Length contraction equation. what am i doing wrong?

Click For Summary

Homework Help Overview

The discussion revolves around the length contraction equation in the context of special relativity, specifically focusing on determining the velocity (v) when given contracted length (L) and proper length (L*).

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to manipulate the length contraction formula but encounters results suggesting velocities exceeding the speed of light, prompting questions about their calculations and assumptions.

Discussion Status

Participants are actively engaging with the original poster's calculations, offering corrections and pointing out errors in the algebraic manipulation of the equations. There is an acknowledgment of mistakes, and some participants provide alternative calculations to clarify the issue.

Contextual Notes

There is a focus on ensuring that the derived velocity does not exceed the speed of light, which is a fundamental constraint in relativity. The original poster expresses urgency for assistance.

Sneil
Messages
18
Reaction score
0
Length contraction equation determining v. what am i doing wrong?

OK,

L=L*(1-v2/c2)1/2 (to the power of 1/2)

L=.75m L*=1.0m

now i need to find v and for the life of me i always end up with a value greater then c which is obviously impossible.

ill simplify to v=

L/L*=(1-v2/c2)1/2
(L/L*)squared = 1-v2/c2
1+(L/L*)squared = v2/c2
(1+(L/L*)squared )to the 1/2=v/c
c((1+(L/L*)squared )to the 1/2)=v

and thus i break the barrier of the speed of light and go back in time i guess...

any help?? asap please :redface:
 
Last edited:
Physics news on Phys.org
Sneil said:
OK,

L=L*(1-v2/c2)1/2 (to the power of 1/2)

L=.75m L*=1.0m

now i need to find v and for the life of me i always end up with a value greater then c which is obviously impossible.

ill simplify to v=

L/L*=(1-v2/c2)1/2
(L/L*)squared = 1-v2/c2
1+(L/L*)squared = v2/c2
(1+(L/L*)squared )to the 1/2=v/c
c((1+(L/L*)squared )to the 1/2)=v

and thus i break the barrier of the speed of light and go back in time i guess...

any help?? asap please :redface:
Your mistake is on line 3. It should go:
[tex](\frac{L}{L*})^2=1-\frac{v^2}{c^2}[/tex]
[tex](\frac{L}{L*})^2-1=-\frac{v^2}{c^2}[/tex]
[tex]\frac{v^2}{c^2}=1-(\frac{L}{L*})^2[/tex]
 
Sneil said:
OK,

L=L*(1-v2/c2)1/2 (to the power of 1/2)

L=.75m L*=1.0m

now i need to find v and for the life of me i always end up with a value greater then c which is obviously impossible.

ill simplify to v=

L/L*=(1-v2/c2)1/2
(L/L*)squared = 1-v2/c2
1+(L/L*)squared = v2/c2
(1+(L/L*)squared )to the 1/2=v/c
c((1+(L/L*)squared )to the 1/2)=v

and thus i break the barrier of the speed of light and go back in time i guess...

any help?? asap please :redface:

Let's see. L/L* is .75

(1 - v^2/c^2)^1/2 -.75
1 - v^2/c^2 = .5625
-v^2/c^2 = .5625 -1 = -.4375
v^2/c^2 = .4375
v/c = sqrt(.4375) = .661437828

So v is about 661/7 % of c.

Your error was in going from your second equation to your third. You got the signs wrong in simplifying. Check it out.
 
ah, great. bad mistake on my part. Thank you very much for the quick replies! :smile: :smile:
 

Similar threads

Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 36 ·
2
Replies
36
Views
4K
  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
11
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 67 ·
3
Replies
67
Views
5K
  • · Replies 5 ·
Replies
5
Views
3K