I Question about length contraction

[PeterDonis]

I am not trying to be smug and argue. I am trying to understand.

Then do the math. You are confusing yourself because you are trying to reason using vague ordinary language instead of precise math. You need to stop doing that.

This is not the same as relatively of simultaneity which states that my frame for me and your frame for me don't always align.

That's not what relativity of simultaneity says. Do the math. Explicitly write down coordinates in one frame for all events of interest. Then use the Lorentz transformation to obtain coordinates for the same events in the other frame. Then look at the coordinates in the two frames to see how time dilation, length contraction, and relativity of simultaneity actually work.

 

[Nathan123]

Then do the math. You are confusing yourself because you are trying to reason using vague ordinary language instead of precise math. You need to stop doing that.
That's not what relativity of simultaneity says. Do the math. Explicitly write down coordinates in one frame for all events of interest. Then use the Lorentz transformation to obtain coordinates for the same events in the other frame. Then look at the coordinates in the two frames to see how time dilation, length contraction, and relativity of simultaneity actually work.

I don't have the background to do the math. I am a novice trying to understand to the best of my ability. I am using common terms to try and keep it simple.

 

[PeterDonis]

I don't have the background to do the math.

Then you don't have the background to analyze the scenario you're posing correctly. The math is really not that difficult; taking the time to learn it will be a more productive use of your time than trying to reason about this using vague ordinary language.

The short answer to the questions you are posing is that you cannot just look at length contraction or time dilation in isolation. A correct analysis requires taking into account length contraction, time dilation, and relativity of simultaneity. Leaving out anyone of those three will mislead you. And trying to take all three of those things into account correctly without using math is going to be a lot harder than just learning the math so you can use it.

 

[Nugatory]

So it's nice to align things from my perspective and say that regular time/regular length = slower time/shorter length. But this does not work for the guy on the ship. He will have regular time and shorter distance and he will view that I have slower time and regular distance.

You are still failing to consider relativity of simultaneity in your steps 3 and 4. When you correct this mistake you will find that both frames work properly and symmetrically.

Using the Earth frame: When the ship is lined up with the end of the Earth ruler, it is one kilometer away from the Earth and the ship moves through that distance to reach the Earth in time $(1 km)/.6c$ seconds according to "regular time" on earth.

Using the ship frame: When the ship is lined up with the end of the Earth ruler, the Earth is .8 kilometers away from the ship and the Earth moves through that distance to reach the ship in $(.8 km)/.6c$ seconds according to a "regular time" on the ship.

This is not the same as relatively of simultaneity which states that my frame for me and your frame for me don't always align.

It is not the same as relativity of simultaneiity, but relativity of simultaneity is essential to understanding this problem. The best way to do this is to do as @PeterDonis suggests above and do the math (when you do, you will see why I chose $v=.6c$ - that value makes the arithmetic particularly easy, which is how I was able to do the 200 meter and 800 meter calculations above in my head).

But before you do that, I can point you to the reason why the relativity of simultaneity matters: You are starting from the correct statement that the point that is one kilometer from Earth in the Earth frame is 800 meters from Earth in the ship frame and assuming (whether you realize it or not), that at the moment in the Earth frame that the ship and the Earth are one kilometer apart they will be 800 meters apart in the ship frame. They aren't.

 

[Nathan123]

You are still failing to consider relativity of simultaneity in your steps 3 and 4. When you correct this mistake you will find that both frames work properly and symmetrically.

Using the Earth frame: When the ship is lined up with the end of the Earth ruler, it is one kilometer away from the Earth and the ship moves through that distance to reach the Earth in time $(1 km)/.6c$ seconds according to "regular time" on earth.

Using the ship frame: When the ship is lined up with the end of the Earth ruler, the Earth is .8 kilometers away from the ship and the Earth moves through that distance to reach the ship in $(.8 km)/.6c$ seconds according to a "regular time" on the ship.

Something seems wrong here. If it is 1 km at .6c, then .8 km takes less time, not the same amount of time. It will be equivalent to .6c of the Earth's regular time, but it will take less of its own time, which happens to be slower.

