B Is this an Unavoidable Collision between 2 spaceships traveling at 0.6c each?

[PeterDonis]

At 1.2C takes 1 year, what will take at 0.88C?

At 0.88C over what distance? You appear to be assuming that distance is 1.2 light years. But it isn't. Distance is frame-dependent, as well as time. Also simultaneity is relative. A correct relativistic calculation must take all three of those things into account.

 

[russ_watters]

Note, for a spaceship with radar, the astronauts don't even need to know relativity to know when the collision is going to happen; the radar is reporting their distance and speed.

The only reason you would need relativity for is for the people on Earth to do a conversion from what they see to what the astronauts see.

 

[Sagittarius A-Star]

At 1.2C takes 1 year, what will take at 0.88C?

Consider the following Minkowski diagram.
For the astronaut in spaceship 1 it takes 0.8 years of proper time (time-dilation!) from event A (passing star 1) to event D (collision). With respect to his restframe, event B (spaceship 2 passing star 2) happened before event A and the spatial distance between the stars is length-contracted.

Source:
https://www.geogebra.org/m/NnrRvA46

Event A: (x=0, t=0), event B: (x=1.2 LY, t=0 Y), event D: (x=0.6 LY, t=1 Y)

Lorentz transformed to the rest frame of spaceship 1:
$x'=\gamma(x-vt)$
$t'=\gamma(t-vx)$

Event A: (x'=0, t'=0), event B: (x'=1.5 LY, t'=-0.9 Y), event D: (x'=0 LY, t'=0.8 Y)

Speed of spaceship 2 with reference to the (primed) restframe of spaceship 1:

$v_2 = |{x'_D-x'_B \over t'_D - t'_B}| \approx 0.8824 c$

 

[Ibix]

A meta point: your problems with relativity, @Acodato, are fairly common among people encountering it for the first time. The reason is that there are some pre-relativistic concepts that are really hard to let go of, and those concepts tend to creep in to your thinking about relativity and lead you to a mental model that's an inconsistent mish-mash of relativistic and pre-relativistic ideas. The problem here is not in relativistic physics, it's in the mish-mash that you think is relativistic physics.

Let's draw a diagram. This is a displacement-time diagram, which is a plot of the position of one or more objects at the time read off the time axis. You may have come across these in high school physics - the only catch here is that in relativistic physics the convention is to draw the time axis vertically where in high school physics it's usually drawn horizontally. Also, by convention, we pick units of years for time and light years for distance, or seconds and light seconds, or nanoseconds and feet - or any other combination so that the speed of light is 1 distance unit per time unit. Here are your two ships travelling towards each other at 0.6c:

The red line gives you the position of one ship and the blue line the position of the other. The lines are one light year apart at the bottom of the page (the start of the experiment), and where they cross is where the ships collide (or if one is displaced a little out of the screen, they pass each other wing mirror-to-wing mirror). If you are finding the diagram difficult to interpret, get a piece of paper and lay it across the diagram with one edge horizontal, then slide it slowly up the screen. Keep your eye on the edge of the paper and where the lines disappear under it, and you'll see them moving towards each other.

Now let's draw the red and blue ships emitting pulses of light at regular intervals, and add the tracks of those light pulses to the diagram. I've rendered those as fine red and blue lines travelling from one ship to the other, equally spaced because the emissions are equally spaced:. Notice that you can see that the lines are on the diagonal of the grid squares, so you can see they're travelling at one distance unit per time unit, which is :

OK so far?

Now let's try to work out what this looks like in the rest frame of the blue ship. Newtonian physics would say that you simply subtract 0.6c from the speeds of everything, and that looks like this:

Note that, in this diagram, the fine lines representing the light paths don't have the same slope going one way as the other way: the speed of light is different depending on direction. If you look at how far they go in unit time you'll see that they travel at 1.6c and 0.4c, and the moving ship is doing 1.2c.

That's obviously not the relativistic picture (that the universe doesn't work that way was demonstrated by Michelson and Morley in 1887, much to their surprise). But here's where I think you are going wrong: I think that you think that relativity is just the picture above, modified with an assumption that (for some reason) light doesn't change speed when everything else does. That gives you this diagram:

In that diagram, the red ship rushes ahead of its own light and the other ship never sees it coming. But this is a wildly inconsistent model. For example, if those light pulses are actually very high power lasers, according to the original frame there's only a collision between wreckage while in the other frame one ship passes through the wreckage of the other before being destroyed by the lasers of the other - and we don't even want to ask how the red ship fired its lasers and somehow snuck around ahead of them without passing through them. There are so many inconsistencies in this model it couldn't last seconds, let alone have been the underlying theory of all physics for over a century.

