I Epstein diagrams and a traveling ship

[Myslius]

TL;DR Summary

Epstein diagrams and a traveling ship

So I've made a simple Epstein diagram. Length of a ship is 1, v is 0.6c. Length contraction is visible, time dilation is visible.

So far so good, let's now look at this (I only added a red line)

How a stationary ship (red line), transforms into a line O1 D1. It feels like four vector is not conserved here. The question is: how do you draw a diagram of how the red line transforms into O1 D1 over time. Both the front of the ship and the back of the ship should travel the same distance in spacetime (not space).

 

[Dale]

Epstein diagrams are not spacetime diagrams. There are no coordinate charts and therefore no transforms. The red line isn’t a thing that has any physical meaning.

 

[Ibix]

What do you mean by "transforms over time"? Transforms change between frames and they apply to a whole frame - transforming over time doesn't make sense. Do you mean, the ship accelerates from rest to 0.6c? If so, what happens depends on the acceleration profile. Is it Rindler acceleration so that the rocket maintains its proper length?

Whatever it is, write down the $x(t)$ coordinates of a family of points along the length of the ship, compute $x(\tau)$, and plot.

 

[Ibix]

Additional: you haven't really explained your graph. What are the four lines meant to represent?

Also, note that the representation of an object on an Epstein diagram depends on its history, not just its current state. I would not necessarily expect the final state of an accelerated ship to look like that of an eternally inertial ship with the same speed.

 

[Myslius]

eternally inertial ship with the same speed looks the same in the diagram, eternally. It's just OB and CD lines extended to the infinity.

There's nothing to explain here. It's an Epstein diagram.
This diagram conserves the length of a four vector or the length of a curve (when a particle slowly accelerates). It works fine for a single particle. I'm just curious why this is not the case for a ship, which can be considered as a bunch or particles. I would like to see an example of such diagram where the gain of the final velocity isn't instantaneous. I feel like such diagram won't conserve a length of a curve. Something accelerating would look like this:

the inner line is shorter.
Ok, maybe the Epstein diagrams doesn't conserve the length of the curves, but yet again the question is why does it conserve for a single particle. Regardless of how particles move in the diagram.


For example, a bunch of particles,

The length of a curve is always the same. You can see how far away each particle got (x axis). And how old each particle is (y axis). How do I do this for a ship?

 

[Myslius]

Also, each particle can for example curve back to the x = 0 (and some y point, different one for each particle).This basically means all of them do meet at a single point in space, starting location in space. And all particles can compare how much they aged.

Or... for example you can you can draw a curve x=-y (to represent a light cone) and offset it to any y you want. You will be able to see what a stationary observer sees at each moment in time. His slice.

I explained the Epstein in case you're not familiar with it.

 

[Dale]

It's an Epstein diagram.

The only things that are physically meaningful on that diagram are OA and OB. Additionally the dashed lines are not physically meaningful but do have an interpretation in terms of Einstein synchronization.

Nothing else on the diagram is meaningful. In particular, CD is not physically meaningful and has no interpretation in terms of Einstein synchronization.

Epstein diagrams are **very** limited

 

[Ibix]

There's nothing to explain here.

No, you need to say what the lines represent.

I think CB is the position of the nose of a moving ship and OD is the tail. I presume their clocks are supposed to be synchronised in their rest frame.

You also seem to have drawn a second set of axes on the chart, which makes no sense. You are plotting proper time vertically and this is an Epstein diagram. Diagonal lines are not the axes of other frames.

Now you want to draw an accelerating rocket. What are the coordinates of some points along its length?

you can you can draw a curve x=-y (to represent a light cone) a

That would not be a lightcone on an Epstein diagram. That would be something moving at $-c/\sqrt{2}$.

 

[Myslius]

You are probably right about the lightcone (well this lightcone differs from Minkowski's), anyhow, the coordinates, well I marked the red line in second picture. I bet you want me to tell you how the clocks are set now?

 

[Ibix]

You are probably right about the lightcone, anyhow, the coordinates, well I marked the red line in second picture.

Yeah, but that's only the start position. You need to write down $x(\tau)$.

 

[Ibix]

Note that I'm on my phone, so I don't absolutely guarantee that the algebra here is correct.

