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Lorentz Transformations, help me grasp them?

by aguycalledwil
Tags: grasp, lorentz, transformations
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aguycalledwil
#1
Jan15-11, 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/√1-v^2/c^2 * (t-vx/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!
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JesseM
#2
Jan15-11, 02:51 PM
Sci Advisor
P: 8,470
Quote Quote by aguycalledwil View Post
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/√1-v^2/c^2 * (t-vx/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?
x and t represent the position and time coordinates of some specific event in the first frame (an 'event' is something localized in position and time, say the place and time a firecracker was set off, or say the place and time a particular clock showed a reading of exactly 3 seconds), so this equation gives the time coordinate of the same physical event in a different frame.

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]

Quote Quote by aguycalledwil View Post
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?
Here the two events are two specific events on the worldline of the observer's watch (say, the event of the watch showing a time of 0 seconds and the later event of the watch showing a time of 3600 seconds), and t (corresponding to my [tex]\Delta t[/tex] above) seems to be from the perspective of the frame where the watch is at rest, so in this case x (corresponding to my [tex]\Delta x[/tex] above) should be the difference in position coordinates from the perspective of the frame where the watch is at rest, which would just be zero. By the way it's a lot simpler to use units where c=1, like seconds and light-seconds, that way v=0.75 (also, the numbers will work out simpler if you use a velocity of v=0.6, that way [tex]\frac{1}{\sqrt{1 - v^2/c^2}} = \frac{1}{\sqrt{0.64}} = 1.25[/tex], a nice easy-to-work-with number).
Quote Quote by aguycalledwil View Post
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.
No need for apologies, asking questions is the way to learn!
aguycalledwil
#3
Jan15-11, 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?

JesseM
#4
Jan15-11, 02:59 PM
Sci Advisor
P: 8,470
Lorentz Transformations, help me grasp them?

Quote Quote by aguycalledwil View Post
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?
Right, although as I said you can also use the same equations for the space and time intervals between a pair of events, like two successive readings on some clock.
aguycalledwil
#5
Jan15-11, 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) * (3600-0.75*10,000)/1^2

And that's it?
JesseM
#6
Jan15-11, 04:07 PM
Sci Advisor
P: 8,470
Quote Quote by aguycalledwil View Post
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) * (3600-0.75*10,000)/1^2
The last "/1^2" seems to mean you are thinking the whole (t - vx) is being divided by c^2, but that's not right--it should be (t - (vx/c^2)). Of course if you set c to 1 it makes no difference, but this would be important if you were using some other units. Also, if you're setting c to 1 then your distances and times must actually be in units where c=1--you can't use seconds for time and meters for distance as you did above! But we can fix this by using light-seconds for distance as I suggested, and say the explosion happened 10,000 light-seconds down the track, not 10,000 meters.
Quote Quote by aguycalledwil View Post
And that's it?
Aside from the minor issues I mentioned, this is the right formula for how many seconds pass in the train's frame between the two events, but you have to be careful with a phrase like "how many seconds tick on the train's clock" because of the relativity of simultaneity. In the train's frame, the event of the explosion happens at this time coordinate:

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 out-of-sync--I can explain this in more detail if you want). I put up some illustrations of this idea in this thread.
aguycalledwil
#7
Jan15-11, 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.
JesseM
#8
Jan15-11, 05:06 PM
Sci Advisor
P: 8,470
Quote Quote by aguycalledwil View Post
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?
No, the question of how much time would pass on the train clock (which I called "clock C") as measured in the stationary frame was what I was addressing with this comment:
But in the stationary frame, the event of the explosion was simultaneous with clock C reading 3600/1.511858 = 2381.176 seconds.
Here I was just using the time dilation formula, which says that if you have an observer measuring a clock moving relative to himself, then if you pick two events on the clock's worldline and say the time between them in the observer's frame is t while the time between them as measured by the clock itself is T, the two times are related by t = T * gamma (or equivalently T = t / gamma), and in this case gamma=1.511858 (another way of putting this is just that as seen in the observer's frame, a moving clock runs slow by a factor of gamma). So if we pick two events on the moving clock C with a separation of t=3600 in the stationary frame (here the second event can be the reading on clock C that happened 'at the same time' as the explosion in the stationary frame), only T = 3600/1.5118 seconds must have elapsed on the clock itself, i.e. 2381.176 seconds.

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 t-coordinate in the ground frame.
aguycalledwil
#9
Jan15-11, 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
JesseM
#10
Jan15-11, 05:52 PM
Sci Advisor
P: 8,470
Quote Quote by aguycalledwil View Post
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?
Yeah, although it's easiest to just think of the Lorentz transformation in terms of giving you each frame's coordinate time rather than any particular clock...but naturally since clock C is at rest in the train frame and shows a time of 0 at T=0 in that frame, its readings will always match up with coordinate time in the train frame. So if the Lorentz transform tells you the explosion happens at a coordinate time of T=-5896.2462 in the train frame, and you also that clock C's readings line up with coordinate time in this frame so the event of clock C reading -5896.2462 also happens at a coordinate time of T=-5896.2462, then from that you can see the event of the explosion and the event of clock C reading -5896.2462 were simultaneous (happened at the same T-coordinate) in the train frame.
Quote Quote by aguycalledwil
And these two values differ because of the relativity of simultaneity?
Right, because of the relativity of simultaneity, different frames have different answers to the question "which event on clock C's worldline happened simultaneously with the explosion?"
Quote Quote by aguycalledwil View Post
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
No, it really was a negative reading (i.e. a reading 5896.2462 seconds prior to showing a reading of zero), because in the train frame the event of clock C passing the observer actually happened after the event of the explosion, that was what I was talking about in this part of post #6:
(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)
On the other hand, as long as there is any possibility that one of two events can have causally influenced the other (meaning either a 'time-like' or 'light-like' separation between the events--see here), then all frames will agree about which event happened earlier.
aguycalledwil
#11
Jan15-11, 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.
JesseM
#12
Jan15-11, 09:23 PM
Sci Advisor
P: 8,470
Glad I could help!


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