DimaMazin Posted February 15, 2018 Posted February 15, 2018 GT= p * dx / KE GT is general time=observer time+traveler time p is momentum KE is kinetic energy
Strange Posted February 16, 2018 Posted February 16, 2018 16 hours ago, DimaMazin said: GT= p * dx / KE So GT = 2 * dx / v ; in other words twice the time taken to travel dx at velocity v. (Assuming v is small enough that relativistic effects can be ignored.) What is the point of that?
DimaMazin Posted April 13, 2018 Author Posted April 13, 2018 On 16.02.2018 at 2:07 PM, swansont said: Evidence that this is true? Expense of general time for crossing of one meter= p /KE = (gamma+1 ) /(gamma*v) It is true for both frames. Do you think that physics law can be useless?
Strange Posted April 13, 2018 Posted April 13, 2018 4 minutes ago, DimaMazin said: Expense of general time for crossing of one meter= p /KE = (gamma+1 ) /(gamma*v) 1. What is "expense of general time"? 2. Please show how you derive this equation.
DimaMazin Posted April 29, 2018 Author Posted April 29, 2018 On 13.04.2018 at 9:36 PM, Strange said: 1. What is "expense of general time"? 2. Please show how you derive this equation. Expense of general time is dt+dt' (dt+dt')/dx=(dx/v+dx/(gamma*v))/dx=(gamma+1)/(gamma*v) p/KE=gamma*v*m/((gamma-1)mc2) = (gamma+1)/(gamma*v) (dt+dt')/dx=p/KE
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