Length Contraction rearrangement

[darkhorror]

mananvpanchal in this thread like in other threads you are making assumptions that you know how things should work, yet they don't work the way you think they do. Go back and look at what people posted with an open mind. I feel you misunderstand much of what people say due to you trying to convert what they say to what you think they say.

 

[darkhorror]

If someone tell us to transform to another frame which is moving relative to us, then we use t_0 as time component of the events, not t_1 and t_2. So, we can say that we have one time component for all space component of the rod if the rod is steady in our frame.

Why do those two clocks being in sync in one frame of reference mean that they need to be in sync in another?

 

[ghwellsjr]

Why do those two clocks being in sync in one frame of reference mean that they need to be in sync in another?

It's very important to realize that the two clocks that are in sync at the two ends of the rod in the rest frame of the rod are coordinate clocks for that frame. In another frame moving with respect to the first rest frame, we don't pay attention to those two clocks anymore. Rather, we have a whole new set of coordinate clocks that have been synchronized together with each other to define the new frame. Now, there are no longer just two new coordinate clocks at the two ends of the rod. Instead, we have a whole bunch of new clocks that are moving with respect to the rod, or we could say the rod is moving with respect to the new set of coordinate clocks. That's why, if we looked at the two coordinate clocks in the new frame when the coordinate clocks in the original frame had the same time on them, the new clocks would have different times on them and their spacing apart is different and we end up with a meaningless result. Instead, we have to look at a new coordinate clock located at one end of the rod and wait until the other end of the rod moves past a new clock with the same time on it and then see how far apart those two new clocks are which tells us the contracted length of the rod in the new frame.

 

[mananvpanchal]

Ok, I accept it that we have to pick two different time components in rest frame to find same time component in moving frame, and with this we can get now length contraction.

Now, please take a look in below image.

This is same as displayed in the link http://en.wikipedia.org/wiki/Length_contraction#Geometrical_representation.

We can say that OA' is length between two events at rest in S'. When we transform O and A', we get O and A in S frame. We can easily say that OA' > OA. So for moving frame S, the length is contracted.

Now, take two events happened at rest in S' frame at O and B' time. When we transform the two events, we get O and B in S frame. We can again easily say that OB' > OB. So, we have to say that time elapsed between two events in S' frame is more than S frame. So time is running faster in moving frame than rest frame.

 

[mananvpanchal]

mananvpanchal in this thread like in other threads you are making assumptions that you know how things should work, yet they don't work the way you think they do. Go back and look at what people posted with an open mind. I feel you misunderstand much of what people say due to you trying to convert what they say to what you think they say.

Can you explain me why I am wrong?

 

[Dale]

Hi mananvpanchal, I have glanced through this thread and noticed something fairly consistent, from your OP to your post 34. You continue to talk about transformation of events. Time dilation can be understood in terms of the transformation of 2 events, but not length contraction. Length contraction should be understood in terms of the transformation of 2 worldlines, specifically the worldlines of the ends of some object. The reason is that you need to measure the distance between the endpoints simultaneously in both frames. That is not possible with a pair of events, but is possible with a pair of worldlines.

 

[Chestermiller]

Can you explain me why I am wrong?

Your problem is not so much with your algebraic application of the Lorentz Transformation. You're actually pretty good at that. Your problem is with confusion over how to use space-time diagrams to reconcile the results. Space-time diagrams are often very confusing at the beginning, and have caused untold frustration for people over the years. My advice to you is to, at least temporarily, stop using space-time diagrams, and study the various aspects of the problems you are working on exclusively using the Lorentz Transformation. It will never lead you astray.

Chet

 

[Chestermiller]

I have a problem for you to solve using the Lorentz Transformation. Consider two inertial frames of reference S and S', with S' moving to the right relative to S at a speed of v. There is an object at rest in the S' frame of reference, with its left end situated at x'=-Δx', and its right end situated at x'=+Δx'. The origin of the S' coordinate grid x'=0 coincides with the origin of the S coordinate grid x=0 when t'=t=0. At t=0 throughout the S frame of reference, what are the left end and right end coordinates of the object as reckoned using the coordinate grid in the S reference frame (i.e., the x coordinates of the left- and right end of the object)? If there were clocks present within the S' frame of reference at the two ends of the object, what values of t' would observers in the S frame of reference see displayed on these clocks when their own clocks all display t = 0? (Of course, express the results algebraically). What is your physical interpretation of these readings on the S' clocks?

