Ekpyrotic Posted December 18, 2008 Posted December 18, 2008 Hey guys. I'm interested in knowing how you get to: [math]t = \gamma t'[/math] ...from: [math]t' = \gamma (t - \frac{vx}{c^2})[/math]
Tom Vose Posted December 19, 2008 Posted December 19, 2008 I could solve it for you (probably), however, i want to know, for my own derivation, what the units you are using. By the way, it seems almost as though someone has replaced mass with time, am i wrong...?
iNow Posted December 19, 2008 Posted December 19, 2008 I could solve it for you (probably), however, i want to know, for my own derivation, what the units you are using. Hi Tom - Just a reminder... Please, please, please don't do the work for the requestor. Ideally, you would explain to them how to work through it, maybe splitting the basics into steps without giving away the answer. More than a policy here at SFN, it's actually a good practice in life.
Obelix Posted December 20, 2008 Posted December 20, 2008 Hey guys. I'm interested in knowing how you get to: [math]t = \gamma t'[/math] ...from: [math]t' = \gamma (t - \frac{vx}{c^2})[/math] To begin with, appart from the transformation you have given here (Lorentz transfirmaton for time) one also needs the transformation for [math]x[/math]. I.e.: [math]x'=\frac{x - \upsilon c}{\sqrt{1-\frac{\upsilon^2}{c^2}}}[/math] Having them both, you can solve a system of 2 equations for 2 unknowns: [math]x[/math] and [math]t[/math]
Tom Vose Posted December 20, 2008 Posted December 20, 2008 Hi Tom - Just a reminder... Please, please, please don't do the work for the requestor. Ideally, you would explain to them how to work through it, maybe splitting the basics into steps without giving away the answer. More than a policy here at SFN, it's actually a good practice in life. Oh my bad.
swansont Posted December 20, 2008 Posted December 20, 2008 By the way, it seems almost as though someone has replaced mass with time, am i wrong...? Yes. These are time dilation equations. Mass does not appear in them.
Tom Vose Posted December 20, 2008 Posted December 20, 2008 I don't recognise the v(x) part, this is why i may have asked this question.
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now