GT= p _* dx / KE

GT is general time=observer time+traveler time

p is momentum

KE is kinetic energy

Evidence that this is true, and useful?

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?

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?

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.

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