Why won't Spaceships at Right Angles Collide in Frame S1?

[daniel_i_l]

In frame S there're two spaceships, s1 and s2. s1 starts at (L,0) with a velocity of (-v,0) and s2 starts at (0,-L) with a velocity of (0,v). Obviously they'll collide in this frame.
But let's look at things from a frame traveling with s1 (S1). In S1, s1 has a position of
(L/G, 0) and a speed of 0 and s2 has a position of (0, -L) and a velocity of (v, v/G). [G is gamma]
In this frame s2 won't hit s1 because s2's velocity isn't in the direction of s1-s2.
How can this be?
Thanks

 

[vin300]

In S1, s1 has a position of
(L/G, 0) and a speed of 0 and s2 has a position of (0, -L) and a velocity of (v, v/G). [G is gamma]
In this frame s2 won't hit s1 because s2's velocity isn't in the direction of s1-s2.
How can this be?
Thanks

That part is wrong.It is (v,v).S1 and S2 collide at a time in S1's frame equivalent to the initial frame, L/v secs.In S1's frame, distance traveled by L2 is L(sqrt.2) and velocity is v(sqrt.2).
An observer does not measure contraction in his own frame

 

[daniel_i_l]

That part is wrong.It is (v,v)

We're talking about relativistic speeds. So both the x and y coordinates of the s2 velocity are different in s1's frame.
http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html

 

[vin300]

The uv/c^2 in the denominator is actually a scalar product of u and v, the vector addition of uand v gives v(sqrt.2) as the magnitude of S2's velocity in S1's frame. How did you get v/G anyway?

 

[daniel_i_l]

The uv/c^2 in the denominator is actually a scalar product of u and v where the vector addition of uand v gives v(sqrt.2) as the magnitude of S2's velocity in S1's frame. How did you get v/G anyway?

Here's the formula from the link I posted:

-------------uy
wy = -------------------
--------(1 + uxvx/c2) γ(vx)

In this case, ux = s2x = 0 and so we're left with uy/G = v/G

 

[jambaugh]

[EDIT: Below are corrections to prior post. (I'm an idgit!)]

You are dealing with two dimensional relativistic addition of velocities. I think that will be easiest if we use matrices and work with space-time vectors.

Firstly we need the relativistic space-time velocities for each ship
(four velocities except we're ignoring z):
(Adopt c = 1 units i.e. v in light-seconds per second and L in light-seconds)

ERROR:$$U_1 =\gamma^{-1} (1;- v,0),\quad U_2 = \gamma^{-1}(1;0,v)$$
EDIT: above should read: $$U_1 = \gamma(1;-v,0), \quad U_2 = \gamma(1;0,v)$$

Let v = tanh(b) to define the boost parameter b and note that ERROR: gamma = sech(b) and v/gamma = sinh(b).
EDIT: Above should read gamma = cosh(b) and gamma v = sinh(b).

The gamma factor for the spatial component is due to $U_x = \frac{dx}{d\tau}$ not $\frac{dx}{dt}$.
Note also that I am separating the time component with a semi-colon and that the time component $\frac{dt}{d\tau}$ is one over the gamma factor for that velocity.


Note then in terms of hyperbolic trig we have:
$$U_1 = (\cosh(b);-\sinh(b),0),\quad U_2 = (\cosh(b);0,\sinh(b))$$
Which means these are time-like unit vectors in Minkowski space-time since:
$$\cosh^2(b) - \sinh^2(b) = 1$$

Now with these four velocities we can apply a Lorentz transformation so that U1=(1;0,0)
i.e. so that s1 is only traveling through time and has no spatial velocity (we're in s1's frame).

