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Galilean transformation - can you show me an example?

  1. Dec 14, 2011 #1
    I'm still having trouble with the basic foundations of relativity so I am taking a look here at the Galilean transformation.

    I know the only thing that changes is

    x' = x-vt

    Now can someone explain what each variable stands for and can show me how you would do an actual example with numbers?
     
  2. jcsd
  3. Dec 14, 2011 #2

    tom.stoer

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    Consider a passenger on a train moving with velocity v and an observer (at rest) at the train station. At t=t'=0 the positions of the passenger and the observer at rest coincides, i.e. x'=x (we neglect y, z coordinates).

    At a later time t'=t>0 the passenger on the train has a position x'(t') = x+vt = +vt, whereas the observer at rest has a position x=0.
     
  4. Dec 14, 2011 #3
    So it's really as simple as saying the train is going 40miles/hour X 5 hours and you get 200 miles.

    Does this equation x'(t') mean the train will be at x position at t time?

    What's the difference in saying x+vt or x-vt? If I used x-vt for my train example or 40miles/hour X 5 hours would I get 0-200 = -200 miles. As in the position is 200 miles in the other direction.

    I have a feeling I am being confused by something simple.
     
  5. Dec 15, 2011 #4

    ghwellsjr

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    Do you understand that x and t are coordinates for a location in one frame of reference and x' and t' are the coordinates for the same location in a second frame of reference moving with respect to the first one?

    So in the example of the passenger on the train traveling at 40 mph in the x direction and an observer at rest at the train station, you have decided to consider two frames, one in which the observer is at rest and one in which the passenger is at rest. You have decided to place the origins of the two frames at the location where the passenger has just passed the observer. You call this location 0 in both frames and the time is 0 in both frames. The unprimed frame is the one in which the observer is at rest and the primed frame is the one in which the passenger is at rest.

    So let's look first at the unprimed frame and write down coordinates for the locations of the observer and the passenger at intervals of 1 hour:
    Code (Text):
    Time Observer Passenger
      t    x[SUB]obs[/SUB]      x[SUB]pas[/SUB]
      0      0        0
      1      0       40
      2      0       80
      3      0      120
      4      0      160
      5      0      200
    Note that everything in this frame is unprimed and includes both the observer and the passenger. So what do these numbers mean? First off, the observer never changes his location in the unprimed frame--that's because he is at rest in this frame. On the other hand, the passenger is continually increasing his distance from the origin as time goes by. You can see that his location is increasing by 40 miles each hour. You should be aware that we have not used the Galilean Transform in filling out this table. Rather it is our description of what is happening: the observer remains at rest and the passenger is traveling at 40 mph.

    Now let's use the Galilean Transform to convert all ten locations into the primed frame using the equation x' = x-vt where v = 40:
    Code (Text):
    Time Observer Passenger
      t'   x'[SUB]obs[/SUB]     x'[SUB]pas[/SUB]
      0      0        0
      1    -40        0
      2    -80        0
      3   -120        0
      4   -160        0
      5   -200        0
    Note again that everything in this frame is primed including both the observer and the passenger. Now you can see that the passenger is at rest because his location is not changing. However, the observer's location is changing and is moving to the left at 40 mph.

    Now this is a rather trivial case because we have arbitrarily chosen to select a scenario in which the two people involved are both inertial, that is, they are not accelerating which means that we can select two frame in which each of them is at rest. But let's change the scenario to one in which the train and the passenger are not inertial. Let's say the train is stopped in the station for one hour and then it starts moving at 10 mph for the first hour, 20 mph for the second hour, and 40 mph for the last two hours. What would our unprimed set of locations look like? Here they are:
    Code (Text):
    Time Observer Passenger
      t    x[SUB]obs[/SUB]      x[SUB]pas[/SUB]
      0      0        0
      1      0        0
      2      0       10
      3      0       30
      4      0       70
      5      0      110
    Remember, these numbers came from our statement of the scenario and have nothing to do with the Galilean Transform.

    Now we will use the Galilean Transform to see the coordinates in the primed frame moving at 40 mph:
    Code (Text):
    Time Observer Passenger
      t'   x'[SUB]obs[/SUB]     x'[SUB]pas[/SUB]
      0      0        0
      1    -40      -40
      2    -80      -70
      3   -120      -90
      4   -160      -90
      5   -200      -90
    Now it's pretty obvious that the passenger is not always at rest with respect to ether frame, but he does eventually come to rest in the primed frame.

    Just remember that every frame contains all observers, objects and clocks.
     
  6. Dec 15, 2011 #5
    Can you show me how you got the numbers in your last example there?

    The thing that is confusing me is that you are saying the train is going 10 mph for the first hour, 20 mph for the second hour, and 40 mph for the last two hours but than you are saying the train is going 40mph in the primed frame?