So what does relativity actually say? Well, it says that the background on which everything happens is not space and time, but spacetime - a single 4d entity. And those diagrams we've been drawing are more than they appeared to be: they're actually maps of spacetime (or at least, two dimensions of it). And, just like with normal maps, you can draw different sets of axes on it - or, to put it another way, there is more than one direction that you can call "time" and more than one direction you can choose to call "space". But you need to pick ones that are orthogonal in spacetime (which won't be orthogonal on our graphs, unfortunately, because the graph is a purely spatial plane).

Here's our original diagram, but I've added the time axis that has the blue ship "only moving in time", and the spatial axis that is orthogonal to it:

Note that the scale on the ship's axes isn't the same as on the original axes. That, and the fact that the axes aren't parallel to the original ones, is the origin of length contraction and time dilation. It's also the origin of the more complicated velocity addition formula that's been quoted several times already in this thread. But look at the light pulses: they all go through the corners of the grid squares of the grid on these axes: light travels one distance unit per time unit in this frame too. That's the invariance of light speed.

Finally, here's the correct form of the diagram drawn in the rest frame of the blue ship:

In this frame, it's the original frame's axes that would be shown not orthogonal (again, that's a limitation of the diagram). But the light rays are still 45 degree lines (i.e., travelling at ) and they still arrive before the collision because the other ship is doing less than . Note that the red ship doesn't start emitting at the same time as the blue ship and it starts emitting much earlier and with much larger gaps between the emission times (that's time dilation) and from way more than one light year away, but is much less than one light year away when the blue ship starts emitting (that's length contraction).

If you want to learn more about this I strongly advise getting a good textbook. Taylor and Wheeler's Spacetime Physics is freely downloadable from Taylor's website, and Morin's Relativity for the Enthusiastic Beginner is a cheap download (the first chapter is free).

 

[robphy]

We will develop a spacetime diagram for the scenario you described.
(This method is based on my PF Insights:
https://www.physicsforums.com/insights/spacetime-diagrams-light-clocks/
https://www.physicsforums.com/insights/relativity-rotated-graph-paper/
https://www.physicsforums.com/insights/relativity-using-bondi-k-calculus/
https://www.physicsforums.com/insights/relativity-on-rotated-graph-paper-a-graphical-motivation/
)

Steps 0 through 3 provide a general setup.
Steps 4 and 5 focus on the specifics of this problem.

---

Step 0:
We will work in 1+1 dimensions (for 1 dimension of space).
Start with a large sheet of graph paper rotated by 45-degrees.
This is a grid of "light-clock diamonds" modeled on
"the light-rays in one tick of a light-clock at rest in this [lab] frame".
(Time runs upwards as in the typical spacetime diagram.)
Use this grid to locate events and make measurements according to the "lab frame".

---

Step 1:

Use the diamonds of the lab frame to construct the slopes (velocities) for
the worldlines of the two incoming ships that meet the lab frame at event Z (the origin).
Note that SZ has velocity $\frac{\Delta x}{\Delta t}=\frac{0-6}{0-(-10)}c=\frac{-6}{10}c$, and, similarly, PZ has velocity (6/10)c.

We also display a radar measurement done by the lab frame.
To reach distant event S, the lab frame sends a light-signal at event Q and it receives a radar echo at event U.
(It might help to think of Q and U as the "intersection of the light-cone of (target) event S and the lab-frame (measurer's) worldline".)

So, the lab records $t_Q=-16$ and $t_U=-4$.
The lab says that event M is simultaneous with the distant event S,
and the lab assigns time coordinate $t_S=t_M=\frac{1}{2}(t_U + t_Q)=-10$ (the mid-time [half the sum]).
Next, the lab assigns "space" coordinate $x_S/c=\frac{1}{2}(t_U - t_Q)=6$ (half the round trip time).
This is in agreement with the diagram.

The lab frame can also make a measurement of event Z.
The radar measurement in that case has equal time-readings for the emission and the reception.
So, $t_Z=0$ and $x_Z/c=0$.

In preparation for the next step, we note the following constructions involving the "square-interval".
Form the "causal diamond of MZ" by intersecting the future light-cone of M with the past light-cone of Z.
The parallelogram formed has the area of 100 light-clock diamonds.
The square root gives the magnitudes along the diagonal: thus 10.
That is, MZ (along the timelike diagonal) has magnitude 10.

In an analogous construction, MS is the spacelike diagonal of a causal diamond with area 36 light-clock diamonds. [Technically, this can be assigned the signed-area -36.] Thus MS (along the spacelike diagonal) has magnitude 6. QS (along a lightlike direction) is the diagonal of causal diamond with area 0 light-clock diamonds. Thus QS has magnitude 0.

