Physics 62 Special Relativity (34 of 43) Relativistic Sample Problem - Length

In this video I will find the speed of of a space ship (event) by an observer on Earth.

Next video in the Special Relativity series can be seen at:

[kathyfrejoles]

Now this is what I'm looking for!! Awesome. I just remembered having my physics class back then with all the derivations on the board making the board looking beautifully messy 😂.

[rithikamedari]

Isn't La supposed to be the proper time, so shouldn't we actually end up with Lb= La *sqrt 1-((v^2)/c^2))???

[zakirhussain-jsku]

What would be the velocity of space rocket if Observer A marks two points in space 100 meters apart. He starts his stopwatch when front of space rocket hits the 1st point & stops the stopwatch when front of rocket hits the 2nd point. Observer B also does the same. To be fair, now both Observers have equal distance to work with.

[musicc]

Why do you assume that v (as measured by the observer A) is the same as u (the actual velocity of the rocketship)? I cannot wrap my head around it..

[mrawesomex]

This feels like it is made longer than needed and more complicated than it is needed to be.

[nishantjangid]

don't we need to change time Ta ti frame s'

[zakirhussain-jsku]

What would be the speed calculated by Observer B on the spaceship.

[shredupchopel]

Thanks for the great video, very informative!

[chavab]

I've watched almost all the videos, studied for hours on Khan Academy, watched the Fermi series, etc. There is still a difficult-to-express question rattling around in my mind. I can manipulate Lorenz transformations, but I don't know how one figures out what is "real". The spaceship sees a length shortened (pancake) 'man' on earth also. I could let this go with being just a perceptual problem, but in the time experiments, there is a way to put the twins back together. They can't BOTH be younger than the other. I get the part about no special reference frame but can't make it reconcile with this. Insight please?

[oremazz]

Be careful!!, Lb must be smaller than La. The energetic observer at b will receive the length contraction and time dilation. For checking this out, divide Lb/Tb = C = La/Ta same Lorentz contraction for both, space and time, which will give the quotient L/T always constant C. You can also have this in mind, Total energy is the orthogonal addition of kinetic energy plus mass-energy, so observer at b will "see" no kinetic energy BUT his mass-energy has been incremented because his mass wavelength has shrunk i.e. with Lorentz length contraction (as well a time dilation). In reverse, the observer at a will "see" space ship with kinetic energy where the mass-energy will be less due to a bigger wavelength. Hope this will clear concepts. Besides this, a very clear presentation, as always, Best Regards