I What would be the form of this world line?

[Dale]

I thought those were rhetorical questions. I see you asked me, "If I physically draw a line on a piece of paper, do you need coordinates to determine if it is straight?" My answer to this is no, if we are assuming a Euclidean space.

Ok, this is good. So, how would you determine, physically, without coordinates, if the line on the piece of paper is straight (assuming a Euclidean space)?

 

[PeterDonis]

I suppose it is sufficient to define what we intend to illustrate by using a graph with the conjecture of what kind of space it's in.

No. No. No. No. No.

Banish the word "graph" from your mind. Get rid of it. Forget it. Abolish it. Eradicate it.

I emphasize this because, in spite of the fact that you have been told many times now to stop thinking in terms of coordinates, you keep thinking in terms of coordinates. "Graph" means "coordinates". You need to stop doing that or you will never get anywhere. And this thread will be closed because there is no point in wasting our time responding to you if you are not going to listen to the responses.

 

[student34]

No. How can the traveler change the ball's trajectory (which is what would have to happen for the ball's worldline to change) by moving himself?

This, right here, is the heart of the issue. This is what length contraction seems to be saying Peter. The relative distance between the traveler and the ball changes. We know that it is in fact the ball that shifts toward the traveler on the x-axis in the diagram - how does this happen - we shouldn't be able to explain it away by distorting a graph that fits the situation. How does this increase in speed bring the ball closer to the traveler if not by accelerating it (even though an accelerometer would probably not read anything)?

 

[jbriggs444]

Ok, this is good. So, how would you determine, physically, without coordinates, if the line on the piece of paper is straight (assuming a Euclidean space)?

If it were me, I'd see if I had a pencil with which to make marks on the line and a piece of thread to stretch between the marks.

Then I'd notice something about the triangle inequality.

 

[student34]

I wanted to address this comment. I have a table, and I have two graphs of the table. Here is the graph using Cartesian coordinates:

View attachment 295335
Here is the graph using spherical coordinates:
View attachment 295336

It is not generally true that if you superimpose a graph onto its corresponding real situation, then it would align with what is actually happening.

The fact that the table legs and table top are all curved in the second graph in no way means that they are physically curved. The physical table is flat and the physical legs are straight, regardless of the fact that the second graph makes it hard to see that fact.

This is really interesting. Thanks for the visual. I will keep this in mind.

 

[PeterDonis]

This is what length contraction seems to be saying

No, it isn't. See below.

The relative distance between the traveler and the ball changes.

The calculated distance to the ball in the traveler's rest frame changes. But there is no direct observable that corresponds with this change. It's a calculated number that depends on your choice of coordinates.

How does this increase in speed bring the ball closer to the traveler

The traveler's change in speed does nothing to the ball. The traveler's decision to choose a new frame of reference when he changes speed changes the distance he calculates to the ball, but that has nothing to do with the ball.

This is what comes of continuing to focus on coordinates despite repeated advice not to. All you're doing is confusing yourself further.

 

[PeterDonis]

We know that it is in fact the ball that shifts toward the traveler on the x-axis in the diagram - how does this happen - we shouldn't be able to explain it away by distorting a graph that fits the situation.

We're not the ones who are distorting things. You are, by continuing to harp on coordinates after being told repeatedly not to do that. There is nothing here to "explain away", because nothing happens to the ball when the traveler changes speed. All that happens is that the traveler calculates a new number for what he calls "distance to the ball" in his new reference frame.

 

[PeterDonis]

This, right here, is the heart of the issue.

No, it isn't. It's a distraction that has nothing whatever to do with how you determine whether a worldline is straight or curved without using coordinates, which is the heart of the issue.

 

[student34]

Ok, this is good. So, how would you determine, physically, without coordinates, if the line on the piece of paper is straight (assuming a Euclidean space)?

I am really not sure.

 

[PeterDonis]

I am really not sure.

Then how were you able to answer "no" to @Dale's question?

 

[student34]

We're not the ones who are distorting things. You are, by continuing to harp on coordinates after being told repeatedly not to do that. There is nothing here to "explain away", because nothing happens to the ball when the traveler changes speed. All that happens is that the traveler calculates a new number for what he calls "distance to the ball" in his new reference frame.

If a ball 5 meters from you suddenly starts moving to 3 meters from you, isn't it logical, by deduction, to say that the ball has moved closer to you?