    Can you show me how you got the numbers in the last example?
     
  7. Dec 15, 2011 #6

    ghwellsjr

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    I said that the train is stopped for the first hour so during that time, it and the observer at the station are going -40 mph in the second frame. Does it now make sense or do you still want me to show you the calculations?
     
  8. Dec 15, 2011 #7
    You are making it confusing to me.

    In my view I see the stationary observer on the platform the x frame and the train and it's passenger as the x' frame.

    I see the equation x' = x-vt.

    All I am asking is what does each variable stand for? I understand t stands for time. I understand that v stands for velocity, but for the velocity of the other reference frame? X to me stands for position from the origin.

    Can it be read like the this,

    the position of the train (x') = the position of the observer (x) - velocity of the train X the time?

    Am I correct in saying that? If I say the position of the observer is at the origin so he is 0 - the velocity of the train (40mph) X 5 seconds I get -200 miles.

    This means that the position of the train is 200 miles from the origin in the stationary FoR?

    I'm just confusing myself.
     
  9. Dec 16, 2011 #8

    haushofer

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    Imagine an event P. Then choose two observers who are looking at it. Observer one uses coordinates x to locate the event P. Observer two uses coordinates x' to locate the event P. If the relative velocity between these two observers is v, then x and x' are related by

    [tex]
    x' = x - vt
    [/tex]

    Notice that there is no t'; time is absolute in Galilean relativity, so t=t': the two observers (after clock synchronization) will always agree on the time.

    Of course, you could place the event P in the origin of one of the observers' coordinate system, say x'. In time the succession of events becomes a worldline.
     
  10. Dec 16, 2011 #9

    ghwellsjr

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    If you look up at the first two tables in my post #4, you will see that the first table contains the x positions in the unprimed frame for the observer in the second column labeled xobs and they're all zero and you will see that the second table contains the x' positions in the primed frame for the passenger in the third column labeled x'pas and they're also all zero. That's because you're using two different frames for the observer and the passenger in which they are each at rest and so their locations aren't changing--they stay where they started, at zero.
    You seem to be thinking that this equation is describing the motion of the passenger but it's not. The equation for the location of the passenger as a function of time is x = s*t, where s is the speed of the train so the equation could be written as x = 40t. Notice that this is x and not x'. The equation for the location of the observer is x = 0. Look at the first table in post #4. For values of t from 0 to 5, you can use these two equations to see that the value for the locations of the observer and the passenger are in the second and third columns.

    Your problem is that you are trying to associate x exclusively to the observer and x' exclusively to the passenger but you should be associating x exclusively to locations in the unprimed frame and x' exclusively to locations in the primed frame which we have not considered yet in this post.
    Yes.
    I can see why you are confused. You are taking x to be the position of the observer which is always 0, correct? But it's not associated with any particular person, it's any location that you want to transform from the unprimed frame to the primed frame. It can be where the observer is, where the passenger is, or any other location in the unprimed frame.
    But you see, you got the wrong answer. The train is not at -200 miles after 5 hours, it's at +200 miles.

    Please go back and read the previous posts and keep in mind that in this post, we have only been talking about the unprimed frame--we have not worked on the transform to the primed frame yet.
     
  11. Dec 18, 2011 #10
    Can I go over something from the beginning?

    I am the observer who is standing on a platform and a train moves by me in the right direction at 40mph. After 5 hours the train is now 200 miles away in my FOR.

    x is considered the point where the train is after 5 hours of traveling at 40mph. The point x is 200 miles from my origin. Am I correct in saying this?

    Now does the formula x' = x-vt mean where the point x is in the FOR of the train?

    So for instance x' (the point of the train in the train FOR) = 200 - 40mphX5hours = 0. Is this correct? That means he is at his own origin.

    Am I making sense here?
     
  12. Dec 19, 2011 #11

    ghwellsjr

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    Yes, you are correct and making sense here.
     
  13. Dec 19, 2011 #12
    Yes this makes sense, but I fear that you are failing to understand the purpose of coordinate transformations. In your example, the point x does NOT have to be defined as the origin of the second observer's frame. x can describe ANY point (along the x-axis) in the first observer's frame. The Galilean transformation relates the coordinates of the first observer to the coordinates of the second.


    For example: Say you're standing on a platform (your origin is taken to be where you are standing). A second person is sitting on a train that moves at 30 m/s relative to the platform (his origin is taken to be his seat). At the moment when the train passes and your origins cross, you start a stopwatch. Ten seconds later you see a small explosion occur 400 meters from you (x=400 m). What was the location of the explosion in the second person's frame of reference (x')?

    You use the Galilean transformation:

    x' = x - vt = (400 m) - (30 m/s)(10 s) = 100 m

    So the location of the explosion in the second person's frame was 100 m from the origin.
     
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