The boost matrix will be:
$$T=\left(\begin{array}{ccc}\cosh(b) & \sinh(b) & 0 \\ \sinh(b) & \cosh(b) & 0 \\ 0 & 0 & 1 \end{array}\right)$$

With column vectors we would multiply on the left by T but with row vectors we can multiply from the right by the transpose (same thing in this case since boost matrices are symmetric):
$$T[U_1]=(\cosh(b);-\sinh(b),0)\left(\begin{array}{ccc}\cosh(b) & \sinh(b) & 0 \\ \sinh(b) & \cosh(b) & 0 \\ 0 & 0 & 1 \end{array}\right) = (\cosh^2(b) - \sinh^2(b); 0,0) = (1;0,0)$$

$$TU_2=(\cosh(b);0,\sinh(b))\left(\begin{array}{ccc}\cosh(b) & \sinh(b) & 0 \\ \sinh(b) & \cosh(b) & 0 \\ 0 & 0 & 1 \end{array}\right) = (\cosh^2(b);\sinh(b)\cosh(b),\sinh(b))$$
The spatial velocity of s2 as seen by s1 is then:
$$v'_2 = \left( \frac{sinh(b)\cosh(b)}{\cosh^2(b)},\frac{\sinh(b)}{\cosh^2(b)}\right)<br /> = \text{ERROR:} (\tanh(b),\gamma \tanh(b) ) = (v,\gamma v)$$

EDIT: last part of above should read $$=(\tanh(b),sech(b)\tanh(b) ) = (v,\gamma^{-1} v)$$

Note that from s1's perspective the (L,0) distance is shortened by gamma so this velocity is from s2's position straight at s1. What is happening is that the y-component of s2's motion as seen in the original frame is slowed due to time dilation. This compensates for the length contraction of the distance to keep the velocities colinear with the collision event.

Working it out this way we can also determine the magnitude of the velocity of s2 as seen by s1 and thus the relative time dilation and the time to collision. It may seem more complex but that's because the problem is complex. With this method you can consider s2 traveling at an arbitrary angle on a collision course with s1.

You can of course also apply the transformation matrix to the position vectors by adding a t=0 component to make them space-time positions. Even better is to set t=-L/v so that at t=0 is when the collision occurs. You'll then have the time to collision automatically in the transformed starting events.

EDIT: I made this gamma <--> 1/gamma error because I generally never use gamma but stick to the hypertrig when doing problems. It is so much easier once you get the hang of it.

 

[jambaugh]

[EDIT! I screwed up mixing up gamma and 1/gamma. Edits below are corrections.]
I should elaborate on the hyperbolic trig identities:
$$\cosh^2(b) - \sinh^2(b)=1$$
divide through by cosh^2 to get:
$$1 - \tanh^2(b) = \text{sech}^2(b)$$
You then have:
$$\beta=v/c = \tanh(b)$$
(which you get by boosting a stationary object by boost matrix T(b) and seeing that its velocity will then be v). Applying hyper-trig identities you get:

$$\gamma=\sqrt{1-v^2/c^2}= \sqrt{1-\tanh^2(b)} = \text{sech}(b)$$
EDIT: Correction
$$\gamma^{-1} = ... = \text{sech}(b)$$
END EDIT

$$\frac{1}{\gamma} = \cosh(b)$$
EDIT: Correction
$$\gamma = \cosh(b)$$
END EDIT

$$\frac{v}{c\gamma} = \tanh(b)\cosh(b) = \sinh(b)$$
EDIT: Correction
$$\frac{\gamma v}{c} = \tanh(b)\cosh(b) = \sinh(b)$$
END EDIT

Choosing c=1 units simplifies things. Using hyperbolic trig is then easy to understand non-linear addition of velocities since it is the boost parameters which add when you compose two boosts.

If observer 1 is traveling at velocity v1 and sees object 2 traveling at v2' both in the x-direction then we see object 2 traveling at v2 where:
$$v_1 = \tanh(b_1)$$
$$v'_2 = \tanh(b_2)$$
$$v_2 = \tanh(b_1 + b_2)$$

When composing velocities in different directions it the same issue as rotations about different axes. You have to compose the actions in matrix form to see what you get.

