How Does Length Contraction Affect Measurements in Special Relativity?

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Homework Help Overview

The discussion revolves around the effects of length contraction in special relativity, specifically regarding a spaceship traveling at a significant fraction of the speed of light. The original poster presents a problem involving the measurement of the spaceship's length from different frames of reference and seeks to determine the proper length and the speed required for the relativistic length to be half of the proper length.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between proper length and relativistic length, questioning the application of the length contraction formula. There are attempts to rearrange equations to isolate variables, particularly the speed of the spaceship.

Discussion Status

Some participants provide guidance on the correct interpretation of the length contraction formula and the concept of the gamma factor. There is a mix of confusion and clarification regarding the calculations and the definitions involved, with ongoing attempts to resolve misunderstandings about the relationships between the variables.

Contextual Notes

Participants express uncertainty about the definitions and calculations related to gamma and the proper length. There is mention of a potential misunderstanding of the original equations provided in the homework statement.

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Homework Statement


A spaceship travels past a planet at a speed of 0.80 c as measured from the planets frame of reference. An observer on the planet measures the length of a moving spaceship to be 40m.

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 of its "proper " length?


Homework Equations


L = Lo sqrt(1 - v^2/c^2)


The Attempt at a Solution


L = Lo sqrt(1 - v^2/c^2)
= 40 m sqrt {1 - (o.80)^2/c^2}
= 40 m (0.6)
= 24 m

b) For this question I am not too sure what to do
this is all i can think of:

L = Lo sqrt(1 - v^2/c^2)
i rearranged the equation t^2 = 4L^2 / c^2 - v^2 to get v^2 = c^2/4L^2

by using this:

v^2 = (3.00 x 10^8 m/s)^2 / 4(20m)^2
= (9.0 x 10^16 m^2/s^2) / 1600 m^2
= 5.625 x 10^13
v = sqrt (5.625 x 10^13)
= 7.5 x 10^6 m/s

I am pretty sure that's not right, however, its all i can think of momentarily. Any suggestions??
 
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For part a, you did it backwards. The proper length, which is the length of an object in its own frame is always the longest length. If the length seen from some observer is 40m, you know the proper length of the spaceship is longer than 40m. So instead of 40/gamma, it is 40*gamma. You are overcomplicating part b. You can use the equation L=Lp/gamma, where Lp is the proper length. Since you are trying to find the speed so that the relativistic length is half the proper length, substitute in 1/2Lp for L and solve for the v term in gamma.
 
alright but what do i do for a? do i divide 40m by .6? and for part b, I am honestly not sure how to rearrange the equation to make it the term v, sadly. I missed that part for rearranging square roots.
 
gamma=1/sqrt(1-v^2/c^2). Gamma is not .6. For a, you should be multiplying the relativistic length by the gamma factor which will be greater than one. For b:

L=Lp/gamma

(1/2)Lp=Lp/gamma

1/gamma=1/2

gamma=2

1/sqrt(1-(v^2/c^2)=2

From here, just solve for v.
 
I honestly did multiply it, and checked the calculations and i get the same answer. what is gamma? I've honestly never heard of that term in physics
 
gamma is the factor by which the classical physics equations must be multiplied or divided by in order for the equations to work at relativistic speeds

http://my.morningside.edu/slaven/Physics/relativity/relativity6.html
 
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really? I've never ever heard of that. but for a I've been multiplying it since the beginning, i keep getting 0.6 for that final answer.
 
I have no idea how you are getting 0.6 for part a. I don't even know what your 0.6 equals. Show what you did step by step.
 
alright, so L = Lp sqrt(1 - v^2/c^2)
= 40m sqrt (1 - (0.80 c)^2 / c^2)
= 40m sqrt (1 - 0.64 c^2 / c^2) (c^2 cancels each other)
= 40 m sqrt (0.36)
= 40m (0.6)
 
  • #10
then multiplying its 24 m
 
  • #11
Okay, so you made the mistake I thought you were making. L does not equal Lp*sqrt(1-v^2/c^2), rather Lp equals L*sqrt(1-v^2/c^2). This way you can find the proper length to be the relativistic length[40m] times the gamma factor[1/.6]. Do you see now?
 
  • #12
but where does the 1/.6 come from?
 
  • #13
i'd get the same answer if i divided L by .6
 
  • #14
thanks for the help by the way
 
  • #15
w3390 said:
L does not equal Lp*sqrt(1-v^2/c^2), rather Lp equals L*sqrt(1-v^2/c^2).

Sorry, I mistyped. Lp should equal L*(1/sqrt(1-(v^2/c^2))). Also, you should know that gamma will always be greater or equal to one, so an answer of 0.6 for gamma makes no sense. You need to divide 1 by this number to get 1.667. This makes sense then when you multiply it by the relativistic length because the proper length will be longer.
 
  • #16
oh okay i understand that, but if you honestly just rearrange the original equation to be L/sqrt(1-(v^2/c^2) you'll end up with the exact same answer, without dividing by 1.
 
  • #17
ohh our equations are different. my original equation from the text is L = Lo sqrt(1 - v^2/c^2 that's why I'm so confused
 
  • #18
You're right, Lp= L*(1/sqrt(1-(v^2/c^2))) is the same thing as Lp=L/sqrt(1-(v^2/c^2)). However, earlier you were saying that it was L times just sqrt(1-(v^2/c^2)). Since the symbol gamma represents a fraction, you have to be careful when substituting it into equations or you will run into problems like this one where you confuse numerators and denominators. More importantly anyway, I hope you understand now. For part b, you just do the same thing but in place of the relativistic length you substitute (1/2)Lp.
 
  • #19
thanks, but i still need to rearrange it right? so that it equals v
 
  • #20
Correct
 
  • #21
cause i got to L^2 = Lp (1 - v^2/c^2) and i know i can divide the whole gamma part, but i don't know where to go from there
 
  • #22
currently I am at L^2/ (1 - v^2/c^2) = Lp
 
  • #23
Wait, you should not have an L term in there. The problem specifies that they want to find v when the relativistic length L is equal to (1/2)Lp. Therefore, starting from the equation L=Lp/gamma, you can substitute into get (1/2)Lp=Lp/gamma. This way you don't even have to worry about any length terms since they cancel to get (1/2)=1/gamma. From here, gamma is equal to 2. Now solve for v.
 
  • #24
oh.. sorry i appreciate the patience, i honestly haven't been able to do physics at all today, normally I'm not this bad. so i rearranged it and i got v^2 = 1 + 2c^2
then v = sqrt (1 + 2c^2).
 
  • #25
i forgot about the sqrt, if I am correct the final equation should be v = sqrt(1+4c^2)
 
  • #26
No. I'll take you to this step.

gamma=2

1/sqrt(1-v^/c^2)=2

(1/2)=sqrt(1-(v^2/c^2))

1-(v^2/c^2)=1/4

From here solve for v.
 
  • #27
is it v = sqrt (1/4c^2 + 1) ?
 
  • #28
No. Everything in the parenthesis is one term you can't just mix around stuff. You take the parenthesis to one side and subtract the 1/4 from 1. Therefore, v^2/c^2=3/4. Follow rules of algebra to solve for v.
 
  • #29
alright thanks, the equation is v = sqrt (3/4 c^2) if that's not it, then honestly just don't worry about it. I'll deal with it some other time
 
  • #30
That is correct. To make it easy to get in terms of c, you can just pull out a c term to find the ratio of the value of v to the speed of light. But either way, that is correct.
 

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