 

[jbriggs444]

If a ball 5 meters from you suddenly starts moving to 3 meters from you, isn't it logical, by deduction, to say that the ball has moved closer to you?

Can you describe the experiment more completely? What did you originally measure? What did you measure later?

It is cheating to say "the ball starts moving" and then conclude "the ball has moved". You've included the conclusion in the problem statement.

 

[student34]

Can you describe the experiment more completely? What did you originally measure? What did you measure later?

I don't know why it is so blurry, but in the diagram below, from Wikipedia, an object moves from rest, at the origin 0,0, to a high speed along the x axis. We will assume that A is a ball at rest with the object O at the origin. Let's say that the acceleration is instantaneous for simplification. Roughly speaking, the ball seems to shift or "move" closer to O, let's say from 5 meters to 4 meters.

 

[jbriggs444]

I don't know why it is so blurry, but in the diagram below, from Wikipedia, an object moves from rest, at the origin 0,0, to a high speed along the x axis. We will assume that A is a ball at rest with the object O at the origin. Let's say that the acceleration is instantaneous for simplification. Roughly speaking, the ball seems to shift or "move" closer to O, let's say from 5 meters to 4 meters.

Not responsive to the question that was posed. What was measured?

No one asked "what coordinates were ascribed?"

 

[student34]

The calculated distance to the ball in the traveler's rest frame changes. But there is no direct observable that corresponds with this change. It's a calculated number that depends on your choice of coordinates.

What about in the case of the muons reaching Earth due to there being less distance that the muon has to travel than if there were no length contraction?

 

[jbriggs444]

What about in the case of the muons reaching Earth due to there being less distance that the muon has to travel than if there were no length contraction?

The elapsed proper time for the muon is an invariant. It is the same for a high speed and time-dilated muon traversing a large distance as for a stationary muon being approached by the Earth's surface through a length-contracted atmosphere. Both descriptions yield the same result for the fraction of muons impacting upon the Earth's surface. No difference in measured results corresponds to the difference in description.

 

[student34]

Not responsive to the question that was posed. What was measured?

No one asked "what coordinates were ascribed?"

Object O was measured to be 5 meters from A when both are at rest with each other, and then measured to be 4 meters from A after acceleration.

 

[jbriggs444]

Object O was measured to be 5 meters from A when both are at rest with each other, and then measured to be 4 meters from A after acceleration.

No. It was not. You are asserting coordinates, not providing actual physical measurements.

What was actually measured? How was the measurement performed?

Example:

We took a five meter stick, lined one end up at the origin, placed a ball at the other end. The ball stayed there. We concluded that the ball was continuously 5 meters away from the origin in the lab frame.

We then took two atomic clocks, synchronized them at the origin and then slowly moved one four meters away on the lab floor.

We launched our original 5 meter stick at 60 percent of the speed of light (details available upon request) rightward from well to the left of the origin so that it would have time for any acceleration-induced stresses (*ahem*, length contraction) to relax away. We glued a ball on the right edge of the stick. We observed the left edge of the stick pass by the origin at a particular time on our atomic clock there. We recorded that time reading. We observed the ball on the stick's right edge pass by our 4 meter clock. We recorded the 4 meter clock's reading.

Our two clock readings matched.

We concluded that this ball was 4 meters away from the origin at the time when the left end of the meter stick passed the origin.

If you prefer, we can re-do the experiment for the scenario in which the lab is accelerated leftward and the meter stick is left alone...

So this time we leave the 5 meter stick on the ground and move the lab (and the origin of our coordinate system) well to the right. We set the lab in motion leftward at 60 percent of the speed of light and wait for any acceleration-induced stresses to relax away.

We set up two atomic clocks, synchronized at our lab origin and slowly move one four meters to the right.

We wait for the ball to pass by our 4 meter clock and for the left edge of the meter stick to pass by our clock at the origin and record both time stamps. We observe that they match and conclude that the ball was 4 meters away from our origin when the left end of the meter stick passed the origin.

None of this involves any coordinates. It involves measurements of distance and of time.

 

[Dale]

I am really not sure.

OK, maybe some other suggestions then. Let’s say that you are building a deck and need to mark a straight line on the deck. How do you physically make a reasonably straight line on a floor? (Assuming the floor is flat)

Spoiler

Here is just one common method How does this simple device make a straight line?

 

[PeterDonis]

If a ball 5 meters from you suddenly starts moving to 3 meters from you

That is not what is happening. Again: nothing about the ball changes when you change speed. Are you going to continue to ignore this or not?

