# Length contraction equation. what am i doing wrong?

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 Last edited:

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LeonhardEuler
Gold Member
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 Your mistake is on line 3. It should go:
$$(\frac{L}{L*})^2=1-\frac{v^2}{c^2}$$
$$(\frac{L}{L*})^2-1=-\frac{v^2}{c^2}$$
$$\frac{v^2}{c^2}=1-(\frac{L}{L*})^2$$

Staff Emeritus
Gold Member
Dearly Missed
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 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!  