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

[Acodato]

TL;DR Summary

An inertial observer measures two spaceships on a collision course each one traveling at 60% the speed of light. From inside the spaceship can they measure how much time they have to avoid a collision?

Relativity always talks in terms of observer but fail to explain common sense problems like this. The fact that no one can measure anything faster than light does not preclude to objects moving towards each other at a higher velocity than an external observer can measure. Move the observer to one of the spaceships and now the limits of observation will render every onboard equipment useless to calculate actual time to impact.

 

[kuruman]

Move the observer to one of the spaceships and now the limits of observation will render every onboard equipment useless to calculate actual time to impact.

How do you figure? Do you understand relativistic addition of velocities?

 

[Acodato]

How do you figure? Do you understand relativistic addition of velocities?

For the observer inside the spaceship there is no relativity. His frame of reference is inert for every purpose. Nothing to add or subtract. Just unable to measure an incoming ship at 120% the speed of light.

 

[PeterDonis]

Just unable to measure an incoming ship at 120% the speed of light.

The incoming ship is not traveling at 120% the speed of light relative to the observer inside the spaceship. That was the point of the question @kuruman asked you in post #2.

 

[kuruman]

Just unable to measure an incoming ship at 120% the speed of light.

But the incoming ship will not be traveling at 120% of the speed of light. It will be traveling at a speed $v=\dfrac{0.6+0.6}{1+0.6^2}c =88\%~$ of the speed of light. That is relativistic addition of velocities.

 

[Dale]

Relativity always talks in terms of observer but fail to explain common sense problems like this

Relativity does not fail to explain this. Your lack of awareness of an explanation doesn’t mean one doesn’t exist.

As others have mentioned, this is explained by relativistic velocity addition. This is a very well known phenomenon.

 

[Acodato]

The incoming ship is not traveling at 120% the speed of light relative to the observer inside the spaceship. That was the point of the question @kuruman asked you in post #2.

If you are inside the spaceship, you agree with me you have no information about the relative motion of the two objects? Your measures will probably say it is moving 99.99% C but it will be wrong. The external observer can actually calculate time to impact. Each ship by itself can't.

 

[Orodruin]

If you are inside the spaceship, you agree with me you have no information about the relative motion of the two objects? Your measures will probably say it is moving 99.99% C but it will be wrong. The external observer can actually calculate time to impact. Each ship by itself can't.

It is clear from your posts here that you simply do not understand how the relativistic addition of velocity works. As you have already been told, you can very well measure the other spaceship's velocity relative to you. The computation was performed for you in post #5 with the result ca 0.88c. A light signal will travel faster than that so there is no issue here. Each ship by itself is perfectly capable of computing the time (in their frames) to impact. You are inventing a problem where there is none. This is extremely well known physics.

 

[PeterDonis]

If you are inside the spaceship, you agree with me you have no information about the relative motion of the two objects?

No. I disagree. If the other ship is emitting light signals at you, you can use their Doppler shift to measure its speed towards you. Which will be, as @kuruman calculated, 88% of the speed of light.

Your measures will probably say it is moving 99.99% C

No, 88%. See above.

it will be wrong.

No, you are wrong. See above.

The external observer can actually calculate time to impact. Each ship by itself can't.

The external observer has to have information about the distance between the ships as well as speed to calculate the time of impact. There's no reason why an observer on board one of the ships could not have distance information as well. And that, combined with the information about relative speed, does indeed allow an observer on board one of the ships to calculate the time to impact by their clock.

In other words, you are simply mistaken about what relativity can and cannot allow people to calculate.

 

[Acodato]

But the incoming ship will not be traveling at 120% of the speed of light. It ill be traveling at a speed $v=\dfrac{0.6+0.6}{1+0.6^2}c =88\%~$ of the speed of light. That is relativistic addition of velocities.

Only the observer outside the spaceships has this information. No one inside can make this calculation: there are no 2 speeds from the inertial frame of reference of the spaceship.

 

[Orodruin]

Only the observer outside the spaceships has this information.

If you don’t allow the crew of the spaceship to make their own measurements, then this setup doesn’t need relativity to result in an impossibility to determine the collision time.

Otherwise your assertion is simply false. If an observer in one of the spaceships makes a measurement of the other spaceship’s velocity relative to themselves, they will obtain the result 0.88c - as described above. Not more, not less.

No one inside can make this calculation: there are no 2 speeds from the inertial frame of reference of the spaceship.

it is impossible to try to parse what you actually mean by this. All objects have easily calculable velocities in all frames. There certainly are two different speeds relevant here, the speed of the first ship relative to the original inertial frame, and the speed of the second relative to the original inertial frame.

But more to the point: They do not need to have this information. They can simply measure the speed of the other ship as described in post #9. The result will be 0.88c.

Which will be, as @Orodruin told you, 88% of the speed of light.

Just to direct credit correctly: The computation here was provided by @kuruman in post #5. I merely quoted the result.

 

[berkeman]

If you are inside the spaceship, you agree with me you have no information about the relative motion of the two objects? Your measures will probably say it is moving 99.99% C but it will be wrong.

Maybe I can help.

-Maverick- "Help me out here Goose!"

-Goose- "I got .88c closure on radar, Mav"

-Maverick- "Fox-3!"

 

[Vanadium 50]

IBTL.

What's magic about 60%? Why not 6%? Or 0.000006%? How can people walking towards each other possibly avoid a collision?

 

[berkeman]

How can people walking towards each other possibly avoid a collision?

Fox-3!

Oops, sorry, that would be anti-social. My bad.

 

[PeterDonis]

Only the observer outside the spaceships has this information.

Why not? Don't the spaceships have sensors?

No one inside can make this calculation

Why not? Don't they know how to do math?

there are no 2 speeds from the inertial frame of reference of the spaceship.

