What is the Transformation of Four-Velocity between Frames?

In summary, the speed of rods in the lab frame is the velocity of an object as measured by an observer in the laboratory or reference frame. It can be calculated by dividing the distance traveled by the object by the time it took to travel that distance. Factors such as the object's mass, applied force, and external forces can affect its speed. The speed of rods in the lab frame differs from the speed of light, which is constant and the fastest speed in the universe. Measuring the speed of rods in the lab frame is important for understanding the behavior of objects in motion and developing theories and laws in physics.
  • #1
Kaguro
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Rod R1 has a rest length `1m and rod R2 has rest length 2m. R1 is moving towards right with velocity v and R2 is moving towards left with velocity v with respect to lab frame. If R2 has length 1m in rest frame of R1, then v/c is what?
-----------------------

I let vel of R1 be u, of R2 be w w.r.t lab and velocity of R2 w.r.t. R1 be w'.
Then u = v, and w = -v.

Then by relativistic velocity addition formula:
w' = (w - u)/(1- (u*w)/c^2)
=> w' = -2v/(1+v^2/c^2) -----------(1)

Now, according to Lorentz contraction:
L/L0 = √(1 - (w'/c)^2 )
But L/L0 = 1/2

Solving, I find w'/c = (√3)/2 ------------(2)

From simplifying (1) and (2):
v^2 + (4/√3) vc + c^2 = 0
So, v = -c/√3 or v = -√3 *c.

But... why is answer negative?
Even if I reject the 2nd one and accept the magnitude of first, I get v = 0.577c...

But the exact answer given is v = 0.6c.

Where did I go wrong?Also, the solution given says:
weird_formula.jpg


Where did this one come from??

Any help will be appreciated.

[Moderator's note: Moved from a technical forum and thus no template.]
 
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  • #2
The negative comes in because you neglected a solution for ##w'##. It could be ##w'=\pm\sqrt 3/2##, and your equation 1 implies that ##w'## is negative if ##v## is positive.

Otherwise I have to say I agree with you.
 
  • #3
Ibix said:
The negative comes in because you neglected a solution for ##w'##. It could be ##w'=\pm\sqrt 3/2##, and your equation 1 implies that ##w'## is negative if ##v## is positive.
Oh yeah.. You're right!
 
  • #4
The expression for length contraction in the photo is weird. You get different behaviour as ##v\rightarrow c## and ##v\rightarrow -c##. That's not right, unless it's a specialisation of something with a hidden assumption that ##v>0##. What book did it come from?
 
  • #5
Ibix said:
What book did it come from?
It is a book for IIT-JAM solved papers.
 
  • #6
So just a book of practice problems? In that case (unless someone else comes up with something we've both missed) I'd just bear in mind that some of the answers might not be perfect.
 
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  • #7
The first equation is not the Lorentz Contraction equation since the right-hand-side is the Doppler factor.
 
  • #8
Ibix said:
So just a book of practice problems? In that case (unless someone else comes up with something we've both missed) I'd just bear in mind that some of the answers might not be perfect.

Yeah i thought so too...

thanks everyone.
 
  • #9
Kaguro said:
Yeah i thought so too...

thanks everyone.

You can also do this with rapidities. In this case, in the frame of R1, we have a rapidity of ##2\theta##, where ##\theta## is the rapidity associated with speed ##v##. And:

##2 = \gamma_1 = \cosh 2\theta = 2\cosh^2 \theta - 1##

Hence ##\gamma^2 = \cosh^2 \theta = 3/2##

Then, we have:

##\frac{v}{c} = \tanh \theta = \sqrt{1 - sech^2 \theta} = \frac{1}{\sqrt{3}}##

Or, more generally, we can relate the gamma factor for ##v## to the gamma factor, ##\gamma_1##, in the reference frame of rod 1:

##\gamma^2 = \cosh^2 \theta = \frac{\cosh 2\theta +1}{2} = \frac{\gamma_1 +1}{2}##

And:

##(\frac{v}{c})^2 = 1 - \frac{1}{\gamma^2} = \frac{\gamma_1 - 1}{\gamma_1 + 1}##

In this case, with ##\gamma_1 = 2##, we get:

##(\frac{v}{c})^2 = \frac13##
 
  • #11
Here's another approach, using transformation of the four-velocity between frames.

In the lab frame, R1 has four-velocity ##(\gamma c, \gamma v)##. The time component when transformed to the frame of R2 is:

##\gamma_2 c = \gamma (\gamma c + \frac{v}{c} \gamma v) = \gamma^2 (1 + v^2/c^2) c##

We want ##\gamma_2 = 2##, hence:

##\frac{1 + v^2/c^2}{1- v^2/c^2} = 2## etc.
 

1. What is the "speed of rods in lab frame"?

The "speed of rods in lab frame" is a term used in physics to describe the velocity of a rod or object in a laboratory setting. This refers to the speed at which the rod is moving relative to its surroundings in the lab.

2. How is the speed of rods in the lab frame measured?

The speed of rods in the lab frame can be measured using various methods, such as using a stopwatch to time the distance traveled by the rod or using advanced equipment like motion sensors or high-speed cameras.

3. What factors can affect the speed of rods in the lab frame?

The speed of rods in the lab frame can be affected by factors such as friction, air resistance, and the material and shape of the rod. Additionally, external forces like gravity or applied force can also impact the speed of the rod.

4. How does the speed of rods in the lab frame relate to other frames of reference?

The speed of rods in the lab frame can be compared to the speed of the same rod in other frames of reference, such as the observer's frame or the frame of the rod itself. This can help determine the relative motion of the rod and its surroundings.

5. What are the practical applications of studying the speed of rods in the lab frame?

Understanding the speed of rods in the lab frame is crucial in various fields such as engineering, transportation, and sports. It helps in designing efficient and safe structures, predicting the behavior of moving objects, and improving performance in activities that involve speed and motion.

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