Jump to content

time does not hAVE a speed


Mitch Bass

Recommended Posts

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 /\?

 

 

\\

Link to comment
Share on other sites

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
Link to comment
Share on other sites

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].

Link to comment
Share on other sites

  • 3 weeks later...

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.

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.