---

Step 2:

We know the lab frame says that the duration of SZ takes 10 ticks (a temporal displacement).
We also know the spatial displacement is (-6) space-ticks ["sticks"].
What does the Green-ship measure for the duration of SZ (using Green's wristwatch along SZ)?
We can use the formula $(SZ)_t{}^2 - (SZ)_x{}^2 =(10)^2-(-6)^2=(8)^2$ to find "8 ticks" along SZ.
But let's try to use the diagram.
Construct the causal diamond of SZ, which has area $uv=(4)(16)=(8)^2$. (Here, $v$ is not velocity $V$.)
So, there are 8 of Green's light-clock diamonds along the diagonal SZ, similar in shape to the causal diamond of SZ. Green's light-clock diamonds have the same area as those of the lab-frame (as expected since a boost has determinant one). [The edge-sizes 4 and 16 here are related to the earlier radar-measurement time-readings.]

Step 3:

So, we can construct all of Green's ticks.
The ratio $\frac{MZ}{SZ}=\frac{10}{8}$ is the time-dilation factor $\gamma=\frac{1}{\sqrt{1-(v/c)^2}}$.

Side note: (3/5)c is an arithmetically-nice value for velocity V that leads to calculations involving rational numbers [fractions]. (4/5)c is nice, but (1/2)c is not nice. It turns out that the Doppler factor $k=\sqrt{\frac{1+V/c}{1-V/c}}$ satisfies $k^2=\frac{u}{v}$. When $k$ is rational, then the associated $V$ is arithmetically nice. Here, $k=\sqrt{\frac{4}{16}}=\frac{1}{2}$. So, Green's diamonds are scaled down by 2 along the "u" direction and scaled up by 2 along the "v" direction.
(Check for yourself that $u/v=1$ for the causal diamond of MZ.)
By the way, since $k=\sqrt{\frac{1+V/c}{1-V/c}}$, then $(V/c)=\frac{k^2-1}{k^2+1}$.
If $V$ were not arithmetically-nice, the diamonds along the diagonal would not be as easy to draw by hand on the rotated graph paper.

Checkpoint:
By the principle of relativity, Green should be able to say similar things about the lab-frame.
If Green followed the lab frame's procedure for making a radar measurement,
Green would get radar-times -16 and -4, as the lab frame did.
Green would make the same coordinate assignments [which could be adjusted for signs when they argee on what is "the positive direction"].
In particular, taking "to the right as positive",
Green says that the lab-frame has velocity +(3/5)c according to Green and has time-dilation factor $\gamma=\frac{10}{8}$ and (compared to Green's grid of diamonds) $k=2$.
Further,
note that while the events on segment MS are simultaneous for the lab-frame,
they are not simultaneous for the Green frame.

---

Step 4:
Green's measurement of Blue's velocity

By trial and error, maybe guided by the fact that "composition of velocities" is encoded in the multiplication of Doppler factors (so $k_{\scriptsize{\rm Blue, wrt\ Green}}=(2)(2)=4$ and thus $V_{\scriptsize{\rm Blue, wrt\ Green}}=\frac{4^2-1}{4^2+1}=\frac{15}{17}\approx 0.882$), we find that
Green can do a radar measurement of Blue using signals JP and PY.
Thus, Green uses radar times $t_J=-32$ and $t_Y=-2$.
Thus, $x_P=\frac{1}{2}((-2)-(-32))=15$(which must be adjusted to $-15$ for the common direction of positive) and $t_P=\frac{1}{2}((-2)+(-32))=-17$.
In the end, Green measures the velocity of Blue to be $V_{\scriptsize{\rm Blue, wrt\ Green}}=+\frac{15}{17}$.
(As a check, the time-dilation factor should be $\frac{17}{8}= 2.125$. Try the textbook time-dilation formula. Also check that $(17)^2-(15)^2=(8)^2$.)
Furthermore, Green says Blue will travel from P to Z with velocity (15/17)c in 17-Green-ticks, covering the 15 space-ticks from event K on Green's worldline and event P on Blue's worldline, which Green says are simultaneous.

---

Step 5:
Blue's measurement of Green's velocity

You can follow the procedure in the preceding step to see that
Blue says essentially the same thing about Green and about the lab-frame.

Note that there are three reference frames with their tickmarks shown on this one spacetime diagram...
and you really don't need a calculator (given the arithmetically-nice velocities).
You just need a sheet of graph paper and practice using and interpreting the methods shown here.