Can 2 ships pass each other twice?

[Zeno]

This is showing 2 ships that are passing each other at the moment that T of ship #1 is passing N of ship #2. Each diagram shows one of the ships at rest and the other ship moving.

Special relativity....

Ship #1 at rest at top:

1 light-year
0sec------------0sec
T--------------------N
N-----------T <-------- .8c
0sec 15,147,296sec
.6 light-year

Ship #2 at rest at bottom:

.6 light-year
0sec -15,147,296sec
T---------------N -------> .8c
N--------------------T
0sec------------0sec
1 light-year

T1 is T of ship #1. N1 is N of ship #1. N2 is N of ship #2. T2 is T of ship #2.

T1, N1, T2, and N2 each have their own clock.

At the moment that T1 is lined up with N2 the clocks at T1 and N1 display 0 seconds in their own rest frame and the clocks at T2 and N2 display 0 seconds in their own rest frame. T1 is displaying 0 seconds and N1 is displaying -15,147,296 seconds when viewed from ship #2. N2 is displaying 0 seconds and T2 is displaying 15,147,296 seconds when viewed from ship #1.

Using the math formula for the rear clock ahead effect....
((distance between clocks)*(velocity of reference frames relative to each other))/(c^2) we get ((3.527*10^12 miles)*(149,025 miles/sec))/(3.47*10^10 miles^2/seconds^2) = 15,147,296 seconds.
So the clock in the rear is 15,147,296 seconds ahead of the clock in the front.

Now to the point of this post. There seems to be a problem here. It seems that T1 is passing N2 twice.

T1 passes N2 when the clock at N1 displays -15,147,296 seconds. T1 passes N2 again when the clock at N1 displays 0 seconds.

Also,
T1 passes N2 when the clock at T2 displays 0 seconds. T1 passes N2 again when the clock at T2 displays 15,147,296 seconds.

What is displayed on the clocks at N1 and T2 when T1 passes N2?

What is the resolution to this problem?

 

[DaveC426913]

Why does one diagram have a negative value for time? Time is positive - unless you are implying a countdown previous to some event (eg. 15Ms before the start of a race, or a rocket liftoff).

You say "T1 is T of ship1". What does it mean for Ship1 to have a T? Is that a specific time on a clock or stopwatch?

You never actually say what N is.

This appears to be some sort of axis labeling:
"0sec------------0sec"
Why is it 0sec at both ends?

At one point, you say:
"N1 is displaying -15,147,296 seconds when viewed from ship #2."
Ask yourself this question: "How can ship #2 know what N1 is displaying?"


Finally, we have been through this before. Historically, your diagrams are very confusing - much of your labels are ambiguously named (which then need lines of explanation) and some parts are not explained at all.

I suggest you read up a bit on how relativistic problems are typically diagrammed and described, so that we can spend less time decrypting your peculiar symbology and more time solving the core problem.

 

[Janus58]

This is showing 2 ships that are passing each other at the moment that T of ship #1 is passing N of ship #2. Each diagram shows one of the ships at rest and the other ship moving.

Special relativity....

Ship #1 at rest at top:

1 light-year
0sec------------0sec
T--------------------N
N-----------T <-------- .8c
0sec 15,147,296sec
.6 light-year

Ship #2 at rest at bottom:

.6 light-year
0sec -15,147,296sec
T---------------N -------> .8c
N--------------------T
0sec------------0sec
1 light-year

T1 is T of ship #1. N1 is N of ship #1. N2 is N of ship #2. T2 is T of ship #2.

T1, N1, T2, and N2 each have their own clock.

At the moment that T1 is lined up with N2 the clocks at T1 and N1 display 0 seconds in their own rest frame and the clocks at T2 and N2 display 0 seconds in their own rest frame. T1 is displaying 0 seconds and N1 is displaying -15,147,296 seconds when viewed from ship #2. N2 is displaying 0 seconds and T2 is displaying 15,147,296 seconds when viewed from ship #1.

Using the math formula for the rear clock ahead effect....
((distance between clocks)*(velocity of reference frames relative to each other))/(c^2) we get ((3.527*10^12 miles)*(149,025 miles/sec))/(3.47*10^10 miles^2/seconds^2) = 15,147,296 seconds.
So the clock in the rear is 15,147,296 seconds ahead of the clock in the front.

Now to the point of this post. There seems to be a problem here. It seems that T1 is passing N2 twice.

T1 passes N2 when the clock at N1 displays -15,147,296 seconds. T1 passes N2 again when the clock at N1 displays 0 seconds.

Also,
T1 passes N2 when the clock at T2 displays 0 seconds. T1 passes N2 again when the clock at T2 displays 15,147,296 seconds.

What is displayed on the clocks at N1 and T2 when T1 passes N2?

What is the resolution to this problem?

There is no problem. Below are two animations showing the scenario as measured from each frame. (I did use different time units in order to keep things neater, but That doesn't change the basic principle involved). In both animations the respective times on clocks are the same according to both frames when they pass each other.




Excuse the jump in the numbers, this is due to losing the minus sign ( the -0.0 refers to times between -1 and 0)

 

[Zeno]

When the Ns are lined up is the upper T showing a 1.0 or a 0.5?
When the Ns are lined up is the lower T showing a 0.9 or a 0.2?

