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What does x-vt mean?

  1. Dec 7, 2011 #1
    It says that x' = gamma (x-vt)

    Now what exactly does that formula mean?

    I take it that x is a point in my coordinate system, let's say 2 meters.

    I take it that v is the speed of the other reference frame or am I not correct in saying that? What exactly does that V refer to?

    I take it that the t = time in my coordinate system.

    So would this make sense? In my coordinate system an event happens at distance of 2 meters and at 2 seconds. The following train is moving at .5c relative to me.

    Is this following equation right?

    2-.5(2) x gamma = the distance in the other coordinate system.

    It appears according to the Interactive Minkowski Diagram when the t and d are the same number in my coordinate system, they will be different but equal numbers in the other coordinate system.
     
  2. jcsd
  3. Dec 8, 2011 #2

    atyy

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    Looks right to me.

    A couple of points:

    1) x and x' are not just any coordinate systems, but special types of coordinate systems called Lorentz inertial frames

    2) v is the velocity of the other frame relative to the first frame

    3) When you call x the distance, it is the distance from the origin. The origin is an event that is labelled (t=0,x=0) in one frame and (t'=0, x'=0) in the other.
     
  4. Dec 8, 2011 #3

    ghwellsjr

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    That formula is one half of the Lorentz Transform. The purpose of the LT is to convert coordinates of an event in one Frame of Reference to the coordinates of the same event in another Frame of Reference moving at speed v with respect to the first one.
    You should use the term "event" rather than point because "point" implies just the location part of the event which includes three components of location (x, y & z) and one component of time (t). You must always include the time components although in many cases we leave y and z out of the nomenclature when they are always equal to zero but it is always understood that they are really still there.
    You are correct, v is the speed along the x axis that the second reference frame is traveling with respect to the first frame.
    Yes, it is the time of an event, the fourth coordinate.
    Yes, if you mean, the event happened 2 meters from the origin along the x-axis at 2 seconds after the origin. It's important to understand that the two frames share a common origin where all four coordinates equal zero; that is, all the location components equal zero and the time component equals zero.
    It depends on what you really meant, but let me make some comments.

    First off, events are relative to the origin of the frame, not to you, even if you were located at the origin of the frame at the beginning of the scenario because one second later, you are no longer located at the origin because the event describing you now has a time value of 1 second.

    Secondly, you have to use consistent units. If your unit for distance is meters and your unit for time is seconds, then you cannot use .5 for the speed if you meant .5c which is a whole lot faster, so I'm going to assume that the train is traveling at .5c.

    Thirdly, things can move in a frame and you describe their motion by a series of events, all within the one frame. So if the train were at the origin at the start of your scenario, its coordinates would be t=0 and x=0 (considering just the front of the train). Two seconds later, it will have moved one half the speed of light along the x-axis. So if we use the unit for time of seconds and the unit for distance of light-seconds, then the coordinates would be t=2 and x = 1. (But remember, I have changed the distance unit from meters to light-seconds.) If you leave the two numbers the same and you're talking about something that moved from the origin to that event, then you are talking about light because it is the only thing that can maintain both the x and t coordinates equal to each other.
    That's true. Let's work it out. First we have to calculate gamma. I'm assuming you know how to do this and you will get a value of 1.1547. Now we can continue calculating for light with both coordinates equal to 2. You will note that I am now using β instead of v because I am using compatible units and I want to make things easier when I do the calculation for t'. But first x':

    x' = gamma (x-βt)
    x' = 1.1547 (2-.5(2))
    x' = 1.1547 (2-1)
    x' = 1.1547 (1)
    x' = 1.1547

    Now we have to calculate t':

    t' = gamma (t-βx)
    t' = 1.1547 (2-.5(2))
    t' = 1.1547 (2-1)
    t' = 1.1547 (1)
    t' = 1.1547

    So it's true and obvious if you look at the calculation that when t and x are the same in one frame, they will be the same in any other frame. But this is just demonstrating that the speed of light is the same in all frames.

    Now I think it is much more interesting to do the calculation for the event of an object (a train) moving at .5c. Remember, at t=2, x=1. Now we have:

    x' = gamma (x-βt)
    x' = 1.1547 (1-.5(2))
    x' = 1.1547 (1-1)
    x' = 1.1547 (0)
    x' = 0

    Now we have to calculate t':

    t' = gamma (t-βx)
    t' = 1.1547 (2-.5(1))
    t' = 1.1547 (2-.5)
    t' = 1.1547 (1.5)
    t' = 1.732

    Now what is this telling us? First off, whenever you calculate the events for a moving object that starts out at the origin and you transform them to a frame that is moving at the same speed as the object, you find that the x' coordinates of the event go to zero. This shows us that the object is at rest in this comoving frame.

    Secondly, we see the time has gone from 2 seconds to 1.732 seconds which is the time divided by gamma. Does this sound familiar? It's showing time dilation.
     
  5. Dec 8, 2011 #4
    If you get the answer tell me, because I am confused too. It is NOT a distance as measured by an observer in the other coordinate system. What is it's physical significance then?

    I looked through about a dozen references and finally resorted to Einstein's original paper. The Lorentz transform is the unique transform under which the laws of physics are invariant. In other words, transforming an equation that represents a law of physics will give you a true expression. He transforms t^2c^2 = x^2 + y^2 + z^2 and gets a valid equality. You can transform Maxwell's equations and get something that works as well, which is the main thrust of the paper. So the input to the transform has to be a true equation that relates x,y,z,and t, as far as I can tell. I hope to stand corrected.
     
  6. Dec 8, 2011 #5
    It's better to start with classical ("Gallilean" or Newtonian") mechanics, and do some exercises with that, as a warm-up.
    Classical mechanics says that x' = (x-vt) and the meaning of the symbols is the same (it's a transformation equation of the relationship between two coordinate systems that are in relative motion).

    See:
    - http://en.wikipedia.org/wiki/Galilean_transformation
    - http://en.wikipedia.org/wiki/Lorentz_transformation
     
  7. Dec 8, 2011 #6

    robphy

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    What is the URL of this "Interactive Minkowski Diagram"?
     
  8. Dec 8, 2011 #7

    ghwellsjr

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    Whether or not an observer in the other coordinate system measures the distance, it's still a distance from the location of the origin. But you also need to include the time coordinate or none of this makes any sense.
    You can use the Lorentz Transform to transform various laws of physics. After transformation, they must remain identical to what they were before transformation. But that is a complicated process to do. It is more common to use the Lorentz Transform on events as I did earlier.
     
  9. Dec 8, 2011 #8
  10. Dec 8, 2011 #9

    ghwellsjr

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    I see that your reference also used compatible units for t, v and x (or d) but they used v instead of the less confusing β.
     
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