# Why is length contracted, and not ELONGATED?

by dreamLord
Tags: contracted, elongated, length
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 Quote by dreamLord To be honest, I like your way more too. It's much more logical. The wikipaedia one (and the one I've found in Kleppner, Berkeley and a few national texts) is, like you said, mathematical manipulation which doesn't make sense to me in a logical manner. Still, if you can manage to extract a logical reason from that method, please do so, because I will have to be writing this method in my examinations. However, how can you say that they are not transforming from one frame to another?
If you just write out the equations for the Lorentz transformation process, are you transforming from one frame to another? Or are you merely establishing a mathematical relationship? In any case, let me restate the derivation from wikipedia with some extra steps and using gamma:

The rod is at rest in frame S' and moving at v in frame S. There are a lot of equations that we can write down but we arbitrarily pick two that relate the end points of the rod in S' to their events in S. Note that we are not concerned with the events in S' because we are ignoring the corresponding time coordinates in frame S'. This is because we realize that we can specify any time coordinates for the events in frame S' and it won't change the calculation of the length of the rod since it is at rest. So we specify the two ends of the rods in frame S' as x1' and x2' and we write:

x1' = γ(x1-vt1) and x2' = γ(x2-vt2)

Now we want t1 to be equal to t2 so we replace both of them with simply t.

x1' = γ(x1-vt) and x2' = γ(x2-vt)

Now we note that since L(0) is equal to x2' - x1' we can write:

L(0) = x2' - x1' = γ(x2 -vt) - γ(x1-vt)
L(0) = γx2 - γvt - γx1 + γvt
L(0) = γx2 - γx1 + γvt - γvt
L(0) = γx2 - γx1
L(0) = γ(x2-x1)

But since we also know that L (the length in frame S) is equal to x2-x1 we can make that substitution and get:

L(0) = γL

And rearranging we get the desired final (and correct) result:

L = L(0)/γ

Note that this derivation is very similar to yours in post #1 where you made the mistake of implicitly setting t = t1' = t2'. Instead, I explicitly set t = t1 = t2. So in effect, you were doing the calculation for the situation where the rod is at rest in frame S and you were calculating its length in frame S'. Remember, the equations are establishing a relationship between the coordinates of events in two different frames. You can use the Lorentz transformation equations to solve for the coordinates in either direction but it is more complicated going in the opposite direction and you have to remember that the sense of v is in the opposite direction.
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 Quote by ghwellsjr I don't know what the standard derivation in most texts looks like. Can you link to an online reference? [..].
I also learned to obtain it directly from the Lorentz transformations; maybe because I was taught that way, it looks simpler to me than your method.
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 Quote by harrylin I also learned to obtain it directly from the Lorentz transformations; maybe because I was taught that way, it looks simpler to me than your method.
Now that I have learned how the derivation works, especially with the added steps and using gamma, I agree. It turns out to be very simple.
P: 846
 Quote by dreamLord Silly question, I assume, but I can't quite figure it out. I've just started reading up on Special Theory of Relativity, and only have knowledge about Lorentz Transformations. Anyway, supposing a rod is at rest in the frame S', which is moving with velocity v with respect to frame S. Then the length of the rod in frame S' is the true length L(0), and is given by x2' - x1', where x2' and x1' are the end points of the rod in frame S'. Now we have to find the length of the rod in frame S, which is given by x2 - x1, where x2 and x1 are the end points of the rod in frame S. x2 is related to x2' by x2 = γ(x2' - vt) and x1 is related to x1' by x1 = γ(x1' - vt) Then length in frame S = x2 - x1 = γ(x2' - x1') Or length L = γL(0) Hence length to an observer in frame S is elongated. May I know where I went wrong?
The math has been handled very well here, so I can't add anything more to that. However, dreamLord, it might be good to keep track of the physical picture implied by the Lorentz transformaions as I'm trying to illustrate with the sketch below. We have red and blue meter sticks moving in opposite directions along the X1 axis of the black rest frame. But, the larger implication of the Lorentz transforms in the context of Minkowski 4-dimensional space is that the red and blue meter sticks are 4-dimensional objects. So,we actually have here the 3-D cross-section views of the 4-D universe at one instant of time for each observer. And the reason for the perceived length contraction is immediately obvious.
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 Quote by bobc2 And the reason for the perceived length contraction is immediately obvious.
It's not immediately obvious to me even after many minutes of trying to figure out what you are trying to show.

