Why is length contracted, and not ELONGATED?

[ghwellsjr]

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):

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.

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.

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 x_B. 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.

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 $L_0<L$.

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?

 

[Fredrik]

I made a mistake in my comments about my spacetime diagram. I will have more to say about that when I've figured out how to draw the diagram in a math program.

Edit: I'm trying to relearn how to do this in Mathematica. That will definitely take a while, and I will have to do something else in about ten minutes. So I will have to return to this later. I have posted a reply to ghwellsjr below, but the diagram and my new comments about it will have to wait.

 

[bobc2]

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'.

 

[Fredrik]

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.

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.

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.

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.

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 x_B. 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.

 

[bobc2]

You may prefer measuring both lengths from the origin (but that's not the way it was presented in the initial post).

 

[PAllen]

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.

 

[ghwellsjr]

This sketch describes the situation originally set up in the post.

I don't think so. My comments about Fredrik's diagram applies to this new one also. The only difference is he moved your pairs of events closer so that X₁ and X'₁ were at the same time and appear at the same place on the diagram. But you are also missing the point of this thread--there are only two events. They are simultaneous in the frame in which the rod is moving and they are not simultaneous in the frame in which the rod is at rest.

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.

Well now this is a confusing statement. First off, the systems are moving with respect to each other. We should never think of one of them at rest and the other one moving. What dreamLord proposed in his OP is that the rod is at rest in S' and moving in S. That seems at odds with your attaching "rest" to S and "moving" to S'.

There is no need for calibration curves for this kind of symmetric diagram (Loedel space-time diagram).

I agree and that's how I ended my previous post, "we don't need hyperbolic calibration curves, we just need marked grid lines". But you left out the marked grid lines. That's what is going to tell the story on a Minkowski diagram. Why didn't you put them in like I requested?

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'.

That's not at all clear and I don't know what you mean by "see" or why you put them in quotes. You have arbitrarily decided to assign the primed coordinates and L₀ to the longer rod just so that it will appear longer on the diagram than the other rod. But your mistake is that you don't realize that I could just as legitimately assign the primed coordinates and L₀ to the rod that is drawn shorter on the diagram. You need to get rid of either one of those rods. You need to draw just one rod and one pair of events in your diagram, it doesn't matter which one, but you can't label the events exclusively to one frame. The same events belong to both frames but with different coordinates for each frame. That's where you need to show the grid lines for each frame so that we can see that those two events are simultaneous in the frame in which the rod is moving and not simultaneous in the frame in which the rod is at rest.

 

[ghwellsjr]

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.

Exactly.

And the only reason I brought up Proper Length in post #4 is because dreamLord incorrectly used the term "true length" in his OP. The "true" length of the rod is L₀ in frame S' where it is at rest and L (which is contracted) in frame S where it is moving. Both are "true" lengths as defined in their respective frames. But I was trying to dissuade him from using the term "true length".

 

[ghwellsjr]

You may prefer measuring both lengths from the origin (but that's not the way it was presented in the initial post).

You see how you have labeled the event at the origin with X'₁ and with X₁? Now you should add two more labels to your other two events so they both say X'₂ and X₂. All events on a Minkowski diagram have two labels, one for each coordinate system. Then you need to delete one of those events and the rod that goes with it, it doesn't matter which one and put in those grid lines and you will see that in whichever frame the events defining the end points of the rod are simultaneous, the distance between the rods is shorter than in the other frame in which the events are not simultaneous but in which the rod is at rest.

The issue is not whether one of the events is at the origin or not, it doesn't matter. You could have done the same thing with your other diagram--one rod, one pair of events.

That is, if you want to illustrate the problem that was presented by dreamLord in this thread. If you are showing a different problem then you need to say so.

 

[Fredrik]

I made a couple of mistakes in this thread. First, there was this one:

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.

As PAllen explained in post #36, we can use the fact that (in my notation) $x'_B=x'_C$ to calculate $L_0$ as $x'_B-x'_A$ instead of as $x'_C-x'_A$. So you can certainly do a calculation that ends with the correct result $L=L_0/\gamma>L_0$ without ever mentioning the event C explicitly.

I don't mind using this trick to make the calculation slightly shorter, but I think it would be a bad idea to use it to hide what I think is the most important thing about this problem: The observers are referring to different slices of the congruence when they talk about "the rod right now".

This is clear in the spacetime diagram I posted. Unfortunately I made a major blunder when I wrote down my comments to the diagram:

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 $L_0<L$.

The diagram is not misleading, at least not in the way I said it would be. That first inequality (the one I said is wrong) is correct, as can be seen from the formula $L=L_0/\gamma$ that I derived in post #17.

I didn't notice that the result I obtained here is different from the one I obtained before. I didn't even look at what I had written long enough to realize that, because I thought I remembered that a naive interpretation of the diagram would give you the wrong idea about what length is greater. But I remembered it wrong. A naive interpretation of the spacetime diagram for time dilation will give you the wrong idea, but this one won't.