Sure there are: zero (the speed of that spaceship) and 88% of the speed of light (the speed of the other spaceship in that spaceship's frame). Which is quite sufficient.

 

[PeterDonis]

@Acodato where are you getting your information on Special Relativity? So far you appear to have nothing but misconceptions.

 

[PeterDonis]

Just to direct credit correctly: The computation here was provided by @kuruman in post #5. I merely quoted the result.

Yes, thanks for the correction! I have edited my post.

 

[Vanadium 50]

Fox-3!

An armed society is a polite society.

So far you appear to have nothing but misconceptions.

While I agree, let's see if the "why 60%" question will illuminate where the difficulty lies,'

 

[FactChecker]

Only the observer outside the spaceships has this information. No one inside can make this calculation: there are no 2 speeds from the inertial frame of reference of the spaceship.

You are missing the point. From either of the two spaceships, no calculation is needed. In the physics of his inertial reference frame, the other spaceship is moving toward him at 0.88 c. If he measures its approach speed, that is what he will get. It's as simple (for him) as that.

 

[kuruman]

While I agree, let's see if the "why 60%" question will illuminate where the difficulty lies,'

I am sure that @Acodato is aware that two cars traveling towards each other at 15 miles an hour can easily avoid a collision. If it is true that two spaceships traveling towards each other at 0.6 c cannot avoid a collision, then it must be true that somewhere between 15 mph and 0.6 c there is a threshold speed that separates collision avoidance from collision non-avoidance. Perhaps @Acodato can tell how this threshold speed is determined.

 

[phinds]

@Acodato you should consider the first rule of holes: when you find yourself in one, stop digging.

 

[Acodato]

IBTL.

What's magic about 60%? Why not 6%? Or 0.000006%? How can people walking towards each other possibly avoid a collision?

My point is on non-relativistic notions just as you mentioned. If one spaceship is 1.2 light years apart from a planet, how long does is take for a 0.6C spaceship to reach the planet? 2 years. Now if another spaceship is launched from the planet at 0.6C how long until they meet (collide)? 1 year. Because it is what it is. Now you do the funky mathematics of relativity, doppler, redshift, whatever, and you say they will meet only after 1.4 years due to relativity, that is exactly the inevitable collision.

Why 0.6C? Because any speed below 0.5C could actually be detected.

 

[PeterDonis]

My point is on non-relativistic notions just as you mentioned.

Non-relativistic physics is not correct; relativistic physics is. That has been shown by many, many experiments over the past century and more. So if you are reasoning from non-relativistic notions, to the extent those notions contradict relativity, you are reasoning from factually wrong premises.

If one spaceship is 1.2 light years apart from a planet, how long does is take for a 0.6C spaceship to reach the planet? 2 years. Now if another spaceship is launched from the planet at 0.6C how long until they meet (collide)? 1 year.

These times are times relative to an observer at rest with respect to the planet.

Now you do the funky mathematics of relativity, doppler, redshift, whatever, and you say they will meet only after 1.4 years due to relativity

Who has said that? Have you done a relativistic calculation? In what frame is this claim supposed to be made? Not in the frame of the observer above (at rest relative to the planet); for that observer the time for the spaceships to meet is 1 year. That is what relativity says.

 

[Acodato]

Non-relativistic physics is not correct; relativistic physics is. That has been shown by many, many experiments over the past century and more. So if you are reasoning from non-relativistic notions, to the extent those notions contradict relativity, you are reasoning from factually wrong premises.


These times are times relative to an observer at rest with respect to the planet.


Who has said that? Have you done a relativistic calculation? In what frame is this claim supposed to be made? Not in the frame of the observer above (at rest relative to the planet); for that observer the time for the spaceships to meet is 1 year. That is what relativity says. But in the frame of either spaceship, it will take less than 1 year for them to meet, not more. That is what relativity says.

If you agree from the observer at rest it will take one year to meet half way then they are traveling at 1.2C. Observer can measure 0.6 and 0.6. The spaceship can't measure 1.2C. That is all light is: a measure.

 

[berkeman]

The spaceship can't measure 1.2C. That is all light is: a measure.

Did you read the link supplied early in your thread?

That is relativistic addition of velocities.

If so, what are your thoughts? If not, why not?

 

[PeterDonis]

If you agree from the observer at rest it will take one year to meet half way then they are traveling at 1.2C.

More precisely, their "closure rate" in the frame in which the planet is at rest is 1.2C.

Observer can measure 0.6 and 0.6.

Yes.

The spaceship can't measure 1.2C.

The relative speed of the other spaceship in that spaceship's frame is not 1.2C. It is 0.88C. The spaceship can measure that, as has already been pointed out.

If all you are going to do is keep repeating the same wrong claims, this thread is going nowhere and will be closed.

 

[Acodato]

More precisely, their "closure rate" in the frame in which the planet is at rest is 1.2C.


Yes.


The relative speed of the other spaceship in that spaceship's frame is not 1.2C. It is 0.88C. The spaceship can measure that, as has already been pointed out.

If all you are going to do is keep repeating the same wrong claims, this thread is going nowhere and will be closed.

I already said, at 0.88C the spaceship calculates about 1.4 years to collide. Can you prove this wrong?

 

[PeterDonis]

I already said, at 0.88C the spaceship calculates about 1.4 years to collide.

How did you come up with this number?

 

[phinds]

But in the frame of either spaceship, it will take less than 1 year for them to meet, not more. That is what relativity says.

I already said, at 0.88C the spaceship calculates about 1.4 years to collide. Can you prove this wrong?

HE JUST DID (well, he didn't show you the math but he could). You aren't listening to what people are telling you. You just can't stop digging, can you?

 

[Acodato]

How did you come up with this number?

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

 

[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.