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Length Contraction rearrangement 
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#1
Mar712, 03:27 AM

P: 215

Hello,
Suppose, There is two points [itex][t_{a1}, x_{a1}][/itex] and [itex][t_{a2}, x_{a2}][/itex] stationary in my frame A. I say [itex]L_a=x_{a2}x_{a1}[/itex]. If I want to find x component of coordinates in other frame B which is moving relative to me with constant speed v. I have to use the equation 1. [itex]x_b=\gamma(x_avt_a)[/itex], where [itex]\gamma > 1[/itex]. So, I can get [itex]x_{b1}=\gamma(x_{a1}vt_{a1})[/itex], [itex]x_{b2}=\gamma(x_{a2}vt_{a2})[/itex]. If I take [itex]t_{a1}=t_{a2}=0[/itex], then [itex]x_{b1}=\gamma x_{a1}[/itex] and [itex]x_{b2}=\gamma x_{a2}[/itex]. Now, how much distance B measures in his coordinate system between this two points is [itex]x_{b2}  x_{b1} = \gamma (x_{a2}  x_{a1})[/itex]. If B says [itex]L_b=x_{b2}  x_{b1}[/itex]. So, I would get [itex]L_b=\gamma L_a[/itex]. Here, [itex]L_b > L_a[/itex]. But, this should not be the case. If I want length contraction then I have to derive it oppositely. (I found this method from http://en.wikipedia.org/wiki/Length_...ion#Derivation and http://www.fourmilab.ch/etexts/einst...www/#SECTION14) I have to pick some other B's coordinates [itex][t_{b1}, x_{b1}][/itex] and [itex][t_{b2}, x_{b2}][/itex] which is stationary in B's frame. B says [itex]L_b=x_{b2}x_{b1}[/itex] I have to use this equation 2. [itex]x_{a}=\gamma(x_{b}vt_{b})[/itex], where [itex]\gamma > 1[/itex]. I would find [itex]x_{a1}=\gamma(x_{b1}vt_{b1})[/itex] and [itex]x_{a2}=\gamma(x_{b2}vt_{b2})[/itex]. If I take [itex]t_{b1}=t_{b2}=0[/itex] then, [itex]x_{a1}=\gamma x_{b1}[/itex] and [itex]x_{a2}=\gamma x_{b2}[/itex]. So, [itex]x_{a2}  x_{a1} = \gamma (x_{b2}  x_{b1})[/itex]. I says [itex]L_a=x_{a2}x_{a1}[/itex]. so, [itex]L_a=\gamma L_b[/itex]. so, [tex]L_b=\frac{L_a}{\gamma}[/tex] so, [itex]L_b < L_a[/itex]. Now, we can say this as length contraction. But, I have started this derivation using B's coordinates. I as A don't know B's coordinates. I only know my coordinates because I can physically define it. I have to calculate B's coordinates to get Length Contraction. I cannot use equation (1) for that, it wouldn't give me length contraction. How can I get B's coordinates using my own? If I as A has a stationary point [itex][t_a, x_a][/itex], I can calculate B's coordinate using my own in Galilean transformation. 


#2
Mar712, 04:52 AM

PF Gold
P: 4,785

Whenever you want to determine the length of an object in a given Frame of Reference, the time coordinates must have the same value. You fulfilled this requirement in the first frame when you said t_{a1}=t_{a2}=0 but when you transform the two events at either end of the object into the second frame, the new time coordinates will not be equal, t_{b1}≠t_{b2}.
One way to find a pair of events that will meet the requirement when the first event is at the origin in both frames is to multiply the xcomponent of the second event in the first frame by the velocity (assuming c=1) and use that as the timecoordinate for the second event in the first frame. Now when you perform the Lorentz Transform on the two events, the time coordinates will be equal and you will get the correct contracted length. If the first event is not at the origin, then you have to take the delta between the xcomponents and multiply that by the velocity and add that to the timecoordinate of the first event and use that as the timecoordinate of the second event. So, in your example, let's make the added assumption that x_{a1}=0 which will put the first event at the origin. The we multiply x_{a2} by v to get vx_{a2} so the second event is [vx_{a2},x_{a2}]. When you transform the timecoordinate for this event you get t_{b2}=γ(vx_{a2}vx_{a2})=0, which is what you want, the t coordinate for the second transformed event equal to the first transformed event (which is at the origin). When you transform the xcoordinate for this event you get x_{b2}=γ(x_{a2}vvx_{a2})=γ(x_{a2})(1v^{2}) and since γ=1/√(1v^{2}), this simplifies to x_{b2}=x√(1v^{2})=x/γ, which is what you want. 