You mentioned a black frame and I think you are showing axes for red and blue frames, so are we supposed to think about three frames at the same time?

You said you have a red and a blue meter stick but I see what appears to be two red and two blue meter sticks. That confuses me.

Plus you show all four meter sticks aligned with lines of simultaneity. The whole point of the preceding posts in this thread is that in the frame in which the rod is moving, we pick a pair of events at the two ends of the stick that are simultaneous but in the rod's rest frame these events are not simultaneous. Is your sketch supposed to be illustrating the process described in the wikipedia article or are you showing some other method to illustrate length contraction?

Since you said that both meter sticks are moving in opposite directions in the black frame and since everything looks symmetrical to me, I'm assuming that they are moving at the same speed, in which case they should be length contracted by the same amount and I would think they should be aligned parallel to the black x-axis but I don't see that.

So it's immediately obvious to me that a sketch like this needs some background training in order for it to communicate to other people.
 PF Patron P: 4,161 Here's another way for an observer to measure the length of a moving rod, or for a moving observer to measure the length of a stationary rod. Let's say the rod is 1 unit in length (using units where c=1) stretching from coordinates [t,x] of [0,0] to [0,1]. The observer is moving in the x direction at 0.8c. In the rest frame of the rod, he will be at the origin at time 0. Since v = d/t and t = d/v, the time in the rod's rest frame that he will reach the other end of the rod will be t = d/v = 1/0.8 = 1.25. The event of him arriving at the far end of the rod in the rod's rest frame is [1.25,1]. Transforming this event into the observer's rest frame we get [0.75,0]. Now he can calculate how long the rod is using d = vt = (0.8)(0.75) = 0.6. This is the same answer we would get using the reciprocal of the Lorentz factor, 1/γ = √(1-v2) = √(1-0.82) = √(1-0.64) = √(0.36) = 0.6. Isn't that fun?
P: 846
I guess I did a pretty poor job of laying out the sketch without a useful explanation. You and the others have done such a good job of explaining the contraction with the Lorentz transformations, I thought I might be able to compliment that with the space-time sketch. It's an old habit started in my first graduate course in special relativity when our prof required us to do so many of the old paradox problems for homework. He required us to first do it with just the math and then repeat it using space-time diagrams. I got in the habit of thinking of the universe as 4-dimensional, although my prof never used the term, "block universe" and never discussed the reality of the 4-dimensional objects.

Does the little tutorial on 4-dimensional space and space-time diagrams in this thread help any?

http://www.physicsforums.com/showthr...54#post4139454

If not I can add in some basic analytic geometry on the front end.

 Quote by ghwellsjr It's not immediately obvious to me even after many minutes of trying to figure out what you are trying to show.
I'm trying to show two 4-dimensional meter sticks. Each one is slanted in the opposite direction relative to the black coordinate system (the vertical black line is the black X4 axis and the horizontal black line is the black X1 axis).

 Quote by ghwellsjr You mentioned a black frame and I think you are showing axes for red and blue frames, so are we supposed to think about three frames at the same time?
Yes. I wanted to show red and blue moving in opposite directions with respect to the black rest frame.

 Quote by ghwellsjr You said you have a red and a blue meter stick but I see what appears to be two red and two blue meter sticks. That confuses me.
The two slanting blue lines represent the front and rear ends of the blue meter stick, displayed as a 4-dimensional object (X2 and X3 coordinates are supressed for ease of viewing). Likewise the two red slanted lines represent front and rear ends of the 4-dimensional red meter stick.