In any case, we keep coming back to 1 and 2 when I am trying to understand 3 and 4. I am hearing that relativity of simultaneity is the answer (and I'm afraid it might get overused because it seems to answer everything). But I am not sure exactly where it comes into play here.

So let's try to stick to 3 and 4 and show me where relativity of simultaneity kicks in.

3. Man on ship sees regular time for himself, no time dilation. He also sees a shorter distance to earth.
4. He compares that to the other frame that has slower time and no distance contraction.

So if he gets there before his year is up because of distance contraction, the other person in his estimation will first take at least a year, because there is no distance contraction, and it will take even longer than that in comparison to him because the other person's clock runs slower.

please tell me where relativity of simultaneity kicks in here.

If the answer is just that it's ok that the simultaneity diverges for 3 and 4 even though it does not for 1 and 2, then just say so.

The reason why saying so bothers me is because it not the standard relativity of simultaneity that applies to 1 and 2 as well. Secondly, I often see 1 and 2 so neatly explained, ignoring that 3 and 4 diverge.

 

[Nathan123]

Nathan123, I think the question you are posing is identical to the famous one about the dilemma of the relativistic muons. The link goes to an examination of this question comparing the different frames of reference.

http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/muon.html

I just read that and it gives a completely different option than the 4 I have been mentioning.
It uses the Earth's frame of reference to have no length contraction, but still uses time dilation. Or it uses the muon's frame to have length contraction, but no time dilation. The second one I understand (that would be 3), but the first one does not jive with other things I have read. Because if you completely use the Earth's frame of reference, then you would not have time dilation either (that's number 1). And if you want to have time dilation, then you have to go by the muon's time. So it's kind of contradictory to go by the muon's time and not by the muon's distance. I get the feeling that very few people really understand how this works.

 

[PeterDonis]

I get the feeling that very few people really understand how this works.

No, you don't understand how this works. Relativity physicists understand it just fine. As I've already said, you are confusing yourself by trying to reason about this using ordinary language instead of math.

I am hearing that relativity of simultaneity is the answer (and I'm afraid it might get overused because it seems to answer everything). But I am not sure exactly where it comes into play here.

That's because you aren't doing the math, and, as I've already said, trying to understand how it all fits together without the math is going to be harder than just learning the math.

Here is a quick summary of the math for the Earth and ship scenario.

First, let's look at things in the Earth frame. Let's consider the following events:

A: The ship is 0.6 light-hours from Earth, and is traveling towards Earth at speed $0.6 \text{c}$. We'll call the time of this event time $t = -1$ in the Earth frame (for reasons which will become apparent below), so the coordinates of this event are $(x, t) = (0.6, -1)$ (we're using light-hours and hours as our units of length and time).

E: This is the Earth (more precisely, a chosen point on the Earth, the one the ship will end up arriving at) at the same time, in the Earth frame, as event A. We are treating this point on the Earth as the spatial origin of the frame, so the coordinates of this event are $(0, -1)$.

B: The ship arrives at Earth (at our chosen point). This event has coordinates $(0, 0)$ in the Earth frame (as you can see from the coordinates of event A and the speed of the ship).

The distance of the ship from Earth at event A is the spatial distance between event A and event E, which, since they both have the same time coordinate in this frame, is just $0.6$.

Now, let's look at things in the ship frame. We apply the Lorentz transformation equations, $x' = \gamma \left( x - v t \right)$, $t' = \gamma \left( t - v x \right)$. The relative velocity between the frames is $v = - 0.6$ (because the ship is moving in the $- x$ direction in the Earth frame), so we have $\gamma = 1 / \sqrt{1 - v^2} = 1 / 0.8 = 1.25$. So the coordinates in the ship frame of our three events turn out to be:

A: $(0, - 0.8)$.

E: $(- 0.75, -1.25)$.

B: $(0, 0)$. (Note that, for the Lorentz transformation equations I gave above to work properly, the event that has coordinates $(0, 0)$ in one frame--i.e., the spacetime origin--must also have those coordinates in the other frame.)

Now, observe:

First, in this frame, events A and E do not happen at the same time (whereas they do in the Earth frame). This is relativity of simultaneity.

Second, the time from event A to event B is 0.8 in this frame (whereas it was 1 in the Earth frame). This is time dilation.

