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Posted

you say this matter has been dealt with. what was the conclusion why has the post before mine asking a question that you said hAS BEEN DEalt I stared this post in response to the post which is asking the question am I right r wrong does time have motion /\?

 

 

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Posted (edited)

you say this matter has been dealt with. what was the conclusion why has the post before mine asking a question that you said hAS BEEN DEalt I stared this post in response to the post which is asking the question am I right r wrong does time have motion /\?

 

The thread is directly under yours in the relativity section at the moment: http://www.scienceforums.net/topic/29821-does-time-have-a-speed/

 

Whether or not time has a "speed" is essentially a question about definitions. What exactly do you mean? By any sensible definition, i.e. rate of time with respect to something else, the closest analogue I can think of would be the rate at which one clock in arbitrary motion in an arbitrary spacetime "ticks" relative to some other clock.

Edited by elfmotat
Posted

time is a means to measure matter in motion

Sure it does. We can take our 3D vectors from Euclidean space and create their analogs in a 4D Minkowski space. So, instead of S=(x,y,z), we have S=(ct,x,y,z). We use the time of the rest frame of the object we are studying, so when we are talking about the speed of time, we're talking about the rate time passes according to one frame with respect to an object's rest frame. So, the four-velocity is [math]U=(\frac{dct}{d\tau}, \frac{dx}{d\tau}, \frac{dy}{d\tau}, \frac{dz}{d\tau})[/math]. So, the velocity in the time direction is given by [math]\frac{dct}{d\tau}[/math]. But the cool thing is that [math]t=\gamma\tau[/math], so the t's cancel out giving the speed of time as [math]{\gamma}c[/math].

  • 3 weeks later...
Posted

Sure it does. We can take our 3D vectors from Euclidean space and create their analogs in a 4D Minkowski space. So, instead of S=(x,y,z), we have S=(ct,x,y,z). We use the time of the rest frame of the object we are studying, so when we are talking about the speed of time, we're talking about the rate time passes according to one frame with respect to an object's rest frame. So, the four-velocity is [math]U=(\frac{dct}{d\tau}, \frac{dx}{d\tau}, \frac{dy}{d\tau}, \frac{dz}{d\tau})[/math]. So, the velocity in the time direction is given by [math]\frac{dct}{d\tau}[/math]. But the cool thing is that [math]t=\gamma\tau[/math], so the t's cancel out giving the speed of time as [math]{\gamma}c[/math].

 

You've still defined the measurement of something changing position (in motion) as the OP claimed.

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