 

[James R]

Zeno:

Why haven't you figured out how to do these kinds of problems in the 25 years you've been trying to do them? What have you been doing for the past 25 years, as far as studying up on relativity goes? Have you made any progress at all?

This is showing 2 ships that are passing each other at the moment that T of ship #1 is passing N of ship #2. Each diagram shows one of the ships at rest and the other ship moving.

Special relativity....

Ship #1 at rest at top:

1 light-year
0sec------------0sec
T--------------------N
N-----------T <-------- .8c
.6 light-year

Ok. So, the rest lengths of the two ships are each 1 light year.

You say the lower ship on this diagram contracts to a length of 0.6 ly, as seen by the upper ship. The lower ship is travelling to the left at a speed of v=-0.8c, relative to the upper ship, which results in a Lorentz factor, gamma, of 1.66666.....

This is all fine, so far.

Next, you (sort of) define the nose (N) and tail (T) of each ship.

T1 is T of ship #1. N1 is N of ship #1. N2 is N of ship #2. T2 is T of ship #2.

You then talk about clocks:

T1, N1, T2, and N2 each have their own clock.

We can assume that clocks T1 and N1 will be synchronised in the frame of ship 1, and clocks T2 and N2 will be synchronised in the frame of ship 2.

Let the (x,t) frame of reference be the rest frame of ship 2, and the (x',t') frame be the rest frame of ship 1. The (x',t') frame moves in the +x direction at v=0.8c in frame (x,t).

Next, you specify some spacetime events:

At the moment that T1 is lined up with N2 the clocks at T1 and N1 display 0 seconds in their own rest frame and the clocks at T2 and N2 display 0 seconds in their own rest frame.

Okay.
Event 1: "T1 is aligned with N2" (i.e. T1 and N2 are at the same horizontal coordinate).

You define t=t'=0 at the instant that Event 1 occurs, in both frames.

Let the spacetime coordinates of Event 1 be (x,t)=(0,0). This implies that T1 is located at x=0 in the frame of ship 2. For convenience, let us also define N2 to be at coordinates (x',t')=(0,0) when Event 1 occurs.

The spatial coordinate definitions here imply that, at the instant that Event 1 occurs, N1 has spacetime coordinates (x,t)=(1 ly, 0).

In the frame of ship 1, ship 2 is length contracted to a length of 0.6 ly. So, at the time that event 1 occurs, N2 is at coordinate x=0 and T2 is at coordinate (x=-0.6 ly).

Next you say:

T1 is displaying 0 seconds and N1 is displaying -15,147,296 seconds when viewed from ship #2.

Define:
Event 2: "Time shown on clock N1 when T1 is aligned with N2".

The alignment occurs at t=t'=0, by definition, so event 2 occurs at t=0.

Since the clock on N1 always reads the time as measured in ship 1 (i.e. it always displays the t coordinate value), it reads zero at the time of Event 2, because the time at which event 2 happens.

Why do you claim that the clock on N1 would read -15,147,296 seconds, when T1 is aligned with N2?

Now to the point of this post. There seems to be a problem here. It seems that T1 is passing N2 twice.

It doesn't.

T1 passes N2 when the clock at N1 displays -15,147,296 seconds.

No. By definition, T1 passes N2 at t=t'=0. Since the clock at N1 displays time t at all times, and since this event happens at t=0, the clock displays zero when T1 passes N2.

What is displayed on the clocks at N1 and T2 when T1 passes N2?

At time t=0, N1 is at x=1 ly. The clock at N1 displays "zero", because it always displays the time t.

Define:
Event 3: "Time shown on clock T2 when T1 is aligned with N2"

At time t=0, T2 is at x=-0.6 ly. The clock at T2 displays the time coordinate t' at all times. We can calculate what time it displays using the Lorentz transformation for the time coordinate:

t'=(gamma)(t-vx/c^2)=(1.666)(0-(0.8c)(-0.6 ly)/c^2)=+0.8 years.

 

[DaveC426913]

N=nose, T=tail
Fer cryin out loud.
Thanks James. Pity the OP couldn't have mentioned that.

 

[Zeno]

"Why do you claim that the clock on N1 would read -15,147,296 seconds, when T1 is aligned with N2?"


"Using the math formula for the rear clock ahead effect....
((distance between clocks)*(velocity of reference frames relative to each other))/(c^2) we get ((3.527*10^12 miles)*(149,025 miles/sec))/(3.47*10^10 miles^2/seconds^2) = 15,147,296 seconds.
So the clock in the rear is 15,147,296 seconds ahead of the clock in the front."

 

[Zeno]

Event 3: "Time shown on clock T2 when T1 is aligned with N2"

At time t=0, T2 is at x=-0.6 ly. The clock at T2 displays the time coordinate t' at all times. We can calculate what time it displays using the Lorentz transformation for the time coordinate:

t'=(gamma)(t-vx/c^2)=(1.666)(0-(0.8c)(-0.6 ly)/c^2)=+0.8 years.

Why is the time at T2 0.8 years and not 15,147,296 seconds?

 

[James R]

It easiest to work in units of years (for time) and lightyears (ly) (for length/distance) here.

Using those units, the speed of light is c=1 ly/yr.

There's no need to mess around converting times to seconds and distances to miles and speeds to miles per hour.