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Lorentz Transformations, help me grasp them? 
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#1
Jan1511, 02:37 PM

P: 37

I'd like to start by mentioning that I have very little in the way of experience on the subject, so forgive me if my confusion is somewhat trivial..
My problem lies with understanding what the fundamental variables in the Lorentz Transformations actually represent. For example, it is to my understanding that the transformation for determining the extent of time dilation is as follows; t1 = 1/√1v^2/c^2 * (tvx/c^2) I find no trouble inputting the values for all the variables, apart from the x. What does x refer to in this case? I know it's the distance along the x axis, but the distance of what exactly? Let's say a train was moving at 75% light speed relative to an observer on the embankment, and I needed to know how long one hour on the observers watch took on a clock on the train. I would use the formula stated above with the following variables.. t = 3600 v = 0.75 * 3*10^8 c = 3*10^8 So what would x be? Is it simply the distance the train has travelled in the x axis relative to the observer on the embankment? Apologies again for my lack of basic understanding. I'm just an interested teenager and without proper education on the subject can really only learn from what I've read. Thanks very much! 


#2
Jan1511, 02:51 PM

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P: 8,470

It also works out that if you want to know the time interval [tex]\Delta t_1[/tex] between a specific pair of events in the second frame, and you know the position and time intervals [tex]\Delta x[/tex] and [tex]\Delta t[/tex] between those same events in the first frame, you can use a basically identical equation: [tex]\Delta t_1 = \frac{1}{\sqrt{1  v^2/c^2}} (\Delta t  v \Delta x / c^2)[/tex] 


#3
Jan1511, 02:57 PM

P: 37

Thank you very much, JesseM! I expected to have to trawl my way through a complicated response that would take hours of thinking to figure out, but you've highlighted my misconception excellently, so thank you! So what you're saying is that the Lorentz Transformations are used to transform from one observers view on an even to another, and I couldn't figure out what x was because I didn't 'set' an event?



#4
Jan1511, 02:59 PM

Sci Advisor
P: 8,470

Lorentz Transformations, help me grasp them?



#5
Jan1511, 03:33 PM

P: 37

Okay, thanks again! :)
So, would this revised problem be more accurate? Lets say the train is again moving passed an observer on the embankment at 0.75% light speed. The two clocks are synchronized just as the train passes the observer. This can be our first event. An explosion occurs 10,000m down the tracks, 3600 seconds after the train passes (relative to the stationary observer). This is the second event. I want to find how many seconds tick on the train's clock between the two events. So.. t1 = 1 / √1(0.75^2/1^2) * (36000.75*10,000)/1^2 And that's it? 


#6
Jan1511, 04:07 PM

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P: 8,470

1.511858*(3600  0.75*10,000) = 5896.2462 So if there was a clock C on board the train that in the stationary frame read t=0 when it was passing by position x=0, then in the train frame the event of the explosion was simultaneous with the event of clock C reading 5896.2462 seconds (the relativity of simultaneity means that two frames can actually disagree on the order of two events as long as the events have a "spacelike" separation, meaning it would be impossible for a signal to travel from one event to the other without moving faster than light). But in the stationary frame, the event of the explosion was simultaneous with clock C reading 3600/1.511858 = 2381.176 seconds. It may help to realize that ideally, each frame defines the time coordinates of events using local readings on a system of multiple clocks that have been synchronized using the Einstein synchronization convention (in which each frame synchronizes their own clocks by assuming light moves at c relative to themselves, which results in each frame thinking the other frame's clocks are outofsyncI can explain this in more detail if you want). I put up some illustrations of this idea in this thread. 


#7
Jan1511, 04:46 PM

P: 37

Thanks again, Jesse for your more than helpful discussion.
Yeah my bad, I did mean for the formula to be (t  (vx/c^2)), I just wasn't thinking when I wrote it haha. I get a little confused when you introduce the relativity of simultaneity. As an isolated concept I understand it but put into the context of the question at hand I lose it. So, when I put the formula into my calculator I get a value of 5896. Is this the amount of seconds the stationary observer would 'see' pass on the clock on the train, whilst his own would tick just 3600 times? Thanks, Will. 


#8
Jan1511, 05:06 PM

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P: 8,470

In the train frame, of course the event of clock C reading 2381.176 seconds does not happen at the same time as the explosion which happened at a time of T=5896.2462 seconds. But you can check that everything is consistent with the Lorentz transform by starting from the coordinates of these events in the train frame and then transforming back to the ground frame. To do this we need the position coordinates of both events in the train frame as well as their time coordinates: clock C is at rest at position X=0 in the train frame, and the explosion happens at position X=11036.5634 in the train frame. And since the ground is moving at 0.75c in the train frame, the transformation back to the ground frame would be: x=1.511858*(X  (0.75)*T) t=1.511858*(T  (0.75)*X) So the event of clock C reading 2381.176 happened at coordinates (X=0, T=2381.176) in the train frame, when you plug this in you find it occurred at t=1.511858*(2381.176 + 0.75*0) = 3600 in the ground frame. And the event of the explosion happened at (X=11036.5634, T=5896.2462) in the train frame, when you plug this in you find the explosion happened at t=1.511858*(5896.2462 + 0.75*11036.5634)=3600 in the ground frame. So, this shows that both of these events did indeed happen at the same tcoordinate in the ground frame. 


#9
Jan1511, 05:37 PM

P: 37

Okay.. I think I just about get it, albeit in a very blurred way. So using the time dilation formula will tell me how many seconds pass on the clock C from the reference frame of the observer, whilst the Lorentz transformation formula I used tells me what clock C will read when the explosion occurs in the reference frame of the train? And these two values differ because of the relativity of simultaneity?
The only query I have left, is the significance of the negative solution to the transformation (5896.2462). Would I just treat that as 5896.2462 seconds having passed on clock C between the train passing the observer and the explosion occurring (from the train's reference frame)? Thanks 


#10
Jan1511, 05:52 PM

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#11
Jan1511, 06:03 PM

P: 37

Aah, I guess I underestimated the effects of the relativity of simultaneity, but of course it makes sense that at .75c the explosion would have happened before the train even passed the observer.
Thank you very much, that seems to have cleared everything up for me! I've been confused about this for ages now and never thought to seek any personal help, but i'm definitely glad I did! Regards, Will. 


#12
Jan1511, 09:23 PM

Sci Advisor
P: 8,470

Glad I could help!



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