Undergrad Time Dilation: Lorentz Transf. & Proper Time Equations

[Dyatlov]

Hello!
Got a bit of an issue with thew two above mentioned equations about time.
From the Lorentz transformation t' = [t - (vx)/c^2]/lorentz factor, we get that the time read by a moving observer for an event in the stationary observer's frame of reference will always be slower (longer) because the denominator will always make the nominator grow when v < c.
Here comes proper time: t’^2- x’^2 = t^2 – x^2. From the moving observer frame of reference (x' = 0) we will get: t’^2= t^2 – x^2; t’^2 = t^2 – (vt)^2;t ’^2= t^2[1 – (v/c)^2]. Here comes the confusing part. This last equation reads that the proper time read by the moving observer for an event in the stationary frame, will be less than the proper time that the stationary observer is reading by a factor of [1 – (v/c)^2]. How can it be less? I thought no matter what frame of reference you relate to, you will always see another observer with dilated time.
Am I mixing things? Proper time is the time measured by a clock moving with the frame of reference, so by definition this time should always be the longest, which would mean the last equation makes sense.
Thanks in advance.

 

[vela]

Hello!
Got a bit of an issue with thew two above mentioned equations about time.
From the Lorentz transformation t' = [t - (vx)/c^2]/lorentz factor, we get that the time read by a moving observer for an event in the stationary observer's frame of reference will always be slower (longer) because the denominator will always make the nominator grow when v < c.

Can you elaborate how you're reaching your conclusion? If you use x=vt, don't you get the same result as the one you found below?

Here comes proper time: t’^2- x’^2 = t^2 – x^2. From the moving observer frame of reference (x' = 0) we will get: t’^2= t^2 – x^2; t’^2 = t^2 – (vt)^2;t ’^2= t^2[1 – (v/c)^2]. Here comes the confusing part. This last equation reads that the proper time read by the moving observer for an event in the stationary frame, will be less than the proper time that the stationary observer is reading by a factor of [1 – (v/c)^2]. How can it be less? I thought no matter what frame of reference you relate to, you will always see another observer with dilated time.

If the moving clock reads less, doesn't that mean it's running slower—that is, it's dilated?

Am I mixing things? Proper time is the time measured by a clock moving with the frame of reference, so by definition this time should always be the longest, which would mean the last equation makes sense.
Thanks in advance.

 

[m4r35n357]

If the moving clock reads less, doesn't that mean it's running slower—that is, it's dilated?

And dilation means "getting bigger", so it's a very misleading term indeed.

 

[Dyatlov]

Yes, that was where my confusion was coming from. Thanks.

 

[m4r35n357]

Yes, that was where my confusion was coming from. Thanks.

As a fellow learner I would advise sticking rigidly to the spacetime interval and diagrams (which you seem OK with). It's the simplest, most direct approach to SR. Then you can just let your eyes glaze over and pretend not to understand (like I do) when folks go off an a tangent talking about "shrinking" rods and trains with their ends measured at different times, or time intervals between two different places ;)