# Angular velocity & impossible situation of relativity

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Interesting problem here. Let's see ...

Bob revolves around Alice observing Alice.

When Bob observes Alice, there is a constant Dobbler blue shift, but despite of that all the frequencies Bob sees are the real unchanged frequencies. They must be the real ones, otherwise there is a problem, as has been pointed out by the OP.

Let's consider some optical fibres on a spinning carousell, fibres are installed radially. At the middle of the carousell there is some photon gas in some container, that gas is steered into the fibres. At the other end of the fibres there are containers into which the photon gas goes.

We are interested about the energy of photon gas in a container in the container frame. The gas was given kinetic energy by the carousell, but in the container frame that energy does not exist. So we conclude that no increase of energy caused by the rotation of the carousell will be observed inside a container. Right?

Edited by Toffo
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It doesn't. Don't project your crackpoterry on the laws of nature. It is time that is affected, GR explains exactly how. I pffered to post the explanation but it involves understanding of basic GR, something that you obviously lack. Besides, you asked for a "touchy-feely" explanation.

You can get the ratio of clock rates relatively easily by knowing GR.

$ds^2=(1-r_s/r) (cdt)^2-\frac{dr^2}{1-r_s/r}-(rd\theta)^2$

Use the fact that

$ds=c d\tau$

Then:

$(c d\tau)^2=(1-r_s/r) (cdt)^2-\frac{dr^2}{1-r_s/r}-(rd\theta)^2$

produces, by division with $(cdt)^2$:

$(\frac{d\tau}{dt})^2=(1-r_s/r)-\frac{(v/c)^2}{1-r_s/r}-(\frac{r \omega}{c})^2$

Write the above for two locations , at radial distances $r_1$ and $r_2$ and you get:

$(\frac{d\tau_1}{dt})^2=(1-r_s/r_1)-\frac{(v_1/c)^2}{1-r_s/r_1}-(\frac{r_1 \omega_1}{c})^2$

$(\frac{d\tau_2}{dt})^2=(1-r_s/r_2)-\frac{(v_2/c)^2}{1-r_s/r_2}-(\frac{r_2 \omega_2}{c})^2$

Divide the two:

$(\frac{d\tau_2}{d \tau_1})^2=\frac{1-r_s/r_2-\frac{(v_2/c)^2}{1-r_s/r_2}-(\frac{r_2 \omega_2}{c})^2}{1-r_s/r_1-\frac{(v_1/c)^2}{1-r_s/r_1}-(\frac{r_1 \omega_1}{c})^2}$

where $v_i= \frac{d r_i}{dt}$ i=1,2. If the orbit is circular, then $v_i=0$ but this is not the usual case.

The above equation relates the proper times of two clocks , located at different radial distances and moving at different angular speeds. It is the foundation of the calculations for setting up the frequencies of the GPS clocks (such that they are synched up).

$\displaystyle \frac{d\tau_2}{d \tau_1}=\sqrt{\frac{1-r_s/r_2-\frac{(v_2/c)^2}{1-r_s/r_2}-(\frac{r_2 \omega_2}{c})^2}{1-r_s/r_1-\frac{(v_1/c)^2}{1-r_s/r_1}-(\frac{r_1 \omega_1}{c})^2}}$

Since the above (exact) formula is rather complicated, the GPS calculations use a power series approximation, making use of the fact that $r_s<<r$. The formula encapsulates the effects in:

1. gravitational potential difference , via the term in $\frac{r_s}{r_i}$

2. radial speed difference (if it exists), via the term in $v_i$

3. angular speed difference, via the term in $r_i \omega_i$

In general, the formula is expressed in terms of ratios of frequencies, not clock rates:

$\displaystyle \frac{f_1}{f_2}= \sqrt{\frac{1-r_s/r_2-\frac{(v_2/c)^2}{1-r_s/r_2}-(\frac{r_2 \omega_2}{c})^2}{1-r_s/r_1-\frac{(v_1/c)^2}{1-r_s/r_1}-(\frac{r_1 \omega_1}{c})^2}}$

No, there is no "compression"

Time passage is a complex function of the gravitational potential and of relative motion. Do you have GPS in your car, mobile phone? How do you think it works? Not through crank misconceptions, that much is sure.

Excuse me xyzt. Is dtau1 a time between photon emission and photon reception in frame of observer1?And is dtau2 a time between the same photon emission and the same photon reception only in observer2 frame?

Edited by DimaMazin

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