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Attempting to obtain the velocity subtraction formula (help required)

  1. Jul 11, 2010 #1
    I have run into problems getting my head around the special theory of relativity, as my understanding of it causes parodoxical outcomes. I would much prefer to have a congruent understanding of the theory, rather than maintain my current perplexed perspective. It would be greatly appreciated, if you could help me escape confusion.

    Based on my assumption that Lorentz transformations affect a moving body by slowing down time and contracting length, in order to maintain the speed of light according to all observers, i have run into a problem. This problem does not exist in a situation where a moving observer, moves from point A to point B and the pulse of light in question that must be maintained to be the speed of light, is traveling in the same direction, that being from point A to point B. Although the relative speed between the pulse of light and the moving observer, is equal to the speed of the pulse of light minus the speed of the moving observer, according to a stationary observer. Lorentz transformations affect the measurement of speed by the moving observer and maintain that the speed of the pulse, is the speed of light. My problem emerges in the situation where the pulse instead of travelling from point A to point B, does the opposite and moves from point B to point A. In this situation a stationary observer would consider the relative speed between the pulse of light and the moving observer to be added together, resulting in a faster than the speed of light velocity. In this situation the problem arises due to the fact that time dilation( the slowing of the rate of time) and length contraction, would speed up the pulse of light, rather than slowing it down, which would be inconsistent with the theory.

    This problem does not arise when you consider the velocity subtraction formula U = (S-V)/(1-(SV/C^2))
    U: The velocity of an object in question relative to a moving observer, according to the moving observer
    S: The velocity of the object in question relative to the stationary observer, according to the stationary observer
    V: The velocity of the moving observer relative to a stationary observer, according to the stationary observer
    (S-V): The velocity of the object in question relative to the moving observer, according to the stationary observer
    1/(1-(SV/C^2): The converting factor, that converts (S-V) into U
    C: The speed of light

    If the object in question is a pulse of light then S=C in situation one and -C in situation two. According to the special theory of relativity U=C in situation one and -C in situation two
    U = (C-V)/(1-(CV/C^2))
    = ((C^3)-(C^2)V)/((C^2)-CV)
    = C((C^2)-CV)/((C^2)-CV)
    = C

    U = (-C-V)/(1-(CV/C^2))
    = ((-C^3)-(C^2)V)/((C^2)-CV)
    = -C((C^2)-CV)/((C^2)-CV)
    = -C

    As it can be seen no problem arrises when this formula is used and as this is official formula, my perspective is wrong. In order to rectify my perspective, attempting to understand how the formula is derived would be a good start, with my approach being a good old fashioned thought experiment.

    The flash and superman are on a high speed train, as they are too lazy to run or fly to a ten pin bowling contest. During this train ride the flash and superman get into an argument, over how fast they can bowl a bowling ball down an alley. Each of them states there case to one another but either one concedes. A scientist sitting across from them, sick of their bickering, suggests that they test the speed of their bowls in the observation carriage of the train, with a devise he has in his luggage. This device consists of laser sensors, spaced one meter apart, the lasers shine perpendicular to reflectors, which allow the sensors to know if something has passed between the sensor and reflector (the same as a shop sensor). When the first beam from the first sensor is broken, a stop watch is turned on. When the second beam from the second sensor is broken the stop watch is turned off. The value recorded on the stop watch in the unit seconds divided by 1, will give a determined speed in the units meters per second. The experiment is established in the observation carriage so that, the flash and superman will bowl from the back end of the carriage and have there balls move up the carriage until they come into contact with the first beam and then the second, allowing them to finally resolve there argument. In my minds eye I am able to witness this experiment, track side as the observation carriage is completely glass, allowing me to view a ruler in between the sensors, that sates the distance apart is one metre and the stop watch.

    As i am able to see the reading on the stop watch and the ruler stating that the distance between the sensors to be one meter. I will be able to determine the speed of the bowling balls, relative to the train according to the observers on the train. Qualitatively according to special relativity, the relative speed between me and the the train should determine the difference in what i consider to be a meter and a second and what the flash, superman and the scientist consider to be one meter and a second. If i was not able to see the meter ruler or the stop watch, i should by logic based on the speed of the bowling balls, relative to me according to me and the train relative to me according to me, what they will determine the relative speed is of the balls compared to the train, according to observers on the train. Without Lorentz transformations, what i would determine the relative speed to be between the balls and the train to be, would be the same for those on board, however Lorentz transformations contract length and slow time. My one meter is longer than the trains one meter and my stop watch runs faster than the one on the train. To obtain the relative speed according to the train, the length it appears to cover to me relative to the train, must be converted into the length it appears to cover to the train and the time it takes to cover that length, according to me must be converted into the time it takes to cover the length according to the train. Without knowing the formulas for length contraction and time dilation, one might come up with this formula:
    U = (S-V)*((M/T)/(B/P)
    U: The velocity of the bowling ball relative to the train, according to the observers on the train
    S: The velocity of the bowling ball relative to trackside me, according to trackside me
    V: The velocity of the moving observer relative to a trackside me, according to trackside me
    (S-V): The velocity of the bowling ball relative to the train, according to trackside me
    M: the length of one meter, according to trackside me
    T: the length of what the train says is one meter, according to trackside me
    B: the duration of what the trains stopwatch considers to be one second, according to trackside me
    P: the duration of one second according to trackside me

