# Basic special relativity problem

• sigma_
In summary, particle 1 moves with a uniform velocity ux along the x-axis and particle 2 moves with a uniform velocity vy along the y-axis in the laboratory frame. To find the velocity of particle 2 in the rest frame of particle 1, the Lorentz transformation must be applied. After solving for the world line of particle 2 in particle 1's rest frame, it can be determined that the velocity of particle 2 with respect to particle 1 is the square root of the sum of the squares of -ux and vy.
sigma_
In the laboratory frame, particle 1 moves along the x-axis with a uniform velocity
ux, and particle 2 moves along the y-axis with a uniform velocity vy. Find the
velocity of particle 2 in the rest frame of particle 1

Since we are looking at particle 2 in the laboratory frame from the rest frame of particle 1, I can't figure out whether or not I need to use the lorentz transform on the velocity of particle 2's speed in the y direction. It would make no sense to, as the frame in which particle 2 is moving has no motion in the y direction, and since particle 2 has no component of velocity in the x or -x direction, its component of velocity in the -x direction would be simply -ux

Vx(p2) = -ux
Vy(p2) = vy

So would not the velocity of particle 2 with respect to particle 1 simply be the square root of the sum of the squares of -ux and vy? This approach seems far too simplistic though.

You need a Lorentz transformation to go from the lab frame to the frame of particle 1.

So would not the velocity of particle 2 with respect to particle 1 simply be the square root of the sum of the squares of -ux and vy?
Simple counterexample: ux=0.9c, vy=0.9c, but the speed of particle 2 in the frame of particle 1 cannot exceed c.

I'm still trying to riddle out question for myself. I'm having trouble finding the correct way to apply the velocity transformation equations. It seems for example, that the equation involving the transformation of the x-component velocity involves the variables $$u_x u'_x v$$ and I find myself in a similar situation while looking at the equation for the y component.
The only three variables I can think of to plug in for these are the velocity of particle one from it's rest frame (0) and then the velocity of particle 2 from particle 1's rest frame ($$-u'_x$$). What do I Use for the v in this equation? they component of the particles velocity in it's own frame? That's the only variable I have left to plug in. I'm confused.

mfb said:
You need a Lorentz transformation to go from the lab frame to the frame of particle 1.

Simple counterexample: ux=0.9c, vy=0.9c, but the speed of particle 2 in the frame of particle 1 cannot exceed c.
So allow me to apply the transformation mathematically, and perhaps tell me where I'm going wrong?

vx' = (vx - V)/(1-(vxV/c^2))

However, since we have no value for vx because particle 2 has no velocity in the x direction with respect to the lab frame, the equation is simplifying to this.

vx' = (0-V)/(1-(0V/c^2)) -> vx' = -V, where V is simply the velocity with which the two frames move apart from each other, which would be -ux if we look at the lab frame from the rest frame of particle 1.

Where am I making my misstep?

vx' looks right, what about vy'?

mfb said:
vx' looks right, what about vy'?
Ahhhh. Thank you very much! I must have misappropriated my variables the first time I ran this calculation. It all works out now.

wouldn't V be bigger than -ux because the particle is moving in the y direction as well as the x direction? isn't V the total velocity with which v2 is moving away from vx? that number is greater than ux because it's moving in the y direction as well?

PsychonautQQ said:
wouldn't V be bigger than -ux because the particle is moving in the y direction as well as the x direction? isn't V the total velocity with which v2 is moving away from vx? that number is greater than ux because it's moving in the y direction as well?
V should equal -ux, as it is the speed at which the lab frame as a whole is moving away from particle 1, which we are defining to be the point of reference of the prime frame. If we defined particle 2 as being the reference point of the lab frame (which would not make sense, considering that the speed vy is the speed of particle 2 as observed from the lab frame), then V would be > -ux. Think of particle 2 as a moving object within a moving frame, not as the frame itself. This is probably not the best explanation possible, but its the best I can come up with right now in between electrodynamics homework problems.

The parametric equations for the world line of particle 2 in the lab frame are
\begin{align*}
t &= t \\
x &= 0 \\
y &= v_y t
\end{align*} Try applying the Lorentz transformations to determine the world line of particle 2 in particle 1's rest frame. Once you have x'(t') and y'(t'), you can see by inspection what the components of particle 2's velocity are.

## 1. What is special relativity?

Special relativity is a theory developed by Albert Einstein that explains how the laws of physics work in reference frames that are moving at a constant velocity relative to each other. It is based on the principles of the constancy of the speed of light and the relativity of motion.

## 2. What are the key concepts of special relativity?

The key concepts of special relativity include the principle of relativity, which states that the laws of physics are the same in all inertial reference frames, and the principle of the constancy of the speed of light, which states that the speed of light is always the same regardless of the observer's frame of reference.

## 3. How does time dilation work in special relativity?

Time dilation in special relativity is the phenomenon where time appears to pass slower for objects that are moving at high speeds relative to an observer. This is due to the constant speed of light and the stretching of space-time.

## 4. What is length contraction in special relativity?

Length contraction is the phenomenon where objects appear to be shorter in the direction of motion when measured by an observer in a different frame of reference. This is also a consequence of the constant speed of light and the stretching of space-time.

## 5. How does special relativity relate to general relativity?

Special relativity is a special case of general relativity, which is a more comprehensive theory that includes the effects of gravity. General relativity expands on the concepts of special relativity to explain how gravity affects the fabric of space-time.

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