What about in the case of the muons reaching Earth due to there being less distance that the muon has to travel than if there were no length contraction?

That's not how we account for it in the Earth frame. In the Earth frame, we account for it by the muons being time dilated; the internal "clocks" inside the muons that determine their decay rate is slowed compared with Earth clocks.

In the muon frame, the distance Earth travels to reach them (note how I phrased that: the muons are at rest in this frame so they don't move, the Earth does) is length contracted, compared to the Earth frame. But neither the muons nor the Earth change speed in this scenario, so it is irrelevant to your question about what happens if you change speed.

Object O was measured to be 5 meters from A when both are at rest with each other, and then measured to be 4 meters from A after acceleration.

Wrong. You have already been told repeatedly that there are no actual measurements that correspond to these numbers.

 

[student34]

OK, maybe some other suggestions then. Let’s say that you are building a deck and need to mark a straight line on the deck. How do you physically make a reasonably straight line on a floor? (Assuming the floor is flat)

Spoiler

Here is just one common method How does this simple device make a straight line?

Use a straight instrument?

 

[PeterDonis]

Use a straight instrument?

How do you know the instrument is straight?

 

[student34]

How do you know the instrument is straight?

I do not know.

 

[PeterDonis]

I do not know.

Then that's where you need to start. This is what you should be asking questions about.

 

[student34]

Then that's where you need to start. This is what you should be asking questions about.

Then I would very much like to know the answer to your question.

 

[Dale]

Use a straight instrument?

Yes, that is one possibility. A straight edge will let you know if your line turns left or right. That is one definition of a straight line, a line which doesn’t turn at any point.

Another definition is “the shortest path between two points is a straight line”. So for drawing a straight line between two points on a floor you can run a string between them and tighten the string to minimize its length. The resulting path is also a straight line.

It is possible to prove that these two definitions, no turning and shortest distance, are equivalent. So once you have a concept of “distance” you can get a concept of “straight”. And with “straight” you can also get “not straight”, “curved” or “angled”. Basically, all of geometry follows from the concept of distance.

Does that make sense?

 

[PeterDonis]

Then I would very much like to know the answer to your question.

The general answer is called "parallel transport", and the property (which has been mentioned before in this thread) that a geodesic--the generalization of "straight line" that we need--parallel transports its own tangent vector along itself.

In the simplest case, compare a line on a plane with a circle on a plane. Imagine an arrow (a vector) at some particular point on the line that points along (tangent to) the line, and an arrow (a vector) at some particular point on the circle that points along (tangent to) the circle.

Now imagine moving along the line or circle and seeing what happens to the arrow (vector) to keep it pointing in the right direction to stay tangent to the line or circle. In the case of the line, nothing at all has to happen to it; the arrows tangent to the line at every point are all parallel to each other, so transporting the arrow at one point along the line parallel to itself just gives you the arrow at any other point on the line. Thus, the tangent vector to the line is transported parallel to itself along the line. That is the invariant sense in which the line is "straight". (The equipment used by surveyors is an actual physical realization of what I have been describing: its intent is precisely to realize "straight lines" in this sense.)

When moving along the circle, however, parallel transporting the arrow at one point along the circle does not give you the arrow at another point. For example, suppose we start at a point on the circle where the arrow is pointing "up" (towards the "top" of the plane). Transporting the arrow parallel to itself means the arrow keeps on pointing "up". But the arrows tangent to the circle at other points on the circle do not point "up"; they point in different directions ("left" or "down" or "right"). So the tangent vector to the circle is not transported parallel to itself along the circle. That is the invariant sense in which the circle is "curved".

There is a lot more mathematics lurking beneath the above, which is needed in order to make everything rigorous and give mathematical conditions that can be checked for much more complicated cases where there is no easy way to visualize what is going on (such as general curved 4D spacetimes). Also, for spacetime in relativity, there is the additional physical rule that the reading on an accelerometer following a worldline indicates the worldline's curvature; in other words, that the accelerometer is the spacetime equivalent of the surveyor's equipment in ordinary space.

 

[PeterDonis]

Yes, that is one possibility. A straight edge will let you know if your line turns left or right.

But that only helps if you already know the straight edge is straight. Which just pushes the problem back to how you determine that.

 

[cianfa72]

Also, for spacetime in relativity, there is the additional physical rule that the reading on an accelerometer following a worldline indicates the worldline's curvature; in other words, that the accelerometer is the spacetime equivalent of the surveyor's equipment in ordinary space.

