I need help understanding length contraction. If the equation is: L = L' * √ (1 - v2/c2), and th length of my

Question

space ship is 300m and my speed is .86...Then I know the length observed by a stationary observer is: 153meters. However, what would the length be for the person inside the space ship and why? Would it appear longer than the 300m? And if so how do you find the exact number?

Answer 1

l = L*sqrt(1 - (v/c)^2); where l is the relative length at velocity v < c and L is the rest length (at v = 0).

sqrt(1 - (v/c)^2) is called the Lorentz transform, T(v). T(v = c) = 0 and T(v = 0) = 1. So, we can see that l = L T(c) = 0; so, theoretically the length of a framework going the speed of light would disappear to an outside observer. But when T(0), we have l = L T(0) = L; so that when the observer is in a frame that is not moving relative to the observed space ship, the relative length equals the rest length.

This is important, v is a relative velocity; so when one is traveling inside a framework going v ~ c relative to the outside, that inside observer is still just going v = 0 relative to the inside. And that is why the observer inside the space ship continues to see rest length, and also rest mass and time. So, to calculate what the inside man sees just solve l = L T(0) because the inside man is traveling v = 0 relative to the spaceship.

The inside observer (e.g., Capt. Kirk) will see no change in length. This follows because all dimensional things in the direction of travel will shrink when seen from the outside, even the measuring stick we use to measure the length.

That is, for example, if Capt. Kirk's desk shrinks to half its length in the direction of travel, the one yard measuring stick will also shrink to half its length. So, relatively speaking (pun intended), the inside observer will see the same length he or she saw before taking off on the star trek.

BTW, this observation relationship also applies to mass and time. That is, an outside observer will see the changes because he sees v --> c, but the inside observer will be aware of only rest mass and rest time because v = 0 relative to the spaceship. To the outside observer, mass will increase and time will slow down on board the Enterprise.

Answer 2

The length of the spaceship, as measured by the person in the spaceship, will be unchanged.

So if your ship was built to be 300 meters long, as you walk around it, you will measure it as being 300 meters long - even though the ship is moving very fast relative to the planet Earth. However, if you look at a spaceship that is just "floating" by Earth (and essentially at rest with respect to it), you will IT as being much shorter than it should be. The same reduction factor that it sees for you, you will see for it.

That's the way it works. If you want to understand it, there's no real way around studying the physics of it. I recommend a book called "Spacetime Physics": It's not too long, and it forces you to think about the "paradoxes".

Answer 3

What is the speed of the spaceship as measured by someone aboard it?

Answer 4

seems like we are in the same boat. im sorry i cant answer your question but can you answer this: at what speed does the kinetic energy of an object equal its rest mass energy?