 

[ghwellsjr]

Ok, I accept it that we have to pick two different time components in rest frame to find same time component in moving frame, and with this we can get now length contraction.

Now, please take a look in below image.

View attachment 45190

This is same as displayed in the link http://en.wikipedia.org/wiki/Length_contraction#Geometrical_representation.

But they're not the same. Your diagram has the S system aligned with the horizontal and vertical axes of the graph while the wiki diagram rotates the S system away from these axes so that it presents a symmetrical viewpoint for each frame:

Thus, the wiki text points out that the Proper Length OB at rest in S' is contracted to OA in S. Similarly, you point out in your diagram:

We can say that OA' is length between two events at rest in S'. When we transform O and A', we get O and A in S frame. We can easily say that OA' > OA. So for moving frame S, the length is contracted.

However, the wiki text goes on to say the symmetrical situation also applies--the Proper Length OD at rest in S is contracted to OC in S'. If you do a similar process on your diagram, you will show that the "contracted" length looks longer. This is because you haven't provide markings on your graph to show the numerical values for the coordinates. The wiki diagram gets away with not having to do this because the two frames are symmetrical and the coordinates have equal values.

Now, take two events happened at rest in S' frame at O and B' time. When we transform the two events, we get O and B in S frame. We can again easily say that OB' > OB. So, we have to say that time elapsed between two events in S' frame is more than S frame. So time is running faster in moving frame than rest frame.

Time on a spacetime diagram is like the time shown on a clock--the shorter the length on the diagram, the less time is on the clock. So when you say that OB' > OB, you're saying that a clock at rest in a frame accumulates more time than one moving in that frame, correct?

 

[Chestermiller]

GHWELLSJR's reply is very pertinent and helpful. I would like to add that the diagram in wiki is not a Minkowski diagram, but rather a Brehme diagram. Brehme diagrams are sometimes much easier to work with than Minkowski diagrams (although they are not without certain limitations). As you can see from ghellsjr's reply, in a Brehme diagram, the ct axis is perpendicular to the x' axis, and the ct' axis is perpendicular to the x axis. This is not the case in a Minkowski diagram. But, this allows you to use identical grid spacings along all the axes, and makes it much easier to compare distances visually.

The Minkowski metric equation reads:

x²-(ct)²=(x')²-(ct')²

The Brehme diagram is based on a rearranged form of this equation reading:

x²+(ct')²=(x')²+(ct)²

This "trick" remaps the Lorentzian coordinates into pseudo Eucidean coordinates. This only works in 2D, and with only two frames of reference.

 

[mananvpanchal]

Hi mananvpanchal, I have glanced through this thread and noticed something fairly consistent, from your OP to your post 34. You continue to talk about transformation of events. Time dilation can be understood in terms of the transformation of 2 events, but not length contraction. Length contraction should be understood in terms of the transformation of 2 worldlines, specifically the worldlines of the ends of some object. The reason is that you need to measure the distance between the endpoints simultaneously in both frames. That is not possible with a pair of events, but is possible with a pair of worldlines.

Can we not say t' and t'1 is world lines of the two points O and A in S' frame? And please explain me how to measure distance between two world lines.

 

[mananvpanchal]

I have a problem for you to solve using the Lorentz Transformation. Consider two inertial frames of reference S and S', with S' moving to the right relative to S at a speed of v. There is an object at rest in the S' frame of reference, with its left end situated at x'=-Δx', and its right end situated at x'=+Δx'. The origin of the S' coordinate grid x'=0 coincides with the origin of the S coordinate grid x=0 when t'=t=0. At t=0 throughout the S frame of reference, what are the left end and right end coordinates of the object as reckoned using the coordinate grid in the S reference frame (i.e., the x coordinates of the left- and right end of the object)? If there were clocks present within the S' frame of reference at the two ends of the object, what values of t' would observers in the S frame of reference see displayed on these clocks when their own clocks all display t = 0? (Of course, express the results algebraically). What is your physical interpretation of these readings on the S' clocks?