This can be seen by actually drawing the invariant hyperbola through C. With v=0.6 and L=1, the diagram with the hyperbola looks like this:

As you can see, the hyperbola intersects the x-axis to the right of the point (0,L), not to the left as I said in the quote above. Since that intersection point is known to have coordinates (0,L₀), this means that $L_0>L$, which is consistent with the formula $L=L_0/\gamma<L_0$.

 

[bobc2]

You see how you have labeled the event at the origin with X'₁ and with X₁? Now you should add two more labels to your other two events so they both say X'₂ and X₂. All events on a Minkowski diagram have two labels, one for each coordinate system.

Certainly not. I did not make up the labels for the rod. In the original post it was clear that he wanted two ends of the rod to be labeled so as to have two events that marked the right and left ends, X1' and X2', as rod ends in a simultaneous 3-D space (identified in my sketch as "Simultaneous Blue") corresponding to the S' frame moving relative to an S rest frame. Labels X1 and X2 were explicitly identified as end labels for a simultaneous space in the S rest frame.

Then you need to delete one of those events and the rod that goes with it...

No. I'm trying to emphasize the physical clarity of the problem by utilizing a picture of the 4-dimensional rod. He is specifically talking about a physical rod at rest in the S' frame that is in motion relative to a rest frame, S. So the rod stays.

it doesn't matter which one and put in those grid lines and you will see that in whichever frame the events defining the end points of the rod are simultaneous...

What you are suggesting only works if your definition of simultaneous space refers to a hyperspace with some kind of hypertime. You should spell it out if you intend something unconventional. We are referring here to a 3-dimensional simultaneity (X2 and X3 are supressed in our diagrams).

I'll put in the grid lines for you later. However, I've shown in the sketch the plane of simultaneity for blue and for red. Only lines parallel to the "Simultaneous Blue" can represent a simultaneous space for blue. And only lines parallel to the "Simultaneous Red" can represent simultaneity for red, unless you reject the Einstein special relativity theory.

...the distance between the rods is shorter than in the other frame in which the events are not simultaneous but in which the rod is at rest.

It is seen directly in the sketch that the cross-section view of the 4-dimensional rod, i.e., the length of the rod, is shorter in the rest S frame (red frame) than it is in the moving blue frame (the blue frame is the one in which the rod is at rest).

The issue is not whether one of the events is at the origin or not, it doesn't matter. You could have done the same thing with your other diagram--one rod, one pair of events.

Of course. That's why I did it that way originally--and because in the original setup he used the Lorentz transformation with the offset displacement rather than a simple boost. I did it this last way because I was getting the impression it might be more clear to you (just using a boost).

That is, if you want to illustrate the problem that was presented by dreamLord in this thread. If you are showing a different problem then you need to say so.

I thought I was giving a physical picture consistent with resolving the problem he was having the way he confused the use of the gamma expression. My sketch goes to his original question, "...where I went wrong?" His analysis was indicating that the rod length in the simultaneous red space was longer than the rod length in the simultaneous blue space--which was backwards. It was hoped that the sketch would at least give him some physical insight and guidance.

 

[Fredrik]

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'.

The statement that S is the "rest system" is misleading, since none of them is at rest in an absolute sense (S' has velocity v in S, and S has velocity -v in S'), and (as your diagram shows) the rod is at rest in S', not S.

Also, the labels $x_1,x_2,x'_1,x'_2$ are a bit odd since each event is assigned coordinates by both coordinate systems. You could e.g. call the events that have S coordinates $x_1$ and $x_2$ 1 and 2 respectively, and the events that have S' coordinates $x'_1$ and $x'_2$ 3 and 4 respectively. Then you can define $L_0=x'_4-x'_3$ and $L=x_2-x_1$.

You may prefer measuring both lengths from the origin (but that's not the way it was presented in the initial post).

I like this better just because it looks less cluttered, but I would still put coordinate-independent labels on the events, since the diagram doesn't favor one coordinate system over the other.

You see how you have labeled the event at the origin with X'₁ and with X₁? Now you should add two more labels to your other two events so they both say X'₂ and X₂. All events on a Minkowski diagram have two labels, one for each coordinate system.

This makes sense, but I would recommend coordinate-independent labels (1,2,3 or A,B,C) instead of one label for each coordinate system ($x_1,x'_1,x_2,x'_2,x_3,x'_3$).

Then you need to delete one of those events and the rod that goes with it, it doesn't matter which one and put in those grid lines and you will see that in whichever frame the events defining the end points of the rod are simultaneous, the distance between the rods is shorter than in the other frame in which the events are not simultaneous but in which the rod is at rest.

He certainly doesn't need to delete an event, and there's just one rod in the diagram (represented by the entire region between the world lines of the end points). What you are referring to are two spacelike curves whose proper lengths are equal to the coordinate lengths of the rod in the two coordinate systems.

You have arbitrarily decided to assign the primed coordinates and L₀ to the longer rod just so that it will appear longer on the diagram than the other rod. But your mistake is that you don't realize that I could just as legitimately assign the primed coordinates and L₀ to the rod that is drawn shorter on the diagram.

