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zenterix

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- TL;DR Summary
- In the thought experiment of three moving railcars in a row, the middle one equidistant from the outer two, in which a light ray is emitted from each of the outer two to reach the middle one, how do we reason about the ##x'=0## and ##t'=0## axes in the moving frame of the first railcar?

I am following the book "Special Relativity and Classical Field Theory: The Theoretical Minimum" by Susskind.

I want to go through a sequence of ideas to try to understand simultaneity.

We have a rest frame A and a moving frame B along the x-axis.

Let the coordinates be ##x## and ##t## in frame A and ##x'## and ##t'## in frame B.

Under Newton's laws, we have the transformation from B coordinates to A coordinates as follows

$$t=t'$$

$$x=x'+vt'$$

Here is a depiction of coordinates in frame A

I've drawn in the coordinates of a light ray that is at the origin at ##t=0## (dark green) and the coordinates of frame B if we assume that it too is at the origin at ##t=0## (light green).

Imagine a moving train with three different railcars each separated by a distance of 1. Art, Maggie, and Lenny are each in one of the railcars.

Each of them has an inertial reference frame, though they are simply offset from each other by a fixed distance. We can consider them all to be frame B.

Let's consider frame A our frame.

Let's use relativistic units (meaning that ##c=1## and we can either think about that number as dimensionless or as, e.g. light-seconds per second).

Suppose Art and Lenny both emit a light ray towards Maggie such that the light rays arrive simultaneously at Maggie.

First suppose that the train is not moving. We have the following situation

Since ##v=0## we have

$$x=x'$$

$$t=t'$$

Consider the light green lines.

For the one starting at the origin we have

$$x=t$$

The other one satisfies

$$x=-t+t_0$$

Since they meet at ##b##, which we know is ##(1,1)## we can easily find the equation

$$x=2-t$$

for the other light green line.

Note that a light ray emitted at ##t=t'=0## from the origin and from ##(2,0)## (in both frames) reaches point ##b(1,1)## simultaneously. The two light blue points are synchronized, as are the two pink points.

In fact, any points on any line ##t=t_1## are synchronized.

Now consider that the train is moving with speed ##v##.

Again, the red lines represent the position in spacetime of the three railcars in frame A.

Assuming that the speed of light is the same in every inertial frame, then we can draw the green lines at 45 degree angles representing emitting light rays from different rail cars. For two light rays to arrive simultaneously at the second rail car at point b, they need to be emitted at the two light blue points (one by Art and one by Lenny).

Notice that in frame A, the times at which they are emitted are no longer the same. The emission of the rays is not synchronized from the point of view of frame A.

Maggie receives both light rays at the same time, and from her frame both Art and Lenny at at the same distance from her so to her the emission of the rays is synchronized. In fact, since the emission by Art occurs at t'=0 then so does the emission by Lenny.

From the point of view of Maggie's frame, all points on the thin blue line are synchronized with each other.

Let's see how Newton would view this situation in graphical terms.

As far as I can tell, we have a situation like

I have drawn the two boxes just to show the symmetry and that in fact the angles are drawn correctly (I think). The point I want to talk about here is the black point.

Notice that I used the gridlines to make the graph precise.

We have here that ##c=1## and ##v=\frac{1}{4}##.

The black line starting at the origin represents the light ray emitted by Art from the point of view of frame A

$$x = (c+v)t=(1+v)t=1.25t\tag{1}$$

In Art's frame we have

$$x'=x-vt=(1+v)t-vt=-t\tag{2}$$

In Lenny's frame we have

$$x'=x-vt=(1+v)t-(2+vt)=-2+t\tag{3}$$

Next, consider the black line starting at ##(2,0)## in frame A.

$$x=(-c+v)t+2=(-1+v)t+2=2-0.75t\tag{4}$$

In Art's frame this is

$$x'=(-1+v)t+2-vt=2-t\tag{5}$$

and in Lenny's frame

$$x'=(-1+v)t+2-(2+vt)=-t\tag{6}$$

(1) and (4) are the black lines in the plot.

They meet at

$$(1+v)t=(-1+v)t+2 \implies t=1 \implies x=1+v=1.25$$

In Art's frame we have

$$t=2-t\implies t=1 \implies x'=1$$

And in Lenny's frame we have

$$-2+t=-t\implies t=1\implies x'=-1$$

All of this makes sense mathematically and graphically.

What it means is that under Newton's transformations the origin is synchronous with ##(2,0)## in all frames, which means that a light ray emitted from these two spacetime points reach the equidistant purple point at the same time.

So why is this problematic?

Because measurements of the speed of light in any of the frames shows is constant: ##c=1##.

Thus, equations (1) and (4) are incorrect since these equations show a speed of light of 1.25 and 0.75, respectively.

When we make the assumption that the speed of light is constant in every frame, we end up with the green lines in the picture above and now the two blue points are synchronous in the moving frames. They are not synchronous in frame A.

At this point my questions arise.

I'm having a hard time interpreting these results further.

We have Maggie's train car and for light to reach her from the two other equidistant train cars we've seen that the light rays must be emitted at different times in frame A.

From the point of view of Maggie, however, since

1) they arrive at the same time

2) they are equidistant

3) light has a fixed speed

then they have the same time coordinate in her frame. In the case of the blue dots, this time would be ##t'=0##. In the case of the pink points, it would be some other time ##t_1##.