If I transform x'=-Δx' and x'=+Δx'. I get x=-Δx and x=Δx. If I compare difference between the two events in different frame I get 2Δx < 2Δx'. At t=0 in S frame if I want to read S' frame's clocks. I get t'=-Δt' and t'=+Δt'. If I get difference between the clocks reading I get 2Δt'. But what is your point with this?

 

[mananvpanchal]

But they're not the same. Your diagram has the S system aligned with the horizontal and vertical axes of the graph while the wiki diagram rotates the S system away from these axes so that it presents a symmetrical viewpoint for each frame:

I know you know this already yet I have to tell you this.
The both diagram is same. In my diagram whole situation looked from S frame where in your diagram the both frame shown moving relative to each other from a point which is guessed at rest.

Time on a spacetime diagram is like the time shown on a clock--the shorter the length on the diagram, the less time is on the clock. So when you say that OB' > OB, you're saying that a clock at rest in a frame accumulates more time than one moving in that frame, correct?

OB' > OB. OB' is time elapsed in S' frame between two events at rest in S' frame. OB is time elapsed in S frame between two events at rest in S' frame.
Suppose, OB=1 and OB'=1.25, so there is more time elapsed in S' frame rather than S frame.
Does this not seems that in S frame time running faster than S' frame between two events which is at rest in S' frame?

 

[ghwellsjr]

But they're not the same. Your diagram has the S system aligned with the horizontal and vertical axes of the graph while the wiki diagram rotates the S system away from these axes so that it presents a symmetrical viewpoint for each frame:

I know you know this already yet I have to tell you this.

You don't know what I know. I know that you don't understand Minkowski diagrams at all and it is very difficult to help you when you won't admit your ignorance in this area. For example, you are making a big mistake when think that transforming event A' gives you event A or B' give you B. You really should read the wikipedia article on them. But I would suggest that you forget about Minkowski diagrams and just use the mathematical Lorentz Transformation as Chet suggested in post #37.

The both diagram is same. In my diagram whole situation looked from S frame where in your diagram the both frame shown moving relative to each other from a point which is guessed at rest.

Since you are so sure that they are the same, then I challenge you to extend your drawing in the same way the wiki drawing is extended to show a rod at rest in the S frame and how it is contracted when moving in the S' frame.

Time on a spacetime diagram is like the time shown on a clock--the shorter the length on the diagram, the less time is on the clock. So when you say that OB' > OB, you're saying that a clock at rest in a frame accumulates more time than one moving in that frame, correct?

OB' > OB. OB' is time elapsed in S' frame between two events at rest in S' frame. OB is time elapsed in S frame between two events at rest in S' frame.
Suppose, OB=1 and OB'=1.25, so there is more time elapsed in S' frame rather than S frame.
Does this not seems that in S frame time running faster than S' frame between two events which is at rest in S' frame?

A clock that has accumulated 1 unit of time on it in the same time that another clock has accumulated 1.25 units of time is running slower.

But before you spend any more time on this last issue, please respond to my challenge. Or if you prefer, drop the Minkowski diagrams and use numbers with the Lorentz Transformation.

 

[mananvpanchal]

Since you are so sure that they are the same, then I challenge you to extend your drawing in the same way the wiki drawing is extended to show a rod at rest in the S frame and how it is contracted when moving in the S' frame.

The rod is at rest in S' frame not in S frame in both diagram. So, please, re-explain me your challenge.

A clock that has accumulated 1 unit of time on it in the same time that another clock has accumulated 1.25 units of time is running slower.

But before you spend any more time on this last issue, please respond to my challenge. Or if you prefer, drop the Minkowski diagrams and use numbers with the Lorentz Transformation.

Yes, we will take a look on this later.