But, third, the time from event A to event B is not "the time it takes the ship to reach Earth" in this frame. In this frame, the ship is motionless; the Earth is moving. So $0.8$ hours is the time it takes Earth to reach the ship. And the question is: how far does the Earth travel to reach the ship in this frame?

To figure that out, we need to find the event on Earth's worldline that is simultaneous with event A in the ship frame--i.e., that has time coordinate $t' = - 0.8$. Since the Earth is traveling at speed $0.6$ in this frame (in the positive $x'$ direction), then this event, which I'll call event C, has coordinates:

C: $(x', t') = (- 0.48, - 0.8)$.

So the Earth takes $0.8$ hours to reach the ship in this frame, and travels a distance of $0.48$ light-hours--whereas, in the Earth frame, the ship traveled a distance of $0.6$ light-hours. And that is length contraction: $0.48 = 0.8 * 0.6$.

So, if you do the math, you see how all three things--length contraction, time dilation, and relativity of simultaneity--work together. But leaving out anyone of them leads you astray.

 

[PeterDonis]

I get the feeling that very few people really understand how this works.

Just as a warning: if you continue to take this attitude, this thread will be closed. You say you are not trying to be smug and argue, but trying to understand. But your understanding has to start with the understanding that you are the one who is making a mistake here. Not all of the relativity physicists. Your efforts need to be focused on understanding what mistake you are making, not on trying to convince anyone else that all of the relativity physicists somehow have it wrong or don't really understand what's going on.

 

[PeroK]

I just read that and it gives a completely different option than the 4 I have been mentioning.
It uses the Earth's frame of reference to have no length contraction, but still uses time dilation. Or it uses the muon's frame to have length contraction, but no time dilation. The second one I understand (that would be 3), but the first one does not jive with other things I have read. Because if you completely use the Earth's frame of reference, then you would not have time dilation either (that's number 1). And if you want to have time dilation, then you have to go by the muon's time. So it's kind of contradictory to go by the muon's time and not by the muon's distance. I get the feeling that very few people really understand how this works.

Your study of SR has started with some fundamental misunderstandings. I suggest you back up and start again with a clear and open mind.

Simply restating your mistakes and assuming you are right and those who have mastered the subject are wrong will lead you nowhere.

1) There is no such thing as absolute motion. Motion is frame dependent. No object can be said to be moving any more than any other object. Any object can be given any desired velocity (less than $c$) by appropriate choice of reference frame.

2) Time dilation between inertial reference frames is entirely symmetric. Clocks moving relative to you run slow in your reference frame. But, equally, your clock runs slow in any reference frame where you are not at rest.

3) Length contraction between inertial reference frames is entirely symmetric. An object moving relative to you will be length contracted in the direction of its relative motion. Equally, you are length contracted in any reference frame where you are not at rest.

Unless you accept these three statements and try to understand why they are true, you are on the road to nowhere.

 

[Nathan123]

A: The ship is 0.6 light-hours from Earth, and is traveling towards Earth at speed $0.6 \text{c}$. We'll call the time of this event time $t = -1$ in the Earth frame (for reasons which will become apparent below), so the coordinates of this event are $(x, t) = (0.6, -1)$ (we're using light-hours and hours as our units of length and time).

E: This is the Earth (more precisely, a chosen point on the Earth, the one the ship will end up arriving at) at the same time, in the Earth frame, as event A. We are treating this point on the Earth as the spatial origin of the frame, so the coordinates of this event are $(0, -1)$.

B: The ship arrives at Earth (at our chosen point). This event has coordinates $(0, 0)$ in the Earth frame (as you can see from the coordinates of event A and the speed of the ship).

The distance of the ship from Earth at event A is the spatial distance between event A and event E, which, since they both have the same time coordinate in this frame, is just $0.6$.

Now, let's look at things in the ship frame. We apply the Lorentz transformation equations, $x' = \gamma \left( x - v t \right)$, $t' = \gamma \left( t - v x \right)$. The relative velocity between the frames is $v = - 0.6$ (because the ship is moving in the $- x$ direction in the Earth frame), so we have $\gamma = 1 / \sqrt{1 - v^2} = 1 / 0.8 = 1.25$. So the coordinates in the ship frame of our three events turn out to be:

A: $(0, - 0.8)$.