Finally, there is also the ##x'=0## axis

I want to go through a sequence of ideas to try to understand simultaneity.

We have a rest frame A and a moving frame B along the x-axis.

Let the coordinates be ##x## and ##t## in frame A and ##x'## and ##t'## in frame B.

Under Newton's laws, we have the transformation from B coordinates to A coordinates as follows

$$t=t'$$

$$x=x'+vt'$$

Here is a depiction of coordinates in frame A

I've drawn in the coordinates of a light ray that is at the origin at ##t=0## (dark green) and the coordinates of frame B if we assume that it too is at the origin at ##t=0## (light green).

Imagine a moving train with three different railcars each separated by a distance of 1. Art, Maggie, and Lenny are each in one of the railcars.

Each of them has an inertial reference frame, though they are simply offset from each other by a fixed distance. We can consider them all to be frame B.

Let's consider frame A our frame.

Let's use relativistic units (meaning that ##c=1## and we can either think about that number as dimensionless or as, e.g. light-seconds per second).

Suppose Art and Lenny both emit a light ray towards Maggie such that the light rays arrive simultaneously at Maggie.

First suppose that the train is not moving. We have the following situation

Since ##v=0## we have

$$x=x'$$

$$t=t'$$

Consider the light green lines.

For the one starting at the origin we have

$$x=t$$

The other one satisfies

$$x=-t+t_0$$

Since they meet at ##b##, which we know is ##(1,1)## we can easily find the equation

$$x=2-t$$

for the other light green line.

Note that a light ray emitted at ##t=t'=0## from the origin and from ##(2,0)## (in both frames) reaches point ##b(1,1)## simultaneously. The two light blue points are synchronized, as are the two pink points.

In fact, any points on any line ##t=t_1## are synchronized.

Now consider that the train is moving with speed ##v##.

Again, the red lines represent the position in spacetime of the three railcars in frame A.

Assuming that the speed of light is the same in every inertial frame, then we can draw the green lines at 45 degree angles representing emitting light rays from different rail cars. For two light rays to arrive simultaneously at the second rail car at point b, they need to be emitted at the two light blue points (one by Art and one by Lenny).

Notice that in frame A, the times at which they are emitted are no longer the same. The emission of the rays is not synchronized from the point of view of frame A.

Maggie receives both light rays at the same time, and from her frame both Art and Lenny at at the same distance from her so to her the emission of the rays is synchronized. In fact, since the emission by Art occurs at t'=0 then so does the emission by Lenny.

From the point of view of Maggie's frame, all points on the thin blue line are synchronized with each other.

Let's see how Newton would view this situation in graphical terms.

As far as I can tell, we have a situation like

I have drawn the two boxes just to show the symmetry and that in fact the angles are drawn correctly (I think). The point I want to talk about here is the black point.

Notice that I used the gridlines to make the graph precise.

We have here that ##c=1## and ##v=\frac{1}{4}##.

The black line starting at the origin represents the light ray emitted by Art from the point of view of frame A

$$x = (c+v)t=(1+v)t=1.25t\tag{1}$$

In Art's frame we have

$$x'=x-vt=(1+v)t-vt=-t\tag{2}$$

In Lenny's frame we have

$$x'=x-vt=(1+v)t-(2+vt)=-2+t\tag{3}$$

Next, consider the black line starting at ##(2,0)## in frame A.

$$x=(-c+v)t+2=(-1+v)t+2=2-0.75t\tag{4}$$

In Art's frame this is

$$x'=(-1+v)t+2-vt=2-t\tag{5}$$

and in Lenny's frame

$$x'=(-1+v)t+2-(2+vt)=-t\tag{6}$$

(1) and (4) are the black lines in the plot.

They meet at

$$(1+v)t=(-1+v)t+2 \implies t=1 \implies x=1+v=1.25$$

In Art's frame we have

$$t=2-t\implies t=1 \implies x'=1$$

And in Lenny's frame we have

$$-2+t=-t\implies t=1\implies x'=-1$$

All of this makes sense mathematically and graphically.

What it means is that under Newton's transformations the origin is synchronous with ##(2,0)## in all frames, which means that a light ray emitted from these two spacetime points reach the equidistant purple point at the same time.

So why is this problematic?

Because measurements of the speed of light in any of the frames shows is constant: ##c=1##.

Thus, equations (1) and (4) are incorrect since these equations show a speed of light of 1.25 and 0.75, respectively.

When we make the assumption that the speed of light is constant in every frame, we end up with the green lines in the picture above and now the two blue points are synchronous in the moving frames. They are not synchronous in frame A.

At this point my questions arise.

I'm having a hard time interpreting these results further.

We have Maggie's train car and for light to reach her from the two other equidistant train cars we've seen that the light rays must be emitted at different times in frame A.

From the point of view of Maggie, however, since

1) they arrive at the same time

2) they are equidistant

3) light has a fixed speed

then they have the same time coordinate in her frame. In the case of the blue dots, this time would be ##t'=0##. In the case of the pink points, it would be some other time ##t_1##.

Finally, there is also the ##x'=0## axis

**What does it mean that the ##x'=0## axis and the ##t'=0## axis are not at a right angle?**