 

[ghwellsjr]

Your diagram has a rod at rest in the S' frame just like the diagram in the wiki article. But the wiki article has a second rod at rest in the S frame:

The Lorentz transformation geometrically corresponds to a rotation in four-dimensional spacetime, and it can be illustrated by a Minkowski diagram: If a rod at rest in S' is given, then its endpoints are located upon the ct' axis and the axis parallel to it. In this frame the simultaneous (parallel to the axis of x') positions of the endpoints are O and B, thus the proper length is given by OB. But in S the simultaneous (parallel to the axis of x) positions are O and A, thus the contracted length is given by OA. On the other hand, if another rod is at rest in S, then its endpoints are located upon the ct axis and the axis parallel to it. In this frame the simultaneous (parallel to the axis of x) positions of the endpoints are O and D, thus the proper length is given by OD. But in S' the simultaneous (parallel to the axis of x') positions are O and C, thus the contracted length is given by OC.

I'm asking you to show on your same diagram, a second rod at rest in your S frame and then use an identical process that you used for your first rod to show that this second rod is contracted in your S' frame.

 

[mananvpanchal]

Your diagram has a rod at rest in the S' frame just like the diagram in the wiki article. But the wiki article has a second rod at rest in the S frame:

I'm asking you to show on your same diagram, a second rod at rest in your S frame and then use an identical process that you used for your first rod to show that this second rod is contracted in your S' frame.

Yes, I am agree on length contraction process for frames S and S'. I have created doubt on getting time dilation by the same process.

Now, take two events happened at rest in S' frame at O and B' time. When we transform the two events, we get O and B in S frame. We can again easily say that OB' > OB. So, we have to say that time elapsed between two events in S' frame is more than S frame. So time is running faster in moving frame than rest frame.

 

[mananvpanchal]

You don't know what I know. I know that you don't understand Minkowski diagrams at all and it is very difficult to help you when you won't admit your ignorance in this area. For example, you are making a big mistake when think that transforming event A' gives you event A or B' give you B. You really should read the wikipedia article on them. But I would suggest that you forget about Minkowski diagrams and just use the mathematical Lorentz Transformation as Chet suggested in post #37.

Please, check below image.

There are x and t axis. Difference between each point is 1. Co-ordinates defined as (x, t). We assume another frame moving at speed 0.6c relative to the frame and we take c=1.
So, if I transform (0, 0), (1, 0), (2, 0), (3, 0), (-1, 0), (-2, 0), (-3, 0), (0, 1), (0, 2), (0, 3), (0, -1), (0, -2), (0, -3) co-ordinates using x&#039;=1.25 (x-0.6t) and t&#039;=1.25 (t-0.6x). I get (0, 0), (1.25, -0.75), (2.5, -1.5), (3.75, -2.25), (-1.25, 0.75), (-2.5, 1.5), (-3.75, 2.25), (-0.75, 1.25), (-1.5, 2.5), (-2.25 3.75), (0.75, -1.25), (1.5, -2.5), (2.25, -3.75) respectively. We can draw x&#039; and t&#039; from this points.

If I re-transform the (0, 0), (1.25, -0.75), (2.5, -1.5), (3.75, -2.25), (-1.25, 0.75), (-2.5, 1.5), (-3.75, 2.25), (-0.75, 1.25), (-1.5, 2.5), (-2.25 3.75), (0.75, -1.25), (1.5, -2.5), (2.25, -3.75) points using x&#039;=1.25 (x+0.6t) and t&#039;=1.25 (t+0.6x). I get (0, 0), (1, 0), (2, 0), (3, 0), (-1, 0), (-2, 0), (-3, 0), (0, 1), (0, 2), (0, 3), (0, -1), (0, -2), (0, -3) respectively. So, how can you say that I cannot get B from B' and A from A'?

 

[Doc Al]

Now, take two events happened at rest in S' frame at O and B' time. When we transform the two events, we get O and B in S frame. We can again easily say that OB' > OB. So, we have to say that time elapsed between two events in S' frame is more than S frame. So time is running faster in moving frame than rest frame.

Leaving aside the issue of how you are interpreting your diagrams, do you think that just because the time interval between two events is greater in one frame than another that that shows that 'time' is running faster in the moving frame?

Once again, what you want is for events in S' to take place at the same position. Then you'll be able to demonstrate the usual time dilation formula.

 

[ghwellsjr]

Yes, I am agree on length contraction process for frames S and S'. I have created doubt on getting time dilation by the same process.

