Measuring relativistic length--how does it work and why?

Question

A few days ago the following question was asked in this forum:

"Your friend flies past you at 75% of the speed of light, traveling in a spaceship which she measures at 50 meters end to end. What length do you measure as she goes by?"

The Best Answer (and only answer) was:

"L' = L √(1 - v²/c²)

50 = L √(1 - (0.7c)²/c²)

L = 70.01 meters"

So the answer given is that the observer (stationary?) measures a greater length L (70m) than the friend (passenger) in the spaceship does (L' = 50m).

[Note: I'm not quibbling about the apparent substitution of .7 for .75 in the answer; possibly just an oversight.]

I know about the relativistic mass equation:

(1) m = γ * m0,

where
m = relativistic mass
m0 = rest mass
γ = gamma, the relativistic expansion factor = 1 / √(1 - v² / c²).

In that form, the answer given above seems to be saying that

(2) L = γ * L';

in other words, that L is the relativistic length and L' (50m) is the rest length.

I guess what confuses me is simply substituting L's for the m's in (1).

I would reason that mass is proportional to volume, and volume is proportional to L^3; so that (2) should read as:

(3) L = (γ^1/3) * L' or

(4) L^3 = γ * L' ^3

Can someone explain to me what's going on?

.

Answer 1

Mass and Length have different transformations



whereas mass increases at high speed, Length contracts.



M'=M*γ



but



L'=L/γ



L' is still the relativistic length.



The reason that mass increases is because you are adding energy into the system, and the relativistic mass= rest mass * kinetic energy/C^2



the reason that length decreases is a lengthier explanation

1) relative velocity is the same, as it is what defines the reference frames

2) time dilates in the observed reference frame (from both perspectives)

3) if less time is observed to pass, a shorter distance must be observed to pass to maintain simultaneity between velocities

Answer 2

The calculation is wrong. If the person in the spaceship measures its length as 50m, that is the 'proper length'; the observed length is smaller.



Observed length = 50 x √(1 - v² / c²).

= 50 x √(1 - 0.75²)..

= 33m



On the other hand, observed mass increases - you divide rest mass by √(1 - v² / c²).



As another answer said, the contraction only applies to the the direction towards/away from the observer.



"Mass is proportional to volume" is not appropriate for relativistic calculations and simply refers to a material of constant density under non-relativistic conditions. If fact the apparent density of an object increases under relativistic conditions because of the mass increase (the extra kinetic energy increases the observed mass because E=mc²) and the length contraction reduces the observed volume.

Answer 3

Relativistic Mass:

m' = m γ



Lorentz Contraction:

L' = L / γ



Helpful Unitless Variables:

β = v/c

γ = √(1 - β²)



The answerer made the mistake of confusing the rest-length and the apparent (relativistic) length of the ship. When objects travel faster, they appear to be shorter. The correct answer for the problem above should be: 33.0718914 m.



Also, you can't cube the length formula. The Lorentz Contraction only applies in the direction of the velocity; whereas the perpendicular dimensions are unaffected.

Answer 4

i'm truly puzzled by potential of this question. enable me say this. If the guy on the planet measures the spacecraft to be 40m long, then its easily length might desire to be LONGER THAN 40m. The relativistic length IS the single MEASURED by potential of THE OBSEVER on the planet no longer the single in the spacecraft!!!! the guy in the spacecraft could degree its everyday length. So its authentic length is 40m MEASURED by potential of the guy in the SPACECRAFT and its relativistic length is 24m MEASURED by potential of the guy on the planet.