    If I now look up what the formulas are for M/T and B/P, i should arrive at the same formula as the official one.
    M/T = 1/(1-(V^2/C^2))^0.5
    B/P= (1-(V^2/C^2))^0.5

    Then the velocity subtraction formula should be U = (S-V)/(1-(V^2/C^2))
    The formula above is not the official velocity subtraction formula and only produces the same outcome when the S=V or the relative speed one is trying to determine according to the moving observer, is the moving observer themselves (U=0)

    As you can see at this point, i was not able to derive the official formula and rectify my perspective. What assumptions are false in the thought experiment and what are the correct ones i should have made?
     
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  3. Jul 11, 2010 #2

    Fredrik

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    It would be less confusing to call these speeds something like "closing speed" and "separation speed" rather than "relative speed", which usually refers to the coordinate speed of one object in the inertial coordinate system associated with the motion of the other object.

    A separation speed >c isn't in any way inconsistent with the theory.

    I didn't read the whole thing, but I don't see any Lorentz transformations, so I'd say that's the problem. It helps to think of this in terms of spacetime diagrams. The motion of object moving with velocity v is represented in a diagram by a line with slope 1/v. To obtain the correct formula for velocity "addition", you need to apply a Lorentz transformation to the points on such a line, and determine the slope of the line that the transformed points are on.
     
  4. Jul 11, 2010 #3
    If you can, please have a full read.

    I used Lorentz trasformations to derive the formula that I arrived at. When leaning special relativity is was emphasised that any pulse of light would be considered to be C. If the separating speed could be determined and it was something that is greater than C, then one could do experiments in an inertial frame of reference and determine absolute speed. Special relativity infers no absolute motion !!!

    Thank you for your reply.
     
  5. Jul 11, 2010 #4

    jtbell

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    Those are not the Lorentz transformation formulas, which look like this:

    [tex]x^\prime = \gamma (x - vt)[/tex]

    [tex]t^\prime = \gamma \left( t - \frac{vx}{c^2} \right)[/tex]

    where x and t are the position and time of an event in one inertial reference frame, x' and t' are the position and time of the same event in another inertial reference frame, and v is the relative velocity of the two frames.

    (In a post in your other thread, I listed the same equations, with c = 1 to make calculations easier.)

    What you have are the length contraction and time dilation formulas, which are not completely equivalent to the Lorentz transformation because they are incomplete. In order to get a complete picture, you need to apply length contraction, time dilation, and relativity of simultaneity.

    One way to describe relativity of simultaneity in an equation is like this. Start with two clocks, both at rest in some inertial reference frame, and therefore at rest respect to each other. They are separated by a distance [itex]\Delta x[/itex] in that reference frame. Synchronize them so that they show the same time in that reference frame.

    In a reference frame in which the two clocks are moving at speed v along the line joining them (i.e. one clock is "chasing" the other through space), the clocks are not synchronized. They both run at a slower rate, as given by the time-dilation equation, and they also do not read the same time at a particular moment in the new reference frame. Instead, the clock that is "being chased" lags behind the other clock by the following amount of time:

    [tex]\Delta t = \frac{v \Delta x} {c^2}[/tex]

    And of course the distance between the two clocks in the new frame is length-contracted, but in the equation above, use the "original" distance between them. (It's simply easier to write the equation that way.)

    The equations for length contraction, time dilation and relativity of simultaneity can all be derived from the Lorentz transformation equations. I suppose it's also possible to go the other way: derive the Lorentz transformation equations from length contraction, time dilation and relativity of simultaneity. I've never happened to see anyone do it, though; probably because the Lorentz transformation is considered to be more "fundamental" than the other equations.
     
    Last edited: Jul 11, 2010
  6. Jul 11, 2010 #5
    It has to be the relativity of simultaneity, that is confusing me. I can't see how it fits into the formula. If your aware of the velocity of the train, then your aware of how much the stop watch dilates on board and what the length of 1m on board is to you. Once you know these particular variables (that are dependent on the velocity of the train), shouldn't one be able to convert U say into S, based on the fact that you know U based on the sensors being 1 m apart and the duration, as marked on the stopped watch (this reading would not be skewed by distance; there would only be a delay between when the watch is stopped and when trackside me would recognise this has happened).