You're never going to understand either length contraction or time dilation using Minkowski diagrams until you realize that you are not doing them correctly. If you will attempt to do what I have asked, it may become obvious to you that you don't understand them. Chet also gave you a challenge which you also ignored. We're trying to help you but if you refuse to cooperate, no progress will be made.

 

[mananvpanchal]

Leaving aside the issue of how you are interpreting your diagrams, do you think that just because the time interval between two events is greater in one frame than another that that shows that 'time' is running faster in the moving frame?

Once again, what you want is for events in S' to take place at the same position. Then you'll be able to demonstrate the usual time dilation formula.

Suppose, in S frame there are O is origin, A is on t axis and B is on x axis. Now If I transform the two points I get A' and B' in S' frame. Now OA < OA' and OB < OB'. Now, please you guys interpret your way the time dilation and length contraction. I don't know how to interpret this. I am eagerly waiting to hear the interpretation from starting of the thread. I know this is the place where I am stuck. Please, interpret it for me. Thanks.

 

[mananvpanchal]

You're never going to understand either length contraction or time dilation using Minkowski diagrams until you realize that you are not doing them correctly. If you will attempt to do what I have asked, it may become obvious to you that you don't understand them. Chet also gave you a challenge which you also ignored. We're trying to help you but if you refuse to cooperate, no progress will be made.

Your challenge is about to length contraction and I have already told that I am agree on this. So I am agree then what do I do after taking the challenge. Your challenge should be about to time dilation. And I cannot get fully chat's challenge so I have requested them to give me detail already. Where am I not co-operating?

 

[Doc Al]

Suppose, in S frame there are O is origin, A is on t axis and B is on x axis.

You've picked two events A and B with x,t coordinates A(0,t) and B(x,0). Is that what you really want?

Now If I transform the two points I get A' and B' in S' frame.

When you transform to the S' frame you get the same events A and B just with different coordinates. But in your diagrams you are creating new and different events A' and B'. Don't do that!

Now OA < OA' and OB < OB'.

So what?

The events OA can represent a time measurement made on a clock in S, so of course Δt' > Δt. That's time dilation. Moving clocks run slow.

The events OB are simultaneous in S, so of course Δx' > Δx. That's length contraction. (Note that the events are not simultaneous in S', which I think is where your problem lies.)

Where's the problem?

My suggestion: Forget about diagrams for the moment. Just stick to the Lorentz transformations, which already includes length contraction and time dilation. Find the coordinates of A and B in S'.

 

[mananvpanchal]

You've picked two events A and B with x,t coordinates A(0,t) and B(x,0). Is that what you really want?

Yes.

When you transform to the S' frame you get the same events A and B just with different coordinates. But in your diagrams you are creating new and different events A' and B'. Don't do that!

I am not creating new events. It is the same events but in S' frame with new values. And I defined the new transformed co-ordinates with A' and B'.

The events OA can represent a time measurement made on a clock in S, so of course Δt' > Δt. That's time dilation. Moving clocks run slow.

Ok.

The events OB are simultaneous in S, so of course Δx' > Δx. That's length contraction. (Note that the events are not simultaneous in S', which I think is where your problem lies.)

Yes, this is the place where my problem lies. With your above explanation of time dilation I understand it. Now, I am trying to understand length contraction by taking same time component, but if I accept it then I cannot understand time dilation now!

Where's the problem?

The problem is I want to understand length contraction and time dilation from Minkowski's diagram or Lorentz transformation. But using both I cannot understand it. If I try to understand time dilation I cannot get length contraction and if I try to understand length contraction I cannot get time dilation.

My suggestion: Forget about diagrams for the moment. Just stick to the Lorentz transformations, which already includes length contraction and time dilation. Find the coordinates of A and B in S'.

Ok, if I find the co-ordinates of A and B as A' and B' in S'. What do I do with that?

In S frame if we want time dilation we start transformation from S frame, but if we want length contraction we have to start transformation from S' frame. Why? This is the thing I cannot accept it. Our transformation process should be clear. We cannot have double standard.

 

[Chestermiller]

This is in response to your reply #42, which was in response to my reply #38.