Let's assume we have a Rindler accelerating rocket of proper length $L$ whose tail accelerates at $\alpha_0$. The proper acceleration of a point a distance $x_0$ in front of the tail is $\alpha(x_0)=\alpha_0/(1+\alpha_0x_0)$.

For a Rindler observer undergoing proper acceleration $\alpha$ starting at $x=x_0$ at $\tau=0$, $x=x_0+\dfrac 1\alpha\sinh(\alpha \tau)$. That means that the $x(\tau)$ of a point $x_0$ in front of the tail follows $x=x_0+\dfrac{1+\alpha_0x_0}{\alpha_0}\sinh\left(\dfrac{\alpha_0}{1+\alpha_0x_0}\tau\right)$.

You can now plot the curves on your Epstein diagram associated with various points along the rocket as it accelerates eternally from rest. If you want to terminate the acceleration at some fixed speed $v$ in the rocket's initial rest frame, note that $\dfrac{dx}{d\tau}=\dfrac{dx}{dt}\dfrac{dt}{d\tau}=\gamma v$. Differentiating the expression in the last paragraph gives us $\frac{dx}{d\tau}=\cosh\left(\dfrac{\alpha_0}{1+\alpha_0x_0}\tau\right)$. Simply set the right hand side equal to $v\gamma$ and solve for the $\tau$ coordinate where the point returns to inertial motion.

 

[Ibix]

Note: if you work in light years and years, a proper acceleration of 1 is within a few percent of one gravity.

 

[Myslius]

<quote>I think CB is the position of the nose of a moving ship and OD is the tail</quote>
The opposite, The nose is CD

 

[Ibix]

<quote>I think CB is the position of the nose of a moving ship and OD is the tail</quote>
The opposite, The nose is CD

Ah yes - transcription error. I swapped B and D.

 

[Dale]

the coordinates

An Epstein diagram doesn’t have coordinates. A coordinate chart is a structure that goes on a manifold and space-proper-time isn’t a manifold.

All you can do on an Epstein diagram is draw two intersecting straight lines and project from one onto the other. It lacks any structure to do more than that.

 

[Dale]

All you can do on an Epstein diagram is draw two intersecting straight lines and project from one onto the other.

You may be able to do lines parallel to these. I didn’t test that when I looked into Epstein diagrams years ago. Just additional lines intersecting the original two

 

[Ibix]

You may be able to do lines parallel to these. I didn’t test that when I looked into Epstein diagrams years ago. Just additional lines intersecting the original two

You can have as many lines as you want on an Epstein diagram, I think. And they don't have to be straight lines, and the Euclidean distance along them is the lapse of coordinate time. The fundamental issue, of course, is that plotting a "private" quantity like proper time on a "public" axis (to borrow Bondi's terminology) inevitably leads to weird results, like crossing lines that don't actually meet in reality.

I think the major issue with this rocket example (which I glossed past in my maths) is what "zero of proper time" means if you don't have all your worldlines of interest meeting at one event. In this case I quietly assumed it was the zero of coordinate time in the initial rest frame of the ship's clocks. That isn't really a part of an Epstein diagram as far as I know. And I don't think there's a "correct" way to do it because you can't model clock synchronisation procedures consistently on the diagram due to the issue with meeting lines that don't meet. Light travels on horizontal lines (noting that we're already generalising from "proper time" to "interval"), so a reflected clock synchronisation pulse doesn't return to its emission point at the same place on the diagram as the return event.

As we've noted before, Epstein diagrams are a dead end. They look kind of attractive because of the "Euclidean distance corresponds to coordinate time" thing, but I think this is the same kind of mistake as relativistic mass - trying to make the theory conform to older ideas because they're familiar instead of accepting the new theory on its own terms.

 

[Dale]

You can have as many lines as you want on an Epstein diagram, I think. And they don't have to be straight lines, and the Euclidean distance along them is the lapse of coordinate time.

No, you cannot. The point of the Epstein diagram is to be able to do the projection (those dotted lines in the OP’s diagrams) to express time dilation. As soon as you bring in a third direction the projections fall apart.

So yes you can draw a figure with proper time and coordinate position for more lines, but there is no meaning to the figure. It is just a bunch of lines on a page. No further information that you can extract from it. Doing so eliminates the whole point of the diagram in the first place.

The fundamental issue, of course, is that plotting a "private" quantity like proper time on a "public" axis (to borrow Bondi's terminology) inevitably leads to weird results

Agreed!