E: $(- 0.75, -1.25)$.

B: $(0, 0)$. (Note that, for the Lorentz transformation equations I gave above to work properly, the event that has coordinates $(0, 0)$ in one frame--i.e., the spacetime origin--must also have those coordinates in the other frame.)

Now, observe:

First, in this frame, events A and E do not happen at the same time (whereas they do in the Earth frame). This is relativity of simultaneity.

Second, the time from event A to event B is 0.8 in this frame (whereas it was 1 in the Earth frame). This is time dilation.

But, third, the time from event A to event B is not "the time it takes the ship to reach Earth" in this frame. In this frame, the ship is motionless; the Earth is moving. So $0.8$ hours is the time it takes Earth to reach the ship. And the question is: how far does the Earth travel to reach the ship in this frame?

To figure that out, we need to find the event on Earth's worldline that is simultaneous with event A in the ship frame--i.e., that has time coordinate $t' = - 0.8$. Since the Earth is traveling at speed $0.6$ in this frame (in the positive $x'$ direction), then this event, which I'll call event C, has coordinates:

C: $(x', t') = (- 0.48, - 0.8)$.

So the Earth takes $0.8$ hours to reach the ship in this frame, and travels a distance of $0.48$ light-hours--whereas, in the Earth frame, the ship traveled a distance of $0.6$ light-hours. And that is length contraction: $0.48 = 0.8 * 0.6$.

So, if you do the math, you see how all three things--length contraction, time dilation, and relativity of simultaneity--work together. But leaving out anyone of them leads you astray.

I will need to take this one step at a time in order to understand, if you will be gracious to be patient.
I am trying to understand:

1. why is -.08 that starting time for the ship. Why can't we have the two starting at the same time (-1)? What would make it that the ship has a different starting time. I understand that it's time will be dilated from my perspective, but why can't the starting time be the same?
2. What causes the Earth to have a non simultaneous starting time only from the ship's perspective?
3. Why is the length -.75 for the ship's frame of reference? I don't know what that even means.

 

[Ibix]

What would make it that the ship has a different starting time.

Because what "at the same time" means is frame-dependant. So "at the same time" in the sentence "what time do the ship's clocks show at the same time as the Earth's clocks read zero" means different things in the ship frame to the Earth frame.

This is the relativity of simultaneity, which we have been telling you about since post #2. Look up Einstein's train thought experiment if you want to see how it follows trivially from the postulates of relativity.

 

[Nathan123]

Your study of SR has started with some fundamental misunderstandings. I suggest you back up and start again with a clear and open mind.

Simply restating your mistakes and assuming you are right and those who have mastered the subject are wrong will lead you nowhere.

1) There is no such thing as absolute motion. Motion is frame dependent. No object can be said to be moving any more than any other object. Any object can be given any desired velocity (less than $c$) by appropriate choice of reference frame.

2) Time dilation between inertial reference frames is entirely symmetric. Clocks moving relative to you run slow in your reference frame. But, equally, your clock runs slow in any reference frame where you are not at rest.

3) Length contraction between inertial reference frames is entirely symmetric. An object moving relative to you will be length contracted in the direction of its relative motion. Equally, you are length contracted in any reference frame where you are not at rest.

Unless you accept these three statements and try to understand why they are true, you are on the road to nowhere.

None of what I have written about has anything to do with disputing those three things. Everything I wrote assumes those three things. I understand relative motion. I understand relative time dilation. I understand length contraction for things you see moving.
I presented 4 calculations.
1. mine for me.
2. yours for me.
3. yours for you.
4. mine for you.

I said that time dilation is all about 2 and 4 (comparing a different frame to the main one). I also said that distance contraction is all about 2 and 3 (for the ship that sees "everything" moving. [note I specifically refer to distance contraction and not the length contraction of the ship.] This difference between time dilation and distance contraction is causing me issues. I am being told that relativity of simultaneity is the solution. I still do not understand fully how it works, and more so don't yet understand how it works here.

 

[Nathan123]

Because what "at the same time" means is frame-dependant. So "at the same time" in the sentence "what time do the ship's clocks show at the same time as the Earth's clocks read zero" means different things in the ship frame to the Earth frame.