 

[daniel_i_l]

Thanks for the reply. I have a few questions.

$$U_1 =\gamma^{-1} (1;- v,0),\quad U_2 = \gamma^{-1}(1;0,v)$$

I don't understand why you need gamma before doing any transformations.

The spatial velocity of s2 as seen by s1 is then:
$$v'_2 = \left( \frac{sinh(b)\cosh(b)}{\cosh^2(b)},\frac{\sinh(b)}{\cosh^2(b)}\right)<br /> = (\tanh(b),\gamma \tanh(b) ) = (v,\gamma v)$$]

How can this be? What if $$v = \frac{c}{\sqrt{}{2}}$$? Then s1 would see s2 moving at a speed greater than c
I haven't learned about boost matricies yet. Are you saying that the transformations on the site I posted are wrong? If so, can you explain why?
Thanks

 

[jambaugh]

Thanks for the reply. I have a few questions.


I don't understand why you need gamma before doing any transformations.

You don't really for the problem but doing so gives you the correct gamma factor for the transformed velocity (as reciprocal of the time component.)

How can this be? What if $$v = \frac{c}{\sqrt{}{2}}$$? Then s1 would see s2 moving at a speed greater than c
I haven't learned about boost matricies yet. Are you saying that the transformations on the site I posted are wrong? If so, can you explain why?
Thanks

Let me triple check my work. I was working it as I typeset it and though I'd double checked.

Hmmm... (again letting c = 1 by choice of units)
EDIT: The following is wrong! I'm using the wrong definition of gamma, which I notice right about ...
$$|(v,\gamma v)|^2= v^2 + \gamma^2 v^2 = v^2(1+ 1-v^2) = v^2(2-v^2)$$

Let $$p=v^2$$
then
$$q = p(2-p) = 2p - p^2$$
defines a downward parabola q=f(p) with vertex at p=1. The vertex point is (1,1) so when v = 1 the magnitude of (v, gamma v) is 1 and when v < 1 the magnitude is less than 1.

With your example: (with c=1)
$$v = \frac{1}{\sqrt{2}}$$
$$\gamma = \sqrt{1 - v^2} = \sqrt{1 - 1/2} = \frac{1}{\sqrt{2}}$$

so the transformed velocity will be:
$$(v,\gamma v) = \left( \frac{1}{\sqrt{2}},\frac{1}{2}\right)$$
The magnitude is:
$$|(v,\gamma v)| = \sqrt{ \frac{1}{2} + \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \simeq 0.866$$
or 86.6 percent the speed of light.
EDIT ... Now!
AhhhhhhhhRRRRRG!

Here is where I screwed up. I'm mixing up gamma and its reciprocal.
$$\gamma = \cosh(b)$$!

Let me add an edit to the original posts. Wherever I said gamma replace with 1/gamma and vis versa!
(My apologies, I use hyper-trig now so much I don't bother with the gammas et al just b and cosh sinh et al)

 

[jambaugh]

Thanks for the reply. I have a few questions.
...
I haven't learned about boost matricies yet. Are you saying that the transformations on the site I posted are wrong? If so, can you explain why?
Thanks

Your transformation is correct. If you don't want to use matrices explicitly you can express it component-wise (in c=1 units):
$$t' = \cosh(b)t + \sinh(b) x$$
$$x' = sinh(b)t + cosh(b)x$$
$$y' = y$$
boosting by b in the x direction.
To reverse replace b with -b remembering sinh(-b) = -sinh(b) and cosh(-b) = cosh(b)

Similarly with $$(U_t,U_x,U_y) and (U'_t,U'_x,U'_y)$$

Note that in terms of (the correct) gammas and including c via t-->ct and v--> v/c:
$$ct' = \gamma ct + (v/c)\gamma x$$
$$x' = (v/c)\gamma ct + \gamma x$$
$$y'=y$$

But matrix form is pretty easy to learn and I suggest you pick it up when you have time. It made SR so much easier to understand once I learned to use matrices and hyper-trig. Boosts are pseudo-rotations of the 4-velocity.