Basically the accelerometer at each point (event) along the given worldline "looks" on a small region around it and check if the tangent vector is actually parallel transported to itself or not.

 

[Dale]

But that only helps if you already know the straight edge is straight. Which just pushes the problem back to how you determine that.

Yes, which is why I included the subsequent sentence describing parallel transport in layman's terms. I wanted to focus on the "shortest distance" definition of straight, and just barely mention parallel transport and the fact that they are equivalent.

 

[student34]

Yes, that is one possibility. A straight edge will let you know if your line turns left or right. That is one definition of a straight line, a line which doesn’t turn at any point.

Another definition is “the shortest path between two points is a straight line”. So for drawing a straight line between two points on a floor you can run a string between them and tighten the string to minimize its length. The resulting path is also a straight line.

It is possible to prove that these two definitions, no turning and shortest distance, are equivalent. So once you have a concept of “distance” you can get a concept of “straight”. And with “straight” you can also get “not straight”, “curved” or “angled”. Basically, all of geometry follows from the concept of distance.

Does that make sense?

Yes, thanks

 

[student34]

The general answer is called "parallel transport", and the property (which has been mentioned before in this thread) that a geodesic--the generalization of "straight line" that we need--parallel transports its own tangent vector along itself.

In the simplest case, compare a line on a plane with a circle on a plane. Imagine an arrow (a vector) at some particular point on the line that points along (tangent to) the line, and an arrow (a vector) at some particular point on the circle that points along (tangent to) the circle.

Now imagine moving along the line or circle and seeing what happens to the arrow (vector) to keep it pointing in the right direction to stay tangent to the line or circle. In the case of the line, nothing at all has to happen to it; the arrows tangent to the line at every point are all parallel to each other, so transporting the arrow at one point along the line parallel to itself just gives you the arrow at any other point on the line. Thus, the tangent vector to the line is transported parallel to itself along the line. That is the invariant sense in which the line is "straight". (The equipment used by surveyors is an actual physical realization of what I have been describing: its intent is precisely to realize "straight lines" in this sense.)

When moving along the circle, however, parallel transporting the arrow at one point along the circle does not give you the arrow at another point. For example, suppose we start at a point on the circle where the arrow is pointing "up" (towards the "top" of the plane). Transporting the arrow parallel to itself means the arrow keeps on pointing "up". But the arrows tangent to the circle at other points on the circle do not point "up"; they point in different directions ("left" or "down" or "right"). So the tangent vector to the circle is not transported parallel to itself along the circle. That is the invariant sense in which the circle is "curved".

There is a lot more mathematics lurking beneath the above, which is needed in order to make everything rigorous and give mathematical conditions that can be checked for much more complicated cases where there is no easy way to visualize what is going on (such as general curved 4D spacetimes). Also, for spacetime in relativity, there is the additional physical rule that the reading on an accelerometer following a worldline indicates the worldline's curvature; in other words, that the accelerometer is the spacetime equivalent of the surveyor's equipment in ordinary space.

Interesting, thank you

 

[Dale]

Yes, thanks

Ok, so once you understand that physical geometry is about distance and with that one concept you can get straight, curved, angles, and all other geometric features, then you can go to coordinates.

My table top has a certain physical size, a length and width and an area. On the top of the table I can draw a grid, and I can tell if the grid lines are straight as follows: I can put a pin in any two points on one of the lines, and tighten a string between the pins. If the string follows the grid line then it is straight. Otherwise it is curved. So it is unambiguous if my grid lines are straight or not.

Furthermore, no amount of drawing or redrawing grid lines changes the length of a string along the edge of the table, or across the diagonal. And it also doesn’t change the area of the table top. Coordinates can be drawn, but they only control the numbers I use to locate objects on my desk, not the physical distance between those objects.

All of this stems from being able to measure the distance along any path. Clear?

 

[student34]

Ok, so once you understand that physical geometry is about distance and with that one concept you can get straight, curved, angles, and all other geometric features, then you can go to coordinates.

My table top has a certain physical size, a length and width and an area. On the top of the table I can draw a grid, and I can tell if the grid lines are straight as follows: I can put a pin in any two points on one of the lines, and tighten a string between the pins. If the string follows the grid line then it is straight. Otherwise it is curved. So it is unambiguous if my grid lines are straight or not.