This is the relativity of simultaneity, which we have been telling you about since post #2. Look up Einstein's train thought experiment if you want to see how it follows trivially from the postulates of relativity.

I am trying to understand the underlying factor of "not at the same time". What makes the ship's starting time from its frame of reference different from the ship's starting time from the Earth's frame of reference? Both occupy the same space.

 

[Ibix]

. [note I specifically refer to distance contraction and not the length contraction of the ship.]

There is no difference between distance and length. You keep implicitly using the distance measure from the Earth's frame as if it is somehow the "real" distance. It isn't. The ship frame's distance measure is just as valid, and the Earth's measure of this distance will be length contracted.

 

[Ibix]

I am trying to understand the underlying factor of "not at the same time". What makes the ship's starting time from its frame of reference different from the ship's starting time from the Earth's frame of reference?

It's a direct consequence of the postulates of relativity. I'm guessing from the speed of your response that you ignored my suggestion to look up Einstein's train?

Both occupy the same space.

Both occupy the same spacetime. They are actually using different definitions of which slicing of that counts as space.

 

[PeterDonis]

why is -.08 that starting time for the ship.

Because that's what you get when you Lorentz transform the coordinates of event A, in the Earth frame, into the ship frame.

Why can't we have the two starting at the same time (-1)?

Because that's not how the math works. If you want things to start at time $t' = -1$ in the ship frame, then you are picking a different event--a different point in spacetime--than the one I called event A. You can make either choice, but you can't make both at once, because they're inconsistent: you can only pick one event--one point in spacetime--to be the "starting" point on the ship's worldline.

What would make it that the ship has a different starting time.

Deciding to pick a different event--point--on the ship's worldline as the "starting" point. You can do that, but that will change the "starting time" in both frames. There is no way to pick one event that will have the same "starting time" in both frames. No such event exists. It's mathematically impossible.

What causes the Earth to have a non simultaneous starting time only from the ship's perspective?

You're mis-stating what I said and misinterpreting what the math is telling you. Here is what the math is telling you:

In the Earth frame, event A is simultaneous with event E.

In the ship frame, event A is simultaneous with event C.

Event E and event C are two different events.

Since we have picked event A as the "starting point" for the ship, then "what time" the Earth starts depends on which frame you pick. In the Earth frame, the Earth starts at event E, so it starts at time $t = -1$. In the ship frame, the Earth starts at event C, so it starts at time $t' = - 0.8$. But both of these starting events are simultaneous with the ship's start--event A--in the frame in which we pick them as the starting events. That's what defines what event counts as the "starting event" for Earth--the event on the Earth's worldline that is simultaneous with event A in that frame. So there is never a "non-simultaneous starting time" for Earth.

Why is the length -.75 for the ship's frame of reference?

It isn't. There is no such thing as "the length of a frame of reference". The concept doesn't even make sense.

If what you really mean is to ask, what is the meaning of the spatial coordinate $x' = - 0.75$ for event E, in the ship frame? then here is the answer: it has no special meaning in the ship frame, given that we've picked event A as the "starting event" for the ship. But, we could have picked a different starting event for the ship; for example, we could pick an event that I'll call event D, whose coordinates in the ship frame are:

D: $(x', t') = (0, -1.25)$.

Do you see what this event is? It is the event that is simultaneous, in the ship frame, with event E. In other words, we could, in the ship frame, view things as "starting" at time $t' = -1.25$, instead of $t' = - 0.8$. And then we would find that the Earth travels a distance of $0.75$ light-hours in $1.25$ hours to meet the ship, in the ship frame, i.e., at a speed of $0.6$, as expected. But then, to see how things look in the Earth frame with this choice of a "starting" event, we would have to find the coordinates of event D in the Earth frame. We would do this by the inverse Lorentz transformation, which just reverses the sign of $v$ in the formulas. This gives:

D: $(x, t) = (-0.9375, -1.5625)$.

So in this version of the scenario, we have everything "starting" at time $t = -1.5625$ in the Earth frame, which means that we now have the ship traveling a distance $0.9375$ in $1.5625$ hours (again, for a speed of $0.6$). And we see that the ship frame time of $1.25$ hours is $0.8$ times the Earth frame time (time dilation), and the ship frame distance of $0.75$ light-hours is $0.8$ times the Earth frame distance (length contraction).

