DimaMazin Posted February 15, 2018 Share Posted February 15, 2018 GT= p * dx / KE GT is general time=observer time+traveler time p is momentum KE is kinetic energy Link to comment Share on other sites More sharing options...
swansont Posted February 16, 2018 Share Posted February 16, 2018 Evidence that this is true, and useful? Link to comment Share on other sites More sharing options...
Strange Posted February 16, 2018 Share 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? Link to comment Share on other sites More sharing options...
DimaMazin Posted April 13, 2018 Author Share 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? Link to comment Share on other sites More sharing options...
Strange Posted April 13, 2018 Share 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. Link to comment Share on other sites More sharing options...
DimaMazin Posted April 29, 2018 Author Share 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 Link to comment Share on other sites More sharing options...
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