Furthermore, no amount of drawing or redrawing grid lines changes the length of a string along the edge of the table, or across the diagonal. And it also doesn’t change the area of the table top. Coordinates can be drawn, but they only control the numbers I use to locate objects on my desk, not the physical distance between those objects.

All of this stems from being able to measure the distance along any path. Clear?

Yes that makes sense.

 

[Dale]

Yes that makes sense.

OK, last one then. Now we just move what we have already learned about geometry, lengths, and coordinates and we apply it to spacetime. For simplicity we will neglect gravity, so spacetime is flat, we are off in intergalactic space far from any significantly gravitating object (but do keep in mind that this all generalizes fairly easily to gravity).

Now, on the table, although you can use any coordinates you like, there do exist coordinates such that the distance in terms of the coordinates is given by the Pythagorean theorem: $ds^2=dx^2+dy^2$. Now, this distance is the same as the distance physically measured by a string or a ruler, and if you transform to any other coordinates the formula will change but $ds$ will stay the same. So $ds$ is what we call an invariant.

If we move to spacetime we have almost the same thing. Now, a point particle becomes a line in spacetime, and just like a line in space this line in spacetime has all sorts of geometric features, such as length, straightness, angles, etc. The only difference is that instead of using the Pythagorean theorem to figure out that geometry, we use the Minkowski metric $ds^2=-c^2 dt^2+dx^2 + dy^2 + dz^2$. This is very similar to the Pythagorean theorem, except that there is an extra term $-c^2 dt^2$ that is negative and is related to time. This is "distance" in spacetime and it is called the spacetime interval. This is the invariant measure of geometry that remains regardless of how you draw your coordinates.

Now, when $ds^2$ is negative we call the interval "timelike", and then $d\tau^2 = -ds^2/c^2$ is called the proper time. This is the time directly measured on a clock that moves along some path through spacetime. In other words, if an object's worldline is a string, then a clock measures the length of the string. Similarly, an accelerometer directly measures if the worldline turns any direction. So an accelerometer is equivalent to a straightedge, allowing us to measure bending of a worldline, and a clock is similar to a ruler, allowing us to measure the length of the worldline. We can use both to determine if a worldline is straight or not, as discussed above. A straight worldline doesn't turn anywhere (the accelerometer reads 0) and a straight worldline is the longest interval between two events. (note that for timelike paths a straight worldline is the longest interval instead of shortest interval because of the - sign in the metric).

 

[student34]

OK, last one then. Now we just move what we have already learned about geometry, lengths, and coordinates and we apply it to spacetime. For simplicity we will neglect gravity, so spacetime is flat, we are off in intergalactic space far from any significantly gravitating object (but do keep in mind that this all generalizes fairly easily to gravity).

Now, on the table, although you can use any coordinates you like, there do exist coordinates such that the distance in terms of the coordinates is given by the Pythagorean theorem: $ds^2=dx^2+dy^2$. Now, this distance is the same as the distance physically measured by a string or a ruler, and if you transform to any other coordinates the formula will change but $ds$ will stay the same. So $ds$ is what we call an invariant.

If we move to spacetime we have almost the same thing. Now, a point particle becomes a line in spacetime, and just like a line in space this line in spacetime has all sorts of geometric features, such as length, straightness, angles, etc. The only difference is that instead of using the Pythagorean theorem to figure out that geometry, we use the Minkowski metric $ds^2=-c^2 dt^2+dx^2 + dy^2 + dz^2$. This is very similar to the Pythagorean theorem, except that there is an extra term $-c^2 dt^2$ that is negative and is related to time. This is "distance" in spacetime and it is called the spacetime interval. This is the invariant measure of geometry that remains regardless of how you draw your coordinates.

Now, when $ds^2$ is negative we call the interval "timelike", and then $d\tau^2 = -ds^2/c^2$ is called the proper time. This is the time directly measured on a clock that moves along some path through spacetime. In other words, if an object's worldline is a string, then a clock measures the length of the string. Similarly, an accelerometer directly measures if the worldline turns any direction. So an accelerometer is equivalent to a straightedge, allowing us to measure bending of a worldline, and a clock is similar to a ruler, allowing us to measure the length of the worldline. We can use both to determine if a worldline is straight or not, as discussed above. A straight worldline doesn't turn anywhere (the accelerometer reads 0) and a straight worldline is the longest interval between two events. (note that for timelike paths a straight worldline is the longest interval instead of shortest interval because of the - sign in the metric).

Yeah thanks for that. I will try to keep all this in mind moving forward.