This is wrong. The diagram shows that the rod is at rest in S'. (The world lines of both endpoints of the rod are parallel to the t' axis). The line segment labeled L₀ (in the diagram involving four events) is drawn along a simultaneity line of S' and it starts and ends on the world lines of the endpoints of the rod, so it's clear that that line segment, not the other one, represents the rest length of the rod.

That is, if you want to illustrate the problem that was presented by dreamLord in this thread. If you are showing a different problem then you need to say so.

The OP asked about length contraction. bobc2's diagram illustrates length contraction. Hence it's not a different problem.

 

[greswd]

How did this thread manage to get so long?

It's simple. Both ends of the rod can be represented by two parallel lines in either frame.

Find the positions of both ends of the rod in the "moving frame" at ONE point in time in the moving frame.

This means we need to consider:

t1-vx1/c^2 = t2 -vx2/c^2


Simplify to show that the length is contracted by a factor of gamma.

 

[ghwellsjr]

This thread got to be so long because it wasn't until post #15 that dreamLord finally disclosed why he rejected all my other attempts to explain length contraction:

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?

Once he stated why he needed help with that one particular derivation, I gave him an explanation in post #19 and he hasn't posted since, so I can only assume that he was satisfied.

But then bobc2 posted what he called an "immediately obvious" sketch explaining length contraction and in my response to him that it was not "immediately obvious" to me, I asked him:

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?

But rather than answer that question, he continued to explain his sketch, all the while I'm trying to relate it to the OP's concern. Now it is becoming obvious that all these added posts are not attempting to explain the OP's concern but to show other methods for illustrating length contraction, which is OK, I just wish when I ask if they are other methods, people will simply answer yes.

 

[bobc2]

This thread got to be so long because it wasn't until post #15 that dreamLord finally disclosed why he rejected all my other attempts to explain length contraction:

Once he stated why he needed help with that one particular derivation, I gave him an explanation in post #19 and he hasn't posted since, so I can only assume that he was satisfied.

But then bobc2 posted what he called an "immediately obvious" sketch explaining length contraction and in my response to him that it was not "immediately obvious" to me, I asked him:

But rather than answer that question, he continued to explain his sketch, all the while I'm trying to relate it to the OP's concern. Now it is becoming obvious that all these added posts are not attempting to explain the OP's concern but to show other methods for illustrating length contraction, which is OK, I just wish when I ask if they are other methods, people will simply answer yes.

You're right, ghwellsjr. Actually, I wasn't trying to show a different derivation of length contraction (what I presented was not a derivation). I thought I was supplementing your excellent mathematical presentation to show in picture form the results of your analysis. You have always presented some of the most useful explanations of Lorentz transformations when people show up on the forum seeking help, and you certainly did a good job on this thread.

I think I assumed too much in thinking it would be immediately obvious--I was obviously wrong about that and for a while didn't realize the additional confusion I was bringing to the thread. But, I think you've got the situation back in proper perspective. Thanks.

And thanks to Fredrik for pointing out details in my sketches that could have been represented in a more correct and consistent fashion.

 

[ghwellsjr]

How did this thread manage to get so long?

It's simple. Both ends of the rod can be represented by two parallel lines in either frame.

Find the positions of both ends of the rod in the "moving frame" at ONE point in time in the moving frame.

This means we need to consider:

t1-vx1/c^2 = t2 -vx2/c^2

Simplify to show that the length is contracted by a factor of gamma.

Another reason this thread is so long is because after the OP's question is sufficiently answered, someone else comes along claiming to have a "simple" answer but doesn't provide enough explanation to support that claim. I have tried to understand your post and I'm afraid I need help. Here's as far as I have gotten:

You start with an equality, the two sides of your equation came from the Lorentz Transformation equation for time:

t' = γ(t-vx/c^2)

So for the two events under consideration I'm assuming you started with:

t1' = γ(t1-vx1/c^2) and t2' = γ(t2-vx2/c^2)

Then you decided* that t1' equals t2', correct? So then:

γ(t1-vx1/c^2) = γ(t2-vx2/c^2)

You divided out gamma to get your starting equation:

t1-vx1/c^2 = t2-vx2/c^2

Now you say "simplify". Alright, I think you mean to calculate x1-x2 which is the length of the rod moving in the frame S. Here's where I go with that:

t1-t2 = vx1/c^2-vx2/c^2 = (v/c^2)(x1-x2)

(c^2/v)(t1-t2) = x1-x2

Now I would have thought that the next step would be to set t1 equal to t2 but that only means the equality holds true when x1 also equals x2 so that the length of the rod moving in frame S is zero which isn't where we want to go.

So I'm stumped. I need help. I can't figure out your "simple" explanation nor your request to "simplify".

* NOTE: You implicitly set t1' equal to t2'. This is exactly what the OP did in his original post and which I pointed out in post #5 and which leads to the conclusion that t1 cannot also be equal to t2 which is what we need to calculate the length of the rod moving in frame S and to compare it to the length of the rod stationary in frame S'.