One final note. You can also rotate with matrices and if you want to boost in an arbitrary direction, rotate that direction to the x direction boost in the x direction then rotate back.
$$B = R^{-1} B_x R$$
Just work out the matrix and you're set to go. At least, hopefully, you can see how in principle you would go about it with matrices.

 

[daniel_i_l]

EDIT: last part of above should read $$=(\tanh(b),sech(b)\tanh(b) ) = (v,\gamma^{-1} v)$$

Note that from s1's perspective the (L,0) distance is shortened by gamma so this velocity is from s2's position straight at s1. What is happening is that the y-component of s2's motion as seen in the original frame is slowed due to time dilation. This compensates for the length contraction of the distance to keep the velocities colinear with the collision event.

If the y component of s2's velocity is less than the x component, and the y component of the distance from s2 to s1 bigger than the x component - then how do they collide?
Thanks

 

[Ich]

EDIT: nonsense deleted, nothing left.

 

[MikeLizzi]

A collision as determined by an observer in one reference frame must always happen when observed from any other. But relativistic calculations in 2 or 3 dimensions are tricky as you can tell by the complex math being posted. For that reason I created a 3D simulation of a common "paradox" so students can "see" the results of the calculations.

If your browser can run a Java Applet, may I suggest you try running it. The tutorial to guide you is at:

http://mysite.verizon.net/mikelizzi/TutorialMeterstickAndHole.htm

Time out. Sun Micosystems apparently changed the location if certain Java certificates. If your browser is set to run only if the applet is certified you will have problems.

 

[jambaugh]

If the y component of s2's velocity is less than the x component, and the y component of the distance from s2 to s1 bigger than the x component - then how do they collide?
Thanks

The y component of the velocity is bigger! v/gamma > v. Remember gamma < 1 when v>0.

From s1's perspective the x direction has been shortened relative to the original setup via length contraction so s1 sees himself as at position $$(\gamma L,0)$$ when he "starts" (as far as he's concerned he just sits there).

The y distance is unchanged via boosting in the x-dir so s1 still sees s2 at $$(0,-L)$$ when s2 starts.

The difference in their spatial coordinates is then:
$$(\gamma L,0)-(0,-L) = (\gamma L,L) = \frac{\gamma L}{v}(v,\gamma^{-1}v)$$

So you see the velocity as seen by s1 is a multiple of the differences in positions as seen by s1.

(I think you might have made my error of mixing up gamma and 1/gamma in terms of "bigger" and "smaller")
[EDIT: Sorry! If I'd parsed your original post more carefully we'd have seen that this is the issue sooner. I assumed you must have made an error in calculation and so I reworked it (with my own error)... so now here we are. ]

 

[vin300]

The velocity addition formula for perpendicular velocities is
S=v +u[sqrt(1-v^2/c^2] {v,u and S vectors}
because u(dot)v is zero and
the velocity of projection of a body (frame 1) from another body(frame 2) moving with velocity v w.r.t a stationary observer is u, but u is measured in frame two, not by the stationary and frame 2 is time dilated according to the stationary observer. Because velocity of frame 1 measured by frame 2 increases due to time dilation of frame two by factor (1-v^2/c^2)^-1/2 according to the stationary, u has to be divided by that factor to get the velocity of projection measured by the stationary,i.e u[sqrt(1-v^2/c^2)]
In the case we're discussing, the velocity S in the above formula is (v, v/Gamma)

 

[daniel_i_l]

The y component of the velocity is bigger! v/gamma > v. Remember gamma < 1 when v>0.

From s1's perspective the x direction has been shortened relative to the original setup via length contraction so s1 sees himself as at position $$(\gamma L,0)$$ when he "starts" (as far as he's concerned he just sits there).

The y distance is unchanged via boosting in the x-dir so s1 still sees s2 at $$(0,-L)$$ when s2 starts.

The difference in their spatial coordinates is then:
$$(\gamma L,0)-(0,-L) = (\gamma L,L) = \frac{\gamma L}{v}(v,\gamma^{-1}v)$$

So you see the velocity as seen by s1 is a multiple of the differences in positions as seen by s1.

