# Relativistic kinematics

1. Aug 12, 2010

### PhMichael

http://img842.imageshack.us/img842/9193/38076308.jpg [Broken]

1. The problem statement, all variables and given/known data
The velocity of B w.r.t A is:

$$\vec{v}=0.6c\left ( \frac{\hat{x}-\hat{y}}{\sqrt{2}} \right )$$.

what is $$\alpha$$ (the 90 degrees of the triangle in A's frame) in B's frame?

2. The attempt at a solution

AC and BC will each be lengthened in B's frame, so that:

$$AC' = \frac{1}{\sqrt{1-(v_{x}/c)^{2}}} \cdot 12 = \frac{60 \sqrt{82}}{41}$$

$$BC' = \frac{1}{\sqrt{1-(v_{y}/c)^{2}}} \cdot 12 = \frac{60 \sqrt{82}}{41}$$

where,

$$v_{x}=v_{y}=\frac{0.6c}{\sqrt{2}}$$

But here i'm stuck and can't proceed ... any guidance will be appreciated :)

Last edited by a moderator: May 4, 2017
2. Aug 12, 2010

### vela

Staff Emeritus
In B's frame, the two legs of the triangle will be length contracted, not lengthened. Moreover, it's not a simple scaling of the entire length of each segment since only lengths in the direction of motion are contracted.

3. Aug 12, 2010

### PhMichael

umm, from Lorentz tranf. i get that they're lengthened and i did use the fact that the only legs who are seen "deformed" (+/-) are the ones in the direction of motion, that's why I used the x component of the velocity for AC and the y component for BC.

By lorentz I get:

$$\Delta x' = \gamma_{x} \Delta x$$ , where, $$\Delta x'=AC'$$
$$\Delta y' = \gamma_{y} \Delta y$$ , where, $$\Delta y'=BC'$$

As: $$\Delta t = 0$$

NO?

4. Aug 12, 2010

### vela

Staff Emeritus
Oh, OK, I see what you were doing. By treating the components of the velocity separately, you're essentially looking at a Lorentz boost in the x-direction combined with a separate boost in the y-direction. That combination, however, doesn't yield the same result as a single boost in the direction of B's velocity. You can see this by considering what happens to AB. In the case of a single boost, its length remains unchanged because it's perpendicular to B's velocity. But when you have two separate boosts, both boosts will cause its length to decrease because it's not perpendicular to either component of the velocity.

The other point of confusion comes from applying the Lorentz transformation to find the length of the segments measured by B. You're calculating the spatial distance between two spacetime events when Δt=0, but from B's point of view, those two events occur at different times. You want to calculate the spatial distance when Δt'=0.

5. Aug 13, 2010

### PhMichael

OK, let me see if i've got you correctly:

$$v=|\vec{v}|=0.6c \to \gamma = 1.25$$

$$AB'=AB=12 \sqrt{2}$$

$$AC'=BC'=\frac{12}{\gamma}=\frac{12}{1.25}=9.6$$

and by the law of cosines, we obtain $$\alpha$$ :

$$(12 \sqrt{2} )^{2} = 2(9.6)^{2} - 2(9.6)^{2}cos \alpha$$

$$\alpha = 124.23^{o}$$

Is this correct?

6. Aug 13, 2010

### vela

Staff Emeritus
No. The segments AC and BC don't contract by the factor of γ because they have a component perpendicular to v. Only the component parallel to v becomes shorter.

7. Aug 13, 2010

### PhMichael

Wow, what a messy question ... So I gusee i'll need to describe each leg in the directions parallel and perpindicular to the velocity direction:

$$AB'=AB=12 \sqrt{2}$$

-----------------

$$AC_{\parallel}=6 \sqrt{2} \to AC_{\perp}=6 \sqrt{2}$$

$$AC'_{\parallel}=\frac{AC_{\parallel}}{\gamma}=6.788$$ , $$AC'_{\perp}=AC_{\perp}=6 \sqrt{2}$$

$$AC' = \sqrt{(AC'_{\parallel})^{2}+(AC_{\perp})^{2}}=10.866$$

-----------------

$$BC_{\parallel}=6 \sqrt{2} \to BC_{\perp}=6 \sqrt{2}$$

$$BC'_{\parallel}=\frac{BC_{\parallel}}{\gamma}=6.788$$ , $$BC'_{\perp}=BC_{\perp}=6 \sqrt{2}$$

$$BC' = \sqrt{(BC'_{\parallel})^{2}+(BC_{\perp})^{2}}=10.866$$

------------------

and then by using the law of cosines, i get:

$$\alpha = 102.68$$

Is it now right?!

Last edited: Aug 13, 2010
8. Aug 13, 2010

### vela

Staff Emeritus
How'd you get 9.6 for the parallel components of AC' and BC'? I don't get that number.

As a sanity check, do you expect the angle α to be bigger or smaller as observed by B given how you expect the triangle to change shape?

9. Aug 13, 2010

### PhMichael

I got it by dividing 12 over $$\gamma = 1.25$$.

I think it should be larger than 90 degrees so there must be something messed up.

10. Aug 13, 2010

### PhMichael

I've screwed up ... i should've divided that parallel component by gamma!!

11. Aug 13, 2010

### PhMichael

I've just edited the calculations ... can you please go over them quickly and let me know if they now seem OK ?

12. Aug 13, 2010

### vela

Staff Emeritus
Looks good!

13. Aug 13, 2010

thanks pal!