 Quote by ghwellsjr Plus you show all four meter sticks aligned with lines of simultaneity. The whole point of the preceding posts in this thread is that in the frame in which the rod is moving, we pick a pair of events at the two ends of the stick that are simultaneous but in the rod's rest frame these events are not simultaneous. Is your sketch supposed to be illustrating the process described in the wikipedia article or are you showing some other method to illustrate length contraction?
If you follow the blue line of simultaneity, you will see the front and rear points on the blue meter stick that are simultaneous. And continuing along that blue line of simultaneity you will see the front and rear points of the red stick that are simultaneous in blues simultaneous space (an instantaneous 3-dimensional cross-section vew, cut across the 4-dimensional universe). So, just working within blue's inertial frame you can directly compare the lengths of the blue and red sticks as they appear in blue's 3-dimensional world. Then you can do the same thing for the red world, using the red simultaneous space.

 Quote by ghwellsjr Since you said that both meter sticks are moving in opposite directions in the black frame and since everything looks symmetrical to me, I'm assuming that they are moving at the same speed,...
That is correct, they are moving along the black X1 axis in opposite directions at the same speed. That's why the 4-dimensional stick objects are slanted in opposite directions with the same slant angle relative to the black X4 axis. Note that as you move along the blue 4-dimensional stick in the blue X4 direction, you are increasing the distance along the black X1 axis. Or, another way of saying it, if you are at rest in the black rest frame, that means you are moving straight forward into time, i.e., moving along the black X4 axis at the speed of light--and as you move along black X4, the blue meter stick increases its distance away from you (black) along the black X1 axis. Also if you are red guy, moving along the red X4 at light speed, you also see the blue meter stick increasing its distance away from you along the red X1 axis.

 Quote by ghwellsjr ... in which case they should be length contracted by the same amount...
As mentioned above, the blue guy "sees" the red guy's stick contracted as compared to his own (just comparing lengths within the blue simultaneous space) and the red guy "sees" the blue guy's stick contracted as compared to his own (comparing lengths just within the red simultaneous space). And the reason I've used the symmetric Loedel space-time diagram is that you are able to compare the lengths directly, since the line lengths in the sketch have exactly the same actual distance calibrations (you can't compare distances directly between black and the moving frames without resorting to the hyperbolic calibration curves).

 Quote by ghwellsjr ...and I would think they should be aligned parallel to the black x-axis but I don't see that.
Take another look at the diagram. If the blue and red 4-dimensional sticks were aligned parallel to the black X4 axis, that would mean that the blue and red sticks were at rest in the black rest system. For example, follow the blue stick along the 4th dimension, assuming the blue stick is parallel to the black X4 axis: In that case, as time passes (as you move along blue's stick along the direction of the X4 axis) you would see that the blue stick front and rear positions along black's X1 axis have not changed.

