t=0. There's where your transform equations come in:
    x' = x - vt
    y' = y
    t' = t
    So in frame S' you have three events now:
    Emission (A): x'=x-vt=0-0.150=0, y'=y=0, t'=t=0
    Reflection (B): x'=x-vt=0-0.1550=-7.5, y'=y=10, t'=t=50
    Detection (C): x'=x-vt=0-0.15100=-15, y'=y=0, t'=t=100
    Last edited by Bigum; Dec 14th 2012 at 09:24 AM. Reason: Got the answer
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