I have two points for you to consider. My first point is that, aside from not actually having substituted the input information into the Lorentz Transformation and showing the quantitative algebraic results (as I was hoping you would do), overall you seem to have analyzed the problem correctly (at least qualitatively)... even without the use of a Minkowski diagram.
My second point is with regard to the physical interpretation of the different clock readings at the ends of the object in the S' frame of reference, when viewed at time t=0 from the S frame of reference. You opted to offer no physical interpretation. Here's a possible interpretation. Observers in the S frame of reference at time t = 0 are seeing different parts of the object at different times in its existence in the S' frame of reference (its rest frame). Relative to the S' observer located at x'=0, observers in the S frame of reference always see one end of the object in his future, and the other end of the object in his past.

 

[Dale]

And please explain me how to measure distance between two world lines.

You pick two simultaneous events on the world lines and measure the distance between them.

 

[mananvpanchal]

This is in response to your reply #42, which was in response to my reply #38.

I have two points for you to consider. My first point is that, aside from not actually having substituted the input information into the Lorentz Transformation and showing the quantitative algebraic results (as I was hoping you would do), overall you seem to have analyzed the problem correctly (at least qualitatively)... even without the use of a Minkowski diagram.
My second point is with regard to the physical interpretation of the different clock readings at the ends of the object in the S' frame of reference, when viewed at time t=0 from the S frame of reference. You opted to offer no physical interpretation. Here's a possible interpretation. Observers in the S frame of reference at time t = 0 are seeing different parts of the object at different times in its existence in the S' frame of reference (its rest frame). Relative to the S' observer located at x'=0, observers in the S frame of reference always see one end of the object in his future, and the other end of the object in his past.

Yes, I understand that. And I also understand that if we try to measure length between two points we get more length which we don't want. So we don't want two see both end at different time. So we see both point at same time, and using the process of taking parallel line to t&#039; we can get length contraction. My problem is I don't get time dilation using the same process. And I want to understand now time dilation using any process you explain me with diagram or transformation.

 

[Chestermiller]

Time Dilation:

You have one guy in the S' frame of reference sweeping past multiple guys in the S frame of reference. As he passes each guy in the S reference frame, he writes down the time on his clock and also the time displayed on the clock of the guy he is currently passing. At some point he stops collecting data, and plots a graph of the time on his clock as the horizontal axis, and the time on the S peoples' clocks as the vertical axis. He then measures the slope, and finds that it is >1. Their clocks seem to be running faster then his. This is time dilation.

The people in the S frame of reference also take some data of their own. They follow the motion of one particular guy in the S' frame of reference as he sweeps past them, and, when he goes by each of them, they record the time on their own clock and also the time displayed on his clock. At some point, they stop taking data, and they assemble in a conference room where they plot a graph of the time on the moving guy's clock as the horizontal axis, and the times on their own clocks when he passed each of them. The graph that the S reference frame guys prepare will be virtually the same as the graph that the guy in the S' frame prepared (assuming the same guy in the S' frame was used in both experiments). In particular, the slope of the line will be >1. Both teams agree that time dilation has occurred, and they agree quantitatively.

Note that, in this example, there is only 1 guy in the S' frame, and multiple guys in the S frame. This is key.

 

[darkhorror]

How about we try and make sure we are on the same page here when something is said.

Lets take frame A, with two clocks x and y, both at rest and synchronized in frame A. These two clocks are 5 light seconds apart in frame A. You also have clock z, which is moving at .5c in frame A. z is moving along so that when z hits clock x both clock x and clock z's time reads 0, and then when clock z hits clock y, clock y reads 10 seconds and clock z reads 8.66 seconds.
Does that make sense?

Next looking from the frame of reference where z is stationary.

 

[Chestermiller]

How about we try and make sure we are on the same page here when something is said.

Lets take frame A, with two clocks x and y, both at rest and synchronized in frame A. These two clocks are 5 light seconds apart in frame A. You also have clock z, which is moving at .5c in frame A. z is moving along so that when z hits clock x both clock x and clock z's time reads 0, and then when clock z hits clock y, clock y reads 10 seconds and clock z reads 8.66 seconds.
Does that make sense?

Next looking from the frame of reference where z is stationary.

Yes. This example makes perfect sense to me. The only change I would make in your wording would be to change the words "which is moving at .5c in frame A" to "which is moving at .5c with respect to frame A."