    "A separation speed >c isn't in any way inconsistent with the theory." (Fredrik)

    Does anyone agree with the above statement by Fredrik?
     
  7. Jul 12, 2010 #6
    Yes when read in context Fredrik intended it to be read. It applies to closing speed or separation speed not relative speed. Fredrik was explaining the difference between them.

    Matheinste.
     
  8. Jul 12, 2010 #7

    Fredrik

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    You may want to take a look at Stevmg's thread about this. This is one of the things I said there:
     
  9. Jul 13, 2010 #8
    In the initial post you will see that when iv define the speed of objects relative to another, i have also slipped in "according to". If I understand the above statement correctly then what is being said in other words is that 'the speed of car A is 100 mph relative to car B, according to an observer stationary relative to the road'.

    My initial problem does not concern what the stationary observer considers the relative speed of light to be, between the moving observer and the pulse, but rather what the moving observer considers the velocity of the pulse to be. To my understanding at least, special relativity states that any observer of em radiation in a vacuum, will consider that em radiation to have a velocity of C relative to the observers frame of reference.

    How can length contraction, time dilation and relativity of simultaneity convert a relative velocity, with a magnitude >C according to the stationary observer to a magnitude that is <C or =C according to the moving observer?
     
  10. Jul 13, 2010 #9

    Fredrik

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    I wouldn't use the word "relative" like that, and you probably shouldn't either, because it gets really confusing.

    The "speed of car A relative to car B" is by definition the speed of car A in an inertial frame in which B is stationary, so it's (50+50)/(1+50*50/c^2) mph (if c is given in mph). (This result can only be found using a Lorentz transformation). The closing speed in the road's rest frame is by definition just the number of miles that the distance (in the road's rest frame) decreases each hour (in the road's rest frame), and that's clearly just 50+50=100.

    That is what we'd call the speed of the pulse relative to the moving observer, so this is one of those situations where it gets really confusing to call closing speeds relative velocities.

    That's right (but I prefer to say "in the observer's rest frame" instead of "relative to the observer's frame"...when we're talking about the values of physical quantities. A lot of people talk about objects being "in" frames, and I really don't like that).

    I assume that what you meant to ask is how two objects can have a closing speed >c in one frame and a closing speed =c in another. As with everything else in SR, the easiest way to understand it is to think in terms of spacetime diagrams. In the diagram that represents the frame in which the closing speed is >c, the world lines of the two objects have slopes -1/|v| and 1. In the other frame, the object that had speed |v| in the first frame has speed 0. In the diagram for that frame, the world line of that object is a vertical line. To be more specific, it's the time axis. And since lines with slope 1 (world lines of "objects" moving at speed c") are left invariant by Lorentz transformations, the world line of the front of the light pulse has slope 1 here too.
     
    Last edited: Jul 13, 2010
  11. Jul 13, 2010 #10
    I'm going to do some more reading (hopefully at least then if there is still a problem, i might have the jargon down).

    Instead of
    (S-V): The velocity of the bowling ball relative to the train, according to trackside me
    Should i have put
    (S-V): the separation speed of the bowling ball from the train, in the trackside frame of reference.
    ???

    I would like to know how relativity of simultaneity fits into the bowling ball situation, but it looks like im going to have to do this one myself, somehow.
     
  12. Jul 13, 2010 #11

    Fredrik

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    Yes, that's a clear way of saying it. (But you may not need to say it at all, since the separation speed isn't very interesting).

    If you use just Lorentz transform the velocities you know, you won't have to worry about relativity of simultaneity.
     
  13. Jul 13, 2010 #12
    I am trying to understand the formula S^2 =X^2 + Y^2 + Z^2 - (CT)^2

    Why is there a subtraction in the formula? I thought that there would exist a greater resultant distance in space time if the observed event occurs at a later time.
     
  14. Jul 14, 2010 #13
    If you a 3 velocities, A, B, and C, such that A is statuionary, B is going 100,000 km/s on the +x axis, and C is going 179,975 km/s on the +x axis.

    Now, according to B, A is going 100,000 km/s on the -x axis, while C is going 100,000 km/s on the +x axis. There is no -c, only different directions, but in this case B can say A is going -100,000 km/s, while C is going +100,000 km/s.

    100,000 km/s + 100,000 km/s = 179,975 km/s
    179,975 km/s - 100,000 km/s = 100,000 km/s

    The addition of velocities is given by:
    [tex]u + v = {u+v \over 1+(uv/c^2)}[/tex]
    Where u and v are the two velocities and c is light speed.