 Quote by ghwellsjr So it's immediately obvious to me that a sketch like this needs some background training in order for it to communicate to other people.
Apparently that is the case. I'll have to give more thought as to how to do a better job of presenting the space-time diagrams. Any suggestions?
 P: 846 Post Script. I probably should just give it up as a fruitless effort. I had forgotten the lengthy question and answer sessions it took to get this concept across to some of my students.
 PF Patron Sci Advisor Emeritus P: 8,837 I think there's just too much information in that diagram. (I didn't try to determine if that information is correct or not). I would prefer one with just the necessary information drawn in. Something like this: If you disregard my poor drawing skills (the slope of the x' axis isn't exactly right), the only problem with this diagram is that a person who doesn't understand that the scales on the axes are fixed by the invariant hyperbolas might interpret it as saying that ##L_0=x'_C-x'_A>x_B-x_A=L##. I would recommend section 1.7 in Schutz to anyone who's struggling with that. In particular fig. 1.11 at the bottom of page 17. To see that the inequality actually goes the other way, what you have to do is to draw a curve of the form ##-t^2+x^2=\text{constant}## with the constant chosen so that the curve goes through C. The point where this curve intersects the x axis will have coordinates (0,L0). This point will be to the left of B on the x axis. So ##L_0
 P: 846 Here's the picture with the hyperbolic calibration curves. Of course these are the time-like curves. I could generate the space-like curves as well but just happened to have these handy, so am specifically illustrating the time dilation rather than the length contraction (a little off point, but I think you get the point about the need for hyperbolic calibration curves).
 P: 846 Here are the calibration curves for time-like and space-like. Red and blue inertial frames are included. I still think it's better to avoid the calibration curves by just showing two observers moving in opposite directions. Both time dilation and length contraction are directly viewed. I insisted my students understand these diagrams--it just took a little more time for some of them.
 PF Patron Sci Advisor P: 4,463 Are you perhaps confusing the Enzyte Theory of Relativity with the Einstein Theory of Relativity? --- Sorry mentors, tried and tried to resist, couldn't any longer (stop it). Do what you may.
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 Quote by Fredrik The fact that you're denoting the S' coordinate of the right endpoint of the rod by x2' suggests that you didn't realize that you need to consider three events, not two.
The above comment makes me ask you the same question I asked bobc2 (which he didn't answer):
 Quote by ghwellsjr The whole point of the preceding posts in this thread is that in the frame in which the rod is moving, we pick a pair of events at the two ends of the stick that are simultaneous but in the rod's rest frame these events are not simultaneous. Is your sketch supposed to be illustrating the process described in the wikipedia article or are you showing some other method to illustrate length contraction?
There are only two events under consideration, not three or more. These are the end points of the moving stick in frame S and they must be simultaneous in that frame. The distance between the two events is L. In the rod's rest frame S', those two events are not simultaneous but since the rod is at rest, the difference in the x-coordinates of the two events is equal to the Proper Length of the rod, L0.
 Quote by Fredrik I think there's just too much information in that diagram. (I didn't try to determine if that information is correct or not). I would prefer one with just the necessary information drawn in. Something like this:
I agree that there's too much information in bobc2's drawing but there's too much in yours too, as far as the number of events goes. The two events we care about are labeled A and B. C should not be shown as an event. Erase that dot and the C. If you try to show that there are two simultaneous events in frame S', then you have missed the whole point of the derivation. Events A and B are not simultaneous in frame S' and that's the way we want it to be.
 Quote by Fredrik If you disregard my poor drawing skills (the slope of the x' axis isn't exactly right), the only problem with this diagram is that a person who doesn't understand that the scales on the axes are fixed by the invariant hyperbolas might interpret it as saying that ##L_0=x'_C-x'_A>x_B-x_A=L##.
But that is (almost) the correct interpretation if you recognize that x'C is equal to x'B so you should have said ##L_0=x'_B-x'_A>x_B-x_A=L##. However, you really need to show the grid lines for the two frames, otherwise, it's impossible to tell what the value of x'B is or that it is larger than xB. The whole purpose of a Minkowski diagram is to show the two sets of coordinates for the two frames so that you can see that they are different for the two frames and it's those numbers that show that L is less than L0, not the fact that two line segments are actually drawn to two different lengths.
 Quote by Fredrik I would recommend section 1.7 in Schutz to anyone who's struggling with that. In particular fig. 1.11 at the bottom of page 17. To see that the inequality actually goes the other way, what you have to do is to draw a curve of the form ##-t^2+x^2=\text{constant}## with the constant chosen so that the curve goes through C. The point where this curve intersects the x axis will have coordinates (0,L0). This point will be to the left of B on the x axis. So ##L_0
You already stated the correct relationship between L and L0 at the end of post #17, why'd you change your mind? This last paragraph is all wrong. We don't need hyperbolic calibration curves, we just need marked grid lines. Why don't you or bobc2 draw a correctly annotated Minkowski diagram that illustrates the derivation of the length contraction that is the subject of this thread?
 P: 846 This sketch describes the situation originally set up in the post. Here we can consider red to be the rest system, S, and blue to be the S' system moving with respect to S at relativistic speed. There is no need for calibration curves for this kind of symmetric diagram (Loedel space-time diagram). It is clear that red will "see" a contracted rod, length L, as compared to the length, L', as "seen" in the blue frame S'.
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 Quote by ghwellsjr You already stated the correct relationship between L and L0 at the end of post #17, why'd you change your mind? This last paragraph is all wrong.
I didn't change my mind, but I agree that the last paragraph is all wrong. I must have made some elementary mistake when I wrote that post. I certainly didn't realize that what I was saying was the opposite of what I had said before.