Just to clarify, in my previous post I was defining $$\gamma$$ as :
$$\gamma = \frac{1}{1 - v^2/c^2}$$
By that definition the transformations give s2's velocity in s1's frame as
$$(v, \gamma^{-1}v)$$ just like vin300 posted.
so the y component in smaller than the x component.

It's impossible for the x component to be v and the y component to be bigger than v because then the speed would bigger than c if v would be bigger than $$c/ \sqrt{2}$$

So the y component is smaller but the y distance is bigger. So I still don't see how they collide.

 

[jambaugh]

OK. I think I've figured out the problem here. I did not appreciate the subtlety of this thought experiment until now.

It appears that the distance from s1 to the origin has grown instead of shrunk as you think it should under Lorentz contraction. But in fact we are considering not spatial points but space-time events.

--There is the origin event at the original spatial origin and at the time that the original observer see's both s1 and s2 pass their flags.
--There is the distinct event where&when s1 passes his starting point (the original observer sees a distance L from the origin),
-- There is the distinct event where&when s2 passes her starting point (the original observer sees a distance L from the origin),
and finally
--There is the collision event.

As I mentioned in an earlier post the best space-time origin to pick should be the one corresponding to the collision. I will rework the problem in that frame but for now let's see if we can figure out what is going on in the frame we used in previous posts. Let us indicate observers O at the origin in the original problem description and the second observer O' we get when we boost to be co-moving with s1 but who is passing the O observer's origin at time t=t' = 0!

Let us add a flag f1 at S1's position and at the time t=0. (in the O frame) While we are at it let's add a flag f2 at S2's position at this time. The flag's are stationary in the O frame but note they are moving in the O' frame.

From the O' observer's perspective at t'=0 he is passing the O observer. However O' sees the flag f1 moving in the x direction at speed v but it has not yet crossed paths with the ship s1.

It is incorrect to naively apply a Lorentz contraction to the distance between the O observer's origin and f1 and assume this to be the distance between the O' observer's origin and s1. Remember the Lorentz contraction applies to the spatial separation of two parallel world-lines (ends of a measuring rod) as seen by two observers. Where it applies here is:

O sees the distance between O' and s1 Lorentz contracted relative to the proper distance O' sees.

O' sees the distance between O and f1 Lorentz contracted relative to the proper distance that O sees.

But note that O sees the Lorentz contracted distance between O' and s1 as equal to the proper distance between O and f1. Thus the proper distance between O' and s1 must be the reciprocal of the Lorentz contraction factor times the proper distance between O and f1.

I'm going to outline the details again and to avoid (my) confusion and I'll not refer to gamma's at all since I keep mixing things up. Where desired you may interject $$\gamma = \cosh(b)$$ and $$\gamma^{-1} = \text{sech}(b)$$.

In the O frame s1 and s2 have respective velocities:
$$V_1=c\cdot(-\tanh(b),0)$$ and $$V_2=c\cdot(0,\tanh(b))$$
In the O' boosting s1 and s2 have respective velocities:
$$V'_1 = (0,0)$$ and $$V'_2 = (v,\text{sech}(b) v)$$
In the O frame the flag crossing space-time events occur at:
$$X_1 = (0;L,0)$$ and $$X_2=(0;0,-L)$$
In the O' frame the flag crossing events occur at:
$$X'_1 = (\sinh(b)L/c;\cosh(b)L,0)$$ and $$X'_2=(0;0,-L)$$

Since in the second frame s1 is not moving his spatial position at any time in this frame will be: $$R'_1 = (\cosh(b)L,0)$$

However the time coordinate for the event when s1 and f1 pass is now bigger than zero. This is because in this moving frame our events are no longer simultaneous. This event as seen in the primed frame occurs after the event where s2 and f2 cross paths and where O' and O cross paths.

