# Relativistic and proper length

• AClass
In summary: So, for a), the 40 m is L, not L0.I don't understand what you are trying to do with the square roots in your second post.You've applied the square root of c2 twice. So you've just taken c.In the equation L= Lo x {\sqrt{1 - \frac{ v^2 }{ c^2 }}}, you have c2 on the bottom.So, if you want to get rid of the square root, you multiply both sides by c2.Lc2= Lo x {\sqrt{c2 - v2}}Okay. Now you have to get rid of the square root on the right side. You do that by squaring both sides of
AClass

## Homework Statement

A spaceship travels past a planet at a speed of 0.80 c as measured from the planet’s frame of reference. An observer on the planet measures the length of a moving spaceship to be 40 m.
a) How long is the spaceship, according to the astronaut?
b) At what speed would the spaceship have to travel for its relativistic length to be half its “proper” length?

## Homework Equations

L= Lo x $${\sqrt{1 - \frac{ v^2 }{ c^2 }}}$$

## The Attempt at a Solution

a)

Lo=40m
v=0.80c

Using the equation above I came to a answer of L = 24m

b)

L= 0.5 Lo

0.5Lo = Lo x $${\sqrt{1 - \frac{ v^2 }{ c^2 }}}$$

v^2 = (0.75)(c^2)
v= Square root [(0.75)(c)]
c= 3.0x10^8
v= 15000 m/s

The speed of the spaceship must be 15000 m/s.

Could someone verify this? The 15,000 m/s seems small.

AClass said:
a)

Lo=40m
v=0.80c

Using the equation above I came to a answer of L = 24m
You're not applying the lengths correctly. The astronaut is the one that will measure the longer length for his own spaceship. Putting that into better words, people on the planet will measure L to be less than L0.

Anytime an object moves relative to a particular observer's frame of reference, that object is always shorter (squished up) according to that particular observer. In other words, it's not squished up in his own frame of reference.

The astronaut will see the entire planet "squished" and even the distances between planets as being shorter (if the planets are along the astronaut's line of motion), i.e. meter sticks on the planet, along the astronaut's line of motion, are shorter than 1 m. But the astronaut doesn't observe his own ship as squished. On his own ship, a meter stick is 1 m long.

People on the planet on the other hand measure a meter stick on their own planet to be exactly 1 m long. But they measure meter sticks in the passing spaceship (meter sticks oriented along the spaceship's line of motion) to be less than 1 m.

Length is always less in the moving frame; where the stationary frame is the frame where the measurements are being performed. There is no length contraction in the stationary frame. (And again, the "stationary frame" is whichever frame of reference is doing the measuring.)*

*This assumes that the frame of reference doing the measuring is an inertial frame -- a frame that is not accelerating or rotating.
b)

L= 0.5 Lo

0.5Lo = Lo x $${\sqrt{1 - \frac{ v^2 }{ c^2 }}}$$

v^2 = (0.75)(c^2)
v= Square root [(0.75)(c)]
You've already taken the square root of c2, so c is not inside of the square root.

Last edited:
I'm not getting another answer for a)

From what I was taught,
L is the relativistic contracted distance measured by an observer moving. [Astronaut]
Lo is the proper distance measured by a observer stationary. [Planet Observer]

For b) I realize where I went wrong.

I used a different technique this time.
If v= X x C where X is a multiple of the speed of light.
Then

0.5Lo= Lo x $${\sqrt{1 - \frac{ X^2 x C^2 }{ c^2 }}}$$

0.25 = 1-X^2
0.75= X^2

X= +/- Root 0.75

V= SQR[0.75] x 3.0x^8

AClass said:
I'm not getting another answer for a)

From what I was taught,
L is the relativistic contracted distance measured by an observer moving. [Astronaut]
Lo is the proper distance measured by a observer stationary. [Planet Observer]
(red emphasis mine)

Fine, but in this problem, you are being asked how long the spaceship is according to the astronaut. The astronaut is not moving relative to his/her own spaceship. The length of the spaceship according to the astronaut is L0.

The planet on the other hand is moving relative to the spaceship when the 40 m figure was measured. That's L, the contracted length.

Your calculations for part a) are correct, the length would be 24m according to the astronaut.

For part b), your calculations are also correct. It may seem like a small speed, but keep in mind that this is a significant fraction of the speed of light. Even a small percentage of the speed of light can result in significant relativistic effects.

## 1. What is the concept of Relativistic length?

Relativistic length is the distance between two points in space as measured by an observer moving at a high velocity. It takes into account the effects of time dilation and length contraction as predicted by Einstein's theory of special relativity.

## 2. How is Relativistic length different from proper length?

Relativistic length is dependent on the velocity of the observer, while proper length is the distance between two points in space as measured by an observer at rest relative to those points. Proper length is considered the "true" length, while relativistic length is relative to the observer's frame of reference.

## 3. Can Relativistic length ever be greater than proper length?

No, relativistic length can never be greater than proper length. According to the theory of relativity, objects moving at high velocities will appear shorter to an observer, which means their relativistic length will be smaller than their proper length.

## 4. What is the formula for calculating Relativistic length?

The formula for calculating relativistic length is L = L0/γ, where L0 is the proper length, and γ is the Lorentz factor, which takes into account time dilation and length contraction. This formula is used to calculate the length of an object as measured by an observer moving at a high velocity.

## 5. How is Relativistic length used in practical applications?

Relativistic length is used in various practical applications, such as in GPS navigation systems. The satellites in the GPS system are moving at high velocities relative to Earth, so their clocks run slower, and their length contracts. This must be taken into account when calculating precise location information on Earth's surface.

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