 

[Nathan123]

There is no difference between distance and length. You keep implicitly using the distance measure from the Earth's frame as if it is somehow the "real" distance. It isn't. The ship frame's distance measure is just as valid, and the Earth's measure of this distance will be length contracted.

It was nicely taught to me that there is quite a difference between the length contraction of the actual ship and the distance contraction between the ship and earth, even though both use the same exact principle. When you see the ship moving, it's length is contracted. When the ship sees "everything" moving, the distance between it and say Earth is contracted. I have always been discussing distance contraction because it specifically pertains to the issue at hand, while the contraction of the ship does not. So I want to make it clear what I am referring to.

 

[Nathan123]

It's a direct consequence of the postulates of relativity. I'm guessing from the speed of your response that you ignored my suggestion to look up Einstein's train?
Both occupy the same spacetime. They are actually using different definitions of which slicing of that counts as space.

It's late. I'll get to it when I have time.

 

[PeroK]

It was nicely taught to me that there is quite a difference between the length contraction of the actual ship and the distance contraction between the ship and earth, even though both use the same exact principle. When you see the ship moving, it's length is contracted. When the ship sees "everything" moving, the distance between it and say Earth is contracted. I have always been discussing distance contraction because it specifically pertains to the issue at hand, while the contraction of the ship does not. So I want to make it clear what I am referring to.

In general, distance is the spatial distance between two events; and, length is the distance between two simultaneous events. This is as true in classical physics as in SR.

For example, a train may travel a distance of 100km in an hour (in some reference frame). The two events are separated in both time and space (in that reference frame).

Whereas, to measure the length of the train, you must have simultaneous measurements for the front and back.

This is why simultaneity is so important in the measurement of length.

Now, if you say that at time $t$ in your reference frame, two objects are a certain distance apart, then that is (For you) a statement about simultaneous events. And the distance you measure is a length.

But, in another reference frame those events may not be simultaneous, so the spatial distance between those events may not be a length.

This is a critical point regarding length contraction.

Also, if we go back to the train example. In the reference frame of someone on the train, the departure and arrival events are at the same place. The distance between those events in that reference frame is zero and length contraction certainly does not apply.

In short, length contraction applies to the measured spatial distance between simultaneous events.

If we go back to the example and imagine that the ship has a long extension in front of, say, 1 light hour. When that reaches the Earth, in the ship's reference frame the ship and the Earth are 1 light hour apart. Let's say that's when the experiment begins for the ship.

But, in the Earth frame, the extension is length contracted. So, when the extension reaches the Earth, the ship is less than one light hour away.

If at that time , the experiment starts for Those on Earth, then we see that the ship and the Earth have different starting distances.

 

[Nathan123]

Because that's what you get when you Lorentz transform the coordinates of event A, in the Earth frame, into the ship frame.
Because that's not how the math works. If you want things to start at time $t' = -1$ in the ship frame, then you are picking a different event--a different point in spacetime--than the one I called event A. You can make either choice, but you can't make both at once, because they're inconsistent: you can only pick one event--one point in spacetime--to be the "starting" point on the ship's worldline.
Deciding to pick a different event--point--on the ship's worldline as the "starting" point. You can do that, but that will change the "starting time" in both frames. There is no way to pick one event that will have the same "starting time" in both frames. No such event exists. It's mathematically impossible.
You're mis-stating what I said and misinterpreting what the math is telling you. Here is what the math is telling you:

In the Earth frame, event A is simultaneous with event E.

In the ship frame, event A is simultaneous with event C.

Event E and event C are two different events.

Since we have picked event A as the "starting point" for the ship, then "what time" the Earth starts depends on which frame you pick. In the Earth frame, the Earth starts at event E, so it starts at time $t = -1$. In the ship frame, the Earth starts at event C, so it starts at time $t' = - 0.8$. But both of these starting events are simultaneous with the ship's start--event A--in the frame in which we pick them as the starting events. That's what defines what event counts as the "starting event" for Earth--the event on the Earth's worldline that is simultaneous with event A in that frame. So there is never a "non-simultaneous starting time" for Earth.
It isn't. There is no such thing as "the length of a frame of reference". The concept doesn't even make sense.