Now parametrically express s2's position over time in the primed frame:
$$R'_2(t') =R'_2(0)+t'\cdot V'_2 = (vt',\text{sech}(b)v t' - L)$$

When $$t' = \cosh(b)L/v$$ s2 will have position:
$$R'_2=\left(v\cdot\cosh(b)L/v,\text{sech}(b)v\cdot \cosh(b)L/v\right)=(\cosh(b)L,0)=R'_1$$.

In short, when $$t' = \cosh(b)L/v$$ BOOM!

Now as to the speed of s2 in the primed frame it is necessarily less than c. In particular the square speed is:
$$|V'_2|^2 = c^2 p(2-p)$$
Where $$p = \frac{v^2}{c^2}=\tanh^2(b)$$.
This is a quadratic function with max value c^2 when p = 1 i.e. when v<c the speed in the primed frame is less than c.

Using the collision event as the origin:
[if you please let me again use c=1 units.]
Let the collision event be the space-time origin for all observers. In the original observer's inertial frame O (and still using the flags above) we now set our clocks so that the collision will occur at t=0 x=0,y=0 which will be a period of L/v after each ship passes their flags as seen by the O observer.

Define flag passing events for each ship:
$$X_1 = (-L/v;L,0)$$ and$$X_2 = (-L/v;0,-L)$$

Boost to an observer O* with origin at the same collision event but co-moving with S1. (He is actually S1 using the collision event as t"=0).

$$X^*_1 = \left( (-L/v) \cosh(b)+L\sinh(b);(-L/v)\cdot\sinh(b) + (L) \cosh(b), 0 \right)$$ and $$X^*_2 = \left((-L/v)\cosh(b); (-L/v)\sinh(b),-L\right)$$

(Note that s2's flag event is now shifted in the x-direction! This is where in the earlier analysis we get that extra distance due to simultaneity issues. This is the distance s2 must travel in the x direction and it is now clearer that its value has nothing to do with Lorentz contraction but rather with relative motion and time.)

Now in c=1 units $$v = \tanh(b)$$ so we get $$\sinh(b)/v = \cosh(b)$$.

$$X^*_1 = L(\sinh(b)- \cosh^2(b)/\sinh(b);-\cosh(b) + \cosh(b), 0)=(-T_1;0,0)$$
and
$$X^*_2 = L(-\cosh^2(b)/\sinh(b); -\cosh(b),-1) (-T_2;-\cosh(b)\cdot L,-L)=$$

Here T1 and T2 are the O* durations between the common collision event and when the ships cross paths with their respective flags.

The velocities in the O* (s1's) frame are the same as earlier in the O' frame since they are co-moving.
$$V^*_1=V'_1 = (0,0)$$ and $$V^*_2=V'_2 = (v,\text{sech}(b) v)$$

Our boosted velocity for s2 is parallel and opposite the prior location of s2 so it is moving right at the origin.

We can parametrize s2's space-time coordinates:
$$X^*_2(\Delta t^*) = \left(\Delta t^*-L\cosh^2(b)/\sinh(b); v\Delta t^* - L\cosh(b),\text{sech}(b) v\Delta t" - L\right)$$
with $$t^* = \Delta t^* - L\cosh^2(b)/\sinh(b)$$ we get...

$$X^*_2(t^*) = \left(t^*; vt^* +vL\cosh^2(b)/\sinh(b) - L\cosh(b),\text{sech}(b) vt^* +\text{sech}(b)L\cosh^2(b)/\sinh(b) - L\right)$$
(v = tanh(b) gives:
$$X^*_2(t^*) = \left(t^*; vt^* +L\cosh(b) - L\cosh(b),\text{sech}(b) vt^* +L - L\right)$$
$$= \left(t^*; vt^*,\text{sech}(b) vt^*\right)$$

and clearly at t* = 0 we go BOOM!

 

[daniel_i_l]

Thank you very much. I understand why they collide. And more importantly, I now understand when to apply length contraction and when not to - the proper length between two points is measured in a frame where both of the points are stationary. In all other frames the length will be shorter.