    So the substraction formula you are looking for is:
    [tex]u - v = {u-v \over 1-(uv/c^2)}[/tex]
     
  15. Jul 14, 2010 #14

    Fredrik

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    It's just a definition that has turned out to be useful. It's possible to show that the set of Lorentz transformations is precisely the set of linear transformations that leave this quantity invariant.

    The 4-vector (CT,X,Y,Z) is said to be spacelike, null and timelike when S2 is positive, zero and negative respectively. If the tangent of the curve is timelike everywhere, the curve is said to be timelike. (Spacelike and null curves are defined similarly). The world line of a massive particle is always timelike. A timelike curve has a property called proper time that can be defined as the integral of [tex]dt^2-dx^2-dy^2-dz^2[/tex] along the curve. What a clock measures is the proper time of the curve in spacetime that represents its motion.

    I think he knows that and just wants to find out why this formula holds (without using the velocity addition formula).
     
  16. Jul 14, 2010 #15
    Iv run into difficulty understanding the formula that posted previously.

    The idea behind the concept that is space-time is that two events are not separated by a length in three dimensions, but also time; causing length to be essentially indistinguishable from a length in the context of space time. This concept can be explained by considering a supernova occurring several thousand light years away and it being observed looking up at the sky. To the observer looking up in the sky, the event of looking up in the sky coincides with the supernova and are to the observer simultaneous. But in fact this occurred several thousand years ago and the event can only be recognised as occurring, once the information has travelled the distance. It might turn out in fact that the supernova wasn't a supernova, but a plane in the sky that has turned on a light. These events are not separated by a massive distance and so it can be considered that the fact that they are simultaneous according to the observer, means that the light only turned on a very marginal time before before it was recognised.

    This has been developed into the concept that is space time, a four vector space (x,y,z,t) where time is indifferent to the other 3 dimensions. The resultant formula for space time is:
    S^2 = X^2 + Y^2 + Z^2 -(CT)^2
    S: the resultant distance between event A and event B
    X: the difference between the x coordinates of event A and event B or (Xb-Xa)
    Y: the difference between the Y coordinates of event A and event B or (Yb-Ya)
    Z: the difference between the Z coordinates of event A and event B or (Zb-Za)
    C: the speed of light in a vacuum (converts the time units into the length units used)
    T: the difference in when event B occurs and when event A occurs, if they were in the same position in space or (Tb-Ta)

    Lets consider the supernova situation. Consider event A to be recognition of event B the supernova, the position of event A to be (0,0,0) and event B (100,0,0) units in light years. If in the same position then event A would occur 100 years after event B; As event A is the recognition of event B we know that the resultant distance in 4 vector space is zero and so a result for S must be zero.

    S^2 = (100 light years)^2 + 0^2+ 0^2 -(C*100 years)^2
    = 10,000-10,000
    = 0
    their for S = 0

    This formula works perfectly in this situation, but in the next situation, it seemingly doesn't. Event A is a bunch of 2012 believers looking to the sky expecting the end of the world and event B is a supernova, that emits massive amounts radiation, which when this radiation hits earth, will kill most lifeforms. If event A and event B are in the same position then event B would occur 10,000 years later. event A has a position of (0,0,0) and event B a position of (10000,0,0). Based on the logic that event A will happen, 10,000 years will pass, the supernova occurs and it takes 10,000 years for the em radiation to hit earth; I expect a result of 20,000 light years.

    S^2 = (10,000 light years)^2 + 0^2+ 0^2 -(C*-10,000 years)^2
    = 1,000,000 - 1,000,000
    = 0
    their for S = 0
    According to the result found using the formula, the 2012 believers are right and my logic is wrong.

    My problem with space-time seems to be a difficulty in recognising time behaves the same as a dimension such as x,y,z as my logic seems to consider time to behave the same as S or the resultant. My logic would have the formula for S to be:
    S = (X^2+Y^2+Z^2)^0.5 -CT
     
  17. Jul 15, 2010 #16

    Fredrik

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    I don't really see what's confusing you here. I assume that you were just making some sort of editing mistake when you said that the event where the star goes nova and the event where the believers see it have the same position. The believers may be in Texas when they see it, but the star certainly isn't.

    The quantity you call S2 is always zero when there's a light ray going from one of the events to the other. I wouldn't call it "distance". There's no name for it that I really like. The best I can think of right now is "the Minkowski square of the difference between the position four-vectors",

    The former events has (t,x,y,z) coordinates (0,0,0,0) and the latter (-100,100,0,0). (I always write t first, and use units such that c=1). Two events are simultaneous "to the observer" if the coordinate system associated with his motion (see e.g. this) assigns the same time coordinate to the two events. In this case, it assigns 0 and -100, so the events are not simultaneous.
     
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