# Speed of two spaceships

1. Aug 12, 2004

### niko2000

Hi,
I am doing exercises regarding relativity. I know how to calculate relative speed in one dimension, but I have some problems with the relative speed in 2D.
If we have two spaceships, one is moving in the x direction with speed 0.7c and one is moving in y direction with speed 0.85c.
What is the speed of second spaceship as it has been seen by first pilot?
I have tried to do a geometric sum, but I don't know how to calculate gamma.
Regards,
Niko

2. Aug 12, 2004

### jcsd

Go back and look at the Lorentz transformation.

Also I should say: the questuon is also slightly uncler: i.e. in which frame of refernce are the two speeds orginally measured?

Last edited: Aug 12, 2004
3. Aug 12, 2004

### jcsd

Infact that might not be the best way to do it, if you can look at the four vector velcoity, but make sure you remember which direction is which.

4. Aug 12, 2004

### zefram_c

Here's another way to do it. It's not particularly insightful, but it will get the job done and it's pretty hard to screw up. Start your work in the frame where both ships are moving, one in the x and another in the y direction. Eg. suppose they both start from the origin at t=0 to simplify things. Now after some time passes (1sec?), find their new positions in this frame. Put that time together with the three spatial coordinates in four vector style, and now you have three points in spacetime (one at the origin,t=0 and one for each spacecraft, wherever, t=1sec). Now act with the Lorentz transform matrix to change to either craft's frame of reference (hereafter 'this craft'). You get new four-vectors. This will net you the (spatial) displacement of the other craft as well as the time elapsed for this craft in its own frame. As a bonus you also get the time elapsed on the other craft as seen by this craft, but to get the speed of the other craft you need to divide its total displacement by the time elapsed in the frame of this craft.

5. Aug 12, 2004

### VantagePoint72

Some time ago I derived the following alteration on w = (u + v)/(u*v/c*c + 1) for the situation you described. I don't believe it's sensitive to whether or not the ships start at the same origin. It uses sine and cosine law and looks like a mess (I had no outside resources for this, so as far as relativity equations go it's rather unorthodox and due to a half-job of condensing it there are some rather arbitrary-looking values), so if you'd like to see the explanation PM me or say so here. I imagine there is much more concise way of doing this, but it works. I couldn't get latex to work for this, so I'll type it out, breaking it into small pieces.

Where u and v are the ships' velocities and angles are expressed in degrees:

w = sqrt(u^2 + v^2) / 1 + xy/c^2
where:
x = u·sin(45) / sin(135-z)
y = v·sin(45) / sin(135-z)
where:
z = arccos( ( u+v - sqrt(u^2 + v^2) ) / 2uv )

Let me know what the result is with that. Anyone want to take the challenge and convert that to one big latex equation?

Last edited: Aug 12, 2004
6. Aug 12, 2004

### VantagePoint72

That should work with non-perpendicular velocities too, as long as you make it even messier with components. And I calculated the result: 0.816c
Someone feel free to check me on that with this or another method.

7. Aug 13, 2004

### Chronos

Since the x and y axis are perpendicular, by definition, the observed velocity is easy to derive. It is the hypoteneuse of a right triangle with a relatavistic conversion factor. 0.816c is correct by my math.

8. Aug 13, 2004

### pervect

Staff Emeritus
I'm not getting the same answer. (I could have screwed something up). But I don't think that the sum should be lower than one of the two original velocities.

What I did is to represent the four-velocity of the ship moving in the y direction as

$$\begin{array}{cccc} \gamma_{y} & 0 & v_{y} \gamma_{y} & 0 \end{array}$$

then did a Lorentz boost in the x direction

$$t' = \gamma_{x}(t - v_{x}x) \hspace{.5 in} x' = \gamma_{x}(x - v_{x}t)$$

to get the resulting 4-velocity

$$\begin{array}{cccc} \gamma_x \gamma_y & -v_{x} \gamma_{x} \gamma_{y} & v_{y} \gamma_{y} & 0 \end{array}$$

The interesting point is that the time coefficient is the product of the two gamma's

If you don't care about the angle, you can solve for the velocity that has this gamma

$$v_{tot} = \sqrt{1-\frac {1} {\gamma_{x}^ 2 \gamma_{y} ^2}$$

Last edited: Aug 13, 2004
9. Aug 13, 2004

### niko2000

The result in a book is equal 0.93c

10. Aug 13, 2004

### DW

The internet really isn't the best place to go look for help with your homework.
$$u'_{y} = \frac{u_{y}}{\gamma (1 - \frac{vu_{x}}{c^2})}$$
$$u'_{y} = \frac{u_{y}\sqrt{1 - \frac{v^2}{c^2}}}{(1 - \frac{vu_{x}}{c^2})}$$

$$u'_{y} = \frac{0.85c\sqrt{1 - (0.7)^{2}}}{(1 - (0.7)(0))}$$
$$u'_{y} = 0.61c$$
$$u'_{x} = \frac{u_{x} - v}{1 - \frac{vu_{x}}{c^2}}$$
$$u'_{x} = \frac{0 - 0.7c}{1 - (0.7)(0)}$$
$$u'_{x} = - 0.7c$$
$$u' = \sqrt{u'_{x}^{2} + u'_{y}^{2}} = \sqrt{(-0.7c)^{2} + (0.61c)^{2}}$$
$$u' = 0.93c$$

11. Aug 13, 2004

### VantagePoint72

My bad, guess I didn't derive that properly. I did notice later that something wasn't quite right since the sum was lower than one of the velocities. Thanks for the corrections.

12. Aug 13, 2004

### pervect

Staff Emeritus
I also get .9265