If what you really mean is to ask, what is the meaning of the spatial coordinate $x' = - 0.75$ for event E, in the ship frame? then here is the answer: it has no special meaning in the ship frame, given that we've picked event A as the "starting event" for the ship.

I still don't understand what exactly is the meaning of the coordinate -.75.

 

[SiennaTheGr8]

Just to explicitly state a fundamental point (not sure if this is the source of confusion here):

Two observers with different rest frames can assign the same spacetime coordinates to ONE event. By convention we choose this event so that it has coordinates (x,t) = (0,0) for both frames (i.e., both frames label it the origin event), because that makes the math easier. Then the observers will disagree on the spacetime coordinates of every other event. For a given frame, the x-coordinate of an event represents the spatial location of that event relative to the origin event (negative value would just mean it's in the direction we define as negative), while the t-coordinate of the event represents the time at which that event occurred relative to the origin event (negative value would just mean that it happened before the origin event).

Maybe that helps? (I'm in a hurry and don't have time to proofread really, so sorry if this was word salad.)

 

[SiennaTheGr8]

(And I was specifically talking about the case of 1 spatial dimension. In the general 3+1D case, the observers agree not only on the spacetime coordinates of the origin event, but also on the coordinates of all events that occur simultaneously with the origin event in the plane that's perpendicular to the frames' axis of relative motion.)

 

[SiennaTheGr8]

(I think.)

 

[A.T.]

Time dilation works differently. It only applies to my comparison of your reference to mine...

I don't see the difference.
Time dilation relates durations measured in different frames.
Length contraction relates lengths measured in different frames.

 

[Nathan123]

Just to explicitly state a fundamental point (not sure if this is the source of confusion here):

Two observers with different rest frames can assign the same spacetime coordinates to ONE event. By convention we choose this event so that it has coordinates (x,t) = (0,0) for both frames (i.e., both frames label it the origin event), because that makes the math easier. Then the observers will disagree on the spacetime coordinates of every other event. For a given frame, the x-coordinate of an event represents the spatial location of that event relative to the origin event (negative value would just mean it's in the direction we define as negative), while the t-coordinate of the event represents the time at which that event occurred relative to the origin event (negative value would just mean that it happened before the origin event).

Maybe that helps? (I'm in a hurry and don't have time to proofread really, so sorry if this was word salad.)

Thanks for explaining the coordinates.

 

[PeterDonis]

I still don't understand what exactly is the meaning of the coordinate -.75.

Then my advice is to ignore most of what I posted about it, and just keep this in mind:

If what you really mean is to ask, what is the meaning of the spatial coordinate $x' = - 0.75$ for event E, in the ship frame? then here is the answer: it has no special meaning in the ship frame, given that we've picked event A as the "starting event" for the ship.

In other words, in the scenario as you posed it and as you are thinking about it, the spatial coordinate $x' = - 0.75$ of event E in the ship frame is just a distraction; it doesn't mean anything useful, and you should just ignore it.

 

[Nathan123]

I don't see the difference.
Time dilation relates durations measured in different frames.
Length contraction relates lengths measured in different frames.

Going back first to math of PeterDonis, I still think that it explains 1 and 2 and not 3 and 4.

I will try to explain here how I see length contraction working very differently than time dillation, which is the basis of this whole thread.

The only reason there is time dilation is because we are comparing one frame of reference to another. If A sees B moving and B sees B not moving, A will say that Bs time slows down.
So
1. Only A = no time dilation
2. A comparing B sees time dilation for B
3. Only B = no time dilation
4. B comparing A sees time dilation for ALength contraction from everything I have read and seen works differently.

Here anything you see moving, you will see it contracted.

So if you see a ship moving, only the ship gets contracted. If you are in a car and everything is moving towards you then everything gets contracted.