#3
Mar712, 12:58 PM

P: 215

Ok, please see the wiki http://en.wikipedia.org/wiki/Length_...ion#Derivation.
Object is in rest in [itex]S'[/itex] frame, so there is always [itex]t'_1=t'_2[/itex]. But, the article says we have to put [itex]t_1=t_2[/itex] to calculate length [itex]L[/itex]. How is that possible? As I understand we cannot get [itex][t'_1, x'_1][/itex] and [itex][t'_2, x'_2][/itex] for any [itex][t_1, x_1][/itex] and [itex][t_2, x_2][/itex] where [itex]t'_1=t'_2[/itex] and [itex]t_1=t_2[/itex]. Please, let me know if we would get this. Suppose, the article not saying that [itex]t_1=t_2[/itex] is a requirement, then we cannot get [itex]x'_2  x'_1 = \gamma (x_2  x_1)[/itex]. 


#4
Mar712, 02:00 PM

Mentor
P: 41,563

Length Contraction rearrangement
Using the reverse Lorentz transformation [itex]\Delta x' = \gamma(\Delta x  v\Delta t)[/itex] is all you need to get the length contraction formula. 


#5
Mar712, 03:32 PM

PF Gold
P: 4,785




#6
Mar912, 02:14 AM

P: 215

This is what I am trying to say. If you make time coordinate same in one frame, you never get same time coordinate in transformed frame. So if we do [itex](x_{a2}x_{a1})[/itex], does this gives us exact contracted length? Other thing I want to point out is in wiki And then wiki says 


#8
Mar912, 08:54 AM

P: 215

This is not the measurement time. When a object is in rest, then the object has same time component value. But when the object is moving relative to a S frame then the object has different time component value for S. (Relativity Of Simultaneity) When some event happen near you at same time and a event happen far from you. But, you see the near event first, and after that the far event. That doesn't mean the event is not simultaneous. Please, read "What the Relativity of Simultaneity is NOT" in the link http://www.pitt.edu/~jdnorton/teachi...ivity_rel_sim/ 


#9
Mar912, 08:57 AM

PF Gold
P: 4,785

If you want, you can cheat by knowing that the moving length should be the rest length divided by gamma and picking a pair of events in the second frame separated by that length and with the times equal, and then transforming those events into the first frame and just changing the times so they are equal but now the length will be gamma times the length in the first frame and you can pretend that you now want to transform that new event to the second frame which, of course will give you another length that is too large, but then you can say, aha, I know a set of times in the first frame that will transform into the second frame and result in two events with the same time on them and then it will give the correct length. This might be a good way to fool someone else when you're showing them how you need to select two times in the first frame that are not equal so that the two times in the second frame will be equal, but like I say, it's cheating because you worked the problem out backwards. So another way is by trial and error. All you have to do is start with the two endpoint events with simultaneous time coordinates in the first frame, transform them to see that the new events are not simultaneous, then tweak either one of the old time coordinates, do the transformations again, see if the new time coordinates got closer together and if so, continue tweaking the same old time coordinate in the same direction until the new time coordinates match. If the new time coordinates get farther apart, go back and tweak the old time coordinate in the other direction and continue until you find a pair of old events that transform into new events that are simultaneous. Or you could do my way. Just remember the important point is that when you are changing the time coordinates of the old events, you must not change the spatial coordinates because that would be determining a different length. 


