# How do I get the velocity out?

1. Jun 18, 2009

### phys121

Alright, im not the best at algebra. So i was wondering if anyone could show me how to get the V out on its own in the lorentz factor equation

l= 1/$$\sqrt{1-(v^2/c^2)}$$

I tried to first times both sides by bottom then manipulate it but i just cant get it right.

Any help would be much appreciated!!!!

2. Jun 18, 2009

### Staff: Mentor

Not exactly sure what you're trying to accomplish, but you can take:
γ = 1/√(1 - v²/c²)

and solve for v in terms of γ.

Cross-multiply and then square both sides.

3. Jun 18, 2009

### phys121

What, you mean like this...

$$\gamma$$($$\sqrt{1-v^2/c^2}$$)=1

then: finally finishing with;

$$\sqrt{(1-1^2/\gamma^2)c^2}$$=v

??

Last edited: Jun 18, 2009
4. Jun 18, 2009

### icystrike

your question was 1/$$\sqrt{1-v^2/c^2}$$=1 ?

5. Jun 18, 2009

### phys121

My question is how do i solve for "v", that l should be an L lower case. as in lorentz factor "L". not a 1. sorry

Like what should the final equation be?

6. Jun 18, 2009

### icystrike

this is wrong.
return to the previous step square both sides, you dont have to multiply gamma into the square root.

7. Jun 19, 2009

### phys121

ah, right i get you now. Thanks!