So
1. A (on earth) has no distance contraction between ship and earth.
2. A comparing B will say that B has distance contraction between ship and Earth because B sees everything moving.
3. B (on ship) will have distance contraction between ship and Earth because it sees everything moving.
4. B comparing A will not have distance contraction for A because A doesn't see eveything moving.
So time dilation is 2 and 4 and length contraction is 2 and 3.
And this is causing me issues that the ship reaching Earth coinside for 1 and 2 but diverge for 3 and 4.

 

[SiennaTheGr8]

I know you're trying to be clear, @Nathan123, but it's difficult to follow what you're writing.

In any case, the distinction between "distance" and "length" is an important one in this context. I suggest carefully reading and re-reading PeroK's post above.

Here's a relevant video (note what he says about length contraction and time dilation not really being two sides of the same coin):

 

[A.T.]

The only reason there is time dilation is because we are comparing one frame of reference to another.

Correct.

Length contraction from everything I have read and seen works differently.

It works exactly analogous. You are comparing lengths measured in two different frames.

Here anything you see moving, you will see it contracted. So if you see a ship moving, only the ship gets contracted.

Contracted compared to its proper length measured in its rest frame. So you are again relating measurements in two different frames (its rest frame, and the frame where it moves).

 

[PeterDonis]

I will try to explain here how I see length contraction working very differently than time dillation

You are failing to take heed of what I said before:

your understanding has to start with the understanding that you are the one who is making a mistake here.

You should not be trying to explain how you see it. You should be discarding how you see it, and trying to learn how everyone else sees it, since your way of seeing it is wrong and ours is right.

Length contraction from everything I have read and seen works differently.

No, it doesn't. Take this:

So
1. Only A = no time dilation
2. A comparing B sees time dilation for B
3. Only B = no time dilation
4. B comparing A sees time dilation for A

And substitute "length contraction" for "time dilation" and you will have four statements about length contraction that are just as valid as the four statements you make here about time dilation. They work the same.

Here anything you see moving, you will see it contracted.

Substitute "time dilated" for "length contracted" here and you will have a true statement about time dilation. They work the same.

if you see a ship moving, only the ship gets contracted

Because only the ship is moving; you have specified, implicitly, that everything else is at rest relative to you. So, again, substitute "time dilated" for "length contracted", and you will have a true statement about time dilation: if the ship is the only thing moving, it is the only thing that gets time dilated, relative to you.

If you are in a car and everything is moving towards you then everything gets contracted.

Why switch from ship to car here? If you are in the ship--that, above, you said was the only thing moving relative to you before--then you have switched yourself to the ship frame, in which, now, the ship is the only thing that is not moving. So of course everything else will get length contracted relative to you. But, again, substitute "time dilated" for "length contracted" here and you will have a true statement about time dilation: everything else will get time dilated relative to you. The two work the same.

1. A (on earth) has no distance contraction between ship and earth.

You are confusing yourself here because you are thinking of "distance" as a single object, and asking yourself whether that object is "length contracted" or not. That's not the right way to look at it. The "distance between the ship and earth" depends on which event you pick as the "starting point" for the ship, and which frame you use. In other words, the "distance" in different frames is a distance between different pairs of events.

In terms of my post #37, in the Earth frame, the distance between the ship and Earth (when the ship starts) is the distance between events A and E. But in the ship frame, the distance between the ship and Earth (when the Earth starts, since in this frame the Earth is what moves) is the distance between events A and C. So there is no one "object" that corresponds to "the distance between the ship and Earth". That ordinary language term refers to different objects--different curves in spacetime--in different frames. This is a key reason why trying to reason about all this in ordinary language is not a good idea.

2. A comparing B will say that B has distance contraction between ship and Earth because B sees everything moving

3. B (on ship) will have distance contraction between ship and Earth because it sees everything moving.

4. B comparing A will not have distance contraction for A because A doesn't see eveything moving.

Your language is very muddled here, and you are confusing yourself with this muddled language. Read my paragraph just above for a better way to look at it.

So time dilation is 2 and 4 and length contraction is 2 and 3.

I have no idea what you mean by this. None of your 1., 2., 3., 4. mention time dilation at all. You are confusing yourself with muddled language.

this is causing me issues that the ship reaching Earth coinside for 1 and 2 but diverge for 3 and 4.

Your analysis is so confused at this point that I can't figure out how you are drawing this conclusion. But however you are doing it, it's wrong.