granpa Posted July 1, 2016 Share Posted July 1, 2016 (edited) What is the magnitude of the proper acceleration of an object moving in a circle at a given (unchanging) relativistic speed? For an object moving in a straight line proper acceleration = acceleration*gamma^3 My question has to do with the relationship between transverse mass and longitudinal Mass Edited July 1, 2016 by granpa Link to comment Share on other sites More sharing options...

elfmotat Posted July 1, 2016 Share Posted July 1, 2016 (edited) I'm not exactly sure what your question has to do with transverse and longitudinal mass. Proper acceleration is the magnitude of four-acceleration. So if you have a path (in this case a circle), you can parametrize it, find four-acceleration, then take its norm to find proper acceleration: [math]x^\mu (t)= (ct,Rcos \, \omega t,Rsin \, \omega t)[/math] [math]u^\mu (t)= \frac{dx^\mu}{d \tau} = \frac{dt}{d \tau} \frac{dx^\mu}{dt} = \frac{dt}{d \tau} (c,-R\omega sin \, \omega t,R\omega cos \, \omega t)[/math] we know that: [math]d\tau^2 = dt^2 -c^{-2}( dx^2 + dy^2)[/math] or: [math]\left (\frac{d\tau}{dt} \right )^2 = 1 -c^{-2} \left ( \left ( \frac{dx}{dt} \right )^2 + \left ( \frac{dy}{dt} \right )^2 \right ) = 1 -c^{-2} \left ( R^2 \omega^2 sin^2 \, \omega t + R^2 \omega^2 cos^2 \, \omega t \right ) = 1-\frac{R^2 \omega^2}{c^2}[/math] So: [math]u^\mu (t) = \frac{1}{\sqrt{1-R^2 \omega^2/c^2}} (c,-R\omega sin \, \omega t,R\omega cos \, \omega t)[/math] Then we can compute four-acceleration: [math]a^\mu (t) = \frac{dt}{d \tau} \frac{du^\mu}{dt} = \frac{1}{1-R^2 \omega^2/c^2} (0,-R\omega^2 cos \, \omega t,-R\omega^2 sin \, \omega t)[/math] And from there we can compute proper acceleration: [math]a_{prop}= \sqrt{a^\mu a_\mu} = \sqrt{\frac{R^2\omega^4 cos^2 \, \omega t+R^2\omega^4 sin^2 \, \omega t}{(1-R^2 \omega^2/c^2)^2}} = \frac{R\omega^2}{1-R^2 \omega^2/c^2}[/math] So, given the circular path's radius and the object's angular velocity, the proper acceleration felt by the object is [math]R\omega^2/ (1-R^2 \omega^2/c^2)[/math]. Note that the object's tangential speed [math]v=R \omega[/math] can also be plugged into this equation to get: [math]a_{prop} = \frac{v^2/R}{1-v^2/c^2} = \frac{\gamma^2 v^2}{R}[/math] EDIT: Whoops, forgot to square the demoninator . Edited July 1, 2016 by elfmotat 2 Link to comment Share on other sites More sharing options...

Mordred Posted July 1, 2016 Share Posted July 1, 2016 (edited) Elfomatat. I'm going to quote your post in another discussion in Speculations. http://www.scienceforums.net/topic/96004-galaxy-rotation-rates-explained-without-dark-matter/page-7#entry929330 Edited July 1, 2016 by Mordred Link to comment Share on other sites More sharing options...

granpa Posted July 1, 2016 Author Share Posted July 1, 2016 (edited) Thank you Elfmotat So for circular motion proper accelleration = (proper velocity)^2/r works for all velocities! F = m*(proper acceleration) So centripetal force should equal m*(proper velocity)^2/r This result makes perfect sense though I don't understand why for an object moving in a straight line proper acceleration = acceleration*gamma^3 And I don't see any way of reconciling it with the concept of transverse mass This is what I am trying to understand https://en.wikipedia.org/wiki/Relativistic_mechanics#Force I'll keep working on it I was expecting everything to reduce to F = m*(proper acceleration) https://en.wikipedia.org/wiki/Proper_acceleration#Acceleration_in_.281.2B1.29D According to https://en.m.wikipedia.org/wiki/Relativistic_mechanics#Force F=dp/dt p = m*gamma*v So I assume F=m*d(gamma*v)/dt It then says that the result is that F = gamma^3*ma for straight line acceleration F = gamma*ma for acceleration at a right angle (like a circle) a= v^2/r for a circle ???? Therefore F = gamma^1*m*v^2/r for a circle Edited July 1, 2016 by granpa Link to comment Share on other sites More sharing options...

Toffo Posted July 2, 2016 Share Posted July 2, 2016 (edited) What is the magnitude of the proper acceleration of an object moving in a circle at a given (unchanging) relativistic speed? Let's consider forces affecting an object swung around using a rope. If the Newtonian centripetal force measured on the rope is F, then the relativistic centripetal force is gamma * F, because the transverse mass of the object is gamma*rest mass. This is the force at the other end of the rope. I mean at the end that is not moving relativistically. At the end that is moving relativistically forces are measured to be gamma times bigger. I like to think that's because of time dilation. The proper acceleration is measured by an accelerometer, which is a force meter. So a force meter attached to the object measures this force: Newtonian centripetal force times gamma squared. So when used as an accelerometer the force meter measures this acceleration: Newtonian centripetal acceleration times gamma squared. What is the magnitude of the proper acceleration of an object moving in a circle at a given (unchanging) relativistic speed? For an object moving in a straight line proper acceleration = acceleration*gamma^3 My question has to do with the relationship between transverse mass and longitudinal Mass Longitudinal acceleration and transverse acceleration are quite different: Longitudinal acceleration changes the longitudinal mass and the transverse mass. While transverse acceleration does not change the longitudinal mass or the transverse mass. So it makes sense that longitudinal mass and transverse mass are quite different. Edited July 2, 2016 by Toffo 1 Link to comment Share on other sites More sharing options...

granpa Posted July 2, 2016 Author Share Posted July 2, 2016 (edited) I'm trying to get away from the idea of relativistic mass and instead use proper velocity since momentum equals mass times proper velocity and force equals the rate of change of momentum Edited July 2, 2016 by granpa Link to comment Share on other sites More sharing options...

Toffo Posted July 2, 2016 Share Posted July 2, 2016 What is this proper velocity thing? If a driver measures his speed relative to road to be 0.5 c, then a cop standing on the road agrees that the car moves at speed 0.5 relative to the road. Link to comment Share on other sites More sharing options...

granpa Posted July 2, 2016 Author Share Posted July 2, 2016 Proper velocity = gamma * v Link to comment Share on other sites More sharing options...

granpa Posted July 3, 2016 Author Share Posted July 3, 2016 (edited) So the rate of change of momentum equals force but does not equal proper acceleration Seems like proper acceleration is badly named Edited July 3, 2016 by granpa Link to comment Share on other sites More sharing options...

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