Lorentz factor and Bondi factor

[peterpang1994]

I am quite confusing about Lorentz factor and Bondi factor. In special relativity, the change in time of a relatively moving frame and a observer frame can be written as

∆t=∆t'/√1-v²/c²

But if we consider the world line of these 2 frames, the change in time of a relatively moving frame and a observer frame can be written as

∆t=K^-1∆t'
∆t=√(c-v)/√(c+v)∆t'

I assume that the frames are leaving each another, that which one correct??

 

[Mentz114]

The way I see it, the first expression is the change of coordinates between the frames, and the second is the Doppler shift which is actually observed. They are different things and the first does not have the same physical status as the second. I mean, how do you measure something using the coordinate system of another frame ?

 

[pervect]

If' I'm understanding you correctly, they're both correct, depending on what question you ask.

If you have two bodies moving away from each other, you will have some relativistic doppler shift between them, which you call K^(-1).

Let t=0 be the instant at which the bodies cross. Then a radar signal emitted from one of the bodies at time T will be received at time Lt, where L = 1/K is a number greater than 1. If it's reflected or retransmitted, it will arrive back at the original observer at time L^2 T

Using the principles of radar, considering the frame of either one of the bodies, we can conclude that at the midpoint of T and (1+L^2)T, i.e at time (1+L^2)/2 the distance was 1/2 the total round trip propagation time times c, i.e. c*(L^2 -1)/2.

This implies v/c = (L^2-1)/(L^2+1), or L = sqrt(c+v)/sqrt(c-v), which, when you consider that L = 1/K, is equivalent to your equation for K.

So, K and L gives the ratio of "time of reception" to "time of transmission", the doppler shift. That's what you'd actually perceive directly.

If you consider a frame in which one body is at rest, you can conclude that time (L^2-1)/2 on the body at rest must be radar-simultaneous with time L on the moving body.

Working this out, you'll see this is your second equation. It's about what you compute from what you actually directly see (the doppler shift) - using the einstein convention, in the frame of one of the observers , what time on the moving observer is simultaneous with 1 unit on the stationary observer. This is usually called the relativistic time dilation.

So K is the relativistic doppler shift, and your other equation is the relativistic time dilation.

 

[robphy]

For simplicity, let them set their clocks to zero at the meeting event.

In this equation,

∆t=∆t'/√1-v²/c²

the events at the tips of the future-timelike-displacement vectors ∆t and ∆t' are simultaneous according to the observer whose proper-time is ∆t. In other words, it's a Minkowski right-triangle with two future-timelike vectors with tips joined by a spacelike-vector orthogonal to the ∆t-leg. Gamma in a Minkowski-right triangle [ie. cosh(rapidity) where tanh(rapidity)=v] is analogous to cosine in a Euclidean right-triangle. As pervect said, this is associated with time-dilation.

In this equation,

∆t=K^-1∆t'
∆t=√(c-v)/√(c+v)∆t'

the events at the tips of these future-timelike-displacement vectors ∆t and ∆t' are lightlike-related. In other words, it's a triangle with two future-timelike vectors with tips joined by a lightlike-vector. K in Minkowski-geometry can be expressed in terms of exp(rapidity) where tanh(rapidity)=v. As pervect said, this is associated with the doppler effect.

 

[peterpang1994]

Thanks a lot, one more question. Does it mean that the Lorentz factor is used to compute the time dilation of the point like events respect to different frames,Bondi factor is used to compute the dropper effect with a continuos time difference due to the motion of the frames?