 Quote by ghwellsjr Why don't you or bobc2 draw a correctly annotated Minkowski diagram that illustrates the derivation of the length contraction that is the subject of this thread?
I can't speak for him, but I drew a primitive diagram in Paint because I use math software so seldom that I forget everything I know about them between each time. Of course, now that I see that I have made a blunder, I will have to relearn some of it to draw an exact diagram that includes the relevant hyperbola. I will have to leave my computer for at least an hour 20 minutes from now, so I probably won't be able to do that for a few hours.

I would say that the subject of this thread is length contraction, not some specific calculation. I didn't even look at the derivation in post #1. I just saw that the result was wrong, and that the notation looked weird to me. Then I posted a correct derivation, and later the spacetime diagram that I had described in words. The diagram is correct, but I made some comments about it that are incorrect.

 Quote by ghwellsjr I agree that there's too much information in bobc2's drawing but there's too much in yours too, as far as the number of events goes. The two events we care about are labeled A and B. C should not be shown as an event. Erase that dot and the C. If you try to show that there are two simultaneous events in frame S', then you have missed the whole point of the derivation. Events A and B are not simultaneous in frame S' and that's the way we want it to be.
We definitely need three events. The motion of the (component parts of) the rod is a congruence of curves in spacetime. I only drew two of them (the world lines of the endpoints). The two observers (or rather, the coordinate systems associated with their motions) are slicing spacetime in different ways, into hypersurfaces that they think of as "space at a specific moment in time". What an observer thinks of as "the rod right now" is the union of the congruence and his simultaneity line through the origin. To the observer who's comoving with the rod (and using coordinate system S'), "the rod right now" is the line AC. To the observer who sees the rod moving at velocity v (and is using coordinate system S), "the rod right now" is the line AB. This disagreement is the reason why there is such a thing as length contraction.

 Quote by ghwellsjr The above comment makes me ask you the same question I asked bobc2 (which he didn't answer): There are only two events under consideration, not three or more. These are the end points of the moving stick in frame S and they must be simultaneous in that frame. The distance between the two events is L. In the rod's rest frame S', those two events are not simultaneous but since the rod is at rest, the difference in the x-coordinates of the two events is equal to the Proper Length of the rod, L0.
"Proper length" is assigned only to spacelike curves. You don't seem to have thought about what curve we're talking about here. The "proper length of the rod" is the proper length of the curve AC. This is the significance of the event C.

 Quote by ghwellsjr However, you really need to show the grid lines for the two frames, otherwise, it's impossible to tell what the value of x'B is or that it is larger than xB. The whole purpose of a Minkowski diagram is to show the two sets of coordinates for the two frames so that you can see that they are different for the two frames and it's those numbers that show that L is less than L0, not the fact that two line segments are actually drawn to two different lengths.
The only thing we need to add is an invariant hyperbola through C, to see where it intersects the x axis.
 P: 846 You may prefer measuring both lengths from the origin (but that's not the way it was presented in the initial post).
 PF Patron Sci Advisor P: 4,463 We've seen plenty of confusion here between terms proper distance, proper length, and rest length. I believe ghwellsjr is following one common usage where proper length (not distance, which applies between a pair of events along a specified simultaneity line) of an effectively rigid body is the same as rest length = length in a frame in which it is at rest. In such a frame, there is no requirement to measure length of a rigid object between simultaneous events, because nothing changes or moves in this frame. You can note the coordinate position of one end of the rigid body at time t1, and the position of the other end at time t2, and subtract, and still get proper length = rest length, because neither end is moving in this frame. With this understanding, only two events are needed (that are simultaneous for the frame in which the rod is moving). This same observation is also the basis of the 'cute' derivation in the wikipedia entry attributed to Born.

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