#10
Mar912, 09:02 AM

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P: 41,563




#11
Mar912, 01:58 PM

P: 89

Using the reverse Lorentz transformation, one obtains t'1≠t'2 for the rod's rest frame S'. But this doesn't matter because in S' you can measure the endpoints whenever you want, since the rod is at rest here. Regards, 


#12
Mar912, 02:35 PM

PF Gold
P: 4,785

There's confusion going on here because when we are talking about events in a Frame of Reference and using the Lorentz Transform to get the coordinates of those events in another FoR, there is no consideration for any measurement taking place and no observer located in the scenario to make any measurement, rather, it is pure calculation.
It is true that when a real observer is making a real measurement with a real ruler of a real object that is at rest with respect to him, that the time over which he makes the measurement is of no consequence and, in fact, there does not need to be any FoR assigned or considered. Making measurements is not an issue with a FoR. Furthermore, a real observer making a real measurement of a real object in motion does not need to do it in an instant of time. He can first measure its speed and then he can measure how long it takes for the object to pass by a particular location and then he can calculate its length. But this thread is not about how an observer measures an object but rather how to correctly transform events from one frame to another in a meaningful way, so let's not confuse the two different ideas. 


#13
Mar912, 11:45 PM

P: 215

But, this is not possible. You can have only one time component for all space components of rod if rod is at rest in your frame. Suppose, a long object is in your frame, and If you put two synchronized clocks at both ends. After any how much time you find the clocks is still synchronized. You cannot get different time components for both ends. Actually you have different space component at different different points of rod, but the space components is surely constant with time, it is not changing with time. When you start moving the long object in your reference frame, now time component of both ends surely changed and you get clocks no longer synchronized. How can you draw a Minkowsky diagram for rest rod with different time components? If you draw a rest rod, it looks like horizontal line parallel to space axis. It has one time component for all space component values. If you change the time component then there is no longer horizontal line, now it look like diagonal line. Which is same as diagram of moving rod in your reference frame. 


#14
Mar1112, 08:00 AM

Mentor
P: 5,375

Lets talk about how you would go about measuring the length of an object that is at rest in the S' frame of reference when you are an observer (or, more precisely, a member of a team of observers) at rest in the S frame of reference. The other members of your team are strung out along the direction of motion x. In the S' frame of reference, the object is always situated between coordinates x' and x'+L', with the trailing edge at x' and the leading edge at x'+L'. One way of measuring the length of the object, as reckoned from your frame of reference, is for the various members of your team to each record on a sheet of paper the times displayed on each of their (synchronized) clocks at which the leading edge and the trailing edge of the object pass their locations, together with their own coordinates x. They then get together afterwards and make a plot of x on the vertical axis vs t on the horizontal axis both for the leading edge and the trailing edge of the object (on the same piece of graph paper). How can they then determine how long the object is from their perspective in the S reference frame? The only way that makes sense is to determine the vertical distance L between the two parallel lines on the graph. But this is the distance at each constant value of the time t as measured on the clocks in your frame of reference. This is why you use constant t in the Lorentz Transformation calculations.



#15
Mar1112, 08:09 AM

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P: 41,563




#16
Mar1112, 09:43 AM

PF Gold
P: 4,785

When you say that we can have only one time component for all space components of a rod, how would you respond if I said, Well that violates the principle that you can't go faster than the speed of light? How do you get from one end of the rod to the other end of the rod at the same instant in time? Obviously, no one is traversing that distance, we are merely assigning coordinates to those different positions and showing that the length of the rod that we previously measured over a period of time is the same as what the spatial coordinates of the events indicate. In other words, we previously measured the length of the rod by placing a ruler next to the rod and actually moving from one end of the rod to line up one end of the ruler and then moving to the other end of the rod and looking to see what the markings on the ruler indicate its length to be. This cannot be done in an instant of time because we cannot travel instantly from one end of the rod to the other end. So I don't see why you are saying you can only have one time component for all space components. Now after having "measured" the length of the rod in this manner, you can also just look at the record of where the two ends were at every point in time and pick a pair of coordinates where the times were the same and subtract the end positions and you will get the same length as you "measured" earlier. This is essentially what Chestermiller described for you in post #14. EDIT: This last process is essentially the trial and error method I mentioned earlier. 


#17
Mar1212, 03:11 AM

P: 215

Again I already got the point that to calculate x component we have to take different time component at both end. I was just trying to say that how rest object can have two time component. If I am agree on this then I have another question. 


#18
Mar1212, 06:11 PM

PF Gold
P: 4,785

I bet I haven't answered your question. If I haven't, you'll have to rephrase it because it really doesn't make much sense to me. 


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